Datasets:
euclaise
/
mathqa_programs

Dataset card Data Studio Files
xet
Community
1
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
a ) 9 and 11 , b ) 19 and 21 , c ) 8 and 10 , d ) 10 and 15 , e ) 10 and 12
b
add(multiply(divide(subtract(multiply(10, const_4), 2), 2), const_100), add(divide(subtract(multiply(10, const_4), 2), 2), 2))
two workers completed the painting of a building in 10 days . if they were to paint the building separately , how many days will it take each worker , if one of them can complete it in approximately 2 days earlier than the other .
work = ( a ) ( b ) / ( a + b ) where a and b are the individual times of each entity . here , we ' re told that ( working together ) the two workers would complete a job in 10 days . this means that ( individually ) each of them would take more than 10 days to do the job . answers e , a and c are illogical , since the individual times must both be greater than 10 days . so we can test the values for answers b and d . using the values for answers b and d . . . answer b : ( 19 ) ( 21 ) / ( 19 + 21 ) = 399 / 40 = 9.97 this is a match ( complete it in approximately 2 days earlier than the other ) this is a match final answer : b
a = 10 * 4 b = a - 2 c = b / 2 d = c * 100 e = 10 * 4 f = e - 2 g = f / 2 h = g + 2 i = d + h
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16
e
add(subtract(multiply(const_2, 15), add(15, 3)), 4)
frank is 15 years younger then john . in 3 years john will be twice as old as frank . how old will frank be in 4 years ?
frank is 15 years younger then john - - > f + 15 = j ; in 5 years john will be twice as old as frank ( in 3 year john will be j + 3 years old and frank will be f + 3 years old ) - - > j + 3 = 2 * ( f + 3 ) - - > ( f + 15 ) + 3 = 2 * ( f + 3 ) - - > f = 12 ; in 4 years frank will be 12 + 4 = 16 years old . answer : e .
a = 2 * 15 b = 15 + 3 c = a - b d = c + 4
a ) 800 , b ) 850 , c ) 900 , d ) 500 , e ) 1000
e
multiply(50, multiply(5, const_10))
last year , for every 100 million vehicles that traveled on a certain highway , 50 vehicles were involved in accidents . if 5 billion vehicles traveled on the highway last year , how many of those vehicles were involved in accidents ? ( 1 billion = 1,000 , 000,000 )
"to solve we will set up a proportion . we know that “ 100 million vehicles is to 50 accidents as 5 billion vehicles is to x accidents ” . to express everything in terms of “ millions ” , we can use 5,000 million rather than 5 billion . creating a proportion we have : 100 / 50 = 5,000 / x cross multiplying gives us : 100 x = 2,000 * 50 x = 20 * 50 = 1000 answer : e"
a = 5 * 10 b = 50 * a
a ) 50 kmph , b ) 82 kmph , c ) 62 kmph , d ) 65 kmph , e ) 75 kmph
b
subtract(multiply(divide(280, 9), const_3_6), 30)
a man sitting in a train which is traveling at 30 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . ?
"relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) * ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 30 ) kmph = 82 kmph . answer : b ."
a = 280 / 9 b = a * const_3_6 c = b - 30
a ) 72 , b ) 80 , c ) 120 , d ) 40 , e ) 150
e
divide(add(72, multiply(72, divide(2, 3))), subtract(const_1, divide(20, const_100)))
in a certain school , 20 % of students are below 8 years of age . the number of students above 8 years of age is 2 / 3 of the number of students of 8 years of age which is 72 . what is the total number of students in the school ?
"explanation : let the number of students be x . then , number of students above 8 years of age = ( 100 - 20 ) % of x = 80 % of x . 80 % of x = 72 + 2 / 3 of 72 80 / 100 x = 120 x = 150 . answer : option e"
a = 2 / 3 b = 72 * a c = 72 + b d = 20 / 100 e = 1 - d f = c / e
a ) 1 / 7 , b ) 16 , c ) 15 , d ) 52 / 7 , e ) 60 / 7
b
multiply(divide(add(const_60.0, 4), add(3, 4)), divide(subtract(multiply(3, divide(add(8, 4), add(3, 4))), 4), 3))
if 4 x + 3 y = 8 and y - 3 x = 8 , then what is the value of x + 4 y ?
"4 x + 3 y = 8 . . . equation 1 - 3 x + y = 8 . . . equation 2 adding both the equations x + 4 y = 16 correct answer option b"
a = const_60 + 0 b = 3 + 4 c = a / b d = 8 + 4 e = 3 + 4 f = d / e g = 3 * f h = g - 4 i = h / 3 j = c * i
a ) 343 , b ) 677 , c ) 800 , d ) 867 , e ) 932
c
divide(add(212, 28), divide(30, const_100))
mike needs 30 % to pass . if he scored 212 marks and falls short by 28 marks , what was the maximum marks he could have got ?
"if mike had scored 28 marks more , he could have scored 30 % therefore , mike required 212 + 28 = 240 marks let the maximum marks be m . then 30 % of m = 240 ( 30 / 100 ) × m = 240 m = ( 240 × 100 ) / 30 m = 24000 / 30 m = 800 answer : c"
a = 212 + 28 b = 30 / 100 c = a / b
a ) 1000 , b ) 900 , c ) 800 , d ) 400 , e ) 1200
c
multiply(divide(multiply(multiply(50, 100), const_2), add(50, 100)), 12)
a man can ride on motorcycle at 50 kmph upward road and 100 kmph downward road . he takes 12 hours to ride motorcycle uphill from lower point a to upper point b and back to a . what is the total distance traveled by him in 12 hours ? he must return back to base point a in remaining time .
upward distance traveled per hour - 50 km , distance traveled after 8 hours = 400 km . he remained with 4 hours travel downward , i . e . distance traveled in remaining 4 hours downward = 400 km so total distance traveled from a to b and back to point a = 400 km upward + 400 km downhill = 800 km ( answer : c )
a = 50 * 100 b = a * 2 c = 50 + 100 d = b / c e = d * 12
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
d
subtract(42, divide(multiply(6, 42), 7))
7 machines at a certain factory operate at the same constant rate . if 6 of these machines , operating simultaneously , take 42 hours to fill a certain production order , how many fewer hours does it take all 7 machines , operating simultaneously , to fill the same production order ?
the total work is 6 * 42 = 252 machine hours the time required for seven machines is 252 / 7 = 36 hours , thus 6 fewer hours . the answer is d .
a = 6 * 42 b = a / 7 c = 42 - b
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
a
subtract(subtract(99, 83), divide(99, add(const_1, const_10)))
what number is obtained by adding the units digits of 734 ^ 99 and 347 ^ 83 ?
the units digit of 734 ^ 99 is 4 because 4 raised to the power of an odd integer ends in 4 . the units digit of 347 ^ 83 is 3 because powers of 7 end in 7 , 9 , 3 , or 1 cyclically . since 83 is in the form 4 n + 3 , the units digit is 3 . then 4 + 3 = 7 . the answer is a .
a = 99 - 83 b = 1 + 10 c = 99 / b d = a - c
a ) s . 9621 , b ) s . 6921 , c ) s . 8400 , d ) s . 6261 , e ) s . 6361
c
multiply(8000, power(add(const_1, divide(5, const_100)), 1))
the amount of principal rs . 8000 at compound interest at the ratio of 5 % p . a . for 1 year is
"c . i = p ( 1 + r / 100 ) ^ n = 8000 ( 1 + 5 / 100 ) = rs 8400 answer : c"
a = 5 / 100 b = 1 + a c = b ** 1 d = 8000 * c
a ) 90 , b ) 120 , c ) 160 , d ) 360 , e ) 600
e
divide(150, subtract(const_1, divide(3, 4)))
the visitors of a modern art museum who watched a certain picasso painting were asked to fill in a short questionnaire indicating whether they had enjoyed looking at the picture and whether they felt they had understood it . according to the results of the survey , all 150 visitors who did not enjoy the painting also did not feel they had understood the painting , and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting . if 3 / 4 of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting , then how many visitors answered the questionnaire ?
