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 ) 300 , b ) 400 , c ) 900 , d ) 600 , e ) 700 | c | power(30, const_2) | what is the sum of the first 30 odd natural numbers | sum of first n odd natural nuber is = n ^ 2 so sum of the first 30 odd natural numbers = ( 30 ) ^ 2 = 900 ans 900 answer : c | a = 30 ** 2
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a ) 36 , b ) 40 , c ) 45 , d ) 32 , e ) 56 | c | add(multiply(divide(2, multiply(54, 2)), 2), multiply(divide(2, multiply(54, 2)), 54)) | the l . c . m . of 2 numbers is 54 . the numbers are in the ratio 2 : 3 . find their sum ? | "let the numbers be 2 x and 3 x l . c . m . = 6 x 6 x = 54 x = 9 the numbers are = 18 and 27 required sum = 18 + 27 = 45 answer is c" | a = 54 * 2
b = 2 / a
c = b * 2
d = 54 * 2
e = 2 / d
f = e * 54
g = c + f
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a ) 11 , b ) 10 , c ) 18 , d ) 16 , e ) 14 | e | multiply(multiply(divide(7, 5), divide(10, 7)), 7) | the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 10 . if p invested the money for 7 months , find for how much time did q invest the money ? | "7 * 5 : 7 * x = 7 : 10 x = 14 answer : e" | a = 7 / 5
b = 10 / 7
c = a * b
d = c * 7
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | subtract(power(2, 2), 2) | if x ^ 2 + 1 / x ^ 2 = √ 12 , what is the value of x ^ 4 + 1 / x ^ 4 ? | "important : i notice that if we square x ² , we get x ⁴ , and if we square 1 / x ² , we get 1 / x ⁴ , so let ' s see what happens if we take the equation x ² + 1 / x ² = √ 12 andsquareboth sides : ( x ² + 1 / x ² ) ² = 12 so , ( x ² + 1 / x ² ) ( x ² + 1 / x ² ) = 12 expand to get : x ⁴ + 1 + 1 + 1 / x ⁴ = 12 simplify : x ⁴ + 1 / x ⁴ = 10 answer : a" | a = 2 ** 2
b = a - 2
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a ) 23 , b ) 27 , c ) 30 , d ) 32 , e ) 34 | b | multiply(divide(3, add(add(5, 3), 4)), 108) | the ratio of 3 numbers is 5 : 3 : 4 and their sum is 108 . the second number of the 3 numbers is ? | 5 : 3 : 4 total parts = 12 12 parts - - > 108 1 part - - - - > 9 the second number of the three numbers is = 3 3 parts - - - - > 27 b ) | a = 5 + 3
b = a + 4
c = 3 / b
d = c * 108
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a ) 33 , b ) 54 , c ) 36 , d ) 17 , e ) 01 | c | multiply(subtract(70, 42), divide(90, 70)) | a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 42 adults have their meal , find the total number of children that can be catered with the remaining food . | explanation : as there is meal for 70 adults and 42 have their meal , the meal left can be catered to 28 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 28 adults = 36 children hence , the meal can be catered to 36 children . answer : c | a = 70 - 42
b = 90 / 70
c = a * b
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a ) 5 min , b ) 6 min , c ) 7 and 1 / 2 min , d ) 8 min , e ) 10 min | c | multiply(subtract(divide(divide(multiply(multiply(subtract(15, 10), divide(15, const_60)), const_60), 10), const_2), const_0_25), 10) | tom and john traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour , respectively . after 15 minutes tom passed john , tom reaches a certain gas station , how many w minutes it takes john to reach the station ? | since the question states “ after 15 minutes ” , we can say tom traveled 15 / 4 km for 15 minutes as he can travel 15 km per hour . hence , using the same logic , we can say john traveled 10 / 4 km as he travels 10 km per hour . so , john has to travel ( 15 / 4 ) - ( 10 / 4 ) km = 5 / 4 km more . since john ’ s speed is 10 km / hour , which means 1 km / 6 minutes . as he has to travel 5 / 4 km more , it is going to take him 6 ( 5 / 4 ) minutes . hence , w = 6 ( 5 / 4 ) = 15 / 2 minutes . the correct answer is c . | a = 15 - 10
b = 15 / const_60
c = a * b
d = c * const_60
e = d / 10
f = e / 2
g = f - const_0_25
h = g * 10
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a ) 5 / 3 , b ) 3 / 5 , c ) 3 / 7 , d ) 5 / 7 , e ) 7 / 9 | a | divide(divide(const_1, const_4), divide(40, const_100)) | if 40 % of a number is equal to 2 / 3 of another number , what is the ratio of first number to the second number ? | "let 40 % of a = 2 / 3 b then , 40 * a / 100 = 2 * b / 3 a / b = 5 / 3 ans - a" | a = 1 / 4
b = 40 / 100
c = a / b
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | inverse(add(divide(const_1, 20), divide(4, 20))) | pipe a fills a tank in 20 minutes . pipe b can fill the same tank 4 times as fast as pipe a . if both the pipes are kept open when the tank is empty , how many minutes will it take to fill the tank ? | "a ' s rate is 1 / 20 and b ' s rate is 1 / 5 . the combined rate is 1 / 20 + 1 / 5 = 1 / 4 the pipes will fill the tank in 4 minutes . the answer is c ." | a = 1 / 20
b = 4 / 20
c = a + b
d = 1/(c)
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a ) 9.45 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | a | multiply(divide(add(divide(multiply(12, 20), const_100), divide(multiply(8, 35), const_100)), add(20, 35)), const_100) | in one alloy there is 12 % chromium while in another alloy it is 8 % . 20 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 20 + 35 = 55 kg alloy is 0.12 * 20 + 0.08 * 35 = 5.2 kg , so the percentage is 5.2 / 55 * 100 = 9.2 % . answer : a ." | a = 12 * 20
b = a / 100
c = 8 * 35
d = c / 100
e = b + d
f = 20 + 35
g = e / f
h = g * 100
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a ) 50 , b ) 25 , c ) 52 , d ) 58 , e ) 60 | a | divide(multiply(42, const_100), subtract(const_100, 16)) | the number which exceeds 16 % of it by 42 is ? | "let the number be x . then , x - 16 % of x = 42 . x - 16 / 100 x = 42 x = ( 42 * 25 ) / 21 = 50 answer : a" | a = 42 * 100
b = 100 - 16
c = a / b
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | d | add(reminder(multiply(reminder(53, const_4), 93), const_10), reminder(35, const_10)) | the units digit of ( 35 ) ^ ( 87 ) + ( 93 ) ^ ( 53 ) is : | the units digit of powers of 3 , cycles in a group of 4 : { 3 , 9 , 7 , 1 } 53 has the form 4 k + 1 , so the units digit of 93 ^ 53 is 3 . the units digit of powers of 5 is always 5 . 3 + 5 = 8 , so the units digit is 8 . the answer is d . | a = reminder * (
b = reminder + (
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a ) 0.66 , b ) 0.72 , c ) 0.78 , d ) 0.84 , e ) 0.9 | a | subtract(const_1, add(multiply(divide(40, const_100), divide(40, const_100)), multiply(divide(30, 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 30 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 + 0.7 * 0.6 x = 0.66 x the answer is a ." | a = 40 / 100
b = 40 / 100
c = a * b
d = 30 / 100
e = 40 / 100
f = 1 - e
g = d * f
h = c + g
i = 1 - h
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a ) 8 , b ) 9 , c ) 12 , d ) 14 , e ) 16 | b | divide(subtract(93, multiply(const_3, 13)), multiply(const_3, const_2)) | a number is doubled and 13 is added . if resultant is trebled , it becomes 93 . what is that number | "explanation : = > 3 ( 2 x + 13 ) = 93 = > 2 x + 13 = 31 = > x = 9 option b" | a = 3 * 13
