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 ) - 14 , b ) - 12 , c ) - 8 , d ) 8 , e ) 12 | a | add(subtract(12, 7), subtract(negate(12), 7)) | if | y + 7 | = 12 , what is the sum of all the possible values of y ? | there will be two cases y + 7 = 12 or y + 7 = - 12 = > y = 5 or y = - 19 sum of both the values will be - 19 + 5 = - 14 answer : a | a = 12 - 7
b = negate - (
c = a + b
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a ) 8.5 , b ) 6.5 , c ) 5.5 , d ) 5.6 , e ) 4.4 | b | subtract(const_100, multiply(multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(15, const_100))), const_100)) | the tax on a commodity is diminished by 15 % and its consumption increased by 10 % . the effect on revenue is ? | "100 * 100 = 10000 85 * 110 = 9350 - - - - - - - - - - - 10000 - - - - - - - - - - - 650 100 - - - - - - - - - - - ? = > 6.5 % decrease answer : b" | a = 10 / 100
b = 1 + a
c = 15 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
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a ) 25 , b ) 50 , c ) 75 , d ) 100 , e ) 125 | d | multiply(multiply(divide(50, 25), divide(50, 25)), const_1000) | positive integer y is 50 percent of 50 percent of positive integer x , and y percent of x equals 25 . what is the value of x ? | "y = 0.5 * 0.5 * x = x / 4 y % * x = 25 ( y / 100 ) * x = 25 ( x / 400 ) * x = 25 x ^ 2 = 25 * 400 x = 100 the answer is d ." | a = 50 / 25
b = 50 / 25
c = a * b
d = c * 1000
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a ) $ 0.00 , b ) $ 5.75 , c ) $ 3.40 , d ) $ 5.00 , e ) $ 6.80 | b | subtract(multiply(78.2, divide(add(const_100, 25), const_100)), divide(78.2, divide(subtract(const_100, 15), const_100))) | cindy has her eye on a sundress but thinks it is too expensive . it goes on sale for 15 % less than the original price . before cindy can buy the dress , however , the store raises the new price by 25 % . if the dress cost $ 78.2 after it went on sale for 15 % off , what is the difference between the original price and the final price ? | "0.85 * { original price } = $ 78.2 - - > { original price } = $ 92 . { final price } = $ 78.2 * 1.25 = $ 97.75 . the difference = $ 97.75 - $ 92 = $ 5.75 answer : b ." | a = 100 + 25
b = a / 100
c = 78 * 2
d = 100 - 15
e = d / 100
f = 78 / 2
g = c - f
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a ) 232 , b ) 242 , c ) 252 , d ) 262 , e ) 180 | e | divide(20, divide(450, 20)) | evaluate 20 % of 450 + 45 % of 200 | "explanation : = ( 20 / 100 ) * 450 + ( 45 / 100 ) * 200 = 180 option e" | a = 450 / 20
b = 20 / a
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a ) 1 cm , b ) 2 cm , c ) 24 cm , d ) 55 cm , e ) 50 cm | a | floor(divide(add(multiply(12, const_100), 34), add(multiply(8, const_100), 98))) | which greatest possible length can be used to measure exactly 12 meter 34 cm , 8 meter 98 cm and 6 meter 65 cm | explanation : convert first all terms into cm . i . e . 1234 cm , 898 cm , 665 cm . now whenever we need to calculate this type of question , we need to find the hcf . hcf of above terms is 1 . option a | a = 12 * 100
b = a + 34
c = 8 * 100
d = c + 98
e = b / d
f = math.floor(e)
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a ) 120 , b ) 180 , c ) 240 , d ) 280 , e ) 340 | c | multiply(divide(4, subtract(divide(60, 20), const_1)), 60) | working together , printer a and printer b would finish the task in 20 minutes . printer a alone would finish the task in 60 minutes . how many pages does the task contain if printer b prints 4 pages a minute more than printer a ? | "20 * a + 20 * b = x pages in 20 mins printer a will print = 20 / 60 * x pages = 1 / 3 * x pages thus in 20 mins printer printer b will print x - 1 / 3 * x = 2 / 3 * x pages also it is given that printer b prints 4 more pages per min that printer a . in 20 mins printer b will print 80 more pages than printer a thus 2 / 3 * x - 1 / 3 * x = 80 = > x = 240 pages answer : c" | a = 60 / 20
b = a - 1
c = 4 / b
d = c * 60
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(add(const_4, const_2), const_1) | a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than 20,000 and at the same time n is a perfect square and a perfect cube ? | "if n is a perfect square and a perfect cube , then n = a ^ 6 for some integer a . the numbers are 1 ^ 6 = 1 , 2 ^ 6 = 64 , 3 ^ 6 = 729 , 4 ^ 6 = 4096 , 5 ^ 6 = 15,625 . the answer is c ." | a = 4 + 2
b = a / 1
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a ) 32 % , b ) 35 % , c ) 38 % , d ) 41 % , e ) 44 % | e | add(20, multiply(divide(30, const_100), 80)) | if 20 liters of chemical x are added to 80 liters of a mixture that is 30 % chemical x and 70 % chemical y , then what percentage of the resulting mixture is chemical x ? | "the amount of chemical x in the solution is 20 + 0.3 ( 80 ) = 44 liters . 44 liters / 100 liters = 44 % the answer is e ." | a = 30 / 100
b = a * 80
c = 20 + b
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a ) 86000 , b ) 79860 , c ) 76300 , d ) 82900 , e ) none of these | b | divide(multiply(add(10, const_100), divide(multiply(add(10, const_100), divide(multiply(60000, add(10, const_100)), const_100)), const_100)), const_100) | the present population of a town is 60000 . the population increases annually at 10 % . find the population after 3 years . | explanation : solution : the required population is 60000 ( 1 + 10 / 100 ) ^ 3 = 60000 * 110 / 100 * 110 / 100 * 110 / 100 = 79860 answer : b | a = 10 + 100
b = 10 + 100
c = 10 + 100
d = 60000 * c
e = d / 100
f = b * e
g = f / 100
h = a * g
i = h / 100
|
a ) $ 30 , b ) $ 25 , c ) $ 35 , d ) $ 40 , e ) $ 45 | b | add(multiply(add(multiply(add(divide(1, 2), add(divide(1, 4), divide(1, 3))), 5), 5), 2), const_10) | sam β s car was fined when he gave joe and peter a ride , so they decided to help sam pay the fine . joe paid $ 4 more than 1 / 4 of the fine and peter paid $ 4 less than 1 / 3 of the fine , leaving pay $ 5 less than 1 / 2 the fine to complete the payment . what fraction of the fine did sam pay ? | "call the fine f . joe paid ( 1 / 4 ) f + 4 and peter paid ( 1 / 3 ) f β 4 , leaving ( 1 / 2 ) f β 5 left . if we add those three up , they should add up to f . f = [ ( 1 / 4 ) f + 4 ] + [ ( 1 / 3 ) f β 4 ] + [ ( 1 / 2 ) f β 5 ] f = ( 1 / 4 ) f + ( 1 / 3 ) f + ( 1 / 2 ) f β 5 multiply all terms by 12 to clear the fractions . 12 f = 3 f + 4 f + 6 f β 60 12 f = 13 f β 60 β f = β 60 f = 60 well , if the fine cost $ 60 , then sam paid the part not covered by joe or peter . half the fine is $ 30 , and sam paid $ 5 less than this : $ 25 . answer = b" | a = 1 / 2
b = 1 / 4
c = 1 / 3
d = b + c
e = a + d
f = e * 5
g = f + 5
h = g * 2
i = h + 10
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a ) 20.5 % , b ) 22.5 % , c ) 21.5 % , d ) 10.5 % , e ) 32.5 % | b | subtract(add(const_12, const_12), divide(subtract(add(multiply(90, const_1), const_3), multiply(90, const_1)), const_2)) | the total car population in a chennai is 900 out of which 500 are made by honda . out of every 100 honda cars , 90 are red in color , but only 60 % of the total car population is red . find the % of non - honda cars that are red | "total honda brand red cars = 90 / 100 * 500 = 450 total red cars = 60 / 100 * 900 = 540 no . of non - honda cars = 900 - 500 = 400 no . of non - honda red cars = 540 - 450 = 90 % of non - honda red cars = 90 / 400 * 100 = 22.5 % answer : b" | a = 12 + 12
