options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 69 m sqaure , b ) 49 m sqaure , c ) 52 m sqaure , d ) 64 m sqaure , e ) none of these | a | add(multiply(const_2, add(multiply(add(divide(40, const_100), 1), 5), multiply(add(divide(40, const_100), 1), 7))), multiply(5, 7)) | a cistern 7 m long and 5 m wide contains water up to a breadth of 1 m 40 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 5 * 1.40 + 7 * 1.40 ) ] + 7 * 5 = 69 m square option a" | a = 40 / 100
b = a + 1
c = b * 5
d = 40 / 100
e = d + 1
f = e * 7
g = c + f
h = 2 * g
i = 5 * 7
j = h + i
|
a ) 2 / 1 , b ) 4 / 1 , c ) 2 / 3 , d ) 3 / 4 , e ) 3 / 2 | a | divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 9)) | in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 9 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | "given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 9 ) = 36 / 18 = 2 / 1 . answer : a ." | a = 4 + 1
b = 45 / a
c = 45 - b
d = 4 + 1
e = 45 / d
f = e + 9
g = c / f
|
a ) 600 cm 2 , b ) 500 cm 2 , c ) 400 cm 2 , d ) 432 cm 2 , e ) 100 cm 2 | d | add(multiply(multiply(divide(const_1, const_2), 27), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(27, 27)))), 27) | find the area of a rhombus one side of which measures 20 cm and one diagonal is 27 cm . | "explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = √ ( 20 ) 2 – ( 12 ) 2 = √ 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 × 27 x 32 ) cm 2 = 432 cm 2 answer : option d" | a = 1 / 2
b = a * 27
c = 20 * 20
d = c * 4
e = 27 * 27
f = d - e
g = math.sqrt(f)
h = b * g
i = h + 27
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | multiply(subtract(divide(power(28, const_2), 56), floor(divide(power(28, const_2), 56))), 56) | on dividing a number by 56 , we get 28 as remainder . on dividing the same number by 8 , what will be the remainder ? | "number = 56 x + 28 ( ∵ since the number gives 28 as remainder on dividing by 56 ) = ( 7 × 8 × x ) + ( 3 × 8 ) + 4 hence , if the number is divided by 8 , we will get 4 as remainder . answer : c" | a = 28 ** 2
b = a / 56
c = 28 ** 2
d = c / 56
e = math.floor(d)
f = b - e
g = f * 56
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | divide(multiply(10.5, const_100), subtract(add(25, const_100), divide(multiply(add(30, const_100), subtract(const_100, 20)), const_100))) | a man sells an article at a profit of 25 % . if he had bought it at 20 % less and sold it for rs . 10.50 less , he would have gained 30 % . find the cost of the article . | let c . p = 100 gain = 25 % s . p = 125 supposed c . p = 80 gain = 30 % s . p = ( 130 * 80 ) / 100 = 104 diff = ( 125 - 104 ) = 21 diff 21 when c . p = 100 then diff 10.50 when c . p = ( 100 * 10.50 ) / 21 = 50 answer : c | a = 10 * 5
b = 25 + 100
c = 30 + 100
d = 100 - 20
e = c * d
f = e / 100
g = b - f
h = a / g
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | subtract(6, const_4) | how many three - element subsets of { 5 , 6,7 , 8,9 } are there that do not contain the pair of elements 6 and 9 ? | "required subsets are = { 5 , 6,7 } , { 5 , 7,8 } , { 5 , 8,9 } , { 6 , 7,8 } , { 7 , 8,9 } = 5 answer : e" | a = 6 - 4
|
a ) 112.5 , b ) 122.5 , c ) 132.5 , d ) 67.5 , e ) 212.5 | d | subtract(divide(multiply(multiply(1500, 11.5), 3), const_100), divide(multiply(multiply(1500, 10), 3), const_100)) | if a lends rs . 1500 to b at 10 % per annum and b lends the same sum to c at 11.5 % per annum then the gain of b in a period of 3 years is ? | "( 1500 * 1.5 * 3 ) / 100 = > 67.5 answer : d" | a = 1500 * 11
b = a * 3
c = b / 100
d = 1500 * 10
e = d * 3
f = e / 100
g = c - f
|
a ) 102 kmph , b ) 106 kmph , c ) 108 kmph , d ) 62 kmph , e ) 72 kmph | c | subtract(multiply(divide(340, 8), const_3_6), 45) | a man sitting in a train which is traveling at 45 kmph observes that a goods train , traveling in opposite direction , takes 8 seconds to pass him . if the goods train is 340 m long , find its speed | "explanation : relative speed = 340 / 8 m / sec = ( ( 340 / 8 ) × ( 18 / 5 ) ) kmph = 153 kmph . speed of goods train = ( 153 - 45 ) kmph = 108 kmph answer : option c" | a = 340 / 8
b = a * const_3_6
c = b - 45
|
a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | multiply(multiply(2, add(const_1, const_1)), add(const_1, const_1)) | if x and y are both odd prime numbers and x < y , how many distinct positive integer q factors does 2 xy have ? | since 2 xy prime q factors are x ^ 1 * y ^ 1 * 2 ^ 1 , its total number or factors must be ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 2 ^ 3 = 8 . thus , i think d would be the correct answer . | a = 1 + 1
b = 2 * a
c = 1 + 1
d = b * c
|
a ) - 1 , b ) 6 , c ) 7 , d ) 10 , e ) 14 | d | add(add(const_4, 5), subtract(5, const_4)) | what is the sum of all possible solutions to | x - 5 | ^ 2 + | x - 5 | = 20 ? | "denote | x - 5 | as y : y ^ 2 + y = 20 - - > y = - 5 or y = 4 . discard the first solution since y = | x - 5 | , so it ' s an absolute value and thus can not be negative . y = | x - 5 | = 4 - - > x = 9 or x = - 1 . the sum = 10 answer : d ." | a = 4 + 5
b = 5 - 4
c = a + b
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | e | multiply(const_10, subtract(const_10, const_2)) | how many integers k greater than 100 and less than 1100 are there such that if the hundreds and the unit digits of k are reversed , the resulting integer is k + 99 ? | "not sure if this is the shortest . . but this is how i did this there are 9 sets of integers with hundreds and units digits exchanged that satisfies k + 99 . 1 . 102 | 201 ( satisfies k + 99 , where k = 102 ) 2 . 203 | 302 ( satisfies k + 99 , where k = 203 ) 3 . . . . 4 . . . . 5 . . . . 6 . . . . 7 . 708 | 807 8 . 809 | 908 9 . 910 | 1001 each set has 10 such numbers . 1 . 102 | 201 ( still k + 99 holds good ) 2 . 112 | 211 3 . 122 | 221 4 . 132 | 231 5 . . . . 6 . . . . 7 . . . . 8 . . . . 9 . 182 | 281 10 . 192 | 291 therefore , 9 sets with 10 such number in each set will give 9 x 10 = 90 integers . e" | a = 10 - 2
b = 10 * a
|
a ) 1590 , b ) 1600 , c ) 1950 , d ) 1630 , e ) 1760 | a | add(subtract(add(multiply(35, 53), multiply(const_3, const_10)), multiply(const_4, const_100)), const_100) | anil had to do a multiplication . instead of taking 35 as one of the multipliers , he took 53 . as a result , the product went up by 540 . what is the new product ? | explanation : let the number that anil wanted to multiply be ' y ' . he was expected to find the value of 35 y . instead , he found the value of 53 y . the difference between the value that he got ( 53 y ) and what he was expected to get ( 35 y ) is 540 . = > 53 y - 35 y = 540 = > ( 53 - 35 ) * y = 540 = > y = 30 therefore , the correct product = 53 * 30 = 1590 answer : a | a = 35 * 53
