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 ) 40 % , b ) 55 % , c ) 54.8 % , d ) 60 % , e ) 62 % | c | multiply(divide(10628, add(add(1136, 7636), 10628)), const_100) | 3 candidates in an election and received 1136 , 7636 and 10628 votes respectively . what % of the total votes did the winning candidate gotin that election ? | "total number of votes polled = ( 1136 + 7636 + 10628 ) = 19400 so , required percentage = 10628 / 19400 * 100 = 54.8 % c" | a = 1136 + 7636
b = a + 10628
c = 10628 / b
d = c * 100
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a ) 124 , b ) 297 , c ) 394 , d ) 421 , e ) 842 | a | add(multiply(multiply(4, 3), const_10), 4) | a is the hundreds digit of the 3 digit integer x , b is the tens digit of x , and c is the units digit of x . 4 a = 2 b = c , and a > 0 . what is the difference between the two greatest possible values of x ? tip : dont stop till you have exhausted all answer choices to arrive at the correct one . | ratio of a : b : c = 1 : 2 : 4 two possible greatest single digit values for c are 8 and 4 if c is 8 , then x = 248 if c is 4 , then x = 124 difference = 248 - 124 = 124 a is the answer | a = 4 * 3
b = a * 10
c = b + 4
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a ) 643 , b ) 652 , c ) 562 , d ) 578 , e ) 693 | c | multiply(divide(567, add(const_1, divide(27, const_100))), add(const_1, divide(26, const_100))) | if albert ’ s monthly earnings rise by 27 % , he would earn $ 567 . if , instead , his earnings rise by only 26 % , how much ( in $ ) would he earn this month ? | "= 567 / 1.27 ∗ 1.26 = 562 = 562 answer is c" | a = 27 / 100
b = 1 + a
c = 567 / b
d = 26 / 100
e = 1 + d
f = c * e
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a ) 5 , b ) 2 , c ) 9 , d ) 5 , e ) 1 | b | subtract(subtract(multiply(1250, power(add(const_1, divide(4, const_100)), 2)), 1250), multiply(multiply(1250, divide(4, const_100)), 2)) | indu gave bindu rs . 1250 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "1250 = d ( 100 / 4 ) 2 d = 2 answer : b" | a = 4 / 100
b = 1 + a
c = b ** 2
d = 1250 * c
e = d - 1250
f = 4 / 100
g = 1250 * f
h = g * 2
i = e - h
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a ) 3 , b ) 4 , c ) 5 , d ) 0 , e ) 7 | d | subtract(4, reminder(4, 8)) | when n is divided by 24 , the remainder is 4 . what is the remainder when 4 n is divided by 8 ? | "let n = 4 ( leaves a remainder of 4 when divided by 24 ) 4 n = 4 ( 4 ) = 16 , which leaves a remainder of 0 when divided by 8 . answer d" | a = 4 - reminder
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a ) 60 , b ) 72 , c ) 84 , d ) 96 , e ) 108 | d | multiply(subtract(divide(multiply(const_2, const_2), subtract(8, multiply(const_2, 3))), divide(const_2, subtract(8, 3))), const_60) | tom and linda stand at point a . linda begins to walk in a straight line away from tom at a constant rate of 3 miles per hour . one hour later , tom begins to jog in a straight line in the exact opposite direction at a constant rate of 8 miles per hour . if both tom and linda travel indefinitely , what is the positive difference , in minutes , between the amount of time it takes tom to cover half of the distance that linda has covered and the amount of time it takes tom to cover twice the distance that linda has covered ? | "d is the answer . . . . d = ts where d = distance , t = time and s = speed to travel half distance , ( 2 + 3 t ) = 8 t = = > t = 2 / 5 = = > 24 minutes to travel double distance , 2 ( 2 + 3 t ) = 8 t = = > 2 = = > 120 minutes difference , 96 minutes d" | a = 2 * 2
b = 2 * 3
c = 8 - b
d = a / c
e = 8 - 3
f = 2 / e
g = d - f
h = g * const_60
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a ) 34 , b ) 32 , c ) 17 , d ) 27 , e ) 28 | b | subtract(35, const_3) | 35 - [ 23 - { 15 - x } ] = 12 × 2 ÷ 1 / 2 | explanation : 35 - [ 23 - { 19 - ( 15 - x ) } ] = 12 × 2 × 2 = 48 = > 35 - 23 + ( 19 - 15 + x ) = 48 = > 12 + 4 + x = 48 = > x = 48 - ( 4 + 12 ) = 32 answer : option b | a = 35 - 3
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a ) 600 m , b ) 200 m , c ) 300 m , d ) 400 m , e ) 100 m | c | divide(multiply(18, multiply(1.2, const_1000)), 48) | amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.2 km ? | "c 300 m distance covered by amar = 18 / 4.8 ( 1.6 km ) = 3 / 8 ( 1200 ) = 300 m answer is c" | a = 1 * 2
b = 18 * a
c = b / 48
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a ) 9 , b ) 0 , c ) 1 , d ) 2 , e ) 4 | a | subtract(subtract(20, add(1, 4)), 6) | if n is an integer , f ( n ) = f ( n - 1 ) - n and f ( 4 ) = 20 . what is the value of f ( 6 ) ? | "since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 20 then f ( 5 ) = 20 - 5 = 15 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 15 - 6 = 9 . answer : a . questions on funtions to practice :" | a = 1 + 4
b = 20 - a
c = b - 6
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a ) 22 hours , b ) 21 hours , c ) 23 hours , d ) 20 hours , e ) 28 hours | d | subtract(divide(multiply(divide(const_1, const_2), 40), subtract(50, 40)), divide(const_1, const_2)) | a thief goes away with a santro car at a speed of 40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 50 kmph when will the owner over take the thief from the start ? | "d 20 hours | - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - | 50 40 d = 20 rs = 50 – 40 = 10 t = 20 / 10 = 2 hours" | a = 1 / 2
b = a * 40
c = 50 - 40
d = b / c
e = 1 / 2
f = d - e
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a ) 11260 , b ) 11860 , c ) 12360 , d ) 12960 , e ) 13560 | d | add(multiply(const_100, const_100), subtract(lcm(lcm(lcm(lcm(5, 15), 32), 45), 54), reminder(multiply(const_100, const_100), lcm(lcm(lcm(lcm(5, 15), 32), 45), 54)))) | what is the smallest 5 digit number that is divisible by 15 , 32 , 45 , and 54 ? | 15 = 3 * 5 32 = 2 ^ 5 45 = 3 ^ 2 * 5 54 = 2 * 3 ^ 3 lcm = 2 ^ 5 * 3 ^ 3 * 5 = 4320 the smallest five - digit number that is a multiple of 4320 is 3 * 4320 = 12,960 the answer is d . | a = 100 * 100
b = math.lcm(5, 15)
c = math.lcm(b, 32)
d = math.lcm(c, 45)
e = math.lcm(d, 54)
f = 100 * 100
g = math.lcm(5, 15)
h = math.lcm(g, 32)
i = math.lcm(h, 45)
j = math.lcm(i, 54)
k = e - reminder
l = a + k
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a ) 15 , b ) 88 , c ) 77 , d ) 20 , e ) 99 | a | subtract(multiply(const_1, const_60), multiply(divide(45, 60), const_60)) | excluding stoppages , the average speed of a bus is 60 km / hr and including stoppages , the average speed of the bus is 45 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 60 km without stoppages and 45 km with stoppages . stoppage time = time take to travel ( 60 - 45 ) km i . e 15 km at 60 km / hr . stoppage time = 15 / 60 hrs = 15 min . answer : a" | a = 1 * const_60
b = 45 / 60
c = b * const_60
d = a - c
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['a ) 14 cms', 'b ) 21 cms', 'c ) 42 cms', 'd ) none of these', 'e ) can not be determined'] | b | multiply(divide(divide(divide(divide(multiply(const_100, 3.78), const_3), const_3), const_3), const_2), const_3) | a rectangular courty 3.78 metres long and 5.25 metres wide is to be paved exactly with square tiles , all of the same size . what is the largest size of the tile which could be used for the purpose ? | solution largest size of the tile . h . c . f of 378 cm and 525 cm = 21 cms . answer b | a = 100 * 3
