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 ) 22 , b ) 38 , c ) 11 , d ) 17 , e ) 91 | d | divide(subtract(multiply(add(14, 1), add(14, 1)), multiply(14, 10)), 5) | the average age of a group of 10 students is 14 years . if 5 more students join the group , the average age rises by 1 year . the average age of the new students is : | explanation : total age of the 10 students = 10 × 14 = 140 total age of 15 students including the newly joined 5 students = 15 × 15 = 225 total age of the new students = 225 − 140 = 85 average age = 85 / 5 = 17 years answer : d | a = 14 + 1
b = 14 + 1
c = a * b
d = 14 * 10
e = c - d
f = e / 5
|
a ) 350 , b ) 370 , c ) 390 , d ) 430 , e ) none | a | add(310, divide(multiply(4, subtract(420, 310)), add(3, 4))) | average expenditure of a person for the first 3 days of a week is rs . 310 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is : | "explanation : assumed mean = rs . 310 total excess than assumed mean = 4 × ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 / 7 = rs . 40 therefore , average expenditure for 7 days = rs . 310 + rs . 40 = rs . 350 correct option : a" | a = 420 - 310
b = 4 * a
c = 3 + 4
d = b / c
e = 310 + d
|
a ) 286 m , b ) 899 m , c ) 300 m , d ) 166 m , e ) 187 m | c | subtract(divide(800, const_2), 100) | if the perimeter of a rectangular garden is 800 m , its length when its breadth is 100 m is ? | "2 ( l + 100 ) = 800 = > l = 300 m answer : c" | a = 800 / 2
b = a - 100
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a ) 1200 , b ) 600 , c ) 750 , d ) 900 , e ) none of these | d | divide(multiply(divide(31.5, subtract(4.5, 4)), const_100), 7) | two equal sums of money were invested , one at 4 % and the other at 4.5 % . at the end of 7 years , the simple interest received from the latter exceeded to that received from the former by 31.50 . each sum was : | difference of s . i . = √ 31.50 let each sum be x . then x × 4 1 / 2 × 7 / 100 − x × 4 × 7 / 100 = 31.50 or 7 ⁄ 100 x × 1 ⁄ 2 = 63 ⁄ 2 or x = 900 answer d | a = 4 - 5
b = 31 / 5
c = b * 100
d = c / 7
|
a ) 65 , b ) 69 , c ) 75 , d ) 85 , e ) 73 | e | divide(add(add(add(add(76, 65), 82), 67), 75), add(const_1, const_4)) | shekar scored 76 , 65 , 82 , 67 and 75 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | "explanation : average = ( 76 + 65 + 82 + 67 + 75 ) / 5 = 365 / 5 = 73 hence average = 73 answer : e" | a = 76 + 65
b = a + 82
c = b + 67
d = c + 75
e = 1 + 4
f = d / e
|
a ) $ 16,830 , b ) $ 17,820 , c ) $ 18,000 , d ) $ 19,800 , e ) $ 21,780 | a | floor(divide(add(multiply(2, subtract(subtract(10000, 100), divide(multiply(subtract(10000, 100), 30), 100))), divide(multiply(subtract(10000, 100), 30), 100)), const_1000)) | tickets for all but 100 seats in a 10000 - seat stadium were sold . of the tickets sold , 30 % were sold at half price and the remaining tickets were sold at the full price of $ 2 . what was the total revenue from ticket sales ? | 10000 seats - - > full price : half price = 7000 : 3000 price when all seats are filled = 14000 + 3000 = 17000 100 seats are unsold - - > loss due to unfilled seats = 30 + 2 * 70 = 170 revenue = 17000 - 170 = 16830 answer : a | a = 10000 - 100
b = 10000 - 100
c = b * 30
d = c / 100
e = a - d
f = 2 * e
g = 10000 - 100
h = g * 30
i = h / 100
j = f + i
k = j / 1000
l = math.floor(k)
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a ) 3 : 2 , b ) 9 : 2 , c ) 18 : 20 , d ) 1 : 2 , e ) 18 : 4 | d | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | a and b started a business investing rs . 10,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "a : b = 10000 : 20000 = 1 : 2 answer : d" | a = 3 * 3
b = 3 * 3
c = b + 1
d = a * c
e = d + 2
f = 3 * 3
g = f + 1
h = 2 * g
i = e / h
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a ) 9 : 3 , b ) 9 : 5 , c ) 5 : 8 , d ) 5 : 4 , e ) 9 : 2 | a | divide(divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(9, 7)), divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(9, 7))) | rs . 6000 is lent out in two parts . one part is lent at 7 % p . a simple interest and the other is lent at 9 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | "let the amount lent at 7 % be rs . x amount lent at 9 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 7 / 100 x + 9 / 100 ( 6000 - x ) = 540 - 2 x / 100 = > 540 - 1 / 50 x = 450 = > x = 4500 amount lent at 10 % = 1500 required ratio = 4500 : 1500 = 9 : 3 answer : a" | a = 450 * 100
b = 6000 * 7
c = a - b
d = 9 - 7
e = c / d
f = 450 * 100
g = 6000 * 7
h = f - g
i = 9 - 7
j = h / i
k = e / j
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a ) 52 kmph . , b ) 62 kmph . , c ) 97 kmph . , d ) 80 kmph . , e ) none | c | subtract(multiply(divide(280, 9), const_3_6), 15) | a man sitting in a train which is travelling at 15 kmph observes that a goods train , travelling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed ? | "solution relative speed = ( 280 / 9 ) m / sec = ( 280 / 9 x 18 / 5 ) = 112 kmph . speed of the train = ( 112 - 15 ) kmph = 97 kmph . answer c" | a = 280 / 9
b = a * const_3_6
c = b - 15
|
a ) 28 , b ) 30 , c ) 32 , d ) 35 , e ) none of these | e | subtract(35, subtract(add(20, 19), 9)) | there are 35 student in a school . 20 if them speek hindi 19 of them speak eng 9 of them speak both , then how many student neithe speak eng nor speak hindi | total number of students = 35 no of hindi speaking students ( h ) = 20 no of english speaking students ( e ) = 19 h intersection e = 9 thus h union e = 20 + 19 - 9 = 30 thus number of student neithe speak eng nor speak hindi = 35 - 30 = 5 answer : e | a = 20 + 19
b = a - 9
c = 35 - b
|
a ) 127 , b ) 100 , c ) 129 , d ) 160 , e ) 200 | a | divide(add(102, 152), 2) | a student chose a number , multiplied it by 2 , then subtracted 152 from the result and got 102 . what was the number he chose ? | "solution : let x be the number he chose , then 2 * x * 152 = 102 2 x = 254 x = 127 correct answer a" | a = 102 + 152
b = a / 2
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a ) 1000 min , b ) 1200 min , c ) 1300 min , d ) 1400 min , e ) 1600 min | a | divide(multiply(divide(80, const_100), multiply(multiply(50, 150), 10)), 60) | the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 50 feet wide by 150 feet long by 10 feet deep and is currently at 80 % capacity , how long will it take to drain the pool ? | volume of pool = 50 * 150 * 10 cu . ft , 80 % full = 50 * 150 * 10 * 0.8 cu . ft water is available to drain . draining capacity = 60 cu . ft / min therefore time taken = 50 * 150 * 10 * 0.8 / 60 min = 1000 min a | a = 80 / 100
b = 50 * 150
c = b * 10
d = a * c
e = d / 60
|
a ) 1 , b ) 1 / 2 , c ) 5 / 6 , d ) 2 / 3 , e ) 4 / 3 | e | subtract(divide(subtract(26, const_1), add(14, const_1)), divide(add(9, const_1), subtract(31, const_1))) | if a is an integer greater than 9 but less than 26 and b is an integer greater than 14 but less than 31 , what is the range of a / b ? | "range of a / b = max ( a / b ) - min ( a / b ) to get max ( a / b ) = > max ( a ) / min ( b ) = 25 / 15 to get min ( a / b ) = > min ( a ) / max ( b ) = 10 / 30 range = 25 / 15 - 10 / 30 = 4 / 3" | a = 26 - 1