"if we exclude those cases and take the question at face value , then it seems straightforward . group # 1 = ( did n ' t like , did n ' t understand ) = 150 group # 2 = ( likeunderstood ) = 3 / 4 ( 1 / 4 ) n = 150 n = 600 answer = ( e )"
a = 3 / 4 b = 1 - a c = 150 / b
a ) 4 ^ 14 , b ) 4 ^ 16 , c ) 4 ^ 18 , d ) 4 ^ 20 , e ) 4 ^ 22
c
multiply(power(4, 16), 16)
if x = 4 ^ 16 and x ^ x = 4 ^ k , what is k ?
"solution : we know that x = 4 ^ 16 which implies x ^ x = ( 4 ^ 16 ) ^ ( 4 ^ 16 ) = 4 ^ ( 16 * 4 ^ 16 ) [ because ( x ^ y ) ^ z = x ^ ( y * z ) ) ] so 4 ^ ( 4 ^ 2 * 4 ^ 16 ) = 4 ^ ( 4 ^ ( 2 + 16 ) ) [ because x ^ a * x ^ b = x ^ ( a + b ) ] therefore x ^ x = 4 ^ ( 4 ^ 18 ) given that x ^ x = 4 ^ k so 4 ^ ( 4 ^ 18 ) = 4 ^ k since the base is same the exponent will also be same therefore k = 4 ^ 18 answer : c"
a = 4 ** 16 b = a * 16
a ) 2.25 , b ) 2.35 , c ) 2.45 , d ) 2.55 , e ) 2.65
b
multiply(divide(multiply(add(5, 1.2), subtract(5, 1.2)), add(add(5, 1.2), subtract(5, 1.2))), const_2)
a man can row 5 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . how far is the place ?
"m = 5 s = 1.2 ds = 5 + 1.2 = 6.2 us = 5 - 1.2 = 3.8 x / 6.2 + x / 3.8 = 1 x = 2.35 . answer : b"
a = 5 + 1 b = 5 - 1 c = a * b d = 5 + 1 e = 5 - 1 f = d + e g = c / f h = g * 2
a ) 4 miles , b ) 4 1 / 4 miles , c ) 4 3 / 4 miles , d ) 5 1 / 2 miles , e ) 6 1 / 2 miles
e
divide(divide(multiply(subtract(4.9, 1.15), const_100), const_3), const_4)
a taxi company charges $ 1.15 for the first quarter of a mile and fifteen cents for each additional quarter of a mile . what is the maximum distance someone could travel with $ 4.90 ?
if we start out with $ 4.90 and have to spend $ 1.15 for the first quarter - mile , we will have $ 3.75 left to spend on quarter - mile intervals . since $ 3.75 / $ 0.15 = 25 , we can buy 25 more quarter - miles , and will travel 26 quarter miles in all : 26 × 1 / 4 = 6 1 / 2 miles . the correct answer is choice ( e ) .
a = 4 - 9 b = a * 100 c = b / 3 d = c / 4
a ) 1 / 140 , b ) 1 / 180 , c ) 71 / 105 , d ) 21 / 113 , e ) 57 / 120
c
add(add(divide(1, 7), divide(1, 3)), divide(1, 5))
in a race where 10 cars are running , the chance that car x will win is 1 / 7 , that y will win is 1 / 3 and that z will win is 1 / 5 . assuming that a dead heat is impossible , find the chance that one of them will win .
"required probability = p ( x ) + p ( y ) + p ( z ) ( all the events are mutually exclusive ) . = 1 / 7 + 1 / 3 + 1 / 5 = 71 / 105 answer : c"
a = 1 / 7 b = 1 / 3 c = a + b d = 1 / 5 e = c + d
a ) 40 , b ) 80 , c ) 70 , d ) 60 , e ) 50
b
subtract(100, 20)
a person decided to build a house in 100 days . he employed 100 men in the beginning and 100 more after 20 days and completed the construction in stipulated time . if he had not employed the additional men , how many days behind schedule would it have been finished ?
"200 men do the rest of the work in 100 - 20 = 80 days 100 men can do the rest of the work in 80 * 200 / 100 = 160 days required number of days = 160 - 80 = 80 days answer is b"
a = 100 - 20
a ) 287 , b ) 132 , c ) 150.9 , d ) 158 , e ) 267
c
multiply(circumface(divide(32, const_2)), 1.50)
find the cost of fencing around a circular field of diameter 32 m at the rate of rs . 1.50 a meter ?
"2 * 22 / 7 * 16 = 100.6 100.6 * 1 1 / 2 = rs . 150.9 answer : c"
a = 32 / 2 b = circumface * (
a ) 3 % , b ) 50 % , c ) 8 % , d ) 10 % , e ) 12 %
b
multiply(multiply(10, 10), subtract(const_1, divide(add(multiply(5, const_60), 20), add(multiply(10, const_60), 40))))
bob wants to run a mile in the same time as his sister . if bob ’ s time for a mile is currently 10 minutes 40 seconds and his sister ’ s time is currently 5 minutes 20 seconds , by what percent does bob need to improve his time in order run a mile in the same time as his sister ?
"bob ' s time = 640 secs . his sis ' time = 320 secs . percent increase needed = ( 640 - 320 / 640 ) * 100 = 320 / 640 * 100 = 50 % . ans ( b ) ."
a = 10 * 10 b = 5 * const_60 c = b + 20 d = 10 * const_60 e = d + 40 f = c / e g = 1 - f h = a * g
a ) 7.9 s , b ) 2.5 s , c ) 5 s , d ) 7.6 s , e ) 7.4 s
c
multiply(divide(divide(100, const_1000), 72), const_3600)
how much time does a train 100 metres long running at 72 km / hr take to pass a pole ?
explanation : 72 km / hr = 72 * 5 / 18 = 20 m / s speed = distance / time ; v = d / t 20 = 100 / t t = 5 s answer : c
a = 100 / 1000 b = a / 72 c = b * 3600
a ) rs . 22000 , b ) rs . 24000 , c ) rs . 26000 , d ) rs . 27000 , e ) none of these
d
add(add(add(12000, 5000), 1000), multiply(divide(add(add(12000, 5000), 1000), const_100), 50))
sahil purchased a machine at rs 12000 , then got it repaired at rs 5000 , then gave its transportation charges rs 1000 . then he sold it with 50 % of profit . at what price he actually sold it .
"explanation : question seems a bit tricky , but it is very simple . just calculate all cost price , then get 150 % of cp . c . p . = 12000 + 5000 + 1000 = 18000 150 % of 18000 = 150 / 100 * 18000 = 27000 option d"
a = 12000 + 5000 b = a + 1000 c = 12000 + 5000 d = c + 1000 e = d / 100 f = e * 50 g = b + f
a ) 277 , b ) 270 , c ) 180 , d ) 266 , e ) 121
c
multiply(450, divide(8, const_100))
find the simple interest on rs . 450 for 8 months at 5 paisa per month ?
"i = ( 450 * 8 * 5 ) / 100 = 180 answer : c"
a = 8 / 100 b = 450 * a
a ) 0 , b ) 12 , c ) 13 , d ) 20 , e ) 28
e
divide(multiply(12, 49), 21)
in a division sum , the remainder is 0 . as student mistook the divisor by 12 instead of 21 and obtained 49 as quotient . what is the correct quotient ?
"12 * 49 = 588 588 % 21 = 28 answer : e"
a = 12 * 49 b = a / 21
a ) 2749 , b ) 5449 , c ) 6749 , d ) 6449 , e ) 6468
a
subtract(multiply(divide(54671, const_100), 18456), multiply(divide(const_1, const_3), multiply(divide(54671, const_100), 18456)))
54671 - 18456 - 33466 = ?
"a if we calculate we will get 2749"
a = 54671 / 100 b = a * 18456 c = 1 / 3 d = 54671 / 100 e = d * 18456 f = c * e g = b - f
a ) 66 % , b ) 64 % , c ) 68 % , d ) 69 % , e ) 52 %
e
subtract(const_100, add(multiply(4, 6), multiply(6, 4)))
uba capital recently bought brand new vehicles for office use . uba capital only went for toyota and honda and bought more of toyota than honda at the ratio of 4 : 6 . if 40 % of the toyota bought and 60 % of the honda bought were suv ã ¢ â ‚ ¬ â „ ¢ s . how many suv ã ¢ â ‚ ¬ â „ ¢ s did uba capital buy in the aforementioned purchase ?
