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 ) 228 , b ) 288 , c ) 27 , d ) 120 , e ) 881 | d | subtract(add(multiply(15000, subtract(const_1, divide(4, const_100))), multiply(8000, add(const_1, divide(9, const_100)))), add(15000, 8000)) | ravi purchased a refrigerator and a mobile phone for rs . 15000 and rs . 8000 respectively . he sold the refrigerator at a loss of 4 percent and the mobile phone at a profit of 9 percent . overall he make a . | let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 9 / 100 ) = 8000 + 720 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 720 = rs . 120 as this is positive , an overall profit of rs . 200 was made . answer : d | a = 4 / 100
b = 1 - a
c = 15000 * b
d = 9 / 100
e = 1 + d
f = 8000 * e
g = c + f
h = 15000 + 8000
i = g - h
|
a ) 22000 , b ) 20000 , c ) 11875 , d ) 12340 , e ) none of these | c | divide(multiply(3800, const_100), 32) | sheela deposits rs . 3800 in bank savings account . if this is 32 % of her monthly income . what is her monthly income in ? | "explanation : 32 % of income = rs . 3800 100 % of income = 3800 x 100 / 32 = rs . 11875 answer : c" | a = 3800 * 100
b = a / 32
|
a ) 35 , b ) 45 , c ) 55 , d ) 65 , e ) 75 | c | divide(multiply(11, subtract(11, const_1)), const_2) | there are 11 teams in a certain league and each team plays each of the other teams exactly once . what is the total number of games played ? | 11 c 2 = 55 the answer is c . | a = 11 - 1
b = 11 * a
c = b / 2
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a ) $ 60 , b ) $ 64 , c ) $ 75 , d ) $ 110 , e ) can not be determined | d | divide(add(140, 80), 2) | if greg buys 6 shirts , 4 trousers and 2 ties , the total cost is $ 80 . if greg buys 4 shirts , 2 trousers and 2 ties , the total cost is $ 140 . how much will it cost him to buy 3 trousers , 5 shirts and 2 ties ? | solution : 6 x + 4 y + 2 z = 80 4 x + 2 y + 2 z = 140 adding both the equations = 10 x + 6 y + 4 z = 220 5 x + 3 y + 2 z = 110 ans d | a = 140 + 80
b = a / 2
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a ) 18 square meters , b ) 19 square meters , c ) 24 square meters , d ) 28 square meters , e ) 30 square meters | b | divide(subtract(subtract(200, 138), 24), const_2) | three rugs have a combined area of 200 square meters . by overlapping the rugs to cover floor area of 138 square meters , the area that is covered by exactly two layers of rug is 24 square meters . what is the area that is covered with three layers of rug ? | "total = rug 1 + rug 2 + rug 3 - { overlap of exactly 2 rugs } - 2 * { overlap of exactly 3 rugs } 138 = 200 - 24 - 2 * { overlap of exactly 2 rugs } - - > { overlap of exactly 3 rugs } = 19 . answer : b ." | a = 200 - 138
b = a - 24
c = b / 2
|
a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 10 , d ) 3 / 50 , e ) 1 / 20 | d | divide(const_3, multiply(5, 10)) | if jenee earns a raise of between 5 % and 10 % , non - inclusive , by what fraction could her salary have increased ? | perhaps it is also easy to just translate the % into fractions : 6 % : 6 / 100 = 3 / 50 7 % : 7 / 100 8 % : 8 / 100 = 4 / 50 = 2 / 25 9 % : 9 / 100 only 3 / 50 exists in the answer choices , so ans d | a = 5 * 10
b = 3 / a
|
a ) 54 m , b ) 64 m , c ) 74 m , d ) 84 m , e ) 94 m | b | divide(add(divide(5300, 26.50), multiply(const_2, 28)), const_4) | length of a rectangular plot is 28 mtr more than its breadth . if the cost of fencin gthe plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 28 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 28 ) + x ] = 200 2 x + 28 = 100 2 x = 72 x = 36 . hence , length = x + 28 = 64 m b" | a = 5300 / 26
b = 2 * 28
c = a + b
d = c / 4
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) it can not be determined from the information given | a | divide(multiply(3, const_3), 3) | if x = 3 y , where y is a prime number greater than 2 , how many different positive even divisors does x have , including x ? | for x to have even divisors , xx has to be an even number . we know that x = 3 y , where yy is a prime number greater than 2 . as 2 is the only even prime number , y has to be an odd prime number . this means that x must also be an odd number , since odd x odd = odd . as x is odd , it can not have any even divisors . therefore , a is the correct answer . | a = 3 * 3
b = a / 3
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a ) 0.36 , b ) 3.6 , c ) 360 , d ) 6000 , e ) 36000 | d | multiply(divide(1, add(const_2, const_4)), multiply(divide(const_3600, const_10), const_100)) | if anangletis defined as 1 percent of 1 degree , then how many anglets are there in a sixth of a circle ? | 1 degree * 1 / 100 = 1 anglet so 1 degree = 100 anglets = > 60 degrees = 6000 anglets answer - d | a = 2 + 4
b = 1 / a
c = 3600 / 10
d = c * 100
e = b * d
|
a ) 100 , b ) 120 , c ) 80 , d ) 90 , e ) none of the above | b | multiply(divide(multiply(20, 3), subtract(multiply(5, 3), multiply(6, 2))), 6) | ratio of girl and boys in class is 5 : 6 . if 20 girl leave the class and ratio become 2 : 3 . calculate the no . of boys ? | girl 5 x and boys 6 x 5 x - 20 / 6 x = 2 / 3,15 x - 60 = 12 x , 3 x = 60 , x = 20 boys = 6 x = 120 answer b | a = 20 * 3
b = 5 * 3
c = 6 * 2
d = b - c
e = a / d
f = e * 6
|
a ) 120 , b ) 200 , c ) 180 , d ) 150 , e ) none of these | c | add(divide(18, divide(subtract(30, 20), const_100)), 20) | if 30 % of a number exceeds 20 % of it by 18 , then find the number ? | "let the number be x . 30 % of x = 30 / 100 * x = 3 x / 10 20 % of x = 20 / 100 * x = x / 5 = 2 x / 10 . given that 3 x / 10 - 2 x / 10 = 18 = > x = 180 . required number is 180 . answer : c" | a = 30 - 20
b = a / 100
c = 18 / b
d = c + 20
|
a ) 310 , b ) 200 , c ) 215 , d ) 285 , e ) 210 | e | divide(multiply(1, 20), const_4) | what is the sum of the numbers between 1 and 20 , inclusive ? | "sol . 1 + 2 + 3 . . . 17 + 18 + 19 + 20 answer . e" | a = 1 * 20
b = a / 4
|
a ) 65,000 , b ) 67,000 , c ) 69,000 , d ) 71,000 , e ) 73,000 | c | subtract(const_100, multiply(const_4, const_2)) | a company recently conducted a survey and found that 30,000 of its customers live in rural areas . if the number of customers who live in urban areas is 130 percent greater than the number of customers who live in rural areas , how many customers live in urban areas ? | "the number of customers in urban areas is 30,000 + 1.3 * 30,000 = 69,000 . the answer is c ." | a = 4 * 2
b = 100 - a
|
