Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 16 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | "sum of 10 numbers = 402 corrected sum of 10 numbers = 402 β 13 + 31 β 16 = 404 hence , new average = 404 β 10 = 40.4 answer b" | a ) 40.2 , b ) 40.4 , c ) 40.6 , d ) 40.8 , e ) none of the above | b | divide(subtract(add(multiply(40.2, 10), add(13, 16)), 31), 10) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#2,n4)|divide(#3,n0)| | general |
rs . 705 is divided amongst a , b , c so that 3 times a ' s share , 5 times b ' s share and 4 times c ' s share are all equal . find b ' s share ? | a + b + c = 705 3 a = 5 b = 4 c = x a : b : c = 1 / 3 : 1 / 5 : 1 / 4 = 20 : 12 : 15 12 / 47 * 705 = rs . 180 answer : c | a ) 177 , b ) 150 , c ) 180 , d ) 716 , e ) 616 | c | divide(705, add(add(divide(4, 3), divide(4, 5)), const_1)) | divide(n3,n1)|divide(n3,n2)|add(#0,#1)|add(#2,const_1)|divide(n0,#3)| | general |
the sale price sarees listed for rs . 175 after successive discount is 20 % and 5 % is ? | "175 * ( 80 / 100 ) * ( 95 / 100 ) = 133 answer : b" | a ) 187 , b ) 133 , c ) 142 , d ) 178 , e ) 175 | b | subtract(subtract(175, divide(multiply(175, 20), const_100)), divide(multiply(subtract(175, divide(multiply(175, 20), const_100)), 5), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
the average salary of all the workers in a workshop is rs . 9500 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is ? | let the total number of workers be x . then , 9500 x = ( 12000 * 7 ) + 6000 ( x - 7 ) = > 3500 x = 42000 = x = 12 . answer : e | a ) a ) 23 , b ) b ) 21 , c ) c ) 52 , d ) d ) 56 , e ) e ) 12 | e | add(7, divide(multiply(7, subtract(12000, 9500)), subtract(9500, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3) | general |
solve x 2 β 3 x β 4 = 0 . | "this one factors easily : x 2 β 3 x β 4 = 0 ( x + 1 ) ( x β 4 ) = 0 x + 1 = 0 or x β 4 = 0 x = β 1 or x = 4 the solution is x = β 1 , 4 answer b - 1,4" | a ) 1,4 , b ) - 1,4 , c ) 2,3 , d ) 2 , - 3 , e ) 3,4 | b | multiply(4, 2) | multiply(n0,n2)| | general |
a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 2 / 5 % his yearly income diminishes by rs . 61.50 . his capital is | "explanation : capital = rs . x , then 3 / 5 x = 61.5 x = 102.50 answer : e ) rs . 102.50" | a ) 223.78 , b ) 37.78 , c ) 246.0 , d ) 278.88 , e ) 102.5 | e | divide(61.50, divide(const_4, 5)) | divide(const_4,n3)|divide(n4,#0)| | gain |
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 | explanation : let sum = rs . x c . i . when compounded half yearly = c . i . when compounded annually = = > x = 20000 answer : b ) 20000 | a ) 200000 , b ) 20000 , c ) 200029 , d ) 288778 , e ) 277789 | b | divide(482, subtract(power(add(divide(divide(20, const_2), const_100), const_1), const_4), power(add(divide(20, const_100), const_1), const_2))) | divide(n1,const_2)|divide(n1,const_100)|add(#1,const_1)|divide(#0,const_100)|add(#3,const_1)|power(#2,const_2)|power(#4,const_4)|subtract(#6,#5)|divide(n2,#7) | gain |
a monkey start climbing up a tree 20 ft tall . each hour it hops 3 ft and slips back 2 ft . how much time would it take the monkey to reach the top . | if monkey hops 3 ft and slips back 2 ft in a hour , it means the monkey hops ( 3 ft - 2 ft ) = 1 ft / hr . similarly in 17 hrs it wil be 17 ft . bt since the height of the tree is 20 ft , so if the monkey hops up the tree in the next hr i . e 18 th hr then it reaches at the top of the tree . hence it takes 18 hrs for m... | a ) 15 hrs , b ) 18 hrs , c ) 19 hrs , d ) 17 hrs , e ) 16 hrs | b | subtract(divide(20, subtract(3, 2)), 2) | subtract(n1,n2)|divide(n0,#0)|subtract(#1,n2) | physics |
1 / 0.06 is equal to | "explanation : 1 / 0.06 = ( 1 * 100 ) / 6 = 100 / 6 = 16.7 option b" | a ) 25.5 , b ) 16.7 , c ) 25 , d ) . 25 , e ) none of these | b | divide(1, 0.06) | divide(n0,n1)| | general |
rs 3600 is divided into two parts such that one part is put out at 3 % and the other at 5 % . if the annual interest earned from both the investments be rs 144 , find the first part . | "explanation : average rate = ( 144 / 3600 ) * 100 = 4 ratio = 1 : 1 so , first part = ( 1 / 2 ) * 3600 = rs 1800 . answer : c" | a ) s 4000 , b ) s 2800 , c ) s 1800 , d ) s 3500 , e ) s 3100 | c | divide(subtract(multiply(3600, divide(5, const_100)), 144), subtract(divide(5, const_100), divide(3, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)| | gain |
without stoppages , a train travels certain distance with an average speed of 200 km / h , and with stoppages , it covers the same distance with an average speed of 160 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 40 km less . time taken to cover 40 km = 40 Γ’ Β β 200 h = 1 Γ’ Β β 5 h = 1 Γ’ Β β 5 Γ£ β 60 min = 12 min answer e" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | subtract(const_60, multiply(divide(160, 200), const_60)) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 56 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 56 ) = 457.8 hence answer will be ( d )" | a ) 320.8 $ , b ) 387.8 $ , c ) 420.8 $ , d ) 457.8 $ , e ) 480.8 $ | d | multiply(multiply(0.65, 56), 12) | multiply(n1,n3)|multiply(n2,#0)| | general |
the price of an item is discounted 2 percent on day 1 of a sale . on day 2 , the item is discounted another 12 percent , and on day 3 , it is discounted an additional 25 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "let initial price be 1000 price in day 1 after 2 % discount = 980 price in day 2 after 12 % discount = 862.4 price in day 3 after 25 % discount = 646.8 so , price in day 3 as percentage of the sale price on day 1 will be = 646.8 / 980 * 100 = > 66 % answer will definitely be ( a )" | a ) 66 % , b ) 67.5 % , c ) 69 % , d ) 70 % , e ) 72 % | a | add(multiply(divide(divide(25, const_100), subtract(1, divide(1, 2))), const_100), 2) | divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4)| | gain |
find the smallest number of four digits exactly divisible by 4 , 5,7 and 8 . | "smallest number of four digits is 1000 . required number must be divisible by l . c . m . of 4,5 , 7,8 i . e 280 , on dividing 1000 by 280 , we get 120 as remainder . therefore , required number = 1000 + ( 280 Γ’ β¬ β 120 ) = 1160 . answer is b ." | a ) 1150 , b ) 1160 , c ) 1140 , d ) 1152 , e ) 1145 | b | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 5,7) | multiply(const_100,const_100)|multiply(#0,const_10)|subtract(#1,const_100)|add(n1,#2)| | general |
a sum fetched a total simple interest of rs . 2700 at the rate of 6 p . c . p . a . in 3 years . what is the sum ? | "sol . principal = rs . [ 100 * 2700 / 6 * 3 ] = rs . [ 270000 / 18 ] = rs . 15000 . answer e" | a ) 25000 , b ) 20000 , c ) 18000 , d ) 14000 , e ) 15000 | e | divide(divide(multiply(2700, const_100), 6), 3) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain |
