Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
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find the unknown term 8 , 86 , x , - 4 , - 12 | 8 , 86 , x , - 4 , - 12 i guess each differ in d range of 2 8 - 8 = 0 8 - 6 = 2 6 - 2 = 4 2 - 6 = - 4 - 4 - 8 = - 12 answer : a | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 9 | a | subtract(divide(12, 4), const_1) | divide(n3,n2)|subtract(#0,const_1) | general |
if a train , travelling at a speed of 360 kmph , crosses a pole in 5 sec , then the length of train is ? | "c c = 360 * 5 / 18 * 5 = 500 m" | a ) 381 m , b ) 112 m , c ) 500 m , d ) 125 m , e ) 433 m | c | multiply(multiply(360, const_0_2778), 5) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
if y is the smallest positive integer such that 1260 multiplied by y is the square of an integer , then y must be | "1260 = 2 ^ 2 * 3 ^ 2 * 5 * 7 to be perfect square , we need to multiply by at least 5 * 7 = 35 . the answer is b ." | a ) 25 , b ) 35 , c ) 60 , d ) 75 , e ) 140 | b | multiply(add(const_2, const_3), const_2) | add(const_2,const_3)|multiply(#0,const_2)| | geometry |
a certain company retirement plan has arule of 70 provision that allows an employee to retire when the employee ' s age plus years of employment with the company total at least 70 . in what year could a female employee hired in 1988 on her 32 nd birthday first be eligible to retire under this provision ? | "she must gain at least 70 points , now she has 32 and every year gives her two more points : one for age and one for additional year of employment , so 32 + 2 * ( # of years ) = 70 - - > ( # of years ) = 19 - - > 1988 + 19 = 2007 . answer : e ." | a ) 2003 , b ) 2004 , c ) 2005 , d ) 2006 , e ) 2007 | e | add(1988, divide(subtract(70, 32), const_2)) | subtract(n0,n3)|divide(#0,const_2)|add(n2,#1)| | general |
a coin is tossed 3 times . what is the probability of getting exactly 2 heads ? | "the number of possible outcomes is 2 ^ 3 = 8 there are 3 ways to get exactly 2 heads . p ( exactly 2 heads ) = 3 / 8 the answer is c ." | a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 8 , d ) 5 / 8 , e ) 3 / 16 | c | multiply(power(divide(const_1, const_2), 2), multiply(choose(3, 2), divide(const_1, const_2))) | choose(n0,n1)|divide(const_1,const_2)|multiply(#0,#1)|power(#1,n1)|multiply(#2,#3)| | probability |
3 , 4.5 , 6.75 , 10.13 , 15.18 , ( . . . ) | "3 ( 3 ã — 3 ) ã · 2 = 4.5 ( 4.5 ã — 3 ) ã · 2 = 6.75 ( 6.75 ã — 3 ) ã · 2 = 10.13 ( 10.18 ã — 3 ) ã · 2 = 15.18 ( 15.18 ã — 3 ) ã · 2 = 60.75 answer is c" | a ) 60 , b ) 60.75 , c ) 22.78 , d ) 32.52 , e ) 88 | c | subtract(negate(10.13), multiply(subtract(4.5, 6.75), divide(subtract(4.5, 6.75), subtract(3, 4.5)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
a certain protective additive increases from 45 days to 60 days the time between required maintenance checks on an industrial vehicle . by what percent is the time between maintenance checks increased by using the additive ? | "general formula for percent increase or decrease , ( percent change ) : percent = change / original ∗ 100 so , the time between maintenance checks increased by 60 − 45 / 45 ∗ 100 = 33 1 / 3 answer : b ." | a ) 25 % , b ) 33 1 / 3 % , c ) 50 % , d ) 66 2 / 3 % , e ) 75 % | b | multiply(divide(subtract(60, 45), 45), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | general |
if x * y = xy + 2 ( x + y ) for all integers x and y , then 2 * ( – 3 ) = | 2 * ( - 3 ) = 2 * ( - 3 ) + 2 ( 2 + ( - 3 ) ) = - 6 - 2 = - 8 option ( c ) | a ) – 16 , b ) – 11 , c ) – 8 , d ) 4 , e ) 16 | c | add(multiply(2, negate(3)), multiply(2, add(2, negate(3)))) | negate(n2)|add(n0,#0)|multiply(n0,#0)|multiply(n0,#1)|add(#2,#3) | general |
a can run 4.5 km distance in 1 min 30 seconds , while b can run this distance in 3 min . by how much distance can a beat b ? | a takes time 1.30 minutes = 90 sec b takes time 3 minutes = 180 sec diffrence = 180 - 90 = 90 sec now we are to find distance covered in 90 sec by b 180 sec = 4500 m 1 sec = 25 m 90 sec = 25 x 90 = 2250 m answer : a | a ) 2250 m , b ) 1000 m , c ) 1250 m , d ) 1800 m , e ) 2000 m | a | multiply(add(multiply(1, const_60), 30), divide(multiply(4.5, const_1000), multiply(3, const_60))) | multiply(n1,const_60)|multiply(n0,const_1000)|multiply(n3,const_60)|add(n2,#0)|divide(#1,#2)|multiply(#3,#4) | physics |
a , b and c rents a pasture for rs . 957 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 957 = 297 answer : c" | a ) 270 , b ) 289 , c ) 297 , d ) 255 , e ) 552 | c | multiply(divide(957, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|multiply(#5,#1)| | general |
a student got 74 % in math and 84 % in history . to get an overall average of 75 % , how much should the student get in the third subject ? | 74 + 84 + x = 3 * 75 x = 67 the answer is d . | a ) 61 % , b ) 63 % , c ) 65 % , d ) 67 % , e ) 69 % | d | subtract(multiply(const_3, 75), add(74, 84)) | add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0) | general |
the sector of a circle has radius of 7 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 7 ) + 2 ( 7 ) = 16.5 + 14 = 30.5 cm answer : d" | a ) 91.5 cm , b ) 92.2 cm , c ) 28.9 cm , d ) 30.5 cm , e ) 98.2 cm | d | multiply(multiply(const_2, divide(multiply(subtract(7, const_3), const_2), add(const_4, const_3))), 7) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
the width of a rectangular hall is ½ of its length . if the area of the hall is 128 sq . m , what is the difference between its length and breadth ? | "let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 128 = x * 1 x / 2 x ² = 256 x = 16 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 16 / 2 = 8 m answer : a" | a ) 8 m , b ) 10 m , c ) 12 m , d ) 15 m , e ) 17 m | a | divide(sqrt(divide(128, divide(const_1, const_2))), const_2) | divide(const_1,const_2)|divide(n0,#0)|sqrt(#1)|divide(#2,const_2)| | geometry |
two trains running in opposite directions cross a man standing on the platform in 47 seconds and 38 seconds respectively and they cross each other in 40 seconds . the ratio of their speeds is : | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 47 x meters , and length of the second train = 38 y meters . ( 47 x + 37 y ) / ( x + y ) = 40 = = > 47 x + 38 y = 40 x + 40 y = = > 7 x = 2 y = = > x / y = 2 / 7 answer : option c" | a ) 3 : 1 , b ) 3 : 2 , c ) 2 : 7 , d ) 3 : 25 , e ) 3 : 4 | c | divide(subtract(47, 40), subtract(40, 38)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 13 km / hr completes one round in 8 minutes , then the area of the park ( in sq . m ) is | "perimeter = distance covered in 8 min . = 18000 x 8 m = 2400 m . 60 let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 2400 or x = 240 . length = 720 m and breadth = 480 m . area = ( 720 x 480 ) m 2 = 345600 m 2 . a" | a ) 345600 , b ) 153400 , c ) 153600 , d ) 154000 , e ) 154200 | a | rectangle_area(divide(divide(multiply(multiply(divide(13, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(13, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
