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 |
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a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | e | multiply(const_2, divide(divide(divide(divide(divide(3150, const_2), add(const_2, const_3)), add(const_2, const_3)), const_3), const_3)) | if g is the smallest positive integer such that 3150 multiplied by g is the square of an integer , then g must be | solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors , every prime factor ' s exponent is an even number . let ’ s start by prime factorizing 3150 . 3150 = 315 x 10 = 5 x 63 x 10 = 5 x 7 x 3 x 3 x 5 x 2 3150 = 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 ( notice that the exponents of both 2 and 7 are not even numbers . this tells us that 3150 itself is not a perfect square . ) we also are given that 3150 multiplied by g is the square of an integer . we can write this as : 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 x g = square of an integer according to our rule , we need all unique prime factors ' exponents to be even numbers . thus , we need one more 2 and one more 7 . therefore , g = 7 x 2 = 14 answer is e . | a = 3150 / 2
b = 2 + 3
c = a / b
d = 2 + 3
e = c / d
f = e / 3
g = f / 3
h = 2 * g
|
a ) a ) 11 , b ) b ) 12 , c ) c ) 13 , d ) d ) 14 , e ) e ) 15 | c | multiply(13, const_1) | the total age of a and b is 13 years more than the total age of b and c . c is how many year younger than a | "explanation : given that a + b = 13 + b + c = > a – c = 13 + b – b = 13 = > c is younger than a by 13 years answer : option c" | a = 13 * 1
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a ) 277 , b ) 36 , c ) 64 , d ) 72 , e ) 30 | e | divide(multiply(108.0, const_100), 360) | ? % of 360 = 108.0 | "? % of 360 = 108.0 or , ? = 108.0 × 100 / 360 = 30 answer e" | a = 108 * 0
b = a / 360
|
a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | c | multiply(1, 7) | two numbers are in the ratio of 1 : 2 . if 7 be added to both , their ratio changes to 3 : 5 . the greater number is | "let the ratio be x : y , given x / y = 1 / 2 , ( x + 7 ) / ( y + 7 ) = 3 / 5 = > x = 14 and y = 28 answer : c" | a = 1 * 7
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a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 9 , 3,32 , d ) 9 , 6,12 , e ) 12 , 8,4 | e | divide(multiply(1, 3), 3) | find the numbers which are in the ratio 3 : 2 : 1 such that the sum of the first and the second added to the difference of the third and the second is 16 ? | "let the numbers be a , b and c . a : b : c = 3 : 2 : 1 given , ( a + b ) + ( c - b ) = 16 = > a + c = 16 = > 3 x + x = 16 = > x = 4 a , b , c are 3 x , 2 x , x a , b , c are 12 , 8 , 4 . answer : e" | a = 1 * 3
b = a / 3
|
a ) 40 , b ) 45 , c ) 49 , d ) 51 , e ) 43 | b | multiply(175, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 175 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 × 7 : 12 × 5 : 15 × 3 = 2 × 7 : 12 × 1 : 3 × 3 = 14 : 12 : 9 amount that c should pay = 175 × ( 9 / 35 ) = 5 × 9 = 45 answer is b ." | a = 15 * 3
b = 10 * 7
c = 12 * 5
d = b + c
e = 15 * 3
f = d + e
g = a / f
h = 175 * g
|
a ) $ 130.00 , b ) $ 145.60 , c ) $ 154.70 , d ) $ 182.00 , e ) $ 210.00 | c | multiply(260, divide(add(30, 15), const_100)) | a discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price . if , during a special sale , an additional 15 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 260 before any discount ? | "since the question is essentially just about multiplication , you can do the various mathstepsin a variety of ways ( depending on whichever method you find easiest ) . we ' re told that the first discount is 10 % to 30 % , inclusive . we ' re told that the next discount is 20 % off of the discounted price . . . . we ' re told to maximize the discount ( thus , 30 % off the original price and then 20 % off of the discounted price ) . thatmathcan be written in a number of different ways ( fractions , decimals , etc . ) : 30 % off = ( 1 - . 3 ) = ( 1 - 30 / 100 ) = ( . 7 ) and the same can be done with the 20 % additional discount . . . the final price of an item that originally cost $ 260 would be . . . . . ( $ 260 ) ( . 7 ) ( . 85 ) = ( $ 260 ) ( . 595 ) = 154.7 final answer : c" | a = 30 + 15
b = a / 100
c = 260 * b
|
a ) 40 , b ) 84 , c ) 90 , d ) 120 , e ) 240 | b | divide(7, subtract(divide(1, 4), divide(0.5, const_3))) | a grocer purchased a quantity of bananas at 3 pounds for $ 0.50 and sold the entire quantity at 4 pounds for $ 1.00 . how many pounds did the grocer purchase if the profit from selling the bananas was $ 7.00 ? | cost price of 1 pound of bananas = 0.5 / 3 = 1 / 6 selling price of 1 pound of bananas = 1 / 4 profit per pound = ( 1 / 4 - 1 / 6 ) = ( 1 / 12 ) total profit is given as 7 ( 1 / 12 ) * x = 7 x = 84 answer : b | a = 1 / 4
b = 0 / 5
c = a - b
d = 7 / c
|
['a ) 9 / 16', 'b ) 8 / 16', 'c ) 7 / 16', 'd ) 6 / 16', 'e ) 5 / 16'] | a | divide(power(3, const_2), subtract(power(5, const_2), power(3, const_2))) | in a triangle abc , point d is on side ab and point e is on side ac , such that bced is a trapezium . de : bc = 3 : 5 . calculate the ratio of the area of triangle ade and the trapezium bced . | area of tri ( ade ) / area of tri ( abc ) = de ^ 2 / bc ^ 2 since tri ( ade ) - tri ( abc ) area of tri ( ade ) / area of tri ( abc ) = 9 / 25 area of tri ( abc ) / area of tri ( ade ) = 25 / 9 area of tri ( abc ) = area of tri ( ade ) + area of trap ( bcde ) ( area of tri ( ade ) + area of trap ( bcde ) ) / area of tri ( ade ) = 25 / 9 1 + area of trap ( bcde ) / area of tri ( ade ) = 25 / 9 take 1 on both sides area of trap ( bcde ) / area of tri ( ade ) = ( 25 / 9 ) - 1 = 16 / 9 since question is asking for area of tri ( ade ) / area of trap ( bcde ) = 9 / 16 answer : a | a = 3 ** 2
b = 5 ** 2
c = 3 ** 2
d = b - c
e = a / d
|
a ) 26799 , b ) 24000 , c ) 26682 , d ) 29973 , e ) 12312 | b | multiply(divide(const_100, 96), 23040) | 96 % of the population of a village is 23040 . the total population of the village is ? | "x * ( 96 / 100 ) = 23040 x = 240 * 100 x = 24000 answer : b" | a = 100 / 96
b = a * 23040
|
a ) 12 , b ) 5 , c ) 10 , d ) 9 , e ) 3 | c | divide(add(87, 4), add(3, 6)) | solve the equation for x : 6 x - 87 + 3 x = 4 + 9 - x | "c 10 9 x + x = 13 + 87 10 x = 100 = > x = 10" | a = 87 + 4
b = 3 + 6
c = a / b
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a ) 100 , b ) 102 , c ) 104 , d ) 106 , e ) 108 | c | multiply(2, divide(117, add(2, 16))) | water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 117 grams of water ? | "( 16 / 18 ) * 117 = 104 grams the answer is c ." | a = 2 + 16
b = 117 / a
c = 2 * b
|
a ) 3600 , b ) 3890 , c ) 3300 , d ) 2789 , e ) 2891 | c | subtract(multiply(add(1200, 100), add(20, const_1)), multiply(1200, 20)) | the average monthly salary of 20 employees in an organisation is rs . 1200 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 1300 * 21 - 1200 * 20 ) = rs . 3300 . answer : c" | a = 1200 + 100
