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 ) a ) 399 , b ) b ) 418 , c ) c ) 519 , d ) d ) 521 , e ) e ) 525 | a | add(subtract(414, 16), const_1) | if the average ( arithmetic mean ) of 16 consecutive odd integers is 414 , then the least of these integers is | "a very helpful rule to know in arithmetic is the rule that in evenly spaced sets , average = median . because the average will equal the median in these sets , then we quickly know that the median of this set of consecutive odd integer numbers is 414 . there are 16 numbers in the set , and in a set with an even number of terms the median is just the average of the two most median terms ( here the 7 th and 8 th numbers in the set ) . this means that numbers 7 and 8 in this set are 413 and 415 . because we know that number 7 is 413 , we know that the smallest number is 7 odd numbers below this , which means that it is 7 * 2 = 14 below this ( every odd number is every other number ) . therefore 413 - 14 = 399 , answer choice a ." | a = 414 - 16
b = a + 1
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a ) 3.5 gallons , b ) 2.7 gallons , c ) 6.7 gallons , d ) 4.5 gallons , e ) 7.5 gallons | c | divide(200, 30) | a car gets 30 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 200 kilometers ? | "each 30 kilometers , 1 gallon is needed . we need to know how many 30 kilometers are there in 200 kilometers ? 200 ã · 30 = 6.7 ã — 1 gallon = 6.7 gallons correct answer is c ) 6.7 gallons" | a = 200 / 30
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a ) 1 / 10 , b ) 3 / 20 , c ) 1 / 5 , d ) 1 / 4 , e ) 9 / 20 | e | subtract(add(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))), multiply(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12)))) | maths , physics and chemistry books are stored on a library shelf that can accommodate 25 books . currently , 20 % of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books . among all the books , 12 books are soft cover and the remaining are hard - cover . if there are a total of 7 hard - cover books among the maths and physics books . what is the probability w , that a book selected at random is either a hard cover book or a chemistry book ? | "first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so , you have the equations : m + p + c = 20 ( since 4 / 5 of the 25 spots are full of books ) m = 2 p p = 4 + c from that , you can use substitution to get everything down to one variable . c = p - 4 m = 2 p p = p then ( p - 4 ) + 2 p + p = 20 , so 4 p = 24 and p = 6 . that means that there are 12 math , 6 physics , and 2 chemistry books on the shelf . with those numbers , you also know that there are 8 total hardcovers , 1 of which is chemistry . so if your goal is to get either a hardcover or a chemistry , there are 9 ways towin - either one of the 7 hardcovers that are n ' t chemistry or the two chemistry books . so out of the 20 total , w = 9 provide the desired outcome , making the answer e ." | a = 2 + 4
b = 2 + a
c = b + 12
d = 2 / c
e = 2 + 4
f = 2 + e
g = 2 + 4
h = 2 + g
i = h + 12
j = f / i
k = d + j
l = 2 + 4
m = 2 + l
n = m + 12
o = 2 / n
p = 2 + 4
q = 2 + p
r = 2 + 4
s = 2 + r
t = s + 12
u = q / t
v = o * u
w = k - v
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a ) 6 , b ) 2 , c ) 3 , d ) 5 , e ) 4 | a | multiply(const_2, sqrt(divide(219, const_2))) | the number 219 can be written as sum of the squares of 3 different positive integers . what is the difference of these 2 different larger integers ? | "sum of the squares of 3 different positive integers = 219 13 ^ 2 + 7 ^ 2 + 1 ^ 2 = 219 now , difference of these 2 different larger integers = 13 - 7 = 6 ans - a" | a = 219 / 2
b = math.sqrt(a)
c = 2 * b
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a ) 99 % , b ) 100.0 % , c ) 102.8 % , d ) 104.5 % , e ) 105.0 % | a | divide(multiply(add(const_100, multiply(const_100, 10)), add(subtract(const_100, 10), const_4)), multiply(const_100, 10)) | this year , mbb consulting fired 10 % of its employees and left remaining employee salaries unchanged . sally , a first - year post - mba consultant , noticed that that the average ( arithmetic mean ) of employee salaries at mbb was 10 % more after the employee headcount reduction than before . the total salary pool allocated to employees after headcount reduction is what percent of that before the headcount reduction ? | "100 employees getting 1000 $ avg , so total salary for 100 ppl = 100000 10 % reduction in employees lead to 90 employees and a salary increase of 10 % of previous avg salary thus the new avg salary is = 10 % ( 1000 ) + 1000 = 1100 so total salary of 90 employees is 90 * 1100 = 99000 now the new salary is more than previous salary by x % . x = ( 99000 / 100000 ) * 100 = 99 % so the answer is a" | a = 100 * 10
b = 100 + a
c = 100 - 10
d = c + 4
e = b * d
f = 100 * 10
g = e / f
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a ) 100 , b ) 200 , c ) 300 , d ) 600 , e ) 900 | c | multiply(subtract(36, 150), 150) | what is the greatest positive integer n such that 3 ^ n is a factor of 36 ^ 150 ? | "36 = 3 ^ 2 * 2 ^ 2 . 36 ^ 150 = 3 ^ 300 * 2 ^ 300 the answer is c ." | a = 36 - 150
b = a * 150
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a ) 39 / 6 , b ) 62 / 11 , c ) 11 / 61 , d ) 21 / 61 , e ) 20 / 61 | a | multiply(divide(divide(multiply(divide(add(subtract(const_100, 30), multiply(subtract(const_100, 30), divide(subtract(const_100, 15), const_100))), const_2), subtract(const_100, 40)), const_100), multiply(subtract(const_100, 30), divide(subtract(const_100, 15), const_100))), const_10) | real - estate salesman z is selling a house at a 30 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 15 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 40 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ? | let the retail price be = x selling price of z = 0.7 x selling price of x = 0.85 * 0.7 x = 0.60 x selling price of y = ( ( 0.7 x + 0.6 x ) / 2 ) * 0.60 = 0.65 x * 0.60 = 0.39 x 0.39 x = k * 0.60 x k = 0.39 / 0.6 = 39 / 6 answer : a | a = 100 - 30
b = 100 - 30
c = 100 - 15
d = c / 100
e = b * d
f = a + e
g = f / 2
h = 100 - 40
i = g * h
j = i / 100
k = 100 - 30
l = 100 - 15
m = l / 100
n = k * m
o = j / n
p = o * 10
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a ) 5 , b ) 8 , c ) 14 , d ) 20 , e ) 28 | a | subtract(40, add(11, 24)) | in a certain alphabet , 11 letters contain a dot and a straight line . 24 letters contain a straight line but do not contain a dot . if that alphabet has 40 letters , all of which contain either a dot or a straight line or both , how many letters contain a dot but do not contain a straight line ? | "we are told that all of the letters contain either a dot or a straight line or both , which implies that there are no letters without a dot and a line ( no line / no dot box = 0 ) . first we find the total # of letters with lines : 11 + 24 = 35 ; next , we find the total # of letters without line : 40 - 35 = 5 ; finally , we find the # of letters that contain a dot but do not contain a straight line : 5 - 0 = 5 . a" | a = 11 + 24
b = 40 - a
