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 |
|---|---|---|---|---|---|---|---|
if a ( a - 9 ) = 10 and b ( b - 9 ) = 10 , where a ≠ b , then a + b = | "i . e . if a = - 1 then b = 10 or if a = 10 then b = - 1 but in each case a + b = - 1 + 10 = 9 answer : option d" | a ) − 48 , b ) − 2 , c ) 2 , d ) 9 , e ) 48 | d | subtract(subtract(subtract(subtract(add(add(9, 10), subtract(9, 10)), const_1), const_1), const_1), const_1) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1)| | general | D |
one pump drains one - half of a pond in 6 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ? | "first pump drains 1 / 2 of the tank in 6 hours so 12 hours it will take to drain the full tank . let , 2 nd pump drains the full tank in a hours so both together can drain ( 1 / 12 + 1 / a ) part in 1 hour son in 1 / 2 hour they drain 1 / 2 * ( 1 / 12 + 1 / a ) part of the tank given that in 1 / 2 hour they drain 1 / 2 of the tank hence we can say 1 / 2 * ( 1 / 12 + 1 / a ) = 1 / 2 solving u get a = 12 / 11 = 1.1 hence answer is b" | a ) 1 hour , b ) 1.1 hour , c ) 3 hours , d ) 5 hours , e ) 6 hours | b | divide(const_1, subtract(const_1, divide(const_1, multiply(6, const_2)))) | multiply(n0,const_2)|divide(const_1,#0)|subtract(const_1,#1)|divide(const_1,#2)| | physics | B |
a tank is filled in 10 hours by 3 pipes a , b and c . pipe a is twice as fast as pipe b , and b is twice as fast as c . how much time will pipe b alone take to fill the tank ? | 1 / a + 1 / b + 1 / c = 1 / 10 ( given ) also given that a = 2 b and b = 2 c = > 1 / 2 b + 1 / b + 2 / b = 1 / 10 = > ( 1 + 2 + 4 ) / 2 b = 1 / 10 = > 2 b / 7 = 10 = > b = 35 hours . answer : c | a ) 56 hours , b ) 28 hours , c ) 35 hours , d ) 66 hours , e ) 47 hours | c | divide(10, divide(const_2, add(3, const_4))) | add(n1,const_4)|divide(const_2,#0)|divide(n0,#1) | physics | C |
an amount of money is to be divided between p , q and r in the ratio of 3 : 7 : 12 . if the difference between the shares of p and q is rs . 2500 , what will be the difference between q and r ' s share ? | "4 - - - 2100 5 - - - ? = > 2625 answer : d" | a ) 2225 , b ) 2325 , c ) 2425 , d ) 2625 , e ) 2525 | d | multiply(subtract(12, 7), divide(2500, subtract(7, 3))) | subtract(n1,n0)|subtract(n2,n1)|divide(n3,#0)|multiply(#2,#1)| | general | D |
if the number of white balls in a box are 6 times the black balls , which of the below is the total number of balls , if there are 8 black balls in the box ? | let the number of black balls = x and the number of white balls = 6 x then , total number of balls in the box = x + 6 x = 7 x i . e . , the total number of balls must be a multiple of 7 from the given choices , only 56 is the multiple of 7 . hence , 56 are the number of balls in the box . answer is a . | a ) a - 56 , b ) b - 57 , c ) c - 58 , d ) d - 59 , e ) e - 60 | a | add(multiply(6, 8), 8) | multiply(n0,n1)|add(n1,#0) | general | A |
solution a is 20 % salt and solution b is 60 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ? | "forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 60 - 50 ) / ( 50 - 20 ) = 1 / 3 so the volume of the two solutions will be equal . answer has to be 1 : 3 e ." | a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 1 : 3 | e | divide(divide(subtract(multiply(50, divide(60, const_100)), multiply(50, divide(50, const_100))), subtract(divide(60, const_100), divide(20, const_100))), subtract(50, divide(subtract(multiply(50, divide(60, const_100)), multiply(50, divide(50, const_100))), subtract(divide(60, const_100), divide(20, const_100))))) | divide(n1,const_100)|divide(n4,const_100)|divide(n0,const_100)|multiply(n4,#0)|multiply(n4,#1)|subtract(#0,#2)|subtract(#3,#4)|divide(#6,#5)|subtract(n4,#7)|divide(#7,#8)| | other | E |
if the sum of 3 consecutive even numbers is 44 more than the average of these numbers , then the largest of these numbers is ? | answer : 24 explanation : let the smallest of these number be x . the other two numbers are ( x + 2 ) and ( x + 4 ) . x + ( x + 2 ) + ( x + 4 ) = ( x + ( x + 2 ) + ( x + 4 ) ) / 3 + 44 3 x + 3 * ( x + 2 ) + 3 * ( x + 4 ) = x + ( x + 2 ) + ( x + 4 ) + 132 9 x + 18 = 3 x + 138 6 x = 120 x = 20 therefore , the largest number is 24 . answer : a | a ) 24 , b ) 66 , c ) 9 , d ) 7 , e ) 01 | a | add(add(divide(subtract(44, const_4), const_2), 3), const_1) | subtract(n1,const_4)|divide(#0,const_2)|add(n0,#1)|add(#2,const_1) | general | A |
if the average ( arithmetic mean ) of a and b is 35 and the average of b and c is 80 , what is the value of c − a ? | "- ( a + b = 70 ) b + c = 160 c - a = 90 c . 90" | a ) 25 , b ) 50 , c ) 90 , d ) 140 , e ) it can not be determined from the information given . | c | subtract(multiply(80, const_2), multiply(35, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general | C |
it is currently 8 : 34 pm . at what time in the morning was it exactly 39,668 minutes ago ? | divide 39,668 by 60 to convert to minutes : 39,668 / 60 = 661 r 8 . that is 661 hours , 8 minutes . all of the answer choices are during the same hour before : 34 , so we can assume the 661 hours takes the time back into the 7 am hour to 7 : 34 am . going back 8 more minutes yields the time 7 : 26 am . c | a ) 7 : 22 , b ) 7 : 24 , c ) 7 : 26 , d ) 7 : 28 , e ) 7 : 30 | c | divide(8, subtract(34, const_3)) | subtract(n1,const_3)|divide(n0,#0)| | physics | C |
the temperature of a certain cup of coffee 15 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 * 2 ^ ( - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 30 minutes after it was poured was how many degrees fahrenheit ? | "first , we have to find a . we know that after t = 15 minutes the temperature f = 120 degrees . hence : 120 = 120 * ( 2 ^ - 15 a ) + 60 60 = 120 * ( 2 ^ - 15 a ) 60 / 120 = 2 ^ - 15 a 1 / 2 = 2 ^ - 15 a 2 ^ - 1 = 2 ^ - 15 a - 1 = - 15 a 1 / 15 = a now we need to find f after t = 30 minutes : f = 120 * ( 2 ^ - 1 / 15 * 30 ) + 60 f = 120 * ( 2 ^ - 2 ) + 60 f = 120 * ( 1 / 2 ^ 2 ) + 60 f = 120 * 1 / 4 + 60 f = 30 + 60 = 90 answer e !" | a ) 65 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | e | add(multiply(power(2, multiply(divide(60, 15), subtract(const_1, 2))), 120), 60) | divide(n4,n0)|subtract(const_1,n3)|multiply(#0,#1)|power(n3,#2)|multiply(n1,#3)|add(n4,#4)| | general | E |
what least number should be subtracted from 642 so that the remainder when divided by 5 , 7 , and 9 will leave in each case the same remainder 4 ? | "the lcm of 5 , 7 , and 9 is 315 . the next multiple is 2 * 315 = 630 . 630 + { remainder } = 630 + 4 = 634 , which is 8 less than 642 . answer : c ." | a ) 5 , b ) 6 , c ) 8 , d ) 9 , e ) 10 | c | subtract(642, add(multiply(multiply(multiply(5, 7), 9), const_2), 4)) | multiply(n1,n2)|multiply(n3,#0)|multiply(#1,const_2)|add(n4,#2)|subtract(n0,#3)| | general | C |
