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 ) 15 , b ) 30 , c ) 120 , d ) 240 , e ) 256 | c | divide(factorial(16), multiply(factorial(subtract(16, const_2)), factorial(const_2))) | there are 16 teams in a tournament . if during the first round , each team plays every other team exactly once , how many games will be played in the first round ? | number of ways to pick 2 teams from a total of 16 = 16 c 2 = 16 ! / 14 ! * 2 ! = 120 . hence c . | a = math.factorial(16)
b = 16 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 72 , b ) 85 , c ) 110 , d ) 98 , e ) 108 | c | subtract(add(subtract(193, 59), 25), 49) | there are 193 items that are members of set u . of these items , 49 are members of set b , 59 are not members of either of set a or set b , and 25 are members of both sets a and b . how many of the members of set u are members of set a ? | you had the answer almost right . the x = 85 refers to only set a . however what ' s being asked is how many members are part of set a . this will include : 1 . only set a 2 . set a and set b so the answer is set a = 85 + set ab = 85 + 25 = 110 c | a = 193 - 59
b = a + 25
c = b - 49
|
a ) 2 / 9 , b ) 4 / 17 , c ) 6 / 29 , d ) 8 / 35 , e ) 10 / 43 | b | multiply(divide(add(const_4, const_4), 18), divide(subtract(add(const_4, const_4), 1), subtract(18, 1))) | 18 balls are numbered 1 to 18 . a ball is drawn and then another ball is drawn without replacement . what is the probability that both balls have even numbers ? | "p ( 1 st ball is even ) = 9 / 18 p ( 2 nd ball is also even ) = 8 / 17 p ( both balls are even ) = 9 / 18 * 8 / 17 = 4 / 17 the answer is b ." | a = 4 + 4
b = a / 18
c = 4 + 4
d = c - 1
e = 18 - 1
f = d / e
g = b * f
|
a ) 1 / 144 , b ) 1 / 36 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 49 | e | multiply(divide(7, power(7, const_2)), divide(7, power(7, const_2))) | what is the probability that the sum of two dice will yield a 8 , and then when both are thrown again , their sum will again yield a 8 ? assume that each die has 7 sides with faces numbered 1 to 7 . | "solution - rolling dices is an independent event . the combinations to get 8 are ( 1,7 ) , ( 7,1 ) , ( 2,6 ) , ( 6,2 ) , ( 3,5 ) , ( 5,3 ) , ( 4 , , 4 ) and total combinations of both dices is 49 . the probability of getting 8 in first attempt is 7 / 49 = 1 / 7 . probability of getting 8 again in second attempt = ( 1 / 7 ) * ( 1 / 7 ) = 1 / 49 . ans e" | a = 7 ** 2
b = 7 / a
c = 7 ** 2
d = 7 / c
e = b * d
|
a ) 270 , b ) 277 , c ) 187 , d ) 257 , e ) 150 | e | multiply(divide(subtract(72, 36), const_3_6), 15) | two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 15 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 15 sec = 15 * 10 = 150 m . the length of the faster train = 150 m . answer : e" | a = 72 - 36
b = a / const_3_6
c = b * 15
|
a ) 1.125 , b ) 1.128 , c ) 1.126 , d ) 1.123 , e ) 1.121 | a | divide(add(divide(45, 12), divide(72, 12)), const_2) | a man swims downstream 72 km and upstream 45 km taking 12 hours each time ; what is the speed of the current ? | "72 - - - 12 ds = 6 ? - - - - 1 45 - - - - 12 us = 3.75 ? - - - - 1 s = ? s = ( 6 - 3.75 ) / 2 = 1.125 answer : a" | a = 45 / 12
b = 72 / 12
c = a + b
d = c / 2
|
a ) s . 768 , b ) s . 329 , c ) s . 595 , d ) s . 600 , e ) s . 725 | c | multiply(850, divide(10, const_100)) | find the simple interest on rs . 850 for 10 months at 7 paisa per month ? | "explanation : i = ( 850 * 10 * 7 ) / 100 = 595 answer : option c" | a = 10 / 100
b = 850 * a
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a ) $ 38 , b ) $ 18 , c ) $ 99 , d ) $ 27 , e ) $ 20 | e | divide(add(1480, 920), add(65, 55)) | sandy bought 65 books for $ 1480 from one shop and 55 books for $ 920 from another shop . what is the average price that sandy paid per book ? | "average price per book = ( 1480 + 920 ) / ( 65 + 55 ) = 2400 / 120 = $ 20 the answer is e ." | a = 1480 + 920
b = 65 + 55
c = a / b
|
a ) 3 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | d | divide(divide(divide(120, 4), const_2), const_3) | how many of the positive divisors d of 120 are also multiples of 4 not including 120 ? | "4 , 8,12 , 20,24 , 40,60 . ( 7 ) is the answer other way : factors of 120 = 2 ^ 3 * 3 * 5 separate 2 ^ 2 ( which means 4 ) now , calculate the number of other factors . d = 2 * 3 * 5 = total positive factors are 2 * 2 * 2 = 8 this 8 factors include 120 so subtract 1 from 8 ans is 7 = d" | a = 120 / 4
b = a / 2
c = b / 3
|
a ) 1601249 , b ) 1607249 , c ) 1701249 , d ) 1708249 , e ) none | d | multiply(1307, 1307) | calculate 1307 x 1307 = ? | "answer 1307 x 1307 = ( 1307 ) 2 = ( 1300 + 7 ) 2 = ( 1300 ) 2 + ( 7 ) 2 + 2 x 1300 x 7 = 1690000 + 49 + 18200 = 1708249 option : d" | a = 1307 * 1307
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a ) 1 / 35 , b ) 3 / 35 , c ) 5 / 37 , d ) 4 / 63 , e ) none | d | divide(80, multiply(6, 210)) | the hcf and lcm of two numbers m and n are respectively 6 and 210 . if m + n = 80 , then 1 / m + 1 / n is equal to | "answer we have , m x n = 6 x 210 = 1260 â ˆ ´ 1 / m + 1 / n = ( m + n ) / mn = 80 / 1260 = 4 / 63 correct option : d" | a = 6 * 210
b = 80 / a
|
a ) $ 0.04 , b ) $ 0.12 , c ) $ 0.05 , d ) $ 0.03 , e ) $ 0.002 | a | subtract(divide(2.99, 130), divide(6.30, 100)) | bottle r contains 100 capsules and costs $ 6.30 . bottle t contains 130 capsules and costs $ 2.99 . what is the difference between the cost per capsule for bottle r and the cost per capsule for bottle t ? | "cost per capsule in r is 6.30 / 100 = 0.063 cost per capsule in t is 2.99 / 130 = 0.023 the difference is 0.004 the answer is a" | a = 2 / 99
b = 6 / 30
c = a - b
|
a ) 140 , b ) 146 , c ) 152 , d ) 162 , e ) 183 | e | multiply(multiply(multiply(250, subtract(1, divide(1, 10))), subtract(1, divide(1, 10))), subtract(1, divide(1, 10))) | in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 250 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ? | "the probability of survival for each of the first 3 months of life is 1 - 1 / 10 = 9 / 10 , so of 250 newborn 250 * 9 / 10 * 9 / 10 * 9 / 10 = ~ 183 is expected to survive . answer : e ." | a = 1 / 10
b = 1 - a
c = 250 * b
d = 1 / 10
e = 1 - d
f = c * e
g = 1 / 10
h = 1 - g
i = f * h
|
