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 ) - 1 , b ) 77 , c ) 8 , d ) 3 , e ) 2 | a | divide(add(17, 21), add(19, 19)) | solve the equation for x : 19 ( x + y ) + 17 = 19 ( - x + y ) - 21 ? | "19 x + 19 y + 17 = - 19 x + 19 y - 21 38 x = - 38 = > x = - 1 answer : a" | a = 17 + 21
b = 19 + 19
c = a / b
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a ) 3377 , b ) 2688 , c ) 2688 , d ) 9125 , e ) 1268 | d | multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100))) | if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 5 years the amount to be paid is ? | "a = 7500 ( 26 / 25 ) ^ 5 = 9125 answer : d" | a = 100 + 4
b = a / 100
c = 100 + 4
d = c / 100
e = b * d
f = 7500 * e
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a ) 22 , b ) 36 , c ) 55 , d ) 77 , e ) 88 | b | subtract(divide(204, add(divide(10, const_100), divide(3, 4))), 204) | a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year 10 % of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals 204 , while 3 / 4 of all his trees are pure fuji . how many of his trees are pure gala ? | "let f = pure fuji , g = pure gala and c - cross pollinated . c = 10 % of x where x is total trees . c = . 1 x also 3 x / 4 = f and c + f = 204 = > . 1 x + 3 / 4 x = 204 = > x = 240 240 - 204 = pure gala = 36 . answer b" | a = 10 / 100
b = 3 / 4
c = a + b
d = 204 / c
e = d - 204
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a ) 54210 , b ) 54250 , c ) 52150 , d ) none of these , e ) 521500 | c | divide(add(32323, 7344), 41330) | ( 32323 + 7344 + 41330 ) - ( 317 x 91 ) = ? | "its speed and accuracy which decides the winner . you can use various methods including vedic mathematics to do complex calculations quickly . ( 32323 + 7344 + 41330 ) - ( 317 x 91 ) = 80997 - 28847 = 52150 answer is c" | a = 32323 + 7344
b = a / 41330
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a ) rs . 390 , b ) rs . 368 , c ) rs . 380 , d ) rs . 360 , e ) rs . 311 | d | multiply(divide(1560, add(add(2, 3), 4)), 4) | an amount of rs . 1560 was divided among a , b and c , in the ratio 1 / 2 : 1 / 3 : 1 / 4 . find the share of c ? | "let the shares of a , b and c be a , b and c respectively . a : b : c = 1 / 2 : 1 / 3 : 1 / 4 let us express each term with a common denominator which is the last number divisible by the denominators of each term i . e . , 12 . a : b : c = 6 / 12 : 4 / 12 : 3 / 12 = 6 : 4 : 3 . share of c = 3 / 13 * 1560 = rs . 360 . answer : d" | a = 2 + 3
b = a + 4
c = 1560 / b
d = c * 4
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a ) 28 , b ) 77 , c ) 36 , d ) 25 , e ) 13 | a | divide(multiply(18, 42), 27) | 42 men can complete a piece of work in 18 days . in how many days will 27 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 27 : 42 : : 18 : x [ please pay attention , we have written 27 : 42 rather than 42 : 27 , in indirect proportion , if you get it then chain rule is clear to you : ) ] { \ color { blue } x = \ frac { 42 \ times 18 } { 27 } } x = 28 so 28 days will be required to get work done by 27 men . answer : a" | a = 18 * 42
b = a / 27
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a ) none , b ) one , c ) two , d ) three , e ) four | a | add(add(1, 1), 1) | how many three - digit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6 ? | "the possible number n can be written as follow : n = multiple of lcm ( 6,10 ) + 1 st such number n = 30 x + 1 possible values = 1 , 31 , 61 , 91 , 121,151 , 181,211 , 241,271 . . . . . 991 answer : none : a" | a = 1 + 1
b = a + 1
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a ) 576 cm 2 , b ) 384 cm 2 , c ) 672 cm 2 , d ) 267 cm 2 , e ) 286 cm 2 | a | multiply(32, 18) | find the area of a parallelogram with base 32 cm and height 18 cm ? | "area of a parallelogram = base * height = 32 * 18 = 576 cm 2 answer : a" | a = 32 * 18
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a ) 8 , b ) 12 , c ) 21 , d ) 27 , e ) 39 | b | multiply(floor(power(divide(91, const_2), divide(const_1, const_3))), power(subtract(91, power(floor(power(divide(91, const_2), divide(const_1, const_3))), const_3)), divide(const_1, const_3))) | the number 91 is equal to the sum of the cubes of two integers . what is the product of those integers ? | "3 ^ 3 + 4 ^ 3 = 91 therefore number is 3 * 4 = 12 b" | a = 91 / 2
b = 1 / 3
c = a ** b
d = math.floor(c)
e = 91 / 2
f = 1 / 3
g = e ** f
h = math.floor(g)
i = h ** 3
j = 91 - i
k = 1 / 3
l = j ** k
m = d * l
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a ) 54 , b ) 48 , c ) 65 , d ) 25 , e ) 64 | e | multiply(divide(subtract(divide(5, 3), divide(3, 5)), divide(5, 3)), const_100) | a student multiplied a number by 3 / 5 instead of 5 / 3 . what is the percentage error in the calculation ? | let the number be x . then , error = 5 / 3 x - 3 / 5 x = 16 / 15 x . error % = ( 16 x / 15 * 3 / 5 x * 100 ) % = 64 % answer e ) 64 % | a = 5 / 3
b = 3 / 5
c = a - b
d = 5 / 3
e = c / d
f = e * 100
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a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 12 | a | multiply(divide(subtract(250, divide(360, const_2)), 70), const_2) | angelina walked 250 meters from her home to the grocery at a constant speed . she then walked 360 meters to the gym at double the speed . she spent 70 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed , in meters per second , from the grocery to the gym ? | "let the speed be x . . . so time taken from home to grocery = 250 / x . . the speed to gym = 2 x . . so time taken = 360 / 2 x = 180 / x . . its given 250 / x - 180 / x = 70 . . 70 / x = 70 . . x = 1 m / secs . . so grocery to gym = 2 * 1 = 2 m / s . . . answer : a" | a = 360 / 2
b = 250 - a
c = b / 70
d = c * 2
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a ) 2 ^ 9 , b ) 2 ^ 10 , c ) 2 ^ 12 , d ) 2 ^ 16 , e ) 2 ^ 20 | a | add(add(add(power(2, 6), add(power(2, 5), add(power(2, 4), add(power(2, 3), add(power(const_2, const_2), add(2, 2)))))), power(2, 7)), power(2, 8)) | 2 + 2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + 2 ^ 5 + 2 ^ 6 + 2 ^ 7 + 2 ^ 8 = | let ' s see that : 2 + 2 is 2 ^ 2 , then : 2 ^ 2 + 2 ^ 2 becomes 2 x 2 ^ 2 that is 2 ^ 3 . you follow with the next : 2 ^ 3 + 2 ^ 3 is equal to 2 x 2 ^ 3 that is 2 ^ 4 . then is assumed right before 2 ^ 8 : 2 ^ 8 x 2 ^ 8 becomes 2 x 2 ^ 8 that is 2 ^ 9 . option a , 2 ^ 9 . | a = 2 ** 6
b = 2 ** 5
c = 2 ** 4
d = 2 ** 3
e = 2 ** 2
f = 2 + 2
g = e + f
h = d + g
i = c + h
j = b + i
k = a + j
l = 2 ** 7
m = k + l
n = 2 ** 8
o = m + n
