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 |
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the school soccer team looks at their record and finds that they win , lose , and tie games in a ratio of 4 : 3 : 1 . how many games did they play if they lost 9 games ? | the ratio is 4 wins : 3 losses : 1 tie . think of ratio as ` ` parts . ' ' divide 9 ( total losses ) by 3 ( ratio losses ) to find 1 ` ` part ' ' of the ratio . 9 / 3 = 3 this means the team tied 3 games . multiply 3 ( 1 ` ` part ' ' of ratio ) by 4 ( ratio wins ) to find total wins . 3 * 4 = 12 . this mean the team won 12 games . add up wins , losses , and ties . 12 + 9 + 3 = 24 the answer is e | a ) 3 , b ) 9 , c ) 12 , d ) 18 , e ) 24 | e | multiply(divide(9, 3), add(add(4, 3), 1)) | add(n0,n1)|divide(n3,n1)|add(n2,#0)|multiply(#2,#1) | other | E |
two boats are heading towards each other at constant speeds of 5 miles / hr and 21 miles / hr respectively . they begin at a distance 20 miles from each other . how far are they ( in miles ) one minute before they collide ? | the question asks : how far apart will they be 1 minute = 1 / 60 hours before they collide ? since the combined rate of the boats is 5 + 21 = 26 mph then 1 / 60 hours before they collide they ' ll be rate * time = distance - - > 26 * 1 / 60 = 13 / 30 miles apart . answer : d . | a ) 1 / 12 , b ) 5 / 12 , c ) 1 / 6 , d ) 13 / 30 , e ) 1 / 5 | d | divide(add(21, 5), const_60) | add(n0,n1)|divide(#0,const_60) | physics | D |
317 x 317 + 283 x 283 = ? | "= ( 317 ) ^ 2 + ( 283 ) ^ 2 = ( 300 + 17 ) ^ 2 + ( 300 - 17 ) ^ 2 = 2 [ ( 300 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 90000 + 289 ] = 2 x 90289 = 180578 answer is d" | a ) 79698 , b ) 80578 , c ) 80698 , d ) 180578 , e ) none of them | d | multiply(317, power(317, 283)) | power(n1,n2)|multiply(n0,#0)| | general | D |
if 15 % of a class averages 100 % on a test , 50 % of the class averages 78 % on the test , and the remainder of the class averages 63 % on the test , what is the overall class average ? ( round final answer to the nearest percent ) . | "this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 15 % - 50 % = 35 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.15 x 100 + 0.50 x 78 + 0.35 x 63 = 76.05 the class average ( rounded ) is 76 % final answer a ) 76 %" | a ) 76 % , b ) 77 % , c ) 78 % , d ) 79 % , e ) 80 % | a | divide(add(add(multiply(15, 100), multiply(50, 78)), multiply(subtract(const_100, add(15, 50)), 63)), const_100) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(const_100,#0)|multiply(n4,#4)|add(#3,#5)|divide(#6,const_100)| | gain | A |
a certain no . when divided by 95 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 95 + 25 = 120 / 15 = 8 ( remainder ) d" | a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 9 | d | subtract(subtract(subtract(95, 25), const_4), const_2) | subtract(n0,n1)|subtract(#0,const_4)|subtract(#1,const_2)| | general | D |
a shopkeeper sold an article at $ 100 with 20 % profit . then find its cost price ? | "cost price = selling price * 100 / ( 100 + profit ) c . p . = 100 * 100 / 120 = $ 83 ( approximately ) answer is d" | a ) $ 120 , b ) $ 100 , c ) $ 91 , d ) $ 83 , e ) $ 69 | d | multiply(100, divide(100, add(100, 20))) | add(n1,n0)|divide(n0,#0)|multiply(n0,#1)| | gain | D |
how many pieces of 0.42 meteres can be cut from a rod 36.5 meteres long | "explanation : we need so simple divide 36.5 / 0.42 , = ( 3650 / 42 ) = 86 option c" | a ) 30 , b ) 40 , c ) 86 , d ) 89 , e ) 95 | c | divide(36.5, 0.42) | divide(n1,n0)| | physics | C |
a certain library assesses fines for overdue books as follows . on the first day that a book is overdue , the total fine is $ 0.20 . for each additional day that the book is overdue , the total fine is either increased by $ 0.30 or doubled , whichever results in the lesser amount . what is the total for a book on the third day it is overdue ? | "1 st day fine - 0.2 2 nd day fine - 0.2 * 2 = 0.4 ( as doubling gives lower value ) 3 rd day fine - 0.4 + 3 = 0.7 ( as doubling gives higher value ) answer : b ." | a ) $ 0.60 , b ) $ 0.70 , c ) $ 0.80 , d ) $ 0.90 , e ) $ 1.00 | b | add(multiply(multiply(multiply(0.20, const_2), const_2), const_2), 0.30) | multiply(n0,const_2)|multiply(#0,const_2)|multiply(#1,const_2)|add(n1,#2)| | general | B |
x and y are positive integers . when x is divided by 6 , the remainder is 3 , and when x is divided by 16 , the remainder is 5 . when y is divided by 9 , the remainder is 5 , and when y is divided by 7 , the remainder is 4 . what is the least possible value of x / y ? | "when x is divided by 6 , the remainder is 3 : so , the possible values of x are : 3 , 9 , 15 , 21 , etc . when x is divided by 16 , the remainder is 5 : so , the possible values of x are : 5,21 . . . stop . since both lists include 21 , the smallest possible value of x is 21 . when y is divided by 9 , the remainder is 5 : so , the possible values of y are : 5 , 14 , 23,32 etc . when y is divided by 7 , the remainder is 4 : so , the possible values of y are : 6 , 14 , . . . stop . since both lists include 14 , the smallest possible value of y is 14 since the smallest possible values of x and y are 21 and 14 respectively , the smallest possible value of x / y is 1.5 . so , e is the correct answer to the original question ." | a ) 2.5 , b ) 3 , c ) 2 , d ) 1.3 , e ) 1.5 | e | subtract(add(multiply(9, const_2), 5), add(3, 6)) | add(n0,n1)|multiply(n4,const_2)|add(n5,#1)|subtract(#2,#0)| | general | E |
if 15 horses eat 15 bags of gram in 15 days , in how many days will one horse eat one bag of grain ? | horse * days = bags 15 * 15 = 15 and 1 * days = 1 ( 15 * 15 ) / ( 1 * days ) = 15 / 1 days = 15 answer : a | a ) 15 days , b ) 1 / 15 days , c ) 1 day , d ) 30 days , e ) 225 days | a | inverse(inverse(15)) | inverse(n0)|inverse(#0) | physics | A |
find the largest 6 digit number which is exactly divisible by 88 ? | largest 6 digit number is 999999 after doing 999999 ÷ 88 we get remainder 55 hence largest 6 digit number exactly divisible by 88 = 999999 - 55 = 999944 c | a ) 998765 , b ) 998907 , c ) 999944 , d ) 999954 , e ) 999990 | c | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(6, const_10)), const_3), 88) | multiply(const_100,const_100)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(n0,const_10)|add(#0,#1)|add(#4,#2)|add(#5,#3)|add(#6,const_3)|multiply(n1,#7) | general | C |
