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 ) 130 , b ) 152 , c ) 225 , d ) 260 , e ) 180 | a | multiply(20, 6.5) | a man performs 3 / 5 of the total journey by rail , 17 / 20 by bus and the remaining 6.5 km on foot . his total journey is ? | "let the total journey be x km then , ( 3 x / 5 ) + ( 7 x / 20 ) + 6.5 = x 12 x + 7 x + 20 * 6.5 = 20 x x = 130 km answer is a" | a = 20 * 6
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a ) 1 / 5 , b ) 2 / 22 , c ) 3 / 4 , d ) 7 / 15 , e ) 5 / 6 | d | multiply(divide(divide(3, 5), divide(6, 7)), divide(2, 3)) | find the fraction which has the same ratio to 2 / 3 that 3 / 5 has to 6 / 7 | "p : 2 / 3 = 3 / 5 : 6 / 7 as the product of the means is equal to the product of the extremes . p * 6 / 7 = 2 / 3 * 3 / 5 p * 6 / 7 = 6 / 15 p = 7 / 15 = > p = 7 / 15 answer : d" | a = 3 / 5
b = 6 / 7
c = a / b
d = 2 / 3
e = c * d
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a ) 3 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | d | divide(divide(divide(120, 4), const_2), const_3) | how many of the positive divisors g of 120 are also multiples of 4 not including 120 ? | "4 , 8,12 , 20,24 , 40,60 . ( 7 ) is the answer other way : factors of 120 = 2 ^ 3 * 3 * 5 separate 2 ^ 2 ( which means 4 ) now , calculate the number of other factors . g = 2 * 3 * 5 = total positive factors are 2 * 2 * 2 = 8 this 8 factors include 120 so subtract 1 from 8 ans is 7 = d" | a = 120 / 4
b = a / 2
c = b / 3
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a ) 5846381256 , b ) 5339179476 , c ) 5346381356 , d ) 5846381406 , e ) 5346381456 | b | multiply(9568422, power(add(const_4, const_1), const_4)) | ( 9568422 x 558 ) = ? | 9568422 x 558 = 5339179476 ans b | a = 4 + 1
b = a ** 4
c = 9568422 * b
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a ) 8 % , b ) 16 % , c ) 20 % , d ) 24 % , e ) 40 % | a | multiply(divide(subtract(multiply(add(add(multiply(50, const_2), 50), multiply(50, const_2)), divide(4, 5)), 70), add(add(multiply(50, const_2), 50), multiply(50, const_2))), multiply(50, const_2)) | in a certain math department , students are required to enroll in either calculus or trigonometry , each of which is offered in beginner and advanced courses . the number of students enrolled in trigonometry is 50 % greater than the number of students enrolled in calculus , and 70 % of calculus students are enrolled in the beginner course . if 4 / 5 of students are in the beginner courses , and one student is selected at random , what is the probability that an advanced trigonometry student is selected ? | "let x be the number of students in calculus . then the number of students in trigonometry is 1.5 x the number of students on beginner calculus is 0.7 x the number of students in beginner trigonometry is 4 / 5 * ( 2.5 x ) - 0.7 x = 1.3 x the number of students in advanced trigonometry is 0.2 x the percentage in advanced trigonometry is 0.2 x / 2.5 x = 8 % the answer is a ." | a = 50 * 2
b = a + 50
c = 50 * 2
d = b + c
e = 4 / 5
f = d * e
g = f - 70
h = 50 * 2
i = h + 50
j = 50 * 2
k = i + j
l = g / k
m = 50 * 2
n = l * m
|
['a ) 50 %', 'b ) 46 %', 'c ) 36 %', 'd ) 26 %', 'e ) 38 %'] | c | divide(multiply(circle_area(60), const_100), circle_area(const_100)) | if the diameter of circle r is 60 % of the diameter of circle s , the area of circle r is what percent of the area of circle s ? | let diameter of circle r , dr = 60 and diameter of circle s , ds = 100 radius of circle r , rr = 30 radius of circle s , rs = 50 area of circle r / area of circle s = ( pi * rr ^ 2 ) / ( pi * rs ^ 2 ) = ( 30 / 50 ) ^ 2 = ( 6 / 10 ) ^ 2 = 36 % answer : c | a = circle_area * (
b = a / 100
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a ) a ) 182 , b ) b ) 208 , c ) c ) 220 , d ) d ) 224 , e ) e ) 302 | e | divide(263, divide(subtract(const_100, 13), const_100)) | a small company reduced its faculty by approximately 13 percent to 263 employees . what was the original number of employees ? | if x is the original number of employees , then after 13 % reduction in employees number is . 87 x but we are given . 87 x = 263 x = 302 so the original number of employees is 302 correct answer - e | a = 100 - 13
b = a / 100
c = 263 / b
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a ) 90 , b ) 65 , c ) 85 , d ) 95 , e ) 80 | a | add(multiply(8, 2.5), 70) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 70 kg . what is the weight of the new person ? | "total increase in weight = 8 × 2.5 = 20 if x is the weight of the new person , total increase in weight = x − 70 = > 20 = x - 70 = > x = 20 + 70 = 90 answer is a ." | a = 8 * 2
b = a + 70
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a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter m of liquid a was contained by the can initially ? | "as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 similarly , for b , you remove ( 5 / 12 ) * 9 - - - > b remaining = 15 - ( 5 / 12 ) * 9 = 45 / 4 and then add 9 more litres of b - - - > 9 + 45 / 4 = 81 / 4 thus a / b ( final ratio ) = ( 45 / 4 ) / ( 81 / 4 ) = 7 : 9 , the same as the final ratio mentioned in the question . hence c is the correct answer . a / b = 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) , where 7 x and 5 x are initial quantities of a and b respectively . thus , 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) - - - > giving you x = 3 . thus a ( original ) m = 7 * 3 = 21 . c" | a = 7 + 9
b = 7 + 5
c = 5 / b
d = c * 9
e = 9 - d
f = a * e
g = 9 * 7
h = 7 * 5
i = g - h
j = f / i
k = 7 * j
|
a ) 200 m , b ) 250 m , c ) 300 m , d ) 350 m , e ) 400 m | e | multiply(divide(multiply(120, const_1000), const_3600), 12) | a train running at the speed of 120 km / hr crosses a pole in 12 seconds . what is the length of the train ? | "speed = ( 120 * 5 / 18 ) m / sec = ( 100 / 3 ) m / sec length of the train = ( speed x time ) = ( 100 / 3 * 12 ) m = 400 m . answer : e" | a = 120 * 1000
b = a / 3600
c = b * 12
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | add(add(2, 2), 3) | the function f ( y ) represents the number of ways that prime numbers can be uniquely summed to form a certain number y such that y = a + b + c + d … where those summed variables are each prime and a ≤ b ≤ c ≤ d . . . for instance f ( 8 ) = 3 and the unique ways are 2 + 2 + 2 + 2 and 2 + 3 + 3 and 3 + 5 . what is f ( 12 ) ? | it is so better to start with 2 and check whether sum of two primes is primes is even . 1 ) 2 ( 6 times ) 2 ) 2 ( 3 times ) + 3 ( 2 times ) 3 ) 2 ( 2 times ) + 3 + 5 4 ) 2 + 3 + 7 5 ) 2 + 5 + 5 6 ) 3 ( 4 times ) 7 ) 5 + 7 answer : d | a = 2 + 2
b = a + 3
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(40, subtract(20, 12)) | a person can swim in still water at 20 km / h . if the speed of water 12 km / h , how many hours will the man take to swim back against the current for 40 km ? | "m = 20 s = 12 us = 20 - 12 = 8 d = 40 t = 40 / 8 = 5 answer : c" | a = 20 - 12
b = 40 / a
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a ) 18 , b ) 92 , c ) 27 , d ) 26 , e ) 19 | a | add(multiply(3, 3), multiply(3, 3)) | a person was asked to state his age in years . his reply was , ` ` take my age 3 years hence , multiply it by 3 and subtract 3 times my age 3 years ago and you will know how old i am . ' ' what was the age of the person ? | "explanation : let the present age of person be x years . then , 3 ( x + 3 ) - 3 ( x - 3 ) = x < = > ( 3 x + 9 ) - ( 3 x - 9 ) = x < = > x = 18 . . answer : a ) 18" | a = 3 * 3
