options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 288 , b ) 279 , c ) 135 , d ) 272 , e ) 150 | c | multiply(divide(multiply(54, const_1000), const_3600), 9) | a train running at the speed of 54 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 54 * 5 / 18 = 15 m / sec length of the train = speed * time = 15 * 9 = 135 m answer : c" | a = 54 * 1000
b = a / 3600
c = b * 9
|
a ) 7 , b ) 9 , c ) 11 , d ) 12 , e ) 15 | b | add(reminder(multiply(6, const_4), const_10), const_1) | one half of a two digit number exceeds its one third by 6 . what is the sum of the digits of the number ? | "explanation : x / 2 – x / 3 = 6 = > x = 6 3 + 6 = 9 b )" | a = 6 * 4
b = reminder + (
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a ) 1 / 190 , b ) 1 / 20 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 92 | c | divide(const_1, subtract(10, const_1)) | a box contains 5 pairs of shoes ( 10 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ? | "the problem with your solution is that we do n ' t choose 1 shoe from 20 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 9 ( as after taking one at random there are 9 shoes left and only one is the pair of the first one ) = 1 / 9 answer : c ." | a = 10 - 1
b = 1 / a
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a ) $ 54,000 , b ) $ 55,000 , c ) $ 56,000 , d ) $ 57,000 , e ) $ 58,000 | e | subtract(divide(multiply(divide(subtract(subtract(multiply(add(add(multiply(multiply(4, 4), add(4, const_1)), const_2), divide(const_1, const_2)), 16), multiply(multiply(16, 4), add(4, const_1))), multiply(multiply(multiply(4, 4), add(4, const_1)), 4)), 4), const_1000), const_1000), 8) | in a company of 16 employees , 8 employees earn $ 38,000 , 4 employees earn $ 42,000 , and the 4 highest - paid employees earn the same amount . if the average annual salary for the 16 employees is $ 44,000 , what is the annual salary for each of the highest - paid employees ? | "8 * 38,000 + 4 * 42,000 + 4 x = 16 * 44,000 4 x = 704,000 - 304,000 - 168,000 4 x = 232,000 x = 58,000 the answer is e ." | a = 4 * 4
b = 4 + 1
c = a * b
d = c + 2
e = 1 / 2
f = d + e
g = f * 16
h = 16 * 4
i = 4 + 1
j = h * i
k = g - j
l = 4 * 4
m = 4 + 1
n = l * m
o = n * 4
p = k - o
q = p / 4
r = q * 1000
s = r / 1000
t = s - 8
|
a ) 25 % , b ) 20 % , c ) 24.2 % , d ) 33.33 % , e ) none of these | c | multiply(subtract(const_1, divide(const_100, add(const_100, 32))), const_100) | if the price of petrol increases by 32 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ? | explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 32 ) i . e 132 . now , his consumption should be reduced to : - = ( 132 − 100 ) / 132 ∗ 100 . hence , the consumption should be reduced to 24.2 % . answer : c | a = 100 + 32
b = 100 / a
c = 1 - b
d = c * 100
|
a ) 9 % less , b ) 1 % less , c ) equal to each other , d ) 1 % more , e ) 2.1 % more | e | divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 7), subtract(const_100, 7)))) | 108 . triangle a ’ s base is 7 % greater than the base of triangle b , and a ’ s height is 7 % 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 = 8 / 7 * base of b height of a = 6 / 7 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 8 / 7 * 6 / 7 * area of b = 48 / 49 * area of b area of a is 2.1 % more than the area of b . answer ( e )" | a = 100 * 100
b = 100 + 7
c = 100 - 7
d = b * c
e = a - d
f = 100 / e
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a ) 0 , b ) 1 , c ) 2 , d ) 5 , e ) 8 | e | divide(64, add(add(2, 2), 1)) | if 2 / z = 2 / ( z + 1 ) + 2 / ( z + 64 ) which of these integers could be the value of z ? | "solving for z algebraically in this problem would not be easy . instead , we can follow the hint in the question ( “ which of these integers … ” ) and test each answer choice : a . 2 / 0 = 2 / 1 + 2 / 64 incorrect ( division by zero ) b . 2 / 1 = 2 / 2 + 2 / 65 incorrect c . 2 / 2 = 2 / 3 + 2 / 66 incorrect d . 2 / 3 = 2 / 4 + 2 / 69 incorrect e . 2 / 4 = 2 / 5 + 2 / 72 correct the correct answer is e , because it contains the only value that makes the equation work . notice how quickly this strategy worked in this case" | a = 2 + 2
b = a + 1
c = 64 / b
|
a ) s . 7500 , b ) s . 8640 , c ) s . 8500 , d ) s . 9000 , e ) s . 6000 | b | divide(divide(3360, subtract(const_1, divide(5, const_12))), divide(2, 3)) | praveen starts business with rs . 3360 and after 5 months , hari joins with praveen as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is hari ’ s contribution in the capital ? | let hari ’ s capital be rs . x . then , 3360 * 12 / 7 x = 2 / 3 = > 14 x = 120960 = > x = 8640 . answer : b | a = 5 / 12
b = 1 - a
c = 3360 / b
d = 2 / 3
e = c / d
|
a ) 25 , b ) 60 , c ) 50 , d ) 30 , e ) 40 | a | subtract(add(add(10, 20), 60), add(add(multiply(5, const_3), 10), 40)) | the average ( arithmetic mean ) of 10 , 20 , and 60 is 5 more than the average of 10 , 40 , and what number ? | "a 1 = 90 / 3 = 30 a 2 = a 1 - 5 = 25 sum of second list = 25 * 3 = 75 therefore the number = 75 - 50 = 25 a" | a = 10 + 20
b = a + 60
c = 5 * 3
d = c + 10
e = d + 40
f = b - e
|
a ) $ 10079.44 , b ) $ 10815.83 , c ) $ 12652.61 , d ) $ 14232.14 , e ) $ 20598.11 | b | multiply(10000, power(add(const_1, divide(divide(3.96, const_100), const_2)), const_4)) | jill invests $ 10000 in an account that pays an annual rate of 3.96 % , compounding semi - annually . approximately how much does she have in her account after two years ? | solution : first of all , notice the magic word “ approximately ” — the test - writer is letting us know estimation is perfectly fine . furthermore , the answer choices are nicely spread out , which will facilitate estimating . ok , get ready for some fast & furious estimation . the interest rate 3.96 % is an ugly number , so i ’ m going to approximate that as 4 % . it compounds semiannually , so that means that there ’ s 2 % every six months , and that happens four times in two years . well , 2 % of $ 10000 is $ 200 . if you get $ 200 , or a little more , on four occasions , that ’ s a little more than $ 800 in interest . we expect an answer slightly higher than $ 10800 , so of course ( b ) is just right . notice , i estimated so that everything up until the last sum was single - digit math . single - digit calculations are a good standard for which to strive when you are practicing estimation . by the way , if you find the bank that will do answer ( e ) , double your money in only two years , that ’ s terrific , but it probably is something wildly illegal , a ponzi scheme or worse ! in the real world , that just doesn ’ t happen . on word problems , especially in financial situations , you should always have your antenna up for what ’ s realistic or unrealistic . answer ( b ) | a = 3 / 96
b = a / 2
c = 1 + b
d = c ** 4
e = 10000 * d
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a ) 700 , b ) 666 , c ) 600 , d ) 400 , e ) 800 | c | divide(multiply(3, multiply(250, 16)), 20) | 250 men work for 16 days and finish a work . how many men would do a job 3 times the previous one in 20 days ? | number of men needed to finish the required work ( variation , proportion method ) = 3 * 16 * 250 / 20 = 600 answer : c | a = 250 * 16
