options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 3200 , b ) 4000 , c ) 3250 , d ) 3825 , e ) 4096 | e | multiply(multiply(48000, subtract(multiply(const_3, 4), 4)), divide(14080, add(add(multiply(36000, multiply(const_3, 4)), multiply(42000, multiply(const_3, 4))), multiply(48000, subtract(multiply(const_3, 4), 4))))) | x and y started a business by investing rs . 36000 and rs . 42000 respectively after 4 months z joined in the business with an investment of rs . 48000 , then find share of z in the profit of rs . 14080 ? | "ratio of investment , as investments is for different time . investment x number of units of time . ratio of investments x : y : z = 36000 : 42000 : 48000 = > 6 : 7 : 8 . x = 6 x 12 months = 72 , y = 7 x 12 = 84 , z = 8 x 8 = 64 = > 18 : 21 : 16 . ratio of investments = > x : y : z = 18 : 21 : 16 . investment ratio = profit sharing ratio . z = 14080 ã — 16 / 55 = rs . 4096 . share of z in the profit is rs . 4096 . option e" | a = 3 * 4
b = a - 4
c = 48000 * b
d = 3 * 4
e = 36000 * d
f = 3 * 4
g = 42000 * f
h = e + g
i = 3 * 4
j = i - 4
k = 48000 * j
l = h + k
m = 14080 / l
n = c * m
|
a ) a ) 15 , b ) b ) 20 , c ) c ) 30 , d ) d ) 40 , e ) e ) 5 | e | divide(subtract(subtract(160, 80), 60), const_4) | 6 ) a marketing firm determined that , of 160 households surveyed , 80 used neither brand a nor brand b soap . 60 used only brand a soap and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 household surveyed used both brands of soap ? | "80 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 120 = > 4 x = 20 = > x = 5 answer - e" | a = 160 - 80
b = a - 60
c = b / 4
|
a ) 15 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | a | multiply(divide(54, add(add(3, 7), 8)), subtract(8, 3)) | every disk in a bag is either blue , yellow or green . the ratio of blue disks to yellow disks to green disks in this bag is 3 : 7 : 8 . if the total number of disks in the bag is 54 , how many more green disks than blue disks are in the bag ? | "let b : y : g = 3 x : 7 x : 8 x . 3 x + 7 x + 8 x = 18 x = 54 - - > x = 3 . g - b = 8 x - 3 x = 5 x = 15 . the answer is a ." | a = 3 + 7
b = a + 8
c = 54 / b
d = 8 - 3
e = c * d
|
a ) 1 / 20 , b ) 3 / 50 , c ) 1 / 75 , d ) 1 / 25 , e ) none of these | b | divide(circle_area(divide(6, const_2)), const_2) | what will be the fraction of 6 % | "explanation : 6 * 1 / 100 = 3 / 50 . option b" | a = 6 / 2
b = circle_area / (
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a ) 155 , b ) 160 , c ) 175 , d ) 190 , e ) 195 | a | subtract(subtract(250, 90), divide(subtract(subtract(add(add(116, 127), 107), 90), 250), const_2)) | among 250 viewers interviewed who watch at least one of the 3 tv channels namely a , bc . 116 watch a , 127 watch c , while 107 watch b . if 90 watch exactly two channels . how many watch exactly one channel ? | 250 = n ( exactly 1 channel ) + n ( exactly 2 channels ) + n ( exactly 3 channels ) 250 = n ( exactly 1 channel ) + 90 + n ( exactly 3 channels ) let ' s find the value of n ( exactly 3 channels ) = x 250 = n ( a ) + n ( b ) + n ( c ) - n ( a and b ) - n ( b and c ) - n ( c and a ) + n ( a and b and c ) note that each of n ( a and b ) is the sum of ' number of people watching exactly two channels a and b ' and ' number of people watching all three channels ' . 250 = 116 + 127 + 107 - n ( exactly 2 channels ) - 3 x + x 250 = 116 + 127 + 107 - 90 - 2 x x = 25 250 = n ( exactly 1 channel ) + 90 + 25 n ( exactly 1 channel ) = 155 answer ( a ) | a = 250 - 90
b = 116 + 127
c = b + 107
d = c - 90
e = d - 250
f = e / 2
g = a - f
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a ) 1 / 30 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | e | divide(subtract(divide(30, const_100), divide(50, const_100)), subtract(divide(25, const_100), divide(50, const_100))) | some of 50 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 30 % . what fraction of the original paint was replaced ? | "let total paint = 1 let amount replaced = x 50 ( 1 - x ) + 25 x = 30 x = 4 / 5 answer : e" | a = 30 / 100
b = 50 / 100
c = a - b
d = 25 / 100
e = 50 / 100
f = d - e
g = c / f
|
a ) 4 cm , b ) 6 cm , c ) 8 cm , d ) 10 cm , e ) 12 cm | c | multiply(divide(divide(divide(divide(multiply(divide(volume_cylinder(divide(8, const_2), 48), const_pi), const_3), const_4), 9), const_4), const_4), const_2) | 9 spheres of the same size are made from melting a solid cylinder of 8 cm diameter and 48 cm height . what is the diameter of each sphere ? | "volume of cylinder = pi * r ^ 2 * h volume of a sphere = 4 * pi * r ^ 3 / 3 9 * 4 * pi * r ^ 3 / 3 = pi * r ^ 2 * h r ^ 3 = r ^ 2 * h / 12 = 64 cm ^ 3 r = 4 cm d = 8 cm the answer is c ." | a = 8 / 2
b = volume_cylinder / (
c = b * math.pi
d = c / 3
e = d / 4
f = e / 9
g = f / 4
h = g * 4
|
a ) 9 , b ) 10 , c ) 12 , d ) 14 , e ) 18 | c | multiply(multiply(6, 4), divide(5, 10)) | working simultaneously and independently at an identical constant rate , 6 machines of a certain type can produce a total of x units of product p in 5 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 4 x units of product p in 10 days ? | "the rate of 6 machines is rate = job / time = x / 5 units per day - - > the rate of 1 machine 1 / 6 * ( x / 5 ) = x / 30 units per day ; now , again as { time } * { combined rate } = { job done } then 10 * ( m * x / 30 ) = 4 x - - > m = 12 . answer : c ." | a = 6 * 4
b = 5 / 10
c = a * b
|
a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 9 | d | divide(log(220), log(add(const_4, const_1))) | for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 220 true ? | "answer d ( a + b ) ( a - b ) = 220 6 cases for ( a + b ) , ( a - b ) 220 , 1 110 , 2 55 , 4 44,5 22 , 10 20,11 answer d" | a = math.log(220)
b = 4 + 1
c = math.log(b)
d = a / c
|
a ) 10 , b ) 20 , c ) 25 , d ) 35 , e ) 15 | a | divide(subtract(divide(100, 4), divide(75, 15)), const_2) | a boat goes 100 km downstream in 4 hours , and 75 km upstream in 15 hours . the speed of the stream is ? | "100 - - - 10 ds = 25 ? - - - - 1 75 - - - - 15 us = 5 ? - - - - - 1 s = ( 25 - 5 ) / 2 = 10 kmph . answer : a" | a = 100 / 4
b = 75 / 15
c = a - b
d = c / 2
|
a ) 0 , b ) 5 , c ) 2 , d ) 14 , e ) 16 | a | subtract(6, 8) | what is 10 - 8 + 6 - 4 + . . . + ( - 10 ) ? | "the expression considers all even numbers between 10 and - 10 with alternate addition and subtraction of the numbers . the numbers to be used are : 10 , 8 , 6 , 4 , 2 , 0 , - 2 , - 4 , - 6 , - 8 , and - 10 now , the first term is positive and the next term is subtracted . so , the required expression becomes , 10 - 8 + 6 - 4 + 2 - 0 + ( - 2 ) - ( - 4 ) + ( - 6 ) - ( - 8 ) + ( - 10 ) = 10 - 8 + 6 - 4 + 2 - 0 - 2 + 4 - 6 + 8 - 10 = 30 - 30 = 0 hence the correct answer choice is a ." | a = 6 - 8
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a ) $ 35,000 , b ) $ 45,000 , c ) $ 55,000 , d ) $ 65,000 , e ) $ 75,000 | a | divide(add(multiply(multiply(multiply(const_3, const_3), const_10), multiply(const_100, const_10)), multiply(multiply(const_3, const_10), multiply(const_100, const_10))), add(75, 15)) | a company has 15 managers and 75 associates . the 15 managers have an average salary of $ 60,000 . the 75 associates have an average salary of $ 30,000 . what is the average salary for the company ? | "another method is to get ratios say 30000 = a and we know the # of people are in 1 : 5 ratio average = ( 2 a * 1 + a * 5 ) / 6 = 7 a / 6 = 35000 answer is a . $ 35,000" | a = 3 * 3
