Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
|---|---|---|---|---|---|---|---|
ages of two persons differ by 16 years . if 6 year ago , the elder one be 3 times as old the younger one , find their present age | "explanation : let the age of younger person is x , then elder person age is ( x + 16 ) = > 3 ( x - 6 ) = ( x + 16 - 6 ) [ 6 years before ] = > 3 x - 18 = x + 10 = > x = 14 . so other person age is x + 16 = 30 answer : option b" | a ) 12,28 , b ) 14,30 , c ) 16,32 , d ) 18,34 , e ) 19,32 | b | subtract(divide(subtract(add(multiply(6, 6), 16), 6), subtract(6, const_1)), const_1) | multiply(n1,n1)|subtract(n1,const_1)|add(n0,#0)|subtract(#2,n1)|divide(#3,#1)|subtract(#4,const_1)| | general | B |
a man is 22 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ? | "let the son ' s present age be x years . then , man ' s present age = ( x + 22 ) years . ( x + 22 ) + 2 = 2 ( x + 2 ) x + 24 = 2 x + 4 = > x = 20 . answer : a" | a ) 20 , b ) 25 , c ) 27 , d ) 22 , e ) 91 | a | divide(subtract(22, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general | A |
at 15 : 00 there were 20 students in the computer lab . at 15 : 03 and every three minutes after that , 3 students entered the lab . if at 15 : 10 and every ten minutes after that 7 students left the lab , how many students were in the computer lab at 15 : 44 ? | "initial no of students + 3 * ( 1 + no of possible 3 minute intervals between 15 : 03 and 15 : 44 ) - 8 * ( 1 + no of possible 10 minute intervals between 15 : 10 and 15 : 44 ) 20 + 3 * 14 - 8 * 4 = 25 c" | a ) 7 , b ) 14 , c ) 25 , d ) 27 , e ) 30 | c | add(subtract(add(multiply(floor(divide(44, 03)), 03), 20), multiply(floor(divide(44, 7)), 7)), 03) | divide(n10,n4)|divide(n10,n8)|floor(#0)|floor(#1)|multiply(n4,#2)|multiply(n8,#3)|add(n2,#4)|subtract(#6,#5)|add(n4,#7)| | physics | C |
6.006 / 3.003 | "answer is 2 , move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is 6006 / 3003 = 2 answer e" | a ) 0.002 , b ) 0.02 , c ) 0.2 , d ) 20 , e ) 2 | e | multiply(divide(6.006, 3.003), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | E |
a certain car uses 12 gallons of gasoline in traveling 420 miles . in order for the car to travel the same distance using 10 gallons of gasoline , by how many miles per gallon must the car β s gas mileage be increased ? | "420 / 10 = 42 the difference is 42 - 35 = 7 . answer d" | a ) 2 , b ) 4 , c ) 6 , d ) 7 , e ) 10 | d | subtract(divide(420, 10), divide(420, 12)) | divide(n1,n2)|divide(n1,n0)|subtract(#0,#1)| | physics | D |
find the area of the quadrilateral of one of its diagonals is 20 cm and its off sets 5 cm and 4 cm ? | "1 / 2 * 20 ( 5 + 4 ) = 90 cm 2 answer : e" | a ) 189 cm 2 , b ) 150 cm 2 , c ) 127 cm 2 , d ) 177 cm 2 , e ) 90 cm 2 | e | multiply(multiply(divide(const_1, const_2), add(4, 5)), 20) | add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2)| | geometry | E |
two pipes can fill a tank in 20 minutes and 15 minutes . an outlet pipe can empty the tank in 10 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 20 + v / 15 - v / 10 = v / 60 per minute the tank will be filled in 60 minutes . the answer is e ." | a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60 | e | subtract(add(divide(const_1, 20), divide(const_1, 15)), divide(const_1, 10)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)| | physics | E |
if 13 = 13 w / ( 1 - w ) , then ( 4 w ) 2 = | "13 - 13 w = 13 w 26 w = 13 w = 1 / 2 4 w = 2 4 w * 2 = 2 * 2 = 4 answer : e" | a ) 1 / 4 , b ) 1 / 2 , c ) 1 , d ) 2 , e ) 4 | e | multiply(divide(13, add(13, 13)), 4) | add(n0,n0)|divide(n0,#0)|multiply(n3,#1)| | general | E |
two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 20 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ? | "same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 10 / speed b since the times are same , 50 pi / speed a = 20 pi / speed b speeda / speed b = 50 pi / 30 pi = 5 / 2 correct option : c" | a ) 3 : 5 , b ) 9 : 25 , c ) 5 : 2 , d ) 25 : 9 , e ) can not be determined from the information provided | c | divide(circumface(divide(50, const_2)), circumface(divide(20, const_2))) | divide(n1,const_2)|divide(n0,const_2)|circumface(#0)|circumface(#1)|divide(#2,#3)| | physics | C |
two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 40 mph between the 2 friends until the friends meet . how many miles does the pigeon travel ? | "c 120 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 40 mph = 120 miles" | a ) 54 , b ) 110 , c ) 120 , d ) 16 , e ) 180 | c | multiply(40, divide(divide(36, 2), 6)) | divide(n1,n3)|divide(#0,n0)|multiply(n2,#1)| | physics | C |
how many integers are divisible by 3 between 10 ! and 10 ! + 20 inclusive ? | "0 ! is divisible by 3 - the way i look factorials is that any number included will also be divisible by the product . 10,9 , 8,7 , 6,5 , 4,3 , 2,1 are all divisors of 10 ! there are 6 numbers between 10 ! and 10 ! + 20 that are divisible by 3 . hence b : 7" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | add(divide(20, 3), const_1) | divide(n3,n0)|add(#0,const_1)| | general | B |
20 beavers , working together in a constant pace , can build a dam in 3 hours . how many b hours will it take 12 beavers that work at the same pace , to build the same dam ? | "c . 5 hrs if there were 10 beavers it could have taken double b = 6 hrs . . so closest to that option is 5 . c" | a ) 2 . , b ) 4 . , c ) b = 5 . , d ) b = 6 . , e ) b = 8 . | c | divide(multiply(3, 20), 12) | multiply(n0,n1)|divide(#0,n2)| | physics | C |
in a certain pond , 70 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ? | "this is a rather straight forward ratio problem . 1 . 70 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 70 / x thus , x = 1750 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . c" | a ) 400 , b ) 625 , c ) 1,750 , d ) 2,500 , e ) 10,000 | c | divide(70, divide(2, 50)) | divide(n2,n1)|divide(n0,#0)| | gain | C |
in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the rate in the remaining 40 overs to reach the target of 292 runs ? | "required run rate = [ 292 - ( 3.2 * 10 ) ] / 40 = 260 / 40 = 6.5 answer : a" | a ) 6.5 , b ) 6.22 , c ) 6.29 , d ) 6.39 , e ) 6.13 | a | divide(subtract(292, multiply(10, 3.2)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain | A |
excluding stoppages , the average speed of a bus is 60 km / hr and including stoppages , the average speed of the bus is 35 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 60 km without stoppages and 35 km with stoppages . stoppage time = time take to travel ( 60 - 35 ) km i . e 25 km at 60 km / hr . stoppage time = 25 / 60 hrs = 24 min . answer : c" | a ) 22 , b ) 88 , c ) 24 , d ) 20 , e ) 99 | c | subtract(multiply(const_1, const_60), multiply(divide(35, 60), const_60)) | divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)| | general | C |
