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 ) 48 , b ) 24.8 , c ) 24.21 , d ) 24.88 , e ) 49 | e | multiply(divide(14, subtract(9, 7)), 7) | sachin is younger than rahul by 14 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "explanation : if rahul age is x , then sachin age is x - 14 , so , 9 x - 126 = 7 x 2 x = 126 x = 63 so sachin age is 63 - 14 = 49 answer : e ) 49" | a = 9 - 7
b = 14 / a
c = b * 7
|
a ) 50000 , b ) 750000 , c ) 850000 , d ) 7 , 500000 , e ) w = 75000,000 | e | divide(reminder(multiply(subtract(800000, multiply(50, 15000)), 15000), multiply(const_100, multiply(const_1000, const_1000))), const_100) | the number x of cars sold each week varies with the price y in dollars according to the equation x = 800000 β 50 y . what would be the total weekly revenue w , in dollars , from the sale of cars priced at $ 15000 ? | number of cars sold = x = 800000 - 50 y y = 15000 x = 800000 - 750000 = 50000 revenue from 50000 cars = 15000 * 50000 = 750000000 e | a = 50 * 15000
b = 800000 - a
c = b * 15000
d = 1000 * 1000
e = 100 * d
f = reminder / (
|
a ) 17 : 73 , b ) 78 : 14 , c ) 45 : 30 , d ) 73 : 17 , e ) 4 : 9 | d | multiply(divide(multiply(divide(5, 1), divide(3, add(3, 2))), multiply(divide(7, 2), divide(2, add(3, 2)))), const_2) | two varieties of steel , a and b , have a ratio of iron to chromium as 5 : 1 and 7 : 2 , respectively . steel c is produced by mixing alloys a and b at a ratio of 3 : 2 . what is the ratio of iron to chromium in c ? | in 6 parts of alloy a , 5 parts are iron , and 1 part is chromium . in 9 parts of alloy b , 7 parts are iron and 2 parts are chromium . first , to compare the two alloys , get the same number of parts in total - we can use 18 . so we have : in 18 parts of alloy a , 15 parts are iron and 3 are chromium . in 18 parts of alloy b , 14 parts are iron and 4 are chromium . so combining 3 of a with 2 of b , we ' d be combining 45 and 28 parts of iron , or 73 parts of iron , with 9 and 8 of chromium , or 17 of chromium , so 73 to 17 is the answer . answer : d | a = 5 / 1
b = 3 + 2
c = 3 / b
d = a * c
e = 7 / 2
f = 3 + 2
g = 2 / f
h = e * g
i = d / h
j = i * 2
|
a ) 6600 , b ) 1300 , c ) 6780 , d ) 1976 , e ) 2448 | a | multiply(choose(11, 2), choose(10, 3)) | there are 11 boys and 10 girls in a class . if three students are selected at random , in how many ways that 3 girl & 2 boys are selected ? | "e = event that 3 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 11 and 3 girl from 10 = 11 c 2 * 10 c 3 = 6600 ans - a" | a = math.comb(11, 2)
b = math.comb(10, 3)
c = a * b
|
a ) 38.4 , b ) 32.6 , c ) 48 , d ) 27.4 , e ) 21 | a | add(45, 30) | two goods trains each 400 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 400 + 400 = 800 m . required time = 800 * 6 / 125 = 38.40 sec . answer : a" | a = 45 + 30
|
a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | c | subtract(multiply(const_10, 150), add(multiply(3, const_100.0), multiply(5, 150))) | a man purchased 3 blankets @ rs . 200 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ? | "10 * 150 = 1500 3 * 200 + 5 * 150 = 1350 1350 β 1050 = 300 answer : c" | a = 10 * 150
b = 3 * 100
c = 5 * 150
d = b + c
e = a - d
|
a ) 4,514 , b ) 4,475 , c ) 4,521 , d ) 4,428 , e ) 4,349 | a | divide(factorial(subtract(add(const_4, 15), const_1)), multiply(factorial(15), factorial(subtract(const_4, const_1)))) | how many positive integers less than 5,000 are evenly divisible by neither 15 nor 22 ? | "integers less than 5000 divisible by 15 5000 / 15 = 333 . something , so 333 integers less than 5000 divisible by 22 5000 / 22 = 238 . # # , so 238 we have double counted some , so take lcm of 15 and 22 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 15 and 22 = 333 + 238 - 47 = 524 now subtract that from 4999 . 4999 - 524 = 4514 answer a ." | a = 4 + 15
b = a - 1
c = math.factorial(b)
d = math.factorial(15)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) s . 1090 , b ) s . 1160 , c ) s . 1190 , d ) s . 1350 , e ) s . 1256 | d | divide(multiply(subtract(const_100, 25), 1800), const_100) | a man buys a cycle for rs . 1800 and sells it at a loss of 25 % . what is the selling price of the cycle ? | "s . p . = 75 % of rs . 1800 = rs . 75 / 100 x 1800 = rs . 1350 answer : d" | a = 100 - 25
b = a * 1800
c = b / 100
|
a ) 49 kmph , b ) 58 kmph , c ) 44 kmph , d ) 47 kmph , e ) 48 kmph | e | divide(add(60, 40), const_2) | a man goes from a to b at a speed of 60 kmph and comes back to a at a speed of 40 kmph . find his average speed for the entire journey ? | "distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 60 ) + ( d / 40 ] = ( 2 d ) / [ 5 d / 120 ) = > 48 kmph . answer : e" | a = 60 + 40
b = a / 2
|
a ) 272258 , b ) 272358 , c ) 289332 , d ) 274258 , e ) 274358 | c | multiply(divide(5358, 54), const_100) | 5358 x 54 = ? | "5358 x 51 = 5358 x ( 50 + 4 ) = 5358 x 50 + 5358 x 4 = 267900 + 21432 = 289332 . c )" | a = 5358 / 54
b = a * 100
|
a ) 50 , b ) 55 , c ) 60 , d ) 75 , e ) 100 | e | subtract(subtract(subtract(subtract(multiply(add(divide(divide(divide(divide(20, 2), 2), 2), 2), add(add(add(add(add(add(20, divide(20, 2)), divide(20, 2)), divide(divide(20, 2), 2)), divide(divide(20, 2), 2)), divide(divide(divide(20, 2), 2), 2)), divide(divide(divide(20, 2), 2), 2))), 2), divide(divide(divide(20, 2), 2), 2)), divide(divide(divide(20, 2), 2), 2)), divide(divide(divide(20, 2), 2), 2)), divide(divide(divide(20, 2), 2), 2)) | a basketball is dropped from a height of 20 feet . if it bounces back up to a height that is exactly half of its previous height , and it stops bouncing after hitting the ground for the fourth time , then how many total feet will the ball have traveled after 2 full bounces . | initial distance = 40 feet first bounce = 20 feet up + 20 feet down = 40 feet second bouche = 10 feet up + 10 feet down = 20 feet total distance covered = 40 + 40 + 20 = 100 answer is e | a = 20 / 2
b = a / 2
c = b / 2
d = c / 2
e = 20 / 2
f = 20 + e
g = 20 / 2
h = f + g
i = 20 / 2
j = i / 2
k = h + j
l = 20 / 2
m = l / 2
n = k + m
o = 20 / 2
p = o / 2
q = p / 2
r = n + q
s = 20 / 2
t = s / 2
u = t / 2
v = r + u
w = d + v
x = w * 2
y = 20 / 2
z = y / 2
A = z / 2
B = x - A
C = 20 / 2
D = C / 2
E = D / 2
F = B - E
G = 20 / 2
H = G / 2
I = H / 2
J = F - I
K = 20 / 2
L = K / 2
M = L / 2
N = J - M
|
a ) 228 , b ) 108 , c ) 1266 , d ) 188 , e ) 80 | e | multiply(divide(200, 9), const_3_6) | a 200 meter long train crosses a man standing on the platform in 9 sec . what is the speed of the train ? | "s = 200 / 9 * 18 / 5 = 80 kmph answer : e" | a = 200 / 9
b = a * const_3_6
|