b = 3 * a
c = b / 20
|
a ) 2.91 , b ) 1.794 , c ) 1.338 , d ) 1.986 , e ) 1.999 | b | add(add(1.35, divide(123, const_1000)), divide(321, const_1000)) | solution for 1.35 + . 123 + . 321 | "1.35 + . 123 + . 321 = 0 0 = 0 - 1.35 - 0.123 - 0.321 0 = - 1.794 answer : b" | a = 123 / 1000
b = 1 + 35
c = 321 / 1000
d = b + c
|
a ) a ) 400 , b ) b ) 500 , c ) c ) 600 , d ) d ) 700 , e ) e ) 800 | e | divide(1150, multiply(add(const_1, divide(25, const_100)), add(const_1, divide(15, const_100)))) | the population of a town increases 25 % and 15 % respectively in two consecutive years . after the growth the present population of the town is 1150 . then what is the population of the town 2 years ago ? | explanation : formula : ( after = 100 denominator ago = 100 numerator ) 1150 * 100 / 125 * 100 / 115 = 800 answer : option e | a = 25 / 100
b = 1 + a
c = 15 / 100
d = 1 + c
e = b * d
f = 1150 / e
|
a ) a . 3 , b ) b . 8 , c ) c . 10 , d ) d . 12 , e ) e . 16 | a | subtract(divide(multiply(16, 45), multiply(divide(const_3, const_4), 45)), 16) | if 16 machine can finish a job in 45 days , then how many more machines would be needed to finish the job in one - third less time ? | "you might think of this in a management context - we can use the principle of ' person - hours ' to solve any problem where we have identical workers . so , using simpler numbers , suppose you know that 6 identical employees , working simultaneously , would finish a job in 5 hours . then that job requires 6 * 5 = 30 total hours of person - work . if instead you wanted the job done in 3 hours , you ' d assign 30 / 3 = 10 employees to do the job , because you want to get a total of 30 hours of work from the employees . we can solve this problem identically . if 16 machines ( identical ones , i assume ) work simultaneously for 45 days , they will do a total of 16 * 45 machine - days of work . so the job requires 16 * 45 days of machine work in total . we instead want the job done in 1 / 3 less time , so in 30 days . so we ' ll need 16 * 36 / 30 = 19.2 ~ = 19 machines , or 3 additional machines . a" | a = 16 * 45
b = 3 / 4
c = b * 45
d = a / c
e = d - 16
|
a ) 56 , b ) 58 , c ) 60 , d ) 62 , e ) 64 | c | multiply(divide(160, 5), divide(160, add(divide(160, 5), divide(160, 3)))) | two trains start simultaneously from opposite ends of a 160 - km route and travel toward each other on parallel tracks . train x , traveling at a constant rate , completes the 160 - km trip in 5 hours . train y , travelling at a constant rate , completes the 160 - km trip in 3 hours . how many kilometers had train x traveled when it met train y ? | if the two trains cover a total distance d , then train x travels ( 3 / 8 ) * d while train y travels ( 5 / 8 ) * d . if the trains travel 160 km to the meeting point , then train x travels ( 3 / 8 ) * 160 = 60 km . the answer is c . | a = 160 / 5
b = 160 / 5
c = 160 / 3
d = b + c
e = 160 / d
f = a * e
|
a ) rs . 6030 , b ) rs . 4050 , c ) rs . 6050 , d ) rs . 7050 , e ) rs . 6080 | c | multiply(divide(6300, add(add(6300, 4200), 10500)), 12100) | a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the total share of a & b in profit of rs . 12100 after a year ? | "6300 : 4200 : 10500 3 : 2 : 5 a ' share = 3 / 10 * 12100 = 3630 b ' s share = 2 / 10 * 12100 = 2420 required total share = 3630 + 2420 = 6050 answer : c" | a = 6300 + 4200
b = a + 10500
c = 6300 / b
d = c * 12100
|
a ) 23 , b ) 38 , c ) 18 , d ) 30 , e ) 28 | c | multiply(divide(200, 40), const_3_6) | an athlete runs 200 metres race in 40 seconds . what is his speed ? | "speed = distance / time = 200 / 40 = 5 m / s = 5 * 18 / 5 = 3618 km / hr answer : c" | a = 200 / 40
b = a * const_3_6
|
a ) 391 , b ) 426 , c ) 410 , d ) 423 , e ) 445 | b | divide(490, add(const_1, divide(15, const_100))) | company p had 15 percent more employees in december than it had in january . if company p had 490 employees in december , how many employees did it have in january ? | "d = number of employees in december j = number of employees in january j x 1.15 = d j x 1.15 = 490 j = 490 / 1.15 j = 49,000 / 115 = 426 thus b is the correct answer ." | a = 15 / 100
b = 1 + a
c = 490 / b
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a ) rs . 1325 , b ) rs . 1300 , c ) rs . 1350 , d ) rs . 1500 , e ) none | a | add(divide(divide(1404, const_2), divide(add(const_100, 8), const_100)), divide(divide(1404, const_2), divide(add(const_100, divide(8, const_2)), const_100))) | the present worth of rs . 1404 due in two equal half - yearly instalments at 8 % per annum simple interest is : | solution required sum = p . w . of rs . 702 due 6 month hence + p . w . of rs . 702 due 1 year hence = rs . [ ( 100 x 702 / 100 + 8 x 1 / 2 ) + ( 100 x 702 / 100 + ( 8 x 1 ) ) ] = rs . ( 675 + 650 ) = rs . 1325 . answer a | a = 1404 / 2
b = 100 + 8
c = b / 100
d = a / c
e = 1404 / 2
f = 8 / 2
g = 100 + f
h = g / 100
i = e / h
j = d + i
|
a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | e | divide(add(1, divide(7, 3)), add(divide(7, 3), divide(5, 3))) | the weight of every type a widget is the same , the weight of every type b widget is the same , and the weight of every type c widget is the same . if the weight of 7 type a widgets is equal to the weight of 3 type b widgets , and the weight of 5 type b widgets is equal to the weight of 7 type c widgets . what is the ratio of the total weight of 1 type a widget and 1 type b widget , to the total weight of 1 type b widget and 1 type c widget ? | "5 b = 7 c and so b = 7 c / 5 7 a = 3 b and so a = 3 b / 7 = 3 c / 5 a + b = 3 c / 5 + 7 c / 5 = 10 c / 5 b + c = 7 c / 5 + c = 12 c / 5 the ratio of a + b : b + c = 10 : 12 = 5 : 6 the answer is e ." | a = 7 / 3
b = 1 + a
c = 7 / 3
d = 5 / 3
e = c + d
f = b / e
|
a ) 34 % , b ) 39 % , c ) 50 % , d ) 35 % , e ) 33 % | c | divide(multiply(4, const_100), 6) | a shopkeeper buys mangoes at the rate of 6 a rupee and sells them at 4 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 24 . if he buys 6 a rupee , his cp = 4 he selling at 4 a rupee , his sp = 6 profit = sp - cp = 6 - 4 = 2 profit percent = 2 / 4 * 100 = 50 % answer : c" | a = 4 * 100
b = a / 6
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a ) a ) 0.9 , b ) b ) 0.6 , c ) c ) 0.45 , d ) d ) 0.3 , e ) e ) 0.25 | b | subtract(const_1, multiply(divide(60, const_100), divide(subtract(const_100, 20), const_100))) | in the graduating class of a certain college , 48 percent of the students are male and 52 percent are female . in this class 60 percent of the male and 20 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | "percent of students who are 25 years old or older is 0.6 * 48 + 0.2 * 52 = ~ 39 , so percent of people who are less than 25 years old is 100 - 39 = 61 . answer : b ." | a = 60 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 1 - d
|
