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 ) s . 3600 , b ) s . 5000 , c ) s . 4500 , d ) s . 4800 , e ) s . 5800 | a | divide(3969, power(add(divide(5, const_100), const_1), 2)) | a sum amounts to rs . 3969 in 2 years at the rate of 5 % p . a . if interest was compounded yearly then what was the principal ? | "ci = 3969 , r = 5 , n = 2 ci = p [ 1 + r / 100 ] ^ 2 = p [ 1 + 5 / 100 ] ^ 2 3969 = p [ 21 / 20 ] ^ 2 3969 [ 20 / 21 ] ^ 2 3600 answer : a" | a = 5 / 100
b = a + 1
c = b ** 2
d = 3969 / c
|
a ) 15 , b ) 20 , c ) 30 , d ) 35 , e ) 45 | a | divide(subtract(const_100, add(35, 20)), const_3) | a polling company surveyed a certain country , and it found that 35 % of that country ’ s registered voters had an unfavorable impression of both of that state ’ s major political parties and that 20 % had a favorable impression only of party r . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b , then what percentage of the country ’ s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? | "s = 100 not ( r and b ) = 35 only r = 20 ( r and b ) / b = 1 / 2 let ( r and b ) = x only b = 2 x so now , 20 + 35 + x + 2 x = 100 x = 15 a ans" | a = 35 + 20
b = 100 - a
c = b / 3
|
a ) 25 , b ) 35 , c ) 45 , d ) 55 , e ) 65 | c | divide(multiply(38.25, const_100), 85) | how many pieces of 85 cm length can be cut from a rod of 38.25 meters long ? | "number of pieces = 3825 / 85 = 45 the answer is c ." | a = 38 * 25
b = a / 85
|
a ) 12 th , b ) 13 th , c ) 14 th , d ) 15 th , e ) 16 th | d | subtract(divide(multiply(5.6, const_1000), 350), const_1) | gretzky street begins at orr street and runs directly east for 5.6 kilometers until it ends when it meets howe street . gretzky street is intersected every 350 meters by a perpendicular street , and each of those streets other than orr street and howe street is given a number beginning at 1 st street ( one block east of orr street ) and continuing consecutively ( 2 nd street , 3 rd street , etc . . . ) until the highest - numbered street one block west of howe street . what is the highest - numbered street that intersects gretzky street ? | 5.6 km / 350 m = 16 . however , the street at the 5.6 - km mark is not 16 th street ; it is howe street . therefore , the highest numbered street is 15 th street . the answer is d . | a = 5 * 6
b = a / 350
c = b - 1
|
a ) 10.6 kg , b ) 10.8 kg , c ) 11 kg , d ) 13 kg , e ) none | d | subtract(multiply(add(19, const_1), 14.9), multiply(19, 15)) | the average weight of 19 students is 15 kg . by the admission of a new student the average weight is reduced to 14.9 kg . the weight of the new student is ? | "answer weight of new student = total weight of all 20 students - total weight of initial 19 students = ( 20 x 14.9 - 19 x 15 ) kg = 13 kg . correct option : d" | a = 19 + 1
b = a * 14
c = 19 * 15
d = b - c
|
a ) 4 , b ) 8 , c ) 16 , d ) 12 , e ) 14 | b | subtract(add(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2)), subtract(multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2), divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)))) | the difference between a two - digit number and the number obtained by interchanging the digits is 36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ? | "since the number is greater than the number obtained on reversing the digits , so the ten ' s digit is greater than the unit ' s digit . let ten ' s and unit ' s digits be 2 x and x respectively . then , ( 10 x 2 x + x ) - ( 10 x + 2 x ) = 36 9 x = 36 x = 4 . required difference = ( 2 x + x ) - ( 2 x - x ) = 2 x = 8 . answer : b" | a = 2 * 10
b = a + 1
c = b - 10
d = c - 2
e = 36 / d
f = 2 * 10
g = f + 1
h = g - 10
i = h - 2
j = 36 / i
k = j * 2
l = e + k
m = 2 * 10
n = m + 1
o = n - 10
p = o - 2
q = 36 / p
r = q * 2
s = 2 * 10
t = s + 1
u = t - 10
v = u - 2
w = 36 / v
x = r - w
y = l - x
|
a ) 15 min , b ) 18 min , c ) 16 min , d ) 20 min , e ) 30 min | e | subtract(multiply(const_1, const_60), multiply(divide(40, 80), const_60)) | excluding stoppages , the average speed of a bus is 80 km / hr and including stoppages , the average speed of the bus is 40 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 80 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 80 - 40 ) km i . e 40 km at 80 km / hr . stoppage time = 40 / 80 hrs = 30 min answer : e" | a = 1 * const_60
b = 40 / 80
c = b * const_60
d = a - c
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | add(floor(divide(12, const_3)), const_1) | what is the smallest integer n for which 25 ^ n > 5 ^ 12 ? | "to solve , we want to get the bases the same . thus we need to break 25 ^ n into prime factors . 25 ^ n = ( 5 ^ 2 ) ^ n = 5 ^ ( 2 n ) ( remember that when we have a power to a power , we multiply the exponents . ) we can use the new value in the given inequality : 5 ^ ( 2 n ) > 5 ^ 12 since we have the same bases on either side of the inequality we can drop the bases and set up an equation involving just the exponents . 2 n > 12 n > 6 because n is greater than 6 , the smallest integer that satisfies the inequality 25 ^ n > 5 ^ 12 is 7 . the answer is b ." | a = 12 / 3
b = math.floor(a)
c = b + 1
|
a ) 26 : 64 , b ) 64 : 9 , c ) 9 : 64 , d ) 9 : 60 , e ) 20 : 64 | c | divide(multiply(multiply(4, 3), 3), multiply(multiply(4, 4), 4)) | a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 4 , and the ratio of soap to salt is five times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is quartered by water evaporation . at that time , what is the ratio of water to salt in the solution ? | "water : soap = 3 : 4 soap : salt = 15 : 20 = > for 15 soap , salt = 20 = > for 4 soap , salt = ( 20 / 15 ) * 4 = 80 / 15 = 16 / 3 so , water : soap : salt = 3 : 4 : 16 / 3 = 9 : 12 : 16 after open container , water : soap : salt = 2.25 : 12 : 16 so , water : salt = 2.25 : 16 = 9 : 64 answer : c" | a = 4 * 3
b = a * 3
c = 4 * 4
d = c * 4
e = b / d
|
a ) 25 % , b ) 30 % , c ) 50 % , d ) 20 % , e ) 10 % | e | divide(multiply(10, const_100), subtract(110, 10)) | by selling an article for $ 110 , a person gains $ 10 . what is the gain % ? | "s . p . = $ 110 gain = $ 10 c . p . = 110 - 10 = 100 gain % = 10 / 100 * 100 % = 10 % answer is e" | a = 10 * 100
b = 110 - 10
c = a / b
|
a ) 64 , b ) 82 , c ) 128 , d ) 512 , e ) 4096 | e | power(4, multiply(2, 3)) | tough and tricky questions : functions . let a be a positive integer . let n # a equal n ^ ( 2 a ) if a is odd and n ^ ( 3 a ) if a is even . then ( 4 # 3 ) + ( 3 # 4 ) – ( 3 # 3 ) is equal to | answer : 4 ^ 6 + 3 ^ 6 - 3 ^ 6 = 4 ^ 6 = 4096 ans e | a = 2 * 3
b = 4 ** a
|
a ) 100011 , b ) 111111 , c ) 101111 , d ) 201111 , e ) 211111 | a | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 111) | find the smallest number of 6 digits which is exactly divisible by 111 . | smallest number of 6 digits is 100000 . on dividing 100000 by 111 , we get 100 as remainder . number to be added = ( 111 - 100 ) - 11 . hence , required number = 100011 option a | a = 100 * 100
b = 10 * a
c = b - 100
d = c + 111
|