"let total no of vehicles bought be 100 , toyota 40 and honda 60 , so total number of suv ' s bought for toyota and honda respectively 40 * 40 / 100 = 16 and 60 * 60 / 100 = 36 so total 52 suv ' s were bought out of 100 vehicles bought . . so required % is 52 % answer : e"
a = 4 * 6 b = 6 * 4 c = a + b d = 100 - c
a ) 16.06 % , b ) 16.07 % , c ) 12.36 % , d ) 6.09 % , e ) 6.19 %
c
add(add(divide(12, const_2), divide(12, const_2)), divide(multiply(divide(12, const_2), divide(12, const_2)), const_100))
the effective annual rate of interest corresponding to a nominal rate of 12 % per annum payable half - yearly is ?
"amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 6 / 100 ) 2 ] = rs . 112.36 effective rate = ( 112.36 - 100 ) = 12.36 % answer : c"
a = 12 / 2 b = 12 / 2 c = a + b d = 12 / 2 e = 12 / 2 f = d * e g = f / 100 h = c + g
a ) 75 % , b ) 58 % , c ) 42 % , d ) 14.5 % , e ) 19.6 %
e
divide(multiply(divide(multiply(525, 60), const_100), const_100), 1600)
an association of mathematics teachers has 1600 members . only 525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 60 percent of the votes cast ?
total umber of members = 1600 number of members that cast votes = 525 since , winning candidate received 60 percent of the votes cast number of votes for winning candidate = ( 60 / 100 ) * 525 = 315 percent of total membership that voted for winning candidate = ( 315 / 1600 ) * 100 = 19.6 % answer e
a = 525 * 60 b = a / 100 c = b * 100 d = c / 1600
a ) 0 , b ) 1 , c ) 2 , d ) 13 , e ) 5
d
subtract(power(add(3, 2), 2), multiply(13, const_4))
if n is a prime number greater than 3 , what is the remainder when n ^ 2 is divided by 13 ?
"there are several algebraic ways to solve this question including the one under the spoiler . but the easiest way is as follows : since we can not have two correct answersjust pick a prime greater than 3 , square it and see what would be the remainder upon division of it by 13 . n = 5 - - > n ^ 2 = 25 - - > remainder upon division 25 by 13 is 12 . answer : d ."
a = 3 + 2 b = a ** 2 c = 13 * 4 d = b - c
a ) $ 2.25 , b ) $ 3.00 , c ) $ 3.25 , d ) $ 3.65 , e ) $ 4.80
a
divide(0.60, subtract(const_1, add(divide(3, 5), multiply(divide(const_1, 3), subtract(const_1, divide(3, 5))))))
having received his weekly allowance , john spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.60 at the candy store . what is john ’ s weekly allowance ?
"x = 3 x / 5 + 1 / 3 * 2 x / 5 + 60 4 x / 15 = 60 x = 225 = $ 2.25 the answer is a ."
a = 3 / 5 b = 1 / 3 c = 3 / 5 d = 1 - c e = b * d f = a + e g = 1 - f h = 0 / 60
a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 385
e
multiply(multiply(11, 3), 7)
a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 11 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 ?
"ans : 385 7 c 1 * 11 c 2 answer e )"
a = 11 * 3 b = a * 7
a ) 11000 , b ) 12000 , c ) 13000 , d ) 14000 , e ) 15000
c
add(add(multiply(5000, add(const_1, divide(20, const_100))), multiply(3000, add(const_1, divide(2000, 3000)))), 2000)
in the year 1990 there are 5000 men 3000 women 2000 boys . in 1994 men are increased by 20 % women are increased by ratio of boys and women ?
total population in year 1990 = 5000 + 3000 + 2000 = 10000 total population in year 1994 = ( ( 5000 * 120 / 100 ) + ( 3000 + ( 3000 * 2 / 3 ) ) + ( 2000 ) ) = 13000 answer : c
a = 20 / 100 b = 1 + a c = 5000 * b d = 2000 / 3000 e = 1 + d f = 3000 * e g = c + f h = g + 2000
a ) 10 , b ) 8 , c ) 6 , d ) 5 , e ) 7
a
divide(multiply(subtract(27, const_2), const_2), add(const_4, const_1))
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 27 , the how old is b
"let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 27 5 x = 25 x = 5 . hence , b ' s age = 2 x = 10 years answer : a"
a = 27 - 2 b = a * 2 c = 4 + 1 d = b / c
a ) 1,890 , b ) 1,960 , c ) 2,200 , d ) 3,780 , e ) 4,400
b
multiply(divide(add(subtract(125, const_3), add(13, const_2)), const_2), add(divide(subtract(subtract(125, const_3), add(13, const_2)), 4), const_1))
what is the sum of the multiples of 4 between 13 and 125 inclusive ?
"the first multiple of 12 between 13 and 125 is 16 while the last is 124 total number of terms is given by 124 = 16 + ( n - 1 ) 4 4 n - 4 = 108 n = 112 / 4 = 28 or , ( 124 - 16 ) / 4 + 1 = 28 so , we have 28 multiples of 12 between 13 and 125 sum = 14 ( 2 * 16 + ( 27 ) 4 ) = 14 ( 140 ) = 1960 answer : b"
a = 125 - 3 b = 13 + 2 c = a + b d = c / 2 e = 125 - 3 f = 13 + 2 g = e - f h = g / 4 i = h + 1 j = d * i
a ) 0.125 , b ) 0.25 , c ) 0.5 , d ) 0.75 , e ) not enough information to determine the rate
e
divide(840, 0.5)
the volume of a rectangular swimming pool is 840 cubic meters and water is flowing into the swimming pool . if the surface level of the water is rising at the rate of 0.5 meters per minute , what is the rate s , in cubic meters per minutes , at which the water is flowing into the swimming pool ?
"the correct answer is e . there are not enough info to answer the question . a 840 cubic meters rectangle is built from : height * length * width . from the question we know the volume of the pool and the filling rate . a pool can have a height of 10 * width 8.4 * length 10 and have a volume of 840 cubic meters , and it can have a height of 1 meter , width of 100 meters and length of 8.4 . in both cases the pool will fill up in a different rate = e"
a = 840 / 0
['a ) 1050 sq . m', 'b ) 3850 sq . m', 'c ) 950 sq . m', 'd ) 1075 sq . m', 'e ) 1065 sq . m']
a
subtract(square_area(70), circle_area(divide(70, const_2)))
four horses are tethered at 4 corners of a square field of side 70 metres so that they just can not reach one another . the area left ungrazed by the horses is :
area ungrazed is given by total area - 4 * area grazed by each horse = 70 * 70 - 4 * ( 90 / 360 ) * pi * ( 70 / 2 ) ^ 2 as the angle made by the horse is 90 degree , so applying the area of the sector , = theta / 360 * pi * radius ^ 2 above = 70 * 70 - pi * ( 70 / 2 ) * ( 70 / 2 ) = 70 * 70 { 1 - pi / 4 } = 70 * 70 { 6 / ( 7 * 4 ) } , expanding pi = 22 / 7 = ( 70 * 70 * 6 ) / ( 7 * 4 ) = 1050 sq m answer : a
a = square_area - (
a ) 80 % , b ) 90 % , c ) 59 % , d ) 40 % , e ) 53 %
b
subtract(multiply(38, add(const_4, const_1)), multiply(25, const_4))
a dealer purchases 15 articles for rs . 25 and sells 12 articles for rs . 38 . find the profit percentage ?
l . c . m of 15 and 12 = 60 cp of 60 articles = rs . 100 ( 25 * 4 ) sp of 60 articles = rs . 190 ( 30 * 5 ) profit percentage = ( 190 - 100 ) / 100 * 100 = 90 % answer : b
a = 4 + 1 b = 38 * a c = 25 * 4 d = b - c
a ) $ 4500 , b ) $ 3500 , c ) $ 5500 , d ) $ 5000 , e ) $ 6300
d
subtract(multiply(subtract(50, 20), 500), 10000)
redo ’ s manufacturing costs for sets of horseshoes include a $ 10000 initial outlay , and $ 20 per set . they can sell the sets $ 50 . if profit is revenue from sales minus manufacturing costs , and the company producessells 500 sets of horseshoes , what was their profit ?
total manufacturing cost = 10000 + 500 * 20 = 20000 total selling cost = 500 * 50 = 25000 profit = 25000 - 20000 = 5000 answer : d
a = 50 - 20 b = a * 500 c = b - 10000
a ) 3 , b ) 2 , c ) 7 , d ) 8 , e ) 9
b
sqrt(50)
from below option 50 is divisible by which one ?