a ) 150 , b ) 200 , c ) 140 , d ) 240 , e ) 145 | b | divide(divide(divide(1900, divide(add(15, 4), const_2)), 4), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 15 m wide at the top and 4 m wide at the bottom and the area of cross - section is 1900 sq m , the depth of cannel is ? | "1 / 2 * d ( 15 + 4 ) = 1900 d = 200 answer : b" | a = 15 + 4
b = a / 2
c = 1900 / b
d = c / 4
e = d / 2
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a ) 14.4 , b ) 180 , c ) 50.4 , d ) 60 , e ) 90 | b | divide(multiply(4, 9), divide(20, const_100)) | 20 percent of andrea ' s living room floor is covered by a carpet that is 4 feet by 9 feet . what is the area of her living room floor ? | "20 % of area of the floor = 4 * 9 square feet = 36 square feet i . e . 100 % area of floor = ( 36 / 20 ) * 100 = 180 square feet answer : option b" | a = 4 * 9
b = 20 / 100
c = a / b
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a ) 86 , b ) 160 , c ) 76 , d ) 140 , e ) 26 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(60, 3)) | a student scored an average of 60 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | "given m + p + c = 60 * 3 = 180 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 180 = 140 answer : d" | a = 90 * 2
b = 70 * 2
c = a + b
d = 60 * 3
e = c - d
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | divide(21, add(2, const_1)) | if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 21 - m ) then m = ? | "2 m = 21 - m 3 m = 21 m = 7 the answer is c ." | a = 2 + 1
b = 21 / a
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a ) 26 , b ) 19 , c ) 11 , d ) 8 , e ) 9 | e | subtract(add(29, 17), subtract(45, 8)) | each of the dogs in a certain kennel is a single color . each of the dogs in the kennel either has long fur or does not . of the 45 dogs in the kennel , 29 have long fur , 17 are brown , and 8 are neither long - furred nor brown . how many long - furred dogs are brown ? | no of dogs = 45 long fur = 29 brown = 17 neither long fur nor brown = 8 therefore , either long fur or brown = 45 - 8 = 37 37 = 29 + 17 - both both = 9 answer e | a = 29 + 17
b = 45 - 8
c = a - b
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a ) 300 , b ) 350 , c ) 400 , d ) 450 , e ) 500 | e | multiply(subtract(62, 42), add(divide(subtract(90, 42), const_2), const_1)) | set a contains all the even numbers between 42 and 90 inclusive . set b contains all the even numbers between 62 and 110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "each term in set b is 20 more than the corresponding term in set a . the difference of the sums = 25 * 20 = 500 . the answer is e ." | a = 62 - 42
b = 90 - 42
c = b / 2
d = c + 1
e = a * d
|
a ) 25 % , b ) 22.2 % , c ) 20 % , d ) 29.4 % , e ) 11.1 % | d | multiply(divide(multiply(divide(1, 3), subtract(1, divide(1, 6))), add(multiply(divide(1, 3), subtract(1, divide(1, 6))), subtract(1, divide(1, 3)))), const_100) | of the 3,600 employees of company x , 1 / 3 are clerical . if the clerical staff were to be reduced by 1 / 6 , what percent of the total number of the remaining employees would then be clerical ? | "welcome , just post the question and the choices let ' s see , the way i did it was 1 / 3 are clerical out of 3600 so 1200 are clerical 1200 reduced by 1 / 6 is 1200 * 1 / 6 so it reduced 200 people , so there is 1000 clerical people left but since 200 people left , it also reduced from the total of 3600 so there are 3400 people total since 1000 clerical left / 3400 people total you get ( a ) 29.4 % answer : d" | a = 1 / 3
b = 1 / 6
c = 1 - b
d = a * c
e = 1 / 3
f = 1 / 6
g = 1 - f
h = e * g
i = 1 / 3
j = 1 - i
k = h + j
l = d / k
m = l * 100
|
a ) 1 / pi , b ) sqrt ( 2 / pi ) , c ) 2 / sqrt ( pi ) , d ) sqrt ( 12.5 / pi ) , e ) pi / 2 | d | sqrt(divide(divide(square_area(5), 2), const_pi)) | an artist wishes to paint a circular region on a square poster that is 5 feet on a side . if the area of the circular region is to be 1 / 2 the area of the poster , what must be the radius of the circular region in feet ? | "area of the poster is 5 x 5 = 25 1 / 2 the area = 12.5 pi * r ^ 2 = 12.5 r ^ 2 = 12.5 / pi r = sqrt ( 12.5 / pi ) answer ( d )" | a = square_area / (
b = a / 2
c = math.sqrt(b)
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a ) 2 % , b ) 8 % , c ) 4.5 % , d ) 6.5 % , e ) 1.5 % | c | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(17, const_100))), const_100)) | the tax on a commodity is diminished by 17 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 83 * 115 = 9545 - - - - - - - - - - - 10000 - - - - - - - - - - - 455 100 - - - - - - - - - - - ? = > 4.55 % decrease answer : c" | a = 15 / 100
b = 1 + a
c = 17 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
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a ) 68 , b ) 70.4 , c ) 96 , d ) 105.6 , e ) 108 | c | add(80, multiply(divide(20, const_100), 80)) | if x is 20 percent greater than 80 , then x = | "x is 20 % greater than 80 means x is 1.2 times 80 ( in other words 80 + 20 / 100 * 80 = 1.2 * 80 ) therefore , x = 1.2 * 80 = 96.00 answer : c" | a = 20 / 100
b = a * 80
c = 80 + b
|
a ) 3 / 7 , b ) 2 / 7 , c ) 12 / 13 , d ) 5 / 11 , e ) 4 / 7 | b | divide(2, 7) | a pet groomer has 7 animals to groom for the day ( 2 cats and 5 dogs ) . if she randomly selects 4 animals to groom before lunch , what is the probability she will finish all the cats before lunch ? | combination probability formula : ncr = n ! / [ r ! ( n - r ) ! ] total possible , select 4 animals from 7 animals = 7 c 4 = 7 ! / [ 4 ! ( 7 - 4 ) ! ] = 35 . to finish all 2 cats there must be 2 dogs , select 2 dogs from 5 = 5 c 2 = 10 . and , select 2 cats from 2 = 2 c 2 = 1 . 2 cats and 2 dogs = ( 1 ) ( 10 ) = 10 probability = ( number outcomes favorable ) / ( total number outcomes ) = 10 / 735 = 2 / 7 answer : b | a = 2 / 7
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a ) 1,108 , b ) 2,100 , c ) 2,108 , d ) 2,628 , e ) 2,256 | d | multiply(divide(250, 22.95), 250) | at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.75 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 250 ? | "to maximize number of hot dogs with 250 $ total number of hot dogs bought in 250 - pack = 22.95 * 10 = 229.5 $ amount remaining = 250 - 229.5 = 20.5 $ total number of hot dogs bought in 20 - pack = 3.05 * 6 = 18.3 $ amount remaining = 20.5 - 18.3 = 2.2 $ total number of hot dogs bought in 8 - pack = 1.55 * 1 = 1.55 $ amount remaining = 2.2 - 1.75 = 0.45 $ this amount is too less to buy any 8 - pack . greatest number of hot dogs one can buy with 250 $ = 250 * 10 + 20 * 6 + 8 * 1 = 2628 answer d" | a = 250 / 22
b = a * 250
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a ) 20 coins , b ) 30 coins , c ) 40 coins , d ) 25 coins , e ) none of these | c | divide(70, add(add(inverse(const_4), inverse(const_2)), const_1)) | a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 70 , how many coins of each type are there ? | "let number of each type of coin = x . then , 1 × x + . 50 × x + . 25 x = 70 ⇒ 1.75 x = 70 ⇒ x = 40 coins answer c" | a = 1/(4)
b = 1/(2)
c = a + b
d = c + 1
e = 70 / d
|