if an investor puts $ 1000 in a savings account that earns 10 percent annual interest compounded semiannually , how much money will be in the account after one year ? | "1.05 * 1.05 * 1000 = $ 1102.50 the answer is e ." | a ) $ 1100.25 , b ) $ 1100.75 , c ) $ 1101.25 , d ) $ 1101.75 , e ) $ 1102.50 | e | multiply(1000, power(add(const_1, divide(divide(10, const_2), const_100)), const_2)) | divide(n1,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)| | gain | E |
there are 2 sections a and b in a class , consisting of 36 and 24 students respectively . if the average weight of section a is 30 kg and that of section b is 30 kg , find the average of the whole class ? | "total weight of 36 + 44 students = 36 * 30 + 24 * 30 = 1800 average weight of the class is = 1800 / 60 = 30 kg answer is a" | a ) 30 kg , b ) 37 kg , c ) 42 kg , d ) 55.12 kg , e ) 29.78 kg | a | divide(add(multiply(36, 30), multiply(24, 30)), add(36, 24)) | add(n1,n2)|multiply(n1,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)| | general | A |
due to construction , the speed limit along an 7 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "7 / 35 - 7 / 55 = 7 / 5 * ( 11 - 7 ) / 77 = 7 / 5 * 4 / 77 * 60 min = 7 * 12 * 4 / 77 = 336 / 77 ~ 4.3 answer - c" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | max(multiply(subtract(add(55, 7), const_1), subtract(divide(7, 35), divide(7, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics | C |
an order was placed for the supply of a carper whose length and breadth were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and breadth were in the ratio 7 : 3 but were was no change in its parameter . find the ratio of the areas of the carpets in both the cases . | "explanation : let the length and breadth of the carpet in the first case be 3 x units and 2 x units respectively . let the dimensions of the carpet in the second case be 7 y , 3 y units respectively . from the data , . 2 ( 3 x + 2 x ) = 2 ( 7 y + 3 y ) = > 5 x = 10 y = > x = 2 y required ratio of the areas of the carpet in both the cases = 3 x * 2 x : 7 y : 3 y = 6 x 2 : 21 y 2 = 6 * ( 2 y ) 2 : 21 y 2 = 6 * 4 y 2 : 21 y 2 = 8 : 7 answer is b" | a ) 4 : 3 , b ) 8 : 7 , c ) 4 : 1 , d ) 6 : 5 , e ) none of these | b | divide(multiply(multiply(3, 2), const_4), multiply(7, 3)) | multiply(n0,n1)|multiply(n0,n2)|multiply(#0,const_4)|divide(#2,#1)| | other | B |
a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 36 km / hr . what is the length of the platform ? | "speed = 36 * 5 / 18 = 10 m / sec . length of the train = 10 * 20 = 200 m . let the length of the platform be x m . then , ( x + 200 ) / 36 = 10 = > x = 160 m . answer : a" | a ) 160 , b ) 240 , c ) 288 , d ) 277 , e ) 221 | a | multiply(20, multiply(36, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics | A |
of the total amount that jill spent on a shopping trip , excluding taxes , she spent 20 percent on clothing , 50 percent on food , and 30 percent on other items . if jill paid a 10 percent tax on the clothing , 15 percent tax on the food , and no tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | "assume she has $ 200 to spend . tax clothing = 20 % = $ 40 = $ 10.00 food = 50 % = $ 100 = $ 15.00 items = 30 % = $ 60 = $ 0.00 total tax = $ 25.00 % of total amount = 25 / 200 * 100 = 12.25 % answer e" | a ) 12.45 , b ) 12.5 , c ) 12.65 , d ) 12.35 , e ) 12.25 | e | multiply(divide(add(multiply(20, divide(10, const_100)), multiply(30, divide(15, const_100))), const_100), const_100) | divide(n3,const_100)|divide(n4,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,const_100)|multiply(#5,const_100)| | general | E |
a certain rectangular crate measures 20 feet by 20 feet by 20 feet . a cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces . what should the radius of the tank be if it is to be of the largest possible volume ? | "for max volume of cylinder ( pi * r ^ 2 * h ) we need to max out r ^ 2 * h . we do n ' t know what the dimensions of the crate refer to . . therefore for max vol base should be 20 x 20 i . e . of radius 20 / 2 = 10 e" | a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | e | sqrt(divide(volume_cylinder(divide(20, const_2), 20), multiply(const_pi, 20))) | divide(n1,const_2)|multiply(n0,const_pi)|volume_cylinder(#0,n0)|divide(#2,#1)|sqrt(#3)| | geometry | E |
how long will a boy take to run round a square field of side 20 meters , if he runs at the rate of 12 km / hr ? | "speed = 12 km / hr = 12 * 5 / 18 = 10 / 3 m / sec distance = 20 * 4 = 80 m time taken = 80 * 3 / 10 = 24 sec answer is d" | a ) 52 sec , b ) 45 sec , c ) 60 sec , d ) 24 sec , e ) 39 sec | d | divide(multiply(20, const_4), multiply(12, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain | D |
50 boys and 100 girls are examined in a test ; 50 % of the boys and 40 % of the girls pass . the percentage of the total who failed is ? | "total number of students = 50 + 100 = 150 number of students passed = ( 50 % of 50 + 40 % of 100 ) = 25 + 40 = 65 number of failures = 85 * 100 / 150 = 56.7 % answer is c" | a ) 42.5 % , b ) 62.1 % , c ) 56.7 % , d ) 39.9 % , e ) 51.4 % | c | multiply(const_100, divide(subtract(add(50, 100), add(multiply(50, divide(50, const_100)), multiply(100, divide(40, const_100)))), add(50, 100))) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100)| | general | C |
a man can do a piece of work in 4 days , but with the help of his daughter , he can do it in 3 days . in what time can his daughter do it alone ? | explanation : daughter β s 1 day β s work = ( 1 / 3 β 1 / 4 ) = 1 / 12 daughter alone can do the work in 12 / 1 = 12 days answer : a | a ) 12 , b ) 28 , c ) 26 , d ) 19 , e ) 112 | a | divide(multiply(4, 3), subtract(4, 3)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1) | physics | A |
the population of a town increases 22 % and 25 % respectively in two consecutive years . after the growth the present population of the town is 1220 . then what is the population of the town 2 years ago ? | "explanation : formula : ( after = 100 denominator ago = 100 numerator ) 1220 * 100 / 122 * 100 / 125 = 800 answer : option c" | a ) a ) 600 , b ) b ) 700 , c ) c ) 800 , d ) d ) 900 , e ) e ) 1000 | c | divide(1220, multiply(add(const_1, divide(22, const_100)), add(const_1, divide(25, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)| | gain | C |
a circle is inscribed in a triangle of side 6 cm . and a square is inscribed in the circle . what is the area of square ? | the radius of the incircle in a equilateral triangle is a / 2 ( sqrt ) 3 where a = side of triangle = 6 cm so the diameter of the circle would be the diagonal of square so diagonal = 2 * a / 2 ( sqrt ) 3 = a / ( sqrt ) 3 so area of square = 1 / 2 * ( diagonal ) ^ 2 = 6 * 6 / 2 * 3 = 6 cm ^ 2 answer : b | ['a ) 4 cm ^ 2', 'b ) 6 cm ^ 2', 'c ) 5 cm ^ 2', 'd ) 7 cm ^ 2', 'e ) 8 cm ^ 2'] | b | sqrt(multiply(6, 6)) | multiply(n0,n0)|sqrt(#0) | geometry | B |