b = 3 * 3
c = a + b
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a ) 80 kg , b ) 63 kg , c ) 70 kg , d ) 75 kg , e ) 85 kg | b | add(48, multiply(10, add(1, divide(const_1, 1)))) | the average weight of 10 men is increased by 1 ½ kg when one of the men who weighs 48 kg is replaced by a new man . what is the weight of the new man ? | "since the average has increased by 1.5 kg , the weight of the man who stepped in must be equal to 48 + 10 x 1.5 48 + 15 = 63 kg ans : ' b '" | a = 1 / 1
b = 1 + a
c = 10 * b
d = 48 + c
|
a ) 5 kmph , b ) 7 kmph , c ) 12 kmph , d ) 8 kmph , e ) 1 kmph | c | divide(subtract(divide(105, 2), divide(45, 2)), const_2) | a man rows his boat 105 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 105 / ( 2 1 / 2 ) = 42 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 42 - 18 ) / 2 = 12 kmph answer : c" | a = 105 / 2
b = 45 / 2
c = a - b
d = c / 2
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a ) 14 days , b ) 16 days , c ) 24 days , d ) 11 days , e ) 19 days | c | add(divide(const_1, 20), divide(const_1, 30)) | a can do a job in 20 days and b can do it in 30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "c 1 / 20 + 1 / 30 = 1 / 12 12 * 2 = 24 days" | a = 1 / 20
b = 1 / 30
c = a + b
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a ) 750 , b ) 752 , c ) 754 , d ) 755 , e ) 756 | d | subtract(multiply(750, const_10), add(add(add(add(add(add(add(add(744, 745), 747), 748), 749), 752), 752), 753), 755)) | if the average of 744 , 745 , 747 , 748 , 749 , 752 , 752 , 753 , 755 and x is 750 , what is the value of x ? | "sum of the deviations of the numbers in the set from the mean is always zero 744 , 745 , 747 , 748 , 749 , 752 , 752 , 753 , 755 mean is 750 so the list is - 6 - 5 - 3 - 2 - 1 + 2 + 2 + 3 + 5 . . . this shud total to zero but this is - 5 , hence we need a number that is 5 more than the mean to get a + 5 and make it zero hence the answer is 750 + 5 = 755 d" | a = 750 * 10
b = 744 + 745
c = b + 747
d = c + 748
e = d + 749
f = e + 752
g = f + 752
h = g + 753
i = h + 755
j = a - i
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a ) 50 , b ) 200 , c ) 400 , d ) 625 , e ) 800 | e | divide(divide(divide(divide(divide(volume_rectangular_prism(100, 100, 50), const_3), const_2), 2), 2), 2) | the ratio , by volume , of soap to alcohol to water in a certain solution is 2 : 50 : 100 . the solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved . if the altered solution will contain 100 cubic centimeters of alcohol , how many cubic centimeters of water will it contain ? | "soap : alcohol initial ratio soap : alcohol : water - - > 2 : 50 : 100 initial soap : alcohol = 2 / 50 = 2 : 50 after doubled soap : alcohol = 2 * 2 / 50 = 4 : 50 initial soap : water = 2 / 100 = 2 : 100 after halved soap : water : 1 / 2 * 2 / 100 = 1 / 100 = 1 : 100 = 4 : 400 after soap : alcohol : water - - > 4 : 50 : 400 given alcohol 100 cumeter . ratio is 8 : 100 : 800 ( 4 : 50 : 400 ) for 100 cu meter of alcohol - - - 800 cu m water is required answer : e" | a = volume_rectangular_prism / (
b = a / 3
c = b / 2
d = c / 2
e = d / 2
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a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | b | multiply(sqrt(divide(72, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer that must divide n is | "possible values of n = 12 , 24 , 36 and 48 . but it is 12 that can divide all the possible values of n . if we consider 48 , it will not divide 12 , 24 and 36 . hence , 12 is the value that must divide n . answer : b" | a = 72 / 2
b = math.sqrt(a)
c = b * 2
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a ) $ 17.05 , b ) $ 17.55 , c ) $ 17.15 , d ) $ 17.35 , e ) $ 17.25 | a | subtract(20.10, add(1.05, add(1.00, 1.00))) | little john had $ 20.10 . he spent $ 1.05 on sweets and gave to his two friends $ 1.00 each . how much money was left ? | "john spent and gave to his two friends a total of 1.05 + 1.00 + 1.00 = $ 3.05 money left 20.10 - 3.05 = $ 17.05 answer : a" | a = 1 + 0
b = 1 + 5
c = 20 - 10
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a ) 1000 , b ) 1250 , c ) 1300 , d ) 1600 , e ) 1625 | e | divide(add(divide(multiply(300, const_100), 30), 300), divide(80, const_100)) | a small pool filled only with water will require an additional 300 gallons of water in order to be filled to 80 % of its capacity . if pumping in these additional 300 gallons of water will increase the amount of water in the pool by 30 % , what is the total capacity of the pool in gallons ? | "since pumping in additional 300 gallons of water will increase the amount of water in the pool by 30 % , then initially the pool is filled with 1,000 gallons of water . so , we have that 1,000 + 300 = 0.8 * { total } - - > { total } = 1,625 . answer : e ." | a = 300 * 100
b = a / 30
c = b + 300
d = 80 / 100
e = c / d
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a ) 35 m 2 , b ) 36 m 2 , c ) 37 m 2 , d ) 38 m 2 , e ) 40 m 2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(30, divide(const_3600, const_10))) | the area of sector of a circle whose radius is 12 metro and whose angle at the center is 30 еў is ? | "30 / 360 * 22 / 7 * 12 * 12 = 37 m 2 answer : c" | a = 12 ** 2
b = 2 * 10
c = b + 2
d = 4 + 3
e = c / d
f = a * e
g = 3600 / 10
h = 30 / g
i = f * h
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a ) 8 , b ) 11 , c ) 12 , d ) 15 , e ) 17 | c | divide(multiply(32, 3), const_4) | there are two positive numbers in the ratio 3 : 11 . if the larger number exceeds the smaller by 32 , then what is the smaller number ? | "let the two positive numbers be 3 x and 11 x respectively . 11 x - 3 x = 32 8 x = 32 = > x = 4 thus , the smaller number = 3 x = 12 . answer : c" | a = 32 * 3
b = a / 4
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['a ) 2', 'b ) 4', 'c ) 1', 'd ) 5', 'e ) 3'] | e | divide(const_1, divide(multiply(divide(const_4, const_3), const_pi), multiply(const_4, const_pi))) | if the volume and surface area of a sphere are numerically the same , then its radius is : ? | 4 / 3 π r ( power 3 ) = 4 π r ( power 2 ) r = 3 answer is e . | a = 4 / 3
b = a * math.pi
c = 4 * math.pi
d = b / c
e = 1 / d
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a ) 300 , b ) t = 400 , c ) t = 500 , d ) 600 , e ) none of these | c | divide(200, divide(2, 5)) | there are 200 female managers in a certain company . find the total number t of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | "{ managers } = { female managers } + { male managers } ; we are told that the total number of managers in the company is 2 / 5 of all the employees , thus { managers } = 2 / 5 ( m + f ) , where m and f are number of female and male employees , respectively . also , we know that 2 / 5 of all male employees are managers : { male managers } = 2 / 5 * mas well as there are total of 200 female managers : { female managers } = 200 ; thus : 2 / 5 ( m + f ) = 200 + 2 / 5 * m - - > f = 500 . answer : c ." | a = 2 / 5
b = 200 / a
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a ) 23 , b ) 78 , c ) 27 , d ) 61 , e ) 81 | d | add(60, divide(multiply(5, 12), divide(180, 3))) | 60 + 5 * 12 / ( 180 / 3 ) = ? | "explanation : 60 + 5 * 12 / ( 180 / 3 ) = 60 + 5 * 12 / ( 60 ) = 60 + ( 5 * 12 ) / 60 = 60 + 1 = 61 . answer : d" | a = 5 * 12