a ) 6 / 22 , b ) 5 / 22 , c ) 7 / 22 , d ) 4 / 22 , e ) 3 / 22 | c | divide(add(4, 3), const_10) | company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 3 / 4 were basic and the rest were deluxe . if it takes 7 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ? | "# of basic stereos was 3 / 4 of total and # of deluxe stereos was 1 / 4 of total , let ' s assume total = 16 , then basic = 12 and deluxe = 4 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 7 / 5 units . total time for basic would be 12 * 1 = 12 and total time for deluxe would be 4 * 7 / 5 = 28 / 5 - - > total time for both of them would be 12 + 28 / 5 = 88 / 5 - - > deluxe / total = 28 / 5 / 88 / 5 = 28 / 88 = 7 / 22 c" | a = 4 + 3
b = a / 10
|
a ) 159202573 , b ) 160902641 , c ) 172502765 , d ) 182302834 , e ) 191102977 | e | divide(add(power(24, 7), 24), 24) | what is ( 24 ^ 7 + 24 ) / 24 ? | "( 24 ^ 7 + 24 ) / 24 = 24 * ( 24 ^ 6 + 1 ) / 24 = 24 ^ 6 + 1 clearly this is a number which ends with a 7 in the units place . the answer is e ." | a = 24 ** 7
b = a + 24
c = b / 24
|
a ) 19 , b ) 21 , c ) 23 , d ) 25 , e ) 27 | a | subtract(multiply(7, 3), const_2) | 3 consecutive even numbers are such that thrice the first number exceeds double the third number by 7 . what is the third number ? | let the three numbers be x , x + 2 , and x + 4 . 3 x = 2 ( x + 4 ) + 7 x = 15 and then the third number is x + 4 = 19 . the answer is a . | a = 7 * 3
b = a - 2
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(add(multiply(factorial(22), factorial(4)), multiply(factorial(22), factorial(3))), 22) | what is the units digit of ( 22 ^ 4 ) ( 16 ^ 3 ) ( 41 ^ 8 ) ? | "the units digit of 22 ^ 4 is the units digit of 2 ^ 4 which is 6 . the units digit of 16 ^ 3 is the units digit of 6 ^ 3 which is 6 . the units digit of 41 ^ 8 is the units digit of 1 ^ 8 which is 1 . the units digit of 6 * 6 * 1 is 6 . the answer is e ." | a = math.factorial(22)
b = math.factorial(4)
c = a * b
d = math.factorial(22)
e = math.factorial(3)
f = d * e
g = c + f
h = g / 22
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | subtract(8, multiply(divide(subtract(8, 5), subtract(1000, 400)), 1000)) | in a factory , each day the expected number of accidents is related to the number of overtime hour by linear equation . suppose that on one day there were 1000 overtime hours logged and 8 accidents reported and on another day there were 400 overtime hours logged and 5 accidents . what is the expected number of accidents when no overtime hours are logged ? | explanation : let , the number of over time hours be x and expected number of accidents be y . we will use the equation y = mx + c , where m and c are constant . now , according to the question , if x = 1000 , then y = 8 i . e 8 = 1000 x + c - - - - ( i ) and , if x = 400 , then y = 5 i . e 5 = 400 x + c . - - - - ( ii ) on solving these equations , we get m = 1 / 200 and c = 3 . hence , in case , no overtime is there i . e x = 0 , then the number of expected accident will be 3 . answer : b | a = 8 - 5
b = 1000 - 400
c = a / b
d = c * 1000
e = 8 - d
|
a ) 150 % , b ) 170 % , c ) 175 % , d ) 160 % , e ) 180 % | d | divide(multiply(40, subtract(const_100, 20)), 20) | last week vartan spent 20 percent of his wages on recreation . this week , his wages are 20 percent less than last week ʼ s wages and he spent 40 percent of his wages on recreation . the amount he spends on recreation this week is what percent of the amount he spent on recreation last week ? | "say vartan ' s wages last week were $ 100 , so he spent 0.20 * 100 = $ 20 on recreation ; this week ' s wages is 0.8 * 100 = $ 80 , so he spends 0.4 * 80 = $ 32 on recreation ; 32 / 20 = 1.6 , hence the amount he spends on recreation this week is 160 % of the amount he spent on recreation last week : 20 * 1.6 = 32 answer : d" | a = 100 - 20
b = 40 * a
c = b / 20
|
a ) 1440 , b ) 10080 , c ) 6200 , d ) 10020 , e ) 12500 | a | subtract(negate(240), multiply(subtract(12, 48), divide(subtract(12, 48), subtract(4, 12)))) | 4 , 12 , 48 , 240 , ( . . . ) | "explanation : go on multiplying the given numbers by 3 , 4 , 5 , 6 , answer : option a" | a = negate - (
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a ) s 400 , b ) s 200 , c ) s 800 , d ) s 500 , e ) s 100 | c | divide(subtract(multiply(3200, divide(5, const_100)), 144), subtract(divide(5, const_100), divide(3, const_100))) | rs 3200 is divided into two parts such that one part is put out at 3 % and the other at 5 % . if the annual interest earned from both the investments be rs 144 , find the first part . | "explanation : average rate = ( 144 / 3200 ) * 100 = 4.5 ratio = 5 : 15 so , first part = ( 5 / 20 ) * 3200 = rs 800 . answer : c" | a = 5 / 100
b = 3200 * a
c = b - 144
d = 5 / 100
e = 3 / 100
f = d - e
g = c / f
|
a ) 55 , b ) 25 , c ) 85 , d ) 45 , e ) 35 | e | multiply(divide(add(const_1, subtract(const_10, const_1)), const_2), subtract(const_10, const_1)) | if you add all the numbers on your mobile phone except 8 and 2 , what is the answer ? | "we have to add 0 , 1,3 , 4,5 , 6,7 & 9 to find the answer . therefore 0 + 1 + 3 + 4 + 5 + 6 + 7 + 9 = 35 answer is e" | a = 10 - 1
b = 1 + a
c = b / 2
d = 10 - 1
e = c * d
|
a ) 50 min , b ) 60 min , c ) 72 min , d ) 80 min , e ) 70 min | c | multiply(const_10, multiply(const_1, 6)) | a pipe takes a hours to fill the tank . but because of a leakage it took 6 times of its original time . find the time taken by the leakage to empty the tank | "pipe a can do a work 60 min . lets leakage time is x ; then 1 / 60 - 1 / x = 1 / 360 x = 72 min answer : c" | a = 1 * 6
b = 10 * a
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a ) 20 , b ) 25 , c ) 10 , d ) 16 , e ) 30 | e | subtract(40, multiply(multiply(10, 2), 2)) | evaluate : 40 - 10 ÷ 2 × 2 = | "according to order of operations , 10 ÷ 2 × 2 ( division and multiplication ) is done first from left to right 10 ÷ 2 × 2 = 5 × 2 = 10 hence 40 - 10 ÷ 2 × 2 = 40 - 10 = 30 correct answer e ) 30" | a = 10 * 2
b = a * 2
c = 40 - b
|
a ) 80 , b ) 85 , c ) 65 , d ) 55 , e ) 50 | a | multiply(6, divide(lcm(add(5, 3), add(6, 4)), add(6, 4))) | the number of stamps that kaye and alberto had were in the ration of 5 : 3 respectively . after kaye gave alberto 10 of her stamps , the ration of the number of kaye had to the number of alberto had was 6 : 4 . as a result of the gift , kaye had how many more stamps than alberto ? | "c k 1 = 5 x a 1 = 3 x k 2 = 5 x - 10 a 2 = 3 x + 10 k 2 / a 2 = 6 / 4 ( 5 x - 10 ) / ( 3 x + 10 ) = 6 / 4 4 * ( 5 x - 10 ) = 6 * ( 3 x + 10 ) 20 x - 40 = 18 x + 60 2 x = 100 x = 50 k 2 = 5 * 50 - 10 = 240 a 2 = 3 * 50 + 10 = 160 k 2 - a 2 = 80 answer : a" | a = 5 + 3
b = 6 + 4
c = math.lcm(a, b)
d = 6 + 4
e = c / d
f = 6 * e
|