if 20 men can build a water fountain 56 metres long in 21 days , what length of a similar water fountain can be built by 35 men in 3 days ? | "explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 21 : 3 : : 56 : x therefore ( 20 x 21 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 420 = 14 hence , the required length is 14 m . answer : b" | a ) 10 m , b ) 14 m , c ) 17 m , d ) 19 m , e ) 20 m | b | multiply(divide(56, multiply(20, 21)), multiply(35, 3)) | multiply(n0,n2)|multiply(n3,n4)|divide(n1,#0)|multiply(#2,#1)| | physics |
a circular garden is surrounded by a fence of negligible width along the boundary . if the length of the fence is 1 / 5 of th area of the garden . what is the radius of the circular garden ? | "as per the question - - width is negligible now , let l be the length of the fence = 2 pir l = 1 / 5 ( pir ^ 2 ) pir ^ 2 = 10 pir r = 10 answer : c" | a ) 1 , b ) 2 , c ) 10 , d ) 8 , e ) 16 | c | multiply(const_2, sqrt(power(5, const_2))) | power(n1,const_2)|sqrt(#0)|multiply(#1,const_2)| | geometry |
the average of 1 st 3 of 4 numbers is 6 and of the last 3 are 3 . if the sum of the first and the last number is 13 . what is the last numbers ? | a + b + c = 18 b + c + d = 9 a + d = 13 a β d = 9 a + d = 13 2 d = 4 d = 2 answer : b | a ) 4 , b ) 2 , c ) 6 , d ) 7 , e ) 8 | b | subtract(subtract(multiply(3, 6), add(subtract(13, 6), 3)), 6) | multiply(n1,n3)|subtract(n6,n3)|add(n1,#1)|subtract(#0,#2)|subtract(#3,n3) | general |
a classroom has equal number of boys and girls . 8 girls left to play kho - kho , leaving twice as many boys as girls in the classroom . what was the total number of girls and boys present initially ? | after 8 girls left remaining 8 girls now boys 16 are twice as many as remaining girls . initially boys = 16 and girls = 16 . answer : c | a ) 16 , b ) 24 , c ) 32 , d ) 48 , e ) 54 | c | multiply(multiply(const_2, 8), const_2) | multiply(n0,const_2)|multiply(#0,const_2) | general |
a person ' s present age is two - fifth of the age of his mother . after 8 years , he will be one - half of the age of his mother . what is the present age of the mother ? | "let present age of the mother = 5 x then , present age of the person = 2 x 5 x + 8 = 2 ( 2 x + 8 ) 5 x + 8 = 4 x + 16 x = 8 present age of the mother = 5 x = 40 answer is d ." | a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | d | divide(subtract(8, add(const_2, const_3)), subtract(divide(const_1, const_2), divide(const_2, add(const_2, const_3)))) | add(const_2,const_3)|divide(const_1,const_2)|divide(const_2,#0)|subtract(n0,#0)|subtract(#1,#2)|divide(#3,#4)| | general |
a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than 1,000 and at the same time n is a perfect square and a perfect cube ? | "given : positive integer n is a perfect square and a perfect cube - - > n is of a form of n = x 6 for some positive integer x - - > 0 < x ^ 6 < 10 ^ 3 - - > 0 < x ^ 2 < 10 - - > x can be 1 , 2 or 3 hence nn can be 1 ^ 6 , 2 ^ 6 or 3 ^ 6 . answer : b ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(add(const_4, const_2), const_1) | add(const_2,const_4)|divide(#0,const_1)| | geometry |
the speed of a boat in still water is 50 kmph and the speed of the current is 20 kmph . find the speed and upstream ? | speed downstream = 50 + 20 = 70 kmph speed upstream = 50 - 20 = 30 kmph answer : a | a ) 30 kmph , b ) 40 kmph , c ) 78 kmph , d ) 15 kmph , e ) 23 kmph | a | subtract(50, 20) | subtract(n0,n1) | physics |
if 5 x = 6 y and xy β 0 , what is the ratio of 1 / 5 * x to 1 / 8 * y ? | "5 x = 6 y = > x / y = 6 / 5 1 / 5 * x to 1 / 8 * y = x / y * 8 / 5 = ( 6 / 5 ) * ( 8 / 5 ) = 48 / 25 ans : a" | a ) 48 / 25 , b ) 36 / 25 , c ) 6 / 5 , d ) 5 / 6 , e ) 25 / 36 | a | multiply(divide(divide(1, const_3.0), divide(1, 5)), divide(6, 5)) | divide(n3,const_3.0)|divide(n3,n0)|divide(n1,n0)|divide(#0,#1)|multiply(#3,#2)| | general |
when a number is divided by 6 & then multiply by 12 the answer is 10 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 10 = > 2 x = 10 = > x = 5 b" | a ) 4.5 , b ) 5 , c ) 5.6 , d ) 5.7 , e ) 6.5 | b | multiply(divide(10, 12), 6) | divide(n2,n1)|multiply(n0,#0)| | general |
a company that ships boxes to a total of 13 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors... | "back - solving is the best way to solve this problem . you basically need 13 combinations ( including single colors ) if we start from option 1 - > 1 = > 4 c 2 + 4 = 10 ( not enough ) 2 = > 5 c 2 + 5 = 15 ( enough ) since the minimum number is asked . it should be 5 . answer - b" | a ) 6 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | b | subtract(divide(factorial(subtract(divide(13, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(13, const_2), const_1)) | divide(n0,const_2)|factorial(const_3)|factorial(const_2)|multiply(#1,#2)|subtract(#0,const_1)|factorial(#4)|divide(#5,#3)|subtract(#6,#4)| | general |
a and b start from house at 10 am . they travel on the mg road at 20 kmph and 40 kmph . there is a junction t on their path . a turns left at t junction at 12 : 00 noon , b reaches t earlier , and turns right . both of them continue to travel till 2 pm . what is the distance between a and b at 2 pm | at 12 a will travel = 20 * 2 = 40 km b will travel this 40 km in 40 / 40 = 1 hr i . e . by 11 am . after t junction for a - distance travelled = 2 * 20 = 40 km for b distance travelled = 40 * 3 = 120 so distance between a & b is = 120 + 40 = 160 km answer : a | a ) 160 km , b ) 170 km , c ) 180 km , d ) 190 km , e ) 150 km | a | add(multiply(20, subtract(12, 10)), multiply(40, const_3)) | multiply(n2,const_3)|subtract(n3,n0)|multiply(n1,#1)|add(#2,#0) | physics |
the circulation for magazine p in 1971 was 4 times the average ( arithmetic mean ) yearly circulation for magazine p for the years 1972 - 1980 . what is the ratio of the circulation in 1971 to the total circulation during 1971 - 1980 for magazine p ? | there are 9 years from 1972 - 1980 , inclusive . let ' s say the average circulation every year between 1972 - 1980 inclusive is x . so the total circulation is 9 x from 1972 - 1980 , inclusive . in 1971 , the circulation is 4 x . so total circulation for 1971 - 1980 is 4 x + 9 x = 13 x . ratio of circulation in 1971 t... | a ) 4 / 7 , b ) 5 / 7 , c ) 3 / 7 , d ) 2 / 7 , e ) 1 / 7 | d | divide(4, add(add(add(subtract(1980, 1972), const_1), 4), const_1)) | subtract(n3,n2)|add(#0,const_1)|add(n1,#1)|add(#2,const_1)|divide(n1,#3) | general |