b = 180 / 3
c = a / b
d = 60 + c
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a ) 20 % , b ) 30 % , c ) 25 % , d ) 15 % , e ) 50 % | b | multiply(divide(multiply(divide(40, const_100), 3), add(3, const_1)), const_100) | to a sugar solution of 3 liters containing 40 % sugar , one liter of water is added . the percentage of sugar in the new solution is ? | quantity of sugar = 40 * 3 / 100 = 1.2 kg new percentage = 1.2 / 4 * 100 = 30 % answer is b | a = 40 / 100
b = a * 3
c = 3 + 1
d = b / c
e = d * 100
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a ) 24 % , b ) 20 % , c ) 17 % , d ) 18 % , e ) none of these | a | subtract(subtract(add(const_100, 20), multiply(add(const_100, 20), divide(5, const_100))), const_100) | a shopkeeper labeled the price of his articles so as to earn a profit of 20 % on the cost price . he then sold the articles by offering a discount of 5 % on the labeled price . what is the actual percent profit earned in the deal ? | "explanation : let the cp of the article = rs . 100 . then labeled price = rs . 120 . sp = rs . 120 - 5 % of 120 = rs . 120 - 6 = rs . 124 . gain = rs . 124 â € “ rs . 100 = rs . 24 therefore , gain / profit percent = 24 % . answer : option a" | a = 100 + 20
b = 100 + 20
c = 5 / 100
d = b * c
e = a - d
f = e - 100
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a ) 8 , b ) 7 , c ) 6 , d ) 11 , e ) 9 | d | multiply(divide(const_1, multiply(add(const_100, 30), divide(const_1, subtract(const_100, 30)))), 11) | by selling 11 pencils for a rupee a man loses 30 % . how many for a rupee should he sell in order to gain 30 % ? | "70 % - - - 12 130 % - - - ? 70 / 130 * 11 = 6 answer : d" | a = 100 + 30
b = 100 - 30
c = 1 / b
d = a * c
e = 1 / d
f = e * 11
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a ) 21 ½ days , b ) 22 ½ days , c ) 33 3 / 4 days , d ) 12 ½ days , e ) none of these | c | add(divide(const_1, 18), divide(const_1, 30)) | a can do a job in 18 days and b can do it in 30 days . a and b working together will finish thrice the amount of work in - - - - - - - days ? | "explanation : 1 / 18 + 1 / 30 = 8 / 90 = 4 / 45 45 / 4 = 45 / 4 * 3 = 33 3 / 4 days answer : c" | a = 1 / 18
b = 1 / 30
c = a + b
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['a ) 300 square meters', 'b ) 400 square meters', 'c ) 500 square meters', 'd ) 600 square meters', 'e ) 700 square meters'] | a | divide(1800, multiply(const_2, const_3)) | the length of a rectangle is increased to 2 times its original size and its width is increased to 3 times its original size . if the area of the new rectangle is equal to 1800 square meters , what is the area of the original rectangle ? | if l and w be the original length and width of the rectangle and its area is given by l ? w after increase the length becomes 2 l and the width becomes 3 w . the area is then given by ( 2 l ) ? ( 3 w ) and is known . hence ( 2 l ) ? ( 3 w ) = 1800 solve the above equation to find l ? w 6 l ? w = 1800 l ? w = 1800 / 6 = 300 square meters , area of original rectangle correct answer a | a = 2 * 3
b = 1800 / a
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a ) 300 , b ) y = 400 , c ) y = 500 , d ) y = 600 , e ) none of these | c | divide(200, divide(2, 5)) | there are 200 female managers in a certain company . find the total number y of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | "{ managers } = { female managers } + { male managers } ; we are told that the total number of managers in the company is 2 / 5 of all the employees , thus { managers } = 2 / 5 ( m + f ) , where m and f are number of female and male employees , respectively . also , we know that 2 / 5 of all male employees are managers : { male managers } = 2 / 5 * mas well as there are total of 200 female managers : { female managers } = 200 ; thus : 2 / 5 ( m + f ) = 200 + 2 / 5 * m - - > f = 500 . answer : c ." | a = 2 / 5
b = 200 / a
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a ) 33361667 , b ) 33391667 , c ) 33331666 , d ) 33331668 , e ) 33331669 | c | subtract(divide(multiply(multiply(multiply(add(add(1000, 2000), 5000), 2), const_4), 2000), add(const_2, const_1)), multiply(2000, 5000)) | a and b r 2 men who enter into business and they invest rs 1000 , rs 2000 respectively . how will they divide the income of rs 5000 | they invested in the ratio 1 : 2 a ' s share = 1 / 3 * 5000 = 1666 b ' s share = 2 / 3 * 5000 = 3333 answer : c | a = 1000 + 2000
b = a + 5000
c = b * 2
d = c * 4
e = d * 2000
f = 2 + 1
g = e / f
h = 2000 * 5000
i = g - h
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a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | a | divide(power(2, divide(power(2, 4), power(2, 2))), power(2, 3)) | if the operation ø is defined for all positive integers x and w by x ø w = ( 2 ^ x ) / ( 2 ^ w ) then ( 4 ø 2 ) ø 3 = ? | 4 ø 2 = 2 ^ 4 / 2 ^ 2 = 4 4 ø 3 = 2 ^ 4 / 2 ^ 3 = 2 the answer is a . | a = 2 ** 4
b = 2 ** 2
c = a / b
d = 2 ** c
e = 2 ** 3
f = d / e
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a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | d | subtract(divide(add(subtract(100, 4), 18), add(const_1, const_2)), 18) | a survey was taken among 100 brainiacs . of those surveyed , twice as many brainiacs likerebus teasers as math teasers . if 18 brainiacs like both rebus teasers and math teasers and 4 like neither kind of teaser , how many brainiacs like math teasers but not rebus teasers ? | let x brainiacs like only rebus teasers and y brainiacs like only math teasers . now , x + y = 100 - 18 - 4 = 78 . also given that , x + 18 = 2 ( y + 18 ) x = 2 y + 18 . 78 - y = 2 y + 18 3 y = 60 y = 20 so 20 brainiacs like math teasers but not rebus teasers answer : d | a = 100 - 4
b = a + 18
c = 1 + 2
d = b / c
e = d - 18
|
a ) 400 , b ) 625 , c ) 3,200 , d ) 4,500 , e ) 10,000 | c | divide(80, divide(2, 80)) | in a certain pond , 80 fish were caught , tagged , and returned to the pond . a few days later , 80 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ? | "total fish = x percentage of second catch = ( 2 / 80 ) * 100 = 2.5 % so , x * 2.5 % = 80 x = 3200 ans c ." | a = 2 / 80
b = 80 / a
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a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | divide(divide(subtract(500, add(2, 3)), 3), 7) | x is a positive integer of t less than 500 . when x is divided by 7 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ? | took me more than 2 mins to solve . that ' s how i did it . x is between 1 - 499 included . smallest number that can be written as 7 n + 1 ( 7 * 0 + 1 ) is 1 largest number that can be written as 7 n + 1 is ( 7 * 71 + 1 ) is 498 so there are total 72 numbers that can be written as 7 n + 1 because x can also be written as 3 m + 2 , we have to see how many numbers that can be written as 7 n + 1 also can be written as 3 m + 2 7 * 0 + 1 can not be written in the form 3 m + 2 7 * 1 + 1 can be written in the form 3 m + 2 7 * 2 + 1 can not be written in the form 3 m + 2 7 * 3 + 1 can not be written in the form 3 m + 2 7 * 4 + 1 can be written in the form 3 m + 2 here we see a sequence that one out of every 4 numbers can be written as 3 m + 1 72 / 4 = 24 numbers can be written as 3 m + 1 d is the answer | a = 2 + 3
b = 500 - a
c = b / 3
d = c / 7