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a ) $ 230.00 , b ) $ 224.00 , c ) $ 263.80 , d ) $ 282.00 , e ) $ 210.00 | b | multiply(400, divide(add(30, 20), const_100)) | a discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price . if , during a special sale , an additional 20 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 400 before any discount ? | "since the question is essentially just about multiplication , you can do the various mathstepsin a variety of ways ( depending on whichever method you find easiest ) . we ' re told that the first discount is 10 % to 30 % , inclusive . we ' re told that the next discount is 20 % off of the discounted price . . . . we ' re told to maximize the discount ( thus , 30 % off the original price and then 20 % off of the discounted price ) . thatmathcan be written in a number of different ways ( fractions , decimals , etc . ) : 30 % off = ( 1 - . 3 ) = ( 1 - 30 / 100 ) = ( . 7 ) and the same can be done with the 20 % additional discount . . . the final price of an item that originally cost $ 400 would be . . . . . ( $ 400 ) ( . 7 ) ( . 8 ) = ( $ 400 ) ( . 56 ) = 224 final answer : b" | a = 30 + 20
b = a / 100
c = 400 * b
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a ) 32 , b ) 34 , c ) 35 , d ) 36 , e ) 37 | a | divide(multiply(add(multiply(0.8, const_4), 0.8), 16), const_2) | logx 16 = 0.8 then what is the value of x ? | x ^ 0.8 = 16 x = ( 2 ^ 4 ) ^ ( 10 / 8 ) x = 2 ^ 5 x = 32 answer : a | a = 0 * 8
b = a + 0
c = b * 16
d = c / 2
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a ) 165 , b ) 170 , c ) 175 , d ) 180 , e ) 200 | e | multiply(divide(const_100, 8), 16) | a 16 % stock yielding 8 % is quoted at ? | "assume that face value = rs . 100 as it is not given to earn rs . 8 , money invested = rs . 100 to earn rs . 16 , money invested = 100 × 16 / 8 = rs . 175 ie , market value of the stock = rs . 200 answer is e ." | a = 100 / 8
b = a * 16
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a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 3 / 5 , e ) 2 / 7 | c | subtract(const_1, divide(1, 3)) | at an elementary school , 70 % of the faculty members are women and 40 % of the faculty members are married . if 1 ⁄ 3 of the men are single , what fraction of the men are married ? | "- - - - - - - - - - - - - - - - - - - - m - - - - - - w - - - - - - - - total marrried - - - - - - - - - - 20 - - - - - 20 - - - - - - - - - 40 not married - - - - - 10 - - - - - 50 - - - - - - - - - 60 total - - - - - - - - - - - - - 30 - - - - - 70 - - - - - - - - 100 need married man / total man , so 20 / 30 = 2 / 3 c" | a = 1 / 3
b = 1 - a
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a ) 1200 , b ) 1992 , c ) 9921 , d ) 2798 , e ) 2789 | a | multiply(multiply(divide(12, multiply(10, 2)), const_100), multiply(10, 2)) | find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 12 of money ? | "p = 12 ( 100 / 10 ) 2 = > p = 1200 answer : a" | a = 10 * 2
b = 12 / a
c = b * 100
d = 10 * 2
e = c * d
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a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | c | divide(subtract(multiply(floor(12.4), 3), 26), subtract(12.4, floor(12.4))) | a man whose bowling average is 12.4 , takes 3 wickets for 26 runs and there by decreases his average by 0.4 . the number of wickets taken by him before his last match is ? | "12.4 * x + 26 = ( 3 + x ) 12 solve equation x = 25 answer : c" | a = math.floor(12, 4)
b = a * 3
c = b - 26
d = math.floor(12, 4)
e = 12 - 4
f = c / e
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a ) 8 % , b ) 11 % , c ) 14 % , d ) 17 % , e ) 20 % | b | multiply(divide(add(multiply(40, divide(5, const_100)), 3.5), add(40, add(3.5, 6.5))), const_100) | a 40 - liter solution of alcohol and water is 5 percent alcohol . if 3.5 liters of alcohol and 6.5 liters of water are added to this solution , what percent of the solution produced is alcohol ? | "the percent of alcohol in the solution is ( 0.05 ( 40 ) + 3.5 ) / 50 = 5.5 / 50 = 11 % the answer is b ." | a = 5 / 100
b = 40 * a
c = b + 3
d = 3 + 5
e = 40 + d
f = c / e
g = f * 100
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a ) sec , b ) sec , c ) sec , d ) sec , e ) sec | d | divide(add(110, 170), multiply(90, const_0_2778)) | how long does a train 110 m long traveling at 90 kmph takes to cross a bridge of 170 m in length ? | "d = 110 + 170 = 280 m s = 90 * 5 / 18 = 50 / 2 t = 280 * 2 / 50 = 11.2 sec answer : d" | a = 110 + 170
b = 90 * const_0_2778
c = a / b
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a ) 127 , b ) 237 , c ) 300 , d ) 450 , e ) 281 | c | divide(multiply(36, 30), divide(multiply(12, 24), 80)) | if 12 men can reap 80 hectares in 24 days , then how many hectares can 36 men reap in 30 days ? | "explanation : let the required no of hectares be x . then men - - - hectares - - - days 12 - - - - - - - - - 80 - - - - - - - - - 24 36 - - - - - - - - - x - - - - - - - - - 30 more men , more hectares ( direct proportion ) more days , more hectares ( direct proportion ) x = 36 / 12 * 30 / 24 * 80 x = 300 answer : c" | a = 36 * 30
b = 12 * 24
c = b / 80
d = a / c
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a ) 36 , b ) 40 , c ) 42 , d ) 54 , e ) 65 | d | divide(add(add(60, 60), 45), const_3) | a martian bat flies at 60 yards per second from its nest to a dry lake . when it arrives there , it immediately continues at 45 yards per second to a nearby cave . if the distance between the lake and the cave is half the distance between the nest and the lake , what is the average speed , in yards per second , of the bat during the whole flight ? | distance from nest to lake = d , speed = 60 , time = d / 60 distance from lake to cave = d / 2 , speed = 45 , time = d / 90 average speed = ( d + d / 2 ) / ( d / 60 + d / 90 ) = ( 3 / 2 ) / ( 1 / 36 ) = 54 answer : d | a = 60 + 60
b = a + 45
c = b / 3
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a ) 0 , b ) - 2 , c ) - 25 , d ) - 49 , e ) - 37 | e | add(add(negate(20), const_1), add(add(negate(20), const_1), const_1)) | the sum of all the integers k such that – 20 < k < 24 is | "- 19 - - - - - - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - - - 23 values upto + 23 cancels outwe are left with only - 19 - 18 sum of which is - 37 . hence option d . e" | a = negate + (
b = a + 1
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a ) 1.25 % , b ) 3.75 % , c ) 6.25 % , d ) 6.67 % , e ) 11.7 % | c | multiply(divide(multiply(multiply(const_100, const_100), divide(5, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume . if 2,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "the amount of sodium chloride is 0.05 * 10,000 = 500 gallons 500 / 8000 = 1 / 16 = 6.25 % the answer is c ." | a = 100 * 100
b = 5 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
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a ) 56 hours , b ) 28 hours , c ) 35 hours , d ) 66 hours , e ) 47 hours | c | divide(10, divide(const_2, add(3, const_4))) | a tank is filled in 10 hours by 3 pipes a , b and c . pipe a is twice as fast as pipe b , and b is twice as fast as c . how much time will pipe b alone take to fill the tank ? | 1 / a + 1 / b + 1 / c = 1 / 10 ( given ) also given that a = 2 b and b = 2 c = > 1 / 2 b + 1 / b + 2 / b = 1 / 10 = > ( 1 + 2 + 4 ) / 2 b = 1 / 10 = > 2 b / 7 = 10 = > b = 35 hours . answer : c | a = 3 + 4
