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 ) 50 % , b ) 75 % , c ) 120 % , d ) 133 1 / 3 % , e ) 160 % | e | add(multiply(50, const_0_33), add(const_100, const_0_33)) | if the price of a certain bond on may 1 st was 1 / 2 the price of the bond on june 1 st and the price of the bond on july 1 st was 50 % greater than the price of the bond on may 1 st . then the price of the bond on june 1 st st was what percent of the average ( arithmetic mean ) price of the bond on may 1 st and july 1 st ? | "the price on june 1 st = 12 ( assume ) ; the price on may 1 st = 1 / 2 * 12 = 6 ; the price on july 1 st = 6 * 1.5 = 9 . the average price of the bond on may 1 st and july 1 st = ( 6 + 9 ) / 2 = 7.5 . the price of the bond on june 1 st ( 12 ) is 8 / 5 times ( 160 % ) the average price of the bond on may 1 st and july 1 st . answer : e ." | a = 50 * const_0_33
b = 100 + const_0_33
c = a + b
|
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 4 | e | add(floor(divide(log(const_3), log(add(const_1, divide(33.34, const_100))))), const_1) | an investment compounds annually at an interest rate of 33.34 % what is the smallest investment period by which time the investment will more than triple in value ? | "assume initial amount is x annual interest is 34.1 % so after 1 year the amount will become x * ( 100 + 33.34 ) / 100 = > x * 4 / 3 now we need to find n for x * ( 4 / 3 ) ^ n = 3 x or in other words n = 4 e" | a = math.log(3)
b = 33 / 34
c = 1 + b
d = math.log(c)
e = a / d
f = math.floor(e)
g = f + 1
|
a ) 4 , b ) 6 , c ) 9 , d ) 11 , e ) 14 | c | divide(add(220, 43), 67) | a no . when divided by 220 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 67 ? | "221 + 43 = 264 / 17 = 9 ( remainder ) c" | a = 220 + 43
b = a / 67
|
a ) 777 , b ) 750 , c ) 789 , d ) 776 , e ) 881 | b | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | a and b ’ s salaries together amount to rs . 3,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is b ’ s salary ? | "( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 3000 4 b = 3000 = > b = 750 answer b" | a = 2 + 3
b = 4 * a
c = 2 * b
d = c * 100
e = 85 / 100
f = 1 - e
g = d * f
h = 95 / 100
i = 1 - h
j = 85 / 100
k = 1 - j
l = i + k
m = g / l
|
a ) 18.38 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) none of these | a | multiply(divide(add(multiply(const_2, multiply(multiply(const_2, add(const_1, const_4)), const_100)), multiply(add(const_1, const_4), const_100)), divide(add(multiply(multiply(const_2, add(const_1, const_4)), const_100), multiply(add(const_2, const_4), const_100)), divide(12.5, const_100))), const_100) | ritesh and co . generated revenue of rs . 1,700 in 2006 . this was 12.5 % of its gross revenue . in 2007 , the gross revenue grew by rs . 2,500 . what is the percentage increase in the revenue in 2007 ? | "explanation : given , ritesh and co . generated revenue of rs . 1,700 in 2006 and that this was 12.5 % of the gross revenue . hence , if 1700 is 12.5 % of the revenue , then 100 % ( gross revenue ) is : = > ( 100 / 12.5 ) × 1700 . = > 13,600 . hence , the total revenue by end of 2007 is rs . 13,600 . in 2006 , revenue grew by rs . 2500 . this is a growth of : = > ( 2500 / 13600 ) × 100 . = > 18.38 % . answer : a" | a = 1 + 4
b = 2 * a
c = b * 100
d = 2 * c
e = 1 + 4
f = e * 100
g = d + f
h = 1 + 4
i = 2 * h
j = i * 100
k = 2 + 4
l = k * 100
m = j + l
n = 12 / 5
o = m / n
p = g / o
q = p * 100
|
a ) 1000 , b ) 3889 , c ) 8908 , d ) 3780 , e ) 2709 | a | divide(1120, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1000 answer : a" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 1120 / f
|
a ) rs 14000 , b ) rs 15000 , c ) rs 16000 , d ) rs 17000 , e ) none of these | b | divide(multiply(5400, const_100), multiply(12, 3)) | a man took a loan at rate of 12 % per annum simple interest . after 3 years he had to pay 5400 interest . the principal amount borrowed by him was . | "explanation : s . i . = p ∗ r ∗ t / 100 = > p = s . i . ∗ 100 / r ∗ t = > p = 5400 ∗ 100 / 12 ∗ 3 = rs 15000 option b" | a = 5400 * 100
b = 12 * 3
c = a / b
|
a ) 139 , b ) 141 , c ) 142 , d ) 143 , e ) 144 | a | power(add(power(multiply(subtract(1, divide(1, 12)), 150), 2), power(subtract(power(add(power(150, 2), power(90, 2)), divide(1, 2)), multiply(subtract(1, divide(1, 12)), 150)), 2)), divide(1, 2)) | in a village there are 150 men and 90 women in present time . if in next year population will be p = ( a ^ 2 + b ^ 2 ) ^ 1 / 2 , and in every year men are reduces 12 % . what is population of after 2 year . | "next year total population = [ 150 ^ 2 + 90 ^ 2 ] ^ . 5 = 174.92 = 175 man decreased by 11 % so total man = 150 * . 88 = 132 women will be = 175 - 132 = 43 so population after two years = [ 135 ^ 2 + 43 ^ 2 ] ^ . 5 = 138.8 = 139 so population after two year = 139 answer : a" | a = 1 / 12
b = 1 - a
c = b * 150
d = c ** 2
e = 150 ** 2
f = 90 ** 2
g = e + f
h = 1 / 2
i = g ** h
j = 1 / 12
k = 1 - j
l = k * 150
m = i - l
n = m ** 2
o = d + n
p = 1 / 2
q = o ** p
|
a ) 42 , b ) 50 , c ) 62 , d ) 72 , e ) 82 | b | subtract(58, multiply(multiply(12, const_2.0), 2)) | evaluate : 58 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 58 - 12 * 3 * 2 = 58 - 8 = 50 correct answer b" | a = 12 * 2
b = a * 2
c = 58 - b
|
a ) 5 minutes , b ) 10 minutes , c ) 15 minutes , d ) 20 minutes , e ) 25 minutes | c | subtract(30, divide(0.75, divide(1.5, 30))) | jenny has recently moved house and is working out her optimal commute . she can take an indirect route which involves taking a bus for 1 mile and then walk 0.75 miles . or she can take a direct route and walk for 1.5 miles and this takes her 30 minutes . what is the most amount time that the bus journey can take for jenny to be indifferent between the two routes assuming travel time is her only concern ? | if jenny can walk 1.5 miles in 30 minutes , then she can talk 0.75 miles in half the time i . e . 15 minutes . for jenny to be indifferent between the two routes , the two journeys must have the same travel time . the direct route has a travel time of 30 minutes ( walking ) . the indirect route has a travel time of 15 minutes ( walking ) plus the bus journey . for the two routes to be the same the bus journey must take 15 minutes ( 30 - 15 ) . answer : c | a = 1 / 5
b = 0 / 75
c = 30 - b
|
a ) 10 , b ) 12 , c ) 16 , d ) 18 , e ) 20 | c | add(15, 1) | seller selling an apple for rs . 15 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | "sp = 15 loss = cp 16 loss = cp − sp = cp − 15 ⇒ cp 16 = cp − 15 ⇒ 15 cp 16 = 15 ⇒ cp 16 = 1 ⇒ cp = 16 c" | a = 15 + 1
|
['a ) x ^ 2', 'b ) x ^ 3', 'c ) x ^ 6', 'd ) 6 x ^ 2', 'e ) 6 x ^ 3'] | d | multiply(multiply(const_2, const_3), multiply(const_1, const_1)) | if the volume of a cube is x ^ 3 cubic units , what is the number of square units in the surface area of the cube ? | the question basically asks you to calculate the surface area of the cube . volume of the cube = x ^ 3 this means each side of the cube = x hence area of each side = x ^ 2 total surfaces in a cube = 6 total surface area = 6 * x 2 correct option : d | a = 2 * 3