the regular price per can of a certain brand of soda is $ 0.30 . if the regular price per can is discounted 15 percent when the soda is purchased in 24 - can cases , what is the price of 72 cans of this brand of soda purchased in 24 - can cases ? | "the discounted price of one can of soda is ( 0.85 ) ( $ 0.30 ) , or $ 0.255 . therefore , the price of 72 cans of soda at the discounted price would be ( 72 ) ( $ 0.255 ) = 18.36 answer : b ." | a ) $ 16.32 , b ) $ 18.36 , c ) $ 21.60 , d ) $ 24.48 , e ) $ 28.80 | b | multiply(divide(subtract(const_100, 15), const_100), multiply(0.30, 72)) | multiply(n0,n3)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#2,#0)| | gain |
a bag contains 12 red marbles . if someone were to remove 2 marbles from the bag , one at a time , and replace the first marble after it was removed , the probability that neither marble would be red is 9 / 16 . how many marbles are in the bag ? | "ok let me see if i can explain what went on in the previous post lets say i have x marbles in the bag in total - - > out of them 12 are red so the probability of pulling a non - red marble is ( x - 12 ) / x now the marble is placed back in the bag and we have x marbles again , of which again 12 are red . so the probab... | a ) 48 , b ) 24 , c ) 60 , d ) 72 , e ) 84 | a | divide(12, subtract(const_1, sqrt(divide(9, 16)))) | divide(n2,n3)|sqrt(#0)|subtract(const_1,#1)|divide(n0,#2)| | other |
the average marks of a class of 26 students is 40 and that of another class of 50 students is 60 . find the average marks of all the students ? | sum of the marks for the class of 26 students = 26 * 40 = 1040 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 76 students = 1040 + 3000 = 4040 average marks of all the students = 4040 / 76 = 53.15 answer : e | a ) 52.2 , b ) 59.5 , c ) 52.8 , d ) 52.5 , e ) 53.1 | e | divide(add(multiply(26, 40), multiply(50, 60)), add(26, 50)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0) | general |
a train 125 m long passes a man , running at 6 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 6 ) km / hr . x - 6 = 45 = = > x = 51 km / hr . answer : a" | a ) 51 , b ) 50 , c ) 88 , d ) 65 , e ) 22 | a | divide(divide(subtract(125, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
tom and john traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour , respectively . after 15 minutes tom passed john , tom reaches a certain gas station , how many l minutes it takes john to reach the station ? | since the question states β after 15 minutes β , we can say tom traveled 15 / 4 km for 15 minutes as he can travel 15 km per hour . hence , using the same logic , we can say john traveled 10 / 4 km as he travels 10 km per hour . so , john has to travel ( 15 / 4 ) - ( 10 / 4 ) km = 5 / 4 km more . since john β s speed i... | a ) 5 min , b ) 6 min , c ) 7 and 1 / 2 min , d ) 8 min , e ) 10 min | c | multiply(subtract(divide(divide(multiply(multiply(subtract(15, 10), divide(15, const_60)), const_60), 10), const_2), const_0_25), 10) | divide(n0,const_60)|subtract(n0,n1)|multiply(#0,#1)|multiply(#2,const_60)|divide(#3,n1)|divide(#4,const_2)|subtract(#5,const_0_25)|multiply(n1,#6) | physics |
pipe a can fill a tank in 8 hours . due to a leak at the bottom , it takes 12 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 8 - 1 / x = 1 / 12 = > 1 / x = 1 / 8 - 1 / 12 = ( 3 - 2 ) / 24 = 1 / 24 = > x = 24 . answer : b" | a ) 5 , b ) 24 , c ) 78 , d ) 90 , e ) 54 | b | divide(multiply(12, 8), subtract(12, 8)) | multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)| | physics |
three 8 faced dice are thrown together . the probability that all the three show the same number on them is ? | "it all 3 numbers have to be same basically we want triplets . 111 , 222 , 333 , 444 , 555 , 666 , 777 and 888 . those are eight in number . further the three dice can fall in 8 * 8 * 8 = 512 ways . hence the probability is 8 / 512 = 1 / 64 answer : d" | a ) 1 / 38 , b ) 1 / 48 , c ) 1 / 56 , d ) 1 / 64 , e ) 1 / 68 | d | multiply(multiply(multiply(divide(const_1, 8), divide(const_1, 8)), divide(const_1, 8)), divide(const_1, 8)) | divide(const_1,n0)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2)| | probability |
an escalator is descending at constant speed . a walks down and takes 50 steps to reach the bottom . b runs down and takes 90 steps in the same time as a takes 10 steps . how many steps are visible when the escalator is not operating ? | lets suppose that a walks down 1 step / min and escalator moves n steps / min it is given that a takes 50 steps to reach the bottom in the same time escalator would have covered 50 n steps so total steps on escalator is 50 + 50 n . again it is given that b takes 90 steps to reach the bottom and time taken by him for th... | a ) 80 , b ) 90 , c ) 100 , d ) 110 , e ) 120 | c | add(50, multiply(50, divide(subtract(90, 50), subtract(90, 50)))) | subtract(n1,n0)|divide(#0,#0)|multiply(n0,#1)|add(n0,#2) | physics |
how many kg of sugar must be added to 30 kg of 2 % solution of sugar and water to increase it to a 10 % solution ? | explanation : solution : amount of sugar in 30 kg solution = ( 2 / 100 * 30 ) kg = 0.6 kg . let x kg of sugar be added . then , ( 0.6 + x ) / ( 30 + x ) = 10 / 100 60 + 100 x = 300 + 10 x x = 240 / 90 = 8 / 3 = 2.6 answer : a | a ) 2.6 kg , b ) 4 kg , c ) 3.4 kg , d ) 1.8 kg , e ) none of these | a | divide(subtract(multiply(30, 10), multiply(multiply(divide(2, const_100), 30), const_100)), multiply(30, const_3)) | divide(n1,const_100)|multiply(n0,n2)|multiply(n0,const_3)|multiply(n0,#0)|multiply(#3,const_100)|subtract(#1,#4)|divide(#5,#2) | general |
each of the integers from 1 to 18 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement , how many cards must be drawn to ensure that the product of all the integers drawn is even | out of the 18 integers : 9 are odd and 9 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : 1 . we will end up picking odd numbers one by one , so we will pick all 9 odd numbers first 2 . 10 th n... | a ) 19 , b ) 12 , c ) 11 , d ) 10 , e ) 3 | d | add(divide(18, const_2), 1) | divide(n1,const_2)|add(n0,#0) | general |
suppose f ( x ) is an odd function for which f ( x + 2 ) = f ( x ) for all x , and f ( x ) = x 2 for x 2 ( 0 , 1 ) . compute f ( - 3 / 2 ) + f ( 1 ) . | because f is periodic , we know that f ( - 3 / 2 ) = f ( 1 / 2 ) = ( 1 / 2 ) 2 = 1 = 4 . because f is odd , we know that f ( 1 ) = - f ( / 1 ) , but because f is periodic , f ( 1 ) = f ( - 1 ) . therefore , f ( 1 ) = 0 and the answer is 1 = 4 . correct answer d | a ) - 1 , b ) - 1 / 2 , c ) - 1 / 4 , d ) 1 / 4 , e ) 1 / 2 | d | divide(1, add(2, 2)) | add(n0,n0)|divide(n4,#0) | general |