a ) 64 / 27 , b ) 9 / 16 , c ) 1 , d ) 6 / 8 , e ) 8 / 6 | c | divide(multiply(3, 4), multiply(4, 3)) | if 3 a = 4 b and ab ≠0 , what is the ratio of a / 4 to b / 3 ? | "a nice fast approach is the first find a pair of numbers that satisfy the given equation : 3 a = 4 b here ' s one pair : a = 4 and b = 3 what is the ratio of a / 9 to b / 8 ? in other words , what is the value of ( a / 4 ) / ( b / 3 ) ? plug in values to get : ( a / 4 ) / ( b / 3 ) = ( 4 / 4 ) / ( 3 / 3 ) = 1 / 1 = 1 c" | a = 3 * 4
b = 4 * 3
c = a / b
|
a ) x = 53186760 , b ) x = 26593379 , c ) x = 73260 , d ) x = 36629 , e ) x = 6666 | d | divide(subtract(lcm(lcm(lcm(33, 44), 55), 666), 2), 2) | what is the smallest integer x such that 2 x + 2 is divisible by 33 , 44 , 55 , and 666 ? | 33 = 3 * 11 44 = 2 * 2 * 11 55 = 5 * 11 666 = 2 * 3 * 3 * 37 so 2 x + 2 = 2 * 2 * 3 * 3 * 5 * 11 * 37 = 73260 , and x = ( 73260 - 2 ) / 2 = 36629 , which is option d . | a = math.lcm(33, 44)
b = math.lcm(a, 55)
c = math.lcm(b, 666)
d = c - 2
e = d / 2
|
a ) 18 days , b ) 38 days , c ) 42 days , d ) 48 days , e ) 44 days | d | divide(multiply(12, 80), 20) | if 12 men do a work in 80 days , in how many days will 20 men do it ? | "12 * 80 = 20 * x x = 48 days answer : d" | a = 12 * 80
b = a / 20
|
a ) 423762 , b ) 433758 , c ) 433298 , d ) 433998 , e ) 437382 | d | multiply(divide(5358, 81), const_100) | 5358 x 81 = ? | "d 273258 5358 x 81 = 5358 x ( 80 + 1 ) = 5358 x 80 + 5358 x 1 = 428640 + 5358 = 433998" | a = 5358 / 81
b = a * 100
|
a ) 3 , b ) 1 / 3 , c ) 2 , d ) 5 , e ) 1 / 5 | d | power(subtract(126, const_1), divide(const_1, 3)) | if the ratio of the sum of the first 6 terms of a g . p . to the sum of the first 3 terms of the g . p . is 126 , what is the common ratio of the g . p ? | "126 = ( a 1 + a 2 + a 3 + a 4 + a 5 + a 6 ) / ( a 1 + a 2 + a 3 ) factorize the same terms 126 = 1 + ( a 4 + a 5 + a 6 ) / ( a 1 + a 2 + a 3 ) write every term with respect to r a 1 = a 1 a 2 = a 1 * r ^ 1 a 3 = a 1 * r ^ 2 . . . . . . . . . 126 = 1 + ( a 1 ( r ^ 3 + r ^ 4 + r ^ 5 ) ) / ( a 1 ( 1 + r ^ 1 + r ^ 2 ) ) 125 = ( r ^ 3 ( 1 + r ^ 1 + r ^ 2 ) ) / ( ( 1 + r ^ 1 + r ^ 2 ) ) 125 = r ^ 3 r = 5 d" | a = 126 - 1
b = 1 / 3
c = a ** b
|
a ) 10 , b ) 20 , c ) 45 , d ) 60 , e ) 120 | c | divide(multiply(6, 15), const_2) | there are 6 chess amateurs playing in villa ' s chess club tournament . if each chess amateur plays with exactly 15 other amateurs , what is the total number of chess games possible to be played in the tournament ? | "each person is one participant of 6 games . so there are in all 6 * 15 = 90 instances of one participant games . but each game has 2 participants so total number of games = 90 / 2 = 45 c" | a = 6 * 15
b = a / 2
|
a ) 37 / 64 , b ) 8 / 27 , c ) 2 / 3 , d ) 19 / 27 , e ) 65 / 81 | a | add(add(divide(const_1, add(3, const_1)), multiply(subtract(const_1, divide(const_1, add(3, const_1))), divide(const_1, add(3, const_1)))), multiply(subtract(const_1, add(multiply(subtract(const_1, divide(const_1, add(3, const_1))), divide(const_1, add(3, const_1))), divide(const_1, add(3, const_1)))), divide(const_1, add(3, const_1)))) | tough and tricky questions : probability . medical analysts predict that a quarter of all people who are infected by a certain biological agent could be expected to be killed for each day that passes during which they have not received an antidote . what fraction of a group of 1000 people could be expected to be killed if infected and not treated for 3 full days ? | at the end of each day we will have 3 / 4 alive . so after 3 days we will have ( 3 / 4 ) ^ 3 people alive . therefore , fraction of dead people will be 1 - ( 3 / 4 ) ^ 3 = 37 / 64 . the correct answer is a . | a = 3 + 1
b = 1 / a
c = 3 + 1
d = 1 / c
e = 1 - d
f = 3 + 1
g = 1 / f
h = e * g
i = b + h
j = 3 + 1
k = 1 / j
l = 1 - k
m = 3 + 1
n = 1 / m
o = l * n
p = 3 + 1
q = 1 / p
r = o + q
s = 1 - r
t = 3 + 1
u = 1 / t
v = s * u
w = i + v
|
a ) 11 , b ) 13 , c ) 14 , d ) 18 , e ) 10 | d | divide(multiply(subtract(add(divide(1, 2), 3), add(divide(1, 3), 3)), 10800), const_100) | if the sales tax reduced from 3 1 / 2 % to 3 1 / 3 % , then what difference does it make to a person who purchases an article with market price of rs . 10800 ? | "required difference = [ 3 ½ % of rs . 10800 ] – [ 3 1 / 3 % of rs . 10800 ] = [ ( 7 / 20 - ( 10 / 3 ) ] % of rs . 10800 = 1 / 6 % of rs . 10800 = rs . [ ( 1 / 6 ) 8 ( 1 / 100 ) * 10800 ] = rs . 18 . answer is d ." | a = 1 / 2
b = a + 3
c = 1 / 3
d = c + 3
e = b - d
f = e * 10800
g = f / 100
|
a ) $ 2.4 million , b ) $ 2.9 million , c ) $ 3.0 million , d ) $ 3.1 million , e ) $ 3.6 million | d | divide(subtract(subtract(const_100, 20), 20), 20) | in 1985 a company sold a brand of shoes to retailers for a fixed price per pair . in 1986 the number of pairs of the shoes that the company sold to retailers decreased by 20 percent , while the price per pair increased by 20 percent . if the company ’ s revenue from the sale of the shoes in 1986 was $ 3.0 million , what was the approximate revenue from the sale of the shoes in 1985 ? | let the number of pairs of shoes sold in 1985 = n let price per pair of shoe sold in 1985 = p in 1986 , number of pairs of shoes sold decreased by 20 % = . 8 n price per pair increased by 20 % = 1.2 p company ' s revenue from sale of shoes in 1986 = 3 million $ = > . 8 n * 1.2 p = 3 = > . 96 np = 3 = > np = 3 / . 96 = 3.125 million $ answer d | a = 100 - 20
b = a - 20
c = b / 20
|
a ) 225 : 121 , b ) 221 : 127 , c ) 225 : 191 , d ) 181 : 161 , e ) 81 : 182 | a | power(divide(3375, 1331), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 3375 : 1331 . what is the ratio of their total surface areas ? | "explanation : ratio of the sides = ³ √ 3375 : ³ √ 1331 = 15 : 11 ratio of surface areas = 15 ^ 2 : 11 ^ 2 = 225 : 121 answer : option a" | a = 3375 / 1331
b = 1 / 3
c = a ** b
|
a ) 3 / 4 , b ) 4 , c ) 18 / 19 , d ) 9 , e ) 12 | c | divide(const_1, multiply(2, add(divide(const_1, 36), divide(const_1, 2)))) | a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 2 hours . if the company used the same number of each type of machine to do job in 12 hours , how many machine r were used ? | yes there is a typo in the question , i got the same ques on my gmat prep last week , and the questions goes as : a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 2 hours . if the company used the same number of each type of machine to do job in 2 hours , how many machine r were used ? so for a job to be done in 2 hours r = 1 / 2 r _ a ( rate of machine r ) = 1 / 36 r _ s ( rate of machine s ) = 1 / 2 lets say x machines are used to attain the desired rate , thus x / 36 + x / 2 = 1 / 2 ( desired r = 1 / 2 i . e . to complete the job in 2 hours ) ( x + 18 x ) / 36 = 1 / 2 19 x / 36 = 1 / 2 x = 18 / 19 . qa = 18 / 19 ( answer c ) | a = 1 / 36