b = 1 * 1
c = a * b
|
a ) 2.5 min , b ) 3 min , c ) 2 min , d ) 4 min , e ) 4.5 min | c | divide(28808, multiply(7.2, const_1000)) | the area of a square field is 28808 sq m . how long will a lady take to cross the field diagonally at the rate of 7.2 km / hr ? | "area of a square field = 28808 sq m let the side of square = a a ^ 2 = 28808 = > a = 169.73 diagonal = ( 2 ) ^ ( 1 / 2 ) * a = 1.414 * 169.73 = 240 speed of lady = 7.2 km / hour = 7200 m / hour = 120 m / min time taken by lady to cross the field diagonally = 240 / 120 = 2 min answer c" | a = 7 * 2
b = 28808 / a
|
a ) a . 36000 , b ) b . 37000 , c ) c . 38000 , d ) d . 39000 , e ) e . 40000 | a | multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3)) | if a town of 35,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ? | 1 % is quite small and hence the answer is a ) | a = 3 + 4
b = 5 * a
c = b * 100
d = 3 + 4
e = d + 3
f = c * e
|
a ) 5 / 36 , b ) 7 / 64 , c ) 9 / 83 , d ) 11 / 111 , e ) 13 / 135 | d | multiply(divide(12, add(add(12, 17), 8)), divide(subtract(12, const_1), subtract(add(add(12, 17), 8), const_1))) | there are 12 slate rocks , 17 pumice rocks , and 8 granite rocks randomly distributed in a certain field . if 2 rocks are chosen at random and without replacement , what is the probability that both rocks will be slate rocks ? | "12 / 37 * 11 / 36 = 11 / 111 the answer is d ." | a = 12 + 17
b = a + 8
c = 12 / b
d = 12 - 1
e = 12 + 17
f = e + 8
g = f - 1
h = d / g
i = c * h
|
a ) 13 / 20 , b ) 11 / 15 , c ) 7 / 10 , d ) 4 / 7 , e ) 2 / 5 | e | multiply(divide(subtract(divide(multiply(4, multiply(multiply(4, 5), 6)), 6), divide(multiply(divide(multiply(4, multiply(multiply(4, 5), 6)), 6), 4), 5)), subtract(multiply(multiply(4, 5), 6), divide(multiply(3, multiply(multiply(4, 5), 6)), 4))), const_100) | one day a car rental agency rented 3 / 4 of its cars , including 5 / 6 of its cars with cd players . if 3 / 5 of its cars have cd players , what percent of the cars that were not rented had cd players ? | "the cars with cd players which were not rented is ( 1 / 6 ) ( 3 / 5 ) = 1 / 10 of all the cars . the cars which were not rented is 1 / 4 of all the cars . the percent of non - rented cars which had cd players is ( 1 / 10 ) / ( 1 / 4 ) = 2 / 5 the answer is e ." | a = 4 * 5
b = a * 6
c = 4 * b
d = c / 6
e = 4 * 5
f = e * 6
g = 4 * f
h = g / 6
i = h * 4
j = i / 5
k = d - j
l = 4 * 5
m = l * 6
n = 4 * 5
o = n * 6
p = 3 * o
q = p / 4
r = m - q
s = k / r
t = s * 100
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | factorial(const_3) | how many words , with or without meaning , can be formed using all letters of the word red using each letter exactly once ? | the word red has exactly 3 letters which are all different . therefore the number of words that can be formed = number of permutations of 3 letters taken all at a time . = p ( 3 , 3 ) = 3 ! = 3 x 2 × 1 = 6 answer : a | a = math.factorial(3)
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a ) 2 , b ) 8 , c ) 10 , d ) 15 , e ) 18 | c | subtract(65, subtract(add(30, 45), 12)) | of the 65 cars on a car lot , 30 have air - conditioning , 45 have air bag , and 12 have both air - conditioning and air bag . how many of the cars on the lot have neither air - conditioning nor air bag ? | "total - neither = all air conditioning + all air bags - both or 65 - neither = 30 + 45 - 12 = 63 . = > neither = 2 , hence c . answer : c" | a = 30 + 45
b = a - 12
c = 65 - b
|
a ) 1.5 days , b ) 4 days , c ) 3 days , d ) 1 day , e ) 2 days | e | inverse(subtract(const_1, divide(1, 2))) | a and b can do a piece of work in 2 days . with the help of c they finish the work in 1 days . c alone can do that piece of work in ? | c = 1 / 1 – 1 / 2 = 1 / 2 = > 2 days answer : e | a = 1 / 2
b = 1 - a
c = 1/(b)
|
a ) 5 , b ) 3 , c ) 4 , d ) 6 , e ) 1 | c | subtract(divide(multiply(20, add(const_4, const_1)), 10), add(const_4, const_1)) | four machines , each working at the same constant rate , together can complete a certain job in 20 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 10 days ? | "rate of one machine = 1 job / ( 20 * 4 ) days let x = number of machines needed to complete the job in 10 days 1 / ( 4 * 20 ) * 10 * x = 1 job x = 8 8 - 4 = 4 answer : c" | a = 4 + 1
b = 20 * a
c = b / 10
d = 4 + 1
e = c - d
|
a ) 84 , b ) 88 , c ) 92 , d ) 96 , e ) 100 | d | subtract(multiply(multiply(divide(32, 80), const_100), 12), multiply(32, 12)) | a car gets 32 miles to the gallon . if it is modified to use a solar panel , it will use only 80 percent as much fuel as it does now . if the fuel tank holds 12 gallons , how many more miles will the car be able to travel , per full tank of fuel , after it has been modified ? | originally , the distance the car could go on a full tank was 12 * 32 = 384 miles . after it has been modified , the car can go 32 / 0.8 = 40 miles per gallon . on a full tank , the car can go 12 * 40 = 480 miles , thus 96 miles more . the answer is d . | a = 32 / 80
b = a * 100
c = b * 12
d = 32 * 12
e = c - d
|
a ) 46 , b ) 47 , c ) 58 , d ) 56 , e ) 54 | b | divide(434.75, 9.25) | john bought 9.25 m of cloth for $ 434.75 . find the cost price per metre . | "cloth bought by john = 9.25 m cost of 9.25 m = $ 434.75 cost per metre = 434.75 ÷ 9.25 the cost of the cloth per metre = $ 47 answers : b" | a = 434 / 75
|
a ) 24 , b ) 25 , c ) 48 , d ) 50 , e ) none | c | divide(add(multiply(60, 40), multiply(40, 60)), add(60, 40)) | the average of 60 results is 40 and the average of other 40 results is 60 . what is the average of all the results ? | "answer sum of 100 result = sum of 60 result + sum of 40 result . = 60 x 40 + 40 x 60 = 4800 / 100 correct option : c" | a = 60 * 40
b = 40 * 60
c = a + b
d = 60 + 40
e = c / d
|
a ) 43 , b ) 53 , c ) 63 , d ) 33 , e ) 78 | d | subtract(negate(6), multiply(subtract(7,12, 18,25), divide(subtract(7,12, 18,25), subtract(3, 7,12)))) | 3 , 7,12 , 18,25 , . . . . . . . . . . . . . . 6 th terms | "3 + 4 = 7 7 + 5 = 12 12 + 6 = 18 18 + 7 = 25 25 + 8 = 33 answer : d" | a = negate - (
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a ) 18 , b ) 92 , c ) 27 , d ) 24 , e ) 19 | d | add(multiply(3, 4), multiply(3, 4)) | a person was asked to state his age in years . his reply was , ` ` take my age 3 years hence , multiply it by 4 and subtract 4 times my age 3 years ago and you will know how old i am . ' ' what was the age of the person ? | explanation : let the present age of person be x years . then , 4 ( x + 3 ) - 4 ( x - 3 ) = x < = > ( 4 x + 12 ) - ( 4 x - 12 ) = x < = > x = 24 . . answer : d | a = 3 * 4
b = 3 * 4
c = a + b
|