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a ) rs . 15000 , b ) rs . 15550 , c ) rs . 15600 , d ) rs . 28875 , e ) none of these | d | multiply(1400, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 1400 per sq . metre . | "solution area of the floor = ( 5.5 × 3.75 ) m 2 = 20.625 m 2 ∴ cost of paving = rs . ( 1400 × 20.625 ) = 28875 . answer d" | a = 5 * 5
b = 1400 * a
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a ) 23 , b ) 32 , c ) 44 , d ) 15 , e ) 64 | e | multiply(multiply(divide(12, 3), divide(14, 7)), divide(16, 2)) | in a rectangular box measuring 12 inches by 14 inches by 16 inches if small rectangular boxes measuring 3 inches by 7 inches by 2 inches are arranged in it , what is the max number of boxes that fit into it ? | the 3 inch side should be aligned to the 12 inch side ( 4 layer ) 7 inch side should be aligned to the 14 inch side . ( 2 layer ) and 2 inch side should be aligned to the 16 inch side . ( 8 layer ) maximum number of rectangles = 4 * 2 * 8 = 64 answer is e | a = 12 / 3
b = 14 / 7
c = a * b
d = 16 / 2
e = c * d
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a ) 30 % , b ) 25 % , c ) 75 % , d ) 20 % , e ) none of these | d | multiply(divide(25, add(const_100, 25)), const_100) | ashis ' s height is 25 % more than babji , by how much percent babji ' s height is less than ashis . | explanation : solution : babji is less than ashis by ( 25 / ( 100 + 25 ) * 100 ) % = 20 % answer : d | a = 100 + 25
b = 25 / a
c = b * 100
|
a ) none of these , b ) 5993 , c ) 994 , d ) 5994 , e ) 995 | b | divide(divide(297832, 7), 7) | what is the difference between the place value and the face value of 7 in the numeral 297832 ? | explanation : place value of 6 = 7000 face value of 6 = 7 difference = 6000 - 7 = 5993 answer : b | a = 297832 / 7
b = a / 7
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a ) 0.9 , b ) 0.72 , c ) 0.45 , d ) 0.35 , e ) 0.28 | b | subtract(const_1, add(multiply(divide(40, const_100), divide(40, const_100)), multiply(divide(20, const_100), subtract(const_1, divide(40, const_100))))) | in a graduating class , 40 percent of the students are male . in this class , 40 percent of the male students and 20 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | "let x be the total number of students . the number students who are younger than 25 is 0.6 * 0.4 x + 0.8 * 0.6 x = 0.72 x the answer is b ." | a = 40 / 100
b = 40 / 100
c = a * b
d = 20 / 100
e = 40 / 100
f = 1 - e
g = d * f
h = c + g
i = 1 - h
|
a ) 1 , b ) 16 , c ) 20 , d ) 71 , e ) 60 | c | subtract(multiply(const_100, add(const_10, multiply(const_3, const_2))), 696) | what is the least number to be subtracted from 696 to make it a perfect square ? | "the numbers less than 696 and are squares of certain numbers are 676 , 625 . the least number that should be subtracted from 696 to make it perfect square = 696 - 676 = 20 . answer : c" | a = 3 * 2
b = 10 + a
c = 100 * b
d = c - 696
|
a ) 0 , b ) 2 , c ) 3 , d ) 6 , e ) 7 | b | subtract(multiply(12, 2), 22) | if the remainder is 12 when the integer n is divided by 22 , what is the remainder when 2 n is divided by 11 ? | "n = 22 k + 12 2 n = 2 ( 22 k + 12 ) = 4 * 11 k + 24 = 4 * 11 k + 2 * 11 + 2 = 11 j + 2 . the answer is b ." | a = 12 * 2
b = a - 22
|
a ) 52.5 , b ) 52.9 , c ) 52.1 , d ) 52.3 , e ) 65 | e | divide(add(multiply(30, 40), multiply(50, 80)), add(30, 50)) | the average marks of a class of 30 students is 40 and that of another class of 50 students is 80 . find the average marks of all the students ? | "sum of the marks for the class of 30 students = 30 * 40 = 1200 sum of the marks for the class of 50 students = 50 * 80 = 4000 sum of the marks for the class of 80 students = 1200 + 4000 = 5200 average marks of all the students = 5200 / 80 = 65 answer : e" | a = 30 * 40
b = 50 * 80
c = a + b
d = 30 + 50
e = c / d
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a ) 12 , b ) 15 , c ) 16 , d ) 20 , e ) 25 | c | multiply(divide(4, add(4, 1)), divide(multiply(10, divide(4, add(4, 1))), subtract(divide(4, add(4, 1)), divide(2, add(2, 3))))) | a jar contains a mixture of a & b in the ratio 4 : 1 . when 10 l of mixture is replaced with liquid b , ratio becomes 2 : 3 . how many liters of liquid a was present in mixture initially . | 10 litres of mixture that is replaced will contain 8 litres of a and 2 litres of b ( as a : b = 4 : 1 ) let the initial volume of the mixture be 4 k + 1 k = 5 k so by condition , [ 4 k - 8 ] / [ k - 2 + 10 ] = 2 / 3 solve for k which is k = 4 so initial volume of liquid a = 4 k = 16 litres answer : c | a = 4 + 1
b = 4 / a
c = 4 + 1
d = 4 / c
e = 10 * d
f = 4 + 1
g = 4 / f
h = 2 + 3
i = 2 / h
j = g - i
k = e / j
l = b * k
|
a ) 1 / 29 , b ) 2 / 29 , c ) 3 / 29 , d ) 4 / 29 , e ) 5 / 29 | e | divide(5, add(multiply(4, 5), add(4, 5))) | at an m & m factory , two types of m & ms are produced , red and blue . the m & ms are transported individually on a conveyor belt . anna is watching the conveyor belt , and has determined that 4 out of every 5 red m & ms are followed by a blue one , while one out of every 6 blue m & ms is followed by a red one . what proportion of the m & ms are red ? | suppose that the fraction of red m & ms is fr and the fraction of blue m & ms is fb . then the probability that anna sees a red one emerge next is fr . this probabil - ity can also be expressed using conditional probabilities as fr = p ( blue ) p ( red after blue ) + p ( red ) p ( red after red ) , or fr = fb ( 1 = 6 ) + fr ( 1 = 5 ) . likewise , we find fb = fr ( 4 = 5 ) + fb ( 5 = 6 ) . solving the system of equations gives fb = 24 29 and fr = 5 / 29 correct answer e | a = 4 * 5
b = 4 + 5
c = a + b
d = 5 / c
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | b | add(divide(multiply(subtract(const_1, divide(40, multiply(50, 50))), 150), 450), divide(add(multiply(divide(50, multiply(50, 50)), 300), multiply(divide(40, multiply(50, 50)), 150)), 450)) | in a certain corporation , there are 300 male employees and 150 female employees . it is known that 50 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 450 employees is chosen at random , what is the probability this employee has an advanced degree or is female ? | "p ( female ) = 150 / 450 = 1 / 3 p ( male with advanced degree ) = 0.5 * 300 / 450 = 150 / 450 = 1 / 3 the sum of the probabilities is 2 / 3 the answer is b ." | a = 50 * 50
b = 40 / a
c = 1 - b
d = c * 150
e = d / 450
f = 50 * 50
g = 50 / f
h = g * 300
i = 50 * 50
j = 40 / i
k = j * 150
l = h + k
m = l / 450
n = e + m
|
a ) 2 : 72 , b ) 5 : 72 , c ) 7 : 72 , d ) 1 : 72 , e ) 3 : 72 | b | divide(5000, 72000) | ravi and kavi start a business by investing â ‚ ¹ 5000 and â ‚ ¹ 72000 , respectively . find the ratio of their profits at the end of year . | "ratio of profit = ratio of investments = 5000 : 72000 = 5 : 72 answer : b" | a = 5000 / 72000
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a ) - 2 , b ) - 5 , c ) - 8 , d ) - 10 , e ) - 12 | a | add(divide(99, const_10), divide(99, divide(99, const_10))) | if a ( a + 2 ) = 99 and b ( b + 2 ) = 99 , where a ≠ b , then a + b = | "i . e . if a = 9 then b = - 11 or if a = - 11 then b = 9 but in each case a + b = - 11 + 9 = - 2 answer : a" | a = 99 / 10
b = 99 / 10
c = 99 / b
d = a + c
|