a ) 6.67 seconds , b ) 18 seconds , c ) 26 seconds , d ) 14.99 seconds , e ) 6 seconds | d | divide(100, multiply(add(18, 6), const_0_2778)) | the speed at which a man can row a boat in still water is 18 kmph . if he rows downstream , where the speed of current is 6 kmph , what time will he take to cover 100 metres ? | "speed of the boat downstream = 18 + 6 = 24 kmph = 24 * 5 / 18 = 6.67 m / s hence time taken to cover 100 m = 100 / 6.67 = 14.99 seconds . answer : d" | a = 18 + 6
b = a * const_0_2778
c = 100 / b
|
a ) 1.42 hr , b ) 2.22 hrs , c ) 3.24 hrs , d ) 5 hrs , e ) 6.02 hrs | a | divide(50, add(10, 25)) | two cyclist start from the same places in opposite directions . one is going towards north at 10 kmph and the other is going towards south 25 kmph . what time will they take to be 50 km apart ? | "to be ( 10 + 25 ) km apart , they take 1 hour to be 50 km apart , they take 1 / 35 * 50 = 1.42 hrs answer is a" | a = 10 + 25
b = 50 / a
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a ) 10,000 , b ) 11,600 , c ) 12,000 , d ) 15,000 , e ) 16,400 | d | add(5, 6) | jerome anticipated that the webweb . com stock price would fall and sold all his webweb . com stocks for $ 5 per stock . he paid $ 10,000 tax on the revenue . a week later , jerome became convinced that the webweb . com stock price would rise , and he used the money that he had gotten from selling the webweb . com stocks to purchase them again , this time at $ 6 per stock . if jerome ended up having 5000 webweb . com stocks fewer than the number he had before selling them , how many webweb . com stocks did he have originally ? | "let the number of shares be x . 5 * x - 10000 ( money paid in taxes ) = 6 ( x - 5000 ) solving for x , we get the shares as 15000 . ans : ( option d )" | a = 5 + 6
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a ) 30 , b ) 32 , c ) 48 , d ) 55 , e ) 64 | c | add(multiply(8, 5), 8) | the ratio of pens to pencils is 5 to 6 . there are 8 more pencils than pens . how many pencils are there ? | "let x equal the number of pens , and y the number of pencils . then , x = 5 / 6 y . we also know that x = y - 8 . and so , we can substitute : y - 8 = 5 / 6 y so to isolate y , we get : 1 / 6 y = 8 and therefore : y = 48 answer : c" | a = 8 * 5
b = a + 8
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a ) 2 : 3 , b ) 3 : 4 , c ) 3 : 20 , d ) 20 : 3 , e ) none | c | divide(multiply(0.06, const_100), multiply(0.4, const_100)) | if 0.4 of a number is equal to 0.06 of another number , the ratio of the numbers i | "sol . 0.4 a = 0.06 b ⇔ a / b = 0.06 / 0.40 = 6 / 40 = 3 / 20 . ∴ a : b = 3 : 20 . answer c" | a = 0 * 6
b = 0 * 4
c = a / b
|
a ) 230 m , b ) 240 m , c ) 260 m , d ) 320 m , e ) 330 m | b | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 260) | a 260 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "relative speed = ( 120 + 80 ) km / hr = ( 200 x ( 5 / 18 ) ) m / sec = ( 500 / 9 ) m / sec . let the length of the other train be x metres . then , ( x + 260 ) / 9 = 500 / 9 x + 260 = 500 x = 240 . b" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 260
|
a ) 3 hrs , b ) 7 hrs , c ) 9 hrs , d ) 2 hrs , e ) 6 hrs | d | divide(8, subtract(subtract(divide(96, 8), 4), 4)) | a boatman can row 96 km downstream in 8 hr . if the speed of the current is 4 km / hr , then find in what time he will be able to cover 8 km upstream ? | explanation : speed in downstream = 96 / 8 = 12 kmph speed of current = 4 km / hr speed of the boatman in still water = 12 â € “ 4 = 8 kmph speed in upstream = 8 â € “ 4 = 4 kmph time taken to cover 8 km upstream = 8 / 4 = 2 hours . answer : d | a = 96 / 8
b = a - 4
c = b - 4
d = 8 / c
|
a ) 8 : 39 , b ) 8 : 38 , c ) 8 : 33 , d ) 8 : 32 , e ) 8 : 31 | c | divide(add(const_4, const_4), add(const_3, multiply(const_10, const_3))) | choose the number pair / group which is different from others ? | explanation : in all other pairs , 2 nd number = ( 1 st number × 3 ) + 1 . answer : c ) 8 : 33 | a = 4 + 4
b = 10 * 3
c = 3 + b
d = a / c
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | a | divide(divide(945, add(power(3, 3), power(2, const_3))), multiply(3, 3)) | the ratio of two numbers is 2 : 3 and the sum of their cubes is 945 . the difference of number is ? | 2 x 3 x 8 x cube + 27 x cube = 945 35 x cube = 945 x cube = 27 = > x = 3 . answer : a | a = 3 ** 3
b = 2 ** 3
c = a + b
d = 945 / c
e = 3 * 3
f = d / e
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a ) s . 5,000 , b ) s . 5,500 , c ) s . 5,700 , d ) s . 6,300 , e ) s . 7,500 | d | multiply(multiply(add(const_4, const_1), const_4), multiply(2, multiply(const_3, const_4))) | a and b started a business in partnership investing rs . 20,000 and rs . 15,000 respectively . after 6 months , c joined them with rs . 20,000 . whatwill be b ' s share in total profit of rs . 21,000 earned at the end of 2 years from the startingof the business ? | "a : b : c = ( 20,000 x 24 ) : ( 15,000 x 24 ) : ( 20,000 x 18 ) = 4 : 3 : 3 . b ' s share = rs . 21000 x 3 / 10 = rs . 6,300 . d" | a = 4 + 1
b = a * 4
c = 3 * 4
d = 2 * c
e = b * d
|
a ) 15 , b ) 18 , c ) 20 , d ) 25 , e ) 30 | c | divide(add(1, add(add(add(power(const_2, const_2), power(const_3, const_2)), add(power(const_4, const_2), power(add(const_2, const_3), const_2))), add(power(add(const_2, const_4), const_2), power(add(const_3, const_4), const_2)))), add(const_3, const_4)) | the mean of 1 ^ 22 ^ 23 ^ 24 ^ 25 ^ 26 ^ 27 ^ 2 is ? | 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + - - - - - + n ^ 2 = n ( n + 1 ) ( 2 n + 1 ) / 6 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + - - - - - + 7 ^ 2 = 7 * 8 * 156 = 140 required average is = 140 / 7 = 20 answer is c | a = 2 ** 2
b = 3 ** 2
c = a + b
d = 4 ** 2
e = 2 + 3
f = e ** 2
g = d + f
h = c + g
i = 2 + 4
j = i ** 2
k = 3 + 4
l = k ** 2
m = j + l
n = h + m
o = 1 + n
p = 3 + 4
q = o / p
|
a ) 872 , b ) 820 , c ) 837 , d ) 840 , e ) 83 | b | divide(subtract(multiply(1500, divide(15, const_100)), 20), divide(25, const_100)) | if 25 % of x is 20 less than 15 % of 1500 , then x is ? | "25 % of x = x / 4 ; 15 % of 1500 = 15 / 100 * 1500 = 225 given that , x / 4 = 225 - 20 = > x / 4 = 205 = > x = 820 . answer : b" | a = 15 / 100
b = 1500 * a
c = b - 20
d = 25 / 100
e = c / d
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a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 30 | e | divide(15, subtract(divide(40, const_60), divide(10, const_60))) | circular gears p and q start rotating at the same time at constant speeds . gear p makes 10 revolutions per minute and gear q makes 40 revolutions per minute . how many seconds after the gears start rotating will gear q have made exactly 15 more revolutions than gear p ? | gear q makes 30 more revolutions each 60 seconds . the time to make 15 = 30 / 2 more revolutions is 60 / 2 = 30 seconds . the answer is e . | a = 40 / const_60