a ) 78,000 , b ) 80,000 , c ) 82,000 , d ) 84,000 , e ) 86,000 | d | subtract(const_100, multiply(const_4, const_2)) | a company recently conducted a survey and found that 35,000 of its customers live in rural areas . if the number of customers who live in urban areas is 140 percent greater than the number of customers who live in rural areas , how many customers live in urban areas ? | "the number of customers in urban areas is 35,000 + 1.4 * 35,000 = 84,000 . the answer is d ." | a = 4 * 2
b = 100 - a
|
a ) $ 80,000 , b ) $ 130,000 , c ) $ 240,000 , d ) $ 290,000 , e ) $ 340,000 | e | divide(subtract(subtract(multiply(multiply(add(60, 20), const_1000), const_100), multiply(20, multiply(add(60, 20), const_1000))), multiply(multiply(multiply(divide(20, const_2), 20), const_1000), 60)), 20) | in plutarch enterprises , 60 % of the employees are marketers , 20 % are engineers , and the rest are managers . marketers make an average salary of $ 50,000 a year , and engineers make an average of $ 80,000 . what is the average salary for managers if the average for all employees is also $ 80,000 ? | "for sake of ease , let ' s say there are 10 employees : 6 marketers , 2 engineers , and 2 manager . average company salary * number of employees = total company salary > > > $ 80,000 * 10 = $ 800,000 subtract the combined salaries for the marketers ( 6 * $ 50,000 ) and the engineers ( 2 * $ 80,000 ) > > > $ 800,000 - $ 300,000 - $ 160,000 = $ 340,000 . the correct answer is e ." | a = 60 + 20
b = a * 1000
c = b * 100
d = 60 + 20
e = d * 1000
f = 20 * e
g = c - f
h = 20 / 2
i = h * 20
j = i * 1000
k = j * 60
l = g - k
m = l / 20
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 8 | e | add(8, divide(subtract(20, 20), add(25, 20))) | two stations a and b are 20 km apart on a straight line . one train starts from a at 7 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ? | "suppose they meet x hours after 7 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 20 45 x = 45 x = 1 . so , they meet at 8 a . m . answer : option e" | a = 20 - 20
b = 25 + 20
c = a / b
d = 8 + c
|
a ) 2 / 7 , b ) 3 / 5 , c ) 3 / 11 , d ) 1 / 4 , e ) 7 / 16 | d | divide(multiply(choose(4, const_2), choose(add(6, 3), const_1)), choose(add(add(6, 3), 4), 6)) | a bag contains 6 red , 3 yellow and 4 green balls . 3 balls are drawn randomly . what is the probability that the balls drawn contain balls of different colours ? | "total number of balls = 6 + 3 + 4 = 13 n ( s ) = 13 c 3 = 286 n ( e ) = 6 c 1 * 3 c 1 * 4 c 1 = 72 probability = 72 / 286 = 1 / 4 answer is d" | a = math.comb(4, 2)
b = 6 + 3
c = math.comb(b, 1)
d = a * c
e = 6 + 3
f = e + 4
g = math.comb(f, 6)
h = d / g
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | a | divide(divide(multiply(multiply(8, 12), 2), 12), 8) | a crate measures 2 feet by 8 feet by 12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius , in feet , of the pillar with the largest volume that could still fit in the crate ? | "we can find the radius of all the three cases of cylinders . the only crux to find the answer faster is that : voulme is pi * r ^ 2 * h . the volume is a function of r ^ 2 . so r has to be the highest to find the largest volume . so r = 2 for the surface 8 * 12 face . volume = 8 pi answer a" | a = 8 * 12
b = a * 2
c = b / 12
d = c / 8
|
a ) 3 : 15 , b ) 1 : 2 , c ) 3 : 20 , d ) 20 : 3 , e ) none | a | divide(multiply(0.06, const_100), multiply(0.3, const_100)) | if 0.3 of a number is equal to 0.06 of another number , the ratio of the numbers i | "sol . 0.3 a = 0.06 b â ‡ ” a / b = 0.06 / 0.30 = 6 / 30 = 3 / 15 . â ˆ ´ a : b = 3 : 15 . answer a" | a = 0 * 6
b = 0 * 3
c = a / b
|
a ) 95 , b ) 91 , c ) 102 , d ) 101 , e ) 85 | b | subtract(multiply(4, const_2), multiply(2, const_2)) | if the average ( arithmetic mean ) of x , x + 2 , and x + 4 is 93 , what is the value of x ? | "am of x , x + 2 , and x + 4 = x + ( x + 2 ) + ( x + 4 ) / 3 = 3 x + 6 / 3 = x + 2 given that x + 2 = 93 x = 91 answer : b" | a = 4 * 2
b = 2 * 2
c = a - b
|
a ) 4.1 , b ) 4.5 , c ) 4.8 , d ) 5.4 , e ) 9 | e | divide(divide(1800, const_1000), divide(multiply(12, const_60), const_3600)) | a person crosses a 1800 m long street in 12 minutes . what is his speed in km per hour ? | "speed = 1800 / ( 12 x 60 ) m / sec = 2.5 m / sec . converting m / sec to km / hr = 2.5 x ( 18 / 5 ) km / hr = 9 km / hr . answer : e" | a = 1800 / 1000
b = 12 * const_60
c = b / 3600
d = a / c
|
a ) 3980 , b ) 3600 , c ) 3700 , d ) 3800 , e ) 3900 | e | divide(multiply(const_100, 1023), 12) | calculate the amount that an investor needs to be invest to earn $ 1023 in interest in 24 months if the investor plans to invest x dollars in a savings account that pays interest at an annual rate of 12 % compounded semi - annually ? | "the approach is substitution , our interest requirement is $ 1023 after 24 months , 2 compounding period . calculate the compound interest on each option and find out the one that yields $ 1023 in 24 months 3900 yielded $ 1023 using the formula a = p ( 1 + r / n ) nt hence answer is e" | a = 100 * 1023
b = a / 12
|
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 c 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 c / 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 ) 33 , b ) 77 , c ) 26 , d ) 28 , e ) 20 | e | divide(multiply(multiply(subtract(9, 1), add(9, 1)), 4), add(add(9, 1), subtract(9, 1))) | a person can row at 9 kmph and still water . he takes 4 1 / 2 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ? | "let the distance between a and b be x km . total time = x / ( 9 + 1 ) + x / ( 9 - 1 ) = 4.5 = > x / 10 + x / 8 = 9 / 2 = > ( 4 x + 5 x ) / 40 = 9 / 2 = > x = 20 km . answer : e" | a = 9 - 1
b = 9 + 1
c = a * b
d = c * 4
e = 9 + 1
f = 9 - 1
g = e + f
h = d / g
|
a ) 5 . , b ) 10 . , c ) 14 . , d ) 15 . , e ) 17.5 | e | multiply(divide(add(35, divide(35, const_2)), 6), const_2) | the distance from steve ' s house to work is 35 km . on the way back steve drives twice as fast as he did on the way to work . altogether , steve is spending 6 hours a day on the roads . what is steve ' s speed on the way back from work ? | time is in the ratio 2 : 1 : : to : fro office therefore , 2 x + 1 x = 6 hrs time take to come back - 2 hrs , distance travelled - 35 km = > speed = 17.5 kmph e | a = 35 / 2
b = 35 + a
c = b / 6
d = c * 2
|
a ) s . 4528 , b ) s . 4520 , c ) s . 4527 , d ) s . 4530 , e ) s . 2718 | e | multiply(subtract(multiply(add(multiply(25, 12), multiply(15, 12)), const_2), add(multiply(3, multiply(4, 3)), multiply(6, 3))), 3) | the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the 4 walls of the room at rs . 3 per square feet if there is one door of dimensions 6 feet * 3 feet and 3 windows of dimensions 4 feet * 3 feet each ? | area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 3 = rs . 2718 answer : e | a = 25 * 12
b = 15 * 12
c = a + b
d = c * 2
e = 4 * 3
f = 3 * e
g = 6 * 3
h = f + g
i = d - h
j = i * 3
|