a ) 941,1009 , b ) 991,1001 , c ) 996,1004 , d ) 791,1009 , e ) 931,1009 | c | divide(999984, add(multiply(const_100, const_10), add(const_3, const_2))) | there are cats got together and decided to kill the mice of 999984 . each cat kills equal number of mice and each cat kills more number of mice than cats there were . then what are the number of cats ? | "999984 can be written as 1000000 â € “ 16 = 10002 â € “ 42 ie of the form a 2 - b 2 = ( a + b ) ( a - b ) = ( 1000 + 4 ) * ( 1000 - 4 ) = ( 1004 ) * ( 996 ) given that number of cats is less than number if mice . so number of cats is 996 and number of mice were 1004 answer c" | a = 100 * 10
b = 3 + 2
c = a + b
d = 999984 / c
|
a ) 100.6 kg , b ) 120.3 kg , c ) 117.5 kg , d ) 110.8 kg , e ) 114 kg | c | add(multiply(divide(100, add(5, 3)), 3), multiply(divide(200, add(2, 3)), 2)) | 100 kg of an alloy a is mixed with 200 kg of alloy b . if alloy a has lead and tin in the ratio 5 : 3 and alloy b has tin and copper in the ratio 2 : 3 , then the amount of tin in the new alloy is ? | "quantity of tin in 100 kg of a = 100 * 3 / 8 = 37.5 kg quantity of tin in 200 kg of b = 200 * 2 / 5 = 80 kg quantity of tin in the new alloy = 37.5 + 80 = 117.5 kg answer is c" | a = 5 + 3
b = 100 / a
c = b * 3
d = 2 + 3
e = 200 / d
f = e * 2
g = c + f
|
a ) 13 , b ) 18 , c ) 12 , d ) 15 , e ) 17 | c | multiply(multiply(divide(30, subtract(const_1, divide(1, 3))), divide(1, 3)), divide(4, 5)) | a certain automobile company ’ s best - selling model is the speedster . the speedster , like all of their other models , comes in coupe and convertible styles . 1 / 3 of the current inventory is speedsters , of which 4 / 5 are convertibles . if there are 30 vehicles that are not speedsters , how many speedster convertibles are there ? | "total vehicle = 1 / 3 of speedster + 2 / 3 of others . speedster convertibles = 1 / 3 total vehicle * 4 / 5 given : 2 / 3 constitutes 30 vehicles . hence 1 / 3 constitutes 15 speedster convertibls = 15 * 4 / 5 = 12 c" | a = 1 / 3
b = 1 - a
c = 30 / b
d = 1 / 3
e = c * d
f = 4 / 5
g = e * f
|
a ) 33 , b ) 54 , c ) 18 , d ) 45 , e ) 01 | d | multiply(subtract(70, 35), divide(90, 70)) | a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 35 adults have their meal , find the total number of children that can be catered with the remaining food . | "explanation : as there is meal for 70 adults and 35 have their meal , the meal left can be catered to 35 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 35 adults = 45 children hence , the meal can be catered to 45 children . answer : d" | a = 70 - 35
b = 90 / 70
c = a * b
|
a ) 43 , b ) 156 , c ) 68 , d ) 87 , e ) 92 | b | multiply(add(const_10, const_2), divide(1300, rectangle_area(10, 10))) | the monthly rent of a shop of dimension 10 feet × 10 feet is rs . 1300 . what is the annual rent per square foot of the shop ? | "sol . monthly rent per square feet = 1300 / ( 10 * 10 ) = 13 & annual rent per square feet = 12 * 13 = 156 answer : b" | a = 10 + 2
b = 1300 / rectangle_area
c = a * b
|
a ) 7500 , b ) 3388 , c ) 2665 , d ) 2888 , e ) 7600 | e | divide(2280, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100))) | a candidate got 35 % of the votes polled and he lost to his rival by 2280 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2280 100 % - - - - - - - - - ? = > 7600 answer : e" | a = 35 / 100
b = 1 - a
c = 35 / 100
d = b - c
e = 2280 / d
|
a ) 5 : 15 , b ) 5 : 26 , c ) 5 : 37 , d ) 5 : 48 , e ) 5 : 59 | b | subtract(divide(multiply(3, 9), 9), const_0_33) | it takes 3 workers a total of 9 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 ? | "3 workers build 1 / 9 of a machine in one hour . 1 worker builds 1 / 27 of a machine in one hour . in the first 3 hours , 3 workers build 3 * ( 1 / 27 ) * 3 = 9 / 27 of a machine from 2 : 00 to 3 : 00 , 4 workers build another 4 / 27 . the total is 13 / 27 . from 3 : 00 to 4 : 00 , 5 workers build another 5 / 27 . the total is 18 / 27 . from 4 : 00 to 5 : 00 , 6 workers build another 6 / 27 . the total is 24 / 27 . to build another 3 / 27 , 7 workers need ( 3 / 7 ) * 60 minutes which is about 26 minutes . the machine is complete at about 5 : 26 . the answer is b ." | a = 3 * 9
b = a / 9
c = b - const_0_33
|
['a ) 6.862 kg', 'b ) 6.838 kg', 'c ) 6.632 kg', 'd ) 6.832 kg', 'e ) 6.834 kg'] | d | divide(volume_cube(const_1), volume_sphere(divide(const_1, const_2))) | the ratio of the volumes of a cube to that of the sphere which will fit inside the cube is ? | 52 * 40 * 29 = 60320 48 * 36 * 27 = 46650 - - - - - - - - - - - - 13664 13664 * 1 / 2 = 6832 = > 6.832 kg answer : d | a = volume_cube / (
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a ) 4 pm , b ) 6 pm , c ) 3 pm , d ) 10 pm , e ) 5 pm | c | add(1, divide(24, add(5, 7))) | a and b start walking towards each other at 1 pm at speed of 5 kmph and 7 kmph . they were initially 24 km apart . at what time do they meet ? | "time of meeting = distance / relative speed = 24 / 7 + 5 = 24 / 12 = 2 hrs after 1 pm = 3 pm answer is c" | a = 5 + 7
b = 24 / a
c = 1 + b
|
a ) 186 m , b ) 176 m , c ) 450 m , d ) 150 m , e ) 765 m | c | multiply(divide(multiply(60, const_1000), const_3600), 27) | a train running at the speed of 60 km / hr crosses a pole in 27 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 27 ) m = 450 m . answer : c" | a = 60 * 1000
b = a / 3600
c = b * 27
|
a ) 40.00 kg , b ) 45.50 kg , c ) 66.67 kg , d ) 80.56 kg , e ) 90.00 kg | c | multiply(divide(divide(multiply(subtract(const_100, 75), 200), const_100), subtract(const_100, 25)), const_100) | fresh grapes contain 75 % by weight while dried grapes contain 25 % water by weight . what is the weight of dry grapes available from 200 kg of fresh grapes ? | "from the question we know : 200 kg * 75 % = 150 kg of water in the fresh grapes 200 kg - 150 kg of water = 50 kg of non - water mass we are looking for the weight of the dry grapes ( x ) . since the question tells us that 25 % of the weight of the dry graps is water and we know that 50 kg is non - water mass we can set up the following equation : x = 1 / 4 ( x ) + 50 kg 3 / 4 ( x ) = 50 kg x = 66.67 kg answer - c" | a = 100 - 75
b = a * 200
c = b / 100
d = 100 - 25
e = c / d
f = e * 100
|
a ) 81 : 121 , b ) 4 : 9 , c ) 729 : 1331 , d ) 27 : 121 , e ) none of these | b | power(divide(1000, 3375), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 1000 : 3375 . what is the ratio of their total surface areas ? | "ratio of the sides = â ³ â ˆ š 1000 : â ³ â ˆ š 3375 = 10 : 15 ratio of surface areas = 10 ^ 2 : 15 ^ 2 = 100 : 225 = 4 : 9 answer : b" | a = 1000 / 3375
b = 1 / 3
c = a ** b
|
