Compound Interest 1 / 15 Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the interest being compounded half-yearly. 524.32 624.32 724.32 824.32 2 / 15 A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.482 more, if the interest was payable half yearly than if it was payable annually . The sum is 10000 20000 30000 40000 3 / 15 Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually 2109 3109 4109 5109 Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]=Rs. [8000 * (23/20) * (23/20) * (21/20)]= Rs. 11109. .:. C.I. = Rs. (11109 - 8000) = Rs. 3109. 4 / 15 The compound interest on rs.30000 at 7% per annum is Rs.4347. The period is 2 years 3 years 4 years 5 years Amount = Rs.(30000+4347) = Rs.34347 let the time be n years Then,30000(1+7/100)^n = 34347 (107/100)^n = 34347/30000 = 11449/10000 = (107/100)^2 n = 2years 5 / 15 The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is: Rs. 400 Rs. 500 Rs. 600 Rs. 700 6 / 15 The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum? 8 10 12 14 7 / 15 The difference between simple interest and compound on Rs. 1200 for one year at 10% per annum reckoned half-yearly is: 2 3 4 5 8 / 15 At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years? 6% 7% 8% 9% 9 / 15 What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.? Rs. 9000.30 Rs. 9720 Rs. 10123.20 Rs. 10483.20 10 / 15 The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is: 2 3 4 5 11 / 15 What is the difference between the compound interests on Rs. 5000 for 1 years at 4% per annum compounded yearly and half-yearly? Rs. 2.04 Rs. 3.06 Rs. 4.80 Rs. 8.30 12 / 15 The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is: 625 630 640 650 13 / 15 A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is: Rs. 120 Rs. 121 Rs. 122 Rs. 123 14 / 15 There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate? Rs. 2160 Rs. 3120 Rs. 3972 Rs. 6240 15 / 15 A sum of money amounts to Rs.6690 after 3 years and to Rs.10,035 after 6 years on compound interest.find the sum. 4360 4460 4560 4660 Let the sum be Rs.P.thenP(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)On dividing,we get (1+R/100)^3=10025/6690=3/2.Substituting this value in (i),we get:P*(3/2)=6690 or P=(6690*2/3)=4460Hence,the sum is rs.4460. Your score isThe average score is 47% 0% Restart quiz TIME AND DISTANCE 1 / 15 A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour? 3.6 7.2 8.4 10 2 / 15 An airplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 hour, it must travel at a speed of: 300 kmph 360 kmph 600 kmph 720 kmph 3 / 15 If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is: 50 km 56 km 70 km 80 km 4 / 15 A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is: 100 kmph 110 kmph 120 kmph 130 kmph 5 / 15 Excluding stoppages, the speed of a bus is 54 kmph, and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour? 9 10 12 20 6 / 15 In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is: 1 hour 2 hours 3 hours 4 hours 7 / 15 A man completes a journey in 10 hours. He travels the first half of the journey at the rate of 21 km/hr and the second half at the rate of 24 km/hr. Find the total journey in km. 220 km 224 km 230 km 234 km 8 / 15 The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is: 70 km/hr 75 km/hr 84 km/hr 87.5 km/hr 9 / 15 A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is: 35.55 km/hr 36 km/hr 71.11 km/hr 71 km/hr 10 / 15 A car travelling with  of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car. 17 (6/7) km/hr 25 km/hr 30 km/hr 35 km/hr 11 / 15 In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is: 5 kmph 6 kmph 6.25 kmph 7.5 kmph 12 / 15 Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.? 8 kmph 11 kmph 12 kmph 14 kmph 13 / 15 It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is: 2 : 3 3 : 2 3 : 4 4 : 3 14 / 15 A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is: 14 km 15 km 16 km 17 km 15 / 15 A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is: 35 36 38 40 Your score isThe average score is 0% 0% Restart quiz Volume and Surface Area 1 / 20 Water flows into a tank 200 m x 160 m through a rectangular pipe of 1.5m x 1.25 m @ 20 kmph. At what time (in minutes) will the water rise by 2 meters? 92 min 93 min 95 min 96 min Volume required in the tank = (200 x 150 x 2) cu.m = 60000 cu.mLength of water column flown in1 min =(20 x 1000)/60 m =1000/3 mVolume flown per minute = 1.5 x 1.25 x (1000/3) cu.m= 625 cu.m.Required time = (60000/625)min = 96min 2 / 20 How many iron rods, each of length 7 mts and diameter 2 cms can be made out of 0.88 cubic metre of iron ? 500 600 400 300 3 / 20 A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes. Find the least possible number of cubes. 30 40 10 20 4 / 20 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 cu.m , then the rise in the water level in the tank will be: 20 cm 25 cm 35 cm 50 cm 5 / 20 If each edge of a cube is increased by 50%, find the percentage increase in Its surface area 125% 110% 150% 175% 6 / 20 A triangle has an area of 34 sq cm and an inradius of 4 cm. Find the perimeter of the triangle. 17 cm 16 cm 15 cm 14 cm Area of the Triangle = Inradius * Semi Perimeter of the Triangle = Inradius * Perimeter/2Perimeter = 2 * Area of the Triangle / InradiusPerimeter = 17 cm 7 / 20 Find the height of a parallelogram if its base and area are  30cm and 540 sq cm, respectively. 6 cm 12 cm 18 cm 24 cm Area = Base × HeightHeight = Area/ Base = 540/30 = 18 cm 8 / 20 Which of the following statements is true if the areas of a circle, a square, and an equilateral triangle are C, S, and T, respectively, and their perimeters are the same? C=S=T S<C<T C>S>T C<S<T Perimeter of Circle= Perimeter of Square= Perimeter of Triangle 2πr= 4a=3s  [where r is the radius of the circle, a is the side of the square, and s is the side length of the triangle]r= 2a/π , r = 1.5s  ——-(i)Area of the Circle = πr2Area of the Square = a2Area of the Triangle = 1/2.s. h, [h= (s√3)/2]Using (i)C : S : T =  πr2  : (πr/2)2  : 1/2 . (r/1.5). [(r/1.5)√3)/2]on Solving,C > S> T 9 / 20 If a cone’s height and base radius are both increased by 100%, the cone’s volume will change by what percentage? 2 4 6 8 Volume of the cone=v = 1/3 πr2hNew Radius = R=r+r=2rNew Height= H= h+h=2hNew Volume =V= 1/3πR2H = 1/3 π(2r)2(2h) = 8 v 10 / 20 A solid spherical ball can displace 5 cubic meters of water. What will the increase in water level be if 40 such balls are submerged in a tank at once?  Dimensions of the tank are 40 m X 20 m X 10 m. 20 cm 25 cm 30 cm 35 cm Total Displaced water = 5* 40 m3 = Length of the Tank * Breadth * Increase in the Water Level5*40 = 40 * 20 * HH= 5/20 m = 25 cm 11 / 20 A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is: 90 cm 1 dm 1 m 1.1 cm 12 / 20 The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the ratio of its diameter to its height. 3 : 7 7 : 3 6 : 7 7 : 6 13 / 20 A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is: 4830 5120 6420 8960 14 / 20 A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is: 49 50 53.5 55 15 / 20 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m3, then the rise in the water level in the tank will be: 20 cm 25 cm 35 cm 50 cm 16 / 20 A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3, then the weight of the pipe is: 3.6 kg 3.696 kg 36 kg 36.9 kg 17 / 20 A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is: 12 kg 60 kg 72 kg 96 kg 18 / 20 66 cubic centimeters of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be: 84 90 168 336 19 / 20 A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is: 720 900 1200 1800 20 / 20 In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is: 75 cu. m 750 cu. m 7500 cu. m 75000 cu. m Your score isThe average score is 0% 0% Restart quiz Problems on H.C.F and L.C.M 1 / 23 'Dandia' is a popular dance of Punjab Gujarat Tamil Nadu Maharashtra No answer description is available. 2 / 23 'Natya - Shastra' the main source of India's classical dances was written by Nara Muni Bharat Muni Abhinav Gupt Tandu Muni No answer description is available. 3 / 23 Rabindranath Tagore's 'Jana Gana Mana' has been adopted as India's National Anthem. How many stanzas of the said song were adopted? Only the first stanza The whole song Third and Fourth stanza First and Second stanza No answer description is available. 4 / 23 Find the highest common factor of 36 and 84. 4 6 12 18 5 / 23 Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is: 75 81 85 89 6 / 23 The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is: 15 cm 25 cm 35 cm 45 cm Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm. 7 / 23 252 can be expressed as a product of primes as: 2 x 2 x 3 x 3 x 7 2 x 2 x 2 x 3 x 7 3 x 3 x 3 x 3 x 7 2 x 3 x 3 x 3 x 7 Clearly, 252 = 2 x 2 x 3 x 3 x 7. 8 / 23 The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is: 1008 1015 1022 1032 Required number = (L.C.M. of 12,16, 18, 21, 28) + 7= 1008 + 7= 1015 9 / 23 The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is: 12 16 24 48 Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.So, the numbers 12 and 16.L.C.M. of 12 and 16 = 48. 10 / 23 The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: 279 283 308 318 11 / 23 A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ? 26 minutes and 18 seconds 42 minutes and 36 seconds 45 minutes 46 minutes and 12 seconds L.C.M. of 252, 308 and 198 = 2772.So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec. 12 / 23 The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is: 1677 1683 2523 3363 L.C.M. of 5, 6, 7, 8 = 840. Required number is of the form 840k + 3Least value of k for which (840k + 3) is divisible by 9 is k = 2. Required number = (840 x 2 + 3) = 1683. 13 / 23 The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: 3 13 23 33 L.C.M. of 5, 6, 4 and 3 = 60.On dividing 2497 by 60, the remainder is 37. Number to be added = (60 - 37) = 23. 14 / 23 The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: 74 94 184 364 L.C.M. of 6, 9, 15 and 18 is 90.Let required number be 90k + 4, which is multiple of 7.Least value of k for which (90k + 4) is divisible by 7 is k = 4. Required number = (90 x 4) + 4  = 364. 15 / 23 The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: 1 2 3 4 Let the numbers 13a and 13b.Then, 13a x 13b = 2028 ab = 12.Now, the co-primes with product 12 are (1, 12) and (3, 4).[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).Clearly, there are 2 such pairs. 16 / 23 The G.C.D. of 1.08, 0.36 and 0.9 is: 0.03 0.9 0.18 0.108 Given numbers are 1.08, 0.36 and 0.90.   H.C.F. of 108, 36 and 90 is 18, H.C.F. of given numbers = 0.18. 17 / 23 Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is: 40 80 120 200 Let the numbers be 3x, 4x and 5x.Then, their L.C.M. = 60x.So, 60x = 2400 or x = 40. The numbers are (3 x 40), (4 x 40) and (5 x 40).Hence, required H.C.F. = 40. 18 / 23 The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: 101 107 111 115 Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111. 19 / 23 The greatest number of four digits which is divisible by 15, 25, 40 and 75 is: 9000 9400 9600 9800 Greatest number of 4-digits is 9999.L.C.M. of 15, 25, 40 and 75 is 600.On dividing 9999 by 600, the remainder is 399. Required number (9999 - 399) = 9600. 20 / 23 Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is: 4 5 6 8 N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)= H.C.F. of 3360, 2240 and 5600 = 1120.Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4 21 / 23 Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ? 4 10 15 16 L.C.M. of 2, 4, 6, 8, 10, 12 is 120.So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together30/2+1 =16 times 22 / 23 The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is: 276 299 322 345 Clearly, the numbers are (23 x 13) and (23 x 14). Larger number = (23 x 14) = 322. 23 / 23 Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. 4 7 9 13 Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)= H.C.F. of 48, 92 and 140 = 4. Your score isThe average score is 0% 0% Restart quiz Boats and Streams 1 / 20 A 500 m long train takes 36 seconds to cross a man walking in the opposite direction at the speed of 10 km/hr. Find the speed of the train. 40 km/hr. 50 km/hr. 60 km/hr. 70 km/hr. Let the speed of the train be T km/hr.=> Relative speed = T + 10 km / hr=> Length of the train = 500 m = 0.5 kmWe know that Distance = Speed x Time=> 0.5 = (T + 10) x (36 / 3600)=> 50 = T + 10=> T = 40 km/hrTherefore, the speed of the train is 40 km/hr. 2 / 20 Two trains 140 m and 160 m long are moving in the same direction on parallel tracks with speeds of 40 km/hr and 50 km/hr respectively. How much time would the faster train require to overtake the slower train? 90 sec 100 sec 108 sec 120 sec Total distance to be covered = 140 + 160 m = 300 mRelative speed = 50 – 40 = 10 km / hr = 10 x (5 / 18) m / sec = 50 / 18 m / secTherefore, the time is taken by the faster train to overtake the slower train = 300 / (50/18) = 108 sec 3 / 20 Two trains 140 m and 160 m long are moving towards each other on parallel tracks with speeds of 40 km/hr and 50 km/hr respectively. How much time would they take to pass each other completely? 9 sec 10 sec 11 sec 12 sec Total distance to be covered = 140 + 160 m = 300 mRelative speed = 40 + 50 = 90 km / hr = 90 x (5 / 18) m / sec = 25 m / secTherefore, time taken to pass each other = 300 / 25 = 12 sec 4 / 20 A man standing near a railway track observes that a train passes him in 80 seconds but to pass by a 180 m long bridge, the same train takes 200 seconds. Find the speed of the train. 1 m/sec 1.5 m/sec 2 m/sec 3 m/sec Let the length of the train be L meters.