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 |
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a ) 270 , b ) 277 , c ) 187 , d ) 180 , e ) 271 | d | multiply(divide(subtract(72, 36), const_3_6), 18) | two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 18 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 18 sec = 18 * 10 = 180 m . the length of the faster train = 180 m . answer : d" | a = 72 - 36
b = a / const_3_6
c = b * 18
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | divide(subtract(1,000, const_100), subtract(550, subtract(550, 150))) | matt gets a $ 1,000 commission on a big sale . this commission alone raises his average commission by $ 150 . if matt ' s new average commission is $ 550 , how many sales has matt made ? | "let , average commission = x no . of items sold = y total commission = xy new commission = xy + 1000 new average = ( xy + 1000 ) / ( y + 1 ) = 150 + x i . e . ( xy + 1000 ) = ( y + 1 ) * ( 150 + x ) i . e . ( xy + 1000 ) = ( xy + x + 150 y + 150 ) i . e . ( 850 ) = ( x + 150 y ) new commission = 550 = 150 + x i . e . x = 400 i . e . y = 3 new sales = y + 1 = 4 answer : option b" | a = 1 - 0
b = 550 - 150
c = 550 - b
d = a / c
|
a ) 234 , b ) 600 , c ) 388 , d ) 278 , e ) 129 | b | multiply(divide(14400, divide(add(20, const_100), const_100)), divide(5, const_100)) | a man invested rs . 14400 in rs . 100 shares of a company at 20 % premium . if his company declares 5 % dividend at the end of the year , then how much does he get ? | explanation : number of shares = = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = = rs . 600 answer : b ) rs . 600 | a = 20 + 100
b = a / 100
c = 14400 / b
d = 5 / 100
e = c * d
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a ) 09 am , b ) 07 am , c ) 11 am , d ) 05 pm , e ) 03 pm | d | add(divide(add(1140, 75), add(60, 75)), 8) | the distance between two cities a and b is 1140 km . a train starts from a at 8 a . m . and travel towards b at 60 km / hr . another train starts from b at 9 a . m and travels towards a at 75 km / hr . at what time do they meet ? | "explanation : suppose they meet x hrs after 8 a . m then , [ distance moved by first in x hrs ] + [ distance moved by second in ( x - 1 ) hrs ] = 1140 . therefore , 60 x + 75 ( x - 1 ) = 1140 . = > x = 9 . so , they meet at ( 8 + 9 ) i . e , 5 pm answer : d" | a = 1140 + 75
b = 60 + 75
c = a / b
d = c + 8
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a ) 500 , b ) 334 , c ) 555 , d ) 536 , e ) 5598 | d | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 15)) | the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 15 % , then the cost price of the article is : | "110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 110 * 560 ) / 115 = rs . 536 answer : option d" | a = 616 * 100
b = 100 + 10
c = a / b
d = 100 + 10
e = c * d
f = 100 + 15
g = e / f
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a ) 16 : 00 , b ) 18 : 00 , c ) 19 : 00 , d ) 20 : 00 , e ) 17 : 00 | e | multiply(divide(const_3, 4), divide(const_1, subtract(divide(const_1, 4), divide(const_1, 6)))) | pipe a fills a swimming pool in 4 hours . pipe b empties the pool in 6 hours . if pipe a was opened at 7 : 00 am and pipe b at 8 : 00 am , at what time will the pool be full ? | "pipe a fills the pool in 4 hrs . 1 hour ' s work : 1 / 4 pipe b empties the pool in 6 hrs . 1 hour ' s work : 1 / 6 together if they work , 1 hour ' s work = 1 / 4 - 1 / 6 = 1 / 12 given : pipe a started at 7 : 00 a . m and pipe b at 8 : 00 a . m pool filled after 1 hour by pipe a : 1 / 4 or 3 / 12 after 8 : 00 a . m pool filled after 1 hour with both the pipes on : 1 / 12 pool filled after 9 hours with both pipes on : 9 / 12 pool filled in 1 hour + pool filled in 9 hours = 3 / 12 + 9 / 12 = 1 therefore , it takes 10 hrs to fill the pool as pipe a started at 7 : 00 a . m , pool is full at 17 : 00 hrs answer : e" | a = 3 / 4
b = 1 / 4
c = 1 / 6
d = b - c
e = 1 / d
f = a * e
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a ) 8 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 21.5 % | e | multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(35, const_100))), divide(multiply(30, const_100), add(35, const_100))), const_100) | selling an kite for rs . 30 , a shop keeper gains 35 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ? | "explanation : marked price = rs . 30 c . p . = 100 / 135 * 30 = rs . 22.22 sale price = 90 % of rs . 30 = rs . 27 required gain % = 4.77 / 22.22 * 100 = 21.5 % . answer : e" | a = 30 / 100
b = 100 - 10
c = a * b
d = 30 * 100
e = 35 + 100
f = d / e
g = c - f
h = 30 * 100
i = 35 + 100
j = h / i
k = g / j
l = k * 100
|
a ) s . 59.45 , b ) s . 56.22 , c ) s . 51.219 , d ) s . 59.18 , e ) s . 51.11 | a | subtract(add(add(divide(multiply(divide(58, multiply(divide(5, const_100), 2)), 5), const_100), divide(58, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(58, multiply(divide(5, const_100), 2)), 5), const_100), divide(58, multiply(divide(5, const_100), 2))), 5), const_100)), divide(58, multiply(divide(5, const_100), 2))) | if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 58 , what is the compound interest on the same sum at the rate and for the same time ? | "explanation : sum = ( 58 * 100 ) / ( 2 * 5 ) = rs . 580 amount = [ 580 * ( 1 + 5 / 100 ) 2 ] = rs . 639.45 c . i . = ( 639.45 - 580 ) = rs . 59.45 answer : a" | a = 5 / 100
b = a * 2
c = 58 / b
d = c * 5
e = d / 100
f = 5 / 100
g = f * 2
h = 58 / g
i = e + h
j = 5 / 100
k = j * 2
l = 58 / k
m = l * 5
n = m / 100
o = 5 / 100
p = o * 2
q = 58 / p
r = n + q
s = r * 5
t = s / 100
u = i + t
v = 5 / 100
w = v * 2
x = 58 / w
y = u - x
|
a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 54 | d | divide(power(105, 3), multiply(multiply(21, 25), 45)) | if a = 105 and a ^ 3 = 21 × 25 × 45 × d , what is the value of d ? | "a = 105 = 3 * 5 * 7 a ^ 3 = 21 × 25 × 45 × d = > a ^ 3 = ( 7 * 3 ) x ( 5 * 5 ) x ( 3 ^ 2 * 5 ) x d = > a ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x d = > ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x d d = 7 ^ 2 = 49 answer d" | a = 105 ** 3
b = 21 * 25
c = b * 45
d = a / c
|
a ) $ 36540 , b ) $ 36100 , c ) $ 29580 , d ) $ 31256 , e ) $ 41250 | b | multiply(40000, power(subtract(const_1, divide(5, const_100)), 2)) | a present value of a machine is $ 40000 . its value depletiation rate is 5 % per annum then find the machine value after 2 years ? | "p = $ 40000 r = 5 % t = 2 years machine value after 3 years = p / ( 1 - r / 100 ) ^ t = 40000 * 19 / 20 * 19 / 20 * 19 / 20 = $ 36100 answer is b" | a = 5 / 100
b = 1 - a
c = b ** 2
d = 40000 * c
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a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 10 , d ) 4 : 10 , e ) 3 : 2 | c | multiply(divide(3, const_3.0), multiply(divide(3, 4), divide(4, 4))) | find the compound ratio of ( 3 : 4 ) , ( 4 : 5 ) and ( 1 : 2 ) is | "required ratio = 3 / 4 * 4 / 5 * 1 / 2 = 3 / 10 = 3 : 10 answer is c" | a = 3 / 3