b = 1 / 2
c = a + b
d = 2 * c
e = 1 / d
|
a ) 13000 , b ) 13500 , c ) 14000 , d ) 14500 , e ) 15000 | b | subtract(divide(multiply(multiply(multiply(multiply(184, 3), 3), 3), 6), 2), const_1000) | the difference between compound and simple interest on a particular sum of money for 3 years at 6 and 2 / 3 % p . a is rs . 184 . what is the sum ? | "p = ( 184 * 10 ^ 6 ) / ( 6 2 / 3 * 6 2 / 3 * ( 300 * 6 2 / 3 ) ) = = > 13500 answer b" | a = 184 * 3
b = a * 3
c = b * 3
d = c * 6
e = d / 2
f = e - 1000
|
a ) 4 , b ) 4 √ 2 , c ) 12 , d ) 8 √ 2 , e ) can not be determined from the information provided | c | sqrt(add(power(multiply(sqrt(36), sqrt(const_2)), const_2), power(multiply(sqrt(36), sqrt(const_2)), const_2))) | triangle xyz is an isosceles right triangle . if side xy is longer than side yz , and the area of the triangle is 36 , what is the measure of side xy ? | "ans c . . 12 . . xy being larger means it is the hyp . . area = ( 1 / 2 ) * ( yz ) ^ 2 = 36 or yz = 3 * \ sqrt { 2 } . . therefore hyp = xy = 12" | a = math.sqrt(36)
b = math.sqrt(2)
c = a * b
d = c ** 2
e = math.sqrt(36)
f = math.sqrt(2)
g = e * f
h = g ** 2
i = d + h
j = math.sqrt(i)
|
a ) 47 , b ) 88 , c ) 28 , d ) 66 , e ) 11 | b | multiply(divide(40, subtract(6, const_1)), subtract(12, const_1)) | at 6 ′ o a clock ticks 6 times . the time between first and last ticks is 40 seconds . how long does it tick at 12 ′ o clock | "explanation : for ticking 6 times , there are 5 intervals . each interval has time duration of 40 / 5 = 8 secs at 12 o ' clock , there are 11 intervals , so total time for 11 intervals = 11 × 8 = 88 secs . answer : b" | a = 6 - 1
b = 40 / a
c = 12 - 1
d = b * c
|
a ) 84 , b ) 87 , c ) 96 , d ) 79 , e ) 74 | e | divide(subtract(35, power(12, const_2)), const_2) | the sum of two numbers is 12 and the product of the numbers is 35 . find sum of the squares of that numbers . | let a and b be the two numbers ( a + b ) ^ 2 = a ^ 2 + 2 ab + b ^ 2 given ( a + b ) = 12 ab = 35 so , 12 ^ 2 = a ^ 2 + b ^ 2 + 2 * 35 144 = a ^ 2 + b ^ 2 + 70 a ^ 2 + b ^ 2 = 74 ans e | a = 12 ** 2
b = 35 - a
c = b / 2
|
a ) 4 % , b ) 8 % , c ) 25 % , d ) 16.66 % , e ) 9 % | a | subtract(add(multiply(multiply(const_4, 20), divide(30, const_100)), multiply(const_4, 20)), const_100) | the balance of a trader weighs 20 % less than it should . still the trader marks up his goods to get an overall profit of 30 % . what is the mark up on the cost price ? | the most natural way to deal with ' weights ' questions is by assuming values . say the trader ' s balance shows 100 gms . it is actually 80 gms because it weighs 20 % less . say , the cost price is $ 80 ( $ 1 / gm ) . since he gets a profit of 30 % , the selling price must be 80 + ( 30 / 100 ) * 80 = $ 104 since the cost price is actually supposed to be $ 100 ( for 100 gms ) and the selling price is $ 104 , the mark up is simply 4 % . ans : a | a = 4 * 20
b = 30 / 100
c = a * b
d = 4 * 20
e = c + d
f = e - 100
|
a ) 4 : 30 , b ) 4 : 45 , c ) 5 : 00 , d ) 5 : 15 , e ) 5 : 30 | c | subtract(divide(multiply(2, 12), 12), const_0_33) | it takes 2 workers a total of 12 hours to build a giant machine with each worker working at the same rate . if 3 workers start to build the machine at 11 : 00 am , and one extra worker per hour is added beginning at 2 : 00 pm , at what time will the machine be complete ? | 2 workers build 1 / 12 of a machine in one hour . 1 worker builds 1 / 24 of a machine in one hour . in the first 3 hours , 3 workers build 3 * ( 1 / 24 ) * 3 = 9 / 24 of a machine from 2 : 00 to 3 : 00 , 4 workers build another 4 / 24 . the total is 13 / 24 . from 3 : 00 to 4 : 00 , 5 workers build another 5 / 24 . the total is 18 / 24 . from 4 : 00 to 5 : 00 , 6 workers build another 6 / 24 . the total is 24 / 24 . the machine is complete at 5 : 00 . the answer is c . | a = 2 * 12
b = a / 12
c = b - const_0_33
|
a ) 120 , b ) 123 , c ) 124 , d ) 125 , e ) 359 | d | multiply(divide(50, const_2), 5) | find the number , difference between number and its 3 / 5 is 50 . | "explanation : let the number = x , then , x - ( 3 / 5 ) x = 50 , = > ( 2 / 5 ) x = 50 = > 2 x = 50 * 5 , = > x = 125 option d" | a = 50 / 2
b = a * 5
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) none of the above | d | divide(subtract(multiply(180, 3), multiply(60, 3)), subtract(add(add(60, 120), 180), multiply(100, const_3))) | in a coconut grove , ( x + 3 ) trees yield 60 nuts per year , x trees yield 120 nuts per year and ( x – 3 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 3 ) × 60 + x × 120 + ( x − 3 ) × 180 / ( x + 3 ) + x + ( x − 3 ) = 100 ⇒ 360 x − 360 / 3 x = 100 ⇒ 60 x = 360 ⇒ x = 6 answer d" | a = 180 * 3
b = 60 * 3
c = a - b
d = 60 + 120
e = d + 180
f = 100 * 3
g = e - f
h = c / g
|
a ) $ 10079.44 , b ) w = $ 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 w does she have in her account after two years ? | "ps . i guess one can use simple interest to solve cause the answer choices are quite spread between you can easily arrive at something near 8 % hence b the answer" | a = 3 / 96
b = a / 2
c = 1 + b
d = c ** 4
e = 10000 * d
|
a ) 230 m , b ) 210 m , c ) 260 m , d ) 320 m , e ) 330 m | b | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 290) | a 290 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "relative speed = ( 120 + 80 ) km / hr = ( 200 x ( 5 / 18 ) ) m / sec = ( 500 / 9 ) m / sec . let the length of the other train be x metres . then , ( x + 290 ) / 9 = 500 / 9 x + 290 = 500 x = 210 . b" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 290
|
a ) 30 , b ) 4 , c ) 9 , d ) 8 , e ) none | c | divide(add(add(add(4, const_1), add(add(4, const_1), const_2)), add(subtract(15, 4), subtract(15, const_2))), const_4) | find the average of all prime numbers between 4 and 15 . | sol . there are four prime numbers between 4 and 15 . they are 5 , 7 , 11 , 13 â ˆ ´ required average = [ 5 + 7 + 11 + 13 / 4 ] = 36 / 4 = 9 answer c | a = 4 + 1
b = 4 + 1
c = b + 2
d = a + c
e = 15 - 4
f = 15 - 2
g = e + f
h = d + g
i = h / 4
|
