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 ) 112 , b ) 77 , c ) 267 , d ) 15 , e ) 99 | d | subtract(divide(714, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7))) | how many numbers between 100 and 714 are divisible by 2 , 3 , and 7 together ? | "explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 714 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 714 : 714 / 42 = 17 hence , 42 * 14 = 714 is the last number divisible by 42 within 714 hence , total numbers divisible by 2 , 3 , 7 together are ( 17 â € “ 2 ) = 15 answer : d" | a = 2 * 3
b = a * 7
c = 714 / b
d = 2 * 3
e = d * 7
f = 100 / e
g = c - f
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a ) 99 , b ) 140 , c ) 26 , d ) 17 , e ) 12 | b | divide(rectangle_area(60, 14), rectangle_area(3, 2)) | how many paying stones , each measuring 3 m * 2 m are required to pave a rectangular court yard 60 m long and 14 m board ? | "60 * 14 = 3 * 2 * x = > x = 140 answer : b" | a = rectangle_area / (
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a ) 301 times , b ) 351 times , c ) 401 times , d ) 451 times , e ) 481 times | a | divide(multiply(divide(const_3600, const_4), const_3), 6) | if a light flashes every 6 seconds , how many times will it flash in 1 / 2 of an hour ? | "in 1 / 2 of an hour there are 30 * 60 = 1800 seconds the number of 6 - second intervals = 1800 / 6 = 300 after the first flash , there will be 300 more flashes for a total of 301 . the answer is a ." | a = 3600 / 4
b = a * 3
c = b / 6
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a ) 24.3 , b ) 25.3 , c ) 26.3 , d ) 28.3 , e ) 34.5 | a | divide(add(multiply(40, 30), multiply(13, 10)), add(30, 10)) | the average runs scored by a batsman in 30 matches is 40 . in the next 10 matches the batsman scored an average of 13 runs . find his average in all the 30 matches ? | "total score of the batsman in 30 matches = 600 . total score of the batsman in the next 10 matches = 130 . total score of the batsman in the 30 matches = 730 . average score of the batsman = 730 / 30 = 24.3 . answer : a" | a = 40 * 30
b = 13 * 10
c = a + b
d = 30 + 10
e = c / d
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | subtract(multiply(divide(36, add(const_1, divide(80, const_100))), const_2), 36) | a retailer bought a shirt at wholesale and marked it up 80 % to its initial price of $ 36 . by how many more dollars does he need to increase the price to achieve a 100 % markup ? | let x be the wholesale price . then 1.8 x = 36 and x = 36 / 1.8 = 20 . to achieve a 100 % markup , the price needs to be $ 40 . the retailer needs to increase the price by $ 4 more . the answer is b . | a = 80 / 100
b = 1 + a
c = 36 / b
d = c * 2
e = d - 36
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a ) 2 , b ) 4 , c ) 8 , d ) 7 , e ) 32 | d | divide(subtract(subtract(add(add(30, 40), 16), 12), 60), const_2) | in the city of san durango , 60 people own cats , dogs , or rabbits . if 30 people owned cats , 40 owned dogs , 16 owned rabbits , and 12 owned exactly two of the 3 types of pet , how many people owned all 3 ? | you are considering a case when cat dogs and rabbit are not exactly equal to 12 . the solution shall be 60 = 30 + 40 + 16 - 12 - 2 x and hence x shall be 7 . answer d | a = 30 + 40
b = a + 16
c = b - 12
d = c - 60
e = d / 2
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a ) 5 % , b ) 18 % , c ) 33 % , d ) 35 % , e ) 46.25 % | e | multiply(subtract(multiply(add(divide(25, const_100), const_1), add(const_1, divide(17, const_100))), const_1), const_100) | a certain company ’ s profit in 1996 was 17 percent greater than its profit in 1995 , and its profit in 1997 was 25 percent greater than its profit in 1996 . the company ’ s profit in 1997 was what percent greater than its profit in 1995 ? | "profit in 1995 - 100 profit in 1996 - 117 % increae profit in 1997 in comparison to 1995 = 17 + 117 * 25 % = 46.25 correct option : e" | a = 25 / 100
b = a + 1
c = 17 / 100
d = 1 + c
e = b * d
f = e - 1
g = f * 100
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a ) 130.12 , b ) 130.43 , c ) 130.22 , d ) 111.0 , e ) 111.12 | b | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 200)), subtract(multiply(const_100, const_10), 80))) | a can give b 80 meters start and c 200 meters start in a kilometer race . how much start can b give c in a kilometer race ? | a runs 1000 m while b runs 920 m and c runs 800 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 800 ) / 920 = 869.56 m b can give c = 1000 - 869.56 = 130.43 m . answer : b | a = 100 * 10
b = 100 * 10
c = 100 * 10
d = c - 200
e = b * d
f = 100 * 10
g = f - 80
h = e / g
i = a - h
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a ) 140 cm , b ) 767 cm , c ) 88 cm , d ) 666 cm , e ) 776 cm | a | multiply(sqrt(divide(58.8, 30)), const_100) | 30 square stone slabs of equal size were needed to cover a floor area of 58.8 sq . m . find the length of each stone slab ? | "area of each slab = 58.8 / 30 m 2 = 1.96 m 2 length of each slab √ 1.96 = 1.4 m = 140 cm" | a = 58 / 8
b = math.sqrt(a)
c = b * 100
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a ) 60 , b ) 70 , c ) 80 , d ) 50 , e ) 40 | a | subtract(divide(multiply(2, multiply(multiply(100, 25), 8)), multiply(25, 10)), 100) | nhai employs 100 men to build a highway of 2 km in 50 days working 8 hours a day . if in 25 days they completed 1 / 3 part of work . than how many more emploies should nhai hire to finish it in time working 10 hours a day | "( m 1 * d 1 * h 1 ) / w 1 = ( m 2 * d 2 * h 2 ) / w 2 ( 100 * 25 * 8 ) / ( 1 / 3 ) = ( ( 100 + a ) * 25 * 10 ) ) / ( 2 / 3 ) 160 = 100 + a a = 60 answer : a" | a = 100 * 25
b = a * 8
c = 2 * b
d = 25 * 10
e = c / d
f = e - 100
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a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | c | add(20, const_1) | the average of first four prime numbers greater than 20 is ? | "23 + 29 + 31 + 37 = 120 / 4 = 30 answer : c" | a = 20 + 1
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a ) 2 / 5 , b ) 1 / 4 , c ) 1 / 3 , d ) 1 / 2 , e ) 2 / 3 | d | add(divide(1, const_2.0), multiply(divide(1, 4), divide(1, 4))) | jar x is 1 / 4 full of water . jar y , which has half the capacity of jar x , is 1 / 2 full of water . if the water in jar y is poured into jar x , then jar x will be filled to what fraction of its capacity ? | "let p be the capacity of jar x . the amount of water in jar y is 1 / 2 * p / 2 = p / 4 then the total amount in jar x is p / 4 + p / 4 = p / 2 the answer is d ." | a = 1 / 2
b = 1 / 4
c = 1 / 4
d = b * c
e = a + d