=> The train covers L meters in 80 seconds and L + 180 meters in 200 seconds, with the same speed.We know that Speed = Distance / Time.=> Speed = L / 80 = (L + 180) / 200=> L / 80 = (L + 180) / 200=> 2.5 L = L + 180=> 1.5 L = 180=> L = 120Thus, speed of the train = 120 / 80 = 1.5 m / sec 5 / 20 A 100 m long train moving at a speed of 60 km/hr passes a man standing on the pavement near a railway track. Find the time taken by the train to pass the man. 3 sec 4 sec 5 sec 6 sec Length of the train = 100 m = 0.1 kmSpeed of the train = 60 km / hrSo, the time taken by the train to pass the man = time taken to cover 0.1 km at the speed of 60 km/hrTherefore, time taken by the train to pass the man = 0.1 / 60 hour = (0.1 / 60) x 3600 sec = 6 sec 6 / 20 A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is: 1 km/hr 1.5 km/hr 2 km/hr 2.5 km/hr 7 / 20 A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is: 2 : 1 3 : 1 3 : 2 4 : 3 8 / 20 The speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is: 16 hours 18 hours 20 hours 24 hours 9 / 20 A man can row three-quarters of a kilometre against the stream in 11 minutes and down the stream in 7 minutes. The speed (in km/hr) of the man in still water is: 2 3 4 5 10 / 20 A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water? 40 minutes 1 hour 1 hr 15 min 1 hr 30 min 11 / 20 A boat covers a certain distance downstream in 1 hour, while it comes back in 1 hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water? 12 kmph 13 kmph 14 kmph 15 kmph 12 / 20 A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place? 2.4 km 2.5 km 3 km 3.6 km 13 / 20 A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is: 2 mph 2.5 mph 3 mph 4 mph 14 / 20 The speed of a boat in still water in 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 12 minutes is: 1.2 km 1.8 km 2.4 km 3.6 km 15 / 20 A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream, it takes 4 hours. What is the speed of the boat in still water? 4 km/hr 6 km/hr 8 km/hr 10 km/hr 16 / 20 In one hour, a boat goes 11 km/hr along the stream and 5 km/hr against the stream. The speed of the boat in still water (in km/hr) is: 3 km/hr 5 km/hr 8 km/hr 9 km/hr Speed in still water =1/2(11 + 5) kmph = 8 kmph. 17 / 20 A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is: 4 5 6 7 18 / 20 A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively? 2 : 1 3 : 2 8 : 3 3 : 8 19 / 20 A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is: 8.5 km/hr 9 km/hr 10 km/hr 12.5 km/hr Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr.Man's rate against the current = (12.5 - 2.5) km/hr = 10 km/hr. 20 / 20 A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream. 2 hours 3 hours 4 hours 5 hours Your score isThe average score is 0% 0% Restart quiz Height and Distance 1 / 20 The stature of a tree is 10m. It is twisted by the wind in a manner that its top touches the ground and makes a point of 60 with the ground. At what range from base did the tree get twisted? (√3=1.73) 4.6m 4.8m 5.2m 5.4m Let AB be the tree bowed at the point C so that part CB takes the position CD.Then ,CD=CB. Let AC=x meters. At that point, CD = CB= (10-X) m and ∠ADC=60°.AC/AC=sin60° => x/(10-x) = √3/2=>2x=10 √3-√3x=> (2+ √3) x= 10 √3=>x=10 √3/ (2+ √3)*(2-√3)/(2-√3)=20 √3-30)m= (20*1.73-30) m=4.6m=> Required height=4.6m. 2 / 20 A kite is flying at a tallness of 75 m from the level of ground, joined to a string slanted at 60° to the level. The string's length is: 50 √2 m 50√3 m 50m/√2 50m/√3 Let AB be the kite and AC be the level groundSo that BC - AC.At that point, ∠BAC=60°and BC=75m. Let AB=x meters.Presently AB/BC=coses60°=2/ √3=> x/75=2/√3 =>x=150/√3 =150* √3/3=50 √3m.∴ Length of the string=50 √3m. 3 / 20 At a moment, the shadow's length of a shaft is √3 times the stature of the shaft. The edge of rise of the sun is: 30° 45° 60° 90° Let AB be the post and AC be its shadow.Let AB =X m. Then, AC= √3Xm.Let∠ACB=θ.AB/AC=tanθ => tanθ = X/√3X = 1/√3 =tan30°.∴ θ = 30°.Hence, the point of rise is 30°. 4 / 20 Two men are inverse sides of a tower. They gauge the edge of the rise of the highest point of the tower as 30° and 45° respectively. On the off chance that the tallness of the tower is 50 m, discover the separation between the two men. (Take √3=1.732) 135.5m 136.5 m 137.5 m 138.5m Let AB be the tower and let C and D be the two's positions men.At that point ∠ACB=30°,∠ADB= 45°and AB= 50 mAC/AB = Cot30°=√3 => AC/50 = √3=>AC=50√3mAD/AB=cot 45°=1 => AD/50=1=> AD=50M.Separation between the two men =CD= (AC+AD)= (50√3+50) m=50(√3+1)=50(1.73+1)m=(50*2.73)m=136.5m. 5 / 20 The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature. 16.32m 17.32 m 18.32 m 19.32 m Let AB be the building and AC be its shadow.Then, AC=20m and ∠ACB=60°.Let AB= x m.Presently AB/AC=tan 60°=√3=>x/10=√3=>x=10√3m= (10*1.732) m=17.32m.∴ Height of the building is 17.32m. 6 / 20 A straight tree is broken due to thunder storm. The broken part is bent in such a way that the peak touches the ground at an angle elevation of 45°. The distance of peak of tree (where it touches the root of the tree is 20 m. Then the height of the tree is 48.28 meters 30.28 meters 24.14 meters 28.14meters Let the total length of the tree be X+Y metersFrom the figure tan 45=X/20 =>X=20cos 45 = 20/Y =>Y=20/cos 45 =20√2X+Y=20+20radic;2=20+2x10x1.414 =48.28 meters 7 / 20 An aeroplane when 750 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 45° and 30&de; respectively. Approximately, how many meters higher is the one than the other? 250(1- √3) 750(3- √3) 250(3- √3) 275(1- √3) Let C and D be the position of the aeroplanes.Given that CB = 900 m,∠CAB = 60°,∠DAB = 45°From the right △ ABC,Tan45=CB/AB=>CB=ABFrom the right △ ADB,Tan30=DB/AB=>DB=ABtan30=CBx(1/√3)=750/√3CB=CD+DB=> Required height CD=CB-DB=750-750/√3=250(3- √3) 8 / 20 A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 10 minutes for the angle of depression to change from 45° to 60°, how soon after this will the car reach the observation tower? 10 min 30 sec 13 min 20 sec 15 min 10 sec 16 min 30 sec Let AB be the tower and C and D be the two positions of the car.Then,from figureAB/AC=tan 60 =√3 => AB=√3ACAB/AD=tan 45=1 => AB=ADAB=AC+CDCD=AB-AC=√3AC - AC=AC (√3-1)CD = AC (√3-1) =>10 min AC=> ?AC/(AC(√3-1)) x 10=?=10/(√3-1)=13.66=13 min 20 sec(approx) 9 / 20 The horizontal distance between two towers is 90 m. The angular depression of the top of the first as seen from the top of the second which is 180 m high is 450.Then the height of the first is 60 m 45 m 90 m 135 m =>(180-h)/90 = Tan(45)=> h =90 m 10 / 20 A man is watching form the top of the tower a boat speeding away from the tower. The boat makes the angle of depression of 60° with the man's eye when at a distance of 75 meters from the tower. After 10 seconds the angle of depression becomes 45°. What is the approximate speed of the boat, assuming that it is running in still water? 20.8 kmph 20 kmph 19 kmph 19.8 kmph Let AB be the tower and C and D be the positions of the boat.Distance travelled by boat = CDFrom the figure 75tan(60)=(75+CD)tan(45)=>75√3 = 75+CD=>CD =55 mSpeed = distance/time=55/10=5.5 m/sec=19.8 kmph 11 / 20 An observer 1.4 m tall is 10√3 away from a tower. The angle of elevation from his eye to the top of the tower is 60°. The heights of the tower is 12.4 m 10.4 m 6.4 m 11.4 m Let AB be the observer and CD be the tower.Then, CE = AB = 1.4 m, BE = AC = 10v3 m.DE/BE=Tan (30) =1/√3DE=10√3/√3=10CD=CE+DE=1.4+10=11.4 m 12 / 20 From a point P on a level ground, the angle of elevation of the top tower is 60°. If the tower is 180 m high, the distance of point P from the foot of the tower is 60√3 40√3 30√3 20√3 From ∠APB = 60° and AB = 180 m.AB/AP= tan 60° =√3AP=AB/√3 =180/√3=60√3 13 / 20 The heights of two towers are 90 meters and 45 meters. The line joining their tops make an angle 450 with the horizontal then the distance between the two towers is 22.5 m 45 m 60 m 30 m Let the distance between the towers be X From the right angled triangle CFD Tan(45)= (90-45)/X => x=45 meters 14 / 20 The angles of elevation of the tops of two vertical towers as seen from the middle point of the lines joining the foot of the towers are 45° & 60°.The ratio of the height of the towers is √3:2 √3:1 2:√3 2:1 Tan(60)=h1/AB=> h1=√3ABTan(45)=h1/BC=> h2=BCh1/ h2=√3/1=> h1:h2=√3:1 15 / 20 On the level ground, the angle of elevation of the top of a tower is 30°.on moving 20 meters nearer, the angle of elevation is 45°.Then the height of the tower is 10 √3 10√3 20√3 Let h be the height of towerFrom figure.20 =h ( cot30 - cot60) 20 =h (√3-1/√3) => 20√3 = h (3-1) => h=10√3. 16 / 20 The angle of elevation of a tower at a point 90 m from it is cot-1(4/5).Then the height of the tower is 45 90 112.5 135 Let cot-1(4/5) = x => cotx = 4/5 => tan(x) = 5/4 From the right angled triangleTan(x) = h/90 => h = 5/4*90 =112.5 m 17 / 20 From a point 375 meters away from the foot of a tower, the top of the tower is observed at an angle of elevation of 45°, then the height (in meters) of the tower is? 375 450 225 250 From the right angled triangleTan(45°)= X/375=> X = 375 m 18 / 20 From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is: 149 m 156 m 173 m 200 m 19 / 20 The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is: 2.3 m 4.6 m 7.8 m 9.2 m Let AB be the wall and BC be the ladder.Then, ACB = 60° and AC = 4.6 m.AC/BC = COS 60°= 1/2  BC= 2 x AC= (2 x 4.6) m= 9.2 m. 20 / 20 Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is: 173 m 200 m 273 m 300 m Your score isThe average score is 0% 0% Restart quiz Square Root and Cube Root 1 / 20 Find the value of a and b such that ab = 512 where a > b and 1 < b < 4. Also, find the value of a1/b. 2 4 6 8 Now, as 512 has 2 in its unit place, then it cannot be a square of any number.But, we see that 83 = 512,∴ a = 8 and b = 3Now, 81/3 = ∛8 = 2. 2 / 20 There is a certain number of rows of chairs in a room. The number of chairs in each row is thrice the total number of rows. Find the number of chairs in each row and the number of rows in the room if the total number of chairs is 2187. 17 & 51 19 & 57 23 & 69 27 & 81 Let there be x rows of chairs in that room.Number of chairs in each row = 3xTotal number of chairs = 2187⇒ x × 3= 2187⇒ x2 = 2187/3 = 729⇒ x = √729 = 27∴ there are 27 rows and 81 chairs in each row. 3 / 20 Find the cube root of 42875 by shortcut method. 25 55 45 35 Step I: Looking at the unit digit of 42875, since it is 5 the cube root will also have 5 in its unit’s place.Step II: Strike out last three digits42875Now, 42 > 33∴ ∛42875 = 35. 4 / 20 Find the smallest number that must be subtracted from 130 to make it a perfect cube number. Also, find the cube root of the number. 5 10 15 25 Now 53 = 125 < 130Thus, if we subtract 5 from 130 we get a perfect cube number, that is 125∛125 = 5. 5 / 20 Check whether 1512 is a perfect cube or not. Also, find the smallest number which must be multiplied by it to make the given number a perfect cube number and find the cube root as well. 22 32 42 52 Prime factorisation of 1512 = 2 × 2 × 2 × 3 × 3 × 3 × 7Thus, 7 × 7 = 49 must be multiplied by 1512, 1512 × 49 = 74088 is a perfect cube number.∛74088 = ∛(2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 7) = 2 × 3 × 7 = 42. 6 / 20 Find the cube root of 21952 by the prime factorisation method. 18 24 28 34 Prime factorisation of 21952 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7Now making pairs of three∛21952 = ∛(2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7) = 2 × 2 × 7 = 28∴ ∛21952 = 28. 7 / 20 Find the square root of 1444 by the prime factorisation method. 28 38 48 58 Prime factorisation of 1444 = 2 × 2 × 19 × 19Now making pairs of two√1444 = √(2 × 2 × 19 × 19) = 2 × 19 = 38∴ √1444 = 38. 8 / 20 The square root of 64009 is: 253 347 363 803 9 / 20 How many two-digit numbers satisfy this property.: The last digit (unit's digit) of the square of the two-digit number is 8 ? 1 2 3 None of these 10 / 20 The least perfect square, which is divisible by each of 21, 36 and 66 is: 213444 214344 214434 231444 L.C.M. of 21, 36, 66 = 2772.Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11To make it a perfect square, it must be multiplied by 7 x 11.So, required number = 22 x 32 x 72 x 112 = 213444 11 / 20 The cube root of .000216 is: 0.6 0.06 77 87 12 / 20 A B C D 13 / 20 10 100 1000 0.1 14 / 20 5 6 7 8 15 / 20 5.59 7.826 8.944 10.062 16 / 20 A B C D 17 / 20 A B C D 18 / 20 A B C D 19 / 20 A B C D 20 / 20 1.05 1.25 1.45 1.55 Your score isThe average score is 65% 0% Restart quiz profit and loss 1 / 20 A trader has 600kgs of rice , a part of which he sells at 15% profit and the remaining quantity at 20% loss. On the whole , he incurs an overall loss of 6% .What is the quantity of rice he sold at 20% loss?ARs 360BRs 960CRs 560DRs 460 A B C D 2 / 20 A dishonest shopkeeper professes to sell pulses at the 20% profit , also he uses a false weight of 800gm for a kg, His actual gain is____%.A25%B15%C20%D50% A B C D Let CP of 1000g item =Rs.1000 CP of 800g item =Rs.800SP of 800g item =Rs.1000Profit by false weight = 1000−800 = 200profit by selling = 20% of 1000 = 200P% = 400/800×100 =50% 3 / 20 On selling 15 balls at Rs 400 there is loss equal to Cost Price of 5 balls. The cost price of a ball is?ARs 30BRs 35CRs 40DRs 45 A B C D Let the cost price of 1 ball be x.According to the problem, the loss incurred on selling 15 balls at Rs 400 is equal to the cost price of 5 balls. This can be written as:Loss = Cost price of 5 balls15x−400 = 5x      10x = 400         x = 40Therefore, the cost price of 1 ball is Rs 40. 