b = 3 * 10
c = a + b
d = 4 * 100
e = c - d
f = e + 100
|
a ) 6.54 % , b ) 4.54 % , c ) 8 . 2 % , d ) 3.03 % , e ) 5.54 % | d | divide(const_100, 33) | at what rate percent of simple interest will a sum of money double itself in 33 years ? | "let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 33 ) = 100 / 33 = 3.03 % answer : d" | a = 100 / 33
|
a ) 229 , b ) 108 , c ) 278 , d ) 216 , e ) 112 | d | multiply(divide(360, 6), const_3_6) | a 360 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | "s = 360 / 6 * 18 / 5 = 216 kmph answer : d" | a = 360 / 6
b = a * const_3_6
|
a ) 118 , b ) 110 , c ) 112 , d ) 113 , e ) 98 | e | multiply(subtract(divide(divide(multiply(subtract(const_100, 30), add(const_100, 40)), const_100), const_100), const_1), const_100) | a trader bought a car at 30 % discount on its original price . he sold it at a 40 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 70 s = 70 * ( 140 / 100 ) = 98 100 - 112 = 2 % answer : e" | a = 100 - 30
b = 100 + 40
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
|
a ) 51 , b ) 20 , c ) 40 , d ) 50 , e ) 55 | d | multiply(subtract(25, 20), const_10) | 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 25 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 = 25 x / 10 = 5 x = 50 answer : d" | a = 25 - 20
b = a * 10
|
a ) 36 seconds , b ) 18 seconds , c ) 40 seconds , d ) 19 seconds , e ) 45 seconds | a | divide(add(310, 140), divide(multiply(45, const_1000), const_3600)) | a train is 310 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 310 + 140 = 500 meter time = distance / speed = 450 * ( 2 / 25 ) = 36 seconds answer : a" | a = 310 + 140
b = 45 * 1000
c = b / 3600
d = a / c
|
a ) 12 , b ) 20 , c ) 88 , d ) 77 , e ) 14 | e | subtract(49, divide(49, add(divide(2, 5), const_1))) | a 49 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ? | "1 : 2 / 5 = 5 : 2 2 / 7 * 49 = 14 answer : e" | a = 2 / 5
b = a + 1
c = 49 / b
d = 49 - c
|
a ) 5 , b ) 21 , c ) 33 , d ) 60 , e ) 6 | b | add(add(add(add(6, subtract(6, 1)), subtract(subtract(6, 1), 1)), subtract(subtract(subtract(6, 1), 1), 1)), 2) | there are 6 boxes numbered 1 , 2 , . . . 6 . each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered . the total number of ways in which this can be done is : | "explanatory answer list down possibilities : from only 1 box all the way to all 6 if only one of the boxes has a green ball , it can be any of the 6 boxes . so , we have 6 possibilities . if two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes . 12 , 23 , 34 , 45 , 56 . if 3 of the boxes have green balls , there are 4 possibilities : 123 , 234 , 345 , 456 . if 4 boxes have green balls , there are 3 possibilities : 1234 , 2345 , 3456 . if 5 boxes have green balls , there are 2 possibilities : 12345 , 23456 . if all 6 boxes have green balls , there is just 1 possibility . total number of possibilities = 6 + 5 + 4 + 3 + 2 + 1 = 21 . choice b" | a = 6 - 1
b = 6 + a
c = 6 - 1
d = c - 1
e = b + d
f = 6 - 1
g = f - 1
h = g - 1
i = e + h
j = i + 2
|
a ) 11 , b ) 15 , c ) 12 , d ) 17 , e ) 18 | e | divide(18, const_1) | the overall age of x and y is 18 year greater than the overall age of y and z . z is how many decades younger that x ? | "e 18 ( x + y ) â € “ ( y + z ) = 18 x â € “ z = 18" | a = 18 / 1
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a ) 4 / 7 , b ) 5 / 7 , c ) 6 / 7 , d ) 1 / 7 , e ) 3 / 7 | a | divide(const_4, add(multiply(const_4, 2), const_1)) | tom , working alone , can paint a room in 8 hours . peter and john , working independently , can paint the same room in 4 hours and 2 hours , respectively . tom starts painting the room and works on his own for two hour . he is then joined by peter and they work together for two hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ? | "let the time when all three were working together be t hours . then : tom worked for t + 4 hour and has done 1 / 8 * ( t + 4 ) part of the job ; peter worked for t + 2 hour and has done 1 / 4 * ( t + 2 ) part of the job ; john worked for t hours and has done 1 / 2 * t part of the job : 1 / 8 * ( t + 4 ) + 1 / 4 * ( t + 2 ) + 1 / 2 * t = 1 - - > multiply by 8 - - > ( t + 4 ) + ( 2 t + 2 ) + 4 t = 8 - - > t = 2 / 7 ; hence peter has done 1 / 4 * ( 2 / 7 + 2 ) = 1 / 4 * 16 / 7 = 16 / 28 = 8 / 14 = 4 / 7 answer : a" | a = 4 * 2
b = a + 1
c = 4 / b
|
a ) 10.2 kmph , b ) 10.4 kmph , c ) 10.8 kmph , d ) 11.8 kmph , e ) none | c | divide(add(10, 12), add(divide(10, 12), divide(12, 10))) | a boy rides his bicycle 10 km at an average sped of 12 km / hr and again travels 12 km at an average speed of 10 km / hr . his average speed for the entire trip is approximately . | sol . total distance travelled = ( 10 + 12 ) km / hr = 22 km / hr . total time taken = [ 10 / 12 + 12 / 10 ] hrs = 61 / 30 hrs ∴ average speed = [ 22 * 30 / 61 ] km / hr = 10.8 km / hr . answer c | a = 10 + 12
b = 10 / 12
c = 12 / 10
d = b + c
e = a / d
|
a ) 8 / 2 , b ) 2 / 4 , c ) 1 / 8 , d ) 2 / 4 , e ) 4 / 6 | a | divide(multiply(4, 2), 2) | ( a / 2 ) / ( b / c ) in the expression above , a , b , and c are different numbers and each is one of the numbers 1 , 2 , or 4 . what is the largest possible value of the expression ? | ( a / 2 ) / ( b / c ) = ( a * c ) / 2 b the expression will have the largest value when numerator ( a * c ) is the largest . = ( 2 * 4 ) / 2 * 1 = 8 / 2 answer a | a = 4 * 2
b = a / 2
|
a ) 562 , b ) 356 , c ) 452 , d ) 494 , e ) 512 | d | add(multiply(divide(8, 5), 190), 190) | in a college the ratio of the numbers of boys to the girls is 8 : 5 . if there are 190 girls , the total number of students in the college is ? | "let the number of boys and girls be 8 x and 5 x then , 5 x = 190 x = 38 total number of students = 13 x = 13 * 38 = 494 answer is d" | a = 8 / 5
b = a * 190
c = b + 190
|
a ) 1 / 2 , b ) 3 / 5 , c ) 5 / 7 , d ) 9 / 11 , e ) 3 / 4 | e | divide(multiply(subtract(const_100, 8), 15), multiply(add(const_100, 15), 16)) | if the numerator of a fraction be increased by 15 % and its denominator be diminished by 8 % , the value of the fraction is 15 / 16 . find the original fraction ? | "let the original fraction be x / y then , 115 % of x / 92 % of y = 15 / 16 115 x / 92 y = 15 / 16 x / y = 3 / 4 answer is e" | a = 100 - 8
b = a * 15
c = 100 + 15
d = c * 16
e = b / d
|
a ) 130 , b ) 124 , c ) 120 , d ) 115 , e ) 112 | e | add(multiply(32, 8), multiply(24, 16)) | simplify : 32 ÷ 8 × 24 + 16 = | "32 ÷ 8 × × 24 + 16 = > 4 × × 24 + 16 = 96 + 16 = 112 option e" | a = 32 * 8
b = 24 * 16
c = a + b
|