a train 200 m long running at 80 kmph crosses a platform in 22 sec . what is the length of the platform ? | "d = 80 * 5 / 18 = 22 = 488 – 200 = 288 answer : d" | a ) 338 , b ) 277 , c ) 350 , d ) 288 , e ) 271 | d | subtract(multiply(22, multiply(80, const_0_2778)), 200) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
a fruit seller sells mangoes at the rate of rs . 9 per kg and thereby loses 20 % . at what price per kg , he should have sold them to make a profit of 5 % ? | "solution 85 : 9 = 105 : x x = ( 9 × 105 / 80 ) = rs . 11.81 hence , s . p per kg = rs . 11.81 answer a" | a ) rs . 11.81 , b ) rs . 12 , c ) rs . 12.25 , d ) rs . 12.31 , e ) none | a | divide(multiply(9, add(const_100, 5)), subtract(const_100, 20)) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,#1)| | gain |
if the simple interest on a certain sum of money for 8 years is one – fifth of the sum , then the rate of interest per annum is | explanation : let the principal ( p ) be x then , simple interest ( si ) = x / 5 time ( t ) = 8 years rate of interest per annum ( r ) = ( 100 × si ) / pt = ( 100 × ( x / 5 ) / ( x × 8 ) = 20 / 8 = 2.5 % answer : option c | a ) 4.1 % , b ) 7 % , c ) 2.5 % , d ) 2.8 % , e ) 3 % | c | divide(divide(const_100, add(const_1, const_4)), 8) | add(const_1,const_4)|divide(const_100,#0)|divide(#1,n0) | gain |
two pipes a and b can fill a tank in 20 and 30 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by a in 1 min = 1 / 20 part filled by b in 1 min = 1 / 30 part filled by ( a + b ) in 1 min = ( 1 / 20 + 1 / 30 ) = 1 / 12 so both pipes can fill the tank in 12 minutes . answer : a" | a ) 12 min , b ) 15 min , c ) 25 min , d ) 50 min , e ) 55 min | a | add(multiply(20, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 30)), const_2)), 20), const_60)) | divide(const_1,n0)|divide(const_1,n1)|multiply(n0,const_100)|add(#0,#1)|multiply(#3,const_2)|subtract(const_1,#4)|multiply(n0,#5)|multiply(#6,const_60)|add(#2,#7)| | physics |
the captain of a cricket team of 11 members is 24 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation : let the average age of the whole team by x years . 11 x â € “ ( 24 + 27 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 42 2 x = 42 x = 21 . so , average age of the team is 21 years . answer b" | a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 24 years | b | divide(subtract(add(24, add(24, 3)), multiply(3, 3)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
the sum of 111 consecutive integers is 11211 . what is the greatest integer in the set ? | "let x be the first integer in the set , then x + 110 is the largest integer . the sum is : x + ( x + 1 ) + ( x + 2 ) + . . . + ( x + 110 ) = 111 x + 110 * 111 / 2 = 111 ( x + 55 ) then x + 55 = 101 x = 46 the largest integer in the set is 46 + 110 = 156 the answer is e ." | a ) 148 , b ) 150 , c ) 152 , d ) 154 , e ) 156 | e | add(add(power(add(add(divide(subtract(subtract(111, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(111, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(111, const_10), const_2), const_4), const_2), power(... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
according to the formula f = 9 / 5 ( c ) + 32 , if the temperature in degrees farenheit ( f ) increases by 28 , by how much does the temperature in degrees celsius ( c ) increase ? | you can plug in values . c = 5 / 9 * ( f - 32 ) f = 32 - - > c = 0 ; f = 32 + 28 = 60 - - > c = 5 / 9 * 28 = 15.55 . increase = 15.55 degrees . answer : b . | a ) 9 , b ) 15.55 , c ) 47 , d ) 48 3 / 5 , e ) 59 | b | divide(multiply(28, 5), 9) | multiply(n1,n3)|divide(#0,n0) | general |
how many positive integers less than 24 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ? | "9 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } 2 odd multiples of 5 : { 5 , 15 } 9 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 } notice , that 5 is in two sets , thus total # of integers satisfying the given ... | a ) 19 , b ) 25 , c ) 24 , d ) 22 , e ) 20 | a | subtract(subtract(subtract(24, 2), const_1), const_1) | subtract(n0,n2)|subtract(#0,const_1)|subtract(#1,const_1)| | general |
without stoppages , a train travels certain distance with an average speed of 250 km / h , and with stoppages , it covers the same distance with an average speed of 125 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 125 km less . time taken to cover 125 km = 125 â „ 250 h = 1 â „ 2 h = 1 â „ 2 ã — 60 min = 30 min answer b" | a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | b | subtract(const_60, multiply(divide(125, 250), const_60)) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
cara took out a charge account at the general store and agreed to pay 5 % simple annual interest . if she charges $ 54 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | 1.05 * $ 54 = $ 56.70 the answer is c . | a ) $ 55.70 , b ) $ 56.20 , c ) $ 56.70 , d ) $ 57.20 , e ) $ 57.70 | c | multiply(add(const_1, divide(5, const_100)), 54) | divide(n0,const_100)|add(#0,const_1)|multiply(n1,#1) | general |
a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter t of liquid a was contained by the can initially ? | "as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 simil... | a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | add(n0,n2)|add(n0,n1)|multiply(n0,n2)|multiply(n0,n1)|divide(n1,#1)|subtract(#2,#3)|multiply(n2,#4)|subtract(n2,#6)|multiply(#0,#7)|divide(#8,#5)|multiply(n0,#9)| | other |
the length of a rectangle is double its width . if the length is diminished by 5 cm and the width is increased by 5 cm then its area is increased by 75 cm square . what is the length of the rectangle ? | sol . according to question ( l - 5 ) ( b + 5 ) - lb = 75 and l = 2 b so , b = 20 , l = 40 ans . ( c ) | ['a ) 20', 'b ) 30', 'c ) 40', 'd ) 56', 'e ) 65'] | c | multiply(const_2, divide(add(multiply(5, 5), 75), 5)) | multiply(n0,n0)|add(n2,#0)|divide(#1,n0)|multiply(#2,const_2) | geometry |