b = a / 3
c = b / 3
d = c / 3
e = d / 2
f = e * 3
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a ) 4 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | b | multiply(multiply(add(inverse(multiply(const_2, const_4)), const_1), const_3), multiply(const_2, const_4)) | working together , wayne and his son can shovel the entire driveway in three hours . if wayne can shovel five times as fast as his son can , how many hours would it take for his son to shovel the entire driveway on his own ? | "w : the time for wyane to do the job s : the time for his son to do the job we have 1 / w + 1 / s = 1 / 5 and w = 5 s then we have 1 / ( 5 * s ) + 1 / s = 1 / 5 < = > 6 / ( 5 * s ) = 1 / 5 < = > s = 6 ans : b" | a = 2 * 4
b = 1/(a)
c = b + 1
d = c * 3
e = 2 * 4
f = d * e
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a ) 240 seconds , b ) 120 seconds , c ) 60 seconds , d ) 180 seconds , e ) 100 seconds | b | divide(960, subtract(20, 12)) | 3 bodies x , y and z start moving around a circular track of length 960 m from the same point simultaneously in the same direction at speeds of 12 m / s , 20 m / s and 36 m / s respectively . when will they meet for the first time after they started moving ? | if they all meet after t seconds , it means they covered the distances 12 t , 20 t , and 36 t respectively . since they all arrive to the same spot , it means that the differences taken pairwise between the distances must be positive integer multiples of the length of the track , which is 960 m . so , 8 t , 16 t , and 24 t must all be multiples of 960 . 8 t multiple of 960 means t multiple of 120 . the smallest t with this property is 120 and is on the list of answers . answer b . | a = 20 - 12
b = 960 / a
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a ) 4 : 2 , b ) 4 : 8 , c ) 4 : 3 , d ) 4 : 0 , e ) 6 : 5 | e | multiply(divide(12, const_100), 10) | a part of certain sum of money is invested at 10 % per annum and the rest at 12 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "12 : 10 = 6 : 5 answer : e" | a = 12 / 100
b = a * 10
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a ) 4 / 3 , b ) 2 / 3 , c ) 1 / 3 , d ) 3 / 4 , e ) 1 / 4 | b | divide(subtract(const_100, 60), 60) | kelly and chris are moving into a new city . both of them love books and thus packed several boxes with books . if chris packed 60 % of the total number of boxes , what was the ratio of the number of boxes kelly packed to the number of boxes chris packed ? | the ratio of the number of boxes kelly packed to the number of boxes chris packed = 40 / 60 = 2 / 3 answer : b | a = 100 - 60
b = a / 60
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a ) 10 / 16 , b ) 9 / 15 , c ) 4 / 16 , d ) 6 / 10 , e ) 4 / 10 | b | divide(divide(subtract(24, 6), add(const_1, const_1)), add(divide(subtract(24, 6), add(const_1, const_1)), 6)) | there are 6 more women than there are men on a local co - ed softball team . if there are a total of 24 players on the team , what is the ratio of men to women ? | "w = m + 6 w + m = 24 m + 6 + m = 24 2 m = 18 m = 9 w = 15 ratio : 9 : 15 ans : b" | a = 24 - 6
b = 1 + 1
c = a / b
d = 24 - 6
e = 1 + 1
f = d / e
g = f + 6
h = c / g
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a ) 0.75 , b ) 1 , c ) 1.25 , d ) 1.5 , e ) 2.0 | e | divide(60, add(13, 17)) | two cars start at the same time from opposite ends of a highway that is 60 miles long . one car is riding at 13 mph and the second car is riding at 17 mph . how long after they begin will they meet ? | "as cars are moving in opposite directions their speeds will be added . so their relative speeds : 17 + 13 = 30 mph total distance to be covered = 60 miles . time taken would be : 60 miles / 30 mph = 2.0 hours e is the answer ." | a = 13 + 17
b = 60 / a
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a ) 2 , b ) 2.2 , c ) 1 , d ) 4 , e ) 5 | c | subtract(divide(42, 6), 6) | ( x + 6 ) is a factor in x ^ 2 - mx - 42 . what is the value of m ? | i solved the second degree equation and found it like this : x ^ 2 - mx - 42 = 0 ( x - 7 ) ( x + 6 ) = 0 x = 7 or x = - 6 substituting both values for x in the equation we find : x ^ 2 - mx - 42 = > ( - 6 ) ^ 2 - m ( - 6 ) = 42 = > 36 + 6 m = 42 = > 6 m = 42 - 36 = 6 = > m = 1 and with 7 , using a similar process we end up with : ( 7 ) ^ 2 - m ( 7 ) = 42 - 7 m = 42 - 49 = - 7 m = - 1 ao , ans c | a = 42 / 6
b = a - 6
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a ) 30 , b ) 35 , c ) 45 , d ) 50 , e ) 40 | e | multiply(10, 8) | local kennel has cats and dogs in the ratio of 6 : 8 . if there are 10 fewer cats than dogs , how many dogs are in the kennel ? | "lets work with the data given to us . we know that there ratio of cats to dogs is 6 : 8 or cats 6 dogs 8 we can write number of cats as 6 x and number of dogs as 8 x and we know that 8 x - 6 x = 10 ( therefore 2 x = 10 = > x = 5 ) then # of dogs = 8 x 5 = 40 answer is e" | a = 10 * 8
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a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | b | add(10, add(const_0_25, add(const_0_33, divide(divide(144, 10), const_2)))) | a rectangular field has a length 10 meters more than it is width . if the area of the field is 144 , what is the length ( in meters ) of the rectangular field ? | "area = l * w = ( l ) * ( l - 10 ) = 171 trial and error : 20 * 10 = 200 ( too high ) 19 * 9 = 171 ( too high ) 18 * 8 = 144 the length is 18 meters . the answer is b ." | a = 144 / 10
b = a / 2
c = const_0_33 + b
d = const_0_25 + c
e = 10 + d
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a ) 9.2 , b ) 10.5 , c ) 11.5 , d ) 12.3 , e ) 15 | a | add(divide(multiply(15, 25), const_100), divide(multiply(12, 45), const_100)) | add 15 % of 25 and 12 % of 45 . | 15 % of 25 + 12 % of 45 25 * 15 / 100 + 45 * 12 / 100 3.8 + 5.4 = 9.2 answer a | a = 15 * 25
b = a / 100
c = 12 * 45
d = c / 100
e = b + d
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a ) 941,1009 , b ) 991,1001 , c ) 991,1009 , d ) 791,1009 , e ) 931,1009 | c | divide(999919, add(multiply(const_100, const_10), add(const_3, const_2))) | there are cats got together and decided to kill the mice of 999919 . each cat kills equal number of mice and each cat kills more number of mice than cats there were . then what are the number of cats ? | "999919 can be written as 1000000 – 81 = 10002 – 92 ie of the form a 2 - b 2 = ( a + b ) ( a - b ) = ( 1000 + 9 ) * ( 1000 - 9 ) = ( 1009 ) * ( 991 ) given that number of cats is less than number if mice . so number of cats is 991 and number of mice were 1009 answer c" | a = 100 * 10
b = 3 + 2
c = a + b
d = 999919 / c
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | divide(divide(multiply(multiply(8, 12), 4), 12), 8) | a crate measures 4 feet by 8 feet by 12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius , in feet , of the pillar with the largest volume that could still fit in the crate ? | "to fit the cylinder with largest radius inside this cuboid , we should make the base of the crate as wide as possible so we will take the base as 12 feet by 8 feet now since the limiting number in the base is 8 feet ; therefore a cylinder { we can visualise that a cylinder ' s width is its diameter } can only fit inside the crate if it is 8 feet or less . therefore the radius of the cylinder will become diameter 2 = = = > 8 / 2 = 4 diameter answer : d" | a = 8 * 12