b = 93 - a
c = 3 * 2
d = b / c
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a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(2020, 2020), add(divide(2020, 2020), divide(2020, 2020))) | we define that k @ j is the product of j number from k in increasing order for positive integers k , j . for example , 6 @ 4 = 6 * 7 * 8 * 9 . if a = 2020 and b = 2120 , what is the value t of the ratio a / b ? | "t - > a / b = 20 * 21 * … … * 39 / 21 * 22 * … . * 39 * 40 = 20 / 40 = 1 / 2 . therefore , the answer is a ." | a = 2020 / 2020
b = 2020 / 2020
c = 2020 / 2020
d = b + c
e = a / d
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a ) 190 , b ) 284.6 , c ) 300 , d ) 268 , e ) 312 | d | multiply(12.2, 22) | a type of extra - large suv averages 12.2 miles per gallon ( mpg ) on the highway , but only 7.6 mpg in the city . what is the maximum distance , in miles , that this suv could be driven on 22 gallons of gasoline ? | "so 12.2 * 22 = 268 . . imo option d is correct answer . ." | a = 12 * 2
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a ) 740 , b ) 750 , c ) 760 , d ) 770 , e ) 775 | b | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 1000), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 700)), 30) | a library has an average of 1000 visitors on sundays and 700 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "explanation : since the month begins with a sunday , so there will be five sundays in the month , required average = ( 1000 * 5 + 700 * 25 ) / 30 = 22500 / 30 = 750 answer : b ) 750" | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 1000
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 700
l = e + k
m = l / 30
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a ) 11 , b ) 25 , c ) 30 , d ) 88 , e ) 34 | b | divide(add(12, 38), const_2) | a man can row upstream at 12 kmph and downstream at 38 kmph , and then find the speed of the man in still water ? | "us = 12 ds = 38 m = ( 12 + 38 ) / 2 = 25 answer : b" | a = 12 + 38
b = a / 2
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a ) 312 , b ) 356 , c ) 452 , d ) 416 , e ) 512 | a | add(multiply(divide(8, 5), 120), 120) | in a college the ratio of the numbers of boys to the girls is 8 : 5 . if there are 120 girls , the total number of students in the college is ? | let the number of boys and girls be 8 x and 5 x then , 5 x = 120 x = 24 total number of students = 13 x = 13 * 24 = 312 answer is a | a = 8 / 5
b = a * 120
c = b + 120
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a ) 36.4543 , b ) 36.2727 , c ) 31.45128 , d ) 36.41288 , e ) 31.41222 | b | multiply(divide(21, 12), 20) | there are 20 poles with a constant distance between each pole . a car takes 21 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 21 seconds 1 pole ( 21 / 11 ) seconds to reach the last ( 20 th ) pole , the car needs to travel 19 poles . 19 pole 19 x ( 21 / 11 ) seconds = 36.2727 seconds answer : b" | a = 21 / 12
b = a * 20
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a ) 15 , b ) 12.5 , c ) 10 , d ) 8 , e ) 7.75 | a | divide(multiply(5, 3), subtract(multiply(const_2, 3), 5)) | noelle walks from point a to point b at an average speed of 3 kilometers per hour . at what speed , in kilometers per hour , must noelle walk from point b to point a so that her average speed for the entire trip is 5 kilometers per hour ? | let ' s suppose that speed while returning was xkm / h since the distance is same , we can apply the formula of avg speed avg speed = 2 s 1 s 2 / s 1 + s 2 5 = 2 * 3 * x / 3 + x x = 15 a is the answer | a = 5 * 3
b = 2 * 3
c = b - 5
d = a / c
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a ) 10 % , b ) ( 8 ) 20 % , c ) 30 % , d ) 50 % , e ) 60 % | e | subtract(add(add(75, 65), 20), const_100) | if 75 percent of a class answered the first question on a certain test correctly , 65 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ? | "{ total } = { first } + { second } - { both } + { neither } 100 % = 75 % + 65 % - { both } + 20 % - - > { both } = 60 % . answer : e ." | a = 75 + 65
b = a + 20
c = b - 100
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a ) 11 , b ) 12 , c ) 10.9 , d ) 14 , e ) 15 | c | divide(multiply(120, const_2), add(speed(120, 12), speed(120, 10))) | two trains of equal lengths take 10 sec and 12 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 12 = 10 m / sec . relative speed = 12 + 10 = 22 m / sec . required time = ( 120 + 120 ) / 22 = 10.9 sec . answer : option c" | a = 120 * 2
b = speed + (
c = a / b
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a ) 2 , b ) 7 , c ) 5 , d ) 3 , e ) 4 | a | divide(add(add(1, 1), 2), 1) | the area of a triangle will be when a = 1 m , b = 1 m , c = 2 m , a , b , c being lengths of respective sides ? | "s = ( 1 + 1 + 2 ) / 2 = 2 answer : a" | a = 1 + 1
b = a + 2
c = b / 1
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a ) 10.6 kg , b ) 10.8 kg , c ) 7 kg , d ) 14.9 kg , e ) none | c | subtract(multiply(add(19, const_1), 14.6), multiply(19, 15)) | the average weight of 19 students is 15 kg . by the admission of a new student the average weight is reduced to 14.6 kg . the weight of the new student is ? | "answer weight of new student = total weight of all 20 students - total weight of initial 19 students = ( 20 x 14.6 - 19 x 15 ) kg = 7 kg . correct option : c" | a = 19 + 1
b = a * 14
c = 19 * 15
d = b - c
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a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | c | multiply(9, const_2) | how many two digit numbers are there such that the product of their digits after reducing it to the smallest form is a prime number ? for example if we take 98 then 9 * 8 = 72 , 72 = 7 * 2 = 14 , 14 = 1 * 4 = 4 . consider only 4 prime no . s ( 23 , 57 ) | we know prime no . are 2 , 3 , 5 , 7 mulply by 1 is prime no 2 = 21 ( 2 * 1 ) = 73 ( 7 * 3 ) 2 = 21 ( 2 * 1 ) = 37 ( 3 * 7 ) 2 = 12 ( 1 * 2 ) = 43 ( 4 * 3 ) or 34 ( 3 * 4 ) or 26 ( 2 * 6 ) or 62 ( 6 * 2 ) . . . similarly for all prime no then u will got it . . . . 12,13 , 15,17 , 21,26 , 31,34 , 35,37 , 43,51 , 53,57 , 62,71 , 73,75 ans is 18 answer : c | a = 9 * 2
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a ) 9 , b ) 12 , c ) 11 , d ) 10 , e ) 3 | a | add(divide(16, const_2), 1) | each of the integers from 1 to 16 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement , how many cards must be drawn to ensure that the product of all the integers drawn is even ? | "out of the 16 integers : 8 are odd and 8 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : 1 . we will end up picking odd numbers one by one , so we will pick all 8 odd numbers first 2 . 9 th number will be the first even number so we need to withdraw at least 9 numbers to make sure that we get one even number and the product of all the integers picked is even . so , answer will be 9 . ( a )" | a = 16 / 2