a ) 498 cm 2 , b ) 384 cm 2 , c ) 430 cm 2 , d ) 128 cm 2 , e ) 448 cm 2 | e | multiply(32, 14) | find the area of a parallelogram with base 32 cm and height 14 cm ? | "area of a parallelogram = base * height = 32 * 14 = 448 cm 2 answer : e" | a = 32 * 14
|
a ) 129.25 , b ) 287.25 , c ) 194.25 , d ) 188.25 , e ) 112.25 | c | add(divide(add(multiply(44, 150), multiply(36, 125)), add(36, 44)), multiply(divide(add(multiply(44, 150), multiply(36, 125)), add(36, 44)), divide(40, const_100))) | raman mixed 44 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | "explanation : cp per kg of mixture = [ 44 ( 150 ) + 36 ( 125 ) ] / ( 44 + 36 ) = rs . 138.75 sp = cp [ ( 100 + profit % ) / 100 ] = 138.75 * [ ( 100 + 40 ) / 100 ] = rs . 194.25 answer : c" | a = 44 * 150
b = 36 * 125
c = a + b
d = 36 + 44
e = c / d
f = 44 * 150
g = 36 * 125
h = f + g
i = 36 + 44
j = h / i
k = 40 / 100
l = j * k
m = e + l
|
a ) 31 , b ) 56 , c ) 41 , d ) 59 , e ) none | d | add(42, const_1) | the average age of 42 students in a group is 16 years . when teacher Γ’ β¬ β’ s age is included to it , the average increases by one . what is the teacher Γ’ β¬ β’ s age in years ? | "sol . age of the teacher = ( 43 Γ£ β 17 Γ’ β¬ β 42 Γ£ β 16 ) years = 59 years . answer d" | a = 42 + 1
|
a ) 330 , b ) 300 , c ) 270 , d ) 250 , e ) 350 | b | divide(multiply(30, 2310), 231) | the l . c . m of two numbers is 2310 and their h . c . f is 30 . if one number is 231 the other is | "the other number = l . c . m * h . c . f / given number = 2310 * 30 / 231 = 300 answer is b ." | a = 30 * 2310
b = a / 231
|
a ) 36 , b ) 198 , c ) 132 , d ) 264 , e ) 364 | b | divide(multiply(12, 396), 24) | hcf and lcm two numbers are 12 and 396 respectively . if one of the numbers is 24 , then the other number is ? | "12 * 396 = 24 * x x = 198 answer : b" | a = 12 * 396
b = a / 24
|
a ) 66 % , b ) 79 % , c ) 86 % , d ) 116 % , e ) 150 % | b | multiply(divide(add(100, 30), add(100, divide(add(100, 30), const_2))), const_100) | company kw is being sold , and both company a and company b were considering the purchase . the price of company kw is 30 % more than company a has in assets , and this same price is also 100 % more than company b has in assets . if companies a and b were to merge and combine their assets , the price of company kw would be approximately what percent of these combined assets ? | let the price of company a ' s assets be 100 price of assets of kw is 30 % more than company a ' s assets which is 130 price of assets of kw is 100 % more than company b ' s assets which means price of company b ' s assets is half the price of kw = 65 a + b = 165 kw = 130 kw / ( a + b ) * 100 = 130 / 165 * 100 = 78.78 % or 79 % b | a = 100 + 30
b = 100 + 30
c = b / 2
d = 100 + c
e = a / d
f = e * 100
|
a ) 288 , b ) 540 , c ) 877 , d ) 278 , e ) 178 | b | add(multiply(divide(60, subtract(21, 24)), 24), multiply(divide(60, subtract(21, 24)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 24 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 24 + 21 = 45 t = 12 d = 45 * 12 = 540 answer : b" | a = 21 - 24
b = 60 / a
c = b * 24
d = 21 - 24
e = 60 / d
f = e * 21
g = c + f
|
a ) 192 , b ) 120 , c ) 168 , d ) 160 , e ) 184 | c | divide(multiply(48, 72), const_4) | what is the sum of the greatest common factor and the lowest common multiple of 48 and 72 ? | "prime factorization of the given numbers 72 = 2 ^ 3 * 3 ^ 2 48 = 2 ^ 4 * 3 greatest common factor = 2 ^ 3 * 3 = 24 lowest common multiple = 2 ^ 4 * 3 ^ 2 = 144 sum = 24 + 144 = 168 answer c" | a = 48 * 72
b = a / 4
|
a ) 21 : 22 , b ) 13 : 17 , c ) 15 : 43 , d ) 11 : 18 , e ) 15 : 23 | d | divide(multiply(2, add(divide(10, const_100), const_1)), multiply(3, add(divide(20, const_100), const_1))) | the ration of the number of boys and girls in a college is 2 : 3 . if the percentage is increase in the number of boys and girls be 10 % and 20 % respectively . what will be the new ration ? | "let the number of boys and girls be 2 x and 3 x their increased number is 110 % of 2 x and 120 % of 3 x 2 x * 110 / 100 and 3 x * 120 / 100 11 x / 5 and 18 x / 5 required ratio = 11 x / 5 : 18 x / 5 = 11 : 18 answer is d" | a = 10 / 100
b = a + 1
c = 2 * b
d = 20 / 100
e = d + 1
f = 3 * e
g = c / f
|
a ) 6 , b ) 5 , c ) 4 , d ) 7 , e ) 3 | d | divide(subtract(100, 1), 13) | how many positive integers between 1 and 100 are there such that they are multiples of 13 ? | "multiples of 13 = 13 , 26,39 , - - - - - 91 number of multiples of 13 = > 13 * 7 = 91 answer is d" | a = 100 - 1
b = a / 13
|
a ) 6 , b ) 2 , c ) - 2 , d ) - 6 , e ) - 5.5 | e | divide(subtract(7, 4), subtract(sqrt(add(9, 7)), sqrt(add(4, 9)))) | in the xy - coordinate plane , the graph of y = - x ^ 2 + 9 intersects line l at ( p , 4 ) and ( t , - 7 ) . what is the least possible value of the slope of line l ? | "we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p , 5 ) . hence , x = p and y = 5 should sactisfy the graph . soliving 5 = - p 2 + 9 p 2 = 4 p = + or - 2 simillarly point ( t , - 7 ) should satisfy the equation . hence x = t and y = - 7 . - 7 = - t 2 + 9 t = + or - 4 considering p = - 2 and t = 4 , the least slope is ( - 7 - 4 ) / ( 4 - 2 ) = - 5.5 imo option e is correct answer ." | a = 7 - 4
b = 9 + 7
c = math.sqrt(b)
d = 4 + 9
e = math.sqrt(d)
f = c - e
g = a / f
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 3 | c | divide(15, subtract(45, 15)) | a chemist mixes one liter of pure water with x liters of a 45 % salt solution , and the resulting mixture is a 15 % salt solution . what is the value of x ? | "concentration of salt in pure solution = 0 concentration of salt in salt solution = 45 % concentration of salt in the mixed solution = 15 % the pure solution and the salt solution is mixed in the ratio of - - > ( 45 - 15 ) / ( 15 - 0 ) = 2 / 1 1 / x = 2 / 1 x = 1 / 2 answer : c" | a = 45 - 15
b = 15 / a
|
a ) 76 kg , b ) 76.5 kg , c ) 90 kg , d ) data inadequate , e ) none of these | c | add(multiply(5, 10.0), 40) | the average weight of 5 person ' s increases by 10.0 kg when a new person comes in place of one of them weighing 40 kg . what might be the weight of the new person ? | "explanation : total weight increased = ( 5 x 10.00 ) kg = 50 kg . weight of new person = ( 40 + 50 ) kg = 90 kg . answer : c" | a = 5 * 10
b = a + 40
|
a ) 10.9 sec , b ) 10.1 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 11.16 sec | e | divide(add(140, 170), multiply(add(60, 40), const_0_2778)) | two trains 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 which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 140 + 170 = 310 m . required time = 310 * 9 / 250 = 11.16 sec . answer : e" | a = 140 + 170