b = 3 / 4
c = 4 / 4
d = b * c
e = a * d
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a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 47 | e | multiply(40, const_1) | at veridux corporation , there are 218 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 135 male associates , how many female managers are there ? | "well , first let â € ™ s take care of the â € œ totals â € . the numbers in the â € œ totals â € row must add up . if 90 are females , the other 218 â € “ 90 = 128 must be males . similarly , the numbers in the â € œ totals â € column must add up . if 40 are managers , then the other 218 â € “ 40 = 178 must be associates . now , in the â € œ associate â € row , 135 + e = 178 , which means e = 43 â € ” the other 43 associates must be female . now , to find b , which is what the question is asking , we need only look at the sum in the â € œ female â € column : b + 43 = 90 , which means b = 47 . there are fifteen female managers in this company . thus , the answer = 47 ( e ) ." | a = 40 * 1
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a ) 25 , b ) 28 , c ) 30 , d ) 34 , e ) 100 | e | add(add(add(add(add(add(add(add(const_2, const_3), add(const_2, const_3)), add(add(const_2, const_3), const_2)), add(9, const_2)), add(add(9, const_2), const_2)), add(add(add(9, const_2), const_2), const_4)), add(add(add(add(9, const_2), const_2), const_4), const_2)), add(add(add(add(add(9, const_2), const_2), const_4), const_2), const_4)) | find a sum for 1 st 9 prime number ' s ? | required sum = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ) = 100 note : 1 is not a prime number option e | a = 2 + 3
b = 2 + 3
c = a + b
d = 2 + 3
e = d + 2
f = c + e
g = 9 + 2
h = f + g
i = 9 + 2
j = i + 2
k = h + j
l = 9 + 2
m = l + 2
n = m + 4
o = k + n
p = 9 + 2
q = p + 2
r = q + 4
s = r + 2
t = o + s
u = 9 + 2
v = u + 2
w = v + 4
x = w + 2
y = x + 4
z = t + y
|
a ) one , b ) two , c ) three , d ) seven , e ) ten | c | add(1, 2) | if d = 1 / ( 2 ^ 3 * 5 ^ 11 ) 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 ^ 11 ) = multiply and divide by 2 ^ 8 = 2 ^ 8 / ( 2 ^ 3 * 2 ^ 8 * 5 ^ 11 ) = 2 ^ 8 / 10 ^ 11 = > non zero digits are 256 = > ans c" | a = 1 + 2
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a ) 11 , b ) 23 , c ) 27 , d ) 22 , e ) 91 | b | divide(subtract(25, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 25 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ? | "let the son ' s present age be x years . then , man ' s present age = ( x + 25 ) years . ( x + 25 ) + 2 = 2 ( x + 2 ) x + 27 = 2 x + 4 = > x = 23 . answer : b" | a = 2 * 2
b = a - 2
c = 25 - b
d = 2 - 1
e = c / d
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a ) 2017 , b ) 2647 , c ) 2147 , d ) 2045 , e ) 4005 | e | multiply(add(89, const_1), divide(89, const_2)) | calculate the sum of first 89 natural numbers . | "solution we know that ( 1 + 2 + 3 + . . . . . + 89 ) = n ( n + 1 ) / 2 therefore ( 1 + 2 + 3 + . . . . + 89 ) = ( 89 × 90 / 2 ) = 4005 . answer e" | a = 89 + 1
b = 89 / 2
c = a * b
|
a ) 76 sec , b ) 67 sec , c ) 98 sec , d ) 46 sec , e ) 23 sec | d | divide(add(340, 120), multiply(subtract(46, 10), divide(divide(const_10, const_2), divide(subtract(46, 10), const_2)))) | a jogger running at 10 km / hr along side a railway track is 340 m ahead of the engine of a 120 m long train running at 46 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 46 - 10 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 340 + 120 = 460 m . time taken = 460 / 10 = 46 sec . answer : d" | a = 340 + 120
b = 46 - 10
c = 10 / 2
d = 46 - 10
e = d / 2
f = c / e
g = b * f
h = a / g
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a ) 2377 , b ) 3750 , c ) 2997 , d ) 2677 , e ) 1987 | b | divide(subtract(multiply(10000, divide(add(9, divide(3, 4)), const_100)), multiply(10000, divide(9, const_100))), subtract(divide(11, const_100), divide(9, const_100))) | an amount of rs . 10000 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 3 / 4 % , then the amount invested at 11 % was ? | let the sum invested at 9 % be rs . x and that invested at 11 % be rs . ( 10000 - x ) . then , ( x * 9 * 1 ) / 100 + [ ( 10000 - x ) * 11 * 1 ] / 100 = ( 10000 * 39 / 4 * 1 / 100 ) ( 9 x + 110000 - 11 x ) / 100 = 3900 / 4 = 975 ( 9 x + 110000 - 11 x ) = 97500 x = 6250 sum invested at 9 % = rs . 6250 sum invested at 11 % = rs . ( 10000 - 6250 ) = rs . 3750 . answer : b | a = 3 / 4
b = 9 + a
c = b / 100
d = 10000 * c
e = 9 / 100
f = 10000 * e
g = d - f
h = 11 / 100
i = 9 / 100
j = h - i
k = g / j
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a ) 100 , b ) 120 , c ) 200 , d ) 220 , e ) 330 | e | add(300, divide(multiply(300, 10), const_100)) | the present population of a town is 300 . population increase rate is 10 % p . a . find the population of town after 1 years ? | p = 300 r = 10 % required population of town = p * ( 1 + r / 100 ) ^ t = 300 * ( 1 + 10 / 100 ) = 300 * ( 11 / 10 ) = 330 answer is e | a = 300 * 10
b = a / 100
c = 300 + b
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a ) 236 , b ) 127.5 , c ) 267 , d ) 268 , e ) 281 | b | subtract(1000, divide(multiply(subtract(1000, 100), subtract(800, 100)), 800)) | in a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 600 m ? | "when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 600 m , the number of meters by which a beats c = ( 600 * 212.5 ) / 1000 = 127.5 m answer : b" | a = 1000 - 100
b = 800 - 100
c = a * b
d = c / 800
e = 1000 - d
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['a ) 168 cm ³', 'b ) 192 cm ³', 'c ) 228 cm ³', 'd ) 236 cm ³', 'e ) none'] | b | subtract(volume_rectangular_prism(8, 8, 14), volume_cylinder(divide(8, const_2), 14)) | a powder tin has a square base with side 8 cm and height 14 cm . another tin has a circular base with diameter 8 cm and height 14 cm . the difference in their capacities is : | sol . difference in capacities = [ 8 * 8 * 14 - 22 / 7 * 4 * 4 * 14 ] cm ³ = 192 cm ³ answer b | a = volume_rectangular_prism - (
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a ) 22 , b ) 20 , c ) 16 , d ) 18 , e ) 14 | c | add(subtract(divide(add(add(40, 12), const_2), const_2), 12), const_1) | today david , who is 40 years old , and his daughter , who is 12 years old , celebrate their birthdays . how many years will pass before david ’ s age is twice his daughter ’ s age ? | forget conventional ways of solving math questions . in ps , ivy approach is the easiest and quickest way to find the answer . after x years passes david ’ s age will be ( 40 + x ) years old , and his daughter ’ s age will be ( 12 + x ) years old . since the david ’ s age is twice his daughter ’ s age ( 40 + x ) = 2 * ( 12 + x ) - - - > 40 + x = 24 + 2 x - - - > x = 16 the answer is ( c ) . | a = 40 + 12