kanul spent $ 5000 in buying raw materials , $ 200 in buying machinery and 30 % of the total amount he had as cash with him . what was the total amount ? | "let the total amount be x then , ( 100 - 20 ) % of x = 5000 + 200 70 % of x = 5200 70 x / 100 = 5200 x = $ 52000 / 7 x = $ 7428.57 answer is c" | a ) $ 7456.00 , b ) $ 7500.55 , c ) $ 7428.57 , d ) $ 7852.56 , e ) $ 7864.00 | c | divide(add(5000, 200), subtract(const_1, divide(30, const_100))) | add(n0,n1)|divide(n2,const_100)|subtract(const_1,#1)|divide(#0,#2)| | gain | C |
find the area of a parallelogram with base 32 cm and height 18 cm ? | "area of a parallelogram = base * height = 32 * 18 = 576 cm 2 answer : a" | a ) 576 cm 2 , b ) 384 cm 2 , c ) 672 cm 2 , d ) 267 cm 2 , e ) 286 cm 2 | a | multiply(32, 18) | multiply(n0,n1)| | geometry | A |
a tradesman sold an article at a loss of 20 % . if the selling price had been increased by $ 100 , there would have been a gain of 5 % . what was the cost price of the article ? | "let c . p . be $ x then , ( 105 % of x ) - ( 80 % of x ) = 100 or 25 % of x = 100 x / 4 = 100 x = 400 c . p . = $ 400 correct option is d" | a ) $ 100 , b ) $ 200 , c ) $ 300 , d ) $ 400 , e ) $ 500 | d | divide(100, divide(add(20, 5), const_100)) | add(n0,n2)|divide(#0,const_100)|divide(n1,#1)| | gain | D |
the diagonals of a rhombus are 17 cm and 20 cm . find its area ? | "1 / 2 * 17 * 20 = 170 answer : d" | a ) 176 , b ) 186 , c ) 150 , d ) 170 , e ) 172 | d | rhombus_area(17, 20) | rhombus_area(n0,n1)| | geometry | D |
three numbers are in the ratio 4 : 5 : 6 and their average is 36 . the largest number is : | "explanation : let the numbers be 4 x , 5 x and 6 x . therefore , ( 4 x + 5 x + 6 x ) / 3 = 36 15 x = 108 x = 7.2 largest number = 6 x = 43.2 . answer c" | a ) 28 , b ) 32 , c ) 43.2 , d ) 42 , e ) 45 | c | add(multiply(multiply(4, 6), const_100), multiply(5, 6)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | general | C |
a student got 72 % in math and 84 % in history . to get an overall average of 75 % , how much should the student get in the third subject ? | "72 + 84 + x = 3 * 75 x = 69 the answer is c ." | a ) 65 % , b ) 67 % , c ) 69 % , d ) 71 % , e ) 73 % | c | subtract(multiply(const_3, 75), add(72, 84)) | add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0)| | general | C |
the area of one square is x ^ 2 + 10 x + 25 and the area of another square is 4 x ^ 2 − 20 x + 25 . if the sum of the perimeters of both squares is 36 , what is the value of x ? | "the areas are ( x + 5 ) ^ 2 and ( 2 x - 5 ) ^ 2 . the lengths of the sides are x + 5 and 2 x - 5 . if we add the two perimeters : 4 ( x + 5 ) + 4 ( 2 x - 5 ) = 36 12 x = 36 x = 3 the answer is a ." | a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | a | divide(subtract(36, subtract(multiply(4, divide(10, 2)), 10)), 10) | divide(n1,n0)|multiply(#0,n3)|subtract(#1,n1)|subtract(n7,#2)|divide(#3,n1)| | general | A |
what is the sum of all remainders obtained when the first 200 natural numbers are divided by 9 ? | "a positive integer can give only the following 9 remainders when divided by 9 : 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , and 0 . 1 divided by 9 gives the remainder of 1 ; 2 divided by 9 gives the remainder of 2 ; . . . 8 divided by 9 gives the remainder of 8 ; 9 divided by 9 gives the remainder of 0 . we ' ll have 11 such blocks , since 99 / 9 = 11 . the last will be : 91 divided by 9 gives the remainder of 1 ; 92 divided by 9 gives the remainder of 2 ; . . . 98 divided by 9 gives the remainder of 8 ; 99 divided by 9 gives the remainder of 0 . the last number , 100 , gives the remainder of 1 when divided by 9 , thus the sum of all remainders will be : 11 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 0 ) + 1 = 401 . answer : b ." | a ) 397 , b ) 401 , c ) 403 , d ) 405 , e ) 399 | b | add(const_3, multiply(divide(subtract(200, reminder(200, 9)), 9), add(add(add(const_1, const_2), const_3), add(add(const_1, const_4), add(const_4, add(add(const_1, const_2), const_3)))))) | add(const_1,const_2)|add(const_1,const_4)|reminder(n0,n1)|add(#0,const_3)|subtract(n0,#2)|add(#3,const_4)|divide(#4,n1)|add(#1,#5)|add(#3,#7)|multiply(#8,#6)|add(#9,const_3)| | general | B |
if a 2 - b 2 = 5 and a * b = 2 , find a 4 + b 4 . | "a 2 - b 2 = 5 : given a 4 + b 4 - 2 a 2 b 2 = 52 : square both sides and expand . a * b = 2 : given a 2 b 2 = 22 : square both sides . a 4 + b 4 - 2 ( 4 ) = 25 : substitute a 4 + b 4 = 33 correct answer e" | a ) 50 , b ) 72 , c ) 25 , d ) 92 , e ) 33 | e | add(power(5, 2), multiply(power(2, 2), 2)) | power(n3,n0)|power(n2,n0)|multiply(#0,n0)|add(#2,#1)| | general | E |
a batsman in his 10 th innings makes a score of 85 , and thereby increases his average by 3 . what is his average after the 10 th innings ? he had never been ’ not out ’ . | "average score before 10 th innings = 85 - 3 × 10 = 50 average score after 10 th innings = > 50 + 3 = 53 answer : b" | a ) 47 , b ) 53 , c ) 39 , d ) 43 , e ) 42 | b | add(subtract(85, multiply(3, 10)), 3) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general | B |
if an article is sold at 18 % profit instead of 11 % profit , then the profit would be $ 63 more . what is the cost price ? | "7 % * cost price = $ 63 1 % * cost price = $ 63 / 7 = $ 9 the cost price is $ 900 . the answer is b ." | a ) $ 600 , b ) $ 900 , c ) $ 1200 , d ) $ 1500 , e ) $ 1800 | b | multiply(divide(63, 11), const_100) | divide(n2,n1)|multiply(#0,const_100)| | gain | B |
an article costing rs . 160 is sold at 20 % discount on a mark - up price . what is the selling price after discount ? | 160 * 80 / 100 = 128 answer : c | a ) 106 , b ) 116 , c ) 128 , d ) 136 , e ) 146 | c | subtract(160, divide(multiply(160, 20), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1) | gain | C |
in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 39 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ? | "working backward from the day it ' s covered : day 39 : fully covered day 38 : half covered so 38 days answer : d" | a ) 36 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47 | d | subtract(39, const_1) | subtract(n0,const_1)| | physics | D |
by investing in 16 % stock at 64 , one earns rs . 1500 . the investment made is : | "explanation : to earn rs . , investment = rs . 64 . to earn rs . 1500 , investment = = rs . 5760 . answer : c ) 5760" | a ) 5760 , b ) 7667 , c ) 4445 , d ) 4566 , e ) 3344 | c | multiply(divide(multiply(multiply(multiply(const_4, const_2), multiply(const_4, const_2)), const_3), divide(const_100, const_2)), subtract(1500, multiply(const_4, const_100))) | divide(const_100,const_2)|multiply(const_2,const_4)|multiply(const_100,const_4)|multiply(#1,#1)|subtract(n2,#2)|multiply(#3,const_3)|divide(#5,#0)|multiply(#6,#4)| | gain | C |