b = 90 * 1
c = b + 3
d = 90 * 1
e = c - d
f = e / 2
g = a - f
|
a ) 1080 , b ) 10080 , c ) 10025 , d ) 11080 , e ) 12080 | b | multiply(add(multiply(const_4, const_2), multiply(const_12, const_4)), lcm(lcm(12, 15), 18)) | find the least number of 5 digits which is exactly divisible by 12 , 15 and 18 ? | lcm = 180 180 ) 10000 ( 55 9900 - - - - - - - - - 100 10000 + 180 - 100 = 10080 answer : b | a = 4 * 2
b = 12 * 4
c = a + b
d = math.lcm(12, 15)
e = math.lcm(d, 18)
f = c * e
|
a ) 4.5 , b ) 3.5 , c ) 3 , d ) 2 , e ) 1.5 | e | add(divide(subtract(1640, 1400), divide(subtract(2000, 1400), subtract(5, 2))), subtract(divide(2, 5), divide(const_1, const_10))) | a certain sum amount to $ 1400 in 2 yrs and $ 2000 in 5 years on certain fixed simple interest . in how many yrs would the amount be $ 1640 had it been put at the rate of interest compounded annually ? | we can form equations as following : initial amount = p , total amount after interest = t , interest per year = i t = p + i given t 1 = p + 2 i = 1400 t 2 = p + 5 i = 2000 from this we get , p = 1000 i = 200 $ now we know that simple interest is < compound interest ( ci = si only after 1 st year if ci is compounded annually ) consider simple interest - after 1 st yer it is 200 $ , after 2 nd years it is 400 $ and after 3 rd year it is 600 $ . to get a 240 $ compound interest the duration have to be > 1 year and < 2 years . only option fitting : option e . | a = 1640 - 1400
b = 2000 - 1400
c = 5 - 2
d = b / c
e = a / d
f = 2 / 5
g = 1 / 10
h = f - g
i = e + h
|
a ) 36.6 , b ) 42.8 , c ) 28.4 , d ) 11.6 , e ) 24.8 | e | multiply(15, 375) | the h . c . f . of two numbers is 15 and their l . c . m . is 620 . if one of the numbers is 375 , then the other is : | "other number = ( 15 x 620 ) / 375 = 24.8 . answer : e" | a = 15 * 375
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a ) 4 / 13 , b ) 1 / 52 , c ) 1 / 4 , d ) 1 / 3 , e ) none of these | a | divide(multiply(const_4, const_3), 52) | from a pack of 52 cards , 1 card is drawn at random . find the probability of a face card drawn . | "explanation : total number of cases = 52 total face cards = 16 [ favourable cases ] so probability = 16 / 52 = 4 / 13 option a" | a = 4 * 3
b = a / 52
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a ) 0.552 , b ) 69.0 , c ) 0.6845 , d ) 0.6859 , e ) 0.69 | a | divide(multiply(92, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 92 % of 3 / 5 ? | "92 % * ( 3 / 5 ) = 0.92 * 0.6 = 0.552 answer : a" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 92 * n
p = o / 100
|
a ) - 4 , b ) - 2 , c ) - 80 , d ) 2 , e ) 4 | c | multiply(negate(multiply(divide(80, 2), 2)), 2) | if 9 a - b = 10 b + 80 = - 12 b - 2 a , what is the value of 2 a + 22 b ? | "this implies 9 a - b = 10 b + 80 , 9 a - b = - 12 b - 2 a , 10 b + 80 = - 12 b - 2 a manipulating the second equation gives us 10 b + 80 = - 12 b - 2 a = = > 2 a + 22 b = - 80 answer is c" | a = 80 / 2
b = a * 2
c = negate * (
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a ) 20 days , b ) 10 days , c ) 6 days , d ) 8 days , e ) 7 days | c | divide(const_1, subtract(divide(const_1, 8), divide(const_1, 24))) | a can do a work in 24 days and b can do it in 8 days . in how many days a and b can do the work ? | "explanation : a ' s 1 day ' s work = 1 / 24 b ' s 1 day ' s work = 1 / 8 they work together = 1 / 24 + 1 / 8 = 6 days answer : option c" | a = 1 / 8
b = 1 / 24
c = a - b
d = 1 / c
|
a ) 25 years , b ) 35 years , c ) 45 years , d ) 55 years , e ) 75 years | b | add(subtract(10, subtract(10, add(const_3, const_2))), multiply(subtract(10, add(const_3, const_2)), const_2)) | the difference between the ages of two persons is 10 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the elder person is : | "solution let their ages be x year and ( x + 10 ) years respectively . then , ( x + 10 ) - 15 = 2 ( x - 15 ) β x - 5 = 2 x - 30 β x = 25 . β΄ present age of the elder person = ( x + 10 ) = 35 years . answer b" | a = 3 + 2
b = 10 - a
c = 10 - b
d = 3 + 2
e = 10 - d
f = e * 2
g = c + f
|
a ) 19 , b ) 21 , c ) 23 , d ) 25 , e ) 27 | d | divide(add(multiply(2, 30), 65), add(2, 3)) | sandy gets 3 marks for each correct sum and loses 2 marks for each incorrect sum . sandy attempts 30 sums and obtains 65 marks . how many sums did sandy get correct ? | "let x be the correct sums and ( 30 - x ) be the incorrect sums . 3 x - 2 ( 30 - x ) = 65 5 x = 125 x = 25 the answer is d ." | a = 2 * 30
b = a + 65
c = 2 + 3
d = b / c
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a ) a ) 452000 , b ) b ) 562000 , c ) c ) 800000 , d ) d ) 400000 , e ) e ) 652000 | d | multiply(divide(40000, subtract(const_100, add(add(multiply(20, 3), 25), 5))), const_100) | a person distributed 20 % of his income to his 3 children each . he deposited 25 % of his income to his wife ' s account . he donated 5 % of remaining amount to an orphan house . finally he has $ 40000 . find his total income ? | "3 children got = 3 * 20 % = 60 % wife got = 25 % orphan house = 5 % total = 60 + 25 + 5 = 90 % remaining = 100 - 90 = 10 % 10 % = 40000 100 % = 40000 * 100 / 10 = $ 400000 answer is d" | a = 20 * 3
b = a + 25
c = b + 5
d = 100 - c
e = 40000 / d
f = e * 100
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a ) 71 kg , b ) 77 kg , c ) 76.5 kg , d ) data inadequate , e ) none of these | a | add(65, multiply(4, 1.5)) | the average weight of 4 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ? | "total weight increases = 4 Γ 1.5 = 6 kg so the weight of new person = 65 + 6 = 71 kg answer a" | a = 4 * 1
b = 65 + a
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a ) 18 / 50 , b ) 16 / 50 , c ) 17 / 50 , d ) 19 / 50 , e ) none | d | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 0.38), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 0.38 in to a vulgar fraction ? | "answer 0.38 = 38 / 100 = 19 / 50 correct option : d" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 0
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
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a ) rs . 20 , b ) rs . 60 . , c ) rs . 40 , d ) rs . 30 , e ) rs . 90 | b | divide(720, subtract(17, 5)) | on selling 17 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is : | ( c . p . of 17 balls ) - ( s . p . of 17 balls ) = ( c . p . of 5 balls ) c . p . of 12 balls = s . p . of 17 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 12 = rs . 60 . answer : b | a = 17 - 5
b = 720 / a
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a ) 1 / 5 , b ) 1 / 10 , c ) 1 / 12 , d ) 1 / 16 , e ) 1 / 32 | d | divide(const_1, power(const_2, const_4.0)) | when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 5 times , what is the probability that it will land on either side all 5 times ? | "must be either 5 heads or 5 tails 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 + 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 1 / 32 + 1 / 32 = 1 / 16 answer : d" | a = 2 ** 4
b = 1 / a