a textile manufacturing firm employees 56 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 56 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs 75000 . if one loom breaks down and remains idle for one month , the decrease in profit is : | "explanation : profit = 5 , 00,000 â ˆ ’ ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 ã — ( 55 / 56 ) â ˆ ’ 150000 ã — ( 55 / 56 ) â ˆ ’ 75000 . = > rs 2 , 68,750 decrease in profit = > 2 , 75,000 â ˆ ’ 2 , 68,750 = > rs . 6,250 answer : b" | a ) 13000 , b ) 6250 , c ) 10000 , d ) 5000 , e ) none of these | b | subtract(subtract(multiply(multiply(5, const_1000), const_100), add(multiply(75000, const_2), 75000)), subtract(subtract(multiply(divide(subtract(56, 1), 56), multiply(multiply(5, const_1000), const_100)), multiply(multiply(75000, const_2), divide(subtract(56, 1), 56))), 75000)) | multiply(n2,const_1000)|multiply(n6,const_2)|subtract(n0,n4)|add(n6,#1)|divide(#2,n0)|multiply(#0,const_100)|multiply(#4,#5)|multiply(#4,#1)|subtract(#5,#3)|subtract(#6,#7)|subtract(#9,n6)|subtract(#8,#10)| | general | B |
if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 34 is subtracted and divided by 10 ? | "explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 34 ) / 10 = ( 54 - 34 ) / 10 = 2 answer : option d" | a ) 4 , b ) 7 , c ) 8 , d ) 2 , e ) 3 | d | divide(subtract(add(multiply(7, 7), 5), 34), 10) | multiply(n1,n1)|add(n0,#0)|subtract(#1,n3)|divide(#2,n4)| | general | D |
the compound ratio of 1 : 2 , 3 : 2 and 4 : 5 ? | "1 / 2 * 3 / 2 * 4 / 5 = 3 / 5 = 3 : 5 answer : d" | a ) 1 : 9 , b ) 1 : 7 , c ) 1 : 2 , d ) 3 : 5 , e ) 1 : 4 | d | divide(divide(multiply(1, 3), multiply(2, 2)), divide(multiply(3, 4), multiply(2, 5))) | multiply(n0,n2)|multiply(n1,n3)|multiply(n2,n4)|multiply(n3,n5)|divide(#0,#1)|divide(#2,#3)|divide(#4,#5)| | other | D |
if t = 5 / 9 * ( k - 32 ) , and if t = 20 , then what is the value of k ? | "k - 32 = 9 t / 5 k = 9 t / 5 + 32 k = 9 ( 20 ) / 5 + 32 = 68 the answer is a ." | a ) 68 , b ) 70 , c ) 72 , d ) 74 , e ) 76 | a | add(divide(multiply(20, 9), 5), 32) | multiply(n1,n3)|divide(#0,n0)|add(n2,#1)| | general | A |
the present worth of rs . 1404 due in two equal half - yearly installments at 8 % per annum simple interest is : | "time = 1 / 2 yr r = 8 % let pw = x si = 1404 - x x * 1 / 2 * 8 / 100 = 1404 - x . ( ref problm 10 of si , rs aggarwal ) x = 1404 * 25 / 26 hence x = 1350 answer : c" | a ) rs . 1325 , b ) rs . 1300 , c ) rs . 1350 , d ) rs . 1500 , e ) rs . 1600 | c | add(divide(divide(1404, const_2), divide(add(const_100, 8), const_100)), divide(divide(1404, const_2), divide(add(const_100, divide(8, const_2)), const_100))) | add(n1,const_100)|divide(n0,const_2)|divide(n1,const_2)|add(#2,const_100)|divide(#0,const_100)|divide(#1,#4)|divide(#3,const_100)|divide(#1,#6)|add(#5,#7)| | gain | C |
arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 13 from the product of their ages a year from today . what is their age today ? | ad = ( a + 1 ) ( d + 1 ) - 13 0 = a + d - 12 a + d = 12 a = d ( as they are twin brothers ) a = d = 6 b is the answer | a ) 2 . , b ) 6 . , c ) 5 . , d ) 7 . , e ) 9 . | b | divide(subtract(13, const_1), const_2) | subtract(n0,const_1)|divide(#0,const_2) | general | B |
population is 20000 . pop increases by 10 % every year , then the pop after 3 years is ? | population after 1 st year = 20000 * 10 / 100 = 2000 = = = > 20000 + 2000 = 22000 population after 2 nd year = 22000 * 10 / 100 = 2200 = = = > 22000 + 2200 = 24200 population after 3 rd year = 24200 * 10 / 100 = 2420 = = = > 24200 + 2420 = 26620 answer : d | a ) 26630 , b ) 26640 , c ) 36620 , d ) 26620 , e ) 26820 | d | add(add(add(20000, multiply(20000, divide(10, const_100))), multiply(add(20000, multiply(20000, divide(10, const_100))), divide(10, const_100))), multiply(add(add(20000, multiply(20000, divide(10, const_100))), multiply(add(20000, multiply(20000, divide(10, const_100))), divide(10, const_100))), divide(10, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|multiply(#2,#0)|add(#2,#3)|multiply(#4,#0)|add(#4,#5) | gain | D |
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 162 grams of water ? | "solution : we are given that the ratio of hydrogen to oxygen in water , by mass , is 2 : 16 . using our ratio multiplier we can re - write this as 2 x : 16 x . we can now use these expressions to determine how much oxygen is in 162 grams of water . 2 x + 16 x = 162 18 x = 162 x = 9 since x is 9 , we know that there are 16 x 9 = 144 grams of oxygen in 162 grams of water . answer e ." | a ) 16 , b ) 72 , c ) 112 , d ) 128 , e ) 144 | e | multiply(2, divide(162, add(2, 16))) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other | E |
a boy goes to his school from his house at a speed of 3 km / hr and returns at a speed of 2 km / hr . if he takes 5 hours in going and coming . the distance between his house and school is ? | "average speed = ( 2 * 3 * 2 ) / ( 3 + 2 ) = 12 / 5 km / hr . distance traveled = 12 / 5 * 5 = 12 km . distance between house and school = 12 / 2 = 6 km . answer : c" | a ) 3 , b ) 4 , c ) 6 , d ) 5 , e ) 7 | c | multiply(divide(5, add(divide(3, 2), const_1)), 3) | divide(n0,n1)|add(#0,const_1)|divide(n2,#1)|multiply(n0,#2)| | physics | C |
the cost of 2 chairs and 3 tables is rs . 1800 . the cost of 3 chairs and 2 tables is rs . 1200 . the cost of each table is more than that of each chair by ? | "explanation : 2 c + 3 t = 1800 - - - ( 1 ) 3 c + 3 t = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - c + t = 600 = > t - c = 600 answer : e" | a ) 228 , b ) 287 , c ) 277 , d ) 188 , e ) 600 | e | subtract(divide(subtract(multiply(3, 1800), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))), divide(subtract(3, multiply(2, divide(subtract(multiply(3, 1800), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))))), 3)) | multiply(n2,const_3)|multiply(n3,const_2)|multiply(n1,const_3)|multiply(n0,const_2)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(n0,#6)|subtract(n3,#7)|divide(#8,n1)|subtract(#6,#9)| | general | E |
a number divided by 44 leaves remainder 12 what is the remainder when same number divided by 12 | "add 44 + 12 = 56 now 56 divided by 12 so we get 8 as reaminder answer : b" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | divide(subtract(12, 12), subtract(12, 12)) | subtract(n1,n2)|divide(#0,#0)| | general | B |