a ) 123 , b ) 127 , c ) 235 , d ) 305 , e ) 505 | b | gcd(subtract(2037, 5), subtract(1657, 6)) | the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively , is : | "explanation : required number = h . c . f . of ( 1657 - 6 ) and ( 2037 - 5 ) = h . c . f . of 1651 and 2032 = 127 . answer : b" | a = 2037 - 5
b = 1657 - 6
c = math.gcd(a, b)
|
a ) 33 , b ) 37 , c ) 54 , d ) 99 , e ) 01 | c | divide(multiply(300, subtract(const_100, add(28, 54))), const_100) | in an examination , 300 students appeared . out of these students ; 28 % got first division , 54 % got second division and the remaining just passed . assuming that no student failed ; find the number of students who just passed . | "the number of students with first division = 28 % of 300 = 28 / 100 × 300 = 8400 / 100 = 84 and , the number of students with second division = 54 % of 300 = 54 / 100 × 300 = 16200 / 100 = 162 therefore , the number of students who just passed = 300 – ( 84 + 162 ) = 54 answer : c" | a = 28 + 54
b = 100 - a
c = 300 * b
d = c / 100
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | e | add(13, const_1) | the average of first 13 even numbers is ? | "sum of 13 even numbers = 13 * 14 = 110 average = 182 / 13 = 14 answer : e" | a = 13 + 1
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a ) 41.1 sec , b ) 20.2 sec , c ) 31.8 sec , d ) 50.4 sec , e ) none of the above | d | divide(add(250, 520), multiply(55, const_0_2778)) | a 250 m long train is running at a speed of 55 km / hr . it crossed a platform of length 520 m in ? | "speed = 55 km / hr ( to convert km / hr in to m / s ) = 55 x 5 / 18 m / s distance = 250 m + 520 m ( if questions is about train crossing a post you need to consider only the length of train , ) = 770 m time = distance / speed = 770 x 18 / ( 5 x 55 ) = 50.4 sec ans is : d" | a = 250 + 520
b = 55 * const_0_2778
c = a / b
|
a ) 7.3 , b ) 8.1 , c ) 9.5 , d ) 10.2 , e ) 11.6 | b | divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 15 percent of 54 ? | "( 15 / 100 ) * 54 = 8.1 the answer is b ." | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 15 * n
p = o / 100
|
a ) 600 , b ) 450 , c ) 650 , d ) 500 , e ) 550 | a | subtract(720, divide(multiply(subtract(1020, 720), 2), 5)) | a sum of money at simple interest amount to rs 720 after 2 years and to rs 1020 after a further period of 5 years . the sum is : | simple interest for 5 year = rs . ( 1020 - 720 ) = rs . 300 . simple interest for 2 year = rs . ( 300 / 5 × 2 ) = rs . 120 . principal = rs . ( 720 - 120 ) = rs . 600 . answer : a | a = 1020 - 720
b = a * 2
c = b / 5
d = 720 - c
|
a ) 23777 , b ) 35000 , c ) 29977 , d ) 26777 , e ) 19871 | b | subtract(100000, divide(subtract(multiply(divide(11, const_100), 100000), multiply(divide(add(9, divide(2, 3)), const_100), 100000)), subtract(divide(11, const_100), divide(9, const_100)))) | an amount of rs . 100000 is invested in two types of shares . the first yields an interest of 9 % p . a and the second , 11 % p . a . if the total interest at the end of one year is 9 2 / 3 % , then the amount invested in each share was ? | let the sum invested at 9 % be rs . x and that invested at 11 % be rs . ( 100000 - x ) . then , ( x * 9 * 1 ) / 100 + [ ( 100000 - x ) * 11 * 1 ] / 100 = ( 100000 * 29 / 3 * 1 / 100 ) ( 9 x + 1100000 - 11 x ) / 100 = 29000 / 3 = 9700 ( 9 x + 1100000 - 11 x ) = 970000 x = 65000 sum invested at 9 % = rs . 65000 sum invested at 11 % = rs . ( 100000 - 65000 ) = rs . 35000 . answer : b | a = 11 / 100
b = a * 100000
c = 2 / 3
d = 9 + c
e = d / 100
f = e * 100000
g = b - f
h = 11 / 100
i = 9 / 100
j = h - i
k = g / j
l = 100000 - k
|
a ) s . 350 , b ) s . 700 , c ) s . 200 , d ) s . 600 , e ) s . 500 | c | divide(100, multiply(divide(5, const_100), 10)) | a sum was put at simple interest at a certain rate for 10 years . had it been put at 5 % higher rate , it would have fetched rs . 100 more . what was the sum ? | "at 5 % more rate , the increase in s . i for 10 years = rs . 100 ( given ) so , at 5 % more rate , the increase in si for 1 year = 100 / 10 = rs . 10 / - i . e . rs . 10 is 5 % of the invested sum so , 1 % of the invested sum = 10 / 5 therefore , the invested sum = 10 × 100 / 5 = rs . 200 answer : c" | a = 5 / 100
b = a * 10
c = 100 / b
|
a ) $ 9,300 , b ) $ 10,400 , c ) $ 11,500 , d ) $ 12,600 , e ) $ 13,700 | c | multiply(multiply(2, const_3), const_100) | a merchant gets a 5 % discount on each meter of fabric he buys after the first 2,000 meters and a 7 % discount on every meter after the next 1,500 meters . the price , before discount , of one meter of fabric is $ 2 , what is the total amount of money the merchant spends on 6,000 meters of fabric ? | "for first 2000 meters he does not get any discount . the price is 2 * 2000 = $ 4000 for next 1500 meters , he gets a 5 % discount . the price is 1.9 * 1500 = $ 2850 for the next 1500 meters , he gets a 7 % discount . the price is 1.86 * 2500 = $ 4650 the total price is $ 4000 + $ 2850 + $ 4650 = $ 11,500 the answer is c ." | a = 2 * 3
b = a * 100
|
a ) 1 / 18 , b ) 1 / 24 , c ) 1 / 30 , d ) 1 / 36 , e ) 1 / 48 | c | divide(choose(4, 3), choose(add(6, 4), 3)) | there are 6 red balls and 4 blue balls in a jar . if 3 balls are selected from the jar , what is the probability that all 3 balls selected are blue balls ? | the number of ways of choosing 3 balls from the jar is 10 c 3 = 120 . the number of ways of choosing 3 blue balls is 4 c 3 = 4 . p ( 3 blue balls ) = 4 / 120 = 1 / 30 . the answer is c . | a = math.comb(4, 3)
b = 6 + 4
c = math.comb(b, 3)
d = a / c
|
a ) $ 80,000 , b ) $ 130,000 , c ) $ 240,000 , d ) $ 370,000 , e ) $ 320,000 | d | divide(subtract(subtract(multiply(multiply(add(70, 10), const_1000), const_100), multiply(const_10, multiply(add(70, 10), const_1000))), multiply(multiply(multiply(divide(const_10, const_2), const_10), const_1000), 70)), 10) | in plutarch enterprises , 70 % of the employees are marketers , 10 % are engineers , and the rest are managers . marketers make an average salary of $ 50,000 a year , and engineers make an average of $ 80,000 . what is the average salary for managers if the average for all employees is also $ 80,000 ? | for sake of ease , let ' s say there are 10 employees : 7 marketers , 1 engineers , and 2 manager . average company salary * number of employees = total company salary > > > $ 80,000 * 10 = $ 800,000 subtract the combined salaries for the marketers ( 7 * $ 50,000 ) and the engineers ( $ 80,000 ) > > > $ 800,000 - $ 350,000 - $ 80,000 = $ 370,000 . the correct answer is d . | a = 70 + 10
b = a * 1000
c = b * 100
d = 70 + 10
e = d * 1000
f = 10 * e
g = c - f
h = 10 / 2
i = h * 10
j = i * 1000
k = j * 70
l = g - k
m = l / 10
|
a ) 4 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | d | multiply(multiply(add(inverse(multiply(const_2, const_4)), const_1), const_3), multiply(const_2, const_4)) | working together , wayne and his son can shovel the entire driveway in three hours . if wayne can shovel eight times as fast as his son can , how many hours would it take for his son to shovel the entire driveway on his own ? | w : the time for wyane to do the job s : the time for his son to do the job we have 1 / w + 1 / s = 1 / 8 and w = 8 s then we have 1 / ( 8 * s ) + 1 / s = 1 / 8 < = > 9 / ( 8 * s ) = 1 / 8 < = > s = 9 ans : d | a = 2 * 4