let the least number of 6 digits , which when divided by 4 , 610 and 15 leaves in each case the same remainder of 2 , be n . the sum of the digits in n is : | solution least number of 6 digital is 100000 . l . c . m . of 4 , 610 and 15 = 60 . on dividing 100000 by 60 , the remainder obtained is 40 . so , least number of 6 digits divisible by 4 , 610 and 15 = 100000 + ( 60 - 40 ) = 100020 . so , n = ( 100020 + 2 ) = 100022 . sum of digits in n = ( 1 + 2 + 2 ) = 5 . answer c | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(add(add(const_1, 2), const_1), const_1) | add(n4,const_1)|add(#0,const_1)|add(#1,const_1) | general | C |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 450 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 450 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 450 x 8 = 3600 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 3700 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 3700 ( answer b )" | a ) 1,108 , b ) 3,700 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 450) | divide(n6,n5)|multiply(n4,#0)| | general | B |
a train passes a station platform in 36 sec and a man standing on the platform in 23 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 23 = 345 m . let the length of the platform be x m . then , ( x + 345 ) / 36 = 15 = > x = 195 m . answer : b" | a ) 288 , b ) 195 , c ) 881 , d ) 1277 , e ) 121 | b | multiply(23, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics | B |
x + ( 1 / x ) = 2.5 find x ^ 2 + ( 1 / x ^ 2 ) | "squaring on both sides ( x + 1 / x ) ^ 2 = 2.5 ^ 2 x ^ 2 + 1 / x ^ 2 = 6.25 - 2 x ^ 2 + 1 / x ^ 2 = 4.25 answer : c" | a ) 2.25 , b ) 3.25 , c ) 4.25 , d ) 5.25 , e ) 6.25 | c | subtract(power(2.5, 2), 2) | power(n1,n2)|subtract(#0,n2)| | general | C |
one night a certain hotel rented 2 / 3 of its rooms , including 2 / 3 of their air conditioned rooms . if 3 / 5 of its rooms were air conditioned , what percent of the rooms that were not rented were air conditioned ? | "the rooms which were not rented is 1 / 3 the ac rooms which were not rented is ( 1 / 3 ) * ( 3 / 5 ) = 1 / 5 the percentage of unrented rooms which were ac rooms is ( 1 / 5 ) / ( 1 / 3 ) = 3 / 5 = 60 % the answer is b ." | a ) 50 % , b ) 60 % , c ) 65 % , d ) 70 % , e ) 75 % | b | multiply(divide(multiply(subtract(const_1, divide(2, 2)), multiply(divide(2, 3), const_100)), subtract(const_100, multiply(divide(2, 3), const_100))), const_100) | divide(n0,n1)|divide(n2,n0)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#2,#3)|subtract(const_100,#2)|divide(#4,#5)|multiply(#6,const_100)| | gain | B |
what will be the difference between simple and compound interest @ 10 % per annum on a sum of $ 1000 after 4 years ? | "s . i . = 1000 * 10 * 4 / 100 = $ 400 c . i . = 1000 * ( 1 + 10 / 100 ) ^ 4 - 1000 = $ 464.10 difference = 464.10 - 400 = $ 64.10 answer is c" | a ) $ 48.12 , b ) $ 52.64 , c ) $ 64.10 , d ) $ 72.85 , e ) $ 60.10 | c | subtract(subtract(multiply(1000, power(add(divide(10, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(10, const_100)), 4)) | divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|multiply(n2,#2)|power(#1,n2)|multiply(n1,#4)|subtract(#5,n1)|subtract(#6,#3)| | gain | C |
a city with a population of 218,000 is to be divided into 10 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? | "the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 218,000 = p + 9 ( 1.1 ) p 10.9 p = 218,000 p = 20,000 the answer is e ." | a ) 19,920 , b ) 19,940 , c ) 19,960 , d ) 19,980 , e ) 20,000 | e | divide(divide(subtract(multiply(const_1000, const_100), subtract(subtract(const_3600, const_100), const_1000)), const_1000), add(multiply(add(const_1, divide(10, const_100)), subtract(10, const_1)), const_1)) | divide(n2,const_100)|multiply(const_100,const_1000)|subtract(const_3600,const_100)|subtract(n1,const_1)|add(#0,const_1)|subtract(#2,const_1000)|multiply(#4,#3)|subtract(#1,#5)|add(#6,const_1)|divide(#7,const_1000)|divide(#9,#8)| | general | E |
what will be the cost of building a fence around a square plot with area equal to 36 sq ft , if the price per foot of building the fence is rs . 58 ? | "let the side of the square plot be a ft . a 2 = 36 = > a = 6 length of the fence = perimeter of the plot = 4 a = 24 ft . cost of building the fence = 24 * 58 = rs . 1392 . answer : e" | a ) 3944 , b ) 2882 , c ) 2999 , d ) 2667 , e ) 1392 | e | multiply(square_perimeter(sqrt(36)), 58) | sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)| | geometry | E |
a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 450 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 450 = > x = 22500 answer : b" | a ) 20029 , b ) 22500 , c ) 20289 , d ) 20027 , e ) 20026 | b | divide(450, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|multiply(#0,#4)|subtract(#6,#5)|divide(n3,#7)| | gain | B |
a couple decides to have 3 children . if they succeed in having 3 children and each child is equally likely to be a boy or a girl , what is the probability that they will have exactly 1 girl and 2 boys ? | sample space = 2 ^ 3 = 8 favourable events = { bgg } , { bgb } , { bbb } , { ggg } , { gbg } , probability = 5 / 8 = 5 / 8 . ans ( c ) . | a ) 3 / 4 , b ) 1 / 2 , c ) 5 / 8 , d ) 1 , e ) 1 / 4 | c | subtract(1, multiply(divide(factorial(3), factorial(2)), power(divide(1, 2), 3))) | divide(n2,n3)|factorial(n0)|factorial(n3)|divide(#1,#2)|power(#0,n0)|multiply(#3,#4)|subtract(n2,#5) | general | C |
the price of lunch for 15 people was $ 211.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ? | "take the initial price before the gratuity is 100 the gratuity is calculated on the final price , so as we assumed the final bill before adding gratuity is 100 so gratuity is 15 % of 100 is 15 so the total price of meals is 115 so the given amount i . e 211 is for 115 then we have to calculate for 100 for 115 211 for 100 x so by cross multiplication we get 115 x = 100 * 211 = > x = 100 * 211 / 110 by simplifying we get x as 191.81 which is the price of lunch before gratuity so the gratuity is 19.19 so as the question ask the average price person excluding gratuity is 191.81 / 15 = 12.78 so our answer is b )" | a ) $ 11.73 , b ) $ 12.78 , c ) $ 13.80 , d ) $ 14.00 , e ) $ 15.87 | b | multiply(multiply(divide(211.00, add(const_100, 15)), const_100), divide(const_1, 15)) | add(n0,const_100)|divide(const_1,n0)|divide(n1,#0)|multiply(#2,const_100)|multiply(#1,#3)| | general | B |
a light flashes every 15 seconds , how many times will it flash in ? of an hour ? | "1 flash = 15 sec for 1 min = 4 flashes so for 1 hour = 4 * 60 = 240 flashes . answer : c" | a ) 550 , b ) 600 , c ) 240 , d ) 700 , e ) 750 | c | divide(const_3600, 15) | divide(const_3600,n0)| | physics | C |
maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 60 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h , what is the distance traveled by brad ? | "time taken = total distance / relative speed total distance = 60 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken = 60 / 10 = 6 hrs distance traveled by brad = brad ' s speed * time taken = 6 * 6 = 36 kms . . . answer - b" | a ) 16 , b ) 36 , c ) 20 , d ) 24 , e ) 30 | b | multiply(4, divide(60, add(4, 6))) | add(n1,n2)|divide(n0,#0)|multiply(n1,#1)| | physics | B |
what could be the range of a set consisting of odd multiples of 11 ? | 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 : 11 * ( 2 n + 1 ) where n is any positive integer . range = 11 * ( 2 m + 1 ) - 11 * ( 2 n + 1 ) = 11 * 2 * ( m - n ) = 22 * ( m - n ) . m , n - any positive integers the answer must be divisible by 22 , which is only 44 . the correct answer is c | a ) 21 , b ) 24 , c ) 44 , d ) 62 , e ) 70 | c | multiply(11, const_4) | multiply(n0,const_4) | general | C |
johnny travels a total of one hour to and from school . on the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 25 miles per hour . how far is it to the school ? | answer : c ) 6.6 miles . average speed for round trip = 2 * a * b / ( a + b ) , where a , b are speeds so , average speed was = 2 * 5 * 25 / ( 5 + 25 ) = 6.6 m / hr the distance between schoolhome should be half of that . ie . 6.6 miles answer c | a ) 2 miles , b ) 4 miles , c ) 6.6 miles , d ) 8 miles , e ) 10 miles | c | multiply(divide(const_1, add(divide(const_1, 5), divide(const_1, 25))), const_1_6) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,const_1_6) | physics | C |
the ratio of 3 numbers is 5 : 3 : 4 and their sum is 108 . the second number of the 3 numbers is ? | 5 : 3 : 4 total parts = 12 12 parts - - > 108 1 part - - - - > 9 the second number of the three numbers is = 3 3 parts - - - - > 27 b ) | a ) 23 , b ) 27 , c ) 30 , d ) 32 , e ) 34 | b | multiply(divide(3, add(add(5, 3), 4)), 108) | add(n0,n1)|add(n3,#0)|divide(n0,#1)|multiply(n4,#2) | other | B |
if it is assumed that 70 percent of those who receive a questionnaire by mail will respond and 300 responses are needed , what is the minimum number of questionnaires that should be mailed ? | "minimum no of mail to be sent for getting 300 responses at 70 % = 300 / 0.7 = 428.5 option b" | a ) 400 , b ) 428.5 , c ) 480 , d ) 500 , e ) 600 | b | divide(300, divide(70, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain | B |
the ratio of the length and the width of a rectangle is 4 : 3 and the area of the rectangle is 6348 sq cm . what is the ratio of the width and the area of the rectangle ? | "let the length and the width be 4 x and 3 x respectively . area = ( 4 x ) ( 3 x ) = 6348 12 x ^ 2 = 6348 x ^ 2 = 529 x = 23 the ratio of the width and the area is 3 x : 12 x ^ 2 = 1 : 4 x = 1 : 92 the answer is d ." | a ) 1 : 96 , b ) 1 : 99 , c ) 1 : 94 , d ) 1 : 92 , e ) 1 : 91 | d | divide(divide(sqrt(multiply(3, 6348)), const_2), 6348) | multiply(n2,n1)|sqrt(#0)|divide(#1,const_2)|divide(#2,n2)| | geometry | D |
at present the ratio between the ages of arun and deepak is 5 : 7 . after 6 years , arun ' s age will be 36 years . what is the age of deepak at present ? | "let the present ages of arun and deepak be 5 x years and 7 x years respectively 5 x + 6 = 36 5 x = 30 x = 6 deepak ' s age = 7 x = 42 years answer is c" | a ) 35 years , b ) 25 years , c ) 42 years , d ) 39 years , e ) 40 years | c | divide(multiply(subtract(36, 6), 7), 5) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other | C |
there are 361 doctors and nurses in a hospital . if the ratio of the doctors to the nurses is 8 : 11 , then how many nurses are there in the hospital ? | "given , the ratio of the doctors to the nurses is 8 : 11 number of nurses = 11 / 19 x 361 = 209 answer : a" | a ) 152 , b ) 209 , c ) 57 , d ) 171 , e ) 181 | a | multiply(multiply(8, subtract(11, 8)), 11) | subtract(n2,n1)|multiply(n1,#0)|multiply(n2,#1)| | other | A |
a cyclist traveled for two days . on the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of 240 miles and spent a total of 12 hours traveling , what was her average speed on the second day ? | solution : d = 280 mi t = 12 hrs Δ Γ’ y 1 time = t 1 d Γ’ y 2 time = t 2 t 2 - t 1 = 4 hrs - - - - - ( i ) t 1 + t 2 = 12 hrs - - - - - ( ii ) adding i and ii , t 2 = 8 hrs and t 1 = 4 hrs d Γ y 1 rate = r 1 d Γ’ y 2 rate = r 2 r 1 - r 2 = 10 mph Γ . αΊΉ . r 1 = 10 + r 2 280 = 8 r 2 + 4 r 1 Γ . αΊΉ . 280 = 8 r 2 + 4 ( 10 + r 2 ) Γ . αΊΉ . r 2 = 20 mph answer : b | a ) 5 mph , b ) 10 mph , c ) 20 mph , d ) 30 mph , e ) 40 mph | b | divide(subtract(240, multiply(divide(subtract(12, 4), const_2), 10)), add(divide(subtract(12, 4), const_2), add(divide(subtract(12, 4), const_2), 4))) | subtract(n3,n0)|divide(#0,const_2)|add(n0,#1)|multiply(n1,#1)|add(#2,#1)|subtract(n2,#3)|divide(#5,#4) | physics | B |
36 welders work at a constant rate they complete an order in 3 days . if after the first day , 12 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | "1 . we need to find out the time taken by 24 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 36 * 3 = 108 days 4 . but on day 1 thirty - six workers had already worked finishing 1 / 3 of the job . so 24 workers have to finish only 2 / 3 of the job . 5 . total time taken by 24 workers can be got from formula used at ( 2 ) . i . e . , 24 * total time taken = 108 . total time taken by 6 workers to finish the complete job is 108 / 24 = 4.5 days . 6 . time taken by 24 workers to finish 2 / 3 of the job is 2 / 3 * 4.5 = 3 days . the answer is choice a" | a ) 3 , b ) 2 , c ) 8 , d ) 4 , e ) 6 | a | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(36, 3)), 36)), subtract(36, 12)), divide(const_1, multiply(36, 3))) | multiply(n0,n1)|subtract(n0,n2)|divide(const_1,#0)|multiply(n0,#2)|subtract(const_1,#3)|divide(#4,#1)|divide(#5,#2)| | physics | A |
a computer manufacturer produces a certain electronic component at a cost of $ 150 per component . shipping costs for delivering the components are $ 1 per unit . further , the manufacturer has costs of $ 19,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 250 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | "$ 19500 is a fixed cost each component is $ 151 ( $ 150 to produce , $ 1 to ship ) manufacturer will be producing and selling 250 components so therefore the equation to find price would be 250 * p = 19500 + ( 250 * 150 ) + ( 250 * 1 ) p = ( 19500 + 37500 + 250 ) / 250 p = 229 answer : e" | a ) 199 , b ) 249 , c ) 233 , d ) 215 , e ) 229 | e | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(1, const_100)), multiply(add(150, 1), 250)), 250) | add(n0,n1)|multiply(const_4,const_4)|multiply(n1,const_100)|multiply(#1,const_1000)|multiply(n3,#0)|add(#3,#2)|add(#5,#4)|divide(#6,n3)| | general | E |