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 2 , b ) 4 , c ) 1 , d ) 8 , e ) 0 | c | add(reminder(multiply(reminder(46, const_4), 10), const_10), reminder(3, const_10)) | the units digit of ( 3 ) ^ ( 44 ) + ( 10 ) ^ ( 46 ) is : | "any power of anything ending in zero always has a units digit of 0 . so the first term has a units digit of 0 . the period is 4 . this means , 3 to the power of any multiple of 4 will have a units digit of 1 . 3 ^ 44 has a units digit of 1 . of course 0 + 1 = 1 c" | a = reminder * (
b = reminder + (
|
a ) β 8 , b ) β 2 , c ) 2 , d ) 46 , e ) 48 | a | add(divide(9, const_10), divide(9, divide(9, const_10))) | if a ( a + 8 ) = 9 and b ( b + 8 ) = 9 , where a β b , then a + b = | "a ( a + 8 ) = 9 = > we have a = 1 or - 9 also b ( b + 8 ) = 9 = > b = 1 or - 9 given a β b 1 ) when a = 1 , b = - 9 and a + b = - 8 1 ) when a = - 9 , b = 1 and a + b = - 8 answer choice a" | a = 9 / 10
b = 9 / 10
c = 9 / b
d = a + c
|
a ) 0 , b ) 1 , c ) 3 , d ) 5 , e ) 6 | c | floor(multiply(const_100, divide(66, 5000))) | what is the thousandths digit in the decimal equivalent of 66 / 5000 ? | "66 / 5000 = 66 / ( 5 * 10 ^ 3 ) = ( 66 / 5 ) * 10 ^ - 3 = 13.2 * 10 ^ - 3 = . 0132 thousandths digit = 3 answer c" | a = 66 / 5000
b = 100 * a
c = math.floor(b)
|
a ) 17.2 , b ) 16.5 , c ) 17.5 , d ) 17.9 , e ) 17.3 | c | divide(multiply(10, 7), subtract(10, 6)) | a contractor undertook to do a piece of work in 6 days . he employed certain number of laboures but 7 of them were absent from the very first day and the rest could finish the work in only 10 days . find the number of men originally employed ? | "let the number of men originally employed be x . 6 x = 10 ( x Γ’ β¬ β 7 ) or x = 17.5 answer c" | a = 10 * 7
b = 10 - 6
c = a / b
|
a ) 2 : 5 , b ) 3 : 7 , c ) 3 : 5 , d ) 4 : 7 , e ) 7 : 3 | b | divide(add(multiply(12, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))), subtract(add(12, 8), add(multiply(12, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))))) | a container holding 12 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 8 ounces of a solution that is 1 part alcohol to 3 parts water . what is the ratio of alcohol to water in the resulting solution ? | "container 1 has 12 ounces in the ratio 1 : 2 or , x + 2 x = 12 gives x ( alcohol ) = 4 and remaining water = 8 container 2 has 8 ounces in the ratio 1 : 3 or , x + 3 x = 8 gives x ( alcohol ) = 2 and remaining water = 6 mixing both we have alcohol = 4 + 2 and water = 8 + 6 ratio thus alcohol / water = 6 / 14 = 3 / 7 answer : b" | a = 1 + 2
b = 1 / a
c = 12 * b
d = 1 + 3
e = 1 / d
f = 8 * e
g = c + f
h = 12 + 8
i = 1 + 2
j = 1 / i
k = 12 * j
l = 1 + 3
m = 1 / l
n = 8 * m
o = k + n
p = h - o
q = g / p
|
['a ) 20 Ο', 'b ) 24 Ο', 'c ) 25 Ο', 'd ) 96 Ο', 'e ) 100 Ο'] | c | max(max(volume_cylinder(divide(4, const_2), 5), volume_cylinder(divide(4, const_2), 6)), volume_cylinder(divide(5, const_2), 4)) | the inside dimensions of a rectangular wooden box are 4 meters by 5 meters by 6 meters . a cylindrical drum is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces . of all such drums that could be used , what is the volume , in cubic meters , of the one that has maximum volume ? | you have three options here . approach is to calculate volumes for all of them and see which one is greatest . 1 . cylinder ' s base rests on a 4 x 5 side of the box . so , height becomes 6 and the maximum possible radius becomes 4 / 2 . so , maximum volume = Ο . ( 4 / 2 ) ^ 2.6 = Ο 24 2 . cylinder ' s base rests on a 5 x 6 side of the box . so , height becomes 4 and the maximum possible radius becomes 5 / 2 . so , maximum volume = Ο . ( 5 / 2 ) ^ 2.4 = Ο 25 3 . cylinder ' s base rests on a 4 x 6 side of the box . so , height becomes 5 and the maximum possible radius becomes 4 / 2 . so , maximum volume = Ο . ( 4 / 2 ) ^ 2.5 = Ο 20 option 2 gives maximum value and so that should be the answer , Ο 25 . answer : c | a = 4 / 2
b = max(volume_cylinder)
c = 4 / 2
d = max(b)
|
a ) 45 , b ) 27 , c ) 28 , d ) 26 , e ) 75 | e | divide(10, multiply(multiply(subtract(1, divide(1, 3)), subtract(1, divide(2, 5))), subtract(1, divide(2, 3)))) | there are some people in party , 1 / 3 rd left the party . then 2 / 5 th of the remaining left the party , then 2 / 3 rd of the remaining left the party . at last 10 were remaining . how many people were in total ? | "sol : 45 if x persons were there in total , then x Γ ( 1 β 1 / 3 ) Γ ( 1 β 2 / 5 ) Γ ( 1 β 2 / 3 ) = 10 x Γ 2 / 3 Γ 3 / 5 Γ 1 / 3 = 10 x = 75 answer : e" | a = 1 / 3
b = 1 - a
c = 2 / 5
d = 1 - c
e = b * d
f = 2 / 3
g = 1 - f
h = e * g
i = 10 / h
|
a ) 462 , b ) 450 , c ) 488 , d ) 480 , e ) 555 | d | multiply(40, 12) | the hcf of two numbers is 40 and the other two factors of their lcm are 11 and 12 . what is the largest number . | "explanation : hcf of the two numbers = 40 hcf will be always a factor of lcm 40 is factor of lcm other two factors are 11 & 12 then the numbers are ( 40 * 11 ) and ( 40 x 12 ) = 440 and 480 answer : option d" | a = 40 * 12
|
a ) 1 : 5 , b ) 1 : 4 , c ) 2 : 3 , d ) 4 : 7 , e ) 1 : 2 | d | divide(divide(const_100, add(const_100, 110)), divide(const_100, add(const_100, 20))) | bert and rebecca were looking at the price of a condominium . the price of the condominium was 110 % more than bert had in savings , and separately , the same price was also 20 % more than rebecca had in savings . what is the ratio of what bert has in savings to what rebecca has in savings . | "suppose bert had 100 so price becomes 210 , this 210 = 1.2 times r ' s saving . . so r ' s saving becomes 175 so required ratio is 100 : 175 = 4 : 7 answer : d" | a = 100 + 110
b = 100 / a
c = 100 + 20
d = 100 / c
e = b / d
|
a ) 3.75 % , b ) 5.93 % , c ) 4.75 % , d ) 5.33 % , e ) 6.33 % | a | multiply(divide(divide(subtract(950, 800), 800), 5), const_100) | at what rate percent on simple interest will rs . 800 amount to rs . 950 in 5 years ? | "150 = ( 800 * 5 * r ) / 100 r = 3.75 % answer : a" | a = 950 - 800
b = a / 800
c = b / 5
d = c * 100
|
a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | c | add(divide(subtract(115, add(add(add(2, add(2, 2)), add(add(2, 2), 2)), add(add(add(2, 2), 2), 2))), 5), add(add(add(2, 2), 2), 2)) | in a school with 5 classes , each class has 2 students less than the previous class . how many students are there in the largest class if the total number of students at school is 115 ? | "let x be the number of students in the largest class . then x + ( x - 2 ) + ( x - 4 ) + ( x - 6 ) + ( x - 8 ) = 115 5 x - 20 = 115 5 x = 135 x = 27 the answer is c ." | a = 2 + 2
b = 2 + a
c = 2 + 2
d = c + 2
e = b + d
f = 2 + 2
g = f + 2
h = g + 2
i = e + h
j = 115 - i
k = j / 5
l = 2 + 2
m = l + 2
n = m + 2
o = k + n
|