b = a * 10
c = 100 * 10
d = b * c
e = 3 * 10
f = 100 * 10
g = e * f
h = d + g
i = 75 + 15
j = h / i
|
a ) 35 , b ) 15 , c ) 29 , d ) 31 , e ) 29 | d | add(multiply(divide(multiply(divide(multiply(4.8, const_1000), multiply(multiply(const_2, const_2), 10)), const_2), multiply(const_4, const_4)), const_2), const_1) | along a road lie an odd number of stones placed at intervals of 10 m . these stones have to be assembled around the middle stone . a person can carry only one stone at a time . a man carried out the job starting with the stone in the middle , carrying stones in succession , thereby covering a distance of 4.8 km . then the number of stones is | explanation : if he travels 10 m for the first stone he has to travel that dist back as well . . . so he travels 2 * 10 m dist for the first stone and so on . = > 2400 = 2 * ( 10 + 20 + 30 . . . . 10 n ) using formula for ap = > 2400 = 2 * n / 2 * ( 2 * 10 + ( n - 1 ) * 10 ) . = > 2400 = 10 n 2 + 10 n . on solving this , we get n = 15 as the positive root , so total numer of stones on both sides = 30 . and including the one which he was at = 31 . answer : d | a = 4 * 8
b = 2 * 2
c = b * 10
d = a / c
e = d * 2
f = 4 * 4
g = e / f
h = g * 2
i = h + 1
|
a ) a ) 31 , b ) b ) 10 , c ) c ) 15 , d ) d ) 26 , e ) e ) 17 | b | subtract(850, divide(1500, 8)) | 850 - ( 1500 ÷ ? ) × 8 / 15 = 770 | "explanation : = > 850 - ( 1500 ÷ ? ) × 8 / 15 = 770 = > 850 - 770 = ( 1500 ÷ ? ) × 8 / 15 = > 80 = ( 1500 ÷ ? ) × 8 / 15 = > 80 × 15 / 8 = ( 1500 ÷ ? ) = 150 = > 1 / ? = 150 / 1500 = 1 / 10 = > ? = 10 answer : option b" | a = 1500 / 8
b = 850 - a
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a ) 210 : 283 , b ) 312 : 433 , c ) 178 : 213 , d ) 197 : 283 , e ) 145 : 256 | d | divide(add(multiply(2, 3), multiply(5, 5)), add(multiply(3, 3), multiply(7, 5))) | the contents of two vessels containing copper and tin in the ratio 2 : 3 and 5 : 7 are mixed in the ratio 3 : 5 . the resulting mixture will have copper and tin in the ratio ? | the ratio of copper and tin the new vessel = ( 2 / 5 * 3 / 8 + 5 / 12 * 5 / 8 ) : ( 3 / 5 * 3 / 8 + 7 / 12 * 5 / 8 ) = 197 / 480 : 283 / 480 = 197 : 283 answer is d | a = 2 * 3
b = 5 * 5
c = a + b
d = 3 * 3
e = 7 * 5
f = d + e
g = c / f
|
a ) 6 days , b ) 8 days , c ) 12 days , d ) 24 days , e ) 25 days | d | add(inverse(subtract(divide(const_1, 8), divide(const_1, 12))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 8 days . a alone can do it in 12 days . what time b will take to do the work alone ? | "explanation : a and b 1 day ' s work = 1 / 8 a alone can do 1 day ' s work = 1 / 12 what time b will take to do the work alone ? b = ( a + b ) - a = ( 1 / 8 ) - ( 1 / 12 ) = 24 days answer : option d" | a = 1 / 8
b = 1 / 12
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
|
a ) 389 , b ) 390 , c ) 388 , d ) 391 , e ) 392 | d | subtract(add(multiply(9, const_2), 5), add(2, 15)) | x and y are positive integers . when x is divided by 15 , the remainder is 2 , and when x is divided by 10 , the remainder is 7 . when y is divided by 9 , the remainder is 5 , and when y is divided by 7 , the remainder is 2 . what is the least possible value of x * y ? | "when x is divided by 15 , the remainder is 2 : so , the possible values of x are : 2 , 17 , 32 , 47 , etc . when x is divided by 10 , the remainder is 7 : so , the possible values of x are : 7,17 . . . stop . since both lists include 17 , the smallest possible value of x is 17 . when y is divided by 9 , the remainder is 5 : so , the possible values of y are : 5 , 14 , 23,32 etc . when y is divided by 7 , the remainder is 2 : so , the possible values of y are : 2 , 9 , 16 , 23 . . . stop . since both lists include 23 , the smallest possible value of y is 23 since the smallest possible values of x and y are 17 and 23 respectively , the smallest possible value of x * y is 391 . so , d is the correct answer to the original question ." | a = 9 * 2
b = a + 5
c = 2 + 15
d = b - c
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a ) 25 % , b ) 50 % , c ) 75 % , d ) 100 % , e ) 200 % | d | divide(multiply(const_2, const_100), const_2) | what is the minimum percentage increase in the mean of set s { - 4 , - 1 , 0 , 6 , 9 } if its two smallest elements are replaced with two different primes ? | mean of s is ( - 4 - 1 + 0 + 6 + 9 ) / 5 = 2 ; in order the increase to be minimal we should replace two smallest elements of s , which are - 4 and - 1 , with two smallest primes , which are 2 and 3 . hence our new set will be { 2 , 3 , 0 , 6 , 9 } - - > new mean is ( 2 + 3 + 0 + 6 + 9 ) / 5 = 4 . percent increase = ( 4 - 2 ) / 2 * 100 = 100 % . answer : d . | a = 2 * 100
b = a / 2
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a ) 5 , b ) 8 , c ) 24 , d ) 15 , e ) 20 | c | divide(4, subtract(divide(20, multiply(add(10, const_1), 20)), divide(10, multiply(add(10, const_1), 20)))) | marla starts running around a circular track at the same time nick starts walking around the same circular track . marla completes 20 laps around the track per hour and nick completes 10 laps around the track per hour . how many minutes after marla and nick begin moving will marla have completed 4 more laps around the track than nick ? | "maria ' s rate - 20 laps per hour - - > 20 / 60 laps / min nick ' s rate - 10 laps per hour - - > 10 / 60 laps / min lets set equations : 20 / 60 * t = 4 ( since maria had to run 4 laps before nick would start ) 10 / 60 * t = 0 ( hick has just started and has n ' t run any lap yet ) ( 20 / 60 - 10 / 60 ) * t = 4 - 0 ( since nick was chasing maria ) t = 24 min needed maria to run 4 laps answer : c" | a = 10 + 1
b = a * 20
c = 20 / b
d = 10 + 1
e = d * 20
f = 10 / e
g = c - f
h = 4 / g
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a ) 129,500 , b ) 129,600 , c ) 129,700 , d ) 129,800 , e ) 129,900 | b | multiply(multiply(subtract(6, 3), const_3600), const_12) | in a renowned city , the average birth rate is 6 people every two seconds and the death rate is 3 people every two seconds . estimate the size of the population net increase that occurs in one day . | "this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 6 - 3 = 3 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 3 = 129,600 . hence b ." | a = 6 - 3
b = a * 3600
c = b * 12
|
a ) 21 , b ) 22 , c ) 17 , d ) 26 , e ) 25 | b | divide(subtract(multiply(const_2, multiply(const_2, const_12)), const_4), const_2) | how many times in a day , are the hands of a clock in straight line but opposite in direction ? | b 22 the hands of a clock point in opposite directions ( in the same straight line ) 11 times in every 12 hours . ( because between 5 and 7 they point in opposite directions at 6 o ' clcok only ) . so , in a day , the hands point in the opposite directions 22 times . | a = 2 * 12
b = 2 * a
c = b - 4
d = c / 2