how many 3 - digit even numbers are possible such that if one of the digits is 5 , the next / succeeding digit to it should be 1 ? | 510 , 512 , 514 , 516 , and 518 , so total 5 . hence option a . | a ) 5 , b ) 305 , c ) 365 , d ) 405 , e ) 495 | a | add(add(3, 1), 1) | add(n0,n2)|add(n2,#0) | general | A |
jim â € ™ s taxi service charges an initial fee of $ 2.25 at the beginning of a trip and an additional charge of $ 0.3 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | let the fixed charge of jim â € ™ s taxi service = 2.25 $ and charge per 2 / 5 mile ( . 4 mile ) = . 3 $ total charge for a trip of 3.6 miles = 2.25 + ( 3.6 / . 4 ) * . 3 = 2.25 + 9 * . 3 = 4.95 $ answer b | a ) $ 3.15 , b ) $ 4.95 , c ) $ 4.80 , d ) $ 5.05 , e ) $ 5.40 | b | add(multiply(divide(3.6, divide(2, 5)), 0.3), 2.25) | divide(n2,n3)|divide(n4,#0)|multiply(n1,#1)|add(n0,#2) | general | B |
if 0.60 : x : : 6 : 2 , then x is equal to | "sol . ( x × 6 ) = ( 0.60 × 2 ) ⇒ x = 0.60 / 6 = 0.2 . answer a" | a ) 0.2 , b ) 0.5 , c ) 2.5 , d ) 1.8 , e ) none | a | divide(multiply(0.60, 2), 6) | multiply(n0,n2)|divide(#0,n1)| | general | A |
an air - conditioning unit costs $ 470 . on december there was a discount for christmas of 16 % . 6 months later , the holiday season was over so the company raised the price of the air - conditioning by 12 % . how much will an air - conditioning unit cost in november ? | if its previous november ( before discount ) then price is $ 470 . but if its november of next year then 16 % discount on $ 470 = 470 ( 1 - 16 / 100 ) = $ 394.8 again a corrected raised price of 12 % over $ 394.8 = 394.8 ( 1 + 12 / 100 ) = 442.176 ~ $ 4442 ans a | a ) $ 442 , b ) $ 470 , c ) $ 472 , d ) $ 484 , e ) $ 491 | a | multiply(add(divide(12, const_100), const_1), multiply(subtract(const_1, divide(16, const_100)), 470)) | divide(n3,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(n0,#3)|multiply(#2,#4) | gain | A |
lilly has 10 fish and rosy has 14 fish . in total , how many fish do they have in all ? | "10 + 14 = 24 the answer is d ." | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | add(10, 14) | add(n0,n1)| | general | D |
if 9 a - b = 10 b + 80 = - 12 b - 2 a , what is the value of 2 a + 22 b ? | "this implies 9 a - b = 10 b + 80 , 9 a - b = - 12 b - 2 a , 10 b + 80 = - 12 b - 2 a manipulating the second equation gives us 10 b + 80 = - 12 b - 2 a = = > 2 a + 22 b = - 80 answer is c" | a ) - 4 , b ) - 2 , c ) - 80 , d ) 2 , e ) 4 | c | multiply(negate(multiply(divide(80, 2), 2)), 2) | divide(n2,n4)|multiply(#0,n4)|negate(#1)|multiply(n5,#2)| | general | C |
set a consists of the integers from 3 to 12 , inclusive , while set b consists of the integers from 6 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 3,4 , 5,6 , 7,8 , 9,10 , 11,12 } b = { 6 , 7,8 , 9,10 , 11,12 . . . } thus we see that there are 7 distinct integers that are common to both . b is the correct answer ." | a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | add(6, 3) | add(n0,n2)| | other | B |
a vessel of capacity 2 litre has 20 % of alcohol and another vessel of capacity 6 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "20 % of 2 litres = 0.4 litres 40 % of 6 litres = 2.4 litres therefore , total quantity of alcohol is 2.8 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 28 % answer : e" | a ) 31 % . , b ) 71 % . , c ) 49 % . , d ) 29 % . , e ) 28 % . | e | multiply(divide(add(multiply(divide(20, const_100), 2), multiply(divide(40, const_100), 6)), 10), const_100) | divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100)| | general | E |
rectangular tile each of size 25 cm by 16 cm must be laid horizontally on a rectangular floor of size 180 cm by 120 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | "area of tile = 25 * 16 = 400 area of floor = 180 * 120 = 21600 no of tiles = 21600 / 400 = 54 so , the no of tile = 54 answer : b" | a ) 52 , b ) 54 , c ) 55 , d ) 65 , e ) 48 | b | divide(multiply(180, 120), multiply(25, 16)) | multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)| | geometry | B |
we have a rectangular metallic piece of paper that covers exactly the area of a cube . the length of the piece of paper is 144 inches and the width is 72 inches . what is the volume of the cube in cubic feet is 1 feet is 12 inches ? | "l = 144 / 12 = 12 ft w = 72 / 12 = 6 ft area of paper = 72 area of cube = 12 * side ^ 2 side of cube = 6 v of cube = 216" | a ) a 216 , b ) b 196 , c ) c 170 , d ) d 140 , e ) e 121 | a | add(cube_edge_by_volume(144), 144) | cube_edge_by_volume(n0)|add(n0,#0)| | geometry | A |
ages of two persons differ by 18 years . if 4 years ago , the elder one was 4 times as old the younger one , find their present age | explanation : let the age of younger person be x , then elder person ' s age is ( x + 18 ) = > 4 ( x - 4 ) = ( x + 18 - 4 ) [ 4 years before ] = > 4 x - 16 = x + 14 = > x = 10 . so elder person ' s age is x + 18 = 28 answer : option c | a ) 6,24 , b ) 9,27 , c ) 10,28 , d ) 18,36 , e ) 19,37 | c | subtract(divide(subtract(add(multiply(4, 4), 18), 4), subtract(4, 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 | C |
a soccer team played 140 games and won 50 percent of them . how many games did it win ? | "50 % of 140 = x 0.50 * 140 = x 70 = x answer : d" | a ) 140 , b ) 94 , c ) 104 , d ) 70 , e ) 80 | d | divide(multiply(50, 140), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain | D |
find the fraction which has the same ratio to 2 / 5 that 3 / 8 has to 1 / 5 | "p : 2 / 5 = 3 / 8 : 1 / 5 as the product of the means is equal to the product of the extremes . p * 1 / 5 = 2 / 5 * 3 / 8 p * 1 / 5 = 6 / 40 p = 3 / 4 = > p = 3 / 4 answer : b" | a ) 1 / 15 , b ) 3 / 4 , c ) 5 / 8 , d ) 9 / 6 , e ) 8 / 7 | b | multiply(divide(divide(3, 8), divide(1, 5)), divide(2, 5)) | divide(n2,n3)|divide(n4,n5)|divide(n0,n1)|divide(#0,#1)|multiply(#3,#2)| | other | B |
if x and y are numbers such that ( x + 4 ) ( y - 4 ) = 0 , what is the smallest possible value of x ^ 2 + y ^ 2 | "from ( x + 4 ) ( y - 4 ) = 0 it follows that either x = - 4 or y = 4 . thus either x ^ 2 = 16 or y ^ 2 = 16 . now , if x ^ 2 = 16 , then the least value of y ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 16 + 0 = 16 . similarly if y ^ 2 = 16 , then the least value of x ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 0 + 16 = 16 . answer : d ." | a ) 0 , b ) 9 , c ) 12 , d ) 16 , e ) 18 | d | power(4, 2) | power(n0,n3)| | general | D |