b = 2 / a
c = 10 / b
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a ) 5100 , b ) 7550 , c ) 10100 , d ) 15500 , e ) 20100 | b | multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1)) | the sum of the first 50 positive even integers is 2550 . what is the sum e of even integers from 102 to 200 inclusive ? | "my solution is : first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : e = 2550 + ( 100 * 50 ) = 7550 . b" | a = 200 + 102
b = a / 2
c = 200 - 102
d = c / 2
e = d + 1
f = b * e
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a ) 270 , b ) 289 , c ) 288 , d ) 255 , e ) 552 | c | multiply(divide(928, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 928 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 928 = 288 answer : c" | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 928 / e
g = 16 * 9
h = f * g
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a ) a ) 47.3 , b ) b ) 45.6 , c ) c ) 44 , d ) d ) 48 , e ) e ) 49 | a | divide(add(multiply(45, subtract(28, add(3, 4))), multiply(95, 4)), 28) | in a class of 28 students in an examination in maths 4 students scored 95 marks each , 3 get zero each and the average of the rest was 45 . what is the average of the whole class ? | "explanation : total marks obtained by a class of 25 students = 4 * 95 + 3 * 0 + 21 * 45 = 1325 average marks of whole class = 1325 / 28 = 47.3 answer : option a" | a = 3 + 4
b = 28 - a
c = 45 * b
d = 95 * 4
e = c + d
f = e / 28
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a ) 27 , b ) 36 , c ) 29 , d ) 10 , e ) 11 | d | subtract(multiply(divide(subtract(90, 65), subtract(const_12, 9)), const_12), 90) | gopi gives rs . 90 plus one turban as salary to his servant for one year . the servant leaves after 9 months and receives rs . 65 and the turban . find the price of the turban . | "let the price of turban be x . thus , for one year the salary = ( 90 + x ) for 9 months he should earn 3434 ( 90 + x ) . now he gets one turban and rs . 65 . thus , 3434 ( 90 + x ) = 65 + x or 270 + 3 x = 260 + 4 x or x = 10 answer : d" | a = 90 - 65
b = 12 - 9
c = a / b
d = c * 12
e = d - 90
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a ) 65 , b ) 120 , c ) 80 , d ) 85 , e ) 90 | b | add(multiply(power(2, multiply(divide(60, 30), subtract(const_1, 2))), 120), 60) | the temperature of a certain cup of coffee 30 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 * 2 ^ ( - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 30 minutes after it was poured was how many degrees fahrenheit ? | "first , we have to find a . we know that after t = 30 minutes the temperature f = 120 degrees . hence : 120 = 120 * ( 2 ^ - 30 a ) + 60 60 = 120 * ( 2 ^ - 30 a ) 60 / 120 = 2 ^ - 30 a 1 / 2 = 2 ^ - 30 a 2 ^ - 1 = 2 ^ - 30 a - 1 = - 30 a 1 / 30 = a now we need to find f after t = 30 minutes : f = 120 * ( 2 ^ - 1 / 30 * 30 ) + 60 f = 120 * ( 2 ^ - 1 ) + 60 f = 120 * ( 1 / 2 ^ 1 ) + 60 f = 120 * 1 / 2 + 60 f = 60 + 60 = 120 answer b !" | a = 60 / 30
b = 1 - 2
c = a * b
d = 2 ** c
e = d * 120
f = e + 60
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a ) 80 , 82,84 , b ) 65 , 68,71 , c ) 45 , 47,42 , d ) 100 , 102,49 , e ) 87 , 88,89 | a | add(add(power(add(add(divide(subtract(subtract(246, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(246, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(246, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(246, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive even numbers is 246 . what are the numbers ? | "first x make the first x second x + 2 even numbers , sowe add 2 to get the next third x + 4 add 2 more ( 4 total ) to get the third f + s + t = 246 summeansaddfirst ( f ) plussecond ( s ) plusthird ( t ) ( x ) + ( x + 2 ) + ( x + 4 ) = 246 replace each f , s , and t withwhatwe labeled them x + x + 2 + x + 4 = 246 here the parenthesis are not needed 3 x + 6 = 246 combine like terms x + x + x and 2 + 4 − 6 − 6 subtract 6 fromboth sides 3 x = 240 the variable ismultiplied by 3 3 3 divide both sides by 3 x = 80 our solution for x first 80 replace x in the origional listwith 80 . second ( 80 ) + 2 = 82 the numbers are 80 , 82 , and 84 . third ( 80 ) + 4 = 84 correct answer a" | a = 246 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 246 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 246 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 246 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
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a ) 90 min , b ) 100 min , c ) 110 min , d ) 120 min , e ) 130 min | d | inverse(subtract(add(inverse(60), inverse(120)), inverse(60))) | two pipes a and b can separately fill a cistern in 60 minutes and 120 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened , then the cistern is full in 60 minutes . in how much time , the third pipe alone can empty the cistern ? | "1 / 60 - ( 1 / 60 + 1 / 120 ) = - 1 / 120 third pipe can empty in 120 minutes answer : d" | a = 1/(60)
b = 1/(120)
c = a + b
d = 1/(60)
e = c - d
f = 1/(e)
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a ) 15 , b ) 20 , c ) 23 , d ) 28 , e ) 25 | d | add(multiply(10, const_2), multiply(subtract(23.5, multiply(10, const_2)), 10)) | if the average of 10 consecutive integers is 23.5 then the 10 th integer is : - | "the average falls between the 5 th and 6 th integers , integer 5 = 23 , integer 6 = 24 . counting up to the tenth integer we get 28 . answer : d" | a = 10 * 2
b = 10 * 2
c = 23 - 5
d = c * 10
e = a + d
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a ) 2.9 hrs , b ) 8.9 hrs , c ) 2.9 hrs , d ) 7.2 hrs , e ) 8.6 hrs | d | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 9))) | a cistern can be filled by a tap in 4 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 / 4 - 1 / 9 ) = 5 / 36 the cistern will be filled in 36 / 5 hrs i . e . , 7.2 hrs . answer : d" | a = 1 / 4
b = 1 / 9
c = a - b
d = 1 / c
|
a ) 3 . , b ) 7 . , c ) 10 . , d ) 12 . , e ) 16 . | e | multiply(5, 5) | the distance between west - town to east - town is 16 kilometers . two birds start flying simultaneously towards one another , the first leaving from west - town at a speed of 5 kilometers per minute and the second bird , leaving from east - town , at a speed of 1 kilometers per minute . what will be the distance , in kilometers , between the meeting point and west - town ? | "time taken by the birds to meet = 16 / ( 5 + 1 ) = 4 mins distance traveled by the bird traveling from west - town = 4 * 4 = 16 answer : e" | a = 5 * 5