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a ) 252 kms , b ) 152 kms , c ) 612 kms , d ) 752 kms , e ) 152 kms | c | multiply(add(multiply(2, 40), multiply(subtract(12, const_1), 2)), divide(12, 2)) | the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 40 kms . what was the total distance traveled in 12 hours ? | "explanation : total distance travelled in 12 hours = ( 40 + 42 + 44 + . . . . . upto 12 terms ) this is an a . p with first term , a = 40 , number of terms , n = 12 , d = 2 . required distance = 12 / 2 [ 2 x 40 + { 12 - 1 ) x 2 ] = 6 ( 102 ) = 612 kms . answer : c" | a = 2 * 40
b = 12 - 1
c = b * 2
d = a + c
e = 12 / 2
f = d * e
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a ) rs . 50000 , b ) rs . 52000 , c ) rs . 57000 , d ) rs . 60000 , e ) rs . 67000 | c | add(add(multiply(multiply(multiply(20, 15), const_2), 60), multiply(multiply(multiply(20, 5), const_2), 60)), multiply(multiply(multiply(15, 5), const_2), 60)) | in a house a hall is 20 m long , 15 m wide and 5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . 60 per square m ? | "length ( l ) = 20 m , breadth ( b ) = 15 m and height ( h ) = 5 m total area of the hall to be covered with mat = 2 ( lb + bh + hl ) = 2 ( 20 * 15 + 15 * 5 + 5 * 20 ) = 2 ( 300 + 75 + 100 ) = 2 * 475 = 950 sq . m total expenditure = 60 * 950 = rs . 57000 c" | a = 20 * 15
b = a * 2
c = b * 60
d = 20 * 5
e = d * 2
f = e * 60
g = c + f
h = 15 * 5
i = h * 2
j = i * 60
k = g + j
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a ) 6.8 , b ) 8.5 , c ) 6.1 , d ) 9.1 , e ) 7.14 | e | divide(divide(5000, const_2), multiply(subtract(50.4, 15.4), divide(5000, 500))) | a small island country of 5000 people has an emigration and immigration statistic per 500 persons of 50.4 and 15.4 respectively . how long will it take for the population to be half of its current number . | annual decrease is ( 500 + 15.4 - 50.4 ) * 10 = 4650 hence every year there is a decrease of 350 for the population to become half 2500 must emigrate therefore 2500 / 350 = 133.3 correct option is e ) 7.14 | a = 5000 / 2
b = 50 - 4
c = 5000 / 500
d = b * c
e = a / d
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a ) 58 kgs , b ) 58.85 kgs , c ) 58.9 kgs , d ) 59 kgs , e ) 59.85 kgs | c | divide(add(multiply(58.4, 20), subtract(66, 56)), 20) | the average weight of a class of 20 boys was calculated to be 58.4 kgs and it was later found that one weight was misread as 56 kg instead of 66 kg . what is the correct weight ? | "actual total weight is ( 20 x 58.4 - 56 + 66 ) = 1178 kgs actual average weight is 1178 / 20 = 58.9 kgs c" | a = 58 * 4
b = 66 - 56
c = a + b
d = c / 20
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a ) 67 % , b ) 80 % , c ) 90 % , d ) 100 % , e ) none of above | a | multiply(divide(subtract(add(const_100, 320), add(const_100, 40)), add(const_100, 320)), const_100) | in a certain store , the profit is 320 % of the cost . if the cost increases by 40 % but the selling price remains constant , approximately what percentage of the selling price is the profit | explanation : let c . p . = rs . 100 . then , profit = rs . 320 , s . p . = rs . 420 . new c . p . = 140 % of rs . 100 = rs . 140 new s . p . = rs . 420 . profit = rs . ( 420 - 140 ) = rs . 280 required percentage = ( 280 / 420 ) * 100 = 66.67 % ( approx ) = 67 % answer : a | a = 100 + 320
b = 100 + 40
c = a - b
d = 100 + 320
e = c / d
f = e * 100
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a ) 299 m , b ) 140 m , c ) 200 m , d ) 167 m , e ) 128 m | b | subtract(divide(480, const_2), 100) | if the perimeter of a rectangular garden is 480 m , its length when its breadth is 100 m is ? | "2 ( l + 100 ) = 480 = > l = 140 m answer : b" | a = 480 / 2
b = a - 100
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a ) $ 1,000 , b ) $ 1,200 , c ) $ 1,400 , d ) $ 1,600 , e ) $ 2,000 | d | subtract(1000, 800) | a family pays $ 800 per year for an insurance plan that pays 60 percent of the first $ 1000 in expenses and 100 percent of all medical expenses thereafter . in any given year , the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total how much ? | assuming the medical expenses are $ 1000 or more , the family pays $ 800 + $ 400 = $ 1200 . the total amount paid by insurance plan for the first $ 1000 of expenses is $ 600 . the insurance will pay another $ 600 when the medical expenses are $ 1600 . the answer is d . | a = 1000 - 800
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a ) 350 , b ) 250 , c ) 450 , d ) 500 , e ) 620 | c | multiply(30, 15) | the h . c . f . of two numbers is 30 and the other two factors of their l . c . m . are 10 and 15 . the larger of the two numbers is : | "explanation : clearly , the numbers are ( 30 x 10 ) and ( 30 x 15 ) . { \ color { blue } \ therefore } larger number = ( 30 x 15 ) = 450 . answer : c ) 450" | a = 30 * 15
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a ) 40 , b ) 44 , c ) 48 , d ) 55 , e ) none of these | d | add(add(multiply(5, divide(10, subtract(multiply(divide(6, 5), 5), 4))), divide(10, subtract(multiply(divide(6, 5), 5), 4))), 10) | in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 10 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 10 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 10 = 4 / 9 ( t - 10 ) 10 t = 990 - 440 = > t = 55 . answer : d" | a = 6 / 5
b = a * 5
c = b - 4
d = 10 / c
e = 5 * d
f = 6 / 5
g = f * 5
h = g - 4
i = 10 / h
j = e + i
k = j + 10
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a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | b | subtract(add(70, 50), subtract(const_100, 20)) | in a particular state , 70 % of the counties received some rain on monday , and 50 % of the counties received some rain on tuesday . no rain fell either day in 20 % of the counties in the state . what percent of the counties received some rain on monday and tuesday ? | "70 + 50 + 20 = 140 % the number is 40 % above 100 % because 40 % of the counties were counted twice . the answer is b ." | a = 70 + 50
b = 100 - 20
c = a - b
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a ) 40 , b ) 50 , c ) 90 , d ) 30 , e ) 20 | c | multiply(multiply(const_0_2778, 54), subtract(16, 10)) | a train passes a platform in 16 seconds . the same train passes a man standing on the platform in 10 seconds . if the speed of the train is 54 km / hr , the length of the platform is | "speed of the train = 54 km / hr = ( 54 × 10 ) / 36 m / s = 15 m / s length of the train = speed × time taken to cross the man = 15 × 10 = 150 m let the length of the platform = l time taken to cross the platform = ( 150 + l ) / 15 = > ( 150 + l ) / 15 = 16 = > 150 + l = 15 × 16 = 240 = > l = 240 - 150 = 90 meter answer is c" | a = const_0_2778 * 54
b = 16 - 10
c = a * b