b = a * 4
c = b / 12
d = c / 8
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a ) 180 cm , b ) 220 cm , c ) 240 cm , d ) 270 cm , e ) 300 cm | d | add(triangle_perimeter(45, 45, 45), triangle_perimeter(45, 45, 45)) | an equilateral triangle t 2 is formed by joining the mid points of the sides of another equilateral triangle t 1 . a third equilateral triangle t 3 is formed by joining the mid - points of t 2 and this process is continued indefinitely . if each side of t 1 is 45 cm , find the sum of the perimeters of all the triangles . | "we have 45 for first triangle , when we join mid - points of first triangle we get the second equilateral triangle then the length of second one is 22.5 and continues . so we have 45 , 22.5 , 11.25 , . . . we have ratio = 1 / 2 , and it is gp type . sum of infinite triangle is a / 1 - r = 45 / 1 - ( 1 / 2 ) = 90 equilateral triangle perimeter is 3 a = 3 * 90 = 270 . so option d ." | a = triangle_perimeter + (
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a ) $ 880 , b ) $ 990 , c ) $ 1,000 , d ) $ 1,100 , e ) $ 1,080 | e | subtract(multiply(120, divide(const_100, 10)), 120) | if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 120 , how much more remains to be paid ? | "10 / 100 p = 120 > > p = 120 * 100 / 10 = 1200 1200 - 120 = 1080 answer : e" | a = 100 / 10
b = 120 * a
c = b - 120
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a ) 27 , b ) 22 , c ) 24 , d ) 26 , e ) 30 | a | subtract(power(3, 3), const_1) | a telephone company needs to create a set of 3 - digit area codes . the company is entitled to use only digits 2 , 4 and 6 , which can be repeated . if the product of the digits in the area code must be even , how many different codes can be created ? | "total # of codes possible is 3 * 3 * 3 = 27 . oit of those 27 codes answer : a" | a = 3 ** 3
b = a - 1
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a ) 14300 , b ) 17280 , c ) 14500 , d ) 14600 , e ) 15400 | b | add(10000, multiply(divide(multiply(10000, 20), const_100), 3)) | the population of a town is 10000 . it increases annually at the rate of 20 % p . a . what will be its population after 3 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 10000 × 120 / 100 ^ 3 = 17280 b )" | a = 10000 * 20
b = a / 100
c = b * 3
d = 10000 + c
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a ) 25 days , b ) 18 days , c ) 21 days , d ) 11 days , e ) 13 days | b | multiply(divide(multiply(6, add(const_2, const_1)), const_2), const_2) | a is twice as good a workman as b and they took 6 days together to do the work b alone can do it in ? | "wc = 2 : 1 2 x + x = 1 / 6 x = 1 / 18 = > 18 days answer : b" | a = 2 + 1
b = 6 * a
c = b / 2
d = c * 2
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a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | add(divide(subtract(301, 149), multiply(15, const_2)), const_1) | how many even multiples of 15 are there between 149 and 301 ? | "150 = 10 * 15 300 = 20 * 15 the even multiples are 15 multiplied by 10 , 12 , 14 , 16 , 18 , and 20 for a total of 6 . the answer is b ." | a = 301 - 149
b = 15 * 2
c = a / b
d = c + 1
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a ) a ) 4 , b ) b ) 1 , c ) c ) 2 , d ) d ) 3 , e ) e ) 5 | d | subtract(24, reminder(1101, 24)) | what least number should be added to 1101 , so that the sum is completely divisible by 24 | "explanation : ( 1056 / 24 ) gives remainder 21 21 + 3 = 24 , so we need to add 3 . answer : option d" | a = 24 - reminder
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a ) 12 / 24 , b ) 48 / 63 , c ) 45 / 56 , d ) 48 / 19 , e ) 28 / 12 | d | divide(const_1, divide(add(add(inverse(6), inverse(2)), inverse(8)), 6)) | a and b can do a work in 6 days , b and c in 2 days and c and a in 8 days . in how many days will the work be completed , if all three of them work together ? | "one day work of a and b = 1 / 6 one day work of b and c = 1 / 2 one day work of c and a = 1 / 8 2 ( a + b + c ) = 1 / 6 + 1 / 2 + 1 / 8 2 ( a + b + c ) = 19 / 24 ( a + b + c ) = 19 / 48 number of days required = 48 / 19 days . answer : d" | a = 1/(6)
b = 1/(2)
c = a + b
d = 1/(8)
e = c + d
f = e / 6
g = 1 / f
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a ) 2 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | a | inverse(add(multiply(divide(const_1, 5), add(divide(10, const_100), const_1)), multiply(divide(const_1, 3), divide(10, const_100)))) | working alone , mary can pave a driveway in 5 hours and hillary can pave the same driveway in 3 hours . when they work together , mary thrives on teamwork so her rate increases by 10 % , but hillary becomes distracted and her rate decreases by 10 % . if they both work together , how many hours will it take to pave the driveway ? | "initial working rates : mary = 1 / 5 per hour hillary = 1 / 3 per hour rate when working together : mary = 1 / 5 + ( 1 / 10 * 1 / 5 ) = 2 / 9 per hour hillary = 1 / 3 - ( 1 / 10 * 1 / 3 ) = 3 / 10 per hour together they work 2 / 9 + 3 / 10 = 1 / 2 per hour so they will need 2 hours to complete the driveway . the correct answer is a ." | a = 1 / 5
b = 10 / 100
c = b + 1
d = a * c
e = 1 / 3
f = 10 / 100
g = e * f
h = d + g
i = 1/(h)
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a ) 24 , b ) 54 , c ) 60 , d ) 84 , e ) 94 | c | multiply(divide(multiply(add(add(subtract(add(const_4, const_4), divide(500, const_100)), const_1), add(const_4, const_4)), divide(500, const_100)), subtract(3, const_1)), const_2) | how many 3 digit no ' s are between 100 - 500 , where sum of two digit is 3 rd ? | total permutations possible = 18 + 16 + 14 + 12 = 60 answer : c | a = 4 + 4
b = 500 / 100
c = a - b
d = c + 1
e = 4 + 4
f = d + e
g = 500 / 100
h = f * g
i = 3 - 1
j = h / i
k = j * 2
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a ) 2 / 1 , b ) 4 / 1 , c ) 6 / 5 , d ) 3 / 4 , e ) 3 / 2 | c | divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 21)) | in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 21 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | "given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 21 ) = 36 / 30 = 6 / 5 . answer : c ." | a = 4 + 1
b = 45 / a
c = 45 - b
d = 4 + 1
e = 45 / d
f = e + 21
g = c / f
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a ) 2 m , b ) 6 m , c ) 4 m , d ) 8 m , e ) 9 m | b | divide(subtract(640, multiply(multiply(80, divide(const_1, const_2)), 10)), multiply(80, divide(const_1, const_2))) | the cross - section of a stream is a trapezium in shape . if the stream is 10 m wide at the top and the area of cross - section is 640 sq m , the depth of stream is 80 m and width at the bottom is ? | 1 / 2 * 80 ( 10 + b ) = 640 b = 6 m answer : b | a = 1 / 2
b = 80 * a
c = b * 10
d = 640 - c
e = 1 / 2
f = 80 * e
g = d / f
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a ) 8 , b ) 5 , c ) 7 , d ) 4 , e ) 6 | c | divide(subtract(multiply(2, 27), 26), subtract(multiply(2, 3), 2)) | the cost of 2 books and 2 magazines is $ 26 . the cost of 1 book and 3 magazines is $ 27 . how much does 1 magazine cost ? | let the cost of 1 book = x , let the cost of 1 magazine = y 2 x + 2 y = 26 x = ( 26 - 2 y ) / 2 x = 13 - y again , ( 13 - y ) + 3 y = 27 13 + 2 y = 27 2 y = 14 y = 7 answer : c | a = 2 * 27