a ) 37.5 , b ) 75 , c ) 100 , d ) 150 , e ) 175 | b | divide(subtract(multiply(10, const_2), 5), subtract(subtract(subtract(1, divide(20, const_100)), multiply(subtract(1, divide(20, const_100)), divide(1, const_4))), multiply(const_2, divide(20, const_100)))) | a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 5 gallons of water and 10 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ? | "nope , 150 . i can only get it by following pr ' s backsolving explanation . i hate that . original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 10 gallons salt is added and 5 gallons of water is added . so salt now becomes - > ( 0.2 x + 10 ) and water - - > ( 0.6 x + 5 ) amount of salt is 33.33 % of total . so amount of water is 66.66 % . so salt is half of the volume of water . so ( 0.2 x + 10 ) = ( 0.6 x + 5 ) / 2 = 0.4 x + 20 = 0.6 x + 5 = > 0.2 x = 15 solving , x = 75 answer : b" | a = 10 * 2
b = a - 5
c = 20 / 100
d = 1 - c
e = 20 / 100
f = 1 - e
g = 1 / 4
h = f * g
i = d - h
j = 20 / 100
k = 2 * j
l = i - k
m = b / l
|
a ) 20 % , b ) 18 % , c ) 15 % , d ) 12 % , e ) 24 % | e | subtract(const_100, divide(multiply(add(const_100, 12), subtract(const_100, 32)), const_100)) | the tax on a commodity is diminished by 32 % but its consumption is increased by 12 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 68 * 112 = 7616 10000 - - - - - - - 2384 100 - - - - - - - ? = 24 % answer : e" | a = 100 + 12
b = 100 - 32
c = a * b
d = c / 100
e = 100 - d
|
a ) 2400 , b ) 2000 , c ) 1904 , d ) 1906 , e ) none of them | a | multiply(106, power(106, 94)) | 106 x 106 - 94 x 94 = ? | "= ( 106 ) ^ 2 - ( 94 ) ^ 2 = ( 106 + 94 ) ( 106 - 94 ) = ( 200 x 12 ) = 2400 answer is a" | a = 106 ** 94
b = 106 * a
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a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 35 | e | subtract(90, add(add(subtract(50, 35), subtract(40, 35)), 35)) | of the 90 house in a development , 50 have a two - car garage , 40 have an in - the - ground swimming pool , and 35 have both a two - car garage and an in - the - ground swimming pool . how many houses in the development have neither a two - car garage nor an in - the - ground swimming pool ? | "neither car nor garage = total - garage - ( swim - common ) = 90 - 50 - ( 40 - 35 ) = 90 - 55 = 35 answer e" | a = 50 - 35
b = 40 - 35
c = a + b
d = c + 35
e = 90 - d
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a ) 4 , b ) 6 , c ) 10 , d ) 8 , e ) 11 | b | divide(multiply(multiply(18, 15), 2), multiply(multiply(5, 6), 3)) | how many cuboids of length 5 m , width 6 m and height 3 m can be farmed from a cuboid of 18 m length , 15 m width and 2 m height . | "( 18 ã — 15 ã — 12 ) / ( 5 ã — 3 ã — 2 ) = 6 answer is b ." | a = 18 * 15
b = a * 2
c = 5 * 6
d = c * 3
e = b / d
|
a ) - 4 , b ) - 2 , c ) 11 , d ) 13 , e ) 22 | d | divide(add(26, 26), add(3, const_1)) | when x is multiplied by 3 , the result is 26 more than the result of subtracting x from 26 . what is the value of x ? | "the equation that can be formed is : 3 x - 26 = 26 - x or , 4 x = 52 or , x = 13 . d answer ." | a = 26 + 26
b = 3 + 1
c = a / b
|
a ) 85 , b ) 94 , c ) 83 , d ) 72 , e ) none | c | add(multiply(divide(add(11, 5), const_2), 11), subtract(11, divide(add(11, 5), const_2))) | the sum of digits of a two digit number is 11 , the difference between the digits is 5 . find the number | "description : = > x + y = 11 , x - y = 5 adding these 2 x = 16 = > x = 8 , y = 3 . thus the number is 83 answer c" | a = 11 + 5
b = a / 2
c = b * 11
d = 11 + 5
e = d / 2
f = 11 - e
g = c + f
|
a ) 15 , b ) 25 , c ) 71 , d ) 35 , e ) 45 | c | add(subtract(divide(multiply(multiply(3, 7), 5), 3), divide(multiply(multiply(3, 7), 5), 7)), divide(multiply(multiply(3, 7), 5), 5)) | if x , y , and z are positive integers and 3 x = 7 y = 5 z , then the least possible value of x + y + z is | "given 3 x = 7 y = 5 z x + y + z in terms of x = x + ( 3 x / 7 ) + ( 3 x / 5 ) = 71 x / 35 now checking with each of the answers and see which value gives a minimum integer value . a x = 35 / 71 * 15 , not an integer b , d , e can be ruled out similarly . c is minimum value as x = 71 * 35 / 71 = 35 answer is c" | a = 3 * 7
b = a * 5
c = b / 3
d = 3 * 7
e = d * 5
f = e / 7
g = c - f
h = 3 * 7
i = h * 5
j = i / 5
k = g + j
|
a ) 180 , b ) 130 , c ) 140 , d ) 160 , e ) 288 | e | multiply(divide(multiply(8, 90), subtract(13, 8)), const_2) | a sports retailer ordered white and yellow tennis balls in equal number but the dispatch clerk of the tennis ball company erred and dispatched 90 extra yellow balls and making ratio of white balls to yellow balls 8 / 13 . how many tennis balls did the retailer order originally . | "white : yellow = x : ( x + 90 ) = 8 : 13 - - > 13 x = 8 x + 720 - - > x = 144 . the total # of balls originally x + x = 144 + 144 = 288 . answer : e ." | a = 8 * 90
b = 13 - 8
c = a / b
d = c * 2
|
a ) 1,000 , b ) 500 , c ) 250 , d ) 50 , e ) 0 | a | divide(divide(10000, 10), divide(10000, 10)) | a combustion reaction forms carbon dioxide . a carbon dioxide molecule contains one carbon and two oxygen atoms . if , over a period of 10 minutes , a combustion reaction creates 10000 molecules of carbon dioxide then approximately how many more atoms of oxygen than carbon are created on average per minute ? | solution : 10,000 carbon dioxide molecules are created over a period of 10 minutes . therefore 10,000 / 10 = 1,000 carbon dioxide molecules are created on average per minute each carbon dioxide molecule contains one carbon atom and two oxygen atoms . so 1,000 carbon dioxide molecules contain 1 × 1,000 = 1,000 carbon atoms and 2 × 1,000 = 2,000 oxygen atoms . the difference is 2,000 – 1,000 = 1,000 . the correct answer is a . | a = 10000 / 10
b = 10000 / 10
c = a / b
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a ) 45 , b ) 55 , c ) 65 , d ) 75 , e ) 85 | d | multiply(add(5, 10), 5) | if two girls starting from same point , walking in the opposite directions with 5 km / hr and 10 km / hr as average speeds respectively . then the distance between them after 5 hours is ? | "explanation : total distance = distance traveled by person a + distance traveled by person b = ( 5 ã — 5 ) + ( 10 ã — 5 ) = 25 + 50 = 75 km answer : d" | a = 5 + 10
b = a * 5
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a ) 4 . , b ) 6 . , c ) 7 . , d ) 8 . , e ) 9 . | e | add(multiply(15, divide(const_1, 1)), const_1) | in the junior basketball league there are 15 teams , 1 / 3 of them are bad and ½ are rich . what ca n ' t be the number of teams that are rich and bad ? | "total teams = 15 bad teams = ( 1 / 3 ) * 15 = 5 rich teams = 8 so maximum value that the both rich and bad can take will be 8 . so e = 9 can not be that value . ans e ." | a = 1 / 1
b = 15 * a
c = b + 1