b = 14 + 1
c = a / b
d = 9 + 1
e = 31 - 1
f = d / e
g = c - f
|
a ) 150 % , b ) 120 % , c ) 130 % , d ) 200 % , e ) none of these | a | multiply(divide(subtract(const_100, 40), 40), const_100) | if the cost price is 40 % of selling price . then what is the profit percent . | "explanation : let the s . p = 100 then c . p . = 40 profit = 60 profit % = ( 60 / 40 ) * 100 = 150 % . answer : a" | a = 100 - 40
b = a / 40
c = b * 100
|
a ) 84 , b ) 98 , c ) 51 , d ) 65 , e ) 100 | e | divide(12, subtract(75.12, floor(75.12))) | when positive integer x is divided by positive integer y , the remainder is 12 . if x / y = 75.12 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 12 - - > x = qy + 12 ; x / y = 75.12 - - > x = 75 y + 0.12 y ( so q above equals to 75 ) ; 0.12 y = 12 - - > y = 100 . answer : e ." | a = math.floor(75, 12)
b = 75 - 12
c = 12 / b
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a ) $ 3.58 , b ) $ 1.67 , c ) $ 2.25 , d ) $ 2.37 , e ) $ 2.50 | a | divide(add(20.00, multiply(1.75, subtract(9, 2))), 9) | the cost to park a car in a certain parking garage is $ 20.00 for up to 2 hours of parking and $ 1.75 for each hour in excess of 2 hours . what is the average ( arithmetic mean ) cost per hour to park a car in the parking garage for 9 hours ? | "total cost of parking for 9 hours = 20 $ for the first 2 hours and then 1.75 for ( 9 - 2 ) hours = 20 + 7 * 1.75 = 32.25 thus the average parking price = 32.25 / 9 = 3.58 $ a is the correct answer ." | a = 9 - 2
b = 1 * 75
c = 20 + 0
d = c / 9
|
a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | a | subtract(multiply(multiply(multiply(611, 704), 912), 261), subtract(multiply(multiply(multiply(611, 704), 912), 261), add(const_4, const_4))) | the unit digit in the product ( 611 * 704 * 912 * 261 ) is : | "explanation : unit digit in the given product = unit digit in ( 1 * 4 * 2 * 1 ) = 2 answer : a" | a = 611 * 704
b = a * 912
c = b * 261
d = 611 * 704
e = d * 912
f = e * 261
g = 4 + 4
h = f - g
i = c - h
|
a ) 56 , b ) 48 , c ) 47 , d ) 30 , e ) 25 | d | multiply(divide(add(1.10, 0.9), add(150, 90)), const_3600) | two trains are moving in opposite directions with speed of 150 km / hr and 90 km / hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds | "explanation : relative speed = 150 + 90 = 240 km / hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 / 240 hr = 1 / 120 hr = 3600 / 120 seconds = 30 seconds answer : option d" | a = 1 + 10
b = 150 + 90
c = a / b
d = c * 3600
|
a ) 7 , b ) 6 , c ) 8 , d ) 2 , e ) 3 | e | divide(60, multiply(add(60, 6), const_0_2778)) | a train 60 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 60 * 3 / 55 = 3 sec . answer e" | a = 60 + 6
b = a * const_0_2778
c = 60 / b
|
a ) 12 , b ) 28 , c ) 160 , d ) 200 , e ) 18 | d | subtract(power(divide(add(20, 10), const_2), const_2), power(subtract(20, divide(add(20, 10), const_2)), const_2)) | if the sum and difference of two numbers are 20 and 10 respectively , then the difference of their square is : | "let the numbers be x and y . then , x + y = 20 and x - y = 8 x 2 - y 2 = ( x + y ) ( x - y ) = 20 * 10 = 200 . answer : d" | a = 20 + 10
b = a / 2
c = b ** 2
d = 20 + 10
e = d / 2
f = 20 - e
g = f ** 2
h = c - g
|
a ) 6 % . , b ) 7.5 % . , c ) 19.6 % . , d ) 10.5 % . , e ) 11 % . | c | multiply(divide(add(subtract(subtract(340, multiply(340, divide(75, const_100))), multiply(340, divide(5, const_100))), 3.2), add(add(add(340, 3.2), 12), 6.8)), const_100) | a 340 - liter solution of kola is made from 75 % water , 5 % concentrated kola and the rest is made from sugar . if 3.2 liters of sugar , 12 liter of water and 6.8 liters of concentrated kola were added to the solution , what percent of the solution is made from sugar ? | "denominator : 340 + 12 + 3.2 + 6.8 = 362 numerator : 340 ( 1 - . 75 - . 05 ) + 3.2 340 ( 0.2 ) + 3.2 68 + 3.2 71.2 ratio : 71.2 / 362 = 0.196 answer : c" | a = 75 / 100
b = 340 * a
c = 340 - b
d = 5 / 100
e = 340 * d
f = c - e
g = f + 3
h = 340 + 3
i = h + 12
j = i + 6
k = g / j
l = k * 100
|
a ) 10 , b ) 40 , c ) 20 , d ) 30 , e ) 25 | c | divide(subtract(multiply(95, 2), multiply(2, 45)), subtract(95, 90)) | the average mark of the students of a class in a particular exam is 90 . if 2 students whose average mark in that exam is 45 are excluded , the average mark of the remaining will be 95 . find the number of students who wrote the exam ? | "let the number of students who wrote the exam be x . total marks of students = 90 x . total marks of ( x - 2 ) students = 95 ( x - 2 ) 90 x - ( 2 * 45 ) = 95 ( x - 2 ) 100 = 5 x = > x = 20 answer : c" | a = 95 * 2
b = 2 * 45
c = a - b
d = 95 - 90
e = c / d
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a ) 2348 , b ) 7250 , c ) 2767 , d ) 1998 , e ) 2771 | b | multiply(add(add(multiply(3, 6), multiply(4, 5)), multiply(5, 4)), divide(250, subtract(multiply(4, 5), multiply(3, 6)))) | a , b and c invest in the ratio of 3 : 4 : 5 . the percentage of return on their investments are in the ratio of 6 : 5 : 4 . find the total earnings , if b earns rs . 250 more than a : | "explanation : a b c investment 3 x 4 x 5 x rate of return 6 y % 5 y % 4 y % return \ inline \ frac { 18 xy } { 100 } \ inline \ frac { 20 xy } { 100 } \ inline \ frac { 20 xy } { 100 } total = ( 18 + 20 + 20 ) = \ inline \ frac { 58 xy } { 100 } b ' s earnings - a ' s earnings = \ inline \ frac { 2 xy } { 100 } = 250 total earning = \ inline \ frac { 58 xy } { 100 } = 7250 answer : b ) rs . 7250" | a = 3 * 6
b = 4 * 5
c = a + b
d = 5 * 4
e = c + d
f = 4 * 5
g = 3 * 6
h = f - g
i = 250 / h
j = e * i
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | subtract(33, reminder(9, 11)) | when positive integer n is divided by 3 , the remainder is 1 . when n is divided by 11 , the remainder is 9 . what is the smallest positive integer k such that k + n is a multiple of 33 ? | "n = 3 p + 1 = 11 q + 9 n + 2 = 3 p + 3 = 11 q + 11 n + 2 is a multiple of 3 and 11 , so it is a multiple of 33 . the answer is a ." | a = 33 - reminder
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a ) 90 , b ) 2,700 , c ) 3,300 , d ) 5,400 , e ) 324,000 | c | divide(multiply(add(35, 75), const_60), const_2) | a copy machine , working at a constant rate , makes 35 copies per minute . a second copy machine , working at a constant rate , makes 75 copies per minute . working together at their respective rates , how many copies do the two machines make in half an hour ? | "together the two machines make 35 + 75 = 110 copies per minute . so , in half an hour they will make 110 * 30 = 3,300 copies . answer : c ." | a = 35 + 75
b = a * const_60
c = b / 2