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a ) 42 , b ) 70 , c ) 140 , d ) 180 , e ) 315 | d | multiply(multiply(10, 3), 4) | a certain university will select 1 of 4 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "1 of 7 will be chosen for the math 2 of 10 will be chosen for the computer none of the 3 chosen people can be in more thn one deparments . we can choose any of the 4 candidates for the math dep . , which gives as 4 selections . we can choose 2 of the 10 candidates for the computer dep . , which gives us 2 selections and 8 rejections . so , the way to find how many different selections of 2 candidates we can have for the computer dep . , we do : 10 ! / 2 ! * 8 ! = ( 9 * 10 ) / 2 = 90 / 2 = 45 . we are multiplying our individual selections : 4 * 45 = 180 in the bolded part , we do n ' t have to multiply all of the numbers , as those in 8 ! are included in 10 ! , so we simplify instead . ans d" | a = 10 * 3
b = a * 4
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a ) 20 , 5 , b ) 10 , 15 , c ) 15 , 10 , d ) 5 , 20 , e ) 12 , 13 | c | subtract(25, divide(subtract(25, divide(5, const_2)), const_2)) | sum of two numbers is 25 . first exceeds by 5 from the other . then the numbers will be ? | "explanation : x + y = 25 x - y = 5 x = 15 y = 10 answer : c" | a = 5 / 2
b = 25 - a
c = b / 2
d = 25 - c
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a ) 546 , b ) 550 , c ) 555 , d ) 532 , e ) 515 | a | multiply(divide(840, const_100), subtract(const_100, 35)) | if 35 % of the 840 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | "we know 35 % people study biology , therefore the no of people not studying = 100 - 35 = 65 % > therefore the people not studying biology out of a total 840 people are = 65 % of 840 > ( 65 / 100 ) * 840 = 546 people a" | a = 840 / 100
b = 100 - 35
c = a * b
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a ) 76 seconds , b ) 18 seconds , c ) 40 seconds , d ) 19 seconds , e ) 44 seconds | e | divide(add(410, 140), divide(multiply(45, const_1000), const_3600)) | a train is 410 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 410 + 140 = 550 meter time = distance / speed = 550 * ( 2 / 25 ) = 44 seconds answer : e" | a = 410 + 140
b = 45 * 1000
c = b / 3600
d = a / c
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a ) 36 , b ) 42 , c ) 28 , d ) 54 , e ) 40 | e | inverse(divide(const_3, multiply(30, const_4))) | if a is thrice as fast as b and together can do a work in 30 days . in how many days a alone can do the work ? | a ’ s one day ’ s work = 1 / x b ’ s one day ’ s work = 1 / 3 x a + b ’ s one day ’ s work = 1 / x + 1 / 3 x = 1 / 30 = 3 + 1 / 3 x = 4 / 3 x = 1 / 30 x = 30 * 4 / 3 = 40 answer : e | a = 30 * 4
b = 3 / a
c = 1/(b)
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a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 12 | a | multiply(divide(subtract(200, divide(300, const_2)), 50), const_2) | angelina walked 200 meters from her home to the grocery at a constant speed . she then walked 300 meters to the gym at double the speed . she spent 50 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 = 200 / x . . the speed to gym = 2 x . . so time taken = 300 / 2 x = 150 / x . . its given 200 / x - 150 / x = 50 . . 50 / x = 50 . . x = 1 m / secs . . so grocery to gym = 2 * 1 = 2 m / s . . . answer : a" | a = 300 / 2
b = 200 - a
c = b / 50
d = c * 2
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a ) 14 year , b ) 8 year , c ) 10 year , d ) none of these , e ) 9 year | a | divide(subtract(divide(40, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | the sum of the ages of 5 children born at the intervals of 3 year each is 40 year . what is the age of the eldest child ? | "solution let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) year . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 40 â ‡ ” 5 x = 10 â ‡ ” x = 2 . â ˆ ´ age of the youngest child = x + 12 = 2 + 12 = 14 years . answer a" | a = 5 / 3
b = 40 / a
c = 5 - 1
d = c * 3
e = b - d
f = e / 3
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a ) 4,514 , b ) 4,480 , c ) 4,521 , d ) 4,428 , e ) 4,349 | b | divide(factorial(subtract(add(const_4, 17), const_1)), multiply(factorial(17), factorial(subtract(const_4, const_1)))) | how many positive integers less than 5,000 are evenly divisible by neither 17 nor 21 ? | "integers less than 5000 divisible by 17 5000 / 17 = 333 . something , so 333 integers less than 5000 divisible by 21 5000 / 21 = 238 . # # , so 238 we have double counted some , so take lcm of 17 and 21 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 17 and 21 = 333 + 238 - 47 = 524 now subtract that from 4999 . 4999 - 524 = 4480 answer b ." | a = 4 + 17
b = a - 1
c = math.factorial(b)
d = math.factorial(17)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
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a ) 30 , b ) 20 , c ) 40 , d ) 60 , e ) 90 | e | divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 2), divide(2, 3))) | 15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 2 hrs . each would it take for 21 women if 3 women do as much work as 2 men ? | "let 1 man does 1 unit / hr of work 15 m in 21 days of 8 hrs will do ( 15 * 21 * 8 ) units 3 w = 2 m 1 w = ( 2 / 3 ) units / hr 21 w with 2 hrs a day will take ( 15 * 21 * 8 ) / ( 21 * 2 * ( 2 / 3 ) ) days = > 90 days answer : e" | a = 15 * 21
b = a * 8
c = 21 * 2
d = 2 / 3
e = c * d
f = b / e
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a ) 14 , b ) 28 , c ) 63 , d ) 196 , e ) 252 | e | add(56, const_1) | if x and y are positive integers and 42 x = 56 y what is the least possible value of xy ? | "42 x = 56 y = > x / y = 4 / 3 = > 3 x = 4 y 3 ( 14 ) = 4 ( 14 ) = > x * y = 196 e" | a = 56 + 1
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a ) 5 sec , b ) 6 sec , c ) 7 sec , d ) 8 sec , e ) 9 sec | b | divide(150, multiply(add(98, 8), const_0_2778)) | a train 150 m long is running with a speed of 98 kmph . in what time will it pass a man who is running at 8 kmph in the same direction in which the train is going | "explanation : speed of the train relative to man = ( 98 - 8 ) kmph = ( 90 × 5 / 18 ) m / sec = 25 m / sec time taken by the train to cross the man i = time taken by it to cover 150 m at 25 m / sec = 150 × 1 / 25 sec = 6 sec answer : option b" | a = 98 + 8
b = a * const_0_2778
c = 150 / b
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a ) 512 , b ) 508 , c ) 451 , d ) 516 , e ) 451 | b | divide(multiply(36, 56), const_4) | what is the sum of the greatest common factor and the lowest common multiple of 36 and 56 ? | "prime factorization of 36 = 2 x 2 x 3 x 3 prime factorization of 56 = 2 x 2 x 2 x 7 gcf = 4 lcm = 504 sum = 508 ans : b" | a = 36 * 56
b = a / 4
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a ) 1287 , b ) 2887 , c ) 2500 , d ) 1129 , e ) 1192 | c | divide(7225, add(add(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), subtract(const_1, divide(10, const_100))), const_1)) | vishal invested 10 % more than trishul . trishul invested 10 % less than raghu . if the total sum of their investments is rs . 7225 , how much amount did raghu invest ? | "let money invested by raghu = rs . x money invested by trishul = 9 / 10 x = 0.9 x money invested by vishal = 9 / 10 x * 110 / 100 = 0.99 x also , x + 0.9 x + 0.99 x = 7225 = x = 7225 / 2.89 = 2500 therefore , amount invested by raghu is rs . 2500 . answer : c" | a = 10 / 100