in a class of 52 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ? | total = english + german - both + neither - - > 52 = english + 22 - 12 + 0 - - > english = 42 - - > only english = english - both = 42 - 12 = 30 answer : a . | a ) 30 , b ) 10 , c ) 18 , d ) 28 , e ) 32 | a | subtract(subtract(add(52, 12), 22), 12) | add(n0,n1)|subtract(#0,n2)|subtract(#1,n1) | other | A |
the radius of a semi circle is 2.1 cm then its perimeter is ? | "36 / 7 r = 2.1 = 10.80 answer : d" | a ) 32.5 , b ) 11.12 , c ) 32.1 , d ) 10.8 , e ) 32.3 | d | add(divide(circumface(2.1), const_2), multiply(2.1, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | physics | D |
300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 39 = 50 / 3 3 ( x + 300 ) = 1950 è x = 350 m . answer : c | a ) 388 , b ) 266 , c ) 350 , d ) 112 , e ) 134 | c | subtract(multiply(speed(300, 18), 39), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0) | physics | C |
the average of first 25 natural numbers is ? | "sum of 25 natural no . = 650 / 2 = 325 average = 325 / 25 = 13 answer : b" | a ) 14 , b ) 13 , c ) 15 , d ) 18 , e ) 12 | b | add(25, const_1) | add(n0,const_1)| | general | B |
the length of the bridge , which a train 160 metres long and travelling at 45 km / hr can cross in 30 seconds , is ? | "speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 160 + x ) / 30 = 25 / 2 = > 2 ( 160 + x ) = 750 = > x = 215 m . answer : e" | a ) 255 , b ) 267 , c ) 245 , d ) 277 , e ) 215 | e | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 160) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics | E |
in a simultaneous throw of a pair of dice , find the probability of getting a total more than 4 | "total number of cases = 3 * 3 = 9 favourable cases = [ ( 2,3 ) , ( 3,2 ) , ( 3,3 ) ] = 3 so probability = 3 / 9 = 1 / 3 answer is e" | a ) 1 / 2 , b ) 7 / 12 , c ) 5 / 13 , d ) 5 / 12 , e ) 1 / 3 | e | divide(subtract(4, multiply(const_2, const_3)), 4) | multiply(const_2,const_3)|subtract(n0,#0)|divide(#1,n0)| | general | E |
a can do a piece of work in 4 days . b can do it in 15 days . with the assistance of c they completed the work in 24 days . find in how many days can c alone do it ? | "c = 1 / 2 - 1 / 24 - 1 / 15 = 0.391 days answer : a" | a ) 0.391 days , b ) 0.491 days , c ) 0.331 days , d ) 1.391 days , e ) 0.301 days | a | divide(multiply(4, 15), divide(subtract(multiply(4, 15), multiply(add(divide(multiply(4, 15), 4), divide(multiply(4, 15), 15)), 24)), 24)) | multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|multiply(n2,#3)|subtract(#0,#4)|divide(#5,n2)|divide(#0,#6)| | physics | A |
the interest on a certain deposit at 5 % per annum is rs . 101.20 in one year . how much will the additional interest in one year be on the same deposit at 6 % per annum ? | explanation : principal , p = 100 × si / rt = 100 × 101.20 / 5 × 1 = 20 × 101.20 = rs . 2024 simple interest for rs . 2024 at 6 % per annum for 1 year , si = 2024 × 6 × 1100 = 121.44 additional interest = rs . 121.44 - rs . 101.20 = rs . 20.24 answer : option c | a ) rs . 20.8 , b ) rs . 19.74 , c ) rs . 20.24 , d ) rs . 19.5 , e ) rs . 19.00 | c | subtract(divide(multiply(divide(multiply(const_100, 101.2), 5), 6), const_100), 101.2) | multiply(n1,const_100)|divide(#0,n0)|multiply(n2,#1)|divide(#2,const_100)|subtract(#3,n1) | gain | C |
the probability that event a occurs is 0.4 , and the probability that events a and b both occur is 0.25 . if the probability that either event a or event b occurs is 0.6 , what is the probability that event b will occur ? | p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.6 = 0.4 + p ( b ) - 0.25 p ( b ) = 0.45 answer : c | a ) 0.05 , b ) 0.15 , c ) 0.45 , d ) 0.5 , e ) 0.55 | c | subtract(add(0.6, 0.25), 0.4) | add(n1,n2)|subtract(#0,n0) | other | C |
a man engaged a servant on the condition that he would pay him rs . 900 and a uniform after 1 year service . he served onlyfor 9 months and received uniform & rs . 650 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b" | a ) rs . 90 , b ) rs . 100 , c ) rs . 150 , d ) rs . 190 , e ) rs . 200 | b | multiply(divide(subtract(multiply(1, 900), multiply(multiply(const_3, const_4), 9)), multiply(multiply(const_3, const_4), const_1)), const_4) | multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4)| | general | B |
what is the maximum number q of 27 cubic centimetre cubes that can fit in a rectangular box measuring 8 centimetre x 9 centimetre x 12 centimetre ? | 27 cubic centimetre cubes gives side = 3 cm so if : l * w * h is 9 * 12 * 8 , then max . cube we can have are 3 * 4 * 2 = 24 l * w * h is 9 * 8 * 12 , then max . cube we can have are 3 * 2 * 4 = 24 l * w * h is 12 * 8 * 9 , then max . cube we can have are 4 * 2 * 3 = 24 l * w * h is 12 * 9 * 8 , then max . cube we can have are 4 * 3 * 2 = 24 l * w * h is 8 * 12 * 9 , then max . cube we can have are 2 * 4 * 3 = 24 l * w * h is 8 * 9 * 12 , then max . cube we can have are 2 * 3 * 4 = 24 in all cases we get q = 24 cubes . ans . c | ['a ) 36', 'b ) 32', 'c ) 24', 'd ) 21', 'e ) 15'] | c | divide(volume_rectangular_prism(multiply(const_2, const_3), 9, 12), 27) | multiply(const_2,const_3)|volume_rectangular_prism(n2,n3,#0)|divide(#1,n0) | geometry | C |
a , b , c , d and e are 5 consecutive points on a straight line . if bc = 3 cd , de = 7 , ab = 5 and ac = 11 , what is the length of ae ? | ac = 11 and ab = 5 , so bc = 6 . bc = 3 cd so cd = 2 . the length of ae is ab + bc + cd + de = 5 + 6 + 2 + 7 = 20 the answer is c . | a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | c | add(add(11, divide(subtract(11, 5), 3)), 7) | subtract(n4,n0)|divide(#0,n1)|add(n4,#1)|add(n2,#2) | physics | C |
a man buy a book in rs 50 & sale it rs 50 . what is the rate of profit ? ? ? | "cp = 50 sp = 50 profit = 50 - 50 = 0 % = 0 / 50 * 100 = 0 % answer : b" | a ) 10 % , b ) 0 % , c ) 30 % , d ) 25 % , e ) 28 % | b | multiply(divide(subtract(50, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain | B |
a man can swim in still water at 7.5 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ? | "m = 7.5 s = x ds = 7.5 + x us = 7.5 - x 7.5 + x = ( 7.5 - x ) 2 7.5 + x = 15 - 2 x 3 x = 7.5 x = 2.5 answer : b" | a ) 3 , b ) 2.5 , c ) 2.25 , d ) 1.5 , e ) 4 | b | divide(7.5, const_3) | divide(n0,const_3)| | general | B |
in a sample of 800 high school students in which all students are either freshmen , sophomores , juniors , or seniors , 26 percent are juniors and 75 percent are not sophomores . if there are 160 seniors , how many more freshmen than sophomores are there among the sample of students ? | "200 are sophomores . the number of freshmen is 600 - 160 - 0.26 ( 800 ) = 232 the answer is a ." | a ) 32 , b ) 40 , c ) 48 , d ) 54 , e ) 60 | a | subtract(divide(multiply(subtract(subtract(subtract(const_100, multiply(divide(160, 800), const_100)), 26), subtract(const_100, 75)), 800), const_100), divide(multiply(subtract(const_100, 75), 800), const_100)) | divide(n3,n0)|subtract(const_100,n2)|multiply(#0,const_100)|multiply(n0,#1)|divide(#3,const_100)|subtract(const_100,#2)|subtract(#5,n1)|subtract(#6,#1)|multiply(n0,#7)|divide(#8,const_100)|subtract(#9,#4)| | gain | A |