b = 1/(a)
c = b + 1
d = c * 3
e = 2 * 4
f = d * e
|
a ) 2 : 3 , b ) 5 : 6 , c ) 4 : 5 , d ) 13 : 1 , e ) 8 : 1 | d | subtract(7, 6) | a boat running up stram takes 6 hours to cover a certain distance , while it takes 7 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ? | "explanation : let speed of boat is x km / h and speed stream is y km / hr 6 ( x + y ) = 7 ( x - y ) 6 x + 6 y = 7 x - 7 y 13 y = x 13 y = x x / y = 13 / 1 13 : 1 answer : option d" | a = 7 - 6
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a ) 7 / 17 , b ) 14 / 31 , c ) 7 / 15 , d ) 17 / 35 , e ) 1 / 2 | e | divide(add(3, 2), const_10) | company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 1.6 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ? | the easiest way for me is to plug in numbers . let the number of basic stereos produced be 40 , and number of delux stereos produced be 20 . total of 60 stereos . if it takes an hour to produce a basic stereo then it will take 1.6 hours to produce a deluxe stereo . 40 basic stereos = 40 hours . 20 delux stereos = 32 hours . total hours = 72 then the fraction would be 32 / 72 = 4 / 9 . therefore answer e | a = 3 + 2
b = a / 10
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a ) 150 , b ) 200 , c ) 250 , d ) 300 , e ) 237.5 | a | divide(subtract(add(60, 50), 75), subtract(divide(add(60, 50), 60), divide(75, 50))) | a car traveled 75 % of the way from town a to town b at an average speed of 50 miles per hour . the car travels at an average speed of s miles per hour for the remaining part of the trip . the average speed for the entire trip was 60 miles per hour . what is s ? | "total distance = 100 miles ( easier to work with % ) 75 % of the distance = 75 miles 25 % of the distance = 25 miles 1 st part of the trip → 75 / 50 = 1.5 2 nd part of the trip → 25 / s = t total trip → ( 75 + 25 ) / 60 = 1.5 + t » 100 / 60 = 1.5 + t » 2.5 = 1.5 + t » t = 0.1667 back to 2 nd part of the trip formula : 25 / s = 0.1667 » s = 150 ans a" | a = 60 + 50
b = a - 75
c = 60 + 50
d = c / 60
e = 75 / 50
f = d - e
g = b / f
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a ) 300 , b ) 600 , c ) 500 , d ) 112 , e ) 620 | b | divide(12, subtract(134.02, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 12 . if n / j = 134.02 , what is value of j ? | "1 ) we know that decimal part of decimal quotient = { remainder / divisor } so 0.02 , the decimal part of the decimal quotient , must equal the remainder , 12 , divided by the divisor j . 0.02 = 12 / j 0.02 * j = 12 j = 12 / 0.02 = 1200 / 2 = 600 / 1 = 600 so j = 600 , answer = b ." | a = 4 * 10
b = a + 2
c = 100 + b
d = 134 - 2
e = 12 / d
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a ) 1 / 21 , b ) 1 / 12 , c ) 1 / 9 , d ) 1 / 6 , e ) 1 / 3 | a | divide(divide(factorial(6), multiply(factorial(2), factorial(5))), factorial(6)) | joshua and jose work at an auto repair center with 5 other workers . for a survey on health care insurance , 2 of the 6 workers will be randomly chosen to be interviewed . what is the probability that joshua and jose will both be chosen ? | "two methods 1 ) probability of chosing josh first = 1 / 7 probability of chosing jose second = 1 / 6 total = 1 / 42 probability of chosing jose first = 1 / 7 probability of chosing josh second = 1 / 6 total = 1 / 42 final = 1 / 42 + 1 / 42 = 1 / 21 a" | a = math.factorial(6)
b = math.factorial(2)
c = math.factorial(5)
d = b * c
e = a / d
f = math.factorial(6)
g = e / f
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a ) 4.5 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | c | multiply(multiply(10, const_4.0), divide(2, 2)) | in the coordinate plane , points ( x , 2 ) and ( 10 , y ) are on line k . if line k passes through the origin and has slope 1 / 2 , then x + y = | "line k passes through the origin and has slope 1 / 2 means that its equation is y = 1 / 2 * x . thus : ( x , 2 ) = ( 4 , 2 ) and ( 10 , y ) = ( 10,5 ) - - > 4 + 5 = 2 + 5 = 9 . answer : c ." | a = 10 * 4
b = 2 / 2
c = a * b
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a ) 26 , b ) 27 , c ) 30 , d ) 35 , e ) 42 | a | divide(add(multiply(2, 20), multiply(3, 30)), 5) | the average score of a cricketer in 2 matches is 20 and in other 3 matches is 30 . then find the average score in all the 5 matches ? | "average in 5 matches = ( 2 * 20 + 3 * 30 ) / 2 + 3 = 40 + 90 / 5 = 130 / 5 = 26 answer is a" | a = 2 * 20
b = 3 * 30
c = a + b
d = c / 5
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a ) 25 , b ) 19 , c ) 39 , d ) 61 , e ) 16 | a | multiply(multiply(4, divide(10, 4)), divide(10, 4)) | 4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 10 mat - weavers in 10 days ? | "let the required number of bottles be x . more weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) wavers 4 : 10 : : 4 : x days 4 : 10 4 * 4 * x = 10 * 10 * 4 x = ( 10 * 10 * 4 ) / ( 4 x 4 ) x = 25 . answer is a ." | a = 10 / 4
b = 4 * a
c = 10 / 4
d = b * c
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a ) 8 , b ) 10 , c ) 9 , d ) 6 , e ) 5 | a | divide(subtract(78, 54), subtract(54, 51)) | rahul played weel in this season . his current batting averagge is 51 . if he score 78 runs in today match . his batting average will become 54 . how many matches had he played in this season . | "51 x + 78 = 54 ( x + 1 ) = > 3 x = 24 = > x = 8 answer : a" | a = 78 - 54
b = 54 - 51
c = a / b
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a ) 80 % , b ) 105 % , c ) 120 % , d ) 112 % , e ) 138 % | d | multiply(divide(multiply(subtract(const_1, divide(20, const_100)), divide(14, const_100)), divide(10, const_100)), const_100) | in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 20 percent , but profits were 14 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,112 r = x / 100 * 0.1 r answer d" | a = 20 / 100
b = 1 - a
c = 14 / 100
d = b * c
e = 10 / 100
f = d / e
g = f * 100
|
a ) 65 days , b ) 30 days , c ) 10 days , d ) 16 days , e ) 18 days | b | divide(multiply(subtract(31, 25), 400), 320) | a garrison of 400 men had a provision for 31 days . after 25 days 320 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? | "400 - - - 31 400 - - - 6 80 - - - ? 400 * 6 = 80 * x = > x = 30 days . answer : b" | a = 31 - 25
b = a * 400
c = b / 320
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a ) 46 , b ) 36 , c ) 18 , d ) 22 , e ) 23 | b | divide(add(240, 120), divide(multiply(subtract(45, 9), const_1000), const_3600)) | a jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . the train is running at 45 kmph in the same direction . how much time does it take for the train to pass the jogger ? | explanation : distance to be covered = 240 + 120 = 360 m relative speed = 36 km / hr = 36 × 10 / 36 = 10 m / s time = distance / speed = 360 / 10 = 36 seconds answer : option b | a = 240 + 120
b = 45 - 9
c = b * 1000
d = c / 3600
e = a / d