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a ) 1 : 5 , b ) 7 : 3 , c ) 5 : 3 , d ) 3 : 1 , e ) 4 : 3 | d | divide(subtract(divide(5, add(1, 7)), divide(1, 7)), subtract(divide(1, 7), divide(1, add(1, 7)))) | an alloy of copper and zinc contains copper and zinc in the ratio 5 : 3 . another alloy of copper and zinc contains copper and zinc in the ratio 1 : 7 . in what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc ? | "alloy - 1 : copper : zinc = 5 : 3 , let quantity = x i . e . copper in alloy - 1 = [ 5 / ( 5 + 3 ) ] * x = ( 5 / 8 ) x and zinc in alloy - 1 = [ 3 / ( 5 + 3 ) ] * x = ( 3 / 8 ) x alloy - 2 : copper : zinc = 1 : 7 , let quantity = y i . e . copper in alloy - 2 = [ 1 / ( 1 + 7 ) ] * y = ( 1 / 8 ) y and zinc in alloy - 1 = [ 7 / ( 1 + 7 ) ] * y = ( 7 / 8 ) y after combining the two alloys copper = ( 5 / 8 ) x + ( 1 / 8 ) y zinc = ( 3 / 8 ) x + ( 7 / 8 ) y now , ( 5 x + y ) / 8 = ( 3 x + 7 y ) / 8 i . e . 2 x = 6 y i . e . x = 3 y i . e . x / y = 3 / 1 answer : option d" | a = 1 + 7
b = 5 / a
c = 1 / 7
d = b - c
e = 1 / 7
f = 1 + 7
g = 1 / f
h = e - g
i = d / h
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a ) 150 , b ) 122 , c ) 30 , d ) 120 , e ) 102 | e | add(add(40, add(30, 30)), 2) | the average weight of 30 boys sitting in a bus had some value . a new person added to them whose weight was 40 kg only . due to his arrival , the average weight of all the boys decreased by 2 kg . find the average weight of first 30 boys ? | 30 x + 40 = 31 ( x â € “ 2 ) x = 102 e | a = 30 + 30
b = 40 + a
c = b + 2
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a ) $ 178.50 , b ) $ 182.50 , c ) $ 185.50 , d ) $ 187.50 , e ) $ 200 | d | subtract(add(200, divide(multiply(200, 25), const_100)), divide(multiply(add(200, divide(multiply(200, 25), const_100)), 25), const_100)) | the original price of a suit is $ 200 . the price increased 25 % , and after this increase , the store published a 25 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 25 % off the increased price , how much did these consumers pay for the suit ? | "0.75 * ( 1.25 * 200 ) = $ 187.50 the answer is d ." | a = 200 * 25
b = a / 100
c = 200 + b
d = 200 * 25
e = d / 100
f = 200 + e
g = f * 25
h = g / 100
i = c - h
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a ) 270 , b ) 198 , c ) 676 , d ) 156 , e ) 122 | b | multiply(divide(638, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 638 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 638 = 198 answer : b" | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 638 / e
g = 16 * 9
h = f * g
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a ) 6 , b ) 4 , c ) 5 , d ) 1 , e ) 2 | a | multiply(add(divide(3, 6), 1), 4) | if x / 4 - x - 3 / 6 = 1 , then find the value of x . | ( x / 4 ) - ( ( x - 3 ) / 6 ) = 1 = > ( 3 x - 2 ( x - 3 ) ) / 12 = 1 = > 3 x - 2 x + 6 = 12 = > x = 6 . answer is a | a = 3 / 6
b = a + 1
c = b * 4
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a ) $ 480 , b ) $ 570 , c ) $ 660 , d ) $ 720 , e ) $ 850 | c | multiply(divide(42, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100))) | bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 42 more . what was his original selling price ? | "let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 42 0.07 = 42 p = 600 the original selling price was 600 * 1.1 = 660 . the correct answer is c ." | a = 10 / 100
b = 1 - a
c = 30 / 100
d = 1 + c
e = b * d
f = 10 / 100
g = 1 + f
h = e - g
i = 42 / h
j = 10 / 100
k = 1 + j
l = i * k
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a ) 3200 , b ) 4000 , c ) 3250 , d ) 4160 , e ) 3985 | d | multiply(multiply(48000, subtract(multiply(const_3, 4), 4)), divide(14300, add(add(multiply(36000, multiply(const_3, 4)), multiply(42000, multiply(const_3, 4))), multiply(48000, subtract(multiply(const_3, 4), 4))))) | x and y started a business by investing rs . 36000 and rs . 42000 respectively after 4 months z joined in the business with an investment of rs . 48000 , then find share of z in the profit of rs . 14300 ? | "ratio of investment , as investments is for different time . investment x number of units of time . ratio of investments x : y : z = 36000 : 42000 : 48000 = > 6 : 7 : 8 . x = 6 x 12 months = 72 , y = 7 x 12 = 84 , z = 8 x 8 = 64 = > 18 : 21 : 16 . ratio of investments = > x : y : z = 18 : 21 : 16 . investment ratio = profit sharing ratio . z = 14300 ã — 16 / 55 = rs . 4160 . share of z in the profit is rs . 4160 . option d" | a = 3 * 4
b = a - 4
c = 48000 * b
d = 3 * 4
e = 36000 * d
f = 3 * 4
g = 42000 * f
h = e + g
i = 3 * 4
j = i - 4
k = 48000 * j
l = h + k
m = 14300 / l
n = c * m
|
a ) 30 and 75 , b ) 35 and 70 , c ) 40 and 65 , d ) 45 and 60 , e ) 55 and 70 | d | divide(180, 15) | hcf of two numbers is 15 and their lcm is 180 . if their sum is 105 , then the numbers are : | "explanation : let the numbers be 15 a and 15 b . then , 15 a + 15 b = 105 or a + b = 7 . . ( i ) lcm = 15 ab = 180 ab = 12 . . ( ii ) solving equations ( i ) and ( ii ) , we get a = 4 , b = 3 so , the numbers are 15 × 4 and 15 × 3 , i . e . , 60 and 45 answer : d" | a = 180 / 15
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a ) 6.25 , b ) 5.5 , c ) 25 , d ) 15 , e ) 6 | c | divide(subtract(282, multiply(10, 3.2)), 10) | in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 10 overs to reach the target of 282 runs ? | "explanation : runs scored in the first 10 overs = 10 × 3.2 = 32 total runs = 282 remaining runs to be scored = 282 - 32 = 250 remaining overs = 10 run rate needed = 250 / 10 = 25 answer : option c" | a = 10 * 3
b = 282 - a
c = b / 10
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a ) 16 , b ) 127 , c ) 12 , d ) 18 , e ) 100 | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 50)), divide(multiply(const_100, 25), 50))) | if the cost price of 50 articles is equal to the selling price of 25 articles , then the gain or loss percent is ? | "percentage of profit = 25 / 25 * 100 = 100 % answer : e" | a = 100 * 25
b = a / 50
c = 100 - b
d = 100 * 25
e = d / 50
f = c / e
g = 100 * f
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a ) 96 days , b ) 48 days , c ) 32 days , d ) 24 days , e ) 28 days | b | inverse(subtract(divide(const_1, 8), add(divide(const_1, 16), divide(const_1, 24)))) | jake can dig a well in 16 days . paul can dig the same well in 24 days . jake , paul and hari together dig the well in 8 days . hari alone can dig the well in | jake 1 day work = 1 / 16 paul 1 day work = 1 / 24 j + p + h 1 ady work = 1 / 8 1 / 16 + 1 / 24 + 1 / x = 1 / 8 1 / x = 1 / 48 x = 48 so , hari alone can dig the well in 48 days answer : b | a = 1 / 8
b = 1 / 16
c = 1 / 24
d = b + c
e = a - d
f = 1/(e)
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a ) 24 , b ) 32 , c ) 36 , d ) 42 , e ) 45 | a | multiply(divide(20, divide(add(add(4, 5), 6), 3)), 6) | 3 numbers are in the ratio 4 : 5 : 6 and their average is 20 . the largest number is : | explanation : let the numbers be 4 x , 5 x and 6 x . therefore , ( 4 x + 5 x + 6 x ) / 3 = 20 15 x = 60 x = 4 largest number = 6 x = 24 . answer a | a = 4 + 5
b = a + 6
c = b / 3
d = 20 / c
e = d * 6
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a ) 14 , b ) 19 , c ) 38 , d ) 33 , e ) 35 | a | subtract(divide(38, const_2), 5) | a father told his son ` ` i was as old as you are at present , at the time of your birth ' ' . if the father is 38 years old now , then what was the son ' s age 5 years ago in years ? | et son ' s present age is = x then 38 - x = x x = 19 son ' s age 5 years back is 19 - 5 = 14 years . answer : a | a = 38 / 2
b = a - 5
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a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 240 | e | divide(subtract(multiply(120, divide(60, const_100)), multiply(120, divide(55, const_100))), subtract(divide(55, const_100), divide(50, const_100))) | a survey of n people in the town of eros found that 50 % of them preferred brand a . another survey of 120 people in the town of angie found that 60 % preferred brand a . in total , 55 % of all the people surveyed together preferred brand a . what is the total number of people surveyed ? | "it is simply a weighted average question . since the given average of 50 % and 60 % is 55 % ( right in the middle ) , it means the number of people surveyed in eros ( n ) is same as the number of people surveyed in angie . so n = 120 total = 120 + 120 = 240 answer ( e )" | a = 60 / 100