a ) 27 , b ) 21 , c ) 24 , d ) 22 , e ) 20 | b | subtract(subtract(subtract(26, 2), const_1), const_1) | how many positive integers less than 26 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ? | 9 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } 3 odd multiples of 5 : { 5 , 15 , 25 } 10 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 } notice , that 5 is in two sets , thus total # of integers satisfying the given conditions is 9 + 3 + 10 - 1 = 21 . answer : b . | a = 26 - 2
b = a - 1
c = b - 1
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a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | d | subtract(50, multiply(divide(50, const_100), 5)) | how many liters of water must be evaporated from 50 liters of a 4 percent sugar solution to get a 5 percent sugar solution ? | "let x be the amount that needs to be evaporated . 0.04 ( 50 ) = 0.05 ( 50 - x ) 0.05 x = 2.5 - 2 x = 0.5 / 0.05 = 10 liters the answer is d ." | a = 50 / 100
b = a * 5
c = 50 - b
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a ) 800 , b ) 1000 , c ) 1200 , d ) 1400 , e ) 1600 | b | divide(520, subtract(const_1, divide(multiply(6, 8), const_100))) | a person lent a certain sum of money at 6 % per annum at simple interest and in 8 years the interest amounted to $ 520 less than the sum lent . what was the sum lent ? | "p - 520 = ( p * 6 * 8 ) / 100 p = 1000 the answer is b ." | a = 6 * 8
b = a / 100
c = 1 - b
d = 520 / c
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a ) 10 , b ) 30 , c ) 435 , d ) 60 , e ) 90 | c | divide(multiply(30, subtract(30, const_1)), const_2) | there are 30 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "30 players are there . two players play one game with one another . so 30 c 2 = 30 * 29 / 2 = 435 so option c is correct" | a = 30 - 1
b = 30 * a
c = b / 2
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | a | divide(multiply(subtract(4.00, 1.50), 100), 1.50) | a wholesaler wishes to sell 100 pounds of mixed nuts at $ 1.50 a pound . she mixes peanuts worth $ 1.50 a pound with cashews worth $ 4.00 a pound . how many pounds of cashews must she use ? | "from the question stem we know that we need a mixture of 100 pounds of peanuts and cashews . if we represent peanuts as x and cashews as y , we get x + y = 100 . since the wholesaler wants to sell the mixture of 100 pounds @ $ 2.50 , we can write this as : $ 1.5 * ( x + y ) = $ 1.5 x + $ 4 y from the equation x + y = 100 , we can rewrite y as y = 100 - x and substitute this into our equation to get : $ 2.5 * ( x + 100 - x ) = $ 1.5 x + $ 4 ( 100 - x ) if you solve for x , you will get x = 60 , and therefore y = 40 . so the wholesaler must use 40 pounds of cashews . you can substitute into the original equation to see that : $ 250 = $ 1.5 ( 60 ) + $ 4 ( 40 ) answer is a ." | a = 4 - 0
b = a * 100
c = b / 1
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | e | divide(90, add(const_4, divide(const_2, const_2))) | a certain number of horses and an equal number of men are going somewhere . half of the owners are on their horses ' back while the remaining ones are walking along leading their horses . if the number of legs walking on the ground is 90 , how many horses are there ? | "legs 18 * 4 = 72 now half on their horses so remaining on the walk so 9 men 9 men has 18 legs so , 18 + 72 = 90 legs walking answer : e" | a = 2 / 2
b = 4 + a
c = 90 / b
|
a ) 200 , b ) 210 , c ) 180 , d ) 288 , e ) 220 | d | add(240, multiply(240, divide(20, const_100))) | 240 is increased by 20 % . find the final number . | explanation final number = initial number + 20 % ( original number ) = 240 + 20 % ( 240 ) = 240 + 48 = 288 . answer d | a = 20 / 100
b = 240 * a
c = 240 + b
|
a ) 25630 yards , b ) 35200 yards , c ) 39520 yards , d ) 42560 yards , e ) 41520 yards | b | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 20), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 20 miles into yards ? | "1 mile = 1760 yards 20 miles = 20 * 1760 = 35200 yards answer is b" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 20
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 18 % , b ) 20 % , c ) 25 % , d ) 23 % , e ) can not be determined | c | add(divide(multiply(10, 4), 2), add(10, 5)) | two kinds of vodka are mixed in the ratio 1 : 2 and 2 : 1 and they are sold fetching the profit 10 % and 25 % 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 : c . | a = 10 * 4
b = a / 2
c = 10 + 5
d = b + c
|
a ) 5 , b ) 5.5 , c ) 6 , d ) 6.5 , e ) 7 | d | divide(subtract(multiply(10, 11), multiply(add(const_4, const_1), 9)), 10) | the average of 10 consecutive integers is 11 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ? | "the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is d ." | a = 10 * 11
b = 4 + 1
c = b * 9
d = a - c
e = d / 10
|
a ) 14 , b ) 20 , c ) 26 , d ) 28 , e ) 30 | a | divide(196, subtract(15, const_1)) | let the number which when multiplied by 15 is increased by 196 . | "solution let the number be x . then , 15 x - x = 196 ‹ = › 14 x = 196 x ‹ = › 14 . answer a" | a = 15 - 1
b = 196 / a
|
a ) 36 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | a | divide(subtract(multiply(multiply(3, 4), 2), multiply(3, 4)), 2) | running at their respective constant rates , machine x takes 2 days longer to produce w widgets than machine y . at these rates , if the two machines together produce 5 / 4 w widgets in 3 days , how many days would it take machine x alone to produce 6 w widgets ? | "let y produce w widgets in y days hence , in 1 day y will produce w / y widgets . also , x will produce w widgets in y + 2 days ( given , x takes two more days ) hence , in 1 day x will produce w / y + 2 widgets . hence together x and y in 1 day will produce { w / y + w / y + 2 } widgets . together x and y in 3 days will produce = 3 * [ { w / y + w / y + 2 } ] widgets . it is given that in 3 days together they produce ( 5 / 4 ) w widgets . equating , 3 * [ { w / y + w / y + 2 } ] = ( 5 / 4 ) w take out w common and move 3 to denominator of rhs w { 1 / y + 1 / ( y + 2 ) } = ( 5 / 12 ) w canceling w from both sides { 1 / y + 1 / ( y + 2 ) } = 5 / 12 2 y + 2 / y ( y + 2 ) = 5 / 12 24 y + 24 = 5 y ^ 2 + 10 y 5 y ^ 2 - 14 y - 24 = 0 5 y ^ 2 - 20 y + 6 y - 24 = 0 5 y ( y - 4 ) + 6 ( y - 4 ) = 0 ( 5 y + 6 ) + ( y - 4 ) = 0 y = - 6 / 5 or y = 4 discarding y = - 6 / 5 as no of days can not be negative y = 4 hence it takes y , 4 days to produce w widgets . therefore , it will take x ( 4 + 2 ) = 6 days to produce w widgets . hence it will take x 6 * 6 = 36 days to produce 2 w widgets . answer : a" | a = 3 * 4
b = a * 2
c = 3 * 4
d = b - c
e = d / 2
|
['a ) 46', 'b ) 81', 'c ) 126', 'd ) 252', 'e ) none of these'] | c | add(add(add(add(37, 37), 37), const_10), const_4) | a rectangular parking space is marked out by painting three of its sides . if the length of the unpainted side is 9 feet , and the sum of the lengths of the painted sides is 37 feet , then what is the area of the parking space in square feet ? | solution clearly , we have : l = 9 and l + 2 b = 37 or b = 14 . ∴ area = ( l × b ) = ( 9 × 14 ) sq . ft . = 126 sq . ft . answer c | a = 37 + 37
b = a + 37
c = b + 10
d = c + 4
|