a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 1980 in all from both of them as interest . the rate of interest per annum is ? | "let the rate be r % p . a . then , ( 5000 * r * 2 ) / 100 + ( 3000 * r * 4 ) / 100 = 1980 100 r + 120 r = 1980 r = 9 % answer : c" | a ) 16 % , b ) 12 % , c ) 9 % , d ) 10 % , e ) 45 % | c | multiply(divide(1980, add(multiply(5000, 2), multiply(3000, 4))), const_100) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(n4,#2)|multiply(#3,const_100)| | gain |
sari and ken climb up a mountain . at night , they camp together . on the day they are supposed to reach the summit , sari wakes up at 06 : 00 and starts climbing at a constant pace . ken starts climbing only at 08 : 00 , when sari is already 600 meters ahead of him . nevertheless , ken climbs at a constant pace of 500... | "both sari and ken climb in the same direction . speed of sari = 600 / 2 = 300 meters / hr ( since she covers 700 meters in 2 hrs ) speed of ken = 500 meters / hr at 8 : 00 , distance between ken and sari is 600 meters . ken needs to cover this and another 50 meters . time he will take = total distance to be covered / ... | a ) 13 : 00 , b ) 11 : 15 , c ) 14 : 00 , d ) 15 : 00 , e ) 15 : 30 | b | add(divide(add(600, 50), subtract(500, divide(600, const_2))), 08) | add(n4,n6)|divide(n4,const_2)|subtract(n5,#1)|divide(#0,#2)|add(n2,#3)| | physics |
if x > 0 , x / 50 + x / 5 is what percent of x ? | "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 50 + 100 / 5 = 2 + 20 = 25 25 is 25 % of 100 = b" | a ) 6 % , b ) 25 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | b | multiply(add(divide(const_1, 50), divide(const_1, 5)), const_100) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|multiply(#2,const_100)| | general |
in what time will a railway train 60 m long moving at the rate of 72 kmph pass a telegraph post on its way ? | "t = 60 / 72 * 18 / 5 = 3 sec answer : b" | a ) 8 sec , b ) 3 sec , c ) 2 sec , d ) 6 sec , e ) 9 sec | b | divide(60, multiply(72, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
a can do a piece of work in 18 days and b alone can do it in 10 days . b works at it for 5 days and then leaves . a alone can finish the remaining work in | "explanation : b ' s 5 days work = 1 / 10 β 5 = 1 / 2 remaining work = 1 β 1 / 2 = 1 / 2 a can finish work = 18 β 1 / 2 = 9 days option e" | a ) 5 days , b ) 6 days , c ) 7.5 days , d ) 8.5 days , e ) 9 days | e | divide(const_1, add(divide(const_1, 18), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
x starts a business with rs . 45000 . y joins in the business after 3 months with rs . 36000 . what will be the ratio in which they should share the profit at the end of the year ? | "ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 Γ 12 : 36000 Γ 9 = 45 Γ 12 : 36 Γ 9 = 15 Γ 12 : 9 Γ 9 = 20 : 9 answer is a ." | a ) 20 : 9 , b ) 2 : 1 , c ) 3 : 2 , d ) 2 : 3 , e ) 5 : 3 | a | divide(multiply(45000, const_12), multiply(36000, add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)| | other |
a car travels first 160 km at 75 km / hr and the next 160 km at 80 km / hr . what is the average speed for the first 320 km of the tour ? | car travels first 160 km at 75 km / hr time taken to travel first 160 km = distancespeed = 160 / 75 car travels next 160 km at 80 km / hr time taken to travel next 160 km = distancespeed = 160 / 80 total distance traveled = 160 + 160 = 2 Γ 160 total time taken = 160 / 75 + 160 / 80 average speed = total distance travel... | a ) 71.11 , b ) 71.12 , c ) 71.1 , d ) 71.17 , e ) 77.42 | e | multiply(320, divide(const_1, add(divide(160, 75), divide(160, 80)))) | divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(const_1,#2)|multiply(n4,#3) | general |
if a β 8 = b , then determine the value of | a - b | - | b - a | . | sol . ( b ) | a - b | = | 8 | = 8 = > | b - a | = | - 8 | = 8 = > | a - b | - | b - a | = 8 - 8 = 0 . answer b | a ) 16 , b ) 0 , c ) 4 , d ) 2 , e ) 1 | b | add(subtract(add(subtract(const_1, const_1), 8), add(subtract(const_1, const_1), 8)), const_1) | subtract(const_1,const_1)|add(n0,#0)|subtract(#1,#1)|add(#2,const_1) | general |
the measure of the side of a square is multiplied by 5 . if x represents the perimeter of the original square , what is the value of the new perimeter ? | original perimeter = x hence original side = x / 4 new side = 5 x / 4 new perimeter = 4 * 5 x / 4 = 5 x correct option : c | ['a ) 3 x', 'b ) 4 x', 'c ) 5 x', 'd ) 12 x', 'e ) 27 x'] | c | square_perimeter(multiply(5, const_4)) | multiply(n0,const_4)|square_perimeter(#0) | geometry |
the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 18 . find the sum : | "sol . ( d ) let the sum be 100 . therefore , si = 100 Γ 5 Γ 2100 = 10100 Γ 5 Γ 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 β 100100 ( 1 + 5100 ) 2 β 100 β΄ = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 difference of ci and si = 41 β 4 - 10 = 1 β 4 if the difference is 1 β 4 , the sum = 100 = > if the ... | a ) 4500 , b ) 7500 , c ) 5000 , d ) 7200 , e ) none of these | d | divide(18, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100))))) | divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|add(#2,const_1)|power(#1,n1)|subtract(#4,#3)|divide(n2,#5)| | gain |
a can contains a mixture of 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 liter w of liquid a was contained by the can initially ? | "as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 simil... | a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | add(n0,n2)|add(n0,n1)|multiply(n0,n2)|multiply(n0,n1)|divide(n1,#1)|subtract(#2,#3)|multiply(n2,#4)|subtract(n2,#6)|multiply(#0,#7)|divide(#8,#5)|multiply(n0,#9)| | other |
a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 6 hours , then the distance covered at 40 kmph is | "let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 6 or , x / 40 + ( 250 - x ) / 60 = 6 or , ( 3 x + 500 - 2 x ) / / 120 = 6 or 500 + x = 720 x = 220 answer : a" | a ) 220 km , b ) 120 km , c ) 100 km , d ) 140 km , e ) none of these | a | multiply(divide(subtract(multiply(60, 6), 250), negate(subtract(40, 60))), 40) | multiply(n2,n3)|subtract(n1,n2)|negate(#1)|subtract(#0,n0)|divide(#3,#2)|multiply(n1,#4)| | physics |
a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 180 m long ? | "speed = 45 km / hr = 45 Γ£ β ( 5 / 18 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 180 = 540 meter time taken to cross the platform = 540 / ( 25 / 2 ) = 540 Γ£ β 2 / 25 = 43.2 seconds answer : c" | a ) 38 sec , b ) 35 sec , c ) 43.2 sec , d ) 40 sec , e ) none of these | c | multiply(divide(add(divide(180, const_1000), divide(360, const_1000)), 45), const_3600) | divide(n2,const_1000)|divide(n0,const_1000)|add(#0,#1)|divide(#2,n1)|multiply(#3,const_3600)| | physics |
if the length of the longest chord of a certain circle is 22 , what is the radius of that certain circle ? | "longest chord of a circle is the diameter of the circle diameter = 2 * radius if diameter of the circle is given as 22 = 2 * 11 so radius of the circle = 11 correct answer - a" | a ) 11 , b ) 5 , c ) 10 , d ) 15 , e ) 20 | a | divide(22, const_2) | divide(n0,const_2)| | geometry |