b = 120 * a
c = 55 / 100
d = 120 * c
e = b - d
f = 55 / 100
g = 50 / 100
h = f - g
i = e / h
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a ) rs . 4480 , b ) rs . 5275 , c ) rs . 6275 , d ) rs . 6725 , e ) none of these | a | divide(5600, add(const_1, divide(25, const_100))) | the owner of a furniture shop charges his customer 25 % more than the cost price . if a customer paid rs . 5600 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 5600 ( 100 / 125 ) = rs . 4480 . answer : a" | a = 25 / 100
b = 1 + a
c = 5600 / b
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a ) 6 : 5 , b ) 4 : 7 , c ) 3 : 5 , d ) 5 : 7 , e ) 8 : 9 | a | divide(multiply(4, 3), multiply(5, 2)) | the marks obtained by vijay and amith are in the ratio 4 : 5 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio o | "4 : 5 3 : 2 - - - - - - - 12 : 15 : 10 12 : 10 = = > 6.5 answer a" | a = 4 * 3
b = 5 * 2
c = a / b
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a ) 78.4 cm , b ) 11.5 cm , c ) 91.8 cm , d ) 92.5 cm , e ) 99.5 cm | a | multiply(multiply(const_2, divide(multiply(subtract(18, const_3), const_2), add(const_4, const_3))), 18) | the sector of a circle has radius of 18 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 18 ) + 2 ( 18 ) = 42.4 + 36 = 78.4 cm answer : a" | a = 18 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 18
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a ) 288 , b ) 320 , c ) 2,880 , d ) 3,200 , e ) 28,800 | c | multiply(96, multiply(3, const_10)) | last year , for every 100 million vehicles that traveled on a certain highway , 96 vehicles were involved in accidents . if 3 billion vehicles traveled on the highway last year , how many of those vehicles were involved in accidents ? ( 1 billion = 1,000 , 000,000 ) | to solve we will set up a proportion . we know that “ 100 million vehicles is to 96 accidents as 3 billion vehicles is to x accidents ” . to express everything in terms of “ millions ” , we can use 3,000 million rather than 3 billion . creating a proportion we have : 100 / 96 = 3,000 / x cross multiplying gives us : 100 x = 3,000 * 96 x = 30 * 96 = 2,880 correct answer is c . | a = 3 * 10
b = 96 * a
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a ) 0.2 , b ) 1.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | d | subtract(6.2, subtract(floor(6.2), const_1)) | for any number y , y * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.2 – 6.2 * ? | "since y * is defined as the greatest positive even integer less than or equal to y , then 6.2 * = 4 ( the greatest positive even integer less than or equal to 6.2 is 4 ) . hence , 6.2 – 6.2 * = 6.2 - 4 = 2.2 answer : d ." | a = math.floor(6, 2)
b = a - 1
c = 6 - 2
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['a ) 0.25 % less', 'b ) 1 % less', 'c ) equal to each other', 'd ) 1 % more', 'e ) 9 % more'] | a | divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 20), subtract(const_100, 20)))) | 108 . triangle a ’ s base is 20 % greater than the base of triangle b , and a ’ s height is 20 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ? | wish the question specified that we are talking about corresponding height . base of a = 21 / 20 * base of b height of a = 19 / 20 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 21 / 20 * 19 / 20 * area of b = 399 / 400 * area of b area of a is 0.25 % less than the area of b . answer ( a ) | a = 100 * 100
b = 100 + 20
c = 100 - 20
d = b * c
e = a - d
f = 100 / e
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['a ) 4 x / 5', 'b ) 75 x', 'c ) 48 x', 'd ) 24 x', 'e ) x / 75'] | a | multiply(multiply(multiply(const_2, const_3), const_10), divide(28, 35)) | a circular rim 28 inches in diameter rotates the same number of inches per second as a circular rim 35 inches in diameter . if the smaller rim makes x revolutions per second , how many revolutions per second does the larger rim makes in terms of x ? | let ' s try the explanation . we have two wheels . one with 28 pi and the other one with 35 pi . they have the same speed . in the smaller wheel it ' s 28 pi * x , which must be equal to the speed of the bigger one ( 35 pi * a number of revolutions ) . they are asking that number of revolutions ( but in minutes , which makes the question even harder ) . anyway , we have 28 pi * x = 35 pi * a . ( 28 pi * x ) / ( 35 pi ) . = 4 x / 5 . ans : a | a = 2 * 3
b = a * 10
c = 28 / 35
d = b * c
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a ) 39 , b ) 33 , c ) 26 , d ) 21 , e ) 10 | b | subtract(divide(11, divide(2, 3)), 11) | a certain lab experiments with white and brown mice only . in one experiment , 2 / 3 of the mice are white . if there are 11 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of white mice = 2 / 3 m number of brown mice = 1 / 3 m = 11 = > m = 33 answer b" | a = 2 / 3
b = 11 / a
c = b - 11
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a ) 18 , b ) 19 , c ) 20 , d ) 22 , e ) 24 | c | multiply(divide(const_3.0, divide(4, 5)), 4) | walking with 5 / 4 of my usual speed , i miss the bus by 5 minutes . what is my usual time ? | "speed ratio = 1 : 5 / 4 = 4 : 5 time ratio = 5 : 4 1 - - - - - - - - 5 4 - - - - - - - - - ? è 20 answer : c" | a = 4 / 5
b = 3 / 0
c = b * 4
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a ) 3 / 80 , b ) 3 / 5 , c ) 4 , d ) 16 , e ) 80 / 3 | d | divide(log(16), log(power(2, 0.25))) | if n = 2 ^ 0.25 and n ^ b = 16 , b must equal | "25 / 100 = 1 / 4 n = 2 ^ 1 / 4 n ^ b = 2 ^ 4 ( 2 ^ 1 / 4 ) ^ b = 2 ^ 4 b = 16 answer : d" | a = math.log(16)
b = 2 ** 0
c = math.log(b)
d = a / c
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a ) 720 , b ) 864 , c ) 900 , d ) 936 , e ) 1296 | d | multiply(multiply(divide(divide(factorial(4), factorial(const_2)), factorial(const_2)), divide(divide(factorial(4), factorial(const_2)), factorial(const_2))), multiply(subtract(multiply(const_2, const_3), const_1), subtract(multiply(const_2, const_3), const_1))) | in how many q ways can a 4 - letter password be chosen , using the letters a , b , c , d , e , and / or f , such that at least one letter is repeated within the password ? | total number of four letter passwords = 6 * 6 * 6 * 6 = 1296 - - - - - - ( 1 ) total number of passwords in which no letter repeats = 6 c 4 * 4 ! = 15 * 24 = 360 - - - - - - ( 2 ) therefore required value q = ( 1 ) - ( 2 ) = 1296 - 360 = 936 . d | a = math.factorial(4)
b = math.factorial(2)
c = a / b
d = math.factorial(2)
e = c / d
f = math.factorial(4)
g = math.factorial(2)
h = f / g
i = math.factorial(2)
j = h / i
k = e * j
l = 2 * 3
m = l - 1
n = 2 * 3
o = n - 1
p = m * o
q = k * p
|
a ) 11 , b ) 77 , c ) 8 , d ) 10 , e ) 12 | c | add(6, divide(150, add(25, 40))) | the distance between delhi and mathura is 150 kms . a starts from delhi with a speed of 25 kmph at 5 a . m . for mathura and b starts from mathura with a speed of 40 kmph at 6 p . m . from delhi . when will they meet ? | d = 150 – 25 = 125 rs = 40 + 25 = 65 t = 125 / 65 = 2 hours 6 a . m . + 2 = 8 a . m . . answer : c | a = 25 + 40
b = 150 / a
c = 6 + b