a train 220 m long is running with a speed of 90 km / hr . in what time will it pass a bus that is running with a speed of 60 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to bus = 90 + 60 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . time taken to pass the bus = 220 * 3 / 125 = 5.28 sec . answer : b" | a ) 4.37 , b ) 5.28 , c ) 6.75 , d ) 8 , e ) 3.1 | b | divide(220, multiply(add(90, 60), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics | B |
a cycle is bought for rs . 1000 and sold for rs . 2000 , find the gain percent ? | "1000 - - - - 1000 100 - - - - ? = > 100 % answer : c" | a ) 11 , b ) 20 , c ) 100 , d ) 77 , e ) 12 | c | multiply(divide(subtract(2000, 1000), 1000), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain | C |
the diagonals of a rhombus are 12 cm and 10 cm . find its area ? | "1 / 2 * 12 * 10 = 60 answer : e" | a ) 158 , b ) 129 , c ) 150 , d ) 123 , e ) 60 | e | rhombus_area(12, 10) | rhombus_area(n0,n1)| | geometry | E |
the average weight of 6 students decreases by 3 kg when one of them weighing 80 kg is replaced by a new student . the weight of the student is | "explanation : let the weight of student be x kg . given , difference in average weight = 3 kg = > ( 80 - x ) / 6 = 3 = > x = 62 answer : a" | a ) 62 kg , b ) 60 kg , c ) 70 kg , d ) 72 kg , e ) none of these | a | subtract(80, multiply(6, 3)) | multiply(n0,n1)|subtract(n2,#0)| | general | A |
a person decided to build a house in 100 days . he employed 100 men in the beginning and 100 more after 25 days and completed the construction in stipulated time . if he had not employed the additional men , how many days behind schedule would it have been finished ? | "200 men do the rest of the work in 100 - 25 = 75 days 100 men can do the rest of the work in 75 * 200 / 100 = 75 days required number of days = 75 - 80 = 5 days answer is b" | a ) 7 , b ) 5 , c ) 2 , d ) 3 , e ) 1 | b | subtract(100, 25) | subtract(n0,n3)| | general | B |
a furniture dealer purchased a desk for $ 240 and then set the selling price equal to the purchase price plus a markup that was 40 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ? | "purchase price = 240 selling price = x 240 + 0.4 * x = x 0.6 * x = 240 x = 400 profit = 400 - 240 = 160 answer : c" | a ) $ 40 , b ) $ 60 , c ) $ 160 , d ) $ 90 , e ) $ 100 | c | divide(multiply(subtract(divide(240, subtract(const_1, divide(40, const_100))), 240), const_100), 240) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|multiply(#3,const_100)|divide(#4,n0)| | gain | C |
bruce purchased 8 kg of grapes at the rate of 70 per kg and 8 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 70 × 8 = 560 . cost of 8 kg of mangoes = 55 × 8 = 440 . total cost he has to pay = 560 + 440 = 1000 c" | a ) a ) 1040 , b ) b ) 1050 , c ) c ) 1000 , d ) d ) 1065 , e ) e ) 1075 | c | add(multiply(8, 70), multiply(8, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain | C |
a , b , c subscribe rs . 50,000 for a business . if a subscribes rs . 4000 more than b and b rs . 5000 more than c , out of a total profit of rs . 37,000 , what will be the amount a receives ? | "total amount invested = 50000 assume that investment of c = x . then investment of b = 5000 + x , investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000 ⇒ 3 x + 14000 = 50000 ⇒ 3 x = 50000 – 14000 = 36000 ⇒ x = 36000 / 3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a , b and c = 21000 : 17000 : 12000 = 21 : 17 : 12 share of a = total profit × 21 / 50 = 37000 × 21 / 50 = 15540 answer is e" | a ) 14700 , b ) 14500 , c ) 14900 , d ) 14300 , e ) 15540 | e | multiply(add(divide(subtract(multiply(5000, const_10), add(add(4000, 5000), 5000)), const_3), add(4000, 5000)), divide(multiply(multiply(const_3, const_12), const_1000), multiply(5000, const_10))) | add(n1,n2)|multiply(n2,const_10)|multiply(const_12,const_3)|add(n2,#0)|multiply(#2,const_1000)|divide(#4,#1)|subtract(#1,#3)|divide(#6,const_3)|add(#0,#7)|multiply(#8,#5)| | general | E |
how many numbers from 10 to 40 are exactly divisible by 3 ? | "12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 . 10 numbers . 10 / 3 = 3 and 40 / 3 = 13 = = > 13 - 3 = 10 . therefore 10 digits d )" | a ) 13 , b ) 15 , c ) 16 , d ) 10 , e ) 18 | d | add(divide(subtract(multiply(floor(divide(40, 3)), 3), multiply(add(floor(divide(10, 3)), const_1), 3)), 3), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general | D |
a shop produces sarongs . the daily average production is given by 5 n + 20 , where n is the number of workers aside from the owner . in the first k days , 500 units are produced , and then 5 workers are added to the team . after another k days , the cumulative total is 800 . how many workers were part of the latter production run ? | "the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 1250 . . thus for the current period = 800 - 500 = 300 ( 5 n + 45 ) k = 300 k = 300 / 5 n + 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 equate 1 and 2 500 / 5 n + 20 = 300 / 5 n + 45 500 ( 5 n + 45 ) = 300 ( 5 n + 20 ) 2500 n + 22500 = 1500 n + 6000 1000 n = - 16500 n = 16 thus n + 5 = 21 hence d" | a ) a ) 6 , b ) b ) 10 , c ) c ) 11 , d ) d ) 21 , e ) e ) 23.5 | d | add(20, 5) | add(n0,n1)| | general | D |
a train running at the speed of 60 km / hr crosses a pole in 6 seconds . what is the length of the train ? | "speed = 60 x 5 / 18 m / sec = 50 / 3 m / sec . length of the train = ( speed x time ) . length of the train = 50 / 3 x 6 m = 100 m . option c" | a ) 120 metres , b ) 180 metres , c ) 100 metres , d ) 228 metres , e ) 150 metres | c | multiply(divide(multiply(60, const_1000), const_3600), 6) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | C |
a no . when divided by 222 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 17 ? | "222 + 43 = 265 / 17 = 10 ( remainder ) d" | a ) 2 , b ) 7 , c ) 9 , d ) 10 , e ) 15 | d | divide(add(222, 43), 17) | add(n0,n1)|divide(#0,n2)| | general | D |
find the area of a cuboid of length 12 cm , breadth 6 cm . and height 10 cm . | "area of a cuboid = lxbxh = 12 cm x 6 cm x 10 cm = 720 cm cube answer : c" | a ) 400 cm cube , b ) 410 cm cube , c ) 720 cm cube , d ) 730 cm cube , e ) 480 cm cube | c | multiply(multiply(12, 6), 10) | multiply(n0,n1)|multiply(n2,#0)| | physics | C |