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a ) 9.5 , b ) 10.5 , c ) 11.5 , d ) 12 , e ) 15 | c | add(divide(multiply(11, 36), const_100), divide(multiply(12, 64), const_100)) | add 11 % of 36 and 12 % of 64 . | "11 % of 36 + 12 % of 64 36 * 11 / 100 + 64 * 12 / 100 3.9 + 7.6 = 11.5 answer c" | a = 11 * 36
b = a / 100
c = 12 * 64
d = c / 100
e = b + d
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a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 50), 55)), divide(multiply(const_100, 50), 55))) | if the cost price of 55 articles is equal to the selling price of 50 articles , then what is the percent profit ? | "let x be the cost price of one article . let y be the selling price of one article . 50 y = 55 x y = 1.1 x the answer is b ." | a = 100 * 50
b = a / 55
c = 100 - b
d = 100 * 50
e = d / 55
f = c / e
g = 100 * f
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a ) 39 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | a | add(divide(360, 10), subtract(subtract(8, 4), const_1)) | in a certain quiz that consists of 8 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 360 points , how many points is the third question worth ? | x x + 4 x + 8 x + 12 x + 16 x + 20 x + 24 x + 28 8 x + 112 = 360 8 x = 248 x = 31 3 rd question = 31 + 8 = 31 + 8 = 39 answer a | a = 360 / 10
b = 8 - 4
c = b - 1
d = a + c
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['a ) 1', 'b ) 2', 'c ) 3', 'd ) 4', 'e ) 5'] | b | subtract(const_2, 0) | line l passes through the points ( - 2,0 ) and ( 0 , a ) . line ll passes through the points ( 4,0 ) and ( 6,2 ) . what value of a makes the two lines parallel ? | find slopes first slope of l = ( a - 0 ) / ( 0 - ( - 2 ) ) = a / 2 slope of ll = ( 2 - 0 ) / ( 6 - 4 ) = 1 for l and ll to be parallel , their slopes must be equal a / 2 = 1 , a = 2 correct answer b | a = 2 - 0
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a ) 20 cs , b ) cs / 2 , c ) 72 cs , d ) ( 2 cs ) / 12 , e ) ( 24 c ) / s | c | multiply(6, const_12) | a certain school implemented a reading program for its students , with the goal of getting each student to read 6 books per month year - round . if the school has c classes made up of s students in each class , how many books will the entire student body read in one year ? | "ans : c solution : simple multiplication s students , c classes , 6 books / month = 72 books a year total number of books = 72 cs" | a = 6 * 12
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a ) 135 , b ) 105 , c ) 95 , d ) 75 , e ) 45 | d | subtract(divide(multiply(divide(multiply(divide(multiply(40, 45), const_60), multiply(40, const_2)), const_100), const_60), 12), 45) | debony normally drives to work in 45 minutes at an average speed of 40 miles per hour . this week , however , she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20 % . if debony averages between 12 and 16 miles per hour when biking , how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives ? | debony normally drives to work in 45 minutes at an average speed of 40 miles per hour . use formula d = rt car : t 1 : 45 min r 1 : 40 mph d 1 : [ ( 40 * 45 ) / 60 ] = 30 miles bike : t 1 : ? r 2 : 12 - 16 mph d 2 : 08 * d 1 = 24 miles t 1 : [ ( 24 * 60 ) / 12 ] = 120 min ( only 12 mph speed yields an answer given in the choices ) therefore , deb has to leave 120 min - 45 min = 75 min early answer : d | a = 40 * 45
b = a / const_60
c = 40 * 2
d = b * c
e = d / 100
f = e * const_60
g = f / 12
h = g - 45
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a ) 25 , b ) 40 , c ) 90 , d ) 140 , e ) it can not be determined from the information given . | b | subtract(multiply(60, const_2), multiply(40, const_2)) | if the average ( arithmetic mean ) of a and b is 40 and the average of b and c is 60 , what is the value of c − a ? | "- ( a + b = 80 ) b + c = 120 c - a = 40 b . 40" | a = 60 * 2
b = 40 * 2
c = a - b
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a ) 2.1 , b ) 2.4 , c ) 2.7 , d ) 3.0 , e ) 3.3 | d | subtract(6, multiply(divide(multiply(6, 25), const_100), const_2)) | a 6 - liter solution is 25 % alcohol . how many liters of pure alcohol must be added to produce a solution that is 50 % alcohol ? | "let x be the amount of pure alcohol required . 0.25 ( 6 ) + x = 0.5 ( x + 6 ) 0.5 x = 3 - 1.5 x = 3 liters the answer is d ." | a = 6 * 25
b = a / 100
c = b * 2
d = 6 - c
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a ) 2 : 5 , b ) 3 : 7 , c ) 3 : 5 , d ) 5 : 12 , e ) 7 : 3 | d | divide(add(multiply(9, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))), subtract(add(9, 8), add(multiply(9, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))))) | a container holding 9 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 8 ounces of a solution that is 1 part alcohol to 3 parts water . what is the ratio of alcohol to water in the resulting solution ? | "container 1 has 9 ounces in the ratio 1 : 2 or , x + 2 x = 9 gives x ( alcohol ) = 3 and remaining water = 6 container 2 has 8 ounces in the ratio 1 : 3 or , x + 3 x = 8 gives x ( alcohol ) = 2 and remaining water = 6 mixing both we have alcohol = 3 + 2 and water = 6 + 6 ratio thus alcohol / water = 5 / 12 answer d" | a = 1 + 2
b = 1 / a
c = 9 * b
d = 1 + 3
e = 1 / d
f = 8 * e
g = c + f
h = 9 + 8
i = 1 + 2
j = 1 / i
k = 9 * j
l = 1 + 3
m = 1 / l
n = 8 * m
o = k + n
p = h - o
q = g / p
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a ) 2 : 3 , b ) 1 : 2 , c ) 3 : 5 , d ) 2 : 5 , e ) 5 : 6 | b | divide(divide(3, 2), 3) | if the ratio of apples to bananas is 3 to 2 and the ratio of bananas to cucumbers is 1 to 3 , what is the ratio of apples to cucumbers ? | "the ratio of bananas to cucumbers is 1 to 3 which equals 2 to 6 . the ratio of apples to bananas to cucumbers is 3 to 2 to 6 . the ratio of apples to cucumbers is 3 to 6 which equals 1 : 2 . the answer is b ." | a = 3 / 2
b = a / 3
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a ) two , b ) three , c ) four , d ) five , e ) six | e | multiply(const_3, const_1) | if k and w are the dimensions of a rectangle that has area 60 , and if k and w are integers such that k > w , what is the total number of possible values of k ? | "kw = 60 = 60 * 1 = 30 * 2 = 15 * 4 = 20 * 3 = 12 * 5 = 10 * 6 - - > k can take 6 values answer : e" | a = 3 * 1
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a ) 110 , b ) 120 , c ) 130 , d ) 140 , e ) 190 | e | divide(add(60, 20), divide(42, const_100)) | a candidate appearing for an examination has to secure 42 % marks to pass paper i . but he secured only 60 marks and failed by 20 marks . what is the maximum mark for paper i ? | "he secured 60 marks nd fail by 20 marks so total marks for pass the examinatn = 80 let toal marks x x * 42 / 100 = 80 x = 190 answer : e" | a = 60 + 20
b = 42 / 100
c = a / b
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a ) $ 8,000 , b ) $ 5,600 , c ) $ 3,200 , d ) $ 2,400 , e ) $ 800 | c | subtract(multiply(multiply(const_100, const_10), multiply(const_2, const_4)), multiply(multiply(multiply(const_100, 30), multiply(const_2, const_4)), multiply(divide(30, const_100), const_2))) | the market value of a certain machine decreased by 30 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 8,000 , what was its market value two years later ? | "market value in 1982 = $ 8000 market value in 1983 = $ 8000 - ( $ 8000 x 30 / 100 ) = 8000 - 2400 = $ 5600 market value in 1984 = market value in 1983 - ( 30 % of $ 8000 ) = 5600 - 2400 = $ 3200 answer : c" | a = 100 * 10