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a ) s . 15,550 , b ) s . 15,600 , c ) s . 16,500 , d ) s . 17,600 , e ) s . 17,900 | c | multiply(750, multiply(5.5, 4)) | the lenght of a room is 5.5 m and width is 4 m . find the cost of paving the floor by slabs at the rate of rs . 750 per sq . metre . | "area of the floor = ( 5.5 ã — 4 ) m 2 = 22 m 2 . cost of paving = rs . ( 750 ã — 22 ) = rs . 16500 . answer : option c" | a = 5 * 5
b = 750 * a
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a ) 22 % , b ) 25 % , c ) 77 % , d ) 5.26 % , e ) 12 % | d | subtract(multiply(divide(const_100, 950), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 950 grams per kg , what is his percent ? | 950 - - - 50 100 - - - ? = > 5.26 % answer : d | a = 100 / 950
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
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a ) rs 2060.808 , b ) rs 2060.801 , c ) rs 2060.804 , d ) rs 2060.802 , e ) rs 2060.805 | c | subtract(multiply(power(add(const_1, divide(divide(2, const_4), const_100)), const_3), multiply(multiply(multiply(const_4, const_4), const_100), sqrt(const_100))), multiply(multiply(multiply(const_4, const_4), const_100), sqrt(const_100))) | find the compound interest on rs . 25000 in 2 years at 4 % per annum . the interest being compounded half yearly . | "explanation : given : principal = rs . 25000 , rate = 4 % per half year , time = 2 years = 4 half years therefore , amount = p ( 1 + ( r / 2 ) / 100 ) 2 n amount = rs . [ 25000 * ( 1 + 2 / 100 ) 4 ] = rs . ( 25000 * 51 / 50 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 27060.804 therefore , c . i . = rs . ( 27060.804 – 25000 ) = rs 2060.804 answer : c" | a = 2 / 4
b = a / 100
c = 1 + b
d = c ** 3
e = 4 * 4
f = e * 100
g = math.sqrt(100)
h = f * g
i = d * h
j = 4 * 4
k = j * 100
l = math.sqrt(100)
m = k * l
n = i - m
|
a ) 10 kmph , b ) 18 kmph , c ) 20 kmph , d ) 26 kmph , e ) 28 kmph | b | multiply(divide(900, 180), const_3_6) | ravi covers a distance of 900 mtrs in 180 secs . find his speed in kmph ? | given distance d = 900 meters and time = 180 seconds speed = distance / time = 900 / 180 = 5 m / s . but speed is asked in kmph , to convert 5 mps to kmph , multiply it by 18 / 5 . speed in kmph = 5 x 18 / 5 = 18 kmph b | a = 900 / 180
b = a * const_3_6
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a ) 2 / 3 , b ) - 1 , c ) 0 , d ) - 2 / 3 , e ) 2 | a | divide(subtract(2, sqrt(subtract(power(2, 3), multiply(5, 5)))), 3) | find the value of x from the below equation ? : 3 x ^ 2 - 5 x + 2 = 0 | "a = 3 , b = - 5 , c = 2 x 1,2 = ( 5 â ± â ˆ š ( ( - 5 ) ^ 2 - 4 ã — 3 ã — 2 ) ) / ( 2 ã — 3 ) = ( 5 â ± â ˆ š ( 25 - 24 ) ) / 6 = ( 5 â ± 1 ) / 6 x 1 = ( 5 + 1 ) / 6 = 6 / 6 = 1 x 2 = ( 5 - 1 ) / 6 = 4 / 6 = 2 / 3 a" | a = 2 ** 3
b = 5 * 5
c = a - b
d = math.sqrt(c)
e = 2 - d
f = e / 3
|
a ) 4 , b ) 8 , c ) 2 , d ) 6 , e ) 10 | b | divide(power(8, 2), 8) | n ^ ( n / 2 ) = 4 is true when n = 4 in the same way what is the value of n if n ^ ( n / 2 ) = 8 ? | n ^ ( n / 2 ) = 8 apply log n / 2 logn = log 8 nlogn = 2 log 8 = log 8 ^ 2 = log 64 logn = log 64 now apply antilog n = 64 / n now n = 8 . answer : b | a = 8 ** 2
b = a / 8
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a ) 3 / 80 , b ) 3 / 5 , c ) 12 , d ) 5 / 3 , e ) 80 / 3 | c | divide(log(8), log(power(2, 0.25))) | if n = 2 ^ 0.25 and n ^ b = 8 , b must equal | "25 / 100 = 1 / 4 n = 2 ^ 1 / 4 n ^ b = 2 ^ 3 ( 2 ^ 1 / 4 ) ^ b = 2 ^ 3 b = 12 answer : c" | a = math.log(8)
b = 2 ** 0
c = math.log(b)
d = a / c
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a ) 0.045388 , b ) 4.5388 , c ) 453.88 , d ) 473.88 , e ) 0.04863 | e | divide(divide(multiply(3.242, 15), 100), const_10) | [ ( 3.242 x 15 ) / 100 ] = ? | "answer multiplying 3.242 x 15 = 4.863 now divide 4.863 by 100 so , 4.863 ÷ 100 = 0.04863 ∴ shift the decimal two places to the left as 100 correct option : e" | a = 3 * 242
b = a / 100
c = b / 10
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a ) 50540 ways , b ) 60540 ways , c ) 50840 ways , d ) 40540 ways , e ) 51540 ways | a | add(multiply(divide(divide(factorial(subtract(41, floor(divide(41, const_2)))), factorial(subtract(4, const_1))), factorial(subtract(subtract(41, floor(divide(41, const_2))), subtract(4, const_1)))), floor(divide(41, const_2))), multiply(divide(divide(factorial(floor(divide(41, const_2))), factorial(subtract(floor(divide(41, const_2)), subtract(4, const_1)))), factorial(subtract(4, const_1))), subtract(41, floor(divide(41, const_2))))) | there are 41 students in a class , number of girls is one more than number of guys . we need to form a team of 4 students . all 4 in the team can not be from same gender . number of girls and guys in the team should not be equal . how many ways can such a team be made ? | boys = 20 and girls = 21 now the combinations are { girl , girl , girl , boy } or { boy , boy , boy , girl } so 21 c 3 * 20 c 1 + 20 c 3 * 21 c 1 = 50540 ways answer : a | a = 41 / 2
b = math.floor(a)
c = 41 - b
d = math.factorial(c)
e = 4 - 1
f = math.factorial(e)
g = d / f
h = 41 / 2
i = math.floor(h)
j = 41 - i
k = 4 - 1
l = j - k
m = math.factorial(l)
n = g / m
o = 41 / 2
p = math.floor(o)
q = n * p
r = 41 / 2
s = math.floor(r)
t = math.factorial(s)
u = 41 / 2
v = math.floor(u)
w = 4 - 1
x = v - w
y = math.factorial(x)
z = t / y
A = 4 - 1
B = math.factorial(A)
C = z / B
D = 41 / 2
E = math.floor(D)
F = 41 - E
G = C * F
H = q + G
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a ) 26 , b ) 22 , c ) 25 , d ) 27 , e ) 29 | e | add(add(multiply(2, 11), 5), 2) | find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 | "( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = 2 ^ 11 x 2 ^ 11 x 7 ^ 5 x 11 ^ 2 = 2 ^ 22 x 7 ^ 5 x 11 ^ 2 total number of prime factors = ( 22 + 5 + 2 ) = 29 . answer is e ." | a = 2 * 11
b = a + 5
c = b + 2
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a ) 9 / 29 , b ) 8 / 23 , c ) 3 / 8 , d ) 3 / 11 , e ) 3 / 4 | d | divide(multiply(3, 3), add(multiply(multiply(4, 3), const_2), multiply(3, 3))) | pipe p can drain the liquid from a tank in 3 / 4 the time that it takes pipe q to drain it and in 3 / 3 the time that it takes pipe r to do it . if all 3 pipes operating simultaneously but independently are used to drain liquid from the tank , then pipe q drains what portion of the liquid from the tank ? | suppose q can drain in 1 hr . so , rq = 1 / 1 = 1 so , rp = 1 / [ ( 3 / 4 ) rq ] = 4 / 3 also , rp = rr / ( 3 / 3 ) = > 4 / 3 = rr / ( 3 / 3 ) = > rr = 4 / 3 let h is the time it takes to drain by running all 3 pipes simultaneously so combined rate = rc = 1 / h = 1 + 4 / 3 + 4 / 3 = 11 / 3 = 1 / ( 3 / 11 ) thus running simultaneously , pipe q will drain 3 / 11 of the liquid . thus answer = d . | a = 3 * 3
b = 4 * 3
c = b * 2
d = 3 * 3
e = c + d
f = a / e
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a ) 40 , b ) 48 , c ) 44 , d ) 36 , e ) 30 | b | add(add(add(12, 12), 12), 12) | xavier starts from p towards q at a speed of 40 kmph and after every 12 mins increases his speed by 20 kmph . if the distance between p and q is 56 km , then how much time does he take to cover the distance ? | "first 12 min = 40 * 12 / 60 = 8 km 2 nd 12 min = 60 * 12 / 60 = 12 km 3 rd 12 min = 80 * 12 / 60 = 16 km 4 th 12 min = 100 * 12 / 60 = 20 km total time 12.4 = 48 min b" | a = 12 + 12