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a ) 50 ° f , b ) 76 ° f , c ) 95 ° f , d ) 110 ° f , e ) 120 ° f | c | add(multiply(divide(35, 100), subtract(212, 32)), 32) | water boils at 212 ° f or 100 ° c and melts at 32 ° f or 0 ° c . if the temperature of the particular day is 35 ° c , it is equal to | let f and c denotes the temparature in fahrenheit anid celcsius respectively . then , ( f - 32 ) / ( 212 - 32 ) = ( c - 0 ) / ( 100 - 0 ) , if c = 35 , then f = 95 . c | a = 35 / 100
b = 212 - 32
c = a * b
d = c + 32
|
a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | a | sqrt(96) | the difference between c . i . and s . i . on an amount of rs . 15,000 for 2 years is rs . 96 . what is the rate of interest per annum ? | "explanation : [ 15000 * ( 1 + r / 100 ) 2 - 15000 ] - ( 15000 * r * 2 ) / 100 = 96 15000 [ ( 1 + r / 100 ) 2 - 1 - 2 r / 100 ] = 96 15000 [ ( 100 + r ) 2 - 10000 - 200 r ] / 10000 = 96 r 2 = ( 96 * 2 ) / 3 = 64 = > r = 8 rate = 8 % answer : a" | a = math.sqrt(96)
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a ) 50 th minute , b ) 41 st minute , c ) 45 th minute , d ) 42 nd minute , e ) 49 th minute | e | add(multiply(multiply(const_4, 2), 2), 1) | a monkey ascends a greased pole 26 meters high . he ascends 2 meters in the first minute and then slips down 1 meter in the alternate minute . if this pattern continues until he climbs the pole , in how many minutes would he reach at the top of the pole ? | "the money is climbing 1 meter in 2 min . this pattern will go on till he reaches 24 meters . i mean this will continue for first 24 * 2 = 48 mins . he would have reached 24 meters . after that he will climb 2 meters and he will reach the pole . so total time taken = 48 + 1 = 49 mins . so , asnwer will be e" | a = 4 * 2
b = a * 2
c = b + 1
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a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | b | add(divide(100, 5), 5) | a man distributed rs . 100 equally among his friends . if there had been 5 more friends , each would have received one rupee less . how many friends had he ? | firstly we have 20 friends and money wiill be distribute each of them 5 rupee then 5 more friends it means total no frnd = 25 each recv 4 rupee which means all recv 1 less rupee answer : b | a = 100 / 5
b = a + 5
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a ) rs . 245.8 , b ) rs . 653.8 , c ) rs . 300 , d ) rs . 400 , e ) rs . 748.5 | b | multiply(1800, divide(inverse(9), add(inverse(12), add(inverse(6), inverse(9))))) | a , b and c can do a work in 6 , 9 and 12 days respectively doing the work together and get a payment of rs . 1800 . what is b ’ s share ? | "wc = 1 / 6 : 1 / 9 : 1 / 12 = > 6 : 4 : 3 4 / 13 * 1800 = 653.8 answer : b" | a = 1/(9)
b = 1/(12)
c = 1/(6)
d = 1/(9)
e = c + d
f = b + e
g = a / f
h = 1800 * g
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a ) 6000 , b ) 2999 , c ) 1000 , d ) 2651 , e ) 1971 | a | divide(6270, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 6270 . what was his original salary ? | x * ( 110 / 100 ) * ( 95 / 100 ) = 6270 x * ( 11 / 10 ) * ( 1 / 100 ) = 66 x = 6000 answer : a | a = 10 / 100
b = 1 + a
c = 5 / 100
d = 1 - c
e = b * d
f = 6270 / e
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a ) 54 , b ) 55 , c ) 56 , d ) 57 , e ) none of these | a | multiply(divide(45, subtract(270, 45)), 270) | the true discount on a bill of rs . 270 is rs . 45 . the banker ' s discount is | solution p . w = rs . ( 270 - 45 ) = rs . 225 s . i on rs . 270 = rs . ( 45 / 225 x 270 ) = rs . 54 . answer a | a = 270 - 45
b = 45 / a
c = b * 270
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a ) 4 : 2 , b ) 4 : 8 , c ) 7 : 3 , d ) 4 : 0 , e ) 4 : 9 | c | multiply(divide(21, const_100), 9) | a part of certain sum of money is invested at 9 % per annum and the rest at 21 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "21 : 9 = 7 : 3 answer : c" | a = 21 / 100
b = a * 9
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['a ) 20', 'b ) 40', 'c ) 60', 'd ) 80', 'e ) 100'] | d | multiply(power(const_2, const_3), 10) | a carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet . if the carpenter made instead a sandbox which was twice as long , twice as wide and twice as high as the original , what would the capacity be of the new larger sandbox ? | explanation : when all the dimensions of a three dimensional object are in fact doubled , then the capacity increases by a factor of 2 x 2 x 2 = 2 ^ 3 = 8 . thus the capacity of the new sandbox is 8 x 10 = 80 cubic feet . answer : option d | a = 2 ** 3
b = a * 10
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a ) 2 / 3 , b ) 2 / 5 , c ) 4 / 9 , d ) 5 / 7 , e ) 4 / 11 | c | divide(multiply(divide(2, 3), 8), 12) | a pipe can empty 2 / 3 rd of a cistern in 12 mins . in 8 mins , what part of the cistern will be empty ? | "2 / 3 - - - - 12 ? - - - - - 8 = = > 4 / 9 c" | a = 2 / 3
b = a * 8
c = b / 12
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a ) 14 , b ) 15 , c ) 16 , d ) 28 , e ) 29 | a | add(multiply(divide(multiply(divide(12.00, multiply(subtract(0.75, 0.30), const_2)), const_2), const_10), const_2), multiply(divide(12.00, multiply(subtract(0.75, 0.30), const_2)), const_2)) | the toll for crossing a certain bridge is $ 0.75 each crossing . drivers who frequently use the bridge may instead purchase a sticker each month for $ 12.00 and then pay only $ 0.30 each crossing during that month . if a particular driver will cross the bridge twice on each of x days next month and will not cross the bridge on any other day , what is the least value of x for which this driver can save money by using the sticker ? | "option # 1 : $ 0.75 / crossing . . . . cross twice a day = $ 1.5 / day option # 2 : $ 0.30 / crossing . . . . cross twice a day = $ 0.6 / day + $ 13 one time charge . if we go down the list of possible answers , you can quickly see that 14 days will not be worth purchasing the sticker . 1.5 x 14 ( 21 ) is cheaper than 0.6 x 14 + 13 ( 21.4 ) . . . it ' s pretty close so let ' s see if one more day will make it worth it . . . if we raise the number of days to 15 , the sticker option looks like a better deal . . . 1.5 x 15 ( 22.5 ) vs 0.6 x 15 + 13 ( 22 ) . answer : a" | a = 0 - 75
b = a * 2
c = 12 / 0
d = c * 2
e = d / 10
f = e * 2
g = 0 - 75
h = g * 2
i = 12 / 0
j = i * 2
k = f + j
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a ) 3 : 4 , b ) 4 : 3 , c ) 5 : 4 , d ) 6 : 7 , e ) 5 : 7 | c | divide(divide(25, const_100), divide(20, const_100)) | if 20 % of a is the same as 25 % of b , then a : b is : | "expl : 20 % of a i = 25 % of b = 20 a / 100 = 25 b / 100 = 5 / 4 = 5 : 4 answer : c" | a = 25 / 100
b = 20 / 100
c = a / b
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a ) 22 , b ) 367 , c ) 27 , d ) 32 , e ) 25 | b | subtract(negate(43), multiply(subtract(7, 16), divide(subtract(7, 16), subtract(4, 7)))) | 4 , 7 , 16 , 43 , 124 , ( . . . ) | "explanation : 4 4 × 3 - 5 = 7 7 × 3 - 5 = 16 16 × 3 - 5 = 43 43 × 3 - 5 = 124 124 × 3 - 5 = 367 answer : option b" | a = negate - (
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a ) 10 % , b ) 15 % , c ) 25 % , d ) 20 % , e ) 30 % | d | subtract(divide(120, divide(100, const_100)), const_100) | a man buys an article for $ 100 . and sells it for $ 120 . find the gain percent ? | "c . p . = $ 100 s . p . = $ 120 gain = $ 20 gain % = 20 / 100 * 100 = 20 % answer is d" | a = 100 / 100
b = 120 / a
c = b - 100
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a ) 45 kg , b ) 50 kg , c ) 53 kg , d ) 55 kg , e ) none of these | a | subtract(multiply(add(35, divide(400, const_1000)), add(24, const_1)), multiply(24, 35)) | the average weight of a class of 24 students is 35 kg . if the weight of the teacher be included , the average rises by 400 g . the weight of the teacher is | "solution weight of the teacher = ( 35.4 × 25 - 35 × 24 ) kg = 45 kg . answer a" | a = 400 / 1000