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a ) 33 , b ) 34 , c ) 35 , d ) 36 , e ) 37 | e | multiply(add(const_10, 2), const_3) | which 2 digit number has its digits reversed when you double it and take 1 ? | reverse it and you get 73 which is 2 * 37 - 1 . answer : e | a = 10 + 2
b = a * 3
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a ) 6.54 , b ) 9.45 , c ) 7.34 , d ) 6.67 , e ) 5.09 | a | multiply(divide(const_1, multiply(add(const_100, 10), divide(const_1, subtract(const_100, 10)))), 8) | by selling 8 pencils for a rupee a man loses 10 % . how many for a rupee should he sell in order to gain 10 % ? | "90 % - - - 8 110 % - - - ? 90 / 110 * 8 = 6.54 answer : a" | a = 100 + 10
b = 100 - 10
c = 1 / b
d = a * c
e = 1 / d
f = e * 8
|
a ) 15 , b ) 11 , c ) 28 , d ) 91 , e ) none of these | b | floor(sqrt(132)) | if the sum of a number and its square is 132 , what is the number ? | let the number be x . then , x + x 2 = 132 ( x + 12 ) ( x - 11 ) = 0 x = 11 answer : b | a = math.sqrt(132)
b = math.floor(a)
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a ) 280 , b ) 320 , c ) 360 , d ) 420 , e ) 480 | e | divide(180, subtract(1, add(divide(1, 2), multiply(divide(1, 2), divide(1, 4))))) | during a car trip , maria stopped to rest after she traveled 1 / 2 of the total distance to her destination . she stopped again after she traveled 1 / 4 of the distance remaining between her first stop and her destination , and then she drove the remaining 180 miles to her detination . what was the total distance , in miles from maria ' s starting point to her destination ? | "easy 480 is the answer . 3 / 4 ( x / 2 ) = 180 x = 180 * 8 / 3 = 480 . answer e" | a = 1 / 2
b = 1 / 2
c = 1 / 4
d = b * c
e = a + d
f = 1 - e
g = 180 / f
|
a ) $ 136.50 , b ) $ 145.60 , c ) $ 163.80 , d ) $ 182.00 , e ) $ 210.00 | a | multiply(260, divide(add(30, 25), const_100)) | a discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price . if , during a special sale , an additional 25 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 260 before any discount ? | since the question is essentially just about multiplication , you can do the various mathstepsin a variety of ways ( depending on whichever method you find easiest ) . we ' re told that the first discount is 10 % to 30 % , inclusive . we ' re told that the next discount is 25 % off of the discounted price . . . . we ' re told to maximize the discount ( thus , 30 % off the original price and then 20 % off of the discounted price ) . thatmathcan be written in a number of different ways ( fractions , decimals , etc . ) : 30 % off = ( 1 - . 3 ) = ( 1 - 30 / 100 ) = ( . 7 ) and the same can be done with the 20 % additional discount . . . the final price of an item that originally cost $ 260 would be . . . . . ( $ 260 ) ( . 7 ) ( . 75 ) = ( $ 260 ) ( . 525 ) = 136.5 final answer : a | a = 30 + 25
b = a / 100
c = 260 * b
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a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 25 | c | divide(multiply(2200, 1.12), 2464) | if 2464 / 1.12 = 2200 , then 246.4 / 11.2 is equal to ? | "answer given expression 246.4 / 11.2 = 2464 / 112 = 2464 / ( 1.12 x 100 ) = 2200 / 100 = 22 correct option : c" | a = 2200 * 1
b = a / 2464
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a ) 3 days , b ) 2 days , c ) 5 days , d ) 6 days , e ) 1 day | b | divide(50, multiply(divide(divide(60, 3), 4), 5)) | if 4 men can paint 60 m long wall in 3 days , then 5 men can paint 50 m long wall in | the length of wall painted by one man in one day = 60 / 4 ã — 3 = 5 m no . of days required to paint 50 m wall by 5 men = 50 / 5 ã — 5 = 2 day . b | a = 60 / 3
b = a / 4
c = b * 5
d = 50 / c
|
a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 22.22 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 10), subtract(const_100, 10)), const_100), subtract(const_100, 10)) | a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 10 % while buying and by 10 % while selling . what is his percentage profit ? | "the owner buys 100 kg but actually gets 110 kg ; the owner sells 100 kg but actually gives 90 kg ; profit : ( 110 - 90 ) / 90 * 100 = ~ 22.22 % answer : d ." | a = 100 + 10
b = 100 - 10
c = a - b
d = c * 100
e = 100 - 10
f = d / e
|
a ) 46.2 % , b ) 54.54 % , c ) 60.52 % , d ) 70 % , e ) none of these | c | multiply(divide(subtract(152, add(multiply(12, const_4), multiply(2, multiply(2, const_3)))), 152), const_100) | a cricketer scored 152 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 152 - ( 12 x 4 + 2 x 6 ) = 152 - ( 60 ) = 92 now , we need to calculate 92 is what percent of 152 . = > 92 / 152 * 100 = 60.52 % answer : c" | a = 12 * 4
b = 2 * 3
c = 2 * b
d = a + c
e = 152 - d
f = e / 152
g = f * 100
|
a ) 4000 , b ) 3000 , c ) 5500 , d ) 4500 , e ) none of these | a | subtract(multiply(5, divide(20000, add(add(2, 3), 5))), multiply(3, divide(20000, add(add(2, 3), 5)))) | a profit of rs . 20000 is to be distributed among a , b , c in the proportion 2 : 3 : 5 . what will be the difference between b ’ s and c ’ s shares | explanation : b ' s share = rs 20000 * ( 3 / 10 ) = rs 6000 c ' s share = rs 20000 * ( 5 / 10 ) = rs 10000 difference in b ' s and c ' s shares = rs 4000 answer : a | a = 2 + 3
b = a + 5
c = 20000 / b
d = 5 * c
e = 2 + 3
f = e + 5
g = 20000 / f
h = 3 * g
i = d - h
|
a ) 2 / 5,25 , b ) 4,5 / 2 , c ) 1,16 , d ) 2,5 , e ) 1,5 | d | divide(divide(divide(250, const_2), multiply(const_0_25, const_100)), const_2) | find the first term and the common ratio of a g . p whose fourth term is 250 and seventh term is 31,250 | a * r ^ 3 = 250 , a * r ^ 6 = 31,250 , r ^ 3 = 31,250 / 250 = 125 r = 5 , a * 125 = 250 , a = 2 answer : d | a = 250 / 2
b = const_0_25 * 100
c = a / b
d = c / 2
|
a ) 25 % , b ) 20 % , c ) 15 % , d ) 10 % , e ) 5 % | a | multiply(divide(subtract(7500, 6000), 6000), const_100) | a sum of money deposited at c . i . amounts to rs . 6000 in 6 years and to rs . 7500 in 7 years . find the rate percent ? | "6000 - - - 1500 100 - - - ? = > 25 % answer : a" | a = 7500 - 6000
b = a / 6000
c = b * 100
|
a ) 90 % , b ) 65 % , c ) 50 % , d ) 85 % , e ) 80 % | a | multiply(divide(10, add(add(5, 4), 2)), const_100) | a dog breeder would like to make the largest profit off of his latest litter of puppies . puppies with more than 4 spots sell for more money . the last litter had 10 puppies ; 6 had 5 spots , 3 had 4 spots , and 1 had 2 spots . what % of puppies can the dog breeder expect to sell for a greater profit ? | number of dogs with profitable spots = ( 6 + 3 ) = 6 + 3 = 9 now , we need to calculate 9 is what percent of 10 . = > 9 / 10 * 100 = 90 % a | a = 5 + 4
b = a + 2
c = 10 / b
d = c * 100
|