|
a ) 2 m , b ) 2.5 m , c ) 1.5 m , d ) 1.25 m , e ) 1 m | e | divide(multiply(10, 5), subtract(rectangle_area(10, 10), rectangle_area(5, 5))) | the dimensions of a field are 10 m by 10 m . a pit 5 m long , 5 m wide and 3 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field . what will be the rise in the height of field as a result of this operation ? | "the volume of the earth removed is 5 * 5 * 3 = 75 m ^ 3 . the remaining area of the field is 10 * 10 - 5 * 5 = 75 m ^ 2 . 75 m ^ 3 of the earth evenly spread over the area of 75 m ^ 2 will rise the height by ( height ) = ( volume ) / ( area ) = 75 / 75 = 1 m . answer : e" | a = 10 * 5
b = rectangle_area - (
c = a / b
|
a ) rs . 9471 , b ) rs . 12,628 , c ) rs . 912.54 , d ) rs . 18,942 , e ) none | c | multiply(multiply(const_0_25, const_100), 10) | suganya and suriya are partners in a business . suganya invests rs . 32,000 for 8 months and suriya invests rs . 28,000 for 10 months . out of a profit of rs . 30,570 . suganya ' s share is | "solution ratio of their shares = ( 32000 ã — 8 ) : ( 28000 ã — 10 ) = 32 : 35 . suganya ' s share = rs . ( 31570 ã — 2 / 67 ) = rs . 912.54 . answer c" | a = const_0_25 * 100
b = a * 10
|
a ) 13 , b ) 59 , c ) 35 , d ) 37 , e ) 12 | d | add(24, 11) | a number when divided by a divisor leaves a remainder of 24 . when twice the original number is divided by the same divisor , the remainder is 11 . what is the value of the divisor ? | "explanatory answer decoding ` ` a number when divided by a divisor leaves a remainder of 24 ' ' let the original number be ' a ' . let the divisor be ' d ' . let the quotient of dividing ' a ' by ' d ' be ' x ' . therefore , we can write the division as a / d = x and the remainder is 24 . i . e . , a = dx + 24 decoding ` ` when twice the original number is divided by the same divisor , the remainder is 11 ' ' twice the original number is divided by d means 2 a is divided by d . we know that a = dx + 24 . therefore , 2 a = 2 ( dx + 48 ) or 2 a = 2 dx + 48 when ( 2 dx + 48 ) is divided by ' d ' the remainder is 11 . 2 dx is divisible by ' d ' and will therefore , not leave a remainder . the remainder of 11 would be the remainder of dividing 48 by d . the question is ` ` what number will leave a remainder of 11 when it divides 48 ? ' ' when 37 divides 48 , the remainder is 11 . hence , the divisor is 37 . choice d" | a = 24 + 11
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a ) 12.50 % , b ) 13.50 % , c ) 14 % , d ) 23.75 % , e ) none | d | multiply(subtract(subtract(add(const_1, divide(65, const_100)), multiply(add(const_1, divide(65, const_100)), divide(25, const_100))), const_1), const_100) | an uneducated retailer marks all his goods at 65 % above the cost price and thinking that he will still make 25 % profit , offers a discount of 25 % on the marked price . what is his actual profit on the sales ? | "sol . let c . p . = rs . 100 . then , marked price = rs . 165 . s . p . = 75 % of rs . 165 = rs . 123.75 . ∴ gain % = 23.75 % . answer d" | a = 65 / 100
b = 1 + a
c = 65 / 100
d = 1 + c
e = 25 / 100
f = d * e
g = b - f
h = g - 1
i = h * 100
|
a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) 8 hours | c | divide(126, add(16, 5)) | a boat can travel with a speed of 16 km / hr in still water . if the rate of stream is 5 km / hr , then find the time taken by the boat to cover distance of 126 km downstream . | explanation : it is very important to check , if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( 16 + 5 ) = 21 kmph time = distance / speed = 126 / 21 = 6 hours option c | a = 16 + 5
b = 126 / a
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a ) 75 , b ) 125 , c ) 675 , d ) 1100 , e ) 1250 | c | multiply(42, multiply(divide(15, 7), divide(15, 2))) | if 7 a = 2 b = 15 , then 42 ab = | "7 a * 2 b = 15 * 15 = 225 14 ab = 225 i . e . 42 ab = 675 answer : option c" | a = 15 / 7
b = 15 / 2
c = a * b
d = 42 * c
|
a ) 130 , b ) 140 , c ) 110 , d ) 120 , e ) none | d | add(180, divide(add(multiply(50, 9), 180), 9)) | 10 friends went to a hotel and decided to pay the bill amount equally . but 9 of them could pay rs . 50 each as a result 10 th has to pay rs . 180 extra than his share . find the amount paid by him . | "explanation : average amount paid by 9 persons = rs . 50 increase in average due to rs . 50 paid extra by the 10 th men = rs . 180 / 9 = rs . 20 therefore , average expenditure of 10 friends = rs . 50 + rs . 20 = rs . 70 therefore , amount paid by the 10 th men = rs . 70 + rs . 50 = rs . 120 correct option : d" | a = 50 * 9
b = a + 180
c = b / 9
d = 180 + c
|
a ) 1.05 , b ) 1.16 , c ) 1.2 , d ) 1.3 , e ) none of these | a | divide(multiply(0.75, 7), 5) | if 0.75 : x : : 5 : 7 , then x is equal to : | "explanation : ( x * 5 ) = ( 0.75 * 7 ) x = 5.25 / 5 = 1.05 answer : a" | a = 0 * 75
b = a / 5
|
a ) 67 / 90 , b ) 29 / 60 , c ) 17 / 30 , d ) 19 / 30 , e ) 11 / 15 | a | multiply(add(multiply(7, 3), 1), multiply(divide(1, 3), divide(1, 5))) | a new tower has just been built at the verbico military hospital ; the number of beds available for patients at the hospital is now 7 times the number available before the new tower was built . currently , 1 / 3 of the hospital ' s original beds , as well as 1 / 5 of the beds in the new tower , are occupied . for the purposes of renovating the hospital ' s original wing , all of the patients in the hospital ' s original beds must be transferred to beds in the new tower . if patients are neither admitted nor discharged during the transfer , what fraction of the beds in the new tower will be unoccupied once the transfer is complete ? | "i think a - 67 / 90 is the correct answer . here goes : lets assume originally the number of beds = x after the new tower , the total combined no of beds = 7 x so old = x , new = 6 x now 1 / 3 of x are occupied and 1 / 5 of 6 x are occupied which simplifies to ( 6 / 5 ) x we are shifting 1 / 3 of x to the new ward so there will now be : 1 / 3 of x plus 6 / 5 of x occupied in the new ward . add them up to get 23 / 15 of x there are 6 x beds in new tower so ratio is : ( 23 / 15 ) x / 6 x = 23 / 90 of x subtract that from 90 / 90 of x and you get the number of un - occupied beds to total capacity of new tower = 67 / 90 . a" | a = 7 * 3
b = a + 1
c = 1 / 3
d = 1 / 5
e = c * d
f = b * e
|
a ) 4 kmph , b ) 5 kmph , c ) 6 kmph , d ) 7 kmph , e ) none of these | b | multiply(const_3_6, multiply(add(divide(750, add(multiply(7, const_60), divide(multiply(const_1, const_60), const_2))), divide(750, 675)), divide(const_1, const_2))) | a man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes . his rowing speed in s ɵ ll water is | explanation : rate upstream = ( 750 / 675 ) = 10 / 9 m / sec rate downstream ( 750 / 450 ) m / sec = 5 / 3 m / sec rate in still water = ( 1 / 2 ) * [ ( 10 / 9 ) + ( 5 / 3 ) ] m / sec . = 25 / 18 m / sec = ( 25 / 18 ) * ( 18 / 5 ) kmph = 5 kmph answer : b | a = 7 * const_60
b = 1 * const_60
c = b / 2
d = a + c
e = 750 / d
f = 750 / 675
g = e + f
h = 1 / 2
i = g * h