b = a + 1
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a ) 7.98 , b ) 8.98 , c ) 9.98 , d ) 6.98 , e ) 8.12 | b | divide(multiply(5000, 3), add(add(divide(5000, 8.5), divide(5000, 9)), divide(5000, 9.5))) | a taxi owner buys diesel at rs . 8.50 , rs . 9 and rs . 9.50 per litre for 3 successive years . what approximately is the average cost per litre of diesel if he spends rs . 5000 each year ? | total cost = 5000 × 3 = 15000 total diesel used = 5000 / 8.50 + 5000 / 9 + 5000 / 9.50 = 588 + 555 + 527 = 1670 average cost per litre of diesel = 15000 / 1670 = 8.98 answer : b | a = 5000 * 3
b = 5000 / 8
c = 5000 / 9
d = b + c
e = 5000 / 9
f = d + e
g = a / f
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a ) rs . 1200 , b ) rs . 1300 , c ) rs . 2500 , d ) rs . 2000 , e ) none of these | c | multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6))) | a man invests rs . 5,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | "explanation : interest on rs . 5000 at 5 % per annum = ( 5000 × 5 × 1 ) / 100 = rs . 250 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x × 8 × 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 5000 + x ) × 6 × 1 / 100 . = > 250 + 2 x / 25 = ( 5000 + x ) × 6 × 1 / 100 . = > 25000 + 8 x = ( 5000 + x ) × 6 . = > 25000 + 8 x = 30000 + 6 x . = > 2 x = 5000 . = > x = 2500 . answer : c" | a = 8 * 1000
b = 6 - 5
c = 8 - 6
d = b / c
e = a * d
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a ) 2.215 , b ) 2.5 . , c ) 2.775 , d ) 3.2 , e ) 3.5 | a | multiply(divide(divide(500, 500), divide(12, const_60)), subtract(const_1, divide(15, const_100))) | an ant walks an average of 500 meters in 12 minutes . a beetle walks 15 % less distance at the same time on the average . assuming the beetle walks at her regular rate , what is its speed in km / h ? | "the ant walks an average of 500 meters in 12 minutes or , 500 meters in 1 / 5 hours the beetle walks 15 % less distance = 500 - 75 = 425 meters in 12 minutes or , 0.425 km in 12 / 60 = 1 / 5 hours speed = 0.425 * 5 = 2.125 km / h ans should be 2.125 answer : a" | a = 500 / 500
b = 12 / const_60
c = a / b
d = 15 / 100
e = 1 - d
f = c * e
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a ) 0 , b ) 50 , c ) 450 , d ) 495 , e ) 500 | d | subtract(multiply(multiply(add(const_3, const_4), const_1000), divide(1, 10)), multiply(divide(divide(1, 10), const_100), multiply(add(const_3, const_4), const_1000))) | when 1 / 10 percent of 5,000 is subtracted from 1 / 10 of 5,000 , the difference is | "1 / 10 of 5,000 is 500 ; 1 / 10 percent of 5,000 is 100 times smaller than the above , so it ' s 5 ; thus the difference is 500 - 5 = 495 . answer : d ." | a = 3 + 4
b = a * 1000
c = 1 / 10
d = b * c
e = 1 / 10
f = e / 100
g = 3 + 4
h = g * 1000
i = f * h
j = d - i
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a ) 3 / 4 , b ) 1 / 5 , c ) 1 / 10 , d ) 1 / 11 , e ) none of above | a | divide(circle_area(divide(75, const_2)), const_2) | what will be the fraction of 75 % | "explanation : it will 75 * 1 / 100 = 3 / 4 option a" | a = 75 / 2
b = circle_area / (
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a ) 26 , b ) 32 , c ) 38 , d ) 46 , e ) 52 | c | add(3, add(4, add(6, add(5, multiply(5, const_2))))) | what is the least number which should be added to 2782 so that the sum is exactly divisible by 5 , 6 , 4 , and 3 ? | "l . c . m . of 5 , 6 , 4 and 3 = 60 . when dividing 2782 by 60 , the remainder is 22 . the number to be added = 60 - 22 = 38 . the answer is c ." | a = 5 * 2
b = 5 + a
c = 6 + b
d = 4 + c
e = 3 + d
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a ) 29 , b ) 776 , c ) 66 , d ) 14 , e ) 99 | d | subtract(multiply(40, divide(35, const_100)), multiply(divide(4, 5), 25)) | how much is 35 % of 40 is greater than 4 / 5 of 25 ? | "( 80 / 100 ) * 40 – ( 4 / 5 ) * 25 35 - 20 = 14 answer : d" | a = 35 / 100
b = 40 * a
c = 4 / 5
d = c * 25
e = b - d
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a ) 1.2 hr , b ) 2.5 hrs , c ) 3 hrs , d ) 5.2 hrs , e ) 6 hrs | b | divide(50, add(10, 10)) | two cyclist start from the same places in opposite directions . one is going towards north at 10 kmph and the other is going towards south 10 kmph . what time will they take to be 50 km apart ? | "to be ( 10 + 10 ) km apart , they take 1 hour to be 50 km apart , they take 1 / 20 * 50 = 2.5 hrs answer is b" | a = 10 + 10
b = 50 / a
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a ) 5 / 3 , b ) 9 / 5 , c ) 41 / 4 , d ) 41 / 12 , e ) none | a | multiply(0.2, add(add(0.2, 0.1), 0.02)) | the value of ( 0.2 × 0.2 + 0.01 ) ( 0.1 × 0.1 + 0.02 ) - 1 is equal to | "solution given expression ( 0.2 × 0.2 + 0.01 ) / ( 0.1 × 0.1 × + 0.02 ) = 0.04 + 0.01 / 0.01 + 0.02 = 0.05 / 0.03 = 5 / 3 . answer a" | a = 0 + 2
b = a + 0
c = 0 * 2
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a ) 600 , b ) 882 , c ) 775 , d ) 662 , e ) 521 | c | add(620, multiply(620, divide(25, const_100))) | a person buys an article at rs . 620 . at what price should he sell the article so as to make a profit of 25 % ? | "cost price = rs . 620 profit = 25 % of 620 = rs . 155 selling price = cost price + profit = 620 + 155 = 775 answer : c" | a = 25 / 100
b = 620 * a
c = 620 + b
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a ) 7 , b ) 17 , c ) 70 , d ) 100 , e ) 107 | c | multiply(10, 7) | the average of 10 numbers is 7 . if each number is multiplied by 10 , then what is the average of the new set of numbers ? | the sum of the ten numbers is 7 * 10 = 70 if each number is multiplied by 10 , the new sum is 10 * ( 70 ) . the average is then 10 * 7 = 70 the answer is c . | a = 10 * 7
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a ) 1 / 2 , b ) 2 , c ) 1 / 3 , d ) 3 , e ) 5 / 6 | e | sqrt(divide(25, 36)) | if xy = 25 , x / y = 36 , for positive numbers x and y , y = ? | "very easy question . 2 variables and 2 easy equations . xy = 25 - - - > x = 25 / y - ( i ) x / y = 36 - - - > replacing ( i ) here - - - > 25 / ( y ^ 2 ) = 36 - - - > y ^ 2 = 25 / 36 - - - > y = 5 / 6 or - 5 / 6 the question states that x and y are positive integers . therefore , y = 5 / 6 is the answer . answer e ." | a = 25 / 36
b = math.sqrt(a)
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a ) 44 % , b ) 48 % , c ) 54 % , d ) 62 % , e ) 70 % | d | divide(add(20, 70), multiply(multiply(const_5, const_5), const_4)) | there is a 20 % chance sandy will visit china this year , while there is a 70 % chance that she will visit malaysia this year . what is the probability that sandy will visit either china or malaysia this year , but not both ? | "p ( china and not malaysia ) = 0.2 * 0.3 = 0.06 p ( malaysia and not china ) = 0.7 * 0.8 = 0.56 total probability = 0.06 + 0.56 = 0.62 = 62 % the answer is d ." | a = 20 + 70
b = 5 * 5
c = b * 4
d = a / c