a person walks at a speed of 4 km / hr and runs at a speed of 8 km / hr . how many hours will the person require to cover a distance of 20 km , if the person completes half of the distance by walking and the other half by running ? | "time = 10 / 4 + 10 / 8 = 30 / 8 = 3.75 hours the answer is c ." | a ) 2.75 , b ) 3.25 , c ) 3.75 , d ) 4.25 , e ) 4.75 | c | add(divide(divide(20, const_2), 4), divide(divide(20, const_2), 8)) | divide(n2,const_2)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)| | physics | C |
jane and thomas are among the 6 people from which a committee of 3 people is to be selected . how many different possible committees of 3 people can be selected from these 6 people if at least one of either jane or thomas is to be selected ? | the total number of ways to choose 3 people from 6 is 6 c 3 = 20 . the number of committees without jane or thomas is 4 c 3 = 4 . there are 20 - 4 = 16 possible committees which include jane and / or thomas . the answer is c . | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | c | add(add(choose(6, const_1), choose(6, const_1)), choose(const_4, const_1)) | choose(n0,const_1)|choose(const_4,const_1)|add(#0,#0)|add(#2,#1) | probability | C |
the denominator of a fraction is 8 greater than the numerator . if the numerator and the denominator are increased by 5 , the resulting fraction is equal to 6 â „ 7 . what is the value of the original fraction ? | "let the numerator be x . then the denominator is x + 8 . x + 5 / x + 13 = 6 / 7 . 7 x + 35 = 6 x + 78 . x = 43 . the original fraction is 43 / 51 . the answer is d ." | a ) 25 / 33 , b ) 31 / 39 , c ) 37 / 45 , d ) 43 / 51 , e ) 53 / 61 | d | divide(divide(subtract(multiply(6, add(5, 8)), 7), subtract(7, 6)), add(divide(subtract(multiply(6, add(5, 8)), 7), subtract(7, 6)), 8)) | add(n0,n1)|subtract(n3,n2)|multiply(n2,#0)|subtract(#2,n3)|divide(#3,#1)|add(n0,#4)|divide(#4,#5)| | general | D |
a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box , what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd ? | the sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls ( odd + odd + odd = odd ) or two even numbered balls and one odd numbered ball ( even + even + odd = odd ) ; p ( ooo ) = ( 1 / 2 ) ^ 3 ; p ( eeo ) = 3 * ( 1 / 2 ) ^ 2 * 1 / 2 = 3 / 8 ( you should multiply by 3 as the scenario of two even numbered balls and one odd numbered ball can occur in 3 different ways : eeo , eoe , or oee ) ; so finally p = 1 / 8 + 3 / 8 = 1 / 2 . answer : c . | a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4 | c | add(power(divide(1, const_2), const_3), multiply(multiply(3, power(divide(1, const_2), const_2)), divide(1, const_2))) | divide(n1,const_2)|power(#0,const_2)|power(#0,const_3)|multiply(n3,#1)|multiply(#0,#3)|add(#4,#2) | general | C |
if x + ( 1 / x ) = 5 , what is the value of e = x ^ 2 + ( 1 / x ) ^ 2 ? | "squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c" | a ) e = 21 , b ) e = 22 , c ) e = 23 , d ) 24 , e ) 27 | c | subtract(power(5, 2), 2) | power(n1,n2)|subtract(#0,n2)| | general | C |
a vendor bought toffees at 6 for a rupee . how many for a rupee must he sell to gain 80 % ? | "c . p . of 6 toffees = re . 1 s . p . of 6 toffees = 480 % of re . 1 = rs . 24 / 5 for rs . 24 / 5 , toffees sold = 6 . for re . 1 , toffees sold = 6 x 24 / 5 = 28.8 . answer c" | a ) 10.8 , b ) 20.8 , c ) 28.8 , d ) 18.8 , e ) 17.8 | c | multiply(6, add(const_1, divide(80, const_100))) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)| | gain | C |
pipe a can fill a tank in 4 hours . due to a leak at the bottom , it takes 8 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 / 4 - 1 / x = 1 / 8 = > 1 / x = 1 / 4 - 1 / 8 = ( 2 - 1 ) / 8 = 1 / 8 = > x = 8 . answer : d" | a ) 13 , b ) 17 , c ) 18 , d ) 8 , e ) 12 | d | divide(multiply(8, 4), subtract(8, 4)) | multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)| | physics | D |
two trains 140 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time ( in seconds ) which they take to cross each other , is : | "explanation : relative speed = ( 60 + 40 ) km / hr = [ 100 x ( 5 / 18 ) ] m / sec = ( 250 / 9 ) m / sec . distance covered in crossing each other = ( 140 + 160 ) m = 300 m . required time = [ 300 x ( 9 / 250 ) ] sec = ( 54 / 5 ) sec = 10.8 sec . answer : d" | a ) 9 , b ) 9.6 , c ) 10 , d ) 10.8 , e ) 7 | d | divide(add(140, 160), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics | D |
p is able to do a piece of work in 5 days and q can do the same work in 10 days . if they can work together for 3 days , what is the fraction of work completed ? | "explanation : amount of work p can do in 1 day = 1 / 5 amount of work q can do in 1 day = 1 / 10 amount of work p and q can do in 1 day = 1 / 5 + 1 / 10 = 3 / 10 amount of work p and q can together do in 3 days = 3 × ( 3 / 10 ) = 9 / 10 answer : option e" | a ) 13 / 10 , b ) 4 / 5 , c ) 11 / 10 , d ) 7 / 10 , e ) 9 / 10 | e | subtract(const_1, multiply(add(divide(const_1, 10), divide(const_1, 5)), 3)) | divide(const_1,n1)|divide(const_1,n0)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics | E |
a contractor is engaged for 30 days on the condition that he receives rs . 25 for eachday he works & is fined rs . 7.50 for each day is absent . he gets rs . 425 in all . for how many days was heabsent ? | "30 * 25 = 750 425 - - - - - - - - - - - 325 25 + 7.50 = 32.5 325 / 32.5 = 10 b" | a ) 5 , b ) 10 , c ) 17 , d ) 25 , e ) 30 | b | subtract(30, divide(add(multiply(30, 7.50), 425), add(25, 7.50))) | add(n1,n2)|multiply(n0,n2)|add(n3,#1)|divide(#2,#0)|subtract(n0,#3)| | physics | B |
what is the least value of x , so that 2 x 5475 is divisible by 9 | "explanation : the sum of the digits of the number is divisible by 9 . then the number is divisible by 9 . 2 + x + 5 + 4 + 7 + 5 = 23 + x least value of x may be ' 4 ' , so that the total 23 + 4 = 27 is divisible by 9 . answer : option c" | a ) 7 , b ) 8 , c ) 4 , d ) 3 , e ) 2 | c | divide(divide(divide(lcm(2, 5475), 5475), const_4), const_4) | lcm(n0,n1)|divide(#0,n1)|divide(#1,const_4)|divide(#2,const_4)| | general | C |
if [ x ] is the greatest integer less than or equal to x , what is the value of [ - 6.2 ] + [ - 3.4 ] + [ 12.7 ] ? | "you are asked what the closest lesser integer value to [ x ] is . [ - 6.2 ] = - 7.0 [ - 3.4 ] = - 4.0 [ 12.7 ] = 12.0 therefore , answer is : - 7.0 - 4.0 + 12.0 = - 1.0 option a ." | a ) - 1 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | a | subtract(add(3.4, 12.7), 6.2) | add(n1,n2)|subtract(#0,n0)| | general | A |
find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1568 ? | "1568 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1400 . answer : d" | a ) 1000 , b ) 2777 , c ) 2889 , d ) 1400 , e ) 2771 | d | divide(1568, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)| | general | D |