b = a + 12
c = b + 12
|
a ) 13 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | d | subtract(add(const_10, divide(200, multiply(6, 4))), const_1) | in the next cricket world cup t - 20 , team w has decided to score 200 runs only through 4 s and 6 s . in how many ways can the team w score these 200 runs ? | team w can score a maximum of 50 fours and a minimum of 2 fours with an interval or spacing of 3 units to accommodate the 6 ' s . so the number of fours scored forms an ap 2 , 5 , 8 , . . . 50 with a common difference of 3 . number of ways of scoring 200 only through 4 ' s and 6 ' s = ( 50 - 2 ) / 3 + 1 = 17 answer : d alternate solution : 4 x + 6 y = 200 - - > 2 x + 3 y = 100 x = ( 100 - 3 y ) / 2 - - > 100 - 3 y should be even . this is possible when 3 y is even . there are 17 even multiples of 3 between 0 and 100 . answer : d | a = 6 * 4
b = 200 / a
c = 10 + b
d = c - 1
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a ) 30 , b ) 10 , c ) 18 , d ) 28 , e ) 32 | d | subtract(subtract(add(50, 12), 22), 12) | in a class of 50 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ? | "total = english + german - both + neither - - > 50 = english + 22 - 12 + 0 - - > english = 40 - - > only english = english - both = 40 - 12 = 28 . answer : d ." | a = 50 + 12
b = a - 22
c = b - 12
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a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 82 kmph | d | divide(divide(subtract(130, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | a train 130 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is | "speed of train relative to man : 130 / 6 * 18 / 5 km / hr = 78 km / hr let speed of train = x therefore x + 6 = 78 x = 78 - 6 x = 72 km / hr answer : d" | a = 6 * const_0_2778
b = a * 6
c = 130 - b
d = c / 6
e = d / const_0_2778
|
a ) 48 , b ) 54 , c ) 60 , d ) 66 , e ) 72 | c | subtract(add(divide(const_1, 30), divide(const_1, 20)), divide(const_1, 15)) | two pipes can fill a tank in 30 minutes and 20 minutes . an outlet pipe can empty the tank in 15 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 30 + v / 20 - v / 15 = v / 60 per minute the tank will be filled in 60 minutes . the answer is c ." | a = 1 / 30
b = 1 / 20
c = a + b
d = 1 / 15
e = c - d
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a ) 15 minutes , b ) 16 minutes , c ) 17 minutes , d ) 18 minutes , e ) none of these | b | multiply(4, subtract(divide(36, const_2), divide(subtract(multiply(divide(36, const_2), 6), 24), 6))) | a barrel full of beer has 2 taps one midway , , which draw a litre in 6 minutes and the other at the bottom , which draws a litre in 4 minutes . the lower tap is lower normally used after the level of beer in the barrel is lower than midway . the capacity of the barrel is 36 litres . a new assistant opens the lower tap when the barrel is full and draws out some beer . as a result the lower tap has been used 24 minutes before the usual time . for how long was the beer drawn out by the new assistant ? | sol . the top tab is operational till 18 litres is drawn out . ∴ time after which the lower tap is usually open = 18 × 6 = 108 minutes ∴ time after which it is open now = 108 – 24 = 84 minutes ∴ litres drawn = 84 / 6 = 14 litres ∴ 18 – 14 = 4 litres were drawn by the new assistant . ∴ time = 4 × 4 = 16 minutes answer b | a = 36 / 2
b = 36 / 2
c = b * 6
d = c - 24
e = d / 6
f = a - e
g = 4 * f
|
a ) 1296 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | a | power(6, multiply(const_4, 1)) | xy = 1 then what is ( 6 ^ ( x + y ) ^ 2 ) / ( 6 ^ ( x - y ) ^ 2 ) | "( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 4 6 ^ 4 = 1296 answer a" | a = 4 * 1
b = 6 ** a
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a ) 3.6 , b ) 6 , c ) 4 , d ) can not be determined , e ) none of these | c | sqrt(divide(multiply(192, const_100), 1200)) | reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 192 as interest at the end of the loan period , what was the rate of interest ? | "let rate = r % and time = r years . then , 1200 x r x r / 100 = 192 12 r 2 = 192 r 2 = 16 r = 4 . answer : c" | a = 192 * 100
b = a / 1200
c = math.sqrt(b)
|
a ) 5 , b ) 14 , c ) 25 , d ) 21 , e ) none of these | c | sqrt(power(25, 2)) | √ ( 25 ) ^ 2 | "explanation √ ( 25 ) ^ 2 = ? or , ? = 25 answer c" | a = 25 ** 2
b = math.sqrt(a)
|
a ) 5 , b ) 3 , c ) 4 , d ) 7 , e ) 8 | a | divide(subtract(const_1, multiply(12, divide(const_1, 20))), divide(const_1, 15)) | x can finish a work in 15 days . y can finish the same work in 20 days . y worked for 12 days and left the job . how many days does x alone need to finish the remaining work ? | "work done by x in 1 day = 1 / 15 work done by y in 1 day = 1 / 20 work done by y in 12 days = 12 / 20 = 3 / 5 remaining work = 1 – 3 / 5 = 2 / 5 number of days in which x can finish the remaining work = ( 1 / 3 ) / ( 1 / 15 ) = 5 a" | a = 1 / 20
b = 12 * a
c = 1 - b
d = 1 / 15
e = c / d
|
a ) 5.6 , b ) 9.5 , c ) 9.1 , d ) 9.8 , e ) 5.2 | b | add(18, const_1) | the average of first 18 natural numbers is ? | "sum of 18 natural no . = 342 / 2 = 171 average = 171 / 18 = 9.5 answer : b" | a = 18 + 1
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a ) - 9 , b ) 9 , c ) - 10 , d ) 10 , e ) 20 | d | add(9, const_1) | what is the sum of 20 consecutive integers from - 9 inclusive , in a increasing order ? | "from - 9 to - 1 - - > 9 nos . zero - - > 1 number from + 1 to + 9 - - > 9 nos . when we add up nos . from - 9 to + 9 sum will be zero . total 19 nos will be added . 20 th number will be 10 . sum of these 20 nos . = 10 . d is the answer ." | a = 9 + 1
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a ) 30 , b ) 28 , c ) 27 , d ) 26 , e ) 40 | e | add(multiply(const_4, 2.9421), divide(log(const_100), log(const_10))) | if log 1087.5 = 2.9421 , then the number of digits in ( 875 ) 10 is ? | "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 2087.5 + 1 ) = 10 ( 2.9421 + 1 ) = 10 ( 3.9421 ) = 39.421 x = antilog ( 39.421 ) therefore , number of digits in x = 40 . answer : e" | a = 4 * 2
b = math.log(100)
c = math.log(10)
d = b / c
e = a + d
|
a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) 100 | b | multiply(divide(subtract(divide(12, multiply(subtract(15, 12), divide(40, const_60))), multiply(subtract(15, 12), divide(40, const_60))), subtract(15, 12)), const_60) | annie and sam set out together on bicycles traveling at 15 and 12 km per hour respectively . after 40 minutes , annie stops to fix a flat tire . if it takes annie 20 minutes to fix the flat tire and sam continues to ride during this time , how many minutes will it take annie to catch up with sam assuming that annie resumes riding at 15 km per hour ? | "annie gains 3 km per hour ( or 1 km every 20 minutes ) on sam . after 40 minutes annie is 2 km ahead . sam rides 1 km every 5 minutes . in the next 20 minutes , sam rides 4 km so sam will be 2 km ahead . it will take annie 40 minutes to catch sam . the answer is b ." | a = 15 - 12