b = 10 / const_60
c = a - b
d = 15 / c
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a ) 4584638125 , b ) 4164647415 , c ) 4584638135 , d ) 4584638140 , e ) 4584638145 | b | multiply(8956231, power(add(const_4, const_1), const_4)) | ( 8956231 x 465 ) = ? | 8956231 x 465 = 4164647415 ans b | a = 4 + 1
b = a ** 4
c = 8956231 * b
|
a ) 26630 , b ) 33275 , c ) 36625 , d ) 26625 , e ) 26825 | b | multiply(25000, add(const_1, divide(multiply(3, 10), const_100))) | population is 25000 . population increases by 10 % every year , then the population after 3 years is ? | population after 1 st year = 25000 * 10 / 100 = 2500 = = = > 25000 + 2000 = 27500 population after 2 nd year = 27500 * 10 / 100 = 2750 = = = > 27500 + 2750 = 30250 population after 3 rd year = 30250 * 10 / 100 = 3025 = = = > 30250 + 3025 = 33275 answer : b | a = 3 * 10
b = a / 100
c = 1 + b
d = 25000 * c
|
a ) 1 / 5 , b ) 2 / 8 , c ) 1 / 3 , d ) 2 / 3 , e ) 7 / 9 | b | divide(subtract(10, 8), 8) | a number , x is chosen at random from the set of positive integers less than 10 . what is the probability that ( 8 / x ) > x ? | number x has to be chosen from numbers 1 - 9 ( 8 / x ) > x = > 8 > x ^ 2 = > x ^ 2 - 8 < 0 x can have 2 values only 1 , 2 therefore , probability = 2 / 8 answer b | a = 10 - 8
b = a / 8
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a ) 3 , b ) 6 , c ) 1 / 9 , d ) 1 / 3 , e ) 9 / 4 | e | divide(multiply(1, power(3, const_2)), power(2, const_2)) | x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 2 will be equal to : | "explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 2 ^ 2 = 9 / 4 answer : e" | a = 3 ** 2
b = 1 * a
c = 2 ** 2
d = b / c
|
a ) 51.1 minutes , b ) 51.3 minutes , c ) 51.5 minutes , d ) 51.7 minutes , e ) 51.9 minutes | c | inverse(add(divide(const_1, multiply(1.5, const_60)), divide(const_1, multiply(2, const_60)))) | it takes 1.5 hours for tim to mow the lawn . linda can mow the same lawn in 2 hours . how long will it take john and linda , work together , to mow the lawn ? | we first calculate the rate of work of john and linda john : 1 / 1.5 and linda 1 / 2 let t be the time for john and linda to mow the lawn . the work done by john alone is given by t * ( 1 / 1.5 ) the work done by linda alone is given by t * ( 1 / 2 ) when the two work together , their work will be added . hence t * ( 1 / 1.5 ) + t * ( 1 / 2 ) = 1 multiply all terms by 6 6 ( t * ( 1 / 1.5 ) + t * ( 1 / 2 ) ) = 6 and simplify 4 t + 3 t = 6 solve for t t = 6 / 7 hours = 51.5 minutes . answer c | a = 1 * 5
b = 1 / a
c = 2 * const_60
d = 1 / c
e = b + d
f = 1/(e)
|
a ) 39 , b ) 40 , c ) 31 , d ) 42 , e ) 43 | c | subtract(multiply(add(20, 5), const_2), 19) | you collect pens . suppose you start out with 5 . mike gives you another 20 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 19 pens . how many pens do you have at the end ? | "solution start with 5 pens . mike gives you 20 pens : 5 + 20 = 25 pens . cindy doubles the number of pens you have : 25 ã — 2 = 50 pens . sharon takes 19 pens from you : 50 - 19 = 31 pens . so you have 31 at the end . correct answer : c" | a = 20 + 5
b = a * 2
c = b - 19
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a ) $ 80,000 , b ) $ 100,000 , c ) $ 120,000 , d ) $ 150,000 , e ) $ 200,000 | a | divide(multiply(const_100, multiply(const_100, add(const_1, const_4))), add(divide(25, const_100), multiply(multiply(divide(25, const_100), subtract(const_1, divide(25, const_100))), const_2))) | the majority owner of a business received 25 % of the profit , with each of 4 partners receiving 25 % of the remaining profit . if the majority owner and two of the owners combined to receive $ 50,000 , how much profit did the business make ? | let p be the total profit . p / 4 + 1 / 2 * ( 3 p / 4 ) = p / 4 + 3 p / 8 = 5 p / 8 = 50000 p = $ 80,000 the answer is a . | a = 1 + 4
b = 100 * a
c = 100 * b
d = 25 / 100
e = 25 / 100
f = 25 / 100
g = 1 - f
h = e * g
i = h * 2
j = d + i
k = c / j
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a ) 7 , b ) 6 , c ) 9 , d ) 4 , e ) 8 | c | subtract(28, add(7, 10)) | in a group of 28 junior high school students , 7 take french , 10 take spanish , and 2 take both languages . the students taking both french and spanish are not counted with the 7 taking french or the 10 taking spanish . how many students are not taking either french or spanish ? | "c 9 add 7 + 10 + 2 to get 19 then subtract 21 from the total students ⇒ 28 – 19 = 9 . answer is c" | a = 7 + 10
b = 28 - a
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a ) 20000 , b ) 14000 , c ) 23000 , d ) 120000 , e ) none of these | d | multiply(multiply(const_60, 10), multiply(20, 10)) | a wheel rotates 10 times every minute and moves 20 cm during each rotation . how many ems does the wheel move in 1 hour ? | explanation : number of wheel moves in 1 hour = 10 * 60 = 600 distance moved = 600 * 200 = 12000 answer : d | a = const_60 * 10
b = 20 * 10
c = a * b
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a ) 150 , b ) 175 , c ) 200 , d ) 225 , e ) 300 | a | subtract(divide(add(200, 400), const_2), divide(add(100, 200), const_2)) | the average ( arithmetic mean ) of the integers from 200 to 400 , inclusive , is how much greater than the average of the integers from 100 to 200 , inclusive ? | "for an ap the mean or average of series is average of first and last term . so , average of numbers between 200 to 400 , inclusive = ( 200 + 400 ) / 2 = 300 average of numbers between 100 to 200 , inclusive = ( 100 + 200 ) / 2 = 150 difference = 300 - 150 = 150 answer is a" | a = 200 + 400
b = a / 2
c = 100 + 200
d = c / 2
e = b - d
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a ) 5.7 , b ) 6.0 , c ) 6.7 , d ) 10.2 , e ) 18.0 | d | divide(add(divide(divide(64, const_3), const_3), divide(multiply(divide(64, const_3), const_2), const_3)), const_2) | the total circumference of two circles is 64 . if the first circle has a circumference that is exactly twice the circumference of the second circle , then what is the approximate sum of their two radii ? | let r = radius of smaller circle . let r = radius of larger circle therefore : 2 π r + 2 π r = 64 where 2 r = r thus : 2 π r + 4 π r = 64 6 π r = 64 r = approx 3.4 π r + 2 r π = 64 3 π r = 64 r = approx 6.8 r + r = approx 10.2 answer d | a = 64 / 3
b = a / 3
c = 64 / 3
d = c * 2
e = d / 3
f = b + e
g = f / 2
|