"50 / 2 = 25 b"
a = math.sqrt(50)
a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 8
a
add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2)
in n is a positive integer less than 200 , and 24 n / 60 is an integer , then n has how many different positive prime factors ?
"( a ) . 24 n / 60 must be an integer . = > 2 n / 5 must be an integer . hence n must be a multiple of 5 . = > n has 1 different prime integers ."
a = 1 + 4 b = 60 / 2 c = b / 2 d = c / 3 e = a / d f = e + 2
a ) 0 , b ) 4 , c ) 5 , d ) 6 , e ) 7
a
subtract(add(const_4, 4), divide(divide(add(42, 14), const_2), 4))
on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 4 cups of tea . last week mo drank a total of 42 cups of tea and hot chocolate together . if during that week mo drank 14 more tea cups than hot chocolate cups , then how many rainy days were there last week ?
"t = the number of cups of tea c = the number of cups of hot chocolate t + c = 42 t - c = 14 - > t = 28 . c = 14 . mo drinks 4 cups of tea a day then number of days that are not rainy = 28 / 4 = 7 so number of rainy days = 7 - 7 = 0 a is the answer ."
a = 4 + 4 b = 42 + 14 c = b / 2 d = c / 4 e = a - d
a ) 1691100843 , b ) 4591100843 , c ) 4691100843 , d ) 3691100843 , e ) none of these
c
multiply(divide(469157, 9999), const_100)
469157 * 9999
"explanation : 469157 * ( 10000 - 1 ) = 4691570000 - 469157 = 4691100843 option c"
a = 469157 / 9999 b = a * 100
a ) 4676 , b ) 4678 , c ) 8888 , d ) 9504 , e ) 9936
e
multiply(floor(divide(power(const_10, 4), 72)), 72)
what is the largest 4 digit number exactly divisible by 72 ?
"largest 4 digit number = 9999 9999 ÷ 72 = 138 , remainder = 63 hence largest 4 digit number exactly divisible by 88 = 9999 - 63 = 9936 answer : e"
a = 10 ** 4 b = a / 72 c = math.floor(b) d = c * 72
a ) 100 , b ) 250 , c ) 125 , d ) 110 , e ) 115
b
multiply(divide(subtract(360, multiply(subtract(33, 30), divide(subtract(420, 360), subtract(36, 33)))), 30), 25)
apple costs l rupees per kilogram for first 30 kgs and q rupees per kilogram for each additional kilogram . if the price of 33 kilograms is 360 and for 36 kgs of apples is 420 then the cost of first 25 kgs of apples is
"ans : by framing equations we get 30 l + 3 q = 360 30 l + 6 q = 420 eliminate q by multiplying the first equation by 2 and subtracting second equation from the first then we get l = 10 cost of 10 kgs of apples = 25 x 10 = 250 answer : b"
a = 33 - 30 b = 420 - 360 c = 36 - 33 d = b / c e = a * d f = 360 - e g = f / 30 h = g * 25
a ) 3 ^ 9 , b ) 3 ^ 12 , c ) 3 ^ 27 , d ) 3 ^ 30 , e ) 3 ^ 33
d
multiply(power(3, 27), 27)
if x = 3 ^ 27 and x ^ x = 3 ^ k , what is k ?
"solution : we know that x = 3 ^ 27 which implies x ^ x = ( 3 ^ 27 ) ^ ( 3 ^ 27 ) = 3 ^ ( 27 * 3 ^ 27 ) [ because ( x ^ y ) ^ z = x ^ ( y * z ) ) ] so 3 ^ ( 3 ^ 3 * 3 ^ 27 ) = 3 ^ ( 3 ^ ( 3 + 27 ) ) [ because x ^ a * x ^ b = x ^ ( a + b ) ] therefore x ^ x = 3 ^ ( 3 ^ 30 ) given that x ^ x = 3 ^ k so 3 ^ ( 3 ^ 30 ) = 3 ^ k since the base is same the exponent will also be same therefore k = 3 ^ 30 answer : d"
a = 3 ** 27 b = a * 27
a ) 9632 , b ) 7896 , c ) 8741 , d ) 1683 , e ) 8523
d
add(3, lcm(5,6, 7,8))
find the least number which when divided by 5,6 , 7,8 leaves a remainder 3 but when divided by 9 leaves no remainder
"l . c . m of 5,6 , 7,8 = 840 required number is of the form of 840 k + 3 least value of k for which ( 840 k + 3 ) is divided by 9 is k = 2 required number = ( 840 * 2 + 3 ) = 1683 answer ( d )"
a = math.lcm(5, 6) b = 3 + a
a ) 1 : 7 , b ) 9 : 16 , c ) 1 : 9 , d ) 3 : 7 , e ) 3 : 4
b
divide(circle_area(3), circle_area(4))
the ratio of the radius of two circles is 3 : 4 , and then the ratio of their areas is ?
"r 1 : r 2 = 3 : 4 î r 1 ^ 2 : î r 2 ^ 2 r 1 ^ 2 : r 2 ^ 2 = 9 : 16 answer : b"
a = circle_area / (
a ) 16 , b ) 26 , c ) 36 , d ) 46 , e ) 56
b
subtract(30, negate(add(negate(10), 6)))
evaluate : 30 - | - x + 6 | for x = 10
substitute x by 10 in the given expression and evaluate 30 - | - ( 10 ) + 6 | = 30 - | - 10 + 6 | = 30 - | - 4 | = 30 - 4 = 26 correct answer b ) 26
a = negate + ( b = 30 - negate
a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) none of these
a
multiply(inverse(add(multiply(6, 3), add(const_1, const_4))), 2300)
in a partnership , a invests 1 ⁄ 6 of the capital for 1 ⁄ 6 of the time , b invests 1 ⁄ 3 of the capital for 1 ⁄ 3 of the time and c , the rest of the capital for whole time . find a ’ s share of the total profit of 2300
remaining capital = 1 - ( 1 ⁄ 6 + 1 ⁄ 3 ) = 1 ⁄ 2 ratio of their profit = 1 ⁄ 6 × [ 1 ⁄ 6 × 12 ] : 1 ⁄ 3 × [ 1 ⁄ 3 × 12 ] : 1 ⁄ 2 × 12 = 1 ⁄ 3 : 4 ⁄ 3 : 6 = 1 : 4 : 18 ∴ a ' s share = 1 / 1 + 4 + 18 × 2300 = 100 answer a
a = 6 * 3 b = 1 + 4 c = a + b d = 1/(c) e = d * 2300
a ) 23 % , b ) 25 % , c ) 26 % , d ) 28 % , e ) 35 %
a
divide(multiply(subtract(add(multiply(divide(multiply(280, 40), const_100), divide(add(const_100, 20), const_100)), multiply(divide(multiply(280, 50), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280)
a shopkeeper has 280 kg of apples . he sells 40 % of these at 20 % profit and remaining 50 % at 30 % profit . find his % profit on total .
"if the total quantity was 100 then 40 x 20 % + 50 x 30 % = 23 this profit will remain same for any total quantity unless the % of products remains the same . hence ' a ' is the answer"
a = 280 * 40 b = a / 100 c = 100 + 20 d = c / 100 e = b * d f = 280 * 50 g = f / 100 h = 100 + 30 i = h / 100 j = g * i k = e + j l = k - 280 m = l * 100 n = m / 280
a ) 105 , b ) 140 , c ) 175 , d ) 210 , e ) 245
e
multiply(35, divide(multiply(divide(48, const_60), 35), subtract(39, 35)))
car x began traveling at an average speed of 35 miles per hour . after 48 minutes , car y began traveling at an average speed of 39 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ?