b = 2 * 4
c = a * b
d = 100 * 30
e = 2 * 4
f = d * e
g = 30 / 100
h = g * 2
i = f * h
j = c - i
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a ) 5100 , b ) 7550 , c ) 10100 , d ) 15500 , e ) 20100 | b | multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1)) | the sum of the first 50 positive even integers is 2550 . what is the sum of even integers from 102 to 200 inclusive ? | first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : 2550 + ( 100 * 50 ) = 7550 ; answer : b | a = 200 + 102
b = a / 2
c = 200 - 102
d = c / 2
e = d + 1
f = b * e
|
a ) 10 , b ) 3 , c ) 5 , d ) 7 , e ) 8 | b | divide(multiply(subtract(26, 22), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 22 years rahul age will be 26 years . what is deepak present age | "explanation : present age is 4 x and 3 x , = > 4 x + 22 = 26 = > x = 1 so deepak age is = 3 ( 1 ) = 3 answer : option b" | a = 26 - 22
b = a * 3
c = b / 4
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a ) 15 , b ) 37 , c ) 55 , d ) 26 , e ) 73 | d | divide(208, multiply(4, const_2)) | which of the options is a two - digit number , which has a unit ' s digit that exceeds its 10 ' s digit by 4 and the product of the given number and the sum of its digits is equal to 208 . | using the elimination method the option that fits this description is 26 6 - 2 = 4 ( unit ' s digit that exceeds its ten ' s digit by 3 ) 26 * 8 = 208 ( the product of the given number and the sum of its digits is equal to 175 ) answer : d | a = 4 * 2
b = 208 / a
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a ) 8 , b ) 5 , c ) 7 , d ) 3 , e ) 12 | e | subtract(negate(8), multiply(subtract(4, 6), divide(subtract(4, 6), subtract(2, 4)))) | 2 , 4 , 6 , 8 , 10 , . . . ? | "each number increases by 2 . 2 + 2 = 4 4 + 2 = 6 6 + 2 = 8 8 + 2 = 10 10 + 2 = 12 answer : e" | a = negate - (
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a ) 1 / 14 , b ) 3 / 35 , c ) 5 / 37 , d ) 2 / 35 , e ) none | a | divide(75, multiply(5, 210)) | the hcf and lcm of two numbers m and n are respectively 5 and 210 . if m + n = 75 , then 1 / m + 1 / n is equal to | "answer we have , m x n = 5 x 210 = 1050 â ˆ ´ 1 / m + 1 / n = ( m + n ) / mn = 75 / 1050 = 1 / 14 correct option : a" | a = 5 * 210
b = 75 / a
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a ) 135 , b ) 277 , c ) 187 , d ) 257 , e ) 271 | a | multiply(divide(subtract(54, 36), const_3_6), 27) | two trains are moving in the same direction at 54 kmph and 36 kmph . the faster train crosses a man in the slower train in 27 seconds . find the length of the faster train ? | "relative speed = ( 54 - 36 ) * 5 / 18 = 5 mps . distance covered in 27 sec = 27 * 5 = 135 m . the length of the faster train = 135 m . answer : a" | a = 54 - 36
b = a / const_3_6
c = b * 27
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a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 5 / 6 , e ) 7 / 8 | a | divide(add(multiply(divide(1, 4), const_12), multiply(divide(const_12, const_2), divide(1, 2))), const_12) | drum x is 1 / 2 full of oil and drum y , which has twice the capacity of drum x , is 1 / 4 full of oil . if all of the oil in drum x is poured into drum y , then drum y will be filled to what capacity ? | ( 1 / 2 ) x = ( 1 / 4 ) y ( 1 / 4 ) y + ( 1 / 4 ) y = ( 1 / 2 ) y the answer is a . | a = 1 / 4
b = a * 12
c = 12 / 2
d = 1 / 2
e = c * d
f = b + e
g = f / 12
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a ) 10 , b ) 12 , c ) 15 , d ) 14 , e ) 9 | d | sqrt(divide(2744, add(power(3, 2), add(power(1, 2), power(2, 2))))) | the ratio of three numbers is 1 : 2 : 3 and the sum of their squares is 2744 . the sum of the numbers is ? | "let the numbers be x , 2 x , 3 x then , x ^ 2 + 4 x ^ 2 + 9 x ^ 2 = 2744 14 x ^ 2 = 2744 x ^ 2 = 196 x = 14 answer is d" | a = 3 ** 2
b = 1 ** 2
c = 2 ** 2
d = b + c
e = a + d
f = 2744 / e
g = math.sqrt(f)
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a ) 3 : 4 , b ) 4 : 3 , c ) 2 : 15 , d ) 2 : 9 , e ) 3 : 8 | c | divide(0.08, 0.6) | 0.6 of a number is equal to 0.08 of another number . the ratio of the numbers will be | let the numbers be x and y 0.6 x = 0.08 y = > x / y = 0.08 / 0.6 = 2 : 15 answer c . | a = 0 / 8
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a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 32 , d ) 1 / 2 , e ) 1 / 64 | e | divide(1, power(2, 6)) | if a coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 6 times , what is the probability that it will land heads up on all 6 flips ? | ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 64 answer : e | a = 2 ** 6
b = 1 / a
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a ) 640 , b ) 277 , c ) 500 , d ) 297 , e ) 435 | a | divide(660, multiply(subtract(78, 1), const_0_2778)) | a train 660 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 660 + x ) / 60 = 65 / 3 x = 640 m . answer : a" | a = 78 - 1
b = a * const_0_2778
c = 660 / b
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a ) 22 sec , b ) 81.6 sec , c ) 48 sec , d ) 18.3 sec , e ) 9.32 sec | b | add(45, 30) | two goods trains each 850 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 850 + 850 = 1700 m . required time = 1700 * 6 / 125 = 81.6 sec . answer : b" | a = 45 + 30
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a ) rs . 6725 , b ) rs . 6727 , c ) rs . 6908 , d ) rs . 6725 , e ) rs . 7000 | e | divide(8400, add(const_1, divide(20, const_100))) | the owner of a furniture shop charges his customer 20 % more than the cost price . if a customer paid rs . 8400 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8400 ( 100 / 120 ) = rs . 7000 . answer : e" | a = 20 / 100
b = 1 + a
c = 8400 / b
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a ) 6.9 , b ) 69.0 , c ) 0.6845 , d ) 0.6859 , e ) 0.69 | e | divide(multiply(92, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 92 % of 3 / 4 ? | "soln : - 120 % = 12 / 10 5 / 8 * 12 / 10 = 60 / 80 = 6 / 8 = 3 / 4 = 0.69 answer : e ." | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 92 * n
p = o / 100
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a ) $ 480 , b ) $ 450 , c ) $ 960 , d ) $ 832 , e ) $ 660 | d | add(divide(multiply(divide(add(832, 448), const_2), 30), const_100), divide(add(832, 448), const_2)) | 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 for making 30 % profit ? | "let c . p . = $ x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 130 % of $ 640 = $ 832 . d" | a = 832 + 448
b = a / 2
c = b * 30
d = c / 100
e = 832 + 448
f = e / 2
g = d + f
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a ) 228 , b ) 108 , c ) 84 , d ) 188 , e ) 211 | c | multiply(divide(140, 6), const_3_6) | a 140 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | "s = 140 / 6 * 18 / 5 = 84 kmph answer : c" | a = 140 / 6
b = a * const_3_6