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a ) a ) 200 , b ) b ) 2000 , c ) c ) 800 , d ) d ) 2100 , e ) e ) 1100 | e | subtract(subtract(multiply(210, 10), multiply(5, 40)), multiply(4, 200)) | the average of 10 numbers is 210 . the average of first 5 numbers is 40 and the average of last 4 numbers is 200 . what is the middle number ? | the total of ten numbers = 10 x 210 = 2100 the total of first 5 and last 4 numbers is = 5 x 40 + 4 x 200 = 1000 so , the middle number is ( 2100 - 1000 ) = 1100 e | a = 210 * 10
b = 5 * 40
c = a - b
d = 4 * 200
e = c - d
|
a ) $ 36.55 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 39.50 , e ) $ 40.60 | a | add(multiply(35, divide(40, const_100)), 35) | a farmer spent $ 35 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 10 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have spent on the chicken feed and goat feed combined ? | "a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 35 = $ 14 on chicken feed , thus he spent the remaining 35 - 14 = $ 21 on goat feed . now , since he bought chicken feed at a 20 % discount then the original price of it was x * 0.9 = $ 14 - - > x = $ 15.55 . therefore if the farmer had paid full price for both the chicken feed and the goat feed , then he would he have spent 15.55 + 21 = $ 36.55 . answer : a ." | a = 40 / 100
b = 35 * a
c = b + 35
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | three numbers are in the ratio 3 : 4 : 5 and their l . c . m . is 240 . their h . c . f is ? | "let the numbers be 3 x , 4 x and 5 x their l . c . m . = 60 x 60 x = 240 x = 4 the numbers are 3 * 4 , 4 * 4 , 5 * 4 hence required h . c . f . = 4 answer is c" | a = 3 * 5
b = a * 100
c = 4 * 5
d = b + c
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a ) 33 , b ) 88 , c ) 40 , d ) 99 , e ) 112 | e | inverse(subtract(divide(const_1, 14), divide(const_1, add(14, const_2)))) | a cistern is normally filled in 14 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full , the leak will empty it in ? | "1 / 14 - 1 / x = 1 / 16 x = 112 answer : e" | a = 1 / 14
b = 14 + 2
c = 1 / b
d = a - c
e = 1/(d)
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a ) 11 , b ) 15 , c ) 12 , d ) 17 , e ) 19 | b | divide(15, const_1) | the overall age of x and y is 15 year greater than the overall age of y and z . z is how many decades younger that x ? | b 15 ( x + y ) Γ’ β¬ β ( y + z ) = 15 x Γ’ β¬ β z = 15 | a = 15 / 1
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a ) 24.58 , b ) 24.5 , c ) 24.3 , d ) 24.9 , e ) 15.5 | e | multiply(divide(5, subtract(9, 7)), 7) | sachin is younger than rahul by 5 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "if rahul age is x , then sachin age is x - 5 , so ( x - 5 ) / x = 7 / 9 = > 9 x - 45 = 7 x = > 2 x = 45 = > x = 22.5 so sachin age is 22.5 - 7 = 15.5 answer : e" | a = 9 - 7
b = 5 / a
c = b * 7
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a ) 22 , b ) 27 , c ) 50 , d ) 99 , e ) 12 | c | add(42, 30) | two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 42 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 42 + 30 = 72 km / hr . 72 * 5 / 18 = 20 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 / 20 = 50 sec . answer : c" | a = 42 + 30
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a ) 640 , b ) 720 , c ) 1080 , d ) 1200 , e ) 1440 | a | divide(128, divide(multiply(const_2, subtract(140, 128)), 128)) | julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ 128 and $ 140 from the simple interest account and the compound interest account respectively . if the interest rates for both accounts were the same , what was the amount of julie ' s initial savings ? | "$ 128 for 2 years = $ 64 per year . extra $ 12 yearned with the compound interest is the percent yearned on percent . so , $ 12 is yearned on $ 64 , which means that the interest = 5 % . this on the other hand means that half of the savings = 64 * 5 = $ 320 . twice of that = $ 640 . answer : a ." | a = 140 - 128
b = 2 * a
c = b / 128
d = 128 / c
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a ) 3 / 20 , b ) 1 / 3 , c ) 5 / 12 , d ) 12 / 17 , e ) 5 / 11 | e | divide(multiply(5, 1), add(3, multiply(5, 5))) | every student in a room is either a junior or a senior . there is at least one junior and at least one senior in the room . if 2 / 5 of the juniors is equal to 1 / 3 of the seniors , what fraction of the students in the room are juniors ? | "let total number of juniors = j total number of seniors = s ( 2 / 5 ) j = ( 1 / 3 ) s = > s = 6 / 5 j total number of students = j + s = ( 11 / 5 ) j fraction of the students in the room are juniors = j / ( j + s ) = j / [ ( 11 / 5 ) j ] = 5 / 11 answer e" | a = 5 * 1
b = 5 * 5
c = 3 + b
d = a / c
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a ) 858 , b ) 1875 , c ) 14667 , d ) 63840 , e ) 146667 | b | add(divide(multiply(subtract(6.25, 2.85), 2000), subtract(5.8, 1.8)), const_100) | dominos charges $ 6.25 for a veg pizza of std size that costs a total of $ 2.85 to make . pizza hut charges $ 5.80 for a similar pizza that costs $ 1.80 to make . if dominos sells 2000 pizzas , how many must pizza hut sell to make at least as much in total gross profit as its competitor does ? | dominos : profit / pizza = 6.25 - 2.85 = 3.75 : no of pizza = 2000 : gross profit = 2000 * 3.75 = 7500 pizza hut : profit / pizza = 4 : gross profit = 7500 : no of cups = 7500 / 4 = 1875 answer b | a = 6 - 25
b = a * 2000
c = 5 - 8
d = b / c
e = d + 100
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a ) 50 % , b ) 40 % , c ) 35 % , d ) 32 % , e ) 17 % | e | multiply(divide(subtract(add(const_1, divide(40, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_100) | a certain company reported that the revenue on sales increased 20 % from 2000 to 2003 , and increased 40 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ? | "assume the revenue in 2000 to be 100 . then in 2003 it would be 120 and and in 2005 140 , so from 2003 to 2005 it increased by ( 140 - 120 ) / 120 = 20 / 120 = 17 % answer : e ." | a = 40 / 100
b = 1 + a
c = 20 / 100
d = 1 + c
e = b - d
f = 20 / 100
g = 1 + f
h = e / g
i = h * 100
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a ) 174.5 , b ) 173.5 , c ) 172.5 , d ) 171.5 , e ) 170.5 | a | divide(subtract(multiply(30, 175), subtract(151, 136)), 30) | the average height of 30 students in a class was calculated as 175 cm . it has later found that the height of one of the students in the class was incorrectly written as 151 cm whereas the actual height was 136 cm . what was the actual average height of the students in the class ? | "the total height was 15 cm too much . the average height should be reduced by 15 cm / 30 = 0.5 cm the answer is a ." | a = 30 * 175
b = 151 - 136
c = a - b
d = c / 30
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a ) 5 , b ) 7 , c ) 9 , d ) 10 , e ) 12 | b | add(subtract(divide(57, 7), 2), const_1) | how many integers are between 2 and 57 / 7 , inclusive ? | "57 / 7 = 8 . xx we are not concerned about the exact value of 57 / 7 as we just need the integers . the different integers between 2 and 57 / 7 would be 2,3 , 4 , 5 , 6 , 7 , 8 . total number of integers = 7 option b" | a = 57 / 7