a ) 21 , b ) 24 , c ) 28 , d ) 30 , e ) 32 | a | subtract(100, add(add(38, const_1), 40)) | 100 people are attending a newspaper conference . 40 of them are writers and more than 38 are editors . of the people at the conference , x are both writers and editors and 2 x are neither . what is the largest possible number of people who are both writers and editors ? | "{ total } = { writers } + { editors } - { both } + { neither } . { total } = 100 ; { writers } = 40 ; { editors } > 38 ; { both } = x ; { neither } = 2 x ; 100 = 40 + { editors } - x + 2 x - - > x = 60 - { editors } . we want to maximize x , thus we should minimize { editors } , minimum possible value of { editors } is 39 , thus x = { both } = 60 - 39 = 21 . answer : a ." | a = 38 + 1
b = a + 40
c = 100 - b
|
a ) 166.69 % , b ) 166.66 % , c ) 166.76 % , d ) 136.66 % , e ) 566.66 % | b | multiply(subtract(divide(multiply(add(6, 2), 2), 6), const_1), const_100) | a milkman purchases the milk at rs . x per litre and sells it at rs . 2 x per litre still he mixes 2 litres water with every 6 litres of pure milk . what is the profit percentage ? | let the cost price of 1 litre pure milk be re . 1 , then \ inline \ begin { bmatrix } 6 & litres ( milk ) & \ rightarrow & cp = rs . 6 \ \ 2 & litres ( water ) & \ rightarrow & cp = rs . 0 \ end { bmatrix } \ rightarrow cp = rs . 6 only and 8 litre mixture \ rightarrow sp \ rightarrow 8 x 2 = rs . 16 profit % = % answer : b | a = 6 + 2
b = a * 2
c = b / 6
d = c - 1
e = d * 100
|
a ) 5 days , b ) 7 5 / 6 days , c ) 10 days , d ) 15 2 / 3 days , e ) none of these | c | divide(const_1, divide(add(add(inverse(12), inverse(15)), inverse(20)), const_2)) | a and b can do a work in 12 days , b and c in 15 days , c and a in 20 days . if a , b and c work together , they will complete the work in : | "( a + b ) ' s 1 day ' s work = 1 / 12 , ( b + c ) ' s 1 day ' s work = 1 / 15 ; ( a + c ) ' s 1 day ' s work = 1 / 20 adding , we get : 2 ( a + b + c ) ' s 1 day ' s work = ( 1 / 12 + 1 / 15 + 1 / 20 ) = 12 / 60 = 1 / 5 ( a + b + c ) ' s 1 day ' s work = 1 / 10 so , a , b and c together can complete the work in 10 days . correct option : c" | a = 1/(12)
b = 1/(15)
c = a + b
d = 1/(20)
e = c + d
f = e / 2
g = 1 / f
|
a ) a ) 266666.7 , b ) b ) 562000 , c ) c ) 800000 , d ) d ) 500000 , e ) e ) 652000 | a | multiply(divide(40000, subtract(const_100, add(add(multiply(15, 3), 30), 10))), const_100) | a person distributed 15 % of his income to his 3 children each . he deposited 30 % of his income to his wife ' s account . he donated 10 % of remaining amount to an orphan house . finally he has $ 40000 . find his total income ? | 3 children got = 3 * 15 % = 45 % wife got = 30 % orphan house = 10 % total = 45 + 30 + 10 = 85 % remaining = 100 - 85 = 15 % 15 % = 40000 100 % = 40000 * 100 / 15 = $ 266666.7 answer is a | a = 15 * 3
b = a + 30
c = b + 10
d = 100 - c
e = 40000 / d
f = e * 100
|
a ) 30 , b ) 31 , c ) 32 , d ) 33 , e ) 34 | c | divide(factorial(subtract(add(const_4, 15), const_1)), multiply(factorial(15), factorial(subtract(const_4, const_1)))) | how many positive integers less than 250 are there such that they are multiples of 15 or multiples of 14 ? | "250 / 15 = 16 ( plus remainder ) so there are 16 multiples of 15 250 / 14 = 17 ( plus remainder ) so there are 17 multiples of 14 we need to subtract 1 because 15 * 14 is a multiple of both so it was counted twice . the total is 16 + 17 - 1 = 32 the answer is c ." | a = 4 + 15
b = a - 1
c = math.factorial(b)
d = math.factorial(15)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 38 sec , b ) 35 sec , c ) 44 sec , d ) 48 sec , e ) none of these | d | multiply(divide(add(divide(240, const_1000), divide(360, const_1000)), 45), const_3600) | a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 240 m long ? | speed = 45 km / hr = 45 ã — ( 5 / 18 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 240 = 600 meter time taken to cross the platform = 600 / ( 25 / 2 ) = 600 ã — 2 / 25 = 48 seconds answer : d | a = 240 / 1000
b = 360 / 1000
c = a + b
d = c / 45
e = d * 3600
|
a ) 32 , b ) 48 , c ) 29 , d ) 27 , e ) 21 | b | divide(divide(add(300, 300), const_1000), divide(45, const_3600)) | a train 300 meters long completely crosses a 300 meters long bridge in 45 seconds . what is the speed of the train is ? | "s = ( 300 + 300 ) / 45 = 600 / 45 * 18 / 5 = 48 answer : b" | a = 300 + 300
b = a / 1000
c = 45 / 3600
d = b / c
|
a ) 28 , b ) 41.25 , c ) 55 , d ) 18 , e ) 12 | b | divide(add(165, 660), multiply(72, const_0_2778)) | how long does a train 165 meters long running at the rate of 72 kmph take to cross a bridge 660 meters in length ? | "t = ( 660 + 165 ) / 72 * 18 / 5 t = 41.25 answer : b" | a = 165 + 660
b = 72 * const_0_2778
c = a / b
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | b | divide(700, const_10) | how many integers from 101 to 700 , inclusive , remains the value unchanged when the digits were reversed ? | "question is asking for palindrome first digit possibilities - 1 through 6 = 6 7 is not possible here because it would result in a number greater than 7 ( i . e 707 , 717 . . ) second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 6 * 10 = 60 answer is b ." | a = 700 / 10
|
a ) 7945 , b ) 8877 , c ) 2877 , d ) 2678 , e ) 1011 | a | divide(8800, power(add(subtract(divide(9261, 8800), const_1), const_1), 2)) | what sum of money put at c . i amounts in 2 years to rs . 8800 and in 3 years to rs . 9261 ? | "8800 - - - - 461 100 - - - - ? = > 5.238 % x * 105.238 / 100 * 105.238 / 100 = 8800 x * 1.1075 = 8800 x = 8800 / 1.1075 = > 7945.70 answer : a" | a = 9261 / 8800
b = a - 1
c = b + 1
d = c ** 2
e = 8800 / d
|
a ) 1 , b ) 5 , c ) 3 , d ) 6 , e ) 7 | c | floor(add(reminder(add(subtract(multiply(divide(47, const_100), 1442), multiply(divide(36, const_100), 1412)), 63), const_10), const_1)) | find the value of x . ( 47 % of 1442 - 36 % of 1412 ) + 63 = x ? | c ) 3 | a = 47 / 100
b = a * 1442
c = 36 / 100
d = c * 1412
e = b - d
f = e + 63
g = reminder + (
h = math.floor(g, 1)
|
a ) 3569 , b ) 1552 , c ) 4573 , d ) 4735 , e ) 5235 | d | add(divide(const_1000, const_2), divide(12345, const_3)) | if in a certain code ` ` range ' ' is coded as 12345 and ` ` random ' ' is coded as 123678 . then the code for the word ` ` gone ' ' would be | g - 4 o - 7 n - 3 e - 5 so for mango the code is 4735 answer : d | a = 1000 / 2
b = 12345 / 3
c = a + b
|
a ) 32 , b ) 40 , c ) 64 , d ) 80 , e ) 96 | a | multiply(4, divide(const_1, subtract(divide(7, 8), divide(3, 4)))) | if henry were to add 4 gallons of water to a tank that is already 3 / 4 full of water , the tank would be 7 / 8 full . how many gallons of water would the tank hold if it were full ? | "7 / 8 x - 3 / 4 x = 4 galls 1 / 8 * x = 4 gallons x = 32 gallons answer a" | a = 7 / 8