"in 48 minutes , car x travels 28 miles . car y gains 4 miles each hour , so it takes 7 hours to catch car x . in 7 hours , car x travels 245 miles . the answer is e ."
a = 48 / const_60 b = a * 35 c = 39 - 35 d = b / c e = 35 * d
a ) 625 , b ) 600 , c ) 500 , d ) 400 , e ) 256
c
multiply(multiply(add(4, 4), add(4, 4)), multiply(add(4, 4), multiply(4, 4)))
how many 4 - digit positive integers are there in which all 4 digits are even ?
"positive integers - 2 , 4,6 , 8,0 let the integers of a four digit positive number be abcd a can take four values ( 2,4 , 6,8 ) b can take five values ( 0 , 2,4 , 6,8 ) c can take five values ( 0 , 2,4 , 6,8 ) d can take five values ( 0 , 2,4 , 6,8 ) the total is 5 * 5 * 5 * 4 the answer according to me is 500 answer : c"
a = 4 + 4 b = 4 + 4 c = a * b d = 4 + 4 e = 4 * 4 f = d * e g = c * f
a ) 8 , b ) 9 , c ) 5 , d ) 7 , e ) 10
c
divide(divide(360, 4), multiply(9, const_2))
benny goes to the market for buying some apples to be distributed between her 9 kids equally . she takes 360 dollars with her . the cost of each apple is 4 dollars . how many apples does she buy to share them equally between her eighteen kids ?
cost of each apple = 4 dollars apples that benny can buy with the amount she has = 360 / 4 = 90 . apples that each kid gets evenly = 90 / 18 = 5 apples . so the answer is c = 5
a = 360 / 4 b = 9 * 2 c = a / b
a ) 13.58 , b ) 13.87 , c ) 14.24 , d ) 14.59 , e ) 14.85
d
subtract(inverse(add(inverse(multiply(add(add(const_2, const_3), const_4), const_60)), inverse(add(multiply(const_3, const_4), const_3)))), divide(subtract(multiply(multiply(const_4, const_4), const_3), const_2), multiply(const_100, const_100)))
it takes nine minutes to load a certain video on a cellphone , and fifteen seconds to load that same video on a laptop . if the two devices were connected so that they operated in concert at their respective rates , how many seconds would it take them to load the video , rounded to the nearest hundredth ?
the laptop can load the video at a rate of 1 / 15 of the video per second . the phone can load the video at a rate of 1 / ( 60 * 9 ) = 1 / 540 of the video per second . the combined rate is 1 / 15 + 1 / 540 = 37 / 540 of the video per second . the time required to load the video is 540 / 37 = 14.59 seconds . the answer is d .
a = 2 + 3 b = a + 4 c = b * const_60 d = 1/(c) e = 3 * 4 f = e + 3 g = 1/(f) h = d + g i = 1/(h) j = 4 * 4 k = j * 3 l = k - 2 m = 100 * 100 n = l / m o = i - n
a ) 0.4 % , b ) 5.0 % , c ) 2.5 % , d ) 7 % , e ) 8 %
c
divide(multiply(const_100, subtract(820, 800)), 800)
a sum of money invested at compound interest to rs . 800 in 3 years and to rs 820 in 4 years . the rate on interest per annum is .
explanation : s . i . on rs 800 for 1 year = 20 rate = ( 100 * 20 ) / ( 800 * 1 ) = 2.5 % answer : c
a = 820 - 800 b = 100 * a c = b / 800
a ) 67 % . , b ) 70 % . , c ) 60 % . , d ) 28.6 % . , e ) 80 % .
d
multiply(divide(1, 7), const_100)
if two positive numbers are in the ratio 1 / 9 : 1 / 7 , then by what percent is the second number more than the first ?
"given ratio = 1 / 9 : 1 / 7 = 7 : 9 let first number be 7 x and the second number be 9 x . the second number is more than first number by 2 x . required percentage = 2 x / 7 x * 100 = 28.6 % . answer : d"
a = 1 / 7 b = a * 100
a ) rs 312 , b ) rs 412 , c ) rs 512 , d ) rs 612 , e ) none of these
d
subtract(add(add(7500, divide(multiply(7500, const_4), const_100)), divide(multiply(add(7500, divide(multiply(7500, const_4), const_100)), 4), const_100)), 7500)
find compound interest on rs . 7500 at 4 % per annum for 2 years , compounded annually
explanation : please apply the formula amount = p ( 1 + r 100 ) nc . i . = amount - p answer : d
a = 7500 * 4 b = a / 100 c = 7500 + b d = 7500 * 4 e = d / 100 f = 7500 + e g = f * 4 h = g / 100 i = c + h j = i - 7500
a ) 30 π , b ) 45 π , c ) 60 π , d ) 90 π , e ) none
c
divide(multiply(multiply(multiply(2, 10), 3), multiply(subtract(const_12, const_1), 2)), add(const_3, const_4))
the front wheels of a wagon are 2 π feet in circumference and the rear wheels are 3 π feet in circumference . when the front wheels have made 10 more revolutions than the rear wheels , how many feet has the wagon travelled ?
solution let the rear wheel make x revolutions . then , the front wheel makes ( x + 10 ) revolutions . ( x + 10 ) x 3 π = x × 2 π ‹ = › 3 x + 30 = 2 x ‹ = › x = 30 . distance travelled by the wagon = ( 2 π x 30 ) ft ‹ = › ( 60 π ) ft . answer c
a = 2 * 10 b = a * 3 c = 12 - 1 d = c * 2 e = b * d f = 3 + 4 g = e / f
a ) 269 , b ) 285 , c ) 300 , d ) 275 , e ) none of these
d
divide(multiply(7700, 11), 308)
lcm of two numbers is 7700 and hcf is 11 . if one number is 308 then other number is
option d
a = 7700 * 11 b = a / 308
a ) 100 m , b ) 440 m , c ) 180 m , d ) 200 m , e ) 250 m
b
multiply(divide(80, const_3_6), 20)
a car is running at a speed of 80 kmph . what distance will it cover in 20 sec ?
speed = 80 kmph = 80 * 5 / 18 = 22 m / s distance covered in 20 sec = 22 * 10 = 440 m answer is b
a = 80 / const_3_6 b = a * 20
a ) rs . 8400 , b ) rs . 11,900 , c ) rs . 13,600 , d ) rs . 14,700 , e ) none
d
subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1)
a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35,000 , a receives
"solution : let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 . = > 3 x = 36000 . = > x = 12000 . a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . so a ' s share = rs . ( 35000 x 21 / 50 ) = rs . 14,700 . answer : option d"
a = 10 * 5000 b = a - 5000 c = 4000 + 5000 d = b - c e = d / 3 f = 4000 + 5000 g = e + f h = 10 * 5000 i = g / h j = 3 + 4 k = j * 5000 l = i * k m = l / 1000 n = math.floor(m) o = n - 1
a ) 50 , b ) 56 , c ) 58 , d ) 62 , e ) 66
c
subtract(multiply(66, add(1, 2)), multiply(70, 2))
a charitable association sold an average of 66 raffle tickets per member . among the female members , the average was 70 raffle tickets . the male to female ratio of the association is 1 : 2 . what was the average number r of tickets sold by the male members of the association
"given that , total average r sold is 66 , male / female = 1 / 2 and female average is 70 . average of male members isx . ( 70 * f + x * m ) / ( m + f ) = 66 - > solving this equation after substituting 2 m = f , x = 58 . ans c ."
a = 1 + 2 b = 66 * a c = 70 * 2 d = b - c
a ) 510 , b ) 540 , c ) 500 , d ) 420 , e ) 589
d
divide(880, multiply(subtract(78, 1), const_0_2778))
a train 880 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 , ( 880 + x ) / 60 = 65 / 3 x = 420 m . answer : option d"
a = 78 - 1 b = a * const_0_2778 c = 880 / b
a ) 33.33 % , b ) 100 % , c ) 75 % , d ) 66.66 % , e ) none of these
b
multiply(subtract(divide(const_100, subtract(const_100, 50)), const_1), const_100)
if x is less than y by 50 % then y exceed x by :
"using formula ( x / ( 100 - x ) * 100 ) where x is percentage decrease ( here it is 25 % ) = > 50 ( 100 - 50 ) * 100 = 100 % answer : b"
a = 100 - 50 b = 100 / a c = b - 1 d = c * 100
a ) 66.3 , b ) 76.3 , c ) 86.3 , d ) 16.3 , e ) 36.3
d
divide(add(add(multiply(30, 11.50), multiply(20, 14.25)), multiply(divide(add(multiply(30, 11.50), multiply(20, 14.25)), const_100), 30)), add(30, 20))
arun purchased 30 kg of wheat at the rate of rs . 11.50 per kg and 20 kg of wheat at the rate of 14.25 per kg . he mixed the two and sold the mixture . approximately what price per kg should be sell the mixture to make 30 % profit ?