j = const_3_6 * i
|
a ) rs . 1200 , b ) rs . 1500 , c ) rs . 2100 , d ) rs . 2000 , e ) none of these | c | multiply(divide(divide(multiply(15, const_2), 10), add(divide(multiply(15, const_2), 10), divide(multiply(15, const_2), 15))), 3500) | a alone can finish a work in 10 days which b alone can finish in 15 days . if they work together and finish it , then out of a total wages of rs . 3500 , a will get : | "explanation : ratio of working days of a : b = 10 : 15 therefore , their wages ratio = reverse ratio = 15 : 10 therefore , a will get 15 units of ratio total ratio = 25 1 unit of ratio = 3500 / 25 = 140 so , a ’ s amount = 120 × 15 = rs . 2100 . answer : option c" | a = 15 * 2
b = a / 10
c = 15 * 2
d = c / 10
e = 15 * 2
f = e / 15
g = d + f
h = b / g
i = h * 3500
|
a ) 17 / 25 , b ) 17 / 27 , c ) 17 / 18 , d ) 17 / 22 , e ) 17 / 09 | a | divide(multiply(subtract(multiply(const_6, const_3), const_1), const_2), 50) | the probability that a number selected at random from the first 50 natural numbers is a composite number is - . | "the number of exhaustive events = ⁵ ⁰ c ₁ = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . required probability = 34 / 50 = 17 / 25 . answer : a" | a = 6 * 3
b = a - 1
c = b * 2
d = c / 50
|
a ) 66 , b ) 67 , c ) 68 , d ) 69 , e ) 70 | a | add(add(41, 23), const_2) | there are certain number of hats and gloves in a box . they are of 41 red , 23 green , 11 orange . power gone but a woman can differentiate between hats and gloves . how many draws are required to obtain a pair of each color ? | the worst case senario will be first take 40 red then take 22 green then take 1 + 1 red + green then take 2 orange . so total 40 + 22 + 2 + 2 = 66 answer : a | a = 41 + 23
b = a + 2
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a ) 1.01252633 , b ) 0.01262533 , c ) 0.12526333 , d ) 0.012793333 , e ) 0.12725333 | d | divide(19.19, 2000) | 19.19 / 2000 is equal to : | "19.19 / 2000 = 2525 / 200000 = 0.012793333 answer : d" | a = 19 / 19
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a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) 59 | b | subtract(multiply(add(20, const_1), 5), 49) | the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 49 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ? | "20 x + 49 = 21 ( x – 5 ) x = 56 answer : b" | a = 20 + 1
b = a * 5
c = b - 49
|
a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 7 , d ) 1 / 2 , e ) 4 / 7 | a | divide(multiply(3, const_1), add(multiply(3, const_1), multiply(3, const_1))) | harold works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 3 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? | "the time from beginning of march to the end of september is 7 months . if x is the average monthly tip for all months other than august then his august month tip will be 3 * x his total tip for the 7 months = 6 * ( average tip for the months other than august ) + 3 x = 9 x august tips as a fraction of total tips = 3 x / 9 x = 1 / 3 . so answer is a" | a = 3 * 1
b = 3 * 1
c = 3 * 1
d = b + c
e = a / d
|
a ) 33 , b ) 88 , c ) 27 , d ) 88 , e ) 99 | c | multiply(power(3, const_2), 3) | how many shots of 1 cm radius can be prepared from a sphere of 3 cm radius ? | "4 / 3 π * 3 * 3 * 3 = 4 / 3 π * 1 * 1 * 1 * x x = 27 answer : c" | a = 3 ** 2
b = a * 3
|
a ) 10.7 sec , b ) 2.7 sec , c ) 11.9 sec , d ) 12.7 sec , e ) 25.7 sec | c | multiply(const_3600, divide(divide(add(190, 190), const_1000), add(65, 50))) | two trains each 190 m in length each , are running on two parallel lines in opposite directions . if one goes at the speed of 65 km / h while the other travels at 50 km / h . how long will it take for them to pass each other completely . | "explanation : d = 190 m + 190 m = 380 m rs = 65 + 50 = 115 * 5 / 18 = 319 / 10 t = 380 * 10 / 319 = 11.9 sec answer : option c" | a = 190 + 190
b = a / 1000
c = 65 + 50
d = b / c
e = 3600 * d
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a ) 288 , b ) 190 , c ) 188 , d ) 188 , e ) 105 | e | divide(add(600, 450), divide(600, 60)) | a 600 m long train crosses a tree in 60 sec , how much time will it take to pass a platform 450 m long ? | "l = s * t s = 600 / 60 s = 10 m / sec . total length ( d ) = 1050 m t = d / s t = 1050 / 10 t = 105 sec answer : e" | a = 600 + 450
b = 600 / 60
c = a / b
|
a ) 5 kg , b ) 15 kg , c ) 25 kg , d ) 30 kg , e ) none | e | divide(const_100, divide(subtract(const_100, 10), 20)) | the price of rice falls by 10 % . how much rice can be bought now with the money that was sufficient to buy 20 kg of rice previously ? | "solution : let rs . 100 be spend on rice initially for 20 kg . as the price falls by 10 % , new price for 20 kg rice , = ( 100 - 10 % of 100 ) = 90 new price of rice = 90 / 20 = rs . 4,5 per kg . rice can bought now at = 100 / 4,5 = 22,22 kg . answer : option e" | a = 100 - 10
b = a / 20
c = 100 / b
|
a ) 448 , b ) 488 , c ) 542 , d ) 1268 , e ) 560 | d | multiply(63, const_10) | the least number , which when divided by 12 , 15 , 20 and 63 leaves in each case a remainder of 8 is : | "required number = ( l . c . m . of 12 , 15 , 20 , 63 ) + 8 = 1260 + 8 = 1268 . answer : d" | a = 63 * 10
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a ) 20 , b ) 30 , c ) 5 , d ) 8 , e ) 10 | e | subtract(50, add(multiply(50, divide(3, 5)), multiply(50, divide(1, 5)))) | in a class of 50 students , 3 / 5 went away do painting , 1 / 5 went to play in the field . how many students are left in the classroom ? | 3 / 5 + 1 / 5 are the students who have left the classroom 3 / 5 + 1 / 5 = ( 3 + 1 ) / 5 , since denominator is same for both fractions = 4 / 5 remaining students in the classroom = 1 - 4 / 5 = 5 / 5 - 4 / 5 = ( 5 - 4 ) / 5 = 1 / 5 1 / 5 of 50 = 10 ans : e | a = 3 / 5
b = 50 * a
c = 1 / 5
d = 50 * c
e = b + d
f = 50 - e
|
a ) 105 kg , b ) 103 kg , c ) 108 kg , d ) 125 kg , e ) 117 kg | d | multiply(subtract(const_1, divide(15, const_100)), multiply(subtract(const_1, divide(30, const_100)), multiply(300, subtract(const_1, divide(30, const_100))))) | a statue is being carved by a sculptor . the original piece of marble weighed 300 kg . in the first week 30 percent is cut away . in the second week 30 percent of the remainder is cut away . in the third week the statue is completed when 15 percent of the remainder is cut away . what is the weight of the final statue ? | "d 125 kg 300 ã — 0.7 ã — 0.7 ã — 0.85 = 125 kg ." | a = 15 / 100
b = 1 - a
c = 30 / 100
d = 1 - c
e = 30 / 100
f = 1 - e
g = 300 * f
h = d * g
i = b * h
|
a ) 9 % , b ) 10 % , c ) 105 / 8 % , d ) 11 % , e ) 12 % | d | multiply(divide(subtract(multiply(add(500, multiply(const_3, 500)), divide(10, const_100)), multiply(500, divide(7, const_100))), multiply(const_3, 500)), const_100) | a $ 500 investment and a $ 1,500 investment have a combined yearly return of 10 percent of the total of the two investments . if the $ 500 investment has a yearly return of 7 percent , what percent yearly return does the $ 1,500 investment have ? | the equation we can form the question : return on total investment = sum of individual investments ( 500 + 1500 ) ( 10 ) = ( 500 â ˆ — 7 ) + ( 1500 x ) , where x is the return on investment of 1500 . solving the equation , we get x = 11 % ( option d ) answer : d | a = 3 * 500