a ) 0.20833 , b ) 0.14544 , c ) 0.21978 , d ) 0.35466 , e ) 0.63435 | c | multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100) | lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 64,000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is : | "total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 364000 x \ therefore required percentage = \ inline \ frac { 800 x } { 364000 } \ times 100 = 0.21978 c" | a = 60 / 100
b = 480 / a
c = 3 * 100
d = c * 1000
e = 4 * 10
f = e + 4
g = f * 1000
h = d + g
i = b / h
j = i * 100
|
a ) 3 / 7 , b ) 34 , c ) 10 / 7 , d ) 2 , e ) 3 | c | add(divide(6, 7), divide(subtract(2, divide(4, 5)), add(2, divide(4, 5)))) | if p / q = 4 / 5 , then the value of 6 / 7 + { ( 2 q - p ) / ( 2 q + p ) } is ? | "answer given exp . = 6 / 7 + { ( 2 q - p ) / ( 2 q + p ) } dividing numerator as well as denominator by q , exp = 6 / 7 + { 2 - p / q ) / ( 2 + p / q ) } = 6 / 7 + { ( 2 - 4 / 5 ) / ( 2 + 4 / 5 ) } = 6 / 7 + 6 / 14 = 6 / 7 + 3 / 7 = 10 / 7 correct option : c" | a = 6 / 7
b = 4 / 5
c = 2 - b
d = 4 / 5
e = 2 + d
f = c / e
g = a + f
|
a ) 45.5 km , b ) 27.250 km , c ) 28.250 km , d ) 29.250 km , e ) 25.250 km | a | add(multiply(6, add(3, divide(const_2, const_4))), multiply(7, add(3, divide(const_2, const_4)))) | two persons start at the same point , walk in opposite directions with 6 km / hr and 7 km / hr respectively . what is the distance separated after 3 and half hrs ? | as the two persons are moving in the opposite direction , so they will be separateed in 1 hour = 6 + 7 = 13 km . they will be separated in 2.5 hours = 13 * 3.5 = 45.5 km answer : a | a = 2 / 4
b = 3 + a
c = 6 * b
d = 2 / 4
e = 3 + d
f = 7 * e
g = c + f
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | c | subtract(add(divide(add(add(86, 50), 53), const_3), const_1), 53) | on day one , a store sold 86 cups . on each of the next d days , the company sold 50 cups . if the average daily sales of cups over this time period ( including day one ) is 53 cups , what is the value of d ? | 86 + 50 d = 53 ( d + 1 ) . 3 d = 33 . d = 11 . the answer is c . | a = 86 + 50
b = a + 53
c = b / 3
d = c + 1
e = d - 53
|
a ) 75 , b ) 80 , c ) 85 , d ) 90 , e ) 95 | d | multiply(divide(subtract(subtract(power(multiply(8, divide(add(const_100, 25), const_100)), const_2), power(multiply(4, divide(subtract(const_100, 25), const_100)), const_2)), subtract(power(8, const_2), power(4, const_2))), subtract(power(8, const_2), power(4, const_2))), const_100) | there are two concentric circles with radii 8 and 4 . if the radius of the outer circle is increased by 25 % and the radius of the inner circle decreased by 25 % , by what percent does the area between the circles increase ? | "the area of a circle is pir ^ 2 , where r is the radius . the area of the big circle is 64 pi . the area of the small circle is 16 pi . the area a 1 between the circles is 48 pi . when the big circle ' s radius increases , the new area is 100 pi . when the small circle ' s radius decreases , the new area is 9 pi . the area a 2 between the circles is 91 pi . the ratio of a 2 / a 1 is 91 / 48 = 1.9 which is an increase of 90 % . the answer is d ." | a = 100 + 25
b = a / 100
c = 8 * b
d = c ** 2
e = 100 - 25
f = e / 100
g = 4 * f
h = g ** 2
i = d - h
j = 8 ** 2
k = 4 ** 2
l = j - k
m = i - l
n = 8 ** 2
o = 4 ** 2
p = n - o
q = m / p
r = q * 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | subtract(divide(60, 5), divide(60, 6)) | alice can bake a pie in 5 minutes . bob can bake a pie in 6 minutes . compute how many more pies alice can bake than bob in 60 minutes . | alice can bake 60 / 5 = 12 pies in 60 minutes . bob can bake 60 / 6 = 10 pies in 60 minutes . alice can therefore bake 12 - 10 = 2 more pies than bob . correct answer b | a = 60 / 5
b = 60 / 6
c = a - b
|
a ) rs . 5870 , b ) rs . 5991 , c ) rs . 6470 , d ) rs . 6850 , e ) none of these | c | subtract(multiply(add(5, const_1), 6100), add(add(add(add(5420, 5660), 6200), 6350), 6500)) | a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6100 ? | "explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6100 * 6 ) – 30,130 ] = rs . ( 36600 – 30,130 ) = rs . 6470 answer c" | a = 5 + 1
b = a * 6100
c = 5420 + 5660
d = c + 6200
e = d + 6350
f = e + 6500
g = b - f
|
a ) 17 / 216 , b ) 19 / 216 , c ) 23 / 216 , d ) 25 / 216 , e ) 35 / 216 | d | subtract(1, divide(const_2, 6)) | when a random experiment is conducted , the probability that event a occurs is 1 / 6 . if the random experiment is conducted 4 independent times , what is the probability that event a occurs exactly twice ? | "one case is : 1 / 6 * 1 / 6 * 5 / 6 * 5 / 6 = 25 / 1296 the total number of possible cases is 4 c 2 = 6 p ( event a occurs exactly twice ) = 6 * ( 25 / 1296 ) = 25 / 216 the answer is d ." | a = 2 / 6
b = 1 - a
|
a ) s . 14,000 , b ) s . 14,200 , c ) s . 4,400 , d ) s . 8,400 , e ) s . 4,800 | d | subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1) | a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35,000 , c receives : | "let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . c ' s share = rs . ( 35000 x 12 / 50 ) = rs . 8,400 . d" | a = 10 * 5000
b = a - 5000
c = 4000 + 5000
d = b - c
e = d / 3
f = 4000 + 5000
g = e + f
h = 10 * 5000
i = g / h
j = 3 + 4
k = j * 5000
l = i * k
m = l / 1000
n = math.floor(m)
o = n - 1
|
a ) 22 , b ) 23 , c ) 28 , d ) 29 , e ) 26 | e | divide(add(add(multiply(20, 35), multiply(15, 20)), multiply(5, 8)), add(add(20, 15), 5)) | a building contractor employs 20 male , 15 female and 5 child workers . to a male worker he pays rs . 35 per day , to a female worker rs . 20 per day and a child worker rs . 8 per day . the average wage per day paid by the contractor is ? | "20 15 5 35 20 8 500 + 300 + 40 = 1040 / 40 = 26 answer : e" | a = 20 * 35
b = 15 * 20
c = a + b
d = 5 * 8
e = c + d
f = 20 + 15
g = f + 5
h = e / g
|
a ) 7 , b ) 18 , c ) 34 , d ) 35 , e ) 36 | a | divide(log(divide(multiply(power(1, 18), multiply(power(10, 35), 2)), power(4, 18))), log(power(5, 5))) | ( ( 1 ^ 5 m ) / ( 5 ^ 5 m ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) what is m ? | ( ( 1 ^ 5 m ) / ( 5 ^ 5 m ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) ( ( 1 / 5 ) ^ 5 m ) * ( ( 1 / 2 ) ^ 36 ) = 1 / ( 2 * ( 2 * 5 ) ^ 35 ) ) 2 ^ 36 will cancel out , since 1 can be written as 1 ^ 35 , so ( 1 / 5 ) ^ 5 m = ( 1 / 5 ) ^ 35 ( ( 1 / 5 ) ^ 5 m ) * ( ( 1 / 2 ) ^ 36 ) = 1 / [ ( 2 ^ 36 ) * ( 5 ^ 35 ) ] so , m = 7 answer a | a = 1 ** 18
b = 10 ** 35
c = b * 2
d = a * c
e = 4 ** 18
f = d / e
g = math.log(f)
h = 5 ** 5
i = math.log(h)
j = g / i
|
a ) 6 . , b ) 2 . , c ) 3 . , d ) 5 . , e ) 4 . | d | divide(subtract(79, multiply(6, const_4)), 11) | eggs are sold in packages of 6 or 11 only . if doris bought 79 eggs exactly , what could be the number of large packs doris bought ? | no strategy involved . simple question demanding fast calculation . 11 x 5 = 55 = > 79 - 55 = 24 = > 24 / 6 is an integer ans d . 5 . good luck | a = 6 * 4
b = 79 - a
c = b / 11
|