a man invests in a 16 % stock at 128 . the interest obtained by him is | by investing rs 128 , income derived = rs . 16 by investing rs . 100 , income derived = = rs . 12.5 interest obtained = 12.5 % answer : c | a ) 22.5 % , b ) 42.5 % , c ) 12.5 % , d ) 62.5 % , e ) 82.5 % | c | multiply(divide(16, 128), const_100) | divide(n0,n1)|multiply(#0,const_100) | gain | C |
the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 2090 . what was his original salary ? | "x * ( 110 / 100 ) * ( 95 / 100 ) = 2090 x * ( 11 / 10 ) * ( 1 / 100 ) = 22 x = 2000 answer : e" | a ) 2277 , b ) 2999 , c ) 1000 , d ) 2651 , e ) 2000 | e | divide(2090, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4)| | gain | E |
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 : d" | a ) 150 , b ) 278 , c ) 179 , d ) 200 , e ) 191 | d | multiply(divide(multiply(60, const_1000), const_3600), 12) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | D |
a tradesman sold an article at a loss of 25 % . if the selling price had been increased by $ 500 , there would have been a gain of 15 % . what was the cost price of the article ? | let c . p . be $ x then 125 % of x - 85 % of x = 500 40 % of x = 500 2 x / 5 = 500 x = $ 1250 answer is e | a ) $ 1000 , b ) $ 1200 , c ) $ 1120 , d ) $ 1350 , e ) $ 1250 | e | divide(500, divide(add(25, 15), const_100)) | add(n0,n2)|divide(#0,const_100)|divide(n1,#1) | gain | E |
mary invested a certain sum of money in a bank that paid simple interest . the amount grew to $ 260 at the end of 2 years . she waited for another 3 years and got a final amount of $ 350 . what was the principal amount that she invested at the beginning ? | what shall be the rate of interest . ? does that is not required for the calculation ? not really ! keep in mind that the interest earned each year will be the same in simple interest . at the end of 2 years , amount = $ 260 at the end of 5 years , amount = $ 350 this means she earned an interest of $ 90 in 3 years . or $ 30 in each year . we know that the interest earned each year will be the same . therefore she must have earned $ 60 in 2 years . hence principal amount = $ 260 - $ 60 = $ 200 option d | a ) $ 220 , b ) $ 230 , c ) $ 240 , d ) $ 200 , e ) $ 250 | d | subtract(260, multiply(divide(subtract(350, 260), 3), 2)) | subtract(n3,n0)|divide(#0,n2)|multiply(n1,#1)|subtract(n0,#2) | gain | D |
a certain box has 12 cards and each card has one of the integers from 1 to 12 inclusive . each card has a different number . if 2 different cards are selected at random , what is the probability that the sum of the numbers written on the 2 cards is less than the average ( arithmetic mean ) of all the numbers written on the 12 cards ? | "the average of the numbers is 6.5 the total number of ways to choose 2 cards from 12 cards is 12 c 2 = 66 . the ways to choose 2 cards with a sum less than the average are : { 1,2 } , { 1,3 } , { 1,4 } , { 1,5 } , { 2,3 } , { 2,4 } the probability is 6 / 66 = 1 / 11 the answer is a ." | a ) 1 / 11 , b ) 2 / 23 , c ) 3 / 34 , d ) 4 / 45 , e ) 5 / 56 | a | divide(const_4, divide(factorial(12), multiply(factorial(2), factorial(subtract(12, 2))))) | factorial(n0)|factorial(n3)|subtract(n0,n3)|factorial(#2)|multiply(#1,#3)|divide(#0,#4)|divide(const_4,#5)| | general | A |
if x + y = 4 , x - y = 36 , for integers of x and y , x = ? | "x + y = 4 x - y = 36 2 x = 40 x = 20 answer is a" | a ) 20 , b ) 15 , c ) 25 , d ) 13 , e ) 42 | a | divide(add(4, 36), const_2) | add(n0,n1)|divide(#0,const_2)| | general | A |
two trains of equal length , running with the speeds of 60 and 20 kmph , take 50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 50 d = 50 * 100 / 18 = 2500 / 9 rs = 60 + 20 = 80 * 5 / 18 t = 2500 / 9 * 18 / 400 = 12.5 sec . answer : a" | a ) 12.5 , b ) 8.01 , c ) 7.5 , d ) 26 , e ) 22 | a | multiply(multiply(multiply(const_0_2778, subtract(60, 20)), 50), inverse(multiply(const_0_2778, add(60, 20)))) | add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5)| | physics | A |
the average of 10 numbers is calculated as 46 . it is discovered later on that while calculating the average , the number 75 was incorrectly read as 25 , and this incorrect number was used in the calculation . what is the correct average ? | "the total sum of the numbers should be increased by 50 . then the average will increase by 50 / 10 = 5 . the correct average is 51 . the answer is b ." | a ) 48 , b ) 51 , c ) 60 , d ) 71 , e ) 96 | b | divide(add(subtract(multiply(46, 10), 25), 75), 10) | multiply(n0,n1)|subtract(#0,n3)|add(n2,#1)|divide(#2,n0)| | general | B |
a shopkeeper sells 200 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ? | sp per metre = 18000 / 200 = rs . 90 loss per metre = rs . 5 cp per metre = 90 + 5 = rs . 95 . answer : e | a ) 12 , b ) 27 , c ) 29 , d ) 55 , e ) 95 | e | add(divide(18000, 200), 5) | divide(n1,n0)|add(n2,#0)| | gain | E |
if x , y , and z are positive real numbers such that x ( y + z ) = 62 , y ( z + x ) = 82 , and z ( x + y ) = 100 , then xyz is | "xy + xz = 62 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 ) yz + yx = 82 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 ) xz + zy = 100 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 3 ) re - writing equation 3 as follows : xz + zy = 82 + 18 xz + zy = yz + yx + 18 xz = yx + 18 . . . . . . . . . . . . . . . ( 4 ) adding ( 1 ) ( 4 ) 2 xz = 80 xz = 40 xyz has to be multiple of 40 , only 120 fits in answer = a" | a ) 120 , b ) 50 , c ) 60 , d ) 70 , e ) 90 | a | subtract(100, subtract(82, 62)) | subtract(n1,n0)|subtract(n2,#0)| | general | A |
a recipe requires 2 1 / 2 ( mixed number ) cups of flour 2 3 / 4 ( mixed number ) cups of sugar and 1 1 / 3 ( mixed number ) cups of milk to make one cake . victor has 15 cups if flour , 16 cups of sugar and 8 cups of milk . what is the greatest number of cakes john can make using this recipe ? | "less work up front : go through each item and see what the greatest number of cakes you can make with each . the lowest of these will be the right answer . flour : 15 cups , we need 2.5 cups each . just keep going up the line to see how many cakes we can make : that means i can make 2 cakes with 5 cups , so 6 cakes overall with 15 cups . i ' ve already got the answer narrowed to either a or b . sugar : 16 cups , we need 2.75 cups each . same principle . i can make 2 cups with 5.5 cups , so to make 6 cakes i ' d need 16.5 cups . i do n ' t have that much sugar , so we ' re limited to 5 cakes . no need to even do milk because we ' re already at 5 . sugar will be the limiting factor . answer is a" | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | min(min(divide(15, add(2, divide(1, 2))), floor(divide(16, add(divide(3, 4), 2)))), divide(8, add(divide(1, 3), 1))) | divide(n1,n4)|divide(n1,n0)|divide(n4,n5)|add(n1,#0)|add(n0,#1)|add(n0,#2)|divide(n11,#3)|divide(n9,#4)|divide(n10,#5)|floor(#8)|min(#7,#9)|min(#6,#10)| | general | A |