a ) 30 , b ) 98 , c ) 900 , d ) 21 , e ) 22 | c | add(multiply(multiply(5, 6), const_100), multiply(4, 6)) | three numbers are in the ratio 5 : 4 : 6 and their average is 750 . the largest number is : | "explanation : let the numbers be 5 x , 4 x and 6 x , then , ( 5 x + 4 x + 6 x ) / 3 = 750 = > 15 x = 750 * 3 = > x = 150 largest number 6 x = 6 * 150 = 900 answer : c" | a = 5 * 6
b = a * 100
c = 4 * 6
d = b + c
|
a ) 8 , b ) 12 , c ) 6 , d ) 4 , e ) 9 | b | multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 18) | by selling 18 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ? | "80 % - - - 18 120 % - - - ? 80 / 120 * 18 = 12 answer : b" | a = 100 + 20
b = 100 - 20
c = 1 / b
d = a * c
e = 1 / d
f = e * 18
|
a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | e | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | find the average of all numbers between 3 and 69 which are divisible by 5 | "explanation : average = ( 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 + 60 + 65 ) / 13 = 455 / 13 = 35 answer : option e" | a = 5 * 3
b = 5 * 2
c = 5 + b
d = a + c
e = 5 * 4
f = d + e
g = 4 + 1
h = g * 5
i = f + h
j = i / 5
|
a ) 1 , b ) 13 , c ) 26 , d ) 25 , e ) 1014 | d | multiply(add(subtract(13, const_10), const_2), add(subtract(13, const_10), const_2)) | if x is a sum of all even integers on the interval 13 . . . 63 and y is their number , what is the gcd ( x , y ) ? | x = 14 + 16 + . . . + 62 = ( largest + smallest ) / 2 * ( # of terms ) = ( 14 + 62 ) / 2 * 25 = 38 * 25 . gcd of 25 and 39 * 25 is 25 . answer : d . | a = 13 - 10
b = a + 2
c = 13 - 10
d = c + 2
e = b * d
|
a ) 6 % , b ) 8 % , c ) 10 % , d ) 12 % , e ) 15 % | c | multiply(divide(divide(subtract(360, 260), 5), subtract(260, multiply(divide(subtract(360, 260), 5), 3))), const_100) | joe invested a certain sum of money in a simple interest bond whose value grew to $ 260 at the end of 3 years and to $ 360 at the end of another 5 years . what was the rate of interest in which he invested his sum ? | "in 5 years , the value grew $ 100 , so the simple interest was $ 20 per year . in 3 years , the total interest was 3 * $ 20 = $ 60 the principal is $ 260 - $ 60 = 200 . the interest rate is $ 20 / $ 200 = 10 % the answer is c ." | a = 360 - 260
b = a / 5
c = 360 - 260
d = c / 5
e = d * 3
f = 260 - e
g = b / f
h = g * 100
|
a ) 19.7 $ . , b ) 23 $ . , c ) 14.7 $ . , d ) 35 $ . , e ) 37.4 $ . | c | add(divide(subtract(74.7, add(add(multiply(1.5, 0.7), 3), subtract(multiply(1.5, add(multiply(1.5, 0.7), 3)), divide(50, const_100)))), add(add(1.5, multiply(1.5, 1.5)), const_1)), 0.7) | a cuban cigar would cost 3 dollar less than 1.5 times a french cigar , had the french cigar cost 0.7 dollar less than it does now . an arabian cigar costs 50 cents more than 1.5 times the cuban cigar . the 3 cigars together cost 74.7 dollars . what is the price of the french cigar ? | the three cigars together cost 74.7 dollars . if each cost the same , they would have cost a little less than 25 dollars each . from the given data we know french cigar < cuban cigar < arabic cigar and each is more expensive 1.5 times . therefore eliminate options a , d , e straightaway . since french cigar is at least 1.5 times cheaper we know b can not be the answer . therefore c is the answer . | a = 1 * 5
b = a + 3
c = 1 * 5
d = c + 3
e = 1 * 5
f = 50 / 100
g = e - f
h = b + g
i = 74 - 7
j = 1 * 5
k = 1 + 5
l = k + 1
m = i / l
n = m + 0
|
a ) 5.25 % increase , b ) 5.25 % decrease , c ) 2.25 % increase , d ) 2.25 % decrease , e ) no change | d | subtract(const_100, subtract(add(15, const_100), divide(multiply(add(15, const_100), 15), const_100))) | the salary of a worker is first increased by 15 % and afterwards reduced by 15 % . what is the net change in the worker ' s salary ? | "let x be the original salary . the final salary is 0.85 ( 1.15 x ) = 0.9775 x the answer is d ." | a = 15 + 100
b = 15 + 100
c = b * 15
d = c / 100
e = a - d
f = 100 - e
|
a ) 2.0 , b ) 4.0 , c ) 5.0 , d ) 6.0 , e ) 8.0 | b | divide(const_100, 25) | a company decreased the price of its main product by 25 % . subsequently , the number of units sold increased such that the total revenue remained unchanged . what was the ratio of the percent increase in the units sold to the percent decrease of the original price for this product ? | "for the total revenue to remain the same when the price is one fourth , the number of products sold must four times . therefore increase in the number of products sold is 100 % = > the required ratio = 100 % / 25 % = 4.0 answer : b" | a = 100 / 25
|
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | b | multiply(divide(35, subtract(multiply(34, 2), 63)), 2) | a firm is comprised of partners and associates in a ratio of 2 : 63 . if 35 more associates were hired , the ratio of partners to associates would be 1 : 34 . how many partners are currently in the firm ? | "the ratio 1 : 34 = 2 : 68 so the ratio changed from 2 : 63 to 2 : 68 . 68 - 63 = 5 which is 1 / 7 of the increase in 35 associates . the ratio changed from 14 : 441 to 14 : 476 . thus the number of partners is 14 . the answer is b ." | a = 34 * 2
b = a - 63
c = 35 / b
d = c * 2
|
a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | d | divide(rectangle_area(12, 15), 9) | carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 12 inches by 15 inches and jordan ' s rectangle is 9 inches long , how wide is jordan ' s rectangle , in inches ? | "area of first rectangle is 12 * 15 = 180 hence area of second would be 9 x = 180 x x = 20 answer is d" | a = rectangle_area / (
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a ) 1 : 2 , b ) 2 : 1 , c ) 3 : 1 , d ) 2 : 3 , e ) 5 : 3 | c | divide(multiply(45000, const_12), multiply(30000, add(const_4, const_3))) | x starts a business with rs . 45000 . y joins in the business after 6 months with rs . 30000 . what will be the ratio in which they should share the profit at the end of the year ? | "ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 × 12 : 30000 × 6 = 45 × 12 : 30 × 6 = 3 × 12 : 2 × 6 = 3 : 1 answer is c" | a = 45000 * 12
b = 4 + 3
c = 30000 * b
d = a / c
|
a ) $ 1542 , b ) $ 1145 , c ) $ 1210 , d ) $ 1642 , e ) $ 1020 | c | multiply(multiply(add(divide(add(const_4, const_1), const_100), const_1), 1000), add(divide(add(const_4, const_1), const_100), const_1)) | sam invested $ 1000 @ 20 % per annum for one year . if the interest is compounded half yearly , then the amount received by sam at the end of the year will be ? | "p = $ 1000 r = 20 % p . a . = 10 % t = 2 half years amount = 1000 * ( 1 + 10 / 100 ) ^ 2 = 1000 * 11 / 10 * 11 / 10 = $ 1210 answer is c" | a = 4 + 1
b = a / 100
c = b + 1
d = c * 1000
e = 4 + 1
f = e / 100
g = f + 1
h = d * g
|
a ) - 8 , b ) - 9 , c ) - 5 , d ) - 4 , e ) 1 | c | divide(negate(add(20, 5)), 3) | solve below question 3 x - 5 = - 20 | 1 . subtract 1 from both sides : 3 x - 5 + 5 = - 20 + 5 2 . simplify both sides : 3 x = - 15 3 . divide both sides by 3 : 4 . simplify both sides : x = - 5 c | a = 20 + 5
b = negate / (
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a ) 87 , b ) 74 , c ) 10 , d ) 76 , e ) 17 | b | subtract(multiply(add(10, const_1), add(4, 30)), multiply(10, 30)) | the average of runs of a cricket player of 10 innings was 30 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "average after 11 innings = 34 required number of runs = ( 34 * 11 ) - ( 30 * 10 ) = 374 - 300 = 74 . answer : b" | a = 10 + 1
b = 4 + 30