a ) 82.1 sec , b ) 12.1 sec , c ) 16.1 sec , d ) 13.1 sec , e ) 16.6 sec | e | divide(add(200, 132), multiply(72, const_0_2778)) | how long does a train 200 m long running at the speed of 72 km / hr takes to cross a bridge 132 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 200 + 132 = 332 m . required time = 332 / 20 = 16.6 sec . answer : e" | a = 200 + 132
b = 72 * const_0_2778
c = a / b
|
a ) 6 , b ) 14 , c ) 16 , d ) 20 , e ) 18 | a | divide(add(125, 125), add(divide(125, 15), divide(125, 10))) | two tains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post . if the length of each train be 125 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | "sol . speed of the first train = [ 150 / 10 ] m / sec = 15 m / sec . speed of the second train = [ 150 / 15 ] m / sec = 10 m / sec . relative speed = ( 15 + 10 ) = m / sec = 25 m / sec . ∴ required time = ( 125 + 125 ) / 25 secc = 6 sec . answer a" | a = 125 + 125
b = 125 / 15
c = 125 / 10
d = b + c
e = a / d
|
a ) 306.06 , b ) 306.02 , c ) 306.04 , d ) 306.09 , e ) 306.12 | c | subtract(multiply(5000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 5000) | find out the c . i on rs . 5000 at 4 % p . a . compound half - yearly for 1 1 / 2 year ? | "a = 5000 ( 51 / 50 ) 3 = 5306.04 5000 - - - - - - - - - - - 306.04 answer : c" | a = 4 / 100
b = 1 + a
c = 3 / 2
d = b ** c
e = 5000 * d
f = e - 5000
|
a ) 50 days , b ) 100 days , c ) 120 days , d ) 110 days , e ) 90 days | a | inverse(subtract(inverse(20), inverse(multiply(inverse(subtract(const_1, multiply(inverse(20), 14))), 10)))) | micheal and adam can do together a piece of work in 20 days . after they have worked together for 14 days micheal stops and adam completes the remaining work in 10 days . in how many days micheal complete the work separately . | "rate of both = 1 / 20 together they do = 1 / 20 * 14 = 7 / 10 left work = 1 - 7 / 10 = 3 / 10 adam completes 3 / 10 work in 10 day so he took 10 * 10 / 3 = 100 / 3 days to complete the left work alone . thus the rate of adam is 3 / 100 rate of micheal = 1 / 20 - 3 / 100 = 1 / 50 thus micheal takes 50 days to complete the whole work . ans . a" | a = 1/(20)
b = 1/(20)
c = b * 14
d = 1 - c
e = 1/(d)
f = e * 10
g = 1/(f)
h = a - g
i = 1/(h)
|
a ) 17 / 30 , b ) 2 / 5 , c ) 7 / 15 , d ) 4 / 15 , e ) 11 / 30 | e | divide(add(floor(divide(30, 3)), divide(30, 17)), 30) | a number is selected at random from the first 30 natural numbers . what is the probability that the number is a multiple of either 3 or 17 ? | "number of multiples of 3 from 1 through 30 = 30 / 3 = 10 number of multiples of 17 from 1 through 30 = 30 / 17 = 1 number of multiples of 3 and 17 both from 1 through 30 = number of multiples of 17 * 3 ( = 51 ) = 0 total favourable cases = 10 + 1 - 0 = 11 probability = 11 / 30 answer : option e" | a = 30 / 3
b = math.floor(a)
c = 30 / 17
d = b + c
e = d / 30
|
a ) s . 416 , b ) s . 480 , c ) s . 429 , d ) s . 128 , e ) s . 419 | a | subtract(divide(multiply(800, const_100), add(35, const_100)), divide(multiply(divide(multiply(800, const_100), add(35, const_100)), 25), const_100)) | by selling an article at rs . 800 , a shopkeeper makes a profit of 35 % . at what price should he sell the article so as to make a loss of 25 % ? | "sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 35 % = 35 % of 640 = rs . 224 sp = cp - loss = 640 - 224 = rs . 416 answer : a" | a = 800 * 100
b = 35 + 100
c = a / b
d = 800 * 100
e = 35 + 100
f = d / e
g = f * 25
h = g / 100
i = c - h
|
a ) $ 900 , b ) $ 810 , c ) $ 915 , d ) $ 715 , e ) $ 729 | e | multiply(900, power(subtract(const_1, divide(10, const_100)), 2)) | a present value of a machine is $ 900 . its value depletion rate is 10 % per annum then find the machine value after 2 years ? | "p = $ 900 r = 10 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 900 * 9 / 10 * 9 / 10 = $ 729 answer is e" | a = 10 / 100
b = 1 - a
c = b ** 2
d = 900 * c
|
a ) 4 , b ) 8 , c ) 12 , d ) 24 , e ) 36 | e | multiply(multiply(6, 3), const_2) | a motorist knows 6 different routes from bristol to birmingham . from birmingham to sheffield he knows 3 different routes and from sheffield to carlisle he knows two different routes . how many routes does he know from bristol to carlisle ? | explanation : total number of routes from bristol to carlisle = ( 6 x 3 x 2 ) = 36 . answer : e | a = 6 * 3
b = a * 2
|
a ) 52 , b ) 54 , c ) 48 , d ) 58 , e ) 60 | c | add(multiply(divide(45, const_3), const_2), divide(54, const_3)) | the grade point average of one third of the classroom is 54 ; the grade point average of the rest is 45 . what is the grade point average of the whole class ? | "let n = total students in class total points for 1 / 3 class = 54 n / 3 = 18 n total points for 2 / 3 class = 45 * 2 n / 3 = 30 n total points for whole class = 18 n + 30 n = 48 n 48 n total class points / n total students = 48 grade point average for total class answer : c" | a = 45 / 3
b = a * 2
c = 54 / 3
d = b + c
|
a ) 2.5 , b ) 5 , c ) 2.4 , d ) 2.8 , e ) 2.1 | b | divide(100, multiply(72, const_0_2778)) | in what time will a train 100 m long cross an electric pole , it its speed be 72 km / hr ? | "speed = 72 * 5 / 18 = 20 m / sec time taken = 100 / 20 = 5 sec . answer : b" | a = 72 * const_0_2778
b = 100 / a
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | add(add(3, 3), 1) | if x is an integer that has exactly 3 positive divisors ( these include 1 and x ) , how many positive divisors does x ^ 3 have ? | since factors of ' 4 ' are { 1 , 2,4 } 4 ^ 3 = 64 number of factors / divisors of 64 = 2 ^ 6 we know that when a number is expressed as a product of the prime factors as below : n = a ^ x * b ^ y * c ^ z then no . of divisors = ( x + 1 ) * ( y + 1 ) * ( z + 1 ) then here ( 6 + 1 ) = 7 answer : d | a = 3 + 3
b = a + 1
|
a ) 1 / 39 , b ) 1 / 36 , c ) 1 / 32 , d ) 1 / 31 , e ) 1 / 30 | b | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | three 6 faced dice are thrown together . the probability that all the three show the same number on them is - | "explanation : it all 3 numbers have to be same basically we want triplets . 111 , 222 , 333 , 444 , 555 and 666 . those are six in number . further the three dice can fall in 6 * 6 * 6 = 216 ways . hence the probability is 6 / 216 = 1 / 36 answer : b" | a = 1 / 6
b = 1 / 6
c = a * b
d = 1 / 6
e = c * d
f = 1 / 6
g = e * f
|
a ) 5600 , b ) 6000 , c ) 2909 , d ) 7200 , e ) 8600 | c | divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 2), multiply(multiply(5, 11), 6)) | how many bricks , each measuring 5 cm x 11 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 2 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 2 ) / ( 5 x 11 x 6 ) = 2909 . answer : option c" | a = 8 * 100
b = 6 * 100
c = a * b
d = c * 2
e = 5 * 11
f = e * 6
g = d / f
|
a ) 6 / 11 , b ) 8 / 13 , c ) 3 / 13 , d ) 9 / 11 , e ) 10 / 13 | c | divide(const_1, add(divide(const_1, 7), divide(const_1, 6))) | a can do a piece of work in 7 days and b can do it in 6 days how long will they both work together to complete the work ? | "explanation : a ’ s one day work = 1 / 5 b ’ s one day work = 1 / 6 ( a + b ) ’ s one day work = 1 / 7 + 1 / 6 = 13 / 42 = > time = 42 / 13 = 3 3 / 13 days answer : option c" | a = 1 / 7