a ) $ 5 , b ) $ 10 , c ) $ 12 , d ) $ 15 , e ) $ 20 | a | subtract(20, 15) | a man goes into a barbershop and gets a $ 15 haircut . he pays the barber with a $ 20 bill , but the barber does n ' t have change because it is an especially slow day . the barber goes to the flower shop next to his own and trades the $ 20 bill for a $ 10 dollar bill and two $ 5 bills . then he gives the man his $ 5 in change . later the lady from the flower shop confronts the barber and tells him that the $ 20 bill he gave her was counterfeit . the barber agrees and he gives the lady a different $ 20 bill . later that day he tries to figure out how much money he lost . what did he lose ? | he lost $ 5 . the lady from the flower shop gave him $ 20 in change , but eventually got $ 20 back . she broke even . the man who got the haircut gave the barber nothing ( except counterfeit money ) but got $ 5 back from the barber . so the barber lost only $ 5 , and gave a free haircut . | a = 20 - 15
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a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | c | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 60), 72)), divide(multiply(const_100, 60), 72))) | if the cost price of 72 articles is equal to the selling price of 60 articles , then what is the percent profit ? | "let x be the cost price of one article . let y be the selling price of one article . 60 y = 72 x y = 1.2 x the answer is c ." | a = 100 * 60
b = a / 72
c = 100 - b
d = 100 * 60
e = d / 72
f = c / e
g = 100 * f
|
a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 8 | e | divide(add(1, const_2.0), multiply(3, 8)) | of all the homes on gotham street , 1 / 3 are termite - ridden , and 5 / 8 of these are collapsing . what fraction of the homes are termite - ridden , but not collapsing ? | "the fraction of homes which are termite - infested but not collapsing is 3 / 8 * 1 / 3 = 1 / 8 the answer is e ." | a = 1 + 2
b = 3 * 8
c = a / b
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a ) rs . 8908.42 , b ) rs . 9720 , c ) rs . 10123.20 , d ) rs . 10483.20 , e ) none | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 22,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 22000 * ( 1 + 12 / 100 ) 3 ] = 22000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 30908.42 c . i . = ( 30908.42 - 22000 ) = rs . 8908.42 answer : a" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
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a ) 300 , b ) 340 , c ) 380 , d ) 420 , e ) 460 | b | divide(68, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | of the votes cast on a certain proposal , 68 more were in favor of the proposal than were against it . if the number of votes against the proposal was 40 percent of the total vote , what was the total number of votes cast ? ( each vote cast was either in favor of the proposal or against it . ) | "let x be the total number of votes cast . 0.6 x = 0.4 x + 68 0.2 x = 68 x = 340 the answer is b ." | a = 40 / 100
b = 1 - a
c = 40 / 100
d = b - c
e = 68 / d
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['a ) 54', 'b ) 62', 'c ) 25', 'd ) 34', 'e ) 60'] | b | add(multiply(const_pi, 12), multiply(const_2, 12)) | a box is in the shape of a semicircle with a radius of 12 . what is the approximate perimeter of the semicircle ? | perimeter of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 12 + 2 * 12 = 61.68 approximately 62 answer b | a = math.pi * 12
b = 2 * 12
c = a + b
|
a ) 1 / 2 , b ) 5 / 8 , c ) 9 / 14 , d ) 16 / 21 , e ) 9 / 10 | c | divide(add(multiply(divide(24, const_2), add(const_2, const_3)), multiply(24, const_2)), multiply(24, add(const_3, const_4))) | a certain electric - company plan offers customers reduced rates for electricity used between 8 p . m . and 8 a . m . weekdays and 24 hours a day saturdays and sundays . under this plan , the reduced rates x apply to what fraction of a week ? | "number of hours between 8 pm to 8 am = 12 number of hours with reduced rates = ( 12 * 5 ) + ( 24 * 2 ) hours with reduced rates x / total number of hours in a week = ( 12 * 5 ) + ( 24 * 2 ) / ( 24 * 7 ) = 108 / ( 24 * 7 ) = 9 / 14 answer : c" | a = 24 / 2
b = 2 + 3
c = a * b
d = 24 * 2
e = c + d
f = 3 + 4
g = 24 * f
h = e / g
|
a ) 30 , b ) 45 , c ) 60 , d ) 90 , e ) 120 | a | multiply(4, inverse(subtract(1, add(divide(1, 5), divide(2, 3))))) | in traveling from a dormitory to a certain city , a student went 1 / 5 of the way by foot , 2 / 3 of the way by bus , and the remaining 4 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "whole trip = distance by foot + distance by bus + distance by car x = 1 / 5 x + 2 / 3 x + 4 x - 13 / 15 x = 4 2 / 15 x = 4 = > so x = ( 15 / 2 ) * 4 = 30 km answer a" | a = 1 / 5
b = 2 / 3
c = a + b
d = 1 - c
e = 1/(d)
f = 4 * e
|
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | a | divide(subtract(power(36, const_2), power(12, const_2)), multiply(36, const_2)) | a tree of height 36 m is on one edge of a road broke at a certain height . it fell in such a way that the top of the tree touches the other edge of the road . if the breadth of the road is 12 m , then what is the height at which the tree broke ? | let the tree was broken at x meters height from the ground and 36 - x be the length of other part of the tree . from the diagram , ( 36 − x ) 2 = x 2 + 122 ( 36 − x ) 2 = x 2 + 122 ⇒ 1296 − 72 x + x 2 = x 2 + 144 ⇒ 1296 − 72 x + x 2 = x 2 + 144 ⇒ 72 x = 1296 − 144 ⇒ 72 x = 1296 − 144 ⇒ x = 16 answer : a | a = 36 ** 2
b = 12 ** 2
c = a - b
d = 36 * 2
e = c / d
|
a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 12 | e | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 3,675 multiplied by y is the square of an integer , then y must be | "i just tried plugging in the numbers and found out that 12 * 3675 = 44 , 100 , which is a square of 210 e" | a = 2 + 3
b = a * 2
|
a ) 50 , b ) 65 , c ) 70 , d ) 61 , e ) 60 | e | multiply(12, divide(20, subtract(16, 12))) | if a person walks at 16 km / hr instead of 12 km / hr , he would have walked 20 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 12 = ( x + 20 ) / 16 x - 60 = > x = 60 km . answer : e" | a = 16 - 12
b = 20 / a
c = 12 * b
|
a ) 14 years , b ) 18 years , c ) 20 years , d ) 27 years , e ) 16 years | d | divide(subtract(29, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 29 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 29 ) + 2 = 2 ( x + 2 ) x + 31 = 2 x + 4 x = 27 . answer : d" | a = 2 * 2
b = a - 2
c = 29 - b
d = 2 - 1
e = c / d
|
a ) 60 , b ) 75 , c ) 90 , d ) 100 , e ) 110 | c | subtract(subtract(multiply(25, 18), multiply(12, 20)), multiply(12, 10)) | the average of 25 results is 18 . the average of first 12 of them is 10 and that of last 12 is 20 . find the 13 th result ? | 13 th result = sum of 25 results - sum of 24 results = 18 * 25 - 12 * 10 - 12 * 20 = 450 - 120 - 240 = 90 answer is c | a = 25 * 18
b = 12 * 20
c = a - b
d = 12 * 10
e = c - d
|
a ) a ) 5 , b ) b ) 8 , c ) c ) 6 , d ) d ) 2 , e ) e ) 1 | a | add(subtract(36, add(30, 1)), 1) | the average weight of a group of boys is 30 kg . after a boy of weight 36 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 36 kg joins the group , total weight of boys = 30 x + 36 so 30 x + 36 = 31 ( x + 1 ) = > x = 5 . answer : a" | a = 30 + 1
b = 36 - a
c = b + 1
|
