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 ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | divide(divide(14, const_2), const_2) | an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 14 - meter platform and caught it after it had traveled 41.7 meters . how many times did the ball bounce ? | "ans : 6 division of total diatance travelled will be 14 + 14 + 7 + 3.5 + 1.75 + 0.9 + 0.5 ans b" | a = 14 / 2
b = a / 2
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a ) 11 , b ) 13 , c ) 17 , d ) 21 , e ) 22 / 5 | e | divide(multiply(add(add(2, const_3), const_2), divide(2, const_2)), add(const_2, divide(const_1, const_2))) | a and b are two partially filled buckets of water . if 2 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 2 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let a contains a , b contains b liters so , ( a - 2 ) / ( b + 2 ) = 1 / 3 . . . . . . . ( 1 ) or 3 a - 6 = b + 2 or 3 a - b = 8 again , ( b - 2 ) / ( a + 2 ) = 1 / 2 . . ( 2 ) or 2 b - 4 = a + 2 or 2 b - a = 6 from ( 1 ) ( 2 ) we find a = 22 / 5 ans : e" | a = 2 + 3
b = a + 2
c = 2 / 2
d = b * c
e = 1 / 2
f = 2 + e
g = d / f
|
a ) 126 , b ) 136 , c ) 146 , d ) 156 , e ) 182 | e | divide(multiply(14, 312), 24) | the reciprocal of the hcf and lcm of two are 1 / 14 and 1 / 312 . if one of the number is 24 then other no . is | "reciprocal of the hcf and lcm of two are 1 / 14 and 1 / 312 so , hcf = 14 , lcm = 312 lcm * hcf = product of two numbers = a * b = > b = lcm * hcf / a so , other = 14 * 312 / 24 = 182 answer : e" | a = 14 * 312
b = a / 24
|
a ) 15 , b ) 30 , c ) 35 , d ) 45 , e ) 55 | b | divide(subtract(multiply(90, 5), multiply(5, 30)), subtract(90, 80)) | the average mark of the students of a class in a particular exam is 80 . if 5 students whose average mark in that exam is 30 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam . | "let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 80 x - ( 5 * 30 ) = 90 ( x - 5 ) 300 = 10 x = > x = 30 answer : b" | a = 90 * 5
b = 5 * 30
c = a - b
d = 90 - 80
e = c / d
|
a ) a ) 100 , b ) b ) 110 , c ) c ) 120 , d ) d ) 130 , e ) e ) 140 | a | subtract(add(200, 350), 450) | a , b and c have rs . 450 between them , a and c together have rs . 200 and b and c rs . 350 . how much does c have ? | "a + b + c = 450 a + c = 200 b + c = 350 - - - - - - - - - - - - - - a + b + 2 c = 550 a + b + c = 450 - - - - - - - - - - - - - - - - c = 100 answer : a" | a = 200 + 350
b = a - 450
|
a ) 1 / 3 , b ) 7 , c ) 1 / 2 , d ) 1 / 7 , e ) 1 / 8 | b | subtract(multiply(divide(4, 3), 4), 3) | at a certain paint store forest green is made by mixing 4 parts blue paint with 3 parts yellow paint . verdant green is made by mixing 4 parts yellow paint with 3 parts blue paint . how many liters of yellow paint must be added to 21 liters of forest green to change it to verdant green ? | "21 liter of forset green have 12 liter of blue and 9 liter of yellow suppose we add x liter of yellow to make it a verdant green so the ratio of blue to yellow in verdant green is ΒΎ so the equation is blue / yellow = 12 / ( 9 + x ) = ΒΎ 27 + 3 x = 48 = > x = 7 answer : b" | a = 4 / 3
b = a * 4
c = b - 3
|
a ) 20 % , b ) 17 % , c ) 4.0 % , d ) 3.3 % , e ) 2.8 % | d | multiply(divide(divide(divide(const_100, const_3), 6), const_100), const_100) | mr . evans will states that each of his children will receive an equal share of his estate and that his grandchildren will split a portion of the estate that is equal to the share received by each of his children . if mr . evans has 4 children and 6 grandchildren , then approximately what percentage of mr . evans estate will each grandchild receive ? | "20 / 6 = 3.33 answer : d" | a = 100 / 3
b = a / 6
c = b / 100
d = c * 100
|
a ) 2 : 5 , b ) 3 : 5 , c ) 4 : 5 , d ) 3 : 7 , e ) 5 : 7 | a | divide(add(multiply(4000, 4), multiply(divide(6000, const_3), multiply(4, 4))), add(multiply(6000, multiply(4, const_3)), multiply(subtract(6000, divide(6000, const_3)), multiply(4, const_3)))) | a and b invests rs . 4000 and rs . 6000 in a business . after 4 months , a withdraws three - fourth of his capital and 4 months later , b withdraws half of his capital . in what ratio should they share the profits at the end of the year ? | "a : b ( 4000 * 4 ) + ( 1000 * 8 ) : ( 6000 * 8 ) + ( 3000 * 4 ) 24000 : 60000 2 : 5 answer : a" | a = 4000 * 4
b = 6000 / 3
c = 4 * 4
d = b * c
e = a + d
f = 4 * 3
g = 6000 * f
h = 6000 / 3
i = 6000 - h
j = 4 * 3
k = i * j
l = g + k
m = e / l
|
a ) 10 % , b ) 10.5 % , c ) 11 % , d ) 12.5 % , e ) none | a | multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), 1), add(const_100, 10))) | a trader marked the selling price of an article at 10 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of : | "sol . let c . p . = rs . 100 . then , marked price = rs . 110 , s . p . = rs . 99 . β΄ discount % = [ 11 / 110 * 100 ] % = 10 % answer a" | a = 2 + 3
b = a * 2
c = b + 1
d = 100 + 10
e = c / d
f = 100 * e
|
a ) 230 m , b ) 270 m , c ) 240 m , d ) 832 m , e ) 270 m | c | multiply(subtract(26, divide(280, multiply(const_0_2778, 72))), multiply(const_0_2778, 72)) | a goods train runs at the speed of 72 kmph and crosses a 280 m long platform in 26 seconds . what is the length of the goods train ? | "speed = ( 72 x 5 / 18 ) m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , x + 280 / 26 = 20 x + 280 = 520 x = 240 . answer : c" | a = const_0_2778 * 72
b = 280 / a
c = 26 - b
d = const_0_2778 * 72
e = c * d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(500, add(40, 60)) | two cars start at the same time from opposite ends of a highway that is 500 miles long . one car is riding at 40 mph and the second car is riding at 60 mph . how long after they begin will they meet ? | as cars are moving in opposite directions their speeds will be added . so their relative speeds : 60 + 40 = 100 mph total distance to be covered = 500 miles . time taken would be : 500 miles / 100 mph = 5 hours e is the answer . | a = 40 + 60
b = 500 / a
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['a ) β 2 / pi', 'b ) 2', 'c ) 3', 'd ) 4', 'e ) 6'] | e | divide(12, const_2) | the area of circle o is added to its diameter . if the circumference of circle o is then subtracted from this total , the result is 12 . what is the radius of circle o ? | pi * r ^ 2 + 2 r - 2 * pi * r = 12 simplifying the equation : pi * r ( r - 2 ) + 2 r = 12 without much algebraic : we can test the answers quickly , then 6 is the only possible answer that will eliminate pi from equation . answer is e | a = 12 / 2
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a ) $ 750 , b ) $ 350 , c ) $ 650 , d ) $ 550 , e ) $ 850 | a | divide(250, subtract(const_1, divide(2, 3))) | linda spent 2 / 3 of her savings on furniture and the rest on a tv . if the tv cost her $ 250 , what were her original savings ? | "if linda spent 2 / 3 of her savings on furniture , the rest 3 / 3 - 2 / 3 = 1 / 3 on a tv but the tv cost her $ 250 . so 1 / 3 of her savings is $ 250 . so her original savings are 3 times $ 250 = $ 750 correct answer a" | a = 2 / 3
b = 1 - a
c = 250 / b
|