b = 1 - a
c = 10 / 100
d = 1 + c
e = b * d
f = 10 / 100
g = 1 - f
h = e + g
i = h + 1
j = 7225 / i
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['a ) 27 , 23', 'b ) 25 , 25', 'c ) 24 , 26', 'd ) 23 , 29', 'e ) 24 , 45'] | a | add(multiply(add(divide(subtract(divide(multiply(250, const_2), 10), 4), const_2), const_4), const_100), divide(subtract(divide(multiply(250, const_2), 10), 4), const_2)) | the difference between two parallel sides of a trapezium is 4 cm . perpendicular distance between them is 10 cm . if the area of the trapezium is 250 cm 2 find the lengths of the parallel side | explanation : let the two parallel sides of the trapezium be a cm and b cm . then , a - b = 4 - - - - - - ( 1 ) and , ( 1 / 2 ) x ( a + b ) x 10 = 475 = > ( a + b ) = ( ( 250 x 2 ) / 10 ) = > a + b = 50 - - - - - - - ( 2 ) solving 1 and 2 , we get : a = 27 , b = 23 so , the two parallel sides are 27 cm and 23 cm . answer : option a | a = 250 * 2
b = a / 10
c = b - 4
d = c / 2
e = d + 4
f = e * 100
g = 250 * 2
h = g / 10
i = h - 4
j = i / 2
k = f + j
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a ) 1 / 12 , b ) 5 / 14 , c ) 2 / 13 , d ) 2 / 15 , e ) 1 / 17 | b | divide(divide(factorial(subtract(8, 2)), multiply(factorial(subtract(subtract(8, 2), 3)), factorial(3))), divide(factorial(8), multiply(factorial(subtract(8, 3)), factorial(3)))) | in a box of 8 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 6 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 8 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 6 c 3 / 8 c 3 = 20 / 56 = 5 / 14 answer is b" | a = 8 - 2
b = math.factorial(a)
c = 8 - 2
d = c - 3
e = math.factorial(d)
f = math.factorial(3)
g = e * f
h = b / g
i = math.factorial(8)
j = 8 - 3
k = math.factorial(j)
l = math.factorial(3)
m = k * l
n = i / m
o = h / n
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a ) 22.48 % , b ) 20.48 % , c ) 24.48 % , d ) 25.48 % , e ) 26.48 % | b | add(20, divide(add(add(add(20, 20), 7), 1), const_100)) | a sum is divided between a and b in the ratio of 1 : 2 . a purchased a car from his part , which depreciates 14 2 / 7 % per annum and b deposited his amount in a bank , which pays him 20 % interest per annum compounded annually . by what percentage will the total sum of money increase after two years due to this investment pattern ( approximately ) . | let the total money = 300 a ' s part = 100 b ' s part = 200 after 2 years a ' s part = 73.46 ( as it depreciates by 14.29 % per annum ) after 2 years b ' s part = 288 ( using the compound interest formula , a = p ( 1 + r / n ) ^ nt ) , where p = 200 , r = 0.2 , n = 1 , t = 2 ) so , after 2 years total amount = 361.46 therefore % increase in money = ( 361.46 - 300 ) * 100 / 300 = 20.48 % answer : b | a = 20 + 20
b = a + 7
c = b + 1
d = c / 100
e = 20 + d
|
a ) 54 sec , b ) 35 sec , c ) 25 sec , d ) 45 sec , e ) 24 sec | b | divide(add(250, 100), multiply(36, const_0_2778)) | how many seconds will a train 100 meters long take to cross a bridge 250 meters long if the speed of the train is 36 kmph ? | "d = 100 + 250 = 350 s = 36 * 5 / 18 = 10 mps t = 350 / 10 = 35 sec answer : b" | a = 250 + 100
b = 36 * const_0_2778
c = a / b
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a ) 18 sec , b ) 70 sec , c ) 13 sec , d ) 20 sec , e ) 19 sec | c | divide(add(100, 160), multiply(add(42, 30), const_0_2778)) | two trains of length 100 m and 160 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 100 + 160 = 260 m . the time required = d / s = 260 / 20 = 13 sec . answer : c" | a = 100 + 160
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 4 mins , b ) 6 mins , c ) 5 mins , d ) 3 mins , e ) 7 mins | c | inverse(add(divide(const_1, 30), divide(5, 30))) | pipe a fills a tank in 30 minutes . pipe b can fill the same tank 5 times as fast as pipe a . if both the pipes were kept open when the tank is empty , how much time will it take for the tank to overflow ? | "let the total capacity of tank be 90 liters . capacity of tank filled in 1 minute by a = 3 liters . capacity of tank filled in 1 minute by b = 15 liters . therefore , capacity of the tank filled by both a and b in 1 minute = 18 liters . hence , time taken by both the pipes to overflow the tank = 90 / 18 = 5 minutes . answer : c" | a = 1 / 30
b = 5 / 30
c = a + b
d = 1/(c)
|
a ) 22 , b ) 23 , c ) 26 , d ) 29 , e ) 31 | c | subtract(multiply(add(15, 1), add(10, 1)), multiply(15, 10)) | the average age of a class of 15 students is 10 years . if the age of the teacher be included , then the average increased by 1 year . find the age of the teacher ? | "total age of 15 persons = 15 * 10 = 150 years average age of 16 persons = 11 years total age of 16 persons = 16 * 11 = 176 years age of the teacher = 176 - 150 = 26 years answer is c" | a = 15 + 1
b = 10 + 1
c = a * b
d = 15 * 10
e = c - d
|
a ) 84 , b ) 94 , c ) 83 , d ) 72 , e ) none | a | add(multiply(divide(add(12, 4), const_2), 12), subtract(12, divide(add(12, 4), const_2))) | the sum of digits of a two digit number is 12 , the difference between the digits is 4 . find the number | "description : = > x + y = 12 , x - y = 4 adding these 2 x = 16 = > x = 8 , y = 4 . thus the number is 84 answer a" | a = 12 + 4
b = a / 2
c = b * 12
d = 12 + 4
e = d / 2
f = 12 - e
g = c + f
|
a ) 18 , b ) 22 , c ) 28 , d ) 26 , e ) 16 | e | divide(multiply(80, 80), const_100) | find number which is 80 % less than 80 . | "explanation : 80 % less is 20 % of the given number therefore , 20 % of 80 is 16 . answer : e" | a = 80 * 80
b = a / 100
|
a ) 60 , b ) 27 , c ) 28 , d ) 26 , e ) 91 | a | divide(8, multiply(divide(3, 5), multiply(divide(2, 3), divide(1, 3)))) | there are some birds in cage , 1 / 3 rd fly away . then 2 / 5 th of the remaining left the cage , then 2 / 3 rd of the remaining left the cage . at last 8 were remaining . how many birds were in total ? | sol : if x birds were there in total , then x × ( 1 – 1 / 3 ) × ( 1 – 2 / 5 ) × ( 1 – 2 / 3 ) = 8 x × 2 / 3 × 3 / 5 × 1 / 3 = 8 x = 60 answer : a | a = 3 / 5
b = 2 / 3
c = 1 / 3
d = b * c
e = a * d
f = 8 / e
|
a ) 11 days , b ) 80 days , c ) 13 days , d ) 15 days , e ) 19 days | b | multiply(add(3, const_1), 20) | x is 3 times as fast as y and working together , they can complete a work in 20 days . in how many days can y alone complete the work ? | x = 3 y x + y = 3 y + y = 4 y these 4 y people together can do the work in 20 days , which means y can alone do the work in 4 * 20 = 80 days . answer : b | a = 3 + 1
b = a * 20
|
a ) 39 , b ) 28 , c ) 70 , d ) 80 , e ) 71 | c | divide(divide(divide(560, divide(add(9, 7), const_2)), 7), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 9 m wide at the top and 7 m wide at the bottom and the area of cross - section is 560 sq m , the depth of cannel is ? | "1 / 2 * d ( 9 + 7 ) = 560 d = 70 answer : c" | a = 9 + 7