b = a - 2
c = b + 1
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a ) 360 , b ) 380 , c ) 400 , d ) 406 , e ) 433 | e | multiply(divide(subtract(divide(360, subtract(const_1, divide(const_1, 10))), 360), 12), 130) | mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 130 mangoes . | "mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 120 mangoes . method 1 : let price per mango = x . let number of mangoes be n . then , nx = 360 . now price = 0.9 x ; number of mangoes = n + 12 . total amount = 0.9 x * ( n + 12 ) = 360 . nx = 0.9 nx + 10.8 x = > 0.1 nx = 10.8 x = > n = 108 = > x = 360 / 108 = 3.33 original price of 130 mangoes = 130 * 3.33 = 433 . answer e" | a = 1 / 10
b = 1 - a
c = 360 / b
d = c - 360
e = d / 12
f = e * 130
|
a ) 3 , b ) 9 , c ) 10 , d ) 12 , e ) 14 | b | subtract(subtract(add(subtract(subtract(subtract(subtract(subtract(subtract(subtract(32, 5), 5), 2), 2), const_1), 3), const_1), 2), 3), 5) | if 3 a β 2 b β 2 c = 32 and β 3 a - β ( 2 b + 2 c ) = 5 , what is the value of a + b + c ? | "when we look at the two equations , we can relize some similarity , so lets work on it . . 3 a β 2 b β 2 c = 32 can be written as β 3 a ^ 2 - β ( 2 b + 2 c ) ^ 2 = 32 { β 3 a - β ( 2 b + 2 c ) } { β 3 a + β ( 2 b + 2 c ) } = 32 . . or 4 * β 3 a + β ( 2 b + 2 c ) = 32 . . or β 3 a + β ( 2 b + 2 c ) = 8 . . now lets work on these two equations 1 ) β 3 a - β ( 2 b + 2 c ) = 4 . . 2 ) β 3 a + β ( 2 b + 2 c ) = 8 . . a ) add the two eq . . β 3 a + β ( 2 b + 2 c ) + β 3 a - β ( 2 b + 2 c ) = 12 . . 2 β 3 a = 12 . . or β 3 a = 6 . . 3 a = 36 . . a = 12 . b ) subtract 1 from 2 . . β 3 a + β ( 2 b + 2 c ) - β 3 a + β ( 2 b + 2 c ) = 4 . . 2 β ( 2 b + 2 c ) = 4 . . β ( 2 b + 2 c ) = 2 . . 2 b + 2 c = 4 . . or b + c = 2 . . from a and b a + b + c = 12 + 2 = 14 . . b" | a = 32 - 5
b = a - 5
c = b - 2
d = c - 2
e = d - 1
f = e - 3
g = f - 1
h = g + 2
i = h - 3
j = i - 5
|
a ) 12 , b ) 11 , c ) 14 , d ) 13 , e ) 15 | e | divide(1000, const_2) | if n is a positive integer and the product of all the integers from 1 to n , inclusive , is a multiple of 1000 , what is the least possible value of n ? | 1000 = 2 * 2 * 2 * 5 * 5 * 5 , so we need 3 factors of 2 and 3 factors of 5 . therefore , the smallest value of n is 15 , so the answer is e . | a = 1000 / 2
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a ) 7 : 20 , b ) 9 : 23 , c ) 11 : 27 , d ) 23 : 54 , e ) 31 : 77 | a | divide(add(1, 1), add(const_3.0, 2)) | in the first m games of a team ' s season , the ratio of the team ' s wins to its losses was 1 : 2 . in the subsequent n games , the ratio of the team Β΄ s wins to losses was 1 : 4 . if m : n = 4 : 5 , what was the ratio of the team ' s wins to its losses for all m + n games ? | "m = 4 / 9 of total games n = 5 / 9 of total games wins = 1 / 3 * 4 / 9 + 1 / 5 * 5 / 9 = 4 / 27 + 3 / 27 = 7 / 27 losses = 1 - 7 / 27 = 20 / 27 the ratio of wins to losses is 7 : 20 . the answer is a ." | a = 1 + 1
b = 3 + 0
c = a / b
|
a ) 10 , b ) 50 , c ) 55 , d ) 35 , e ) 75 | e | add(subtract(80, 120), subtract(50, 35)) | in an it company , there are a total of 120 employees including 50 programmers . the number of male employees is 80 , including 35 male programmers . how many employees must be selected to guaranty that we have 3 programmers of the same sex ? | "you could pick 70 non - programmers , 2 male programmers , and 2 female programmers , and still not have 3 programmers of the same sex . but if you pick one more person , you must either pick a male or a female programmer , so the answer is 75 . e" | a = 80 - 120
b = 50 - 35
c = a + b
|
a ) 48 days , b ) 55 days , c ) 24 days , d ) 33 days , e ) 40 days | a | divide(multiply(6, const_3), subtract(divide(add(divide(multiply(6, const_3), 12), add(divide(multiply(6, const_3), 6), divide(multiply(6, const_3), 8))), const_2), divide(multiply(6, const_3), 6))) | a and b can do a piece of work in 6 days , b and c in 8 days , c and a in 12 days . how long will c take to do it ? | "2 c = 1 / 8 + 1 / 12 β 1 / 6 = 1 / 24 c = 1 / 48 = > 48 days answer : a" | a = 6 * 3
b = 6 * 3
c = b / 12
d = 6 * 3
e = d / 6
f = 6 * 3
g = f / 8
h = e + g
i = c + h
j = i / 2
k = 6 * 3
l = k / 6
m = j - l
n = a / m
|
a ) 28 km , b ) 12 km , c ) 16 km , d ) 30 km , e ) 20 km | b | divide(5, add(divide(1, add(5, 1)), divide(1, subtract(5, 1)))) | john can row at 5 kmph and still water . he takes 5 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ? | let the distance between a and b be x km . total time = x / ( 5 + 1 ) + x / ( 5 - 1 ) = 5 = > x / 6 + x / 4 = 4 = > ( 2 x + 3 x ) / 12 = 5 = > x = 12 km . answer : b | a = 5 + 1
b = 1 / a
c = 5 - 1
d = 1 / c
e = b + d
f = 5 / e
|
a ) 58 , b ) 62 , c ) 66 , d ) 70 , e ) 74 | b | add(divide(circumface(12), const_2), multiply(const_2, 12)) | a semicircle has a radius of 12 . what is the approximate perimeter of the semicircle ? | "the perimeter of a circle is 2 * pi * r . the perimeter of a semicircle is 2 * pi * r / 2 + 2 r = pi * r + 2 r the perimeter is pi * 12 + 2 * 12 which is about 62 . the answer is b ." | a = circumface / (
b = a + 2
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a ) 328 , b ) 262 , c ) 72 , d ) 205 , e ) 267 | a | multiply(divide(64, 40), add(add(const_100, 65), 40)) | a certain sum of money is divided among a , b and c so that for each rs . a has , b has 65 paisa and c 40 paisa . if c ' s share is rs . 64 , find the sum of money ? | a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 64 41 - - - - ? = > rs . 328 answer : a | a = 64 / 40
b = 100 + 65
c = b + 40
d = a * c
|
a ) 19,872 , b ) 19,780 , c ) 19,680 , d ) 19,380 , e ) none of these | a | floor(divide(divide(multiply(add(multiply(multiply(add(multiply(add(const_1, const_4), const_2), 8), const_100), multiply(add(const_1, const_4), const_2)), multiply(add(const_1, const_4), const_100)), add(const_100, 8)), const_100), multiply(multiply(add(const_1, const_4), const_100), const_2))) | sonika bought a v . c . r . at the list price of 18,400 . if the rate of sales tax was 8 % , find the amount she had to pay for purchasing the v . c . r . | "sol . list price of v . c . r . = 18,400 rate of sales tax = 8 % β΄ sales tax = 8 % of 18,400 = 8 β 100 Γ 18400 = 1472 so , total amount which sonika had to pay for purchasing the v . c . r . = 18,400 + 1472 = 19,872 . answer a" | a = 1 + 4
b = a * 2
c = b + 8
d = c * 100
e = 1 + 4
f = e * 2
g = d * f
h = 1 + 4
i = h * 100
j = g + i
k = 100 + 8
l = j * k
m = l / 100
n = 1 + 4
o = n * 100
p = o * 2
q = m / p
r = math.floor(q)
|
a ) 1361 , b ) 1481 , c ) 1601 , d ) 1721 , e ) 1841 | b | add(multiply(divide(add(60, 80), const_2), add(subtract(80, 60), const_1)), add(divide(subtract(80, 60), const_2), const_1)) | if x is equal to the sum of the integers from 60 to 80 , inclusive , and y is the number of even integers from 60 to 80 , inclusive , what is the value of x + y ? | "x = 60 + 61 + . . . + 80 = 21 ( 70 ) = 1470 y = 11 x + y = 1481 the answer is b ." | a = 60 + 80
b = a / 2
c = 80 - 60
d = c + 1
e = b * d
f = 80 - 60
g = f / 2
h = g + 1
i = e + h
|