the average age of an adult class is 40 years . 15 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 year . find what was theoriginal strength of class ? | "let original strength = y then , 40 y + 15 x 32 = ( y + 15 ) x 36 â ‡ ’ 40 y + 480 = 36 y + 540 â ‡ ’ 4 y = 60 â ˆ ´ y = 15 c" | a ) 8 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | c | divide(subtract(multiply(15, subtract(40, 4)), multiply(15, 32)), 4) | multiply(n1,n2)|subtract(n0,n3)|multiply(n1,#1)|subtract(#2,#0)|divide(#3,n3)| | general | C |
two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 15 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 15 sec = 15 * 10 = 150 m . the length of the faster train = 150 m . answer : e" | a ) 270 , b ) 277 , c ) 187 , d ) 257 , e ) 150 | e | multiply(divide(subtract(72, 36), const_3_6), 15) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics | E |
what is 25 % of 4 / 12 of 520 ? | 25 % = 25 / 100 = 1 / 4 of 4 / 12 = 1 / 4 * 4 / 12 = 1 / 12 of 520 = 1 / 12 * 520 = 43.33 = = 43 ans - d | a ) 49 , b ) 56 , c ) 60 , d ) 43 , e ) 70 | d | divide(multiply(25, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain | D |
a fruit drink is made of orange , watermelon , and grape juice , where 25 percent of the drink is orange juice and 40 percent is watermelon juice . if the drink is made with 105 ounces of grape juice , how many ounces is the drink total ? | "let the total number of ounces in the drink be x . % of orange = 25 % % of watermelon = 40 % % of grape = 100 % - 65 % = 35 % 0.35 x = 105 x = 300 therefore there are a total of 300 ounces in the drink . the answer is d ." | a ) 220 , b ) 250 , c ) 280 , d ) 300 , e ) 340 | d | add(add(105, divide(multiply(105, 25), subtract(const_100, add(40, 25)))), divide(multiply(105, 40), subtract(const_100, add(40, 25)))) | add(n0,n1)|multiply(n0,n2)|multiply(n1,n2)|subtract(const_100,#0)|divide(#1,#3)|divide(#2,#3)|add(n2,#4)|add(#6,#5)| | general | D |
shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 50 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 40 minutes after shannon . if maxine rider her bike home at the rate of r per hour and shannon still drives at a rate of 2 r miles per hour , shannon arrives home 2 hours before maxine . how far does maxine live from work ? | "nice question + 1 we have that x / 24 - ( 60 - x ) / 2 r = 40 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 46 hence answer is e" | a ) 20 , b ) 34 , c ) 38 , d ) 40 , e ) 46 | e | divide(subtract(multiply(const_3, 50), 50), const_3) | multiply(n0,const_3)|subtract(#0,n0)|divide(#1,const_3)| | physics | E |
the breath of a rectangle is three - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 2025 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the length is 10 units ? | given that the area of the square = 2025 sq . units = > side of square = √ 2025 = 45 units the radius of the circle = side of the square = 45 units breath of the rectangle = 3 / 5 * 45 = 27 units given that length = 10 units area of the rectangle = lb = 27 * 10 = 270 sq . units answer : b | ['a ) 370 sq . units', 'b ) 270 sq . units', 'c ) 170 sq . units', 'd ) 470 sq . units', 'e ) 570 sq . units'] | b | rectangle_area(10, divide(multiply(square_edge_by_area(2025), const_3), add(const_4, const_1))) | add(const_1,const_4)|square_edge_by_area(n0)|multiply(#1,const_3)|divide(#2,#0)|rectangle_area(n1,#3) | geometry | B |
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in 40 min , then the slower pipe alone will be able to fill the tank in ? | "let the slower pipe alone fill the tank in x min . then , faster pipe will fill it in x / 3 min . 1 / x + 3 / x = 1 / 40 4 / x = 1 / 40 = > x = 160 min . answer : a" | a ) 160 , b ) 787 , c ) 144 , d ) 128 , e ) 121 | a | multiply(add(const_1, const_4), 40) | add(const_1,const_4)|multiply(n0,#0)| | physics | A |
how many pieces of 85 cm length can be cut from a rod of 17 meters long ? | number of pieces = 1700 / 85 = 20 the answer is d . | a ) 50 , b ) 40 , c ) 30 , d ) 20 , e ) 10 | d | divide(multiply(17, const_100), 85) | multiply(n1,const_100)|divide(#0,n0) | physics | D |
find the number of shares that can be bought for rs . 8200 if the market value is rs . 28 each with brokerage being 2.5 % . | "explanation : cost of each share = ( 28 + 2.5 % of 28 ) = rs . 28.7 therefore , number of shares = 8200 / 28.7 = 285.7 answer : d" | a ) 237 , b ) 270 , c ) 177 , d ) 285.7 , e ) 111 | d | floor(divide(8200, add(28, divide(2.5, const_100)))) | divide(n2,const_100)|add(n1,#0)|divide(n0,#1)|floor(#2)| | gain | D |
if 25 % of x is 15 less than 12 % of 1500 , then x is ? | "25 % of x = x / 4 ; 12 % of 1500 = 12 / 100 * 1500 = 180 given that , x / 4 = 180 - 15 = > x / 4 = 165 = > x = 660 . answer : a" | a ) 660 , b ) 738 , c ) 837 , d ) 840 , e ) 83 | a | divide(subtract(multiply(1500, divide(12, const_100)), 15), divide(25, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#2,n1)|divide(#3,#1)| | general | A |
what is rate of interest if principal . amount be 400 , simple interest 140 and time 2 year . | "s . i = ( p * r * t ) / 100 140 = 800 r / 100 r = 140 / 8 = 17.5 % answer b" | a ) 10 , b ) 17.5 , c ) 25 , d ) 12 , e ) 14.5 | b | multiply(divide(140, multiply(400, 2)), const_100) | multiply(n0,n2)|divide(n1,#0)|multiply(#1,const_100)| | gain | B |
a cistern is filled by pipe a in 10 hours and the full cistern can be leaked out by an exhaust pipe b in 15 hours . if both the pipes are opened , in what time the cistern is full ? | "time taken to full the cistern = ( 1 / 10 - 1 / 15 ) hrs = 1 / 30 = 30 hrs answer : b" | a ) 50 hrs , b ) 30 hrs , c ) 70 hrs , d ) 80 hrs , e ) 90 hrs | b | divide(const_1, subtract(divide(const_1, 10), divide(const_1, 15))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics | B |