b = 20 + 1
c = a * b
d = 1200 * 20
e = c - d
|
a ) 27 , b ) 48 , c ) 45 , d ) 72 , e ) 18 | b | subtract(multiply(multiply(5, 5), divide(141, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(141, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5))))) | the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 141 , what is the difference between the ages of patrick and monica ? | "ages of p and mi = 3 x + 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 47 x = 141 x = 3 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 3 = 48 answer b" | a = 5 * 5
b = 3 * 3
c = 3 * 5
d = b + c
e = 5 * 5
f = d + e
g = 141 / f
h = a * g
i = 3 * 3
j = 3 * 3
k = 3 * 5
l = j + k
m = 5 * 5
n = l + m
o = 141 / n
p = i * o
q = h - p
|
a ) 31 , b ) 32 , c ) 35 , d ) 30 , e ) 29 | e | divide(add(408, 288), multiply(multiply(multiply(const_2, const_2), const_2), const_3)) | there are 408 boys and 288 girls in a school which are to be divided into equal sections of either boys or girls alone . find the total number of sections thus formed . | explanation : hcf ( 408 , 288 ) = 24 the number of boys or girls that can be placed in a section = 24 . thus the total number of sections is given by 408 / 24 + 288 / 24 = 17 + 12 = 29 answer : e | a = 408 + 288
b = 2 * 2
c = b * 2
d = c * 3
e = a / d
|
a ) 0 , b ) 3 , c ) 4 , d ) 8 , e ) 6 | d | add(add(const_4, const_3), const_2) | what is the units digit of the expression 14 ^ 7 − 17 ^ 4 ? | i think answer on this one should be d too . since we know that 14 ^ 7 > 17 ^ 4 , as will said one should always check if the number is positive . | a = 4 + 3
b = a + 2
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a ) 0 and 13 , b ) 0 and 14 , c ) 1 and 10 , d ) 1 and 9 , e ) 2 and 8 | b | divide(subtract(multiply(42, 2), 55), 2) | a certain experimental mathematics program was tried out in 2 classes in each of 42 elementary schools and involved 55 teachers . each of the classes had 1 teacher and each of the teachers taught at least 1 , but not more than 3 , of the classes . if the number of teachers who taught 3 classes is n , then the least and greatest possible values of n , respectively , are | "one may notice that greatest possible values differ in each answer choice in contrast to the least values , which repeat . to find out the greatest value you should count the total classes ( 42 * 2 = 84 ) , then subtract the total # of teachers since we know from the question that each teacher taught at least one class ( 84 - 55 = 29 ) . thus we get a number of the available extra - classes for teachers , and all that we need is just to count how many teachers could take 2 more classes , which is 29 / 2 = 14.5 . so the greatest possible value of the # of teachers who had 3 classes is 14 . only answer b has this option ." | a = 42 * 2
b = a - 55
c = b / 2
|
a ) 0 and 1 , b ) 1 and 2 , c ) 2 and 3 , d ) 3 and 4 , e ) 4 and 5 | d | add(multiply(multiply(4, 4), const_2), const_2) | if a ^ 4 + b ^ 4 = 100 , then the greatest possible value of b is between | for the greatest possible value of b ^ 4 , we must minimize the value of b ^ 4 i . e . lets say a ^ 4 = 0 then we need to find a number b such that b ^ 4 < 100 . 3 ^ 4 = 81 and 4 ^ 4 = 256 so we can say that the maximum possible value of b can be a little more than 3 hence answer = between 3 and 4 hence d | a = 4 * 4
b = a * 2
c = b + 2
|
a ) 33 , b ) 20 , c ) 88 , d ) 66 , e ) 21 | b | divide(const_100, 5) | in how many years will a sum of money doubles itself at 5 % per annum on simple interest ? | "p = ( p * 5 * r ) / 100 r = 20 % answer : b" | a = 100 / 5
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a ) 400 , b ) 300 , c ) 200 , d ) 100 , e ) 40 | c | sqrt(divide(20, add(power(2, 5), add(power(2, 5), power(5, 5))))) | the ratio of 2 numbers is 5 : 2 and their h . c . f . is 20 . their l . c . m . is ? | "let the numbers be 5 x and 2 x their h . c . f . = 20 so the numbers are 5 * 20 , 2 * 20 = 100 , 40 l . c . m . = 200 answer is c" | a = 2 ** 5
b = 2 ** 5
c = 5 ** 5
d = b + c
e = a + d
f = 20 / e
g = math.sqrt(f)
|
a ) 80 cm , b ) 50 cm , c ) 40 cm , d ) 10 cm , e ) 100 cm | a | multiply(20, const_4) | what is the greatest possible length which can be used to measure exactly the lengths 180 m , 500 m and 15 m 20 cm ? | required length = hcf of 18000 cm , 50000 cm , 1520 cm = 80 cm answer is a | a = 20 * 4
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a ) 9 : 10 , b ) 19 : 20 , c ) 29 : 30 , d ) 39 : 40 , e ) 49 : 50 | e | divide(rectangle_area(3, 2), rectangle_area(divide(divide(rectangle_perimeter(3, 2), 2), add(4, 3)), multiply(divide(divide(rectangle_perimeter(3, 2), 2), add(4, 3)), 4))) | an order was placed for a carpet whose length and width were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and width were in the ratio 4 : 3 but were was no change in its perimeter . what is the ratio of the areas of the carpets ? | "let the length and width of one carpet be 3 x and 2 x . let the length and width of the other carpet be 4 y and 3 y . 2 ( 3 x + 2 x ) = 2 ( 4 y + 3 y ) 5 x = 7 y ( 5 / 7 ) * x = y the ratio of the areas of the carpet in both cases : = 3 x * 2 x : 4 y * 3 y = 6 x ^ 2 : 12 y ^ 2 = x ^ 2 : 2 y ^ 2 = x ^ 2 : 2 * ( 25 / 49 ) * x ^ 2 = 49 : 50 the answer is e ." | a = rectangle_area / (
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['a ) 2 : 1', 'b ) 1 : 2', 'c ) 2 : 3', 'd ) 1 : 4', 'e ) 4 : 1'] | e | divide(multiply(inverse(const_2), const_4), inverse(const_2)) | the ratio of the areas of two squares , one having double its diagonal then the other is : | lenth of the diagonals be 2 x and x units . areas are 1 / 2 × ( 2 x ) power 2 and ( 1 / 2 × ( xpower 2 ) ) required ratio = 1 / 2 × 4 x 2 : 1 / 2 x 2 = 4 : 1 answer is e . | a = 1/(2)
b = a * 4
c = 1/(2)
d = b / c
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(subtract(subtract(15, 7), 3), 2) | a group of people participate in some curriculum , 25 of them practice yoga , 15 study cooking , 8 study weaving , 2 of them study cooking only , 7 of them study both the cooking and yoga , 3 of them participate all curriculums . how many people study both cooking and weaving ? | both cooking and weaving = 15 - ( 2 + 3 + 7 ) = 3 so , the correct answer is c . | a = 15 - 7
b = a - 3
c = b - 2
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a ) 71.11 , b ) 71.12 , c ) 71.1 , d ) 74.66 , e ) 71.13 | d | multiply(320, divide(const_1, add(divide(160, 70), divide(160, 80)))) | a car travels first 160 km at 70 km / hr and the next 160 km at 80 km / hr . what is the average speed for the first 320 km of the tour ? | "car travels first 160 km at 70 km / hr time taken to travel first 160 km = distancespeed = 160 / 70 car travels next 160 km at 80 km / hr time taken to travel next 160 km = distancespeed = 160 / 80 total distance traveled = 160 + 160 = 2 × 160 total time taken = 160 / 70 + 160 / 80 average speed = total distance traveled / total time taken = 320 / ( 160 / 70 + 160 / 80 ) = 74.66 km / hr answer : d" | a = 160 / 70