"explanation : c . p . of 50 kg wheat = ( 30 * 11.50 + 20 * 14.25 ) = rs . 630 . s . p . of 50 kg wheat = 130 % of rs . 630 = 130 / 100 * 630 = rs . 819 . s . p . per kg = 819 / 50 = rs . 16.38 = 16.30 . answer : d"
a = 30 * 11 b = 20 * 14 c = a + b d = 30 * 11 e = 20 * 14 f = d + e g = f / 100 h = g * 30 i = c + h j = 30 + 20 k = i / j
a ) 6200 , b ) 3000 , c ) 3000 , d ) 2000 , e ) 1000
a
divide(3100, subtract(const_1, divide(multiply(5, 10), const_100)))
what was the principal , if at 5 % per annum the interest after 10 years amounted to rs . 3100 less than the sum lent ?
"p - 2336 = ( p * 10 * 5 ) / 100 p = 6200 answer : a"
a = 5 * 10 b = a / 100 c = 1 - b d = 3100 / c
a ) 3.23 % , b ) 4.23 % , c ) 5.23 % , d ) 7.23 % , e ) 8.23 %
e
multiply(divide(subtract(subtract(68, multiply(68, divide(10, const_100))), 56.16), subtract(68, multiply(68, divide(10, const_100)))), const_100)
the list price of an article is rs . 68 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ?
"68 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 56.16 x = 8.23 % answer : e"
a = 10 / 100 b = 68 * a c = 68 - b d = c - 56 e = 10 / 100 f = 68 * e g = 68 - f h = d / g i = h * 100
a ) rs . 850 , b ) rs . 860 , c ) rs . 876 , d ) rs . 886 , e ) none of these
c
multiply(divide(surface_cube(8), 16), 36.50)
the cost of the paint is rs . 36.50 per kg . if 1 kg of paint covers 16 square feet , how much will it cost to paint outside of a cube having 8 feet each side .
"explanation : we will first calculate the surface area of cube , then we will calculate the quantity of paint required to get answer . here we go , surface area = 6 a 2 = 6 ∗ 8 ( 2 ) = 384 sq feet quantity required = 38416 = 24 kg cost of painting = 36.50 ∗ 24 = rs . 876 option c"
a = surface_cube / ( b = a * 16
a ) 41.4 , b ) 34.1 , c ) 13.4 , d ) 12.4 , e ) 10.8
a
add(inverse(subtract(divide(const_1, 12.5), divide(const_1, 30))), inverse(subtract(divide(const_1, 7.5), divide(const_1, 12))))
two consultants can type up a report in 12.5 hours and edit it in 7.5 hours . if mary needs 30 hours to type the report and jim needs 12 hours to edit it alone , how many w hours will it take if jim types the report and mary edits it immediately after he is done ?
"break down the problem into two pieces : typing and editing . mary needs 30 hours to type the report - - > mary ' s typing rate = 1 / 30 ( rate reciprocal of time ) ( point 1 in theory below ) ; mary and jim can type up a report in 12.5 and - - > 1 / 30 + 1 / x = 1 / 12.5 = 2 / 25 ( where x is the time needed for jim to type the report alone ) ( point 23 in theory below ) - - > x = 150 / 7 ; jim needs 12 hours to edit the report - - > jim ' s editing rate = 1 / 12 ; mary and jim can edit a report in 7.5 and - - > 1 / y + 1 / 12 = 1 / 7.5 = 2 / 15 ( where y is the time needed for mary to edit the report alone ) - - > y = 20 ; how many w hours will it take if jim types the report and mary edits it immediately after he is done - - > x + y = 150 / 7 + 20 = ~ 41.4 answer : a ."
a = 1 / 12 b = 1 / 30 c = a - b d = 1/(c) e = 1 / 7 f = 1 / 12 g = e - f h = 1/(g) i = d + h
a ) 160 minutes , b ) 120 minutes , c ) 100 minutes , d ) 76 minutes , e ) 77 minutes
c
divide(3500, 35)
a scuba diver descends at a rate of 35 feet per minute . a diver dive from a ship to search for a lost ship at the depth of 3500 feet below the sea level . . how long will he take to reach the ship ?
"time taken to reach = 3500 / 35 = 100 minutes answer : c"
a = 3500 / 35
a ) 20 , b ) 26 , c ) 32 , d ) 39 , e ) 51
d
subtract(42, 4)
a snail , climbing a 42 feet high wall , climbs up 4 feet on the first day but slides down 2 feet on the second . it climbs 4 feet on the third day and slides down again 2 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ?
"total transaction in two days = 4 - 2 = 2 feet in 38 days it will climb 38 feet on the 39 th day , the snail will climb 4 feet , thus reaching the top therefore , total no of days required = 39 d"
a = 42 - 4
a ) 1.11 % , b ) 5.93 % , c ) 4.33 % , d ) 5.33 % , e ) 6.33 %
a
multiply(divide(divide(subtract(950, 900), 900), 5), const_100)
at what rate percent on simple interest will rs . 900 amount to rs . 950 in 5 years ?
"50 = ( 900 * 5 * r ) / 100 r = 1.11 % answer : a"
a = 950 - 900 b = a / 900 c = b / 5 d = c * 100
a ) 17 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) none
a
multiply(subtract(const_1, divide(multiply(const_1, 10), 12)), const_100)
if the price of sugar rises from rs . 10 per kg to rs . 12 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by
"sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 10 = x × 12 = x = 83 kg . ∴ reduction in consumption = 17 % . answer a"
a = 1 * 10 b = a / 12 c = 1 - b d = c * 100
a ) 12 , b ) 14 , c ) 15 , d ) 24 , e ) 36
d
multiply(2, subtract(divide(multiply(5, subtract(14, 2)), 4), 3))
initially , the men and women in a room were in the ratio of 4 : 5 . then , 2 men entered the room and 3 women left the room . then , the number of women doubled . now there are 14 men in the room . how many e women are currently in the room ?
"the number of women doubled means that they have become 24 from 12 . . and we have to tell the current strength so 24 is the answer . . let the number be 4 x and 5 x . . given 4 x + 2 = 14 . . so x = 3 . . women number = 5 * 3 - 3 = 12 , then doubled = 24 . . ans d"
a = 14 - 2 b = 5 * a c = b / 4 d = c - 3 e = 2 * d
a ) 1 : 2 , b ) 2 : 3 , c ) 1 : 8 , d ) 4 : 5 , e ) 3 : 2
c
multiply(divide(2, const_3.0), multiply(divide(2, 3), divide(3, 3)))
find the compound ratio of ( 2 : 3 ) , ( 3 : 4 ) and ( 1 : 4 ) is
"required ratio = 2 / 3 * 3 / 4 * 1 / 4 = 1 / 8 = 1 : 8 answer is c"
a = 2 / 3 b = 2 / 3 c = 3 / 3 d = b * c e = a * d
a ) 30 , b ) 35 , c ) 20 , d ) 18 , e ) 10
b
subtract(subtract(add(subtract(90, 11), 16), 44), 16)
in a neighborhood having 90 households , 11 did not have either a car or a bike . if 16 households had a both a car and a bike and 44 had a car , how many had bike only ?
"{ total } = { car } + { bike } - { both } + { neither } - - > 90 = 44 + { bike } - 16 + 11 - - > { bike } = 51 - - > # those who have bike only is { bike } - { both } = 51 - 16 = 35 . answer : b ."
a = 90 - 11 b = a + 16 c = b - 44 d = c - 16
a ) 10 % , b ) 25 % , c ) 20 % , d ) 50 % , e ) 45 %
c
multiply(divide(subtract(150, 120), 150), const_100)
a bag marked at $ 150 is sold for $ 120 . the rate of discount is ?