b = 40 / const_60
c = a * b
d = 12 / c
e = 15 - 12
f = 40 / const_60
g = e * f
h = d - g
i = 15 - 12
j = h / i
k = j * const_60
|
a ) $ 12700 , b ) $ 13000 , c ) $ 23000 , d ) $ 11500 , e ) $ 33000 | d | add(10000, multiply(10000, divide(15, 100))) | last year the range of the annual yield on equal amount of investments in 100 mutual funds was $ 10000 . if the annual yield this year for each of the 100 mutual funds has improved by 15 percent this year than it was last year , what is the range of the annual yield on mutual funds this year ? | let the lowest yield be x . therefore , highest yield is x + 10000 . now yield of each mutual fund investment is improved by 10 % . therefore the yields will remain arranged in the same order as before . or lowest yield = 1.15 x and highest = 1.15 * ( x + 10000 ) or range = highest - lowest = 1.15 * ( x + 10000 ) - 1.15 x = 11500 , hence , d | a = 15 / 100
b = 10000 * a
c = 10000 + b
|
a ) $ 2.00 , b ) $ 2.15 , c ) $ 2.30 , d ) $ 2.40 , e ) $ 2.65 | d | divide(1.80, add(multiply(const_0_25, const_2), multiply(const_0_33, const_1))) | a customer purchased a package of ground beef at a cost of $ 1.80 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 25 percent less than the package of ground beef . what was the cost per pound of the steak ? | "for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.80 . let x be the price per pound for the steak . then 0.75 x = 180 x = 180 / 0.75 = $ 2.40 the answer is d ." | a = const_0_25 * 2
b = const_0_33 * 1
c = a + b
d = 1 / 80
|
a ) 40 , b ) 60 , c ) 150 , d ) 80 , e ) 100 | c | multiply(100, sqrt(divide(9, 4))) | two trains a and b starting from two points and travelling in opposite directions , reach their destinations 9 hours and 4 hours respectively after meeting each other . if the train a travels at 100 kmph , find the rate at which the train b runs . | "if two objects a and b start simultaneously from opposite points and , after meeting , reach their destinations in ‘ a ’ and ‘ b ’ hours respectively ( i . e . a takes ‘ a hrs ’ to travel from the meeting point to his destination and b takes ‘ b hrs ’ to travel from the meeting point to his destination ) , then the ratio of their speeds is given by : sa / sb = √ ( b / a ) i . e . ratio of speeds is given by the square root of the inverse ratio of time taken . sa / sb = √ ( 4 / 9 ) = 2 / 3 this gives us that the ratio of the speed of a : speed of b as 2 : 3 . since speed of a is 100 kmph , speed of b must be 100 * ( 3 / 2 ) = 150 kmph answer c" | a = 9 / 4
b = math.sqrt(a)
c = 100 * b
|
a ) 15360 , b ) 16010 , c ) 18180 , d ) 14930 , e ) 16075 | c | subtract(add(6000, 15000), add(multiply(6000, divide(12, const_100)), multiply(divide(14, const_100), 15000))) | a soft drink company had 6000 small and 15000 big bottles in storage . if 12 % of small 14 % of big bottles have been sold , then the total bottles remaining in storage is | "6000 + 15000 - ( 0.12 * 6000 + 0.14 * 15000 ) = 18180 . answer : c ." | a = 6000 + 15000
b = 12 / 100
c = 6000 * b
d = 14 / 100
e = d * 15000
f = c + e
g = a - f
|
a ) 1 / 12 , b ) 7 / 15 , c ) 2 / 13 , d ) 2 / 15 , e ) 1 / 17 | b | divide(divide(factorial(subtract(10, 2)), multiply(factorial(subtract(subtract(10, 2), 3)), factorial(3))), divide(factorial(10), multiply(factorial(subtract(10, 3)), factorial(3)))) | in a box of 10 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | "first , there are 8 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 10 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 8 c 3 / 10 c 3 = 56 / 120 = 7 / 15 answer is b" | a = 10 - 2
b = math.factorial(a)
c = 10 - 2
d = c - 3
e = math.factorial(d)
f = math.factorial(3)
g = e * f
h = b / g
i = math.factorial(10)
j = 10 - 3
k = math.factorial(j)
l = math.factorial(3)
m = k * l
n = i / m
o = h / n
|
a ) 300 , b ) 400 , c ) 500 , d ) 600 , e ) 700 | c | divide(subtract(multiply(3.25, multiply(const_100, const_1000)), multiply(3, multiply(const_100, const_1000))), 50) | company workers decided to raise rs . 3 lakhs by equal contribution from each . had they contributed rs . 50 each extra , the contribution would have been rs . 3.25 lakhs . how many workers were they ? | explanation : n * 50 = ( 325000 - 300000 ) = 25000 n = 25000 / 50 = 500 option c | a = 100 * 1000
b = 3 * 25
c = 100 * 1000
d = 3 * c
e = b - d
f = e / 50
|
a ) a ) 100 , b ) b ) 122 , c ) c ) 120 , d ) d ) 125 , e ) e ) 145 | a | multiply(20, 5) | the average of 7 numbers is 20 . if each number be multiplied by 5 . find the average of new set of numbers ? | "explanation : average of new numbers = 20 * 5 = 100 answer : option a" | a = 20 * 5
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['a ) 5', 'b ) 10', 'c ) 15', 'd ) 20', 'e ) 25'] | e | multiply(triangle_area(sqrt(5), sqrt(5)), 10) | right triangle pqr is the base of the prism in the figure above . if pq = pr = â ˆ š 5 and the height of the prism is 10 , what is the volume of the prism ? | volume of prism = area of base * height = 1 / 2 * ( square root of 5 ) * ( square root of 5 ) * 10 = 25 answer : e | a = math.sqrt(5)
b = math.sqrt(5)
c = triangle_area * (
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a ) 18695 , b ) 18325 , c ) 18365 , d ) 18395 , e ) 18485 | d | add(add(add(add(multiply(9, const_1000), multiply(add(3, 4), const_100)), multiply(add(4, const_1), const_10)), 3), add(add(add(multiply(multiply(4, const_2), const_1000), multiply(multiply(3, const_2), const_100)), multiply(4, const_10)), const_2)) | each of the positive integers a and c is a 4 - digit integer . if each of the digits 0 through 9 appears in one of these 3 integers , what is the maximum possible value of the sum of a and c ? | according to the stem we should use the digits 0 through 9 to construct 2 four - digit integers , so that their sum is as big as possible . to maximize the sum , maximize the thousands digits of a and c , so make them 9 and 8 . next , maximize hundreds digits . make them 7 and 6 . next maximize 10 ' s digit place by 5 and 4 use the remaining digits ( 3 , and 2 ) for units digits . so , a would be 9753 , and c would be 8642 . 9753 + 8642 = 18395 answer : d . | a = 9 * 1000
b = 3 + 4
c = b * 100
d = a + c
e = 4 + 1
f = e * 10
g = d + f
h = g + 3
i = 4 * 2
j = i * 1000
k = 3 * 2
l = k * 100
m = j + l
n = 4 * 10
o = m + n
p = o + 2
q = h + p
|
a ) 250 % , b ) 300 % , c ) 500 % , d ) 650 % , e ) 700 % | b | multiply(subtract(multiply(const_2, const_3), const_1), const_10) | the length of a rectangle is doubled while its width is doubled . what is the % change in area ? | "the original area is l * w the new area is 2 l * 2 w = 4 * l * w = l * w + 3 * l * w the area increased by 300 % . the answer is b ." | a = 2 * 3