a ) 430 , b ) 438 , c ) 436 , d ) 470 , e ) 422 | d | divide(188, divide(subtract(70, subtract(const_100, 70)), const_100)) | in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 188 votes . find the total number of valid votes ? | "let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 188 = > 4 x / 10 = 188 = > 4 x = 1880 = > x = 470 . answer : d" | a = 100 - 70
b = 70 - a
c = b / 100
d = 188 / c
|
a ) 5 , b ) 6 , c ) 7 , d ) 12 , e ) 14 | a | divide(add(12, 8), const_4) | in a classroom , 12 students brought apples and 8 students brought bananas . if exactly 10 students brought only one of the two types of fruits , how many students brought both types of fruits ? | say x students brought both fruits . ( 12 - x ) + ( 8 - x ) = 10 - - > x = 5 . answer : a . | a = 12 + 8
b = a / 4
|
a ) 40 , b ) 15 , c ) 35 , d ) 20 , e ) 25 | c | subtract(divide(add(140, 10), const_2), divide(add(60, 20), const_2)) | the average ( arithmetic mean ) of the even integers from 20 to 60 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 10 to 140 inclusive ? | so , according to a mean of a set of even numbers from 20 to 60 = ( 20 + 60 ) / 2 = 40 and mean of a set of even numbers from 10 to 140 = ( 10 + 140 ) / 2 = 75 difference = 75 - 40 = 35 answer : c . | a = 140 + 10
b = a / 2
c = 60 + 20
d = c / 2
e = b - d
|
a ) 14 , b ) 15 , c ) 20 , d ) 22 , e ) 24 | b | divide(multiply(subtract(26, 6), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 26 years . what is deepak present age . | "explanation : present age is 4 x and 3 x , = > 4 x + 6 = 26 = > x = 5 so deepak age is = 3 ( 5 ) = 15 option b" | a = 26 - 6
b = a * 3
c = b / 4
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a ) 205 , b ) 245 , c ) 285 , d ) 325 , e ) 365 | c | multiply(divide(multiply(30, const_2), subtract(38, 30)), 38) | train a leaves the station traveling at 30 miles per hour . two hours later train Π² leaves the same station traveling in the same direction at 38 miles per hour . how many miles from the station was train a overtaken by train b ? | after two hours , train a is ahead by 60 miles . train b can catch up at a rate of 8 miles per hour . the time to catch up is 60 / 8 = 7.5 hours . in 7.5 hours , train a travels another 30 * 7.5 = 225 miles for a total of 285 miles . the answer is c . | a = 30 * 2
b = 38 - 30
c = a / b
d = c * 38
|
a ) 12.25 , b ) 44 , c ) 18 , d ) 16 , e ) 14 | a | multiply(sqrt(divide(divide(100, 2), const_3)), const_3) | the length of a rectangular floor is more than its breadth by 200 % . if rs . 100 is required to paint the floor at the rate of rs . 2 per sq m , then what would be the length of the floor ? | let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 100 / 2 = 50 sq m l b = 50 i . e . , l * l / 3 = 150 l 2 = 150 = > l = 12.25 answer : a | a = 100 / 2
b = a / 3
c = math.sqrt(b)
d = c * 3
|
a ) 20 , b ) 40 , c ) - 20 , d ) 0 , e ) - 40 | c | add(10, 10) | if | x + 10 | = 10 what is the sum of all the values of x . | "there will be two cases x + 10 = 10 and x + 10 = - 10 solve for x = > x = 10 - 10 = > x = 0 or x = - 10 - 10 = > x = - 20 the sum of both values will be 0 + - 20 = - 20 answer is c" | a = 10 + 10
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a ) 25 , b ) 34 , c ) 39 , d ) 21 , e ) 11 | b | add(add(12, 12), 10) | anne bought doughnuts for a class breakfast party . she bought 12 chocolate doughnuts , 12 coconut doughnuts , and 10 jam - filled doughnuts . how many doughnuts did anne buy in all ? | "add the numbers of doughnuts . 12 + 12 + 10 = 34 . answer is b ." | a = 12 + 12
b = a + 10
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a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 7 % , e ) 40 % | d | multiply(divide(multiply(50, 0.014), 10), const_100) | a bowl was filled with 10 ounces of water , and 0.014 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | "total amount of water evaporated each day during a 50 - day period = . 014 * 50 = . 014 * 100 / 2 = 1.4 / 2 = . 7 percent of the original amount of water evaporated during this period = ( . 7 / 10 ) * 100 % = 7 % answer d" | a = 50 * 0
b = a / 10
c = b * 100
|
a ) 15 / 16 , b ) 1 / 16 , c ) 7 / 8 , d ) 9 / 8 , e ) none of these | a | divide(subtract(16, 1), 16) | a and b play a game where each is asked to select a number from 1 to 16 . if the two numbers match , both of them win a prize . find the probability that they will not win a prize in a single trial . | explanation : a and b win a prize , only if both a and b select the number . so , if a chooses a number , b has to select the same number . hence , it is problem on dependent event . let a select a number . , then p ( a ) = 1 / 16 = probability of a . now , b selects the same number after a has already selected that number . then , p ( b / a ) = 1 / 1 = dependent probability of b given a has occurred . p ( ab ) = probability of winning a prize by both a and b is : - = > p ( a ) x p ( ab ) . = > ( 1 / 16 ) x 1 = 1 / 16 . hence , the probability of not winning a prize is : - = 1 - ( 1 / 16 ) . = 15 / 16 . answer : a | a = 16 - 1
b = a / 16
|
a ) $ 1,000 , b ) $ 1,200 , c ) $ 1,500 , d ) $ 1,800 , e ) $ 2,200 | c | subtract(1,000, 700) | a family pays $ 700 per year for an insurance plan that pays 60 percent of the first $ 1,000 in expenses and 100 percent of all medical expenses thereafter . in any given year , the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total how much ? | "assuming the medical expenses are $ 1000 or more , the family pays $ 700 + $ 400 = $ 1100 . the total amount paid by insurance plan for the first $ 1000 of expenses is $ 600 . the insurance will pay another $ 500 when the medical expenses are $ 1500 . the answer is c ." | a = 1 - 0
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a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014 | d | add(2001, divide(add(divide(90, const_100), subtract(4.50, 4.20)), subtract(divide(30, const_100), subtract(4.50, 4.20)))) | the price of commodity x increases by 30 cents every year , while the price of commodity y increases by 20 cents every year . if in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 4.50 , in which year will commodity x cost 90 cents more than the commodity y ? | "the cost of commodity x increases by 10 cents per year relative to commodity y . the price of x must gain 30 + 90 = $ 1.20 cents on commodity y , which will take 12 years . the answer is d ." | a = 90 / 100
b = 4 - 50
c = a + b
d = 30 / 100
e = 4 - 50
f = d - e
g = c / f
h = 2001 + g
|
a ) 34.4543 , b ) 34.5455 , c ) 34.45128 , d ) 34.51288 , e ) 34.41222 | b | multiply(divide(20, 12), 20) | there are 20 poles with a constant distance between each pole . a car takes 20 second to reach the 12 th pole . how much will it take to reach the last pole . | "assuming the car starts at the first pole . to reach the 12 th pole , the car need to travel 11 poles ( the first pole does n ' t count , as the car is already there ) . 11 poles 20 seconds 1 pole ( 20 / 11 ) seconds to reach the last ( 20 th ) pole , the car needs to travel 19 poles . 19 pole 19 x ( 20 / 11 ) seconds = 34.5455 seconds answer : b" | a = 20 / 12
b = a * 20