b = 160 / 80
c = a + b
d = 1 / c
e = 320 * d
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a ) 330000 , b ) 340000 , c ) 347000 , d ) 356000 , e ) 357000 | e | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(75, const_100)) | in an election , candidate a got 75 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate . | "total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 75 % therefore , the number of valid votes polled in favour of candidate a = 75 % of 476000 = 75 / 100 × 476000 = 35700000 / 100 = 357000 e )" | a = 15 / 100
b = 1 - a
c = 560000 * b
d = 75 / 100
e = c * d
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a ) 20 , b ) 25 , c ) 55 , d ) 75 , e ) 80 | d | subtract(multiply(90, 4), multiply(95, const_3)) | joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 90 . he was allowed to drop his lowest score . after doing so , his average test score improved to 95 . what is the lowest test score that was dropped ? | "the arithmetic mean of 4 equally weighted tests was 90 . so what we can assume is that we have 4 test scores , each 90 . he dropped his lowest score and the avg went to 95 . this means that the lowest score was not 90 and other three scores had given the lowest score 5 each to make it up to 90 too . when the lowest score was removed , the other 3 scores got their 5 back . so the lowest score was 3 * 5 = 15 less than 90 . so the lowest score = 90 - 15 = 75 answer ( d )" | a = 90 * 4
b = 95 * 3
c = a - b
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a ) 55 , b ) 45 , c ) 65 , d ) 78 , e ) 64 | a | add(40, 15) | john found that the average of 15 numbers is 40 . if 15 is added to each number then the mean of number is ? | "( x + x 1 + . . . x 14 ) / 15 = 40 55 option a" | a = 40 + 15
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a ) 429 : 94 , b ) 201 : 21 , c ) 138 : 38 , d ) 93 : 39 , e ) 95 : 39 | a | multiply(divide(357, 73), const_100) | 357 : 73 : : ? | "357 : 73 to get 2 nd number , middle digit is removed , 1 st & last interchanged their position of 1 st no . only option a ) match this . answer : a" | a = 357 / 73
b = a * 100
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a ) 68 , b ) 72 , c ) 78 , d ) 88 , e ) 312 | a | subtract(multiply(subtract(78, 2), add(4, const_1)), multiply(4, 78)) | scott ' s average ( arithmetic mean ) golf score on his first 4 rounds was 78 . what score does he need on his fifth round to drop his average score by 2 points ? | average score for first 4 rounds = 78 average score after 5 rounds = 76 score scott needs on his 5 th round to drop average score by 2 points = 78 - ( 2 * 5 ) = 68 answer a alternatively , sum of scores of first 4 rounds = 78 * 4 = 312 sum of scores after 5 rounds = 76 * 5 = 380 score scott needs on his 5 th round to drop average score by 2 points = 380 - 312 = 68 answer a | a = 78 - 2
b = 4 + 1
c = a * b
d = 4 * 78
e = c - d
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a ) 61 , b ) 78 , c ) 71 , d ) 68 , e ) 94 | c | add(subtract(107, multiply(15, 4)), multiply(6, 4)) | in 15 year ’ s time the combined age of my 4 brothers will be 107 . what will it be in the 6 year ’ s time ? | c 71 combined age in 15 years = 107 . 4 × 15 = 60 , therefore combined age now is 107 – 60 = 47 . in 6 years time , therefore , combined age will be 47 + 24 ( 4 x 6 ) = 71 . | a = 15 * 4
b = 107 - a
c = 6 * 4
d = b + c
|
a ) 70 , b ) 90 , c ) 120 , d ) 100 , e ) 140 | d | subtract(add(250, 450), 600) | a , b and c have rs . 600 between them , a and c together have rs . 250 and b and c rs . 450 . how much does c have ? | "a + b + c = 600 a + c = 250 b + c = 450 - - - - - - - - - - - - - - a + b + 2 c = 700 a + b + c = 600 - - - - - - - - - - - - - - - - c = 100 answer : d" | a = 250 + 450
b = a - 600
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a ) 76.5 , b ) 80.9 , c ) 81.0 , d ) 81.1 , e ) 81.9 | a | multiply(multiply(divide(subtract(const_100, 15), const_100), divide(subtract(const_100, 10), const_100)), const_100) | a store reduced the price of all items in the store by 15 % on the first day and by another 10 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ? | "consider price of the all items as $ 100 after a initial reduction of 15 % price becomes = 0.85 * 100 = $ 85 after the final reduction of 10 % price becomes = 0.9 * 85 = $ 76.5 price of all items on second day is 76.5 % of price on first day correct answer option a" | a = 100 - 15
b = a / 100
c = 100 - 10
d = c / 100
e = b * d
f = e * 100
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['a ) 189 cm 2', 'b ) 300 cm 2', 'c ) 127 cm 2', 'd ) 177 cm 2', 'e ) 187 cm 2'] | b | multiply(multiply(divide(const_1, const_2), add(6, 9)), 40) | find the area of the quadrilateral of one of its diagonals is 40 cm and its off sets 9 cm and 6 cm ? | 1 / 2 * 40 ( 9 + 6 ) = 300 cm 2 answer : b | a = 1 / 2
b = 6 + 9
c = a * b
d = c * 40
|
a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 3 | b | subtract(15, add(multiply(const_4, 2), multiply(const_4, 1.5))) | an equal number of desks and bookcases are to be placed along a library wall that is 15 meters long . each desk is 2 meters long , and each bookshelf is 1.5 meters long . if the maximum possible number of desks and bookcases are to be placed along the wall , then the space along the wall that is left over will be how many meters k long ? | "let x be the number of desks and bookcases that are placed along the library wall . 2 x + 1.5 x < 15 3.5 x < 15 since x is a non negative integer , the largest number x can be is 4 . when x is 4 , the desks and bookcases take up 3.5 * 4 = 14 m = k , leaving 1 m of empty space . thus , i believe the answer is b ) 1" | a = 4 * 2
b = 4 * 1
c = a + b
d = 15 - c
|
a ) a ) 6.7 , b ) b ) 1.3 , c ) c ) 9.6 , d ) d ) 20 , e ) e ) 7.9 | d | divide(subtract(25, 10), subtract(const_1, divide(25, 100))) | how many kg of pure salt must be added to 100 kg of 10 % solution of salt and water to increase it to a 25 % solution ? | "amount salt in 100 kg solution = 10 * 100 / 100 = 10 kg let x kg of pure salt be added then ( 10 + x ) / ( 100 + x ) = 25 / 100 40 + 4 x = 100 + x 3 x = 60 x = 20 answer is d" | a = 25 - 10
b = 25 / 100
c = 1 - b
d = a / c
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a ) 80 , b ) 90 , c ) 85 , d ) 84 , e ) 89 | d | multiply(divide(140, add(const_10, const_10)), const_12) | my grandson is about as many days as my son in weeks , and my grandson is as many months as i am in years . my grandson , my son and i together are 140 years . can you tell me my age in years ? | let the age of the man = x years so according to the given condition age of my grandson = x / 12 years and age of my son = x 30 / 52 = 15 x / 26 and x + ( x / 12 ) + ( 15 x / 26 ) = 140 or ( 156 x + 13 x + 90 x ) / 156 = 140 or 259 x = 156 * 140 or x = 156 * 140 / 259 = 84 answer : d | a = 10 + 10
b = 140 / a