b = a + 2
c = b / 2
d = c - 12
e = d + 1
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a ) 2.29 , b ) 2.75 , c ) 4.25 , d ) 4.5 , e ) none of these | d | multiply(17, 17) | ( 17 ) 3.5 x ( 17 ) ? = 178 | "solution let ( 17 ) 3.5 * ( 17 ) x = 178 . then , ( 17 ) 3.5 + x = ( 17 ) 8 . ∴ 3.5 + x = 8 ⇔ x = ( 8 - 3.5 ) ⇔ x = 4.5 answer d" | a = 17 * 17
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a ) 2 minutes , b ) 5 minutes , c ) 7 minutes , d ) 3 minutes , e ) 4 minutes | a | divide(120, add(36, divide(36, divide(const_3, const_2)))) | earl can stuff advertising circulars into envelopes at the rate of 36 envelopes per minutes and ellen requires a minutes and half to stuff the same number of envelops . working together , how long will it take earl and ellen to stuff 120 envelopes | "earl takes 1 min . for 36 envelopes . ellen takes 3 / 2 mins for the same . so ellen can stuff ( ( 36 ) / ( 3 / 2 ) ) in 1 min . i . e . , 24 envelopes a min . so both of them when work together can stuff 36 + 24 = 60 envelopes in 1 min . for 120 envelopes they will take 120 / 60 mins . i . e . , 2 mins . answer : a" | a = 3 / 2
b = 36 / a
c = 36 + b
d = 120 / c
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a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | e | divide(divide(multiply(400, 7000), 28000), 100) | if ( 400 ) ( 7000 ) = ( 28000 ) ( 100 ^ x ) , what is the value of x ? | ( 400 ) ( 7,000 ) = ( 2,800 ) ( 100 ^ x ) = > ( 400 ) ( 7,000 ) / 2,800 = 100 ^ x = > 100 = 100 ^ x = > 100 ^ 1 = 100 ^ x since , base is same so powers will be same too . so , x = 1 answer will be e | a = 400 * 7000
b = a / 28000
c = b / 100
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a ) $ 2.40 , b ) $ 3.00 , c ) $ 3.20 , d ) $ 3.60 , e ) $ 4.80 | a | divide(0.64, subtract(const_1, add(divide(3, 5), multiply(divide(const_1, 3), subtract(const_1, divide(3, 5)))))) | having received his weekly allowance , john spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.64 at the candy store . what is john ’ s weekly allowance ? | "total allowance = x amount spent at the arcade = 3 / 5 x amount remaining = 2 / 5 x amount spent at the toy store = 2 / 5 * 1 / 3 x = 2 / 15 x amount remaining = 2 / 5 x - 2 / 15 x = 4 / 15 x now , 4 / 15 x = $ 0.64 therefore , x = $ 2.40 . answer a" | a = 3 / 5
b = 1 / 3
c = 3 / 5
d = 1 - c
e = b * d
f = a + e
g = 1 - f
h = 0 / 64
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a ) 13 , b ) 12 , c ) 15 , d ) 16 , e ) 17 | a | divide(41, 3) | a number of 41 marbles is to be divided and contain with boxes . if each box is to contain 3 , 4 , or 5 marbles , what is the largest possible number of boxes ? | "to maximize # of boxes we should minimize marbles per box : 12 * 3 + 1 * 5 = 41 - - > 12 + 1 = 13 . answer : a ." | a = 41 / 3
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a ) and 25 , b ) and 24 , c ) and 22 , d ) and 29 , e ) of these | a | subtract(41, divide(add(41, 7), const_3)) | the sum of the present age of henry and jill is 41 . what is their present ages if 7 years ago henry was twice the age of jill ? | let the age of jill 7 years ago be x , age of henry be 2 x x + 7 + 2 x + 7 = 41 x = 9 present ages will be 16 and 25 answer : a | a = 41 + 7
b = a / 3
c = 41 - b
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a ) 4 , b ) 10 , c ) 5 , d ) 9 , e ) 8 | c | add(subtract(12, 0), 0) | set x consists of the integers from 0 to 12 , inclusive , while set y consists of the integers from 6 to 10 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "x = { 0 , 1,2 , 3,4 , 5,6 , 7 , 8 , 9 , 10 , 11 , 12 } y = { 6 , 7 , 8 , 9 , 10 } common elements = { 6 , 7 , 8 , 9 , 10 } = 5 elements answer : option c ." | a = 12 - 0
b = a + 0
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a ) 1000 , b ) 2998 , c ) 2500 , d ) 2788 , e ) 2991 | c | divide(multiply(500, const_100), subtract(add(const_100, 10), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 500 more , there would have been a gain of 10 % . what is the cost price ? | "90 % 110 % - - - - - - - - 20 % - - - - 500 100 % - - - - ? = > rs . 2500 answer : c" | a = 500 * 100
b = 100 + 10
c = 100 - 10
d = b - c
e = a / d
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a ) 287 , b ) 269 , c ) 338 , d ) 200 , e ) 230 | c | divide(square_area(26), const_2) | what is the area of a square field whose diagonal of length 26 m ? | "d 2 / 2 = ( 26 * 26 ) / 2 = 338 answer : c" | a = square_area / (
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | multiply(divide(subtract(624, 600), 600), const_100) | a book is bought for $ 600 and sold for $ 624 . what is the profit in percentage ? | "624 / 600 = 1.04 the answer is b ." | a = 624 - 600
b = a / 600
c = b * 100
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a ) $ 14,755 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 12,285 , e ) $ 17,155 | d | multiply(divide(const_3, const_4), const_1000) | a store owner estimates that the average price of type a products will increase by 35 % next year and that the price of type b products will increase by 20 % next year . this year , the total amount aid for type a products was $ 3500 and the total price paid for type b products was $ 6300 . according to the store owner ' s estimate , and assuming the number of products purchased next year remains the same as that of this year , how much will be spent for both products next year ? | "cost of type a products next year = 1.35 * 3500 = 4725 cost of type b products next year = 1.2 * 6300 = 7560 total 4725 + 7560 = 12285 answer : d" | a = 3 / 4
b = a * 1000
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a ) 11 , b ) 18 , c ) 13 , d ) 17 , e ) 12 | e | multiply(divide(75, add(add(divide(2, 3), divide(5, 2)), 2)), 5) | a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 3 and b : c = 2 : 5 . if the total runs scored by all of them are 75 , the runs scored by a are ? | "a : b = 2 : 3 b : c = 2 : 5 a : b : c = 4 : 6 : 15 4 / 25 * 75 = 12 answer : e" | a = 2 / 3
b = 5 / 2
c = a + b
d = c + 2
e = 75 / d
f = e * 5
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a ) 270 , b ) 288 , c ) 100 , d ) 299 , e ) 126 | c | multiply(divide(subtract(72, 36), const_3_6), 10) | two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 10 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 27 sec = 10 * 10 = 100 m . the length of the faster train = 100 m . answer : c" | a = 72 - 36
b = a / const_3_6
c = b * 10
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a ) 22 , b ) 75 , c ) 60 , d ) 58 , e ) 11 | d | subtract(divide(6788, 78), 29) | a trader sells 78 meters of cloth for rs . 6788 at the profit of rs . 29 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 6788 / 78 = rs . 87 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 87 - rs . 29 = rs . 58 . answer : d" | a = 6788 / 78