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a ) 11 , b ) 17 , c ) 26 , d ) 101 , e ) 1322 | c | sqrt(divide(2028, const_3)) | the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 2028 sq m , then what is the breadth of the rectangular plot ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 2028 3 b 2 = 2028 b 2 = 676 = 26 ( b > 0 ) b = 26 m . answer : c" | a = 2028 / 3
b = math.sqrt(a)
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a ) 1683 , b ) 1684 , c ) 1685 , d ) 1582 , e ) 1584 | a | add(multiply(lcm(lcm(lcm(5, multiply(const_2, const_3)), add(const_3, const_4)), 8), const_2), 3) | find the least number which when divided by 5 , 67 , and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder | explanation : l . c . m . of 5,6 , 7,8 = 840 . required number is of the form 840 k + 3 least value of k for which ( 840 k + 3 ) is divisible by 9 is k = 2 . required number = ( 840 x 2 + 3 ) = 1683 answer is a | a = 2 * 3
b = math.lcm(5, a)
c = 3 + 4
d = math.lcm(b, c)
e = math.lcm(d, 8)
f = e * 2
g = f + 3
|
a ) 1 / 32 , b ) 7 / 8 , c ) 9 / 64 , d ) 5 / 64 , e ) 3 / 64 | a | multiply(multiply(divide(1, 6), divide(1, 2)), subtract(1, divide(5, 8))) | xavier , yvonne , and zelda each try independently to solve a problem . if their individual probabilities for success are 1 / 6 , 1 / 2 and 5 / 8 , respectively , what is the probability that xavier and yvonne , but not zelda , will solve the problem ? | "p ( xavier will solve ) = 1 / 6 p ( yvonne will solve ) = 1 / 2 p ( zelda will not solve ) = 1 - 5 / 8 = 3 / 8 . now , we need to multiply all this ps to find an answer : p = ( 1 / 6 ) * ( 1 / 2 ) * ( 3 / 8 ) = 1 / 32 . ans . a ." | a = 1 / 6
b = 1 / 2
c = a * b
d = 5 / 8
e = 1 - d
f = c * e
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a ) 0 , b ) 0 , c ) 27 , d ) 54 , e ) can not be determined | b | divide(multiply(multiply(multiply(3, 2), multiply(3, 2)), multiply(3, 2)), const_4) | for any positive number x , the function [ x ] denotes the greatest integer less than or equal to x . for example , [ 1 ] = 1 , [ 1.267 ] = 1 and [ 1.999 ] = 1 . if k is a positive integer such that k ^ 2 is divisible by 45 and 80 , what is the units digit of k ^ 3 / 4000 ? | "k = [ lcm of 80 and 45 ] * ( any integer ) however minimum value of k is sq . rt of 3 ^ 2 * 4 ^ 2 * 5 ^ 2 = 60 * any integer for value of k ( 60 ) * any integer unit value will be always zero . b" | a = 3 * 2
b = 3 * 2
c = a * b
d = 3 * 2
e = c * d
f = e / 4
|
a ) 74 , b ) 166 , c ) 86 , d ) 92 , e ) 98 | b | add(multiply(add(multiply(6, const_3), 22), divide(add(multiply(6, const_3), 22), 5)), 6) | in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 22 to the thrice of the remainder . the dividend is | "divisor = ( 6 * 3 ) + 22 = 40 5 * quotient = 40 quotient = 8 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 8 ) + 6 = 166 . b )" | a = 6 * 3
b = a + 22
c = 6 * 3
d = c + 22
e = d / 5
f = b * e
g = f + 6
|
a ) 8925 , b ) 8032.5 , c ) 7302.3 , d ) 8900 , e ) none of these | c | divide(multiply(const_100, 4016.25), multiply(11, 5)) | a sum fetched total simple interest of 4016.25 at the rate of 11 p . c . p . a . in 5 years . what is the sum ? | "let the sums be p . now , 55 % of p = 4016.25 or , p = 7302.3 answer c" | a = 100 * 4016
b = 11 * 5
c = a / b
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a ) 18 sec , b ) 12 sec , c ) 15 sec , d ) 20 sec , e ) 13.5 sec | e | divide(subtract(multiply(12, 15), 15), divide(multiply(12, 15), 18)) | a train consists of 12 boggies , each boggy 15 metres long . the train crosses a telegraph post in 18 seconds . due to some problem , three boggies were detached . the train now crosses a telegraph post in | "length of train = 12 ã — 15 = 180 m . then , speed of train = 180 â „ 18 = 10 m / s now , length of train = 9 ã — 15 = 135 m â ˆ ´ required time = 135 â „ 10 = 13.5 sec . answer e" | a = 12 * 15
b = a - 15
c = 12 * 15
d = c / 18
e = b / d
|
a ) 92.9 , b ) 96.3 , c ) 92.2 , d ) 96.7 , e ) 92.8 | c | add(subtract(100, 8), divide(1, 5)) | what is the cp of rs 100 stock at 8 discount , with 1 / 5 % brokerage ? | "explanation : use the formula , cp = 100 â € “ discount + brokerage % cp = 100 - 8 + 1 / 5 92.2 thus the cp is rs 92.2 . answer : c" | a = 100 - 8
b = 1 / 5
c = a + b
|
a ) 58 kg , b ) 60 kg , c ) 64 kg , d ) 70 kg , e ) none | b | add(multiply(divide(11, 9), 27), 27) | zinc and copper are melted together in the ratio 9 : 11 . what is the weight of melted mixture , if 27 kg of zinc has been consumed in it ? | "sol . for 9 kg zinc , mixture melted = ( 9 + 11 ) kg . for 27 kg zinc , mixture , melted = [ 20 / 9 x 27 ] kg = 60 kg . answer b" | a = 11 / 9
b = a * 27
c = b + 27
|
a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | c | subtract(subtract(9, 1), 1) | if x is an integer such that 3 < x < 10 , 5 < x < 18 , 9 > x > – 2 , 8 > x > 0 , and x + 1 < 9 , then x is | "3 < x < 10 , 6 < x < 18 , - 2 < x < 9 0 < x < 8 x < 8 from above : 6 < x < 8 - - > x = 7 . answer : c ." | a = 9 - 1
b = a - 1
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a ) 2 , b ) 11 , c ) 13 , d ) 17 , e ) 211 | e | divide(add(multiply(multiply(multiply(multiply(11, add(const_4, const_3)), add(const_4, const_1)), const_3), const_2), 11), 11) | if n is a positive integer greater than 1 , then p ( n ) represents the product of all the prime numbers less than or equal to n . the second smallest prime factor of p ( 12 ) + 11 is | p ( 12 ) + 11 = 2 ∗ 3 ∗ 5 ∗ 7 ∗ 11 + 11 = 11 ( 2 ∗ 3 ∗ 5 ∗ 7 + 1 ) = 11 ∗ 211 . both 11 and 211 are primes : 11 is the smallest prime of p ( 12 ) + 11 and 211 is the second smallest prime of p ( 12 ) + 11 . answer : e . | a = 4 + 3
b = 11 * a
c = 4 + 1
d = b * c
e = d * 3
f = e * 2
g = f + 11
h = g / 11
|
a ) a ) 35 , b ) b ) 38 , c ) c ) 90 , d ) d ) 88 , e ) e ) 40 | e | divide(add(140, 60), add(4, 1)) | in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 140 marks , the no of questions he attempts correctly is : | "let the number of correct answers be x . number of incorrect answers = ( 60 – x ) . 4 x – ( 60 – x ) = 140 = > 5 x = 200 = > x = 40 answer : e" | a = 140 + 60
b = 4 + 1
c = a / b
|
a ) 1 , b ) 19 , c ) 29 , d ) 13 , e ) 33 | d | subtract(subtract(subtract(multiply(70, 4), 90), subtract(90, const_1)), subtract(90, const_2)) | the average ( arithmetic mean ) of 4 different integers is 70 . if the largest integer is 90 , what is the least possible value of the smallest integer ? | "total of integers = 70 * 4 = 280 lowest of the least possible integer is when the middle 2 intergers are at the maximum or equal to the highest possible integer . but all integers are distinct . so if the largest integer is 90 , then the middle 2 will be 88 and 89 lowest of least possible integer = 280 - ( 90 + 89 + 88 ) = 280 - 267 = 13 answer : d" | a = 70 * 4