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | b | multiply(10, const_1) | a mixture contains alcohol and water in the ratio 2 : 5 . if 10 liters of water is added to the mixture , the ratio becomes 2 : 7 . find the quality of alcohol in the given mixture . | let the quantity of alcohol and water be 2 x and 5 x 2 x / ( 5 x + 10 ) = 2 / 7 20 x = 4 ( 3 x + 5 ) x = 5 quantity of alcohol = 2 * 5 = 10 liters . answer is b | a = 10 * 1
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a ) 100 , b ) 200 , c ) 500 , d ) 1000 , e ) 5000 | c | divide(multiply(power(10, 6), 5.4), 10800) | in the biology lab of ` ` jefferson ' ' high school there are 5.4 * 10 ^ 6 germs , equally divided among 10800 petri dishes . how many germs live happily in a single dish ? | 5.4 * 10 ^ 6 / 10800 = = > 54 * 10 ^ 3 / 108 = = > 54000 / 108 = 500 answer : c | a = 10 ** 6
b = a * 5
c = b / 10800
|
a ) 88.8 , b ) 91.0 , c ) 88.0 , d ) 70.9 , e ) 71.2 | a | add(80, multiply(divide(11, const_100), 80)) | if x is 11 percent greater than 80 , then x = | "11 % of 80 = ( 80 * 0.11 ) = 8.8 11 % greater than 80 = 80 + 8.8 = 88.8 answer is clearly a ." | a = 11 / 100
b = a * 80
c = 80 + b
|
a ) s . 109 , b ) s . 108 , c ) s . 110 , d ) s . 103 , e ) s . 102 | c | multiply(5, divide(1628, add(add(multiply(6, 3), multiply(9, 4)), multiply(4, 5)))) | a , b and c completed a piece of work , a worked for 6 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was rs . 1628 ? | "3 x 4 x 5 x 6 9 4 18 x + 36 x + 20 x = 1628 74 x = 1628 = > x = 22 5 x = 110 rs . answer : c" | a = 6 * 3
b = 9 * 4
c = a + b
d = 4 * 5
e = c + d
f = 1628 / e
g = 5 * f
|
a ) $ 314.16 , b ) $ 314.17 , c ) $ 314.18 , d ) $ 314.19 , e ) $ 314.20 | a | add(314.12, divide(const_3, const_100)) | 8 people decided to split the restaurant bill evenly . if the bill was $ 314.12 dollars , how much money did they 1 cent is the smallest unit ? | "if the last three digits of a whole number are divisible by 8 , then the entire number is divisible by 8 the last 3 digit 412 not divisible by a hence , we need to add 4 to this number for it to be divisible by 8 correct option : a" | a = 3 / 100
b = 314 + 12
|
a ) $ 1,800 , b ) $ 1,500 , c ) $ 1,300 , d ) $ 1,600 , e ) $ 1,100 | d | subtract(subtract(subtract(multiply(1200, 4), multiply(subtract(4, const_1), 1000)), divide(subtract(multiply(1200, 4), multiply(subtract(4, const_1), 1000)), add(divide(50, const_100), const_1))), const_100) | the average ( arithmetic mean ) monthly income of 4 workers is $ 1000 . after one worker β s income increases by 50 percent the new average income is $ 1200 . what was the original income of the worker whose monthly income increased ? | increase in total income was 200 * 4 = $ 800 , we know that this increase was 50 % ( 1 / 2 ) of the workers original income , thus his / her original income was 800 * 2 = $ 1,600 . answer : d | a = 1200 * 4
b = 4 - 1
c = b * 1000
d = a - c
e = 1200 * 4
f = 4 - 1
g = f * 1000
h = e - g
i = 50 / 100
j = i + 1
k = h / j
l = d - k
m = l - 100
|
a ) 3 / 4 , b ) 1 / 4 , c ) 1 / 2 , d ) 1 , e ) none of them | a | divide(multiply(subtract(const_100, 8), 15), multiply(add(const_100, 15), 16)) | if the numerator of a fraction be increased by 15 % and its denominator be diminished by 8 % , the value of the fraction is 15 / 16 . find the original fraction . | "let the original fraction be x / y . then ( 115 % of x ) / ( 92 % of y ) = 15 / 16 = > ( 115 x / 92 y ) = 15 / 16 = ( ( 15 / 16 ) * ( 92 / 115 ) ) = 3 / 4 answer is a ." | a = 100 - 8
b = a * 15
c = 100 + 15
d = c * 16
e = b / d
|
a ) 16 % , b ) 25 % , c ) 32 % , d ) 40 % , e ) 50 % | e | multiply(divide(subtract(72, 36), 72), const_100) | in town x , 72 percent of the population are employed , and 36 percent of the population are employed males . what percent of the employed people in town x are females ? | "total employed people 72 % , out of which 36 are employed males , hence 36 % are employed females . ( employed females ) / ( total employed people ) = 36 / 72 = 1 / 2 = 50 % answer : e ." | a = 72 - 36
b = a / 72
c = b * 100
|
a ) 140 m , b ) 141 m , c ) 142 m , d ) 143 m , e ) 144 m | b | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 100), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 100 resolutions . | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 100 resolutions . = 100 * 2 * 22 / 7 * 22.4 = 14100 cm = 141 m answer : b" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 100
i = h / 100
|
a ) 92 hours , b ) 24 hours , c ) 84 hours , d ) 48 hours , e ) none | c | add(divide(210, add(9, 6)), divide(210, subtract(9, 6))) | speed of a boat in standing water is 9 kmph and the speed of the stream is 6 kmph . a man rows to place at a distance of 210 km and comes back to the starting point . the total time taken by him is : | "sol . speed upstream = 3 kmph ; speed downstream = 9 kmph . β΄ total time taken = [ 210 / 3 + 210 / 9 ] hours = 84 hours . answer c" | a = 9 + 6
b = 210 / a
c = 9 - 6
d = 210 / c
e = b + d
|
a ) 49 m 2 , b ) 50 m 2 , c ) 53.5 m 2 , d ) 68.5 m 2 , e ) 57 m 2 | d | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 9))), multiply(4, 9)) | a cistern 9 m long and 4 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is : | "area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 9 x 1.25 ) + 9 x 4 ] m 2 = 68.5 m 2 . answer : option d" | a = 25 / 100
b = a + 1
c = b * 4
d = 25 / 100
e = d + 1
f = e * 9
g = c + f
h = 2 * g
i = 4 * 9
j = h + i
|
a ) 55 % , b ) 60 % , c ) 50 % , d ) 45 % , e ) 66 % | b | multiply(divide(multiply(divide(2,500, 2), add(1, divide(1, 5))), 2,500), const_100) | at a contest with 2,500 participants , 1 / 2 of the people are aged 10 to 12 . next year , the number of people aged 10 to 12 will increase by 1 / 5 . after this change , what percentage of the total 2,500 people will the 10 - to 12 - year - olds represent ? | "i just wanted to mention a couple of things here : * this is a pure ratio question ; the number 2,500 is completely irrelevant , and you can ignore it if you like . when we increase something by 1 / 5 , we are multiplying it by 1 + 1 / 5 = 6 / 5 , so the answer here must be ( 1 / 2 ) * ( 6 / 5 ) = 3 / 5 = 60 % . answer : b" | a = 2 / 500
b = 1 / 5
c = 1 + b
d = a * c
e = d / 2
f = e * 100
|
a ) 3 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | c | divide(140, add(24, 4)) | a boat can travel with a speed of 24 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 140 km downstream | "explanation : speed of the boat in still water = 24 km / hr speed of the stream = 4 km / hr speed downstream = ( 22 + 5 ) = 28 km / hr distance travelled downstream = 140 km time taken = distance / speed = 140 / 28 = 5 hours answer : option c" | a = 24 + 4
b = 140 / a
|
a ) 90 , b ) 85 , c ) 75 , d ) 95 , e ) 82 | a | add(choose(10, 2), choose(10, 2)) | there are 10 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed . | first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 10 and for second = 9 so total methods = 10 * 9 = 90 answer : a | a = math.comb(10, 2)
b = math.comb(10, 2)
c = a + b
|
a ) a ) 442 , b ) b ) 490 , c ) c ) 380 , d ) d ) 450 , e ) e ) 690 | a | multiply(17, 26) | a large box contains 17 small boxes and each small box contains 26 chocolate bars . how many chocolate bars are in the large box ? | the only thing you have to do is a simple multiply we have : 17 * 26 = 442 correct answer is : a ) 442 | a = 17 * 26