a ) 25 , b ) 35 , c ) 50 , d ) 20 , e ) 15 | c | multiply(divide(subtract(subtract(add(const_1, divide(20, const_100)), multiply(add(const_1, divide(10, const_100)), divide(75, const_100))), subtract(const_1, divide(75, const_100))), subtract(const_1, divide(75, const_100))), const_100) | paulson spends 75 % of his income . his income is increased by 20 % and he increased his expenditure by 10 % . find the percentage increase in his savings . | let the original income = $ 100 . then , expenditure = $ 75 and savings = $ 25 new income = $ 120 , new expenditure = $ ( ( 110 / 100 ) * 75 ) = rs . 165 / 2 new savings = $ ( 120 - ( 165 / 2 ) ) = $ 75 / 2 increase in savings = $ ( ( 75 / 2 ) - 25 ) = $ 25 / 2 increase % = ( ( 25 / 2 ) * ( 1 / 25 ) * 100 ) % = 50 % . answer c ) 50 . | a = 20 / 100
b = 1 + a
c = 10 / 100
d = 1 + c
e = 75 / 100
f = d * e
g = b - f
h = 75 / 100
i = 1 - h
j = g - i
k = 75 / 100
l = 1 - k
m = j / l
n = m * 100
|
a ) 4.5 hr , b ) 4.8 hr , c ) 4.9 hr , d ) 5.5 hr , e ) 6.5 hr | a | add(divide(7, const_2), divide(2, const_2)) | on the average mr . jackson has 2 hours to plan for an long hour meeting he talk to his assistant 7 hours to prepare the paper work needed . how much hours should jackson and assistant has t plan for 30 min meeting ? | for a 1 hour talk he took total of 9 hr time 1 hr = 9 hr ( 2 + 7 ) 30 min = 1 / 2 hr = ? so 9 * 1 / 2 div by 1 4.5 hr answer : a | a = 7 / 2
b = 2 / 2
c = a + b
|
a ) a ) 10 , b ) b ) 12 , c ) c ) 24 , d ) d ) 50 , e ) e ) 100 | d | multiply(factorial(4), factorial(2)) | there are 4 red chips and 2 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ? | using anagram method : 6 _ 5 _ 4 _ 3 _ 2 _ 1 r _ r _ r _ r _ b _ b so . . 6 ! / number of repeated letters ( 4 ! ) ( 2 ! ) = 5 ans : d | a = math.factorial(4)
b = math.factorial(2)
c = a * b
|
a ) 5100 , b ) 7550 , c ) 10100 , d ) 15500 , e ) 20100 | b | multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1)) | the sum of the first 50 positive even integers is 2550 . what is the sum t of even integers from 102 to 200 inclusive ? | "my solution is : first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : t = 2550 + ( 100 * 50 ) = 7550 . b" | a = 200 + 102
b = a / 2
c = 200 - 102
d = c / 2
e = d + 1
f = b * e
|
['a ) 44', 'b ) 45', 'c ) 46', 'd ) 47', 'e ) 49'] | e | subtract(multiply(add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3)), add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3))), 4440) | what should be added to 4440 so that it may become a perfect square ? | 67 x 67 = 4489 4489 - 4440 = 49 if added to 49 get perfect square answer = e | a = 10 + 10
b = 10 + 10
c = a + b
d = 10 + 10
e = c + d
f = 4 + 3
g = e + f
h = 10 + 10
i = 10 + 10
j = h + i
k = 10 + 10
l = j + k
m = 4 + 3
n = l + m
o = g * n
p = o - 4440
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 3 | a | divide(10, subtract(50, 10)) | a chemist mixes one liter of pure water with x liters of a 50 % salt solution , and the resulting mixture is a 10 % salt solution . what is the value of x ? | concentration of salt in pure solution = 0 concentration of salt in salt solution = 50 % concentration of salt in the mixed solution = 10 % the pure solution and the salt solution is mixed in the ratio of - - > ( 50 - 10 ) / ( 10 - 0 ) = 4 / 1 1 / x = 4 / 1 x = 1 / 4 answer : a | a = 50 - 10
b = 10 / a
|
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 10 | e | multiply(3, const_2) | how many positive even integers less than 100 contain digits 3 or 9 ? | "two digit numbers : 3 at tens place : 30 , 32,34 , 36,38 9 at tens place : 90 , 92,94 , 96,98 if 3 and 7 is at units place , the number cant be even total : 5 + 5 = 10 answer e" | a = 3 * 2
|
a ) 25 , b ) 21 , c ) 20 , d ) 10 , e ) 15 | b | multiply(sqrt(9), 7) | a can contains a mixture of two liquids a and b in the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many litres of liquid a was contained by the can initially ? | suppose the can initially contains 7 x and 5 x litres mixtures a and b respectively quantity of a in mix . left = [ 7 x – 7 / 12 * 9 ] = [ 7 x – 21 / 4 ] quantity of b in mix . left = [ 5 x – 5 / 12 * 9 ] = [ 5 x - 15 / 4 ] therefore [ 7 x - 21 / 4 ] = 7 / 9 or 28 x - 21 = 7 / 9 [ 5 x - 15 / 4 ] + 9 20 x + 21 252 x - 189 = 140 x + 147 or 112 x = 336 , x = 3 . quantity of a in the can initially = 7 * 3 = 21 answer : b | a = math.sqrt(9)
b = a * 7
|
a ) $ 60,000 , b ) $ 65,000 , c ) $ 70,000 , d ) $ 75,000 , e ) $ 80,000 | c | subtract(divide(multiply(divide(subtract(subtract(multiply(add(add(multiply(multiply(const_3, const_3), add(const_4, const_1)), const_2), divide(const_1, const_2)), 12), multiply(multiply(12, 3), add(const_4, const_1))), multiply(multiply(multiply(const_3, const_3), add(const_4, const_1)), 4)), 3), const_1000), const_1000), 5) | in a company of 12 employees , 5 employees earn $ 36,000 , 4 employees earn $ 45,000 , and the 3 highest - paid employees earn the same amount . if the average annual salary for the 12 employees is $ 47,500 , what is the annual salary for each of the highest - paid employees ? | 5 * 36,000 + 4 * 45,000 + 3 x = 12 * 47,500 3 x = 570,000 - 180,000 - 180,000 3 x = 210,000 x = 70,000 the answer is c . | a = 3 * 3
b = 4 + 1
c = a * b
d = c + 2
e = 1 / 2
f = d + e
g = f * 12
h = 12 * 3
i = 4 + 1
j = h * i
k = g - j
l = 3 * 3
m = 4 + 1
n = l * m
o = n * 4
p = k - o
q = p / 3
r = q * 1000
s = r / 1000
t = s - 5
|
a ) 30.6 % , b ) 34.4 % , c ) 38.7 % , d ) 42.9 % , e ) 46.5 % | d | multiply(divide(75, add(75, const_100)), const_100) | if y is 75 % greater than x , than x is what % less than y ? | "y = 1.75 x x = y / 1.75 = 100 y / 175 = 4 y / 7 x is 3 / 7 less which is about 42.9 % less than y . the answer is d ." | a = 75 + 100
b = 75 / a
c = b * 100
|
a ) 41.1 sec , b ) 20.2 sec , c ) 36 sec , d ) 50.4 sec , e ) none of the above | c | divide(add(250, 300), multiply(55, const_0_2778)) | a 250 m long train is running at a speed of 55 km / hr . it crossed a platform of length 300 m in ? | "speed = 55 km / hr ( to convert km / hr in to m / s ) = 55 x 5 / 18 m / s distance = 250 m + 300 m ( if questions is about train crossing a post you need to consider only the length of train , ) = 550 m time = distance / speed = 550 x 18 / ( 5 x 55 ) = 36 sec ans is : c" | a = 250 + 300
b = 55 * const_0_2778
c = a / b
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(subtract(multiply(multiply(multiply(const_4, const_100), const_100), const_12), multiply(multiply(multiply(const_2, const_100), const_100), const_12)), multiply(multiply(40, const_100), const_10)) | arjun started a business with rs . 20,000 and is joined afterwards by anoop with rs . 40 , 000 . after how many months did anoop join if the profits at the end of the year are divided equally ? | suppose anoop joined after 3 months . then , 20000 * 12 = 40000 * ( 12 – x ) = > = > x = 6 . answer : d | a = 4 * 100
b = a * 100
c = b * 12
d = 2 * 100
e = d * 100
f = e * 12
g = c - f
h = 40 * 100
i = h * 10
j = g / i