b = a - 90
c = 90 - 1
d = b - c
e = 90 - 2
f = d - e
|
a ) 0.03 , b ) 0.0005 , c ) 0.25 , d ) 0.005 , e ) none of these | a | multiply(divide(divide(6, const_100), 6), const_2) | double of quarter of 6 percent written as a decimal is : | "explanation : solution : ( 2 ) * ( 1 / 4 ) * 6 % = 0.03 answer : a" | a = 6 / 100
b = a / 6
c = b * 2
|
a ) 5 days , b ) 10 days , c ) 14 days , d ) 22 days , e ) 26 days | b | add(4, divide(subtract(const_1, divide(4, 20)), add(inverse(20), inverse(12)))) | p and q can complete a work in 20 days and 12 days respectively . p alone started the work and q joined him after 4 days till the completion of the work . how long did the work last ? | "explanation : work done by p in 1 day = 1 / 20 work done by q in 1 day = 1 / 12 work done by p in 4 days = 4 × ( 1 / 20 ) = 1 / 5 remaining work = 1 – 1 / 5 = 4 / 5 work done by p and q in 1 day = 1 / 20 + 1 / 12 = 8 / 60 = 2 / 15 number of days p and q take to complete the remaining work = ( 4 / 5 ) / ( 2 / 15 ) = 6 total days = 4 + 6 = 10 answer : option b" | a = 4 / 20
b = 1 - a
c = 1/(20)
d = 1/(12)
e = c + d
f = b / e
g = 4 + f
|
a ) 18 kmph , b ) 15.5 kmph , c ) 12.5 kmph , d ) 13.5 kmph , e ) 22.5 kmph | c | divide(add(16, 9), const_2) | if in one hour , a canoe rows at 16 km / hr downstream and 9 km / hr upstream , calculate the speed of a boat in still water ( in km / hr ) . | speed in still water = ( 16 + 9 ) 1 / 2 kmph = 12.5 kmph . answer : c | a = 16 + 9
b = a / 2
|
a ) rs . 259 , b ) rs . 252 , c ) rs . 258 , d ) rs . 251 , e ) rs . 252 | c | divide(multiply(multiply(subtract(add(55, 35), 4), 4), 75), const_100) | a rectangular lawn 55 m by 35 m has two roads each 4 m wide running in the middle of it . one parallel to the length and the other parallel to breadth . the cost of graveling the roads at 75 paise per sq meter is ? | "area of cross roads = 55 * 4 + 35 * 4 - 4 * 4 = 344 sq m cost of graveling = 344 * ( 75 / 100 ) = rs . 258 answer : c" | a = 55 + 35
b = a - 4
c = b * 4
d = c * 75
e = d / 100
|
a ) 32500 , b ) 2500 , c ) 22500 , d ) 30000 , e ) 40000 | a | divide(multiply(add(31100, const_100), divide(const_100, const_4)), add(const_12, const_12)) | a salesman ` s commission is 5 % on all sales upto rs . 10000 and 4 % on all sales exceeding this . he remits rs . 31100 to his parent company after deducing his commission . find the total sales . | let his total sales be rs . x . now ( total sales ) – ( commission ) = rs . 31100 x - [ ( 5 % of 10000 + 4 % of ( x - 10000 ) ] = 31100 x - [ ( ( 5 / 100 ) * 10000 + ( 4 / 100 ) * ( x - 10000 ) ] = 31100 x - 500 - ( ( x - 10000 ) / 25 ) = 31100 x - ( x / 25 ) = 31200 24 x / 25 = 31200 x = [ ( 31200 * 25 ) / 24 ) = 32500 . total sales = rs . 32500 answer a | a = 31100 + 100
b = 100 / 4
c = a * b
d = 12 + 12
e = c / d
|
a ) 669 , b ) 660 , c ) 360 , d ) 720 , e ) 1,440 | a | divide(divide(1, 2), divide(multiply(add(divide(add(divide(add(divide(add(divide(multiply(const_3, const_2), const_10), const_1), const_10), const_4), const_10), const_1), const_10), const_3), 15), multiply(add(add(multiply(add(multiply(add(multiply(multiply(const_3, const_2), const_10), const_3), const_10), const_3), const_10), const_3), const_2), const_10))) | approximately how many revolutions will be made by a car tire with a 15 - inch diameter if the car travels 1 ⁄ 2 mile ? | diameter of car tire = 15 inches radius of car tire = 7.5 inches = 7.5 * 2.54 cm = 19.05 cm circumference of the car tire = 2 * pi * 19.05 = 2 * 3.14 * 19.05 = 119.63 cm distance traveled by car = . 5 mile = . 5 * 1.6 km = . 8 km = 800 m = 8 * 10 ^ 4 cm number of revolutions = distance traveled / circumference of tire = 8 * 10 ^ 4 / 119.63 = 668.73 since the answer choices are far apart we should use estimation in this question . ideally , in the gmat the conversion between units would be provided . answer a | a = 1 / 2
b = 3 * 2
c = b / 10
d = c + 1
e = d / 10
f = e + 4
g = f / 10
h = g + 1
i = h / 10
j = i + 3
k = j * 15
l = 3 * 2
m = l * 10
n = m + 3
o = n * 10
p = o + 3
q = p * 10
r = q + 3
s = r + 2
t = s * 10
u = k / t
v = a / u
|
a ) 190 , b ) 200 , c ) 210 , d ) 220 , e ) 240 | e | multiply(16, subtract(16, const_1)) | 16 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | 2 * 16 c 2 = 2 * 120 = 240 the answer is e . | a = 16 - 1
b = 16 * a
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a ) 44 , b ) 41 , c ) 38 , d ) 35 , e ) 32 | e | add(multiply(divide(33, const_3), const_2), divide(30, const_3)) | the grade point average of one third of the classroom is 30 ; the grade point average of the rest is 33 . what is the grade point average of the whole class ? | "let n = total students in class total points for 1 / 3 class = 30 n / 3 = 10 n total points for 2 / 3 class = 33 * 2 n / 3 = 22 n total points for whole class = 10 n + 22 n = 32 n 32 n total class points / n total students = 32 grade point average for total class answer : e" | a = 33 / 3
b = a * 2
c = 30 / 3
d = b + c
|
['a ) 21 .', 'b ) 28 .', 'c ) 5 .', 'd ) 49 .', 'e ) 52 .'] | c | add(const_3, const_2) | in a rectangular axis system , what is the area of a parallelogram with the coordinates : ( 4,4 ) , ( 7,4 ) , ( 5,9 ) , ( 8,9 ) ? | delta x will give us the dimension of one side of the parallelogram = 5 - 4 = 1 unit delta y will give us the dimension of the other side of parallelogram = 9 - 4 = 5 unit area of parallelogram = 1 * 5 = 5 answer is c | a = 3 + 2
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a ) 54 , b ) 66 , c ) 80 , d ) 36 , e ) 96 | a | multiply(18, divide(divide(36, 2), 6)) | two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 18 mph between the 2 friends until the friends meet . how many miles does the pigeon travel ? | "a 54 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 18 mph = 54 miles" | a = 36 / 2
b = a / 6
c = 18 * b
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a ) 350 , b ) 370 , c ) 390 , d ) 400 , e ) none | d | add(360, divide(multiply(4, subtract(420, 360)), add(3, 4))) | average expenditure of a person for the first 3 days of a week is rs . 360 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is : | "explanation : assumed mean = rs . 360 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 . 360 + rs . 40 = rs . 400 correct option : d" | a = 420 - 360
b = 4 * a
c = 3 + 4
d = b / c
e = 360 + d
|
a ) 237 , b ) 278 , c ) 278 , d ) 252 , e ) 271 | d | add(add(add(add(add(const_10, const_4), const_4), const_1), divide(add(add(36, const_10), const_12), const_100)), add(63, subtract(multiply(divide(47, const_100), 1442), multiply(divide(36, const_100), 1412)))) | what approximate value will come in place of the question mark ( ? ) in the below question ? ( 47 % of 1442 - 36 % of 1412 ) + 63 = ? | explanation : since there are 5 cups of each kind , prepared with milk or tea leaves added first , are identical hence , total number of different people ways of presenting the cups to the expert is 10 ! / ( 5 ! x 5 ! ) = 252 answer : d ) 252 | a = 10 + 4