4 / 20 If selling price is doubled then, the profit triples. What is profit per cent?A20%B75%C100%D50% A B C D Let the C.P be Rs.100 and S.P be Rs.x, Then The profit is (x−100) Now the S.P is doubled,  then the new S.P is 2x New profit is (2x−100)Now as per the given condition;=> 3(x−100) = 2x−100By solving, we getx = 200Then the Profit percent = (200−100)/100                               = 100Hence the profit percentage is 100% 5 / 20 A and B, there are two companies, selling the packs of cold-drinks. For the same selling price A gives two successive discounts of 10% and 25%. While B sells it by giving two successive discounts of 15% and 20%. What is the ratio of their marked price?A. 143 : 144B. 19 : 11C. 136 : 135D. 73 : 77 A B C D A = 90/100*75/100=> 0.675B = 85/100*80/100=> 0.68=> 680 : 675=> 136 : 135 6 / 20 A person sold a pen at Rs. 96 in such a way that his percentage profit is same as the cost price of the watch. If he sells it at twice the percentage profit of its previous percentage profit then new selling price will be?A. Rs. 132B. Rs. 150C. Rs. 180D. Rs. 192 A B C D Rs. 132Explanation: CP = xProfit Percentage = x%SP = x(100 + x)/100x(100 + x)/100 = 96x = 60Profit Percentage = 60%New SP = 60 * 220 / 100 = 132 7 / 20 An article is sold at 25% profit. If the CP and the SP of the article are increased by Rs 60 and Rs 30 respectively, the profit% decreases by 15%. Find the cost price of the article.A. Rs. 240B. Rs. 270C. Rs. 285D. Rs. 305 A B C D CP = x, then SP = (125/100)*x = 5x/4New CP = (x+60), new SP = (5x/4 + 30), new profit% = 25-15 = 10So (5x/4 + 30) = (110/100) * (x+60)Solve, x = 240 8 / 20 A shopkeeper bought 150 pen drives at the rate of Rs. 500 per pen drive. He spent Rs. 500 on transportation and packing. If the marked price of pen drive is Rs. 520 per pen drive and the shopkeeper gives a discount of 5% on the marked price then what will be the percentage profit gained by the shopkeeper?A. 1.2%B. 3.8%C. 4.5%D. 5.5% A B C D C.P. of 150 calculators = 150 * 500 = Rs. 75000∴ Total C.P. = 75000 + 500 = Rs. 75500Marked price of 150 calculator = 150 * 570 = Rs. 82500Selling price after discount = 82500 * 95 / 100 = Rs. 78375∴ percentage profit = [(78375 – 75500) / 75500] * 100 = 3.8% 9 / 20 If an article is sold at a loss of 66 2/3%, what is the loss in terms of the selling price?A. 50%B. 100%C. 150%D. 200% A B C D Let the C. P. = 100∴ Amount of loss = 66 2/3 or 200/3⇒ S. P = 100 – 66 2/3 = 33 1/3 or 100/3∴ Loss expressed in terms of S. P. = 100 × (200/3)/(100/3) 100 = 200% 10 / 20 Bhajan Singh purchased 120 reams of paper at Rs. 80 per ream. He spent Rs. 280 on transportation. Paid octrai at the rate of 40 paise per ream and paid Rs. 72 to the coolie. If he wants to have a gain of 8 %. What must be the selling price per ream?A. Rs. 86B. Rs. 88C. Rs. 90D. Rs. 92 A B C D C.P of 120 reams = Rs. (120 × 80 + 280 + 72 + 120 × 40) = Rs. 10000C.P of 1 ream = (10000/120) = Rs. (250/3)S.P of 1 ream = Rs. (108/100 × 250/3) = Rs. 90 11 / 20 If two mixers and one T.V cost Rs. 7000. While two T.V s and one mixer cost Rs. 9800. The value of one T.V is:A. Rs. 3,200B. Rs. 3,600C. Rs. 4,200D. Rs. 4,600 A B C D 2X + Y = 7000X + 2Y = 9800Solving (i) and (ii), we get Y = 4200 12 / 20 Profit after selling a commodity for Rs. 425 is same as loss after selling it for Rs.355. The cost of the commodity is:A. Rs. 385B. Rs. 390C. Rs. 395D. Rs. 400 A B C D Let C.P = Rs. X425 – X = X – 355 (or)2X = 780 or X = 390 13 / 20 A man sold two houses for Rs. 7.8 lakhs each. On the one, he gained 5% and on the other, he lost 5%. What percent is the effect of the sale on the whole?A. 0.25 % lossB. 0.25 % gainC. 25 % lossD. 25 % gain A B C D Loss% = (5/10)^2 = 1/4% = 0.25% 14 / 20 Amit bought 160 shirts at the rate of Rs. 225/shirt. The transport expenditure was Rs. 1,400. He paid an octroi at the rate of Rs. 1.75/shirt and labour charges were Rs. 320. What should be the selling price of 1 shirt, if he wants a profit of 20%?A. Rs. 260B. Rs. 275C. Rs. 280D. Rs. 285 A B C D Total CP per shirt = 225 + 1400/160 + 1.75 + 320/160 = Rs. 237.5SP = CP[(100 + profit%)/100]= 237.5 * [(100 + 20)/100] = Rs. 285 15 / 20 By selling an article at Rs. 800, a shopkeeper makes a profit of 25%. At what price should he sell the article so as to make a loss of 25%?A. Rs. 460B. Rs. 500C. Rs. 480D. Rs. 520 A B C D  SP = 800, Profit = 25%CP = (SP)*[100/(100+P)]= 800 * [100/125] = 640Loss = 25% = 25% of 640 = Rs. 160SP = CP – Loss = 640 – 160 = Rs. 480 16 / 20 A, B and C enter into a partnership. A invests some money at the beginning, B invests double the amount after 6 months, and C invests thrice the amount after 8 months. If the annual gain be Rs. 18,000. A's share is?A. Rs. 5,750B. Rs. 6,000C. Rs. 7,200D. Rs. 7,500 A B C D  x*12 : 2x*6: 3x*41:1:11/3 * 18000 = 6000 17 / 20 A trader sells 85 m of cloth for Rs. 8,925 at the profit of Rs. 15/m of cloth. What is the cost price of 1 m of cloth?A. Rs. 84B. Rs. 90C. Rs. 96D. Rs. 102 A B C D SP of 1m of cloth = 8925/85 = Rs. 105CP of 1m of cloth = SP of 1m of cloth – profit on 1m of cloth= Rs. 105 – Rs. 15 = Rs. 90 18 / 20 A person buys a horse for 15 pounds. After one year, he sells it for 20 pounds. After one year, again he buys the same horse at 30 pounds and sells it for 40 pounds. What is the overall profit percent for that person over both the transactions?A. 30.33%B. 33.33%C. 35.33%D. 40.33% A B C D Total C.P. = 45Total S.P. = 60Profit% = (15/45) *100 = 33.33% 19 / 20 David sold a bicycle at Rs 2100 and made a loss of 25%. At what price should he sell the bicycle if he wants to make a profit of 15%?ARs 2500BRs 2990CRs 3220DRs 3500 A B C D When David sold the bicycle for Rs 2100 with a loss of 25%, it means that his selling price was only 75% of the original cost price. Let the original cost price of the bicycle be x.Selling price with a loss of 25% = 0.75xGiven that the selling price is Rs 2100, we can write:0.75x = 2100x = 2800So, the original cost price of the bicycle is Rs 2800.Now, David wants to sell the bicycle at a profit of 15%. Let the selling price be y.Selling price with a profit of 15% = 1.15xWe know that x = 2800, so:Selling price with a profit of 15% = 1.15(2800)= 3220Therefore, David should sell the bicycle for Rs 3220 if he wants to make a profit of 15% 20 / 20 A vendor bought 6 oranges for Re 10 and sold them at 4 for Re 6. Find his loss or gain percent.A - 8% gainB - 10% gainC - 8% lossD - 10% loss A B C D Suppose, number of oranges bought = LCM of 6 and 4 = 12∴CP = Re (10/6 * 12) = Re 20 and SP = Re (6/4 * 12) = Re 18∴Loss% = (2/20 * 100)% = 10% Your score isThe average score is 0% 0% Restart quiz Time and Work 1 / 15 Two brothers(younger and older) can do a piece of work in 70 and 60 days respectively. They began to work together, but the younger brother leaves after some days and the older brother finished the remaining work in 47 days. After how many days did the younger brother leave?A) 14 daysB) 16 daysC) 11 daysD) 7 days A B C D The Older brother would have done 47/60 work in 47 days.The remaining work = (1 -47/60) = 13/60 must have done by them together.Two brothers 1 day work = (1/70 + 1/60) = 13/420So, they would have done 13/60 work in 420/13 × 13/60 = 7 days.Therefore, The younger brother left the work after 7 days. 2 / 15  Akash can reap a field in 45 days and Vishal is 200% more expert than Akash to reap the field, then find total time taken to reap the field by Vishal.A) 15B) 20C) 22D) 25 A B C D According to the question, Vishal is more efficient than Akash. So, he takes less days to complete the work.Let, Vishal takes time = x daysSo, Akash takes time = 3x daysAccording to the question,3x = 45x = 15Thus, Vishal takes time 15 days. 3 / 15 If 12 men or 18 boys can build a wall in 48 days, then how long will 6 men or 9 boys build the wall?A) 48 daysB) 36 daysC) 64 daysD) 96 days A B C D 12 men = 18 boys2 men = 3 boysNow, 12 men + 2×18/3 men = 24 MenAnd, 6 men + 2× 9/3 men = 12 menHere, M1 = 24, M2 = 12, D1 = 48, D2 =?According to the formula,M1 × D1 = M2 × D2=> 24 × 48 = 12 × D2=> D2 = 96 4 / 15  E and F are two friends working together to finish a work in 24 days and E alone can do the same work in 36 days. In how many days can F alone complete the work?A) 36 daysB) 24 daysC) 72 daysD) 48 days A B C D (E+F)’s 1 days work = 1/24E’s 1-day work = 1/36F’s 1 day work = (1/24 – 1/36)  = 1/72Thus, the time is taken by F to finish the work alone= 72 days 5 / 15 P and Q can do a job in 3 days. Q and R can do the same job in 9 days, while R and P can do it in 12 days. In how many days the job will be finished when P, Q and R working together.A) 72/19 daysB) 83/10 days.C) 61/3 days.D) 67/4 days A B C D Here, (P + Q)’s 1 day’s work = 1/3(Q + R )’s 1 day’s work= 1/9and (R + P)’s 1 day’s work = 1/12Now, 2 (P + Q + R)’s 1 day’s work = (1/3+1/9+1/12)= 19/36(P + Q + R)’s 1 day’s work = 19/(36 ×2)=19/72Hence, (P + Q + R) complete the work in= 72/19 days 6 / 15 M did a piece of work in 5 days. That piece of work was done by N in 9 days. If M and N worked together, they got total wages of Rs 4200. Find the share of N.A) 1500B) 2000C) 1000D) 1200 A B C D M  :  NTime  =      5   :  9Efficiency=    9   :  5(Time and efficiency are inversely proportional)N gets = 4200 ×  5/14= 1500Thus, N gets the wages of Rs 1500. 7 / 15 P, Q and R together can complete a work in 16 days and R alone complete the work in 20 days. If P, Q and R started the work together and after 10 days P and Q left the work, in how many days R alone complete the remaining work?A) 12½ daysB) 20½ daysC) 4 daysD) 7½ days A B C D P + Q + R = 16 daysR =20 daysTotal work (LCM of 16 and 20) = 80( P + Q +R) ‘s work= 80 /16= 5 unitWork done by R = 80 /20 = 4 unit(P + Q + R) ‘s 10 days work= 5 × 10= 50 unitRemaining work= (80 – 50)= 30 unitRemaining work done by R= 30/4= 7½ days 8 / 15 In a factory, 20 people can make 20 toys in 15 days working 10 hours per day. Then, in how many days can 25 persons make 30 toys working 20 hr per day?A) 6 daysB) 9 daysC) 10 daysD) 12 days A B C D Here, M1 = 20, M2 = 25, D1 = 15, D2 =? , T1 = 10 , T2= 20, W1 = 20 and W2 = 30.We know,M1 × D1 × T1 × W2 = M2 × D2 × T2 × W1=> 20 × 15 × 10 × 30 = 25 × D2 × 20 × 20=> D2 = 9.Thus, the required day = 9 days 9 / 15 Karan completes 1/15 part of a certain job in 1 day. In how many, he will complete the full job.A) 30 daysB) 15 daysC) 20 daysD) 5 days A B C D Here, 1/n = 1/15So, n = 15Thus, the required days = 15 10 / 15 Vicky completes a job in 45/2 days. What part of the job will he do in 2 days?A. 4/45B. 1/45C. 2/45D. 8/45 A B C D We know, if a person does a job in n days, then his 1-day work = 1/nHere, n = 45/2Vicky’s 1-day work = 2/45Thus, Vicky’s 2 days work = 2× 2/45 = 4/45 11 / 15 Ankit alone can do a piece of work in 6 days and Bishal alone in 8 days. Ankit and Bishal undertook to do it for Rs. 4800. With the help of Dinesh, they completed the work in 3 days. How much is to be paid to Dinesh?A. Rs. 1375B. Rs. 1400C. Rs. 1600D. Rs. 1800 A B C D Ankit’s 1day work = 1/6Bishal’s 1 day work = 1/8(Ankit + Bishal + Dinesh)’s 1 day work =1/3Dinesh’s 1 day work = 1/3 – (1/6+1/8) = 1/24Ankit’s wages : Bishal’s wages : Dinesh’s wages=1/6 : 1/8 : 1/24.Here the (LCM of 6,8 and 24) is 24. Now, taking LCM we get ratio= 4 : 3 : 1Dinesh’s share (for 3 days) = 3 × 1/8 × 4800 = 1800 12 / 15 M’s efficiency is three times N’s efficiency. M can finish a job in 60 days less than N. If they work together, then in how many days the job will be done.A. 20 daysB. 22.5 daysC. 25 daysD. 30 days A B C D The ratio of times taken by M and N = 1 : 3  (Since the efficiency of M is three times to N)The time difference is (3 – 1) = 2 days, while N take 3 days and M takes 1 day.2 units = 60 days1 unit = 30 daysSo, M takes 30 days to do the job.And N takes (30×3) = 90 days to do the job.M’s 1 day’s work = 1/30N’s 1 day’s work = 1/90(M + N)’s 1 day’s work =( 1/30 + 1/90) = 2/45M and N together can do the job in 45/2 days = 22.5 days. 13 / 15 P, Q and R can do a job in 20, 30 and 60 days respectively. In how many days can P do the job if he is assisted by Q and R every third day?A. 11 daysB. 15 daysC. 17 daysD. 16 days A B C D P’s 2 day’s work = 2/20 = 1/10(P + Q + R)’s 1 day’s work= (1/20 + 1/30 + 1/60)= 6/60  = 1/10Job done in 3 days = (1/10 + 1/10) = 1/5Now, 1/5 jobs is done in 3 daysWhole job  will be done in (3 x 5) = 15 days. 14 / 15 L can finish a work in  16 days and M can do the same work in 12 days. With help of N, they did the work in 4 days only. Then, N alone can do the work in how many days.A. 48/5 daysB. 48/7 daysC. 48/11 daysD. 10 days A B C D (L + M + N)’s 1 day’s work =1/4L’s 1 day’s work = 1/16M’s 1 day’s work = 1/12Therefore, N’s 1 day’s work= 1/4 – (1/16 + 1/12)= 1/4 – 7/48= 5/48So, N alone can do the work in 48/5 days. 15 / 15 A man can do a work in 20 days and a woman in 15 days. If they work on it together for 5 days, then the fraction of the work that is left is :A. 1/12B. 1/10C. 5/12D. 7/15 A B C D Man’s 1 day’s work = 1/20Woman’s 1 day’s work = 1/15(Man + woman)’s 1 day’s work = (1/20 + 1/15)  = 7/60(Man + woman)’s 5 day’s work = (7/60 × 5) = 7/12Thus, Remaining work = 1 – 7/12 = 5/12 Your score isThe average score is 0% 0% Restart quiz Problems on Trains 1 / 20 20.A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train? 69.5 km/hr 70 km/hr 79 km/hr 79.2 km/hr Explanation:Let the length of the train be x metres and its speed by y m/sec. Then,x= 8        x = 8yy Now,x + 264= y20  8y + 264 = 20y y = 22.  