a ) 8 , b ) 10 , c ) 2 , d ) 17 , e ) 19 | c | subtract(30, divide(add(multiply(7.50, 30), 685), add(7.50, 25))) | a contractor is engaged for 30 days on the condition thathe receives rs . 25 for each day he works & is fined rs . 7.50 for each day is absent . he gets rs . 685 in all . for how many days was he absent ? | "30 * 25 = 750 685 - - - - - - - - - - - 65 25 + 7.50 = 32.5 62 / 32.5 = 2 c" | a = 7 * 50
b = a + 685
c = 7 + 50
d = b / c
e = 30 - d
|
a ) 17 , b ) 18 , c ) 54 , d ) 108 , e ) 162 | e | divide(multiply(multiply(72, 48), 24), volume_cube(divide(24, const_2))) | a box measuring 72 inches long by 48 inches wide by 24 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 . 8 is the biggest number that could divide all three , 72 , 48 and 24 . thus side of cube must be 8 , and total number of cubes = 72 / 8 * 48 / 8 * 24 / 8 = 162 ans e it is ." | a = 72 * 48
b = a * 24
c = 24 / 2
d = b / volume_cube
|
a ) 900 , b ) 840 , c ) 750 , d ) 860 , e ) 3530 | e | divide(multiply(add(80, divide(370, 860)), 1550), const_100) | 80 370 860 1550 2440 ? | "10 ^ 2 - 20 = 80 20 ^ 2 - 30 = 370 30 ^ 2 - 40 = 860 40 ^ 2 - 50 = 1550 50 ^ 2 - 60 = 2440 60 ^ 2 - 70 = 3530 . answer : e" | a = 370 / 860
b = 80 + a
c = b * 1550
d = c / 100
|
a ) 30 % gain , b ) 30 % loss , c ) 40 % gain , d ) 40 % loss , e ) 52 % loss | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 12)), divide(multiply(const_100, 25), 12))) | if the cost price of 12 articles is same as the selling price of 25 articles . find the gain or loss percentage ? | "explanation : 12 cp = 25 sp cost price cp = 25 selling price sp = 12 formula = ( sp - cp ) / cp * 100 = ( 12 - 25 ) / 25 * 100 = 52 % loss answer : option e" | a = 100 * 25
b = a / 12
c = 100 - b
d = 100 * 25
e = d / 12
f = c / e
g = 100 * f
|
a ) 330 , b ) 200 , c ) 150 , d ) 100 , e ) 91 | d | divide(multiply(200, 50), const_100) | the cost of an article is decreased by 50 % . if the original cost is $ 200 , find the decrease cost . | original cost = $ 200 decrease in it = 50 % of $ 200 = 50 / 100 ã — 200 = 10000 / 100 = $ 100 therefore , decrease cost = $ 200 - $ 100 = $ 100 answer : d | a = 200 * 50
b = a / 100
|
a ) 0 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | a | divide(1, 37) | what is the 16 th digit to the right of the decimal point in the decimal expansion of 1 / 37 ? | "1 / 37 = 0.027027 . . . so , we have a repeating cycle of 027 . every third digit ( 2 nd , 4 th , 6 th , . . . ) to the right of the decimal point is alternating by 2 , 0 , 7 thus 16 th digit is 0 . answer : a ." | a = 1 / 37
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a ) 90 , b ) 82 , c ) 80 , d ) 45 , e ) 36 | c | subtract(subtract(subtract(const_1000, 700), const_1), add(multiply(multiply(subtract(const_10, divide(700, const_100)), subtract(const_10, const_1)), subtract(const_10, const_2)), const_3)) | of the 3 - digit integers greater than 700 , how many have two digits t that are equal to each other and the remaining digit different from the other two ? | three - digit integers greater than 700 : 701 to 999 , inclusive . possible values for hundreds - digit - - > 7 , 89 possible values for tens - digit and ones - digit - - > 0 , 1 , 23 , 45 , 67 , 89 when hundreds - digit and tens - digit are the same : ( 3 x 1 x 10 ) - 3 = 27 - - - > we minus three to exclude 777 , 888 and 999 when hundreds - digit and ones - digit are the same : ( 3 x 10 x 1 ) - 3 = 27 - - - > we minus three to exclude 777 , 888 and 999 when tens - digit and hundreds - digit are the same : [ ( 3 x 10 x 1 ) - 3 ] - 1 = 26 - - - > we minus three to exclude 777 , 888 and 999 ; we minus one to exclude 700 t = 27 + 27 + 26 = 80 answer : c | a = 1000 - 700
b = a - 1
c = 700 / 100
d = 10 - c
e = 10 - 1
f = d * e
g = 10 - 2
h = f * g
i = h + 3
j = b - i
|
a ) 30 , 10 , b ) 25 , 5 , c ) 29 , 9 , d ) 50 , 30 , e ) 20,10 | a | add(divide(add(multiply(5, 5), subtract(20, 5)), subtract(5, const_1)), 20) | the ages of two person 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 | "let their ages be x and ( x + 20 ) years . then , 5 ( x - 5 ) = ( x + 20 - 5 ) = > 4 x = 40 = > x = 10 their present ages are 30 years and 10 year . answer : a" | a = 5 * 5
b = 20 - 5
c = a + b
d = 5 - 1
e = c / d
f = e + 20
|
a ) 20 % , b ) 28 % , c ) 27 % , d ) 35 % , e ) 70 % | d | subtract(divide(subtract(const_100, 10), divide(2, 3)), const_100) | what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 10 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 90 2 / 3 sp 1 = 90 sp 1 = 135 100 - - - 35 = > 35 % answer : d" | a = 100 - 10
b = 2 / 3
c = a / b
d = c - 100
|
a ) 5 , b ) 10 , c ) 15 , d ) 25 , e ) 35 | a | subtract(20, subtract(40, add(10, 15))) | out of 40 applicants to a law school , 15 majored in political science , 20 had a grade point average higher than 3.0 , and 10 did not major in political science and had a gpa equal to or lower than 3.0 . how many w applicants majored in political science and had a gpa higher than 3.0 ? | "total applicants = 40 political science = 15 and non political science = 40 - 15 = 25 gpa > 3.0 = 20 and gpa < = 3.0 = 20 10 non political science students had gpa < = 3.0 - - > 15 non political science students had gpa > 3.0 gpa > 3.0 in political science = total - ( gpa > 3.0 in non political science ) w = 20 - 15 = 5 answer : a" | a = 10 + 15
b = 40 - a
c = 20 - b
|
a ) 80 kg , b ) 73 kg , c ) 70 kg , d ) 75 kg , e ) 85 kg | b | add(58, multiply(10, add(1, divide(const_1, 1)))) | the average weight of 10 men is increased by 1 ½ kg when one of the men who weighs 58 kg is replaced by a new man . what is the weight of the new man ? | "since the average has increased by 1.5 kg , the weight of the man who stepped in must be equal to 58 + 10 x 1.5 58 + 15 = 73 kg ans : ' b '" | a = 1 / 1
b = 1 + a
c = 10 * b
d = 58 + c
|
a ) 1500 , b ) 200 , c ) 1700 , d ) 2000 , e ) 2500 | a | multiply(500, const_3) | the price of a car and ac are in the ratio 3 : 2 . if the scooter costs $ 500 more than the ac . then the price of the ac ? | let the price of the car and ac is 3 x and 2 x 3 x - 2 x = 500 x = 500 price of ac = 3 * 500 = $ 1500 answer is a | a = 500 * 3
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a ) 24 , b ) 28 , c ) 30 , d ) 32 , e ) 50 | e | subtract(divide(multiply(add(const_100, 44), const_100), subtract(const_100, 4)), const_100) | a shopkeeper sold an article offering a discount of 4 % and earned a profit of 44 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 144 let marked price be rs . x . then , 96 / 100 x = 144 x = 14400 / 96 = rs . 150 now , s . p . = rs . 150 , c . p . = rs . 100 profit % = 50 % . answer : e" | a = 100 + 44
b = a * 100
c = 100 - 4
d = b / c
e = d - 100
|