b = a - 1
c = b * 10
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a ) 25 kms , b ) 11 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms | b | multiply(speed(add(18, 24), const_60), 15) | riya and priya set on a journey . riya moves eastward at a speed of 18 kmph and priya moves westward at a speed of 24 kmph . how far will be priya from riya after 15 minutes | total eastward distance = 18 kmph * 1 / 4 hr = 4.5 km total westward distance = 24 kmph * 1 / 4 hr = 6 km total distn betn them = 4.5 + 6 = 10.5 m ans 11 km answer : b | a = 18 + 24
b = speed * (
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a ) 12 , b ) 16 , c ) 13 , d ) 18 , e ) 19 | a | divide(multiply(multiply(36, 12), 5), multiply(30, 6)) | 36 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 36 : : 12 : x working hours / day 6 : 5 30 x 6 x x = 36 x 5 x 12 x = ( 36 x 5 x 12 ) / ( 30 x 6 ) x = 12 answer a" | a = 36 * 12
b = a * 5
c = 30 * 6
d = b / c
|
a ) 200 , b ) 278 , c ) 100 , d ) 202 , e ) 270 | c | multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 10)), 1000), multiply(divide(const_100, add(const_100, 10)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 10)), 1000), multiply(divide(const_100, add(const_100, 10)), 1000))), const_100) | a shopkeeper buys two articles for rs . 1000 each and then sells them , making 10 % profit on the first article and 10 % loss on second article . find the net profit or loss percent ? | profit on first article = 10 % of 1000 = 100 . this is equal to the loss he makes on the second article . that , is he makes neither profit nor loss . answer : c | a = 100 - 10
b = 100 / a
c = b * 1000
d = 100 + 10
e = 100 / d
f = e * 1000
g = c + f
h = 1000 + 1000
i = g - h
j = i * 100
k = 100 - 10
l = 100 / k
m = l * 1000
n = 100 + 10
o = 100 / n
p = o * 1000
q = m + p
r = j / q
s = r * 100
|
a ) 70 m , b ) 60 m , c ) 80 m , d ) 65 m , e ) 84 m | e | add(multiply(4, divide(divide(63, 4), subtract(4, const_1))), 63) | a can run 4 times as fast as b and gives b a start of 63 m . how long should the race course be so that a and b might reach in the same time ? | "speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 63 m he gains 63 * ( 4 / 3 ) i . e 84 m answer : e" | a = 63 / 4
b = 4 - 1
c = a / b
d = 4 * c
e = d + 63
|
a ) 2.5 % , b ) 15 % , c ) 25 % , d ) 35 % , e ) 250 % | e | multiply(divide(divide(25, const_100), divide(10, const_100)), const_100) | if c is 25 % of a and 10 % of b , what percent of a is b ? | "c is 25 % of a - - > c = a / 4 ; c is 10 % of b - - > c = b / 10 ; thus a / 4 = b / 10 - - > b = 5 / 2 * a = 2.5 a . therefore , b is 250 % of a . answer : e" | a = 25 / 100
b = 10 / 100
c = a / b
d = c * 100
|
a ) 205 , b ) 220 , c ) 240 , d ) 247 , e ) 294 | e | multiply(35, divide(multiply(divide(72, const_60), 35), subtract(40, 35))) | car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 40 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? | "car y began travelling after 72 minutes or 1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so , 35 ( t + 1.2 ) = 40 t t = 8.4 distance traveled by car x from the time car y began traveling until both cars stopped is 35 x 8.4 = 294 miles answer : - e" | a = 72 / const_60
b = a * 35
c = 40 - 35
d = b / c
e = 35 * d
|
a ) 4 : 5 , b ) 1 : 3 , c ) 2.5 : 1 , d ) 3.5 : 1 , e ) 3 : 2 | c | divide(2.5, divide(2.5, 2.5)) | what is the ratio between perimeters of two squares one having 2.5 times the diagonal then the other ? | "d = 2.5 d d = d a √ 2 = 2.5 d a √ 2 = d a = 2.5 d / √ 2 a = d / √ 2 = > 2.5 : 1 answer : c" | a = 2 / 5
b = 2 / 5
|
a ) 8 , b ) 5 , c ) 7 , d ) 6 , e ) 9 | b | subtract(63, multiply(29, const_2)) | a number when divided by 899 gives a remainder 63 . what remainder will be obtained by dividing the same number by 29 | 63 / 29 , thereforerequired number is : 5 , correct answer ( b ) | a = 29 * 2
b = 63 - a
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a ) 16 , b ) 32 , c ) 64 , d ) 96 , e ) none of these | d | divide(subtract(const_1, multiply(divide(add(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 25)), const_2), 4)), subtract(divide(add(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 25)), const_2), divide(const_1, 15))) | a and b can do a piece of work in 10 days , while b and c can do the same work in 15 days and c and a in 25 days . they started working together , after 4 days a left . after another 4 days b left . in how many days c can finish the remaining work ? | "let the rates of a , b and c be a , b , and c respectively . a and b can do a piece of work in 10 days : a + b = 1 / 10 ; b and c can do the same work in 15 days : b + c = 1 / 15 ; c and a can do the same work in 25 days : c + a = 1 / 25 . sum the above 3 equations : 2 ( a + b + c ) = 31 / 150 - - > a + b + c = 31 / 300 subtract a + b = 1 / 10 from above to get c = 1 / 300 . for 4 days all 3 worked and completed 4 * ( a + b + a ) = 124 / 300 of the work . for the next 4 days b and c worked and they completed 4 ( b + c ) = 4 / 15 = 80 / 300 of the work . so , by the time c is left alone 1 - ( 124 / 300 + 80 / 300 ) = 96 / 300 of the work is left to be completed by c alone . time = job / rate = ( 96 / 300 ) / ( 1 / 300 ) = 96 days . answer : d" | a = 1 / 10
b = 1 / 15
c = a + b
d = 1 / 25
e = c + d
f = e / 2
g = f * 4
h = 1 - g
i = 1 / 10
j = 1 / 15
k = i + j
l = 1 / 25
m = k + l
n = m / 2
o = 1 / 15
p = n - o
q = h / p
|
a ) 34 , b ) 87 , c ) 30 , d ) 99 , e ) 77 | a | add(floor(divide(multiply(multiply(21, 8), multiply(17, 3)), multiply(multiply(21, 2), 6))), const_1) | 17 men take 21 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ? | "3 w = 2 m 17 m - - - - - - 21 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 17 * 21 * 8 = 14 * x * 6 x = 34 answer : a" | a = 21 * 8
b = 17 * 3
c = a * b
d = 21 * 2
e = d * 6
f = c / e
g = math.floor(f)
h = g + 1
|
a ) 1 / 2 , b ) 2 , c ) 1 / 3 , d ) 3 , e ) 1 / 6 | a | sqrt(divide(9, 36)) | if xy = 9 , x / y = 36 , for positive numbers x and y , y = ? | very easy question . 2 variables and 2 easy equations . xy = 9 - - - > x = 9 / y - ( i ) x / y = 36 - - - > replacing ( i ) here - - - > 9 / ( y ^ 2 ) = 36 - - - > y ^ 2 = 9 / 36 = 1 / 4 - - - > y = 1 / 2 or - 1 / 2 the question states that x and y are positive integers . therefore , y = 1 / 2 is the answer . answer a . | a = 9 / 36
b = math.sqrt(a)
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a ) 10000 , b ) 8000 , c ) 1988 , d ) 1277 , e ) 2081 | a | divide(divide(9600, subtract(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))) | in one year , the population , of a village increased by 20 % and in the next year , it decreased by 20 % . if at the end of 2 nd year , the population was 9600 , what was it in the beginning ? | x * 120 / 100 * 80 / 100 = 9600 x * 0.96 = 9600 x = 9600 / 0.96 = > 10000 answer : a | a = 20 / 100
b = 1 - a
c = 9600 / b
d = 20 / 100
e = 1 + d
f = c / e
|