a ) 80 m , b ) 180 m , c ) 100 m , d ) 60 m , e ) of these | a | multiply(divide(100, subtract(add(6, 4), 5)), 4) | the length of 3 ropes are in the ratio 4 : 5 : 6 . if the sum of the weights of the longest and the shortest rope is 100 metres more than the length of the third rope , what is the length of the shortest rope ? | let the lengths of the three ropes be 4 k , 5 k and 6 k respectively . 4 k + 6 k = 5 k + 100 = > 5 k = 100 = > k = 20 therefore the weight of the lightest boy = 4 k = 4 ( 20 ) = 80 m answer : a | a = 6 + 4
b = a - 5
c = 100 / b
d = c * 4
|
a ) 3 / 4 , b ) 7 / 8 , c ) 15 / 16 , d ) 31 / 32 , e ) 63 / 64 | e | subtract(const_1, power(divide(const_1, 2), 6)) | on a ranch , a rancher can place a loop of rope , called a lasso , once in every 2 throws around a cow β s neck . what is the probability that the rancher will be able to place a lasso around a cow β s neck at least once in 6 attempts ? | "p ( missing all 6 ) = ( 1 / 2 ) ^ 6 = 1 / 64 p ( success on at least one attempt ) = 1 - 1 / 64 = 63 / 64 the answer is e ." | a = 1 / 2
b = a ** 6
c = 1 - b
|
a ) 6.12 hr , b ) 8 hr , c ) 8.5 hr , d ) 10 hr , e ) none of these | a | inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 50))) | two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 50 hr . if all pipes are opened simultaneously , then the cistern will be filled in | solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 β 1 / 50 = 8 / 49 hence , tank will be filled in 49 / 8 = 6.12 hour option ( a ) | a = 1 / 10
b = 1 / 12
c = a + b
d = 1 / 50
e = c - d
f = 1/(e)
|
a ) $ 514.16 , b ) $ 514.17 , c ) $ 514.18 , d ) $ 514.19 , e ) $ 514.20 | b | add(514.16, divide(const_3, const_100)) | 9 people decided to split the restaurant bill evenly . if the bill was $ 514.16 dollars , how much money did they 1 cent is the smallest unit ? | "this is equivalent to finding the first number that is divisible by 9 that occurs after 51416 . in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 51416 = 17 and 18 is divisible by 9 . hence , we need to add 1 to this number for it to be divisible by 9 correct option : b" | a = 3 / 100
b = 514 + 16
|
a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0 | e | multiply(add(add(26, 5), 5), divide(5, const_60)) | a certain car increased its average speed by 5 miles per hour in each successive 5 - minute interval after the first interval . if in the first 5 - minute interval its average speed was 26 miles per hour , how many miles did the car travel in the third 5 - minute interval ? | in the third time interval the average speed of the car was 22 + 5 + 5 = 36 miles per hour ; in 5 minutes ( 1 / 12 hour ) at that speed car would travel 36 * 1 / 12 = 3 miles . answer : e . | a = 26 + 5
b = a + 5
c = 5 / const_60
d = b * c
|
a ) 120 , b ) 480 , c ) 780 , d ) 720 , e ) 800 | c | multiply(choose(13, 2), choose(10, 1)) | there are 13 boys and 10 girls in a class . if three students are selected at random , in how many ways that 1 girl & 2 boys are selected ? | n ( s ) = sample space = 23 c 3 = 1771 e = event that 1 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 13 and 1 girl from 10 = 13 c 2 * 10 c 1 = 780 ans - c | a = math.comb(13, 2)
b = math.comb(10, 1)
c = a * b
|
a ) 76 , b ) 77 , c ) 78 , d ) 79 , e ) 80 | e | divide(subtract(multiply(84, const_4), 96), subtract(const_4, const_1)) | the grade point average of the entire class is 84 . if the average of one fourth of the class is 96 , what is the average of the rest of the class ? | "let x be the number of students in the class . let p be the average of the rest of the class . 84 x = ( 1 / 4 ) 96 x + ( 3 / 4 ) ( p ) x 336 = 96 + 3 p 3 p = 240 p = 80 . the answer is e ." | a = 84 * 4
b = a - 96
c = 4 - 1
d = b / c
|
a ) 1 kg , b ) 10.8 kg , c ) 11 kg , d ) 14.9 kg , e ) none | a | subtract(multiply(add(19, const_1), 14.2), 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 14.2 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 14.2 - 19 x 15 ) kg = 1 kg . correct option : a" | a = 19 + 1
b = a * 14
c = 19 * 15
d = b - c
|
a ) 82 , b ) 84 , c ) 86 , d ) 88 , e ) 90 | e | subtract(132, divide(subtract(132, 6), const_3)) | if a student loses 6 kilograms , he will weigh twice as much as his sister . together they now weigh 132 kilograms . what is the student ' s present weight in kilograms ? | "let x be the weight of the sister . then the student ' s weight is 2 x + 6 . x + ( 2 x + 6 ) = 132 3 x = 126 x = 42 kg then the student ' s weight is 90 kg . the answer is e ." | a = 132 - 6
b = a / 3
c = 132 - b
|
a ) 2.215 . , b ) 2.5 . , c ) 2.6 , d ) 2.7 , e ) 3.5 . | d | multiply(divide(divide(600, const_1000), divide(10, const_60)), subtract(const_1, divide(25, const_100))) | an ant walks an average of 600 meters in 10 minutes . a beetle walks 25 % less distance at the same time on the average . assuming the beetle walks at her regular rate , what is its speed in km / h ? | "the ant walks an average of 600 meters in 10 minutes 600 meters in 5 / 30 hours the beetle walks 25 % less distance = 600 - 150 = 450 meters in 10 minutes 0.450 km in 5 / 30 = 5 / 30 hours speed = 0.450 * 30 / 5 = 2.7 km / h i guess option d should be 2.7" | a = 600 / 1000
b = 10 / const_60
c = a / b
d = 25 / 100
e = 1 - d
f = c * e
|
a ) 8 m , b ) 10 m , c ) 12 m , d ) 15 m , e ) 17 m | e | divide(sqrt(divide(578, divide(const_1, const_2))), const_2) | the width of a rectangular hall is Β½ of its length . if the area of the hall is 578 sq . m , what is the difference between its length and breadth ? | "let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 578 = x * 1 x / 2 x Β² = 1156 x = 34 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 34 / 2 = 17 m answer : e" | a = 1 / 2
b = 578 / a
c = math.sqrt(b)
d = c / 2
|
a ) 28 , b ) 27 , c ) 33 , d ) 18 , e ) 12 | c | divide(add(165, 660), multiply(90, const_0_2778)) | how long does a train 165 meters long running at the rate of 90 kmph take to cross a bridge 660 meters in length ? | "t = ( 660 + 165 ) / 90 * 18 / 5 t = 33 answer : c" | a = 165 + 660
b = 90 * const_0_2778
c = a / b
|
a ) 12 kg , b ) 60 kg , c ) 72 kg , d ) 96 kg , e ) none | b | multiply(multiply(multiply(3, 2), divide(1, const_100)), const_1000) | a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of man is | "solution volume of water displaced = ( 3 x 2 x 0.01 ) m 3 = 0.06 m 3 . mass of man = volume of water displaced Γ density of water = ( 0.06 Γ 1000 ) kg = 60 kg . answer b" | a = 3 * 2
b = 1 / 100
c = a * b
d = c * 1000
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | add(divide(12, 3), const_1) | how many different pairs of numbers ( s , t ) such that s = 3 t can be obtained if s and t are selected from the set of number { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 } | given s = 3 t t can take 0,1 , 2,3 , 4 and so s = 0,3 , 6,9 , 12 4 such pairs can be formed . answer e | a = 12 / 3