b = 3 / 4
c = a - b
d = 1 / c
e = 4 * d
|
a ) 110 , b ) 130 , c ) 120 , d ) 125 , e ) 135 | b | subtract(multiply(15, multiply(60, const_0_2778)), 120) | a train 120 m long crosses a platform with a speed of 60 kmph in 15 sec ; find the length of the platform ? | d = 120 + x t = 15 s = 120 + x / 15 * 18 / 5 = 60 kmph . = > ( 120 + x ) * 18 = 4500 = > 120 + x = 250 = > x = 130 m answer : b | a = 60 * const_0_2778
b = 15 * a
c = b - 120
|
a ) 12021 , b ) 14520 , c ) 16065 , d ) 18925 , e ) 12887 | c | divide(divide(multiply(4016.25, const_100), 5), 5) | a sum fetched a total simple interest of rs . 4016.25 at the rate of 5 % p . a . in 5 years . what is the sum ? | "principal = ( 100 * 4016.25 ) / ( 5 * 5 ) = rs . 16065 . answer : c" | a = 4016 * 25
b = a / 5
c = b / 5
|
a ) 6 kmph , b ) 5 kmph , c ) 9 kmph , d ) 8 kmph , e ) 1 kmph | c | divide(subtract(divide(84, 2), divide(48, 2)), const_2) | a man rows his boat 84 km downstream and 48 km upstream , taking 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 84 / ( 2 ) = 42 kmph speed upstream = d / t = 48 / ( 2 ) = 24 kmph the speed of the stream = ( 42 - 24 ) / 2 = 9 kmph answer : c" | a = 84 / 2
b = 48 / 2
c = a - b
d = c / 2
|
a ) 130 , b ) 100 , c ) 200 , d ) 150 , e ) 120 | c | multiply(10, 20) | what number has a 20 : 1 ratio to the number 10 ? | "20 : 1 = x : 10 x = 10 * 20 x = 150 answer : c" | a = 10 * 20
|
a ) 27 , b ) 48 , c ) 45 , d ) 72 , e ) 18 | b | subtract(multiply(multiply(5, 5), divide(146, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(146, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5))))) | the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 146 , what is the difference between the ages of patrick and monica ? | "ages of p and mi = 3 x + 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 47 x = 146 x = 3 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 3 = 48 answer b" | a = 5 * 5
b = 3 * 3
c = 3 * 5
d = b + c
e = 5 * 5
f = d + e
g = 146 / f
h = a * g
i = 3 * 3
j = 3 * 3
k = 3 * 5
l = j + k
m = 5 * 5
n = l + m
o = 146 / n
p = i * o
q = h - p
|
a ) $ 1.75 , b ) $ 1.85 , c ) $ 2.00 , d ) $ 2.12 , e ) $ 2.30 | d | add(divide(subtract(8.25, multiply(divide(multiply(subtract(6, const_1), 6), const_2), 0.3)), 6), multiply(subtract(6, const_1), 0.3)) | mark bought a set of 6 flower pots of different sizes at a total cost of $ 8.25 . each pot cost 0.3 more than the next one below it in size . what was the cost , in dollars , of the largest pot ? | "this question can be solved with a handful of different algebra approaches ( as has been shown in the various posts ) . since the question asks for the price of the largest pot , and the answers are prices , we can test the answers . we ' re told that there are 6 pots and that each pot costs 30 cents more than the next . the total price of the pots is $ 8.25 . we ' re asked for the price of the largest ( most expensive ) pot . since the total price is $ 8.25 ( a 30 - cent increment ) and the the difference in sequential prices of the pots is 30 cents , the largest pot probably has a price that is a 30 - cent increment . from the answer choices , i would then test answer d if . . . . the largest pot = $ 2.125 the other 5 pots are . . . . 0.625 0.925 1.225 1.525 1.825 2.125 total = $ 8.25 so this must be the answer . d" | a = 6 - 1
b = a * 6
c = b / 2
d = c * 0
e = 8 - 25
f = e / 6
g = 6 - 1
h = g * 0
i = f + h
|
a ) 5 liters , b ) 10 liters , c ) 15 liters , d ) 8 liters , e ) 6 liters | a | subtract(15, divide(multiply(15, 40), 60)) | what quantity of water should taken out to concentrate 15 liters of 40 % acidic liquid to 60 % acidic liquid ? | required answer is = 15 ( 60 - 40 ) / 60 = 5 liters answer is a | a = 15 * 40
b = a / 60
c = 15 - b
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(divide(5, const_2), multiply(9, 9)) | what is the remainder when 9 ^ 19 is divided by 5 ? | "9 ^ 0 = 1 9 ^ 1 = 9 9 ^ 2 = 81 9 ^ 3 = 729 9 ^ 4 = . . . 1 9 ^ 5 = . . . . . 9 and so on in a distinct last - digit pattern of 9,1 , 9,1 . . . after ignoring the first last - digit of 1 19 / 2 = 9 ( with a remainder of 1 ) so the last number is a 9 . alternatively , 19 is an odd number and hence the last digit is a 9 when 9 ^ 19 is divided by 5 the remainder is 4 , ans = e" | a = 5 / 2
b = 9 * 9
c = a - b
|
a ) 675 , b ) 750 , c ) 1,125 , d ) 5,900 , e ) 7,500 | c | subtract(multiply(5, 360), multiply(5, divide(multiply(5, 360), add(const_4, const_4)))) | eight identical machines can produce 360 aluminum cans per hour . if all of the machines work at the same constant rate , how many cans could 5 such machines produce in 5 hours ? | "8 machines / 360 cans = 5 machines / x cans 8 x = 1800 x = 225 ( 225 ) ( 5 hours ) = 1125 cans . the answer is c ." | a = 5 * 360
b = 5 * 360
c = 4 + 4
d = b / c
e = 5 * d
f = a - e
|
a ) 22 , b ) 99 , c ) 26 , d ) 66 , e ) 887 | c | divide(multiply(65, 3), add(divide(65, 65), divide(multiply(2, 65), 20))) | a trained covered x km at 65 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km . | "total time taken = x / 65 + 2 x / 20 hours = 3 x / 26 hours average speed = 3 x / ( 3 x / 26 ) = 26 kmph answer : c" | a = 65 * 3
b = 65 / 65
c = 2 * 65
d = c / 20
e = b + d
f = a / e
|
a ) $ 118.00 , b ) $ 120.00 , c ) $ 169.40 , d ) $ 122.00 , e ) $ 140.00 | c | multiply(140, power(add(const_1, divide(divide(20, const_100), const_2)), const_2)) | ms . lopez deposits $ 140 in an account that pays 20 % interest , compounded semiannually . how much money will there be in the account at the end of one year ? | using formula - a = p ( 1 + r / n ) ^ nt given p = 140 n = 2 t = 1 r = 0.2 substituting value in formula a = 140 ( 1 + 0.2 / 2 ) ^ 2 a = 169.40 $ = c | a = 20 / 100
b = a / 2
c = 1 + b
d = c ** 2
e = 140 * d
|
a ) 286 , b ) 278 , c ) 255 , d ) 180 , e ) 287 | d | multiply(divide(multiply(72, const_1000), const_3600), 9) | a train running at the speed of 72 km / hr crosses a pole in 9 seconds . what is the length of the train ? | "speed = ( 72 * 5 / 18 ) m / sec = ( 20 ) m / sec length of the train = ( speed x time ) = ( 20 * 9 ) m = 180 m . answer : d" | a = 72 * 1000
b = a / 3600
c = b * 9
|