b = a - 29
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a ) s . 429 , b ) s . 480 , c ) s . 320 , d ) s . 128 , e ) s . 419 | c | subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 50), const_100)) | by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 50 % ? | "sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 50 % = 50 % of 640 = rs . 320 sp = cp - loss = 640 - 320 = rs . 320 answer : c" | a = 800 * 100
b = 25 + 100
c = a / b
d = 800 * 100
e = 25 + 100
f = d / e
g = f * 50
h = g / 100
i = c - h
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a ) 3 . , b ) 4 . , c ) 5 . , d ) 6 . , e ) 9 . | e | power(subtract(power(8, 2), 17), divide(17, 8)) | a ( 8 , w ^ 2 ) is the ( x , y ) coordinate of point located on the parabola y = x ^ 2 + 17 . what is the value of w ? | "y = x ^ 2 + 17 w ^ 2 = 8 ^ 2 + 17 w ^ 2 = 81 w = 9 answer e" | a = 8 ** 2
b = a - 17
c = 17 / 8
d = b ** c
|
a ) 40 % , b ) 50 % , c ) 60 % , d ) 80 % , e ) 90 % | d | divide(40, subtract(const_1, divide(50, const_100))) | on a certain transatlantic crossing , 40 percent of a ship ’ s passengers held round - trip tickets and also took their cars abroad the ship . if 50 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ’ s passengers held round - trip tickets ? | let t be the total number of passengers . let x be the number of people with round trip tickets . 0.4 t had round trip tickets and took their cars . 0.5 x had round trip tickets and took their cars . 0.5 x = 0.4 t x = 0.8 t the answer is d . | a = 50 / 100
b = 1 - a
c = 40 / b
|
a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 56 | b | multiply(0.30103, subtract(2128, multiply(const_2, const_1000))) | if log 2 = 0.30103 , the number of digits in 2128 is . | log ( 2128 ) = 128 log ( 2 ) = 128 × 0.30103 ≈ 38.4 ie , its characteristic = 38 hence , number of digits in 2128 = 38 + 1 = 39 answer : option b | a = 2 * 1000
b = 2128 - a
c = 0 * 30103
|
a ) 8 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | a | divide(40, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 40 white marbles . how many red marbles could be in bag a ? | "6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 3 let w in bag b be 4 k w = 30 = 6 k + 4 k = > k = 40 / 10 = 4 total red ' s in bag a will be 2 k = 8 a" | a = 3 * 2
b = a + 4
c = 40 / b
|
a ) 14 , b ) 20 , c ) 18 , d ) 24 , e ) 21 | d | multiply(4, divide(6, subtract(5, 4))) | if a person walks at 5 km / hr instead of 4 km / hr , he would have walked 6 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 4 = ( x + 6 ) / 5 x - 24 = > x = 24 km . answer : d" | a = 5 - 4
b = 6 / a
c = 4 * b
|
a ) $ 30.60 , b ) $ 60.60 , c ) $ 72.96 , d ) $ 40.60 , e ) $ 50.60 | c | add(60.8, divide(multiply(60.8, 20), const_100)) | if tim had lunch at $ 60.80 and he gave 20 % tip , how much did he spend ? | the tip is 20 % of what he paid for lunch . hence tip = 20 % of 60.80 = ( 20 / 100 ) * 60.80 = $ 12.16 total spent 60.80 + 12.16 = $ 72.96 correct answer c | a = 60 * 8
b = a / 100
c = 60 + 8
|
a ) 2 pm , b ) 9 pm , c ) 3 pm , d ) 8 pm , e ) 6 pm | e | subtract(multiply(const_2, const_12), divide(multiply(1, const_12), add(divide(1, 3), const_1))) | when asked what the time is , a person answered that the amount of time left is 1 / 3 of the time already completed . what is the time . | "a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 − x = 1 / 3 x ⇒ x = 18 time is 6 pm answer : e" | a = 2 * 12
b = 1 * 12
c = 1 / 3
d = c + 1
e = b / d
f = a - e
|
a ) 12.5 , b ) 15 , c ) 18 , d ) 19 , e ) 25 | a | sqrt(subtract(power(18, const_2), power(subtract(sqrt(subtract(power(18, const_2), power(6, const_2))), 4), const_2))) | a ladder 18 feet long is leaning against a wall that is perpendicular to level ground . the bottom of the ladder is 6 feet from the base of the wall . if the top of the ladder slips down 4 feet , how many feet will the bottom of the ladder slip ? | "18 ^ 2 - 6 ^ 2 = 288 it means that the height is equal to 16.9 . since the top of the ladder slips down 4 feet , then the height of the wall = 16.9 - 4 = 12.9 the bottom = sqrt ( 18 ^ 2 - 12.9 ^ 2 ) = sqrt ( 324 - 166.41 ) = 12.5 ans is a" | a = 18 ** 2
b = 18 ** 2
c = 6 ** 2
d = b - c
e = math.sqrt(d)
f = e - 4
g = f ** 2
h = a - g
i = math.sqrt(h)
|
a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | d | divide(power(2, 2), multiply(choose(4, 2), choose(2, 1))) | if x is to be chosen at random from the set { 1 , 2 , 3 , 4 } and y is to be chosen at random from the set { 5 , 6 , 7 } , what is the probability that xy will be even ? | "in order to make even nos . by multiplication , we should have even * odd , odd * even or even * even total even nos . possible by multiplying nos . from the 2 sets : ( 1 * 6 ) ; 2 * any of the three from set b ; 3 * 6 ; & 4 * any of the three from set b 1 + 3 + 1 + 3 = 8 total possibilities = 4 * 3 = 12 p ( event ) = 8 / 12 or 2 / 3 ans d" | a = 2 ** 2
b = math.comb(4, 2)
c = math.comb(2, 1)
d = b * c
e = a / d
|
a ) $ 30 , b ) $ 54 , c ) $ 26 , d ) $ 12 , e ) $ 9.60 | c | multiply(3.25, 8) | johnny makes $ 3.25 per hour at his work . if he works 8 hours , how much money will he earn ? | 3.25 * 8 = 26 . answer is c . | a = 3 * 25
|
a ) 28 % , b ) 29 % , c ) 16 % , d ) 26 % , e ) 32 % | b | subtract(const_100, divide(multiply(900, const_100), 1280)) | an article is bought for rs . 1280 and sold for rs . 900 , find the loss percent ? | "1280 - - - - 380 100 - - - - ? = > 29 % answer : b" | a = 900 * 100
b = a / 1280
c = 100 - b
|
a ) 5 % , b ) 6.6 % , c ) 9 % , d ) 2 % , e ) 4 % | b | divide(multiply(const_100, 160), multiply(800, 3)) | what is the rate percent when the simple interest on rs . 800 amount to rs . 160 in 3 years ? | "160 = ( 800 * 3 * r ) / 100 r = 6.6 % answer : b" | a = 100 * 160
b = 800 * 3
c = a / b
|
a ) - 8 , b ) - 9 , c ) 9 , d ) 8 , e ) - 7 | b | divide(negate(add(17, 1)), 2) | solve below question 2 x + 1 = - 17 | "1 . subtract 1 from both sides : 2 x + 1 - 1 = - 17 - 1 2 . simplify both sides : 2 x = - 18 3 . divide both sides by 2 : 4 . simplify both sides : x = - 9 b" | a = 17 + 1
b = negate / (
|
a ) 6084 , b ) 3788 , c ) 2077 , d ) 8082 , e ) 1812 | d | subtract(add(add(add(multiply(multiply(9, const_100), const_10), multiply(const_4.0, const_100)), multiply(3, const_10)), 1), add(add(add(const_1000, multiply(3, const_100)), multiply(2, const_10)), 9)) | what is the difference between the largest number and the least number written with the digits 9 , 3 , 1 , 2 ? | "explanation : 1239 9321 - - - - - - - - - - - - 8082 answer : d" | a = 9 * 100
b = a * 10
c = 4 * 0
d = b + c
e = 3 * 10
f = d + e
g = f + 1
h = 3 * 100
i = 1000 + h
j = 2 * 10
k = i + j
l = k + 9
m = g - l
|