b = 35 + a
c = 24 + 1
d = b * c
e = 24 * 35
f = d - e
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a ) 0.25 % , b ) 4 % , c ) 25 % , d ) 27.5 % , e ) 250 % | d | multiply(divide(200, 55), const_100) | what percent of 200 is 55 ? | "200 * x / 100 = 55 x = 27.5 ans : d" | a = 200 / 55
b = a * 100
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a ) 600 , b ) 887 , c ) 625 , d ) 654 , e ) 712 | c | add(500, multiply(500, divide(25, const_100))) | a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 25 % ? | "cost price = rs . 500 profit = 25 % of 500 = rs . 125 selling price = cost price + profit = 500 + 125 = 625 answer : c" | a = 25 / 100
b = 500 * a
c = 500 + b
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a ) 12 , b ) 14 , c ) 8 , d ) 25 , e ) 18 | e | divide(545, 31) | if 28 a + 30 b + 31 c = 545 . then a + b + c = ? . a , b , c are natural numbers | have a look on your calender since , we all knw a year consist of 365 days february is d only month which has 28 days 4 months in a year has 30 days and , rest 7 months has 31 days . . so , following d given eq . we can write 28 * 1 + 30 * 10 + 31 * 7 . . hence values of a , b and c are 1 , 10 and 7 respectively . . a + b + c = 18 answer : e | a = 545 / 31
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a ) 5 : 1 , b ) 17 : 3 , c ) 5 : 6 , d ) 17 : 7 , e ) 3 : 8 | a | divide(75000, 15000) | p and q started a business investing rs 75000 and rs 15000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 75000 : 15000 = 75 : 15 = 5 : 1 option a" | a = 75000 / 15000
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a ) 47 , b ) 52 , c ) 32 , d ) 12 , e ) 22 | a | subtract(55, multiply(multiply(12, const_2.0), 2)) | evaluate : 55 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 55 - 12 * 3 * 2 = 55 - 8 = 47 correct answer a" | a = 12 * 2
b = a * 2
c = 55 - b
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a ) 56.8 m 2 , b ) 82.9 m 2 , c ) 52.8 m 2 , d ) 17.9 m 2 , e ) 72.9 m 2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(42, divide(const_3600, const_10))) | the area of sector of a circle whose radius is 12 metro and whose angle at the center is 42 ° is ? | "42 / 360 * 22 / 7 * 12 * 12 = 52.8 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 = 42 / g
i = f * h
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a ) z / 4 , b ) z / 2 , c ) z , d ) 2 z , e ) 4 z | a | divide(subtract(divide(multiply(2, const_100), const_2), const_2), add(divide(multiply(2, const_100), const_2), const_2)) | if 2 x = 4 y = z , what is x - y , in terms of z ? | "2 x = 4 y = z i . e . x = z / 2 and y = z / 4 x - y = z / 2 - z / 4 = z / 4 answer : option a" | a = 2 * 100
b = a / 2
c = b - 2
d = 2 * 100
e = d / 2
f = e + 2
g = c / f
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | add(subtract(40, add(35, 1)), const_1) | the average weight of a group of boys is 35 kg . after a boy of weight 40 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | let the number off boys in the group originally be x . total weight of the boys = 35 x after the boy weighing 40 kg joins the group , total weight of boys = 35 x + 40 so 35 x + 40 = 36 ( x + 1 ) = > x = 5 . answer : b | a = 35 + 1
b = 40 - a
c = b + 1
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a ) 100 , b ) 610 , c ) 729 , d ) 900 , e ) 9000000 | e | multiply(multiply(multiply(multiply(13, const_10), const_10), const_10), const_10) | how many 13 - digits number are palindromic numbers ? a palindromic number reads the same forward and backward , example 1234567654321 . | "take the task of building palindromes and break it intostages . stage 1 : select the 13 th digit we can choose 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 1 in 9 ways stage 2 : select the 12 th 11 th 10 th , 9 th , 8 th , 7 th , we can choose 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 2 in 10 ways important : at this point , the remaining digits are alreadylocked in . stage 3 : select the 6 th , 5 th , 4 th , 3 rd , 2 nd , 1 st digit so , we can complete this stage in 1 way . by thefundamental counting principle ( fcp ) , we can complete all 5 stages ( and thus build a 13 - digit palindrome ) in ( 9 ) ( 10 ) ( 10 ) ( 10 ) ( 10 ) ( 10 ) ( 10 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) ways ( = 9000000 ways ) answer : e" | a = 13 * 10
b = a * 10
c = b * 10
d = c * 10
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a ) 266015 , b ) 266014 , c ) 266016 , d ) 266116 , e ) 266226 | c | multiply(divide(5216, 51), const_100) | 5216 × 51 = ? | "normal way of multiplication may take time . here is one alternative . 5216 × 51 = ( 5216 × 50 ) + 5216 = ( 5216 × 1002 ) + 5216 = 5216002 + 5216 = 260800 + 5216 = 266016 answer is c ." | a = 5216 / 51
b = a * 100
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a ) 90 , b ) 85 , c ) 85.5 , d ) 70 , e ) 90 | a | multiply(divide(const_3, add(divide(const_2, divide(100, 75)), const_1)), 75) | points x , y , and z lie , in that order , on a straight railroad track . the distance from point x to point y is twice the distance from point y to point z . a train traveled from point x to point z without stopping . the train ' s average speed when traveling from point x to point y was 100 miles per hour and the train ' s average speed when traveling from point y to point z was 75 miles per hour . what was the train ' s average speed , in miles per hour , when traveling from point x to point z ? | "average speed = distance / time because we are looking for average speed we can pick a distance for the variable d . speed x - y = 100 speed y - z = 75 average speed = total distance / total rate rate = distance / time x = = = = = = = = = = = = = = = = = = = = y = = = = = = = = = = z if x - y is twice the length of y - z then let x - y = 2 d and let y - z = d average speed = 3 d / ( 2 d / 100 ) + ( d / 75 ) 3 d / ( 6 d / 300 ) + ( 4 d / 300 ) 3 d / ( 10 d / 300 ) 900 d / 10 d average speed = 90 answer : a" | a = 100 / 75
b = 2 / a
c = b + 1
d = 3 / c
e = d * 75
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a ) 72 min , b ) 62 min , c ) 70 min , d ) 32.72727 min , e ) 76 min | d | multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 20), multiply(const_2, 20)))), 20) | if there are 200 questions in a 3 hr examination . among these questions are 20 type a problems , which requires twice as much as time be spent than the rest of the type b problems . how many minutes should be spent on type a problems ? | "x = time for type b prolems 2 x = time for type a problem total time = 3 hrs = 180 min 180 x + 20 * 2 x = 180 x = 180 / 220 x = 0.818182 time taken for type a problem = 20 * 2 * 0.818182 = 32.72727 min answer : d" | a = 3 * const_60
b = 200 - 20
c = 2 * 20
d = b + c
e = a / d
f = 2 * e
g = f * 20
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a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | c | multiply(divide(divide(100, 2), 100), 2) | an integer n between 1 and 100 , inclusive , is to be chosen at random . what is the probability that n ( n + 2 ) will be divisible by 2 ? | "n ( n + 2 ) to be divisible by 2 either n or n + 2 must be a multiples of 2 . in each following group of numbers : { 1 , 2 , 3 , 4 } , { 5 , 6 , 7 , 8 } , . . . , { 97 , 98 , 99 , 100 } there are exactly 2 numbers out of 4 satisfying the above condition . for example in { 1 , 2 , 3 , 4 } n can be : 2 , or 4 . thus , the overall probability is 2 / 4 = 1 / 2 . answer : c ." | a = 100 / 2