b = a - 26
c = 2 * 3
d = c - 2
e = b / d
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a ) 22 , b ) 35 , c ) 27 , d ) 32 , e ) 161 | e | subtract(negate(81), multiply(subtract(21, 41), divide(subtract(21, 41), subtract(6,11, 21)))) | 6,11 , 21 , 41 , 81 , ( . . . ) | "explanation : 6 6 × 2 - 1 = 11 11 × 2 - 1 = 21 21 × 2 - 1 = 41 41 × 2 - 1 = 81 81 × 2 - 1 = 161 answer : option e" | a = negate - (
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a ) 45 % , b ) 56 % , c ) 64 % , d ) 75 % , e ) 80 % | c | multiply(divide(multiply(20, const_4), add(const_100, 25)), const_100) | the organizers of a fair projected a 25 percent increase in attendance this year over that of last year , but attendance this year actually decreased by 20 percent . what percent of the projected attendance was the actual attendance ? | last year attendance = 100 ( assume ) ; projected attendance = 125 ; actual attendance = 80 . the actual attendance therefore was ( actual ) / ( project ) = 80 / 125 * 100 = 64 $ of the projected attendance . answer : c . | a = 20 * 4
b = 100 + 25
c = a / b
d = c * 100
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a ) 5 , b ) 35 , c ) 34 , d ) 36 , e ) 38 | a | sqrt(85) | the length of the longest tape in cm which can be used to measure exactly , the length 10 m ; 3 m 85 cm ; and 11 m 50 cm is : | "the three lengths in cm are 1000 , 385 & 1150 . hcf of 700 , 385 & 1295 is 5 . hence , the answer is 5 cm . answer : a" | a = math.sqrt(85)
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a ) 200 , b ) 235 , c ) 50 , d ) 115 , e ) 150 | b | divide(add(270, 200), const_2) | if x + y = 270 , x - y = 200 , for integers of x and y , y = ? | "x + y = 270 x - y = 200 2 x = 70 x = 35 y = 235 answer is b" | a = 270 + 200
b = a / 2
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a ) 12 days , b ) 15 days , c ) 66 / 5 days , d ) 21 days , e ) 22 / 5 days | c | inverse(subtract(inverse(6), inverse(11))) | a and b together can do a piece of work in 6 days and a alone can do it in 11 days . in how many days can b alone can do it ? | "explanation : a and b can do work 1 / 6 in 1 day a alone can do 1 / 11 work in 1 day b alone can do ( 1 / 6 - 1 / 11 ) = 5 / 66 work in 1 day = > complete work can be done in 66 / 5 days by b answer : option c" | a = 1/(6)
b = 1/(11)
c = a - b
d = 1/(c)
|
a ) 48.89 , b ) 42.25 , c ) 50 , d ) 51.25 , e ) 52.25 | a | divide(add(multiply(22, 50.25), multiply(8, 45.15)), add(22, 8)) | the average weight of 22 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg . find the average weights of all the boys in the class . | explanation : average weight of 22 boys = 50.25 total weight of 22 boys = 50.25 × 22 average weight of remaining 8 boys = 45.15 total weight of remaining 8 boys = 45.15 × 8 total weight of all boys in the class = ( 50.25 × 16 ) + ( 45.15 × 8 ) total boys = 22 + 8 = 30 average weight of all the boys = ( ( 50.25 × 22 ) + ( 45.15 × 8 ) ) / 30 = 48.89 answer : option a | a = 22 * 50
b = 8 * 45
c = a + b
d = 22 + 8
e = c / d
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a ) $ 150 , b ) $ 900 , c ) $ 600 , d ) $ 450 , e ) none | b | add(add(add(divide(300, 1), divide(300, 2)), divide(300, 2)), divide(300, 1)) | profits in the partnership of bess , bill and bob are shared in the ratio 1 : 2 : 3 . if bill ' s share of the profits is $ 300 , what is bob ' s share ? | ans is d given profit ratio , bess : bill : bob = 1 : 2 : 3 to make bill ' s portion $ 300 , multiply by $ 150 = > bess : bill : bob = $ 150 : $ 300 : $ 450 = > bob ' s share = $ 450 | a = 300 / 1
b = 300 / 2
c = a + b
d = 300 / 2
e = c + d
f = 300 / 1
g = e + f
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a ) 24 , b ) 12 , c ) 8 , d ) 4 , e ) 2 | a | subtract(204, multiply(60, 3)) | a sporting goods store sold 60 frisbees in one week , some for $ 3 and the rest for $ 4 each . if receipts from frisbee sales for the week totaled $ 204 , what is the fewest number of $ 4 frisbees that could have been sold ? | "in this question however , because we are told that exactly 64 frisbees have been sold and revenue was exactly $ 204 , there is only one possible solution for the number of $ 3 and $ 4 frisbees sold . to solve , we have 2 equations and 2 unknowns let x = number of $ 3 frisbees sold let y = number of $ 4 frisbees sold x + y = 60 3 x + 4 y = 204 x = 60 - y 3 ( 60 - y ) + 4 y = 204 180 - 3 y + 4 y = 204 y = 24 answer : a" | a = 60 * 3
b = 204 - a
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a ) 0.75 , b ) 0.8 , c ) 1 , d ) 1.2 , e ) 1.44 | e | inverse(divide(90, add(90, 40))) | patrick purchased 90 pencils and sold them at a loss equal to the selling price of 40 pencils . the cost of 90 pencils is how many times the selling price of 90 pencils ? | "say the cost price of 90 pencils was $ 90 ( $ 1 per pencil ) and the selling price of 1 pencil was p . selling at a loss : 90 - 90 p = 40 p - - > p = 9 / 13 . ( cost price ) / ( selling price ) = 1 / ( 9 / 13 ) = 13 / 9 = 1.44 . answer : e ." | a = 90 + 40
b = 90 / a
c = 1/(b)
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a ) a ) 1040 , b ) b ) 1045 , c ) c ) 1055 , d ) d ) 1060 , e ) e ) 1235 | e | add(multiply(8, 70), multiply(9, 75)) | tom purchased 8 kg of apples at the rate of 70 per kg and 9 kg of mangoes at the rate of 75 per kg . how much amount did he pay to the shopkeeper ? | cost of 8 kg apples = 70 × 8 = 560 . cost of 9 kg of mangoes = 75 × 9 = 675 . total cost he has to pay = 560 + 675 = 1235 . e ) | a = 8 * 70
b = 9 * 75
c = a + b
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a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | b | divide(6, subtract(divide(multiply(const_2, 6), 8), 1)) | it takes joey the postman 1 hours to run a 6 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 8 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 6 miles an hour let the other half be x miles an hour now , avg speed = 8 mile an hour 2 * 6 * x / 6 + x = 8 12 x = 8 x + 48 = > x = 12 b" | a = 2 * 6
b = a / 8
c = b - 1
d = 6 / c
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a ) 4 . , b ) 8 . , c ) 12 . , d ) 16 . , e ) 36 . | e | power(subtract(6, divide(add(18, 6), 2)), 2) | if ( a - b - c + d = 18 ) and ( a + b - c - d = 6 ) , what is the value of ( b - d ) ^ 2 ? | "eq 1 : a - b - c + d = 18 eq 2 : a + b - c - d = 6 ( 1 ) subtract eq 1 from eq 2 a - b - c + d = 18 - a + b - c - d = 6 - - - - - - - - - - - - - - - - - - - - - - - - - 2 b + 2 d = 12 ( 2 ) simplify - b + d = 6 b - d = - 6 ( b - d ) ^ 2 = ( - 6 ) ^ 2 = 36 my answer : e" | a = 18 + 6
b = a / 2
c = 6 - b
d = c ** 2
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a ) 70 , b ) 72 , c ) 74 , d ) 75 , e ) 55 | e | divide(1, divide(add(multiply(const_3600, divide(1, 60)), 5), const_3600)) | a car traveling at a certain constant speed takes 5 seconds longer to travel 1 km than it would take to travel 1 km at 60 km / hour . at what speed , in km / hr , is the car traveling ? | "time to cover 1 kilometer at 80 kilometers per hour is 1 / 60 hours = 3,600 / 60 seconds = 60 seconds ; time to cover 1 kilometer at regular speed is 60 + 5 = 65 seconds = 65 / 3,600 hours = 1 / 55 hours ; so , we get that to cover 1 kilometer 1 / 55 hours is needed - - > regular speed 55 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : e" | a = 1 / 60