a ) 2 , b ) 3 , c ) 15 , d ) 30 , e ) 45 | c | floor(power(multiply(1800, const_2), const_0_33)) | if m and n are positive integer , and 1800 m = n ^ 3 , what is the least possible value of m ? | 1800 * m = n ^ 3 m = n ^ 3 / 1800 = n ^ 3 / ( 3 ^ 2 * 5 ^ 2 * 2 ^ 3 ) so n has to be 3 * 5 * 2 so that n ^ 3 is divisible by 1800 . so m = ( 2 * 3 * 5 ) ^ 3 / ( 3 ^ 2 * 5 ^ 2 * 2 ^ 3 ) m = 15 answer : c | a = 1800 * 2
b = a ** const_0_33
c = math.floor(b)
|
a ) 60 , b ) 61.2 , c ) 3,744 , d ) 3,600 , e ) 3,672 | c | multiply(1.04, const_3600) | the moon revolves around the earth at a speed of approximately 1.04 kilometers per second . this approximate speed is how many kilometers per hour ? | "moon revolves around the earth at a speed of 1.02 kilometers per second . one hour equal to 60 minutes . one minute equals to 60 seconds . so one hour equals to 3600 seconds . so one hour , speed = 1.04 * 3600 = 3744 kilometers per hour . option c is correct" | a = 1 * 4
|
a ) $ 690 , b ) $ 620 , c ) $ 650 , d ) $ 640 , e ) $ 770 | a | add(add(multiply(100, 6), multiply(30, 3)), multiply(multiply(10, 3), const_2)) | rates for having a manuscript typed at a certain typing service are $ 6 per page for the first time a page is typed and $ 3 per page each time a page is revised . if a certain manuscript has 100 pages , of which 30 were revised only once , 10 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ? | "for 100 - 30 - 10 = 60 pages only cost is 6 $ per page for the first time page is typed - 60 * 5 = 300 $ ; for 30 pages the cost is : first time 6 $ + 3 $ of the first revision - 30 * ( 6 + 3 ) = 270 $ ; for 10 pages the cost is : first time 5 $ + 3 $ of the first revision + 3 $ of the second revision - 10 ( 6 + 3 + 3 ) = 120 $ ; total : 300 + 270 + 120 = 690 $ . answer : a ." | a = 100 * 6
b = 30 * 3
c = a + b
d = 10 * 3
e = d * 2
f = c + e
|
a ) rs . 90 , b ) rs . 76 , c ) rs . 98 , d ) rs . 18 , e ) rs . 02 | c | subtract(divide(8925, 85), 15) | a trader sells 85 meters of cloth for rs . 8925 at the profit of rs . 15 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 8925 / 85 = rs . 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 15 = rs . 90 . answer : c" | a = 8925 / 85
b = a - 15
|
a ) 150 , b ) 288 , c ) 1687 , d ) 997 , e ) 121 | a | divide(120, divide(120, const_100)) | 20 % of a number is added to 120 , the result is the same number . find the number ? | "( 20 / 100 ) * x + 120 = x 2 x = 300 x = 150 answer : a" | a = 120 / 100
b = 120 / a
|
a ) 333 , b ) 200 , c ) 288 , d ) 276 , e ) 220 | e | subtract(divide(divide(5830, 26.50), const_2), multiply(const_2, 20)) | the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 5830 , what is the length of the plot in metres ? | "let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 5830 [ 4 l - 40 ] = 5830 / 26.50 = 220 4 l = 260 l = 260 / 4 = 65 meters . answer : e" | a = 5830 / 26
b = a / 2
c = 2 * 20
d = b - c
|
a ) 300 km , b ) 325 km , c ) 375 km , d ) 425 km , e ) 450 km | c | multiply(divide(multiply(divide(multiply(divide(480, 4), 8), 3), 15), 16), 5) | there is a train and car . the ratio between the speed of a train & a car is 16 : 15 respectively . also , a bus covered a distance of 480 km in 8 hours . the speed of the bus is 3 / 4 th of the speed of the train . how much distance will the car cover in 5 hours ? | "the speed of the bus is 480 / 8 = 60 km / hr the speed of the train is ( 60 * 4 ) / 3 = 80 km / hr the speed of the car is 80 / 16 * 15 = 75 km / hr the distance covered by the car in 5 hours is 75 × 5 = 375 km the answer is c ." | a = 480 / 4
b = a * 8
c = b / 3
d = c * 15
e = d / 16
f = e * 5
|
a ) a . 66000 , b ) b . 67000 , c ) c . 68000 , d ) d . 69000 , e ) e . 70000 | a | multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3)) | if a town of 65,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ? | 1 % is quite small and hence the answer is a ) | a = 3 + 4
b = 5 * a
c = b * 100
d = 3 + 4
e = d + 3
f = c * e
|
a ) 20 , b ) 22.5 , c ) 30 , d ) 37.5 , e ) 45 | c | multiply(divide(2, add(3, 2)), 75) | two trains , a and b , started simultaneously from opposite ends of a 75 - mile route and traveled toward each other on parallel tracks . train a , traveling at a constant rate , completed the 75 - mile trip in 3 hours ; train b , traveling at a constant rate , completed the 75 - mile trip in 2 hours . how many miles had train a traveled when it met train b ? | "as the ratio of the rates of a and b is 2 to 3 then the distance covered at the time of the meeting ( so after traveling the same time interval ) would also be in that ratio , which means that x would cover 2 / ( 2 + 3 ) = 2 / 5 of 75 miles : 75 * 2 / 5 = 30 miles . answer : c ." | a = 3 + 2
b = 2 / a
c = b * 75
|
a ) 9.4 % , b ) 9.33 % , c ) 9.6 % , d ) 9.8 % , e ) 10 % | b | multiply(divide(add(divide(multiply(12, 15), const_100), divide(multiply(8, 30), const_100)), add(15, 30)), const_100) | in one alloy there is 12 % chromium while in another alloy it is 8 % . 15 kg of the first alloy was melted together with 30 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 + 30 = 45 kg alloy is 0.12 * 15 + 0.08 * 30 = 4.2 kg , so the percentage is 4.2 / 45 * 100 = 9.33 % . answer : b ." | a = 12 * 15
b = a / 100
c = 8 * 30
d = c / 100
e = b + d
f = 15 + 30
g = e / f
h = g * 100
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(divide(const_100.0, const_2), multiply(135, 135)) | what is the remainder when 135 ^ 77 is divided by 7 ? | "( 15 ^ 77 * 9 ^ 77 ) / 7 ; ( 15 ^ 77 ) / 7 * ( 9 ^ 77 ) / 7 ; 1 * 2 ^ 77 / 7 ; ( ( 2 ^ 3 ) ^ 25 * 2 ^ 2 ) / 7 ; 1 * 4 = 4 answer : e" | a = 100 / 0
b = 135 * 135
c = a - b
|
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | b | divide(45, subtract(const_10, const_1)) | the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 45 . what is the difference between the two digits of that number ? | "sol . let the ten ’ s digit be x and unit ’ s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 45 ⇔ 9 ( x - y ) = 45 ⇔ x - y = 5 answer b" | a = 10 - 1
b = 45 / a
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | c | add(divide(multiply(subtract(const_0_25, divide(const_1, const_100)), 50), divide(subtract(42.30, subtract(const_0_25, divide(const_1, const_100))), 14)), divide(subtract(42.30, subtract(const_0_25, divide(const_1, const_100))), 14)) | if 42.30 = k ( 14 + m / 50 ) , where k and m are positive integers and m < 50 , then what is the value of k + m ? | "42.30 = 14 k + km / 50 . . . we can rewrite the number as follows : 42 + 0.30 = 14 k + km / 50 . . . . . . . . since k is integer , then 42 = 14 k . . . . . . . . . . k = 3 0.30 = km / 50 . . . . . . 40 / 100 = 3 m / 50 . . . . . . m = 5 k + m = 3 + 5 = 8 answer : c" | a = 1 / 100
b = const_0_25 - a
c = b * 50
d = 1 / 100
e = const_0_25 - d
f = 42 - 30
g = f / 14