b = a + 4
c = b + 1
d = 36 + 10
e = d + 12
f = e / 100
g = c + f
h = 47 / 100
i = h * 1442
j = 36 / 100
k = j * 1412
l = i - k
m = 63 + l
n = g + m
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | a | add(reminder(multiply(reminder(47, const_4), 93), const_10), reminder(35, const_10)) | the units digit of ( 35 ) ^ ( 87 ) + ( 93 ) ^ ( 47 ) is : | "the units digit of powers of 3 , cycles in a group of 4 : { 3 , 9 , 7 , 1 } 47 has the form 4 k + 3 , so the units digit of 93 ^ 47 is 7 . the units digit if powers of 5 is always 5 . 7 + 5 = 12 , so the units digit is 2 . the answer is a ." | a = reminder * (
b = reminder + (
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a ) 200 , b ) 216 , c ) 300 , d ) 400 , e ) 2,500 | c | multiply(multiply(subtract(6, const_4), const_10), const_10) | a “ palindromic integer ” is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 400,000 ? | "first digit = last digit = 4 or 6 or 8 so three possibilities . . second = fifth = any of ten digits third = fourth = any of ten digits . . so total integers = 3 * 10 * 10 = 300 answer : c" | a = 6 - 4
b = a * 10
c = b * 10
|
a ) 200 , b ) 300 , c ) 400 , d ) 500 , e ) 600 | c | add(sqrt(divide(9375, 15)), divide(9375, sqrt(divide(9375, 15)))) | multiplication of 2 numbers is 9375 and the quotient , when the larger one is divided by the smalleris 15 . find the sum of the numbers ? | let the numbers be x and y . then , xy = 9375 and x = 15 . y xy = 9375 ( x / y ) 15 y 2 = 625 . y = 25 . x = 15 y = ( 15 x 25 ) = 375 . sum of the numbers = x + y = 375 + 25 = 400 . c | a = 9375 / 15
b = math.sqrt(a)
c = 9375 / 15
d = math.sqrt(c)
e = 9375 / d
f = b + e
|
a ) 452000 , b ) 562000 , c ) 800000 , d ) 500000 , e ) 652000 | c | multiply(divide(40000, subtract(const_100, add(add(multiply(20, 3), 30), 5))), const_100) | a person distributed 20 % of his income to his 3 children each . he deposited 30 % of his income to his wife ' s account . he donated 5 % of remaining amount to an orphan house . finally he has $ 40000 . find his total income ? | "3 children got = 3 * 20 % = 60 % wife got = 30 % orphan house = 5 % total = 60 + 30 + 5 = 95 % remaining = 100 - 95 = 5 % 5 % = 40000 100 % = 40000 * 100 / 5 = $ 800000 answer is c" | a = 20 * 3
b = a + 30
c = b + 5
d = 100 - c
e = 40000 / d
f = e * 100
|
a ) 7 / 19 , b ) 6 / 19 , c ) 5 / 19 , d ) 4 / 19 , e ) none of these | a | subtract(const_1, divide(choose(subtract(20, 4), const_2), choose(20, const_2))) | a box contains 20 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is | "explanation : please remember that maximum portability is 1 . so we can get total probability of non defective bulbs and subtract it form 1 to get total probability of defective bulbs . so here we go , total cases of non defective bulbs 16 c 2 = 16 ∗ 15 / 2 ∗ 1 = 120 total cases = 20 c 2 = 20 ∗ 19 / 2 ∗ 1 = 190 probability = 120 / 190 = 12 / 19 p of at least one defective = 1 − 12 / 19 = 7 / 19 option a" | a = 20 - 4
b = math.comb(a, 2)
c = math.comb(20, 2)
d = b / c
e = 1 - d
|
a ) 1667 , b ) 6789 , c ) 1300 , d ) 6151 , e ) 1421 | c | divide(988, subtract(const_1, divide(24, const_100))) | after decreasing 24 % in the price of an article costs rs . 988 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 988 cp = 13 * 100 = > cp = 1300 answer : c" | a = 24 / 100
b = 1 - a
c = 988 / b
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a ) s . 738 , b ) s . 638 , c ) s . 650 , d ) s . 762 , e ) s . 735 | d | subtract(825, divide(multiply(subtract(846, 825), 3), 4)) | a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 846 in 4 years . the sum is ? | "s . i . for 1 year = ( 846 - 825 ) = rs . 29 s . i . for 3 years = 21 * 3 = rs . 63 principal = ( 825 - 63 ) = rs . 762 . answer : d" | a = 846 - 825
b = a * 3
c = b / 4
d = 825 - c
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a ) 230 , b ) 245 , c ) 260 , d ) 275 , e ) 290 | e | divide(58, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | of the votes cast on a certain proposal , 58 more were in favor of the proposal than were against it . if the number of votes against the proposal was 40 percent of the total vote , what was the total number of votes cast ? ( each vote cast was either in favor of the proposal or against it . ) | "let x be the total number of votes cast . 0.6 x = 0.4 x + 58 0.2 x = 58 x = 290 the answer is e ." | a = 40 / 100
b = 1 - a
c = 40 / 100
d = b - c
e = 58 / d
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a ) 2197 , b ) 1267 , c ) 1750 , d ) 2267 , e ) 1400 | e | divide(divide(subtract(multiply(4000, power(add(const_1, divide(10, const_100)), 2)), 4000), 2), multiply(3, divide(10, const_100))) | the s . i . on a certain sum of money for 3 years at 10 % per annum is half the c . i . on rs . 4000 for 2 years at 10 % per annum . the sum placed on s . i . is ? | "explanation : c . i . = [ 4000 * ( 1 + 10 / 100 ) 2 - 4000 ] = ( 4000 * 11 / 10 * 11 / 10 - 4000 ) = rs . 840 . sum = ( 420 * 100 ) / ( 3 * 10 ) = rs . 1400 answer : e" | a = 10 / 100
b = 1 + a
c = b ** 2
d = 4000 * c
e = d - 4000
f = e / 2
g = 10 / 100
h = 3 * g
i = f / h
|
a ) 0.45 , b ) 0.5 , c ) 6.45 , d ) 0.25 , e ) 6.0 | a | subtract(6.45, floor(6.45)) | for any number z , z * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.45 – 6.45 * ? | "since z * is defined as the greatest positive even integer less than or equal to z , then 6.45 * = 6 ( the greatest positive even integer less than or equal to 6.45 is 6 ) . hence , 6.45 – 6.45 * = 6.45 - 6 = 0.45 answer : a ." | a = math.floor(6, 45)
b = 6 - 45
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a ) 16 , b ) 18 , c ) 20 , d ) 24 , e ) 30 | d | divide(add(54, 6), add(4, 6)) | maxwell leaves his home and walks toward brad ' s house . one hour later , brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 54 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h . what is the distance traveled by maxwell ? | "after the 1 st hour , maxwell has travelled 4 km . and brad just left home . so the distance between them = 54 - 4 = 50 . relative speed = 4 + 6 = 10 kmph . now to cover these 50 kms , they will take 50 / 10 = 5 hrs . so in 5 hrs , maxwell walks 5 * 4 = 20 kms . so total distance travelled by maxwell = 4 + 20 = 24 . answer : d" | a = 54 + 6
b = 4 + 6
c = a / b