b = 3600 * a
c = b + 5
d = c / 3600
e = 1 / d
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a ) 1 / 7 , b ) 2 / 7 , c ) 3 / 7 , d ) 4 / 7 , e ) 5 / 7 | b | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting 53 mondays in a leap year ? | "in a leap year contains 52 monday ' s since 52 weeks are present and 2 days extra it may be sunday and monday 2 . monday and tuesday 3 . tuesday and thursday 4 . thursday and friday similary 7 alternatives are possible . among this only 2 are possible cases so 2 / 7 answer : b" | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
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a ) 3125 , b ) 625 , c ) 5 , d ) 25 , e ) 125 | e | multiply(power(5, 5), 5) | if a and b are positive integers and ( 5 ^ a ) ^ b = 5 ^ 2 , what is the value of 2 ^ a * 2 ^ b ? | "5 ^ ab = 5 ^ 2 therefore ab = 2 either a = 1 or 2 or b = 2 or 1 therefore 5 ^ a * 5 ^ b = 5 ^ ( a + b ) = 5 ^ 3 = 125 e" | a = 5 ** 5
b = a * 5
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a ) 111 , b ) 28 , c ) 160 , d ) 213 , e ) 107 | d | subtract(add(floor(divide(subtract(332, 10), 3)), divide(subtract(332, 10), 2)), floor(divide(subtract(332, 10), multiply(2, 3)))) | if w is the set of all the integers between 10 and 332 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 330 , the smallest is 12 ) : 330 - 12 = 318 step 2 . divide by 3 : 318 / 3 = 106 step 3 . add 1 : 106 + 1 = 107 . so there are 107 multiples of 3 within the range : examples are 51 , 54 , 57 , 60 , etc . number of multiples of 2 step 1 . subtract the extreme multiples of 2 within the range ( the greatest is 330 , the smallest is 12 ) : 330 - 12 = 318 step 2 . divide by 2 : 318 / 2 = 159 step 3 . add 1 : 159 + 1 = 160 . so there are 160 multiples of 2 within the range : examples are 50 , 52 , 54 , 56 , 58 , 60 etc . add the 107 multiples of 3 and the 160 multiples of 2 : 107 + 160 = 267 . however , by adding the multiples of 2 and the multiples of 3 , we are effectively counting several numbers twice : for example , 54 and 60 are parts of both the lists above . so we ca n ' t just take 107 + 160 = 267 . find the number of multiples of 6 ( which are the double counted , as 6 is divisible by both 2 and 3 ) , and subtract it from 25 : step 1 . subtract the extreme multiples of 6 within the range ( the greatest is 72 , the smallest is 54 ) : 330 - 12 = 318 step 2 . divide by 6 : 318 / 6 = 53 step 3 . add 1 : 53 + 1 = 54 . so there are 54 multiples of 6 within the range : we counted 54 numbers twice . subtract the 54 multiples of 6 from the sum of the multiples of 2 and 3 : = 107 + 160 - 54 = 267 - 54 = 213 therefore , the final number of multiples of 2 , 3 or 6 is 213 . hence , this is the correct answer . ( d )" | a = 332 - 10
b = a / 3
c = math.floor(b)
d = 332 - 10
e = d / 2
f = c + e
g = 332 - 10
h = 2 * 3
i = g / h
j = math.floor(i)
k = f - j
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['a ) 2345', 'b ) 3456', 'c ) 4567', 'd ) 5678', 'e ) 6789'] | b | divide(multiply(30, power(12, const_3)), subtract(add(12, 6), 3)) | a tank with a volume of 30 cubic feet has one inlet pipe and 2 outlet pipes . the inlet pipe fills water into the tank at the rate of 3 cubic inches / min and the 2 outlet pipes empty it out at the rates of 12 cubic inches / min and 6 cubic inches / min respectively . if all 3 pipes are opened when the tank is full , how many minutes does it take to empty the tank ? ( 1 foot = 12 inches ) | the tank is emptied at this rate : 12 + 6 - 3 = 15 cubic inches / min the tank has a volume of 30 * 12 * 12 * 12 = 51840 cubic inches . the time it takes to empty the tank is 51840 / 15 = 3456 minutes . the answer is b . | a = 12 ** 3
b = 30 * a
c = 12 + 6
d = c - 3
e = b / d
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a ) $ 28,300 , b ) $ 30,800 , c ) $ 31,300 , d ) $ 32,500 , e ) $ 35,100 | b | multiply(divide(231, divide(9, multiply(const_3, const_4))), const_100) | an investment yields an interest payment of $ 231 each month . if the simple annual interest rate is 9 % , what is the amount of the investment ? | "let the principal amount = p simple annual interest = 9 % simple monthly interest = ( 9 / 12 ) = ( 3 / 4 ) % ( 3 / 4 ) * ( p / 100 ) = 230 = > p = ( 231 * 4 * 10 ^ 2 ) / 3 = 77 * 4 * 10 ^ 2 = 308 * 10 ^ 2 = 30800 answer b" | a = 3 * 4
b = 9 / a
c = 231 / b
d = c * 100
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a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) none | b | multiply(subtract(const_1, divide(multiply(const_1, 6), 7.50)), const_100) | if the price of sugar rises from rs . 6 per kg to rs . 7.50 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 6 = x × 7.50 = x = 80 kg . ∴ reduction in consumption = 20 % . answer b" | a = 1 * 6
b = a / 7
c = 1 - b
d = c * 100
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a ) 941,1009 , b ) 991,1001 , c ) 995,1005 , d ) 791,1009 , e ) 931,1009 | c | divide(999975, add(multiply(const_100, const_10), add(const_3, const_2))) | there are cats got together and decided to kill the mice of 999975 . each cat kills equal number of mice and each cat kills more number of mice than cats there were . then what are the number of cats ? | 999975 can be written as 1000000 â € “ 25 = 10002 â € “ 52 ie of the form a 2 - b 2 = ( a + b ) ( a - b ) = ( 1000 + 5 ) * ( 1000 - 5 ) = ( 1005 ) * ( 995 ) given that number of cats is less than number if mice . so number of cats is 995 and number of mice were 1005 answer c | a = 100 * 10
b = 3 + 2
c = a + b
d = 999975 / c
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a ) 10 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | a | divide(factorial(subtract(add(const_4, 01), const_1)), multiply(factorial(01), factorial(subtract(const_4, const_1)))) | how many positive integers less than 50 have a reminder 01 when divided by 5 ? | "1 also gives the remainder of 1 when divided by 5 . so , there are total of 10 numbers . answer : a ." | a = 4 + 1
b = a - 1
c = math.factorial(b)
d = math.factorial(1)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
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a ) 2.125 , b ) 2.375 , c ) 2.625 , d ) 2.675 , e ) 2.825 | c | divide(divide(multiply(add(4, 2), 3.5), 2), 4) | natasha climbs up a hill , and descends along the same way she went up . it takes her 4 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 3.5 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "let the distance to the top be x , so the total distance traveled by natasha is 2 x . the total time is 4 + 2 = 6 hours the average speed = total distance / total time taken = 2 x / 6 = x / 3 the average speed of the complete journey is 3.5 km / hour x / 3 = 3.5 x = 10.5 km the average speed while climbing = distance / time = 10.5 / 4 = 2.625 km / h the answer is c ." | a = 4 + 2
b = a * 3
c = b / 2
d = c / 4