a ) 16 seconds , b ) 76 seconds , c ) 26 seconds , d ) 9.4 seconds , e ) 18 seconds | d | divide(60, multiply(add(20, 3), const_0_2778)) | the speed at which a man can row a boat in still water is 20 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 60 metres ? | "speed of the boat downstream = 20 + 3 = 23 kmph = 23 * 5 / 18 = 115 / 18 m / s hence time taken to cover 60 m = 60 * 18 / 115 = 9.4 seconds . answer : d" | a = 20 + 3
b = a * const_0_2778
c = 60 / b
|
a ) s . 83.33 , b ) s . 110 , c ) s . 114 , d ) s . 120 , e ) s . 140 | c | multiply(divide(const_100, 14), 16) | a 16 % stock yielding 14 % is quoted at : | "income of rs 14 on investment of rs 100 income of rs 16 on investment of ? = ( 16 * 100 ) / 14 = 114 answer : c" | a = 100 / 14
b = a * 16
|
a ) 0 , b ) 1 / 12 , c ) 1 , d ) 1 / 5 , e ) 1 / 3 | d | divide(const_3, 15) | the value of ( ( x – y ) ³ + ( y - z ) ³ + ( z – x ) ³ ) / ( 15 ( x – y ) ( y – z ) ( z – x ) ) is equal to : | since ( x – y ) + ( y – z ) + ( z – x ) = 0 , so ( x – y ) ³ + ( y – z ) ³ + ( z – x ) ³ = 3 ( x – y ) ( y – z ) ( z – x ) . ( 3 ( x – y ) ( y – z ) ( z – x ) ) / ( 15 ( x – y ) ( y – z ) ( z – x ) ) = 1 / 5 . answer : d | a = 3 / 15
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(add(divide(50, divide(const_10, const_2)), const_2), const_2) | a is the product of each integer from 1 to 50 , inclusive and b = 100 ^ k , where k is an integer . what is the greatest value of k for which y is a factor of a ? | the number of trailing zeros in the decimal representation of n ! , the factorial of a non - negative integer n , can be determined with this formula : n 5 + n 52 + n 53 + . . . + n 5 k , where k must be chosen such that 5 k ≤ n x = 1 * 2 * 3 . . . . * 50 = 50 ! no . of trailing zeros in 50 ! = 50 / 5 + 50 / 5 ^ 2 = 10 + 2 = 12 100 ^ k = 10 ^ 2 k → k = 12 / 2 = 6 a | a = 10 / 2
b = 50 / a
c = b + 2
d = c / 2
|
a ) 3.6 , b ) 6 , c ) 18 , d ) can not be determined , e ) none of these | b | sqrt(divide(multiply(432, const_100), 1200)) | reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 432 as interest at the end of the loan period , what was the rate of interest ? | "let rate = r % and time = r years . then , 1200 x r x r / 100 = 432 12 r 2 = 432 r 2 = 36 r = 6 . answer : b" | a = 432 * 100
b = a / 1200
c = math.sqrt(b)
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | subtract(divide(10, const_2), const_1) | if n is the greatest positive integer for which 4 ^ n is a factor of 10 ! , then n = ? | 10 ! = 3628800 e . 4 ^ 10 = 1048576 ( 3628800 / 1048576 ) - this is not a factor of 10 ! d . 4 ^ 8 = 65536 ( 3628800 / 65536 ) - this is not a factor of 10 ! c . 4 ^ 6 = 4096 ( 3628800 / 4096 ) - this is not a factor of 10 ! b . 4 ^ 4 = 256 ( 3628800 / 256 ) - this is a factor of 10 ! b is answer | a = 10 / 2
b = a - 1
|
a ) 360 , b ) 3150 , c ) 540 , d ) 640 , e ) 720 | b | divide(divide(1400, subtract(1, divide(1, 3))), subtract(1, divide(1, 3))) | an art gallery has only paintings and sculptures . currently , 1 / 3 of the pieces of art are displayed , and 1 / 6 of the pieces on display are sculptures . if 1 / 3 of the pieces not on display are paintings , and 1400 sculptures are not on display , how many pieces of art does the gallery have ? | "too many words and redundant info there . ( i ) 1 / 3 of the pieces of art are displayed , hence 2 / 3 of the pieces of art are not displayed . ( ii ) 1 / 6 of the pieces on display are sculptures , hence 5 / 6 of the pieces on display are paintings . ( iii ) 1 / 3 of the pieces not on display are paintings , hence 2 / 3 of the pieces not on display are sculptures . 1400 sculptures are not on display , so according to ( iii ) 2 / 3 * { not on display } = 1400 - - > { not on display } = 2100 . according to ( i ) 2 / 3 * { total } = 2100 - - > { total } = 3150 . answer : b ." | a = 1 / 3
b = 1 - a
c = 1400 / b
d = 1 / 3
e = 1 - d
f = c / e
|
a ) 55 , b ) 56 , c ) 64 , d ) 65 , e ) 66 | c | add(multiply(divide(66, const_3), const_2), divide(60, const_3)) | the grade point average of one third of the classroom is 60 ; the grade point average of the rest is 66 . what is the grade point average of the whole class ? | "let n = total students in class total points for 1 / 3 class = 60 n / 3 = 20 n total points for 2 / 3 class = 66 * 2 n / 3 = 44 n total points for whole class = 20 n + 44 n = 64 n 64 n total class points / n total students = 64 grade point average for total class answer : c" | a = 66 / 3
b = a * 2
c = 60 / 3
d = b + c
|
a ) 0.33 , b ) 33 , c ) 0.033 , d ) 0.0033 , e ) 3.3 e - 05 | b | multiply(divide(multiply(multiply(multiply(10, 10), subtract(multiply(10, 10), const_1)), divide(33, subtract(multiply(10, 10), const_1))), const_1000), 10) | if the digits 33 in the decimal 0.00033 repeat indefinitely , what is the value of ( 10 ^ 5 - 10 ^ 3 ) ( 0.00033 ) ? | 99 * 0.33 = 32.67 approx . 33 answer : b | a = 10 * 10
b = 10 * 10
c = b - 1
d = a * c
e = 10 * 10
f = e - 1
g = 33 / f
h = d * g
i = h / 1000
j = i * 10
|
a ) a ) 7 , b ) b ) 9 , c ) c ) 11 , d ) d ) 13 , e ) e ) 15 | a | divide(add(multiply(10, 3), 5), add(const_2, 3)) | sum of two numbers is 10 . two times of the first exceeds by 5 from the 3 times of the other . then the numbers will be ? | explanation : x + y = 10 2 x – 3 y = 5 x = 7 y = 3 a ) | a = 10 * 3
b = a + 5
c = 2 + 3
d = b / c
|
a ) 3 : 4 , b ) 3 : 5 , c ) 4 : 5 , d ) 5 : 7 , e ) none | a | divide(divide(subtract(divide(72, const_2), const_10), const_2), add(divide(72, const_2), const_10)) | two whole numbers whose sum is 72 can not be in the ratio | "sol . the sum of the ratio terms must divide 72 . so , the ratio can not be 3 : 4 . answer a" | a = 72 / 2
b = a - 10
c = b / 2
d = 72 / 2
e = d + 10
f = c / e
|
a ) 15 , b ) 22 , c ) 27 , d ) 29 , e ) 18 | b | divide(multiply(2, 22), subtract(22, 20)) | a group of men decided to do a work in 20 days , but 2 of them became absent . if the rest of the group did the work in 22 days , find the original number of men ? | "original number of men = 2 * 22 / ( 22 - 20 ) = 22 answer is b" | a = 2 * 22
b = 22 - 20
c = a / b
|
a ) 1.5 seconds , b ) 9.56 seconds , c ) 28.56 seconds , d ) 30.56 seconds , e ) 3.5 seconds | b | divide(add(280, 210), multiply(const_60.0, const_0_2778)) | a train 280 m long is running at 70 kmph . in how much time will it pass a platform 210 m long ? | distance travelled = 280 + 210 m = 490 m speed = 70 * 5 / 8 = 205 / 4 m time = 490 * 4 / 205 = 9.56 seconds answer : b . | a = 280 + 210
b = const_60 * 0
c = a / b
|