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a ) 233 : 97 , b ) 235 : 98 , c ) 236 : 97 , d ) 265 : 98 , e ) 234 : 97 | a | divide(add(divide(9, add(9, 2)), add(divide(3, add(3, 2)), divide(7, add(7, 3)))), add(divide(2, add(9, 2)), add(divide(2, add(3, 2)), divide(3, add(7, 3))))) | find out the ratio of milk to that of water , if a mixture is formed after mixing milk and water of 3 different vessels containning the mixture of milk to that of water in the proportion 3 : 2 , 7 : 3 , 9 : 2 respectively . | 3 + 2 = 57 + 3 = 109 + 2 = 11 now , 3 / 5 + 7 / 10 + 9 / 11 : 2 / 5 + 3 / 10 + 2 / 11 = > 233 / 110 : 97 / 110 = > 233 : 97 answer : a | a = 9 + 2
b = 9 / a
c = 3 + 2
d = 3 / c
e = 7 + 3
f = 7 / e
g = d + f
h = b + g
i = 9 + 2
j = 2 / i
k = 3 + 2
l = 2 / k
m = 7 + 3
n = 3 / m
o = l + n
p = j + o
q = h / p
|
a ) 86 , b ) 16 , c ) 76 , d ) 65 , e ) 26 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(85, 3)) | a student scored an average of 85 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | given m + p + c = 85 * 3 = 255 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 255 = 65 answer : d | a = 90 * 2
b = 70 * 2
c = a + b
d = 85 * 3
e = c - d
|
a ) a ) 11 , b ) b ) 18 , c ) c ) 20 , d ) d ) 21 , e ) e ) 22 | c | divide(subtract(100, 1), 2) | if you write down all the numbers from 1 to 100 , then how many times do you write 2 ? | "explanation : clearly , from 1 to 100 , there are ten numbers with 2 as the unit ' s digit - 2 , 12 , 22 , 32 , 42 , 52 , 62 , 72 , 82 , 92 ; and ten numbers with 2 as the ten ' s digit - 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 . so , required number = 10 + 10 = 20 . answer : c" | a = 100 - 1
b = a / 2
|
a ) 5 , b ) 10 , c ) 12 , d ) 15 , e ) 20 | b | sqrt(divide(add(subtract(power(20, const_2), power(subtract(20, divide(subtract(20, 10), const_2)), const_2)), power(divide(subtract(20, 10), const_2), const_2)), const_2)) | abce is an isosceles trapezoid and acde is a rectangle . ab = 10 and ec = 20 . what is the length of ae . | as abce is an isosceles trapezoid and ec is twice times ab , we can consider the given trapezoid as half of a regular hexagon ( as length of diagonal in regular hexagon is always double the side ) . thus , sides ab = ae = bc . = > ae = 10 . answer : b | a = 20 ** 2
b = 20 - 10
c = b / 2
d = 20 - c
e = d ** 2
f = a - e
g = 20 - 10
h = g / 2
i = h ** 2
j = f + i
k = j / 2
l = math.sqrt(k)
|
a ) 631.344 , b ) 622.443 , c ) 623.434 , d ) 632.423 , e ) 632.443 | b | divide(24.39, divide(562.093, 24.39)) | evaluate 24.39 + 562.093 + 35.96 | "24.39 + 562.093 + 35.96 = 622.443 option b" | a = 562 / 93
b = 24 / 39
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a ) s 4000 , b ) s 1800 , c ) s 800 , d ) s 500 , e ) s 1300 | e | divide(subtract(multiply(3400, divide(5, const_100)), 144), subtract(divide(5, const_100), divide(3, const_100))) | rs 3400 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 / 3400 ) * 100 = 4.24 ratio = 7.65 : 12.35 so , first part = ( 7.65 / 20 ) * 3400 = rs 1300 . answer : e" | a = 5 / 100
b = 3400 * a
c = b - 144
d = 5 / 100
e = 3 / 100
f = d - e
g = c / f
|
a ) 10 % , b ) 12 % , c ) 14 % , d ) 16 % , e ) 18 % | a | divide(multiply(divide(multiply(40, 50), const_100), 50), const_100) | at a certain college , 40 percent of the total number of students are freshmen . if 50 percent of the fresh - men are enrolled in the school of liberal arts and , of these , 50 percent are psychology majors , what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? | "let ' s say there is a total of 100 students at this college . 40 percent of the total number of students are freshmen . # of freshmen = 40 % of 100 = 40 50 percent of the fresh - men are enrolled in the school of liberal arts . . . number of liberal arts freshmen = 50 % of 40 = 20 . . . and , of these , 50 percent are psychology majors . . . number of liberal arts freshmen who are psychology majors = 50 % of 20 = 10 what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? 10 / 100 = 10 % answer : a" | a = 40 * 50
b = a / 100
c = b * 50
d = c / 100
|
a ) 125 , b ) 150 , c ) 225 , d ) 250 , e ) 500 | b | sqrt(divide(multiply(90, const_100), divide(40, const_100))) | 90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 40 % of the total school population of x students , what is x ? | "90 = ( x / 100 ) * ( # boys ) # boys = . 4 x substitute and solve : 90 = ( x / 100 ) * ( . 4 x ) 9000 = . 4 ( x ^ 2 ) 25000 = x ^ 2 x = 150 answer : b" | a = 90 * 100
b = 40 / 100
c = a / b
d = math.sqrt(c)
|
a ) 30 , b ) 20 , c ) 25 , d ) 40 , e ) 50 | c | subtract(multiply(subtract(60, 20), const_3), multiply(subtract(60, 30), const_3)) | in x game of billiards , x can give y 20 points in 60 and he can give z 30 points in 60 . how many points can y give z in x game of 100 ? | "c 25 x scores 60 while y score 40 and z scores 30 . the number of points that z scores when y scores 100 = ( 100 * 30 ) / 40 = 25 * 3 = 75 . in x game of 100 points , y gives ( 100 - 75 ) = 25 points to c ." | a = 60 - 20
b = a * 3
c = 60 - 30
d = c * 3
e = b - d
|
a ) 3451.21 , b ) 3029.0 , c ) 2993.03 , d ) 2984.36 , e ) none of these | a | subtract(3500, divide(1000, 20.5)) | 3500 β ( 1000 Γ· 20.50 ) is equal to : | explanation : given exp . = 3500 - ( ( 1000 Γ· 2050 ) * 100 ) = 3451.21 answer a | a = 1000 / 20
b = 3500 - a
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a ) 3 kmph , b ) 2.5 kmph , c ) 13 kmph , d ) 6.5 kmph , e ) 7 : 3 kmph | b | divide(add(divide(40, 4), divide(60, 4)), const_2) | a man swims downstream 60 km and upstream 40 km taking 4 hours each time ; what is the speed of the current ? | "explanation : 60 - - - 4 ds = 15 ? - - - - 1 40 - - - - 4 us = 10 ? - - - - 1 s = ? s = ( 15 - 10 ) / 2 = 2.5 answer : option b" | a = 40 / 4
b = 60 / 4
c = a + b
d = c / 2
|
a ) 24 , 22 , b ) 32 , 16 , c ) 30 , 18 , d ) 28 , 14 , e ) none of these | b | subtract(multiply(divide(add(120, 120), 10), 2), divide(add(120, 120), 15)) | two trains of equal length 120 metres move in the same direction . the faster train completely overtakes the slower one in 15 seconds . if the slower train were to move at half its speed , the over taking would take in 10 seconds . at what speeds are the 2 trains moving ( faster and slower respectively in m / s ) | the total distance covered for over taking = length of the two trains = 120 + 120 = 240 speed of faster train be x m / s and slower train be y m / s in the first case relative speed = x - y relative speed = distance / time taken = x - y = 240 / 15 = 16 m / s β¦ β¦ β¦ ( 1 ) in the second case , the slower train moves at half its speed relative speed = x - 0.5 y = 240 / 10 = 24 m / s β¦ β¦ β¦ β¦ β¦ β¦ β¦ . . ( 2 ) solving equation 1 & 2 we get x = 32 and y = 16 . : . the speed of the faster and slower trains are 32 m / s and 16 m / s respectively answer : b | a = 120 + 120
b = a / 10
c = b * 2
d = 120 + 120
e = d / 15
f = c - e