b = a + 1
|
a ) 9.0 % , b ) 9.4 % , c ) 9.2 % , d ) 8.8 % , e ) 8.6 % | e | multiply(divide(add(divide(multiply(10, 15), const_100), divide(multiply(8, 35), const_100)), add(15, 35)), const_100) | in one alloy there is 10 % chromium while in another alloy it is 8 % . 15 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 15 + 35 = 50 kg alloy is 0.10 * 15 + 0.08 * 35 = 4.3 kg , so the percentage is 4.3 / 50 * 100 = 8.6 % . answer : e ." | a = 10 * 15
b = a / 100
c = 8 * 35
d = c / 100
e = b + d
f = 15 + 35
g = e / f
h = g * 100
|
a ) 15 , b ) 16 , c ) 18 , d ) 20 , e ) 24 | c | divide(log(divide(multiply(const_3, const_10), add(const_4, const_1))), log(power(divide(multiply(const_2, const_10), add(const_4, const_1)), divide(const_1, 12)))) | on a certain date , pat invested $ 10,000 at x percent annual interest , compounded annually . if the total value of the investment plus interest at the end of 12 years will be $ 40,000 , in how many years , the total value of the investment plus interest will increase to $ 80,000 ? | "x - interest rate 80.000 = 10.000 ( 1 + x ) ^ year = > 8 = ( 1 + x ) ^ year 40.000 = 10.000 . ( 1 + x ) ^ 12 = > 4 = ( 1 + x ) ^ 12 = > 2 = ( 1 + x ) ^ 6 = > 8 = ( 1 + x ) ^ 18 so , after 18 years , the total value of the investment plus interest will increase to $ 80,000 . answer : c" | a = 3 * 10
b = 4 + 1
c = a / b
d = math.log(c)
e = 2 * 10
f = 4 + 1
g = e / f
h = 1 / 12
i = g ** h
j = math.log(i)
k = d / j
|
a ) 10 hours , b ) 12 hours , c ) 14 hours , d ) 16 hours , e ) none of these | b | add(divide(32, subtract(6, 2)), divide(32, add(6, 2))) | in a river flowing at 2 km / hr , a boat travels 32 km upstream and then returns downstream to the starting point . if its speed in still water be 6 km / hr , find the total journey time . | "explanation : speed of the boat = 6 km / hr speed downstream = ( 6 + 2 ) = 8 km / hr speed upstream = ( 6 - 2 ) = 4 km / hr distance travelled downstream = distance travelled upstream = 32 km total time taken = time taken downstream + time taken upstream = ( 32 / 8 ) + ( 32 / 4 ) = ( 32 / 8 ) + ( 64 / 8 ) = ( 96 / 8 ) = 12 hr . answer : option b" | a = 6 - 2
b = 32 / a
c = 6 + 2
d = 32 / c
e = b + d
|
a ) 16 , b ) 20 , c ) 36 , d ) 44 , e ) 48 | c | divide(subtract(15, multiply(divide(20, const_100), 66)), subtract(divide(25, const_100), divide(20, const_100))) | a bowl of nuts is prepared for a party . brand p mixed nuts are 20 % almonds and brand q ' s deluxe nuts are 25 % almonds . if a bowl contains a total of 66 ounces of nuts , representing a mixture of both brands , and 15 ounces of the mixture are almonds , how many ounces of brand q ' s deluxe mixed nuts are used ? | "lets say x ounces of p is mixed with q . = > 66 - x ounces of q is present in the mixture ( as the total = 66 ounces ) given total almond weight = 15 ounces ( 20 x / 100 ) + ( 25 / 100 ) ( 66 - x ) = 15 = > x = 30 = > 66 - 30 = 36 ounces of q is present in the mixture . answer is c ." | a = 20 / 100
b = a * 66
c = 15 - b
d = 25 / 100
e = 20 / 100
f = d - e
g = c / f
|
a ) 1 , b ) 0 , c ) 2 , d ) 3 , e ) 4 | b | divide(add(multiply(factorial(6), factorial(4)), multiply(factorial(6), factorial(5))), 6) | what is the units digit of ( 6 ! * 4 ! + 6 ! * 5 ! ) / 6 ? | ( 6 ! * 4 ! + 6 ! * 5 ! ) / 6 = 6 ! ( 4 ! + 5 ! ) / 6 = 720 ( 24 + 120 ) / 6 = ( 720 * 144 ) / 6 = 720 * 24 units digit of the above product will be equal to 0 answer b | a = math.factorial(6)
b = math.factorial(4)
c = a * b
d = math.factorial(6)
e = math.factorial(5)
f = d * e
g = c + f
h = g / 6
|
a ) 6 , b ) 16 , c ) 22 , d ) 30 , e ) 174 | a | add(subtract(10, 6), subtract(8, 6)) | if x and y are sets of integers , x # y denotes the set of integers that belong to set x or set y , but not both . if x consists of 8 integers , y consists of 10 integers , and 6 of the integers are in both x and y , then x # y consists of how many integers ? | "the number of integers that belong to set x only is 8 - 6 = 2 ; the number of integers that belong to set y only is 10 - 6 = 4 ; the number of integers that belong to set x or set y , but not both is 2 + 4 = 6 . answer : a ." | a = 10 - 6
b = 8 - 6
c = a + b
|
a ) 36.54 , b ) 36.58 , c ) 36.23 , d ) 36.14 , e ) 36.81 | a | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 21)), 50) | the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 21 . the corrected new mean is : | "explanation : correct sum = ( 36 * 50 + 48 - 21 ) = 1827 . correct mean = = 1827 / 50 = 36.54 answer : a ) 36.54" | a = 36 * 50
b = 50 - 2
c = b - 21
d = a + c
e = d / 50
|
a ) 60000 , b ) 30000 , c ) 18000 , d ) 72000 , e ) 22200 | d | multiply(multiply(const_3, const_60), const_60) | if an object travels at ten feet per second , how many feet does it travel in two hours ? | "if an object travels at 10 feet per second it covers 10 x 60 feet in one minute , and 10 x 60 x 2 x 60 feet in two hour . answer = 72000 answer : d" | a = 3 * const_60
b = a * const_60
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(divide(multiply(3, 24), 18), 3) | 3 medical experts , working together at the same constant rate , can write an anatomy textbook in 24 days . how many additional experts , working together at this same constant rate , are needed to write the textbook in 18 days ? | each expert can write 1 / 72 of the book per day . to complete the book in 18 days , we need 72 / 18 = 4 experts , thus 1 more expert is needed . the answer is a . | a = 3 * 24
b = a / 18
c = b - 3
|
a ) 2 , b ) 1.15 , c ) 2.05 , d ) 2.15 , e ) 2.35 | d | divide(divide(multiply(multiply(34.31, 0.473), 1.567), multiply(multiply(7.57, 23.25), 0.0673)), const_10) | the value of ( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) is close to | "( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) = 25.4303 / 11.845 = 2.15 answer : d" | a = 34 * 31
b = a * 1
c = 7 * 57
d = c * 0
e = b / d
f = e / 10
|
a ) 6 meters , b ) 7 meters , c ) 8 meters , d ) 9 meters , e ) 10 meters | a | divide(multiply(divide(multiply(36, const_100), 30), divide(75, const_100)), 15) | the cost of carpeting a room 15 meters long with a carpet 75 cm wide at 30 paise per meter is rs . 36 . the breadth of the room is ? | "length of carpet = total cost / rate = 3600 / 30 = 120 m area of carpet = ( 120 x 75 ) / 100 m 2 = 90 m 2 β΄ area of the room = 90 m 2 breadth of the room = area / length = 90 / 15 m = 6 m answer : a" | a = 36 * 100
b = a / 30
c = 75 / 100
d = b * c
e = d / 15
|