b = 500 + a
c = 10 / 100
d = b * c
e = 7 / 100
f = 500 * e
g = d - f
h = 3 * 500
i = g / h
j = i * 100
|
a ) 640 , b ) 780 , c ) 380 , d ) 540 , e ) 680 | e | divide(factorial(subtract(18, const_1)), multiply(factorial(subtract(const_4, const_1)), factorial(subtract(subtract(18, const_1), subtract(const_4, const_1))))) | the number of positive integer solutions for the equation x + y + z + t = 18 is | "the number of positive integer solutions for the equatio fx 1 + x 2 + ⋯ + xn = k ( k - 1 ) c ( n - 1 ) - where k is the number and n is number of variable in the equation . 18 - 1 c 4 - 1 = 17 c 3 = 680 answer : e" | a = 18 - 1
b = math.factorial(a)
c = 4 - 1
d = math.factorial(c)
e = 18 - 1
f = 4 - 1
g = e - f
h = math.factorial(g)
i = d * h
j = b / i
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | divide(subtract(37, 17), subtract(17, 15)) | the average age of a group of n people is 15 years old . one more person aged 37 joins the group and the new average is 17 years old . what is the value of n ? | "15 n + 37 = 17 ( n + 1 ) 2 n = 20 n = 10 the answer is a ." | a = 37 - 17
b = 17 - 15
c = a / b
|
a ) 87.0 , b ) 87.5 , c ) 87.2 , d ) 87.1 , e ) 87.4 | b | multiply(divide(divide(400, 4), 8), 7) | the ratio between the speeds of two trains is 7 : 8 . if the second train runs 400 km in 4 hours , what is the the speed of the first train ? | speed and time are inversely proportional ( when distance is constant ) ⇒ speed ∝ 1 time ( when distance is constant ) here distance is constant and hence speed and time are inversely proportionalspeed ∝ 1 time ⇒ speed 1 speed 2 = time 2 time 1 ⇒ 78 = 4 time 1 ⇒ time 1 = 4 × 87 hr ⇒ speed of the first train = distancetime 1 = 400 ( 4 × 87 ) = 100 × 78 = 12.5 × 7 = 87.5 km / hr answer : b | a = 400 / 4
b = a / 8
c = b * 7
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | divide(multiply(factorial(8), factorial(8)), multiply(factorial(8), factorial(3))) | what is the unit digit of ( 8 ! * 4 ! / 8 ! * 3 ! ) ? | "( 8 ! * 4 ! / 8 ! * 3 ! ) = ( 4 ! / 3 ! ) = 24 / 6 = 4 units digit of the above product will be equal to 4 answer e" | a = math.factorial(8)
b = math.factorial(8)
c = a * b
d = math.factorial(8)
e = math.factorial(3)
f = d * e
g = c / f
|
['a ) x / 2', 'b ) x', 'c ) 2 x', 'd ) 4 x', 'e ) 3 x'] | c | multiply(power(const_1, const_2), const_2) | area of square with side x is equal to the area of a triangle with base x . the altitude of the triangle is | x ^ 2 = 1 / 2 ã ƒ â € ” x ã ƒ â € ” h or h = 2 x ^ 2 / x = 2 x answer : c | a = 1 ** 2
b = a * 2
|
a ) 28 % , b ) 30 % , c ) 32 % , d ) 36 % , e ) 64 % | e | subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(60, const_100))), const_100)) | a baseball card decreased in value 60 % in its first year and 10 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | let the initial value of baseball card = 100 after first year , value of baseball card = ( 1 - 60 / 100 ) * 100 = 40 after second year , value of baseball card = ( 1 - 10 / 100 ) * 40 = 36 total percent decrease of the card ' s value over the two years = ( 100 - 36 ) / 100 * 100 % = 64 % answer e | a = 10 / 100
b = 1 - a
c = 60 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 75 , b ) 120 , c ) 210 , d ) 246 , e ) 252 | d | subtract(choose(10, 5), choose(6, 5)) | there are 10 books on a shelf , of which 4 are paperbacks and 6 are hardbacks . how many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback ? | "paperbacks - 4 , hardbacks - 6 5 books in total and at least 1 from each . total combinations for 5 books = ( 1 pb , 4 hb ) + ( 4 pb , 1 hb ) + ( 3 pb , 2 hb ) + ( 2 pb , 3 hb ) 1 pb , 4 hb = 4 c 1 * 6 c 4 = 60 4 pb , 1 hb = 4 c 4 * 6 c 1 = 6 3 pb , 2 hb = 4 c 3 * 6 c 2 = 60 2 pb , 3 hb = 4 c 2 * 6 c 3 = 120 total combinations of 5 books = 60 + 6 + 60 + 120 = > 246 ans d ." | a = math.comb(10, 5)
b = math.comb(6, 5)
c = a - b
|
a ) 30 m , b ) 20 m , c ) 35 m , d ) 40 m , e ) 38 m | a | add(12, sqrt(subtract(power(20, const_2), power(8, const_2)))) | a walk 20 m towards north east . b walk towards east 8 m and then 12 m south . now calculate distance between a and b ? | its coming 30.4 m since 1 st triangle we get 18.31 and extra 12 so 18.31 + 12 = 30.31 = 30 m answer : a | a = 20 ** 2
b = 8 ** 2
c = a - b
d = math.sqrt(c)
e = 12 + d
|
a ) rs . 3500 , b ) rs . 1500 , c ) rs . 2500 , d ) rs . 3000 , e ) none | e | multiply(multiply(132, 20), 20) | if the difference between compound interest ( interest compounded yearly ) and simple interest on a sum for 3 years at 20 % p . a . is rs . 132 , then the sum is | "c . i - s . i = 132 c . i = a - p a = p ( 1 + ( r / 100 ) ) ^ n a = x ( 1 + ( 20 / 100 ) ) ^ 3 a = x ( 6 / 5 ) ^ 3 a = 216 / 125 ( x ) c . i = ( 91 / 125 ) x s . i = ( pnr ) / 100 = ( x * 3 * 20 ) / 100 = ( 6 / 10 ) x diff = 132 diff = ( 91 / 25 ) x - ( 6 / 10 ) x = ( 182 x - 150 x ) / 250 = 32 x / 250 32 x / 250 = 132 x = ( 132 * 250 ) / 32 x = 1031.25 answer : e" | a = 132 * 20
b = a * 20
|
a ) 0.8 , b ) 1.42 , c ) 8.0 , d ) 12.5 , e ) 80.0 | b | subtract(divide(power(const_100, const_3), multiply(700, 1,000)), const_2) | the mass of 1 cubic meter of a substance is 700 kilograms under certain conditions . what is the volume , in cubic centimeters , of 1 gram of this substance under these conditions ? ( 1 kilogram = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | "density is mass divided by volume . so density of the given substance will be mass / volume = 700 kg / 1 m ^ 3 = 700 kg / m ^ 3 or 1 g / 1.42 cm ^ 3 = 0.7 g / cm ^ 3 . next , ask yourself if 700,000 g is equivalent to 1 , 000,000 cubic centimeters then 1 g is equivalent to how many cubic centimeters ? - - > 1 g - 1 , 000,000 / 700,000 = 10 / 8 = 1.42 cubic centimeters . answer is b" | a = 100 ** 3
b = 700 * 1
c = a / b
d = c - 2
|
['a ) 7207', 'b ) 7206', 'c ) 7203', 'd ) 7200', 'e ) 7201'] | d | add(add(triangle_area(40, 30), triangle_area(50, 120)), multiply(40, subtract(120, 30))) | a plot abcd is as shown in figure , where af = 30 m , ce = 40 m , ed = 50 m , ae = 120 m . find the area of the plot abcd ? | area of plot abcd = area of ade + area of afb + area of bcef = 1 / 2 * 50 * 120 + 1 / 2 * 40 * 30 + 40 * 90 = 3000 + 600 + 3600 = 7200 sq . m answer : d | a = triangle_area + (
b = a + triangle_area
|
a ) 8.5 seconds , b ) 2.86 seconds , c ) 3.5 seconds , d ) 2.5 seconds , e ) 2.6 seconds | b | divide(100, multiply(126, const_0_2778)) | in what time will a train 100 meters long cross an electric pole , if its speed is 126 km / hr | "first convert speed into m / sec speed = 126 * ( 5 / 18 ) = 35 m / sec time = distance / speed = 100 / 35 = 2.86 seconds answer : b" | a = 126 * const_0_2778
b = 100 / a
|