b = 1 / 6
c = a + b
d = 1 / c
|
a ) 38400 , b ) 24000 , c ) 24936 , d ) 25640 , e ) none | a | multiply(divide(const_100, 60), 23040) | 60 % of the population of a village is 23040 . the total population of the village is ? | answer ∵ 60 % of p = 23040 ∴ p = ( 23040 x 100 ) / 60 = 38400 correct option : a | a = 100 / 60
b = a * 23040
|
a ) 96 , b ) 75 , c ) 36 , d ) 25 , e ) 12 | c | divide(9, subtract(96.25, floor(96.25))) | when positive integer x is divided by positive integer y , the remainder is 9 . if x / y = 96.25 , what is the value of y ? | "guys , one more simple funda . 5 / 2 = 2.5 now . 5 x 2 = 1 is the remainder 25 / 4 = 6.25 now . 25 x 4 = 1 is the remainder 32 / 5 = 6.4 now . 4 x 5 = 2 is the remainder given x / y = 96.25 and remainder is 9 so . 25 x y = 9 hence y = 36 ans c" | a = math.floor(96, 25)
b = 96 - 25
c = 9 / b
|
a ) 2 / 3 , b ) 3 / 4 , c ) 1 / 2 , d ) 2 / 8 , e ) none of these | a | multiply(divide(378, 252), const_100) | 378 × ? = 252 | "explanation : 378 × ? = 252 ? = 252 / 378 = 126 / 189 = 14 / 21 = 2 / 3 answer is a" | a = 378 / 252
b = a * 100
|
a ) 75 , b ) 65 , c ) 85 , d ) 40 , e ) 80 | d | add(multiply(8, 2.5), 20) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 20 kg . what is the weight of the new person ? | total increase in weight = 8 × 2.5 = 20 if x is the weight of the new person , total increase in weight = x − 20 = > 20 = x - 20 = > x = 20 + 20 = 40 answer is d . | a = 8 * 2
b = a + 20
|
a ) 50 mph , b ) 45 mph , c ) 48 mph , d ) 52 mph , e ) 40 mph | e | add(20, multiply(5, const_4)) | liam is pulled over for speeding just as he is arriving at work . he explains to the police officer that he could not afford to be late today , and has arrived at work only four minutes before he is to start . the officer explains that if liam had driven 5 mph slower for his whole commute , he would have arrived at work exactly on time . if liam ' s commute is 20 miles long , how fast was he actually driving ? ( assume that liam drove at a constant speed for the duration of his commute . ) | "let t be the number of hours he would need to reach office on time . when he is driving with over speed , he reached office 4 min earlier ! so the equation for this is s ( t - 4 / 60 ) = 20 where s is the speed and 20 is the distance . if he decreases his speed by 5 mph then he would have reached his office on time : ( s - 5 ) t = 20 if you solve above equations , you will arrive at t = 2 / 3 hr and s = 40 mph therefore answer is e" | a = 5 * 4
b = 20 + a
|
a ) s . 528 , b ) s . 542 , c ) s . 780 , d ) s . 540 , e ) s . 549 | c | multiply(4, divide(1690, add(add(4, 2), const_3))) | rs . 1690 is divided so that 4 times the first share , thrice the 2 nd share and twice the third share amount to the same . what is the value of the third share ? | "a + b + c = 1690 4 a = 3 b = 2 c = x a : b : c = 1 / 4 : 1 / 3 : 1 / 2 = 3 : 4 : 6 6 / 13 * 1690 = rs . 780 answer : c" | a = 4 + 2
b = a + 3
c = 1690 / b
d = 4 * c
|
a ) 6 , b ) 2 , c ) 5 , d ) 0 , e ) 9 | d | add(const_2, const_3) | if 41 / 88 = 0.46590 , what is the 77 th digit to the right of the decimal point of the fraction ? | "we are not concerned what 41 / 88 means . . we have to look at the decimal . . 0.6590 means 0.465909090 . . . . so leaving girst , second and third digit to the right of decimal , all odd numbered are 0 and all even numbered are 9 . . here 77 is odd , so ans is 0 answer is d" | a = 2 + 3
|
a ) 300 , b ) 400 , c ) 500 , d ) 600 , e ) 700 | c | multiply(divide(multiply(100, const_1000), const_3600), 18) | a train running at the speed of 100 km / hr crosses a pole in 18 seconds . what is the length of the train ? | speed = ( 100 * 5 / 18 ) m / sec = ( 250 / 9 ) m / sec length of the train = ( speed x time ) = ( 250 / 9 * 18 ) m = 500 m . answer : c | a = 100 * 1000
b = a / 3600
c = b * 18
|
a ) 12 % , b ) 15 % , c ) 18 % , d ) 21 % , e ) 24 % | d | multiply(subtract(const_1, divide(const_100, add(add(const_100, 15), divide(multiply(add(const_100, 15), 10), const_100)))), const_100) | there has been successive increases of 15 % and then 10 % in the price of gas from the previous month . by what percentage should a driver reduce gas consumption so that the expenditure does not change ? | "let p be the original price per unit of gas . let x be the original gas consumption . let y be the reduced gas consumption . y * 1.1 * 1.15 * p = x * p y = x / ( 1.1 * 1.15 ) which is about 0.79 x which is a decrease of about 21 % . the answer is d ." | a = 100 + 15
b = 100 + 15
c = b * 10
d = c / 100
e = a + d
f = 100 / e
g = 1 - f
h = g * 100
|
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | a | multiply(subtract(add(30, const_1), 30), const_10) | two numbers are less than a third number by 30 % and 37 % . how much percent is the second number is less than the first ? | let the third number be x first number = 70 % of x = 7 x / 10 ; second number = 63 % of x = 63 x / 100 ; difference = 7 x / 10 - 63 x / 100 = 7 x / 100 ; required percentage = 7 x / 100 * 10 / 7 x * 100 = 10 % answer is a | a = 30 + 1
b = a - 30
c = b * 10
|
a ) 600 , b ) 900 , c ) 1200 , d ) 1350 , e ) 1500 | e | multiply(15, const_100) | there are 300 seniors at morse high school , and 40 % of them have cars . of the remaining grades ( freshmen , sophomores , and juniors ) , only 10 % of them have cars . if 15 % of all the students at morse have cars , how many students are in those other 3 lower grades ? | let the total no of students be n . the no of seniors having cars is 40 % of 300 i . e . 120 . the rest of the students with other three grades who have cars is 10 % of ( n - 300 ) . the total no of students with cars is 15 % of n . thus , 0.15 n = 120 + 0.1 ( n - 300 ) on solving this , we get n = 1800 . hence , the no of students with other three grades is 1800 - 300 , i . e . 1500 ( e ) . | a = 15 * 100
|
a ) a ) 20 , b ) b ) 38 , c ) c ) 90 , d ) d ) 88 , e ) e ) 37 | a | divide(add(280, 60), add(6, 1)) | in an examination , a student scores 6 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 280 marks , the no of questions he attempts correctly is : | "let the number of correct answers be x . number of incorrect answers = ( 60 – x ) . 6 x – ( 60 – x ) = 280 = > 7 x = 340 = > x = 20 answer : a" | a = 280 + 60
b = 6 + 1
c = a / b
|
a ) 17 , b ) 18 , c ) 50 , d ) 108 , e ) 864 | c | divide(multiply(multiply(90, 45), 9), volume_cube(divide(9, const_2))) | a box measuring 90 inches long by 45 inches wide by 9 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ? | "least number of cubes will be required when the cubes that could fit in are biggest . 9 is the biggest number that could divide all three , 90 , 45 and 9 . thus side of cube must be 9 , and total number of cubes = 90 / 9 * 45 / 9 * 9 / 9 = 50 ans c it is ." | a = 90 * 45
b = a * 9
c = 9 / 2
d = b / volume_cube
|