h = c / g
i = 1 / 100
j = const_0_25 - i
k = 42 - 30
l = k / 14
m = h + l
|
a ) 12 , b ) 8 , c ) 6 , d ) 10 , e ) 16 | a | multiply(const_4, power(27, divide(const_1, const_3))) | a cube is painted red on all faces . it is then cut into 27 equal smaller cubes . how many j cubes are painted on only 2 faces ? | "1 ) draw a simple cube 2 ) draw 9 squares on each face of the cube ( so that it looks like a rubik ' s cube ) - this is what the cube will look like when it ' s cut into 27 equal smaller cubes . 3 ) remember that the outside of the cube is the part that ' s painted . . . . the mini - cubes with 2 painted sides are all on the edge of the cube , in themiddleof the edge . there are 4 in front , 4 in back and 4 more on thestripthat runs around the left / top / right / bottom of the cube . j = 4 + 4 + 4 = 12 . answer a" | a = 1 / 3
b = 27 ** a
c = 4 * b
|
a ) 35 years , b ) 25 years , c ) 42 years , d ) 39 years , e ) 40 years | c | divide(multiply(subtract(36, 6), 7), 5) | at present the ratio between the ages of arun and deepak is 5 : 7 . after 6 years , arun ' s age will be 36 years . what is the age of deepak at present ? | "let the present ages of arun and deepak be 5 x years and 7 x years respectively 5 x + 6 = 36 5 x = 30 x = 6 deepak ' s age = 7 x = 42 years answer is c" | a = 36 - 6
b = a * 7
c = b / 5
|
a ) 18 years , b ) 16 years , c ) 14 years , d ) 12 years , e ) 10 years | e | subtract(multiply(15, 15), add(multiply(5, 14), multiply(10, 16))) | the average age of 15 students of a class is 16 years . out of these , the average age of 5 students is 14 years and that of the other 10 students is 16 years . tee age of the 15 th student is ? | "age of the 15 th student = [ 15 * 16 - ( 14 * 5 + 16 * 10 ) ] = ( 240 - 230 ) = 10 years . answer : e" | a = 15 * 15
b = 5 * 14
c = 10 * 16
d = b + c
e = a - d
|
a ) 15 , b ) 24 , c ) 17 , d ) 19 , e ) none of these | b | multiply(const_60, divide(subtract(75, 45), 75)) | excluding stoppages , the speed of a bus is 75 kmph and including stoppages , it is 45 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 30 km less . time taken to cover 30 km = ( ( 30 / 75 ) ã — 60 ) = 24 min . option ( b ) is correct" | a = 75 - 45
b = a / 75
c = const_60 * b
|
a ) 90 , b ) 120 , c ) 160 , d ) 360 , e ) 520 | e | divide(130, subtract(const_1, divide(3, 4))) | the visitors of a modern art museum who watched a certain picasso painting were asked to fill in a short questionnaire indicating whether they had enjoyed looking at the picture and whether they felt they had understood it . according to the results of the survey , all 130 visitors who did not enjoy the painting also did not feel they had understood the painting , and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting . if 3 / 4 of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting , then how many visitors answered the questionnaire ? | if we exclude those cases and take the question at face value , then it seems straightforward . group # 1 = ( did n ' t like , did n ' t understand ) = 130 group # 2 = ( like understood ) = 3 / 4 ( 1 / 4 ) n = 130 n = 520 answer = ( e ) | a = 3 / 4
b = 1 - a
c = 130 / b
|
['a ) 1296', 'b ) 1369', 'c ) 1681', 'd ) 1764', 'e ) 2500'] | a | multiply(multiply(multiply(const_3, const_3), const_4), multiply(multiply(const_3, const_3), const_4)) | when a certain perfect square is increased by 148 , the result is another perfect square . what is the value of the original perfect square ? | let ’ s call the two perfect squares x ^ 2 and y ^ 2 , respectively . then the given information translates as x ^ 2 + 148 = y ^ 2 . subtracting x ^ 2 gives 148 = y ^ 2 − x ^ 2 , a difference of squares . this , in turn , factors as ( y + x ) ( y − x ) = 148 . the next step is tricky . it begins with factoring 148 , which breaks down as 2 ∗ 2 ∗ 37 . since we ’ re dealing with perfect squares , x and y are positive integers , and ( y + x ) and ( y − x ) must be paired integer factors of 148 . the options are 148 ∗ 1,74 ∗ 2 , and 37 ∗ 4 . but our number properties establish that ( y + x ) and ( y − x ) must be either both odd or both even , so only 74 ∗ 2 is an actual possibility . and because for any positive integers ( y + x ) > ( y − x ) , we can conclude that y + x = 74 and y – x = 2 . solving by elimination , 2 y = 76 , y = 38 , and x = 36 . finally , we just need to square 36 . but rather than multiplying it out , note that 36 ^ 2 ends in 6 – as does only one answer , choice a . this answer must be the one we want . | a = 3 * 3
b = a * 4
c = 3 * 3
d = c * 4
e = b * d
|
a ) 4 hours , b ) 2 hours , c ) 2 hours 30 minutes , d ) 5 hours , e ) 3 hours 20 minutes | d | divide(add(13, 17), subtract(43, 37)) | train a is 13 miles ahead of train b , which is traveling the same direction along the same route as train a . if train a is traveling at an average speed of 37 miles per hour and train b is traveling at an average speed of 43 miles per hour , how long will it take train b to overtake and drive 17 miles ahead of train a ? | relative speed = 43 - 37 = 6 miles per hour dist required = 13 + 17 = 30 miles time taken to overtake = 30 / 6 = 5 hours . d is the answer . | a = 13 + 17
b = 43 - 37
c = a / b
|
a ) 1 / 20 , b ) 1 / 12 , c ) 2 / 15 , d ) 3 / 10 , e ) 1 / 7 | b | multiply(divide(const_1, const_2), divide(const_1, const_4)) | x , y , and z are all unique numbers . if x is chosen randomly from the set { 9 , 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ? | "p ( x is prime ) = 1 / 3 p ( y is prime ) = 1 / 4 if y is prime , then z is not prime since y and z are unique . then the probability is 1 / 3 * 1 / 4 = 1 / 12 the answer is b ." | a = 1 / 2
b = 1 / 4
c = a * b
|
a ) 4 , b ) 5 , c ) 56 , d ) 2 , e ) 7 | a | subtract(add(const_100, 30), add(divide(multiply(add(const_100, 30), 20), const_100), const_100)) | on increasing the price of t . v . sets by 30 % , their sale decreases by 20 % . what is the effect on the revenue receipts of the shop ? | "explanation : let the price be = rs . 100 , and number of units sold = 100 then , sale value = rs . ( 100 × 100 ) = rs . 10000 new sale value = rs . ( 130 × 80 ) = rs . 10400 increase % = 4001000040010000 × 100 = 4 % answer : a" | a = 100 + 30
b = 100 + 30
c = b * 20
d = c / 100
e = d + 100
f = a - e
|
a ) $ 374 , b ) $ 330 , c ) $ 385 , d ) $ 392 , e ) $ 399 | b | add(divide(348, add(const_1, divide(16, const_100))), multiply(divide(10, const_100), divide(348, add(const_1, divide(16, const_100))))) | if sharon ' s weekly salary increased by 16 percent , she would earn $ 348 per week . if instead , her weekly salary were to increase by 10 percent , how much would she earn per week ? | "( 348 / 116 ) 110 = 330 in this case long division does not take much time . ( 348 / 116 ) = 3 3 * 110 = 330 ( 300 + 30 ) answer b" | a = 16 / 100