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a ) 31 , b ) 22 , c ) 25 , d ) 27 , e ) 29 | a | add(add(multiply(2, 11), 7), 2) | find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 7 x ( 11 ) ^ 2 | "( 4 ) ^ 11 x ( 7 ) ^ 7 x ( 11 ) ^ 2 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 7 x ( 11 ) ^ 2 = 2 ^ 11 x 2 ^ 11 x 7 ^ 7 x 11 ^ 2 = 2 ^ 22 x 7 ^ 7 x 11 ^ 2 total number of prime factors = ( 22 + 7 + 2 ) = 31 . answer is a ." | a = 2 * 11
b = a + 7
c = b + 2
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(subtract(multiply(48, 10), multiply(45, 10)), subtract(60, 45)) | at a certain fruit stand , the price of each apple is 40 cents and the price of each orange is 60 cents . mary selects a total of 10 apples and oranges from the fruit stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 48 cents . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 45 cents ? | "let number of apples = a number of oranges = b a + b = 10 - - - 1 . 48 = ( . 4 a + . 6 b ) / 10 = > 48 = 4 a + 6 b - - - - 2 solving 1 and 2 , we get a = 6 b = 4 let the number of oranges put back = c 45 * ( 10 - c ) = 40 * 6 + 60 ( 4 - c ) = > c = 2 answer b" | a = 48 * 10
b = 45 * 10
c = a - b
d = 60 - 45
e = c / d
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a ) 400 , b ) 1550 , c ) 1575 , d ) 1600 , e ) 1625 | c | add(rectangle_area(200, 5), rectangle_area(120, 5)) | a rectangular lawn of length 200 m by 120 m has two roads running along its center , one along the length and the other along the width . if the width of the roads is 5 m what is the area q covered by the two roads ? | "area covered by road along the length = 5 * 200 = 1000 square meter area covered by road along the width = 5 * 120 = 600 square meter common area in both roads ( where the roads intersect ) = square with side 5 meter = 5 * 5 = 25 total area of the roads q = 1000 + 600 - 25 = 1575 answer : option c" | a = rectangle_area + (
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a ) 73 % , b ) 56 % , c ) 41 % , d ) 37 % , e ) 29 % | b | multiply(divide(subtract(5, divide(4, 5)), divide(4, 5)), const_100) | by approximately what percent is x greater than 4 / 5 if ( 4 / 5 ) ( x ) = 1 ? | 4 / 5 = 80 % , 5 / 4 = 125 % increase by 45 percentage points 45 / 80 = 56 % , hence b | a = 4 / 5
b = 5 - a
c = 4 / 5
d = b / c
e = d * 100
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a ) 22 , b ) 48.2 , c ) 99 , d ) 288 , e ) 12 | b | divide(divide(subtract(120, multiply(multiply(10, const_0_2778), 10)), 5), const_0_2778) | a train 120 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | speed of the train relative to man = ( 120 / 10 ) m / sec = ( 12 ) m / sec . [ ( 12 ) * ( 18 / 5 ) ] km / hr = 43.2 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 43.2 = = > x = 48.2 km / hr . answer : b | a = 10 * const_0_2778
b = a * 10
c = 120 - b
d = c / 5
e = d / const_0_2778
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a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 43.75 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 15), subtract(const_100, 20)), const_100), subtract(const_100, 20)) | a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 15 % while buying and by 20 % while selling . what is his percentage profit ? | "the owner buys 100 kg but actually gets 115 kg ; the owner sells 100 kg but actually gives 80 kg ; profit : ( 115 - 80 ) / 80 * 100 = ~ 43.75 % answer : d ." | a = 100 + 15
b = 100 - 20
c = a - b
d = c * 100
e = 100 - 20
f = d / e
|
a ) 0.5 metre , b ) 0.3 metre , c ) 0.8 metre , d ) 0.6 metre , e ) 0.7 metre | a | divide(multiply(multiply(25, 20), 4), subtract(multiply(90, 50), multiply(25, 20))) | a field is 90 metre long and 50 metre broad . a 25 metre long , 20 metre broad and 4 metre deep tank dug in the field and the earth taken out is spread evenly over the remaining field . how much the level of field will rise ? | area of field = 90 x 50 = 4500 m 2 area of field dug out = 25 x 20 = 500 m 2 therefore , area of remaining field = 4500 m 2 - 500 m 2 = 4000 m 2 volume of the earth dug out = 25 x 20 x 4 = 2000 m 3 therefore , field will rise by 2000 / 4000 = 0.5 metre answer a | a = 25 * 20
b = a * 4
c = 90 * 50
d = 25 * 20
e = c - d
f = b / e
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | a | add(10, const_4) | how many odd numbers between 10 and 800 are the squares of integers ? | "the square of an odd number is an odd number : 10 < odd < 1,000 10 < odd ^ 2 < 1,000 3 . something < odd < 31 . something ( by taking the square root ) . so , that odd number could be any odd number from 5 to 31 , inclusive : 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 , 29 , and 31 . 12 numbers . answer : a ." | a = 10 + 4
|
a ) 8 % , b ) 47 % , c ) 45 % , d ) 52 % , e ) 56 % | b | multiply(divide(3, 20), const_100) | a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of and then $ 9 million in royalties on the next $ 106 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ? | "( 9 / 106 ) / ( 3 / 20 ) = 30 / 54 = 56,6 % it means that 9 / 106 represents only 56,6 % . therefore a decrease of 47 % . answer b" | a = 3 / 20
b = a * 100
|
a ) 11.73 , b ) 12 , c ) 13.8 , d ) 14 , e ) 15.87 | b | divide(divide(multiply(207, subtract(const_100, 15)), const_100), 15) | the price of lunch for 15 people was $ 207 including a 15 % gratuity for service . what was the average price per person , excluding the gratuity ? | "x - > price without gratuity 207 = x + 0.15 x = > x = 207 / 1.15 = 180 so average price per person , excluding the gratuity = 180 / 15 = 12 the answer is b ." | a = 100 - 15
b = 207 * a
c = b / 100
d = c / 15
|
a ) 15360 , b ) 16010 , c ) 15060 , d ) 14930 , e ) 18540 | e | subtract(add(6000, 15000), add(multiply(6000, divide(11, const_100)), multiply(divide(12, const_100), 15000))) | a soft drink company had 6000 small and 15000 big bottles in storage . if 11 % of small 12 % of big bottles have been sold , then the total bottles remaining in storage is | "6000 + 15000 - ( 0.11 * 6000 + 0.12 * 15000 ) = 18540 . answer : e ." | a = 6000 + 15000
b = 11 / 100
c = 6000 * b
d = 12 / 100
e = d * 15000
f = c + e
g = a - f
|
a ) 4 , b ) 5 , c ) 3 , d ) 9 , e ) 6 | a | divide(add(27, 4), add(6, 6)) | solve the equation for x : 6 x - 27 + 3 x = 4 + 9 - x ? | "9 x + x = 13 + 27 10 x = 40 = > x = 4 answer : a" | a = 27 + 4
b = 6 + 6
c = a / b
|
a ) 647 , b ) 798 , c ) 654 , d ) 847 , e ) 976 | b | subtract(840, divide(multiply(subtract(854, 840), 3), 4)) | a sum of money at simple interest amounts to rs . 840 in 3 years and to rs . 854 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 854 - 840 ) = rs . 14 . s . i . for 3 years = rs . ( 14 x 3 ) = rs . 42 . principal = rs . ( 840 - 42 ) = rs . 798 . answer : b" | a = 854 - 840
b = a * 3
c = b / 4
d = 840 - c
|