b = a / 2
c = 560 / b
d = c / 7
e = d / 2
|
a ) a ) 12 , b ) b ) 14 , c ) c ) 16 , d ) d ) 18 , e ) e ) 22 | d | floor(divide(199, 11)) | on dividing 199 by a number , the quotient is 11 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 199 - 1 ) / 11 = 198 / 11 = 18 d )" | a = 199 / 11
b = math.floor(a)
|
a ) 70 , b ) 96 , c ) 108 , d ) 120 , e ) 150 | d | divide(subtract(32, 20), divide(10, const_100)) | of the diplomats who attended a summit conference : 20 spoke french , 32 did not speak hindi and 20 % of the diplomats spoke neither french nor hindi . if 10 % of the diplomats spoke both french and hindi , then how many diplomats attended the conference ? | "2 x 2 matrix will be the easiest way to calculate this . text in black : given statements text in red : calculated values thus d = 120 is the correct answer" | a = 32 - 20
b = 10 / 100
c = a / b
|
a ) 320.20 $ , b ) 380.20 $ , c ) 421.20 $ , d ) 451.20 $ , e ) 481.20 $ | e | multiply(multiply(0.65, 59), 12) | in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 59 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 59 ) = 481.20 hence answer will be ( e )" | a = 0 * 65
b = a * 12
|
a ) 2289 , b ) 2310 , c ) 2333 , d ) 2456 , e ) 2309 | b | add(2304, reminder(2304, 21)) | what is the nearest no . to 2304 which is divisible by 21 ? | remainder when 2304 is divided by 21 = 15 so , no . should added to make 15 to 21 = 6 no . = 2304 + 6 = 2310 answer b | a = 2304 + reminder
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a ) 14000 , b ) 14400 , c ) 14500 , d ) 15625 , e ) 14700 | d | add(10000, multiply(divide(multiply(10000, 25), const_100), const_2)) | the population of a town is 10000 . it increases annually at the rate of 25 % p . a . what will be its population after 2 years ? | formula : 10000 × 125 / 100 × 125 / 100 = 15625 answer : d | a = 10000 * 25
b = a / 100
c = b * 2
d = 10000 + c
|
a ) 2 pm , b ) 9 pm , c ) 4 pm , d ) 8 pm , e ) 6 pm | c | subtract(multiply(const_2, const_12), divide(multiply(const_2, const_12), add(divide(4, 8), const_1))) | when asked what the time is , a person answered that the amount of time left is 4 / 8 of the time already completed . what is the time . | "a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 − x = 4 / 8 x ⇒ x = 20 time is 4 pm answer : c" | a = 2 * 12
b = 2 * 12
c = 4 / 8
d = c + 1
e = b / d
f = a - e
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | c | subtract(24, 20) | if x and y are integers , what is the least positive number of 24 x + 20 y ? | 24 x + 20 y = 4 ( 6 x + 5 y ) which will be a minimum positive number when 6 x + 5 y = 1 . 6 ( 1 ) + 5 ( - 1 ) = 1 then 4 ( 6 x + 5 y ) can have a minimum positive value of 4 . the answer is c . | a = 24 - 20
|
a ) 14 , b ) 10 , c ) 9 , d ) 7 , e ) 5 | b | add(subtract(add(14, 18), subtract(25, 3)), subtract(18, 14)) | of 25 applicants for a job , 14 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "set a : people with more than 4 years exp set b : people with degree aub = total - ( less than 4 exp and no degree ) aub = 25 - 3 = 22 aub = a + b - aib aib = 18 + 14 - 22 = 10 answer b" | a = 14 + 18
b = 25 - 3
c = a - b
d = 18 - 14
e = c + d
|
a ) 100 cm , b ) 85 cm , c ) 75 cm , d ) 95 cm , e ) 110 cm | a | multiply(multiply(const_2, divide(multiply(subtract(28, const_3), const_2), add(const_4, const_3))), 28) | the sector of a circle has radius of 28 cm and central angle 90 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 90 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 28 ) = 44 + 56 = 100 cm answer : a" | a = 28 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 28
|
a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(10, subtract(subtract(const_100, 80), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? | "let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . [ ( 2 - 1 ) / 1 ] * 100 % = 100 % answer : c" | a = 100 - 80
b = a - 10
c = 10 / b
d = c * 100
|
a ) 125 , b ) 126 , c ) 130 , d ) 148 , e ) 151 | a | add(add(add(multiply(3, 11), multiply(3, 12)), multiply(3, 15)), 11) | 3 * 11 + 3 * 12 + 3 * 15 + 11 = ? | "we have : 3 * 11 = 33 3 * 12 = 36 3 * 15 = 45 all of this is : ( 33 + 36 + 45 ) + 11 = 114 + 11 = 125 correct answer a" | a = 3 * 11
b = 3 * 12
c = a + b
d = 3 * 15
e = c + d
f = e + 11
|
a ) 1 hour , b ) 1.1 hour , c ) 3 hours , d ) 5 hours , e ) 6 hours | b | divide(const_1, subtract(const_1, divide(const_1, multiply(8, const_2)))) | one pump drains one - half of a pond in 8 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ? | first pump drains 1 / 2 of the tank in 8 hours so 16 hours it will take to drain the full tank . let , 2 nd pump drains the full tank in a hours so both together can drain ( 1 / 16 + 1 / a ) part in 1 hour son in 1 / 2 hour they drain 1 / 2 * ( 1 / 16 + 1 / a ) part of the tank given that in 1 / 2 hour they drain 1 / 2 of the tank hence we can say 1 / 2 * ( 1 / 16 + 1 / a ) = 1 / 2 solving u get a = 16 / 15 = 1.1 hence answer is b | a = 8 * 2
b = 1 / a
c = 1 - b
d = 1 / c
|
a ) a ) 10 , b ) b ) 12 , c ) c ) 24 , d ) d ) 60 , e ) e ) 100 | d | multiply(factorial(2), factorial(2)) | there are 2 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 : 4 _ 3 _ 2 _ 1 r _ r _ b _ b so . . 4 ! / number of repeated letters ( 2 ! ) ( 2 ! ) = 6 ans : d" | a = math.factorial(2)
b = math.factorial(2)
c = a * b
|
a ) 79 kmph , b ) 85 kmph , c ) 55 kmph , d ) 23 kmph , e ) 14 kmph | c | divide(add(50, 60), const_2) | the speed of a car is 50 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 55 + 60 ) / 2 = 55 kmph answer : c" | a = 50 + 60
b = a / 2
|
a ) 64.67 m , b ) 66.67 m , c ) 68.67 m , d ) 56.67 m , e ) 46.67 m | b | subtract(100, multiply(const_0_33, const_100)) | the top and bottom of a tower were seen to be at angles of depression 30 ° and 60 ° from the top of a hill of height 100 m . find the height of the tower | by making a diagram ac represents the hill and de represents the pole given that ac = 100 m anglexad = angleadb = 30 ° ( ∵ ax | | bd ) anglexae = angleaec = 60 ° ( ∵ ax | | ce ) let de = h then , bc = de = h , ab = ( 100 - h ) ( ∵ ac = 100 and bc = h ) , bd = ce tan 60 ° = acce = > 3 √ = 100 ce = > ce = 1003 √ - - - ( 1 ) tan 30 ° = abbd = > 13 √ = 100 − hbd = > 13 √ = 100 − h ( 1003 √ ) ( ∵ bd = ce and substituted the value of ce from equation 1 ) = > ( 100 − h ) = 13 √ × 1003 √ = 1003 = 33.33 = > h = 100 − 33.33 = 66.67 m i . e . , the height of the pole = 66.67 m answer : b | a = const_0_33 * 100
b = 100 - a
|