c = a * b
d = 10 * 30
e = c - d
|
a ) 4 : 7 , b ) 6 : 5 , c ) 1 : 2 , d ) 2 : 3 , e ) 5 : 6 | e | divide(divide(const_1, const_4), divide(30, const_100)) | if 30 % of a number is equal to one - fourth of another number , what is the ratio of first number to the second number ? | let 30 % of a = 1 / 4 b . then , 30 a / 100 = 1 b / 4 = > 3 a / 10 = 1 b / 4 a / b = ( 1 / 4 * 10 / 3 ) = 5 / 6 a : b = 5 : 6 . answer : e | a = 1 / 4
b = 30 / 100
c = a / b
|
a ) 12 minutes , b ) 12.5 min , c ) 14.4 min , d ) 10.2 min , e ) 14.66 min | c | divide(multiply(multiply(12, const_3), multiply(8, const_3)), add(multiply(12, const_3), multiply(8, const_3))) | a cistern is two - third full of water . pipe a can fill the remaining part in 12 minutes and pipe b in 8 minutes . once the cistern is emptied , how much time will they take to fill it together completely ? | as pipe a takes 12 min . to fill remaining one - third it takes 36 min . to fill completely . similarly pipe b takes 24 min . to fill completely so , total time taken by both together is reciprocal of : ( 1 / 36 ) + ( 1 / 24 ) = 5 / 72 ans : 14.4 min . answer : c | a = 12 * 3
b = 8 * 3
c = a * b
d = 12 * 3
e = 8 * 3
f = d + e
g = c / f
|
a ) 20 , b ) 23 , c ) 26 , d ) 27 , e ) 29 | a | subtract(multiply(add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3)), add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3))), 5021) | what should be added to 5021 so that it may become a perfect square ? | "71 x 71 = 5041 5041 - 5021 = 20 if added to 20 get perfect square answer = a" | a = 10 + 10
b = 10 + 10
c = a + b
d = 10 + 10
e = c + d
f = 4 + 3
g = e + f
h = 10 + 10
i = 10 + 10
j = h + i
k = 10 + 10
l = j + k
m = 4 + 3
n = l + m
o = g * n
p = o - 5021
|
a ) 5.5 kg , b ) 11 kg , c ) 30 kg , d ) 36.5 kg , e ) 7 kg | e | subtract(44, divide(subtract(add(multiply(30, subtract(44, add(30, 1))), 44), multiply(subtract(subtract(44, add(30, 1)), 1), 30)), const_2)) | when a student joe , weighing 44 kg , joins a group of students whose average weight is 30 kg , the average weight goes up by 1 kg . subsequently , if two students , excluding joe , leave the group the average weight comes back to 30 kg . what is the difference between the average weight of the two students who left and the weight of joe ? | "after two persons leave the group the average remains the same . that means the weight of the two persons = 44 + 30 = 74 so , the average the two persons = 37 that gives the answer 44 - 37 = 7 answer e" | a = 30 + 1
b = 44 - a
c = 30 * b
d = c + 44
e = 30 + 1
f = 44 - e
g = f - 1
h = g * 30
i = d - h
j = i / 2
k = 44 - j
|
a ) 35 % , b ) 40 % , c ) 45 % , d ) 50 % , e ) 56 % | e | multiply(divide(70, 125), const_100) | 70 is what percent of 125 ? | 70 / 125 × 100 = 56 % answer : e | a = 70 / 125
b = a * 100
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a ) 12000 , b ) 24000 , c ) 16000 , d ) 14000 , e ) 16200 | c | divide(add(525, 275), 4) | ( 525 + 275 ) ã — 4 ã — ( 3 + 2 ) = ? | "( 525 + 275 ) ã — 4 ã — ( 3 + 2 ) = ? or , ? = 800 ã — 4 ã — 5 = 16000 answer c" | a = 525 + 275
b = a / 4
|
a ) 62 , b ) 72 , c ) 82 , d ) 92 , e ) 62 | b | add(power(8, 2), multiply(power(2, 2), 2)) | if a 2 - b 2 = 8 and a * b = 2 , find a 4 + b 4 . | "a 2 - b 2 = 8 : given a 4 + b 4 - 2 a 2 b 2 = 82 : square both sides and expand . a * b = 2 : given a 2 b 2 = 22 : square both sides . a 4 + b 4 - 2 ( 4 ) = 82 : substitute a 4 + b 4 = 72 correct answer b" | a = 8 ** 2
b = 2 ** 2
c = b * 2
d = a + c
|
a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 12 | c | divide(5.76, subtract(96.12, floor(96.12))) | when positive integer x is divided by positive integer y , the remainder is 5.76 . if x / y = 96.12 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 5.76 - - > x = qy + 5.76 ; x / y = 96.12 - - > x = 96 y + 0.12 y ( so q above equals to 96 ) ; 0.12 y = 5.76 - - > y = 48 . answer : c ." | a = math.floor(96, 12)
b = 96 - 12
c = 5 / 76
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a ) 1 / 3 , b ) 9 / 5 , c ) 4 , d ) 9 , e ) 12 | b | multiply(divide(6, multiply(const_4, subtract(36, 6))), 36) | timothy leaves home for school , riding his bicycle at a rate of 6 miles per hour . fifteen minutes after he leaves , his mother sees timothy ’ s math homework lying on his bed and immediately leaves home to bring it to him . if his mother drives at 36 miles per hour , how far ( in terms of miles ) must she drive before she reaches timothy ? i think is a 700 level problem but i tag it as 600 / 700 , let me know . either way i hope in an explanationthanks | in 15 mins , timothy travels = 6 / 4 miles . now , let his mother takes x hours to reach him , traveling at 36 mph . so , 36 x = 6 x + 6 / 4 x = 1 / 20 hrs . thus , the distance traveled by his mother to reach = 36 * 1 / 20 = 9 / 5 miles . ans b | a = 36 - 6
b = 4 * a
c = 6 / b
d = c * 36
|
a ) 5 , b ) 6 , c ) 7 , d ) 10 , e ) 12 | d | subtract(20, const_4) | in a group of cows and hens , the number of legs are 20 more than twice the number of heads . the number of cows is : | "let no of cows be x , no of hens be y . so heads = x + y legs = 4 x + 2 y now , 4 x + 2 y = 2 ( x + y ) + 20 2 x = 20 x = 10 . answer : d" | a = 20 - 4
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a ) 10 sec , b ) 16 sec , c ) 13 sec , d ) 67 sec , e ) 11 sec | e | multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 55), inverse(multiply(const_0_2778, add(60, 40)))) | two trains of equal length , running with the speeds of 60 and 40 kmph , take 55 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 55 d = 55 * 100 / 18 = 2750 / 9 rs = 60 + 40 = 100 * 5 / 18 t = 2750 / 9 * 18 / 500 = 11 sec answer : e" | a = 60 - 40
b = const_0_2778 * a
c = b * 55
d = 60 + 40
e = const_0_2778 * d
f = 1/(e)
g = c * f
|
a ) 1 / 12 , b ) 1 / 9 , c ) 2 / 3 , d ) 5 / 9 , e ) 2 1 / 9 | d | subtract(add(divide(5, 9), divide(1, 3)), divide(8, 24)) | the instructions state that cheryl needs 5 / 9 square yards of one type of material and 1 / 3 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 8 / 24 square yards left that she did not use . what is the total amount of square yards of material cheryl used ? | "total bought = 5 / 9 + 1 / 3 left part 8 / 24 - - - > 1 / 3 so used part 5 / 9 + 1 / 3 - 1 / 3 = 5 / 9 answer : d" | a = 5 / 9
b = 1 / 3
c = a + b
d = 8 / 24
e = c - d
|