a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 600 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 1800 m t = d / s t = 1800 / 10 t = 180 sec answer : b" | a ) 266 sec , b ) 180 sec , c ) 776 sec , d ) 166 sec , e ) 997 sec | b | divide(add(1200, 600), divide(1200, 120)) | add(n0,n2)|divide(n0,n1)|divide(#0,#1)| | physics | B |
the average of 6 observations is 15 . a new observation is included and the new average is decreased by 1 . the seventh observation is ? | "let seventh observation = x . then , according to the question we have = > ( 90 + x ) / 7 = 14 = > x = 8 hence , the seventh observation is 8 answer : b" | a ) 1 , b ) 8 , c ) 5 , d ) 6 , e ) 7 | b | subtract(multiply(subtract(15, 1), add(6, 1)), multiply(15, 6)) | add(n0,n2)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#2)|subtract(#3,#1)| | general | B |
when positive integer n is divided by 2 , the remainder is 1 . when n is divided by 7 , the remainder is 5 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 10 ? | "when positive integer n is divided by 2 , the remainder is 1 i . e . , n = 2 x + 1 values of n can be one of { 1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 , 17,19 , . . . . . . . . . . . . . . 31 , 33,35 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 7 , the remainder is 5 . . i . e . , n = 7 y + 5 values of n can be one of { 5 , 12 , 19 , 26 , 33 , 40 , 47 , 54 , 61 . . . . . . . . } combining both the sets we get n = { 19 , 52 , . . . . . . . . . . . } what is the smallest positive integer p , such that ( n + p ) is a multiple of 21 or 21 x in case of n = 5 p = 5 so for min value of p , we take min value of n . c is the answer ." | a ) 1 , b ) 2 , c ) 5 , d ) 19 , e ) 20 | c | subtract(10, reminder(5, 7)) | reminder(n3,n2)|subtract(n4,#0)| | general | C |
i bought two books ; for rs . 480 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | x * ( 85 / 100 ) = ( 480 - x ) 119 / 100 x = 280 answer : b | a ) rs . 28 , b ) rs . 280 , c ) rs . 140 , d ) rs . 70 , e ) rs . 80 | b | divide(multiply(480, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | add(n2,const_100)|subtract(const_100,n1)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2) | gain | B |
what percentage of numbers from 1 to 60 have 1 or 9 in the unit ' s digit ? | "clearly , the numbers which have 1 or 9 in the unit ' s digit , have squares that end in the digit 1 . such numbers from 1 to 60 are 1 , 9 , 11 , 19 , 21 , 29 , 31 , 39 , 41 , 49 , 51 , 59 number of such number = 11 answer : c" | a ) 12 , b ) 14.11 , c ) 18.33 , d ) 20 , e ) 21 | c | multiply(divide(multiply(const_2, divide(60, const_10)), 60), const_100) | divide(n1,const_10)|multiply(#0,const_2)|divide(#1,n1)|multiply(#2,const_100)| | gain | C |
sheela deposits rs . 4500 in bank savings account . if this is 28 % of her monthly income . what is her monthly income in ? | "explanation : 28 % of income = rs . 4500 100 % of income = 4500 x 100 / 28 = rs . 16071 answer : c" | a ) 22000 , b ) 27000 , c ) 16071 , d ) 16789 , e ) none of these | c | divide(multiply(4500, const_100), 28) | multiply(n0,const_100)|divide(#0,n1)| | gain | C |
lionel left his house and walked towards walt ' s house , 48 miles away . two hours later , walt left his house and ran towards lionel ' s house . if lionel ' s speed was 3 miles per hour and walt ' s 3 miles per hour , how many miles had lionel walked when he met walt ? | "in the first 2 hours lionel at the rate of 3 miles per hour covered distance = rate * time = 3 * 2 = 6 miles . so , the distance between him and walt was 48 - 6 = 42 miles when walt left his house . now , their combined rate to cover this distance was 3 + 3 = 6 miles per hour , hence they will meet ( they will cover that distance ) in time = distance / rate = 42 / 6 = 7 hours . total time that lionel was walking is 2 + 7 = 9 hours , which means that he covered in that time interval distance = rate * time = 3 * 9 = 27 miles . answer : d ." | a ) 12 , b ) 16 , c ) 20 , d ) 27 , e ) 28 | d | multiply(3, add(divide(subtract(48, multiply(3, 3)), add(3, const_2.0)), 3)) | add(n1,n2)|multiply(n1,n1)|subtract(n0,#1)|divide(#2,#0)|add(#3,const_2)|multiply(n1,#4)| | physics | D |
following an increase in prices , the price of a candy box was 10 pounds and the price of a can of soda was 15 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ? | price of candy before price increase = 10 / 1.25 = 8 price of soda before price increase = 15 / 1.5 = 10 total price = 8 + 10 = 18 d is the answer | a ) 11 . , b ) 12 . , c ) 13 . , d ) 18 . , e ) 14.5 | d | add(divide(multiply(10, const_100), add(const_100, 25)), divide(multiply(15, const_100), add(const_100, 50))) | add(n2,const_100)|add(n3,const_100)|multiply(n0,const_100)|multiply(n1,const_100)|divide(#2,#0)|divide(#3,#1)|add(#4,#5) | general | D |
solve : 12.05 * 5.4 + 0.6 | = 12.05 * ( 5.4 / 0.6 ) = ( 12.05 * 9 ) = 108.45 answer is a . | a ) 108.45 , b ) 110.45 , c ) 106.45 , d ) 109.45 , e ) none of them | a | multiply(12.05, divide(5.4, 0.6)) | divide(n1,n2)|multiply(n0,#0) | general | A |
the standard deviation of a normal distribution of data is 2 , and 3 standard deviations below the mean is greater than 47 . what is a possible value for the mean of the distribution ? | the standard deviation ( { sd } ) = 2 ; 3 standard deviations below the mean is greater than 47 : { mean } - 3 * { sd } > 47 ; { mean } - 6 > 47 ; { mean } > 53 . answer : b . | a ) 46 , b ) 54 , c ) 48 , d ) 49 , e ) 50 | b | add(add(47, multiply(3, 2)), const_1) | multiply(n0,n1)|add(n2,#0)|add(#1,const_1) | general | B |
if 5 < x < 12 and y = x + 3 , what is the greatest possible integer value of x + y ? | "x + y = x + x + 3 = 2 x + 3 we need to maximize this value and it needs to be an integer . 2 x is an integer when the decimal of x is . 0 or . 5 the largest such value is 11.5 then x + y = 11.5 + 14.5 = 26 . the answer is d ." | a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | d | add(add(3, const_10), const_10) | add(n2,const_10)|add(#0,const_10)| | general | D |
a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability c that out of the two , one is white and one is blue ? | "the required probability c = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer ." | a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14 | d | divide(multiply(3, 5), divide(multiply(add(3, 5), add(5, const_2)), const_2)) | add(n0,n1)|add(n1,const_2)|multiply(n0,n1)|multiply(#0,#1)|divide(#3,const_2)|divide(#2,#4)| | probability | D |