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a ) 60 % , b ) 65 % , c ) 70 % , d ) 75 % , e ) 80 % | a | subtract(add(add(80, 75), 5), const_100) | if 80 percent of the test takers taking an old paper and pencil gmat exam answered the first question on a given math section correctly , and 75 percent of the test takers answered the second question correctly , and 5 percent of the test takers answered neither question correctly , what percent answered both correctly ? | "{ total } = { first correctly } + { second correctly } - { both correctly } + { neither correctly } 100 = 80 + 75 - { both correctly } + 5 { both correctly } = 60 . answer : a ." | a = 80 + 75
b = a + 5
c = b - 100
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a ) 67 , b ) 26 , c ) 87 , d ) 26 , e ) 68 | e | divide(add(add(add(add(70, 63), 80), 63), 65), divide(const_10, const_2)) | dacid obtained 70 , 63 , 80 , 63 and 65 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 70 + 63 + 80 + 63 + 65 ) / 5 = 341 / 5 = 68.2 . answer : e" | a = 70 + 63
b = a + 80
c = b + 63
d = c + 65
e = 10 / 2
f = d / e
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a ) 50 min , b ) 33 min , c ) 35 min , d ) 25 min , e ) 20 min | b | add(20, 13) | the jogging track in a sports complex is 1000 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at 20 km / hr and 13 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 1000 m apart to be 20 + 13 = 33 km apart , they take 1 hour to be 1000 m apart , they take 33 * 1000 / 1000 = 33 min . answer is b" | a = 20 + 13
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a ) 134 , b ) 178 , c ) 199 , d ) 166 , e ) 194 | c | subtract(divide(add(979, add(add(53, 79), 105)), 4), 105) | at a recent small town election for mayor a total of 979 votes were cast for the 4 candidates , the winner exceeding his opponents by 53 , 79 and 105 votes , respectively . how many votes were cast for the candidate in fourth place ? | the number of votes the winning candidate received was 979 + 53 + 79 + 105 / 4 = 304 . the second received 304 – 53 = 251 the fourth place received 304 – 105 = 199 . c | a = 53 + 79
b = a + 105
c = 979 + b
d = c / 4
e = d - 105
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a ) 32 kmph , b ) 48 kmph , c ) 34 kmph , d ) 43 kmph , e ) 40 kmph | b | divide(divide(add(100, 300), const_1000), divide(30, const_3600)) | a train 100 meters long completely crosses a 300 meters long bridge in 30 seconds . what is the speed of the train is ? | "s = ( 100 + 300 ) / 30 = 400 / 30 * 18 / 5 = 48 answer : b" | a = 100 + 300
b = a / 1000
c = 30 / 3600
d = b / c
|
a ) 1410 , b ) 1420 , c ) 1430 , d ) 1440 , e ) 1540 | d | divide(multiply(subtract(const_100, 10), 1600), const_100) | a man buys a cycle for rs . 1600 and sells it at a loss of 10 % . what is the selling price of the cycle ? | "s . p . = 90 % of rs . 1600 = 90 / 100 x 1600 = rs . 1440 answer : d" | a = 100 - 10
b = a * 1600
c = b / 100
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a ) 6400 , b ) 6410 , c ) 5600 , d ) 6500 , e ) 6800 | c | divide(multiply(multiply(multiply(7, const_100), multiply(6, const_100)), 22.5), multiply(multiply(25, 11.25), 6)) | how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 7 m x 6 m x 22.5 cm ? | "number of bricks = volume of wall / volume of bricks = 700 x 600 x 22.5 / 25 x 11.25 x 6 = = 5600 answer : c" | a = 7 * 100
b = 6 * 100
c = a * b
d = c * 22
e = 25 * 11
f = e * 6
g = d / f
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a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28 | e | divide(7, subtract(divide(5, 4), const_1)) | walking with 4 / 5 of my usual speed , i arrive at the bus stop 7 minutes later than normal . how many minutes does it take to walk to the bus stop at my usual speed ? | let t = usual time = distance / usual speed t + 7 = distance / ( 4 * usual speed / 5 ) = ( 5 * distance ) / ( 4 * usual speed ) = 5 t / 4 t = 28 the answer is e . | a = 5 / 4
b = a - 1
c = 7 / b
|
a ) 50 sec , b ) 29 sec , c ) 26 sec , d ) 27 sec , e ) 34 sec | a | divide(360, multiply(subtract(36, 140), const_0_2778)) | a train 360 m long is running at a speed of 36 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 36 * 5 / 18 = 10 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 1 / 10 = 50 sec answer : a" | a = 36 - 140
b = a * const_0_2778
c = 360 / b
|
a ) 2.1 , b ) 1.3 , c ) 5.1 , d ) 2.3 , e ) 1.1 | e | add(const_1, divide(const_1, const_10)) | a primitive village follows a strange custom . they have babies only to have a boy . whenever a boy is born in a family , they stop having babies and whenever a girl is born , they have a baby again and again till a boy is born . the probability of having a boy is same as the probability of having a girl . what will be the proportion of boys to girls in the village after some time ? | e 1 : 1 ( approx ) explanation : we know that the probability of having a boy or having a girl is same and thus , half of the couples will stop after having a boy child . half of the others will have a girl and will have a baby again . out of those half of the couples , half will have a boy and will stop and half will have a girl again . this will keep on going like this . now , if there are x number of couples , there will be x boys . 1 / 2 have a boy and stop : 0 girls 1 / 4 have a girl , then a boy : x / 4 girls 1 / 8 have 2 girls , then a boy : 2 * x / 8 girls 1 / 16 have 3 girls , then a boy : 3 * x / 16 girls 1 / 32 have 4 girls , then a boy : 4 * x / 32 girls … total : x boys and 1 x 2 x 3 x 4 x – + – + – + — + … = ~ x therefore , the proportion of boys to girls will be extremely close to 1 : 1 | a = 1 / 10
b = 1 + a
|
a ) 57 , b ) 67 , c ) 77 , d ) 87 , e ) 44 | e | sqrt(multiply(19.36, const_100)) | a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 19.36 , the number of the member is the group is : | "money collected = ( 19.36 x 100 ) paise = 1936 paise . number of members = square root of 1936 = 44 answer : option e" | a = 19 * 36
b = math.sqrt(a)
|
a ) 54 , b ) 162 , c ) 250 , d ) 270 , e ) 322 | d | divide(multiply(324, 10), add(2, 10)) | compound x contains elements a and b at an approximate ratio , by weight , of 2 : 10 . approximately how many grams of element b are there in 324 grams of compound x ? | "total number of fractions = 2 + 10 = 12 element b constitutes = 10 out of 12 parts of x so in 324 gms of x have 324 * 10 / 12 = 270 gms of b and 324 - 270 = 54 gms of a . cross check : - a / b = 54 / 270 = 2 / 10 ( as given ) ans d" | a = 324 * 10
b = 2 + 10
c = a / b
|
a ) 15 % , b ) 23 % , c ) 25 % , d ) 60 % , e ) 80 % | b | multiply(divide(subtract(77.95, 59.95), 77.95), const_100) | a $ 77.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ? | "listed selling price of chair = 77.95 $ discounted selling price of chair = 59.95 $ discount = 77.95 - 59.95 = 18 $ % decrease in price of chair = ( 18 / 77.95 ) * 100 % = 23 % approx answer b" | a = 77 - 95
b = a / 77
c = b * 100
|