a and b go around a circular track of length 1500 m on a cycle at speeds of 36 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 1500 / ( 36 * 5 / 18 ) , 1500 / ( 54 * 5 / 18 ) } = lcm ( 150 , 100 ) = 300 sec . answer : c" | a ) 120 sec , b ) 165 sec , c ) 300 sec , d ) 167 sec , e ) 168 sec | c | divide(1500, subtract(multiply(54, const_0_2778), multiply(36, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)| | physics |
a train 300 m long takes 9 sec to cross a man walking at 3 kmph in a direction opposite to that of the train . find the speed of the train ? | "let the speed of the train be x kmph speed of the train relative to man = x + 3 = ( x + 3 ) * 5 / 18 m / sec 300 / [ ( x + 3 ) * 5 / 18 ] = 9 9 ( x + 3 ) = 1080 x = 117 kmph answer is d" | a ) 100 kmph , b ) 90 kmph , c ) 120 kmph , d ) 117 kmph , e ) 125 kmph | d | subtract(divide(divide(300, 9), const_0_2778), 3) | divide(n0,n1)|divide(#0,const_0_2778)|subtract(#1,n2)| | physics |
the perimeters of 2 squares are 40 cm and 32 cm . find the perimeter of a third square whose area is equal to the difference of the areas of the two squares ? | "side of the 1 st square = 40 / 4 = 10 cm side of the 2 nd square = 32 / 4 = 8 cm area of the 3 rd square = 10 ^ 2 - 8 ^ 2 = 36 cm ^ 2 side of the 3 rd square = 6 cm required perimeter = 6 * 4 = 24 cm answer is b" | a ) 20 cm , b ) 24 cm , c ) 32 cm , d ) 28 cm , e ) 30 cm | b | square_perimeter(sqrt(subtract(square_area(divide(40, const_4)), square_area(divide(32, const_4))))) | divide(n1,const_4)|divide(n2,const_4)|square_area(#0)|square_area(#1)|subtract(#2,#3)|sqrt(#4)|square_perimeter(#5)| | geometry |
if there are 210 workers in a factory , and on a certain day , 198 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) . | 198 / 210 * 100 = 94.28 94.3 % correct answer a | a ) 94.3 % , b ) 95.3 % , c ) 93.3 % , d ) 92.3 % , e ) 91.3 % | a | multiply(divide(198, 210), const_100) | divide(n1,n0)|multiply(#0,const_100) | physics |
a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 210 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 210 kilometers ? 210 / 40 = 5.25 * 1 gallon = 5.25 gallons correct answer d" | a ) 8.25 gallons , b ) 7.5 gallons , c ) 6.55 gallons , d ) 5.25 gallons , e ) 4.5 gallons | d | divide(210, 40) | divide(n1,n0)| | physics |
on a certain farm the ratio of horses to cows is 4 : 1 . if the farm were to sell 15 horses and buy 15 cows , the ratio of horses to cows would then be 7 : 3 . after the transaction , how many more horses than cows would the farm own ? | "originally , there were 4 k horses and k cows . 3 ( 4 k - 15 ) = 7 ( k + 15 ) 12 k - 7 k = 105 + 45 5 k = 150 k = 30 the difference between horses and cows is ( 4 k - 15 ) - ( k + 15 ) = 3 k - 30 = 60 the answer is d ." | a ) 20 , b ) 40 , c ) 50 , d ) 60 , e ) 80 | d | subtract(subtract(multiply(4, divide(add(multiply(15, 7), multiply(15, 3)), subtract(multiply(4, 3), 7))), 15), add(divide(add(multiply(15, 7), multiply(15, 3)), subtract(multiply(4, 3), 7)), 15)) | multiply(n2,n4)|multiply(n2,n5)|multiply(n0,n5)|add(#0,#1)|subtract(#2,n4)|divide(#3,#4)|add(n2,#5)|multiply(n0,#5)|subtract(#7,n2)|subtract(#8,#6)| | other |
two pipes p and q can fill a cistern in 11 and 15 minutes respectively . both are opened together , but at the end of 3 minutes the first is turned off . how much longer will the cistern take to fill ? | "3 / 11 + x / 15 = 1 x = 10 10 / 11 answer : d" | a ) 1 / 8 , b ) 1 / 4 , c ) 2 / 4 , d ) 10 / 11 , e ) 1 / 4 | d | multiply(subtract(const_1, multiply(add(divide(const_1, 11), divide(const_1, 15)), 3)), 15) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n1,#4)| | physics |
rani bought more apples than oranges . she sells apples at ₹ 23 apiece and makes 15 % profit . she sells oranges at ₹ 10 apiece and makes 25 % profit . if she gets ₹ 653 after selling all the apples and oranges , find her profit percentage q . | given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the valu... | a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(1... | add(n1,const_2)|divide(n3,const_100)|multiply(n2,const_2)|add(#2,const_1)|add(#1,const_1)|divide(n2,#4)|multiply(#3,#2)|multiply(#0,#5)|add(#6,#7)|subtract(n4,#8)|divide(#9,#8)|multiply(#10,const_100) | gain |
at a summer camp with 1,800 participants , 1 / 2 of the campers are aged 8 to 12 . next year , the number of campers aged 8 to 12 will increase by 1 / 3 . after this change , what percentage of the total 1,800 campers will the 8 - to 12 - year - olds represent ? | total - 1,800 participants campers are aged 8 to 12 = ( 1 / 2 ) * 1800 = 900 next year , campers are aged 8 to 12 = ( 4 / 3 ) * 900 = 1200 percentage = ( 1200 / 1800 ) * 100 = 66 2 / 3 % answer : option c | a ) 68 % , b ) 66 % , c ) 66 2 / 3 % , d ) 60 % , e ) 70 % | c | add(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), const_2), add(divide(const_1, const_3), const_1)), add(const_1000, multiply(8, const_100))), const_100), const_3), divide(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), const_2), add(divide(const_1, ... | divide(const_1,const_3)|multiply(n3,const_100)|add(#0,const_1)|add(#1,const_1000)|divide(#3,const_2)|multiply(#2,#4)|divide(#5,#3)|multiply(#6,const_100)|multiply(#7,const_3)|divide(#8,const_10)|add(#9,#8) | general |
village p ’ s population is 800 greater than village q ' s population . if village q ’ s population were reduced by 500 people , then village p ’ s population would be 3 times as large as village q ' s population . what is village q ' s current population ? | "p = q + 800 . p = 3 ( q - 500 ) . 3 ( q - 500 ) = q + 800 . 2 q = 2300 . q = 1150 . the answer is b ." | a ) 1100 , b ) 1150 , c ) 1200 , d ) 1250 , e ) 1300 | b | divide(add(800, multiply(500, 3)), const_3) | multiply(n1,n2)|add(n0,#0)|divide(#1,const_3)| | general |
the sale price of a trolley bag including the sale tax is rs . 420 . the rate of sale tax is 12 % . if the shopkeeper has made a profit of 25 % , the cost price of the trolley bag is : | explanation : 112 % of s . p . = 420 s . p . = rs . ( 420 x 100 / 112 ) = rs . 375 . c . p . = rs ( 100 / 125 x 375 ) = rs 300 answer : b | a ) rs 625 , b ) rs 300 , c ) rs 350 , d ) rs 375 , e ) none of these | b | divide(subtract(420, multiply(420, divide(12, const_100))), add(divide(25, const_100), const_1)) | divide(n1,const_100)|divide(n2,const_100)|add(#1,const_1)|multiply(n0,#0)|subtract(n0,#3)|divide(#4,#2) | gain |