a ) 47 , b ) 25 , c ) 37 , d ) 33 , e ) 29 | b | add(subtract(multiply(sqrt(169), const_2), multiply(const_2, 1)), 1) | what is the value of n if the sum of the consecutive odd intergers q from 1 to n equals 169 ? | "before you tackle this question you must first understand that the question is comprised of two key parts , 1 st is finding out how manytermsis in that sequence and 2 nd whatactual number valuethat term is . in an arithmetic progression , in this case consecutive odd integers 1 , 3 , 5 , . . . . , there are two set of rules . rule # 1 ( arithmetic sequence ) : xn = a + d ( n - 1 ) identifies what the actual # in the sequence would be . each number in the sequence has a term such as 1 ( is the first term ) , 3 ( is the second term ) and so on . so if i were to ask you to find out what the 10 th term is of that sequence you would use that formula to find that value . a = 1 ( first term ) d = 2 ( the common difference ) remember in the sequence 1 , 3 , 5 , 7 the common difference is always 2 * on a side note we use n - 1 because we do n ' t have d in the first term , therefore if we were solving for the first term we would get 0 as n - 1 and 0 times d would give us 0 , leaving only the first term . this works regardless what your first term is in any sequence . but remember the question askswhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169 ? which means we first need a consecutive sequence that sums up to 169 and than find what the value of the n is , in this case it would be the last number in that sequence . in order to find that we first need to knowhow many terms ( how many of the n there is ) in order to be able to plug n in this formula given we know what the sum is . for that to happen we need to use rule # 2 . rule # 2 ( summing an arithmetic series ) : 169 = n / 2 ( 2 a + ( n - 1 ) d ) . given the question gives us what the sum is ( 169 in this case ) we would simply use this formula to solve for n . once we solve for n ( 13 in this case ) we can simply plug n into the first formula ( rule 1 ) and find the value . it feels very confusing and difficult at first , but once you identify the steps all you need to do is plug and play . we have the sum ( 169 ) of a sequence , the number of terms in that sequence is ( unknown ) . rule # 2 tells us how many numbers there are in that sequence and rule # 1 gives us what that last term is ." | a = math.sqrt(169)
b = a * 2
c = 2 * 1
d = b - c
e = d + 1
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a ) a ) 6.7 , b ) b ) 1.3 , c ) c ) 9.6 , d ) d ) 12.5 , e ) e ) 28.6 | e | divide(subtract(30, 10), subtract(const_1, divide(30, 100))) | how many kg of pure salt must be added to 100 kg of 10 % solution of salt and water to increase it to a 30 % solution ? | "amount salt in 100 kg solution = 10 * 100 / 100 = 10 kg let x kg of pure salt be added then ( 10 + x ) / ( 100 + x ) = 30 / 100 100 + 10 x = 300 + 3 x 7 x = 200 x = 28.6 answer is e" | a = 30 - 10
b = 30 / 100
c = 1 - b
d = a / c
|
a ) 2299 , b ) 2699 , c ) 3000 , d ) 6725 , e ) 2601 | c | divide(3600, 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 . 3600 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 3600 ( 100 / 120 ) = rs . 3000 answer : c" | a = 20 / 100
b = 1 + a
c = 3600 / b
|
a ) 13 , b ) 13.5 , c ) 14 , d ) 14.5 , e ) 15.3 | b | divide(add(multiply(11, const_10), divide(multiply(11, const_10), const_2)), add(divide(multiply(11, const_10), 12), divide(divide(multiply(11, const_10), const_2), 11))) | a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 12 miles per gallon , and from town b to town c , the car averaged 11 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through town b to town c ? | "ans is b given d _ ab = 2 * d _ bc let d _ ab = d and d _ bc = x so d = 2 x for average miles per gallon = ( d + x ) / ( ( d / 12 ) + ( x / 11 ) ) = 15.3 ( formula avg speed = total distance / total time )" | a = 11 * 10
b = 11 * 10
c = b / 2
d = a + c
e = 11 * 10
f = e / 12
g = 11 * 10
h = g / 2
i = h / 11
j = f + i
k = d / j
|
a ) 32.5 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 22.5 % | e | multiply(const_100, divide(add(divide(30, const_100), multiply(3, divide(20, const_100))), add(const_1, 3))) | because he ’ s taxed by his home planet , mork pays a tax rate of 30 % on his income , while mindy pays a rate of only 20 % on hers . if mindy earned 3 times as much as mork did , what was their combined tax rate ? | say morks income is - 100 so tax paid will be 30 say mindys income is 3 * 100 = 300 so tax paid is 20 % * 300 = 60 total tax paid = 30 + 60 = 90 . combined tax % will be 90 / 100 + 300 = 22.5 % | a = 30 / 100
b = 20 / 100
c = 3 * b
d = a + c
e = 1 + 3
f = d / e
g = 100 * f
|
a ) rs . 22500 , b ) rs . 24500 , c ) rs . 26500 , d ) rs . 28500 , e ) none of these | a | add(add(add(9000, 5000), 1000), multiply(divide(add(add(9000, 5000), 1000), const_100), 50)) | sahil purchased a machine at rs 9000 , then got it repaired at rs 5000 , then gave its transportation charges rs 1000 . then he sold it with 50 % of profit . at what price he actually sold it . | "explanation : question seems a bit tricky , but it is very simple . just calculate all cost price , then get 150 % of cp . c . p . = 9000 + 5000 + 1000 = 15000 150 % of 15000 = 150 / 100 * 15000 = 22500 option a" | a = 9000 + 5000
b = a + 1000
c = 9000 + 5000
d = c + 1000
e = d / 100
f = e * 50
g = b + f
|
a ) 525 , b ) 500 , c ) 510 , d ) 530 , e ) 550 | a | subtract(divide(multiply(multiply(3500, 15), 3), const_100), divide(multiply(multiply(3500, 10), 3), const_100)) | if a lends rs . 3500 to b at 10 % per annum and b lends the same sum to c at 15 % per annum then the gain of b in a period of 3 years is ? | "( 3500 * 5 * 3 ) / 100 = > 525 answer : a" | a = 3500 * 15
b = a * 3
c = b / 100
d = 3500 * 10
e = d * 3
f = e / 100
g = c - f
|
a ) 21 , b ) 23 , c ) 25 , d ) 27 , e ) 30 | c | sqrt(add(power(multiply(7.5, const_2), const_2), power(multiply(10, const_2), const_2))) | two twins sisters sita and geeta were standing back to back and suddenly they started running in opposite directions for 10 km each . then they turned left and ran for another 7.5 km . what is the distance ( in kilometers ) between the the two twins when they stop ? | the distance between them is the hypotenuse of a right angle triangle with sides 15 km and 20 km . the hypotenuse = sqrt ( 15 ^ 2 + 20 ^ 2 ) = 25 the answer is c . | a = 7 * 5
b = a ** 2
c = 10 * 2
d = c ** 2
e = b + d
f = math.sqrt(e)
|
a ) 20 m , b ) 70 m , c ) 50 m , d ) can not be determined , e ) none of these | c | multiply(divide(multiply(30, const_1000), const_3600), 6) | a train running at the speed of 30 km / hr crosses a pole in 6 sec . what is the length of the train ? | "speed = 30 * 5 / 18 = 25 / 3 m / sec length of the train = speed * time = 25 / 3 * 6 = 50 m answer : c" | a = 30 * 1000
b = a / 3600
c = b * 6
|
a ) 10 , b ) 8 , c ) 5 , d ) 4 , e ) 2 | e | subtract(divide(divide(15, const_1), const_3), const_3) | in how many ways can the integer 15 be expressed as a product of two different positive integers ? | 15 = 3 * 5 since 15 is not a perfect square , no of ways = 2 answer e | a = 15 / 1
b = a / 3
c = b - 3
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | divide(add(3, 17), subtract(10, 5)) | if 5 x + 3 = 10 x – 17 , what is the value of x ? | "solve for x : 5 x + 3 = 10 x - 17 20 = 5 x 4 = x correct answer d ) 4" | a = 3 + 17
b = 10 - 5
c = a / b
|
a ) 12 kg , b ) 60 kg , c ) 108 kg , d ) 96 kg , e ) none | c | multiply(multiply(multiply(3, 2), divide(1.8, const_100)), const_1000) | a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1.8 cm when a man gets on it . the mass of man is | "solution volume of water displaced = ( 3 x 2 x 0.018 ) m 3 = 0.108 m 3 . mass of man = volume of water displaced × density of water = ( 0.108 × 1000 ) kg = 108 kg . answer c" | a = 3 * 2
b = 1 / 8
c = a * b
d = c * 1000
|
a ) 766 min , b ) 656 min , c ) 144 min , d ) 877 min , e ) 555 min | c | multiply(add(const_1, const_4), 36) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in 36 min , then the slower pipe alone will be able to fill the tank in ? | "let the slower pipe alone fill the tank in x min . then , faster pipe will fill it in x / 3 min . 1 / x + 3 / x = 1 / 36 4 / x = 1 / 36 = > x = 144 min . answer : c" | a = 1 + 4
b = a * 36
|