a ) 140 cm , b ) 767 cm , c ) 88 cm , d ) 666 cm , e ) 776 cm | a | multiply(sqrt(divide(98, 50)), const_100) | 50 square stone slabs of equal size were needed to cover a floor area of 98 sq . m . find the length of each stone slab ? | "area of each slab = 98 / 50 m 2 = 1.96 m 2 length of each slab √ 1.96 = 1.4 m = 140 cm" | a = 98 / 50
b = math.sqrt(a)
c = b * 100
|
a ) 4236 , b ) 4306 , c ) 4336 , d ) 5336 , e ) none | d | multiply(multiply(5, 46), divide(multiply(5, 46), 10)) | in a division sum , the divisor is 10 times the quotient and 5 times the remainder . if the remainder is 46 , the divident is | "solution divisor = ( 5 × 46 ) = 230 . also . 10 × q = 230 ‹ = › q = 23 . and , r = 46 . dividend = ( 230 × 23 + 46 ) = 5336 . answer d" | a = 5 * 46
b = 5 * 46
c = b / 10
d = a * c
|
a ) 4.8 days , b ) 3.8 days , c ) 3.65 days , d ) 3.77 days , e ) 5.75 days | a | inverse(add(inverse(8), inverse(12))) | a and b complete a work in 8 days . a alone can do it in 12 days . if both together can do the work in how many days ? | "1 / 8 + 1 / 12 = 5 / 24 24 / 5 = 4.8 days answer : a" | a = 1/(8)
b = 1/(12)
c = a + b
d = 1/(c)
|
a ) 1,104 , b ) 1,200 , c ) 1,240 , d ) 1,280 , e ) 1,320 | a | floor(divide(multiply(add(6, 10), add(add(14, 10), multiply(9, add(const_4, const_1)))), const_1000)) | gary ’ s gas station serves an average of 14 cars per hour on saturdays , 10 cars per hour on sundays , and 9 cars per hour on all other days of the week . if the station is open from 6 a . m . to 10 p . m . every day , how many cars does gary ’ s station serve over the course of a typical week ? | "6 a . m . to 10 p . m . = 16 hours number of cars serviced on weekdays = ( 16 * 9 * 5 ) number of cars serviced on saturday = ( 16 * 14 ) number of cars serviced on sunday = ( 16 * 10 ) number of cars served in a week = 16 ( 45 + 14 + 10 ) = 16 * 69 = 1104 answer : a" | a = 6 + 10
b = 14 + 10
c = 4 + 1
d = 9 * c
e = b + d
f = a * e
g = f / 1000
h = math.floor(g)
|
a ) 55 , b ) 60 , c ) 73 , d ) 82 , e ) 91 | e | subtract(100, divide(subtract(100, 73), const_3)) | a teacher grades students ’ tests by subtracting twice the number of incorrect responses from the number of correct responses . if student a answers each of the 100 questions on her test and receives a score of 73 , how many questions did student a answer correctly ? | let the number of correct responses be x then the number of incorrect responses = 100 - x according to question x - 2 ( 100 - x ) = 73 ( subtracting twice of incorrect from correct ) 3 x = 273 x = 91 answer : e | a = 100 - 73
b = a / 3
c = 100 - b
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | divide(add(120, 1), 11) | find the value of 3 / 11 × [ ( 1 + 1 / 3 ) ( 1 + 1 / 4 ) … … … … . ( 1 + 1 / 120 ) ] is | 3 / 11 × [ ( 1 + 1 / 3 ) ( 1 + 1 / 4 ) . . . . . . . ( 1 + 1 / 120 ) ] 3 / 11 × [ 4 / 3 + 5 / 4 × . . . . . . . . 121 / 120 ] 3 / 11 × 4 / 3 × 5 / 4 × . . . . 121 / 120 121 / 11 = 11 . a | a = 120 + 1
b = a / 11
|
a ) 5 / 2 , b ) 2 √ 5 , c ) 5 , d ) 4 √ 2 , e ) none of these | a | divide(10, const_4) | 10 meters of wire is available to fence off a flower bed in the form of a circular sector . what must the radius of the circle in meters be , if we wish to have a flower bed with the greatest possible surface area ? | "area of sector , a = x / 360 * pi * r ^ 2 circumference of the sector = 10 = > x / 360 * 2 * pi * r + 2 r = 20 = > 2 a / r + 2 r = 10 or 2 a + 2 r ^ 2 = 10 r = > a = r 5 - r ^ 2 = r 5 - r ^ 2 we will now max using derivations i . e 5 - 2 r = 0 r = 5 / 2 a" | a = 10 / 4
|
a ) 1776 , b ) 1887 , c ) 1998 , d ) 2332 , e ) 2554 | c | multiply(add(add(multiply(add(add(4, 3), 2), const_100), multiply(add(add(const_4.0, 3), 2), const_10)), add(add(4, 3), 2)), 3) | what is the sum of all possible 3 - digit numbers that can be constructed using the digits 2 , 3 , and 4 if each digit can be used only once in each number ? | "there are 6 possible arrangements of the three numbers . then each number will be in the hundreds , tens , and ones place two times each . the sum is 2 ( 222 ) + 2 ( 333 ) + 2 ( 444 ) = 1998 the answer is c ." | a = 4 + 3
b = a + 2
c = b * 100
d = 4 + 0
e = d + 2
f = e * 10
g = c + f
h = 4 + 3
i = h + 2
j = g + i
k = j * 3
|
a ) 30 % , b ) 33 1 / 3 % , c ) 37 1 / 2 % , d ) 40 % , e ) 50 % | c | multiply(const_100, divide(add(multiply(8, divide(30, const_100)), multiply(2, divide(30, const_100))), add(add(multiply(8, divide(30, const_100)), multiply(2, divide(30, const_100))), add(subtract(multiply(8, divide(70, const_100)), 2), multiply(2, divide(70, const_100)))))) | solution y is 30 percent liquid x and 70 percent water . if 2 kilograms of water evaporate from 8 kilograms of solution y and 2 kilograms of solution y are added to the remaining 6 kilograms of liquid , what percent of this new solution is liquid x ? | "initial solution = > 8 kg ( x : w : : 2.4 : 5.6 ) after evaporation initial solution = > 6 kg ( x : w : : 2.4 : 3.6 ) new solution = > 2 kg ( x : w : : 0.6 : 1.4 ) mixing ( initial solution + new solution ) = > 6 kg ( 2.4 : 3.6 ) + 2 kg ( 0.6 : 1.4 ) = = 8 kg [ ( 2.4 + 0.6 ) : ( 3.6 + 1.4 ) ] = 8 kg ( x : water : : 3 kg : 5 kg ) hence % of x = 3 / 8 = 37.5 % answer : c" | a = 30 / 100
b = 8 * a
c = 30 / 100
d = 2 * c
e = b + d
f = 30 / 100
g = 8 * f
h = 30 / 100
i = 2 * h
j = g + i
k = 70 / 100
l = 8 * k
m = l - 2
n = 70 / 100
o = 2 * n
p = m + o
q = j + p
r = e / q
s = 100 * r
|
a ) $ 600 , b ) $ 820 , c ) $ 850 , d ) $ 980 , e ) $ 1,140 | b | multiply(add(add(subtract(subtract(48, 20), 10), multiply(subtract(const_1, divide(40, const_100)), 10)), multiply(subtract(const_1, divide(15, const_100)), 20)), 20) | tickets to a certain concert sell for $ 20 each . the first 10 people to show up at the ticket booth received a 40 % discount , and the next 20 received a 15 % discount . if 48 people bought tickets to the concert , what was the total revenue from ticket sales ? | "price of 1 ticket = 20 $ revenue generated from sales of first 10 tickets = 10 * ( 60 / 100 * 20 ) = 10 * 12 = 120 revenue generated from sales of next 20 tickets = 20 * ( 85 / 100 * 20 ) = 20 * 17 = 340 revenue generated from sales of last 18 tickets = 20 * 18 = 360 revenue generated from sales of 48 tickets = 120 + 340 + 360 = 820 $ answer b" | a = 48 - 20
b = a - 10
c = 40 / 100
d = 1 - c
e = d * 10
f = b + e
g = 15 / 100
h = 1 - g
i = h * 20
j = f + i
k = j * 20
|
a ) 2888 , b ) 2669 , c ) 2500 , d ) 2762 , e ) 8227 | c | divide(multiply(200, const_100), 2) | a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 200 when the rate to profit rises from 5 to 7 percent . find the capital of b ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 200 b capital = 15000 * 1 / 6 = 2500 answer : c" | a = 200 * 100
b = a / 2