how many bricks each measuring 20 cm x 13.25 cm x 8 cm , will be needed to build a wall 7 m x 8 m x 15.5 m | "explanation : no . of bricks = volume of the wall / volume of 1 brick = ( 700 x 800 x 15.5 ) / ( 20 x 13.25 x 8 ) = 4094 answer : c" | a ) 5600 , b ) 4600 , c ) 4094 , d ) 7200 , e ) none of these | c | divide(multiply(multiply(7, 8), 15.5), divide(divide(multiply(multiply(20, 13.25), 8), const_100), const_100)) | multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#3,const_100)|divide(#4,const_100)|divide(#2,#5)| | physics |
one - tenth of the students at a nursery school are 4 years old or older . if 20 students have not yet reached their third birthday , and a total of 50 students are not between 3 years old and 4 years old , how many children are in the nursery school ? | x / 10 students are > 4 yrs 20 students are < 3 yrs x / 10 + 20 = 50 x / 10 = 30 x = 300 answer : b | a ) 150 , b ) 300 , c ) 450 , d ) 490 , e ) 620 | b | multiply(subtract(50, 20), const_10) | subtract(n2,n1)|multiply(#0,const_10) | general |
if p / q = 3 / 4 then 3 p + 4 q = ? | 3 p + 4 q = ? divided by q , 3 ( p / q ) + 4 = x 3 * ( 3 / 4 ) + 4 = 25 / 4 answer : a | a ) 25 / 4 , b ) 25 / 2 , c ) 20 / 4 , d ) 35 / 4 , e ) 25 / 3 | a | divide(add(multiply(multiply(3, divide(4, 3)), multiply(3, divide(4, 3))), multiply(3, 3)), multiply(3, divide(4, 3))) | divide(n1,n0)|multiply(n0,n0)|multiply(n0,#0)|multiply(#2,#2)|add(#3,#1)|divide(#4,#2) | general |
x and y started a business by investing rs . 36000 and rs . 42000 respectively after 4 months z joined in the business with an investment of rs . 48000 , then find share of z in the profit of rs . 13970 ? | "ratio of investment , as investments is for different time . investment x number of units of time . ratio of investments x : y : z = 36000 : 42000 : 48000 = > 6 : 7 : 8 . x = 6 x 12 months = 72 , y = 7 x 12 = 84 , z = 8 x 8 = 64 = > 18 : 21 : 16 . ratio of investments = > x : y : z = 18 : 21 : 16 . investment ratio = ... | a ) 3200 , b ) 4064 , c ) 3250 , d ) 3825 , e ) 3985 | b | multiply(multiply(48000, subtract(multiply(const_3, 4), 4)), divide(13970, add(add(multiply(36000, multiply(const_3, 4)), multiply(42000, multiply(const_3, 4))), multiply(48000, subtract(multiply(const_3, 4), 4))))) | multiply(const_3,n2)|multiply(n0,#0)|multiply(n1,#0)|subtract(#0,n2)|add(#1,#2)|multiply(n3,#3)|add(#4,#5)|divide(n4,#6)|multiply(#7,#5)| | gain |
the banker â € ™ s discount of a certain sum of money is rs . 42 and the true discount on the same sum for the same time is rs . 36 . the sum due is | "sol . sum = b . d . * t . d . / b . d . - t . d . = rs . [ 42 * 36 / 42 - 36 ] = rs . [ 42 * 36 / 6 ] = rs . 252 answer a" | a ) 252 , b ) 280 , c ) 360 , d ) 450 , e ) none | a | divide(multiply(42, 36), subtract(42, 36)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | gain |
the cost of 20 pens and 12 pencils is rs . 400 and the cost of 6 pens and 4 pencils is rs . 100 . find the cost of each pen ? | "20 p + 12 q = 400 - - - ( 1 ) 6 p + 4 q = 100 8 p + 8 q = 192 - - - ( 2 ) ( 1 ) - ( 2 ) = > 8 p = 160 = > p = 50 answer : a" | a ) rs . 50 , b ) rs . 100 , c ) rs . 150 , d ) rs . 25 , e ) rs . 125 | a | multiply(multiply(const_3, const_4), divide(400, add(20, 4))) | add(const_3.0,n4)|multiply(n0,const_4)|divide(n2,#0)|multiply(#2,#1)| | general |
two trains of equal lengths take 10 sec and 15 sec , respectively , to cross a telegraph post . if the length of each train is 120 m , in what time will they cross each other , travelling in opposite directions ? | speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 5 = 8 m / sec . relative speed = 12 + 8 = 20 m / sec . required time = ( 120 + 120 ) / 20 = 12 sec . answer : b | a ) 17 sec , b ) 12 sec , c ) 18 sec , d ) 19 sec , e ) 52 sec | b | divide(add(120, 120), add(divide(120, 10), divide(120, 15))) | add(n2,n2)|divide(n2,n0)|divide(n2,n1)|add(#1,#2)|divide(#0,#3) | physics |
if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 5 children have two children of the same sex and one of the opposite sex ? | "no of ways of selecting a gender - 2 c 1 no of ways of selecting any 2 children out of 5 = 5 c 2 total possible outcomes - 2 ^ 5 ( each child can be either a girl or a boy ) probability = 2 c 1 * 5 c 2 / 2 ^ 5 = 2 * 5 / 2 * 2 * 2 * 2 * 2 = 10 / 32 = 5 / 16 ans = d" | a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 5 / 16 , e ) 3 / 5 | d | divide(const_2, 5) | divide(const_2,n0)| | general |
2 men and 3 women are lined up in a row . what is the number of cases where they stand with each other in turn ? ( the number of cases in which men ( or women ) do not stand next to each other ) | "first arrange 3 women . 3 women can be arranged in 3 places in 3 ! ways . now w blank w blankw we have to fill the two blanks with 2 men . we can arrange 2 persons in 2 places in 2 ! ways . 3 ! * 2 ! = 12 ways . a is the answer" | a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | a | multiply(factorial(2), factorial(3)) | factorial(n0)|factorial(n1)|multiply(#0,#1)| | general |
a constructor estimates that 3 people can paint mr khans house in 4 days . if he uses 4 people instead of 3 , how long will they take to complete the job ? | explanation : use formula for a work members × days = constant 3 × 4 = 4 × a a = 3 so answer is 3 days answer : d | a ) 2 , b ) 8 , c ) 1 , d ) 3 , e ) 4 | d | divide(const_1, multiply(divide(const_1, multiply(4, 3)), 4)) | multiply(n0,n1)|divide(const_1,#0)|multiply(n1,#1)|divide(const_1,#2) | physics |
( 0.82 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.82 ) ( power 2 ) + 0.082 + ( 0.1 ) ( power 2 ) is : | "given expression = ( 0.82 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.82 ) ( power 2 ) + ( 0.82 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.82 - 0.1 ) = 0.72 answer is e" | a ) 0.86 , b ) 0.68 , c ) 0.96 , d ) 0.69 , e ) 0.72 | e | divide(subtract(power(0.82, 3), power(0.1, 3)), add(add(power(0.82, 2), 0.082), power(0.1, 2))) | power(n0,n1)|power(n2,n1)|power(n0,n5)|power(n2,n5)|add(n6,#2)|subtract(#0,#1)|add(#4,#3)|divide(#5,#6)| | general |
for every x , the action [ x ] is defined : [ x ] is the greatest integer less than or equal to x . what is the value of [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.6 ? | [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.6 = 6 * 0 + 2 * 7.2 + 8 - 6.6 = 0 + 14.4 + 1.4 15.8 answer c | a ) 12.6 . , b ) 14.4 . , c ) 15.8 . , d ) 16.2 . , e ) 16.4 | c | subtract(add(multiply(2, 7.2), subtract(8.4, divide(const_4, const_10))), 6.6) | divide(const_4,const_10)|multiply(n1,n4)|subtract(n5,#0)|add(#1,#2)|subtract(#3,n6) | general |