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a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | e | multiply(subtract(add(5, 4), 3), 4) | present ages of abi and suji are in the ratio of 5 : 4 respectively . 3 years hence , the ratio of their ages will become 11 : 9 respectively . what is suji ' s present age in years ? | present ages = 5 x : 4 x three hence = 5 x + 3 : 4 x + 3 5 x + 3 : 4 x + 3 = 11 : 9 weget x = 6 suji age is = 4 x = 4 * 6 = 24 answer : e | a = 5 + 4
b = a - 3
c = b * 4
|
a ) 8 , b ) 8 / 11 , c ) 8 / 13 , d ) 8 / 15 , e ) none | d | subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 4)) | a can do a work in 15 days and b in 20 days . if they work on it together for 4 days , then the fraction of the work that is left is ? | a ' s 1 day ' s work = 1 ; 15 b ' s 1 day ' s work = 1 ; 20 ( a + b ) ' s 1 day ' s work = ( 1 + 1 ) = 7 . 15 20 60 ( a + b ) ' s 4 day ' s work = ( 7 x 4 ) = 7 . 60 15 therefore , remaining work = ( 1 - 7 ) = 8 . 15 15 option d | a = 1 / 15
b = 1 / 20
c = a + b
d = c * 4
e = 1 - d
|
a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | a | multiply(divide(2, 4), multiply(const_4, 3)) | the ratio of flour to water to sugar in a recipe is 8 : 4 : 3 . the ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar . if the new recipe calls for 2 cups of water , how much sugar is required ? | "the ratio of flour to water is 16 : 4 = 4 : 1 . the ratio of flour to sugar is 4 : 3 . the new ratio of flour to water to sugar is 4 : 1 : 3 if we need 2 cups of water , then we need 6 cups of sugar . the answer is a ." | a = 2 / 4
b = 4 * 3
c = a * b
|
a ) 5 , b ) 6 , c ) 9 , d ) 8 , e ) 5 | c | multiply(const_3600, divide(divide(165, const_1000), add(60, 6))) | a train 165 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 165 * 3 / 55 ] sec = 9 sec answer : c" | a = 165 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 6 , b ) 8 , c ) 10 , d ) 5 , e ) 13 | d | divide(log(add(divide(multiply(12100, const_2), 100), const_1)), log(const_3)) | joan took out a mortgage from hel local bank . each monthly mortgage payment she makes must be triple the amount of the previous month ' s payment . if her first payment is $ 100 , and the total amount she must pay back is $ 12100 , how many months will it take joan to pay back her mortgage ? | "joan starts off with 100 $ . . which is to be tripled every month her monthly payments look like this : 100 , 300 , 900 , 2700 . . . . . . . . . upto 12100 this can be re written as : 100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 . . . . . . 100 x 121 so we have 1 , 3 , 9 , 27 . . . . . 12100 in gp we know that a = 1 , and r = 3 ( its easy to figure it out by looking at the question , but regardless of it being mentioned in the question we can still compute the value of r using the formula tn = a 3 ^ n - 1 . . . ) therefore to find the sum of n terms of a gp we use this formula : sn = a ( 1 - r ^ n ) / 1 - r 121 = 1 - 3 ^ n / 1 - 3 ; 1 - 3 ^ n / - 2 cross multiplying we get 121 x - 2 = 1 - 3 ^ n - 242 = 1 - 3 ^ n - 243 = - 3 ^ n 243 = 3 ^ n ( negatives cancel out ) 243 can also be re written as 3 ^ 5 therefore ; 3 ^ 5 = 3 ^ n thus n = 5 ( d )" | a = 12100 * 2
b = a / 100
c = b + 1
d = math.log(c)
e = math.log(3)
f = d / e
|
['a ) 4 : 25', 'b ) 4 : 15', 'c ) 3 : 25', 'd ) 3 : 15', 'e ) none of these'] | a | divide(const_4, power(5, const_2)) | the diagonals of two squares are in the ratio of 2 : 5 . find the ratio of their areas . | explanation : let the diagonals of the squares be 2 x and 5 x . then ratio of their areas will be area of square = 1 / 2 β diagonal 2 1 / 2 β 2 x 2 : 1 / 2 β 5 x 2 4 x 2 : 25 x 2 = 4 : 25 option a | a = 5 ** 2
b = 4 / a
|
a ) q = 350 , b ) q = 353 , c ) q = 354 , d ) q = 356 , e ) 357 | b | add(add(add(add(const_60, const_4), 4), multiply(add(add(const_60, const_4), 4), 4)), add(multiply(3, 1), multiply(multiply(3, 2), 2))) | s ( n ) is a n - digit number formed by attaching the first n perfect squares , in order , into one integer . for example , s ( 1 ) = 1 , s ( 2 ) = 14 , s ( 3 ) = 149 , s ( 4 ) = 14916 , s ( 5 ) = 1491625 , etc . how many digits q are in s ( 99 ) ? | focus on the points where the number of digits q in squares change : 1 , 2 , 3 - single digit squares . first 2 digit number is 10 . 4 , 5 , . . . 9 - two digit squares . to get 9 , the last number with two digit square , think that first 3 digit number is 100 which is 10 ^ 2 . so 9 ^ 2 must be the last 2 digit square . 10 , 11 , 12 , . . . 31 - three digit squares . to get 31 , think of 1000 - the first 4 digit number . it is not a perfect square but 900 is 30 ^ 2 . 32 ^ 2 = 2 ^ 10 = 1024 , the first 4 digit square . 32 - 99 - four digit squares . to get 99 , think of 10000 - the first 5 digit number which is 100 ^ 2 . so number of digits in s ( 99 ) = 3 * 1 + 6 * 2 + 22 * 3 + 68 * 4 = 3 + 12 + 66 + 272 = 353 . b | a = const_60 + 4
b = a + 4
c = const_60 + 4
d = c + 4
e = d * 4
f = b + e
g = 3 * 1
h = 3 * 2
i = h * 2
j = g + i
k = f + j
|
a ) 2287 , b ) 140 , c ) 128 , d ) 210 , e ) 123 | d | multiply(subtract(divide(300, divide(30, const_100)), 300), divide(30, const_100)) | bhanu spends 30 % of his income on petrol on scooter 30 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ? | "given 30 % ( income ) = 300 β β income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 30 % ( 700 ) = rs . 210 answer : d" | a = 30 / 100
b = 300 / a
c = b - 300
d = 30 / 100
e = c * d
|
['a ) 30 pi m ^ 2', 'b ) 40 pi m ^ 2', 'c ) 60 pi m ^ 2', 'd ) 80 pi m ^ 2', 'e ) 100 pi m ^ 2'] | c | multiply(10, sqrt(subtract(power(10, const_2), power(8, const_2)))) | the slant height of a right circular cone is 10 m and its height is 8 m . find the area of its curved surface . | l = 10 m h = 8 m r = root ( i ^ 2 - h ^ 2 ) = root ( 10 ^ 2 - 8 ^ 2 ) = 6 m curved surface area = pi ri = 60 pi m ^ 2 answer c | a = 10 ** 2
b = 8 ** 2
c = a - b
d = math.sqrt(c)
e = 10 * d
|
a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | d | multiply(40, const_1) | at veridux corporation , there are 250 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 150 male associates , how many female managers are there ? | "250 employees : 90 male , 160 female 40 managers , 210 associates 150 male associates implies 60 female associates which means the remaining 30 females must be managers d . 30" | a = 40 * 1
|
a ) 178.27 cm , b ) 181.29 cm , c ) 978.29 cm , d ) 178.89 cm , e ) 176.29 cm | b | floor(divide(add(subtract(multiply(35, 183), 166), 106), 35)) | the average height of 35 boys in a class was calculated as 183 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . ? | "calculated average height of 35 boys = 183 cm . wrong total height of 35 boys = 183 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 183 cm - ( 166 cm - 106 cm ) / 35 = 183 cm - 60 / 35 cm = 183 cm - 1.71 cm = 181.29 cm . answer : b" | a = 35 * 183
b = a - 166
c = b + 106
d = c / 35
e = math.floor(d)
|
a ) 22 , b ) 77 , c ) 29 , d ) 66 , e ) 21 | d | add(65, divide(multiply(5, 12), divide(180, 3))) | 65 + 5 * 12 / ( 180 / 3 ) = ? | "65 + 5 * 12 / ( 180 / 3 ) = 65 + 5 * 12 / ( 60 ) = 65 + ( 5 * 12 ) / 60 = 65 + 1 = 66 . answer : d" | a = 5 * 12
b = 180 / 3
c = a / b
d = 65 + c
|