c = b * 12
|
a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 120,000 | e | divide(multiply(multiply(add(const_2, const_3), const_1000), 12), const_2) | if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 15,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years later , when luke is ready for college ? | answer equals e in 48 years . i thought by 50 th year it would reach 120,000 . options should have been separated more widely for clarity . | a = 2 + 3
b = a * 1000
c = b * 12
d = c / 2
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a ) 152 , b ) 209 , c ) 57 , d ) 171 , e ) 181 | a | multiply(multiply(8, subtract(11, 8)), 11) | there are 361 doctors and nurses in a hospital . if the ratio of the doctors to the nurses is 8 : 11 , then how many nurses are there in the hospital ? | "given , the ratio of the doctors to the nurses is 8 : 11 number of nurses = 11 / 19 x 361 = 209 answer : a" | a = 11 - 8
b = 8 * a
c = b * 11
|
a ) 33.33 , b ) 34.01 , c ) 26.32 , d ) 28.33 , e ) 19.21 | d | subtract(20, divide(subtract(780, multiply(20, 10)), subtract(22, 10))) | 20 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780 . the bigger part is : | "explanation : let the two parts be ( 20 - x ) and x . then , 10 ( 20 - x ) + 22 x = 780 = > 12 x = 580 = > x = 48.33 . bigger part = ( 20 - x ) = 28.33 . answer : d ) 28.3" | a = 20 * 10
b = 780 - a
c = 22 - 10
d = b / c
e = 20 - d
|
a ) 1276.28 , b ) 1276.22 , c ) 1276.29 , d ) 1276.21 , e ) 1276.23 | a | multiply(power(add(divide(5, const_100), const_1), 5), 1000) | a chartered bank offers a 5 - year escalator guaranteed investment certificate . in successive years it pays annual interest rates of 4 % , 4.5 % , 5 % , 5.5 % , and 6 % , respectively , compounded at the end of each year . the bank also offers regular 5 - year gics paying a fixed rate of 5 % compounded annually . calculate and compare the maturity values of $ 1000 invested in each type of gic . ( note that 5 % is the average of the 5 successive one - year rates paid on the escalator gic . ) | explanation : fv = $ 1000 ( 1.04 ) ( 1.045 ) ( 1.05 ) ( 1.055 ) ( 1.06 ) = $ 1276.14 the maturity value of the regular gic is fv = $ = $ 1276.28 answer : a ) 1276.28 | a = 5 / 100
b = a + 1
c = b ** 5
d = c * 1000
|
a ) 26 sec , b ) 65 sec , c ) 55 sec , d ) 19 sec , e ) 72 sec | e | divide(1200, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 1200 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 1200 * 3 / 50 = 72 sec . answer : e" | a = 63 - 3
b = a * const_0_2778
c = 1200 / b
|
a ) 3 / 4 , b ) 8 / 9 , c ) 18 / 19 , d ) 23 / 24 , e ) it can not be determined from the information given | e | divide(add(3, 4), add(4, 4)) | the ratio of two quantities is 3 to 4 . if each of the quantities is increased 4 and 3 respectively , what is the ratio of these two new quantities ? | "3 / 4 = 3 x / 4 x we need to find out ( 3 x + 4 ) / ( 4 x + 3 ) off course we can not solve this to arrive at any rational number hence e ." | a = 3 + 4
b = 4 + 4
c = a / b
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a ) 399 , b ) 272 , c ) 865.8 , d ) 277 , e ) 311 | c | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 1.30), 3) | the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 1.30 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "answer : option c explanation : a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 â € “ 6 = 666 * 1.3 = 865.8 answer : c" | a = math.sqrt(3136)
b = a * 4
c = 2 * 1
d = b - c
e = d * 1
f = e * 3
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a ) 8 days , b ) 6 days , c ) 7 days , d ) 9 days , e ) 10 days | e | inverse(add(divide(5, multiply(25, 7)), divide(25, multiply(25, 14)))) | 25 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ? | "1 women ' s 1 day work = 1 / 175 1 child ' s 1 day work = 1 / 140 ( 5 women + 10 children ) ' s 1 day work = ( 5 / 175 + 10 / 140 ) = 1 / 10 5 women and 10 children will complete the work in 10 days . answer : e" | a = 25 * 7
b = 5 / a
c = 25 * 14
d = 25 / c
e = b + d
f = 1/(e)
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a ) 300 kmph , b ) 360 kmph , c ) 600 kmph , d ) 720 kmph , e ) none | d | divide(divide(multiply(240, 5), add(const_1, divide(const_2, const_3))), const_2) | an aeroplane covers a certain distance at a speed of 240 kmph in 5 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of : | "explanation : distance = ( 240 x 5 ) = 1200 km . speed = distance / time speed = 1200 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = 1200 x 3 / 5 km / hr = 720 km / hr . answer : option d" | a = 240 * 5
b = 2 / 3
c = 1 + b
d = a / c
e = d / 2
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a ) 43.0 , b ) 45.5 , c ) 46.0 , d ) 49.0 , e ) 50.0 | a | divide(divide(divide(multiply(const_2, 43000), const_2), const_100), const_10) | in arun ' s company 60 % of the employees earn less than $ 50000 a year , 60 % of the employees earn more than $ 40000 a year , 11 % of the employees earn $ 43000 a year and 5 % of the employees earn $ 49000 a year . what is the median salary for the company ? | in arun ' s company 60 % of the employees earn less than $ 50,000 a year , 50 and 51 employee will be each 43 k . hence median = ( 43 k + 43 k ) / 2 | a = 2 * 43000
b = a / 2
c = b / 100
d = c / 10
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a ) 120 cm , b ) 767 cm , c ) 88 cm , d ) 666 cm , e ) 776 cm | a | multiply(sqrt(divide(72, 50)), const_100) | 50 square stone slabs of equal size were needed to cover a floor area of 72 sq . m . find the length of each stone slab ? | "area of each slab = 72 / 50 m 2 = 1.44 m 2 length of each slab √ 1.44 = 1.2 m = 120 cm" | a = 72 / 50
b = math.sqrt(a)
c = b * 100
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a ) 50 % , b ) 36 % , c ) 30 % , d ) 32 % , e ) 29 % | b | multiply(divide(subtract(add(const_1, divide(90, const_100)), add(const_1, divide(40, const_100))), add(const_1, divide(40, const_100))), const_100) | a certain company reported that the revenue on sales increased 40 % from 2000 to 2003 , and increased 90 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ? | "assume the revenue in 2000 to be 100 . then in 2003 it would be 140 and and in 2005 190 , so from 2003 to 2005 it increased by ( 190 - 140 ) / 140 = 50 / 140 = ~ 29 % . answer : b" | a = 90 / 100
b = 1 + a
c = 40 / 100
d = 1 + c
e = b - d
f = 40 / 100
g = 1 + f
h = e / g
i = h * 100
|
a ) 5 sec , b ) 9 sec , c ) 12 sec , d ) 18 sec , e ) 19 sec | d | divide(300, multiply(subtract(68, 8), const_0_2778)) | a train 300 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 300 m at 50 / 3 m / sec = 300 * 3 / 50 sec = 18 sec answer : d ." | a = 68 - 8
b = a * const_0_2778
c = 300 / b
|