a ) 15 , b ) 23 , c ) 27 , d ) 46 , e ) 32 | c | subtract(add(multiply(divide(divide(add(add(const_12, const_1), add(subtract(const_10, const_1), const_10)), const_4), const_2), const_10), const_1), add(const_10, divide(divide(add(add(const_12, const_1), add(subtract(const_10, const_1), const_10)), const_4), const_2))) | the age of the daughter is same as the digit in the mother ' s age , but in reverse order . in thirteen years mother will be twice as old as daughter will be then . what is the difference in their current age ? | daughter ' s age = 10 x + y so mother ' s age = 10 y + x . . after 13 years , 10 x + y + 13 = 2 * ( 10 y + x + 13 ) . . . . so 13 + 19 y = 8 x . . . check for odd values of y , y = 1 satisfies the eqn with x = 4 . . . so ages are 41 and 14 and ans as found correctly by u is c . . 27 | a = 12 + 1
b = 10 - 1
c = b + 10
d = a + c
e = d / 4
f = e / 2
g = f * 10
h = g + 1
i = 12 + 1
j = 10 - 1
k = j + 10
l = i + k
m = l / 4
n = m / 2
o = 10 + n
p = h - o
|
a ) 4 , b ) 6 , c ) 5 , d ) 9 , e ) 12 | c | multiply(multiply(add(inverse(multiply(const_2, const_4)), const_1), const_3), multiply(const_2, const_4)) | working together , wayne and his son can shovel the entire driveway in three hours . if wayne can shovel four times as fast as his son can , how many hours would it take for his son to shovel the entire driveway on his own ? | "w : the time for wyane to do the job s : the time for his son to do the job we have 1 / w + 1 / s = 1 / 4 and w = 4 s then we have 1 / ( 4 * s ) + 1 / s = 1 / 4 < = > 5 / ( 4 * s ) = 1 / 4 < = > s = 5 ans : c" | a = 2 * 4
b = 1/(a)
c = b + 1
d = c * 3
e = 2 * 4
f = d * e
|
a ) 100 , b ) 600 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 12) | the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 12 , find the other ? | "other number = 12 * 600 / 12 = 600 answer is b" | a = 12 * 12
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a ) 1621 , b ) 1711 , c ) 1611 , d ) 1511 , e ) 1311 | c | add(multiply(add(8, 8), const_100), divide(88, 8)) | if below equation are true 24 $ 6 = 64 35 $ 5 = 87 63 $ 7 = 99 by using same logic what is the value of the below question 88 $ 8 = ? | correct answer : c the number on the right side is formed as ( sum of digits of the 1 st number ) no 1 / no 2 ex : 35 $ 5 = 87 ( 3 + 5 = 8 ) ( 35 / 5 = 7 ) 63 $ 7 = 99 ( 6 + 3 = 9 ) ( 63 / 7 = 9 ) | a = 8 + 8
b = a * 100
c = 88 / 8
d = b + c
|
a ) 76 kg , b ) 80 kg , c ) 85 kg , d ) 97 kg , e ) none | d | add(65, multiply(10, 3.2)) | the average weight of 10 persons increases by 3.2 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "solution total weight increased = ( 10 x 3.2 ) kg = 32 kg . weight of new person = ( 65 + 32 ) kg = 97 kg . answer d" | a = 10 * 3
b = 65 + a
|
a ) 76 , b ) 81 , c ) 85 , d ) 87 , e ) 89 | b | subtract(multiply(add(10, const_1), add(4, 37)), multiply(10, 37)) | the average of runs of a cricket player of 10 innings was 37 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "explanation : average = total runs / no . of innings = 37 so , total = average x no . of innings = 37 x 10 = 370 . now increase in avg = 4 runs . so , new avg = 37 + 4 = 41 runs total runs = new avg x new no . of innings = 41 x 11 = 451 runs made in the 11 th inning = 451 - 370 = 81 answer : b" | a = 10 + 1
b = 4 + 37
c = a * b
d = 10 * 37
e = c - d
|
a ) 3 / 5 , b ) 3 / 4 , c ) 3 / 2 , d ) 3 / 7 , e ) 3 / 8 | b | divide(circle_area(divide(0.75, const_2)), const_2) | what will be the vulgar fraction of 0.75 | "explanation : 0.75 = 75 / 100 = 3 / 4 option b" | a = 0 / 75
b = circle_area / (
|
a ) 1.33 , b ) 2.33 , c ) 3.33 , d ) 4.33 , e ) 1.45 | a | divide(const_100, 75) | a company decreased the price of its main product by 75 % . subsequently , the number of units sold increased such that the total revenue remained unchanged . what was the ratio of the percent increase in the units sold to the percent decrease of the original price for this product ? | "for the total revenue to remain the same when the price is 3 / 4 , the number of products sold must 3 / 4 times . therefore increase in the number of products sold is 100 % = > the required ratio = 100 % / 75 % = 1.33 answer : a" | a = 100 / 75
|
a ) 35 % , b ) 48 % , c ) 52 % , d ) 42 % , e ) 46 % | a | multiply(subtract(const_1, multiply(divide(subtract(const_100, 25), const_100), divide(13, 15))), const_100) | a ' s speed is 15 / 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that b beats a by 25 % of the length of the race ? | first calculate the distance , b has covered with his speed in the time , in which a reached 80 % of the race . then add the remaining distance as head start for b to win the race . its best to apply ratios concept here . since a ' s speed is 15 / 13 of b , therefore , b ' s speed is 13 / 15 of a distance covered by b = speed x time = ( 13 / 15 ) x ( 0.75 ) = 0.65 % ( which means b would have covered 0.65 of the race length during the time in which a has covered 0.75 of the race length . therefore to win , b needs a headstart of ( 1 - 0.65 = ) 0.35 of the race length . so the answer is a | a = 100 - 25
b = a / 100
c = 13 / 15
d = b * c
e = 1 - d
f = e * 100
|
a ) 12000 , b ) 11000 , c ) 10000 , d ) 13000 , e ) 15000 | b | subtract(multiply(20, 1200), add(multiply(subtract(10, const_1), 1000), multiply(400, 10))) | company c sells a line of 20 products with an average retail price of $ 1200 . if none of these products sells for less than $ 400 , and exactly 10 of the products sell for less than $ 1000 , what is the greatest possible selling price of the most expensive product ? | the average price of 20 products is $ 1,200 means that the total price of 20 products is 20 * 1,200 = $ 24,000 . next , since exactly 10 of the products sell for less than $ 1,000 , then let ' s make these 10 items to be at $ 400 each ( min possible ) . now , the remaining 9 items can not be priced less than $ 1,000 , thus the minimum possible price of each of these 9 items is $ 1,000 . thus the minimum possible value of 19 products is 10 * 400 + 9 * 1,000 = $ 13,000 . therefore , the greatest possible selling price of the most expensive product is $ 24,000 - $ 13,000 = $ 11,000 . answer : b . | a = 20 * 1200
b = 10 - 1
c = b * 1000
d = 400 * 10
e = c + d
f = a - e
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a ) 1 / 2 , b ) 1 / 5 , c ) 2 / 3 , d ) 2 / 5 , e ) 1 / 7 | b | divide(divide(factorial(subtract(6, 2)), multiply(factorial(subtract(subtract(6, 2), 3)), factorial(3))), divide(factorial(6), multiply(factorial(subtract(6, 3)), factorial(3)))) | in a box of 6 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 4 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 6 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 4 c 3 / 6 c 3 = 4 / 20 = 1 / 5 answer is b" | a = 6 - 2
b = math.factorial(a)
c = 6 - 2
d = c - 3
e = math.factorial(d)
f = math.factorial(3)
g = e * f
h = b / g
i = math.factorial(6)
j = 6 - 3
k = math.factorial(j)
l = math.factorial(3)
m = k * l
n = i / m
o = h / n
|