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a ) 10,100 , b ) 20,200 , c ) 22,650 , d ) 40,200 , e ) 45,150 | b | add(divide(subtract(subtract(301, const_1), add(99, const_1)), const_2), 1) | for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of all the even integers between 99 and 301 ? | firstly calculate the number of even integers between 99 and 301 : no of even integers = ( 300 - 100 ) / 2 + 1 = 101 avg of the even integers is ( 100 + 300 ) / 2 = 200 . the sum would be = no of even integers * avg . = = > 101 * 202 = 20,200 answer b . | a = 301 - 1
b = 99 + 1
c = a - b
d = c / 2
e = d + 1
|
a ) 0.25 % , b ) 4 % , c ) 25 % , d ) 40 % , e ) 250 % | c | multiply(divide(500, 125), const_100) | what percent of 500 is 125 ? | "500 * x / 100 = 125 x = 125 / 5 x = 25 ans : c" | a = 500 / 125
b = a * 100
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a ) 19 % , b ) 18 % , c ) 17 % , d ) 13 % , e ) 16 % | e | divide(multiply(add(40, const_100), subtract(const_100, 40)), const_100) | a number is increased by 40 % and then decreased by 40 % . find the net increase or decrease per cent . | "let the number be 100 . increase in the number = 40 % = 40 % of 100 = ( 40 / 100 Γ 100 ) = 40 therefore , increased number = 100 + 40 = 140 this number is decreased by 40 % therefore , decrease in number = 40 % of 140 = ( 40 / 100 Γ 140 ) = 5600 / 100 = 56 therefore , new number = 140 - 56 = 84 thus , net decreases = 100 - 84 = 16 hence , net percentage decrease = ( 16 / 100 Γ 100 ) % = ( 1600 / 100 ) % = 16 % answer : e" | a = 40 + 100
b = 100 - 40
c = a * b
d = c / 100
|
a ) 25 % , b ) 50 % , c ) 20 % , d ) 15 % , e ) 30 % | a | subtract(divide(25, divide(20, const_100)), const_100) | a man buys an article for $ 20 . and sells it for $ 25 . find the gain percent ? | "c . p . = $ 20 s . p . = $ 25 gain = $ 5 gain % = 5 / 20 * 100 = 25 % answer is a" | a = 20 / 100
b = 25 / a
c = b - 100
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a ) 9.5 , b ) 23 , c ) 25 , d ) 22 , e ) 15 | b | add(divide(multiply(20, 40), const_100), divide(multiply(25, 60), const_100)) | add 20 % of 40 and 25 % of 60 . | "20 % of 40 + 25 % of 60 40 * 20 / 100 + 60 * 25 / 100 8 + 15 = 23 answer b" | a = 20 * 40
b = a / 100
c = 25 * 60
d = c / 100
e = b + d
|
a ) 80 , b ) 110 , c ) 160 , d ) 800 , e ) 400 | d | divide(40, subtract(const_1, add(add(divide(1, 5), divide(1, 4)), divide(1, 2)))) | of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 40 grades are d ' s . what is the number of students in the course ? | we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next , we can use variable t in an equation that we translate from the given information . we are given that , of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 40 grades are d ' s . since this represents all the grades in the class , it represents all the students in the class . thus we know : # a β s + # b β s + # c β s + # d β s = total number of students in the class 1 / 5 ( t ) + ΒΌ ( t ) + Β½ ( t ) + 40 = t we can multiply the entire equation by 20 to cancel out the denominators of the fractions and we have : 4 t + 5 t + 10 t + 800 = 20 t 19 t + 800 = 20 t 800 = t there are a total of 800 students in the math class . answer is d . | a = 1 / 5
b = 1 / 4
c = a + b
d = 1 / 2
e = c + d
f = 1 - e
g = 40 / f
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(floor(divide(subtract(subtract(40, 8), subtract(20, 5)), const_2)), const_1) | the scoring system in a certain football competition goes as follows : 3 points for victory , 1 point for a draw , and 0 points for defeat . each team plays 20 matches . if a team scored 8 points after 5 games , what is the least number of the remaining matches it has to win to reach the 40 - point mark by the end of the tournament ? | to get 40 points as end of season we need another 32 points or more from remaining 15 matches : option a = 6 * 3 + 9 * 1 = 27 option b = 7 * 3 + 8 * 1 = 29 option c = 8 * 3 + 7 * 1 = 31 option d = 9 * 3 + 6 * 1 = 33 hence option d - 9 | a = 40 - 8
b = 20 - 5
c = a - b
d = c / 2
e = math.floor(d)
f = e + 1
|
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | subtract(427398, multiply(floor(divide(427398, 12)), 12)) | what least number must be subtracted from 427398 so that remaining no . is divisible by 12 | explanation : on dividing 427398 by 12 we get the remainder 6 , so 6 should be subtracted answer : option c | a = 427398 / 12
b = math.floor(a)
c = b * 12
d = 427398 - c
|
a ) $ 41.60 , b ) $ 45.00 , c ) $ 50.75 , d ) $ 55.75 , e ) $ 61.25 | a | subtract(add(add(575, multiply(divide(4, const_100), 575)), 82.5), add(add(530, multiply(divide(3, const_100), 530)), 93)) | charges made by two companies for the same a / c company xcompany y price $ 575 $ 530 surcharge as a percent of price 4 % 3 % installation charge $ 82.50 $ 93.00 the table above shows the various charges made by two companies for the same air conditioner . what is the total amount that can be saved on the purchase and installation of the air conditioner by dealing with the company that offers the lower total charge ? | looking at the values / percentage , its clear that company x costlier than company y just observed the oa ; they have 4 distinct numbers after decimal point total calculation of company x gives . 00 + . 00 + . 50 = . 50 at the decimal total calculation of company y gives . 00 + . 90 + . 00 = . 90 at the decimal there difference provides . 60 at the decimal ; only 41.60 fits in answer = a | a = 4 / 100
b = a * 575
c = 575 + b
d = c + 82
e = 3 / 100
f = e * 530
g = 530 + f
h = g + 93
i = d - h
|
a ) 4 : 9 , b ) 5 : 7 , c ) 4 : 5 , d ) 4 : 1 , e ) 4 : 2 | b | divide(sqrt(25), sqrt(49)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 49 hours and 25 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = β b : β a = β 25 : β 49 = 5 : 7 answer : b" | a = math.sqrt(25)
b = math.sqrt(49)
c = a / b
|
a ) $ 30.14 , b ) 40.44 , c ) 34.66 , d ) 32.29 , e ) 33.16 | b | divide(add(211.00, divide(multiply(15, 211.00), const_100)), 6) | total dinning bill for 6 people was $ 211.00 . if they add 15 % tip and divided the bill evenly , approximate . what was each persons find share | "211 * 15 = 3165 / 100 = 31.65 211 + 31.65 = 242.65 242.65 / 6 = 40.44 answer : b" | a = 15 * 211
b = a / 100
c = 211 + 0
d = c / 6
|
a ) 90 , b ) 120 , c ) 250 , d ) 720 , e ) 910 | d | subtract(1000, multiply(add(add(multiply(add(1, 7), divide(10, const_2)), divide(10, const_2)), 1), 10)) | there are 7 baskets numbered from 1 to 7 and filled with apples . 10 children are asked to pick apples one by one from each basket such that the number of apples picked by each child from each basket is equal to the number marked on the basket . if there were 1000 apples in total and the baskets were filled in such a way that none of the basket became empty in the process , how many apples were left in the end ? | "each child takes a total of 1 + 2 + 3 + 4 + 5 + 6 + 7 = 7 * 8 / 2 = 28 apples . the total number of apples taken by 10 children is 10 * 28 = 280 the number of apples left in the end is 1000 - 280 = 720 apples . the answer is d ." | a = 1 + 7