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a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26 | e | subtract(add(const_100, const_1), subtract(add(add(add(add(divide(const_100, const_2), const_1), add(divide(subtract(const_100, const_1), const_3), const_1)), add(divide(const_100, 5), const_1)), add(divide(subtract(const_100, const_10), multiply(multiply(3, 5), 2)), const_1)), add(add(add(divide(subtract(const_100, multiply(const_3, const_2)), multiply(const_3, const_2)), const_1), add(divide(subtract(const_100, const_10), multiply(3, 5)), const_1)), add(divide(const_100, const_10), const_1)))) | how many positive integers d between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ? | 1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the integers ending in 5 ( already took out those ending in 0 ) . take out 10 integers ( 2 ? 5 , ? = 0 to 9 ) , leaving us with 40 integers . 4 ) now the painstaking part . we have to remove the remaining numbers that are multiples of 3 . those are 201 , 207 , 213 , 219 , 231 , 237 , 243 , 249 , 261 , 267 , 273 , 279 , 291 , and 297 . . . a total of 14 numbers . 26 numbers left ! 6 ) answer choice e . | a = 100 + 1
b = 100 / 2
c = b + 1
d = 100 - 1
e = d / 3
f = e + 1
g = c + f
h = 100 / 5
i = h + 1
j = g + i
k = 100 - 10
l = 3 * 5
m = l * 2
n = k / m
o = n + 1
p = j + o
q = 3 * 2
r = 100 - q
s = 3 * 2
t = r / s
u = t + 1
v = 100 - 10
w = 3 * 5
x = v / w
y = x + 1
z = u + y
A = 100 / 10
B = A + 1
C = z + B
D = p - C
E = a - D
|
a ) 63 , b ) 64.28 , c ) 65 , d ) 66 , e ) 68 | b | divide(const_100, add(add(add(divide(const_1, multiply(3, 3)), divide(const_1, multiply(3, 3))), const_1), const_0_33)) | in a certain flower shop , which stocks 4 types of flowers , there are 1 / 3 as many violets as carnations , and 1 / 3 as many tulips as violets . if there are equal numbers of roses and tulips , what percent of the flowers in the shop are carnations ? | given : - violets - c / 3 carnations - c tulip - c / 9 rose - c / 9 total flowers in terms of c = c / 3 + c + c / 9 + c / 9 = 14 c / 9 percentage of carnations = c / 14 c / 9 * 100 = 64.28 % answer b | a = 3 * 3
b = 1 / a
c = 3 * 3
d = 1 / c
e = b + d
f = e + 1
g = f + const_0_33
h = 100 / g
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a ) 45 % , b ) 500 / 11 , c ) 42.22 % , d ) 55 % , e ) 35 % | c | multiply(divide(subtract(90, add(multiply(4, 6), multiply(6, 4))), 90), const_100) | a batsman scored 90 runs which included 4 boundaries and 6 sixes . what percent of his total score did he make by running between the wickets ? | "explanation : number of runs made by running , = > 90 − ( 4 × 4 + 6 × 6 ) . = > 90 − 52 = > 38 hence , the required percentage is : - = > 38 / 90 * 100 = > 42.22 % answer : c" | a = 4 * 6
b = 6 * 4
c = a + b
d = 90 - c
e = d / 90
f = e * 100
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a ) 11 : 00 , b ) 11 : 30 , c ) 12 : 00 , d ) 12 : 30 , e ) 1 : 00 | c | divide(add(70, multiply(70, divide(const_1, const_2))), subtract(84, 70)) | a train sets off at 9 : 00 am at the speed of 70 km / h . another train starts at 9 : 30 am in the same direction at the rate of 84 km / h . at what time will the second train catch the first train ? | "in thirty minutes the first train travels 35 km . the second train catches the first train at a rate of 84 km / h - 70 km / h = 14 km / h . the second train will catch the first train in 35 / 14 = 2.5 hours , so at 12 : 00 noon . the answer is c ." | a = 1 / 2
b = 70 * a
c = 70 + b
d = 84 - 70
e = c / d
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a ) 2 / 35 , b ) 2 / 30 , c ) 2 / 63 , d ) 2 / 29 , e ) 2 / 10 | a | multiply(divide(1, 5), divide(2, 7)) | two brother x and y appeared for an exam . the probability of selection of x is 1 / 5 and that of b is 2 / 7 . find the probability that both of them are selected . | "explanation : let a be the event that x is selected and b is the event that y is selected . p ( a ) = 1 / 5 , p ( b ) = 2 / 7 . let c be the event that both are selected . p ( c ) = p ( a ) ã — p ( b ) as a and b are independent events : = ( 1 / 5 ) ã — ( 2 / 7 ) = 2 / 35 answer : a ) 2 / 35" | a = 1 / 5
b = 2 / 7
c = a * b
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a ) 6 : 5 , b ) 6 : 9 , c ) 6 : 2 , d ) 2 : 2 , e ) 2 : 8 | a | divide(divide(1, 30), power(divide(1, 2), 2)) | the volumes of two cones are in the ratio 1 : 30 and the radii of the cones are in the ratio of 1 : 2 . what is the length of the wire ? | "the volume of the cone = ( 1 / 3 ) π r 2 h only radius ( r ) and height ( h ) are varying . hence , ( 1 / 3 ) π may be ignored . v 1 / v 2 = r 1 ^ 2 . h 1 / r 2 ^ 2 . h 2 = > 1 / 30 = ( 1 ) ^ 2 h 1 / ( 2 ) ^ 2 h 2 = > h 1 / h 2 = 6 / 5 i . e . h 1 : h 2 = 6 : 5 answer : a" | a = 1 / 30
b = 1 / 2
c = b ** 2
d = a / c
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a ) 87 , b ) 86 , c ) 28 , d ) 76 , e ) 80 | b | subtract(multiply(add(10, const_1), add(4, 42)), multiply(10, 42)) | the average of runs of a cricket player of 10 innings was 42 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "average after 11 innings = 46 required number of runs = ( 46 * 11 ) - ( 42 * 10 ) = 506 - 420 = 86 . answer : b" | a = 10 + 1
b = 4 + 42
c = a * b
d = 10 * 42
e = c - d
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a ) 27 % , b ) 21 % , c ) 19 % , d ) 18 % , e ) 16 % | c | add(subtract(subtract(const_100, 40), multiply(divide(3, 4), subtract(const_100, 40))), subtract(40, multiply(divide(9, 10), 40))) | in a survey of parents , exactly 9 / 10 of the mothers and 3 / 4 of the fathers held full - time jobs . if 40 percent of the parents surveyed were women , what percent of the parents did not hold full - time jobs ? | "fathers without full - time jobs are 1 / 4 * 3 / 5 = 3 / 20 of all the parents surveyed . mothers without full - time jobs are 1 / 10 * 2 / 5 = 2 / 50 of all the parents surveyed . the percent of parents without full - time jobs is 3 / 20 + 2 / 50 = 19 / 100 = 19 % the answer is c ." | a = 100 - 40
b = 3 / 4
c = 100 - 40
d = b * c
e = a - d
f = 9 / 10
g = f * 40
h = 40 - g
i = e + h
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a ) 2 : 9 , b ) 2 : 7 , c ) 1 : 2 , d ) 1 : 4 , e ) 1 : 3 | c | divide(subtract(4, 2), subtract(8, 4)) | cereal a is 8 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 4 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "( 8 / 100 ) a + ( 2 / 100 ) b = ( 4 / 100 ) ( a + b ) 4 a = 2 b = > a / b = 1 / 2 answer is c ." | a = 4 - 2
b = 8 - 4
c = a / b
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a ) 30 , b ) 50 , c ) 70 , d ) 80 , e ) 90 | a | subtract(divide(multiply(divide(120, 2), 5), 2), 120) | the ratio of boarders to day students at a school was originally 2 to 5 . however , after a number of new boarders join the initial 120 boarders , the ratio changed to 1 to 2 . if no boarders became day students and vice versa , and no students left the school , how many new boarders joined the school ? | "let x be the number of new boarders . the ratio changed from 2 : 5 = 4 : 10 up to 1 : 2 = 5 : 10 . 120 / ( 120 + x ) = 4 / 5 x = 30 the answer is a ." | a = 120 / 2
b = a * 5
c = b / 2
d = c - 120
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a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3 | e | divide(divide(subtract(2, multiply(divide(0.375, 5), 20)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(20, divide(subtract(2, multiply(divide(0.375, 5), 20)), subtract(divide(1.375, 10), divide(0.375, 5))))) | a 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ? | "the bar lost certain percentage of its weight . we do n ' t know how much tin was lost and how much silver was lost but in all 2 kg was lost with is 10 % of its overall weight . tin loses 1.375 kg in 10 kg so 13.75 % of its weight when it is put in water . silver loses . 375 kg in 5 kg so . 375 / 5 * 100 = 7.5 % of its weight in water . now , we just need to use weighted averages : wt / ws = ( 7.5 - 10 ) / ( 10 - 13.75 ) = 2.5 / 3.75 = 2 / 3 answer ( e )" | a = 0 / 375