company c produces toy trucks at a cost of $ 5.00 each for the first 100 trucks and $ 3.50 for each additional truck . if 500 toy trucks were produced by company c and sold for $ 20.00 each , what was company c β s gross profit ? | "cost of 500 trucks : ( 100 * 5 ) + ( 400 * 3.5 ) = 500 + 1400 = $ 1900 revenue : 500 * 20 = $ 10000 profit : 10000 - 1900 = $ 8100 option c is correct" | a ) $ 7,250 , b ) $ 8,500 , c ) $ 8,100 , d ) $ 8,250 , e ) $ 8,600 | c | floor(divide(subtract(subtract(multiply(500, 20.00), multiply(5.00, 100)), multiply(subtract(500, 100), 3.50)), const_1000)) | multiply(n3,n4)|multiply(n0,n1)|subtract(n3,n1)|multiply(n2,#2)|subtract(#0,#1)|subtract(#4,#3)|divide(#5,const_1000)|floor(#6)| | gain |
if y is 70 % greater than x , than x is what % less than y ? | y = 1.7 x x = y / 1.7 = 10 y / 17 x is 7 / 17 less which is about 41.2 % less than y . the answer is d . | a ) 18.7 , b ) 25.5 , c ) 33.3 , d ) 41.2 , e ) 54.4 | d | multiply(divide(70, add(70, const_100)), const_100) | add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100) | general |
in a room filled with 7 people , 3 people have exactly 1 sibling in the room and 4 people have exactly 2 siblings in the room . if two individuals are selected from the room at random , what is the probability that those two individuals are not siblings ? | there are suppose a b c d e f g members in the room 4 people who have exactly one sibling . . . . a b c d . . . . ( a is bs β sssibl β g β ssand β ssviceversa ) β ss ( c β ssis β ssds β sssibl β g β ssand β ssviceversa ) β ss ( c β ssis β ssdssibl β gandviceversa ) ( cisds sibling and viceversa ) ( c is ds sibling and ... | a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 13 / 21 | e | add(multiply(divide(3, 7), divide(subtract(7, 2), subtract(7, const_1))), multiply(divide(3, 7), divide(subtract(7, 3), subtract(7, const_1)))) | divide(n1,n0)|subtract(n0,n4)|subtract(n0,const_1)|subtract(n0,n1)|divide(#1,#2)|divide(#3,#2)|multiply(#0,#4)|multiply(#0,#5)|add(#6,#7) | general |
mohit sold an article for $ 27000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ? | "let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 27000 - 10 / 100 ( 27000 ) = 27000 - 2700 = $ 22500 = > 1.08 x = 22500 = > x = 22500 b" | a ) 16000 , b ) 22500 , c ) 15000 , d ) 18000 , e ) 17000 | b | multiply(divide(subtract(27000, divide(multiply(27000, 10), const_100)), add(const_100, 8)), const_100) | add(n2,const_100)|multiply(n0,n1)|divide(#1,const_100)|subtract(n0,#2)|divide(#3,#0)|multiply(#4,const_100)| | gain |
if a farmer wants to plough a farm field on time , he must plough 260 hectares a day . for technical reasons he ploughed only 85 hectares a day , hence he had to plough 2 more days than he planned and he still has 40 hectares left . what is the area of the farm field and how many days the farmer planned to work initial... | "let x be the number of days in the initial plan . therefore , the whole field is 260 Γ’ βΉ β¦ x hectares . the farmer had to work for x + 2 days , and he ploughed 85 ( x + 2 ) hectares , leaving 40 hectares unploughed . then we have the equation : 260 x = 85 ( x + 2 ) + 40 175 x = 210 x = 1.2 so the farmer planned to hav... | a ) 312 , b ) 490 , c ) 720 , d ) 435 , e ) 679 | a | multiply(divide(add(40, multiply(2, 85)), subtract(260, 85)), 260) | multiply(n1,n2)|subtract(n0,n1)|add(n3,#0)|divide(#2,#1)|multiply(n0,#3)| | physics |
on my sister ' s 15 th birthday , she was 159 cm in height , having grown 6 % since the year before . how tall was she the previous year ? | explanation : given that height on 15 th birthday = 159 cm and growth = 6 % let the previous year height = x then height on 15 th birthday = x Γ ( 100 + 6 ) / 100 = x Γ 106 / 100 β 159 = x Γ 106 / 100 β x = 159 Γ 100 / 106 = 1.5 Γ 100 = 150 cm answer : option a | a ) 150 cm , b ) 160 cm , c ) 170 cm , d ) 140 cm , e ) 150 cm | a | divide(159, add(const_1, divide(6, const_100))) | divide(n2,const_100)|add(#0,const_1)|divide(n1,#1) | physics |
jim Γ’ β¬ β’ s taxi service charges an initial fee of $ 2.35 at the beginning of a trip and an additional charge of $ 0.35 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim Γ’ β¬ β’ s taxi service = 2.35 $ and charge per 2 / 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.35 + ( 3.6 / . 4 ) * . 35 = 2.35 + 9 * . 35 = 5.5 $ answer d" | a ) $ 3.15 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.50 , e ) $ 5.40 | d | add(multiply(divide(3.6, divide(2, 5)), 0.35), 2.35) | divide(n2,n3)|divide(n4,#0)|multiply(n1,#1)|add(n0,#2)| | general |
a special municipal payroll tax charges not tax on a payroll less than $ 200,000 and only 0.2 % on a company β s payroll above $ 200,000 . if belfried industries paid $ 400 in this special municipal payroll tax , then they must have had a payroll of | "answer : b , ( with different approach ) : the 400 paid is 0.2 % of the additional amount above 200,000 . let it be x now 0.2 % of x = 400 therefore x = 200,000 total = 200,000 + x = 400,000" | a ) $ 180,000 , b ) $ 400,000 , c ) $ 220,000 , d ) $ 402,000 , e ) $ 2 , 200,000 | b | multiply(multiply(400, const_100), const_10) | multiply(n3,const_100)|multiply(#0,const_10)| | general |
the speed of a boat in still water in 20 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 24 minutes is : | "speed downstream = ( 20 + 3 ) = 23 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time Γ speed = 2 / 5 Γ 23 = 9.20 km answer is c ." | a ) 9.6 , b ) 10.2 , c ) 9.2 , d ) 10.4 , e ) 9.0 | c | multiply(add(20, 3), divide(24, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
what is the sum of all possible solutions to | x - 3 | ^ 2 + | x - 3 | = 42 ? | "denote | x - 3 | as y : y ^ 2 + y = 42 - - > y = - 7 or y = 6 . discard the first solution since y = | x - 3 | , so it ' s an absolute value and thus can not be negative . y = | x - 3 | = 6 - - > x = 9 or x = - 4 . the sum = 5 . answer : c ." | a ) 4 , b ) 3 , c ) 5 , d ) 6 , e ) 2 | c | add(add(const_4, 3), subtract(3, const_4)) | add(n0,const_4)|subtract(n0,const_4)|add(#0,#1)| | general |
what is remainder of the division ( 1225 * 1227 * 1229 ) / 12 ? | "remainder will be number / 100 here as the divisor is two digit number = 12 . hence checking for the last two digits = 5 * 7 * 9 = 15 thus remainder = 3 . answer : c" | a ) 1 , b ) 0 , c ) 3 , d ) 2 , e ) 4 | c | subtract(multiply(multiply(1225, 1227), 1229), subtract(multiply(multiply(1225, 1227), 1229), const_3)) | multiply(n0,n1)|multiply(n2,#0)|subtract(#1,const_3)|subtract(#1,#2)| | general |