b = 10 / 2
c = a * b
d = 10 / 2
e = c + d
f = e + 1
g = f * 10
h = 1000 - g
|
a ) 800 , b ) 900 , c ) 1000 , d ) 1100 , e ) none of these | c | divide(400, divide(2, 5)) | there are 400 female managers in a certain company . find the total number of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | as per question stem 2 / 5 m ( portion of men employees who are managers ) + 400 ( portion of female employees who are managers ) = 2 / 5 t ( portion of total number of employees who are managers ) , thus we get that 2 / 5 m + 400 = 2 / 5 t , or 2 / 5 ( t - m ) = 400 , from here we get that t - m = 1000 , that would be total number of female employees and the answer ( c ) | a = 2 / 5
b = 400 / a
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a ) 11 , b ) 7 , c ) 8 , d ) 20 , e ) 25 | b | add(inverse(subtract(divide(const_1, 6), divide(const_1, 42))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 6 days . if a alone can do it in 42 days . in how many days can b alone do it ? | "b 7 1 / 6 Γ’ β¬ β 1 / 42 = 1 / 7 = > 7" | a = 1 / 6
b = 1 / 42
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 14 | b | divide(divide(const_1, const_4), divide(divide(const_1, add(const_2, const_3)), 8)) | a bucket full of nuts was discovered by the crow living in the basement . the crow eats a fifth of the total number of nuts in 8 hours . how many hours in total will it take the crow to finish a quarter of the nuts ? | in one hour , the crow eats 1 / 40 of the nuts . ( 1 / 4 ) / ( 1 / 40 ) = 10 hours the answer is b . | a = 1 / 4
b = 2 + 3
c = 1 / b
d = c / 8
e = a / d
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a ) 87 , b ) 267 , c ) 68 , d ) 26 , e ) 21 | a | subtract(100, 11) | if all the numbers between 11 and 100 are written on a piece of paper . how many times will the number 4 be used ? | we have to consider the number of 4 ' s in two digit numbers . _ _ if we fix 4 in the 10 th place , unit place be filled with 10 ways . if we fix 4 in units place , 10 th place be filled with 9 ways ( 0 is not allowed ) so total 19 ways . alternatively : answer : a | a = 100 - 11
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['a ) 82 %', 'b ) 78 %', 'c ) 80 %', 'd ) 90 %', 'e ) none of these'] | a | multiply(subtract(multiply(add(const_1, divide(40, const_100)), add(const_1, divide(30, const_100))), const_1), const_100) | the length and breadth of a square are increased by 40 % and 30 % respectively . the area of the rectangle so formed exceeds the area of the square by ? | since side 1 x side 2 = area therefore , net % change in area = ( x + y + xy / 100 ) % = [ 40 + 30 + ( 40 x 30 ) / 100 ] % or 82 % . therefore , area is increased by 82 % . answer : a | a = 40 / 100
b = 1 + a
c = 30 / 100
d = 1 + c
e = b * d
f = e - 1
g = f * 100
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a ) 2 pm , b ) 9 pm , c ) 3 pm , d ) 6 pm , e ) 7 pm | d | subtract(multiply(const_2, const_12), divide(multiply(const_2, const_12), add(divide(2, 6), const_1))) | when asked what the time is , a person answered that the amount of time left is 2 / 6 of the time already completed . what is the time . | a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 β x = 2 / 6 x β x = 18 time is 6 pm answer : d | a = 2 * 12
b = 2 * 12
c = 2 / 6
d = c + 1
e = b / d
f = a - e
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a ) 100 km , b ) 140 km , c ) 50 km , d ) 120 km , e ) 200 km | b | multiply(7, 20) | a walks at 10 kmph and 7 hours after his start , b cycles after him at 20 kmph . how far from the start does b catch up with a ? | "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 7 hours . x / 10 - x / 20 = 7 x = 140 km answer is b" | a = 7 * 20
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['a ) 15 m', 'b ) 22.5 m', 'c ) 25 m', 'd ) 30 m', 'e ) none'] | c | subtract(divide(add(sqrt(add(multiply(750, const_4), power(5, const_2))), 5), const_2), 5) | the length of a rectangualr hll is 5 m more than its breadth . the area of the hall is 750 m . the length of the hall is | solution let the length = xmetres . then length = ( x + 5 ) metres . then , x ( x + 5 ) = 750 = x Β² + 5 x - 750 = 0 ( x + 30 ) ( x - 25 ) = 0 x = 25 . answer c | a = 750 * 4
b = 5 ** 2
c = a + b
d = math.sqrt(c)
e = d + 5
f = e / 2
g = f - 5
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a ) 15 , b ) 16 , c ) 12 , d ) 18 , e ) 19 | c | add(multiply(3, 2), const_1) | how many distinct integer values of n satisfy the inequality | | n - 3 | + 2 | β€ 12 ? | "so i can write this as in - 3 i + 2 < = 12 or in - 3 i < = 10 so n can have - 7 to 13 = 12 true values . . . . c" | a = 3 * 2
b = a + 1
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a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 54 | a | divide(power(105, 3), multiply(multiply(21, 35), 45)) | if a = 105 and a ^ 3 = 21 * 35 * 45 * b , what is the value of b ? | "first step will be to break down all the numbers into their prime factors . 105 = 3 * 5 * 7 21 = 7 * 3 35 = 7 * 5 45 = 3 * 3 * 5 so , ( 105 ) ^ 3 = 3 * 7 * 7 * 5 * 3 * 3 * 5 * b therefore ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 2 * 7 ^ 2 * b therefore , b = 3 ^ 3 * 5 ^ 3 * 7 ^ 3 / 3 ^ 3 * 5 ^ 2 * 7 ^ 2 b = 5 * 7 = 35 correct answer a ." | a = 105 ** 3
b = 21 * 35
c = b * 45
d = a / c
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a ) 350 m , b ) 200 m , c ) 400 m , d ) 700 m , e ) none of them | d | divide(multiply(175, 8), subtract(10, 8)) | a thief is spotted by a policeman from a distance of 175 meters . when the policeman starts the chase , the thief also starts running . if the speed of the thief be 8 km / hr and that of the policeman 10 km / hr , how far the thief will have run before he is overtaken ? | relative speed of the policeman = ( 10 - 8 ) km / hr = 2 km / hr . time taken by police man to cover ( 175 m / 1000 ) x 1 / 2 hr = 7 / 80 hr . in 7 / 80 hrs , the thief covers a distance of 8 x 7 / 80 km = 7 / 10 km = 700 m answer is d . | a = 175 * 8
b = 10 - 8
c = a / b
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a ) 1200 , b ) 750 , c ) 800 , d ) 600 , e ) 900 | c | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(60, const_100))), 2000) | in an election between two candidates , one got 60 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 2000 , the number of valid votes that the other candidate got , was : | "number of valid votes = 80 % of 2000 = 1600 . valid votes polled by other candidate = 40 % of 2000 = ( 40 / 100 ) x 2000 = 800 answer = c" | a = 20 / 100
b = 1 - a
c = 60 / 100
d = 1 - c
e = b * d
f = e * 2000
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a ) 50 , b ) 25 , c ) 86 , d ) 75 , e ) 102 | e | add(divide(circumface(20), const_2), multiply(const_2, 20)) | a semicircle has a radius of 20 . what is the approximate perimeter of the semicircle ? | "circumference of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 20 + 2 * 20 = 102.8 approximately 102 answer e" | a = circumface / (
b = a + 2
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a ) 375 m , b ) 200 m , c ) 300 m , d ) 400 m , e ) 100 m | a | divide(multiply(18, multiply(1.5, const_1000)), 48) | amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.5 km ? | "a 375 m distance covered by amar = 18 / 4.8 ( 1.5 km ) = 3 / 8 ( 1500 ) = 375 m" | a = 1 * 5