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a ) 1 / 10 , b ) 3 / 10 , c ) 1 / 2 , d ) 7 / 25 , e ) 9 / 10 | d | divide(subtract(100, subtract(add(50, 40), 18)), 100) | a certain manufacturer of cake , muffin , and bread mixes has 100 buyers , of whom 50 purchases cake mix , 40 purchase muffin mix , and 18 purchase both cake mix and muffin mix . if a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix nor muffin mix ? | "c + m + b - cm - mb - cb - 2 cmb = 100 c - cake buyers , m - muffin and b - bread buyers . cm , mb , cb and cmb are intersecting regions . the question asks for people who have bought only bread mixes = b - cb - mb - 2 cmb has to be found out . 50 + 40 + b - cb - mb - 18 - 2 cmb = 100 b - cb - mb - 2 cmb = 28 hence the probability = 28 / 100 = 7 / 25 . d" | a = 50 + 40
b = a - 18
c = 100 - b
d = c / 100
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a ) 22.8 kg , b ) 25.6 kg , c ) 19 kg , d ) 26.5 kg , e ) none of these | c | divide(multiply(5, 42.75), 11.25) | if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 5 m of the same rod ? | "explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 5 : : 42.75 : x = > 11.25 x x = 5 x 42.75 = > x = ( 5 x 42.75 ) / 11.25 = > x = 19 answer : c" | a = 5 * 42
b = a / 11
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['a ) 20', 'b ) 27', 'c ) 26', 'd ) 188', 'e ) 11'] | a | sqrt(divide(460, add(const_1, divide(15, const_100)))) | the area of a rectangle is 460 square metres . if the length is 15 % more than the breadth , what is the breadth of the rectangular field ? | explanation : let breadth = x meters . then , length = meters given that , = > x = 20 breadth = 20 answer : a ) 20 m | a = 15 / 100
b = 1 + a
c = 460 / b
d = math.sqrt(c)
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a ) 8000 , b ) 16000 , c ) 12000 , d ) 18000 , e ) 360000 | e | multiply(multiply(6, const_1000), 6) | find the product of the local value and absolute value of 6 in 564823 | "explanation : place value = local value face value = absolute value the place value of 6 in 564823 is 6 x 10000 = 60000 the face value of 6 in 564823 is nothing but 6 . = > 60000 x 6 = 360,000 answer : option e" | a = 6 * 1000
b = a * 6
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a ) $ 645.56 , b ) $ 824.32 , c ) $ 954.26 , d ) $ 745.69 , e ) $ 8243.21 | e | subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 100000), 100000) | find the compound interest on $ 100000 in 2 years at 4 % per annum , the interest being compounded half - yearly ? | "principle = $ 10000 rate = 2 % half yearly = 4 half years amount = 100000 * ( 1 + 2 / 100 ) ^ 4 = 100000 * 51 / 50 * 51 / 50 * 51 / 50 * 51 / 50 = $ 108243.21 c . i . = 108243.21 - 100000 = $ 8243.21 answer is e" | a = 4 / 100
b = a / 2
c = b + 1
d = c ** 4
e = d * 100000
f = e - 100000
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a ) 2333 , b ) 2777 , c ) 2688 , d ) 1500 , e ) 2871 | d | divide(subtract(135, divide(multiply(6, 2500), const_100)), subtract(divide(5, const_100), divide(6, const_100))) | rs . 2500 is divided into two parts such that if one part be put out at 5 % simple interest and the other at 6 % , the yearly annual income may be rs . 135 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 2500 - x ) * 6 * 1 ] / 100 = 135 x = 1500 answer : d" | a = 6 * 2500
b = a / 100
c = 135 - b
d = 5 / 100
e = 6 / 100
f = d - e
g = c / f
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a ) 0.86 , b ) 0.56 , c ) 0.96 , d ) 0.69 , e ) 0.76 | b | divide(subtract(power(0.66, 3), power(0.1, 3)), add(add(power(0.66, 2), 0.066), power(0.1, 2))) | ( 0.66 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.66 ) ( power 2 ) + 0.066 + ( 0.1 ) ( power 2 ) is : | "given expression = ( 0.66 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.66 ) ( power 2 ) + ( 0.66 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.66 - 0.1 ) = 0.56 answer is b" | a = 0 ** 66
b = 0 ** 1
c = a - b
d = 0 ** 66
e = d + 0
f = 0 ** 1
g = e + f
h = c / g
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a ) 80 % , b ) 65 % , c ) 28.33 % , d ) 19.109 % , e ) 50 % | b | subtract(multiply(divide(add(const_100, 50), const_100), add(const_100, 10)), const_100) | a trader cheats both his supplier and customer by using faulty weights . when he buys from the supplier , he takes 10 % more than the indicated weight . when he sells to his customer , he gives the customer a weight such that 50 % of that is added to the weight , the weight claimed by the trader is obtained . if he charges the cost price of the weight that he claims , find his profit percentage . | "anyways , one can infer that he ' steals ' 10 % from suppliers and then charges 50 % extra to customers so basically 1.1 * 1.5 = 1.65 given that 1 is start point , we get 65 % more hence answer is b" | a = 100 + 50
b = a / 100
c = 100 + 10
d = b * c
e = d - 100
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(subtract(add(add(7, multiply(const_2, const_4)), multiply(const_2, const_4)), add(multiply(const_2, const_3), const_3)), 11) | when the no . 7 y 86038 is exactly divisible by 11 , then the smallest whole no . in place of y ? | the given number = 7 y 86038 sum of the odd places = 8 + 0 + 8 + 7 = 23 sum of the even places = 3 + 6 + y ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) 23 - ( 9 + y ) = divisible by 11 14 � y = divisible by 11 . y must be 3 , to make given number divisible by 11 . b | a = 2 * 4
b = 7 + a
c = 2 * 4
d = b + c
e = 2 * 3
f = e + 3
g = d - f
h = g - 11
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a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | c | power(3, 0) | set i consists of integers { 3 , - 8 , y , 19 , - 6 } and set b consists of integers { k , - 3 , 0 , 16 , - 5 , 9 } . number l represents the median of set i , number m represents mode of set b and number z = l ^ m . if y is an integer greater than 21 , for what value of k will z be a divisor of 26 ? | i = { 3 , - 8 , y , 19 , - 6 } b = { k , - 3 , 0,16 , - 5,9 } y > 21 so l = median of i = 3 m = mode of set b z = ( 3 ) ^ m if z is a divisor of 26 , ( 3 ) ^ m = 1 because 26 does not have 3 as a factor = > m = 0 hence k = 0 , as m is mode and 0 will be the most frequently occuring number in set b . answer - c | a = 3 ** 0
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a ) 25 , b ) 23 , c ) 22 , d ) 20 , e ) 10 | e | divide(rectangle_area(5, 24), 12) | carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 5 inches by 24 inches and jordan ' s rectangle is 12 inches long , how wide is jordan ' s rectangle , in inches ? | "area of carol ' s rectangle = 24 * 5 = 120 let width of jordan ' s rectangle = w since , the areas are equal 12 w = 120 = > w = 10 answer e" | a = rectangle_area / (
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['a ) 80 feet', 'b ) 85 feet', 'c ) 83 feet', 'd ) 88 feet', 'e ) 90 feet'] | d | subtract(rectangle_perimeter(20, divide(680, 20)), 20) | a rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 680 sq . feet , how many feet of fencing will be required ? | given that area of the field = 680 sq . feet = > lb = 680 sq . feet length ( l ) = 20 feet = > 20 × b = 680 = > b = 680 / 20 = 34 feet required length of the fencing = l + 2 b = 20 + ( 2 × 34 ) = 88 feet answer is d . | a = 680 / 20