the sum of three consecutive multiples of 3 is 99 . what is the largest number ? | "let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 99 9 x = 90 x = 10 largest number = 3 x + 6 = 36 answer : a" | a ) 36 , b ) 33 , c ) 30 , d ) 27 , e ) 24 | a | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(di... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | general |
what number should replace the question mark ? 2 , 16 , 4 , 14 , 6 , 12 , - - - ? | answer : a 2 , 16 , 4 , 14 , 6 , 12 , 8 ? there are two alternate sequences : + 2 and - 2 . | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | add(add(add(2, 2), 2), const_2) | add(n0,n0)|add(n0,#0)|add(#1,const_2) | general |
a garrison of 400 men had a provision for 31 days . after 28 days 280 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? | "400 - - - 31 400 - - - 3 120 - - - ? 400 * 3 = 120 * x = > x = 10 days . answer : c" | a ) 65 days , b ) 45 days , c ) 10 days , d ) 16 days , e ) 18 days | c | divide(multiply(subtract(31, 28), 400), 280) | subtract(n1,n2)|multiply(n0,#0)|divide(#1,n3)| | other |
a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 62 now , the son age 5 years back was | "let the son ' s present age be x years . then , ( 62 - x ) = x x = 31 . son ' s age 5 years back = ( 31 - 5 ) = 26 years answer : e" | a ) 14 , b ) 17 , c ) 11 , d ) 19 , e ) 26 | e | subtract(divide(62, const_2), 5) | divide(n0,const_2)|subtract(#0,n1)| | general |
calculate the average of all the numbers between 7 and 49 which are divisible by 6 . | "explanation : numbers divisible by 6 are 12 , 18,24 , 30,36 , 42,48 , average = ( 12 + 18 + 24 + 30 + 36 + 42 + 48 , ) / 7 = 210 / 7 = 30 answer : e" | a ) 20 , b ) 18 , c ) 19 , d ) 15 , e ) 30 | e | multiply(divide(add(add(floor(divide(7, 6)), const_1), floor(divide(49, 6))), const_2), 6) | divide(n0,n2)|divide(n1,n2)|floor(#0)|floor(#1)|add(#2,const_1)|add(#4,#3)|divide(#5,const_2)|multiply(n2,#6)| | general |
a certain sum of money is divided among a , b and c such that a gets one - third of what b and c together get and b gets two - seventh of what a and c together get . if the amount received by a is $ 35 more than that received by b , find the total amount shared by a , b and c . | "a = 1 / 3 ( b + c ) = > c = 3 a - b - - - ( 1 ) b = 2 / 7 ( a + c ) = > c = 3.5 b - a - - ( b ) a - b = $ 35 a = 35 + b ( 1 ) = = = > c = 105 + 3 b - b = 2 b + 105 = = > 2 b - c = - 105 - - - ( 3 ) ( 2 ) = = = > c = 3.5 b - b - 35 = 2.5 b - 35 = = > 2.5 b - c = 35 - - - ( 4 ) from ( 4 ) and ( 3 ) 0.5 b = 140 b = $ 280... | a ) $ 1360 , b ) $ 1250 , c ) $ 1260 , d ) $ 1270 , e ) $ 1280 | c | add(add(add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)), add(add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)), 35)), subtract(multiply(add(add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)), 35), const_3), add(multiply(multiply(const_3, 35), const_2), multiply(... | multiply(n0,const_3)|multiply(n0,const_2)|multiply(#0,const_2)|add(#2,#1)|add(n0,#3)|add(#3,#4)|multiply(#4,const_3)|subtract(#6,#3)|add(#5,#7)| | general |
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 ? | "explanation : speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) x ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be ' s ' km / hr . then , relative speed = ( s - 5 ) km / hr . s - 5 = 45 = > s = 50 km / hr . answer is d" | a ) 45 kmph , b ) 25 kmph , c ) 30 kmph , d ) 50 kmph , e ) none of these | d | divide(divide(subtract(125, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
the number which exceeds 16 % of it by 21 is : | "solution solution let the number be x . x - 16 % of x = 21 x - 16 / 100 x = 21 x - 4 / 25 x = 21 21 / 25 x = 21 x = ( 21 x 25 / 21 ) = 25 answer a" | a ) 25 , b ) 52 , c ) 58 , d ) 60 , e ) 62 | a | divide(multiply(21, const_100), subtract(const_100, 16)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
excluding stoppages , the speed of a bus is 60 kmph and including stoppages , it is 50 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 10 km less . time taken to cover 10 km = ( 10 / 60 x 60 ) min = 10 min answer : a" | a ) 10 min , b ) 15 min , c ) 12 min , d ) 8 min , e ) 16 min | a | multiply(const_60, divide(subtract(60, 50), 60)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 4960 , r will receive | "explanation : let p ' s capital = p , q ' s capital = q and r ' s capital = r then 4 p = 6 q = 10 r = > 2 p = 3 q = 5 r = > q = 2 p / 3 r = 2 p / 5 p : q : r = p : 2 p / 3 : 2 p / 5 = 15 : 10 : 6 r ' s share = 4960 * ( 6 / 31 ) = 160 * 6 = 960 . answer : option d" | a ) 600 , b ) 700 , c ) 800 , d ) 960 , e ) none of these | d | multiply(4960, divide(6, add(add(add(10, add(4, const_1)), 10), 6))) | add(n0,const_1)|add(n2,#0)|add(n2,#1)|add(n1,#2)|divide(n1,#3)|multiply(n3,#4)| | general |
an isosceles triangle those sides are 13 cm , 13 cm , 10 cm long inscribed in a circle . find the radius of the circle | a , b , c are sides of triangle , ad is the median , ad ^ 2 = ab ^ 2 - bd ^ 2 then ad = 12 , the centriod median divides into 2 : 1 ratio so 12 will be divide into 2 : 1 ratio so 8 is radius . so ans is 8 cm answer : a | ['a ) 8 cm', 'b ) 9 cm', 'c ) 10 cm', 'd ) 11 cm', 'e ) 7 cm'] | a | add(divide(add(subtract(power(13, const_2), power(divide(10, const_2), const_2)), power(divide(10, const_2), const_2)), multiply(sqrt(subtract(power(13, const_2), power(divide(10, const_2), const_2))), const_2)), const_1) | divide(n2,const_2)|power(n0,const_2)|power(#0,const_2)|subtract(#1,#2)|add(#2,#3)|sqrt(#3)|multiply(#5,const_2)|divide(#4,#6)|add(#7,const_1) | geometry |
n is an 80 - digit positive integer ( in the decimal scale ) . all digits except the 44 th digit ( from the left ) are 2 . if n is divisible by 13 , find the 26 th digit ? | detailed solution to begin with , the question should read ` ` find the 44 th digit ' ' . any number of the form abcabc is a multiple of 1001 . 1001 is 7 * 11 * 13 . so , any number of the form abcabc is a multiple of 13 . so , a number comprising 42 2 ' s would be a multiple of 13 , so would a number comprising 36 2 '... | a ) 5 , b ) 6 , c ) 1 , d ) 2 , e ) 3 | b | divide(divide(const_12, const_2), const_1) | divide(const_12,const_2)|divide(#0,const_1) | general |
60 boys can complete a work in 40 days . how many men need to complete twice the work in 20 days | "one man can complete the work in 40 * 60 = 2400 days = one time work to complete the work twice it will be completed in let m be the no . of worker assign for this therefore the eqn becomes m * 20 = 2 * 2400 m = 240 workers answer : e" | a ) 160 , b ) 170 , c ) 180 , d ) 190 , e ) 240 | e | divide(multiply(60, multiply(40, const_2)), 20) | multiply(n1,const_2)|multiply(n0,#0)|divide(#1,n2)| | physics |