a ) 195,143 , b ) 185,133 , c ) 175,123 , d ) 165,113 , e ) none of them | a | multiply(15, 13) | two numbers are in the ratio of 15 : 11 . if their h . c . f . is 13 , find the numbers . | let the required numbers be 15 . x and llx . then , their h . c . f . is x . so , x = 13 . the numbers are ( 15 x 13 and 11 x 13 ) i . e . , 195 and 143 . answer is a . | a = 15 * 13
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a ) 2 / 15 , b ) 2 / 21 , c ) 7 / 40 , d ) 3 / 29 , e ) 4 / 27 | c | divide(choose(7, 2), choose(add(add(7, 5), 4), 2)) | a bag contains 7 red , 5 blue and 4 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 7 c 216 c 2 = 7 c 216 c 2 = 21 / 120 = 7 / 40 c" | a = math.comb(7, 2)
b = 7 + 5
c = b + 4
d = math.comb(c, 2)
e = a / d
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a ) 5 hours , b ) 4 hours , c ) 3 hours , d ) 2 hours , e ) 8 hours | e | divide(216, add(22, 5)) | a boat can travel with a speed of 22 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 216 km downstream | "explanation : speed of the boat in still water = 22 km / hr speed of the stream = 5 km / hr speed downstream = ( 22 + 5 ) = 27 km / hr distance travelled downstream = 216 km time taken = distance / speed = 216 / 27 = 8 hours . answer : option e" | a = 22 + 5
b = 216 / a
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a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3 | a | divide(subtract(add(const_100, 7), add(const_100, 1)), subtract(add(const_100, 1), subtract(const_100, 11))) | a certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio w of revenue from car sales in 1996 to revenue from truck sales in 1996 ? | a . . i have probably solved this question 3 - 4 times by now . . remember the answer . . 1 : 2 | a = 100 + 7
b = 100 + 1
c = a - b
d = 100 + 1
e = 100 - 11
f = d - e
g = c / f
|
a ) 18.6 , b ) 9.3 , c ) 1.2 , d ) 3.1 , e ) 1.6 | c | divide(divide(37.3, 3), multiply(5, 2)) | in 1979 approximately 1 / 3 of the 37.3 million airline passengers traveling to or from the united states used kennedy airport . if the number of such passengers that used miami airport was 1 / 2 the number that used kennedy airport and 5 times the number that used logan airport , approximately how many millions of these passengers used logan airport that year ? | "number of passengers using kennedy airport = 37 / 3 = ~ 12.43 passengers using miami airport = 12.43 / 2 = ~ 6.2 passengers using logan airport = 6.2 / 5 = ~ 1.24 so c" | a = 37 / 3
b = 5 * 2
c = a / b
|
a ) 15.0 , b ) 13.0 , c ) 13.9 , d ) 10.0 , e ) 6.0 | a | subtract(multiply(12.8, 5), multiply(6.8, 5)) | the average of 5 numbers is 6.8 . if one of the numbers is multiplied by a factor of 3 , the average of the numbers increases to 12.8 . what number is multiplied by 3 ? | "the average of 5 numbers is 6.8 the sum of 5 numbers will be 6.8 x 5 = 34 the average of 5 number after one of the number is multiplied by 3 is 12.8 the sum of the numbers will now be 12.8 x 5 = 64 so the sum has increased by 64 - 34 = 30 let the number multiplied by 3 be n then , 3 n = n + 30 or 2 n = 30 or n = 15 answer : - a" | a = 12 * 8
b = 6 * 8
c = a - b
|
a ) 3 kmph , b ) 6 kmph , c ) 6.25 kmph , d ) 7.5 kmph , e ) 7.8 kmph | a | divide(subtract(18, divide(18, 2)), add(1, 2)) | in covering a distance of 18 km , abhay takes 2 hours more than sameer . if abhay doubles his speed , then he would take 1 hour less than sameer . abhay ' s speed is : | let abhay ' s speed be x km / hr . then , 18 / x - 18 / 2 x = 3 6 x = 18 x = 3 km / hr . answer : option a | a = 18 / 2
b = 18 - a
c = 1 + 2
d = b / c
|
a ) 3 / 4 , b ) 1 / 4 , c ) 3 / 4 , d ) 2 / 5 , e ) 1 / 2 | e | divide(subtract(50, 55), subtract(50, 60)) | a portion of the 50 % solution of chemicals was replaced with an equal amount of 60 % solution of chemicals . as a result , 55 % solution of chemicals resulted . what part of the original solution was replaced ? | "this is a weighted average question . say x % of the solution was replaced - - > equate the amount of chemicals : 0.5 ( 1 - x ) + 0.6 * x = 0.55 - - > x = 1 / 2 . answer : e ." | a = 50 - 55
b = 50 - 60
c = a / b
|
a ) 475 , b ) 600 , c ) 550 , d ) 500 , e ) 525 | d | add(450, multiply(add(const_3, const_2), const_10)) | ram get 450 marks in his exam which is 90 % of total marks . what is the total marks ? | x * ( 90 / 100 ) = 450 x = 5 * 100 x = 500 answer : d | a = 3 + 2
b = a * 10
c = 450 + b
|
a ) 54 , b ) 8 , c ) 16 , d ) 24 , e ) 34 | a | divide(multiply(power(subtract(power(125, divide(const_1, const_3)), const_2), const_2), const_60), const_10) | all the faces of a cube are painted with blue colour . then it is cut into 125 small equal cubes . how many small cubes will be formed having only one face coloured ? | no . of small cubes will have only one face painted = ( x - 2 ) 2 * 6 here x = side of the small cube = 5 therfore ( x - 2 ) 2 * 6 = 54 answer : a | a = 1 / 3
b = 125 ** a
c = b - 2
d = c ** 2
e = d * const_60
f = e / 10
|
a ) 54 , b ) 25 , c ) 45 , d ) 35 , e ) 52 | d | multiply(divide(7, 8), 40) | in a certain country 1 / 7 of 8 = 5 . assuming the same proportion , what would be the value of 1 / 5 of 40 ? | "d 35" | a = 7 / 8
b = a * 40
|
a ) 10 % , b ) 12 % , c ) 15 % , d ) 29 % , e ) 20 % | d | multiply(divide(subtract(divide(50, const_100), divide(30, const_100)), subtract(const_1, divide(30, const_100))), const_100) | mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 30 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | let me try a simpler one . lets assume that candidate got 30 % votes and total votes is 100 . candidate won = 30 remaining = 70 to get 50 % , candidate requires 20 votes from 100 which is 20 % and 20 votes from 70 . 20 / 70 = 2 / 7 = . 285 = 28.5 % which is approx 29 % . hence the answer is d . | a = 50 / 100
b = 30 / 100
c = a - b
d = 30 / 100
e = 1 - d
f = c / e
g = f * 100
|
a ) 4 . , b ) 6 . , c ) 7 . , d ) 8 . , e ) 10 | e | add(multiply(18, divide(const_1, const_2)), const_1) | in the junior basketball league there are 18 teams , 2 / 3 of them are bad and ½ are rich . what ca n ' t be the number of teams that are rich and bad ? | otal teams = 18 bad teams = ( 2 / 3 ) * 18 = 12 rich teams = 9 so maximum value that the both rich and bad can take will be 9 . so e = 10 can not be that value . answer : e | a = 1 / 2
b = 18 * a
c = b + 1
|
a ) 1.5 , b ) 2 , c ) 2.4 , d ) 3 , e ) 3.6 | e | inverse(add(inverse(6), inverse(multiply(divide(const_3, const_2), 6)))) | working alone , john finishes cleaning half the house in a third of the time it takes nick to clean the entire house alone . john alone cleans the entire house in 6 hours . how many hours will it take nick and john to clean the entire house if they work together ? | answer is 3.6 hours . john does the complete house in 6 hours while nick does it in 9 hours . 1 / ( 1 / 6 + 1 / 9 ) = 3.6 answer is e | a = 1/(6)
b = 3 / 2
c = b * 6
d = 1/(c)
e = a + d
f = 1/(e)
|