|
a ) 3 , b ) 4 , c ) 12 , d ) 32 , e ) 5 | e | subtract(36, reminder(3, 7)) | when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 36 ? | "n = 5 p + 1 = 6,11 , 16,21 , 26,31 n = 7 q + 3 = 3 , 10,17 , 24,31 = > n = 36 m + 31 to get this , we need to take lcm of co - efficients of p and q and first common number in series . so we need to add 5 more to make it 36 m + 36 answer - e" | a = 36 - reminder
|
['a ) 48', 'b ) 40', 'c ) 24', 'd ) 18', 'e ) 12'] | b | multiply(5, multiply(const_2, const_4)) | a cubical block of metal weighs 5 pounds . how much will another cube of the same metal weigh if its sides are twice as long ? | for example our cube have a side 1 meter , so we have 1 cubical meter in this cube and this cubical meter weigth 5 pounds if we take cube with side 2 meters we will have 8 cubical meters in this cube 8 meters * 5 pounds = 40 pounds so answer is b and similar but more theoretical approach : if we have sides a and b than they have equal ration with their areas : a / b = a ^ 2 / b ^ 2 and they have equal ration with their volumes : a / b = a ^ 3 / b ^ 3 we have two sides 1 / 2 so their volume will be in ratio 1 / 8 weight of one cube * volume of another cube 5 * 8 = 40 so answer is b | a = 2 * 4
b = 5 * a
|
a ) 1 / 4 , b ) 1 / 2 , c ) 1 / 2 , d ) 2 , e ) 2 / 3 | d | divide(add(3, 3), 3) | a box contains 3 red balls and 3 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ? | "my complicated version of your simple approach let the 3 black balls be bbb and 3 red balls be rrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 6 ! ) / ( 3 ! x 3 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 5 slots and 2 black ( bb ) and 3 red ( rrr ) they can be arranged in ( 5 ! ) / ( 2 ! x 3 ! ) hence required probability = [ ( 6 ! ) / ( 3 ! x 3 ! ) ] / [ ( 5 ! ) / ( 2 ! x 3 ! ) ] = 2 d" | a = 3 + 3
b = a / 3
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a ) 40 % , b ) 55 % , c ) 57 % , d ) 63.2 % , e ) 62 % | d | multiply(divide(11628, add(add(1136, 5636), 11628)), const_100) | 3 candidates in an election and received 1136 , 5636 and 11628 votes respectively . what % of the total votes did the winning candidate gotin that election ? | "total number of votes polled = ( 1136 + 5636 + 11628 ) = 18400 so , required percentage = 11628 / 18400 * 100 = 63.2 % d" | a = 1136 + 5636
b = a + 11628
c = 11628 / b
d = c * 100
|
a ) 4.37 % , b ) 5 % , c ) 6 % , d ) 8.75 % , e ) none | b | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | the population of a town increased from 1 , 75,000 to 2 , 62,500 in a decade . the average percent increase of population per year is | "solution increase in 10 years = ( 262500 - 175000 ) = 87500 . increase % = ( 87500 / 175000 × 100 ) % = 50 % . required average = ( 50 / 10 ) % = 5 % . answer b" | a = 10 * 1000
b = a * 10
c = b - 1000
d = c - 1000
e = 2 + 3
f = e * 100
g = d - f
h = 3 + 4
i = h * 10
j = 2 + 3
k = i + j
l = k * 1000
m = g / l
n = m - 1
o = n / 10
p = o * 100
q = p + 4
|
a ) 87 , b ) 91 , c ) 95 , d ) 99 , e ) 103 | b | divide(factorial(14), multiply(factorial(subtract(14, const_2)), factorial(const_2))) | there are 14 teams in a soccer league and each team plays each of the other teams once . how many games will be played in total ? | "14 c 2 = 91 the answer is b ." | a = math.factorial(14)
b = 14 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 10 ^ 9 , b ) 10 ^ 10 , c ) 10 ^ 11 , d ) 10 ^ 12 , e ) 10 ^ 13 | c | divide(0.1, power(0.0001, 3)) | the decimal 0.1 is how many times greater than the decimal ( 0.0001 ) ^ 3 ? | "0.1 = 10 ^ - 1 ( 0.0001 ) ^ 3 = ( 10 ^ - 4 ) ^ 3 = 10 ^ - 12 10 ^ 11 * 10 ^ - 12 = 10 ^ - 1 the answer is c ." | a = 0 ** 1
b = 0 / 1
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a ) 5 , b ) 6 , c ) 4 , d ) 3 , e ) 8 | a | divide(52.416, 0.168) | when 52416 is divided by 312 , the quotient is 168 . what will be the quotient when 52.416 is divided by 0.168 ? | "for the 1 st no . there are 2 digits after decimal for the 2 nd no . there are 4 digits after decimal total no . of decimals = 6 req . no . of digits = ( n - 1 ) = ( 6 - 1 ) = 5 answer : a" | a = 52 / 416
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a ) 130 , b ) 140 , c ) 150 , d ) 170 , e ) 180 | e | divide(450, add(divide(150, const_100), const_1)) | two employees a and b are paid a total of rs . 450 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 450 but x = 150 % of y = 150 y / 100 = 15 y / 10 ∴ 15 y / 10 + y = 450 ⇒ y [ 15 / 10 + 1 ] = 450 ⇒ 25 y / 10 = 450 ⇒ 25 y = 4500 ⇒ y = 4500 / 25 = rs . 180 e" | a = 150 / 100
b = a + 1
c = 450 / b
|
a ) 1 / 2 , b ) 3 / 8 , c ) 1 / 4 , d ) 1 / 5 , e ) 5 / 8 | c | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting only 1 head in a single throw of four fair coins ? | "one possible case is httt . p ( httt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 there are 4 c 1 = 4 possible cases . p ( 1 head ) = 4 * 1 / 16 = 1 / 4 the answer is c ." | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
|
a ) 3.6 sec , b ) 18 sec , c ) 36 sec , d ) 72 sec , e ) none | c | divide(add(240, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 120 = 360 m . time taken = 360 / 10 = 36 sec answer : c" | a = 240 + 120
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
a ) 0.2 % , b ) 8 % , c ) 10 % , d ) 20 % , e ) 500 % | b | multiply(divide(4, 50), const_100) | 4 is what percent of 50 ? | "4 = x * 50 / 100 x = 8 % ans ; b" | a = 4 / 50
b = a * 100
|
['a ) 1 : 7', 'b ) 2 : 9', 'c ) 4 : 9', 'd ) 3 : 7', 'e ) 3 : 4'] | c | divide(circle_area(2), circle_area(3)) | the ratio of the radius of two circles is 2 : 3 , and then the ratio of their areas is ? | r 1 : r 2 = 2 : 3 î r 1 ^ 2 : î r 2 ^ 2 r 1 ^ 2 : r 2 ^ 2 = 4 : 9 answer : c | a = circle_area / (
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a ) $ 480 , b ) $ 450 , c ) $ 992 , d ) $ 870 , e ) $ 660 | c | add(divide(multiply(divide(add(832, 448), const_2), 55), const_100), divide(add(832, 448), const_2)) | the profit earned by selling an article for $ 832 is equal to the loss incurred when the same article is sold for $ 448 . what should be the sale price for making 55 % profit ? | "c let c . p . = $ x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 155 % of $ 640 = 150 / 100 * 640 = $ 992 ." | a = 832 + 448
b = a / 2
c = b * 55
d = c / 100
e = 832 + 448
f = e / 2
g = d + f
|
a ) 0 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | e | add(add(const_4, const_3), const_2) | what is the units digit of the expression 14 ^ 7 − 18 ^ 4 ? | "units digit of 14 ^ 7 − 18 ^ 4 : consider the following unit digits : 4 ^ 1 : 4 4 ^ 2 : 6 4 ^ 3 : 4 4 ^ 4 : 6 ⇒ unit digit of 4 ^ 7 is : 4 the unit digit of 14 ^ 7 will also be 4 also : 8 ^ 1 : 8 8 ^ 2 : 4 8 ^ 3 : 2 8 ^ 4 : 6 ⇒ unit digit of 8 ^ 4 is : 6 the unit digit of 18 ^ 4 will also be 6 finally , 4 - 6 ( or 14 - 6 ) = 8 answer : e )" | a = 4 + 3