a ) 50.0 % , b ) 40.0 % , c ) 30.0 % , d ) 20.0 % , e ) 10.0 % | b | multiply(divide(subtract(divide(50, 60), divide(20, 40)), divide(50, 60)), const_100) | at the store opening , larry ' s grocery had 50 lemons and 60 oranges . by closing , the store at 20 lemons and 40 oranges left . by approximately what percent did the ratio of lemons to orange decrease from opening to closing . | opening : lemons / oranges = 50 / 60 = 100 / 120 closing : lemons / oranges = 20 / 40 = 60 / 120 aside : it ' s useful to write both ratios with the same denominator . this allows us to ignore the denominator and focus solely on the numerators . so , our ratio went from 100 / 120 to 60 / 120 ignoring the denominators , we went from 100 to 60 the percent change = 100 ( difference in values ) / ( original value ) = ( 100 ) ( 100 - 60 ) / 100 = ( 100 ) ( 40 ) / 100 ) = 40 answer : b | a = 50 / 60
b = 20 / 40
c = a - b
d = 50 / 60
e = c / d
f = e * 100
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | divide(5, 11) | what is the 27 th digit to the right of the decimal point in the decimal form of 5 / 11 ? | "5 / 11 = 0.45454545 . . . the odd numbered positions in the decimal expansion are all 4 . the answer is b ." | a = 5 / 11
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a ) 4 / 99 , b ) 2 / 25 , c ) 8 / 99 , d ) 49 / 100 , e ) 86 / 99 | e | divide(subtract(add(multiply(divide(const_100, 4), const_2), multiply(divide(const_100, 5), const_2)), 4), subtract(const_100, 1)) | if x is a positive integer with fewer than 3 digits , what is the probability that z x * ( x + 1 ) is a multiple of either 4 or 5 ? | "interesting question ! also one that we should be able to answer very quickly be keeping an eye on our best friends , the answer choices . we know that x belongs to the set { 1 , 2 , 3 , . . . , 99 } . we want to know the probability z that x ( x + 1 ) is a multiple of either 4 or 5 . when will this happen ? if either x or ( x + 1 ) is a multiple of 4 or 5 . since 4 * 5 is 20 , let ' s look at the first 20 numbers to get a rough idea of how often this happens . out of the numbers from 1 to 20 : 4 , 5 , 6 , 8 , 9 , 10 , 11 , 12 , 13 , 15 , 16 , 17 , 20 so , 14 out of the first 20 numbers match our criteria . since : probability = ( # of desired outcomes ) / ( total # of possibilities ) , we guesstimate the answer to be 14 / 20 . since ( e ) is the only answer greater than 1 / 2 , we go with ( e ) ." | a = 100 / 4
b = a * 2
c = 100 / 5
d = c * 2
e = b + d
f = e - 4
g = 100 - 1
h = f / g
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a ) 123 , b ) 106 , c ) 108 , d ) 156 , e ) 240 | c | subtract(108.25, divide(1, 4)) | the cash realised on selling a 14 % stock is rs . 108.25 , brokerage being 1 / 4 % is | "explanation : cash realised = rs . ( 108.25 - 0.25 ) = rs . 108 . answer : c" | a = 1 / 4
b = 108 - 25
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a ) 24.4 % , b ) 28.6 % , c ) 32.3 % , d ) 36.5 % , e ) 40.9 % | b | multiply(divide(40, add(40, const_100)), const_100) | if y is 40 % greater than x , than x is what % less than y ? | "y = 1.4 x x = y / 1.4 = 10 y / 14 = 5 y / 7 x is 2 / 7 less which is 28.6 % less than y . the answer is b ." | a = 40 + 100
b = 40 / a
c = b * 100
|
a ) 5 , b ) 6 , c ) 4 , d ) 7 , e ) 8 | a | multiply(divide(subtract(subtract(41, const_1), multiply(subtract(16, const_1), const_2)), subtract(multiply(subtract(16, const_1), const_4), multiply(subtract(16, const_1), const_2))), subtract(16, const_1)) | a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and one - legged captain ) to a nearby port . altogether , these passengers have 16 heads and 41 legs . how many cats does the ship host ? | sa ' s + co + ca + cats = 16 . sa ' s + 1 + 1 + cats = 16 or sa ' s + cats = 14 . sa ' s ( 2 ) + 2 + 1 + cats * 4 = 41 sa ' s * 2 + cats * 4 = 38 or sa ' s + cats * 2 = 19 or 14 - cats + cat * 2 = 19 then cats = 5 a | a = 41 - 1
b = 16 - 1
c = b * 2
d = a - c
e = 16 - 1
f = e * 4
g = 16 - 1
h = g * 2
i = f - h
j = d / i
k = 16 - 1
l = j * k
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a ) 320 $ , b ) 420 $ , c ) 582 $ , d ) 650 $ , e ) 780 $ | c | multiply(multiply(0.85, 55), 12) | in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.85 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | 12 * 1.75 + 0.85 * 12 * 55 = 582 hence - c | a = 0 * 85
b = a * 12
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a ) 337 , b ) 500 , c ) 266 , d ) 288 , e ) 536 | e | divide(75, divide(multiply(4, add(3, divide(1, 2))), const_100)) | what sum of money will produce rs . 75 as simple interest in 4 years at 3 1 / 2 percent ? | "75 = ( p * 4 * 7 / 2 ) / 100 p = 536 answer : e" | a = 1 / 2
b = 3 + a
c = 4 * b
d = c / 100
e = 75 / d
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a ) 3600 , b ) 3607 , c ) 3200 , d ) 3602 , e ) 3603 | c | subtract(16000, multiply(divide(4, 5), 16000)) | income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 16000 , then find his savings ? | "let the income and the expenditure of the person be rs . 5 x and rs . 4 x respectively . income , 5 x = 16000 = > x = 3200 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 3200 . answer : c" | a = 4 / 5
b = a * 16000
c = 16000 - b
|
a ) 1 / 14 , b ) 1 / 18 , c ) 1 / 21 , d ) 1 / 25 , e ) 1 / 28 | e | divide(1, 33) | a certain fraction has the same ratio to 1 / 33 , as 3 / 4 does to 7 / 11 . what is this certain fraction ? | "x / ( 1 / 33 ) = ( 3 / 4 ) / ( 7 / 11 ) x = 3 * 11 * 1 / 33 * 4 * 7 = 1 / 28 the answer is e ." | a = 1 / 33
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a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 4.0 | e | multiply(add(add(38, 5), 5), divide(5, const_60)) | a certain car increased its average speed by 5 miles per hour in each successive 5 - minute interval after the first interval . if in the first 5 - minute interval its average speed was 38 miles per hour , how many miles did the car travel in the third 5 - minute interval ? | "in the third time interval the average speed of the car was 38 + 5 + 5 = 30 miles per hour ; in 5 minutes ( 1 / 12 hour ) at that speed car would travel 48 * 1 / 12 = 4 miles . answer : e ." | a = 38 + 5
b = a + 5
c = 5 / const_60
d = b * c
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a ) 1 / 35 , b ) 2 / 3 , c ) 7 / 35 , d ) 5 / 7 , e ) 7 / 5 | a | multiply(divide(1, 7), divide(1, 5)) | two brothers ram and ravi appeared for an exam . the probability of selection of ram is 1 / 7 and that of ravi is 1 / 5 . find the probability that both of them are selected . | "let a be the event that ram is selected and b is the event that ravi is selected . p ( a ) = 1 / 7 p ( b ) = 1 / 5 let c be the event that both are selected . p ( c ) = p ( a ) x p ( b ) as a and b are independent events : = 1 / 7 x 1 / 5 = 1 / 35 answer : a" | a = 1 / 7
b = 1 / 5
c = a * b