the average salary of 16 people in the shipping department at a certain firm is $ 20,000 . the salary of 5 of the employees is $ 25,000 each and the salary of 4 of the employees is $ 16,000 each . what is the average salary of the remaining employees ? | total salary . . . 16 * 20 k = 320 k 5 emp @ 25 k = 125 k 4 emp @ 16 k = 64 k remaing 7 emp sal = 320 k - 125 k - 64 k = 131 k average = 131 k / 7 = 18714 ans : a | a ) $ 18,714 , b ) $ 18,500 , c ) $ 18,000 , d ) $ 15,850 , e ) $ 12,300 | a | divide(subtract(subtract(multiply(multiply(5, 4), multiply(const_4, const_4)), multiply(multiply(5, 5), 5)), multiply(4, 16)), add(const_2, 5)) | add(n2,const_2)|multiply(n2,n4)|multiply(const_4,const_4)|multiply(n2,n2)|multiply(n0,n4)|multiply(#1,#2)|multiply(n2,#3)|subtract(#5,#6)|subtract(#7,#4)|divide(#8,#0) | general | A |
one night 40 percent of the female officers on a police force were on duty . if 240 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 240 number of female officers on duty on that night = 240 / 2 = 120 ( 40 / 100 ) * f = 120 = > f = 300 answer b" | a ) 90 , b ) 300 , c ) 270 , d ) 500 , e ) 1,000 | b | divide(divide(240, const_2), divide(40, const_100)) | divide(n1,const_2)|divide(n0,const_100)|divide(#0,#1)| | gain | B |
the area of one square is x ^ 2 + 6 x + 9 and the area of another square is 4 x ^ 2 β 20 x + 25 . if the sum of the perimeters of both squares is 52 , what is the value of x ? | "the areas are ( x + 3 ) ^ 2 and ( 2 x - 5 ) ^ 2 . the lengths of the sides are x + 3 and 2 x - 5 . if we add the two perimeters : 4 ( x + 3 ) + 4 ( 2 x - 5 ) = 52 12 x = 60 x = 5 the answer is c ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(subtract(52, subtract(multiply(const_4, divide(6, 2)), 20)), 6) | divide(n5,n0)|multiply(#0,n3)|subtract(#1,n5)|subtract(n7,#2)|divide(#3,n1)| | general | C |
cereal a is 10 % sugar by weight , whereas healthier but less delicious cereal b is 3 % sugar by weight . to make a delicious and healthy mixture that is 5 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "ratio of a / ratio of b = ( average wt of mixture - wt of b ) / ( wt of a - average wt of mixture ) = > ratio of a / ratio of b = ( 5 - 3 ) / ( 10 - 5 ) = 2 / 5 so they should be mixed in the ratio 2 : 5 answer - a" | a ) 2 : 5 , b ) 2 : 7 , c ) 1 : 6 , d ) 1 : 4 , e ) 1 : 3 | a | divide(subtract(5, 3), subtract(10, 5)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general | A |
xy = 2 then what is ( 2 ^ ( x + y ) ^ 2 ) / ( 2 ^ ( x - y ) ^ 2 ) | ( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 8 2 ^ 8 = 256 answer e | a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 256 | e | power(2, multiply(const_4, 2)) | multiply(n0,const_4)|power(n0,#0) | general | E |
due to construction , the speed limit along an 10 - mile section of highway is reduced from 55 miles per hour to 30 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 10 miles stretch = 10 * 60 / 55 = 10 * 12 / 11 = 10.9 new time in minutes to cross 10 miles stretch = 10 * 60 / 30 = 10 * 2 = 20 time difference = 9.1 ans : d" | a ) a ) 6.24 , b ) b ) 8 , c ) c ) 10 , d ) d ) 9.1 , e ) e ) 24 | d | max(multiply(subtract(add(55, 10), const_1), subtract(divide(10, 30), divide(10, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics | D |
it takes joey the postman 1 hours to run a 2 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 5 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 2 miles an hour let the other half be x miles an hour now , avg speed = 5 mile an hour 2 * 2 * x / 2 + x = 5 4 x = 5 x + 10 = > x = 10 a" | a ) 10 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | divide(2, subtract(divide(multiply(const_2, 2), 5), 1)) | multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2)| | physics | A |
a can do a piece of work in 10 days and b can do it in 15 days and c can do it 20 days . they started the work together and a leaves after 2 days and b leaves after 4 days from the beginning . how long will work lost ? | "b 10 2 / 3 2 / 10 + 4 / 15 + x / 20 = 1 x = 32 / 3 = 10 2 / 3" | a ) 09 2 / 3 , b ) 10 2 / 3 , c ) 11 2 / 3 , d ) 12 2 / 3 , e ) 16 2 / 3 | b | add(divide(subtract(const_1, add(multiply(subtract(4, 2), add(inverse(15), inverse(20))), multiply(add(inverse(20), add(inverse(10), inverse(15))), 2))), inverse(15)), 4) | inverse(n1)|inverse(n2)|inverse(n0)|subtract(n4,n3)|add(#0,#1)|add(#2,#0)|add(#5,#1)|multiply(#4,#3)|multiply(n3,#6)|add(#7,#8)|subtract(const_1,#9)|divide(#10,#0)|add(n4,#11)| | physics | B |
a train 2500 m long can cross an electric pole in 50 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 2500 / 50 s = 50 m / sec speed = 50 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 180 kmph answer : b" | a ) 178 , b ) 180 , c ) 182 , d ) 184 , e ) 186 | b | divide(divide(2500, const_1000), divide(50, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics | B |
if 5 men undertook a piece of construction work and finished half the job in 15 days . if two men drop out , then the job will be completed in | that is , half the work done = 5 * 15 * 1 / 2 then , 5 * 15 * 1 / 2 = 3 * ? * 1 / 2 i . e . 5 * 15 = 3 * ? = 5 * 15 / 3 = 25 day d | a ) 5 days , b ) 15 days , c ) 10 days , d ) 25 days , e ) 20 days | d | divide(multiply(15, 5), subtract(5, const_2)) | multiply(n0,n1)|subtract(n0,const_2)|divide(#0,#1) | physics | D |
what is the smallest integer that is multiple of 8,7 and 20 | correct answer : e it is the lcm of 8,7 and 20 which is 280 | a ) 141 , b ) 180 , c ) 130 , d ) 122 , e ) 280 | e | lcm(lcm(multiply(const_2, const_4), add(const_3, const_4)), 20) | add(const_3,const_4)|multiply(const_2,const_4)|lcm(#0,#1)|lcm(n1,#2) | general | E |
how many figures are required to number the pages the pages of a book containing 1000 pages ? | 1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 999 = 900 * 3 = 2700 1000 = 4 - - - - - - - - - - - 2893 answer : d | a ) 2793 , b ) 2333 , c ) 2993 , d ) 2893 , e ) 2693 | d | add(add(subtract(divide(divide(1000, const_10), const_10), const_1), subtract(subtract(divide(1000, const_10), const_1), subtract(divide(divide(1000, const_10), const_10), const_1))), multiply(subtract(subtract(1000, const_1), subtract(divide(1000, const_10), const_1)), const_3)) | divide(n0,const_10)|subtract(n0,const_1)|divide(#0,const_10)|subtract(#0,const_1)|subtract(#2,const_1)|subtract(#1,#3)|multiply(#5,const_3)|subtract(#3,#4)|add(#4,#7)|add(#8,#6) | general | D |
a powder tin has a square base with side 8 cm and height 14 cm . another tin has a circular base with diameter 8 cm and height 14 cm . the difference in their capacities is : | sol . difference in capacities = [ 8 * 8 * 14 - 22 / 7 * 4 * 4 * 14 ] cm Β³ = 192 cm Β³ answer b | ['a ) 168 cm Β³', 'b ) 192 cm Β³', 'c ) 228 cm Β³', 'd ) 236 cm Β³', 'e ) none'] | b | subtract(volume_rectangular_prism(8, 8, 14), volume_cylinder(divide(8, const_2), 14)) | divide(n0,const_2)|volume_rectangular_prism(n0,n0,n1)|volume_cylinder(#0,n1)|subtract(#1,#2) | geometry | B |