a ) 48 kmph , b ) 50 kmph , c ) 52 kmph , d ) 56 kmph , e ) 60 kmph | c | divide(469, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 469 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distence = 469 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 469 / 9 = 52 kmph c" | a = 6 * 3
b = a / 2
c = 469 / b
|
a ) 70 min , b ) 16 min , c ) 7 min , d ) 6 min , e ) 40 min | d | multiply(const_60, divide(subtract(62, 56), 62)) | excluding stoppages , the speed of a bus is 62 kmph and including stoppages , it is 56 kmph . for how many minutes does the bus stop per hour ? | "d 10 min due to stoppages , it covers 6 km less . time taken to cover 6 km = ( 6 / 62 x 60 ) min = 6 min" | a = 62 - 56
b = a / 62
c = const_60 * b
|
a ) 23 , b ) 25 , c ) 26 , d ) 28 , e ) 30 | d | divide(subtract(3450, 6), gcd(3560, 3450)) | the greatest number that divides 3560 and 3450 leaving remainder 4 and 6 respectively is : | "explanation : 3560 - 4 = 3556 , 3450 - 6 = 3444 highest number that can divide 3556 and 3444 is hcf of numbers hcf of 3556 and 3444 = 28 answer : d" | a = 3450 - 6
b = math.gcd(3560, 3450)
c = a / b
|
a ) 30.33 , b ) 31.33 , c ) 32.33 , d ) 33.33 , e ) 34.33 | d | multiply(divide(subtract(const_100, 75), 75), const_100) | if the cost price is 75 % of selling price then what is the profit percentage . | "selling price = rs 100 : then cost price = rs 75 : profit = rs 25 . profit = { ( 25 / 75 ) * 100 } % = 33.33 % answer is d ." | a = 100 - 75
b = a / 75
c = b * 100
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a ) $ 250 , b ) $ 300 , c ) $ 500 , d ) $ 150 , e ) $ 100 | d | multiply(500, divide(3, const_100)) | find the simple interest on $ 500 for 3 years at 10 % per annum ? | "si = ptr / 100 = 500 * 3 * 10 / 100 = $ 150 answer is d" | a = 3 / 100
b = 500 * a
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | add(add(const_4, 4), subtract(4, const_4)) | what is the sum of all possible solutions to | x - 4 | ^ 2 + | x - 4 | = 12 ? | "denote | x - 4 | as y : y ^ 2 + y = 12 - - > y = - 4 or y = 3 . discard the first solution since y = | x - 4 | , so it ' s an absolute value and thus can not be negative . y = | x - 4 | = 3 - - > x = 7 or x = 0 . the sum = 7 . answer : e ." | a = 4 + 4
b = 4 - 4
c = a + b
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a ) 10 hours , b ) 20 hours , c ) 30 hours , d ) 40 hours , e ) 12 hours | e | inverse(subtract(divide(const_1, 4), divide(const_1, add(4, 2)))) | a cistern is normally filled in 4 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ? | "if leakage / hour = 1 / x , then 1 / 4 - 1 / x = 1 / 6 , solving 1 / x = 1 / 12 so in 12 hours full cistern will be empty . answer : e" | a = 1 / 4
b = 4 + 2
c = 1 / b
d = a - c
e = 1/(d)
|
a ) s . 28500 , b ) s . 50000 , c ) s . 57000 , d ) s . 60000 , e ) s . 62000 | a | add(add(multiply(multiply(multiply(20, 15), const_2), 30), multiply(multiply(multiply(20, 5), const_2), 30)), multiply(multiply(multiply(15, 5), const_2), 30)) | in a house a hall is 20 m long , 15 m wide and 5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . 30 per square m ? | "length ( l ) = 20 m , breadth ( b ) = 15 m and height ( h ) = 5 m total area of the hall to be covered with mat = 2 ( lb + bh + hl ) = 2 ( 20 * 15 + 15 * 5 + 5 * 20 ) = 2 ( 300 + 75 + 100 ) = 2 * 475 = 950 sq . m total expenditure = 30 * 950 = rs . 28,500 a" | a = 20 * 15
b = a * 2
c = b * 30
d = 20 * 5
e = d * 2
f = e * 30
g = c + f
h = 15 * 5
i = h * 2
j = i * 30
k = g + j
|
a ) 24 , b ) 26 , c ) 30 , d ) 16 , e ) 40 | d | floor(16) | an old man distributed all the gold coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 16 times the difference between the two numbers . how many coins did the old man have ? | "let the number of coins one son got be x and the number of coins another got be y . total = x + y . x ^ 2 - y ^ 2 = 16 ( x - y ) - - > x + y = 16 . answer : d ." | a = math.floor(16)
|
a ) 4.166 , b ) 3.1 , c ) 2.1 , d ) 1.256 , e ) 3.759 | a | multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 960), 960), const_100) | a shopkeeper sells his goods at cost price but uses a faulty meter that weighs 960 grams . find the profit percent . | "explanation : ( 100 + g ) / ( 100 + x ) = true measure / faulty measure x = 0 true measure = 1000 faulty measure = 960 100 + g / 100 + 0 = 1000 / 960 100 + g = 100 / 96 * 100 g = 4.166 answer : a" | a = 4 + 1
b = 4 + 1
c = a + b
d = c * 100
e = d - 960
f = e / 960
g = f * 100
|
a ) 48 , b ) 54 , c ) 72 , d ) 60 , e ) 27 | d | divide(multiply(220, const_3), add(const_10, const_1)) | the sum of the numbers is 220 . if the first number be twice the second and third number be one - third of the first , then the second number is : | "let the second number be x . then , first number = 2 x and third number = 2 x / 3 . 2 x + x + 2 x / 3 = 220 11 x / 3 = 220 x = 60 answer : d" | a = 220 * 3
b = 10 + 1
c = a / b
|
a ) 960 , b ) 1060 , c ) 1,200 , d ) 920 , e ) none of these | a | multiply(divide(add(832, 448), const_2), add(const_1, divide(50, const_100))) | the profit earned by selling an article for 832 is equal to the loss incurred when the same article is sold for 448 . what should be the sale price of the article for making 50 per cent profit ? | "let the profit or loss be x and 832 – x = 448 + x or , x = 384 ⁄ 2 = 192 \ cost price of the article = 832 – x = 448 + x = 640 \ sp of the article = 640 × 150 ⁄ 100 = 960 answer a" | a = 832 + 448
b = a / 2
c = 50 / 100
d = 1 + c
e = b * d
|
a ) 2 / 7 , b ) 4 / 9 , c ) 16 / 25 , d ) 7 / 1 , e ) 7 / 2 | b | divide(power(2, add(2, 2)), power(6, 2)) | for what value of â € œ k â € will the equation ( 2 kx 2 + 6 kx + 2 ) = 0 have equal roots ? | "for a 2 nd degree equation ax 2 + bx _ c = 0 has equal roots the condition is b 2 - 4 ac = 0 in the given equation ( 6 k ) ^ 2 - 4 * 2 k * 2 = 0 by solving this equation we get k = 0 , k = 4 / 9 answer : b" | a = 2 + 2
b = 2 ** a
c = 6 ** 2
d = b / c
|
a ) 13 % , b ) 16 % , c ) 17 % , d ) 78 % , e ) 28 % | a | multiply(divide(subtract(1500, 1305), 1500), const_100) | the cost price of a radio is rs . 1500 and it was sold for rs . 1305 , find the loss % ? | "1500 - - - - 195 100 - - - - ? = > 13 % answer : a" | a = 1500 - 1305
b = a / 1500
c = b * 100
|
a ) $ 122 , b ) $ 150 , c ) $ 176 , d ) $ 200 , e ) $ 250 | a | multiply(divide(366, add(divide(1, 2), const_1)), divide(1, 2)) | $ 366 is divided amongst a , b and c so that a may get 1 / 2 as much as b and c together , b may get 2 / 3 as much as a and c together , then the share of a is | "a : ( b + c ) = 1 : 2 a ' s share = 366 * 1 / 3 = $ 122 answer is a" | a = 1 / 2
b = a + 1
c = 366 / b
d = 1 / 2
e = c * d
|