a ) 12 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | a | multiply(subtract(subtract(subtract(11, const_4), const_4), 1), const_3) | you collect baseball cards . suppose you start out with 11 . maria takes half of one more than the number of baseball cards you have . since you ' re nice , you give peter 1 baseball card . since his father makes baseball cards , paul decides to triple your baseball cards . how many baseball cards do you have at the end ? | "solution start with 11 baseball cards . maria takes half of one more than the number of baseball cards you have . so maria takes half of 11 + 1 which is 6 , so you ' re left with 11 - 6 = 5 . peter takes 1 baseball card from you : 5 - 1 = 4 baseball cards . paul triples the number of baseball cards you have : 4 ã — 3 = 12 baseball cards . so you have 12 at the end . correct answer : a" | a = 11 - 4
b = a - 4
c = b - 1
d = c * 3
|
a ) 23 sec , b ) 13 sec , c ) 12 sec , d ) 11 sec , e ) 16 sec | b | divide(130, divide(multiply(add(50, 10), const_1000), const_3600)) | a bullet train 130 m long is running with a speed of 50 kmph . in what time will it pass a man who is running at 10 kmph in the direction opposite to that in which the bullet train is going ? | "b 13 sec speed of the bullet train relative to man = ( 50 + 10 ) kmph = 60 * 5 / 18 m / sec = 30 / 3 m / sec . time taken by the bullet train to cross the man = time taken by it to cover 130 m at ( 30 / 3 ) m / sec = ( 130 * 3 / 30 ) sec = 13 sec" | a = 50 + 10
b = a * 1000
c = b / 3600
d = 130 / c
|
a ) 10101 , b ) 11000 , c ) 10110 , d ) 10111 , e ) 10100 | e | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 33,66) | find the smallest number of five digits exactly divisible by 22 , 33,66 and 44 . | "smallest number of five digits is 10000 . required number must be divisible by l . c . m . of 22,33 , 66,44 i . e 132 , on dividing 10000 by 132 , we get 32 as remainder . therefore , required number = 10000 + ( 132 â € “ 32 ) = 10100 . answer is e ." | a = 100 * 100
b = 10 * a
c = b - 100
d = c + 33
|
a ) 16 , b ) 20 , c ) 24 , d ) 25 , e ) 28 | b | multiply(const_60, divide(multiply(divide(5, const_60), subtract(25, 5)), 5)) | a man walking at a constant rate of 5 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 25 miles per hour . the woman stops to wait for the man 5 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ? | "when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 5 * 25 / 60 = 25 / 12 miles and the man walks the distance mm , which is 5 * 5 / 60 = 5 / 12 mile . the difference of 25 / 12 - 5 / 12 = 1 / 3 miles ( mw ) will be covered by the man in ( 4 / 3 ) / 4 = 1 / 3 of an hour , which is 20 minutes . answer b ." | a = 5 / const_60
b = 25 - 5
c = a * b
d = c / 5
e = const_60 * d
|
a ) 20,20 , b ) 20,10 , c ) 25,15 , d ) 30,10 , e ) none of these | d | subtract(add(divide(multiply(20, 5), subtract(5, const_1)), 5), 20) | the ages of two persons differ by 20 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | explanation : let their ages be x and ( x + 20 ) years . 5 ( x - 5 ) = ( x + 20 - 5 ) or 4 x = 40 or x = 10 . their present ages are 30 years and 10 years option d | a = 20 * 5
b = 5 - 1
c = a / b
d = c + 5
e = d - 20
|
a ) $ 5 , b ) $ 10 , c ) $ 15 , d ) $ 20 , e ) $ 25 | a | subtract(divide(300, divide(10, subtract(divide(const_3, const_2), const_1))), divide(300, add(divide(10, subtract(divide(const_3, const_2), const_1)), 10))) | p and q are the only two applicants qualified for a short - term research project that pays 300 dollars in total . candidate p has more experience and , if hired , would be paid 50 percent more per hour than candidate q would be paid . candidate q , if hired , would require 10 hours more than candidate p to do the job . candidate p ’ s hourly wage is how many dollars greater than candidate q ’ s hourly wage ? | "let q ' s hourly wage be x , then p ' s hourly wage is 1.5 x let t be the number of hours that q needs , then p needs t - 10 hours to do the job . since they both are paid an equal total amount of $ 300 : x * t = 1.5 x * ( t - 10 ) t = 30 hours and q ' s hourly wage is 300 / 30 = $ 10 p ' s hourly wage is 300 / ( t - 10 ) = $ 15 which is $ 5 per hour more . the answer is a ." | a = 3 / 2
b = a - 1
c = 10 / b
d = 300 / c
e = 3 / 2
f = e - 1
g = 10 / f
h = g + 10
i = 300 / h
j = d - i
|
['a ) 4', 'b ) 6', 'c ) 12', 'd ) 24', 'e ) 36'] | e | multiply(12, const_3) | the average length of the sides of triangle abc is 12 . what is the perimeter of triangle abc ? | ( average ) = ( perimeter ) / 3 ; 12 = ( perimeter ) / 3 ; ( perimeter ) = 36 . answer : e . | a = 12 * 3
|
a ) 105 , b ) 155 , c ) 125 , d ) 185 , e ) 285 | e | multiply(const_100, divide(subtract(subtract(1, divide(1, const_100)), divide(36, 140)), divide(36, 140))) | a retailer buys 140 pens at the market price of 36 pens from a wholesaler , if he sells these pens giving a discount of 1 % , what is the profit % ? | "let the market price of each pen be $ 1 then , cost price of 140 pens = $ 36 selling price of 140 pens = 99 % of $ 140 = $ 138.60 profit % = ( ( 102.60 * 100 ) / 36 ) % = 285 % answer e" | a = 1 / 100
b = 1 - a
c = 36 / 140
d = b - c
e = 36 / 140
f = d / e
g = 100 * f
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | power(multiply(4, power(5, 2)), 3) | if 3 log ( 4 * 5 ^ 2 ) = x , find x | "3 ( log 2 ^ 2 * 5 ^ 2 ) = x 3 log ( 5 * 2 ) ^ 2 = x 3 * 2 log ( 5 * 2 ) = x 6 log 10 = x log 10 base 10 = 1 so 6 * 1 = x x = 6 answer : b" | a = 5 ** 2
b = 4 * a
c = b ** 3
|
a ) 11 , b ) 9 , c ) 13 , d ) 14 , e ) 12.8 | e | divide(multiply(120, const_2), add(speed(120, 18), speed(120, 10))) | two trains of equal lengths take 10 sec and 18 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 / 18 = 6.7 m / sec . relative speed = 12 + 6.7 = 18.7 m / sec . required time = ( 120 + 120 ) / 18.7 = 12.8 sec . answer : option e" | a = 120 * 2
b = speed + (
c = a / b
|
a ) 35 , b ) 25 , c ) 45 , d ) 54 , e ) 52 | d | subtract(add(multiply(15, 3), const_12), const_3) | if one - third of one fourth of a number is 15 , then 3 - tenth of that number is : | let the number be x . then , 1 / 3 of 1 / 4 of x = 15 x = 15 * 12 = 180 so , required number = ( 3 / 10 * 180 ) = 54 . answer : d | a = 15 * 3
b = a + 12
c = b - 3
|
a ) none , b ) two , c ) four , d ) five , e ) seven | e | subtract(const_4, const_1) | r is the set of positive even integers less than 201 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "r is the set of positive even integers less than 201 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 2,4 , 6,8 , 10,12 . . . s = 4,16 , 36,64 . . . numbers : 4 , 16 , 36 , 64 , 100 , 144 , and 196 are even integers ( less than 201 ) that are in both sets . solution : seven answer : e" | a = 4 - 1
|
a ) 2 , b ) 7 , c ) 6 , d ) 8 , e ) 1 | c | subtract(multiply(multiply(multiply(722, 774), 889), 223), subtract(multiply(multiply(multiply(722, 774), 889), 223), add(const_4, const_4))) | the unit digit in the product ( 722 * 774 * 889 * 223 ) is : | "explanation : unit digit in the given product = unit digit in ( 2 * 4 * 9 * 3 ) = 6 answer : c" | a = 722 * 774
b = a * 889
c = b * 223
d = 722 * 774
e = d * 889
f = e * 223
g = 4 + 4
h = f - g
i = c - h
|
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