b = rectangle_perimeter - (
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a ) rs . 6000 , b ) rs . 7935 , c ) rs . 8265.63 , d ) rs . 3967.5 , e ) none | c | divide(multiply(divide(divide(add(divide(multiply(6000, 15), const_100), divide(multiply(add(6000, divide(multiply(6000, 15), const_100)), 15), const_100)), 2), 2), const_100), 12) | simple interest on a certain sum of money for 2 years at 12 % per annum is half the compound interest on rs . 6000 for 2 years at 15 % per annum . the sum placed on simple interest is | "solution c . i . = rs [ 6000 x ( 1 + 15 / 100 ) â ² - 4000 ] rs . ( 4000 x 115 / 100 x 115 / 100 - 6000 ) = rs . 7935 . sum = rs . [ 3967.5 x 100 / 4 x 12 ] = rs . 8265.63 . answer c" | a = 6000 * 15
b = a / 100
c = 6000 * 15
d = c / 100
e = 6000 + d
f = e * 15
g = f / 100
h = b + g
i = h / 2
j = i / 2
k = j * 100
l = k / 12
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a ) 9 days , b ) 10 days , c ) 11 days , d ) 12 days , e ) 18 days | a | divide(divide(multiply(12, 24), 4), 8) | 10 men working 12 hours daily finish a work in 24 days . in how many days will they finish 1 / 4 th of the work working 8 hours daily ? | the number of days needed to finish 1 / 4 of the work = 1 / 4 * 12 / 8 * 24 = 9 days . answer : a | a = 12 * 24
b = a / 4
c = b / 8
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a ) 100 , b ) 230 , c ) 150 , d ) 180 , e ) 199 | b | divide(multiply(multiply(divide(747.50, const_100), 20), const_100), 65) | 65 % of x = 20 % of 747.50 . find the value of x ? | "65 % of x = 20 % of 747.50 then , 65 / 100 * x = 20 / 100 * 7475 / 10 x = 230 answer is b" | a = 747 / 50
b = a * 20
c = b * 100
d = c / 65
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(divide(5, const_2), multiply(2, 2)) | what is the remainder wen 2 ^ 35 is divided by 5 ? | "2 ^ 35 / 5 = > ( 2 ^ 4 ) ^ 8 * 2 ^ 3 / 5 = > ( 5 * 3 + 1 ) ^ 8 * ( 5 + 3 ) / 5 = > ( 1 * 3 ) / 5 = > 3 / 5 = > remainder = 3 answer : c" | a = 5 / 2
b = 2 * 2
c = a - b
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a ) 52 , b ) 65 , c ) 78 , d ) 91 , e ) 104 | b | divide(multiply(divide(multiply(40, 3.8), 1.6), 3.9), 5.7) | a certain car can travel 40 kilometers on a liter of fuel . if the fuel tank ’ s contents decrease by 3.9 gallons over a period of 5.7 hours as the car moves at a constant speed , how fast is the car moving , in miles per hour ? ( 1 gallon = 3.8 liters ; 1 mile = 1.6 kilometers ) | "fuel used 3.9 gallons ; convert to liters - - > 3.9 x 3.8 liters time = 5.7 hours 1 mile = 1.6 kilometers ; convert to miles - - > 1 km = 1 / 1.6 mile speed ( km / hour ) = d / t = 40 ( km * ) x 3.9 x 3.8 / 5.7 replace ( km * ) to miles ; multiply by 1 / 1.6 mile speed ( miles / hour ) = 40 x 3.9 x 3.8 / 5.7 x 1.6 = 78 miles / hour ans : c ps : i felt the factors were easy to cancel out , so did n ' t require much rounding off = 40 x 3.9 x 3.8 / 5.7 x 1.6 = 65 b" | a = 40 * 3
b = a / 1
c = b * 3
d = c / 5
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a ) 300 % , b ) 400 % , c ) 500 % , d ) 200 % , e ) 600 % | a | multiply(subtract(multiply(const_2, const_2), const_1), 100) | each side of a rectangle is increased by 100 % . by what percentage does the area increase ? | original area = a * b where a and b are sides after increase in side increase in area will be = ( ( ( 2 a * 2 b ) - ( a * b ) ) * 100 ) / ( a * b ) = 300 % answer : a | a = 2 * 2
b = a - 1
c = b * 100
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a ) 3600 , b ) 3890 , c ) 88798 , d ) 3400 , e ) 2891 | d | subtract(multiply(add(1300, 100), add(20, const_1)), multiply(1300, 20)) | the average monthly salary of 20 employees in an organisation is rs . 1300 . 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 . ( 1400 * 21 - 1300 * 20 ) = rs . 3400 . answer : d" | a = 1300 + 100
b = 20 + 1
c = a * b
d = 1300 * 20
e = c - d
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a ) 130 ares . , b ) 160 ares . , c ) 180 ares . , d ) 200 ares . , e ) 250 ares . | d | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 2.0), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 2.0 hectares in ares | "2.0 hectares in ares 1 hectare = 100 ares therefore , 2.0 hectares = 2.0 × 100 ares = 200 ares . answer - d" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 2
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
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a ) 10 , b ) 12 , c ) 20 , d ) 18 , e ) 21 | c | add(19, 1) | seller selling an apple for rs . 19 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | sp = 19 loss = cp 20 loss = cp − sp = cp − 19 ⇒ cp 20 = cp − 19 ⇒ 19 cp 20 = 19 ⇒ cp 20 = 1 ⇒ cp = 20 c | a = 19 + 1
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a ) 6893 , b ) 6993 , c ) 6093 , d ) 693 , e ) none | d | subtract(add(add(add(multiply(multiply(7, const_100), const_10), multiply(7, const_100)), multiply(7, const_10)), 0), add(add(add(const_1000, multiply(7, const_100)), multiply(7, const_10)), 7)) | what is the difference between the largest and the smallest number written with 7 , 7 , 0 , 7 ? | "7770 7077 - - - - - - - - - - - - - 693 answer : d" | a = 7 * 100
b = a * 10
c = 7 * 100
d = b + c
e = 7 * 10
f = d + e
g = f + 0
h = 7 * 100
i = 1000 + h
j = 7 * 10
k = i + j
l = k + 7
m = g - l
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a ) 5 / 3 , b ) 5 / 4 , c ) 10 / 3 , d ) 15 / 4 , e ) 15 / 2 | d | divide(1, divide(add(divide(2, 5), multiply(divide(2, 5), divide(1, 3))), const_2)) | if a certain toy store ' s revenue in november was 2 / 5 of its revenue in december and its revenue in january was 1 / 3 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "n = 2 d / 5 j = n / 3 = 2 d / 15 the average of november and january is ( n + j ) / 2 = 8 d / 15 / 2 = 4 d / 15 d is 15 / 4 times the average of november and january . the answer is d ." | a = 2 / 5
b = 2 / 5
c = 1 / 3
d = b * c
e = a + d
f = e / 2
g = 1 / f
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a ) 7 days , b ) 8 days , c ) 9 days , d ) 10 days , e ) 11 days | a | add(multiply(divide(2, 5), 5), 5) | a man bring 2 cats to his home to eliminate all the mice . they do a half of the work in 5 days , then the man brings 3 more cats . if they hunt at the same rate , how many days will take to hunt all the mice , since the first day ? | we have that : 2 cats - - - - - 5 days , ( 2 + 3 ) c - - - - - x d , that is : ( x d / 2 c ) = ( 5 d / 5 p ) , then : x d = ( 5 d / 5 c ) 2 c = 2 d . is to say , the 5 cats hunt the same quantity in 2 days ; as it is only 1 / 2 of the work , all the mice would be hunt in : m = 5 d + 2 d = 7 days . answer a . | a = 2 / 5
b = a * 5
c = b + 5
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a ) 600 , b ) 288 , c ) 2768 , d ) 287 , e ) 191 | a | subtract(720, multiply(divide(subtract(1020, 720), 5), 2)) | a sum of money lent out at s . i . amounts to rs . 720 after 2 years and to rs . 1020 after a further period of 5 years . the sum is ? | "s . i for 5 years = ( 1020 - 720 ) = rs . 300 . s . i . for 2 years = 300 / 5 * 2 = rs . 120 . principal = ( 720 - 120 ) = rs . 600 . answer : a" | a = 1020 - 720