b = 18 * a
c = b / 48
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a ) $ 100 , b ) $ 220 , c ) $ 280 , d ) $ 300 , e ) $ 360 | a | subtract(subtract(400, divide(multiply(400, 2), 5)), multiply(subtract(subtract(20, divide(multiply(20, 3), 5)), const_1), 20)) | a prize of $ 400 is to be distributed among 20 winners , each of whom must be awarded at least $ 20 . if 2 / 5 of the prize will be distributed to 3 / 5 of the winners , what is the greatest possible individual award ? | "total value of the prize = $ 400 number of people = 20 2 / 5 of 400 ( = $ 160 ) should be distributed among 3 / 5 of 20 ( = 12 people ) with each getting $ 20 each . remaining money = 400 - 160 = $ 240 . now in order to ' maximize ' 1 prize , we need to minimise the others and we have been given that each should get $ 20 . thus , minimising the remaining 7 people ( = 20 - 12 - 1 . ' - 1 ' to exclude 1 that needs to be maximised ) = 7 * 20 = 140 . thus the maximum award can be = 240 - 140 = $ 100 , hence a is the correct answer ." | a = 400 * 2
b = a / 5
c = 400 - b
d = 20 * 3
e = d / 5
f = 20 - e
g = f - 1
h = g * 20
i = c - h
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a ) 178.27 cm , b ) 179.29 cm , c ) 978.29 cm , d ) 178.89 cm , e ) 176.29 cm | b | floor(divide(add(subtract(multiply(35, 181), 166), 106), 35)) | the average height of 35 boys in a class was calculated as 181 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . ? | "calculated average height of 35 boys = 181 cm . wrong total height of 35 boys = 181 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 181 cm - ( 166 cm - 106 cm ) / 35 = 181 cm - 60 / 35 cm = 181 cm - 1.71 cm = 179.29 cm . answer : b" | a = 35 * 181
b = a - 166
c = b + 106
d = c / 35
e = math.floor(d)
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a ) 220 , b ) 210 , c ) 200 , d ) 240 , e ) 260 | d | multiply(divide(multiply(20, const_3), 10), 40) | if 10 typists can type 20 letters in 20 minutes , then how many letters will 40 typists working at the same rate complete in 1 hour ? | "no . of letters typing by 10 typists in 20 minutes = 20 no . of letters typing by 10 typists in 60 minutes = 20 * 3 = 60 no . of letters typing by 40 typists in 60 minutes = 60 / 10 * 40 = 240 answer : d" | a = 20 * 3
b = a / 10
c = b * 40
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a ) 13 seconds , b ) 17 seconds , c ) 36 seconds , d ) 34 seconds , e ) 51 seconds | c | divide(add(divide(multiply(3.7, add(16, 2)), subtract(4.2, 3.7)), add(16, 2)), 4.2) | john and steve are speed walkers in a race . john is 16 meters behind steve when he begins his final push . john blazes to the finish at a pace of 4.2 m / s , while steve maintains a blistering 3.7 m / s speed . if john finishes the race 2 meters ahead of steve , how long was john β s final push ? | "let t be the time that john spent for his final push . thus , per the question , 4.2 t = 3.7 t + 16 + 2 - - - > 0.5 t = 18 - - - > t = 36 seconds . c is the correct answer ." | a = 16 + 2
b = 3 * 7
c = 4 - 2
d = b / c
e = 16 + 2
f = d + e
g = f / 4
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a ) 107 , b ) 108 , c ) 109 , d ) 110 , e ) 111 | b | divide(multiply(multiply(9, 12), 3), 3) | a lady builds 9 cm length , 12 cm width , and 3 cm height box using 3 cubic cm cubes . what is the minimum number of cubes required to build the box ? | "number of cubes required = volume of box / volume of cube = 9 * 12 * 3 / 3 = 108 cubes answer : b" | a = 9 * 12
b = a * 3
c = b / 3
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a ) 242 , b ) 353 , c ) 464 , d ) 575 , e ) 686 | c | divide(multiply(58, 32), 4) | dan β s car gets 32 miles per gallon . if gas costs $ 4 / gallon , then how many miles can dan β s car go on $ 58 of gas ? | "58 / 4 = 14.5 gallons 14.5 * 32 = 464 miles the answer is c ." | a = 58 * 32
b = a / 4
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a ) 24 , b ) 9 , c ) 72 , d ) 144 , e ) 216 | b | multiply(multiply(subtract(const_10, const_1), subtract(const_10, const_2)), const_1) | of the two - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? | of the two - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? a . 24 b . 36 c . 72 d . 144 e . 216 choosing the digit for x - 9 ways ; # of permutations of 3 digits in xx - 2 ! / 2 ! total : 9 * 2 ! / 2 ! = 9 . answer : b . | a = 10 - 1
b = 10 - 2
c = a * b
d = c * 1
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a ) 362.33 , b ) 399.6 , c ) 702.33 , d ) 702 , e ) none of them | b | add(add(add(33, 333), 33.3), divide(3, const_10)) | . 3 + 33 + 333 + 33.3 = ? | ". 3 33 333 33.3 - - - - - - - - - - 399.6 - - - - - - - - - - answer is b" | a = 33 + 333
b = a + 33
c = 3 / 10
d = b + c
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a ) 3 , b ) 5 , c ) 6 , d ) 22 , e ) 18 | d | add(5, 17) | if a certain number x is divided by 82 , the reminder is 5 . what is the reminder when x + 17 is divided by 41 ? | "x can be written as 82 k + 5 or x = 5 , 87,169 , etc . x + 17 = 82 k + 5 + 17 = 82 k + 22 or x + 17 = 22,104 , 186 etc . when divided by 41 , we will get the remainder 22 . d" | a = 5 + 17
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a ) 1 , b ) 2 , c ) 3 , d ) 8 , e ) 9 | c | divide(subtract(multiply(48, 15), multiply(45, 15)), subtract(60, 45)) | at a certain fruit stand , the price of each apple is 40 cents and the price of each orange is 60 cents . mary selects a total of 15 apples and oranges from the fruit stand , and the average ( arithmetic mean ) price of the 15 pieces of fruit is 48 cents . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 45 cents ? | let number of apples = a number of oranges = b a + b = 15 - - - 1 . 48 = ( . 4 a + . 6 b ) / 15 = > 48 = 4 a + 6 b - - - - 2 solving 1 and 2 , we get a = 9 b = 6 let the number of oranges put back = c 45 * ( 15 - c ) = 40 * 9 + 60 ( 6 - c ) = > c = 3 answer c | a = 48 * 15
b = 45 * 15
c = a - b
d = 60 - 45
e = c / d
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a ) 100 , b ) 200 , c ) 10 ^ 4 , d ) 10 ^ 5 , e ) 10 ^ 6 | a | power(const_10, subtract(5, 3)) | on the richter scale , which measures the total amount of energy released during an earthquake , a reading of x - 1 indicates one - tenth the released energy as is indicated by a reading of x . on that scale , the frequency corresponding to a reading of 5 is how many times as great as the frequency corresponding to a reading of 3 ? | if richter scale reading goes from x - 1 to x it will be 10 if richter scale reading goes from 3 to 4 it will be 10 if richter scale reading goes from 4 to 5 it will be 10 so it will from 3 to 5 i . e 4,5 = 10 * 10 = 10 ^ 2 answer is a | a = 5 - 3
b = 10 ** a