a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 10 days before the project is completed , in how many days will the project be completed ? | "answer a can complete a project in 20 days . so , a will complete 1 / 20 th of the project in a day . b can complete a project in 30 days . so , b will complete 1 / 30 th of the project in a day . let the total number of days taken to complete the project be x days . b worked for all x days . however , a worked for ( x - 10 ) days because a quits 10 days before the project is completed . in a day , a completes 1 / 20 th of the project . therefore , a would have completed x β 10 / 20 th of the project in ( x - 10 ) days . in a day , b completes 130130 th of the project . therefore , b would have completed x / 30 h of the project in x days . β΄ x β 10 / 20 + x / 30 = 1 or x = 18 . choice a" | a ) 18 days , b ) 27 days , c ) 26.67 days , d ) 16 days , e ) 12 days | a | add(divide(subtract(const_1, multiply(divide(const_1, 30), 10)), add(divide(const_1, 20), divide(const_1, 30))), 10) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)| | physics | A |
a waiter ' s salary consists of his salary and tips . during one week his tips were 3 / 4 of his salary . what fraction of his income came from tips ? | "income = salary ( s ) + tips = s + s * 3 / 4 = s * 7 / 4 tips = s * 3 / 4 fraction of his income came from tips = ( s * 3 / 4 ) / ( s * 7 / 4 ) = 3 / 7 answer : a" | a ) 3 / 7 , b ) 5 / 4 , c ) 5 / 8 , d ) 5 / 9 , e ) 6 / 9 | a | divide(divide(3, 4), add(divide(3, 4), const_1)) | divide(n0,n1)|add(#0,const_1)|divide(#0,#1)| | general | A |
how many positive integers less than 600 can be formed using the numbers 1 , 2 , 3 and 5 for the digits ? | "notice that we can find the number of 2 and 3 digit numbers by just assuming the first digit can also be zero : 0 1 1 1 2 2 2 3 3 3 5 5 5 5 5 number of possibilities = 5 * 4 * 4 = 80 . then , just add up the number of 1 digits numbers = 4 , so total is 80 + 4 = 84 . answer : e" | a ) 48 , b ) 52 , c ) 66 , d ) 68 , e ) 84 | e | divide(factorial(subtract(add(const_4, 1), const_1)), multiply(factorial(1), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | probability | E |
3 * 11 + 3 * 12 + 3 * 15 + 11 = ? | "we have : 3 * 11 = 33 3 * 12 = 36 3 * 15 = 45 all of this is : ( 33 + 36 + 45 ) + 11 = 114 + 11 = 125 correct answer a" | a ) 125 , b ) 126 , c ) 130 , d ) 148 , e ) 151 | a | add(add(add(multiply(3, 11), multiply(3, 12)), multiply(3, 15)), 11) | multiply(n0,n1)|multiply(n0,n3)|multiply(n0,n5)|add(#0,#1)|add(#3,#2)|add(n6,#4)| | general | A |
a certain number when divided by 50 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 50 + 25 = 75 / 15 = 5 ( remainder ) b" | a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | b | reminder(25, 15) | reminder(n1,n2)| | general | B |
5 drainage pipes , each draining water from a pool at the same constant rate , together can drain a certain pool in 16 days . how many additional pipes , each draining water at the same constant rate , will be needed to drain the pool in 4 days ? | this is an inverse proportional problem . . . . . . 5 pipes in 16 days ; so for 4 days , it will be = 16 x 5 / 4 = 20 so , 20 - 5 = 15 answer e | a ) 6 , b ) 9 , c ) 10 , d ) 12 , e ) 15 | e | subtract(multiply(4, 5), 5) | multiply(n0,n2)|subtract(#0,n0) | general | E |
from a pack of 52 cards , two cards are drawn at random together at random what is the probability of both the cards being kings ? | "let s be the sample space then n ( s ) = 52 c 2 = ( 52 * 51 ) / ( 2 * 1 ) = 1326 let e be the event of getting 2 kings out of 4 n ( e ) = 4 c 2 = ( 4 * 3 ) / ( 2 * 1 ) = 6 p ( e ) = n ( e ) / n ( s ) = 6 / 1326 = 1 / 221 answer ( a )" | a ) 1 / 221 , b ) 8 / 221 , c ) 4 / 589 , d ) 4 / 587 , e ) 7 / 654 | a | divide(multiply(divide(52, const_4), divide(52, const_4)), choose(52, const_2)) | choose(n0,const_2)|divide(n0,const_4)|multiply(#1,#1)|divide(#2,#0)| | probability | A |
a rectangular block 8 cm by 24 cm by 56 cm is cut into an exact number of equal cubes . find the least possible number of cubes ? | "volume of the block = 8 * 24 * 56 = 10752 cm ^ 3 side of the largest cube = h . c . f of 8 , 24,56 = 8 cm volume of the cube = 8 * 8 * 8 = 512 cm ^ 3 number of cubes = 10752 / 512 = 21 answer is e" | a ) 6 , b ) 10 , c ) 15 , d ) 40 , e ) 21 | e | divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(56, const_10))), divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(56, const_10))), const_10)) | multiply(const_1000,const_3)|multiply(n2,const_10)|multiply(const_2,const_3)|rectangle_area(const_1000,const_360)|add(#0,#1)|divide(#3,const_10)|multiply(#2,const_1000)|subtract(#5,#6)|add(#4,#7)|divide(#8,const_10)|divide(#8,#9)| | geometry | E |
the speed at which a man can row a boat in still water is 6 km / hr . if he rows downstream , where the speed of current is 3 km / hr , how many seconds will he take to cover 110 meters ? | "the speed of the boat downstream = 6 + 3 = 9 km / hr 9 km / hr * 5 / 18 = 2.5 m / s the time taken to cover 110 meters = 110 / 2.5 = 44 seconds . the answer is b ." | a ) 40 , b ) 44 , c ) 48 , d ) 52 , e ) 56 | b | divide(110, multiply(add(6, 3), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics | B |
1240 men have provisions for 12 days . if 300 more men join them , for how many days will the provisions last now ? | "1240 * 12 = 1540 * x x = 9.7 answer : c" | a ) 12.9 , b ) 12.0 , c ) 9.7 , d ) 8.6 , e ) 12.1 | c | divide(multiply(12, 1240), add(1240, 300)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics | C |
the measurement of a rectangular box with lid is 25 cmx 24 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of Ο cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) | "d = 24 , r = 12 ; volume of the largest sphere = 4 / 3 Ο r 3 = 4 / 3 * Ο * 12 * 12 * 12 = 2304 Ο cm 3 answer : c" | a ) 288 , b ) 2302 , c ) 2304 , d ) 8640 , e ) 964 | c | multiply(divide(const_4, 3), power(3, 3)) | divide(const_4,n3)|power(n3,n3)|multiply(#0,#1)| | geometry | C |
a welder received an order to make a 1 million liter cube - shaped tank . if he has only 4 x 2 meter sheets of metal that can be cut , how many metal sheets will be required for this order r ? ( 1 cubic meter = 1000 liters ) | i get 75 . a cube with 1 million liters cube would be a cube with the dimensions of 100 * 100 * 100 . 4 * 2 covers 8 sq liters so 100 / 8 = 12.5 . r = 12.5 * 6 = 75 . e | a ) 92 , b ) 90 , c ) 82 , d ) 78 , e ) 75 | e | divide(multiply(power(power(1000, divide(const_1, const_3)), const_2), add(const_4, const_2)), multiply(const_4, const_2)) | add(const_2,const_4)|divide(const_1,const_3)|multiply(const_2,const_4)|power(n4,#1)|power(#3,const_2)|multiply(#0,#4)|divide(#5,#2) | general | E |
56 : 65 : : 68 : ? | "ans 86 reverse of 68 answer : c" | a ) 49 , b ) 84 , c ) 86 , d ) 64 , e ) 56 | c | multiply(68, divide(56, 65)) | divide(n0,n1)|multiply(n2,#0)| | general | C |