b = a / 5
c = b * 2
d = 720 - c
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(15, divide(180, 15)) | there are 15 members in a family . it takes 180 days for a women to complete the work , however man completes in 120 days . they take 17 days to complete the work if men work on alternate days and women work every third day . if all of them started together on the 1 st day , then how many women are there in the family | let the number of men be m and so the number of women = 15 - m out of 17 days , men and women work together on 1,4 , 7,10 , 13,16 th days ( total of 6 days ) from the start . men work on alternate days , that is 1 , 3,5 , 7,9 , 11,13 , 15,17 th days ( total of 9 days ) . let the total work be 360 units ( lcm of 180 and 120 ) . 1 man does 360 / 120 = 3 units per day 1 woman does 360 / 180 = 2 units per day based on above facts , total work by men + women in these 17 days = 360 units m * 9 * 3 + ( 15 - m ) * 6 * 2 = 360 , 27 m + 180 - 12 m = 360 , solving m = 12 . num of men = 12 and number of women = 15 - 12 = 3 answer : c | a = 180 / 15
b = 15 - a
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a ) 91 , b ) 51 , c ) 81 , d ) 71 , e ) 61 | c | multiply(multiply(subtract(4, const_1), subtract(4, const_1)), multiply(subtract(4, const_1), subtract(4, const_1))) | how many integers between 1000 and 10000 have no digits other than 4 , 5 or 6 ? | any number between 1000 and 10000 is of 4 digits . the unit ’ s place can be filled up by 45 or 6 , that is , in 3 ways . similarly , the ten ’ s place can be filled up by 4 or 5 or 6 , that is in 3 ways . the hundred ’ s place can be filled up by 4 , 5 or 6 , that is in 3 ways and the thousand ’ s place can . be filled up by 4 or 5 or 6 , that is , in 3 ways . hence the required numbers = 3 x 3 x 3 x 3 = 81 answer : c | a = 4 - 1
b = 4 - 1
c = a * b
d = 4 - 1
e = 4 - 1
f = d * e
g = c * f
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a ) 6 days , b ) 5.33 days , c ) 10 days , d ) 12 days , e ) 14 days | b | divide(const_1, add(divide(const_1, 8), divide(const_1, 16))) | john completes a piece of work in 8 days , rose completes the same work in 16 days . if both of them work together , then the number of days required to complete the work is ? | if a can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y / x + y days . that is , the required no . of days = 8 × 16 / 24 = 5.33 days answer : b | a = 1 / 8
b = 1 / 16
c = a + b
d = 1 / c
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | add(divide(subtract(const_1, multiply(divide(const_1, 30), 5)), add(divide(const_1, 20), divide(const_1, 30))), 5) | a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and b quits 5 days before the project is completed , in how many days total will the project be completed ? | "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 5 days , a can do 1 / 4 of the project . thus a and b must complete 3 / 4 of the project , which takes 9 days . the total number of days is 9 + 5 = 14 . the answer is c ." | a = 1 / 30
b = a * 5
c = 1 - b
d = 1 / 20
e = 1 / 30
f = d + e
g = c / f
h = g + 5
|
a ) 13.6 sec , b ) 12.1 sec , c ) 17.9 sec , d ) 16.8 sec , e ) 14.9 sec | a | divide(add(140, 132), multiply(72, const_0_2778)) | how long does a train 140 m long running at the speed of 72 km / hr takes to cross a bridge 132 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 140 + 132 = 272 m . required time = 272 / 20 = 13.6 sec . answer : a" | a = 140 + 132
b = 72 * const_0_2778
c = a / b
|
a ) 11 , b ) 22 , c ) 77 , d ) 33 , e ) 88 | d | add(add(10, 15), 8) | in kaya ' s teacher ' s desk there are 10 pink highlighters , 15 yellow highlighters , and 8 blue highlighters . how many highlighters are there in all ? | "add the numbers of highlighters . 10 + 15 + 8 = 33 . answer is d ." | a = 10 + 15
b = a + 8
|
a ) rs . 3600 , b ) rs . 3603 , c ) rs . 2000 , d ) rs . 3632 , e ) rs . 3602 | c | subtract(18000, multiply(divide(8, 9), 18000)) | income and expenditure of a person are in the ratio 9 : 8 . if the income of the person is rs . 18000 , then find his savings ? | "let the income and the expenditure of the person be rs . 9 x and rs . 8 x respectively . income , 9 x = 18000 = > x = 2000 savings = income - expenditure = 9 x - 8 x = x so , savings = rs . 2000 answer : c" | a = 8 / 9
b = a * 18000
c = 18000 - b
|
a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | e | divide(subtract(multiply(40, divide(60, const_100)), multiply(40, divide(40, const_100))), subtract(divide(80, const_100), divide(60, const_100))) | a gambler has won 40 % of his 40 poker games for the week so far . if , all of a sudden , his luck changes and he begins winning 80 % of the time , how many more games must he play to end up winning 60 % of all his games for the week ? | "let x be the number of additional games the gambler needs to play . 0.4 ( 40 ) + 0.8 x = 0.6 ( x + 40 ) 0.2 x = 8 x = 40 the answer is e ." | a = 60 / 100
b = 40 * a
c = 40 / 100
d = 40 * c
e = b - d
f = 80 / 100
g = 60 / 100
h = f - g
i = e / h
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | divide(subtract(8, subtract(multiply(const_4, const_2), multiply(const_2, const_3))), const_3) | x , y , and z are consecutive numbers and x > y > z . also , 2 x + 3 y + 3 z = 5 y + 8 . what is the value of z ? | if x , y , and z are consecutive numbers and x > y > z , then y = z + 1 and x = z + 2 . 2 x + 3 y + 3 z = 5 y + 8 2 z + 4 + 3 z + 3 + 3 z = 5 z + 5 + 8 3 z = 6 z = 2 the answer is a . | a = 4 * 2
b = 2 * 3
c = a - b
d = 8 - c
e = d / 3
|
a ) 52 , b ) 46 , c ) 60 , d ) 50 , e ) 35 | c | multiply(multiply(10, 2), divide(6, 2)) | in the coordinate plane , points ( x , 6 ) and ( 10 , y ) are on line k . if line k passes through the origin and has slope 1 / 2 , then x * y = | "line k passes through the origin and has slope 1 / 2 means that its equation is y = 1 / 2 * x . thus : ( x , 6 ) = ( 12 , 6 ) and ( 10 , y ) = ( 10,5 ) - - > x * y = 12 * 5 = 60 . answer : c" | a = 10 * 2
b = 6 / 2
c = a * b
|
a ) 31 % . , b ) 71 % . , c ) 27.5 % . , d ) 29 % . , e ) 51 % . | c | multiply(divide(add(multiply(divide(25, const_100), 3), multiply(divide(40, const_100), 5)), 10), const_100) | a vessel of capacity 3 litre has 25 % of alcohol and another vessel of capacity 5 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "25 % of 3 litres = 0.75 litres 40 % of 5 litres = 2 litres therefore , total quantity of alcohol is 2.75 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 27.5 % c" | a = 25 / 100
b = a * 3
c = 40 / 100
d = c * 5
e = b + d
f = e / 10
g = f * 100
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | add(3, multiply(subtract(18, 3), divide(1, 5))) | a jar contains 18 balls . 3 blue balls are removed from the jar and not replaced . now the probability of getting a blue ball is 1 / 5 then how many blue balls jar contains initially ? | x / 15 = 1 / 5 x = 3 3 + 3 ( removed 3 blue balls ) = 6 answer : e | a = 18 - 3
b = 1 / 5
c = a * b
d = 3 + c
|
