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 profi