of the 13 employees in a certain department , 1 has an annual salary of 38,000 , 2 have an annual salary of 45,800 each , 2 have an annual salary of 42,500 each , 3 have an annual salary of 40,000 each and 5 have an annual salary of 48,500 each . what is the median annual salary for the 13 employees ? | "median is just the value in the middle when you arrange all values in the ascending order in this question , the 7 th value would be the median ( since there are 13 employees ) 38 , 40 , 40 , 40 , 42.5 , 42.5 , 45.8 so , answer is d ." | a ) 38,000 , b ) 40,000 , c ) 42,500 , d ) 45,800 , e ) 48,500 | d | multiply(multiply(const_100, const_100), const_4) | multiply(const_100,const_100)|multiply(#0,const_4)| | general | D |
the length of the bridge , which a train 250 metres long and travelling at 72 km / hr can cross in 30 seconds , is : | "speed = [ 72 x 5 / 18 ] m / sec = 20 m / sec time = 30 sec let the length of bridge be x metres . then , ( 250 + x ) / 30 = 20 = > 250 + x = 600 = > x = 350 m . answer : d" | a ) 200 m , b ) 225 m , c ) 245 m , d ) 350 m , e ) 240 m | d | subtract(multiply(divide(multiply(72, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 250) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics | D |
from the digits 2 , 3 , 4 , 5 , 6 and 7 , how many 5 - digit numbers can be formed that have distinct digits and are multiples of 12 ? | detailed solution any multiple of 12 should be a multiple of 4 and 3 . first , let us look at the constraint for a number being a multiple of 3 . sum of the digits should be a multiple of 3 . sum of all numbers from 2 to 7 is 27 . so , if we have to drop a digit and still retain a multiple of 3 , we should drop either 3 or 6 . so , the possible 5 digits are 2 , 4 , 5 , 6 , 7 or 2 , 3 , 4 , 5 , 7 . when the digits are 2 , 4 , 5 , 6 , 7 . the last two digits possible for the number to be a multiple of 4 are 24 , 64 , 52 , 72 , 56 , 76 . for each of these combinations , there are 6 different numbers possible . so , with this set of 5 digits we can have 36 different numbers . when the digits are 2 , 3 , 4 , 5 , 7 . the last two digits possible for the number to be a multiple of 4 are 32 , 52 , 72 , 24 . for each of these combinations , there are 6 different numbers possible . so , with this set of 5 digits we can have 24 different numbers . overall , there are 60 different 5 - digit numbers possible . correct answer : b | a ) 36 , b ) 60 , c ) 84 , d ) 72 , e ) 86 | b | add(multiply(factorial(subtract(5, const_2)), multiply(const_2, const_3)), multiply(const_4, factorial(subtract(5, const_2)))) | multiply(const_2,const_3)|subtract(n3,const_2)|factorial(#1)|multiply(#2,#0)|multiply(#2,const_4)|add(#3,#4) | general | B |
convert 300 miles into meters ? | "1 mile = 1609.34 meters 300 mile = 300 * 1609.34 = 482802 meters answer is d" | a ) 784596 , b ) 845796 , c ) 804670 , d ) 482802 , e ) 864520 | d | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 300), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics | D |
if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 12 ) , what is the value of x ? | i am guessing the question is : ( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 12 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 12 hence x = 14 . answer is e | a ) 9 , b ) 11 , c ) 13 , d ) 15 , e ) 14 | e | add(12, 2) | add(n0,n5) | general | E |
if 14 lions can kill 14 deers in 14 minutes how long will it take 100 lions to kill 100 deers ? | "we can try the logic of time and work , our work is to kill the deers so 14 ( lions ) * 14 ( min ) / 14 ( deers ) = 100 ( lions ) * x ( min ) / 100 ( deers ) hence answer is x = 14 answer : b" | a ) 1 minutes , b ) 14 minute , c ) 100 minutes , d ) 10000 minutes , e ) 1000 minutes | b | multiply(divide(14, 14), 14) | divide(n0,n0)|multiply(n0,#0)| | physics | B |
the product of x and y is a constant . if the value of x is increased by 20 % , by what percentage must the value of y be decreased ? | "x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.2 x ( 120 ) i . e . 120 * new ' y ' = 10000 i . e . new ' y ' = 10000 / 120 = 83.33 i . e . y decreases from 100 to 83.33 i . e . decrease of 16.66 % answer : option a" | a ) 16.66 % , b ) 33.33 % , c ) 44.44 % , d ) 55.55 % , e ) 19.92 % | a | multiply(subtract(const_1, divide(const_100, add(const_100, 20))), const_100) | add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | general | A |
alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 1021 . find the cost of each ice - cream cup ? | "let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1021 96 + 225 + 490 + 6 x = 1021 6 x = 210 = > x = 35 . answer : a" | a ) 35 , b ) 66 , c ) 77 , d ) 99 , e ) 91 | a | divide(subtract(subtract(subtract(1021, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6) | multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)| | general | A |
a sum of money at simple interest amounts to rs . 1717 in 1 year and to rs . 1734 in 2 years . the sum is : | "s . i . for 1 year = rs . ( 1734 - 1717 ) = rs . 17 . principal = rs . ( 1717 - 17 ) = rs . 1700 . answer : option d" | a ) rs . 1200 , b ) rs . 1690 , c ) rs . 1600 , d ) rs . 1700 , e ) rs . 1500 | d | subtract(1717, divide(multiply(subtract(1734, 1717), 1), 2)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain | D |
in one hour , a boat goes 11 km along the stream and 7 km against the stream . the sped of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 11 + 7 ) km / hr = 9 kmph . answer d" | a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | d | divide(add(11, 7), const_2) | add(n0,n1)|divide(#0,const_2)| | gain | D |
three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 288 . what is their h . c . f . ? | "let the numbers be 2 x , 3 x , and 4 x . lcm of 2 x , 3 x and 4 x is 12 x . 12 x = 288 x = 24 hcf of 2 x , 3 x and 4 x = x = 24 the answer is b ." | a ) 18 , b ) 24 , c ) 36 , d ) 42 , e ) 48 | b | multiply(2, 4) | multiply(n0,n2)| | other | B |
which is the smallest number divides 2800 and gives a perfect square . | 7 is the smallest number which divides 2800 and gives a perfect square . as 2800 = 2 * 2 * 2 * 2 * 5 * 5 * 7 and 7 is not in a pair which gives 400 ( a perfect square of 20 ) on dividing 2800 . answer : c | ['a ) 5', 'b ) 6', 'c ) 7', 'd ) 8', 'e ) 9'] | c | divide(divide(divide(divide(divide(divide(2800, const_2), const_2), const_2), const_2), add(const_2, const_3)), add(const_2, const_3)) | add(const_2,const_3)|divide(n0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(#4,#0)|divide(#5,#0) | geometry | C |