in a certain supermarket , a triangular display of cans is arranged in 10 rows , numbered 1 through 10 from top to bottom . each successively numbered row contains 3 more cans than the row immediately above it . if there are fewer than 150 cans in the entire display , how many cans are in the sixth row ? | "let x be the number of cans in row 1 . the total number of cans is x + ( x + 3 ) + . . . + ( x + 27 ) = 10 x + 3 ( 1 + 2 + . . . + 9 ) = 10 x + 3 ( 9 ) ( 10 ) / 2 = 10 x + 135 since the total is less than 150 , x must equal 1 . the number of cans in the 6 th row is 1 + 3 ( 5 ) = 16 the answer is b ." | a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | b | add(multiply(add(3, 3), 3), floor(divide(subtract(150, multiply(divide(multiply(3, subtract(10, 1)), const_2), 10)), 10))) | add(n3,n3)|subtract(n0,n1)|multiply(n3,#1)|multiply(n3,#0)|divide(#2,const_2)|multiply(n0,#4)|subtract(n4,#5)|divide(#6,n0)|floor(#7)|add(#8,#3)| | general | B |
the average monthly salary of 8 workers and one supervisor in a factory was 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . whenthesupervisor , whosesalarywas 430 . when the supervisor , whose salary was 870 per month , retired , a new person was appointed and then the average salary of 9 people was $ $ 410 per month . the salary of the new supervisor is : | "explanation : total salary of 8 workers and supervisor together = 9 Γ£ β 430 = 3870 now total salary of 8 workers = 3870 Γ’ Λ β 870 = 3000 total salary of 9 workers including the new supervisor = 9 Γ£ β 410 = 3690 salary of the new supervisor = 3690 Γ’ Λ β 3000 = 690 answer : b" | a ) 233 , b ) 690 , c ) 287 , d ) 771 , e ) 191 | b | subtract(multiply(410, 9), subtract(multiply(430, 9), 870)) | multiply(n6,n7)|multiply(n1,n6)|subtract(#1,n5)|subtract(#0,#2)| | general | B |
jane started baby - sitting when she was 20 years old . whenever she baby - sat for a child , that child was no more than half her age at the time . jane is currently 32 years old , and she stopped baby - sitting 10 years ago . what is the current age of the oldest person for whom jane could have baby - sat ? | check two extreme cases : jane = 20 , child = 10 , years ago = 32 - 20 = 12 - - > child ' s age now = 10 + 12 = 22 ; jane = 22 , child = 11 , years ago = 32 - 22 = 10 - - > child ' s age now = 11 + 10 = 21 . answer : c . | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | add(divide(20, const_2), subtract(32, 20)) | divide(n0,const_2)|subtract(n1,n0)|add(#0,#1) | general | C |
having received his weekly allowance , a student spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.40 at the candy store . what is this student β s weekly allowance ? | "let x be the value of the weekly allowance . ( 2 / 3 ) ( 2 / 5 ) x = 40 cents ( 4 / 15 ) x = 40 x = $ 1.50 the answer is b ." | a ) $ 1.20 , b ) $ 1.50 , c ) $ 1.80 , d ) $ 2.00 , e ) $ 2.50 | b | divide(multiply(multiply(3, 5), 0.40), const_4) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_4)| | general | B |
if the cost price of 17 articles is equal to the selling price of 16 articles , what is the percentage of profit or loss that the merchant makes ? | "explanation : let cost price of 1 article be re . 1 . therefore , cost price of 17 articles = rs . 17 . selling price of 16 articles = rs . 17 therefore , selling price of 17 articles is : - = > 17 / 16 Γ£ β 17 = > 18.06 therefore , profit = selling price - cost price . = > 18.06 Γ’ Λ β 17 = 1.06 hence , the percentage of profit = profit x 100 / c . p . = > 1.06 / 17 Γ£ β 100 . = > 6.25 % profit . answer : b" | a ) 20 % loss , b ) 6.25 % profit , c ) 33.33 % loss , d ) 30.33 % loss , e ) none of these | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 16), 17)), divide(multiply(const_100, 16), 17))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain | B |
if albert β s monthly earnings rise by 36 % , he would earn $ 495 . if , instead , his earnings rise by only 25 % , how much ( in $ ) would he earn this month ? | "= 495 / 1.36 β 1.25 = 454 = 454 answer is c" | a ) 643 , b ) 652 , c ) 454 , d ) 460 , e ) 490 | c | multiply(divide(495, add(const_1, divide(36, const_100))), add(const_1, divide(25, const_100))) | divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n1,#3)|multiply(#2,#4)| | gain | C |
a batsman in his 17 th innings makes a score of 85 and their by increasing his average by 3 . what is his average after the 17 th innings ? | "explanation : 16 x + 85 = 17 ( x + 3 ) x = 34 + 3 = 37 answer : d" | a ) 28 , b ) 27 , c ) 12 , d ) 37 , e ) 01 | d | add(subtract(85, multiply(17, 3)), 3) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general | D |
a swimming pool 9 m wide and 12 m long is 1 m deep on the shallow side and 4 m deep on the deeper side . its volume is ? | "volume will be length * breadth * height , but in this case two heights are given so we will take average , volume = ( 12 β 9 β ( 1 + 42 ) ) m 312 β 9 β 2.5 m 3 = 270 m 3 answer : c" | a ) 870 m 3 , b ) 277 m 3 , c ) 270 m 3 , d ) 220 m 3 , e ) 170 m 3 | c | multiply(quadrilateral_area(12, 1, 4), 9) | quadrilateral_area(n1,n2,n3)|multiply(n0,#0)| | physics | C |
in Ξ΄ pqs above , if pq = 7 and ps = 8 , then | "there are two ways to calculate area of pqs . area remains same , so both are equal . 7 * 8 / 2 = pr * 9 / 2 pr = 56 / 9 c" | a ) 9 / 4 , b ) 12 / 5 , c ) 56 / 9 , d ) 15 / 4 , e ) 20 / 3 | c | divide(8, 7) | divide(n1,n0)| | general | C |
at an elementary school , 70 % of the faculty members are women and 60 % of the faculty members are married . if 2 β 3 of the men are single , what fraction of the men are married ? | "- - - - - - - - - - - - - - - - - - - - m - - - - - - w - - - - - - - - total marrried - - - - - - - - - - 10 - - - - - 50 - - - - - - - - - 60 not married - - - - - 20 - - - - - 20 - - - - - - - - - 40 total - - - - - - - - - - - - - 30 - - - - - 70 - - - - - - - - 100 need married man / total man , so 10 / 30 = 1 / 3 c" | a ) 5 β 7 , b ) 7 β 10 , c ) 1 β 3 , d ) 7 β 30 , e ) 5 β 70 | c | subtract(const_1, divide(2, 3)) | divide(n2,n3)|subtract(const_1,#0)| | gain | C |
the effective annual rate of interest corresponding to a nominal rate of 10 % per annum payable half - yearly is ? | "amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 5 / 100 ) 2 ] = rs . 110.25 effective rate = ( 110.25 - 100 ) = 10.25 % answer : b" | a ) 10.06 % , b ) 10.25 % , c ) 10.35 % , d ) 16.09 % , e ) 16.19 % | b | add(add(divide(10, const_2), divide(10, const_2)), divide(multiply(divide(10, const_2), divide(10, const_2)), const_100)) | divide(n0,const_2)|add(#0,#0)|multiply(#0,#0)|divide(#2,const_100)|add(#1,#3)| | gain | B |
what is the value of β 16 % ? | "explanation : br > β 16 % = β 14 / β 100 = 4 / 10 = 40 / 100 = 40 % correct answer is c ) 40 %" | a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | c | circle_area(divide(16, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | gain | C |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.