a ) 1780 , b ) 1716 , c ) 1890 , d ) 1980 , e ) 1450 | b | divide(multiply(590, 8), divide(add(multiply(4, 2), 3), 4)) | an aeroplane covers a certain distance of 590 kmph in 8 hours . to cover the same distance in 2 3 / 4 hours , it must travel at a speed of | speed of aeroplane = 590 kmph distance travelled in 8 hours = 590 * 8 = 4720 km speed of aeroplane to acver 4720 km in 11 / 4 = 4720 * 4 / 11 = 1716 km answer b . | a = 590 * 8
b = 4 * 2
c = b + 3
d = c / 4
e = a / d
|
a ) 12 , b ) 18 , c ) 16 , d ) 20 , e ) 24 | c | inverse(add(divide(6, multiply(12, 24)), divide(6, multiply(12, 12)))) | a work can be completed by 12 boys in 24 days and 12 girls in 12 days . in how many days would the 6 boys and 6 girls working together complete the work ? | in 1 day , 12 boys does 1 / 24 of the total work . = > in 1 day , 1 boy does 1 / ( 24 * 12 ) of the total work = > in 1 days , 6 boys do 6 / ( 24 * 12 ) = 1 / 48 of the total work in 1 day , 12 girls do 1 / 12 of the total work . = > in 1 day , 1 girl does 1 / ( 12 * 12 ) of the total work = > in 1 day , 6 girls do 6 / ( 12 * 12 ) = 1 / 24 of the total work in 1 day , 6 boys and 6 girls do 1 / 48 + 1 / 24 of the total work = > in 1 days , 6 boys and 6 girls do 3 / 48 = 1 / 16 of the total work so , 6 boys and 6 girls complete all work in 16 days . hence , the answer is c | a = 12 * 24
b = 6 / a
c = 12 * 12
d = 6 / c
e = b + d
f = 1/(e)
|
a ) 30 , b ) 36 , c ) 42 , d ) 48 , e ) 72 | e | divide(24, subtract(divide(const_1, divide(75, const_100)), const_1)) | walking at 75 % of his usual speed a man takes 24 minutes more to cover a distance . what is his usual time to cover this distance ? | "speed is inversly proprtional to time walking at 75 % of speed meand 3 / 4 s takes 4 / 3 t . it takes 24 minutes extra to cover the distance . then 4 / 3 t = t + 24 4 t = 3 t + 72 t = 72 option e is correct" | a = 75 / 100
b = 1 / a
c = b - 1
d = 24 / c
|
a ) 22 , b ) 88 , c ) 81 , d ) 66 , e ) 22 | c | multiply(subtract(100, 10), inverse(divide(100, subtract(100, 10)))) | anusha , banu and esha run a running race of 100 meters . anusha is the fastest followed by banu and then esha . anusha , banu and esha maintain constant speeds during the entire race . when anusha reached the goal post , banu was 10 m behind . when banu reached the goal post esha was 10 m behind . how far was behind anusha when the latter reached the goal post . | by that time anusha covered 100 m , bhanu covered 90 m . so ratio of their speeds = 10 : 9 by that time bhanu reached 100 m , esha covered 90 m . so ratio of their speeds = 10 : 9 ratio of the speed of all the three = 100 : 90 : 81 by that time anusha covered 100 m , esha covers only 81 . answer : c | a = 100 - 10
b = 100 - 10
c = 100 / b
d = 1/(c)
e = a * d
|
a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | c | divide(multiply(subtract(add(add(const_100, 10), multiply(add(const_100, 10), divide(30, const_100))), const_100), const_100), add(add(const_100, 10), multiply(add(const_100, 10), divide(30, const_100)))) | the output of a factory was increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output was increased by 30 % . by approximately what percent would the output now have to be decreased in order to restore the original output ? | "the original output increases by 10 % and then 30 % . total % change = a + b + ab / 100 total % change = 10 + 30 + 10 * 30 / 100 = 43 % now , you want to change it to 0 , so , 0 = 43 + x + 43 x / 100 x = - 43 ( 100 ) / 143 = 30 % approximately answer is c" | a = 100 + 10
b = 100 + 10
c = 30 / 100
d = b * c
e = a + d
f = e - 100
g = f * 100
h = 100 + 10
i = 100 + 10
j = 30 / 100
k = i * j
l = h + k
m = g / l
|
a ) 8.35 , b ) 9.35 , c ) 10.35 , d ) 11.35 , e ) 12.35 | e | multiply(divide(subtract(const_100, 89), 89), const_100) | if the cost price is 89 % of selling price then what is the profit percentage . | "selling price = rs 100 : then cost price = rs 89 : profit = rs 11 . profit = { ( 11 / 89 ) * 100 } % = 12.35 % answer is e ." | a = 100 - 89
b = a / 89
c = b * 100
|
a ) 2,200 , b ) 2,000 , c ) 2,100 , d ) 2,250 , e ) 2,560 | b | divide(540, 540) | the rate of interest on a sum of money is 2 % p . a . for the first 3 years , 4 % p . a . for the next 4 years , and 5 % for the period beyond 7 years . if the s . i , occured on the sum for the total period of 8 years is rs . 540 / - , the sum is | "explanation : i 1 = ( p x 3 x 2 ) / 100 = 3 p / 50 i 2 = ( p x 4 x 4 ) / 100 = 4 p / 25 i 3 = ( p x 1 x 5 ) / 100 = p / 20 3 p / 50 + 4 p / 25 + p / 20 = 540 the l . c . m of 50 , 25 , 20 = 100 ( 6 p + 16 p + 5 p ) / 100 = 540 27 p / 100 = 540 27 p = 54000 p = 54000 / 27 p = 2000 answer : option b" | a = 540 / 540
|
a ) 4 % , b ) 7 % , c ) 9 % , d ) 3 % , e ) 14 % | e | divide(multiply(divide(7, 5), const_100), 10) | at what rate percent per annum will the simple interest on a sum of money be 7 / 5 of the amount in 10 years ? | "let sum = x . then , s . i . = 7 x / 5 , time = 10 years . rate = ( 100 * 7 x ) / ( x * 5 * 10 ) = 14 % answer : e" | a = 7 / 5
b = a * 100
c = b / 10
|
a ) 99 , b ) 277 , c ) 48 , d ) 96 , e ) 108 | e | subtract(divide(multiply(2.9, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.6, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 2.9 km and 1.6 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 2.9 + 1.6 = 4.5 km = 4500 m . required time = 4500 * 3 / 125 = 108 sec . answer : e" | a = 2 * 9
b = 60 * 1000
c = b / 3600
d = a / c
e = 1 * 6
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) 1339 , b ) 2300 , c ) 4200 , d ) 1340 , e ) 2414 | a | subtract(458,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit ? | "there is one number in hundred with 2 in th tens digit and 1 in the units digit : 21 , 121 , 221 , 321 , . . . the difference between 324,700 and 458,600 is 458,600 - 324,700 = 133,900 - one number per each hundred gives 133,900 / 100 = 1,339 numbers . answer : 1,339 a" | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 458 - 600
|
a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 123 | e | subtract(119, subtract(add(144, 119), 242)) | in a graduating class of 242 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ? | "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 242 = 21 greatest - smallest is 144 - 21 = 123 so answer must be e . 123" | a = 144 + 119
b = a - 242
c = 119 - b
|
a ) 15 , b ) 45 , c ) 35 , d ) 25 , e ) 55 | d | divide(subtract(multiply(95, 5), multiply(5, 20)), subtract(95, 80)) | the average mark of the students of a class in a particular exam is 80 . if 5 students whose average mark in that exam is 20 are excluded , the average mark of the remaining will be 95 . find the number of students who wrote the exam . | let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 95 ( x - 5 ) 80 x - ( 5 * 20 ) = 95 ( x - 5 ) 375 = 15 x = > x = 25 answer : d | a = 95 * 5