b = a + 2
|
a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | d | divide(divide(subtract(58, 46), 6), 3) | for 6 - digit integer 46 x , y 12 , when x and y are drawn at random from { 3 , 58 } , what is the probability w that the integer drawn is divisible by 8 ? * a solution will be posted in two days . | in general , the remainder divided by integer n is the same as the remainder that up to tens of n is divided by 8 . thus , in order for 46 , x , y 12 to be divided by 8 , y 12 should be divided by 8 . amongst 3 , 5,8 , 3 and 5 can be divided by 8 . therefore , w = two out of three can be divided and the answer is d . | a = 58 - 46
b = a / 6
c = b / 3
|
a ) 48 , b ) 51 , c ) 60 , d ) 71 , e ) 96 | b | divide(add(subtract(multiply(46, 10), 25), 75), 10) | the average of 10 numbers is calculated as 46 . it is discovered later on that while calculating the average , the number 75 was incorrectly read as 25 , and this incorrect number was used in the calculation . what is the correct average ? | "the total sum of the numbers should be increased by 50 . then the average will increase by 50 / 10 = 5 . the correct average is 51 . the answer is b ." | a = 46 * 10
b = a - 25
c = b + 75
d = c / 10
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a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 40 % | b | divide(divide(multiply(40, 60), multiply(80, 60)), add(divide(multiply(40, 60), multiply(80, 60)), 8)) | a certain car can travel 40 minutes on a gallon of gasoline at 60 miles per hour . if the car had started with a full tank and had 8 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 80 miles at 60 mph ? | "otal time for travelling 80 miles @ 60 mph = 80 / 60 = 4 / 3 hour = 80 minutes . given , the car uses 1 gallon for every 40 minutes of driving @ 60 mph . thus in 80 minutes it will use = 2 gallons . thus , full tank = 2 + 8 = 10 gallons - - - > 2 / 10 = 20 % of the fuel used . b is the correct answer ." | a = 40 * 60
b = 80 * 60
c = a / b
d = 40 * 60
e = 80 * 60
f = d / e
g = f + 8
h = c / g
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a ) 50 , b ) 45 , c ) 60 , d ) 55 , e ) 53 | c | multiply(subtract(40, 10), const_2) | all the students of class are told to sit in circle shape . here the boy at the 10 th position is exactly opposite to 40 th boy . total number of boys in the class ? | "as half the circle shape consist of 40 - 10 = 30 boys , so total number of boys in full circle = 2 * 30 = 60 answer : c" | a = 40 - 10
b = a * 2
|
a ) 7 sec , b ) 6 sec , c ) 8 sec , d ) 9 sec , e ) 5 sec | b | divide(120, multiply(72, const_0_2778)) | in what time will a train 120 metres long cross an electric pole , if its speed be 72 km / hr ? | "solution speed = ( 72 x 5 / 18 ) m / sec = 20 m / sec time taken = ( 120 / 20 ) sec = 6 sec . answer b" | a = 72 * const_0_2778
b = 120 / a
|
a ) s . 125 , b ) s . 108 , c ) s . 100 , d ) s . 103 , e ) s . 102 | a | multiply(5, divide(1850, add(add(multiply(6, 3), multiply(9, 4)), multiply(4, 5)))) | a , b and c completed a piece of work , a worked for 6 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was rs . 1850 ? | "3 x 4 x 5 x 6 9 4 18 x + 36 x + 20 x = 1850 74 x = 1850 = > x = 25 5 x = 125 rs . answer : a" | a = 6 * 3
b = 9 * 4
c = a + b
d = 4 * 5
e = c + d
f = 1850 / e
g = 5 * f
|
a ) $ 22 , b ) $ 24 , c ) $ 28 , d ) $ 32 , e ) $ 36 | c | divide(42, add(const_1, divide(50, const_100))) | a worker ' s daily wage is increased by 50 % and the new wage is $ 42 per day . what was the worker ' s daily wage before the increase ? | "let x be the daily wage before the increase . 1.5 x = $ 42 x = $ 28 the answer is c ." | a = 50 / 100
b = 1 + a
c = 42 / b
|
a ) 17.25 % , b ) 19 , c ) 19.5 % , d ) 21.5 % , e ) 22 % | c | multiply(const_100, divide(multiply(divide(26, const_100), 15), add(5, 15))) | 5 ltr of water is added with 15 ltr of a solution containing 26 % of alcohol in the water . the % of alcohol in the new mixture is ? | we have a 15 litre solution containing 26 % alcohol in water . = > quantity of alcohol in the solution = 15 ã — 26 / 100 now 5 litre of water is added to the solution . = > total quantity of the new solution = 15 + 5 = 20 percentage of alcohol in the new solution = 15 ã — 26 / 100 20 ã — 100 = 15 ã — 26 / 10020 ã — 100 = 15 ã — 1.3 / 100 = 19.5 % c | a = 26 / 100
b = a * 15
c = 5 + 15
d = b / c
e = 100 * d
|
a ) 177 , b ) 166 , c ) 350 , d ) 198 , e ) 161 | c | subtract(multiply(divide(300, 18), 39), 300) | a 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 39 = 50 / 3 3 ( x + 300 ) = 1950 è x = 350 m . answer : c" | a = 300 / 18
b = a * 39
c = b - 300
|
a ) 41 , b ) 36 , c ) 39 , d ) 21 , e ) 11 | a | add(add(15, 12), 14) | anne bought doughnuts for a class breakfast party . she bought 15 chocolate doughnuts , 12 coconut doughnuts , and 14 jam - filled doughnuts . how many doughnuts did anne buy in all ? | add the numbers of doughnuts . 15 + 12 + 14 = 41 . answer is a . | a = 15 + 12
b = a + 14
|
a ) 11 . , b ) 12 , c ) 13 , d ) 14 , e ) 14.5 | b | add(divide(multiply(10, const_100), add(const_100, 25)), divide(multiply(6, const_100), add(const_100, 50))) | following an increase in prices , the price of a candy box was 10 pounds and the price of a can of soda was 6 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ? | "let the candy and soda before the increase be c and s respectively so increased price of can = c ( 1 + 25 / 100 ) = 10 - - - - - - - - - - - > c = 8 increase price of soda : s ( 1 + 1 / 2 ) = 6 - - - - - - - - - - - > s = 4 hence total = 8 + 4 = 12 . . hence answer : b" | a = 10 * 100
b = 100 + 25
c = a / b
d = 6 * 100
e = 100 + 50
f = d / e
g = c + f
|
a ) 70 , 30 , b ) 30 , 70 , c ) 70 , 70 , d ) 30 , 30 , e ) 80 , 50 | a | divide(add(divide(240, 6), divide(500, 5)), const_2) | if a boat is rowed downstream for 500 km in 5 hours and upstream for 240 km in 6 hours , what is the speed of the boat and the river ? | "explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x â € “ y x = ( ds + us ) / 2 y = ( ds â € “ us ) / 2 in the above problem ds = 100 ; us = 40 x = ( 100 + 40 ) / 2 = 140 / 2 = 70 km / hr y = ( 100 - 40 ) / 2 = 60 / 2 = 30 km / hr answer : a" | a = 240 / 6
b = 500 / 5
c = a + b
d = c / 2
|
a ) 7297 , b ) 6625 , c ) 2871 , d ) 6725 , e ) 2981 | b | divide(8215, add(const_1, divide(24, const_100))) | the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 8215 for a computer table , then what was the cost price of the computer table ? | "explanation : cp = sp * ( 100 / ( 100 + profit % ) ) = 8215 ( 100 / 124 ) = rs . 6625 . answer : b" | a = 24 / 100
b = 1 + a
c = 8215 / b
|
a ) 12 , b ) 13 , c ) 16.6 , d ) 17 , e ) 23 | a | add(10, const_2) | set d consists of all the prime numbers between 10 and 25 . what is the range of set d ? | answer of set d is 12 = a | a = 10 + 2