"rate of discount = 30 / 150 * 100 = 20 % answer is c"
a = 150 - 120 b = a / 150 c = b * 100
a ) 1 , b ) 4 , c ) 2 , d ) 6 , e ) 8
c
divide(add(divide(30, 10), divide(70, 10)), const_2)
a man swims downstream 70 km and upstream 30 km taking 10 hours each time ; what is the speed of the current ?
"70 - - - 10 ds = 7 ? - - - - 1 30 - - - - 10 us = 3 ? - - - - 1 s = ? s = ( 7 - 3 ) / 2 = 2 answer : c"
a = 30 / 10 b = 70 / 10 c = a + b d = c / 2
a ) 42 , b ) 70 , c ) 140 , d ) 196 , e ) 315
d
multiply(multiply(8, 3), 7)
a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 8 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 ?
"ans : 196 7 c 1 * 8 c 2 answer d )"
a = 8 * 3 b = a * 7
a ) 3488 , b ) 10800 , c ) 12788 , d ) 1800 , e ) 2881
b
multiply(multiply(const_3, const_60), const_60)
if an object travels at three feet per second , how many feet does it travel in one hour ?
explanation : if an object travels at 2 feet per second it covers 3 x 60 feet in one minute , and 3 x 60 x 60 feet in one hour . answer = 10800 answer : b ) 10800
a = 3 * const_60 b = a * const_60
a ) 16 cm , b ) 18 cm , c ) 15 cm , d ) 20 cm , e ) 23 cm
a
divide(const_100, const_3)
the length of a rectangle is twice its breadth . if its length is decreased by 5 cm and breadth is increased by 5 cm , the area of the rectangle is increased by 55 sq . cm . find the length of the rectangle .
"explanation : let breadth = x . then , length = 2 x . then , ( 2 x - 5 ) ( x + 5 ) - 2 x * x = 55 = > 5 x - 25 = 55 = > x = 16 . length of the rectangle = 16 cm . answer : option a"
a = 100 / 3
['a ) 5', 'b ) 4', 'c ) 3', 'd ) 2', 'e ) 1']
c
subtract(divide(volume_cylinder(divide(6, const_2), 8), volume_cylinder(2, 5)), divide(add(const_4, const_2), const_10))
a cylindrical container with 6 meters diameter and a height of 8 meters is filled to capacity with water . if the water is then used to fill several smaller cylinders ( 2 meters radius and 5 meters height ) , how many smaller cylinders can be filled to capacity before the larger cylinder becomes empty ?
calculate the volume of the larger cylinder and divide it by the volume of the smaller cylinder . volume of cylinder = π r 2 h larger cylinder volume = 226.19 smaller cylinder volume = 62.83 therefore the number of smaller cylinders can be filled to capacity = 226.19 / 62.83 = 3.6 answer is c only 3 smaller cylinders can be filled to capacity .
a = 6 / 2 b = volume_cylinder / ( c = b - volume_cylinder
['a ) 12', 'b ) 32', 'c ) 42', 'd ) 52', 'e ) 58']
c
subtract(power(multiply(7, const_2), const_2), multiply(power(7, const_2), const_pi))
four circular cardboard pieces , each of radius 7 cm are placed in such a way that each piece touches two other pieces . the area of the space encosed by the four pieces is. four circular cardboard pieces , each of radius 7 cm are placed in such a way that each piece touches two other pieces . the area of the space encosed by the four pieces is
required area = 14 * 14 - ( 4 * 1 / 4 * 22 / 7 * 7 * 7 ) sq cm = 196 - 154 = 42 sq cm . answer : c
a = 7 * 2 b = a ** 2 c = 7 ** 2 d = c * math.pi e = b - d
a ) 38 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 48 .
e
subtract(multiply(sqrt(divide(720, 5)), 5), sqrt(divide(720, 5)))
the roof of an apartment building is rectangular and its length is 5 times longer than its width . if the area of the roof is 720 feet squared , what is the difference between the length and the width of the roof ?
"answer is e : 48 let w be the width , so length is 5 w . therefore : w * 5 w = 720 , solving for , w = 12 , so 5 w - w = 4 w = 4 * 12 = 48"
a = 720 / 5 b = math.sqrt(a) c = b * 5 d = 720 / 5 e = math.sqrt(d) f = c - e
a ) 237.5 , b ) 234 , c ) 289.5 , d ) 345 , e ) none of these
a
divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4)
jaclyn buys $ 10 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter
explanation : i = ( p x r x t ) / 100 = 10000 * 9.5 / 100 * ( 18 / 12 ) ^ 1 / 6 = 237.5 answer : a
a = 100 * 100 b = a * 9 c = b / 100 d = c / 4
a ) 15 sec , b ) 19 sec , c ) 12 sec , d ) 10 sec , e ) 11 sec
a
divide(250, multiply(80, const_0_2778))
two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 40 kmph respectively . in what time will they cross each other completely ?
"explanation : d = 250 m + 250 m = 500 m rs = 80 + 40 = 120 * 5 / 18 = 100 / 3 t = 500 * 3 / 100 = 15 sec answer : option a"
a = 80 * const_0_2778 b = 250 / a
a ) 33 , b ) 34 , c ) 35 , d ) 39 , e ) 37
d
subtract(add(15, 25), const_1)
at garage sale , all of the prices of the items sold were different . if the price of a radio sold at the garage sale was both the 15 th highest price and the 25 th lowest price among the prices of the items sold , how many items were sold at the garage sale ?
"14 + 24 + 1 = 39 answer : d"
a = 15 + 25 b = a - 1
a ) 1375 , b ) 1376 , c ) 1875 , d ) 1365 , e ) 1345
a
add(multiply(20, 66), 55)
in a division sum , the quotient is 20 , the divisor 66 and the remainder 55 , find the dividend ?
"explanation : 20 * 66 + 55 = 1375 answer : a"
a = 20 * 66 b = a + 55
a ) 66 / 5 , b ) 13 , c ) 17 , d ) 21 , e ) 23
a
divide(multiply(add(add(6, const_3), const_2), divide(6, const_2)), add(const_2, divide(const_1, const_2)))
a and b are two partially filled buckets of water . if 6 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 6 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ?
let bucket a be a and bucket b be b scenario 1 a - 6 = 1 / 3 ( b + 6 ) - - - - > 3 a - 18 = b + 6 scenario 2 b - 6 = 1 / 2 ( a + 6 ) - - - - - > 2 b - 12 = a + 6 from scenario 1 , b = 3 a - 24 substitute b with this information in stmt 2 2 ( 3 a - 24 ) - 12 = a + 6 - - - - - - > 6 a - 48 - 12 = a + 6 - - - - - - > 6 a - a = 60 + 6 - - - > 5 a = 66 a = 66 / 5 , answer choice a
a = 6 + 3 b = a + 2 c = 6 / 2 d = b * c e = 1 / 2 f = 2 + e g = d / f
a ) $ 3.36 , b ) $ 6.85 , c ) $ 8.46 , d ) $ 10.08 , e ) $ 12.46
e
add(multiply(2, 2.49), multiply(4, 1.87))
what is the total cost of 2 sandwiches at $ 2.49 each and 4 sodas at $ 1.87 each ?
"answer = e 2 * 2.49 + 4 * 1.87 = 2 ( 2.50 - 0.01 ) + 4 ( 2.00 - 0.13 ) = 5 + 8 - 0.02 - 0.52 = 13 - 0.54 = 12.46"
a = 2 * 2 b = 4 * 1 c = a + b
a ) 8 , b ) 11 , c ) 10 , d ) 16 , e ) 17
b
divide(add(add(add(add(const_2.0, const_4), add(1, const_4)), add(const_4, const_4)), 38), 4)
the sum of ages of 4 children born 1 years different each is 38 years . what is the age of the elder child ?
"let the ages of children be x , ( x + 1 ) , ( x + 2 ) , ( x + 3 ) years . then , x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) = 38 4 x = 32 x = 8 . x + 3 = 8 + 3 = 11 answer : b"
a = 2 + 0 b = 1 + 4 c = a + b d = 4 + 4 e = c + d f = e + 38 g = f / 4
['a ) 2 * 3 * 6', 'b ) 3 * 2 * 5', 'c ) 5 * 3 * 3', 'd ) 2 * 3', 'e ) 3 * 5 * 4']
b
subtract(power(add(floor(sqrt(595)), const_1), const_2), 595)
what is the positive integer that can be added by 595 to make it a perfect square ?