b = a / 100
c = b * 2
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a ) 16 % , b ) 32 % , c ) 68 % , d ) 84 % , e ) 92 % | c | subtract(const_100, divide(subtract(const_100, 36), const_2)) | a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 36 percent of the distribution lies within one standard deviation d of the mean , what percent of the distribution is less than m + d ? | "this is easiest to solve with a bell - curve histogram . m here is equal to µ in the gaussian normal distribution and thus m = 50 % of the total population . so , if 36 % is one st . dev , then on either side of m we have 36 / 2 = 18 % . so , 18 % are to the right and left of m ( = 50 % ) . in other words , our value m + d = 50 + 18 = 68 % goingfrom the mean m , to the right of the distributionin the bell shaped histogram . . this means that 68 % of the values are below m + d . like i said , doing it on a bell - curve histogram is much easier to fullygethow this works , or you could apply gmat percentile jargon / theory to it c" | a = 100 - 36
b = a / 2
c = 100 - b
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a ) 13 / 20 , b ) 11 / 3 , c ) 5 / 13 , d ) 12 / 17 , e ) 17 / 20 | c | divide(multiply(5, const_2.0), add(2, multiply(5, 5))) | every student in a room is either a junior or a senior . there is at least one junior and at least one senior in the room . if 4 / 5 of the juniors is equal to 1 / 2 of the seniors , what fraction of the students in the room are juniors ? | "let total number of juniors = j total number of seniors = s ( 4 / 5 ) j = ( 1 / 2 ) s = > s = 8 / 5 j total number of students = j + s = ( 13 / 5 ) j fraction of the students in the room are juniors = j / ( j + s ) = j / [ ( 13 / 5 ) j ] = 5 / 13 answer c" | a = 5 * 2
b = 5 * 5
c = 2 + b
d = a / c
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a ) 9 : 8 , b ) 8 : 3 , c ) 3 : 2 , d ) 2 : 3 , e ) 1 : 2 | b | divide(divide(multiply(const_4, 1), multiply(1, 1)), divide(multiply(1, const_4), multiply(2, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 2 . the ratio of economy cars to sport utility vehicles is 4 : 1 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 2 - - > e : l = 3 : 2 = 12 : 8 . the ratio of economy cars to sport utility vehicles is 4 : 1 - - > e : s = 4 : 1 = 12 : 3 . thus , l : s = 8 : 3 . answer : b ." | a = 4 * 1
b = 1 * 1
c = a / b
d = 1 * 4
e = 2 * 4
f = d / e
g = c / f
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a ) s . 41.00 , b ) s . 51.22 , c ) s . 51.219 , d ) s . 41.18 , e ) s . 51.11 | a | subtract(add(add(divide(multiply(divide(40, multiply(divide(5, const_100), 2)), 5), const_100), divide(40, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(40, multiply(divide(5, const_100), 2)), 5), const_100), divide(40, multiply(divide(5, const_100), 2))), 5), const_100)), divide(40, multiply(divide(5, const_100), 2))) | if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 40 , what is the compound interest on the same sum at the rate and for the same time ? | "explanation : sum = ( 40 * 100 ) / ( 2 * 5 ) = rs . 400 amount = [ 400 * ( 1 + 5 / 100 ) 2 ] = rs . 441 c . i . = ( 441 - 400 ) = rs . 41 answer : a" | a = 5 / 100
b = a * 2
c = 40 / b
d = c * 5
e = d / 100
f = 5 / 100
g = f * 2
h = 40 / g
i = e + h
j = 5 / 100
k = j * 2
l = 40 / k
m = l * 5
n = m / 100
o = 5 / 100
p = o * 2
q = 40 / p
r = n + q
s = r * 5
t = s / 100
u = i + t
v = 5 / 100
w = v * 2
x = 40 / w
y = u - x
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a ) rs . 3600 , b ) rs . 3629 , c ) rs . 3279 , d ) rs . 3629 , e ) rs . 3283 | a | subtract(18000, multiply(divide(4, 5), 18000)) | income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 18000 , then find his savings ? | "let the income and the expenditure of the person be rs . 5 x and rs . 4 x respectively . income , 5 x = 18000 = > x = 3600 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 3600 . answer : a" | a = 4 / 5
b = a * 18000
c = 18000 - b
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a ) 68.8 , b ) 73.6 , c ) 75.2 , d ) 76.8 , e ) 79.36 | e | multiply(add(add(7.5, 8.8), 8.5), 3.2) | in a certain diving competition , 5 judges score each dive on a scale from 1 to 10 . the point value of the dive is obtained by dropping the highest score and the lowest score and multiplying the sum of the remaining scores by the degree of difficulty . if a dive with a degree of difficulty of 3.2 received scores of 7.5 , 8.8 , 9.0 , 6.0 , and 8.5 , what was the point value of the dive ? | "degree of difficulty of dive = 3.2 scores are 6.0 , 7.5 , 8.8 , 8.5 and 9.0 we can drop 6.0 and 9.0 sum of the remaining scores = ( 7.5 + 8.8 + 8.5 ) = 24.8 point of value of the dive = 24 * 3.2 = 79.36 answer e" | a = 7 + 5
b = a + 8
c = b * 3
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a ) 430 , b ) 432 , c ) 440 , d ) 442 , e ) 446 | a | add(subtract(310, divide(112, const_2)), subtract(232, divide(112, const_2))) | at the faculty of aerospace engineering , 310 students study random - processing methods , 232 students study scramjet rocket engines and 112 students study them both . if every student in the faculty has to study one of the two subjects , how many students are there in the faculty of aerospace engineering ? | "310 + 232 - 112 ( since 112 is counted twice ) = 430 a is the answer" | a = 112 / 2
b = 310 - a
c = 112 / 2
d = 232 - c
e = b + d
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a ) 1 : 2 , b ) 3 : 7 , c ) 4 : 1 , d ) 5 : 2 , e ) 6 : 5 | b | divide(subtract(12, 9), subtract(19, 12)) | gold is 19 times as heavy as water and copper is 9 times as heavy as water . in what ratio should these be mixed to get an alloy 12 times as heavy as water ? | "g = 19 w c = 9 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 19 w + 9 wx = x + 1 * 12 w 19 + 9 x = 12 ( x + 1 ) x = 7 / 3 ratio of gold with copper = 1 : 7 / 3 = 3 : 7 answer is b" | a = 12 - 9
b = 19 - 12
c = a / b
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a ) 80 % , b ) 112 % , c ) 120 % , d ) 124.2 % , e ) 138 % | b | multiply(divide(multiply(14, 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 14 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 = 0,14 = 14 / 100 x = 80 * 14 / 100 x = 11.2 11.2 / 10 = 1,12 = 112 % , answer b" | a = 20 / 100
b = 1 - a
c = 14 * b
d = c / 10
e = d * 100
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a ) 4 , b ) 3 , c ) 2 , d ) 7 , e ) 6 | a | floor(divide(divide(add(multiply(factorial(3), factorial(5)), multiply(factorial(4), factorial(3))), 2), 100)) | what is the 100 th digit of ( 3 ! * 5 ! + 4 ! * 3 ! ) / 2 ? | ( 3 ! * 5 ! + 4 ! * 3 ! ) / 2 = 3 ! ( 5 ! + 4 ! ) / 2 = 6 ( 120 + 24 ) / 2 = 432 100 th digit of the above product will be equal to 4 answer a | a = math.factorial(3)
b = math.factorial(5)
c = a * b
d = math.factorial(4)
e = math.factorial(3)
f = d * e
g = c + f
h = g / 2
i = h / 100
j = math.floor(i)
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['a ) rs . 55.50', 'b ) rs . 67.50', 'c ) rs . 86.50', 'd ) rs . 98.00', 'e ) none of these'] | d | divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(9408, multiply(3, 4)))), multiply(4, sqrt(divide(9408, multiply(3, 4))))), 25), const_100) | the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 9408 sq . m , the cost of fencing the field @ 25 paise per metre is | solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 9408 ⇔ 12 x 2 = 9408 ⇔ x 2 = 784 ⇔ x = 28 . so , length = 84 m and breadth = 112 m . perimeter = [ 2 ( 84 + 112 ) ] m = 392 m . ∴ cost of fencing = rs . ( 0.25 × 392 ) = rs . 98.00 . answer d | a = 3 * 4