b = 5 * 20
c = a - b
d = 95 - 80
e = c / d
|
a ) 12 / 13 , b ) 11 / 15 , c ) 11 / 17 , d ) 11 / 14 , e ) 11 / 7 | d | multiply(divide(divide(multiply(divide(add(subtract(const_100, 30), multiply(subtract(const_100, 30), divide(subtract(const_100, 30), const_100))), const_2), subtract(const_100, 30)), const_100), multiply(subtract(const_100, 30), divide(subtract(const_100, 30), const_100))), const_10) | real - estate salesman z is selling a house at a 30 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 30 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 30 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ? | "let the retail price be = x selling price of z = 0.70 x selling price of x = 0.70 * 0.80 x = 0.56 x selling price of y = ( ( 0.70 x + 0.56 x ) / 2 ) * 0.70 = 0.63 x * 0.70 = 0.44 x 0.44 x = k * 0.56 x k = 0.44 / 0.56 = 44 / 56 = 11 / 14 answer : d" | a = 100 - 30
b = 100 - 30
c = 100 - 30
d = c / 100
e = b * d
f = a + e
g = f / 2
h = 100 - 30
i = g * h
j = i / 100
k = 100 - 30
l = 100 - 30
m = l / 100
n = k * m
o = j / n
p = o * 10
|
a ) 400 , b ) 300 , c ) 500 , d ) 610 , e ) 175 | a | divide(add(92, 40), divide(33, const_100)) | a student has to obtain 33 % of the total marks to pass . he got 92 marks and failed by 40 marks . the maximum marks are ? | "let the maximum marks be x then , 33 % of x = 92 + 40 33 x / 100 = 132 x = 400 answer is a" | a = 92 + 40
b = 33 / 100
c = a / b
|
a ) 50 days , b ) 75 days , c ) 120 days , d ) 150 days , e ) 80 days | c | subtract(multiply(const_4, 50), multiply(divide(150, const_100), 60)) | p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 150 days , in how many days can r alone complete the remaining work ? | "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 250 days and hence r alone in 400 days p works for 60 days - - > 60 / 200 work done = > 30 % q works for 150 days - - > 150 / 250 work done = > 60 % 30 % work left . . . r alone will take 30 % * 400 = 120 days answer is ( c )" | a = 4 * 50
b = 150 / 100
c = b * 60
d = a - c
|
a ) 500 , b ) 600 , c ) 700 , d ) 300 , e ) 400 | d | multiply(divide(add(subtract(72, const_3), add(48, const_2)), const_2), add(divide(subtract(subtract(72, const_3), add(48, const_2)), 6), const_1)) | what is the sum of the multiples of 6 from 48 to 72 , inclusive ? | "the formula we want to use in this type of problem is this : average * total numbers = sum first , find the average by taking the sum of the f + l number and divide it by 2 : a = ( f + l ) / 2 second , find the total numbers in our range by dividing our f and l numbers by 7 and add 1 . ( 72 / 6 ) - ( 48 / 6 ) + 1 multiply these together so what we show average * total numbers = sum ( 48 + 72 ) / 2 * ( 72 / 6 ) - ( 48 / 6 ) + 1 = sum 60 * 5 = 300 d" | a = 72 - 3
b = 48 + 2
c = a + b
d = c / 2
e = 72 - 3
f = 48 + 2
g = e - f
h = g / 6
i = h + 1
j = d * i
|
a ) one , b ) two , c ) three , d ) seven , e ) ten | c | add(const_1, const_2) | if d = 1 / ( 2 ^ 3 * 5 ^ 10 ) is expressed as a terminating decimal , how many nonzero digits will d have ? | another way to do it is : we know x ^ a * y ^ a = ( x * y ) ^ a given = 1 / ( 2 ^ 3 * 5 ^ 10 ) = multiply and divide by 2 ^ 7 = 2 ^ 7 / ( 2 ^ 3 * 2 ^ 7 * 5 ^ 10 ) = 2 ^ 7 / 10 ^ 10 = > non zero digits are 128 = > ans c | a = 1 + 2
|
a ) 1100 , b ) 800 , c ) 1400 , d ) 600 , e ) none of them | d | multiply(3300, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 3300 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a β s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 3300 x ( 2 / 11 ) ) = rs . 600 answer is d ." | a = 2 * 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 3
f = 2 / e
g = 3300 * f
|
a ) 75 % , b ) 80 % , c ) 100 % , d ) 110 % , e ) 125 % | d | multiply(multiply(power(divide(11, 10), const_2), divide(10, 11)), const_100) | tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 10 meters and a circumference of 11 meters , and the interior of tank b has a height of 11 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "the radius of tank a is 11 / ( 2 * pi ) . the capacity of tank a is 10 * pi * 121 / ( 4 * pi ^ 2 ) = 605 / ( 2 * pi ) the radius of tank b is 10 / ( 2 * pi ) . the capacity of tank b is 11 * pi * 100 / ( 4 * pi ^ 2 ) = 550 / ( 2 * pi ) tank a / tank b = 605 / 550 = 11 / 10 = 110 % the answer is d ." | a = 11 / 10
b = a ** 2
c = 10 / 11
d = b * c
e = d * 100
|
a ) 18 , b ) 16 , c ) 26 , d ) 17 , e ) 11 | e | multiply(divide(subtract(1500, 1335), 1500), const_100) | the cost price of a radio is rs . 1500 and it was sold for rs . 1335 , find the loss % ? | "explanation : 1500 - - - - 165 100 - - - - ? = > 11 % answer : e" | a = 1500 - 1335
b = a / 1500
c = b * 100
|
a ) 1 / 5 , b ) 1 / 4 , c ) 1 / 3 , d ) 2 / 5 , e ) 3 / 7 | c | divide(subtract(0.6, 0.5), subtract(0.8, 0.5)) | each of the products produced yesterday was checked by worker x or worker y . 0.5 % of the products checked by worker x are defective and 0.8 % of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is 0.6 % , what fraction of the products was checked by worker y ? | "x : 0.5 % is 0.1 % - points from 0.6 % . y : 0.8 % is 0.2 % - points from 0.6 % . therefore the ratio of products checked by y : x is 1 : 2 . thus , worker y checked 1 / 3 of the products . the answer is c ." | a = 0 - 6
b = 0 - 8
c = a / b
|
a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | b | multiply(divide(multiply(subtract(const_1, divide(30, const_100)), divide(15, 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 30 percent , but profits were 15 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,105 r = x / 100 * 0.1 r answer b" | a = 30 / 100
b = 1 - a
c = 15 / 100
d = b * c
e = 10 / 100
f = d / e
g = f * 100
|
a ) 12 , b ) 27 , c ) 29 , d ) 50 , e ) 65 | e | add(divide(18000, 300), 5) | a shopkeeper sells 300 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ? | sp per metre = 18000 / 300 = rs . 60 loss per metre = rs . 5 cp per metre = 60 + 5 = rs . 65 . answer : e | a = 18000 / 300
b = a + 5
|
a ) 277 , b ) 36 , c ) 64 , d ) 40 , e ) none of these | d | divide(multiply(144, const_100), 360) | ? % of 360 = 144 | "? % of 360 = 144 or , ? = 144 Γ 100 / 360 = 40 answer d" | a = 144 * 100
b = a / 360
|
a ) $ 9000 , b ) $ 3000 , c ) $ 6000 , d ) $ 7000 , e ) $ 8000 | b | divide(500, subtract(const_1, divide(5, 6))) | linda spent 5 / 6 of her savings on furniture and the rest on a tv . if the tv cost her $ 500 , what were her original savings ? | "if linda spent 5 / 6 of her savings on furniture , the rest 6 / 6 - 5 / 6 = 1 / 6 on a tv but the tv cost her $ 500 . so 1 / 6 of her savings is $ 500 . so her original savings are 6 times $ 500 = $ 3000 correct answer b" | a = 5 / 6