596 is added to a number that gives a perfect square nearest perfect square is 625 . so , 625 - 596 = 30 30 = 2 * 3 * 5 option b is answer
a = math.sqrt(595) b = math.floor(a) c = b + 1 d = c ** 2 e = d - 595
a ) 40 % , b ) 48 % , c ) 54 % , d ) 58 % , e ) 65 %
a
multiply(divide(220, divide(add(110, 220), divide(60, const_100))), const_100)
after a storm deposits 110 billion gallons of water into the city reservoir , the reservoir is 60 % full . if the original contents of the reservoir totaled 220 billion gallons , the reservoir was approximately what percentage full before the storm ?
when the storm deposited 110 billion gallons , volume of water in the reservoir = 220 + 110 = 360 billion gallons if this is only 60 % of the capacity of the reservoir , the total capacity of the reservoir = 330 / 0.6 = 550 billion gallons therefore percentage of reservoir that was full before the storm = ( 220 / 550 ) * 100 = 40 % option a
a = 110 + 220 b = 60 / 100 c = a / b d = 220 / c e = d * 100
a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8
b
multiply(power(const_60.0, 4.5), multiply(power(6, 6.5), power(36, 4.5)))
( 6 ) 6.5 × ( 36 ) 4.5 ÷ ( 216 ) 4.5 = ( 6 ) ?
"explanation : ( 6 ) 6.5 × ( 36 ) 4.5 ÷ ( 216 ) 4.5 = ( 6 ) 6.5 × [ ( 6 ) 2 ] 4.5 ÷ [ ( 6 ) 3 ] 4.5 = ( 6 ) 6.5 × ( 6 ) 9 ÷ ( 6 ) 13.5 = ( 6 ) ( 6.5 + 9 - 13.5 ) = ( 6 ) 2 answer : option b"
a = const_60 ** 0 b = 6 ** 6 c = 36 ** 4 d = b * c e = a * d
a ) s . 1350 , b ) s . 1327 , c ) s . 1328 , d ) s . 1364 , e ) s . 1927
d
multiply(subtract(rectangle_area(add(75, multiply(3.2, 2)), add(55, multiply(3.2, 2))), rectangle_area(75, 55)), 2)
a rectangular grass field is 75 m * 55 m , it has a path of 3.2 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ?
"area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 3.2 * 2 ) 2 * 2.5 = > 682 682 * 2 = rs . 1364 answer : d"
a = 3 * 2 b = 75 + a c = 3 * 2 d = 55 + c e = rectangle_area - ( f = e * rectangle_area
a ) 5 min , b ) 10 min , c ) 15 min , d ) 20 min , e ) 25 min
d
divide(20, 1)
a fill pipe can fill 1 / 2 of cistern in 20 minutes . in how many minutes , it can fill 1 / 2 of the cistern ?
required time = 20 * 2 * 1 / 2 = 20 minutes answer is d
a = 20 / 1
a ) 419 , b ) 551 , c ) 601 , d ) 590 , e ) 721
d
divide(add(add(add(add(add(add(1115, 130), 140), 510), 520), 530), 1115), add(const_3, const_4))
what is the average of 120 , 130 , 140 , 510 , 520 , 530 , 1115 , 1120 , and 1125 ?
"add 120 , 130 , 140 , 510 , 520 , 530 , 1115 , 1120 , and 1125 grouping numbers together may quicken the addition sum = 5310 5310 / 9 = 590 . d"
a = 1115 + 130 b = a + 140 c = b + 510 d = c + 520 e = d + 530 f = e + 1115 g = 3 + 4 h = f / g
a ) 60 , b ) 37 , c ) 26 , d ) 28 , e ) 11
a
multiply(50, add(divide(subtract(multiply(divide(30, const_60), multiply(509509, const_3_6)), multiply(50, divide(42, const_60))), subtract(multiply(509509, const_3_6), 50)), divide(42, const_60)))
a bus travels from town a to town b . if the bus ' s speed is 50 km / hr , it will arrive in town b 42 min later than scheduled . if the bus increases its speed by 509509 m / sec , it will arrive in town b 30 min earlier than scheduled . find : a ) the distance between the two towns ; b ) the bus ' s scheduled time of arrival in b ; c ) the speed of the bus when it ' s on schedule .
first we will determine the speed of the bus following its increase . the speed is increased by 509509 m / sec = 50 ⋅ 60 ⋅ 6091000 = 50 ⋅ 60 ⋅ 6091000 km / hr = 20 = 20 km / hr . therefore , the new speed is v = 50 + 20 = 70 v = 50 + 20 = 70 km / hr . if xx is the number of hours according to the schedule , then at the speed of 50 km / hr the bus travels from a to b within ( x + 4260 ) ( x + 4260 ) hr . when the speed of the bus is v = 70 v = 70 km / hr , the travel time is x − 3060 x − 3060 hr . then 50 ( x + 4260 ) = 70 ( x − 3060 ) 50 ( x + 4260 ) = 70 ( x − 3060 ) 5 ( x + 710 ) = 7 ( x − 12 ) 5 ( x + 710 ) = 7 ( x − 12 ) 72 + 72 = 7 x − 5 x 72 + 72 = 7 x − 5 x 2 x = 72 x = 7 x = 72 x = 72 hr . so , the bus is scheduled to make the trip in 33 hr 3030 min . the distance between the two towns is 70 ( 72 − 12 ) = 70 ⋅ 3 = 21070 ( 72 − 12 ) = 70 ⋅ 3 = 210 km and the scheduled speed is 210 / 7 / 2 = 60 km / hr . answer : a
a = 30 / const_60 b = 509509 * const_3_6 c = a * b d = 42 / const_60 e = 50 * d f = c - e g = 509509 * const_3_6 h = g - 50 i = f / h j = 42 / const_60 k = i + j l = 50 * k
a ) 900 , b ) 11110 , c ) 1100 , d ) 1200 , e ) 1400
b
add(divide(subtract(multiply(floor(divide(100000, 9)), 9), multiply(add(floor(divide(10, 9)), const_1), 9)), 9), const_1)
how many numbers from 10 to 100000 are exactly divisible by 9 ?
"10 / 9 = 1 and 100000 / 9 = 11111 = = > 11111 - 1 = 11110 . answer : b"
a = 100000 / 9 b = math.floor(a) c = b * 9 d = 10 / 9 e = math.floor(d) f = e + 1 g = f * 9 h = c - g i = h / 9 j = i + 1
a ) rs . 48.66 , b ) rs . 51.03 , c ) rs . 54.17 , d ) rs . 55.33 , e ) none of the above
a
divide(add(multiply(10, 40), multiply(5, 66)), add(10, 5))
if 10 litres of an oil of rs . 40 per litres be mixed with 5 litres of another oil of rs . 66 per litre then what is the rate of mixed oil per litre ?
"40 * 10 = 400 66 * 5 = 330 730 / 15 = 48.66 answer : a"
a = 10 * 40 b = 5 * 66 c = a + b d = 10 + 5 e = c / d
a ) 8 hours , b ) 48 hours , c ) 31 hours , d ) 18 hours , e ) 28 hours
c
floor(divide(add(add(25.35, 70.69), 85.96), multiply(6.5, subtract(const_1, divide(const_1, const_10)))))
tom wants to buy items costing $ 25.35 , $ 70.69 , and $ 85.96 . he earns $ 6.50 an hour doing odd jobs . if ten percent of his income is put aside for other purposes , how many hours must he work to earn the money he needs for his purchases ? round your answer to the nearest whole hour .
$ 6.50 x . 10 = $ . 65 is 10 % of his hourly income $ 6.50 - . 65 = $ 5.85 hourly amount available to spend $ 25.35 + $ 70.69 + $ 85.96 = $ 182 total needed $ 182 ÷ $ 5.85 = 31.11 . . . rounds to 31 hours correct answer c
a = 25 + 35 b = a + 85 c = 1 / 10 d = 1 - c e = 6 * 5 f = b / e g = math.floor(f)