a ) 110 , b ) 154 , c ) 130 , d ) 140 , e ) 150 | b | divide(add(45, 35), divide(52, const_100)) | a candidate appearing for an examination has to secure 52 % marks to pass paper i . but he secured only 45 marks and failed by 35 marks . what is the maximum mark for paper i ? | "he secured 45 marks nd fail by 35 marks so total marks for pass the examinatn = 80 let toal marks x x * 52 / 100 = 80 x = 154 answer : b" | a = 45 + 35
b = 52 / 100
c = a / b
|
a ) 432 miles , b ) 651 miles , c ) 387 miles , d ) 954 miles , e ) 556 miles | a | subtract(1200, 768) | jim drove 768 miles of a 1200 miles journey . how many more miles does he need to drive to finish his journey ? | "the number of miles to drive to finish his journey is given by 1200 - 768 = 432 miles correct answer a" | a = 1200 - 768
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a ) - 45 , b ) 50 , c ) - 62 , d ) 35 , e ) - 30 | e | subtract(subtract(subtract(110, 10), add(110, 10)), 10) | if | 20 x - 10 | = 110 , then find the product of the values of x ? | | 20 x - 10 | = 110 20 x - 10 = 110 or 20 x - 10 = - 110 20 x = 120 or 20 x = - 100 x = 6 or x = - 5 product = - 5 * 6 = - 30 answer is e | a = 110 - 10
b = 110 + 10
c = a - b
d = c - 10
|
a ) 125 , b ) 100 , c ) 75 , d ) 150 , e ) 50 | a | divide(10, subtract(205.08, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 10 . if n / j = 205.08 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 205.08 here 205 is the quotient . given that remainder = 10 so , 205.08 = 205 + 10 / j so , j = 125 answer : a" | a = 4 * 10
b = a + 2
c = 100 + b
d = 205 - 8
e = 10 / d
|
a ) 14 , b ) 10 , c ) 8 , d ) 6 , e ) 4 | d | subtract(add(18, 10), subtract(add(18, divide(add(18, 10), const_2)), 10)) | in a certain kindergarten , half of the children are allergic to peanuts , 10 children are not allergic to cashew nuts , 10 children are allergic to both peanuts and cashew nuts , and 18 children are allergic to cashew nuts . how many of the kindergarten ' s children are not allergic to peanuts and are not allergic to cashew nuts ? | we are given children allergic to cashew nuts as 18 and not allergic to cashew nuts as 10 , so total = 10 + 18 = 28 . . . children allergic to both are 10 . . . 18 are allergic to cashew nuts , so and half or 28 / 2 = 14 are allergic to peanuts . . so children allergic to at least one of the two are 18 + 14 - 10 = 22 . . therefore children not allergic to any of the two are 28 - 22 = 6 answer : d | a = 18 + 10
b = 18 + 10
c = b / 2
d = 18 + c
e = d - 10
f = a - e
|
a ) 288 m , b ) 256 m , c ) 200 m , d ) 278 m , e ) 287 m | c | divide(1100, multiply(subtract(78, 1), const_0_2778)) | a train 1100 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 1100 + x ) / 60 = 65 / 3 x = 200 m . answer : c" | a = 78 - 1
b = a * const_0_2778
c = 1100 / b
|
a ) 50 , b ) 40 , c ) 60 , d ) 100 , e ) 75 | e | divide(300, add(divide(300, const_100), const_1)) | the sum of number of boys and girls in a school is 300 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ? | "we have x + x % of 300 = 300 x + x / 100 * 300 = 300 4 * x = 300 x = 75 answer is e" | a = 300 / 100
b = a + 1
c = 300 / b
|
a ) $ 015 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.55 | e | divide(multiply(add(const_1, const_1), 0.82), add(const_1, const_2)) | on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.82 per glass on the first day , what was the price per glass on the second day ? | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.82 = 3 * x - - > x = $ 0.55 . answer : e ." | a = 1 + 1
b = a * 0
c = 1 + 2
d = b / c
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 45 % , e ) 72 % | d | multiply(divide(162, divide(const_3600, const_10)), const_100) | the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 162 ° of the circle , what percentage of megatek employees are in manufacturing ? | answer : d 162 ° divided by 360 ° equals 0.45 , therefore the sector is equal to 45 % of the total | a = 3600 / 10
b = 162 / a
c = b * 100
|
a ) 63.3 % , b ) 52.4 % , c ) 81.2 % , d ) 75.4 % , e ) 69.1 % | a | multiply(const_100, divide(subtract(add(1100, 700), add(multiply(1100, divide(42, const_100)), multiply(700, divide(30, const_100)))), add(1100, 700))) | 1100 boys and 700 girls are examined in a test ; 42 % of the boys and 30 % of the girls pass . the percentage of the total who failed is ? | total number of students = 1100 + 700 = 1800 number of students passed = ( 42 % of 1100 + 30 % of 700 ) = 462 + 210 = 672 number of failures = 1128 * 100 / 1800 = 190 / 3 % = 63.3 % answer is a | a = 1100 + 700
b = 42 / 100
c = 1100 * b
d = 30 / 100
e = 700 * d
f = c + e
g = a - f
h = 1100 + 700
i = g / h
j = 100 * i
|
a ) 16 % , b ) 16.66 % , c ) 17 % , d ) 17.2 % , e ) 18 % | b | multiply(divide(subtract(70, 60), 60), const_100) | john makes $ 60 a week from his job . he earns a raise and nowmakes $ 70 a week . what is the % increase ? | "increase = ( 10 / 60 ) * 100 = ( 1 / 6 ) * 100 = 16.66 % . b" | a = 70 - 60
b = a / 60
c = b * 100
|
a ) 120 , b ) 130 , c ) 150 , d ) 320 , e ) 450 | a | add(90, multiply(divide(subtract(690, 90), add(90, 80)), subtract(90, 80))) | two heavily loaded sixteen - wheeler transport trucks are 690 kilometers apart , sitting at two rest stops on opposite sides of the same highway . driver a begins heading down the highway driving at an average speed of 90 kilometers per hour . exactly one hour later , driver b starts down the highway toward driver a , maintaining an average speed of 80 kilometers per hour . how many kilometers farther than driver b , will driver a have driven when they meet and pass each other on the highway ? | "i ' ve been reading the website for a while and i ' m always keen to see different approaches so i would like to share one that works for me : short version : truck a travels for an hour . distance remaining = 690 - 90 = 600 k ratio of speeds 9 : 8 - > 510 / 17 = 30 truck a = 90 + 30 * 9 = 360 truck b = 30 * 8 = 240 delta = 120 km answer a" | a = 690 - 90
b = 90 + 80
c = a / b
d = 90 - 80
e = c * d
f = 90 + e
|
a ) 14.83 , b ) 14.81 , c ) 14.8 , d ) 14.82 , e ) 18.3 | e | divide(multiply(42, 136), 312) | on a map the distance between two mountains is 312 inches . the actual distance between the mountains is 136 km . ram is camped at a location that on the map is 42 inch from the base of the mountain . how many km is he from the base of the mountain ? | "explanation : since 312 inch = 136 km so 1 inch = 136 / 312 km so 42 inch = ( 136 ã — 42 ) / 312 = 18.3 km answer : e" | a = 42 * 136
b = a / 312
|