a ) $ 4 , b ) $ 0.3 , c ) $ 1 , d ) $ 3 , e ) $ 1.65 | b | multiply(divide(add(multiply(divide(add(100, 100), 100), divide(3, const_2)), multiply(divide(3, const_2), divide(subtract(100, 70), 100))), 3.45), divide(subtract(100, 70), 100)) | the cost per pound of green tea and coffee were the same in june . in july , the price of coffee shot up by 100 % and that of green tea dropped by 70 % . if in july , a mixture containing equal quantities of green tea and coffee costs $ 3.45 for 3 lbs , how much did a pound of green tea cost in july ? | lets assume price of coffee in june = 100 x price of green tea in june = 100 x price of coffee in july = 200 x ( because of 100 % increase in price ) price of green tea in july = 30 x ( because of 70 % decrease in price ) price of 1.5 pound of coffee 1.5 pound of green tea in july will be = 300 x + 45 x = 345 x as per question 345 x = 3.45 $ x = 0.01 s so the price of tea in july = 30 x = 30 x 0.01 = 0.3 $ / pound answer b | a = 100 + 100
b = a / 100
c = 3 / 2
d = b * c
e = 3 / 2
f = 100 - 70
g = f / 100
h = e * g
i = d + h
j = i / 3
k = 100 - 70
l = k / 100
m = j * l
|
a ) 69 , b ) 72 , c ) 75 , d ) 71 , e ) none of them | d | add(multiply(1.8, 10), 53) | the average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew , who weighs 53 kg is replaced by a new man . find the weight of the new man . | "total weight increased = ( 1.8 x 10 ) kg = 18 kg . : . weight of the new man = ( 53 + 18 ) kg = 71 kg . answer is d" | a = 1 * 8
b = a + 53
|
a ) $ 770 , b ) $ 660 , c ) $ 700 , d ) $ 1100 , e ) $ 840 | a | multiply(divide(49, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100))) | bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 49 more . what was his original selling price ? | let the original purchase price be x so original selling price at 10 % profit = 1.1 x if product is purchased at 10 % less of original = 0.9 x profit of 30 % on this price = 1.3 ( 0.9 x ) he would have received $ 49 more in second scenario = > 1.3 ( 0.9 x ) - 1.1 x = 49 = > 0.07 x = 49 = > x = $ 700 original purchase price = $ 700 hence , original selling price ( at 10 % of profit ) = 1.1 ( 700 ) = $ 770 option a | a = 10 / 100
b = 1 - a
c = 30 / 100
d = 1 + c
e = b * d
f = 10 / 100
g = 1 + f
h = e - g
i = 49 / h
j = 10 / 100
k = 1 + j
l = i * k
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a ) 3 % , b ) 4 % , c ) 5 % , d ) 6 % , e ) 7 % | b | divide(multiply(const_100, subtract(612, 600)), divide(600, 2)) | on a sum of money , the s . i . for 2 years is $ 600 , while the c . i . is $ 612 , the rate of interest being the same in both the cases . the rate of interest is ? | difference in c . i . and s . i for 2 years = $ 612 - $ 600 = $ 12 s . i for one year = $ 300 s . i . on $ 300 for 1 year = $ 12 rate = ( 100 * 12 ) / ( 300 ) = 4 % the answer is b . | a = 612 - 600
b = 100 * a
c = 600 / 2
d = b / c
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a ) 81 , b ) 82 , c ) 85 , d ) 90 , e ) 95 | c | add(add(divide(551, gcd(551, 1073)), gcd(551, 1073)), divide(1073, gcd(551, 1073))) | 3 numbers which are co - prime to each other are such that the product of the first two is 551 and that of the last two is 1073 . the sum of the 3 numbers is : | since the numbers are co - prime , they contain only 1 as the common factor . also , the given two products have the middle number in common . so , middle number = h . c . f of 551 and 1073 = 29 ; first number = 551 / 29 = 19 third number = 1073 / 29 = 37 . required sum = 19 + 29 + 37 = 85 . answer : c | a = math.gcd(551, 1073)
b = 551 / a
c = math.gcd(551, 1073)
d = b + c
e = math.gcd(551, 1073)
f = 1073 / e
g = d + f
|
a ) 26 km / hr , b ) 17 km / hr , c ) 60 km / hr , d ) 77 km / hr , e ) 46 km / hr | c | divide(multiply(multiply(divide(add(100, 100), multiply(8, add(const_1, const_2))), const_2), const_3600), const_1000) | two trains , each 100 m long , moving in opposite directions , cross other in 8 sec . if one is moving twice as fast the other , then the speed of the faster train is ? | "let the speed of the slower train be x m / sec . then , speed of the train = 2 x m / sec . relative speed = ( x + 2 x ) = 3 x m / sec . ( 100 + 100 ) / 8 = 3 x = > x = 25 / 3 . so , speed of the faster train = 50 / 3 = 50 / 3 * 18 / 5 = 60 km / hr . answer : c" | a = 100 + 100
b = 1 + 2
c = 8 * b
d = a / c
e = d * 2
f = e * 3600
g = f / 1000
|
a ) 198 mph , b ) 110 mph , c ) 88 mph , d ) 100 mph , e ) 99 mph | e | divide(add(110, 88), const_2) | i flew my tiny seaplane to visit my mother . on the flight up , i flew at 110 mph . on the way home , i flew 88 mph . what was my average speed for the trip ? | "( 110 mph + 88 mph ) / 2 = 99 mph correct answer is : e" | a = 110 + 88
b = a / 2
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a ) 205 km , b ) 195 km , c ) 295 km , d ) 115 km , e ) 215 km | b | multiply(65, 3) | a person is traveling at 65 km / hr and reached his destiny in 3 hr then find the distance ? | "t = 3 hrs d = t * s = 65 * 3 = 195 km answer is b" | a = 65 * 3
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a ) 96 cm 2 , b ) 104 cm 2 , c ) 78 cm 2 , d ) 86 cm 2 , e ) 76 cm 2 | a | multiply(12, 8) | find the area of a parallelogram with base 12 cm and height 8 cm ? | "area of a parallelogram = base * height = 12 * 8 = 96 cm 2 answer : a" | a = 12 * 8
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a ) 2480 , b ) 3490 , c ) 6785 , d ) 8235 , e ) 9255 | d | subtract(divide(multiply(multiply(30, add(30, 1)), add(multiply(2, 30), 1)), add(3, 3)), 1260) | the sum of the squares of the first 15 positive integers ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 15 ^ 2 ) is equal to 1260 . what is the sum of the squares of the second 15 positive integers ( 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 ) ? | "you ' ll never need a formula for the sums of squares on the actual gmat . you do n ' t need to use that formula here , though it ' s not all that straightforward to solve without one . two different approaches : 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 = ( 15 + 1 ) ^ 2 + ( 15 + 2 ) ^ 2 + ( 15 + 3 ) ^ 2 + . . . + ( 15 + 15 ) ^ 2 now we can expand each square ; they are all in the ( x + y ) ^ 2 = x ^ 2 + 2 xy + y ^ 2 pattern . = ( 15 ^ 2 + 2 * 15 + 1 ^ 2 ) + ( 15 ^ 2 + 4 * 15 + 2 ^ 2 ) + ( 15 ^ 2 + 6 * 15 + 3 ^ 2 ) + . . . + ( 15 ^ 2 + 30 * 15 + 15 ^ 2 ) now we have fifteen 15 ^ 2 terms , so adding these gives 15 * 15 ^ 2 = 15 ^ 3 = 3375 . we also have the entire sum 1 ^ 2 + 2 ^ 2 + . . . + 15 ^ 2 , which we know is equal to 1240 . finally adding the middle terms , we have : 2 * 15 + 4 * 15 + 6 * 15 + . . . + 30 * 15 = 15 ( 2 + 4 + 6 + . . . . + 30 ) = 15 * 2 * ( 1 + 2 + 3 + . . . + 15 ) = 15 * 2 * 8 * 15 = 3600 so the sum must be 3375 + 1240 + 3600 = 8215 alternatively , we can use a different factoring pattern . we want to find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 . well if we subtract 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 from this , the answer will be 1240 less than what we want to find . so if we can find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 - ( 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 ) then we can add 1240 to get the answer . now grouping the terms above to get differences of squares , we have = ( 30 ^ 2 - 15 ^ 2 ) + ( 29 ^ 2 - 14 ^ 2 ) + . . . + ( 16 ^ 2 - 1 ^ 2 ) and factoring each of these using x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) , we have = 45 * 15 + 43 * 15 + 41 * 15 + . . . + 17 * 15 = 15 ( 45 + 43 + 41 + . . . + 17 ) in brackets we have an equally spaced sum with fifteen terms , which we can evaluate using the familiar formula . so the above equals 15 * 15 * 62 / 2 = 6975 and adding back the 1260 , we get the answer of 8235 . ( ans d )" | a = 30 + 1