b = a * 20
c = 2 - b
d = 1 / 375
e = 0 / 375
f = d - e
g = c / f
h = 0 / 375
i = h * 20
j = 2 - i
k = 1 / 375
l = 0 / 375
m = k - l
n = j / m
o = 20 - n
p = g / o
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a ) 2200 , b ) 5500 , c ) 3300 , d ) 3771.4 , e ) 4400 | d | multiply(circumface(10), 12) | the radius of a cylinder is 10 m , height 12 m . the volume of the cylinder is : | "cylinder volume = ï € r ( power 2 ) h = 22 / 7 ã — 10 ã — 10 ã — 12 = 3771.4 m ( power 3 ) answer is d ." | a = circumface * (
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a ) $ 6980 , b ) $ 7070 , c ) $ 7120 , d ) $ 7260 , e ) $ 7340 | d | multiply(6000, power(add(const_1, divide(10, const_100)), const_2)) | what amount does an investor receive if the investor invests $ 6000 at 10 % p . a . compound interest for two years , compounding done annually ? | a = ( 1 + r / 100 ) ^ n * p ( 1.1 ) ^ 2 * 6000 = 1.21 * 6000 = 7260 the answer is d . | a = 10 / 100
b = 1 + a
c = b ** 2
d = 6000 * c
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a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | e | multiply(divide(multiply(multiply(50, divide(50, 40)), const_2), 50), const_2) | the racing magic takes 50 seconds to circle the racing track once . the charging bull makes 40 rounds of the track in an hour . if they left the starting point together , how many minutes will it take for them to meet at the starting point for the second time ? | "time taken by racing magic to make one circle = 50 seconds time taken bycharging bullto make one circle = 60 mins / 40 = 1.5 mins = 90 seconds lcm of 90 and 50 seconds = 450 seconds time taken for them to meet at the starting point for the second time = 450 * 2 = 900 seconds = 15 mins answer e" | a = 50 / 40
b = 50 * a
c = b * 2
d = c / 50
e = d * 2
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a ) 2 hours , b ) 4 hours , c ) 3 hours , d ) 5 hours , e ) 6 hours | e | add(add(2, 2), 2) | three pipes , a , b , & c are attached to a tank . a & b can fill it in 20 & 30 minutes respectively while c can empty it in 15 minutes . if a , b & c are kept open successively for 2 minute each , how soon will the tank be filled ? | "in three minute 1 / 20 + 1 / 30 - 1 / 15 = 1 / 60 part is filled 6 min - - - - - - - - 1 / 60 parts x min - - - - - - - - - 1 part ( full ) x = 360 min = 6 hours answer : e" | a = 2 + 2
b = a + 2
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a ) 38 sec , b ) 35 sec , c ) 44 sec , d ) 40 sec , e ) 56 | e | multiply(divide(add(divide(340, const_1000), divide(360, const_1000)), 45), const_3600) | a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 340 m long ? | "speed = 45 km / hr = 45 ã — ( 5 / 18 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 340 = 700 meter time taken to cross the platform = 700 / ( 25 / 2 ) = 700 ã — 2 / 25 = 56 seconds answer : e" | a = 340 / 1000
b = 360 / 1000
c = a + b
d = c / 45
e = d * 3600
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a ) 15 , b ) 20 , c ) 18 , d ) 16.5 , e ) 25 | d | multiply(divide(subtract(add(25, add(const_0_25, const_0_25)), 5), add(const_100, subtract(add(25, add(const_0_25, const_0_25)), 5))), const_100) | in a certificate by mistake a candidate gave his height as 25 % more than actual height . in the interview panel , he clarified that his height was 5 feet 6 nches . find the % correction made by the candidate from his stated height to his actual height ? | his height was = 5 feet 6 inch = 6 + 60 = 66 inch . required % correction = 66 * ( 1.25 - 1 ) = 16.5 d | a = const_0_25 + const_0_25
b = 25 + a
c = b - 5
d = const_0_25 + const_0_25
e = 25 + d
f = e - 5
g = 100 + f
h = c / g
i = h * 100
|
a ) 1 / 25 , b ) 12 / 49 , c ) 1 / 4 , d ) 24 / 49 , e ) 1 / 2 | d | multiply(multiply(divide(subtract(divide(50, 2), const_1), subtract(50, const_1)), divide(divide(50, 2), 50)), 2) | there are 2 available positions and 50 candidates , one half of whom are democrats and another half are republicans . if it was decided that the positions would be filled at random , then what is the probability q that the both positions will be taken by members of just one party ? | "q probability of one party having both spots : ( 1 / 2 ) * ( 24 / 49 ) = 12 / 49 ( 1 / 2 ) or ( 25 / 50 ) because it does not matter which party or which person gets the first spot . ( 24 / 49 ) because after one person from a particular party is chosen , there are 24 members of the same party left out of 49 total candidates . since this result can happen for both parties , ( 12 / 49 ) + ( 12 / 49 ) = ( 24 / 49 ) answer : d" | a = 50 / 2
b = a - 1
c = 50 - 1
d = b / c
e = 50 / 2
f = e / 50
g = d * f
h = g * 2
|
a ) 3 , b ) 9 , c ) 15 , d ) 25 , e ) 63 | a | add(const_3, const_4) | what is the smallest positive integer k such that the product of 3675 x k is a perfect square ? | "a perfect square , is just an integer that can be written as the square of some other integer . for example 16 = 4 ^ 2 , is a perfect square . now , 3675 = 5 ^ 2 * 7 ^ 2 * 3 , so if k = 3 then 3675 k = ( 5 * 7 * 3 ) ^ 2 , which is a perfect square ( basically the least positive value of k must complete only the power of 7 to even power as powers of other primes are already even ) . answer : a ." | a = 3 + 4
|
a ) 144 mins , b ) 140 mins , c ) 136 mins , d ) 156 minw , e ) none of these | d | multiply(add(const_1, const_4), 39) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 39 minutes , then the slower pipe alone will be able to fill the tank in | "explanation : let the slower pipe alone fill the tank in x minutes then faster will fill in x / 3 minutes . part filled by slower pipe in 1 minute = 1 / x part filled by faster pipe in 1 minute = 3 / x part filled by both in 1 minute = 1 / x + 3 / x = 1 / 39 = > 4 / x = 1 / 39 x = 39 ∗ 4 = 156 mins option d" | a = 1 + 4
b = a * 39
|
a ) 22 , b ) 38 , c ) 35 , d ) 29 , e ) 18 | c | multiply(7, add(const_4, const_1)) | there are 7 non - collinear points . how many triangles can be drawn by joining these points ? | explanation : a triangle is formed by joining any three non - collinear points in pairs . there are 7 non - collinear points the number of triangles formed = \ inline { \ color { black } 7 c _ { 3 } } = 35 answer : c ) 35 | a = 4 + 1
b = 7 * a
|
a ) 7.6 % , b ) 7.7 % , c ) 24.82 % , d ) 13.6 % , e ) 27.82 % | c | multiply(const_100, divide(subtract(multiply(58, subtract(const_100, 1)), multiply(46, const_100)), multiply(46, const_100))) | a man buys 58 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ? | "explanation : let marked price be re . 1 each c . p . of 58 pens = rs . 46 s . p . of 58 pens = 99 % of rs . 58 = rs . 57.42 profit % = ( profit / c . p . ) x 100 profit % = ( 11.42 / 46 ) x 100 = 24.82 % answer : c" | a = 100 - 1
b = 58 * a
c = 46 * 100
d = b - c
e = 46 * 100
f = d / e
g = 100 * f
|