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a ) $ 1.60 , b ) $ 32.00 , c ) $ 96.00 , d ) $ 108.00 , e ) $ 196.00 | b | multiply(divide(3.20, 60), surface_cube(10)) | if paint costs $ 3.20 per quart , and a quart covers 60 square feet , how much will it cost to paint the outside of a cube 10 feet on each edge ? | "total surface area = 6 a ^ 2 = 6 * 10 * 10 = 600 each quart covers 20 sqr ft thus total number of quarts = 600 / 60 = 10 cost will be 10 * 3.2 = $ 32 ans : b" | a = 3 / 20
b = a * surface_cube
|
a ) 15 , b ) 30 , c ) 51 , d ) 63 , e ) 76 | c | subtract(divide(subtract(subtract(160, 10), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(160, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | how many even number in the range between 10 to 160 inclusive are not divisible by 3 | "we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 10 , 12,14 . . . 160 using ap formula , we can say 160 = 10 + ( n - 1 ) * 2 or n = 76 . for 6 , 12,18 , . . . 156 using ap formula , we can say 156 = 12 + ( n - 1 ) * 6 or n = 25 . hence , only divisible by 2 but not 3 = 76 - 25 = 51 . hence , answer c" | a = 160 - 10
b = a - 2
c = b / 2
d = 160 - 2
e = 3 * 4
f = d - e
g = f - 3
h = g - 3
i = h / 3
j = i / 2
k = c - j
|
a ) 250 , b ) 240 , c ) 230 , d ) 245 , e ) 235 | b | divide(multiply(100, 24), const_10) | the delhi express runs @ 144 kilometers per hours and it reaches destination 20 minute ahead of schedule . if it averages 100 km / hour , it reaches 24 minute behind schedule . what is the average speed it should maintain in order to reach the destination on schedule ? | if the destination is d kilometer and the speed in order to reach on schedule is s km / hour , the time taken to reach on schedule is d / s hours thus we get two equations : d / 144 = d / s - 1 / 3 and d / 100 = d / s + 2 / 5 and subtracting the first equation from the second we get : 44 d / 14400 = 11 / 15 thus d = 14400 x 11 / 44 x 15 = 240 answer : b | a = 100 * 24
b = a / 10
|
a ) $ 200 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1000 | a | subtract(divide(negate(subtract(multiply(40, 8), multiply(45, 8))), divide(20, const_100)), subtract(multiply(40, 8), multiply(45, 8))) | at a florist shop on a certain day , all corsages sold for either $ 40 or $ 45 . if 8 of the corsages that sold for $ 30 had instead sold for $ 20 , then the store ' s revenue from corsages that day would have been reduced by 20 percent . what was the store ' s actual revenue from corsages that day ? | "i am doing it elaborately , hope it will help you . let , no . of corsages @ $ 40 = x , no . of corsages @ $ 45 = y and revenue = r so , 40 x + 45 y = r . . . . . . . . . ( 1 ) now , given the situation , 40 ( x + 8 ) + 45 ( y - 8 ) = r - . 2 r = > 40 x + 320 + 45 y - 360 = . 8 r = > 40 x + 45 y = . 8 r + 40 . . . . . . . . . . . . ( 2 ) so , r = . 8 r + 40 = > r = 200 the answer is a ." | a = 40 * 8
b = 45 * 8
c = a - b
d = negate / (
e = 20 / 100
f = d - e
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a ) 33,000 , b ) 34,000 , c ) 35,000 , d ) 36,000 , e ) 37,000 | a | multiply(divide(multiply(multiply(multiply(add(const_3, const_4), add(const_3, const_4)), const_100), multiply(add(const_3, const_2), const_2)), subtract(const_100, 10)), add(const_100, 10)) | when a laptop is sold for rs . 27000 , the owner loses 10 % . at what price must that laptop be sold in order to gain 10 % ? | "90 : 27000 = 110 : x x = ( 27000 x 110 ) / 90 = 33000 . hence , s . p . = rs . 33,000 . answer : option a" | a = 3 + 4
b = 3 + 4
c = a * b
d = c * 100
e = 3 + 2
f = e * 2
g = d * f
h = 100 - 10
i = g / h
j = 100 + 10
k = i * j
|
a ) 44 % , b ) 125 % , c ) 144 % , d ) 40 % , e ) none of the above | b | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(50, const_100)), const_2)), const_100) | if the radius of a circle is increased by 50 % then the area is increased by : | "initially a = pi * r 2 now r 2 = 150 / 100 r then area = pi * ( 150 r / 100 ) 2 area = 225 r / 100 that is area increases from 100 to 225 = increase in area = 125 % answer : b" | a = 100 / 100
b = 100 / 100
c = 50 / 100
d = b - c
e = d ** 2
f = a - e
g = f * 100
|
a ) 50 % , b ) 40 % , c ) 14 % , d ) 32 % , e ) 29 % | c | multiply(divide(subtract(add(const_1, divide(60, const_100)), add(const_1, divide(40, const_100))), add(const_1, divide(40, const_100))), const_100) | a certain company reported that the revenue on sales increased 40 % from 2000 to 2003 , and increased 60 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ? | "assume the revenue in 2000 to be 100 . then in 2003 it would be 140 and and in 2005 160 , so from 2003 to 2005 it increased by ( 160 - 140 ) / 140 = 20 / 140 = 1 / 7 = ~ 14 % . answer : c" | a = 60 / 100
b = 1 + a
c = 40 / 100
d = 1 + c
e = b - d
f = 40 / 100
g = 1 + f
h = e / g
i = h * 100
|
a ) 3629 , b ) 3608 , c ) 3600 , d ) 3613 , e ) 6192 | c | divide(144, subtract(power(add(divide(20, const_100), const_1), 2), add(multiply(divide(20, const_100), 2), const_1))) | the difference between the compound interest compounded annually and simple interest for 2 years at 20 % per annum is rs . 144 . find the principal ? | "p = 144 ( 100 / 5 ) 2 = > p = 3600 answer : c" | a = 20 / 100
b = a + 1
c = b ** 2
d = 20 / 100
e = d * 2
f = e + 1
g = c - f
h = 144 / g
|
a ) 1 / 25 , b ) 1 / 6 , c ) 1 / 5 , d ) 5 , e ) 6 | c | divide(divide(6, 5), 5) | if xy > 0 , 1 / x + 1 / y = 6 , and 1 / xy = 5 , then ( x + y ) / 6 = ? | "( 1 / x + 1 / y ) = 6 canbe solved as { ( x + y ) / xy } = 5 . substituting for 1 / xy = 5 , we get x + y = 6 / 5 = = > ( x + y ) / 6 = 6 / ( 5 * 6 ) = 1 / 5 . c" | a = 6 / 5
b = a / 5
|
a ) 32 kmph , b ) 76 kmph , c ) 120 kmph , d ) 43 kmph , e ) 40 kmph | c | divide(divide(add(100, 300), const_1000), divide(12, const_3600)) | a train 100 meters long completely crosses a 300 meters long bridge in 12 seconds . what is the speed of the train is ? | "s = ( 100 + 300 ) / 12 = 400 / 12 * 18 / 5 = 120 answer : c" | a = 100 + 300
b = a / 1000
c = 12 / 3600
d = b / c
|
a ) 7.5 , b ) 8.99 , c ) 8.0 , d ) 9.2 , e ) 8.57 | e | multiply(divide(divide(multiply(divide(30, const_100), 200), 10), multiply(divide(30, const_100), 200)), const_100) | a reduction of 30 % in the price of salt enables a lady to obtain 10 kgs more for rs . 200 , find the original price per kg ? | "answer : 200 * ( 30 / 100 ) = 60 - - - 10 ? - - - 1 = > rs . 6 200 - - - 140 ? - - - 6 = > rs . 8.57 . answer : e" | a = 30 / 100
b = a * 200
c = b / 10
d = 30 / 100
e = d * 200
f = c / e
g = f * 100
|
a ) 40 , b ) 44 , c ) 45 , d ) 46 , e ) 50 | e | multiply(10, add(const_2, const_3)) | if a is a positive integer and 10 ^ a – 74 in decimal notation has digits whose sum is 440 , what is the value of a ? | answer e = 50 we have ( 10 ^ a ) - 74 , where the sum of the digits is 440 . 100 - 74 = 26 , so we already have 2 + 6 = 8 from the 440 . because we have a power of 10 , we will have numbers like 100 , 1000 , 10000 and so on . this minus 74 rests 26 and a lot of 9 s . e . g : 1000 - 74 = 9926 . so dividing the 432 / 9 = 48 , that ` s the number of 9 s . and we have 2 other numbers ( 2 and 6 ) wich were 00 before the subtraction . so we have 48 + 2 as an a = 50 | a = 2 + 3