a ) 40 % , b ) 50 % , c ) 65 % , d ) 70 % , e ) 75 % | e | multiply(divide(subtract(150, add(multiply(3, 8), multiply(8, 3))), 150), const_100) | a batsman scored 150 runs which included 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets | "number of runs made by running = 150 - ( 3 x 4 + 8 x 6 ) = 150 - ( 60 ) = 90 now , we need to calculate 60 is what percent of 120 . = > 90 / 120 * 100 = 75 % e" | a = 3 * 8
b = 8 * 3
c = a + b
d = 150 - c
e = d / 150
f = e * 100
|
a ) 60 % , b ) 70 % , c ) 100 % , d ) 90 % , e ) 80 % | c | divide(multiply(subtract(divide(150, subtract(const_1, divide(50, const_100))), 150), const_100), 150) | a furniture dealer purchased a desk for $ 150 and then set the selling price equal to the purchase price plus a markup that was 50 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ? | "anyway , in this question , there is no discount but the mark up is given as 50 % of the selling price . so it is not 50 % of $ 150 but instead , 50 % of selling price which is obtained by adding mark up to $ 150 . so if selling price is s , 150 + 50 % of s = s s = 300 profit = 150 which is calculated on cost price in % terms . so 150 / 150 * 100 = 100 % is profit . c" | a = 50 / 100
b = 1 - a
c = 150 / b
d = c - 150
e = d * 100
f = e / 150
|
a ) 17 : 1 , b ) 17 : 6 , c ) 17 : 9 , d ) 17 : 2 , e ) 17 : 3 | e | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | p and q started a business investing rs . 85,000 and rs . 15,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 85000 : 15000 = 17 : 3 . answer : e" | a = 2 + 3
b = a + 3
c = 2 + 3
d = c * 2
e = b * d
f = 2 + 3
g = e + f
h = 2 + 3
i = h * 2
j = 3 * i
k = 2 + 3
l = j + k
m = g / l
|
a ) 28 / 3 , b ) 27 / 3 , c ) 26 / 3 , d ) 25 / 3 , e ) 24 / 3 | a | divide(add(multiply(const_2, const_4), multiply(const_10, const_2)), const_3) | 1 / 216 , 415 , 356 , _ ? | take alternative no . 1 / 2 , 43 16 / 4 = 4 15 / 5 = 3 56 / 6 = 28 / 3 hence ans is 28 / 3 answer : a | a = 2 * 4
b = 10 * 2
c = a + b
d = c / 3
|
a ) 276 , b ) 299 , c ) 322 , d ) 345 , e ) 368 | e | multiply(23, 16) | the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 13 and 16 . the larger of the two numbers is : | "clearly , the numbers are ( 23 x 13 ) and ( 23 x 16 ) . larger number = ( 23 x 16 ) = 368 . answer : option e" | a = 23 * 16
|
a ) 1.2 sec , b ) 2.4 sec , c ) 4.8 sec , d ) 9.6 sec , e ) none | d | multiply(multiply(100, inverse(multiply(add(45, 30), const_0_2778))), const_2) | two good train each 100 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 . | "sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 100 + 100 ) m = 1000 m . required time = ( 200 x 6 / 125 ) sec = 9.6 sec . answer d" | a = 45 + 30
b = a * const_0_2778
c = 1/(b)
d = 100 * c
e = d * 2
|
a ) 200 , b ) 320 , c ) 480 , d ) 500 , e ) 600 | b | subtract(subtract(800, 1), subtract(add(add(divide(subtract(subtract(800, const_2), const_2), const_2), 1), add(divide(subtract(subtract(800, divide(800, const_100)), divide(800, const_100)), divide(800, const_100)), 1)), add(divide(subtract(subtract(800, const_10), const_10), const_10), 1))) | what is the total number of positive integers that are less than 800 and that have no positive factor in common with 800 other than 1 ? | "since 800 = 2 ^ 5 * 5 ^ 2 then a number can not have 2 and / or 5 as a factor . the odd numbers do not have 2 as a factor and there are 400 odd numbers from 1 to 800 . we then need to eliminate the 80 numbers that end with 5 , that is 5 , 15 , 25 , . . . , 795 . there are a total of 400 - 80 = 320 such numbers between 1 and 800 . the answer is b ." | a = 800 - 1
b = 800 - 2
c = b - 2
d = c / 2
e = d + 1
f = 800 / 100
g = 800 - f
h = 800 / 100
i = g - h
j = 800 / 100
k = i / j
l = k + 1
m = e + l
n = 800 - 10
o = n - 10
p = o / 10
q = p + 1
r = m - q
s = a - r
|
a ) 35 hours , b ) 40 hours , c ) 36 hours , d ) 38 hours , e ) 42 hours | b | subtract(multiply(divide(120, add(add(1, 2), 3)), 3), multiply(divide(120, add(add(1, 2), 3)), 1)) | the amount of time that three people worked on a special project was in the ratio of 1 to 2 to 3 . if the project took 120 hours , how many more hours did the hardest working person work than the person who worked the least ? | "let the persons be a , b , c . hours worked : a = 1 * 120 / 6 = 20 hours b = 2 * 120 / 6 = 40 hours c = 3 * 120 / 6 = 60 hours c is the hardest worker and a worked for the least number of hours . so the difference is 60 - 20 = 40 hours . answer : b" | a = 1 + 2
b = a + 3
c = 120 / b
d = c * 3
e = 1 + 2
f = e + 3
g = 120 / f
h = g * 1
i = d - h
|
a ) 540 km , b ) 767 km , c ) 276 km , d ) 450 km , e ) 176 km | d | multiply(add(20, 25), divide(50, subtract(25, 20))) | two trains start at same time from two stations and proceed towards each other at the rate of 20 km / hr and 25 km / hr respectively . when they meet , it is found that one train has traveled 50 km more than the other . what is the distance between the two stations ? | explanation : let us assume that trains meet after ' x ' hours distance = speed * time distance traveled by two trains = 20 x km and 25 x km resp . as one train travels 50 km more than the other , 25 x â € “ 20 x = 50 5 x = 50 x = 10 hours as the two trains are moving towards each other , relative speed = 20 + 25 = 45 km / hr therefore , total distance = 45 * 10 = 450 km . answer : d | a = 20 + 25
b = 25 - 20
c = 50 / b
d = a * c
|
a ) 19 , b ) 21 , c ) 23 , d ) 25 , e ) 27 | c | add(add(divide(subtract(150, 40), 5), 1), 1) | a test has 150 questions . each question has 5 options , but only 1 option is correct . if test - takers mark the correct option , they are awarded 1 point . however , if an answer is incorrectly marked , the test - taker loses 0.25 points . no points are awarded or deducted if a question is not attempted . a certain group of test - takers attempted different numbers of questions , but each test - taker still received the same net score of 40 . what is the maximum possible number of such test - takers ? | "a correct answers get you 1 point , an incorrect answer gets you minus 1 / 4 point and a skipped question gets you 0 points . since there are 200 total questions , there are a variety of ways to get a total of 40 points . let c be the number of correct answers and let i be the number of incorrect answers . to get 40 points , a test taker must have at least 40 correct answers . then c = > 40 . for every correct question above 40 , the test taker has 4 incorrect answers . then , the i = 4 * ( c - 40 ) . also , i + c < = 150 . thus 5 c < = 310 and so c < = 62 . then 40 < = c < = 62 and c can have 23 possible values . the answer is c ." | a = 150 - 40
b = a / 5
c = b + 1
d = c + 1
|
a ) 347.4 , b ) 987.4 , c ) 877.2 , d ) 1027.2 , e ) 1667.2 | d | add(add(multiply(4, multiply(divide(24, 10), divide(multiply(24, 6), 2))), multiply(3, divide(multiply(24, 6), 2))), multiply(5, 24)) | the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 24 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ? | "let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 24 = 2 r a = 12 / 5 r and r = 72 a = 144 required total cost = 4 * 144 + 3 * 72 + 5 * 24 = 691.2 + 216 + 120 = $ 1027.20 d" | a = 24 / 10
b = 24 * 6
c = b / 2
d = a * c
e = 4 * d
f = 24 * 6
g = f / 2
h = 3 * g
i = e + h
j = 5 * 24
k = i + j
|
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