john had a stock of 1400 books in his bookshop . he sold 62 on monday , 62 on tuesday , 60 on wednesday , 48 on thursday and 40 on friday . what percentage of the books were not sold ? | "let n be the total number of books sold . hence n = 62 + 62 + 60 + 48 + 40 = 272 let m be the books not sold m = 1400 - n = 1400 - 272 = 1128 percentage books not sold / total number of books = 1128 / 1200 = 0.81 = 80.57 % correct answer e" | a ) 81.57 % , b ) 36.5 % , c ) 80.67 % , d ) 56.5 % , e ) 80.57 % | e | multiply(divide(subtract(1400, add(add(add(62, 62), add(60, 48)), 40)), 1400), const_100) | add(n1,n2)|add(n3,n4)|add(#0,#1)|add(n5,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)| | gain | E |
the s . i . on a certain sum of money for 4 years at 12 % per annum is half the c . i . on rs . 6000 for 2 years at 15 % per annum . the sum placed on s . i . is ? | c . i . = [ 6000 * ( 1 + 15 / 100 ) 2 - 6000 ] = ( 6000 * 23 / 20 * 23 / 20 - 6000 ) = rs . 1935 . sum = ( 967.5 * 100 ) / ( 4 * 12 ) = rs . 2015.625 answer : d | a ) 2019.625 , b ) 2017.625 , c ) 2013.625 , d ) 2015.625 , e ) 2011.625 | d | divide(divide(subtract(multiply(6000, power(add(const_1, divide(15, const_100)), 2)), 6000), 2), multiply(4, divide(12, const_100))) | divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3)| | gain | D |
a participated in cycling contest and he drove the lap at the rate of 6 kmph , 12 kmph , 18 kmph , 24 kmph . . what is his average speed ? | = 4 x / ( x / 6 + x / 12 + x / 18 + x / 24 ) = 11.52 answer : d | a ) 15 kmph , b ) 18 kmph , c ) 14.25 kmph , d ) 11.52 kmph , e ) 16 kmph | d | divide(subtract(multiply(const_100, const_3), const_12), add(24, const_1)) | add(n3,const_1)|multiply(const_100,const_3)|subtract(#1,const_12)|divide(#2,#0) | physics | D |
if the sum of two numbers is 22 and the sum of their squares is 460 , then the product of the numbers is | "according to the given conditions x + y = 22 and x ^ 2 + y ^ 2 = 460 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 22 ^ 2 = 460 + 2 xy so xy = 24 / 2 = 12 answer : c" | a ) 40 , b ) 44 , c ) 12 , d ) 88 , e ) 48 | c | divide(subtract(power(22, const_2), 460), const_2) | power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)| | general | C |
a milk man has 20 liters of milk . if he mixes 25 liters of water , which is freely available , in 20 liters of pure milk . if the cost of pure milk is rs . 18 per liter , then the profit of the milkman , when he sells all the mixture at cost price is : | "explanation : when the water is freely available and all the water is sold at the price of the milk , then the water gives the profit on the cost of 20 liters of milk . therefore , profit percentage = 45 % . answer : b" | a ) 20 % , b ) 45 % , c ) 33.33 % , d ) 18 % , e ) none of these | b | multiply(divide(subtract(multiply(add(add(add(20, 25), const_10), add(divide(25, const_2), const_3)), 18), multiply(20, 18)), multiply(20, 18)), const_100) | add(n1,n2)|divide(n1,const_2)|multiply(n0,n3)|add(#0,const_10)|add(#1,const_3)|add(#3,#4)|multiply(n3,#5)|subtract(#6,#2)|divide(#7,#2)|multiply(#8,const_100)| | gain | B |
one hour after yolanda started walking from x to y , a distance of 52 miles , bob started walking along the same road from y to x . if yolanda ' s walking rate was 3 miles per hour and bob т ' s was 4 miles per hour , how many miles had bob walked when they met ? | "when b started walking y already has covered 3 miles out of 52 , hence the distance at that time between them was 52 - 3 = 49 miles . combined rate of b and y was 3 + 4 = 7 miles per hour , hence they would meet each other in 49 / 7 = 7 hours . in 6 hours b walked 7 * 4 = 28 miles . answer : a ." | a ) 28 , b ) 23 , c ) 22 , d ) 21 , e ) 19.5 | a | multiply(divide(subtract(52, 3), add(3, 4)), 4) | add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n2,#2)| | physics | A |
of the 120 passengers on flight 750 , 55 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 / 3 of the passengers in first class are male , how many females are there in coach class ? | number of passengers on flight = 120 number of female passengers = . 5 * 120 = 66 number of passengers in first class = ( 10 / 100 ) * 120 = 12 number of passengers in coach class = ( 90 / 100 ) * 120 = 108 number of male passengers in first class = 1 / 3 * 12 = 4 number of female passengers in first class = 12 - 4 = 8 number of female passengers in coach class = 66 - 8 = 58 answer e | a ) 44 , b ) 48 , c ) 50 , d ) 52 , e ) 58 | e | subtract(multiply(120, divide(55, const_100)), subtract(multiply(120, divide(10, const_100)), divide(multiply(120, divide(10, const_100)), 3))) | divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|divide(#3,n5)|subtract(#3,#4)|subtract(#2,#5) | gain | E |
in cliff ’ s impressive rock collection , there are half as many igneous rocks as sedimentary rocks . of the igneous rocks , 3 / 5 are shiny and the rest are matte , while 1 / 5 of the sedimentary rocks are shiny . if there are 30 shiny igneous rocks , how many total rocks does cliff have ? | "we can start with the known quantity and then go on to find the others . shiny igneous ricks are 30 . these are ( 3 / 5 ) of total igneous rocks . ( 3 / 5 ) * total igneous rocks = 30 total igneous rocks = 30 * ( 5 / 3 ) = 50 total sedimentary rocks = 2 * total igneous rocks = 2 * 50 = 100 total number of rocks = 50 + 100 = 150 answer ( d )" | a ) 30 , b ) 45 , c ) 60 , d ) 150 , e ) 135 | d | divide(multiply(divide(multiply(multiply(3, 5), 30), 3), 5), 3) | multiply(n0,n1)|multiply(n4,#0)|divide(#1,n0)|multiply(n1,#2)|divide(#3,n0)| | general | D |
company kw is being sold , and both company a and company b were considering the purchase . the price of company kw is 60 % more than company a has in assets , and this same price is also 100 % more than company b has in assets . if companies a and b were to merge and combine their assets , the price of company kw would be approximately what percent of these combined assets ? | "let the price of company a ' s assets be 100 price of assets of kw is 60 % more than company a ' s assets which is 160 price of assets of kw is 100 % more than company b ' s assets which means price of company b ' s assets is half the price of kw = 80 a + b = 180 kw = 160 kw / ( a + b ) * 100 = 160 / 180 * 100 = 88.88 % or 89 % c" | a ) 66 % , b ) 75 % , c ) 89 % , d ) 116 % , e ) 150 % | c | multiply(divide(add(100, 60), add(100, divide(add(100, 60), const_2))), 100) | add(n0,n1)|divide(#0,const_2)|add(n1,#1)|divide(#0,#2)|multiply(#3,n1)| | gain | C |