Speed = 22 m/sec =22 x18km/hr = 79.2 km/hr.5 2 / 20 19.A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is: 50 m 150 m 200 m Data inadequate Explanation:Let the length of the train be x metres and its speed by y m/sec. Then,x= 15        y =x.y15 x + 100=x2515  15(x + 100) = 25x 15x + 1500 = 25x 1500 = 10x x = 150 m. 3 / 20 18.A 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform? 320 m 350 m 650 m Data inadequate Explanation: Speed =300m/sec =50m/sec.183 Let the length of the platform be x metres. Then,x + 300=50393  3(x + 300) = 1950 x = 350 m. 4 / 20 17.A train 800 metres long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 minute, then the length of the tunnel (in meters) is: 130 360 500 540 Explanation: Speed =78 x5m/sec=65m/sec.183 Time = 1 minute = 60 seconds.Let the length of the tunnel be x metres. Then,800 + x=65603  3(800 + x) = 3900 x = 500. 5 / 20 16.A train travelling at a speed of 75 mph enters a tunnel 3 miles long. The train is  mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges? 2.5 min 3 min 3.2 min 3.5 min Explanation: Total distance covered=7+1miles24=15miles.4  Time taken=15hrs4 x 75=1hrs20=1x 60min.20= 3 min. 6 / 20 15.A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going? 5 sec 6 sec 7 sec 10 sec Explanation:Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.   =66 x5m/sec18   =55m/sec.3  Time taken to pass the man =110 x3sec = 6 sec.55 7 / 20 14.Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is: 9 9.6 10 10.8 Explanation: Relative speed = (60 + 40) km/hr =100 x5m/sec=250m/sec.189 Distance covered in crossing each other = (140 + 160) m = 300 m. Required time =300 x9sec=54sec = 10.8 sec.2505 8 / 20 13.Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is: 30 km/hr 45 km/hr 60 km/hr 75 km/hr Explanation:Let the speed of the slower train be x m/sec.Then, speed of the faster train = 2x m/sec.Relative speed = (x + 2x) m/sec = 3x m/sec. (100 + 100)= 3x8  24x = 200  x =25.3 So, speed of the faster train =50m/sec3   =50x18km/hr35 = 60 km/hr. 9 / 20 12.A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train? 230 m 240 m 260 m 270 m Explanation: Speed =72 x5m/sec= 20 m/sec.18 Time = 26 sec.Let the length of the train be x metres. Then,x + 250= 2026  x + 250 = 520 x = 270. 10 / 20 11.A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? 230 m 240 m 260 m 320 m None of these Explanation:Relative speed = (120 + 80) km/hr   =200 x5m/sec18   =500m/sec.9 Let the length of the other train be x metres. Then,x + 270=50099  x + 270 = 500 x = 230. 11 / 20 10.A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger? 3.6 sec 18 sec 36 sec 72 sec Explanation:Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr.   =36 x5m/sec18 = 10 m/sec.Distance to be covered = (240 + 120) m = 360 m. Time taken =360sec= 36 sec.10 12 / 20 9.Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is: 36 45 48 49 Explanation:Relative speed = (60+ 90) km/hr   =150 x5m/sec18   =125m/sec.3 Distance covered = (1.10 + 0.9) km = 2 km = 2000 m. Required time =2000 x3sec = 48 sec.125 13 / 20 8.A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long? 40 sec 42 sec 45 sec 48 sec Explanation: Formula for converting from km/hr to m/s:   X km/hr =X x5m/s.18 Therefore, Speed =45 x5m/sec=25m/sec.182 Total distance to be covered = (360 + 140) m = 500 m. Formula for finding Time =DistanceSpeed  Required time =500 x 2sec= 40 sec.25 14 / 20 7.Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is: 50 m 72 m 80 m 82 m Explanation:Let the length of each train be x metres.Then, distance covered = 2x metres.Relative speed = (46 - 36) km/hr   =10 x5m/sec18   =25m/sec9 2x=25369  2x = 100 x = 50. 15 / 20 6.A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long? 65 sec 89 sec 100 sec 150 sec Explanation: Speed =240m/sec = 10 m/sec.24  Required time =240 + 650sec = 89 sec.10 16 / 20 5.A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? 120 m 240 m 300 m None of these Explanation: Speed =54 x5m/sec = 15 m/sec.18 Length of the train = (15 x 20)m = 300 m.Let the length of the platform be x metres. Then,x + 300= 1536  x + 300 = 540 x = 240 m. 17 / 20 4.Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is: 1 : 3 3 : 2 3 : 4 None of these Explanation:Let the speeds of the two trains be x m/sec and y m/sec respectively.Then, length of the first train = 27x metres,and length of the second train = 17y metres. 27x + 17y= 23x+ y  27x + 17y = 23x + 23y 4x = 6y x=3.y2 18 / 20 3.The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is: 200 m 225 m 245 m 250 m Explanation: Speed =45 x5m/sec=25m/sec.182 Time = 30 sec.Let the length of bridge be x metres. Then,130 + x=25302  2(130 + x) = 750 x = 245 m.Explanation: Speed =45 x5m/sec=25m/sec.182 Time = 30 sec.Let the length of bridge be x metres. Then,130 + x=25302  2(130 + x) = 750 x = 245 m. 19 / 20 2.A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is: 45 km/hr 50 km/hr 54 km/hr 55 km/hr Explanation: Speed of the train relative to man =125m/sec10   =25m/sec.2   =25x18km/hr25 = 45 km/hr.Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 45        x = 50 km/hr. 20 / 20 1.A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train? 120 metres 180 metres 324 metres 150 metres Explanation: Speed =60 x5m/sec=50m/sec.183 Length of the train = (Speed x Time).  Length of the train =50x 9m = 150 m.3 Your score isThe average score is 0% 0% Restart quiz Permutation and Combination 1 / 14 14.In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together? 120 720 4320 2160 Explanation:The word 'OPTICAL' contains 7 different letters.When the vowels OIA are always together, they can be supposed to form one letter.Then, we have to arrange the letters PTCL (OIA).Now, 5 letters can be arranged in 5! = 120 ways.The vowels (OIA) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720. 2 / 14 13.In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together? 10080 4989600 120960 None of these Explanation:In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.Thus, we have MTHMTCS (AEAI).Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.  Number of ways of arranging these letters =8!= 10080.(2!)(2!) Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters =4!= 12.2!  Required number of words = (10080 x 12) = 120960. 3 / 14 12.How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed? 40 400 5040 2520 Explanation:'LOGARITHMS' contains 10 different letters. Required number of words= Number of arrangements of 10 letters, taking 4 at a time.= 10P4= (10 x 9 x 8 x 7)= 5040. 4 / 14 11.In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women? 63 90 126 45 Explanation: Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) =7 x 6x 3= 63.2 x 1 5 / 14 10.In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions? 32 48 36 60 Explanation:There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.Let us mark these positions as under:(1) (2) (3) (4) (5) (6)Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.Number of ways of arranging the vowels = 3P3 = 3! = 6.Also, the 3 consonants can be arranged at the remaining 3 positions.Number of ways of these arrangements = 3P3 = 3! = 6.Total number of ways = (6 x 6) = 36. 6 / 14 9.A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw? 32 48 64 96 Explanation:We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).  Required number of ways= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)=3 x6 x 5+3 x 2x 6+ 12 x 12 x 1= (45 + 18 + 1)= 64. 7 / 14 8.In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women? 266 5040 11760 86400 Explanation:Required number of ways= (8C5 x 10C6)= (8C3 x 10C4)= 8 x 7 x 6x10 x 9 x 8 x 73 x 2 x 14 x 3 x 2 x 1= 11760. 8 / 14 7.How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? 5 10 15 20 Explanation:Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it. Required number of numbers = (1 x 5 x 4) = 20. 9 / 14 6.In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? 159 194 205 209 Explanation:We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).  Required numberof ways= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)= (6 x 4) +6 x 5x4 x 3+6 x 5 x 4x 4+6 x 52 x 12 x 13 x 2 x 12 x 1= (24 + 90 + 80 + 15)= 209. 10 / 14 5.In how many ways can the letters of the word 'LEADER' be arranged? 72 144 360 720 Explanation:The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.  Required number of ways =6!= 360.(1!)(2!)(1!)(1!)(1!) 11 / 14 4.Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? 210 1050 25200 21400 Explanation:Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3 x 4C2)=7 x 6 x 5x4 x 33 x 2 x 12 x 1= 210. Number of groups, each having 3 consonants and 2 vowels = 210.Each group contains 5 letters. Number of ways of arranging5 letters among themselves= 5!= 5 x 4 x 3 x 2 x 1= 120.  Required number of ways = (210 x 120) = 25200. 12 / 14 3.In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? 810 1440 2880 50400 Explanation:In the word 'CORPORATION', we treat the vowels OOAIO as one letter.Thus, we have CRPRTN (OOAIO).This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different. Number of ways arranging these letters =7!= 2520.2! Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in5!= 20 ways.3!  Required number of ways = (2520 x 20) = 50400.Video Explanation: https://youtu.be/o3fwMoB0duw 13 / 14 2.In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? 360 480 720 5040 Explanation:The word 'LEADING' has 7 different letters.When the vowels EAI are always together, they can be supposed to form one letter.Then, we have to arrange the letters LNDG (EAI).Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720. 14 / 14 1.From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done? 564 645 735 756 None of these Explanation:We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).  Required number of ways= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)=7 x 6 x 5x6 x 5+ (7C3 x 6C1) + (7C2)3 x 2 x 12 x 1= 525 +7 x 6 x 5x 6+7 x 63 x 2 x 12 x 1= (525 + 210 + 21)= 756. Your score isThe average score is 0% 0% Restart quiz problems on number 1 / 15 14.The sum of two number is 25 and their difference is 13. Find their product. 104 114 315 325 Explanation:Let the numbers be x and y.Then, x + y = 25 and x - y = 13.4xy = (x + y)2 - (x- y)2= (25)2 - (13)2= (625 - 169)= 456 xy = 114. 2 / 15 15.What is the sum of two consecutive even numbers, the difference of whose squares is 84? 34 38 42 46 Explanation:Let the numbers be x and x + 2.Then, (x + 2)2 - x2 = 84 4x + 4 = 84 4x = 80 x = 20. The required sum = x + (x + 2) = 2x + 2 = 42. 3 / 15 13.A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is: 145 253 370 352 Explanation:Let the middle digit be x.Then, 2x = 10 or x = 5. So, the number is either 253 or 352.Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.Hence, required number = 253. 4 / 15 12.The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is: 20 23 169 None of these Explanation:Let the numbers be x and y.Then, xy = 120 and x2 + y2 = 289. (x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529 x + y = 529 = 23. 5 / 15 11.The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is: 380 395 400 425 Explanation:Let the numbers be x and y. Then, xy = 9375 andx= 15.y xy=9375(x/y)15  y2 = 625. y = 25. x = 15y = (15 x 25) = 375. Sum of the numbers = x + y = 375 + 25 = 400. 6 / 15 10.Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number. 3 10 17 20 Explanation:Let the number be x. Then, x + 17 =60x  x2 + 17x - 60 = 0 (x + 20)(x - 3) = 0 x = 3. 7 / 15 9.In a two-digit, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is: 24 26 42 46 Explanation:Let the ten's digit be x.Then, unit's digit = x + 2.Number = 10x + (x + 2) = 11x + 2.Sum of digits = x + (x + 2) = 2x + 2. (11x + 2)(2x + 2) = 144 22x2 + 26x - 140 = 0 11x2 + 13x - 70 = 0 (x - 2)(11x + 35) = 0 x = 2.Hence, required number = 11x + 2 = 24. 8 / 15 8.A number consists of two digits. If the digi 3 5 9 11 Explanation:Let the ten's digit be x and unit's digit be y.Then, number = 10x + y.Number obtained by interchanging the digits = 10y + x. (10x + y) + (10y + x) = 11(x + y), which is divisible by 11. 9 / 15 7.The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is: 20 30 40 None of these Explanation:Let the numbers be a, b and c.Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400. (a + b + c) = 400 = 20. 10 / 15 6.The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number? 68 78 96 Cannot be determined Explanation:Let the ten's digit be x and unit's digit be y.Then, x + y = 15 and x - y = 3   or   y - x = 3.Solving x + y = 15  and   x - y = 3, we get: x = 9, y = 6.Solving x + y = 15  and   y - x = 3, we get: x = 6, y = 9.So, the number is either 96 or 69.Hence, the number cannot be determined. 11 / 15 5.A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is: 18 24 42 81 Explanation: Let the ten's and unit digit be x and8respectively.x Then,10x +8+ 18 = 10 x8+ xxx  10x2 + 8 + 18x = 80 + x2 9x2 + 18x - 72 = 0 x2 + 2x - 8 = 0 (x + 4)(x - 2) = 0 x = 2. 12 / 15 4.The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ? 4 8 16 None of these Explanation:Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.Let ten's and unit's digits be 2x and x respectively.Then, (10 x 2x + x) - (10x + 2x) = 36 9x = 36 x = 4. Required difference = (2x + x) - (2x - x) = 2x = 8. 13 / 15 3.The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number? 3 4 9 Cannot be determined Explanation:Let the ten's digit be x and unit's digit be y.Then, (10x + y) - (10y + x) = 36 9(x - y) = 36 x - y = 4. 14 / 15 2.Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is: 9 11 13 15 Explanation:Let the three integers be x, x + 2 and x + 4.Then, 3x = 2(x + 4) + 3      x = 11. Third integer = x + 4 = 15. 15 / 15 1.If one-third of one-fourth of a number is 15, then three-tenth of that number is: 35 36 45 54 Explanation:Let the number be x. Then,1of1of x = 15      x = 15 x 3 x 4 = 180.34 So, required number =3x 180= 54.10 Your score isThe average score is 0% 0% Restart quiz simplifications 1 / 10 10.To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present ? 