b = 30 * a
c = 2 * 30
d = c + 1
e = b * d
f = 3 + 3
g = e / f
h = g - 1260
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a ) 15265 , b ) 16458 , c ) 25600 , d ) 17576 , e ) 14785 | d | multiply(multiply(divide(add(4, const_100), const_100), 15625), divide(add(4, const_100), const_100)) | if the annual increase in the population of a town is 4 % and the present number of people is 15625 , what will the population be in 3 years ? | "the required population is = 15625 ( 1 + 4 / 100 ) ^ 3 = 15625 * 26 / 25 * 26 / 25 * 26 / 25 = 17576 answer is d" | a = 4 + 100
b = a / 100
c = b * 15625
d = 4 + 100
e = d / 100
f = c * e
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a ) 180 , b ) 130 , c ) 140 , d ) 160 , e ) 224 | e | multiply(divide(multiply(8, 70), subtract(13, 8)), const_2) | a sports retailer ordered white and yellow tennis balls in equal number but the dispatch clerk of the tennis ball company erred and dispatched 70 extra yellow balls and making ratio of white balls to yellow balls 8 / 13 . how many tennis balls did the retailer order originally . | "white : yellow = x : ( x + 70 ) = 8 : 13 - - > 13 x = 8 x + 560 - - > x = 112 . the total # of balls originally x + x = 112 + 112 = 224 . answer : e ." | a = 8 * 70
b = 13 - 8
c = a / b
d = c * 2
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a ) 7500 , b ) 8880 , c ) 6200 , d ) 5200 , e ) none of them | b | multiply(80, power(100, 80)) | 80 x 100 + 80 x 11 = ? | "= 80 x ( 100 + 11 ) ( by distributive law ) = 80 x 111 = 8880 answer is b" | a = 100 ** 80
b = 80 * a
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a ) 237 , b ) 270 , c ) 177 , d ) 166 , e ) 397.2 | e | floor(divide(8200, add(20, divide(3.2, const_100)))) | find the number of shares that can be bought for rs . 8200 if the market value is rs . 20 each with brokerage being 3.2 % . | "explanation : cost of each share = ( 20 + 3.2 % of 20 ) = rs . 20.64 therefore , number of shares = 8200 / 20.64 = 397.2 answer : e" | a = 3 / 2
b = 20 + a
c = 8200 / b
d = math.floor(c)
|
a ) a ) 125 , b ) b ) 627 , c ) c ) 153 , d ) d ) 721 , e ) e ) 272 | e | divide(multiply(850, subtract(const_100, add(add(44, 14), 10))), const_100) | in a school of 850 boys , 44 % of muslims , 14 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ? | "44 + 14 + 10 = 68 % 100 – 68 = 32 % 850 * 32 / 100 = 272 answer : e" | a = 44 + 14
b = a + 10
c = 100 - b
d = 850 * c
e = d / 100
|
a ) 40 , b ) 50 , c ) 51 , d ) 79 , e ) 80 | c | divide(divide(subtract(125, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | a train 125 m long passes a man , running at 6 km / hr in the same direction in which the train is going , in 10 sec . the speed of the train is ? | "speed of the train relative to man = 125 / 10 = 25 / 2 m / sec . = 25 / 2 * 18 / 5 = 45 km / hr let the speed of the train be x km / hr . then , relative speed = ( x - 6 ) km / hr . x - 6 = 45 = > x = 51 km / hr . answer : option c" | a = 6 * const_0_2778
b = a * 6
c = 125 - b
d = c / 6
e = d / const_0_2778
|
a ) 21 , b ) 24 , c ) 44 , d ) 62 , e ) 70 | c | multiply(11, const_4) | what could be the range of a set consisting of odd multiples of 11 ? | range = the difference between the greatest and the smallest numbers in the sequence . our sequence is odd and is a multiple of 7 . every number in that sequence can be represented like this : 11 * ( 2 n + 1 ) where n is any positive integer . range = 11 * ( 2 m + 1 ) - 11 * ( 2 n + 1 ) = 11 * 2 * ( m - n ) = 22 * ( m - n ) . m , n - any positive integers the answer must be divisible by 22 , which is only 44 . the correct answer is c | a = 11 * 4
|
a ) 177 , b ) 168 , c ) 120 , d ) 882 , e ) 180 | e | divide(18, subtract(divide(18, 10), 6)) | a train covers a distance of 18 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 18 / 10 * 60 ) km / hr = ( 108 * 5 / 18 ) m / sec = 30 m / sec . length of the train = 30 * 6 = 180 m . answer : e" | a = 18 / 10
b = a - 6
c = 18 / b
|
a ) 3 litres , b ) 2 litres , c ) 4 litres , d ) 4.5 litres , e ) 5 litres | d | subtract(divide(multiply(multiply(5, divide(45, const_100)), const_100), 25), 5) | what quantity of water should be added to reduce 5 litres of 45 % acidic liquid to 25 % acidic liquid ? | "quantity of acid in 5 litres = 45 * 5 / 100 = 2.25 litres let x litres of water is added to the solution , then there is 2.25 litres of acid in ( 5 + x ) litres of liquid . 25 % of ( 5 + x ) = 2.25 , 25 / 100 * ( 5 + x ) = 2.25 x = 225 - 125 / 25 = 4 answer : d" | a = 45 / 100
b = 5 * a
c = b * 100
d = c / 25
e = d - 5
|
a ) $ 60,000 , b ) $ 25,000 , c ) $ 35,000 , d ) $ 40,000 , e ) $ 45,000 | a | multiply(divide(multiply(2.55, multiply(10, 7)), 425), const_100) | a certain company had a total annual expenditure of 2.55 ∗ 10 ^ 7 on employee salaries last year . if the company employed 425 people , what was the average employee salary ? | "given : total annual expenditure of 2.55 ∗ 10 ^ 7 on employee salaries total employees = 425 observe that 425 * 6 = 2550 therefore try to bring the numerator in terms of 2550 average salary = ( 2550 * 10 ^ 4 ) / 425 = 6 * 10 ^ 4 = 60,000 option a" | a = 10 * 7
b = 2 * 55
c = b / 425
d = c * 100
|
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | e | divide(add(242, 100), 18) | a no . when divided by 242 gives a remainder 100 , what remainder will be obtainedby dividingthe same no . 18 ? | "242 + 100 = 342 / 18 = 19 ( remainder ) e" | a = 242 + 100
b = a / 18
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | add(const_4, const_4) | if each year the population of the country grows by 40 % , how many years will elapse before the population of the country doubles ? | "till year 2000 , population is 100 . year 2001 : population becomes 140 . . . . . . . . . . . . . 1 year elapsed year 2002 : population becomes 196 . . . . . . . . . . . . . 2 year elapsed year 2003 : population > 200 . . . . . . . . . . . . . . . . . . 3 year elapsed answer : a" | a = 4 + 4
|
a ) 1 / 15 , b ) 1 / 16 , c ) 1 / 11 , d ) 1 / 10 , e ) 1 / 12 | e | subtract(add(divide(const_1, 18), divide(const_1, 20)), divide(const_1, 45)) | two pipes can fill a tank in 18 minutes and 20 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "part of the filled by all the three pipes in one minute = 1 / 18 + 1 / 21 - 1 / 45 = 1 / 12 so , the tank becomes full in 12 minutes . answer : e" | a = 1 / 18