an outlet pipe empties a tank which is full in 8 hours . if the inlet pipe is kept open , which lets water in at the rate of 8 litres / min then outlet pipe would take 4 hours longer . find the capacity of the tank . | "let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 8 litres / min , or 8 * 60 = 480 liters / hour ; net outflow rate when both pipes operate would be x - 480 liters / hour . capacity of the tank = x * 8 hours = ( x - 480 ) * ( 8 + 4 ) hours 8 x = ( x - 480 ) * 12 - - > x = 1440 - - > capacity = 8 x... | a ) 8600 litres , b ) 200 litres , c ) 12800 litres , d ) 11520 litres , e ) 13200 litres | d | multiply(divide(multiply(multiply(8, const_60), add(8, 4)), 4), 8) | add(n0,n2)|multiply(n1,const_60)|multiply(#0,#1)|divide(#2,n2)|multiply(n0,#3)| | physics |
how should we mix two rice packets to get one 1 kg packet priced at rs 8 when the prices of 1 kg packets of the two types are rs 7.1 and rs 9.2 . | explanation : no explanation is available for this question ! answer : a | a ) 4 : 3 , b ) 4 : 7 , c ) 4 : 9 , d ) 4 : 2 , e ) 4 : 1 | a | divide(subtract(9.2, 8), subtract(8, 7.1)) | subtract(n4,n1)|subtract(n1,n3)|divide(#0,#1) | general |
car a runs at the speed of 35 km / hr & reaches its destination in 9 hr . car b runs at the speed of 43 km / h & reaches its destination in 10 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 35 Γ 9 = 315 km distance travelled by car b = 43 Γ 10 = 430 km ratio = 315 / 430 = 63 : 86 e" | a ) 65 : 69 , b ) 56 : 45 , c ) 89 : 71 , d ) 68 : 23 , e ) 63 : 86 | e | divide(multiply(35, 9), multiply(43, 10)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
a and b β s salaries together amount to rs . 3,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is a β s salary ? | "( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 3000 4 b = 3000 = > b = 750 a = 2250 answer b" | a ) 2777 , b ) 2250 , c ) 2789 , d ) 2776 , e ) 2881 | b | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | add(const_2,const_3)|divide(n2,const_100)|divide(n1,const_100)|multiply(#0,const_4)|subtract(const_1,#1)|subtract(const_1,#2)|add(#5,#4)|multiply(#3,const_2)|multiply(#7,const_100)|multiply(#8,#4)|divide(#9,#6)| | gain |
the average weight of 3 packets a , b and c is 84 kg . another packet d added into the group and the average now becomes 80 kg . if another packet e , whose weight is 3 kg more than that of d , replaces a then the average weight of b , c , d and e becomes 79 kg . the weight of a is : | explanation : a + b + c = ( 84 x 3 ) = 252 kg , a + b + c + d = ( 80 x 4 ) = 320 kg . d = ( 320 - 252 ) = 68 , e = ( 68 + 3 ) = 7 l . b + c + d + e = ( 79 x 4 ) = 316 . ( a + b + c + d ) - ( b + c + d + e ) = ( 320 - 316 ) kg = 4 kg . a - e = 4 = a = ( 4 + e ) = 75 kg answer : c | a ) 70 kg , b ) 72 kg , c ) 75 kg , d ) 80 kg , e ) none of these | c | subtract(multiply(3, 84), subtract(multiply(const_4, 79), add(subtract(multiply(const_4, 80), multiply(3, 84)), add(subtract(multiply(const_4, 80), multiply(3, 84)), 3)))) | multiply(n0,n1)|multiply(n4,const_4)|multiply(n2,const_4)|subtract(#2,#0)|add(n0,#3)|add(#4,#3)|subtract(#1,#5)|subtract(#0,#6) | general |
the consumption of diesel per hour of a bus varies directly as square of its speed . when the bus is travelling at 60 kmph its consumption is 1 litre per hour . if each litre costs $ 30 and other expenses per hous is $ 20 , then what would be the minimum expenditure required to cover a distance of 600 km ? | 60 kmph consumption is 1 lt / hr so 600 km will take 10 hrs and the consumption is 10 lt for entire distance . 1 lt costs $ 30 so 10 lt costs $ 300 extra expenses for 1 hr - $ 20 10 hrs - $ 200 total expense - $ 300 + $ 200 = $ 500 answer : a | a ) 500 , b ) 450 , c ) 600 , d ) 650 , e ) 750 | a | add(multiply(divide(600, 60), 30), multiply(divide(600, 60), 20)) | divide(n4,n0)|multiply(n2,#0)|multiply(n3,#0)|add(#1,#2) | physics |
if the average of 10 consecutive integers is 19.5 then the 10 th integer is : - | "the average falls between the 5 th and 6 th integers , integer 5 = 19 , integer 6 = 20 . counting up to the tenth integer we get 24 . answer : d" | a ) 15 , b ) 20 , c ) 23 , d ) 24 , e ) 25 | d | add(multiply(10, const_2), multiply(subtract(19.5, multiply(10, const_2)), 10)) | multiply(n0,const_2)|subtract(n1,#0)|multiply(n0,#1)|add(#0,#2)| | general |
on the first day of her vacation , louisa traveled 100 miles . on the second day , traveling at the same average speed , she traveled 175 miles . if the 100 - mile trip took 3 hours less than the 175 - mile trip , what was the average speed , in miles per hour ? | "( time ) * ( rate ) = ( distance ) - - > ( rate ) = ( distance ) / ( time ) - - > given : ( rate ) = 100 / t = 175 / ( t + 3 ) - - > 4 / t = 7 / ( t + 3 ) - - > 4 t + 12 = 7 t - - - - > 3 t = 12 . t = 4 - - - - > ( rate ) = 100 / 4 = 25 answer : c" | a ) 27 , b ) 28 , c ) 25 , d ) 30 , e ) 35 | c | divide(subtract(175, 100), 3) | subtract(n1,n0)|divide(#0,n3)| | physics |
at a certain paint store forest green is made by mixing 4 parts blue paint with 3 parts yellow paint . verdant green is made by mixing 4 parts yellow paint with 3 parts blue paint . how many liters of yellow paint must be added to 28 liters of forest green to change it to verdant green ? | "28 liter of forset green have 16 liter of blue and 12 liter of yellow suppose we add x liter of yellow to make it a verdant green so the ratio of blue to yellow in verdant green is ΒΎ so the equation is blue / yellow = 16 / ( 12 + x ) = ΒΎ 36 + 3 x = 64 = > x = 28 / 3 answer : a" | a ) 28 / 3 , b ) 13 / 6 , c ) 3 , d ) 4 , e ) 14 / 3 | a | subtract(multiply(divide(4, 3), 4), 3) | divide(n0,n1)|multiply(n0,#0)|subtract(#1,n1)| | general |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 8 minutes , then what is the area of the park ( in sq . m ) ? | "let length = 3 x km , breadth = 2 x km distance travelled by the man at the speed of 12 km / hr in 8 minutes = 2 ( 3 x + 2 x ) = 10 x therefore , 12 Γ 860 = 10 xx = 425 km = 160 m area = 3 x Γ 2 x = 6 x 2 = 6 Γ 1602 = 153600 m 2 answer is c ." | a ) 153500 , b ) 153650 , c ) 153600 , d ) 153700 , e ) 153750 | c | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
in a office work is distribute between p persons . if 1 / 5 members are absent then work increased for each person is ? | "let total % of work is 100 % total person = p 1 / 5 person are absent of total person . so absent person is 1 / 5 p ie p / 5 . left person is , p - p / 5 = 4 p / 5 . p person do the work 100 % 1 person do the work 100 * p % 4 p / 5 person do the work ( 100 * p * 5 ) / 4 p % = 125 % work increased for each person is = ... | a ) 13.28 % , b ) 14.28 % , c ) 15.28 % , d ) 16.28 % , e ) 25 % | e | subtract(divide(multiply(const_100, 5), subtract(5, 1)), const_100) | multiply(n1,const_100)|subtract(n1,n0)|divide(#0,#1)|subtract(#2,const_100)| | general |