b = 1 - a
c = 500 / b
|
a ) 1400 , b ) 2400 , c ) 4000 , d ) 7000 , e ) 3400 | c | divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(40, 11.25), 6)) | how many bricks , each measuring 40 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 40 x 11.25 x 6 ) = 4000 answer : c" | a = 8 * 100
b = 6 * 100
c = a * b
d = c * 22
e = 40 * 11
f = e * 6
g = d / f
|
a ) s . 8500 , b ) s . 8000 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500 | b | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | find the simple interest on rs . 64,000 at 16 2 / 3 % per annum for 9 months . | "p = rs . 64000 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 64,000 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8000 answer is b ." | a = 2 * 3
b = a * 100
c = b * 100
d = 3 * 3
e = d * 100
f = 3 + 2
g = f * 2
h = e * g
i = c + h
j = 16 * 3
k = j + 2
l = k / 3
m = i * l
n = 3 * 3
o = 2 * 3
p = 2 * o
q = n / p
r = m * q
s = 1 / 100
t = r * s
|
a ) 132 cms , b ) 141 cms , c ) 142 cms , d ) 152 cms , e ) 161 cms | b | divide(add(multiply(138, 15), multiply(142, const_10)), 60) | the average height of 15 girls out of a class of 60 is 138 cm . and that of the remaining girls is 142 cm . the average height of the whole class is : | "explanation : average height of the whole class = ( 15 Γ 138 + 45 Γ 142 / 60 ) = 141 cms answer b" | a = 138 * 15
b = 142 * 10
c = a + b
d = c / 60
|
a ) 2.57 hrs , b ) 5 hrs , c ) 6.57 hrs , d ) 7.2 hrs , e ) 9.27 hrs | a | divide(const_1, subtract(divide(const_1, 2), divide(const_1, 9))) | a cistern can be filled by a tap in 2 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 2 - 1 / 9 = 7 / 18 therefore the cistern will be filled in 18 / 7 hours or 2.57 hours . answer : a" | a = 1 / 2
b = 1 / 9
c = a - b
d = 1 / c
|
a ) 1 , b ) 6 , c ) 7 , d ) 8 , e ) 2 | e | subtract(multiply(const_4, const_3), divide(divide(multiply(45000, multiply(const_4, const_3)), 27000), 2)) | a began business with rs . 45000 and was joined afterwards by b with rs . 27000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ? | "45 * 12 : 27 * x = 2 : 1 x = 10 12 - 10 = 2 answer : e" | a = 4 * 3
b = 4 * 3
c = 45000 * b
d = c / 27000
e = d / 2
f = a - e
|
a ) 0 , b ) 2 / 15 , c ) 2 / 5 , d ) 9 / 20 , e ) 5 / 6 | d | divide(9, 20) | 1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 - 9 / 20 = | "we need to determine the result of 1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 let β s add the given fractions in two groups . in the group of the first three fractions , notice that 1 / 3 and 1 / 2 share a common denominator of 6 with 5 / 6 . 1 / 2 + 1 / 3 = 3 / 6 + 2 / 6 = 5 / 6 thus , 5 / 6 β 5 / 6 = 0 looking at the 2 nd group of the fractions ( 1 / 5 , 1 / 4 , and 9 / 20 ) , notice that 1 / 5 and 1 / 4 share a common denominator of 20 with 9 / 20 . 1 / 5 + 1 / 4 = 4 / 20 + 5 / 20 = 9 / 20 thus , 9 / 20 β 9 / 20 = 0 . thus , the result of 1 / 3 + 1 / 2 β 5 / 6 + 1 / 5 + 1 / 4 β 9 / 20 is 9 / 20 . answer : d" | a = 9 / 20
|
a ) 9 : 10 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 1 : 2 | a | divide(divide(multiply(const_4, const_3.0), multiply(5, 5)), divide(multiply(5, const_4), multiply(3, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 4 : 3 . the ratio of economy cars to sport utility vehicles is 6 : 5 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 4 : 3 - - > e : l = 4 : 3 = 24 : 18 . the ratio of economy cars to sport utility vehicles is 6 : 5 - - > e : s = 6 : 5 = 24 : 20 . thus , l : s = 18 : 20 = 9 : 10 . answer : a ." | a = 4 * 3
b = 5 * 5
c = a / b
d = 5 * 4
e = 3 * 4
f = d / e
g = c / f
|
a ) 4,0 , b ) 0,4 , c ) 4,4 , d ) 0,0 , e ) none | a | divide(divide(multiply(lcm(8, 5), const_2), const_10), const_2) | which digits should come in place of @ and # if the number 62684 @ # is divisible by both 8 and 5 ? | explanation : since the given number is divisible by 5 , so 0 or 5 must come in place of # . but , a number ending with 5 is never divisible by 8 . so , 0 will replace # . now , the number formed by the last three digits is 4 @ 0 , which becomes divisible by 8 , if @ is replaced by 4 . hence , digits in place of @ and # are 4 and 0 respectively . answer : a | a = math.lcm(8, 5)
b = a * 2
c = b / 10
d = c / 2
|
a ) $ 100 , b ) $ 75 , c ) $ 25 , d ) $ 120 , e ) $ 50 | c | multiply(100, subtract(const_1, divide(divide(200, const_2), add(300, divide(200, const_2))))) | a invested $ 300 in a business after 6 months b invested $ 200 in the business . end of the year if they got $ 100 as profit . find b ' s shares ? | "a : b = 300 * 12 : 200 * 6 a : b = 1 : 1 b ' s share = 100 * 1 / 4 = $ 25 answer is c" | a = 200 / 2
b = 200 / 2
c = 300 + b
d = a / c
e = 1 - d
f = 100 * e
|
a ) 38 , b ) 27 , c ) 99 , d ) 17 , e ) 86 | e | subtract(multiply(76, 6), multiply(74, 5)) | ashok secured average of 76 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 76 therefore total marks in 6 subjects = 76 * 6 = 456 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 456 β 370 = 86 answer : e" | a = 76 * 6
b = 74 * 5
c = a - b
|
a ) 106 , b ) 301 , c ) 309 , d ) 400 , e ) 450 | b | add(multiply(lcm(lcm(lcm(3, 4), 5), 6), 5), const_1) | a man has a certain number of small boxes to pack into parcles . if he packs 3 , 4 , 5 or 6 in a parcel , he is left with one over ; if he packs 7 in a parcle , none is left over . what is the number of boxes , he may have to pack ? | explanation : clearly , the required number would be such that it leaves a remainder of 1 when divided by 3 , 4 , 5 , or 6 and no remainder when divided by 7 . thus , the number must be of the form ( l . c . m of 3 , 4 , 5 , 6 ) x + 1 i . e . , ( 60 x + 1 ) and a multiple of 7 . clearly , for x = 5 , the number is a multiple of 7 . so the number is 301 . answer : b ) 301 | a = math.lcm(3, 4)
b = math.lcm(a, 5)
c = math.lcm(b, 6)
d = c * 5
e = d + 1
|
a ) - 10 , b ) - 20 , c ) - 30 , d ) - 40 , e ) - 50 | b | multiply(2, multiply(2, 4)) | find b and c so that the parabola with equation y = 4 x 2 - bx - c has a vertex at ( 2 , 4 ) ? | h = b / 8 = 2 : formula for x coordinate of vertex b = 16 : solve for b y = 4 for x = 2 : the vertex point is a solution to the equation of the parabola 4 ( 2 ) 2 - 16 ( 2 ) - c = 4 c = - 20 : solve for c correct answer b | a = 2 * 4
b = 2 * a
|
a ) 18 , b ) 27 , c ) 98 , d ) 27 , e ) 66 | e | multiply(subtract(divide(30, 8), const_1), 24) | pipe a can fill a tank in 8 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ? | let the pipe b be closed after x minutes . 30 / 8 - x / 24 = 1 = > x / 24 = 30 / 8 - 1 = 11 / 4 = > x = 11 / 4 * 24 = 66 . answer : e | a = 30 / 8
b = a - 1
c = b * 24
|
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