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a ) 0 , b ) 1 , c ) 4 , d ) 5 , e ) 6 | a | multiply(multiply(2, const_3), add(const_2, const_3)) | if a and b are distinct positive integers . the units digit of a ^ 2 is equal to the units digit of a , and the units digit of b ^ 2 is equal to the units digit of b . if the units digit of a · b is equal to neither the units digit of a nor the units digit of b , then what is the units digit of a · b ? | there are four digits which when squared give the same units digits : 0 , 1 , 5 , 6 0 ^ 2 = 0 1 ^ 2 = 1 5 ^ 2 = 25 6 ^ 2 = 36 so a and b must take 2 of these values . they are positive integers so 0 is not possible . if one of the digits is 1 , when the other is multiplied by 1 , you will get the same digit . 1 * 5 = 5 so 1 is out of the picture . then you have only two digits left 5 and 6 . so out of a and b , one must be 5 and the other 6 . 5 * 6 = 30 so units digit of their product = 0 answer ( a ) | a = 2 * 3
b = 2 + 3
c = a * b
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a ) 8847 , b ) 9837 , c ) 11,497 , d ) 15,627 , e ) 19,307 | d | subtract(subtract(subtract(multiply(627, const_10), const_100), 3), 3) | what is the next number : 3 , 27 , 627 , __ | "5 ^ 0 + 2 = 3 5 ^ 2 + 2 = 27 5 ^ 4 + 2 = 627 5 ^ 6 + 2 = 15,627 the answer is d ." | a = 627 * 10
b = a - 100
c = b - 3
d = c - 3
|
a ) 6 : 5 , b ) 1 : 4 , c ) 4 : 3 , d ) 2 : 3 , e ) 5 : 4 | e | divide(divide(5, 2), 2) | if the ratio of apples to bananas is 5 to 2 and the ratio of bananas to cucumbers is 1 to 2 , what is the ratio of apples to cucumbers ? | "the ratio of bananas to cucumbers is 1 to 2 which equals 2 to 4 . the ratio of apples to bananas to cucumbers is 5 to 2 to 4 . the ratio of apples to cucumbers is 5 to 4 . the answer is e ." | a = 5 / 2
b = a / 2
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a ) – 2 , b ) – 1 , c ) 0 , d ) 1 , e ) 2 | e | multiply(add(1, const_1), const_2) | what is the least integer greater than 1 + 0.5 ? | "this question is just about doing careful arithmetic and remembering what makes a numberbiggerorsmallercompared to another number . first , let ' s take care of the arithmetic : ( 1 ) + ( 0.5 ) = 1.5 on a number line , since we ' re adding + . 5 to a number , the total moves to the right ( so we ' re moving from 1 to 1.5 ) . next , the question asks for the least integer that is greater than 1.5 again , we can use a number line . numbers become greater as you move to the right . the first integer to the right of 1.5 is 2 . final answer : e" | a = 1 + 1
b = a * 2
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a ) 11 , b ) 14 , c ) 18 , d ) 23 , e ) none of these | b | subtract(add(6, 13), 5) | when 242 is divided by a certain divisor the remainder obtained is 6 . when 698 is divided by the same divisor the remainder obtained is 13 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 5 . what is the value of the divisor ? | "let that divisor be x since remainder is 6 or 13 it means divisor is greater than 13 . now 242 - 6 = 236 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 13 = 685 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 236 + 685 ) + 6 + 13 = x ( k + l ) + 19 when we divide this number by x then remainder will be equal to remainder of ( 19 divided by x ) = 5 hence x = 19 - 5 = 14 hence b" | a = 6 + 13
b = a - 5
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a ) 10 hours , b ) 12 hours , c ) 14 hours , d ) 21 hours , e ) none of these | d | add(divide(56, subtract(6, 2)), divide(56, add(6, 2))) | in a river flowing at 2 km / hr , a boat travels 56 km upstream and then returns downstream to the starting point . if its speed in still water be 6 km / hr , find the total journey time . | "explanation : speed of the boat = 6 km / hr speed downstream = ( 6 + 2 ) = 8 km / hr speed upstream = ( 6 - 2 ) = 4 km / hr distance travelled downstream = distance travelled upstream = 56 km total time taken = time taken downstream + time taken upstream = ( 56 / 8 ) + ( 56 / 4 ) = 21 hr . answer : option d" | a = 6 - 2
b = 56 / a
c = 6 + 2
d = 56 / c
e = b + d
|
a ) 6 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | a | sqrt(divide(volume_cylinder(divide(12, const_2), 8), multiply(const_pi, 8))) | a certain rectangular crate measures 8 feet by 12 feet by 14 feet . a cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces . what should the radius of the tank be if it is to be of the largest possible volume ? | "ans is ` ` b ' ' . for max volume of cylinder ( pi * r ^ 2 * h ) we need to max out r ^ 2 * h . we do n ' t know what the dimensions of the crate refer to . so to maximize the above eqn , radius may be of 9 , 10,12 one of the base area ( 8 x 12 , 14 x 12 or 8 x 14 ) r is maximum for base 14 x 12 and 12 can be maximum value so r = 10 / 2 = 6 a" | a = 12 / 2
b = volume_cylinder / (
c = math.pi * 8
d = math.sqrt(b)
|
a ) 320 , b ) 345 , c ) 375 , d ) 380 , e ) 400 | e | multiply(divide(640, 5), 3) | there are 640 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | "in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 640 . hence 3 x + 5 x = 640 solve for x 8 x = 640 x = 80 number of boys 3 x = 3 × 80 = 240 number of girls 5 x = 5 × 80 = 400 e" | a = 640 / 5
b = a * 3
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a ) 60 , b ) 45 , c ) 20 , d ) 15 , e ) none | c | multiply(divide(80, multiply(5, 4)), 5) | a ratio between two numbers is 5 : 4 and their l . c . m . is 80 . the first number is | "sol . let the required numbers be 5 x and 4 x . then , their l . c . m . is 20 x . ∴ 20 x = 80 ⇔ x = 4 . hence , the first number is 20 . answer c" | a = 5 * 4
b = 80 / a
c = b * 5
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a ) rs . 610.83 , b ) rs . 810.83 , c ) rs . 617 , d ) rs . 618 , e ) rs . 610.83 | e | subtract(add(add(divide(multiply(divide(570, multiply(divide(7, const_100), const_2.0)), 7), const_100), divide(570, multiply(divide(7, const_100), 3))), divide(multiply(add(divide(multiply(divide(570, multiply(divide(7, const_100), 3)), 7), const_100), divide(570, multiply(divide(7, const_100), 3))), 7), const_100)), divide(570, multiply(divide(7, const_100), 3))) | if the simple interest on a sum of money for 3 years at 7 % per annum is rs . 570 , what is the compound interest on the same sum at the rate and for the same time ? | "sum = ( 570 * 100 ) / ( 7 * 3 ) = rs . 2 , 714.29 amount = [ 2 , 714.29 * ( 1 + 7 / 100 ) 3 ] = rs . 3 , 325.12 c . i . = ( 3 , 325.12 - 2 , 714.29 ) = rs . 610.83 . answer : e" | a = 7 / 100
b = a * 2
c = 570 / b
d = c * 7
e = d / 100
f = 7 / 100
g = f * 3
h = 570 / g
i = e + h
j = 7 / 100
k = j * 3
l = 570 / k
m = l * 7
n = m / 100
o = 7 / 100
p = o * 3
q = 570 / p
r = n + q
s = r * 7
t = s / 100
u = i + t
v = 7 / 100
w = v * 3
x = 570 / w
y = u - x
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(subtract(add(const_1, floor(divide(1055, 23))), divide(1055, 23)), 23) | what is the least number that should be added to 1055 , so the sum of the number is divisible by 23 ? | "( 1055 / 23 ) gives a remainder 20 so we need to add 3 . the answer is c ." | a = 1055 / 23
b = math.floor(a)
c = 1 + b
d = 1055 / 23
e = c - d
f = e * 23
|