a ) 3 : 6 , b ) 1.5 : 1 , c ) 3 : 1.5 , d ) 2 : 1 , e ) 3 : 2 | b | divide(1.5, divide(1.5, 1.5)) | what is the ratio between perimeters of two squares one having 1.5 times the diagonal then the other ? | "d = 1.5 d d = d a β 2 = 1.5 d a β 2 = d a = 1.5 d / β 2 a = d / β 2 = > 1.5 : 1 answer : b" | a = 1 / 5
b = 1 / 5
|
a ) 1 , b ) 2 , c ) 4 , d ) 8 , e ) 12 | e | multiply(const_2, sqrt(power(6, const_2))) | a circular garden is surrounded by a fence of negligible width along the boundary . if the length of the fence is 1 / 6 of th area of the garden . what is the radius of the circular garden ? | "as per the question - - width is negligible now , let l be the length of the fence = 2 pir l = 1 / 6 ( pir ^ 2 ) pir ^ 2 = 12 pir r = 12 answer : e" | a = 6 ** 2
b = math.sqrt(a)
c = 2 * b
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(4, 11) | what is the 28 th digit to the right of the decimal point in the decimal form of 4 / 11 ? | "4 / 11 = 0.363636 . . . the even numbered positions in the decimal expansion are all 6 . the answer is d ." | a = 4 / 11
|
a ) 125 , b ) 250 , c ) 300 , d ) 330 , e ) none of these | a | divide(divide(multiply(const_100, multiply(const_1000, 66)), multiply(const_60, const_1)), multiply(multiply(const_2, 140), add(const_3, divide(add(const_2, multiply(const_3, const_4)), power(add(const_2, multiply(const_4, const_2)), const_2))))) | the radius of the wheel of a bus is 140 cms and the speed of the bus is 66 km / h , then the r . p . m . ( revolutions per minutes ) of the wheel is | "radius of the wheel of bus = 140 cm . then , circumference of wheel = 2 Γ― β¬ r = 280 Γ― β¬ = 880 cm distance covered by bus in 1 minute = 66 Γ’ Β β 60 Γ£ β 1000 Γ£ β 100 cms distance covered by one revolution of wheel = circumference of wheel = 880 cm Γ’ Λ Β΄ revolutions per minute = 6600000 / 60 Γ£ β 880 = 125 answer a" | a = 1000 * 66
b = 100 * a
c = const_60 * 1
d = b / c
e = 2 * 140
f = 3 * 4
g = 2 + f
h = 4 * 2
i = 2 + h
j = i ** 2
k = g / j
l = 3 + k
m = e * l
n = d / m
|
a ) 5 , b ) 5 1 / 2 , c ) 6 , d ) 8 , e ) none of these | a | divide(multiply(multiply(divide(const_1, 15), 10), 15), const_2) | a can finish a work in 15 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ? | "b ' s 10 day ' s work = ( 1 / 15 * 10 ) = 2 / 3 remaining work = ( 1 - 2 / 3 ) = 1 / 3 now , 1 / 18 work is done by a in 1 day 1 / 3 work is done by a in ( 15 * 1 / 3 ) = 5 days . correct option : a" | a = 1 / 15
b = a * 10
c = b * 15
d = c / 2
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | subtract(9, add(divide(subtract(50, add(18, 17)), 3), const_2)) | in the county of veenapaniville , there are a total of 50 high schools , of 3 kinds : 25 public schools , 16 parochial schools , and 9 private independent schools . these 50 schools are divided between 3 districts : a , b , and c . district a has 18 high schools total . district b has 17 high schools total , and only two of those are private independent schools . if district c has an equal number of each of the 3 kinds of schools , how many private independent schools are there in district a ? | β in the county of veenapaniville , there are a total of 50 high schools , of three kinds : 25 public schools , 16 parochial schools , and 9 private independent schools . β β district a has 18 high schools total . β β district b has 17 high schools total , and only two of those are private independent schools . β β district c has an equal number of each of the three kinds of schools . β hmm . that β s not a simple number . let β s put that on hold for a moment . so far , from the matrix it looks like we could figure out the missing number in the β totals β row . that number must be 15 , because 18 + 17 + 15 = 50 . now , go back to that last fact : β district c has an equal number of each of the three kinds of schools . β now that makes sense . district c has 15 schools , so it must have 5 of each kind : the question is asking : β how many private independent schools are there in district a ? β well , now that β s just a matter of adding across the β private independent β row . we know 2 + 2 + 5 = 9 , so the missing number must be 2 . there β s still a good deal of information we can β t figure out from the information given in the question , but we are able to answer the specific question asked here . there are 2 private independent schools in district a , so the answer = ( a ) . | a = 18 + 17
b = 50 - a
c = b / 3
d = c + 2
e = 9 - d
|
a ) 400 meters , b ) 120 meters , c ) 1777 meters , d ) 60 meters , e ) none of these | b | multiply(multiply(subtract(100, 64), const_0_2778), 12) | a train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 12 seconds . what is the length of the train in meters ? | "train overtakes a bike means that we are talking about total length of the train . ( train ' s head is close to bike when it started and its tail crosses the bike when it overtakes the bike ) relative speed = 100 - 64 = 36 km / h = 36000 m / h time = 12 seconds distance = speed * time 36000 * 12 / 3600 = 120 meters . b is the answer ." | a = 100 - 64
b = a * const_0_2778
c = b * 12
|
a ) 17 , b ) 13 , c ) 12 , d ) 11 , e ) 10 | a | subtract(subtract(subtract(22, 2), const_4), const_1) | how many positive integers less than 22 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3 ? | "the number = 2 a + 3 b < 20 when a = 1 , b = 1 , 2 , 3 , 4 , 5 , 6 - > 2 a = 2 ; 3 b = 3 , 6 , 9 , 12 , 15 , 18 - > the number = 5 , 8 , 11 , 14 , 17 , 20 - - > 6 numbers when a = 2 , b = 1 , 2,3 , 4,5 , 6 - > . . . . - - > 6 numbers when a = 3 , b = 1,2 , 3,4 , 5 - - > . . . . - - > 5 numbers total number is already 17 . look at the answer there is no number greater than 17 - - > we dont need to try any more answer must be a" | a = 22 - 2
b = a - 4
c = b - 1
|
a ) 10.6 kg , b ) 3 kg , c ) 11 kg , d ) 14.9 kg , e ) none | b | subtract(multiply(add(19, const_1), 14.4), multiply(19, 15)) | the average weight of 19 students is 15 kg . by the admission of a new student the average weight is reduced to 14.4 kg . the weight of the new student is ? | "answer weight of new student = total weight of all 20 students - total weight of initial 19 students = ( 20 x 14.4 - 19 x 15 ) kg = 3 kg . correct option : b" | a = 19 + 1
b = a * 14
c = 19 * 15
d = b - c
|
a ) 2 , b ) 4 / 5 , c ) 7 / 9 , d ) 10 , e ) 84 / 25 | e | divide(const_1, add(add(divide(const_1, 21), divide(const_1, add(const_4, const_2))), divide(const_1, multiply(const_2, add(const_4, const_2))))) | a , b , c can complete a piece of work in 21 , 6,12 days . working together , they complete the same work in how many days ? | "a + b + c 1 day work = 1 / 21 + 1 / 6 + 1 / 12 = 25 / 84 a , b , c together will complete the job in 84 / 25 days answer is e" | a = 1 / 21
b = 4 + 2
c = 1 / b
d = a + c
e = 4 + 2
f = 2 * e
g = 1 / f
h = d + g
i = 1 / h
|
a ) 14 , b ) 15 , c ) 20 , d ) 22 , e ) 23 | b | divide(multiply(subtract(26, 6), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 26 years . what is deepak present age | "explanation : present age is 4 x and 3 x , = > 4 x + 6 = 26 = > x = 5 so deepak age is = 3 ( 5 ) = 15 answer : option b" | a = 26 - 6
b = a * 3
c = b / 4
|
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