a ) 2 , b ) 8 , c ) 1 , d ) 3 , e ) 5 | e | divide(multiply(5, 6), 6) | a constructor estimates that 5 people can repair mr rohan ' s house in 6 days . if he uses 6 people instead of 5 , how long will they take to complete the job ? | explanation : use formula for a work members ã — days = constant 5 ã — 6 = 6 ã — a a = 5 so answer is 3 days answer : e | a = 5 * 6
b = a / 6
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a ) 41 , b ) 42 , c ) 43 , d ) 66 , e ) 45 | d | divide(factorial(12), multiply(factorial(subtract(12, const_2)), factorial(const_2))) | if 12 boys meet at a reunion and each boy shakes hands exactly once with each of the others , then what is the total number of handshakes | "n ( n - 1 ) / 2 = 12 * 11 / 2 = 66 answer : d" | a = math.factorial(12)
b = 12 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 150 % , b ) 200 % , c ) 300 % , d ) 350 % , e ) none of these | c | multiply(divide(subtract(const_100, 25), 25), const_100) | if the cost price is 25 % of selling price . then what is the profit percent . | "explanation : let the s . p = 100 then c . p . = 25 profit = 75 profit % = ( 75 / 25 ) * 100 = 300 % . answer : c" | a = 100 - 25
b = a / 25
c = b * 100
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a ) 15 , b ) 25 , c ) 30 , d ) 45 , e ) 60 | e | multiply(divide(divide(multiply(7, 16), subtract(11, 7)), 7), 15) | in a certain animal shelter , the ratio of the number of dogs to the number of cats is 15 to 7 . if 16 additional cats were to be taken in by the shelter , the ratio of the number of dogs to the number of cats would be 15 to 11 . how many dogs are in the shelter ? | "this ratio question can be solved in a couple of different ways . here ' s an algebraic approach . . . we ' re told that the ratio of the number of dogs to the number of cats is 15 : 7 . we ' re then told that 16 more cats are added to this group and the ratio becomes 15 : 11 . we ' re asked for the number of dogs . algebraically , since the number of dogs is a multiple of 15 and the number of cats is a multiple of 7 , we can write this initial relationship as . . . 15 x / 7 x when we add the 12 cats and factor in the ' ending ratio ' , we have an equation . . . . 15 x / ( 7 x + 16 ) = 15 / 11 here we have 1 variable and 1 equation , so we can solve for x . . . . ( 15 x ) ( 11 ) = ( 7 x + 16 ) ( 15 ) ( x ) ( 11 ) = ( 7 x + 16 ) ( 1 ) 11 x = 7 x + 16 4 x = 16 x = 4 with this x , we can figure out the initial number of dogs and cats . . . initial dogs = 15 x = 15 ( 4 ) = 60 final answer : e" | a = 7 * 16
b = 11 - 7
c = a / b
d = c / 7
e = d * 15
|
['a ) 64', 'b ) 68', 'c ) 72', 'd ) 76', 'e ) 80'] | d | divide(multiply(add(add(3, 7), add(2, 7)), subtract(12, 4)), const_2) | in a rectangular coordinate system , what is the area of a quadrilateral whose vertices have the coordinates ( 4 , - 3 ) , ( 4 , 7 ) , ( 12 , 2 ) , ( 12 , - 7 ) ? | by graphing the points , we can see that this figure is a trapezoid . a trapezoid is any quadrilateral that has one set of parallel sides , and the formula for the area of a trapezoid is : area = ( 1 / 2 ) × ( base 1 + base 2 ) × ( height ) , where the bases are the parallel sides . we can now determine the area of the quadrilateral : area = 1 / 2 × ( 10 + 9 ) × 8 = 76 . the answer is d . | a = 3 + 7
b = 2 + 7
c = a + b
d = 12 - 4
e = c * d
f = e / 2
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | c | divide(5, subtract(divide(40, const_60), divide(10, const_60))) | circular gears p and q start rotating at the same time at constant speeds . gear p makes 10 revolutions per minute and gear q makes 40 revolutions per minute . how many seconds after the gears start rotating will gear q have made exactly 5 more revolutions than gear p ? | "gear q makes 30 more revolutions each 60 seconds . the time to make 5 = 30 / 6 more revolutions is 60 / 6 = 10 seconds . the answer is c ." | a = 40 / const_60
b = 10 / const_60
c = a - b
d = 5 / c
|
a ) 21.2 % , b ) 21.6 % , c ) 22.0 % , d ) 22.4 % , e ) 22.8 % | d | multiply(divide(add(multiply(8, divide(25, const_100)), multiply(divide(12, const_100), multiply(8, divide(25, const_100)))), add(8, 2)), const_100) | 8 liters of a 25 percent solution of alcohol in water are mixed with 2 liters of a 12 percent alcohol in water solution . what is the percentage of alcohol in the new solution ? | the total amount of alcohol is 0.25 ( 8 ) + 0.12 ( 2 ) = 2.24 liters . the percentage is 2.24 / 10 = 22.4 / 100 which is 22.4 % the answer is d . | a = 25 / 100
b = 8 * a
c = 12 / 100
d = 25 / 100
e = 8 * d
f = c * e
g = b + f
h = 8 + 2
i = g / h
j = i * 100
|
a ) 42 , b ) 270 , c ) 140 , d ) 165 , e ) 315 | b | multiply(multiply(10, 3), 6) | a certain university will select 1 of 6 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "1 of 6 will be chosen for the math 2 of 10 will be chosen for the computer none of the 3 chosen people can be in more than one departments . we can choose any of the 6 candidates for the math dep . , which gives as 6 selections . we can choose 2 of the 10 candidates for the computer dep . , which gives us 2 selections and 8 rejections . so , the way to find how many different selections of 2 candidates we can have for the computer dep . , we do : 10 ! / 2 ! * 8 ! = ( 9 * 10 ) / 2 = 90 / 2 = 45 . we are multiplying our individual selections : 6 * 45 = 270 in the bolded part , we do n ' t have to multiply all of the numbers , as those in 8 ! are included in 10 ! , so we simplify instead . ans b" | a = 10 * 3
b = a * 6
|
a ) 5 : 4 , b ) 2 : 3 , c ) 1 : 4 , d ) 3 : 7 , e ) 2 : 5 | a | divide(divide(20, const_100), divide(1, 4)) | in a school 20 % of the boys are same in number as 1 / 4 th of the girls . what is the ratio of boys to the girls in the school ? | "20 % of b = 1 / 4 g 20 b / 100 = g / 4 b = 5 g / 4 b / g = 5 / 4 b : g = 5 : 4 answer is a" | a = 20 / 100
b = 1 / 4
c = a / b
|
a ) 11 , b ) 14 , c ) 18 , d ) 20 , e ) 29 | a | subtract(5264, multiply(floor(divide(5264, 17)), 17)) | find the least number must be subtracted from 5264 so that remaining number is divisible by 17 . | when dividing 5264 by 17 , we get the remainder 11 , so 11 should be subtracted . the answer is a . | a = 5264 / 17
b = math.floor(a)
c = b * 17
d = 5264 - c
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a ) $ 150 , b ) $ 245 , c ) $ 365 , d ) $ 315 , e ) $ 730 | d | subtract(divide(subtract(multiply(500, 7), add(add(add(406, 413), add(436, 420)), 395)), const_2), 400) | a salesman ' s income consists of a commission and a base salary of $ 400 per week . over the past 5 weeks , his weekly income totals have been $ 406 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) commission be per week over the next two weeks so that his average weekly income is $ 500 over the 7 - week period ? | "total weekly income over 5 weeks = $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070 for avg weekly income to be $ 500 over 7 weeks , we need total weekly income over 7 weeks = $ 3500 now , $ 3500 - $ 2070 = $ 1430 from this , we subtract base salary for 2 weeks i . e $ 350 * 2 = $ 700 therefore , commission = $ 1430 - $ 800 = $ 630 for 2 weeks avg weekly commission = $ 315 answer d" | a = 500 * 7