['a ) 10', 'b ) 11', 'c ) 12', 'd ) 9', 'e ) 8'] | a | divide(subtract(multiply(11, const_3), const_3), const_3) | an entry in an unshaded square is obtained by adding the entries connected to it from the row above ( 11 is one such number ) . write the value of n ? | see the values row wise 5 6 n 7 11 6 + n 7 + n 11 + 6 + n 6 + n + 7 + n 60 hence 11 + 6 + n + 6 + n + 7 + n = 60 n = 10 answer : a | a = 11 * 3
b = a - 3
c = b / 3
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a ) 0 , b ) 10 , c ) 15 , d ) 20 , e ) 30 | e | multiply(add(const_3, const_4), add(const_2, const_3)) | in the above number , a and b represent the tens and units digits , respectively . if the above number is divisible by 65 , what is the greatest possible value of b x a ? | "i also was confused when i was looking forabove number : d as far as i understood , 65 is a factor of ab . in other words , the values of b ( units digits can be 5 or 0 . better to have option for 5 in this case to havebigger result ) . now let ' s try 65 x 1 ( a = 6 , b = 5 respectively we have = 30 ) . this is the greatest possible value of b x a . imo e ." | a = 3 + 4
b = 2 + 3
c = a * b
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a ) 2.25 . , b ) 3.125 . , c ) 4.5 . , d ) 5.225 . , e ) 6.25 . | a | divide(divide(3024, 28), multiply(subtract(28, const_4), const_2)) | a computer factory produces 3024 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | "number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 3024 / ( 28 * 24 * 2 ) = 2.25 answer a" | a = 3024 / 28
b = 28 - 4
c = b * 2
d = a / c
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a ) 45 , b ) 60 , c ) 75 , d ) 90 , e ) 95 | c | multiply(divide(400, add(add(const_1, 5), 10)), const_3) | a man has rs . 400 in the denominations of one - rupee notes , 5 - rupee notes and 10 rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? | let number of notes of each denomination be x . then x + 5 x + 10 x = 400 16 x = 400 x = 25 . hence , total number of notes = 3 x = 75 answer : option c | a = 1 + 5
b = a + 10
c = 400 / b
d = c * 3
|
a ) 7 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | subtract(add(add(17, 17), 2), 30) | in a sports club with 30 members , 17 play badminton and 17 play tennis and 2 do not play either . how many members play both badminton and tennis ? | "let x play both badminton and tennis so 17 - x play only badminton and 17 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 17 - x ) + x + 2 = 30 36 - 2 x + x = 30 36 - x = 30 x = 6 so 6 members play both badminton and tennis . b" | a = 17 + 17
b = a + 2
c = b - 30
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a ) 392 , b ) 229 , c ) 753 , d ) 750 , e ) 540 | d | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5000) | 3 / 4 of 1 / 2 of 2 / 5 of 5000 = ? | "d 750 ? = 5000 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 750" | a = 3 / 4
b = 1 / 2
c = a * b
d = 2 / 5
e = c * d
f = e * 5000
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a ) y / z , b ) 20 y ² / z ² , c ) y ² z ² , d ) 5 y ² / z ² , e ) yz | b | divide(add(2, 5), subtract(2, 5)) | if x / y = 2 / z , then 5 x ^ 2 = | "this question is most easily solved by isolating y in the equation and substituting into the expression 5 x ² : x / y = 2 / z x = 2 y / z if we substitute 2 y / z into the expression for x , we get : 5 ( 2 y / z ) ² = 5 ( 4 y ² / z ² ) = 20 y ² / z ² . the correct answer is choice ( b )" | a = 2 + 5
b = 2 - 5
c = a / b
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a ) 35 m east , b ) 35 m north , c ) 30 m west , d ) 45 m west , e ) 55 m east | e | add(20, 30) | sandy walked 20 meters towards south . then sandy turned to her left and walked 25 meters . she then turned to her left and walked 20 meters . she then turned to her right and walked 30 meters . what distance is she from the starting point and in which direction ? | "the net distance is 25 + 30 = 55 meters to the east . the answer is e ." | a = 20 + 30
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a ) 0 , b ) 25 , c ) 50 , d ) 75 , e ) 100 | c | subtract(75, 25) | in a sample of associates at a law firm , 25 percent are second - year associates , and 75 percent are not first - year associates . what percentage of the associates at the law firm have been there for more than two years ? | "75 percent are not first - year associates - - > this means 75 percent associates are either two or more than two years already , we know % of second - year associates = 25 % so , % of associates there for more than two years = 75 % - 25 % = 50 % . . = > ( c ) . . is it correct ?" | a = 75 - 25
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a ) 9 , b ) 11 , c ) 13 , d ) 12 , e ) 17 | d | add(10, 2) | if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 10 ) , what is the value of x ? | "( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 10 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 10 hence x = 12 . answer is d" | a = 10 + 2
|
a ) $ 180,000 , b ) $ 202,000 , c ) $ 230,000 , d ) $ 400,000 , e ) $ 2 , 200,000 | c | multiply(multiply(200, const_100), const_10) | a special municipal payroll tax charges not tax on a payroll less than $ 300,000 and only 1 % on a company ’ s payroll above $ 300,000 . if belfried industries paid $ 200 in this special municipal payroll tax , then they must have had a payroll of ? | "answer : c , ( with different approach ) : the 300 paid is 1 % of the additional amount above 200,000 . let it be x now 1 % of x = 300 therefore x = 30,000 total = 200,000 + x = 230,000" | a = 200 * 100
b = a * 10
|
a ) rs . 18 , 828.80 , b ) rs . 19 , 828.80 , c ) rs . 18 , 028.80 , d ) rs . 17 , 828.80 , e ) none of these | b | add(divide(2828.80, subtract(power(add(const_1, divide(8, const_100)), const_2), const_1)), 2828.80) | the compound interest earned by sunil on a certain amount at the end of two years at the rate of 8 % p . a . was rs . 2828.80 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned . | "let the sum be rs . p p { [ 1 + 8 / 100 ] 2 - 1 } = 2828.80 p ( 8 / 100 ) ( 2 + 8 / 100 ) = 2828.80 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 2828.80 / ( 0.08 ) ( 2.08 ) = 1360 / 0.08 = 17000 principal + interest = rs . 19828.80 answer : b" | a = 8 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = 2828 / 80
f = e + 2828
|
a ) 1 / 11 , b ) 1 / 12 , c ) 1 / 13 , d ) 3 / 19 , e ) 3 / 43 | d | divide(divide(divide(1, 8), divide(2, 3)), add(divide(divide(1, 8), divide(2, 3)), 1)) | in a certain large company , the ratio of college graduates with a graduate degree to non - college graduates is 1 : 8 , and ratio of college graduates without a graduate degree to non - college graduates is 2 : 3 . if one picks a random college graduate at this large company , what is the probability r this college graduate has a graduate degree ? | "in believe the answer is d . please see below for explanation . 0 ) we are told the following ratios cgd - college graduate with degree ncg - non college graduate cgn - college graduate no degree cgd ncg cgn 1 8 3 2 in order to make cgd and cgn comparable we need to find the least common multiple of 8 and 3 and that is 24 multiplying the first ratio by 3 and the second ratio by 8 we get cgd ncg cgn 3 24 16 if one picks a random college graduate at this large company , what is the probability this college graduate has a graduate degree ? nr of cgd = 3 nr of cg = 3 + 16 = 19 probability r of cgd / ( cg ) - > 3 / 19 answer d" | a = 1 / 8