average expenditure of a person for the first 3 days of a week is rs . 350 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is : | "explanation : assumed mean = rs . 350 total excess than assumed mean = 4 × ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 / 7 = rs . 40 therefore , average expenditure for 7 days = rs . 350 + rs . 40 = rs . 390 correct option : c" | a ) 350 , b ) 370 , c ) 390 , d ) 430 , e ) none | c | add(350, divide(multiply(4, subtract(420, 350)), add(3, 4))) | add(n0,n2)|subtract(n3,n1)|multiply(n2,#1)|divide(#2,#0)|add(n1,#3)| | general | C |
when 6 is added to half of one - third of one - fifth of a number , the result is one - fifteenth of the number . find the number ? | "explanation : let the number be 26 + 1 / 2 [ 1 / 3 ( a / 5 ) ] = a / 15 = > 6 = a / 30 = > a = 180 answer : b" | a ) 32 , b ) 180 , c ) 60 , d ) 27 , e ) 11 | b | divide(const_3.0, divide(divide(const_1, multiply(const_3, add(const_2, 6))), const_2)) | add(const_2,const_3)|multiply(#0,const_3)|divide(const_1,#1)|divide(#2,const_2)|divide(n0,#3)| | general | B |
find the value of m 72519 x 9999 = m ? | "72519 x 9999 = 72519 x ( 10000 - 1 ) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481 d" | a ) 724533811 , b ) 353654655 , c ) 545463251 , d ) 725117481 , e ) 477899932 | d | multiply(subtract(9999, const_4), 72519) | subtract(n1,const_4)|multiply(#0,n0)| | general | D |
a rectangular block 10 cm by 20 cm by 35 cm is cut into an exact number of equal cubes . find the least possible number of cubes ? | "volume of the block = 10 * 20 * 35 = 7000 cm ^ 3 side of the largest cube = h . c . f of 10 , 20,35 = 5 cm volume of the cube = 5 * 5 * 5 = 125 cm ^ 3 number of cubes = 7000 / 125 = 56 answer is b" | a ) 6 , b ) 56 , c ) 15 , d ) 40 , e ) 22 | b | 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(35, 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(35, 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 | B |
you have a 6 - sided cube and 6 cans of paint , each a different color . you may not mix colors of paint . how many distinct ways can you paint the cube using a different color for each side ? ( if you can reorient a cube to look like another cube , then the two cubes are not distinct . ) | paint one of the faces red and make it the top face . 5 options for the bottom face . now , four side faces can be painted in ( 4 - 1 ) ! = 3 ! = 6 ways ( circular arrangements of 4 colors ) . total = 5 * 6 = 30 . answer : b . | a ) 24 , b ) 30 , c ) 48 , d ) 60 , e ) 120 | b | multiply(factorial(divide(6, const_2)), subtract(6, const_1)) | divide(n0,const_2)|subtract(n0,const_1)|factorial(#0)|multiply(#2,#1) | general | B |
the average speed of a car decreased by 3 miles per hour every successive 8 - minutes interval . if the car traveled 4.4 miles in the fifth 8 - minute interval , what was the average speed of the car , in miles per hour , in the first 8 minute interval ? | "( 4.4 miles / 8 minutes ) * 60 minutes / hour = 33 mph let x be the original speed . x - 4 ( 3 ) = 33 x = 45 mph the answer is a ." | a ) 45 , b ) 48 , c ) 51 , d ) 54 , e ) 57 | a | add(add(add(add(divide(4.4, divide(8, const_60)), 3), 3), 3), 3) | divide(n1,const_60)|divide(n2,#0)|add(n0,#1)|add(n0,#2)|add(n0,#3)|add(n0,#4)| | physics | A |
sale of rs 6835 , rs . 9927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs , 6800 ? | "total sale for 5 months = rs . ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs . 34009 . required sale = rs . [ ( 6800 x 6 ) - 34009 ] = rs . ( 40800 - 34009 ) = rs . 6791 answer : b" | a ) 4966 , b ) 6791 , c ) 2877 , d ) 2676 , e ) 1881 | b | multiply(subtract(divide(add(add(add(add(6835, 9927), 6855), 7230), 6562), 5), 6800), 5) | add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|divide(#3,n5)|subtract(#4,n6)|multiply(n5,#5)| | general | B |
p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 130 days , in how many days can r alone complete the remaining work ? | "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 250 days and hence r alone in 380 days p works for 60 days - - > 60 / 200 work done = > 30 % q works for 130 days - - > 130 / 250 work done = > 52 % 22 % work left . . . r alone will take 22 % * 380 = 83.6 days answer is ( b )" | a ) 50 days , b ) 83.6 days , c ) 100 days , d ) 150 days , e ) 80 days | b | subtract(multiply(const_4, 50), multiply(divide(130, const_100), 60)) | divide(n4,const_100)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#1,#2)| | physics | B |
the food in a camp lasts for 40 men for 45 days . if fifteen more men join , how many days will the food last ? | one man can consume the same food in 40 * 45 = 1800 days . 15 more men join , the total number of men = 55 the number of days the food will last = 1800 / 55 = 33 days . answer : b | a ) 80 days , b ) 33 days , c ) 35 days , d ) 16 days , e ) 15 days | b | divide(multiply(45, 40), add(add(const_12, const_3), 40)) | add(const_12,const_3)|multiply(n0,n1)|add(n0,#0)|divide(#1,#2) | physics | B |
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in 36 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 / 36 4 / x = 1 / 36 = > x = 144 min . answer : c" | a ) 766 min , b ) 656 min , c ) 144 min , d ) 877 min , e ) 555 min | c | multiply(add(const_1, const_4), 36) | add(const_1,const_4)|multiply(n0,#0)| | physics | C |
the angle between the minute hand and the hour hand of a clock when the time is 8.30 , is : | "angle traced by hour hand in 17 / 2 hrs = ( 360 / 12 x 17 / 2 ) º = 255 . angle traced by min . hand in 30 min . = ( 360 / 60 x 30 ) º = 180 . required angle = ( 255 - 180 ) º = 75 º . answer b" | a ) 10 º , b ) 75 º , c ) 180 º , d ) 270 º , e ) 360 º | b | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | add(const_1,const_4)|divide(const_60,const_2)|multiply(const_3,const_4)|subtract(#2,const_1)|divide(#2,#3)|multiply(#0,#3)|multiply(#4,#5)|subtract(#6,#1)|multiply(#7,#3)|divide(#8,const_2)| | physics | B |
a cylinder of height h is 2 / 3 of water . when all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder , the new cylinder is 3 / 5 full . the height of the new cylinder is what percent of h ? | "basically we can disregard the radius is 25 % information , as we are only asked about the height of the original and the new cylinder . this is becausethe new cylinder is 3 / 5 fullmeans the same as that it ' s height is 3 / 5 . original cylinder 2 / 3 new cylinder 3 / 5 so 3 / 5 / 2 / 3 = 3 / 5 * 3 / 2 = 0.90 or 90 % . answer e" | a ) 25 % , b ) 50 % , c ) 60 % , d ) 80 % , e ) 90 % | e | multiply(divide(divide(3, 5), divide(2, 3)), const_100) | divide(n3,n4)|divide(n0,n1)|divide(#0,#1)|multiply(#2,const_100)| | gain | E |
of 30 applicants for a job , 12 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "d . 7 30 - 3 = 27 27 - 12 - 18 = - 7 then 7 are in the intersection between 4 years experience and degree . answer d" | a ) 14 , b ) 13 , c ) 9 , d ) 7 , e ) 5 | d | add(subtract(add(12, 18), subtract(30, 3)), subtract(18, 12)) | add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)| | general | D |