a ) 6 , b ) 1 , c ) 5 , d ) 4 , e ) 7 | b | multiply(divide(divide(multiply(divide(20, const_100), 100), 25), multiply(divide(20, const_100), 100)), const_100) | a reduction of 20 % in the price of salt enables a lady to obtain 25 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 20 / 100 ) = 20 - - - 25 ? - - - 1 = > rs . 0.8 100 - - - 80 ? - - - 0.8 = > rs . 1 answer : b" | a = 20 / 100
b = a * 100
c = b / 25
d = 20 / 100
e = d * 100
f = c / e
g = f * 100
|
a ) 1 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | a | divide(divide(30, 10), 3) | in how many no . between 10 and 30 exactly two of the digits is 3 ? | it ' s simple can be solved by elimination of answer choices . option b and c are too large , not possible . even de are large to have correct choice . ans : a | a = 30 / 10
b = a / 3
|
a ) a ) 125 , b ) b ) 439 , c ) c ) 225 , d ) d ) 375 , e ) e ) 524 | b | divide(multiply(divide(208, const_100), 1265), 6) | ( 208 % of 1265 ) ÷ 6 = ? | "explanation : ? = ( 208 x 1265 / 100 ) ÷ 6 = 263120 / 600 = 439 answer : option b" | a = 208 / 100
b = a * 1265
c = b / 6
|
a ) s . 96 , b ) s . 45 , c ) s . 120 , d ) s . 180 , e ) s . 189 | a | divide(multiply(120, const_100), add(const_100, 25)) | by selling an article at rs . 120 , a profit of 25 % is made . find its cost price ? | "sp = 120 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 120 * [ 100 / ( 100 + 25 ) ] = 120 * [ 100 / 125 ] = rs . 96 answer : a" | a = 120 * 100
b = 100 + 25
c = a / b
|
a ) 25 , b ) 22 , c ) 30 , d ) 35 , e ) 42 | b | divide(add(multiply(2, 40), multiply(3, 10)), 5) | the average score of a cricketer in 2 matches is 40 and in other 3 matches is 10 . then find the average score in all the 5 matches ? | "average in 5 matches = ( 2 * 40 + 3 * 10 ) / 2 + 3 = 80 + 30 / 5 = 110 / 5 = 22 answer is b" | a = 2 * 40
b = 3 * 10
c = a + b
d = c / 5
|
a ) 69 , b ) 62 , c ) 61 , d ) 63 , e ) 64 | a | multiply(330, subtract(add(floor(divide(8679921, 330)), const_1), divide(8679921, 330))) | which number need to add to 8679921 to get a number exactly divisible by 330 ? | 8679921 / 330 = 26302 and reminder = 261 . 330 - 261 = 69 so , the next number divisible by 339 is 69 places in front of 8679921 which means 69 + 8679921 = 8679990 69 should be added to 8679921 a | a = 8679921 / 330
b = math.floor(a)
c = b + 1
d = 8679921 / 330
e = c - d
f = 330 * e
|
a ) 4 % , b ) 8 % , c ) 4.2 % , d ) 4.8 % , e ) 5 % | d | divide(multiply(divide(multiply(60, 40), const_100), 20), const_100) | at a certain college , 60 percent of the total number of students are freshmen . if 40 percent of the fresh - men are enrolled in the school of liberal arts and , of these , 20 percent are psychology majors , what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? | "let ' s say there is a total of 100 students at this college . 60 percent of the total number of students are freshmen . # of freshmen = 60 % of 100 = 60 40 percent of the fresh - men are enrolled in the school of liberal arts . . . number of liberal arts freshmen = 40 % of 60 = 24 . . . and , of these , 20 percent are psychology majors . . . number of liberal arts freshmen who are psychology majors = 20 % of 24 = 4.8 what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? 4.8 / 100 = 4.8 % answer : d" | a = 60 * 40
b = a / 100
c = b * 20
d = c / 100
|
a ) 300 , b ) 345 , c ) 325 , d ) 335 , e ) none of them | a | multiply(multiply(const_100.0, divide(12, 1008)), 3) | what annual installment will discharge a debt of rs . 1008 due in 3 years at 12 % simple interest ? | "let each installment be rs . x then , ( x + ( ( x * 12 * 1 ) / 100 ) ) + ( x + ( ( x * 12 * 2 ) / 100 ) ) + x = 1008 = ( ( 28 x / 25 ) + ( 31 x / 25 ) + x ) = 1008 ï ƒ › ( 28 x + 31 x + 25 x ) = ( 1008 * 25 ) x = ( 1008 * 25 ) / 84 = rs . 300 . therefore , each installment = rs . 300 . answer is a ." | a = 12 / 1008
b = 100 * 0
c = b * 3
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | add(floor(divide(16, const_3)), const_1) | what is the smallest integer k for which 64 ^ k > 4 ^ 16 ? | "64 ^ k > 4 ^ 16 4 ^ ( 3 k ) > 4 ^ 16 3 k > 16 k = 6 the answer is c ." | a = 16 / 3
b = math.floor(a)
c = b + 1
|
a ) 220 sec , b ) 190 sec , c ) 167 sec , d ) 197 sec , e ) 179 sec | a | divide(add(1200, 1000), divide(1200, 120)) | a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 1000 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 2200 m t = d / s t = 2200 / 10 t = 220 sec answer : a" | a = 1200 + 1000
b = 1200 / 120
c = a / b
|
a ) 116 , b ) 119 , c ) 122 , d ) 125 , e ) 128 | c | add(divide(multiply(50, 9), 5), 32) | if t = 5 / 9 * ( k - 32 ) , and if t = 50 , then what is the value of k ? | "k - 32 = 9 t / 5 k = 9 t / 5 + 32 k = 9 ( 50 ) / 5 + 32 = 122 the answer is c ." | a = 50 * 9
b = a / 5
c = b + 32
|
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 7 | c | add(divide(7, 7), const_2) | let q represent a set of 7 distinct prime numbers . if the sum of the numbers in q is even and x is a member of q , then what is the least possible value that x can be ? | q = p 1 + p 2 + p 3 + p 4 + p 5 + p 6 + p 7 = even ( and all primes are distinct ) if the least prime is 2 then we have sum of q = even . ans . c . 3 | a = 7 / 7
b = a + 2
|
a ) 22 , b ) 20 , c ) 19 , d ) 18 , e ) 17 | d | subtract(multiply(10, 2), const_2) | how many digits are in the number 50 ^ 8 × 8 ^ 3 × 11 ^ 2 x 10 ^ 4 ? | whenever you are asked to find the number of digits , try to bring the number in multiples of 10 . this way , we can wasily calculate the umber of 0 ' s through the powers of 10 50 ^ 8 × 8 ^ 3 × 11 ^ 2 x 10 ^ 4 = ( 5 ^ 2 * 2 ) ^ 8 * 2 ^ 9 * 11 ^ 2 * 10 ^ 4 = 5 ^ 16 * 2 ^ 17 * 11 ^ 2 * 10 ^ 4 = 2 * 11 ^ 2 * 10 ^ 16 * 10 ^ 4 = 242 * 10 ^ 16 x 10 ^ 4 hence we would have 16 trailing 0 ' s and the three digits from 242 total digits = 3 + 20 = 23 correct option : d | a = 10 * 2
b = a - 2
|
a ) 2 , b ) 8 , c ) 9 , d ) 11 , e ) 12 | e | divide(432, add(multiply(8, const_3), multiply(6, const_2))) | sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 432 per week . how much does she earn in dollars per hour ? | "explanation : total hours worked = 8 x 3 + 6 x 2 = 36 total earned = 432 . hourly wage = 432 / 36 = 12 answer : e" | a = 8 * 3
b = 6 * 2
c = a + b
d = 432 / c
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(subtract(11, subtract(multiply(const_4, 2), multiply(2, 3))), 3) | x , y , and z are consecutive numbers and x > y > z . also , 2 x + 3 y + 3 z = 5 y + 11 . what is the value of z ? | "if x , y , and z are consecutive numbers and x > y > z , then y = z + 1 and x = z + 2 . 2 x + 3 y + 3 z = 5 y + 11 2 z + 4 + 3 z + 3 + 3 z = 5 z + 5 + 11 3 z = 9 z = 3 the answer is b ." | a = 4 * 2