b = 2 / 3
c = a / b
d = 1 / 8
e = 2 / 3
f = d / e
g = f + 1
h = c / g
|
a ) 7 , b ) 6 , c ) 3 , d ) 5 , e ) 4 | e | divide(add(multiply(4, 5), multiply(3, 4)), subtract(multiply(4, 5), multiply(3, 4))) | if a / b = 5 / 4 , then ( 4 a + 3 b ) / ( 4 a - 3 b ) = ? | "answer dividing numerator as well as denominator by b , we get given exp . = ( 4 a + 3 b ) / ( 4 a - 3 b ) = ( 4 a / b + 3 ) / ( 4 a / b - 3 ) since a / b = 5 / 4 this implies that = [ ( 4 * 5 ) / 4 + 3 ] / [ ( 4 * 5 ) / 4 - 3 ) ] = ( 5 + 3 ) / ( 5 - 3 ) = 4 option : e" | a = 4 * 5
b = 3 * 4
c = a + b
d = 4 * 5
e = 3 * 4
f = d - e
g = c / f
|
a ) 60 , b ) 61 , c ) 62 , d ) 63 , e ) 64 | e | add(63, const_1) | total 63 matches are conducted in knockout match type . how many players will be participated in that tournament ? | "64 players answer : e" | a = 63 + 1
|
a ) 2013 , b ) 2088 , c ) 270 , d ) 1881 , e ) 1781 | a | add(1, 2012) | if f ( f ( n ) ) + f ( n ) = 2 n + 3 , f ( 0 ) = 1 then f ( 2012 ) = ? | "f ( f ( 0 ) ) + f ( 0 ) = 2 ( 0 ) + 3 ⇒ ⇒ f ( 1 ) = 3 - 1 = 2 , f ( 1 ) = 2 f ( f ( 1 ) ) + f ( 1 ) = 2 ( 1 ) + 3 ⇒ ⇒ f ( 2 ) = 5 - 2 = 3 , f ( 2 ) = 3 f ( f ( 2 ) ) + f ( 2 ) = 2 ( 2 ) + 3 ⇒ ⇒ f ( 3 ) = 7 - 3 = 4 , f ( 3 ) = 4 . . . . . . . . . . . . . . f ( 2012 ) = 2013 ans : a" | a = 1 + 2012
|
a ) 82216043 , b ) 83786095 , c ) 84316108 , d ) 85766122 , e ) 86426237 | d | divide(add(power(21, 7), 21), 21) | what is ( 21 ^ 7 + 21 ) / 21 ? | ( 21 ^ 7 + 21 ) / 21 = 21 * ( 21 ^ 6 + 1 ) / 21 = 21 ^ 6 + 1 clearly this is a number which ends with a 2 in the units place . the answer is d . | a = 21 ** 7
b = a + 21
c = b / 21
|
a ) a ) 3 , b ) b ) 2 , c ) c ) 6 , d ) d ) 9 , e ) e ) 1 / 5 | e | divide(subtract(divide(multiply(102, const_100), const_2), const_2), add(divide(multiply(102, const_100), const_2), const_2)) | if 102 x = 25 , then 10 - x equals : | "102 x = 25 ( 10 x ) 2 = 52 10 x = 5 1 / 10 x = 1 / 5 10 - x = 1 / 5 answer : e" | a = 102 * 100
b = a / 2
c = b - 2
d = 102 * 100
e = d / 2
f = e + 2
g = c / f
|
a ) a ) 540 , b ) b ) 400 , c ) c ) 800 , d ) d ) 650 , e ) e ) 840 | c | divide(multiply(520, const_100), subtract(const_100, 35)) | in an examination 35 % of the students passed and 520 failed . how many students appeared for the examination ? | "let the number of students appeared be x then , 65 % of x = 520 65 x / 100 = 520 x = 520 * 100 / 65 = 800 answer is c" | a = 520 * 100
b = 100 - 35
c = a / b
|
a ) 70 , b ) 88 , c ) 167 , d ) 197 , e ) 161 | a | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(170, divide(20, divide(15, const_3)))) | a train crosses a platform of 170 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be ‘ x ’ x + 170 / 15 = x + 250 / 20 20 x + 3400 = 15 x + 3750 5 x = 350 x = 70 m answer : a" | a = 15 / 3
b = 15 / a
c = 250 * b
d = 15 / 3
e = 20 / d
f = 170 * e
g = c - f
|
a ) a . 0.6 , b ) b . 1 , c ) c . 2.1 , d ) d . 3 , e ) e . 5.4 | d | subtract(6, multiply(const_2, multiply(divide(25, const_100), 6))) | a 6 litre sol is 25 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ? | "25 % of 6 = 1.5 50 % of 6 = 3 shortage is 1.5 so we need to have 1.5 / 50 % to get 50 % alcohol content . = 3 d" | a = 25 / 100
b = a * 6
c = 2 * b
d = 6 - c
|
a ) s . 5000 , b ) s . 2400 , c ) s . 4000 , d ) s . 3200 , e ) s . 4200 | a | multiply(subtract(multiply(divide(2000, 2), 3), 2000), 5) | the ratio of incomes of two person p 1 and p 2 is 5 : 4 and the ratio of their expenditures is 3 : 2 . if at the end of the year , each saves rs . 2000 , then what is the income of p 1 ? | "let the income of p 1 and p 2 be rs . 5 x and rs . 4 x respectively and let their expenditures be rs . 3 y and 2 y respectively . then , 5 x – 3 y = 2000 … ( i ) and 4 x – 2 y = 2000 … … . . ( ii ) on multiplying ( i ) by 2 , ( ii ) by 3 and subtracting , we get : 2 x = 2000 - > x = 1000 p 1 ’ s income = rs 5 * 1000 = rs . 5000 answer : a" | a = 2000 / 2
b = a * 3
c = b - 2000
d = c * 5
|
a ) 64 , b ) 72 , c ) 86 , d ) 98 , e ) 99 | e | subtract(power(add(const_1, add(const_1, const_3)), const_3), 26) | cubes with each side one inch long are glued together to form a larger cube . the larger cube ' s face is painted with red color and the entire assembly is taken apart . 26 small cubes are found with no paints on them . how many of unit cubes have at least one face that is painted red ? | "use the options . the options which after getting added to 26 shows a cube of a number could be right . here 64 + 26 = 90 72 + 26 = 98 86 + 26 = 112 98 + 26 + 124 99 + 26 = 125 - - - ( 5 * 5 * 5 ) so we have 99 as the answer ! e" | a = 1 + 3
b = 1 + a
c = b ** 3
d = c - 26
|
a ) 39 , b ) 24 , c ) 22 , d ) 20 , e ) 18 | c | add(divide(320, 10), subtract(subtract(10, 4), const_1)) | in a certain quiz that consists of 10 questions , each question after the first is worth 4 points more than the preceding question . if the 10 questions on the quiz are worth a total of 320 points , how many points is the third question worth ? | "x x + 4 x + 8 x + 12 x + 16 x + 20 x + 24 x + 28 x + 32 x + 36 10 x + 180 = 320 10 x = 140 x = 14 3 rd question = x + 8 = 14 + 8 = 22 answer c" | a = 320 / 10
b = 10 - 4
c = b - 1
d = a + c
|
a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | a | subtract(30, log(32)) | it takes 30 days to fill a laboratory dish with bacteria . if the size of the bacteria doubles each day , how long did it take for the bacteria to fill 1 / 32 of the dish ? | "the bacteria doubles each day , so after 29 days , the dish was half full . after 28 days , the dish was one quarter full . after 27 days , the dish was one eighth full . after 26 days , the dish was one sixteenth full . after 25 days , the dish was 1 / 32 full . the answer is a ." | a = math.log(32)
b = 30 - a
|
a ) s . 17 , b ) s . 18 , c ) s . 12 , d ) s . 11 , e ) s . 20 | e | divide(add(1500, 340), add(32, 60)) | rahim bought 32 books for rs . 1500 from one shop and 60 books for rs . 340 from another . what is the average price he paid per book ? | "average price per book = ( 1500 + 340 ) / ( 32 + 60 ) = 1840 / 92 = rs . 20 answer : e" | a = 1500 + 340
b = 32 + 60
c = a / b
|
a ) 0.64202 , b ) 0.64204 , c ) 0.642022 , d ) 0.642025 , e ) none | d | divide(add(multiply(0.64204, 0.64203), multiply(0.64204, 0.64202)), 0.64204) | the average of numbers 0.64204 , 0.64203 , 0.64202 and 0.64201 is ? | "answer average = ( 0.64204 + 0.64203 + 0.64202 + 0.64201 ) / 4 = 2.5681 / 4 = 0.642025 correct option : d" | a = 0 * 64204
b = 0 * 64204
c = a + b
d = c / 0
|
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