|
a ) 2788 , b ) 3500 , c ) 7282 , d ) 2782 , e ) 2729 | b | multiply(subtract(12, 7), divide(2800, subtract(7, 3))) | an amount of money is to be divided between p , q and r in the ratio of 3 : 7 : 12 . if the difference between the shares of p and q is rs . 2800 , what will be the difference between q and r ' s share ? | "4 - - - 2800 5 - - - ? = > 3500 answer : b" | a = 12 - 7
b = 7 - 3
c = 2800 / b
d = a * c
|
a ) 10 % , b ) 15 % , c ) 17.5 % , d ) 25 % , e ) 12.5 % | e | multiply(subtract(multiply(divide(add(const_100, 25), const_100), divide(subtract(const_100, 10), const_100)), const_1), const_100) | the price of a mobile was increased by 25 percent . the new price was then decreased by 10 percent . a single increase of what percent is equivalent to these two successive changes ? | "consider base price - $ 100 25 % increase = 1.25 * 100 = $ 125 then a 10 % decrease on new price = 0.9 * 125 = $ 112.5 so final price of radio - $ 112.5 therefore a 12.5 % increase correct option - e" | a = 100 + 25
b = a / 100
c = 100 - 10
d = c / 100
e = b * d
f = e - 1
g = f * 100
|
a ) rs . 1 , 10,000 , b ) rs . 1 , 20,000 , c ) rs . 1 , 22,000 , d ) rs . 1 , 21,000 , e ) none of these | c | add(multiply(multiply(subtract(1, divide(20, const_100)), subtract(1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100))) | the value of a machine depreciates at 20 % per annum . if its present value is rs . 1 , 50,000 , at what price should it be sold after two years such that a profit of rs . 26,000 is made ? | "the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = rs . 96,000 sp such that a profit of rs . 24,000 is made = 96,000 + 26,000 = rs . 1 , 22,000 answer : c" | a = 20 / 100
b = 1 - a
c = 20 / 100
d = 1 - c
e = b * d
f = 100 * 100
g = math.sqrt(100)
h = f * g
i = math.sqrt(100)
j = i / 2
k = j * 100
l = k * 100
m = h + l
n = e * m
o = 20 + 2
p = o * 100
q = math.sqrt(100)
r = p * q
s = n + r
|
a ) 29 , b ) 10 , c ) 20 , d ) 40 , e ) 45 | a | divide(subtract(64.50, 50), subtract(1.50, 1.00)) | caleb spends $ 64.50 on 50 hamburgers for the marching band . if single burgers cost $ 1.00 each and double burgers cost $ 1.50 each , how many double burgers did he buy ? | "solution - lets say , single hamburgersxand double hamburgersy given that , x + y = 50 and 1 x + 1.5 y = 64.50 . by solving the equations y = 29 . ans a ." | a = 64 - 50
b = 1 - 50
c = a / b
|
a ) 150 , b ) 180 , c ) 190 , d ) 210 , e ) 105 | e | add(add(add(add(add(add(add(4, 10), add(4, const_2)), add(4, const_1)), 4), 10), const_2), const_1) | if two integers x , y ( x > y ) are selected from - 10 to 4 ( inclusive ) , how many possible cases are there ? | "if two integers x , y ( x > y ) are selected from - 10 to 9 ( inclusive ) , how many possible cases are there ? a . 150 b . 180 c . 190 d . 210 e . 240 - - > 15 c 2 = 15 * 14 / 2 = 105 . therefore , the answer is e ." | a = 4 + 10
b = 4 + 2
c = a + b
d = 4 + 1
e = c + d
f = e + 4
g = f + 10
h = g + 2
i = h + 1
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | floor(sqrt(divide(30800, multiply(multiply(multiply(11, 7), 5), 2)))) | in a certain archery competition , points were awarded as follows : the first place winner receives 11 points , the second place winner receives 7 points , the third place winner receives 5 points and the fourth place winner receives 2 points . no other points are awarded . john participated several times in the competition and finished first , second , third , or fourth each time . the product of all the points he received was 30800 . how many times did he participate in the competition ? | "30800 = 2 * 2 * 2 * 2 * 5 * 5 * 7 * 11 john participated 8 times . the answer is d ." | a = 11 * 7
b = a * 5
c = b * 2
d = 30800 / c
e = math.sqrt(d)
f = math.floor(e)
|
a ) 46 m , b ) 36 m , c ) 26 m , d ) 56 m , e ) 25 m | b | divide(multiply(add(40, 50), const_2), 5) | a rectangular plot measuring 40 meters by 50 meters is to be enclosed by wire fencing . if the poles of the fence are kept 5 meters apart . how many poles will be needed ? | "perimeter of the plot = 2 ( 40 + 50 ) = 180 m no of poles = 180 / 5 = 36 m answer : b" | a = 40 + 50
b = a * 2
c = b / 5
|
a ) 7 , b ) 141 , c ) 29 , d ) 349 , e ) 359 | d | add(subtract(multiply(2.52, const_100), const_2), subtract(const_100, const_1)) | the no 2.52 , when written as a fraction and reduced to lowest terms , the sum of the numeratr n denominator is : | the actual question has bar on decimal part . hence the number is 2.5252525252 . . . . . . so let x = 2.525252 . . then 100 x = 252.525252 100 x - x = 250 x = 250 / 99 so ans = num + den = 349 answer : d | a = 2 * 52
b = a - 2
c = 100 - 1
d = b + c
|
a ) rs . 540 , b ) rs . 580 , c ) rs . 560 , d ) rs . 620 , e ) rs . 650 | c | add(divide(575, add(const_1, divide(15, const_100))), multiply(divide(12, const_100), divide(575, add(const_1, divide(15, const_100))))) | if sharon ' s weekly salary increased by 15 percent , she would earn $ 575 per week . if instead , her weekly salary were to decreased by 12 percent , how much would she earn per week ? | "soln : - ( 575 / 115 ) 112 = 560 in this case long division does not take much time . ( 23 / 5 ) = 4.6 4.6 * 112 = rs . 560 answer : c" | a = 15 / 100
b = 1 + a
c = 575 / b
d = 12 / 100
e = 15 / 100
f = 1 + e
g = 575 / f
h = d * g
i = c + h
|
a ) 7000 , b ) 6800 , c ) 8000 , d ) 5400 , e ) 7100 | a | divide(630, multiply(divide(subtract(subtract(const_100, 20), 70), const_100), divide(90, const_100))) | of the land owned by a farmer , 90 percent was cleared for planting . of the cleared land , 20 percent was planted with potato and 70 percent of the cleared land was planted with tomato . if the remaining 630 acres of cleared land was planted with corn , how many acres did the farmer own ? | 20 % od 90 % = 18 % , 70 % 0 f 90 % = 63 % so the remaining 90 - 18 - 63 = 9 % = 630 acres or 10 % of 90 % = 9 % - - > 630 / 9 * 100 = 7000 acres answer ( a ) | a = 100 - 20
b = a - 70
c = b / 100
d = 90 / 100
e = c * d
f = 630 / e
|
a ) 7 days , b ) 8 days , c ) 9 days , d ) 10 days , e ) 11 days | b | divide(55, divide(add(add(divide(55, 11), divide(55, 20)), add(divide(55, 11), divide(55, 55))), const_2)) | a , band c can do a piece of work in 11 days , 20 days and 55 days respectively , working alone . how soon can the work be done if a is assisted by band c on alternate days ? | "( a + b ) ' s 1 day ' s work = 1 / 11 + 1 / 20 = 31 / 220 ( a + c ) ' s 1 day ' s work = 1 / 11 + 1 / 55 = 6 / 55 work done in 2 day ' s = 31 / 220 + 6 / 55 = 55 / 220 = 1 / 4 1 / 4 th work done in 2 days work done = 4 * 2 = 8 days answer : b" | a = 55 / 11
b = 55 / 20
c = a + b
d = 55 / 11
e = 55 / 55
f = d + e
g = c + f
h = g / 2
i = 55 / h
|
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