b = 10 * a
|
a ) − 100 , b ) 0.001 , c ) 0.25 , d ) 4 , e ) 8 | b | divide(divide(2, const_2), const_1000) | if x = 7 and y = − 2 , what is the value of ( x − 2 y ) ^ y ? | quickly we can spot that answer is neither integer nor negative . eliminate a , de by inversing and squaring 0.001 answer : b | a = 2 / 2
b = a / 1000
|
a ) 18 , b ) 21.6 , c ) 20 , d ) 22.5 , e ) 27 | c | multiply(multiply(15, divide(add(const_100, 10), const_100)), divide(add(const_100, 20), const_100)) | a distributor sells a product through an on - line store , which take a commission of 20 % of the price set by the distributor . the distributor obtains the product from a producer at the price of $ 15 per item . what is the price that the buyer observers on - line if the distributor wants to maintain a 10 % profit on the cost of the item ? | producer price = $ 15 ; the distributor wants to maintain a 20 % profit on the cost of the item , thus he must get $ 15 * 1.2 = $ 18 after the store takes a commission of 10 % of the final price - - > ( final price ) * 0.9 = $ 18 - - > ( final price ) = $ 20 . answer : c . | a = 100 + 10
b = a / 100
c = 15 * b
d = 100 + 20
e = d / 100
f = c * e
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a ) 32 , b ) 7 , c ) 11 , d ) 13 , e ) 17 | a | divide(add(divide(subtract(multiply(floor(divide(610, 5)), 5), multiply(floor(divide(300, 5)), 5)), 5), const_1), const_2) | if integer k is equal to the sum of all even multiples of 5 between 300 and 610 , what is the greatest prime factor of k ? | "if we break down what the stem is asking what is the sum of all mult of 10 between 300 and 610 . using arithmetic progression to find n : 610 = 300 + ( n - 1 ) 10 310 + 10 = 10 n 320 = 10 n = > n = 32 the sum would be : 32 * mean mean = [ 610 + 300 ] / 2 = 455 32 * 455 a" | a = 610 / 5
b = math.floor(a)
c = b * 5
d = 300 / 5
e = math.floor(d)
f = e * 5
g = c - f
h = g / 5
i = h + 1
j = i / 2
|
a ) 1760 , b ) 1700 , c ) 2350 , d ) 2500 , e ) 8000 | a | divide(multiply(multiply(13, 2200), 16), add(multiply(16, 16), 4)) | one ton has 2200 pounds , and one pound has 16 ounces . how many packets containing wheat weighing 16 pounds and 4 ounces each would totally fill a gunny bag of capacity 13 tons ? | 16 pounds and 4 ounces = 16 * 16 + 4 = 260 ounces . 13 tons = 13 * 2200 pound = 13 * 2200 * 16 ounces . hence the answer is ( 13 * 2200 * 16 ) / 260 = 1760 . answer : a . | a = 13 * 2200
b = a * 16
c = 16 * 16
d = c + 4
e = b / d
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a ) 1 , b ) 8 , c ) 12 , d ) 16 , e ) none of these | a | divide(subtract(62, 25), const_2) | find the greatest number which leaves the same remainder when it divides 25 , 62 and 105 . | "105 - 62 = 43 62 - 25 = 37 105 - 25 = 80 the h . c . f of 37 , 43 and 80 is 1 . answer : a" | a = 62 - 25
b = a / 2
|
a ) 2 , b ) 4 , c ) 8 , d ) 10 , e ) 12 | b | divide(16, const_2) | if 16 ^ y = 2 ^ 16 , what is y ? | "16 ^ y = 2 ^ 4 y = 2 ^ 16 4 y = 16 y = 4 the answer is b ." | a = 16 / 2
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a ) 1305 , b ) 1545 , c ) 1375 , d ) 1345 , e ) 1455 | d | add(add(multiply(multiply(25, divide(4, 5)), subtract(subtract(subtract(60, multiply(60, divide(25, const_100))), multiply(60, divide(25, const_100))), 10)), multiply(25, 10)), add(multiply(25, multiply(60, divide(25, const_100))), multiply(25, multiply(60, divide(25, const_100))))) | an automobile parts supplier charges $ 25 per package of gaskets . when a customer orders more than 10 packages of gaskets , the supplier charges 4 / 5 the price for each package in excess of 10 . during a certain week , the supplier sold 60 packages of gaskets . if 25 percent of the gaskets went to company x , 15 percent to company y , and the rest to company z , what was the total amount , in dollars , that the parts supplier received in payment for the gaskets ? | "$ 25 per packet of gasket in case a customer orders less than 10 in case a customer orders > 10 price per gasket = 25 * 4 / 5 = 20 a certain week the supplier sold 60 gasket 1 . he sold 25 % of the gaskets to x = 15 gaskets = 25 * 10 + 20 * 5 = 250 + 100 = 350 2 . he sold 15 % of the gaskets to y = 9 gaskets = 25 * 9 = 225 3 . he sold remaining 60 % to z = 36 gaskets = 25 * 10 = 250 + 20 * 26 = 770 thus , total money earned 350 + 225 + 770 = 1345 answer is d" | a = 4 / 5
b = 25 * a
c = 25 / 100
d = 60 * c
e = 60 - d
f = 25 / 100
g = 60 * f
h = e - g
i = h - 10
j = b * i
k = 25 * 10
l = j + k
m = 25 / 100
n = 60 * m
o = 25 * n
p = 25 / 100
q = 60 * p
r = 25 * q
s = o + r
t = l + s
|
a ) 18 % , b ) 20 % , c ) 21 % , d ) 23 % , e ) can not be determined | a | add(divide(multiply(20, 4), 2), add(20, 5)) | two kinds of vodka are mixed in the ratio 1 : 2 and 2 : 1 and they are sold fetching the profit 20 % and 20 % respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4 / 3 and 5 / 3 times respectively , then the mixture will fetch the profit of | answer : a . | a = 20 * 4
b = a / 2
c = 20 + 5
d = b + c
|
a ) 3277 , b ) 2977 , c ) 9492 , d ) 6077 , e ) 17112 | c | multiply(multiply(4, multiply(const_100, const_100)), power(divide(3, 4), 5)) | the value of a scooter depreciates in such a way that its value of the end of each year is 3 / 4 of its value of the beginning of the same year . if the initial value of the scooter is rs . 40,000 , what is the value at the end of 5 years ? | "explanation : 40,000 * ( 3 / 4 ) ^ 5 answer : c" | a = 100 * 100
b = 4 * a
c = 3 / 4
d = c ** 5
e = b * d
|
a ) 211 , b ) 219 , c ) 227 , d ) 235 , e ) 243 | c | subtract(multiply(multiply(45, const_0_2778), 30), 148) | what is the length of a bridge ( in meters ) , which a train 148 meters long and travelling at 45 km / h can cross in 30 seconds ? | "speed = 45 km / h = 45000 m / 3600 s = 25 / 2 m / s in 30 seconds , the train can travel 25 / 2 * 30 = 375 meters 375 = length of train + length of bridge length of bridge = 375 - 148 = 227 meters the answer is c ." | a = 45 * const_0_2778
b = a * 30
c = b - 148
|
a ) 7 / 10 , b ) 2 / 5 , c ) 3 / 5 , d ) 4 / 5 , e ) 5 / 8 | a | divide(multiply(7, 4), add(multiply(7, 4), multiply(3, 4))) | at a loading dock , each worker on the night crew loaded 3 / 4 as many boxes as each worker on the day crew . if the night crew has 4 / 7 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "method : x = no . of boxes loaded by day crew . boxes by night crew = 3 / 4 * 4 / 7 x = 3 / 7 x % loaded by day crew = x / ( x + 3 / 7 x ) = 7 / 10 answer a" | a = 7 * 4
b = 7 * 4
c = 3 * 4
d = b + c
e = a / d
|
a ) 80 % , b ) 105 % , c ) 88 % , d ) 124.2 % , e ) 138 % | c | multiply(divide(multiply(11, subtract(const_1, divide(20, const_100))), 10), const_100) | in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 11 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,11 = 11 / 100 x = 80 * 11 / 100 x = 8.8 8.8 / 10 = 0.88 = 88 % , answer c" | a = 20 / 100