b = 9408 / a
c = math.sqrt(b)
d = 3 * c
e = 3 * 4
f = 9408 / e
g = math.sqrt(f)
h = 4 * g
i = rectangle_perimeter * (
j = i / 25
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a ) 25 , b ) 37.5 , c ) 50 , d ) 62.5 , e ) 75 | c | add(subtract(const_100, 70), subtract(90, 70)) | at a certain university , 70 % of the professors are women , and 70 % of the professors are tenured . if 90 % of the professors are women , tenured , or both , then what percent of the men are tenured ? | "answer is 75 % total women = 70 % total men = 40 % total tenured = 70 % ( both men and women ) therefore , women tenured + women professors + men tenured = 90 % men tenured = 20 % but question wants to know the percent of men that are tenured 20 % / 40 % = 50 % c" | a = 100 - 70
b = 90 - 70
c = a + b
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a ) a . 0 , b ) b . 1 , c ) c . 3 , d ) d . 4 , e ) e . 5 | b | subtract(6, 5) | if p and t are positive integers such that p > t > 1 , what is the remainder when 92 p × 5 p + t + 11 t × 6 pt is divided by 10 ? | 92 p × 5 p + t + 11 t × 6 pt - - 1 = b | a = 6 - 5
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a ) 1 / 2 , b ) 1 , c ) 3 / 4 , d ) 2 , e ) 3 | b | multiply(const_60, divide(multiply(40, divide(15, const_60)), subtract(50, 40))) | mary passed a certain gas station on a highway while traveling west at a constant speed of 40 miles per hour . then , 15 minutes later , paul passed the same gas station while traveling west at a constant speed of 50 miles per hour . if both drivers maintained their speeds and both remained on the highway for at least 2 hours , how long after he passed the gas station did paul catch up with mary ? | "d = rt m : r = 40 mph , t = t + 1 / 4 hr d = 40 ( t + 1 / 4 ) p : r = 50 , t = t d = 50 t since they went the same distance : 40 t + 40 / 4 = 50 t 10 t = 10 t = 1 hr , b" | a = 15 / const_60
b = 40 * a
c = 50 - 40
d = b / c
e = const_60 * d
|
a ) 455 , b ) 277 , c ) 269 , d ) 261 , e ) 281 | a | add(350, multiply(350, divide(30, const_100))) | a person buys an article at rs . 350 . at what price should he sell the article so as to make a profit of 30 % ? | "cost price = rs . 350 profit = 30 % of 350 = rs . 105 selling price = cost price + profit = 350 + 105 = 455 answer : a" | a = 30 / 100
b = 350 * a
c = 350 + b
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a ) 10.9 sec , b ) 10.1 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 12.24 sec | e | divide(add(140, 200), multiply(add(60, 40), const_0_2778)) | two trains 140 m and 200 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 140 + 200 = 340 m . required time = 340 * 9 / 250 = 12.24 sec . answer : e" | a = 140 + 200
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 1770 cm , b ) 8000 cm , c ) 7860 cm , d ) 6170 cm , e ) 1870 cm | b | divide(volume_cube(1), volume_cube(divide(5, const_100))) | how many cubes of 5 cm edge can be put in a cubical box of 1 m edge . | "number of cubes = 100 â ˆ — 100 â ˆ — 100 / 5 * 5 * 5 = 8000 note : 1 m = 100 cm answer : b" | a = volume_cube / (
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a ) 9 , b ) 16 , c ) 15 , d ) 8 , e ) 10 | b | add(4, const_1) | the average of first seven multiples of 4 is : | "explanation : ( 4 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) / 7 = 4 x 28 / 7 = 16 answer : b" | a = 4 + 1
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a ) 10 , b ) 20 , c ) 40 , d ) 60 , e ) 120 | a | divide(multiply(5, 4), const_2) | there are 5 chess amateurs playing in villa ' s chess club tournament . if each chess amateur plays with exactly 4 other amateurs , what is the total number of chess games possible to be played in the tournament ? | "method 1 : take the first amateur . he plays a game with each of the other four i . e . 4 games . now take the second one . he has already played a game with the first one . he plays 3 games with the rest of the 3 amateurs i . e . 3 more games are played . now take the third amateur . he has already played a game each with the first and the second amateur . now he plays 2 games with the remaining 2 amateurs so 2 more games are played . now go on to the fourth amateur . he has already played 3 games with the first 3 amateurs . he just needs to play a game with the last one i . e . 1 more game is played . the last amateur has already played 4 games . total no of games = 4 + 3 + 2 + 1 = 10 method 2 : each person is one participant of 4 games . so there are in all 4 * 5 = 20 instances of one participant games . but each game has 2 participants so total number of games = 20 / 2 = 10 answer : a ." | a = 5 * 4
b = a / 2
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a ) 23 % , b ) 17 % , c ) 11 % , d ) 10 % , e ) 25 % | e | subtract(const_100, add(add(add(subtract(const_100, 75), subtract(const_100, 90)), subtract(const_100, 85)), subtract(const_100, 75))) | in a urban village of india named ` ` owlna ' ' , 75 % people have refrigerator , 90 % people have television , 85 % people got computers and 75 % got air - conditionor . how many people ( minimum ) got all these luxury . | "e 10 % 100 - [ ( 100 - 75 ) + ( 100 - 90 ) + ( 100 - 85 ) + ( 100 - 75 ) ] = 100 - ( 25 + 10 + 15 + 25 ) = 100 - 75" | a = 100 - 75
b = 100 - 90
c = a + b
d = 100 - 85
e = c + d
f = 100 - 75
g = e + f
h = 100 - g
|
a ) 16 , b ) 20 , c ) 32 , d ) 44 , e ) 48 | c | divide(subtract(15, multiply(divide(20, const_100), 67)), subtract(divide(25, const_100), divide(20, const_100))) | a bowl of nuts is prepared for a party . brand p mixed nuts are 20 % almonds and brand q ' s deluxe nuts are 25 % almonds . if a bowl contains a total of 67 ounces of nuts , representing a mixture of both brands , and 15 ounces of the mixture are almonds , how many ounces of brand q ' s deluxe mixed nuts are used ? | "lets say x ounces of p is mixed with q . = > 67 - x ounces of q is present in the mixture ( as the total = 67 ounces ) given total almond weight = 15 ounces ( 20 x / 100 ) + ( 25 / 100 ) ( 67 - x ) = 15 = > x = 35 = > 67 - 35 = 32 ounces of q is present in the mixture . answer is c ." | a = 20 / 100
b = a * 67
c = 15 - b
d = 25 / 100
e = 20 / 100
f = d - e
g = c / f
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a ) $ 4 , b ) $ 8 , c ) $ 12 , d ) $ 16 , e ) $ 432 | b | multiply(8, const_1) | if $ 5,000 is invested in an account that earns 8 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ? | "solution amount ( ci ) = p + ( 1 + r / n ) ^ nt = 5000 + ( 1 + 0.08 / 2 ) ^ 2 = 5408 amount ( si ) = p + ptr / 100 = 5000 + ( 5000 * 1 * 8 / 100 ) = 5400 difference = 5408 - 5400 = 8 $ answer : b" | a = 8 * 1
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a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | b | sqrt(divide(675, const_3)) | the length of a rectangular garden is three times its width . if the area of the rectangular garden is 675 square meters , then what is the width of the rectangular garden ? | "let x be the width of the garden . 3 x ^ 2 = 675 x ^ 2 = 225 x = 15 the answer is b ." | a = 675 / 3
b = math.sqrt(a)
|