a ) 10 : 17 , b ) 2 : 5 , c ) 5 : 16 , d ) 25 : 7 , e ) 2 : 3 | e | divide(const_10, add(add(multiply(8, const_2), const_2), const_1)) | the dimensions of a rectangular solid are 4 inches , 6 inches , and 8 inches . if a cube , a side of which is equal to one of the dimensions of the rectangular solid , is placed entirely within thespherejust large enough to hold the cube , what the ratio of the volume of the cube to the volume within thespherethat is not occupied by the cube ? | answer : e . | a = 8 * 2
b = a + 2
c = b + 1
d = 10 / c
|
a ) 3.625 , b ) 3 , c ) 4.5 , d ) 4 , e ) 3.5 | e | add(divide(subtract(divide(51, 12), 2.75), const_2), 2.75) | annika hikes at a constant rate of 12 minutes per kilometer . she has hiked 2.75 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in 51 minutes . if annika continues east , then turns around and retraces her path to reach the start of the trail in exactly 51 minutes , for how many kilometers total did she hike east ? | "set up two r x t = d cases . 1 . 1 / 12 km / min x t = 2.75 from which t = 33 mins . we know total journey time now is 51 + 33 = 84 . the rate is the same ie 1 / 12 km / min . set up second r x t = d case . 1 / 12 km / min x 84 = 7 km now the total journey would be halved as distance would be same in each direction . 7 / 2 = 3.5 e ." | a = 51 / 12
b = a - 2
c = b / 2
d = c + 2
|
a ) 23 , b ) 51 , c ) 38 , d ) 35 , e ) 37 | d | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 1,260 multiplied by y is the square of an integer , then y must be | "1260 = 2 * 2 * 3 * 3 * 7 * 5 , so we need one 7 and one 5 to make it a square of a number . so 7 * 5 = 35 ans : d" | a = 2 + 3
b = a * 2
|
a ) $ 374 , b ) $ 382 , c ) $ 385 , d ) $ 392 , e ) $ 375 | e | add(divide(348, add(const_1, divide(16, const_100))), multiply(divide(25, const_100), divide(348, add(const_1, divide(16, const_100))))) | if sharon ' s weekly salary increased by 16 percent , she would earn $ 348 per week . if instead , her weekly salary were to increase by 25 percent , how much would she earn per week ? | "( 348 / 116 ) 125 = 375 in this case long division does not take much time . ( 348 / 116 ) = 3 3 * 125 = 375 ( 300 + 75 ) answer e" | a = 16 / 100
b = 1 + a
c = 348 / b
d = 25 / 100
e = 16 / 100
f = 1 + e
g = 348 / f
h = d * g
i = c + h
|
a ) 25 , b ) can not be determined , c ) 10 , d ) 15 , e ) 20 | b | multiply(4500, const_1) | how many books each of volume 250 meter cube can be packed into a crate of volume 4500 meter cube | "gud question with a simple concept . in geo if we want to insert one shape into another we need to know the dimensions of the two shapes . in above with volume given , we can come up with different shapes , so we cant know the answer for ex : 4500 m 3 can be 450 * 10 or 45 * 100 or just 4500 * 1 we do n ' t know , so we cant calculate answer : b" | a = 4500 * 1
|
a ) sec , b ) sec , c ) sec , d ) sec , e ) sec | d | divide(add(90, 170), multiply(60, const_0_2778)) | how long does a train 90 m long traveling at 60 kmph takes to cross a bridge of 170 m in length ? | "d = 90 + 170 = 260 m s = 60 * 5 / 18 = 50 / 3 t = 260 * 3 / 50 = 15.6 sec answer : d" | a = 90 + 170
b = 60 * const_0_2778
c = a / b
|
a ) 4 , b ) 16 , c ) 5 , d ) 14 , e ) 11 | b | multiply(divide(400, 50), 2) | a certain hospital has a policy that for every 50 patients there should be 2 doctors . if the number of patients visiting the hospital every day is 400 , how many doctors are there in the hospital ? | 2 / 50 = x / 400 x = 16 answer is b | a = 400 / 50
b = a * 2
|
a ) rs 10123.20 , b ) rs 12147.84 , c ) rs 10123.40 , d ) rs 10123.50 , e ) none of these | b | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on rs . 30000 after 3 years at the rate of 12 % per annum | "explanation : ( 30000 × ( 1 + 12 / 100 ) 3 ) = > 30000 × 28 / 25 × 28 / 25 × 28 / 25 = > 42147.84 so compound interest will be 42147.84 - 30000 = rs 12147.84 option b" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) 2988 , b ) 2776 , c ) 4400 , d ) 2871 , e ) 3162 | e | floor(divide(2553, multiply(divide(subtract(const_100, 5), const_100), divide(subtract(const_100, 15), const_100)))) | 5 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 2553 , how much was it in the beginning ? | "x * ( 95 / 100 ) * ( 85 / 100 ) = 2553 x = 3162 answer : e" | a = 100 - 5
b = a / 100
c = 100 - 15
d = c / 100
e = b * d
f = 2553 / e
g = math.floor(f)
|
a ) 20 , b ) 25 , c ) 40 , d ) 50 , e ) 35 | c | 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(15, 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(15, const_10))), const_10)) | a rectangular block 6 cm by 12 cm by 15 cm is cut into an exact number of equal cubes . find the least possible number of cubes ? | "volume of the block = 6 * 12 * 15 = 1080 cm ^ 3 side of the largest cube = h . c . f of 6 , 12,15 = 3 cm volume of the cube = 3 * 3 * 3 = 27 cm ^ 3 number of cubes = 1080 / 27 = 40 answer is c" | a = rectangle_area / (
b = a - 10
c = 3 * 2
d = 1000 * c
e = b + d
f = 3 * 1000
g = 15 * 10
h = f + g
i = e / h
|
a ) 49 m 2 , b ) 50 m 2 , c ) 52 m 2 , d ) 62 m 2 , e ) 55 m 2 | d | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 8), multiply(add(divide(25, const_100), 1), 4))), multiply(8, 4)) | a cistern 4 m long and 8 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is : | "area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 8 x 1.25 + 4 x 1.25 ) + 8 x 4 ] m 2 = 62 m 2 . answer : d" | a = 25 / 100
b = a + 1
c = b * 8
d = 25 / 100
e = d + 1
f = e * 4
g = c + f
h = 2 * g
i = 8 * 4
j = h + i
|
a ) 3.5 % , b ) 6.4 % , c ) 3 % , d ) 5 % , e ) 2.6 % | b | subtract(subtract(12, 5), divide(multiply(12, 5), const_100)) | in measuring the sides of a rectangle , one side is taken 12 % in excess and other 5 % in deficit . find the error percentage in the area calculated from these measurements . | "say both sides of the rectangle are equal to 100 ( so consider that we have a square ) . in this case the area is 100 * 100 = 10,000 . now , the area obtained with wrong measurements would be 112 * 95 = 10,640 , which is 6.4 % greater than the actual area . answer : b ." | a = 12 - 5
b = 12 * 5
c = b / 100
d = a - c
|
a ) 1888 , b ) 2999 , c ) 2834 , d ) 2509 , e ) 2991 | d | divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 7720), divide(multiply(15, 7720), const_100)), const_2) | in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 7720 votes were polled , then how many valid votes did b get ? | let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 7720 32.5 k = 32.5 k / 100 k * 7720 = 2509 the number of valid votes polled in favour of b is 2834 . answer : d | a = 20 / 100
b = 1 - a
c = b * 7720
d = 15 * 7720
e = d / 100
f = c - e
g = f / 2
|
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