b = 406 + 413
c = 436 + 420
d = b + c
e = d + 395
f = a - e
g = f / 2
h = g - 400
|
a ) 125 miles , b ) 225 miles , c ) 260 miles , d ) 425 miles , e ) 525 miles | c | multiply(65, 4) | a car travels at a speed of 65 miles per hour . how far will it travel in 4 hours ? | during each hour , the car travels 65 miles . for 4 hours it will travel 65 + 65 + 65 + 65 = 4 × 65 = 260 miles correct answer is c ) 260 miles | a = 65 * 4
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a ) 111.12 , b ) 111.67 , c ) 111.64 , d ) 111.11 , e ) 176.47 | e | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 300)), subtract(multiply(const_100, const_10), 150))) | a can give b 150 meters start and c 300 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "a runs 1000 m while b runs 850 m and c runs 700 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 700 ) / 850 = 823.53 m . b can give c = 1000 - 823.53 = 176.47 m . answer : e" | a = 100 * 10
b = 100 * 10
c = 100 * 10
d = c - 300
e = b * d
f = 100 * 10
g = f - 150
h = e / g
i = a - h
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | subtract(divide(subtract(100, 50), subtract(55, 50)), const_1) | for the past n days , the average ( arithmetic mean ) daily production at a company was 50 units . if today ' s production of 100 units raises the average to 55 units per day , what is the value of n ? | "the daily production was raised by 5 units for n days , which has a weighting of 5 n . 5 n = 100 - 55 = 45 n = 9 the answer is d ." | a = 100 - 50
b = 55 - 50
c = a / b
d = c - 1
|
a ) 21 , b ) 20 , c ) 2 , d ) 30 , e ) 45 | c | subtract(divide(multiply(add(divide(60, const_2), 5), 5), 5), 5) | 5 is added to a certain number , the sum is multiplied by 5 , the product is divided by 5 and 5 is subtracted from the quotient . the remainder left is half of 60 . what is the number ? | "let number is x . when 5 added to it , = ( x + 5 ) 5 multiplied to sum , = 5 * ( x + 5 ) now , = [ { 5 * ( x + 5 ) } / 5 ] and , = [ { 5 * ( x + 5 ) } / 5 ] - 5 according to question , [ { 5 * ( x + 5 ) } / 5 ] - 5 = half of 60 [ ( 5 x + 25 ) / 5 ) = 30 + 5 5 x + 25 = 35 * 5 x + 5 = 7 * 1 x + 5 = 7 x = 7 - 5 x = 2 so , required number is : 2 . answer : c" | a = 60 / 2
b = a + 5
c = b * 5
d = c / 5
e = d - 5
|
a ) one , b ) 4 , c ) 5 , d ) 8 , e ) 9 | c | divide(7, 7) | how many points ( x , y ) lie on the line segment between ( 32 , 12 2 / 3 ) and ( 7 , 17 2 / 3 ) such that x and y are both integers ? | "slope = ( 17 2 / 3 - 12 2 / 3 ) / ( 7 - 32 ) = - 1 / 5 y = mx + b = > 12 2 / 3 = - 22 / 3 + b = > b = 19 y = - x / 5 + 19 only integer values work , and the only multiples of 5 between 7 and 32 for x values are 10 , 15 , 20 , 25 and 30 , thus 5 points . c" | a = 7 / 7
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a ) 50 m , b ) 200 m , c ) 250 m , d ) 270 m , e ) 300 m | a | subtract(multiply(15, multiply(72, const_0_2778)), 250) | a train 250 m long running at 72 kmph crosses a platform in 15 sec . what is the length of the platform ? | "d = 72 * 5 / 18 = 15 = 300 â € “ 250 = 50 m answer : a" | a = 72 * const_0_2778
b = 15 * a
c = b - 250
|
a ) 1 / 10 , b ) 1 / 5 , c ) 3 / 10 , d ) 2 / 5 , e ) 1 / 2 | c | divide(choose(2, 2), choose(add(2, 3), 2)) | a bag holds 2 red marbles and 3 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be green ? | "given : 2 r and 3 g marbles required : probability that 2 marbles removed without replacement are both red initially we have to pick one red from a total of 2 red and 3 green marbles after one red has been picked , we need to pick 1 green from a total of 2 green and 2 red marbles . p ( both green ) = ( 3 / 5 ) * ( 2 / 4 ) = 3 / 10 option c" | a = math.comb(2, 2)
b = 2 + 3
c = math.comb(b, 2)
d = a / c
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a ) 28 % , b ) 30 % , c ) 32 % , d ) 36 % , e ) 44 % | e | subtract(const_100, multiply(multiply(subtract(const_1, divide(30, const_100)), subtract(const_1, divide(20, const_100))), const_100)) | a baseball card decreased in value 20 % in its first year and 30 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | "consider the initial value of the baseball card as $ 100 after first year price = 100 * 0.8 = 80 after second year price = 80 * 0.7 = 56 final decrease = [ ( 100 - 56 ) / 100 ] * 100 = 44 % correct answer - e" | a = 30 / 100
b = 1 - a
c = 20 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 39 , b ) 140 , c ) 42 , d ) 40.5 , e ) 41.5 | b | add(subtract(300, multiply(17, 10)), 10) | a batsman makes a score of 300 runs in the 17 th inning and thus increases his averages by 10 . what is his average after 17 th inning ? | "let the average after 16 th inning = x then total run after 16 th inning = 16 x then total run after 17 th inning = 16 x + 300 then average run after 17 th inning = ( 16 x + 300 ) / 17 ( 16 x + 300 ) / 17 = x + 10 = > 16 x + 300 = 17 x + 170 x = 130 average after 17 th inning = 130 + 10 = 140 answer : b" | a = 17 * 10
b = 300 - a
c = b + 10
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a ) 0.46 , b ) 0.56 , c ) 0.64 , d ) 0.65 , e ) none of them | a | multiply(0.6, divide(0.6, 0.6)) | solve : 0.6 * 0.6 + 0.6 * 0.6 | "= 0.6 * 0.6 + ( 0.6 * 6 ) = 0.36 + 0.1 = 0.46 answer is a ." | a = 0 / 6
b = 0 * 6
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a ) 74 , b ) 78 , c ) 86 , d ) 126 , e ) 98 | d | add(multiply(add(multiply(6, const_3), 12), divide(add(multiply(6, const_3), 12), 5)), 6) | in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 12 to the thrice of the remainder . the dividend is | "divisor = ( 6 * 3 ) + 12 = 30 5 * quotient = 30 quotient = 6 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 6 ) + 6 = 126 . d )" | a = 6 * 3
b = a + 12
c = 6 * 3
d = c + 12
e = d / 5
f = b * e
g = f + 6
|
a ) 15 days , b ) 16 days , c ) 19.4 days , d ) 17.2 days , e ) 18 days | c | divide(const_1, add(divide(const_1, 30), divide(const_1, 55))) | a can do a piece of work 30 days . b can do work in 55 days . in how many days they will complete the work together ? | "lcm = 330 , ratio = 30 : 45 = 6 : 11 no of days = 330 / ( 6 + 11 ) = 330 / 17 = 19.4 days answer : c" | a = 1 / 30
b = 1 / 55
c = a + b
d = 1 / c
|