what could be the range of a set consisting of odd multiples of 5 ? | "range = the difference between the greatest and the smallest numbers in the sequence . our sequence is odd and is a multiple of 7 . every number in that sequence can be represented like this : 5 * ( 2 n + 1 ) where n is any positive integer . range = 5 * ( 2 m + 1 ) - 5 * ( 2 n + 1 ) = 5 * 2 * ( m - n ) = 10 * ( m - n... | a ) 21 , b ) 24 , c ) 35 , d ) 62 , e ) 60 | e | multiply(5, const_4) | multiply(n0,const_4)| | general |
a book is bought for $ 60 and sold for $ 75 . what is the profit in percentage ? | "75 / 60 = 1.25 the answer is d ." | a ) 12 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | d | multiply(divide(subtract(75, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
how many multiples of 5 are there between 5 to 85 ? | "explanation : as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the units place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 . answer : a" | a ) 15 , b ) 18 , c ) 17 , d ) none of these , e ) can not be determined | a | add(divide(subtract(85, 5), 5), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
the difference between a number and its two - fifth is 510 . what is 10 % of that number ? | "explanation : let the number be x . then , x - 2 / 5 x = 510 x = ( 510 * 5 ) / 3 = 850 10 % of 850 = 85 . answer is b" | a ) 12.75 , b ) 85 , c ) 204 , d ) 250 , e ) none of these | b | divide(multiply(510, add(const_4, const_1)), add(const_1, const_2)) | add(const_1,const_4)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)| | general |
in a recent election , james received 0.5 percent of the 2,000 votes cast . to win the election , a candidate needed to receive more than 45 percent of the vote . how many additional votes would james have needed to win the election ? | "james = ( 0.5 / 100 ) * 2000 = 10 votes to win = ( 45 / 100 ) * total votes + 1 = ( 45 / 100 ) * 2000 + 1 = 901 remaining voted needed to win election = 901 - 10 = 891 answer : option d" | a ) 901 , b ) 989 , c ) 990 , d ) 891 , e ) 1,001 | d | subtract(add(const_1000, const_1000), multiply(add(const_1000, const_1000), 0.5)) | add(const_1000,const_1000)|multiply(n0,#0)|subtract(#0,#1)| | general |
the value of a machine depreciates at 23 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ? | "the value of the machine after two years = 0.77 * 0.77 * 1 , 50,000 = $ 88,935 sp such that a profit of $ 24,000 is made = 88,935 + 24,000 = $ 1 , 12,935 d" | a ) $ 252935 , b ) $ 432935 , c ) $ 122935 , d ) $ 112935 , e ) $ 152935 | d | add(multiply(multiply(subtract(1, divide(23, const_100)), subtract(1, divide(23, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(23, const_2), const_100), sqrt(const_100))) | add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(n1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10)| | gain |
a train travels 240 km in 3 hours and 450 km in 5 hours . find the average speed of train . | "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 240 + 450 = 690 km thus , total speed = 8 hrs or , average speed = 690 / 8 or , 86 kmph . answer : c" | a ) 80 kmph , b ) 60 kmph , c ) 86 kmph , d ) 90 kmph , e ) none of these | c | divide(add(240, 450), add(3, 5)) | add(n0,n2)|add(n1,n3)|divide(#0,#1)| | physics |
a gambler bought $ 3,000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 16 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could ha... | in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that ... | a ) $ 2,040 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 1,920 , e ) $ 1,400 | b | subtract(multiply(const_3, const_1000), add(multiply(divide(add(16, 2), 2), 20), multiply(subtract(divide(add(16, 2), 2), 2), 100))) | add(n3,n5)|multiply(const_1000,const_3)|divide(#0,n5)|multiply(n1,#2)|subtract(#2,n5)|multiply(n2,#4)|add(#3,#5)|subtract(#1,#6) | general |
a train running at the speed of 60 km / hr crosses a pole in 12 seconds . find the length of the train ? | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 12 = 200 meter answer : b" | a ) 150 meter , b ) 200 meter , c ) 156 meter , d ) 168 meter , e ) 154 meter | b | multiply(divide(multiply(60, const_1000), const_3600), 12) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
two trains of equal lengths take 12 sec and 24 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 12 = 10 m / sec . speed of the second train = 120 / 24 = 5 m / sec . relative speed = 10 + 5 = 15 m / sec . required time = ( 120 + 120 ) / 15 = 16 sec . answer : e" | a ) 18 sec , b ) 12 sec , c ) 17 sec , d ) 21 sec , e ) 16 sec | e | divide(multiply(120, const_2), add(speed(120, 24), speed(120, 12))) | multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)| | physics |
set a contains 4 different positive odd integers and 6 different positive even integers . set b contains two different positive odd integers and 3 different positive even integers . if one integer from set a and one integer from set b are chosen at random , what is the probability that the product of the chosen integer... | there can be an odd product only if both integers are odd . p ( odd product ) = 4 / 10 * 2 / 5 = 4 / 25 p ( even product ) = 1 - 4 / 25 = 21 / 25 the answer is d . | a ) 3 / 5 , b ) 4 / 5 , c ) 17 / 25 , d ) 21 / 25 , e ) 43 / 50 | d | subtract(const_1, multiply(divide(const_2, add(const_2, const_3)), divide(4, add(4, 6)))) | add(const_2,const_3)|add(n0,n1)|divide(const_2,#0)|divide(n0,#1)|multiply(#2,#3)|subtract(const_1,#4) | probability |
a and b can do a piece of work in 16 days , b and c in 24 days , c and a in 12 days . how long will c take to do it ? | "2 c = 1 / 24 + 1 / 12 β 1 / 16 = 1 / 16 c = 1 / 32 = > 32 days answer : d" | a ) 16 days , b ) 55 days , c ) 24 days , d ) 32 days , e ) 22 days | d | divide(multiply(16, const_3), subtract(divide(add(divide(multiply(16, const_3), 12), add(divide(multiply(16, const_3), 16), divide(multiply(16, const_3), 24))), const_2), divide(multiply(16, const_3), 16))) | multiply(n0,const_3)|divide(#0,n0)|divide(#0,n1)|divide(#0,n2)|add(#1,#2)|add(#4,#3)|divide(#5,const_2)|subtract(#6,#1)|divide(#0,#7)| | physics |
population of a city in 20004 was 1000000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrement of 35 % and in 2007 there is an increment of 45 % , then find the population of city at theend of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1083875 e" | a ) 2456789 , b ) 3458009 , c ) 3890043 , d ) 3455048 , e ) 1083875 | e | multiply(1000000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6)| | gain |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.