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a ) 40 % , b ) 19 % , c ) 25 % , d ) 16.66 % , e ) 9 % | b | subtract(add(multiply(multiply(const_4, 8), divide(30, const_100)), multiply(const_4, 8)), const_100) | the balance of a trader weighs 8 % less than it should . still the trader marks up his goods to get an overall profit of 30 % . what is the mark up on the cost price ? | "the most natural way to deal with ' weights ' questions is by assuming values . say the trader ' s balance shows 100 gms . it is actually 92 gms because it weighs 8 % less . say , the cost price is $ 92 ( $ 1 / gm ) . since he gets a profit of 30 % , the selling price must be 92 + ( 30 / 100 ) * 90 = $ 119 since the cost price is actually supposed to be $ 100 ( for 100 gms ) and the selling price is $ 119 , the mark up is simply 19 % ans : b ." | a = 4 * 8
b = 30 / 100
c = a * b
d = 4 * 8
e = c + d
f = e - 100
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a ) 12 liters , b ) 32 liters , c ) 46 liters , d ) 50 liters , e ) 34 liters | c | multiply(divide(115, add(3, 2)), 2) | 115 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ? | "milk = 3 / 5 * 115 = 69 liters water = 46 liters 69 : ( 46 + p ) = 3 : 4 138 + 3 p = 276 = > p = 46 46 liters of water are to be added for the ratio become 3 : 4 . answer : c" | a = 3 + 2
b = 115 / a
c = b * 2
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a ) s . 109 , b ) s . 108 , c ) s . 100 , d ) s . 95 , e ) s . 102 | d | multiply(5, divide(1406, add(add(multiply(6, 3), multiply(9, 4)), multiply(4, 5)))) | a , b and c completed a piece of work , a worked for 6 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was rs . 1406 ? | "3 x 4 x 5 x 6 9 4 18 x + 36 x + 20 x = 1406 74 x = 1406 = > x = 19 5 x = 95 rs . answer : d" | a = 6 * 3
b = 9 * 4
c = a + b
d = 4 * 5
e = c + d
f = 1406 / e
g = 5 * f
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a ) 6 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | a | add(sqrt(divide(subtract(power(10, const_2), power(const_2, const_2)), const_2)), const_2) | efrida and frazer who live 10 miles apart , meet at a restaurant that is directly north of efrida ' s home and directly east of frazer ' s home . if the restaurant is four miles closer to efrida ' s home , than to frazer ' s home , how many miles is the restaurant from frazer ' s home ? | "it ' s a simple geometry problem . look at the diagram below : answer : a ." | a = 10 ** 2
b = 2 ** 2
c = a - b
d = c / 2
e = math.sqrt(d)
f = e + 2
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a ) 0.004 , b ) 0.05555 , c ) 2.775 , d ) 3.6036 , e ) 36.036 | b | inverse(add(divide(2, 0.03), divide(2, 0.37))) | 2 / [ ( 1 / 0.03 ) + ( 1 / 0.37 ) ] = ? | "approximate . 1 / . 03 = 100 / 3 = 33 1 / . 37 = 100 / 37 = 3 denominator becomes 33 + 3 = 36 2 / 36 = . 05 something answer ( b )" | a = 2 / 0
b = 2 / 0
c = a + b
d = 1/(c)
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a ) 7 / 25 , b ) 3 / 25 , c ) 8 / 35 , d ) 2 / 13 , e ) 3 / 17 | c | multiply(divide(4, 7), divide(2, 5)) | if p ( a ) = 4 / 7 and p ( b ) = 2 / 5 , find p ( a n b ) if a and b are independent events . | p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = 4 / 7 . 2 / 5 p ( a n b ) = 8 / 35 . c | a = 4 / 7
b = 2 / 5
c = a * b
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a ) $ 0.25 , b ) $ 0.12 , c ) $ 0.05 , d ) $ 0.003 , e ) $ 0.005 | e | subtract(divide(3.0, 100), divide(6.25, 250)) | bottle r contains 250 capsules and costs $ 6.25 . bottle t contains 100 capsules and costs $ 3.0 . what is the difference between the cost per capsule for bottle r and the cost per capsule for bottle t ? | "cost per capsule in r is 6.25 / 250 = 0.625 / 25 = 0.025 cost per capsule in t is 3.00 / 100 = 0.03 the difference is 0.005 the answer is e" | a = 3 / 0
b = 6 / 25
c = a - b
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a ) 2 / 7 , b ) 9 / 4 , c ) 16 / 25 , d ) 7 / 1 , e ) 7 / 2 | c | divide(multiply(multiply(multiply(2, 2), 2), 2), multiply(5, 5)) | for what value of β k β will the equation ( 2 kx 2 + 5 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 ( 5 k ) ^ 2 - 4 * 2 k * 2 = 0 by solving this equation we get k = 0 , k = 16 / 25 answer : c | a = 2 * 2
b = a * 2
c = b * 2
d = 5 * 5
e = c / d
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a ) w = 16 , b ) w = 32 , c ) 64 , d ) 128 , e ) 512 | b | multiply(8, 4) | the weight of a hollow sphere is directly dependent on its surface area . the surface area of a sphere is 4 Ο Β· r ^ 2 , where r is the radius of the sphere . if a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams , a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams w ? | "weight directly proportional to 4 pi r ^ 2 now , 4 pi is constant , so , weight is directly proportional to r ^ 2 . when radius = 0.15 , weight = 8 , so ( 0.15 ) ^ 2 proportional to 8 ; ( 0.15 ) ^ 2 * 4 proportional to 8 * 4 , solving further ( 0.15 ) ^ 2 * 2 ^ 2 = ( 0.15 * 2 ) ^ 2 = 0.3 ^ 2 ; so answer = 32 ( b )" | a = 8 * 4
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a ) 16 days , b ) 26 days , c ) 27 days , d ) 28 days , e ) 29 days | b | subtract(30, const_4) | it takes 30 days to fill a laboratory dish with bacteria . if the size of the bacteria doubles each day , how long did it take for the bacteria to fill one sixteenth of the dish ? | the bacteria doubles each day , so after 29 days , the dish was half full . after 28 days , the dish was one quarter full . after 27 days , the dish was one eighth full . after 26 days , the dish was one sixteenth full . the answer is b . | a = 30 - 4
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a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | a | multiply(multiply(8, divide(const_1, const_2)), divide(const_1, const_2)) | some persons can do a piece of work in 8 days . two times the number of these people will do half of that work in ? | "8 / ( 2 * 2 ) = 2 days answer : a" | a = 1 / 2
b = 8 * a
c = 1 / 2
d = b * c
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['a ) 600', 'b ) 400', 'c ) 200', 'd ) 500', 'e ) 100'] | c | divide(multiply(1000, 210), multiply(35, 30)) | rectangular tile each of size 35 cm by 30 cm must be laid horizontally on a rectangular floor of size 1000 cm by 210 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | area of tile = 35 * 30 = 1050 area of floor = 1000 * 210 = 210000 no of tiles = 210000 / 1050 = 200 so , the no of tile = 200 answer : c | a = 1000 * 210
b = 35 * 30
c = a / b
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a ) 10 , b ) 100 , c ) 1000 , d ) 10000 , e ) none of these | c | multiply(1000, 10) | ( 1000 ) 7 Γ· ( 10 ) 18 = ? | "explanation : = ( 103 ) 7 / ( 10 ) 18 = ( 10 ) 21 / ( 10 ) 18 = 10 ( 3 ) = 1000 option c" | a = 1000 * 10
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a ) 80 , b ) 90 , c ) 100 , d ) 70 , e ) 110 | c | add(multiply(divide(subtract(100, const_4), multiply(2, const_4)), 3), 2) | the length of a rectangular garden is 2 feet longer than 3 times its width . if the perimeter of the garden is 100 feet , find the width and the length of the garden . | "let l and w be the length and width of the garden . the statement ` ` the length of a rectangular garden is 2 feet longer than 3 times its width ' ' may be formulated by l = 2 + 3 w the formula for the perimeter is given by p = 2 l + 2 w substitute p and l in the above equation by 100 and 2 + 3 w respectively to obtain 100 = 2 ( 2 + 3 w ) + 2 w solve for w and l w = 12 and l = 2 + 3 w = 38 . check that the perimeter of the rectangular garden is 100 p = 2 l + 2 w = 76 + 24 = 100 answer c" | a = 100 - 4