['a ) 40 m 2', 'b ) 44 m 2', 'c ) 48 m 2', 'd ) 37.7 m 2', 'e ) none of these'] | d | multiply(5, multiply(multiply(multiply(const_2, divide(22, 7)), divide(1.2, const_2)), 2)) | the diameter of a garden roller is 1.2 m and it is 2 m long . how much area will it cover in 5 revolutions ? ( use ï € = 22 â „ 7 ) | required area covered in 5 revolutions = 5 ã — 2 ï € rh = 5 ã — 2 ã — 22 â „ 7 ã — 0.6 ã — 2 = 37.7 m 2 answer d | a = 22 / 7
b = 2 * a
c = 1 / 2
d = b * c
e = d * 2
f = 5 * e
|
a ) s 802 , b ) s 208 , c ) s 960 , d ) s 690 , e ) s 906 | c | multiply(3, divide(2080, add(add(3, const_2.0), const_3))) | rs . 2080 is divided so that 3 times the first share , thrice the 2 nd share and four times the third share amount to the same . what is the value of the second share ? | "a + b + c = 2080 3 a = 2 b = 4 c = x a : b : c = 1 / 3 : 1 / 2 : 1 / 4 = 4 : 6 : 3 6 / 13 * 2080 = rs 960 answer : c" | a = 3 + 2
b = a + 3
c = 2080 / b
d = 3 * c
|
a ) 3 / 2 , b ) 3 / 9 , c ) 3 / 6 , d ) 3 / 1 , e ) 3 / 5 | a | divide(subtract(67, 73), subtract(73, 17)) | two trains running in opposite directions cross a man standing on the platform in 67 seconds and 17 seconds respectively and they cross each other in 73 seconds . the ratio of their speeds is : | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 23 = = > 27 x + 17 y = 23 x + 23 y = = > 4 x = 6 y = = > x / y = 3 / 2 . answer : a" | a = 67 - 73
b = 73 - 17
c = a / b
|
a ) 357 , b ) 834 , c ) 500 , d ) 512 , e ) 216 | d | power(subtract(10, const_2), const_3) | you make a cube that is made of 10 x 10 x 10 smaller cubes which sums up to a total of 1000 smaller cubes . now , if someone takes one layer of cubes , how many will remain ? | d 512 the cube that remains will be 8 x 8 x 8 . this will make a total of 512 . this is because , when a layer is taken , it will be taken from all sides of the cube which will reduce the dimension be two and not one . | a = 10 - 2
b = a ** 3
|
a ) a ) 7 , b ) b ) 12 , c ) c ) 15 , d ) d ) 18 , e ) e ) 20 | a | divide(subtract(30, power(4, 2)), 2) | if a - b = 4 and a 2 + b 2 = 30 , find the value of ab . | "explanation : 2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 30 - 16 = 14 ab = 7 answer : a" | a = 4 ** 2
b = 30 - a
c = b / 2
|
a ) 160 , b ) 175 , c ) 180 , d ) 185 , e ) 195 | b | add(multiply(multiply(8, 5), 4), divide(subtract(432, multiply(multiply(multiply(8, 2.4), 5), 4)), 3.2)) | in a regular week , there are 5 working days and for each day , the working hours are 8 . a man gets rs . 2.40 per hour for regular work and rs . 3.20 per hours for overtime . if he earns rs . 432 in 4 weeks , then how many hours does he work for ? a . 160 b . 175 c . 180 d . 195 explanation : | suppose the man works overtime for x hours . now , working hours in 4 weeks = ( 5 x 8 x 4 ) = 160 . 160 x 2.40 + x x 3.20 = 432 3.20 x = 432 - 384 = 48 x = 15 . hence , total hours of work = ( 160 + 15 ) = 175 . answer : option b | a = 8 * 5
b = a * 4
c = 8 * 2
d = c * 5
e = d * 4
f = 432 - e
g = f / 3
h = b + g
|
a ) 50 , b ) 70 , c ) 100 , d ) 130 , e ) 150 | b | subtract(200, subtract(add(90, 60), 20)) | of the 200 stamps in a collection , 90 are foreign and 60 are more than 10 years old . if 20 stamps are both foreign and more than 10 years old , how many stamps are neither foreign nor more than 10 years old ? | "20 stamps are both foreign and more than 10 years old . 70 stamps are foreign only . 40 stamps are 10 years old only . the number of remaining stamps is 200 - ( 20 + 70 + 40 ) = 70 the answer is b ." | a = 90 + 60
b = a - 20
c = 200 - b
|
['a ) 7 cm', 'b ) 8.5 cm', 'c ) 9 cm', 'd ) 9.5 cm', 'e ) none'] | c | power(divide(multiply(volume_cylinder(12, 6.75), const_3), multiply(const_4, const_pi)), divide(const_1, const_3)) | a cylindrical cube of radius 12 cm contains water upto a depth of 20 cm . a spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm . the radius of the ball is : | sol . let the radius of the ball be r cm . volume of ball = volume of water displaced by it . ∴ 4 / 3 ∏ r ³ = ∏ * 12 * 12 * 6.75 ⇒ r ³ = 9 * 9 * 9 ⇒ r = 9 cm . answer c | a = volume_cylinder * (
b = a / 3
c = 4 * math.pi
d = b ** c
|
a ) 42 , b ) 43 , c ) 70 , d ) 45 , e ) 46 | c | add(subtract(92, multiply(12, 2)), 2) | a batsman in his 12 th innings makes a score of 92 and thereby increases his average by 2 runs . what is his average after the 12 th innings if he had never been ‘ not out ’ ? | "let ‘ x ’ be the average score after 12 th innings ⇒ 12 x = 11 × ( x – 2 ) + 92 ∴ x = 70 answer c" | a = 12 * 2
b = 92 - a
c = b + 2
|
a ) 12 , b ) 14 , c ) 15 , d ) 20 , e ) 21 | c | subtract(36, 8) | a snail , climbing a 36 feet high wall , climbs up 8 feet on the first day but slides down 4 feet on the second . it climbs 8 feet on the third day and slides down again 4 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ? | "total transaction in two days = 8 - 4 = 4 feet in 15 days it will climb 36 feet thus reaching the top therefore , total no of days required = 15 c" | a = 36 - 8
|
a ) 21 , b ) 42 , c ) 24 , d ) 40 , e ) 45 | b | divide(16800, multiply(multiply(const_10, const_2), multiply(const_10, const_2))) | what is the least number by which 16800 must be divided to get a number perfect square | 16,800 / 42 = 400 400 = ( 20 ) ^ 2 ans - 42 answer : b | a = 10 * 2
b = 10 * 2
c = a * b
d = 16800 / c
|
a ) 70 , b ) 45 , c ) 50 , d ) 46 , e ) 30 | e | multiply(divide(add(multiply(divide(30, const_100), 300), multiply(subtract(const_1, divide(70, const_100)), 240)), add(300, 240)), const_100) | addison high school ’ s senior class has 300 boys and 240 girls . if 30 % of the boys did not attend the college then 70 % of the total class attended college , what percentage of the girls class did not attend college ? | "% of boy attending the college = 100 - 30 = 70 % number of boys attending the college : 0.70 * 300 = 210 total people = 300 + 240 = 540 total number of students attending the college : 0.70 * 540 = 378 number of girls attending the college = 378 - 210 = 168 number of girls not attending the college = 240 - 168 = 72 % of girls not attending = 72 / 240 = 30 % . e is the correct answer" | a = 30 / 100
b = a * 300
c = 70 / 100
d = 1 - c
e = d * 240
f = b + e
g = 300 + 240
h = f / g
i = h * 100
|
a ) − 220 , b ) − 100 , c ) 100 , d ) 135 , e ) 80 | e | subtract(multiply(const_60.0, const_2), multiply(140, const_2)) | if the average ( arithmetic mean ) of a and b is 140 , and the average of b and c is 100 , what is the value of a − c ? | "a + b = 280 b + c = 200 a - c = 80 . imo option e ." | a = const_60 * 0
b = 140 * 2
c = a - b
|
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