a ) 72 , b ) 88 , c ) 112 , d ) 122 , e ) 144 | e | divide(multiply(divide(factorial(subtract(const_10, const_1)), factorial(subtract(subtract(const_10, const_1), subtract(3, const_1)))), divide(subtract(500, const_100), const_100)), const_2) | how many odd 3 - digit integers smaller than 500 are there such that all their digits are different ? | case 1 : numbers starting with 1 1 _ _ the unit digit can take 4 numbers ( 3,5 , 7,9 ) to be odd so 4 ways . the tens digit can take remaining 8 numbers left after using 2 numbers so 8 ways . total = 8 * 4 = 32 case 2 : numbers starting with 2 2 _ _ the unit digit can take 5 numbers ( 1 , 3,5 , 7,9 ) to be odd so 5 ways . the tens digit can take remaining 8 numbers left after using 2 numbers so 8 ways . total = 8 * 5 = 40 case 3 : numbers starting with 3 3 _ _ the unit digit can take 4 numbers ( 1,5 , 7,9 ) to be odd so 4 ways . the tens digit can take remaining 8 numbers left after using 2 numbers so 8 ways . total = 8 * 4 = 32 case 4 : numbers starting with 4 4 _ _ the unit digit can take 5 numbers ( 1 , 3,5 , 7,9 ) to be odd so 5 ways . the tens digit can take remaining 8 numbers left after using 2 numbers so 8 ways . total = 8 * 5 = 40 hence 32 + 40 + 32 + 40 = 144 , correct answer is e | a = 10 - 1
b = math.factorial(a)
c = 10 - 1
d = 3 - 1
e = c - d
f = math.factorial(e)
g = b / f
h = 500 - 100
i = h / 100
j = g * i
k = j / 2
|
a ) 50 , b ) 52 , c ) 54 , d ) 56 , e ) 58 | e | divide(435, multiply(divide(3, 2), 5)) | a van takes 5 hours to cover a distance of 435 km . what speed in kph should the van maintain to cover the same distance in 3 / 2 of the previous time ? | "( 3 / 2 ) * 5 = 7.5 hours 435 / 7.5 = 58 kph the answer is e ." | a = 3 / 2
b = a * 5
c = 435 / b
|
a ) 20 % increase , b ) 10 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | a | subtract(divide(multiply(subtract(const_100, 20), add(const_100, 50)), const_100), const_100) | if the price of a tv is first decreased by 20 % and then increased by 50 % , then the net change in the price will be : | "explanation : solution : let the original price be rs . 100 . new final price = 150 % of ( 80 % of 100 ) = rs . 150 / 100 * 80 / 100 * 100 = rs . 120 . . ' . increase = 20 % answer : a" | a = 100 - 20
b = 100 + 50
c = a * b
d = c / 100
e = d - 100
|
a ) 44.1 , b ) 32.4 , c ) 66.5 , d ) 75.2 , e ) 53.7 | e | divide(multiply(divide(multiply(68.7, 680), const_100), 18), const_100) | 18 % of 680 - ? = 68.7 | "e 53.7 ( 18 * 680 ) / 100 - ? = 68.7 ? = 122.4 - 68.7 ? = 53.7" | a = 68 * 7
b = a / 100
c = b * 18
d = c / 100
|
a ) 12200 , b ) 11897 , c ) 18799 , d ) 10000 , e ) 12782 | d | add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1)) | a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 241 more if the interest were payable half - yearly , then it were pay able yearly . find the sum . | "p ( 11 / 10 ) ^ 4 - p ( 6 / 5 ) ^ 2 = 241 p = 10000 answer : d" | a = 20 + 1
b = 20 + 1
c = a * b
d = 20 + 1
e = c + d
|
a ) 11 , b ) 5 , c ) 70 , d ) 11 , e ) 13 | d | add(subtract(subtract(24, subtract(40, 30)), 4), const_1) | a , b and c can do a piece of work in 24 , 30 and 40 days respectively . they start the work together but c leaves 4 days before the completion of the work . in how many days is the work done ? | x / 24 + x / 30 + x / 40 = 1 x = 11 days answer : d | a = 40 - 30
b = 24 - a
c = b - 4
d = c + 1
|
a ) 44 , b ) 48 , c ) 46 , d ) 52 , e ) 56 | c | subtract(multiply(120, divide(45, const_100)), subtract(multiply(120, divide(10, const_100)), divide(multiply(120, divide(10, const_100)), 3))) | of the 120 passengers on flight 750 , 45 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 / 3 of the passengers in first class are male , how many females are there in coach class ? | "number of passengers on flight = 120 number of female passengers = . 45 * 120 = 54 number of passengers in first class = ( 10 / 100 ) * 120 = 12 number of passengers in coach class = ( 90 / 100 ) * 120 = 108 number of male passengers in first class = 1 / 3 * 12 = 4 number of female passengers in first class = 12 - 4 = 8 number of female passengers in coach class = 54 - 8 = 46 answer c" | a = 45 / 100
b = 120 * a
c = 10 / 100
d = 120 * c
e = 10 / 100
f = 120 * e
g = f / 3
h = d - g
i = b - h
|
a ) 4 , b ) 6 , c ) 36 , d ) 12 , e ) 18 | c | add(const_3, const_4) | what is the smallest positive integer x , such that 6000 x is a perfect cube ? | "take out the factors of 6000 that will come 2 * 3 * 10 ^ 3 . for perfect cube you need every no . raise to the power 3 . for 6000 x to be a perfect cube , need two 2 and 2 3 that means 36 or 6 ^ 2 c is the answer ." | a = 3 + 4
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a ) 171 , b ) 175 , c ) 180 , d ) 190 , e ) 200 | a | add(multiply(divide(add(10, 20), const_2), add(subtract(20, 10), const_1)), add(divide(subtract(20, 10), const_2), const_1)) | if x is equal to the sum of the integers from 10 to 20 , inclusive , and y is the number of even integers from 10 to 20 , inclusive , what is the value of x + y ? | sum s = n / 2 { 2 a + ( n - 1 ) d } = 11 / 2 { 2 * 10 + ( 11 - 1 ) * 1 } = 11 * 15 = 165 = x number of even number = ( 20 - 10 ) / 2 + 1 = 6 = y x + y = 165 + 6 = 171 a | a = 10 + 20
b = a / 2
c = 20 - 10
d = c + 1
e = b * d
f = 20 - 10
g = f / 2
h = g + 1
i = e + h
|
a ) 4 , b ) 5 , c ) 7 , d ) 9 , e ) 8 | d | multiply(49, 49) | 49 * 49 * 49 * 49 = 7 ? | "sol . 49 * 49 * 49 * 49 = ( 72 * 72 * 72 * 72 ) = 7 ( 2 + 2 + 2 + 2 ) = 78 . so , the correct answer is 8 . answer d" | a = 49 * 49
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a ) 266 , b ) 268 , c ) 270 , d ) 272 , e ) 274 | a | add(2, lcm(33, 8)) | find the least number which when divided by 33 and 8 leaves a remainder of 2 in each case . | "the least number which when divided by different divisors leaving the same remainder in each case = lcm ( different divisors ) + remainder left in each case . hence the required least number = lcm ( 33 , 8 ) + 2 = 266 . answer : a" | a = math.lcm(33, 8)
b = 2 + a
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a ) 240 , b ) 288 , c ) 1687 , d ) 997 , e ) 121 | a | divide(120, divide(120, const_100)) | 50 % of a number is added to 120 , the result is the same number . find the number ? | "( 50 / 100 ) * x + 120 = x x = 240 answer : a" | a = 120 / 100
b = 120 / a
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a ) 4.46 , b ) 4.39 , c ) 4.42 , d ) 4.49 , e ) 4.33 | a | divide(add(4.5, 4.3), const_2) | there are 4 runners on a track team who run an average of 4.5 seconds per 40 yards . if another runner joins their team who runs 4.3 seconds per 40 yards , what will the new average 40 yard time be ? | ( sum of the 4 times ) / 4 = 4.5 sum of the 4 times = 18 new sum = 18 + 4.3 = 22.3 new average = 22.3 / 5 = 4.46 ans : a | a = 4 + 5
b = a / 2
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a ) 1 , b ) 1.5 , c ) 2 , d ) 2.5 , e ) 3 | c | divide(multiply(multiply(30, divide(4, 5)), inverse(subtract(const_1, divide(4, 5)))), const_60) | a train is moving at 4 / 5 of its usual speed . the train is 30 minutes too late . what is the usual time ( in hours ) for the train to complete the journey ? | "new time = d / ( 4 v / 5 ) = 5 / 4 * usual time 30 minutes represents 1 / 4 of the usual time . the usual time is 2 hours . the answer is c ." | a = 4 / 5
b = 30 * a
c = 4 / 5
d = 1 - c
e = 1/(d)
f = b * e
g = f / const_60
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a ) 252 kms , b ) 152 kms , c ) 732 kms , d ) 752 kms , e ) 152 kms | c | multiply(add(multiply(2, 50), multiply(subtract(12, const_1), 2)), divide(12, 2)) | the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 50 kms . what was the total distance traveled in 12 hours ? | "explanation : total distance travelled in 12 hours = ( 50 + 52 + 54 + . . . . . upto 12 terms ) this is an a . p with first term , a = 50 , number of terms , n = 12 , d = 2 . required distance = 12 / 2 [ 2 x 50 + { 12 - 1 ) x 2 ] = 6 ( 122 ) = 732 kms . answer : c" | a = 2 * 50
b = 12 - 1
c = b * 2
d = a + c
e = 12 / 2
f = d * e
|
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