a ) 18 π r 2 , b ) 12 π r 2 , c ) 3 π r 2 , d ) 6 π r 2 , e ) 9 π r 2 | b | multiply(circumface(2), 3) | the radius of a cylinder is 2 r units and height is 3 r units . find the curved surface ? | "explanation : 2 * π * 2 r * 3 r = 12 π r 2 answer : b" | a = circumface * (
|
a ) 10 ^ 2 , b ) 10 ^ 3 , c ) 10 ^ 4 , d ) 10 ^ 5 , e ) 10 ^ 6 | a | power(const_10, subtract(7, 5)) | on the richter scale , which measures the total amount of energy released during an earthquake , a reading of x - 1 indicates one - tenth the released energy as is indicated by a reading of x . on that scale , the frequency corresponding to a reading of 7 is how many times as great as the frequency corresponding to a reading of 5 ? | "if richter scale reading goes from x - 1 to x it will be 10 if richter scale reading goes from 5 to 6 it will be 10 similarly if richter scale reading goes from 6 to 7 it will be 10 so it will from 5 to 7 i . e 6,7 = 10 * 10 = 10 ^ 2 answer is a" | a = 7 - 5
b = 10 ** a
|
a ) 238 , b ) 277 , c ) 278 , d ) 200 , e ) 300 | e | divide(subtract(multiply(344, 25), multiply(71, const_100)), subtract(25, 20)) | the total of 344 of 20 paise and 25 paise make a sum of rs . 71 . the no of 20 paise coins is | explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 344 - x ) . 0.20 * ( x ) + 0.25 ( 344 - x ) = 71 = > x = 300 . . answer : e ) 300 | a = 344 * 25
b = 71 * 100
c = a - b
d = 25 - 20
e = c / d
|
a ) s . 1090 , b ) s . 1160 , c ) s . 680 , d ) s . 520 , e ) s . 700 | c | divide(multiply(subtract(const_100, 15), 800), const_100) | a man buys a cycle for rs . 800 and sells it at a loss of 15 % . what is the selling price of the cycle ? | "s . p . = 85 % of rs . 800 = rs . 85 / 100 x 800 = rs . 680 answer : c" | a = 100 - 15
b = a * 800
c = b / 100
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | subtract(subtract(divide(divide(multiply(26, 8), const_10), const_2), const_0_25), const_0_25) | a math teacher has 26 cards , each of which is in the shape of a geometric figure . half of the cards are rectangles , and a third of the cards are rhombuses . if 8 cards are squares , what is the maximum possible number of cards that re circles . | "a square is a special kind of rhombus ( sides are perpendicular ) a square is a special kind of rectangles ( sides with same length ) among the 26 cards with have : 15 rectangles 10 rhombus 8 squares among the 15 rectangles , there could be 8 special ones ( with sides of same length ) that are squares . that lets at least 7 rectangles that are not square . among the 10 rectangles , there could be 8 special ones ( with sides perpendicular ) that are squares . that lets at least 2 rhombus that are not square . we have 8 squares . so the minimum different cards that represent a square , a rhombus or a rectangle is 2 + 7 + 8 = 17 which means that the maximum number of circles that you could have is 26 - 17 = 9 answer ( a )" | a = 26 * 8
b = a / 10
c = b / 2
d = c - const_0_25
e = d - const_0_25
|
a ) 8 , b ) 5 , c ) 11 , d ) 3 , e ) 1.5 | e | divide(sqrt(multiply(add(6, 2), add(15, 3))), add(7, 1)) | if a * b * c = ( √ ( a + 2 ) ( b + 3 ) ) / ( c + 1 ) , find the value of 6 * 15 * 7 . | "6 * 15 * 3 = ( √ ( 6 + 2 ) ( 15 + 3 ) ) / ( 7 + 1 ) = ( √ 8 * 18 ) / 8 = ( √ 144 ) / 8 = 12 / 8 = 1.5 answer is e" | a = 6 + 2
b = 15 + 3
c = a * b
d = math.sqrt(c)
e = 7 + 1
f = d / e
|
a ) z / 2 , b ) 2 z , c ) z / 3 , d ) 3 z / 5 , e ) z / 9 | a | divide(subtract(divide(multiply(3, const_100), const_2), const_2), add(divide(multiply(3, const_100), const_2), const_2)) | if 3 x = 6 y = z , what is x + y , in terms of z ? | "3 x = 6 y = z x = z / 3 and y = z / 6 x + y = z / 3 + z / 6 = z / 2 answer is a" | a = 3 * 100
b = a / 2
c = b - 2
d = 3 * 100
e = d / 2
f = e + 2
g = c / f
|
a ) 12028 , b ) 12000 , c ) 20000 , d ) 12197 , e ) 12012 | c | divide(600, divide(5, const_100)) | a salesman ’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1400 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ? | "[ 1400 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 20000 answer : c" | a = 5 / 100
b = 600 / a
|
a ) 4955 , b ) 4899 , c ) 4650 , d ) 7200 , e ) none of these | a | divide(multiply(multiply(9, 5), 18.5), divide(divide(multiply(multiply(21, 10), 8), const_100), const_100)) | how many bricks each measuring 21 cm x 10 cm x 8 cm , will be needed to build a wall 9 m x 5 m x 18.5 m | "explanation : no . of bricks = volume of the wall / volume of 1 brick = ( 900 x 500 x 18.5 ) / ( 21 x 10 x 8 ) = 4955 answer : a" | a = 9 * 5
b = a * 18
c = 21 * 10
d = c * 8
e = d / 100
f = e / 100
g = b / f
|
a ) 62.5 , b ) 62.8 , c ) 62.1 , d ) 62.9 , e ) 32.5 | a | divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 5)) | at what price must an article costing rs . 47.50 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "explanation : cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 95 / 100 ) = 59.375 mp = 62.5 answer : a" | a = 47 * 50
b = a / 100
c = 47 + 50
d = c * 100
e = 100 - 5
f = d / e
|
a ) 10 a , b ) 12 a , c ) 14 a , d ) 16 a , e ) 18 a | a | floor(divide(64, add(5, const_1))) | during a certain two - week period , 64 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 5 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedies , in terms of a , were rented during that two - week period ? | total movies = 100 . comedies = 64 . action + drama = 36 . since there were 5 times as many dramas as action movies , then action + 5 * action = 36 - - > action = a = 6 . comedies = 60 = 10 a . a | a = 5 + 1
b = 64 / a
c = math.floor(b)
|
a ) 388 , b ) 282 , c ) 378 , d ) 292 , e ) 281 | e | subtract(subtract(multiply(15, 24), multiply(5, 15)), const_4) | find the greatest 4 digit number which leaves respective remainders of 2 and 5 when divided by 15 and 24 . a . 9974 | explanation : since the difference between the divisors and the respective remainders is not constant , back substitution is the convenient method . none of the given numbers is satisfying the condition . answer : e | a = 15 * 24
b = 5 * 15
c = a - b
d = c - 4
|
a ) $ 45 , b ) $ 47 , c ) $ 49 , d ) $ 51 , e ) $ 53 | c | divide(35, subtract(1, multiply(divide(1, 7), const_2))) | p has $ 35 more than what q and r together would have had if both b and c had 1 / 7 of what p has . how much does p have ? | "p = ( 2 / 7 ) * p + 35 ( 5 / 7 ) * p = 35 p = 49 the answer is c ." | a = 1 / 7
b = a * 2
c = 1 - b
d = 35 / c
|
a ) 1 / 3 , b ) 2 / 3 , c ) 4 / 3 , d ) 5 / 3 , e ) 7 / 3 | b | subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 30)), 4)) | a can do a work in 20 days and b in 30 days . if they work on it together for 4 days , then the fraction of the work that is left is : | "ans is : b a ' s 1 day ' s work = 1 / 20 b ' s 1 day ' s work = 1 / 30 ( a + b ) ' s 1 day ' s work = ( 1 / 20 + 1 / 30 ) = 1 / 12 ( a + b ) ' s 4 day ' s work = ( 1 / 12 * 4 ) = 1 / 3 therefore , remaining work = ( 1 - 1 / 3 ) = 2 / 3" | a = 1 / 20
b = 1 / 30
c = a + b
d = c * 4
e = 1 - d
|
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