b = 2 * 4
c = a / b
d = c * 3
e = d + 2
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a ) 3 / 2 , b ) 5 / 4 , c ) 10 / 3 , d ) 20 / 7 , e ) 25 / 9 | d | divide(const_1, divide(add(divide(3, 5), multiply(divide(3, 5), divide(1, 6))), const_2)) | if a certain toy store ' s revenue in november was 3 / 5 of its revenue in december and its revenue in january was 1 / 6 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 = 3 d / 5 j = n / 6 = d / 10 the average of november and january is ( n + j ) / 2 = 7 d / 10 / 2 = 7 d / 20 d is 20 / 7 times the average of november and january . the answer is d . | a = 3 / 5
b = 3 / 5
c = 1 / 6
d = b * c
e = a + d
f = e / 2
g = 1 / f
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a ) 25.6 % , b ) 21 % , c ) 20 % , d ) 19 % , e ) none of these | a | divide(multiply(subtract(multiply(540, divide(add(const_100, 15), const_100)), 462), const_100), multiply(540, divide(add(const_100, 15), const_100))) | mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 462 ? | "cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 462 , discount = 621 - 462 = 159 discount % = 159 / 621 * 100 = 25.6 % answer : a" | a = 100 + 15
b = a / 100
c = 540 * b
d = c - 462
e = d * 100
f = 100 + 15
g = f / 100
h = 540 * g
i = e / h
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a ) 31.25 % , b ) 35.5 % , c ) 37.5 % , d ) 39.5 % , e ) 30.5 % | a | multiply(subtract(multiply(divide(const_100, 64), divide(subtract(const_100, 16), const_100)), const_1), const_100) | the cost price of an book is 64 % of the marked price . calculate the gain percent after allowing a discount of 16 % ? | "let marked price = $ 100 . then , c . p . = $ 64 , s . p . = $ 84 gain % = 20 / 64 * 100 = 31.25 % . a" | a = 100 / 64
b = 100 - 16
c = b / 100
d = a * c
e = d - 1
f = e * 100
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a ) β 0.3 , b ) 0.8 , c ) 0.3 , d ) 1.08 , e ) 2.46 | b | subtract(multiply(divide(divide(subtract(power(3, 2), power(1.8, 0.8)), const_1000), const_1000), 3), divide(divide(subtract(power(3, 2), power(1.8, 0.8)), const_1000), const_1000)) | what is the value of 3 x ^ 2 β 1.8 x + 0.8 for x = 0.6 ? | "3 x ^ 2 - 1.8 x + 0.8 for x = 0.6 = 3 ( 0.6 * 0.6 ) - 3 * 0.6 * ( 0.6 ) + 0.8 = 0 + 0.8 = 0.8 answer : b" | a = 3 ** 2
b = 1 ** 8
c = a - b
d = c / 1000
e = d / 1000
f = e * 3
g = 3 ** 2
h = 1 ** 8
i = g - h
j = i / 1000
k = j / 1000
l = f - k
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a ) 24,602 , b ) 28,888 , c ) 25,800 , d ) 24,628 , e ) 24,6012 | c | divide(64.50, divide(const_4, 4)) | a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 3 / 4 % his yearly income diminishes by rs . 64.50 , his capital is ? | "let the capital be rs . x . then , ( x * 8 * 1 ) / 100 - ( x * 31 / 4 * 1 / 100 ) = 64.50 32 x - 31 x = 6450 * 4 x = 25,800 . answer : c" | a = 4 / 4
b = 64 / 50
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a ) 212 , b ) 222 , c ) 232 , d ) 242 , e ) 252 | c | multiply(add(divide(subtract(subtract(27, 10), const_2), const_2), 10), divide(add(subtract(27, 10), const_2), const_2)) | what is the sum of all digits for the number 10 ^ 27 - 48 ? | "10 ^ 27 is a 28 - digit number : 1 followed by 27 zeros . 10 ^ 27 - 48 is a 27 - digit number : 25 9 ' s and 52 at the end . the sum of the digits is 25 * 9 + 5 + 2 = 232 . the answer is c ." | a = 27 - 10
b = a - 2
c = b / 2
d = c + 10
e = 27 - 10
f = e + 2
g = f / 2
h = d * g
|
a ) 0 , b ) 1 , c ) 4 , d ) 3 , e ) 6 | d | divide(add(multiply(factorial(493), factorial(913)), multiply(factorial(493), factorial(756))), 493) | what is the units digit of ( 493 ) ( 913 ) ( 381 ) ( 756 ) ( 29 ) | "just multiply the digits in the units place for each term and you will get the answer . it should be 0 . you got a 5 as a unit digit and an even number term . so the multiplication of this will definitely yield a 0 . answer has to be 0 . i also tried it using the calculator and the answer is 3 . imo d ." | a = math.factorial(493)
b = math.factorial(913)
c = a * b
d = math.factorial(493)
e = math.factorial(756)
f = d * e
g = c + f
h = g / 493
|
a ) 123 , b ) 127 , c ) 2 , d ) 305 , e ) 505 | c | gcd(subtract(2037, 5), subtract(1557, 7)) | the greatest number which on dividing 1557 and 2037 leaves remainders 7 and 5 respectively , is : | "explanation : required number = h . c . f . of ( 1557 - 7 ) and ( 2037 - 5 ) = h . c . f . of 1550 and 2032 = 2 . answer : c" | a = 2037 - 5
b = 1557 - 7
c = math.gcd(a, b)
|
['a ) 24 cm', 'b ) 36 * 2 ^ 1 / 2 cm', 'c ) 36 cm', 'd ) 54 cm', 'e ) 42 cm'] | e | add(18, add(18, 6)) | there is a right angle triangle inside a circle of diameter 18 cm . inside this triangle , there is another circle of diameter 6 cm . find the perimeter of the triangle . | 1 ) the radius of the incircle of a right triangle with legs a and b and hypotenuse c is r = ( a + b - c ) / 2 2 ) the radius of the circumcircle is half the length of the hypotenuse r = c / 2 3 ) thus the sum of the circumradius and the inradius is half the sum of the legs r + r = ( a + b ) / 2 from 2 nd rule c = 18 cm from 3 rd rule a + b = 2 * ( 9 + 3 ) = 24 cm hence perimeter a + b + c = 24 + 18 = 42 cm answer : e | a = 18 + 6
b = 18 + a
|
a ) 20 % , b ) 29 % , c ) 13 1 / 30 % , d ) 12 % , e ) 13 % | b | subtract(divide(subtract(const_100, 14), divide(2, 3)), const_100) | what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 14 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 86 2 / 3 sp 1 = 86 sp 1 = 129 100 - - - 29 = > 29 % answer : b" | a = 100 - 14
b = 2 / 3
c = a / b
d = c - 100
|
a ) 5 days , b ) 10 days , c ) 14 days , d ) 22 days , e ) 26 days | a | add(2, divide(subtract(const_1, divide(2, 10)), add(inverse(10), inverse(6)))) | p and q can complete a work in 10 days and 6 days respectively . p alone started the work and q joined him after 2 days till the completion of the work . how long did the work last ? | "explanation : work done by p in 1 day = 1 / 10 work done by q in 1 day = 1 / 6 work done by p in 2 days = 2 Γ£ β ( 1 / 10 ) = 1 / 5 remaining work = 1 Γ’ β¬ β 1 / 5 = 4 / 5 work done by p and q in 1 day = 1 / 10 + 1 / 6 = 4 / 15 number of days p and q take to complete the remaining work = ( 4 / 5 ) / ( 4 / 15 ) = 3 total days = 2 + 3 = 5 answer : option a" | a = 2 / 10
b = 1 - a
c = 1/(10)
d = 1/(6)
e = c + d
f = b / e
g = 2 + f
|
a ) 12 , b ) 14 , c ) 18 , d ) 24 , e ) 23 | a | subtract(add(18, 6), multiply(const_2, 6)) | david is 18 years older than his daughter rosy . in 6 years david will be twice as old as rosy . what is rosy current age . | now : rosy = x , david = x + 18 in 6 years rosy = x + 6 , david = x + 18 + 6 or 2 ( x + 6 ) x + 24 = 2 x + 12 x = 12 rosy is 12 years old answer : a | a = 18 + 6
b = 2 * 6
c = a - b
|