b = 1 + a
c = 348 / b
d = 10 / 100
e = 16 / 100
f = 1 + e
g = 348 / f
h = d * g
i = c + h
|
a ) $ 160 , b ) $ 350 , c ) $ 282 , d ) $ 274 , e ) $ 286 | b | add(multiply(18.00, add(const_3, const_4)), multiply(14.00, subtract(23, add(const_3, const_4)))) | if the charge of staying in a student youth hostel $ 18.00 / day for the first week , and $ 14.00 / day for each additional week , how much does it cost to stay for 23 days ? | "total number of days of stay = 23 charge of staying in first week = 18 * 7 = 126 $ charge of staying for additional days = ( 23 - 7 ) * 14 = 16 * 14 = 224 $ total charge = 126 + 224 = 350 $ answer b" | a = 3 + 4
b = 18 * 0
c = 3 + 4
d = 23 - c
e = 14 * 0
f = b + e
|
a ) 360 , b ) 380 , c ) 400 , d ) 416 , e ) 412 | d | multiply(divide(subtract(divide(360, subtract(const_1, divide(const_1, 10))), 360), 12), 125) | mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 125 mangoes . | "mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 120 mangoes . method 1 : let price per mango = x . let number of mangoes be n . then , nx = 360 . now price = 0.9 x ; number of mangoes = n + 12 . total amount = 0.9 x * ( n + 12 ) = 360 . nx = 0.9 nx + 10.8 x = > 0.1 nx = 10.8 x = > n = 108 = > x = 360 / 108 = 3.33 original price of 125 mangoes = 125 * 3.33 = 416 . answer d" | a = 1 / 10
b = 1 - a
c = 360 / b
d = c - 360
e = d / 12
f = e * 125
|
a ) 90 hours , b ) 80 hours , c ) 50 hours , d ) 120 hours , e ) 110 hours | e | divide(multiply(10, 11), subtract(divide(multiply(10, 11), 10), divide(multiply(10, 11), 11))) | calculate the time it will take for a full tank of water to become completely empty due to a leak given that the tank could be filled in 10 hours , but due to the leak in its bottom it takes 11 hours to be filled ? | part filled without leak in 1 hour = 1 / 10 part filled with leak in 1 hour = 1 / 11 work done by leak in 1 hour = 1 / 10 â ˆ ’ 1 / 11 = 110 hours answer : e | a = 10 * 11
b = 10 * 11
c = b / 10
d = 10 * 11
e = d / 11
f = c - e
g = a / f
|
a ) 16.5 , b ) 13.2 , c ) 10.6 , d ) 11.2 , e ) 7.25 | c | divide(subtract(400, multiply(25, 5.4)), 25) | in the first 25 overs of a cricket game , the run rate was only 5.4 . what should be the run rate in the remaining 25 overs to reach the target of 400 runs ? | required run rate = 400 - ( 5.4 x 25 ) / 25 = 265 / 25 = 10.6 option c | a = 25 * 5
b = 400 - a
c = b / 25
|
a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 13 | d | add(divide(add(power(3, 3), sqrt(add(power(power(3, 3), const_2), power(subtract(279, power(3, 3)), const_2)))), multiply(power(3, const_2), const_2)), subtract(divide(add(power(3, 3), sqrt(add(power(power(3, 3), const_2), power(subtract(279, power(3, 3)), const_2)))), multiply(power(3, const_2), const_2)), 3)) | if one positive integer is greater than another positive integer by 3 , and the difference of their cubes is 279 , what is their sum ? | "1 ^ 3 = 1 2 ^ 3 = 8 3 ^ 3 = 27 4 ^ 3 = 64 5 ^ 3 = 125 6 ^ 3 = 216 7 ^ 3 = 343 the two numbers are 4 and 7 . the answer is d ." | a = 3 ** 3
b = 3 ** 3
c = b ** 2
d = 3 ** 3
e = 279 - d
f = e ** 2
g = c + f
h = math.sqrt(g)
i = a + h
j = 3 ** 2
k = j * 2
l = i / k
m = 3 ** 3
n = 3 ** 3
o = n ** 2
p = 3 ** 3
q = 279 - p
r = q ** 2
s = o + r
t = math.sqrt(s)
u = m + t
v = 3 ** 2
w = v * 2
x = u / w
y = x - 3
z = l + y
|
a ) 3630 , b ) 3877 , c ) 2667 , d ) 2977 , e ) 2378 | a | multiply(divide(6300, add(add(6300, 4200), 10500)), 12100) | a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 12100 after a year ? | "explanation : 6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12100 = 3630 answer : a" | a = 6300 + 4200
b = a + 10500
c = 6300 / b
d = c * 12100
|
a ) 35 years , b ) 45 years , c ) 51 years , d ) 52 years , e ) none of these | d | add(37, const_1) | the average age of 37 students in a group is 14 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ? | "explanation : age of the teacher = ( 38 * 15 - 37 * 14 ) years = 52 years . answer : d" | a = 37 + 1
|
a ) 36 , b ) 87 , c ) 40 , d ) 37 , e ) 86 | a | divide(multiply(18, 12), subtract(18, 12)) | pipe a can fill a tank in 12 hours . due to a leak at the bottom , it takes 18 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 12 - 1 / x = 1 / 18 = > 1 / x = 1 / 12 - 1 / 18 = ( 3 - 2 ) / 36 = 1 / 36 = > x = 36 . answer : a" | a = 18 * 12
b = 18 - 12
c = a / b
|
a ) 11 , b ) 8 , c ) 37 , d ) 15 , e ) 18 | e | divide(multiply(9, 3), divide(const_3, const_2)) | if one and a half women , drink one and a half tea in one and a half minutes . how many tea can 9 women drink in 3 minutes ? | e 18 explanation : more minutes implies more tea . more women implies more tea . the time become twice ( 3 minutes / 1.5 minutes ) . women become six times ( 9 / 1.5 ) . number of tea = 2 * 6 * original ( 1.5 ) | a = 9 * 3
b = 3 / 2
c = a / b
|
a ) 900 m , b ) 500 m , c ) 120 m , d ) 180 m , e ) 190 m | b | subtract(multiply(2, const_1000), multiply(divide(multiply(2, const_1000), add(multiply(3, const_60), 5)), add(multiply(2, const_60), 20))) | a can run 2 km distance in 2 min 20 seconds , while b can run this distance in 3 min 5 sec . . by how much distance can a beat b ? | a takes time 2.20 minutes = 140 sec b takes time 3 minutes = 185 sec difference = 185 - 140 = 45 sec now we are to find distance covered in 40 sec by b 180 sec = 2000 m 1 sec = 100 / 9 m 45 sec = 45 x 100 / 9 = 500 m answer : b | a = 2 * 1000
b = 2 * 1000
c = 3 * const_60
d = c + 5
e = b / d
f = 2 * const_60
g = f + 20
h = e * g
i = a - h
|
a ) rs . 3900 , b ) rs . 3990 , c ) rs . 4000 , d ) rs . 4100 , e ) rs . 4200 | a | add(multiply(divide(4000, add(multiply(15, divide(4, 10)), 2)), 3), multiply(multiply(divide(4, 10), divide(4000, add(multiply(15, divide(4, 10)), 2))), 12)) | the price of 10 chairs is equal to that of 4 tables . the price of 15 chairs and 2 tables together is rs . 4000 . the total price of 12 chairs and 3 tables is : | "let the cost of a chair and that of a table be rs . x and rs . y respectively . then , 10 x = 4 y or y = 5 x . 2 15 x + 2 y = 4000 15 x + 2 x 5 x = 4000 2 20 x = 4000 x = 200 . so , y = 5 x 200 = 500 . 2 hence , the cost of 12 chairs and 3 tables = 12 x + 3 y = rs . ( 2400 + 1500 ) = rs . 3900 . a" | a = 4 / 10
b = 15 * a
c = b + 2
d = 4000 / c
e = d * 3
f = 4 / 10
g = 4 / 10
h = 15 * g
i = h + 2
j = 4000 / i
k = f * j
l = k * 12
m = e + l
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(multiply(15, 2), 28) | if the remainder is 15 when the integer n is divided by 28 , what is the remainder when 2 n is divided by 14 ? | "n = 28 k + 15 2 n = 2 ( 28 k + 15 ) = 4 k * 14 + 30 = 4 k * 14 + 2 * 14 + 2 = 14 j + 2 the answer is c ." | a = 15 * 2
b = a - 28