the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 40 years , was replaced by a new person . find the age of the new person ? | "initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 40 + q = > q = 10 answer : e" | a ) 18 , b ) 56 , c ) 12 , d ) 17 , e ) 10 | e | subtract(40, multiply(10, 3)) | multiply(n0,n1)|subtract(n2,#0)| | general | E |
the population of a town increased from 1 , 75,000 to 2 , 62,500 in a decade . the average percent increase of population per year is | "solution increase in 10 years = ( 262500 - 175000 ) = 87500 . increase % = ( 87500 / 175000 × 100 ) % = 50 % . required average = ( 50 / 10 ) % = 5 % . answer b" | a ) 4.37 % , b ) 5 % , c ) 6 % , d ) 8.75 % , e ) none | b | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,const_4)| | general | B |
what positive number , when squared , is equal to the cube of the positive square root of 12 ? | "let the positive number be x x ^ 2 = ( ( 12 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 16 = > x = 4 answer d" | a ) 64 , b ) 32 , c ) 8 , d ) 4 , e ) 2 | d | sqrt(power(power(12, divide(const_1, const_2)), const_3)) | divide(const_1,const_2)|power(n0,#0)|power(#1,const_3)|sqrt(#2)| | geometry | D |
how many 5 digit number contain number 3 ? | "total 5 digit no . = 9 * 10 * 10 * 10 * 10 = 90000 not containing 3 = 8 * 9 * 9 * 9 * 9 = 52488 total 5 digit number contain 3 = 90000 - 52488 = 37512 answer : e" | a ) 31512 , b ) 32512 , c ) 33512 , d ) 34512 , e ) 37512 | e | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(5, 5)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general | E |
a train running at the speed of 72 km / hr crosses a pole in 9 seconds . what is the length of the train ? | "speed = ( 72 * 5 / 18 ) m / sec = ( 20 ) m / sec length of the train = ( speed x time ) = ( 20 * 9 ) m = 180 m . answer : d" | a ) 286 , b ) 278 , c ) 255 , d ) 180 , e ) 287 | d | multiply(divide(multiply(72, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | D |
working alone , tom finishes cleaning half the house in a third of the time it takes nick to clean the entire house alone . tom alone cleans the entire house in 6 hours . how many hours will it take nick and tom to clean the entire house if they work together ? | answer is 3.6 hours . tom does the complete house in 6 hours while nick does it in 9 hours . 1 / ( 1 / 6 + 1 / 9 ) = 3.6 answer is e | a ) 1.5 , b ) 2 , c ) 2.4 , d ) 3 , e ) 3.6 | e | divide(multiply(6, const_3), add(const_3, const_2)) | add(const_2,const_3)|multiply(n0,const_3)|divide(#1,#0) | physics | E |
if 2 ^ ( 2 w ) = 8 ^ ( w − 3 ) , what is the value of w ? | "2 ^ ( 2 w ) = 8 ^ ( w − 3 ) 2 ^ ( 2 w ) = 2 ^ ( 3 * ( w − 3 ) ) 2 ^ ( 2 w ) = 2 ^ ( 3 w - 9 ) let ' s equate the exponents as the bases are equal . 2 w = 3 w - 9 w = 9 the answer is c ." | a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | c | divide(multiply(const_12, log(2)), log(2)) | log(n0)|multiply(#0,const_12)|divide(#1,#0)| | general | C |
compound interest of rs . 5000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) . | "10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 5000 [ 1 + ( 5 / 100 ) ] ^ 3 - 5000 = 5000 [ ( 21 / 20 ) ^ 3 - 1 ] = 5000 ( 9261 - 8000 ) / 8000 = 5 * 1261 / 8 = 788 answer : d" | a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 788 , e ) none of the above | d | subtract(multiply(5000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 5000) | divide(n1,n4)|divide(n2,n4)|add(n2,#1)|divide(#0,const_100)|add(#3,n2)|multiply(#2,n4)|power(#4,#5)|multiply(n0,#6)|subtract(#7,n0)| | gain | D |
after getting 2 successive discounts , a shirt with a list price of rs 150 is available at rs 105 . if the second discount is 12.55 , find the first discount . | "let the first discount be x % then , 87.5 % of ( 100 - x ) % of 150 = 105 = 87.5 / 100 * ( 100 - x ) / 100 * 450 = 150 = > 105 = > 100 - x = ( 105 * 100 * 100 ) / ( 150 * 87.5 ) = 80 x = ( 100 - 80 ) = 20 first discount = 20 % answer is e ." | a ) 22 % , b ) 18 % , c ) 24 % , d ) 17 % , e ) 20 % | e | divide(multiply(subtract(150, divide(multiply(105, const_100), subtract(const_100, 12.55))), const_100), 150) | multiply(n2,const_100)|subtract(const_100,n3)|divide(#0,#1)|subtract(n1,#2)|multiply(#3,const_100)|divide(#4,n1)| | gain | E |
sony and johnny caught 40 fishes . sony caught 4 times as many as johnny . how many fishes did johnny catch ? | s + j = 40 s = 4 * j j = 8 answer : e | a ) 12 , b ) 16 , c ) 34 , d ) 38 , e ) 8 | e | divide(40, add(4, const_1)) | add(n1,const_1)|divide(n0,#0) | general | E |
what percent is 2 minutes 24 seconds of an hour ? | "2 mt 24 s = 144 s ( 144 * 100 ) / ( 60 * 60 ) = 4 % answer : c" | a ) 6 % , b ) 8 % , c ) 4 % , d ) 2 % , e ) 9 % | c | multiply(divide(2, 24), const_100) | divide(n0,n1)|multiply(#0,const_100)| | physics | C |
a volume of 11248 l water is in a container of sphere . how many hemisphere of volume 4 l each will be required to transfer all the water into the small hemispheres ? | "a volume of 4 l can be kept in 1 hemisphere therefore , a volume of 11248 l can be kept in ( 11248 / 4 ) hemispheres ans . 2812 answer : a" | a ) 2812 , b ) 8231 , c ) 2734 , d ) 4222 , e ) 4254 | a | divide(11248, 4) | divide(n0,n1)| | geometry | A |
a baseball team played 10 games and won 5 . what is the ratio of the number of games played to the number of losses ? | the number of games played is 10 . the number of games won is 5 , so 10 - 5 = 5 = number of game losses . answer = ! 0 : 5 = 2 : 1 answer is e | a ) 5 : 1 , b ) 1 : 1 , c ) 20 : 4 , d ) 10 : 3 , e ) 2 : 1 | e | divide(10, 5) | divide(n0,n1) | other | E |
in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what q percent of all the people in smithtown are left - handed women ? | "looking at the ratio we can take total number of people q = 20 . . ans 5 / 20 or 25 % c" | a ) 50 % , b ) 40 % , c ) 25 % , d ) 20 % , e ) 10 % | c | multiply(divide(subtract(multiply(2, divide(add(3, 1), add(3, 2))), subtract(3, multiply(3, divide(add(3, 1), add(3, 2))))), add(3, 1)), const_100) | add(n0,n1)|add(n0,n3)|divide(#0,#1)|multiply(n3,#2)|multiply(n0,#2)|subtract(n0,#4)|subtract(#3,#5)|divide(#6,#0)|multiply(#7,const_100)| | general | C |
he average weight of 8 persons increases by 3.5 kg when a new person comes in place of one of them weighing 62 kg . what might be the weight of the new person ? | "explanation : total weight increased = ( 8 x 3.5 ) kg = 28 kg . weight of new person = ( 62 + 28 ) kg = 90 kg . answer : e" | a ) 75 kg , b ) 55 kg , c ) 45 kg , d ) 85 kg , e ) 90 kg | e | add(multiply(3.5, 8), 62) | multiply(n0,n1)|add(n2,#0)| | general | E |