a ) 680 rs , b ) 540 rs , c ) 480 rs , d ) 640 rs , e ) 280 rs | a | multiply(divide(80, const_100), subtract(multiply(110, 65), multiply(subtract(110, 5), subtract(65, 5)))) | a rectangular grassy plot 110 m by 65 cm has a gravel path . 5 cm wide all round it on the inside . find the cost of gravelling the path at 80 paise per sq . mt ? | area of theplot = 110 * 65 = 7150 sq m area of the plot excluding the path = ( 110 - 5 ) * ( 65 - 5 ) = 6300 sq m area of the path = 7150 - 6300 = 850 sq m cost of gravelling the path = 850 * 80 / 100 = 680 rs answer : a | a = 80 / 100
b = 110 * 65
c = 110 - 5
d = 65 - 5
e = c * d
f = b - e
g = a * f
|
a ) 1 : 2 , b ) 1 : 7 , c ) 1 : 9 , d ) 1 : 5 , e ) 2 : 5 | e | divide(60, divide(600, subtract(subtract(subtract(divide(600, 60), 5), inverse(const_2)), inverse(const_2)))) | a motorcyclist goes from nagpur to delhi , a distance of 600 kms at an average of 60 kmph speed . another man starts from nagpur by car 5 â ½ hours after the first , and reaches delhi â ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | t = 600 / 60 = 10 h t = 10 - 6 = 4 time ratio = 10 : 4 = 5 : 2 speed ratio = 2 : 5 answer : e | a = 600 / 60
b = a - 5
c = 1/(2)
d = b - c
e = 1/(2)
f = d - e
g = 600 / f
h = 60 / g
|
a ) 120 , b ) 150 , c ) 180 , d ) 240 , e ) 600 | d | add(120, multiply(3, const_10)) | according to the directions on a packet of smoothie mix , 1 3 - ounce packet of smoothie mix is to be combined with 12 ounces of water to make a smoothie . how many 3 - ounce packets of smoothie mix are required to prepare 120 12 - ounce smoothies ? | "this question was n ' t particularly grueling , but i think it ' s the first where i had the opportunity to solve it via theory andinspectionthat many on this board suggest as strategy on the gmat . it actually came to me by accident . basically , if we thought that the 3 packets of powder were included in the 12 ounces of water , that would mean we would need 150 packets of smoothie mix ( along with 12 ( 150 ) ounces of water for a total of 150 packets . however , we know , after a more careful reading of the stimulus , that the 3 ounces are not included in the 12 ounces . as such , the answer has to be less than 150 packets , since 150 would be too much powder considering you already have 150 ( 12 ) ounces of water and need less packets than water to make a smoothie . as such , the only answer less than 150 is 120 , a . does this make sense ? or am i way off base ? d" | a = 3 * 10
b = 120 + a
|
a ) 12 . , b ) 9 . , c ) 8 . , d ) 7 . , e ) 6 . | a | add(divide(subtract(20, 2), 5), 2) | there are 20 balls in a jar . you take out 2 blue balls without putting them back inside , and now the probability of pulling out a blue ball is 1 / 5 . how many blue balls were there in the beginning ? | "12 = 3 blue balls + 18 / 2 answer : a" | a = 20 - 2
b = a / 5
c = b + 2
|
a ) 1 / 4 , b ) 1 / 3 , c ) 5 / 12 , d ) 4 / 9 , e ) 1 / 2 | c | subtract(divide(add(multiply(const_2, const_3), multiply(const_3, const_2)), power(add(1, 4), const_2)), divide(const_2, power(add(1, 4), const_2))) | a = { 20 , 1 , - 3 , 6 , - 8 } b = { - 1 , 2 , - 4 , 7 , - 620 } if a is a number that is randomly selected from set a , and b is a number that is randomly selected from set b , what is the probability that ab > 0 ? | for the product of 2 numbers to be positive either both of them must be positive or both of them must be negative : p ( positive , positive ) = 3 / 6 * 3 / 6 = 9 / 36 ; p ( negative , negative ) = 2 / 6 * 3 / 6 = 6 / 36 . p = 9 / 36 + 6 / 36 = 15 / 36 = 5 / 12 . answer : c . | a = 2 * 3
b = 3 * 2
c = a + b
d = 1 + 4
e = d ** 2
f = c / e
g = 1 + 4
h = g ** 2
i = 2 / h
j = f - i
|
a ) 68 , b ) 70.4 , c ) 123.2 , d ) 105.6 , e ) 108 | c | add(88, multiply(divide(40, const_100), 88)) | if x is 40 percent greater than 88 , then x = | "x = 88 * 1.4 = 123.2 so the answer is c ." | a = 40 / 100
b = a * 88
c = 88 + b
|
a ) 2.82 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(multiply(4, sqrt(2)), 2) | the perimeter of an isosceles right triangle is 4 + 4 sq rt 2 . what is the length of the hypotenuse of the triangle ? | "side of triangle is a then perimeter = a + a + a . sqrt 2 ( right angle and pythagorus ) = 2 a + a . sqrt 2 = 4 + 4 sqrt 2 or , a . ( 2 + sqrt 2 ) = 4 ( 1 + sqrt 2 ) a = 4 . ( 1 + sqrt 2 ) / 2 + sqrt 2 = 4 * 2.414 / 3.414 = then hypotenuse = 2.82 a" | a = math.sqrt(2)
b = 4 * a
c = b / 2
|
a ) 4 , b ) 6 , c ) 7 , d ) 9 , e ) 11 | d | add(power(const_2, const_2), const_2) | how many different positive integers are factors of 36 ? | "6 × 6 = 3 ^ 2 × 2 ^ 2 so total factors = ( 2 + 1 ) ( 2 + 1 ) = 9 answer : d" | a = 2 ** 2
b = a + 2
|
a ) 8 % , b ) 7 % , c ) 10 % , d ) 12 % , e ) 2 % | e | multiply(divide(subtract(subtract(65, multiply(65, divide(10, const_100))), 57.33), subtract(65, multiply(65, divide(10, const_100)))), const_100) | the list price of an article is rs . 65 . a customer pays rs . 57.33 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | "option e explanation : 65 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 57.33 x = 2 %" | a = 10 / 100
b = 65 * a
c = 65 - b
d = c - 57
e = 10 / 100
f = 65 * e
g = 65 - f
h = d / g
i = h * 100
|
a ) 60 , b ) 45 , c ) 90 , d ) 180 , e ) 120 | b | divide(subtract(multiply(const_3, subtract(const_100, const_10)), multiply(subtract(const_100, const_10), const_2)), const_2) | the supplementary of an angle is thrice its complimentary . find the angle . | let angle = x ; by question - - 180 - x = 3 ( 90 - x ) x = 45 answer : b | a = 100 - 10
b = 3 * a
c = 100 - 10
d = c * 2
e = b - d
f = e / 2
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(log(multiply(9, 9)), log(const_10)) | 9 log 9 ( 2 ) = ? | exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 2 ) = 2 correct answer b | a = 9 * 9
b = math.log(a)
c = math.log(10)
d = b / c
|
a ) 1 : 3 , b ) 1 : 4 , c ) 1 : 5 , d ) 1 : 6 , e ) 1 : 7 | b | divide(multiply(divide(5, add(25, 5)), 6), divide(subtract(25, 5), 5)) | there are 2 vessels a & b containing 25 liters each of pure milk and pure water respectively . 5 liters of milk from a is taken and poured into b , then 6 liters of mixture from b is taken and poured in a . what is ratio of water a and b respectively . | at first a cotain milk = 25 water = 0 b cotain milk = 00 water = 25 5 litres of milk taken from a and poured in b then a cotain milk = 20 water = 0 b cotain milk = 5 water = 25 6 litres ( 1 lit milk & 5 lit water ) of mixture from b and poured in a . so , finally a cotain milk = 21 water = 5 b cotain milk = 4 water = 20 ratio of water in a & b is 5 : 20 = 1 : 4 answer : b | a = 25 + 5
b = 5 / a
c = b * 6
d = 25 - 5
e = d / 5
f = c / e
|
a ) 30 miles . , b ) 35 miles . , c ) 45 miles . , d ) 90 miles . , e ) 100 miles . | e | multiply(divide(10, add(5, 10)), 150) | tim and é lan are 150 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and é lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets é lan ? | "tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice e meet the 2 : 1 ( tim : elan = 100 : 50 ) ratio of maintaining total distance traveled 150 miles socorrect answer e" | a = 5 + 10
b = 10 / a
c = b * 150
|
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