a ) 39 , b ) 40 , c ) 65 , d ) 42 , e ) 43 | c | subtract(multiply(add(22, 20), const_2), 19) | you collect pens . suppose you start out with 20 . mike gives you another 22 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 20 pens . mike gives you 22 pens : 20 + 22 = 42 pens . cindy doubles the number of pens you have : 42 Γ£ β 2 = 84 pens . sharon takes 19 pens from you : 84 - 19 = 65 pens . so you have 65 at the end . correct answer : c | a = 22 + 20
b = a * 2
c = b - 19
|
a ) 6 , b ) 6.5 , c ) 7 , d ) 7.5 , e ) 5 | a | divide(210, add(30, add(10, 20))) | a train travels at the rate of 10 miles / hr for the first hour of a trip , at 20 miles / hr for the second hour , at 30 miles / hr for the third hour and so on . how many hours will it take the train to complete a 210 - mile journey ? assume that the train makes no intermediate stops . | "a train travels at the rate of 10 miles / hr for the first hour of a trip , at 20 miles / hr for the second hour , at 30 miles / hr for the third hour and so on . how many hours will it take the train to complete a 210 - mile journey ? assume that the train makes no intermediate stops . i think the easiest way to solve this problem would be simply to count the number of miles it travels per hour ( and in total ) hour miles / hour total miles 1 10 10 2 20 30 3 30 60 4 40 100 5 50 150 6 60 210 it takes a total of nine hours to cover the 210 mile distance . answer : a 6" | a = 10 + 20
b = 30 + a
c = 210 / b
|
a ) 140 , b ) 340 , c ) 210 , d ) 245 , e ) 280 | b | subtract(30, subtract(40, 30)) | sarah is driving to the airport . after driving at 40 miles per hour for one hour , she realizes that if she continues at that same average rate she will be an hour late for her flight . she then travels 50 miles per hour for the rest of the trip , and arrives 30 minutes before her flight departs . how many miles did she drive in all ? | "after driving at 40 miles per hourfor one hour , this distance left to cover is d - 40 . say this distance is x miles . now , we know that the difference in time between covering this distance at 40 miles per hour and 50 miles per hour is 1 + 1 / 2 = 3 / 2 hours . so , we have that x / 40 - x / 50 = 3 / 2 - - > 5 x / 200 - 4 x / 200 = 3 / 2 - - > x / 200 = 3 / 2 - - > x = 300 . total distance = x + 40 = 340 miles . answer : b" | a = 40 - 30
b = 30 - a
|
a ) 121 , b ) 133 , c ) 145 , d ) 150 , e ) 165 | b | add(subtract(131, 3), subtract(add(3, 3), 1)) | the sum of the 1 st and 3 rd of 3 consecutive odd integers is 131 less than 3 times the second integer . find the 3 integers ? | let x , x + 2 and x + 4 be three integers . the sum of the first x and third x + 4 is given by x + ( x + 4 ) 131 less than three times the second 3 ( x + 2 ) is given by 3 ( x + 2 ) - 131 ` ` the sum of the first and third is 131 less than three times the second ' ' gives x + ( x + 4 ) = 3 ( x + 2 ) - 131 solve for x and find all three numbers x = 129 , x + 2 = 131 , x + 4 = 133 as an exercise , check that the sum of the first and third is 131 less than three times b | a = 131 - 3
b = 3 + 3
c = b - 1
d = a + c
|
a ) 100 % , b ) 150 % , c ) 200 % , d ) 250 % , e ) 300 % | c | subtract(subtract(divide(125000, const_100), const_1000), add(multiply(const_4, const_10), const_10)) | a local business made a $ 125000 profit on its first $ 1 million in sales and a $ 80000 profit on its next $ 2 million in sales . by approximately what percent did the ratio of profit to sales decrease from the first $ 1 million in sales to the next $ 2 million in sales ? | solution : this is a percent decrease problem . we will use the formula : percent change = ( new β old ) / old x 100 to calculate the final answer . we first set up the ratios of profits to sales . the first ratio will be for the first 1 million in sales , and the second ratio will be for the next 2 million in sales . because all of the sales are in millions , we do not have to express all the trailing zeros in our ratios . first 1 million profit / sales = 125 / 1000 = 1 / 8 next 2 million profit / sales = 8 / 200 = 2 / 50 we can simplify each ratio by multiplying each by the lcm of the two denominators , which is 400 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first 1 million royalties / sales = ( 1 / 8 ) x 400 = 50 next 2 million royalties / sales = ( 2 / 50 ) x 400 = 16 we can plug 5 and 6 into our percent change formula : ( new β old ) / old x 100 [ ( 16 β 50 ) / 16 ] x 100 = - 34 / 16 x 100 so approximately a 200 % decrease . answer c . | a = 125000 / 100
b = a - 1000
c = 4 * 10
d = c + 10
e = b - d
|
a ) 16 % , b ) 55 % , c ) 29 % , d ) 40 % , e ) 52 % | c | multiply(divide(subtract(64, 44), 64), const_100) | in town x , 64 percent of the population are employed , and 44 percent of the population are employed males . what percent of the employed people in town x are females ? | "answer b male - employed - 64 % female - employed - 20 % total employed 64 % means total unemployed = 36 % therefore d and e are not correct because female unemployed has to be less than 36 % female - unemployed = 32 % male unemployed = 4 % 48 % + 4 % = 52 % 16 % + 32 % = 48 % plug in a and b in place of 32 % and the sum is not 100 % 20 / 64 = 29 c" | a = 64 - 44
b = a / 64
c = b * 100
|
a ) 74 , b ) 76 , c ) 78 , d ) 80 , e ) 82 | a | divide(add(multiply(58, 3), multiply(98, 2)), add(2, 3)) | for a certain exam , a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean . what was the mean score t for the exam ? | "a score of 58 was 2 standard deviations below the mean - - > 58 = mean - 2 d a score of 98 was 3 standard deviations above the mean - - > 98 = mean + 3 d solving above for mean t = 74 . answer : a ." | a = 58 * 3
b = 98 * 2
c = a + b
d = 2 + 3
e = c / d
|
a ) 0.05 , b ) 0.25 , c ) 0.5 , d ) 2.5 , e ) 25 | b | divide(multiply(multiply(sqrt(5), sqrt(5)), 5), const_100) | β 5 percent of 5 β 5 = | "β / 1005 / 100 * 55 β 5 = ( 5 * 5 β 5 β β β 5 β 5 ) / 100 = ( 5 * 5 ) / 100 = 25 / 100 = . 25 answer : b" | a = math.sqrt(5)
b = math.sqrt(5)
c = a * b
d = c * 5
e = d / 100
|
a ) 13 , b ) 15 , c ) 16 , d ) 7 , e ) 18 | d | add(divide(subtract(multiply(floor(divide(31, 3)), 3), multiply(add(floor(divide(10, 3)), const_1), 3)), 3), const_1) | how many numbers from 10 to 31 are exactly divisible by 3 ? | "12 , 15 , 18 , 21 , 24 , 27 , 30 . 7 numbers . 10 / 3 = 3 and 31 / 3 = 10 = = > 10 - 3 = 7 . therefore 7 digits d )" | a = 31 / 3
b = math.floor(a)
c = b * 3
d = 10 / 3
e = math.floor(d)
f = e + 1
g = f * 3
h = c - g
i = h / 3
j = i + 1
|
a ) 28 , b ) 25 , c ) 35 , d ) 45 , e ) 55 | a | subtract(add(add(20, 40), 60), add(add(multiply(4, const_3), 10), 70)) | the average ( arithmetic mean ) of 20 , 40 , and 60 is 4 more than the average of 10 , 70 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 4 = 36 sum of second list = 36 * 3 = 108 therefore the number = 108 - 80 = 28 answer : a" | a = 20 + 40
b = a + 60
c = 4 * 3
d = c + 10
e = d + 70
f = b - e
|
a ) 5 , b ) 8 , c ) 4 , d ) 6 , e ) 7 | c | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 221) | what is the least number . which should be added to 221 to make it a perfect square ? | "221 + 4 = 225 15 ^ 2 answer : c" | a = 12 + 4
b = a + 2
c = b / 100
d = 12 + 4
e = d + 2
f = e / 100
g = c * f
h = g - 221
|
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