b = 2 * 3
c = a - b
d = 11 - c
e = d / 3
|
['a ) 1 : 8 .', 'b ) 1 : 2 .', 'c ) 1 : 4 .', 'd ) 1 : 25 .', 'e ) 1 : 6 .'] | d | divide(multiply(const_pi, const_1), multiply(const_pi, power(5, const_2))) | a and b are two circles . the radius of a is 5 times as large as the radius of b . what is the ratio between the areas of the circles ? | given : the radius of a is 5 times as large as the radius of b . = > r ( a ) = 5 r ( b ) . the radius are in ratio of 1 : 5 thus the area will be in the ratio of square of radius . 1 : 25 . hence d . | a = math.pi * 1
b = 5 ** 2
c = math.pi * b
d = a / c
|
a ) 0.045388 , b ) 4.5388 , c ) 0.03242 , d ) 473.88 , e ) none of these | c | divide(divide(multiply(3.242, 10), 100), const_10) | [ ( 3.242 x 10 ) / 100 ] = ? | "answer multiplying 3.242 x 10 = 3.242 now divide 3.242 by 100 so , 3.242 ÷ 100 = 0.03242 ∴ shift the decimal two places to the left as 100 correct option : c" | a = 3 * 242
b = a / 100
c = b / 10
|
a ) 12 hrs , b ) 6 hrs , c ) 8 hrs , d ) 10 hrs , e ) 24 hrs | a | divide(subtract(12, 8), inverse(3)) | pipe a can fill a pool in 12 hours at a constant rate . if pipe a does the job for 8 hours and pipe b does the rest of the job , which works at 1 / 3 constant rate of pipe a . how long will it take for pipe b alone to do the rest of the job ? | pipe a will do 8 / 12 in 8 hrs , so a does 2 / 3 of the work . . therefore , b will do the remaining 1 / 3 rd work alone . . as the speed of b is 1 / 3 rate of a , b will do the 1 / 3 rd work in same time that a takes to complete full job . . . ans 12 a | a = 12 - 8
b = 1/(3)
c = a / b
|
a ) rs . 340 , b ) rs . 400 , c ) rs . 450 , d ) rs . 475 , e ) none | b | add(add(divide(subtract(1210, add(add(5, 5), 4)), const_3), 4), 5) | rs . 1210 were divided among a , b , c so that a : b = 5 : 4 and b : c = 9 : 10 . then , c gets : | "solution a : b = 5 : 4 , b : c = 9 : 10 = ( 9 x 4 / 9 ) : ( 10 x 4 / 90 = 4 : 40 / 9 ∴ a : b : c = 5 : 4 : 40 / 9 = 45 : 36 : 40 . sum of ratio terms = ( 45 + 36 + 40 ) = 121 ∴ c ’ s share = rs . ( 1210 x 40 / 121 ) = rs . 400 . answer b" | a = 5 + 5
b = a + 4
c = 1210 - b
d = c / 3
e = d + 4
f = e + 5
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | c | subtract(subtract(divide(40, add(const_1, divide(60, const_100))), 1), subtract(40, subtract(divide(40, add(const_1, divide(60, const_100))), 1))) | two sacks together contained 40 kilograms of sugar . if after 1 kilogram of sugar was taken from the first sack and poured into the second the weight of the sugar in the first sack became 60 % the weight of the sugar in the second , what was the original difference in the weights of the sacks ? | say the weight of the second sack after change is x kilograms , then the weight of the first sack after change would be 0.6 x . since the total weight of sugar in both sacks remained the same then x + 0.6 x = 40 , which gives x = 25 . now , if the weight of the second sack after change is 25 kilograms then initially it was 25 − 1 = 24 kilograms and the initial weight of the first sack was 40 − 24 = 16 kilograms , so the difference was 24 − 16 = 8 kilograms . answer : c | a = 60 / 100
b = 1 + a
c = 40 / b
d = c - 1
e = 60 / 100
f = 1 + e
g = 40 / f
h = g - 1
i = 40 - h
j = d - i
|
a ) 5 , b ) 6 , c ) 8 , d ) 7 , e ) 9 | c | divide(subtract(sqrt(add(multiply(128, const_4), power(8, const_2))), 8), const_2) | find a positive number , which when increased by 8 is equal to 128 times reciprocal of the number . | explanation : let the number be x . then , x + 8 = 128 x ( 1 / x ) = > x < sup > 2 < / sup > + 8 x - 128 = 0 = > ( x + 16 ) ( x - 8 ) = 0 = > x = 8 answer : option c | a = 128 * 4
b = 8 ** 2
c = a + b
d = math.sqrt(c)
e = d - 8
f = e / 2
|
a ) 7 , b ) 5 , c ) 6 , d ) 4 , e ) 2 | d | divide(add(7, 2), power(2, 2)) | find the value of x : 2 a ^ 2 - 7 a + 10 = 14 | "using the elimination method substitute options for x and find the correct option . answer : d" | a = 7 + 2
b = 2 ** 2
c = a / b
|
a ) 22666 , b ) 20000 , c ) 25000 , d ) 22235 , e ) none of these | a | divide(multiply(3400, const_100), 15) | sheela deposits rs . 3400 in bank savings account . if this is 15 % of her monthly income . what is her monthly income in ? | "explanation : 15 % of income = rs . 3400 100 % of income = 3400 x 100 / 15 = rs . 22666 answer : a" | a = 3400 * 100
b = a / 15
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(divide(subtract(39, 15), add(2, 1)), 1) | when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of r ? | "a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a person aged 39 is added to a group of n people , the average age increases by 2 . r * 23 + 39 = 25 * ( r + 1 ) r = 7 ans . ( a )" | a = 39 - 15
b = 2 + 1
c = a / b
d = c - 1
|
a ) 10 , b ) 18 , c ) 13 , d ) 21 , e ) 27 | e | divide(108, 4) | a number when multiplied by 3 is odd and divisible by 9 . this number when multiplied by 4 is 108 . what is the original number ? | working backward , 108 divided by 4 is 27 . this number fits the rules of three and nine , where the digits add up to a sum divisible by nine . answer is e . | a = 108 / 4
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(subtract(subtract(5, 5), 1), 1) | in the coordinate plane , line a has a slope of - 1 and an x - intercept of 1 . line b has a slope of 5 and a y - intercept of - 5 . if the two lines intersect at the point ( a , b ) , what is the sum a + b ? | "the equation of line a is y = - x + 1 the equation of line b is y = 5 x - 5 5 x - 5 = - x + 1 x = 1 y = 0 the point of intersection is ( 1,0 ) and then a + b = 1 . the answer is b ." | a = 5 - 5
b = a - 1
c = b / 1
|
a ) rs . 50 , b ) rs . 45 , c ) rs . 40.5 , d ) rs . 48.5 , e ) rs . 49.5 | b | subtract(divide(multiply(divide(405, divide(4.5, const_100)), 5), const_100), 405) | the interest on a certain deposit at 4.5 % p . a . is rs . 405 in one year . how much will the additional interest in one year be on the same deposit at 5 % p . a . ? | s . i . = rs . 405 r = 4.5 % t = 1 year principal = ( 100 * 405 ) / ( 4.5 * 1 ) = rs . 9000 s . i at 5 % interest = ( 9000 * 5 * 1 ) / 100 = rs . 450 difference in interest = 450 – 405 = rs . 45 answer : b | a = 4 / 5
b = 405 / a
c = b * 5
d = c / 100
e = d - 405
|
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