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | b | divide(multiply(divide(25, 2), 16), 4) | rice weighing 25 / 2 pounds was divided equally and placed in 4 containers . how many ounces of rice were in each container ? ( note that 1 pound = 16 ounces ) | "25 / 2 ÷ 4 = 25 / 8 pounds in each container 25 / 8 pounds * 16 ounces / pound = 50 ounces in each container the answer is b ." | a = 25 / 2
b = a * 16
c = b / 4
|
a ) s . 600 , b ) s . 800 , c ) s . 500 , d ) s . 900 , e ) s . 1000 | e | divide(21000, add(18, 3)) | a man sold 18 toys for rs . 21000 , gaining thereby the cost price of 3 toy find the cost price of a toy | "let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 21000 . gain = sp â € “ cp 3 x = 21000 â € “ 18 x 21 x = 21000 x = rs . 1000 . answer : option e" | a = 18 + 3
b = 21000 / a
|
a ) 109 , b ) 119 , c ) 129 , d ) 139 , e ) 149 | b | subtract(700, add(add(multiply(divide(45, const_100), 700), multiply(divide(23, const_100), 700)), multiply(divide(15, const_100), 700))) | in a school of 700 students , 45 % wear blue shirts , 23 % wear red shirts , 15 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ? | 45 + 23 + 15 = 83 % 100 – 83 = 17 % 700 * 17 / 100 = 119 the answer is b . | a = 45 / 100
b = a * 700
c = 23 / 100
d = c * 700
e = b + d
f = 15 / 100
g = f * 700
h = e + g
i = 700 - h
|
a ) 2 / 69 , b ) 2 / 60 , c ) 2 / 63 , d ) 2 / 29 , e ) 2 / 10 | c | multiply(divide(1, 7), divide(2, 9)) | two brother x and y appeared for an exam . the probability of selection of x is 1 / 7 and that of b is 2 / 9 . find the probability that both of them are selected . | "explanation : let a be the event that x is selected and b is the event that y is selected . p ( a ) = 1 / 7 , p ( b ) = 2 / 9 . let c be the event that both are selected . p ( c ) = p ( a ) × p ( b ) as a and b are independent events : = ( 1 / 7 ) × ( 2 / 9 ) = 2 / 63 answer : c ) 2 / 63" | a = 1 / 7
b = 2 / 9
c = a * b
|
a ) 10 , b ) 15 , c ) 14 , d ) 16 , e ) 17 | b | divide(150, add(subtract(11, 2), const_1)) | 150 metres long yard , 11 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "11 trees have 10 gaps between them , required distance ( 150 / 10 ) = 15 b" | a = 11 - 2
b = a + 1
c = 150 / b
|
a ) 150 , b ) 120 , c ) 154 , d ) 160 , e ) 210 | c | divide(231, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100)))) | a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 231 for it , the cost price of the cricket bat for a is : | 125 % of 120 % of a = 231 125 / 100 * 120 / 100 * a = 231 a = 231 * 2 / 3 = 154 . answer c | a = 20 / 100
b = 1 + a
c = 25 / 100
d = 1 + c
e = b * d
f = 231 / e
|
a ) 2 : 3 , b ) 5 : 6 , c ) 4 : 5 , d ) 3 : 1 , e ) 8 : 1 | d | subtract(8, 4) | a boat running up stram takes 4 hours to cover a certain distance , while it takes 8 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 4 ( x + y ) = 8 ( x - y ) 4 x + 4 y = 8 x - 8 y 12 y = 4 x 3 y = x x / y = 3 / 1 3 : 1 answer : option d" | a = 8 - 4
|
a ) 188 , b ) 258 , c ) 376 , d ) 470 , e ) 517 | c | multiply(divide(add(multiply(const_1000, const_1), add(multiply(const_10, const_3), 4)), add(add(5, 2), 4)), 4) | a farmer used 1,034 acres of land for beans , wheat , and corn in the ratio of 5 : 2 : 4 , respectively . how many w acres were used for corn ? | "consider 5 x acres of land used for bean consider 2 x acres of land used for wheat consider 4 x acres of land used for corn total given is 1034 acres 11 x = 1034 x = 94 land used for corn w = 4 * 94 = 376 correct option - c" | a = 1000 * 1
b = 10 * 3
c = b + 4
d = a + c
e = 5 + 2
f = e + 4
g = d / f
h = g * 4
|
a ) 1240 , b ) 1600 , c ) 1320 , d ) 1150 , e ) 1100 | e | multiply(multiply(power(const_3, const_3), multiply(power(const_2, const_3), power(add(const_4, const_1), const_2))), divide(divide(divide(divide(divide(44, const_2), const_2), const_3), add(const_4, const_1)), add(const_4, const_1))) | find the l . c . m of 22 , 25 , 44 and 20 . | "explanation : 2 x 2 x 11 x 5 x 5 = 1100 answer : option e" | a = 3 ** 3
b = 2 ** 3
c = 4 + 1
d = c ** 2
e = b * d
f = a * e
g = 44 / 2
h = g / 2
i = h / 3
j = 4 + 1
k = i / j
l = 4 + 1
m = k / l
n = f * m
|
a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 13 | c | subtract(divide(add(add(19, 17), 15), 3), subtract(3, const_1)) | 3 models ( f , g , and h ) of cars are distributed among 3 showrooms . the number of cars in each showrooms must be equal and each model must be represented by at least one car in every showroom . there are 19 cars of model f , 17 cars of model g , and 15 cars of model h . what is the maximum number of cars of model f in any showroom ? | the total number of cars is 51 . so each showroom has 17 cars ( since the number of cars in each showrooms should be equal 51 / 3 = 17 ) . moreover that the number of model f is maximum means that the numbers of model g and h should be minimum . since each model must be represented by at least one car in every showroom that minimum number should be 1 . so maximum number of model f is 17 - 2 = 15 . the answer is ( c ) | a = 19 + 17
b = a + 15
c = b / 3
d = 3 - 1
e = c - d
|
a ) 56 minutes , b ) 45 minutes , c ) 30 minutes , d ) 48 minutes , e ) 44 minutes | a | divide(multiply(5.00, const_60), 5) | the timing of a college is from 12 p . m to 5.00 p . m . five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students . find the duration of each lecture . | "explanation : total time a student spends in college = 5 hours 00 minutes = 300 minutes as there are 5 lectures , the number of breaks between lectures is 4 . total time of the break = 20 minutes hence , the duration of each lecture is = ( 300 â € “ 20 ) / 5 = 56 minutes answer a" | a = 5 * 0
b = a / 5
|
a ) 0 and 3 , b ) 3 and 6 , c ) 6 and 9 , d ) 9 and 12 , e ) 12 and 15 | b | add(multiply(const_10, const_3), add(4, const_2)) | if q = x ^ 4 + y ^ 4 = 100 , then the greatest possible value of x is between | my attempt : if q = x ^ 4 + y ^ 4 = 100 , then the greatest possible value of x would be when y is minimum . let y ^ 4 be 0 . now x ^ 4 = 100 . x should be definitely greater than 3 but less than 4 . the only option that fits this range is b hence answer is - - b ) 3 and 6 . | a = 10 * 3
b = 4 + 2
c = a + b
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | b | add(add(const_4, const_3), const_2) | what is the units digit of 29 ! + 50 ! + 3 ! + 3 ! ? | "for all n greater than 4 , the units digit of n ! is 0 . the sum of the four units digits is 0 + 0 + 6 + 6 = 12 the units digit is 2 . the answer is b ." | a = 4 + 3
b = a + 2
|
a ) 19 sec , b ) 15 sec , c ) 16 sec , d ) 20 sec , e ) none | a | multiply(divide(285, multiply(54, const_1000)), const_3600) | a train 285 m long , running with a speed of 54 km / hr will pass a tree in | "sol . speed = ( 54 x 5 / 18 ) m / sec . = 15 m / sec . time taken = ( 285 x 1 / 15 ) sec = 19 sec answer a" | a = 54 * 1000
b = 285 / a
c = b * 3600
|
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