a train traveling at 144 kmph crosses a platform in 30 seconds and a man standing on the platform in 12 seconds . what is the length of the platform in meters ? | "answer distance covered by the train when crossing a man and when crossing a platform when a train crosses a man standing on a platform , the distance covered by the train is equal to the length of the train . however , when the same train crosses a platform , the distance covered by the train is equal to the length o... | a ) 240 meters , b ) 360 meters , c ) 480 meters , d ) 600 meters , e ) can not be determined | c | subtract(multiply(divide(multiply(144, const_1000), const_3600), 30), multiply(divide(multiply(144, const_1000), const_3600), 12)) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)| | physics |
the cost of registration at a professional association meeting was $ 50 per person ; a lunch for registrants only was available for an additional $ 22 per person . if the number of registrants who paid for lunch was 40 more than the number who did not , and if receipts for registration and lunch totaled $ 39,480 , how ... | "hope this might be useful to you . let the number of people who have opted only to register = x now since the registration cost is 50 $ per person , the total amount sums to = 50 x $ as per the information given in the question , the number of registrants who paid for lunch was 40 more than the number who did not . th... | a ) 700 , b ) 800 , c ) 300 , d ) 1,500 , e ) 1,800 | c | multiply(const_1, const_1) | multiply(const_1,const_1)| | general |
at 1 : 00 pm , there were 10.0 grams of bacteria . the bacteria increased to x grams at 4 : 00 pm , and 25.6 grams at 7 : 00 pm . if the amount of bacteria present increased by the same fraction during each of the 3 - hour periods , how many grams of bacteria were present at 4 : 00 pm ? | let x be the factor by which the bacteria increases every three hours . at 4 : 00 pm , the amount of bacteria was 10 x and at 7 : 00 pm it was 10 x ^ 2 . 10 x ^ 2 = 25.6 x ^ 2 = 2.56 x = 1.6 at 4 : 00 pm , the amount of bacteria was 10 ( 1.6 ) = 16 grams . the answer is d . | a ) 15.1 , b ) 15.4 , c ) 15.7 , d ) 16.0 , e ) 16.3 | d | multiply(sqrt(divide(25.6, const_10)), const_10) | divide(n5,const_10)|sqrt(#0)|multiply(#1,const_10) | general |
a bag contains 6 red , 4 blue and 2 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 6 c 212 c 2 = 6 c 212 c 2 = 15 / 66 = 5 / 22 d" | a ) 1 / 13 , b ) 2 / 23 , c ) 5 / 26 , d ) 5 / 22 , e ) 3 / 23 | d | divide(choose(6, 2), choose(add(add(6, 4), 2), 2)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)| | other |
the true discount on a bill of rs . 2560 is rs . 360 . what is the banker ' s discount ? | "explanation : f = rs . 2560 td = rs . 360 pw = f - td = 2560 - 360 = rs . 2200 true discount is the simple interest on the present value for unexpired time = > simple interest on rs . 2200 for unexpired time = rs . 360 banker ' s discount is the simple interest on the face value of the bill for unexpired time = simple... | a ) rs . 432 , b ) rs . 422 , c ) rs . 419 , d ) rs . 442 , e ) none of these | c | multiply(divide(360, subtract(2560, 360)), 2560) | subtract(n0,n1)|divide(n1,#0)|multiply(n0,#1)| | gain |
a man took loan from a bank at the rate of 4 % p . a . s . i . after 10 years he had to pay rs . 2000 interest only for the period . the principal amount borrowed by him was ? | "principal = ( 100 * 2000 ) / ( 4 * 10 ) = rs . 5000 answer : a" | a ) 5000 , b ) 4500 , c ) 3200 , d ) 6000 , e ) 1592 | a | divide(divide(multiply(2000, const_100), 10), 4) | multiply(n2,const_100)|divide(#0,n1)|divide(#1,n0)| | gain |
the mean of 50 observations was 36 . it was found later that an observation 46 was wrongly taken as 23 . the corrected new mean is | "solution correct sum = ( 36 x 50 + 46 - 23 ) = 1823 . â ˆ ´ correct mean = 1823 / 50 = 36.46 . answer d" | a ) 35.24 , b ) 36.16 , c ) 36.24 , d ) 36.46 , e ) none | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
a fruit seller had some oranges . he sells 40 % oranges and still has 300 oranges . how many oranges he had originally ? | "explanation : he sells 40 % of oranges and still there are 300 oranges remaining = > 60 % of oranges = 300 ⇒ ( 60 × total oranges ) / 100 = 300 ⇒ total oranges / 100 = 5 ⇒ total oranges = 5 × 100 = 500 answer : option d" | a ) 420 , b ) 700 , c ) 220 , d ) 500 , e ) none of these | d | add(300, multiply(300, divide(40, const_100))) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | gain |
average weight of 15 boys in a class is 48 kgs . the average weight of the class of 25 students is 45 kgs . what is the average weight of the 15 girls in the class ? | total weight of boys in class 15 x 48 = 720 kg total weight of all student in the class room 25 x 45 = 1125 kg total no of girls in the class 1125 - 720 = 405 kg average weight of girls 405 / 10 = 40.5 kg correct ans is e | a ) 45 , b ) 47.3 , c ) 39.8 , d ) 45.5 , e ) 40.5 | e | divide(subtract(multiply(25, 45), multiply(15, 48)), const_10) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,const_10) | general |
the sum of two numbers is 84 , and one of them is 12 more than the other . what are the two numbers ? | 2 x = 84 − 12 = 72 . x = 72 / 2 = 36 . this is the first number . therefore the other number is x + 12 = 36 + 12 = 48 . the sum of 36 + 48 is 84 . answer is b . | a ) 48 , 36 , b ) 36 , 48 , c ) 32 , 48 , d ) 48 , 32 , e ) 32 , 42 | b | divide(subtract(84, 12), const_2) | subtract(n0,n1)|divide(#0,const_2) | general |
12.5 * 3.2 * 6.8 = ? | "b 272 ? = 12.5 * 3.2 * 6.8 = 352" | a ) 352 , b ) 272 , c ) 224 , d ) 646 , e ) 742 | b | multiply(12.5, 3.2) | multiply(n0,n1)| | general |
a man started driving at a constant speed , from the site of a blast , the moment he heard the blast . he heard a second blast after a time of 30 mins and 25 seconds . if the second blast occurred exactly 30 mins after the first , how many meters was he from the site when he heard the second blast ? ( speed of sound = ... | "the distance the sound traveled to the man is 25 * 330 = 8250 meters the answer is a ." | a ) 8250 , b ) 9350 , c ) 10,450 , d ) 11,550 , e ) 12,650 | a | multiply(330, 25) | multiply(n1,n3)| | physics |
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 |