a ) $ 5 , b ) $ 15 , c ) $ 23 , d ) $ 100 , e ) $ 105 | c | subtract(subtract(multiply(500, power(add(const_1, divide(10, const_100)), 2)), 500), subtract(multiply(power(add(const_1, divide(5, const_100)), 2), 800), 800)) | on a certain day , tim invested $ 500 at 10 percent annual interest , compounded annually , and lana invested 800 at 5 percent annual interest , compounded annually . the total amount of interest earned by tim ’ s investment in the first 2 years was how much greater than the total amount of interest earned by lana ’ s investment in the first 2 years ? | "compounded annually means that the interest is applied once per year . one can have 10 % annual interest compounded monthly - in this case 10 % / 12 would be applied each month , or 10 % annual interest compounded daily etc . with respect to the problem at hand , at the end of two years , tim would have 500 ( 1.10 ) ^ 2 = 500 ( 1.21 ) = 605 and lana would have 800 ( 1.05 ) ^ 2 = 800 ( 1.1025 ) = 882 thus , tim earned 105 dollars , while lana earned 82 dollars the difference is $ 23 and the answer is c ." | a = 10 / 100
b = 1 + a
c = b ** 2
d = 500 * c
e = d - 500
f = 5 / 100
g = 1 + f
h = g ** 2
i = h * 800
j = i - 800
k = e - j
|
a ) 279 , b ) 283 , c ) 308 , d ) 318 , e ) 328 | c | multiply(11, 275) | the h . c . f of two numbers is 11 and their l . c . m is 7700 . if one of the numbers is 275 , then the other is : | "other number = ( 11 * 7700 ) / 275 = 308 . answer : c" | a = 11 * 275
|
a ) 3.5 % , b ) 2.4 % , c ) 3 % , d ) 5 % , e ) 2.6 % | e | subtract(subtract(8, 5), divide(multiply(8, 5), const_100)) | in measuring the sides of a rectangle , one side is taken 8 % in excess and other 5 % in deficit . find the error percentage in the area calculated from these measurements . | "let the sides of the rectangle be a and b . actual area of the rectangle without error = ab new side a ( 8 % in excess ) = a + 0.08 a = 1.08 a new side b ( 5 % in deficit ) = b - 0.05 b = 0.95 b area of the new rectangle = ( 1.08 a ) ( 0.95 b ) = ( 1.0260 ) ab say a = 1 and b = 1 area of the original rectangle = 1 area of the new rectangle = 1.026 change in the area calculation = . 026 error percentage in the area calculated = 2.6 % answer : e" | a = 8 - 5
b = 8 * 5
c = b / 100
d = a - c
|
a ) 8100 , b ) 3388 , c ) 7767 , d ) 4050 , e ) 2288 | d | add(add(3200, multiply(divide(1, 8), 3200)), multiply(divide(1, 8), add(3200, multiply(divide(1, 8), 3200)))) | every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 3200 ? | "3200 * 9 / 8 * 9 / 8 = 4050 answer : d" | a = 1 / 8
b = a * 3200
c = 3200 + b
d = 1 / 8
e = 1 / 8
f = e * 3200
g = 3200 + f
h = d * g
i = c + h
|
a ) 30 , b ) 45 , c ) 50 , d ) 80 , e ) 100 | b | divide(add(160, 20), add(add(const_1, const_1), const_2)) | on a certain test , bill scored 20 more points than john , but half as many points as sue . if the sum of the scores of the 3 students was 160 points , how many points did bill receive ? | consider john ' s points as x so bill ' s points are x + 20 so sue ' s points are 2 x + 40 total of all three students is 4 x + 60 , but this total is given as 160 x = 25 so bill ' s points = 25 + 20 = 45 correct answer option b | a = 160 + 20
b = 1 + 1
c = b + 2
d = a / c
|
a ) 99 : 61 , b ) 99 : 160 , c ) 61 : 160 , d ) 61 : 99 , e ) 99 : 160 | a | power(2, 3) | two vessels having volumes in the ratio 3 : 5 are filled with water and milk solutions . the ratio of milk and water in the two vessels are 2 : 3 and 3 : 1 respectively . if the contents of both the vessel are empties into a larger vessel , find the ratio of milk and water in the larger vessel . | "vessel a = 300 gallons - - > milk = 120 , water = 180 ; vessel b = 500 gallons - - > milk = 375 , water = 125 ; vessel a + b = 800 gallons - - > milk = 495 , water 305 . the ratio = 495 / 305 - - > reduce by 5 = 99 / 61 . answer : a ." | a = 2 ** 3
|
a ) 2 , b ) 8 , c ) 9 , d ) 10 , e ) 10.5 | e | divide(add(divide(18, 3), divide(45, 3)), const_2) | a man swims downstream 45 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ? | "45 - - - 3 ds = 15 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 15 + 6 ) / 2 = 10.5 answer : e" | a = 18 / 3
b = 45 / 3
c = a + b
d = c / 2
|
a ) 34 , b ) 36 , c ) 33 , d ) 38 , e ) 31 | c | sqrt(1122) | the product of two successive numbers is 1122 . which is the smaller of the two numbers ? | "c 33 from the given alternatives , 33 × 34 = 1122 ∴ smaller number = 33" | a = math.sqrt(1122)
|
a ) 250 , b ) 305 , c ) 280 , d ) 285 , e ) none | b | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 630), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | a library has an average of 630 visitors on sundays and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "since the month begins with sunday , to there will be five sundays in the month average required = ( 630 x 5 + 240 x 25 / 30 ) = 305 answer : option b" | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 630
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 240
l = e + k
m = l / 30
|
a ) rs 1000 , b ) rs 1515 , c ) rs 1525 , d ) rs 1250 , e ) none of these | a | divide(subtract(1232, multiply(1232, divide(10, const_100))), add(divide(12, const_100), const_1)) | the sale price of a trolley bag including the sale tax is rs . 1232 . the rate of sale tax is 10 % . if the shopkeeper has made a profit of 12 % , the cost price of the trolley bag is : | "explanation : 110 % of s . p . = 1232 s . p . = rs . ( 1232 x 100 / 110 ) = rs . 1120 . c . p . = rs ( 100 / 112 x 1120 ) = rs 1000 answer : a" | a = 10 / 100
b = 1232 * a
c = 1232 - b
d = 12 / 100
e = d + 1
f = c / e
|
a ) rs 4000 , b ) rs 2800 , c ) rs 3800 , d ) rs 3500 , e ) rs 3100 | b | divide(subtract(multiply(4000, divide(5, const_100)), 144), subtract(divide(5, const_100), divide(3, const_100))) | rs 4000 is divided into two parts such that one part is put out at 3 % and the other at 5 % . if the annual interest earned from both the investments be rs 144 , find the first part . | "explanation : average rate = ( 144 / 4000 ) * 100 = 3.6 ratio = 14 : 6 = 7 : 3 so , first part = ( 7 / 10 ) * 4000 = rs 2800 . answer : b" | a = 5 / 100
b = 4000 * a
c = b - 144
d = 5 / 100
e = 3 / 100
f = d - e
g = c / f
|
a ) 4 , b ) 3 , c ) 5 , d ) 6 , e ) 9 | d | add(divide(multiply(add(24, const_1), 60), const_1000), 4.5) | a train of 24 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 4.5 km long ? | "d = 25 * 60 + 4500 = 6000 m t = 6000 / 60 * 18 / 5 = 360 sec = 6 mins answer : d" | a = 24 + 1
b = a * 60
c = b / 1000
d = c + 4
|
a ) 51 / 145 , b ) 9 / 25 , c ) 3 / 5 , d ) 94 / 145 , e ) 5 / 6 | a | divide(choose(18, const_2), choose(add(add(4, 8), 18), const_2)) | the only items of clothing at michelle ’ s boutique are 4 shirts , 8 pairs of pants and 18 ties . what is the probability of randomly selecting 2 ties out of all the items in michelle ’ s boutique ? | we have 4 shirts , 8 pairs of pants and 18 ties . i . e 30 items total ways of picking two items out of all = 30 c 2 = 435 total ways of picking two items out of all ties = 18 c 2 = 153 required probability = 153 / 435 = 51 / 145 answer : option a | a = math.comb(18, 2)
b = 4 + 8
c = b + 18
d = math.comb(c, 2)
e = a / d
|
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