b = 1 / 20
c = a + b
d = 1 / 45
e = c - d
|
a ) s . 7000 , b ) s . 9000 , c ) s . 14000 , d ) s . 17000 , e ) s . 42000 | e | divide(840, divide(multiply(subtract(14, 12), const_2), const_100)) | a certain sum is invested at simple interest at 14 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 840 . find the sum ? | "let the sum be rs . x . ( x * 14 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 840 = > 28 x / 100 - 24 x / 100 = 840 = > 4 x / 100 = 840 = > x = 42000 . answer : e" | a = 14 - 12
b = a * 2
c = b / 100
d = 840 / c
|
a ) 4 / 19 , b ) 7 / 19 , c ) 12 / 37 , d ) 21 / 95 , e ) none | c | subtract(const_1, divide(choose(subtract(23, 4), const_2), choose(23, const_2))) | a box contains 23 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is | solution p ( none is defective ) = 19 c 2 / 23 c 2 = 25 / 37 . p ( at least one is defective ) = ( 1 - 25 / 37 ) = 12 / 37 . answer c | a = 23 - 4
b = math.comb(a, 2)
c = math.comb(23, 2)
d = b / c
e = 1 - d
|
a ) 6 days , b ) 18 days , c ) 21 days , d ) 3 days , e ) 15 days | e | multiply(const_3, 5) | aarti can do a piece of work in 5 days . in how many days will she complete three time of work of same type ? | we have the important relation , more work , more time ( days ) a piece of work can be done in 5 days . three times of work of same type can be done in 5 x 3 = 15 days answer e | a = 3 * 5
|
['a ) 2,000', 'b ) 1,800', 'c ) 1,900', 'd ) 1,700', 'e ) none of these'] | c | multiply(subtract(add(multiply(70, 5), multiply(30, 5)), multiply(5, 5)), 4) | a rectangular lawn 70 m × 30 m has two roads each 5 metres wide , running in the middle of it , one parallel to the length and the other parallel to the breadth . find the cost of gravelling the road at the rate of 4 per square metre . | total area of road = area of road which parallel to length + area of road which parallel to breadth – overlapped road = 70 × 5 + 30 × 5 – 5 × 5 = 350 + 150 – 25 = 500 – 25 = 475 m 2 ∴ cost of gravelling the road = 475 × 4 = 1900 answer c | a = 70 * 5
b = 30 * 5
c = a + b
d = 5 * 5
e = c - d
f = e * 4
|
a ) 52 , b ) 92 , c ) 95 , d ) 88 , e ) 66 | b | add(multiply(add(15, subtract(15, 3)), 3), 11) | in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 11 people all together . how many people can sit in a bus ? | "right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 11 total people can seat = 81 + 11 = 92 answer : b" | a = 15 - 3
b = 15 + a
c = b * 3
d = c + 11
|
a ) 79 , b ) 77 , c ) 85 , d ) 83 , e ) 61 | e | add(multiply(multiply(divide(25, const_100), 20), multiply(divide(25, const_100), 20)), divide(subtract(300, 20), 20)) | in a 300 member association consisting of men and women , exactly 20 % of men and exactly 25 % women are homeowners . what is the least number of members who are homeowners ? | "solution simple out of 300 20 % are male i . e 60 and 25 % are female i . e 75 , so total homeowner is 135 . now min number homeowner is 60 and max is 135 so question ask us to find least and 61 has least value among all option . so ans is 61 . answer : e" | a = 25 / 100
b = a * 20
c = 25 / 100
d = c * 20
e = b * d
f = 300 - 20
g = f / 20
h = e + g
|
a ) 2 : 3 , b ) 4 : 3 , c ) 3 : 6 , d ) 4 : 5 , e ) none | a | divide(sqrt(4), sqrt(9)) | two cars , one from austin to dallas and the other from dallas to austin , start simultaneously . after they meet , the cars reach their destinations after 4 hours and 9 hours respectively the ratio of their speeds is | solution let us name the cars as a and b . = ( car a ' s speed ) : ( car b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 4 : â ˆ š 9 = 2 : 3 answer a | a = math.sqrt(4)
b = math.sqrt(9)
c = a / b
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(multiply(multiply(150, 0.09), 0.3), const_2) | chlorination is used in industry for the purification of water ; 1 gm solution of chlorine will remove half of the germs in 1 liter of water . if water containing less than 0.09 % of germs is considered clean , what is the minimum amount of chlorine required to purify 150 liter of water containing 0.3 liter of harmful substances ? | assuming that it ' s supposed to state ` ` 0.3 ' ' liters of harmful substances , here ' s how you can answer this question . first , we have to figure out what . 09 % of 150 liters is . . . 1 % = . 01 . 1 % = . 001 . 09 % = . 0009 so , . 0009 ( 150 ) = . 135 liters we need enough doses of chlorine to get the 0.3 liters down to less than . 135 liters . each ` ` dose ' ' of chlorine will remove half of the harmful substances . . . start = 0.3 liters after 1 st dose = . 15 liters after 2 nd dose = . 075 liters we ' re now below the . 135 liters that we need to be less than . final answer : b | a = 150 * 0
b = a * 0
c = b / 2
|
a ) 49 , b ) 50 , c ) 12 , d ) 13 , e ) 68 | e | divide(divide(subtract(175, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 175 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | "speed of the train relative to man = ( 175 / 10 ) m / sec = ( 35 / 2 ) m / sec . [ ( 35 / 2 ) * ( 18 / 5 ) ] km / hr = 63 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 63 = = > x = 68 km / hr . answer : option e" | a = 5 * const_0_2778
b = a * 5
c = 175 - b
d = c / 5
e = d / const_0_2778
|
a ) 18.3 % , b ) 15.3 % , c ) 21.3 % , d ) 22.3 % , e ) 25.3 % | c | multiply(divide(subtract(add(multiply(10, 12), 27), multiply(10, 12)), 127), const_100) | on a purchase of $ 127 , a store offered a payment plan consisting of a $ 27 down payment and 12 monthly payments of $ 10 each . what percent of the purchase price , to the nearest tenth of a percent , did the customer pay in interest by using this plan ? | "12 * 10 + 27 = 147 ( 27 / 127 ) * 100 = 21.3 answer : c" | a = 10 * 12
b = a + 27
c = 10 * 12
d = b - c
e = d / 127
f = e * 100
|
['a ) 56', 'b ) 80', 'c ) 100', 'd ) 120', 'e ) 144'] | e | multiply(factorial(subtract(5, const_1)), 6) | in how many r ways can 5 people from a group of 6 people be seated around a circular table | r = 6 c 5 * ( 5 - 1 ) ! ( select 5 out of 6 and arrange them in circular manner ) = 6 * 4 ! = 6 * 24 = 144 answer - e | a = 5 - 1
b = math.factorial(a)
c = b * 6
|
a ) 50 , b ) 100 , c ) 250 , d ) 300 , e ) 400 | c | divide(multiply(50, 100), 20) | 50 men do a work in 100 days . how many men are needed to finish the work in 20 days ? | men required to finish the work in 2 days = 50 * 100 / 20 = 250 answer is c | a = 50 * 100
b = a / 20
|
a ) 4.44 , b ) 5.44 , c ) 6.44 , d ) 7.44 , e ) 8.44 | a | divide(multiply(const_100, 160), multiply(900, 4)) | what is the rate percent when the simple interest on rs . 900 amount to rs . 160 in 4 years ? | "160 = ( 900 * 4 * r ) / 100 r = 4.44 % answer : a" | a = 100 * 160
b = 900 * 4
c = a / b
|
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