if you multiply all the numbers on your mobile phone except 0 and 9 , what is the answer ? | "we have to multiply 1 to 8 to find the answer . therefore 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 = 40320 answer is c" | a ) 25869 , b ) 48586 , c ) 40320 , d ) 58964 , e ) 45698 | c | multiply(const_1, subtract(const_1, const_1)) | subtract(const_1,const_1)|multiply(#0,const_1)| | general |
a train 250 m long is running at a speed of 68 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ? | "speed of the train relative to man = ( 68 - 8 ) kmph = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec time taken by the train to cross the man = time taken by it to cover 250 m at 50 / 3 m / sec = 250 * 3 / 50 sec = 15 sec answer : a ." | a ) 15 sec , b ) 9 sec , c ) 12 sec , d ) 15 sec , e ) 18 sec | a | divide(250, multiply(subtract(68, 8), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 30 percent , but profits were 20 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,14 r = x / 100 * 0.1 r answer c" | a ) 80 % , b ) 105 % , c ) 140 % , d ) 124.2 % , e ) 138 % | c | multiply(divide(multiply(subtract(const_1, divide(30, const_100)), divide(20, const_100)), divide(10, const_100)), const_100) | divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(#0,#3)|divide(#4,#2)|multiply(#5,const_100)| | gain |
the sum of two numbers is 96 , and one of them is 12 more than the other . what are the two numbers ? | "in this problem , we are asked to find two numbers . therefore , we must let x be one of them . let x , then , be the first number . we are told that the other number is 12 more , x + 12 . the problem states that their sum is 96 : word problem = 96 the line over x + 12 is a grouping symbol called a vinculum . it saves... | a ) 36 - 48 , b ) 50 - 34 , c ) 60 - 24 , d ) 42 - 54 , e ) 21 - 63 | d | divide(subtract(96, 12), const_2) | subtract(n0,n1)|divide(#0,const_2)| | general |
in a certain game , each player scores either 2 points or 5 points . if n players score 2 points and m players score 5 points , and the total number of points scored is 50 , what is the least possible positive difference q between n and m ? | "we have equation 2 n + 5 m = 50 we have factor 2 in first number and we have factor 5 in second number . lcm ( 2 , 5 ) = 10 so we can try some numbers and we should start from 5 because it will be less list than for 2 2 * 5 = 10 and n should be equal 20 4 * 5 = 20 and n should be equal 15 6 * 5 = 30 and n should be eq... | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(5, 2) | subtract(n1,n0)| | general |
a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 20 miles per hour . if the ride from marker a to marker b lasted 3 t... | a - b = 120 miles b - c = 60 miles avg speed = 20 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) / total time 20 = 180 / 4 t t = 135 b - c = 135 mph answer b | a ) 40 , b ) 135 , c ) 50 , d ) 55 , e ) 60 | b | multiply(divide(divide(add(divide(120, const_2), 120), 20), const_4), divide(120, const_2)) | divide(n0,const_2)|add(n0,#0)|divide(#1,n1)|divide(#2,const_4)|multiply(#3,#0) | physics |
the difference between the place values of 9 and 4 in the number 529435 is | sol . = ( place value of 9 ) – ( place value of 4 ) = ( 9000 - 400 ) = 8600 answer c | a ) 8500 , b ) 8900 , c ) 8600 , d ) 6970 , e ) none | c | subtract(multiply(const_10, 9), 9) | multiply(n0,const_10)|subtract(#0,n0)| | general |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 450 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 450 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . ... | a ) 1,108 , b ) 3,700 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 450) | divide(n6,n5)|multiply(n4,#0)| | general |
from a total of 3 boys and 4 girls , how many 4 - person committees can be selected if the committee must have exactly 2 boys and 2 girls ? | "answer = c = 18 no of 4 person committees that can be formed = 3 c 2 * 4 c 2 = 18 answer a" | a ) 18 , b ) 24 , c ) 60 , d ) 120 , e ) 240 | a | multiply(multiply(const_3, 3), multiply(3, 2)) | multiply(n0,const_3)|multiply(n0,n3)|multiply(#0,#1)| | general |
mary is expected to pay back a total sum of $ 819200 to her bank as mortgage . if she is expected to make a payment which is double of the previous months amount , how long will it take her to fully repay the loan if her initial payment was $ 400 . | first payment - - $ 400 total amount to be repaid - - $ 819200 first month payment - - $ 400 second month payment - - $ 400 * 2 = $ 800 third month payment - - $ 800 * 2 = $ 1600 fourth month payment - - $ 1600 * 2 = $ 3200 fifth month payment - - $ 3200 * 2 = $ 6400 sixth month payment - - $ 6400 * 2 = $ 12800 seventh... | a ) 13 , b ) 15 , c ) 12 , d ) 10 , e ) 11 | e | subtract(400, add(subtract(subtract(subtract(subtract(divide(divide(divide(divide(819200, const_100), const_2), const_2), const_4), const_10), const_100), const_10), const_4), const_1)) | divide(n0,const_100)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_4)|subtract(#3,const_10)|subtract(#4,const_100)|subtract(#5,const_10)|subtract(#6,const_4)|add(#7,const_1)|subtract(n1,#8) | general |
irin , ingrid and nell bake chocolate chip cookies in the ratio of 9.18 : 5.17 : 2.05 . if altogether they baked a batch of 148 cookies , what percent of the cookies did nell bake ? | "9.18 x + 5.17 x + 2.05 x = 16.4 x = 148 cookies x = 148 / 16.4 = 9 ( approx ) so , tell baked 9 * 2.05 cookies or 19 cookies ( approx ) % share = 19 / 148 = 12.5 approx hence , answer is c ." | a ) 0.125 % , b ) 1.25 % , c ) 12.5 % , d ) 125 % , e ) 0.152 % | c | multiply(divide(divide(multiply(148, 5.17), add(add(9.18, 5.17), 2.05)), 148), const_100) | add(n0,n1)|multiply(n1,n3)|add(n2,#0)|divide(#1,#2)|divide(#3,n3)|multiply(#4,const_100)| | other |
a jar contains a mixture of ab in the ratio 4 : 1 . when 30 l of mixture is replaced with liquid b , ratio becomes 2 : 3 . how many liters of liquid a was present in mixture initially . | "30 litres of mixture that is replaced will contain 24 litres of a and 6 litres of b ( as a : b = 4 : 1 ) let the initial volume of the mixture be 4 k + 1 k = 5 k so by condition , [ 4 k - 24 ] / [ k - 6 + 30 ] = 2 / 3 = > 12 k - 72 = 2 k - 12 + 60 = > 10 k = 120 solve for k which is k = 12 so initial volume of liquid ... | a ) 12 , b ) 48 , c ) 16 , d ) 20 , e ) 25 | b | multiply(divide(multiply(add(multiply(3, 2), multiply(2, 2)), divide(30, subtract(multiply(3, 2), 1))), add(4, 1)), 4) | add(n0,n1)|multiply(n4,n3)|multiply(n3,n3)|add(#1,#2)|subtract(#1,n1)|divide(n2,#4)|multiply(#3,#5)|divide(#6,#0)|multiply(n0,#7)| | other |