if a is a positive integer , and if the units digit of a ^ 2 is 9 and the units digit of ( a + 1 ) ^ 2 is 4 , what is the units d digit of ( a + 2 ) ^ 2 ? | "i also got a . by punching in numers : d . . . 7 ^ 2 = . . . 9 . . . 8 ^ 2 = . . . 4 . . . 9 ^ 2 = . . . 1 . a" | a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) c . 14 | a | power(add(multiply(9, 2), 2), 2) | multiply(n0,n1)|add(n0,#0)|power(#1,n0)| | general | A |
a certain number when divided by 110 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 110 + 25 = 135 / 15 = 9 ( remainder ) e" | a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | e | reminder(25, 15) | reminder(n1,n2)| | general | E |
if 22 lines are drawn in a plane such that no two of them are parallel and no three are concurrent , then in how many points do they intersect ? | "if two lines are not parallel , then they will intersect in exactly one point . lines can be extended infinitely on both ends so somewhere they will intersect with each other if they are not parallel . we are also given that no three lines are concurrent . this means that no three lines intersect at the same point . so every pair of two lines we select will have a unique point of intersection which they will not share with any third line . the number of ways to select 2 lines from 22 lines is 22 c 2 = 231 the answer is c ." | a ) 125 , b ) 187 , c ) 231 , d ) 375 , e ) 460 | c | divide(factorial(22), multiply(factorial(subtract(22, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | physics | C |
three 6 faced dice are thrown together . the probability that exactly two dice show the same number on them is - . | "using question number 11 and 12 , we get the probability as 1 - ( 1 / 36 + 5 / 9 ) = 5 / 12 answer a" | a ) 5 / 12 , b ) 4 / 13 , c ) 5 / 17 , d ) 3 / 12 , e ) 8 / 9 | a | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | divide(const_1,n0)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2)| | general | A |
set a consists of the integers from 5 to 10 , inclusive , while set b consists of the integers from 1 to 8 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 5,6 , 7 , 8 , 9 , 10 } b = { 1,2 , 3,4 , 5,6 , 7,8 } common elements = { 5,6 , 7 , 8 } = 4 elements answer : option e ." | a ) 5 , b ) 10 , c ) 8 , d ) 2 , e ) 4 | e | add(1, 5) | add(n0,n2)| | other | E |
the area of a square is equal to five times the area of a rectangle of dimensions 36 cm * 20 cm . what is the perimeter of the square ? | "area of the square = s * s = 5 ( 36 * 20 ) = > s = 60 = 60 cm perimeter of the square = 4 * 60 = 240 cm . answer : d" | a ) 289 cm , b ) 800 cm , c ) 829 cm , d ) 240 cm , e ) 289 cm | d | multiply(sqrt(multiply(rectangle_area(36, 20), divide(20, const_2))), const_4) | divide(n1,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry | D |
the cost price of a radio is rs . 2400 and it was sold for rs . 2100 , find the loss % ? | "2400 - - - - 300 100 - - - - ? = > 12.5 % answer : a" | a ) 12.5 % , b ) 11 % , c ) 13 % , d ) 15 % , e ) 12.8 % | a | multiply(divide(subtract(2400, 2100), 2400), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain | A |
kiran has 85 currency notes in all , some of which were of rs . 100 denomination and the remaining of rs . 50 denomination . the total amount of all these currency notes was rs . 5000 . how much amount did she have in the denomination of rs . 50 ? a . b . c . none of these d . answer : option b explanation : | let the number of 50 – rupee notes be x . then , the number of 100 - rupee notes = ( 85 – x ) 50 x + 100 ( 85 – x ) = 5000 x + 2 ( 85 – x ) = 100 = x = 70 so , required amount = rs . ( 50 x 70 ) = rs . 3500 answer : b | a ) 1900 , b ) 3500 , c ) 4000 , d ) 2000 , e ) 2500 | b | multiply(subtract(85, divide(subtract(5000, multiply(50, 85)), 50)), 50) | multiply(n0,n2)|subtract(n3,#0)|divide(#1,n2)|subtract(n0,#2)|multiply(n2,#3) | general | B |
a man can row his boat with the stream at 14 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 14 us = 4 s = ? s = ( 14 - 4 ) / 2 = 5 kmph answer : e" | a ) 1 kmph , b ) 3 kmph , c ) 8 kmph , d ) 7 kmph , e ) 5 kmph | e | divide(subtract(14, 4), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain | E |
walking at 5 / 6 th of its usual speed a cab is 8 mnts late . find its usual time to cover the journey ? | "new speed = 5 / 6 th of usual speed new time = 6 / 5 th of usual time 6 / 5 ut - ut = 8 m ut / 5 = 8 m ut = 40 m answer is c" | a ) 25 m , b ) 45 m , c ) 40 m , d ) 50 m , e ) 62 m | c | multiply(5, 8) | multiply(n0,n2)| | physics | C |
a baker makes chocolate cookies and peanut cookies . his recipes allow him to make chocolate cookie in batches of 7 and peanut cookies in batches of 6 . if he makes exactly 96 cookies , what is the minimum number of chocolate chip cookies he makes ? | "7 c + 6 p = 96 we need to maximize p to minimize c so that the eq is also satisfied try substitution for cp to solve so that eqn is satisfied the least value of c for which equation gets satisfied is 5 i . e . 7 * 6 + 6 * 9 = 42 + 54 = 96 hence e is the answer" | a ) 7 , b ) 14 , c ) 21 , d ) 28 , e ) 42 | e | multiply(divide(subtract(96, reminder(96, add(7, 6))), add(7, 6)), 7) | add(n0,n1)|reminder(n2,#0)|subtract(n2,#1)|divide(#2,#0)|multiply(n0,#3)| | general | E |
if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 55 , how much more remains to be paid ? | "90 % remains to be paid so the remaining amount is 9 * 55 = $ 495 . the answer is b ." | a ) $ 455 , b ) $ 495 , c ) $ 525 , d ) $ 550 , e ) $ 585 | b | subtract(multiply(55, divide(const_100, 10)), 55) | divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)| | general | B |
x varies inversely as square of y . given that y = 2 for x = 1 . the value of x for y = 6 will be equal to ? | "given x = k / y 2 , where k is a constant . now , y = 2 and x = 1 gives k = 4 . x = 4 / y 2 = > x = 4 / 62 , when y = 6 = > x = 4 / 36 = 1 / 9 . answer : d" | a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 0 , d ) 1 / 9 , e ) 1 / 5 | d | divide(multiply(1, power(2, const_2)), power(6, const_2)) | power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1)| | general | D |
a pipe can empty 2 / 3 rd of a cistern in 12 mins . in 8 mins , what part of the cistern will be empty ? | "2 / 3 - - - - 12 ? - - - - - 8 = = > 4 / 9 c" | a ) 2 / 3 , b ) 2 / 5 , c ) 4 / 9 , d ) 5 / 7 , e ) 4 / 11 | c | divide(multiply(divide(2, 3), 8), 12) | divide(n0,n1)|multiply(n3,#0)|divide(#1,n2)| | physics | C |
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