10 35 62.5 Cannot be determined Explanation:Let the capacity of 1 bucket = x.Then, the capacity of tank = 25x. New capacity of bucket =2x5  Required number of buckets =25x(2x/5) = 25xx52x =1252 = 62.5 2 / 10 9.Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by: 1 /7 1/ 8 1 /9 1/2 Explanation: Original share of 1 person =18 New share of 1 person =17 Increase =1-1=17856  Required fraction =(1/56)=1x8=1(1/8)5617 3 / 10 8.A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed: 30 birds 60 birds 72 birds 90 birds Explanation:Let the total number of shots be x. Then, Shots fired by A =5x8 Shots fired by B =3x8 Killing shots by A =1of5x=5x3824 Shots missed by B =1of3x=3x2816 3x= 27 or x =27 x 16= 144.163 Birds killed by A =5x=5x 144= 30.2424 4 / 10 7.One-third of Rahul's savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ? Rs. 30,000 Rs. 50,000 Rs. 60,000 Rs. 90,000 Explanation:Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 - x) respectively. Then, 1x =1(150000 - x)32 x+x= 7500032 5x= 750006  x =75000 x 6= 900005  Savings in Public Provident Fund = Rs. (150000 - 90000) = Rs. 60000 5 / 10 6.A sum of Rs. 1360 has been divided among A, B and C such that A gets  of what B gets and B gets  of what C gets. B's share is: Rs. 120 Rs. 160 Rs. 240 Rs. 300 Explanation:Let C's share = Rs. x Then, B's share = Rs.x,  A's share = Rs.2xx= Rs.x4346 x+x+ x = 136064 17x= 136012  x =1360 x 12= Rs. 96017 Hence, B's share = Rs.960= Rs. 240.4 6 / 10 5.The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ? Rs. 1200 Rs. 2400 Rs. 4800 Cannot be determined Explanation:Let the price of a saree and a shirt be Rs. x and Rs. y respectively.Then, 2x + 4y = 1600 .... (i)and x + 6y = 1600 .... (ii) 7 / 10 4.If a - b = 3 and a2 + b2 = 29, find the value of ab. 10 12 15 18 Explanation:2ab = (a2 + b2) - (a - b)2= 29 - 9 = 20ab = 10. 8 / 10 3.The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is: Rs. 3500 Rs. 3750 Rs. 3840 Rs. 3900 Explanation:Let the cost of a chair and that of a table be Rs. x and Rs. y respectively. Then, 10x = 4y   or   y =5x.2  15x + 2y = 4000  15x + 2 x5x = 40002  20x = 4000 x = 200. So, y =5x 200= 500.2 Hence, the cost of 12 chairs and 3 tables = 12x + 3y= Rs. (2400 + 1500)= Rs. 3900. 9 / 10 2.There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is: 20 80 100 200 Explanation:Let the number of students in rooms A and B be x and y respectively.Then, x - 10 = y + 10      x - y = 20 .... (i)and x + 20 = 2(y - 20)      x - 2y = -60 .... (ii)Solving (i) and (ii) we get: x = 100 , y = 80. The required answer A = 100. 10 / 10 1.A man has Rs.480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ? 45 60 75 90 Explanation:Let number of notes of each denomination be x.Then x + 5x + 10x = 480 16x = 480 x = 30.Hence, total number of notes = 3x = 90. Your score isThe average score is 0% 0% Restart quiz Surds and Indices 1 / 15 15.If x = 3 + 22, then the value ofx-1is:x 1 2 5 7 Explanation: x-12= x +1- 2xx   = (3 + 22) +1- 2(3 + 22)   = (3 + 22) +1x(3 - 22)- 2(3 + 22)(3 - 22) = (3 + 22) + (3 - 22) - 2= 4. x-1= 2.x 2 / 15 14.xb(b + c - a).xc(c + a - b).xa(a + b - c)= ?xcxaxb xabc 1 xab + bc + ca xa + b + c Explanation:Given Exp.= x(b - c)(b + c - a) . x(c - a)(c + a - b) . x(a - b)(a + b - c)= x(b - c)(b + c) - a(b - c)  .  x(c - a)(c + a) - b(c - a).  x(a - b)(a + b) - c(a - b)= x(b2 - c2 + c2 - a2 + a2 - b2)  .   x-a(b - c) - b(c - a) - c(a - b)= (x0 x x0)= (1 x 1) = 1. 3 / 15 13.If m and n are whole numbers such that mn = 121, the value of (m - 1)n + 1 is: 1 10 121 1000 Explanation:We know that 112 = 121.Putting m = 11 and n = 2, we get:(m - 1)n + 1 = (11 - 1)(2 + 1) = 103 = 1000. 4 / 15 12.1+1= ?1 + a(n - m)1 + a(m - n) 0 1/2 1 Explanation: 1+1=1 +11 +anam1 +aman1 + a(n - m)1 + a(m - n)   =am+an(am + an)(am + an)   =(am + an)(am + an) = 1. 5 / 15 11.(243)n/5 x 32n + 1= ?9n x 3n - 1 1 2 9 3 Explanation: Given Expression=(243)(n/5) x 32n + 19n x 3n - 1=(35)(n/5) x 32n + 1(32)n x 3n - 1=(35 x (n/5) x 32n + 1)(32n x 3n - 1)=3n x 32n + 132n x 3n - 1=3(n + 2n + 1)3(2n + n - 1)=33n + 133n - 1= 3(3n + 1 - 3n + 1)   = 32   = 9. 6 / 15 10.(0.04)-1.5 = ?25 25 125 250 625 Explanation: (0.04)-1.5 =4-1.5100   =1-(3/2)25 = (25)(3/2)= (52)(3/2)= (5)2 x (3/2)= 53= 125. 7 / 15 9.(25)7.5 x (5)2.5 ÷ (125)1.5 = 5? 8.5 13 16 17.5 Explanation:Let (25)7.5 x (5)2.5 ÷ (125)1.5 = 5x. Then,(52)7.5 x (5)2.5= 5x(53)1.5 5(2 x 7.5) x 52.5= 5x5(3 x 1.5) 515 x 52.5= 5x54.5  5x = 5(15 + 2.5 - 4.5) 5x = 513 x = 13. 8 / 15 8.1 +1+1= ?1 + x(b - a) + x(c - a)1 + x(a - b) + x(c - b)1 + x(b - c) + x(a - c) O 1 xa - b - c Explanation: Given Exp. =1 +1 +11 +xb+xcxaxa1 +xa+xcxbxb1 +xb+xaxcxc   =xa+xb+xc(xa + xb + xc)(xa + xb + xc)(xa + xb + xc)   =(xa + xb + xc)(xa + xb + xc) = 1. 9 / 15 7.The value of [(10)150 ÷ (10)146] 1000 10000 100000 106 Explanation: (10)150 ÷ (10)146 =1015010146 = 10150 - 146= 104= 10000. 10 / 15 6.(256)0.16 x (256)0.09 = ? 4 16 64 256.25 Explanation:(256)0.16 x (256)0.09 = (256)(0.16 + 0.09)= (256)0.25= (256)(25/100)= (256)(1/4)= (44)(1/4)= 44(1/4)= 41= 4 11 / 15 5.If 3(x - y) = 27 and 3(x + y) = 243, then x is equal to: 0 2 4 6 Explanation:3x - y = 27 = 33        x - y = 3 ....(i)3x + y = 243 = 35        x + y = 5 ....(ii)On solving (i) and (ii), we get x = 4. 12 / 15 4.If 5a = 3125, then the value of 5(a - 3) is: 25 125 625 1625 Explanation:5a = 3125       5a = 55 a = 5. 5(a - 3) = 5(5 - 3) = 52 = 25. 13 / 15 3.Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to: 1.45 1.88 2.9 3.7 Explanation:xz = y2       10(0.48z) = 10(2 x 0.70) = 101.40 0.48z = 1.40  z =140=35= 2.9 (approx.)4812 14 / 15 2.Ifax - 1=bx - 3, then the value of x is:ba 1/2 1 2 7/2 Explanation: Given ax - 1=bx - 3ba ax - 1=a-(x - 3) =a(3 - x)bbb  x - 1 = 3 - x 2x = 4 x = 2. 15 / 15 1.(17)3.5 x (17)? = 178 2.29 2.75 4.25 4.5 Explanation:Let (17)3.5 x (17)x = 178.Then, (17)3.5 + x = 178. 3.5 + x = 8 x = (8 - 3.5) x = 4.5 Your score isThe average score is 0% 0% Restart quiz RATIO AND PROPORTION 1 / 12 12.If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number? 2 : 5 3 : 7 5 : 3 7 : 3 Explanation: Let 40% of A =2B3 Then,40A=2B1003 2A=2B53 A=2x5=5B323  A : B = 5 : 3. 2 / 12 11.The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries? 3 : 3 : 10 10 : 11 : 20 23 : 33 : 60 Cannot be determined Explanation:Let A = 2k, B = 3k and C = 5k. A's new salary =115of 2k =115x 2k=23k10010010 B's new salary =110of 3k =110x 3k=33k10010010 C's new salary =120of 5k =120x 5k= 6k100100  New ratio23k:33k: 6k= 23 : 33 : 601010 3 / 12 10.If Rs. 782 be divided into three parts, proportional to  :  : , then the first part is: Rs. 182 Rs. 190 Rs. 196 Rs. 204 Explanation:Given ratio =  :  :  = 6 : 8 : 9.  1st part = Rs.782 x6= Rs. 20423 4 / 12 9.The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is: 20 30 48 58 Explanation:Let the three parts be A, B, C. Then, A : B = 2 : 3 and B : C = 5 : 8 =5 x3:8 x3= 3 :24555  A : B : C = 2 : 3 :24= 10 : 15 : 245  B =98 x15= 30.49 5 / 12 8.If 0.75 : x :: 5 : 8, then x is equal to: 1.12 1.2 1.25 1.30 Explanation: (x x 5) = (0.75 x 8)    x =6= 1.205 6 / 12 7.Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit's salary? Rs. 17,000 Rs. 20,000 Rs. 25,500 Rs. 38,000 Explanation:Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively. Then,2x + 4000=403x + 400057  57(2x + 4000) = 40(3x + 4000) 6x = 68,000 3x = 34,000Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000. 7 / 12 6.The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio? 8 : 9 17 : 18 21 : 22 Cannot be determined Explanation:Originally, let the number of boys and girls in the college be 7x and 8x respectively.Their increased number is (120% of 7x) and (110% of 8x). 120x 7xand110x 8x100100 42xand44x55  The required ratio =42x:44x= 21 : 22.55 8 / 12 5.In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is: 20 litres 30 litres 40 litres 60 litres Explanation: Quantity of milk =60 x2litres = 40 litres.3 Quantity of water in it = (60- 40) litres = 20 litres.New ratio = 1 : 2Let quantity of water to be added further be x litres. Then, milk : water =40.20 + x Now,40=120 + x2  20 + x = 80 x = 60. Quantity of water to be added = 60 litres. 9 / 12 4.Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats? 2 : 3 : 4 6 : 7 : 8 6 : 8 : 9 None of these Explanation:Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x). 140x 5x,150x 7xand175x 8x100100100  7x,21xand 14x.2  The required ratio = 7x :21x: 14x2  14x : 21x : 28x 2 : 3 : 4. 10 / 12 3.A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share? Rs. 500 Rs. 1500 Rs. 2000 Explanation:Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.Then, 4x - 3x = 1000 x = 1000. B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000. 11 / 12 2.Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is 2:5 3:5 4:5 6:5 Explanation:Let the third number be x. Then, first number = 120% of x =120x=6x1005 Second number = 150% of x =150x=3x1002  Ratio of first two numbers =6x:3x= 12x : 15x = 4 : 5.52 12 / 12 1.A and B together have Rs. 1210. If  of A's amount is equal to  of B's amount, how much amount does B have? Rs. 460 Rs. 484 Rs. 550 Rs. 664 Your score isThe average score is 0% 0% Restart quiz Pipes and Cistern 1 / 15 15.Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is: 10 12 14 16 Explanation: Part filled in 2 hours =2=163 Remaining part =1 -1=2.33  (A + B)'s 7 hour's work =23 (A + B)'s 1 hour's work =221  C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }   =1-2=162114  C alone can fill the tank in 14 hours. 2 / 15 14.Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in: 6 hours 6 2 hours 3 7 hours 7 1 hours 2 Explanation: (A + B)'s 1 hour's work =1+1=9=3.12156020 (A + C)'s hour's work =1+1=8=2.12206015 Part filled in 2 hrs =3+2=17.201560 Part filled in 6 hrs =3 x17=17.6020 Remaining part =1 -17=3.2020 Now, it is the turn of A and B and3part is filled by A and B in 1 hour.20 3 / 15 13.A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely? 3 hrs 15 min 3 hrs 45 min 4 hrs 4 hrs 15 min Explanation:Time taken by one tap to fill half of the tank = 3 hrs. Part filled by the four taps in 1 hour =4 x1=2.63 Remaining part =1 -1=1.22 2:1:: 1 : x32  x =1x 1 x3=3hours i.e., 45 mins.224 So, total time taken = 3 hrs. 45 mins. 4 / 15 12.A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half? 15 min 20 min 27.5 min 30 min Explanation: Part filled by (A + B) in 1 minute =1+1=1.604024 Suppose the tank is filled in x minutes. Then,x1+1= 122440 xx1= 1215  x = 30 min. 5 / 15 11.One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in: 81 min. 108 min. 144 min. 192 min. Explanation:Let the slower pipe alone fill the tank in x minutes. Then, faster pipe will fill it inxminutes.3 1+3=1xx36 4=1x36  x = 144 min. 6 / 15 10.Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank? 10 min. 20 sec. 11 min. 45 sec. 12 min. 30 sec. 14 min. 40 sec. Explanation: Part filled in 4 minutes = 41+1=7.152015 Remaining part =1 -7=8.1515 Part filled by B in 1 minute =120 1:8:: 1 : x2015 x =8x 1 x 20= 102min = 10 min. 40 sec.153  The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec. 7 / 15 9.Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank? 12 min 15 min 25 min 50 min Explanation: Part filled by A in 1 min =1.20 Part filled by B in 1 min =1.30 Part filled by (A + B) in 1 min =1+1=1.203012  Both pipes can fill the tank in 12 minutes. 8 / 15 8.Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately? 1 hour 2 hours 6 hours 8 hours Explanation:Let the cistern be filled by pipe A alone in x hours.Then, pipe B will fill it in (x + 6) hours. 1+1=1x(x + 6)4 x + 6 + x=1x(x + 6)4  x2 - 2x - 24 = 0 (x -6)(x + 4) = 0 x = 6.   [neglecting the negative value of x] 9 / 15 7.A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank? 20 hours 25 hours 35 hours Cannot be determined Explanation:Suppose pipe A alone takes x hours to fill the tank. Then, pipes B and C will takexandxhours respectively to fill the tank.24 1+2+4=1xxx5 7=1x5  x = 35 hrs. 10 / 15 6.Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is: 60 gallons 100 gallons 120 gallons 180 gallons Explanation: Work done by the waste pipe in 1 minute =1-1+1152024   =1-1115120   = -1.  [-ve sign means emptying]40  Volume of1part = 3 gallons.40 Volume of whole = (3 x 40) gallons = 120 gallons. 11 / 15 5.A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is: 6 hours 10 hours 15 hours 30 hours Explanation:Suppose, first pipe alone takes x hours to fill the tank .Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank. 1+1=1x(x - 5)(x - 9) x - 5 + x=1x(x - 5)(x - 9)  (2x - 5)(x - 9) = x(x - 5) x2 - 18x + 45 = 0(x - 15)(x - 3) = 0 x = 15.   [neglecting x = 3] 12 / 15 4.Two pipes A and B can fill a cistern in 37 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after: 5 min. 9 min. 10 min. 15 min. Explanation:Let B be turned off after x minutes. Then,Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1.  x2+1+ (30 - x).2= 1754575 11x+(60 -2x)= 122575  11x + 180 - 6x = 225. x = 9. 13 / 15 3.A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in: 4 1 hours 3 7 hours 8 hours 14 hours Explanation: Work done by the leak in 1 hour =1-3=1.2714  Leak will empty the tank in 14 hrs. 14 / 15 2.Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in: 1 13 hours 17 2 8 hours 11 3 9 hours 17 4 1 hours 2 Explanation: Net part filled in 1 hour1+1-1=17.561260  The tank will be full in60hours i.e., 39hours.1717 15 / 15 1.Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes? 5/11 6/11 7/11 8/11 Explanation: Part filled by (A + B + C) in 3 minutes = 31+1+1=3 x11=11.3020106020 Part filled by C in 3 minutes =3.10  Required ratio =3x20=6.101111 Your score isThe average score is 0% 0% Restart quiz Alligation or Mixture 1 / 15 15.A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is: 400 kg 560 kg 600 kg 640 kg Explanation:By the rule of alligation, we have: Profit on 1st partProfit on 2nd part8%Mean Profit14%18%46 Ration of 1st and 2nd parts = 4 : 6 = 2 : 3  Quantity of 2nd kind =3x 1000kg= 600 kg.5 2 / 15 14.8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally? 18 litres 24 litres 32 litres 42 litres Explanation:Let the quantity of the wine in the cask originally be x litres. Then, quantity of wine left in cask after 4 operations =x1 -84 litres.x x(1 - (8/x))4=16x81 1 -84=24x3 x - 8=2x3  3x - 24 = 2x x = 24. 3 / 15 13.The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs. 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is: Rs. 18 Rs. 18.50 Rs. 19 Rs. 19.50 Explanation:Let the price of the mixed variety be Rs. x per kg.By rule of alligation, we have: Cost of 1 kg of Type 1 riceCost of 1 kg of Type 2 riceRs. 15Mean PriceRs. xRs. 20(20 - x)(x - 15) (20 - x)=2(x - 15)3  60 - 3x = 2x - 30 5x = 90 x = 18. 4 / 15 12.In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%? 3:2 3:4 3:5 4:5 Explanation:S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%. C.P. of 1 kg of the mixture = Rs.100x 68.20= Rs. 62.110 By the rule of alligation, we have: Cost of 1 kg tea of 1st kind.Cost of 1 kg tea of 2nd kind.Rs. 60Mean PriceRs. 62Rs. 6532  Required ratio = 3 : 2. 5 / 15 11.Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg. 1:3 2:3 3:4 4:5 Explanation:By the rule of alligation: Cost of 1 kg of 1st kindCost of 1 kg of 2nd kind720 pMean Price630 p570 p6090  Required ratio = 60 : 90 = 2 : 3. 6 / 15 10.In what ratio must water be mixed with milk to gain 16% on selling the mixture at cost price? 1:6 6:1 2:3 4:3 Explanation:Let C.P. of 1 litre milk be Re. 1. S.P. of 1 litre of mixture = Re.1, Gain =50%.3  C.P. of 1 litre of mixture =100 x3x 1=63507 By the rule of alligation, we have: C.P. of 1 litre of waterC.P. of 1 litre of milk0Mean PriceRe.67Re. 11767  Ratio of water and milk =1:6= 1 : 6.77 7 / 15 9.A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is: 1/3 2/3 2/5 3/5 Explanation:By the rule of alligation, we have: Strength of first jarStrength of 2nd jar40%MeanStrength26%19%714 So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2  Required quantity replaced =23 8 / 15 8.A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container? 26.34 litres 27.36 litres 28 litres 29.16 litres Explanation: Amount of milk left after 3 operations =401 -43 litres40   =40 x9x9x9= 29.16 litres.101010 9 / 15 7.How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg? 36 kg 42 kg 54 kg 63 kg Explanation:S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.  C.P. of 1 kg of mixture = Rs.100x 9.24= Rs. 8.40110 By the rule of allilation, we have: C.P. of 1 kg sugar of 1st kindCost of 1 kg sugar of 2nd kindRs. 9Mean PriceRs. 8.40Rs. 71.400.60  Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind.Then, 7 : 3 = x : 27  x =7 x 27= 63 kg.3 10 / 15 6.A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is: 4% 6% 20% 25% Explanation:Let C.P. of 1 litre milk be Re. 1Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%. C.P. of 1 litre mixture = Re.100x 1=41255 By the rule of alligation, we have: C.P. of 1 litre of milkC.P. of 1 litre of waterRe. 1Mean PriceRe.4504515  Ratio of milk to water =4:1= 4 : 1.55 Hence, percentage of water in the mixture =1x 100%= 20%.5 11 / 15 5.In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg? 3:7 5:7 7:3 7:5 Explanation:By the rule of alligation: Cost of 1 kg pulses of 1st kindCost of 1 kg pulses of 2nd kindRs. 15Mean PriceRs. 16.50Rs. 203.501.50  Required rate = 3.50 : 1.50 = 7 : 3. 12 / 15 4.A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5? 4 litres, 8 litres 6 litres, 6 litres 5 litres, 7 litres 7 litres, 5 litres Explanation:Let the cost of 1 litre milk be Re. 1 Milk in 1 litre mix. in 1st can =3litre, C.P. of 1 litre mix. in 1st can Re.344 Milk in 1 litre mix. in 2nd can =1litre, C.P. of 1 litre mix. in 2nd can Re.122 Milk in 1 litre of final mix. =5litre, Mean price = Re.588 By the rule of alligation, we have: C.P. of 1 litre mixture in 1st can  C.P. of 1 litre mixture in 2nd can34Mean Price58121818  Ratio of two mixtures =1:1= 1 : 1.88 So, quantity of mixture taken from each can =1x 12= 6 litres.2 13 / 15 3.A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially? 10 20 21 25 Explanation:Suppose the can initially contains 7x and 5x of mixtures A and B respectively. Quantity of A in mixture left =7x -7x 9litres =7x -21 litres.124 Quantity of B in mixture left =5x -5x 9litres =5x -15 litres.124 7x -214=75x -15 + 949 28x - 21=720x + 219  252x - 189 = 140x + 147 112x = 336 x = 3.So, the can contained 21 litres of A. 14 / 15 2.Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be: Rs. 169.50 Rs. 170 Rs. 175.50 Rs. 180 Explanation:Since first and second varieties are mixed in equal proportions. So, their average price = Rs.126 + 135= Rs. 130.502 So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.By the rule of alligation, we have: Cost of 1 kg of 1st kindCost of 1 kg tea of 2nd kindRs. 130.50Mean PriceRs. 153Rs. x(x - 153)22.50 x - 153= 122.50  x - 153 = 22.50 x = 175.50 15 / 15 1.A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup? 1/3 1/4 1/5 1/7 Explanation:Suppose the vessel initially contains 8 litres of liquid.Let x litres of this liquid be replaced with water. Quantity of water in new mixture =3 -3x+ xlitres8 Quantity of syrup in new mixture =5 -5xlitres8 3 -3x+ x=5 -5x88  5x + 24 = 40 - 5x 10x = 16  x =8.5 So, part of the mixture replaced =8x1=1.585 Your score isThe average score is 0% 0% Restart quiz probability 1 / 14 14.One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)? 1/13 3/13 1/4 9/52 Explanation:Clearly, there are 52 cards, out of which there are 12 face cards.  P (getting a face card) =12=3.5213 2 / 14 13.Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is: 3/20 29/24 47/100 13/102 Explanation:Let S be the sample space. Then, n(S) = 52C2 =(52 x 51)= 1326.(2 x 1) Let E = event of getting 1 spade and 1 heart.  n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13= (13C1 x 13C1)= (13 x 13)= 169.  P(E) =n(E)=169=13.n(S)1326102 3 / 14 12.A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is: 1/22 3/22 2/91 2/77 Explanation:Let S be the sample space. Then, n(S)= number of ways of drawing 3 balls out of 15= 15C3=(15 x 14 x 13)(3 x 2 x 1)= 455. Let E = event of getting all the 3 red balls.  n(E) = 5C3 = 5C2 =(5 x 4)= 10.(2 x 1)  P(E) =n(E)=10=2.n(S)45591 4 / 14 11.A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is: 1/13 2/13 1/26 1/52 Explanation:Here, n(S) = 52.Let E = event of getting a queen of club or a king of heart.Then, n(E) = 2.  P(E) =n(E)=2=1.n(S)5226 5 / 14 10.Two dice are tossed. The probability that the total score is a prime number is: 1/6 5/12 1/2 7/9 Explanation:Clearly, n(S) = (6 x 6) = 36.Let E = Event that the sum is a prime number. Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),(5, 2), (5, 6), (6, 1), (6, 5) }  n(E) = 15.  P(E) =n(E)=15=5.n(S)3612 6 / 14 9.From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings? 1/15 25/57 35/256 1/121 Explanation:Let S be the sample space. Then, n(S) = 52C2 =(52 x 51)= 1326.(2 x 1) Let E = event of getting 2 kings out of 4.  n(E) = 4C2 =(4 x 3)= 6.(2 x 1)  P(E) =n(E)=6=1.n(S)1326221 7 / 14 8.In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize? 1/10 2/5 2/7 5/7 Explanation:P (getting a prize) =10=10=2.(10 + 25)357 8 / 14 7.In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is 21/46 25/117 1/50 3/25 Explanation:Let S be the sample space and E be the event of selecting 1 girl and 2 boys. Then, n(S)= Number ways of selecting 3 students out of 25= 25C3 `=(25 x 24 x 23)(3 x 2 x 1)= 2300. n(E)= (10C1 x 15C2)=10 x(15 x 14)(2 x 1)= 1050.  P(E) =n(E)=1050=21.n(S)230046 9 / 14 6.Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even? 1/2 3/4 3/8 5/16 Explanation:In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36. Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}  n(E) = 27.  P(E) =n(E)=27=3.n(S)364 10 / 14 5.Three unbiased coins are tossed. What is the probability of getting at most two heads? 3/4 1/4 3/8 7/8 Explanation:Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}Let E = event of getting at most two heads.Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.  P(E) =n(E)=7.n(S)8 11 / 14 4.What is the probability of getting a sum 9 from two throws of a dice? 1/6 1/8 1/9 1/12 Explanation:In two throws of a dice, n(S) = (6 x 6) = 36.Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.  P(E) =n(E)=4=1.n(S)369 12 / 14 3.In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green? 1/3 3/4 7/19 8/21 Explanation:Total number of balls = (8 + 7 + 6) = 21. Let E= event that the ball drawn is neither red nor green= event that the ball drawn is blue.  n(E) = 7.  P(E) =n(E)=7=1.n(S)213 13 / 14 2.A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? 10/21 11/21 2/7 5/7 Explanation:Total number of balls = (2 + 3 + 2) = 7.Let S be the sample space. Then, n(S)= Number of ways of drawing 2 balls out of 7= 7C2 `=(7 x 6)(2 x 1)= 21. Let E = Event of drawing 2 balls, none of which is blue.  n(E)= Number of ways of drawing 2 balls out of (2 + 3) balls.= 5C2=(5 x 4)(2 x 1)= 10.  P(E) =n(E)=10.n(S)21 14 / 14 1.Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5? 1/2 2/5 8/15 9/20 Explanation:Here, S = {1, 2, 3, 4, ...., 19, 20}.Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.  P(E) =n(E)=9.n(S)20 Your score isThe average score is 0% 0% Restart quiz SIMPLE INTEREST 1 / 20 A person takes a loan of Rs. 200 at 5% simple interest. He returns Rs.100 at the end of one year. In order clear his dues at the end of 2 years, he would pay : Rs. 100 Rs. 105 Rs. 110 Rs. 115 2 / 20 A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is: 5 8 12 15 3 / 20 A sum was put at simple interest at a certain rate for 10 years . Had it been put at 5% higher rate , it would have fetched Rs.600 more. What was the Sum? Rs.1300 Rs.1500 Rs.1200 Rs.1400 At 5% more rate, the increase in S.I for 10 years = Rs.600  (given)So, at 5% more rate, the increase in SI for 1 year = 600/10 = Rs.60/-i.e. Rs.60 is 5% of the invested sumSo, 1% of the invested sum = 60/5Therefore, the invested sum = 60 × 100/5 = Rs.1200 4 / 20 The interest on a certain deposit at 4.5% p.a. is Rs. 202.50 in one year. How much will the additional interest in one year be on the same deposit at 5% p.a.  ? Rs. 22.5 Rs. 23 Rs. 23.5 Rs. 24 We know that I = PTR/100=> P = 20250/4.5 = 4500Now, new Interest at 5% = 4500x1x5/100 = 225Now the additional amount = 225 - 202.5 = Rs. 22.5 5 / 20 How long will it take for a sum of money to grow from Rs.1250 to Rs.10,000, if it is invested at 12.5% p.a simple interest? 55years 56years 57years 58years Simple interest is given by the formula SI = (pnr/100), where p is the principal, n is the numberof years for which it is invested, r is the rate of interest per annum In this case, Rs. 1250 has become Rs.10,000. Therefore, the interest earned = 10,000 – 1250 = 8750. 8750 = [(1250 x n x 12.5)/100] => n = 700 / 12.5 = 56 years. 6 / 20 A man borrowed Rs 24000 from two money lenders. For one loan, he paid 15% per annum and for the other 18% per annum. At the end of one year, he paid Rs 4050. How much did he borrow at each rate ? Rs.16000 Rs.12000 Rs.15000 Rs.13000 Let the sum at 15% be Rs x and that at 18% be Rs (24000 - x).{(x * 15 * 1)/100 } + { [(24000 – x) * 18 * 1]/100 } = 4050or 15 x + 432000 - 18x = 405000 or x = 9000.Money borrowed at 15% = Rs 9000 .Money borrowed at 18% = Rs 15000. 7 / 20 The difference between the simple interest received from two different sources on Rs.1500 for 3 years is Rs.13.50. The difference between their rates of interest is 0.1% 0.2% 0.3% 0.4% (1500 x R1 x 3)/100=> 4500 (R1-R2) = 1350=> (R1-R2)= 1350/4500 = 0.3 % 8 / 20 A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest 3 4 5 6 9 / 20 A Certain sum of money an amounts to Rs 2500 in a span Of 5 years and further to Rs.3000 in a span of 7 years at simple interest The sum is ? Rs. 1500 Rs. 1450 Rs. 1300 Rs. 1250 2500 in 5th year and 3000 in 7th yearSo in between 2 years Rs. 500 is increased => for a year 500/2 = 250So, per year it is increasing Rs.250 then in 5 years => 250 x 5 = 1250Hence, the initial amount must be 2500 - 1250 = Rs. 1250 10 / 20 Find: S.l. on Rs 6250 at 14% per annum for 146 days. Rs.250 Rs.300 Rs.350 Rs.400 11 / 20 A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest? 3% 3.46% 3.5% 3.75% 12 / 20 At what rate of compound interest per annum will a sum of rs.1200 becomes rs.1348.32 in 2 years 6% 6.5% 7% 7.5% 13 / 20 In how many years will a sum of Rs.800 at 10% per annum compounded semi annually become Rs.926.10 2.5 3.5 1.5 4.5 14 / 20 A father left a will of Rs.35 lakhs between his two daughters aged 8.5 and 16 such that they may get equal amounts when each of them reach the age of 21 years. The original amount of Rs.35 lakhs has been instructed to be invested at 10% p.a. simple interest. How much did the elder daughter get at the time of the will? 17.5 lakhs 21 lakhs 20 lakhs 15 lakhs 15 / 20 A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs 360 more. Find the sum. Rs.4000 Rs.5000 Rs.9000 Rs.6000 Let sum = P and original rate = R. Then[(P * (R+2) * 3)/100] - [ (P * R * 3)/100] = 3603P*(R+2) - 3PR = 360003PR + 6P - 3PR = 360006P = 36000P = 6000 16 / 20 What annual instalment will discharge a debt of Rs 1092 due in 3 years at 12% simple interest? Rs.325 Rs.350 Rs.375 Rs.400 Let each instalment be Rs.x . 