two trains are moving at 50 kmph and 70 kmph in opposite directions . their lengths are 150 m and 100 m respectively . the time they will take to pass each other completely is ? | "70 + 50 = 120 * 5 / 18 = 100 / 3 mps d = 150 + 100 = 250 m t = 250 * 3 / 100 = 15 / 2 = 7 1 / 2 sec answer : d" | a ) 7 1 / 2 sec , b ) 7 1 / 7 sec , c ) 7 8 / 2 sec , d ) 7 1 / 9 sec , e ) 7 2 / 2 sec | d | divide(add(150, 100), multiply(add(50, 70), const_0_2778)) | add(n2,n3)|add(n0,n1)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
if there are 56 laborers in a crew , and on a certain day , 30 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) . | 30 / 56 * 100 = 53.57 53.6 % correct answer a | a ) 53.6 % , b ) 52.6 % , c ) 54.6 % , d ) 55.6 % , e ) 59.6 % | a | divide(multiply(30, const_100), 56) | multiply(n1,const_100)|divide(#0,n0) | physics |
how many of the positive factors of 24 are not factors of 30 | "factors of 24 - 1 , 2 , 3 , 4 , 6 , 8 , 12,24 factors of 30 - 1 , 2 , 3 , 5 , 6 , 10 , 15 and 30 . comparing both , we have four factors of 24 which are not factors of 30 - 4,8 , 12,24 answer ( c )" | a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5 | c | divide(30, 24) | divide(n1,n0)| | other |
if f ( x ) = 12 - x ^ 2 / 2 and f ( 2 k ) = 5 k , what is one possible value for k ? | "first of all , see thisgmat blog postand check the related lesson linked below for some background on function notation . we can plug anything in for x and get a result . you can find f ( 1 ) , for example , by plugging in 1 where x is , and you would get 12 - 1 / 2 = 11.5 . or we could find f ( 2 ) , which would be 1... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 8 | d | add(divide(power(2, 2), 2), const_1) | power(n1,n1)|divide(#0,n1)|add(#1,const_1)| | general |
the average runs scored by a batsman in 20 matches is 40 . in the next 10 matches the batsman scored an average of 13 runs . find his average in all the 30 matches ? | total score of the batsman in 20 matches = 800 . total score of the batsman in the next 10 matches = 130 . total score of the batsman in the 30 matches = 930 . average score of the batsman = 930 / 30 = 31 . answer : a | a ) 31 , b ) 67 , c ) 88 , d ) 69 , e ) 12 | a | divide(add(multiply(40, 20), multiply(13, 10)), add(20, 10)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0) | general |
p , q and r have rs . 9000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "let the amount with r be rs . r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 9000 - r ) = > 3 r = 18000 - 2 r = > 5 r = 18000 = > r = 3600 . answer : b" | a ) rs . 3000 , b ) rs . 3600 , c ) rs . 2400 , d ) rs . 4000 , e ) none of these | b | multiply(divide(9000, add(const_1, divide(const_2, const_3))), divide(const_2, const_3)) | divide(const_2,const_3)|add(#0,const_1)|divide(n0,#1)|multiply(#2,#0)| | general |
a car gets 60 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 170 kilometers ? | "each 60 kilometers , 1 gallon is needed . we need to know how many 60 kilometers are there in 170 kilometers ? 60 ã · 170 = 2.83 ã — 1 gallon = 2.83 gallons correct answer is e ) 2.83 gallons" | a ) 3.5 gallons , b ) 2.37 gallons , c ) 5.7 gallons , d ) 4.25 gallons , e ) 2.83 gallons | e | divide(170, 60) | divide(n1,n0)| | physics |
if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 293 pounds . what is jake ’ s present weight , in pounds ? | "j + k = 293 and so k = 293 - j j - 8 = 2 k j - 8 = 2 ( 293 - j ) 3 j = 594 j = 198 the answer is e ." | a ) 182 , b ) 186 , c ) 190 , d ) 194 , e ) 198 | e | add(multiply(divide(subtract(293, 8), const_3), const_2), 8) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)| | general |
the radius of a circle is 4 / π cm then its perimeter is ? | perimeter = 2 π r 2 π * 4 / π = 8 cm answer : c | ['a ) 10', 'b ) 9', 'c ) 8', 'd ) 12', 'e ) 15'] | c | multiply(const_2, multiply(const_pi, divide(4, const_pi))) | divide(n0,const_pi)|multiply(#0,const_pi)|multiply(#1,const_2) | physics |
the effective annual rate of interest corresponding to a nominal rate of 6 % per annum payable half - yearly is ? | "amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 3 / 100 ) 2 ] = rs . 106.09 effective rate = ( 106.09 - 100 ) = 6.09 % answer : d" | a ) 6.99 % , b ) 6.89 % , c ) 6.08 % , d ) 6.09 % , e ) 6.19 % | d | add(add(divide(6, const_2), divide(6, const_2)), divide(multiply(divide(6, const_2), divide(6, const_2)), const_100)) | divide(n0,const_2)|add(#0,#0)|multiply(#0,#0)|divide(#2,const_100)|add(#1,#3)| | gain |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 5 cm . | "explanation : area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 5 ) = 95 cm 2 answer : option e" | a ) 187 cm 2 , b ) 185 cm 2 , c ) 85 cm 2 , d ) 85 cm 2 , e ) 95 cm 2 | e | quadrilateral_area(5, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
from a pack of 52 cards , 1 card is drawn at random . what is the probability that a heart is drawn ? | "the total number of cards is 52 . the number of hearts is 13 . p ( heart ) = 13 / 52 = 1 / 4 the answer is c ." | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 13 , e ) 3 / 26 | c | divide(multiply(const_4, const_3), 52) | multiply(const_3,const_4)|divide(#0,n0)| | probability |
if each year the population of the country grows by 50 % , how many years will elapse before the population of the country doubles ? | "till year 2000 , population is 100 . year 2001 : population becomes 150 . . . . . . . . . . . . . 1 year elapsed year 2002 : population > 200 . . . . . . . . . . . . . 2 year elapsed answer : c" | a ) 3 , b ) 4 , c ) 2 , d ) 6 , e ) 7 | c | add(const_4, const_4) | add(const_4,const_4)| | gain |
for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 7 + 3 and * 7 + 7 , inclusive ? | "7 ! = 720 * 7 = 5040 # in question = 5043 , 5044 , 5045 , 5046 , 5047 none of these are prime answer - a" | a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(subtract(add(multiply(multiply(multiply(7, 3), const_2), const_4), 7), add(multiply(multiply(multiply(7, 3), const_2), const_4), 3)), 1) | multiply(n1,n2)|multiply(#0,const_2)|multiply(#1,const_4)|add(n1,#2)|add(n2,#2)|subtract(#3,#4)|subtract(#5,n0)| | general |