b = 1 - a
c = 11 * b
d = c / 10
e = d * 100
|
a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 63 | a | multiply(multiply(30, const_2), divide(3, divide(12, const_3))) | in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the ` ` sport ' ' formulation , the ratio of flavoring to corn syrup is 3 times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the ` ` sport ' ' formulation contains 3 ounces of corn syrup , how many ounces of water does it contain ? | f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 3 * 180 / 12 = 45 ounces of water answer : a | a = 30 * 2
b = 12 / 3
c = 3 / b
d = a * c
|
a ) 5 : 3 , b ) 16 : 3 , c ) 11 : 3 , d ) 11 : 5 , e ) 9 : 4 | c | divide(add(24, divide(add(24, subtract(4, const_1)), subtract(4, const_1))), divide(add(24, subtract(4, const_1)), subtract(4, const_1))) | one year before the age of pramila was 4 times than her daughter swati . after 6 years the age of pramila will be 24 year greater than the age of swati . then calculate the ratio of ages of pramila and swati ? | let 1 year before age of swati = x years and age of pramila = 4 x years swati present age = ( x + 1 ) and pramila present age = ( 4 x + 1 ) after 6 year age of swati = ( x + 7 ) , age of pramila = ( 4 x + 7 ) so , 4 x + 7 - ( x + 7 ) = 243 x = 24 , x = 8 . ratio = ( 4 x + 7 ) : ( x + 1 ) = 33 : 9 = 11 : 3 answer c | a = 4 - 1
b = 24 + a
c = 4 - 1
d = b / c
e = 24 + d
f = 4 - 1
g = 24 + f
h = 4 - 1
i = g / h
j = e / i
|
a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | b | divide(subtract(96, add(add(add(add(2, 3), 4), add(2, 3)), 2)), 4) | somehow or other i got talked into buying something on the installment plan . i ' m notsure i got a good deal . the payments to date , according to my checkbook , have reached rs . 96 . the second year cost rs . 2.00 more than the first year ; the third year cost rs . 3.00 more thanthe second ; and the fourth year cost me rs . 4.00 more than the third . what were mypayments the first year ? | let payments for the first year = x given , x + x + 2 + x + 5 + x + 9 = 96 4 x + 16 = 96 x = 20 answer : b | a = 2 + 3
b = a + 4
c = 2 + 3
d = b + c
e = d + 2
f = 96 - e
g = f / 4
|
a ) 1 , b ) 21 , c ) 26 , d ) 52 , e ) 1014 | b | multiply(add(subtract(23, const_10), const_2), add(subtract(23, const_10), const_2)) | if x is a sum of all even integers on the interval 23 . . . 65 and y is their number , what is the gcd ( x , y ) ? | "x = 24 + 16 + . . . + 64 = ( largest + smallest ) / 2 * ( # of terms ) = ( 24 + 64 ) / 2 * 21 = 44 * 16 . gcd of 21 and 44 * 21 is 21 . answer : b ." | a = 23 - 10
b = a + 2
c = 23 - 10
d = c + 2
e = b * d
|
a ) 0 , b ) 1 / 12 , c ) 5 / 12 , d ) 7 / 18 , e ) 4 / 9 | c | multiply(add(const_12, const_3), power(divide(1, 6), const_2)) | a cube with its sides numbered 1 through 6 is rolled twice , first landing on a and then landing on b . if any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6 , what is the probability w that a + b is prime ? | "total # of outcomes is 6 * 6 = 36 ; favorable outcomes : a - b - - > prime 1 - 1 - - > 2 ; 1 - 2 - - > 3 ; 2 - 1 - - > 3 ; 1 - 4 - - > 5 ; 4 - 1 - - > 5 ; 2 - 3 - - > 5 ; 3 - 2 - - > 5 ; 1 - 6 - - > 7 ; 6 - 1 - - > 7 ; 2 - 5 - - > 7 ; 5 - 2 - - > 7 ; 3 - 4 - - > 7 ; 4 - 3 - - > 7 ; 6 - 5 - - > 11 ; 5 - 6 - - > 11 . total of 15 favorable outcomes w = 15 / 36 . answer : c ." | a = 12 + 3
b = 1 / 6
c = b ** 2
d = a * c
|
a ) 277 m , b ) 704 m , c ) 278 m , d ) 288 m , e ) 1056 m | e | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 750), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 750 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 750 resolutions . = 750 * 2 * 22 / 7 * 22.4 = 105600 cm = 1056 m answer : e" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 750
i = h / 100
|
a ) 370 , b ) 365 , c ) 360 , d ) 380 , e ) 350 | c | divide(1, divide(add(multiply(const_3600, divide(1, 450)), 2), const_3600)) | a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 450 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "c 450 * t = 1 km = > t = 1 / 450 km / h v * ( t + 2 / 3600 ) = 1 v ( 1 / 450 + 2 / 3600 ) = 1 = > v = 360 km / h" | a = 1 / 450
b = 3600 * a
c = b + 2
d = c / 3600
e = 1 / d
|
a ) 1642 , b ) 1640 , c ) 1720 , d ) 2795 , e ) 6386 | c | subtract(2795, divide(multiply(multiply(3, 5), 2795), add(multiply(3, 5), multiply(8, 3)))) | a sum of rs . 2795 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2795 - x ) * 3 * 5 ) / 100 24 x / 100 = 41925 / 100 - 15 x / 100 39 x = 41925 = > x = 1075 second sum = 2795 – 1075 = 1720 answer : c" | a = 3 * 5
b = a * 2795
c = 3 * 5
d = 8 * 3
e = c + d
f = b / e
g = 2795 - f
|
a ) 7787 , b ) 8000 , c ) 1988 , d ) 1277 , e ) 20000 | e | divide(divide(18750, subtract(const_1, divide(25, const_100))), add(const_1, divide(25, const_100))) | in one year , the population , of a village increased by 25 % and in the next year , it decreased by 25 % . if at the end of 2 nd year , the population was 18750 , what was it in the beginning ? | "x * 125 / 100 * 75 / 100 = 18750 x * 0.9375 = 18750 x = 18750 / 0.9375 = > 20000 answer : e" | a = 25 / 100
b = 1 - a
c = 18750 / b
d = 25 / 100
e = 1 + d
f = c / e
|
a ) 1000 , b ) 1250 , c ) 1300 , d ) 1529.41 , e ) 1625 | d | divide(add(divide(multiply(300, const_100), 30), 300), divide(85, const_100)) | a small pool filled only with water will require an additional 300 gallons of water in order to be filled to 85 % 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 ? | "300 gallons of water increases capacity by 30 % that means 30 % is 300 gallons , so 100 % would be = 300 * 100 / 30 = 1000 gallons now 1000 + 300 gallons is 85 % capacity of tank . so 100 % capacity would be = 1300 * 100 / 85 = 1529.41 d is the answer" | a = 300 * 100
b = a / 30
c = b + 300
d = 85 / 100
e = c / d
|
a ) 80 , b ) 100 , c ) 70.5 , d ) 90 , e ) none of these | c | divide(36, multiply(divide(85, const_100), divide(3, 5))) | if 85 % of 3 / 5 of a number is 36 , then the number is ? | "let the number be x . then 85 % of 3 / 5 of x = 36 85 / 100 * 3 / 5 * x = 36 x = ( 36 * 100 / 51 ) = 70.5 required number = 70.5 . correct option : c" | a = 85 / 100
b = 3 / 5
c = a * b
d = 36 / c
|
a ) 18 , b ) 15 , c ) 16 , d ) 20 , e ) 25 | a | multiply(divide(multiply(add(multiply(3, 2), multiply(2, 2)), divide(10, subtract(multiply(3, 2), 2))), add(3, 2)), 3) | a jar contains a mixture of ab in the ratio 3 : 2 . 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 6 litres of a and 4 litres of b ( as a : b = 3 : 2 ) let the initial volume of the mixture be 3 k + 2 k = 5 k so by condition , [ 3 k - 6 ] / [ 2 k - 4 + 10 ] = 2 / 3 9 k - 18 = 4 k - 8 + 20 5 k = 12 + 18 k = 6 solve for k which is k = 6 so initial volume of liquid a = 3 k = 18 litres answer : a" | a = 3 * 2
b = 2 * 2
c = a + b
d = 3 * 2
e = d - 2
f = 10 / e
g = c * f
h = 3 + 2
i = g / h
j = i * 3
|
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