a ) 15.8 sec . , b ) 12.8 sec . , c ) 11.16 sec . , d ) 10.8 sec . , e ) 08.8 sec . | c | divide(add(140, 170), multiply(add(60, 40), const_0_2778)) | two bullet train s 140 m and 170 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time ( in seconds ) which they take to cross each other , is : | "relative speed = ( 60 + 40 ) km / hr = 100 x 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = ( 140 + 170 ) m = 310 m . required time = 310 x 9 / 250 = 11.16 sec . answer c" | a = 140 + 170
b = 60 + 40
c = b * const_0_2778
d = a / c
|
['a ) 12 square meters', 'b ) 5 square meters', 'c ) 44 square meters', 'd ) 60 square meters', 'e ) 22 square meters'] | d | multiply(12, 5) | a rectangular garden is 12 m by 5 m , what is its area ? | area of a rectangle : a = w × h d ) 60 square meters | a = 12 * 5
|
a ) s . 10 , b ) s . 2 , c ) s . 5 , d ) s . 3 , e ) s . 4 | c | subtract(subtract(multiply(3125, power(add(const_1, divide(4, const_100)), 2)), 3125), multiply(multiply(3125, divide(4, const_100)), 2)) | indu gave bindu rs . 3125 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "3125 = d ( 100 / 4 ) 2 d = 5 answer : c" | a = 4 / 100
b = 1 + a
c = b ** 2
d = 3125 * c
e = d - 3125
f = 4 / 100
g = 3125 * f
h = g * 2
i = e - h
|
a ) 4 / 15 , b ) 1 / 3 , c ) 2 / 5 , d ) 4 / 5 , e ) 7 / 15 | e | divide(multiply(7, 1), multiply(3, 5)) | if the ratio of a to b is 7 to 3 and the ratio of b to c is 1 to 5 , what is the ratio of a to c ? | "a : b = 7 : 3 - - 1 b : c = 1 : 5 = > b : c = 3 : 15 - - 2 from 1 and 2 , we get a : c = 7 : 15 answer e" | a = 7 * 1
b = 3 * 5
c = a / b
|
a ) 19 , b ) 18 , c ) 16 , d ) 11 , e ) 17 | a | add(18, const_1) | the average of first 18 even numbers is ? | "sum of 18 even numbers = 18 * 19 = 342 average = 342 / 18 = 19 answer : a" | a = 18 + 1
|
a ) 228 , b ) 350 , c ) 289 , d ) 500 , e ) 821 | b | divide(238, subtract(const_1, divide(multiply(4, 8), const_100))) | a person lent a certain sum of money at 4 % per annum at simple interest and in 8 years the interest amounted to rs . 238 less than the sum lent . what was the sum lent ? | "p - 238 = ( p * 4 * 8 ) / 100 p = 350 answer : b" | a = 4 * 8
b = a / 100
c = 1 - b
d = 238 / c
|
a ) 98 , b ) 100 , c ) 0 , d ) 99 , e ) 97 | a | subtract(99, 1) | for every integer n ≥ 3 , the function g ( n ) is defined as the product of all the odd integers from 1 to n , inclusive . what is the value of g ( 99 ) – g ( 97 ) ? | g ( 99 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 g ( 97 ) = 1 * 3 * 5 * 7 * 9 * . . . * 97 g ( 99 ) - g ( 97 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 - 1 * 3 * 5 * 7 * 9 * . . . * 97 = 1 * 3 * 5 * 7 * 9 * . . . * 97 * ( 99 - 1 ) = 1 * 3 * 5 * 7 * 9 * . . . * 97 * 98 hence : a . | a = 99 - 1
|
a ) 6.25 , b ) 6.28 , c ) 6.31 , d ) 6.19 , e ) 6.21 | c | divide(subtract(200, multiply(5, 2.1)), 30) | in the first 5 overs of a cricket game , the run rate was only 2.1 . what should be the rate in the remaining 30 overs to reach the target of 200 runs ? | "required run rate = [ 200 - ( 2.1 * 5 ) ] / 30 = 189.50 / 30 = 6.31 answer : c" | a = 5 * 2
b = 200 - a
c = b / 30
|
a ) 122821 , b ) 202950 , c ) 281199 , d ) 122850 , e ) 128111 | b | divide(multiply(1, 901), const_4) | what is the sum of all even numbers from 1 to 901 ? | "explanation : 900 / 2 = 450 450 * 451 = 202950 answer : b" | a = 1 * 901
b = a / 4
|
a ) 100 , b ) 12 , c ) 35 , d ) 466 , e ) 56 | a | divide(multiply(100, 10), 110) | the lcm and hcf of two numbers are 100 and 10 respectively . find the larger of the two numbers if their sum is 110 . | "there are 2 approaches in solving this . methode 1 . hcf * lcm = the actual number . 100 * 10 = 1000 so the answer which we are looking for has to be a factor of 1000 . so among the options shortlist the answers by eliminating those numbers which is not divisible by 1000 . and then take the highest number as the answer as the question asks abt the highest number . answer is a" | a = 100 * 10
b = a / 110
|
a ) 4 , b ) 3 , c ) 2 , d ) 1 , e ) 1 / 2 | e | subtract(3, multiply(divide(factorial(const_3.0), factorial(const_2.0)), power(divide(3, 1), 4))) | a couple decides to have 4 children . if they succeed in having 4 children and each child is equally likely to be a boy or a girl , what is the probability that they will have exactly 3 girls and 1 boy ? | "sample space = 2 ^ 4 = 16 . favourable events = { bbgg } , { bgbg } , { bggb } , { ggbb } , { gbgb } , { gbbg } . { bggg } , { gbbb ) probability = 8 / 16 = 1 / 2 . ans ( e ) ." | a = math.factorial(3)
b = math.factorial(2)
c = a / b
d = 3 / 1
e = d ** 4
f = c * e
g = 3 - f
|
a ) s : 10123.19 , b ) s : 10123.29 , c ) s : 10123.20 , d ) s : 10123.28 , e ) s : 16197.12 | e | 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 . 40,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 40000 * ( 1 + 12 / 100 ) 3 ] = 40000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 56197.12 c . i . = ( 56197.12 - 40000 ) = rs : 16197.12 answer : e" | 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
|
a ) 90 , b ) 190 , c ) 120 , d ) 130 , e ) 220 | b | subtract(divide(subtract(multiply(12, 370), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8)), subtract(370, divide(subtract(multiply(12, 370), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8)))) | a theater charges $ 12 for seats in the orchestra and $ 8 for seats in the balcony . on a certain night , a total of 370 tickets were sold for a total cost of $ 3,320 . how many more tickets were sold that night for seats in the balcony than for seats in the orchestra ? | orchestra seats - a balcony seats - b a + b = 370 and 12 a + 8 b = 3320 solving equations simultaneously ( multiply equation 1 with 8 and subtract from second equation ) 4 a = 3320 - 8 * 370 = 3320 - 2960 = 360 i . e . a = 90 and b = 370 - 90 = 280 more seats in balcony than orchestra = b - a = 280 - 90 = 190 answer : option b | a = 12 * 370
b = 3 * 1000
c = 3 * 100
d = b + c
e = 2 * 10
f = d + e
g = a - f
h = 12 - 8
i = g / h
j = 12 * 370
k = 3 * 1000
l = 3 * 100
m = k + l
n = 2 * 10
o = m + n
p = j - o
q = 12 - 8
r = p / q
s = 370 - r
t = i - s
|
a ) $ 10 , b ) $ 12 , c ) $ 13.20 , d ) $ 21 , e ) $ 16.80 | d | divide(subtract(multiply(81, 3), multiply(69, 2)), subtract(multiply(3, 3), multiply(2, 2))) | if bill can buy 3 pairs of jeans and 2 shirts for $ 69 or 2 pairs of jeans and 3 shirts for $ 81 , how much does one shirt cost ? | "3 j + 2 s = 69 2 j + 3 s = 81 - - - - - - - - - - - - - - - - 5 j + 5 s = 150 - - - - ( divide by 5 ) - - - > j + s = 30 3 j + 2 s = j + 2 ( j + s ) = j + 60 = 69 - - - > j = 9 3 * 9 + 2 s = 69 27 + 2 s = 69 2 s = 42 s = 21 answer : d" | a = 81 * 3
b = 69 * 2
c = a - b
d = 3 * 3
e = 2 * 2
f = d - e
g = c / f
|
a ) 10 feet , b ) 12 feet , c ) 27 feet , d ) 15 feet , e ) 18 feet | c | multiply(divide(27, 3), 3) | a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 3 feet . how many feet does the squirrel travels if the post is 27 feet tall and 3 feet in circumference ? | "total circuit = 27 / 3 = 9 total feet squirrel travels = 9 * 3 = 27 feet answer : c" | a = 27 / 3
b = a * 3
|
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