1st year =  [x + (x * 12 * 2)/100]2nd year = [ x + (x *12 * 1)/100]3rd year = xThen, [x + (x * 12 * 2)/100] + [ x + (x *12 * 1)/100] + x =10923x + ( 24x/100 ) + ( 12x/100 )  = 1092336x =109200Therefore, x = 325Each instalment = Rs. 325 17 / 20 A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is: 650 690 698 700 S.I. for 1 year =  Rs. (854 - 815) = Rs. 39.S.I. for 3 years = Rs.(39 x 3) = Rs. 117.Principal = Rs. (815 - 117) = Rs. 698 18 / 20 How much time will take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest  ? 4 years 2.5 years 3.5 years 2 years Time = (100 x 81) / (450 x 4.5) = 4 years. 19 / 20 At what rate percent per annum will a sum of money double in 8 years. 12.5% 13.5% 11.5% 14.5% Let principal = P, Then, S.I.= P and Time = 8 years We know that S.I. = PTR/100 Rate = [(100 x P)/ (P x 8)]% = 12.5% per annum. 20 / 20 A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was: Rs. 2000 Rs. 10000 Rs. 15000 Rs. 20000 Your score isThe average score is 0% 0% Restart quiz PERCENTAGE 1 / 20 45% of 750  -  25% of 480 = ? 337.50 217.50 376.21 213.21 Given expression = (45 x 750/100) - (25 x 480/100) = (337.50 - 120) = 217.50 2 / 20 A housewife saved Rs. 2.50 in buying an item on sale. If she spent Rs. 25 for the item, approximately how much percent she saved in the transaction ? 8 9 10 11 3 / 20 What percent of 7.2 kg is 18 gms ? 0.25% 0.50% 0.75% 1% Required percentage = (18/7200 * 100)% =  1/4% = 0.25% 4 / 20 A number is decreased by 10% and then increased by 10%.  The number so obtained is 10 less than the original number. What was the oiginal number ? 1000 2000 3000 4000 Let the original number be x. Final number obtained = 110% of (90% of x) =(110/100 * 90/100 * x) = (99/100)x. x-(99/100)x=10=> x =1000 5 / 20 In an election only two candidates contested 20% of the voters did not vote and 120 votes were declared as invalid. The winner got 200 votes more than his opponent  thus he secured 41% votes of the total voters on the voter list. Percentage votes of the defeated candidate out of the total votes casted is: 30% 35% 40% 45% Let there be x voters and k votes goes to loser then 0.8x - 120 = k + (k + 200) k+200 = 0.41x => k = 1440 and  (k + 200) =1640 Therefore 1440/3200 x 100 = 45% 6 / 20 Gaurav spends 30% of his monthly income on food articles, 40% of the remaining on conveyance and clothes and saves 50% of the remaining. If his monthly salary is Rs. 18,400, how much money does he save every month ? 3864 4903 5849 6789 Saving  = 50% of (100 - 40)% of (100 - 30)% of Rs. 18,400 = Rs. (50/100 * 60/100 * 70/100 * 18400) = Rs. 3864. 7 / 20 405 sweets were distributed equally among children in such a way that the number of sweets received by each child is 20% of the total number of children. How many sweets did each child recieve ? 9 10 11 12 8 / 20 If 75% of a number is added to 75, then the result is the number itself. The number is : 100 200 300 400 Let the number be x, Then75% of x + 75 = x=> x - 75x/100 =75=> x =  300 . 9 / 20 A salesman gets commission on total sales at 9%. If the sale is exceeded Rs.10,000 he gets an additional commission as bonus of 3% on the excess of sales over Rs.10,000. If he gets total commission of Rs.1380, then the bonus he received is: 180 120 150 160 10 / 20 25% of 25% Is Equal To 0.0625 0.0210 0.03145 0.3210 25% of 25% means 25/100 of 25/100 i.e, 1/4 of 1/4 or in other words 1/4 x 1/4=1/16 Hence, the answer is 1/16 = 0.0625. 11 / 20 A's salary is 40% of B's salary which is 25% of C's salary. What percentage of C's salary is A's salary ? 10 20 30 40 A's Salary  = 40% of B = 40% of (25% of C) = 40/100[(52/100)*100] % of C = 10% of C. 12 / 20 The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8748, its purchase price was : 10000 12000 14000 16000 13 / 20 A man spends 35% of his income on food, 25% on children's education and 80% of the remaining on house rent. What percent of his income he is left with ? 6 % 8 % 10 % 12 % Let the total income be x.Then, income left = (100 - 80)% of [100 -  (35 + 25)] % of x = 20% of 40% of x = [(20/100) * (40/100) * 100] % of x = 8 % of x. 14 / 20 Fresh fruit contains 68% water and dry fruit contains 20% water. How much dry fruit can be obtained from 100 kg of fresh fruits ? 20 30 40 50 The fruit content in both the fresh fruit and dry fruit is the same. Given, fresh fruit has 68% water.so remaining 32% is fruit content. weight of fresh fruits is 100kg Dry fruit has 20% water.so remaining 80% is fruit content.let weight if dry fruit be y kg. Fruit % in freshfruit = Fruit% in dryfruit Therefore, (32/100) x 100 = (80/100 ) x y we get, y = 40 kg. 15 / 20 if the price of a book is first decreased by 25% and then increased by 20%, then the net change in the price will be  : 10 20 30 40 Let the original price be Rs. 100.New final price  = 120 %  of (75 % of Rs. 100) = Rs. [(120/100) * (75/100) * 100] = Rs. 90.Decrease = 10% 16 / 20 A student has to obtain 33% of the total marks to pass. He got 125 marks and failed by 40 marks. The maximum marks are : 500 600 800 1000 Given that the student got 125 marks and still he failed by 40 marks => The minimum pass mark = 125 + 40 = 165 Given that minimum pass mark = 33% of the total mark => total mark =33/100 =165 => total mark = 16500/33 = 500 17 / 20 A student multiplied a number by 3/5 instead of 5/3, What is the percentage error in the calculation ? 54 % 64 % 74 % 84 % Let the number be x.Then, ideally he should have multiplied by  x by 5/3. Hence Correct result was x * (5/3)= 5x/3. By mistake he multiplied x by 3/5 . Hence the result with error  = 3x/5Then, error = (5x/3 - 3x/5) = 16x/15Error %  = (error/True vaue) * 100 = [(16/15) * x/(5/3) * x] * 100 = 64 % 18 / 20 If 20% of a = b, then b% of 20 is the same as : 4% of a 6% of a 8% of a 10% of a 20% of a = b=> (20/100) * a = bb% of 20 =(b/100) x 20 = [(20a/100) / 100] x 20= 4a/100 = 4% of a. 19 / 20 In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was : 2500 2700 2900 3100 Total number of votes = 7500Given that 20% of Percentage votes were invalid=> Valid votes = 80%Total valid votes = 7500*(80/100)1st candidate got 55% of the total valid votes.Hence the 2nd candidate should have got 45% of the total valid votes=> Valid votes that 2nd candidate got = total valid votes x (45/100)7500*(80/100)*(45/100) = 2700 20 / 20 John got 30% of the maximum marks in an examination and failed by 10 marks. However, Paul who took the same examination got 40% of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination? 75 80 85 90 Maximum mark be mpassing marks = (30/100)m +10marks obtained by paul = (40/100)m = [(30/100)m +10] + 1540m/100 = 30m/100 + 2510m/100 =25m = 2500/10m=250Thus passing marks = 85 Your score isThe average score is 0% 0% Restart quiz PROBLEMS ON AGE 1 / 20 0.1 decates ago, Vijaya was quadrice as old as her daughter Simran. 0.6 decades hence, Vijaya’s age will beat her daughter’s age by 0.9 decades. The proportion of the current ages of Vijaya and Simran is : [ 0.1 Decates =1 Year; quadrice = 4 times] 8:1 4:9 11:7 13:4 Let, Ages of Vijaya and Simran – 1 year ago was 4A and A years correspondingly.Then, [(4A + 1) + 6] – [(A + 1) + 6] = 93A = 9  => A = 3 Needed Ratio = (4A + 1) : (A + 1) = 13 : 4 2 / 20 A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is 20 21 22 23 Let the son's present age be x years. Then, man's present age = (x + 24) years=> (x + 24) + 2 = 2(x + 2)=> x + 26 = 2x + 4So, x = 22 3 / 20 If Raj was one-third as old as Rahim 5 years back and Raj is 17 years old now, How old is Rahim now? 40 41 42 43 Raj’s age today = 17 decades,Hence, 5 decades back, he must be 12 years old.Rahim must be 36 years old, Because (3×12).5 years back Rahim must be 41 years old today. Because (36+5). 4 / 20 Sum of the ages of Rajesh, Suresh, Mahesh and Dinesh is 76 years. 7 years hence, their age ratio is 7:6:5:8. Find Mahesh’s present age ? 17 years 13 years 14 years 16 years Given R + S + M + D = 76=> R+7 : S+7 : M+7 : D+7 = 7:6:5:8=> Let the ratio = x=> M+7 = 5x=> M = 5x-7=> R = 7x-7=> S = 6x-7=> D = 8x-7=> R + S + M + D = 76=> 7x-7 + 6x-7 + 5x-7 + 8x-7 = 76=> 26x - 28 = 76=> 26x = 104=> x = 4=> Mahesh's present age = 5x-7 = 20 - 7 = 13 years. 5 / 20 The age of a person is thrice the total ages of his 2 daughters. 0.5 decades hence, his age will be twice of the total ages of his daughters. Then what is the father’s current age? [0.5 Decades = 5 Years] 35 years 40 years 45 years 50 years Let, Total of current ages of the 2 daughters is A years.Then, father’s current age = 3A years.(3A + 5) = 2 (A +10)3A + 5 = 2A + 20A = 15Therefore, father’s current age = 45 years. 6 / 20 Six years ago Anita was P times as old as Ben was. If Anita is now 17 years old, how old is Ben now in terms of P ? 11/P + 6 P/11 +6 17 - P/6 17/P Let Ben’s age now be BAnita’s age now is A.(A - 6) = P(B - 6)But A is 17 and therefore 11 = P(B - 6)11/P = B-6(11/P) + 6 = B 7 / 20 Raju got married 8years ago. His present age is 6/5 times his age at the time of his marriage Raju's sister was 10years younger to him at the time of his marriage. The present age of Raju's sister is  ? 30 32 38 40 Let Raju's age at the time of his marriage = X yearsThen from given data=> X + 8 = 6/5 XX = 40.Since got married 8 years ago, his present age  = 48His sister is 10 years younger => 48 - 10 = 38 years. 8 / 20 The ratio of the ages of Maala and Kala is 4 : 3. The total of their ages is 2.8 decades. The proportion of their ages after 0.8 decades will be [1 Decade = 10 years] 4:3 12:11 7:4 6:5 Let, Maala’s age = 4A and Kala’s age = 3AThen 4A + 3A = 28A = 4Maala’s age = 16 yearsand Kala’s age = 12 yearsProportion of their ages after 8 is = (16 + 8) : (12 + 8)= 24 : 20= 6 : 5 9 / 20 Mother’s age today is thrice as her daughter’s. After 10 years it would be just double. How old is the daughter today ? 8 years 9 years 10 years 11 years Let the age of the daughter today be ‘x’ years.Mother’s age today = 3x yearsAfter 10 years, we have3x + 10 = 2(x + 10)x = 10 years. 10 / 20 Ratio of the ages of Mahesh and Nilesh is 5 : x. Mahesh is 18 years younger to Ramesh. After nine years Ramesh will be 47 years old. If the difference between the ages of Mahesh and Nilesh is same as the age of Ramesh, what is the value of x ? 11.8 12.9 13.7 14.5 Let the present ages of Mahesh, Nilesh and Ramesh be k, l and m respectively.k/l = 5/x ------ (1)k = m - 18 ------ (2)m + 9 = 47 ------ (3)k - l = m ----- (4)(3) => m = 47 - 9 = 38 years(2) => k = 38 -18 = 20 years(1) => 20/l = 5/x => l = 4x(4) => 4x - 20 = 38=> 4x = 58 => x = 14.5 11 / 20 The ages of X and Y are in the proportion of 6:5 and total of their ages is 44 years. The proportion of their ages after 8 years will be 3:4 4:3 8:7 7:9 Let current ages of X and Y correspondingly, is 6A & 5AGiven: 6A + 5A = 44=> A = 4Proportion of ages after 0.8 decades will be6A + 8 : 5A + 8   32:28 (or) 8:7 12 / 20 The present average age of a family of five members is 26 years. If the present age of the youngest member in the family is ten years, then what was the average age of the family at the time of the birth of the youngest member ? (Assume no death occurred in the family since the birth of the youngest) 19 years 16 years 18 years 20 years 13 / 20 The ages of Krish and Vaibhav are in the proportion of 3 : 5. After 9 years, the proportion of their ages will be 3 : 4. Then the current age of Vaibhav is: 10 13 15 18 Krish’s age = 3A and Vaibhav’s age = 5A(3A+9)/(5A+9) = 3/4=> 4 (3A + 9) = 3 (5A + 9)=> A = 3Therefore, Vaibhav’s age = 15 years. 14 / 20 The total age of A and B is 12 years more than the total age of B and C. C is how many year younger than A 11 12 13 14 Given that A+B = 12 + B + C=> A – C = 12 + B – B = 12=> C is younger than A by 12 years 15 / 20 Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages 35 34 45 25 Let the present age of P and Q be 3x and 4x respectively.Ten years ago, P was half of Q in age=> 2(3x – 10) = (4x – 10)=> 6x – 20 = 4x – 10=> 2x = 10=> x = 5 16 / 20 The sum of the ages of a father and son is 45 years. Five years ago, the product of their ages was four times the fathers age at that time. The present age of father and son 34, 11 35, 10 36, 9 40, 5 Let sons age = x years.Then fathers age = (45 - x)years.(x—5)(45—x—5) = 4(45- x - 5)hence (x—5) = 4 so x = 9sons age = 9 years.fathers age = (45 - x)years.fathers age = (45 - 9)years.Their ages are 36 years and 9 years. 17 / 20 Ages of two persons differ by 16 years. If 6 year ago, the elder one be 3 times as old the younger one, find their present age 12, 28 14, 30 16, 32 18, 34 Let the age of younger person is x,Then elder person age is (x+16)=> 3(x-6) = (x+16-6) [6 years before]=> 3x-18 = x+10=> x = 14.So other person age is x + 16 = 30 18 / 20  The ratio between the present ages of P and Q is 6:7. If Q is 4 years old than P, what will be the ratio of the ages of P and Q after 4 years 7:8 7:9 3:8 5:8 Let P age and Q age is 6x years and 7x years.Then 7x - 6x = 4 <=> x = 4So required ratio will be (6x+4): (7x+4) => 28:32 => 7:8 19 / 20 Sachin is younger than Rahul by 7 years. If the ratio of their ages is 7:9, find the age of Sachin 23.5 24.5 12.5 14.5 If Rahul age is x, then Sachin age is x-7,so (x-7)/x = 7/9=> 9x-63 = 7x=> 2x = 63=> x = 31.5So Sachin age is 31.5 - 7 = 24.5 20 / 20 Raju age after 15 years will be 5 times his age 5 years back, What is the present age of Raju 15 14 10 8 Clearly,x+15 = 5(x-5)<=> 4x = 40 => x = 10 Your score isThe average score is 0% 0% Restart quiz AVERAGE 1 / 20 The average age of a husband and his wife was 23 years at the time of their marriage. After  five years they have a one-year old child. The average age of the family now is : 25 23 19 17 Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) + 1 = 57 years.Required average = (57/3) = 19 years. 