what is the units digit of ( 22 ^ 4 ) ( 16 ^ 3 ) ( 41 ^ 8 ) ? | "the units digit of 22 ^ 4 is the units digit of 2 ^ 4 which is 6 . the units digit of 16 ^ 3 is the units digit of 6 ^ 3 which is 6 . the units digit of 41 ^ 8 is the units digit of 1 ^ 8 which is 1 . the units digit of 6 * 6 * 1 is 6 . the answer is e ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(add(multiply(factorial(22), factorial(4)), multiply(factorial(22), factorial(3))), 22) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general | E |
20 % of major airline companies equip their planes with wireless internet access . 70 % of major airlines offer passengers free on - board snacks . what is the greatest possible percentage of major airline companies that offer both wireless internet and free on - board snacks ? | to maximize the percentage of companies offering both , let ' s assume that all 20 % of companies which offer wireless internet also offer snacks . the answer is a . | a ) 20 % , b ) 30 % , c ) 40 % , d ) 70 % , e ) 90 % | a | multiply(20, const_1) | multiply(n0,const_1) | general | A |
find large number from below question the difference of 2 no ' s is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 15 5 x = 1350 x = 270 large number = 270 + 1365 = 1635 e" | a ) 1236 , b ) 1356 , c ) 1470 , d ) 1556 , e ) 1635 | e | multiply(divide(subtract(2, 6), subtract(1365, const_1)), 1365) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general | E |
if 2 / z = 2 / ( z + 1 ) + 2 / ( z + 25 ) which of these integers could be the value of z ? | solving for z algebraically in this problem would not be easy . instead , we can follow the hint in the question ( “ which of these integers … ” ) and test each answer choice : a . 2 / 0 = 2 / 1 + 2 / 25 incorrect ( division by zero ) b . 2 / 1 = 2 / 2 + 2 / 26 incorrect c . 2 / 2 = 2 / 3 + 2 / 27 incorrect d . 2 / 3 = 2 / 4 + 2 / 28 incorrect e . 2 / 4 = 2 / 5 + 2 / 30 correct the correct answer is e , because it contains the only value that makes the equation work . notice how quickly this strategy worked in this case | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | e | divide(25, add(add(2, 2), 1)) | add(n0,n0)|add(n2,#0)|divide(n4,#1) | general | E |
each week , harry is paid x dollars per hour for the first 30 hours and 2 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 41 hours if harry and james were paid the same amount last week , how many hours did harry work last week ? | "james worked for 41 hours hence he earned 40 * x + 1 * 2 x = 42 x dollars ; we know that harry also earned the same 42 x dollars , out of which he earned 30 x dollars for thefirst 30 hoursplus 12 x additional dollars . since for each additional hour he gets 2 x dollars then he worked for 12 x / 2 x = 6 additional hours , so harry worked for total of 30 + 6 = 36 hours . answer : b ." | a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | b | add(divide(subtract(add(40, 2), 30), 2), 30) | add(n2,n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)| | general | B |
what is x if x + 2 y = 10 and y = 4 ? | "x = 10 - 2 y x = 10 - 8 . x = 2 answer : e" | a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2 | e | subtract(10, multiply(2, 4)) | multiply(n0,n2)|subtract(n1,#0)| | general | E |
a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 800 when the rate to profit rises from 5 to 7 percent . find the capital of b ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 800 b capital = 60000 * 1 / 6 = 10000 . answer : b" | a ) 3999 , b ) 10000 , c ) 2500 , d ) 2772 , e ) 2912 | b | divide(multiply(800, const_100), 2) | multiply(n2,const_100)|divide(#0,n0)| | general | B |
while driving from a - ville to b - town , harriet drove at a constant speed of 105 kilometers per hour . upon arriving in b - town , harriet immediately turned and drove back to a - ville at a constant speed of 145 kilometers per hour . if the entire trip took 5 hours , how many minutes did it take harriet to drive from a - ville to b - town ? | "5 hr = 300 min . if harriet spend equal hrs on each leg she will spend 150 min on each . since speed a - b is less than speed b - a and distance on each leg is the same , time spent on a - b is more than 150 min , which mean we can eliminate ans . a , b and c . now let plug in ans . d or e and verify which one give same distance on each leg . e . t = 174 min * leg a - b - - - > d = 105.174 / 60 = 18270 / 60 * leg b - a - - - - > d = 145 * 126 / 60 = 18270 / 60 so the correct ans . ise" | a ) 138 , b ) 148 , c ) 150 , d ) 162 , e ) 174 | e | multiply(subtract(5, divide(multiply(105, 5), add(105, 145))), const_60) | add(n0,n1)|multiply(n0,n2)|divide(#1,#0)|subtract(n2,#2)|multiply(#3,const_60)| | physics | E |
a theater box office sold an average ( arithmetic mean ) of 64 tickets per staff member to a particular movie . among the daytime staff , the average number sold per member was 76 , and among the evening staff , the average number sold was 60 . if there are no other employees , what was the ratio of the number of daytime staff members to the number of evening staff members ? | "deviation from the mean for the daytime staff = 76 - 64 = 12 . deviation from the mean for the evening staff = 64 - 60 = 4 . thus , the ratio of the number of daytime staff members to the number of evening staff members is 4 : 12 = 1 : 3 . the answer is c ." | a ) 2 : 5 , b ) 1 : 4 , c ) 1 : 3 , d ) 15 : 19 , e ) 64 : 76 | c | divide(subtract(64, 60), subtract(76, 64)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general | C |
a flagstaff 17.5 m high casts a shadow of length 40.25 m . what will be the height of a building , which casts a shadow of length 28.75 m under similar conditions ? | explanation : let the required height of the building be x meter more shadow length , more height ( direct proportion ) hence we can write as ( shadow length ) 40.25 : 28.75 : : 17.5 : x ⇒ 40.25 × x = 28.75 × 17.5 ⇒ x = 28.75 × 17.5 / 40.25 = 2875 × 175 / 40250 = 2875 × 7 / 1610 = 2875 / 230 = 575 / 46 = 12.5 answer : option a | a ) 12.5 m , b ) 10.5 m , c ) 14 , d ) 12 , e ) 10 | a | multiply(28.75, divide(17.5, 40.25)) | divide(n0,n1)|multiply(n2,#0) | physics | A |
the number 0.8 is how much greater than 1 / 2 ? | "let x be the difference then . 8 - 1 / 2 = x 8 / 10 - 1 / 2 = x x = 3 / 10 ans b" | a ) ½ , b ) 3 / 10 , c ) 1 / 50 , d ) 1 / 500 , e ) 2 / 500 | b | subtract(0.8, divide(1, 2)) | divide(n1,n2)|subtract(n0,#0)| | general | B |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.