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 12 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 12 sec = 27 * 10 = 270 m . the length of the faster train = 120 m . answer : d" | a ) 270 , b ) 288 , c ) 266 , d ) 120 , e ) 126 | d | multiply(divide(subtract(72, 36), const_3_6), 12) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
two trains of equal are running on parallel lines in the same direction at 44 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 44 - 36 = 8 km / hr . = 8 * 5 / 18 = 20 / 9 m / sec . 2 x / 36 = 20 / 9 = > x = 40 . answer : a" | a ) 40 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 84 m | a | divide(multiply(36, divide(multiply(subtract(44, 36), const_1000), const_3600)), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general |
a man sold 20 articles for $ 60 and gained 20 % . how many articles should he sell for $ 50 to incur a loss 20 % ? | production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 required production costs for a loss of 20 % : $ 50 * ( 100 % + 20 % ) = $ 60 number of articles to be sold for $ 60 to incur a 20 % loss : $ 60 / $ 2.40 = 25 thus , solution a is correct . | a ) 25 , b ) 36 , c ) 40 , d ) 50 , e ) 48 | a | divide(original_price_before_loss(20, 50), divide(original_price_before_gain(20, 60), 20)) | original_price_before_gain(n0,n1)|original_price_before_loss(n0,n3)|divide(#0,n0)|divide(#1,#2) | gain |
each of the 59 members in lourdes school class is required to sign up for a minimum of one and a maximum of 3 academic clubs . the 3 clubs to choose from are the poetry club , the history club , and the writing club . a total of 22 students sign up for the poetry club , 27 students for the history club , and 28 student... | each of the 59 members in lourdes school class is required to sign up for a minimum of one and a maximum of three academic clubs . total = g 1 + g 2 + g 3 - ( # in exactly 2 ) - 2 * ( # in 3 sets ) 59 = 22 + 27 + 28 - ( 6 ) - 2 x so , # in 3 sets = 6 = c | a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | c | divide(subtract(subtract(add(add(22, 27), 28), 59), 6), const_2) | add(n3,n4)|add(n5,#0)|subtract(#1,n0)|subtract(#2,n6)|divide(#3,const_2) | general |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 22 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 6 ? - - - - - - 60 12 h rs = 16 + 22 = 38 t = 12 d = 38 * 12 = 456 answer : d" | a ) 565 , b ) 444 , c ) 676 , d ) 456 , e ) 663 | d | add(multiply(divide(60, subtract(22, 16)), 16), multiply(divide(60, subtract(22, 16)), 22)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
bruce and bhishma are running on a circular track of length 600 m . speed of bruce is 30 m / s and that of bhishma is 20 m / s . they start fro the same point at the same time in the same direction . when will they meet again for the first time ? | actually arjun has to make a lead of 600 m , because when arjun will be 600 m ahead of bhishma , they will be together again as a person when completes the total length it starts retracing he same path and thus arjun and bhishma can be together again . since , they make a difference of 10 m in 1 second . so , he will c... | a ) 60 sec , b ) 70 sec , c ) 80 sec , d ) 90 sec , e ) 100 sec | d | add(divide(600, subtract(30, 20)), divide(divide(600, subtract(30, 20)), const_2)) | subtract(n1,n2)|divide(n0,#0)|divide(#1,const_2)|add(#1,#2) | physics |
if a certain computer is capable of printing 4700 monthly credit card bills per hour , while a new model is capable of printing at a rate of 6800 per hour , the old model will take approximately how much longer than the new model to print 10000 bills ? | old model ' s 1 hr work = 4700 new model ' s 1 hr work = 6800 old model ' time taken to print 10000 cards for 1 hr - - 4700 cards x - - 10000 x = 10000 / 4700 = 100 / 47 new model ' time taken to print 10000 cards for 1 hr - - 6800 cards y - - 10000 y = 10000 / 6800 = 100 / 68 x = 10000 / 4700 = 100 / 47 = 2 6 / 47 hr ... | a ) 19 mins , b ) 29 mins , c ) 39 mins , d ) 38 mins , e ) 37 mins | c | multiply(subtract(divide(10000, 4700), divide(10000, 6800)), const_60) | divide(n2,n0)|divide(n2,n1)|subtract(#0,#1)|multiply(#2,const_60) | general |
average between two sets of numbers is closer to the set withmore no ' s ? | "if on a test three people answered 90 % of the questions correctly and two people answered 80 % correctly , then the average for the group is not 85 % but rather 3 × 90 + 2 × 805 = 4305 = 86.3 × 90 + 2 × 805 = 4305 = 86 . here , 90 has a weight of 3 = > it occurs 3 times . whereas 80 has a weight of 2 = > it occurs 2 ... | a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | b | multiply(multiply(const_2, const_4), const_10) | multiply(const_2,const_4)|multiply(#0,const_10)| | general |
in a graduating class of 232 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number w and the smallest possible number of students that could have taken both geometry and biology ? | "official solution : first of all , notice that since 144 took geometry and 119 took biology , then the number of students who took both geometry and biology can not be greater than 119 . { total } = { geometry } + { biology } - { both } + { neither } ; 232 = 144 + 119 - { both } + { neither } ; { both } = 31 + { neith... | a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | d | subtract(119, subtract(add(144, 119), 232)) | add(n1,n2)|subtract(#0,n0)|subtract(n2,#1)| | other |
rani bought more apples than oranges . she sells apples at ₹ 23 apiece and makes 15 % profit . she sells oranges at ₹ 10 apiece and makes 25 % profit . if she gets ₹ 653 after selling all the apples and oranges , find her profit percentage . | "given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the val... | a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(1... | add(n1,const_2)|divide(n3,const_100)|multiply(n2,const_2)|add(#2,const_1)|add(#1,const_1)|divide(n2,#4)|multiply(#3,#2)|multiply(#0,#5)|add(#6,#7)|subtract(n4,#8)|divide(#9,#8)|multiply(#10,const_100)| | gain |
a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 70 points . to the nearest tenth of a point , what was the student ’ s average grade for the entire two - year period ? | "let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 70 * 5 = 350 student ' s average = ( 600 + 350 ) / 11 = 86.36 answer d" | a ) 79 , b ) 89 , c ) 95 , d ) 86.36 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 70)), add(6, 5))) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|floor(#4)| | general |
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