2 / 20 Of the four numbers, whose average is 60, the first is one-fourth of the sum of the last three. The first number is : 45 46 47 48 Let the first number be x,Then, sum of the four numbers = x + 4x = 5x.so,  5x/4 = 60 or x = (60 * 4) / 5 = 48. 3 / 20 After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member ? 12 13 14 15 i) Let the ages of the five members at present be a, b, c, d & e years.And the age of the new member be f years.ii) So the new average of five members' age = (a + b + c + d + f)/5 ------- (1)iii) Their corresponding ages 3 years ago = (a-3), (b-3), (c-3), (d-3) & (e-3) yearsSo their average age 3 years ago = (a + b + c + d + e - 15)/5 = x ----- (2)==> a + b + c + d + e = 5x + 15==> a + b + c + d = 5x + 15 - e ------ (3)iv) Substituting this value of a + b + c + d = 5x + 15 - e in (1) above,The new average is: (5x + 15 - e + f)/5Equating this to the average age of x years, 3 yrs, ago as in (2) above,(5x + 15 - e + f)/5 = x==> (5x + 15 - e + f) = 5xSolving e - f = 15 years.Thus the difference of ages between replaced and new member = 15 years. 4 / 20 A team of 8 persons joins in a shooting competition. The best marksman scored 85 points. If he had scored 92 points, the average score for the team would have been 84. The number of points, the team scored was : 660 665 670 675 Let the total score be x.(x + 92 - 85) / 8 = 84.So,  x + 7 = 672  => x = 665. 5 / 20 A car owner buys petrol at Rs 7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year ? 5 6 7 8 Total quantity of petrol consumed in 3 years =(4000/7.50+4000/8+4000/8.50)  liters = 4000(2/15+1/8+2/17) liters = 76700/51 liters Total amount spent = Rs. (3 x 4000) = Rs. 12000. Average cost = Rs. (12000*51/76700) = Rs. 7.98.≈ 8 6 / 20 The average age of 8 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men. The average age of the two new men is : 20 30 40 50 Total age increased = (8 * 2) years = 16 years.Sum of ages of two new men = (21 + 23 + 16) years = 60 yearsAverage age of two new men = (60/2) years = 30 years. 7 / 20 10 years ago, the average age of a family of 4 members was 24 years. Two children having been born (with age diference of 2 years), the present average age of the family is the same. The present age of the youngest child is : 1 2 3 4 Total age of 4 members, 10 years ago = (24 x 4) years = 96 years. Total age of 4 members now = [96 + (10 x 4)] years = 136 years. Total age of 6 members now = (24 x 6) years = 144 years. Sum of the ages of 2 children = (144 - 136) years = 8 years. Let the age of the younger child be x years. Then, age of the elder child = (x+2) years. So, x+(x+2) =8 <=> x=3 Age of younger child  =  3 years. 8 / 20 The average temperature for Monday, Tuesday, Wednesday and Thursday was 48 degrees and for Tuesday, Wednesday, Thursday and Friday was 46 degrees. If the temperature on Monday was 42 degrees. Find the temperature on Friday ? 34 36 38 40 M + Tu + W + Th = 4 x 48 = 192Tu + W + Th + F = 4 x 46 = 184M = 42Tu + W + Th = 192 - 42 = 150F = 184 – 150 = 34 9 / 20 The average of 11 numbers is 10.9. If the average of the first six numbers is 10.5 and that of the last six numbers is 11.4, then the middle number is : 10.5 11.5 12.5 13.5 Middle numbers  =  [(10.5 x 6 + 11.4 x 6) - 10.9 x 11] = (131.4 - 119-9) = 11.5. 10 / 20 The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is : 36 36.5 37 37.5 Correct Sum = (36 * 50 + 48 - 23) = 1825.Correct mean = = 1825/50 = 36.5 11 / 20 A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is : 275 280 285 290 Since the month begins with a Sunday, so there will be five Sundays in the month,Required average = (510 * 5 + 240 * 25) / 30 = 8550/30 = 285. 12 / 20 The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is : 46 48 45 47 Weight of the teacher = (35.4 x 25 - 35 x 24) kg = 45 kg. 13 / 20 The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers : 16 12 8 4 Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.Then  [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.or 5x + 20 = 305 or x = 57.So, required difference = (57 + 8) - 57  = 8 14 / 20 A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The number of pupils in the class is : 45 40 39 37 Let there be x pupils in the class.Total increase in marks = (x * 1/2) = x/2.x/2 = (83 - 63) => x/2 = 20 => x = 40. 15 / 20 The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person ? 70 kg 75 kg 80 kg 85 kg Total weight increased = (8 x 2.5) kg = 20 kg.Weight of new person = (65 + 20) kg = 85 kg. 16 / 20 The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero ? -1 0 1 2 verage of 20 numbers = 0. Sum of 20 numbers = (0 * 20) = 0. 17 / 20 The average of 7 consecutive numbers is 20. The largest of these numbers is : 21 22 23 24 Let the numbers be x, x + 1, x + 2,  x + 3, x + 4, x + 5 and x + 6,Then  (x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)) / 7 = 20.or  7x + 21 = 140 or 7x = 119 or x =17.Latest number = x + 6 = 23. 18 / 20 A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning? 38 39 40 41 Let the average after 7th inning = x Then average after 16th inning = x - 3 16(x-3)+87 = 17x x = 87 - 48 = 39 19 / 20 The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4 ? 76 79 85 87 Average = total runs / no.of innings = 32So, total = Average x no.of innings = 32 x 10 = 320. Now increase in avg = 4runs. So, new avg = 32+4 = 36runsTotal runs = new avg x new no. of innings = 36 x 11 = 396Runs made in the 11th inning = 396 - 320 = 76 20 / 20 A grocer has a sale of Rs 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs, 6500 ? 4991 4995 5000 5050 Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.Required sale = Rs.[(6500 x 6) - 34009]= Rs. (39000 - 34009)= Rs.  4991. Your score isThe average score is 0% 0% Restart quiz Area 1 / 18 The dimensions of a room are 10m x 7m x 5m. There are 2 doors and 3 windows in the room. The dimensions of the doors are 1m x 3m. One window is of size 2m x 1.5m and the other 2 windows are of size 1m x 1.5m. The cost of painting the walls at Rs. 3 per sq m is 500 484 474 450 Area of 4 walls = 2(l+b)h=2(10+7) x 5 = 170 sq mArea of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq marea to be planted = 170 -12 = 158 sq mCost of painting = Rs. 158 x 3 = Rs. 474 2 / 18 The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle? 50 60 70 80 3 / 18 One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field? 110 120 130 140 Other side = [(17 x 17) - (15 x 15)] = (289 - 225) = 8mArea = 15 x 8 =120 sq. m 4 / 18 Find the ratio of the areas of the incircle and circumcircle of a square. 1:1 1:2 3:1 3:4 5 / 18 A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be 7744 7674 7524 7934 6 / 18 The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres? 100 200 300 400 Let length of plot = L meters, then breadth = L - 20 metersand perimeter = 2[L + L - 20] = [4L - 40] meters[4L - 40]  * 26.50 = 5300[4L - 40] = 5300 / 26.50 = 2004L = 240L = 240/4= 60 meters. 7 / 18 The length of a rectangular hall is 5m more than its breadth. The area of the hall is 750 sq.m. The length of the hall is 20 25 30 35 8 / 18 The length of a rectangle is twice its breadth if its length is decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is increased by 75 sq cm. Find the length of the rectangle. 30 40 50 60 9 / 18 A room of 5m 44cm long and 3m 74cm broad is to be paved with squre tiles. Find the least number of squre tiles required to cover the floor. 177 176 175 174 area of the room = 544 x 374 sq.cmsize of largest square tile = H.C.F of 544cm and 374 cm= 34 cmarea of 1 tile = 34x34 sq cmno. of tiles required = (544 x 374) / (34 x 34) = 176 10 / 18 An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is: 4.04 4.24 3.90 3.50 11 / 18 A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is 230 236 336 330 12 / 18 The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area? 20% 30% 40% 50% 13 / 18 A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is 258 358 458 558 14 / 18 The area of the largest circle that can be drawn inside a rectangle with sides 18cm by 14cm is 49 154 378 1078 15 / 18 If the radius of a circle is decreased by 50%, find the percentage decrease in its area. 55% 65% 75% 85% 16 / 18 A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges? 30 40 50 60 17 / 18 The percentage increase in the area of a rectangle, if each of its sides is increased by 20%. 22% 33% 44% 55% 18 / 18 If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle. 110 120 130 140 Your score isThe average score is 0% 0% Restart quiz Partnership 1 / 14 14.Simran started a software business by investing Rs. 50,000. After six months, Nanda joined her with a capital of Rs. 80,000. After 3 years, they earned a profit of Rs. 24,500. What was Simran's share in the profit? Rs. 9,423 Rs. 10,250 Rs. 12,500 Rs. 10,500 Explanation:Simran : Nanda = (50000 x 36) : (80000 x 30) = 3 : 4. Simran's share = Rs.24500 x3= Rs. 10,500.7 2 / 14 13.Arun, Kamal and Vinay invested Rs. 8000, Rs. 4000 and Rs. 8000 respectively in a business. Arun left after six months. If after eight months, there was a gain of Rs. 4005, then what will be the share of Kamal? Rs. 890 Rs. 1335 Rs. 1602 Rs. 1780 Explanation:Arun : Kamal : Vinay = (8,000 x 6) : (4,000 x 8) : (8,000 x 8)= 48 : 32 : 64= 3 : 2 : 4.  Kamal's share = Rs.4005 x2= Rs. 890.9 3 / 14 12.Aman started a business investing Rs. 70,000. Rakhi joined him after six months with an amount of Rs.. 1,05,000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman, Rakhi and Sagar respectively, 3 years after Aman started the business? 7 : 6 : 10 12 : 15 : 16 42 : 45 : 56 Cannot be determined Explanation:Aman : Rakhi : Sagar = (70,000 x 36) : (1,05,000 x 30) : (1,40,000 x 24) = 12 : 15 : 16. 4 / 14 11.A began a business with Rs. 85,000. He was joined afterwards by B with Rs. 42,500. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1? 4 months 5 months 6 months 8 months Explanation:Suppose B joined for x months. Then, Then,85000 x 12=342500 x x1  x =85000 x 12= 8.42500 x 3 So, B joined for 8 months. 5 / 14 10.A and B started a business in partnership investing Rs. 20,000 and Rs. 15,000 respectively. After six months, C joined them with Rs. 20,000. What will be B's share in total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business? Rs. 7500 Rs. 9000 Rs. 9500 Rs. 10,000 Explanation:A : B : C = (20,000 x 24) : (15,000 x 24) : (20,000 x 18) = 4 : 3 : 3.  B's share = Rs.25000 x3= Rs. 7,500.10 6 / 14 9.A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175, how much must C pay as his share of rent? Rs. 45 Rs. 50 Rs. 55 Rs. 60 Explanation:A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9.  C's rent = Rs.175 x9= Rs. 45.35 7 / 14 8.A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C? 3 : 5 : 2 3 : 5 : 5 6 : 10 : 5 Data inadequate Explanation:Let the initial investments of A and B be 3x and 5x.A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5. 8 / 14 7.A and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew  of his capital and B withdrew  of his capital. The gain at the end of 10 months was Rs. 760. A's share in this profit is: Rs. 330 Rs. 360 Rs. 380 Rs. 430 Explanation: A : B =4x x 3 +4x -1x 4xx 7:5x x 3 +5x -1x 5xx 745 = (12x + 21x) : (15x + 28x)= 33x :43x= 33 : 43.  A's share = Rs.760 x33= Rs. 330.76 9 / 14 6.A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B's contribution in the capital? Rs. 7500 Rs. 8000 Rs. 8500 Rs. 9000 Explanation:Let B's capital be Rs. x. Then,3500 x 12=27x3  14x = 126000 x = 9000. 10 / 14 5.Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments? 5 : 7 : 8 20 : 49 : 64 38 : 28 : 21 None of these Explanation:Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively.Then, 14x : 8y : 7z = 5 : 7 : 8. Now,14x=5      98x = 40y        y =49x8y720 And,14x=5      112x = 35z        z =112x =16x.7z8355  x : y : z = x :49x:16x= 20 : 49 : 64.205 11 / 14 4.A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives: Rs. 8400 Rs. 11,900 Rs. 13,600 Rs. 14,700 Explanation:Let C = x.Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000.So, x + x + 5000 + x + 9000 = 50000 3x = 36000 x = 12000A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.  A's share = Rs.35000 x21= Rs. 14,700.50 12 / 14 3.A, B and C enter into a partnership in the ratio  :  : . After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 21,600, then B's share in the profit is: Rs. 2100 Rs. 2400 Rs. 3600 Rs. 4000 Explanation: Ratio of initial investments =7:4:6= 105 : 40 : 36.235 Let the initial investments be 105x, 40x and 36x.  A : B : C =105x x 4 +150x 105x x 8: (40x x 12) : (36x x 12)100 = 1680x : 480x : 432x = 35 : 10 : 9. Hence, B's share = Rs.21600 x10= Rs. 4000.54 13 / 14 2.A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit. Rs. 1900 Rs. 2660 Rs. 2800 Rs. 2840 Explanation:For managing, A received = 5% of Rs. 7400 = Rs. 370.Balance = Rs. (7400 - 370) = Rs. 7030.Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)= 39000 : 42000 : 30000= 13 : 14 : 10  B's share = Rs.7030 x14= Rs. 2660.37 14 / 14 1.A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A's share is Rs. 855, the total profit is: Rs. 1425 Rs. 1500 Rs. 1537.50 Rs. 1576 Explanation:Let the total profit be Rs. 100. After paying to charity, A's share = Rs.95 x3= Rs. 57.5 If A's share is Rs. 57, total profit = Rs. 100. If A's share Rs. 855, total profit =100x 855= 1500.57 Your score isThe average score is 0% 0% Restart quiz