a ) 187 , b ) 169 , c ) 172 , d ) 178 , e ) 171 | e | subtract(subtract(200, divide(multiply(200, 10), const_100)), divide(multiply(subtract(200, divide(multiply(200, 10), const_100)), 5), const_100)) | the sale price sarees listed for rs . 200 after successive discount is 10 % and 5 % is ? | "200 * ( 90 / 100 ) * ( 95 / 100 ) = 171 answer : e" | a = 200 * 10
b = a / 100
c = 200 - b
d = 200 * 10
e = d / 100
f = 200 - e
g = f * 5
h = g / 100
i = c - h
|
a ) $ 8 , b ) $ 9 , c ) $ 27 , d ) $ 32 , e ) $ 36 | d | multiply(divide(68, add(add(const_1, divide(const_4, const_3)), divide(const_1, const_2))), divide(const_4, const_3)) | if josh , doug , and brad have a total of $ 68 between them , and josh has two times as much money as brad but only 3 - fourths as much as doug , how much money does doug have ? | josh + doug + brad = 68 ; josh = 2 brad , josh = 3 / 4 doug josh + 1 / 2 josh + 4 / 3 josh = 68 ( substituted the given values ) josh = 24 . 24 = 3 / 4 doug = > doug = 32 answer is d . | a = 4 / 3
b = 1 + a
c = 1 / 2
d = b + c
e = 68 / d
f = 4 / 3
g = e * f
|
a ) 715 , b ) 716 , c ) 718 , d ) 720 , e ) 722 | d | divide(multiply(120, 144), subtract(144, 120)) | the bankers discount of a certain sum of money is rs . 144 and the true discount on the same sum for the same time is rs . 120 . the sum due is : | sum = ( b . d * t . d ) / ( b . d - t . d ) ( 144 * 120 ) / 144 - 120 ; 720 answer : d | a = 120 * 144
b = 144 - 120
c = a / b
|
a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 6 | c | add(divide(subtract(52, divide(subtract(divide(add(14, 2), 2), 2), 2)), add(subtract(divide(add(14, 2), 2), 2), divide(add(14, 2), 2))), divide(const_1, 2)) | tammy climbed a mountain in two days . she spent a total of 14 hours climbing the mountain . on the second day , she walked at an average speed that was half a kilometer per hour faster , but 2 hours less than what she walked on the first day . if the total distance she climbed during the two days is 52 kilometers , how many t kilometers per hour did tammy walk on the second day ? | "ans : c total time = 14 hrs let time traveled during 1 st day = x let time traveled during 2 nd day = x - 2 total time = 14 x + x - 2 = 14 x = 8 speed * time = distance s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 solving s = 4.5 now speed for 2 nd day is 0.5 less than the 1 st day which is 4.5 thus speed for 2 nd day = 4 its simple algebra for s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 but for some reason im getting 3.5 and not 4.5 . 8 s + 6 s + 3 = 52 14 s = 49 s = 3.5" | a = 14 + 2
b = a / 2
c = b - 2
d = c / 2
e = 52 - d
f = 14 + 2
g = f / 2
h = g - 2
i = 14 + 2
j = i / 2
k = h + j
l = e / k
m = 1 / 2
n = l + m
|
a ) 22 , b ) 12 , c ) 67 , d ) 20 , e ) 81 | d | divide(add(120, 280), multiply(add(42, 30), const_0_2778)) | two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 120 + 280 = 400 m . the time required = d / s = 400 / 20 = 20 sec . answer : d" | a = 120 + 280
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 57 minutes , b ) 14 minutes , c ) 39 minutes , d ) 40 minutes 20 seconds , e ) none of these | a | multiply(divide(950, subtract(add(40, 30), 20)), const_3) | pipe a fills a tank of capacity 950 liters at the rate of 40 liters a minute . another pipe b fills the same tank at the rate of 30 liters a minute . a pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute . if pipe a is kept open for a minute and then closed and pipe b is open for a minute and then closed and then pipe c is open for a minute and then closed and the cycle is repeated , when will the tank be full ? | in one cycle they fill 40 + 30 - 20 = 50 liters 950 = 50 * n = > n = 19 here n = number of cycles . total time = 19 * 3 = 57 as in one cycle there are 3 minutes . thus 57 minutes answer : a | a = 40 + 30
b = a - 20
c = 950 / b
d = c * 3
|
a ) 9 % , b ) 10 % , c ) 11 % , d ) 14 % , e ) 90 % | d | divide(multiply(12, const_100), subtract(const_100, 12)) | during a sale , the price of a pair of shoes is marked down 12 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | "assume the price = 100 price during sale = 88 price after sale = 100 percent increase = 12 / 88 * 100 = 14 % approx . correct option : d" | a = 12 * 100
b = 100 - 12
c = a / b
|
a ) 12 , b ) 14 , c ) 13 , d ) 15 , e ) 16 | b | add(multiply(5, 2), 4) | if p / q = 5 / 4 , then 2 p + q = ? | "let p = 5 , q = 4 then 2 * 5 + 4 = 14 so 2 p + q = 14 . answer : b" | a = 5 * 2
b = a + 4
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | c | divide(0.06, divide(0.75, const_100)) | find the missing figures : 0.75 % of ? = 0.06 | "let 0.75 % of x = 0.06 . then , 0.75 * x / 100 = 0.06 x = [ ( 0.06 * 100 ) / 0.75 ] = 8 . answer is c ." | a = 0 / 75
b = 0 / 6
|
a ) 1 : 25 , b ) 2 : 25 , c ) 3 : 25 , d ) 4 : 25 , e ) 5 : 25 | c | divide(multiply(divide(30, const_100), divide(20, const_100)), divide(50, const_100)) | m is 30 % of q . q is 20 % of p . n is 50 % of p . find m : n ratio | let p be 100 n = 50 % of 100 ( p = 100 ) = 50 q = 20 % of 100 ( p = 100 ) = 20 m = 30 % of 20 ( q = 20 ) = 6 m : n = 6 : 50 m : n = 3 : 25 answer : c | a = 30 / 100
b = 20 / 100
c = a * b
d = 50 / 100
e = c / d
|
a ) s . 375 , b ) s . 425 , c ) s . 625 , d ) s . 800 , e ) s . 850 | c | multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 5000), 3) | a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 5000 . with the help of c , they completed the work in 3 days . how much is to be paid to c ? | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 3 - 7 / 24 = 1 / 24 a ' s wages : b ' s wages : c ' s wages = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 c ' s share ( for 3 days ) = rs . ( 3 * 1 / 24 * 5000 ) = rs . 625 answer = c" | a = 1/(3)
b = 1/(8)
c = 1/(6)
d = b + c
e = a - d
f = e * 5000
g = f * 3
|
a ) 1 / 190 , b ) 1 / 17 , c ) 1 / 19 , d ) 1 / 10 , e ) 1 / 9 | b | divide(const_1, subtract(18, const_1)) | a box contains 9 pairs of shoes ( 18 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ? | "the problem with your solution is that we do n ' t choose 1 shoe from 18 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 17 ( as after taking one at random there are 17 shoes left and only one is the pair of the first one ) = 1 / 17 answer : b ." | a = 18 - 1
b = 1 / a
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) x = 8 | e | divide(240, 30) | a marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each . there can be no less than 8 musicians per row and no more than 30 musicians per row . how many different rectangular formations x are possible ? | the combinations could be { ( 1,240 ) , ( 2,120 ) , ( 3,80 ) , ( 4,60 ) , ( 5,48 ) , ( 6,40 ) , ( 8,30 ) , ( 10,24 ) , ( 12,20 ) , ) 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) , ( 40,6 ) , ( 48,5 ) , ( 60,4 ) , ( 80,3 ) , ( 120,2 ) , ( 240,1 ) } of these we are told 8 < = t < = 30 so we can remove these pairs , and we are left only with . { ( 8,30 , ( 10,24 ) , ( 12,20 ) , ( 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) } hence 8 . e | a = 240 / 30
|
a ) 420 , b ) 840 , c ) 1,260 , d ) 2,520 , e ) 5,040 | a | lcm(1, 7) | what is the lowest positive integer that is divisible by each of the integers 1 through 7 , inclusive ? | "the integer should be divisible by : 2 , 3 , 4 ( = 2 ^ 2 ) , 5 , 6 ( = 2 * 3 ) , and 7 . the least common multiple of these integers is lcm = 2 ^ 2 * 3 * 5 * 7 = 420 . answer : a ." | a = math.lcm(1, 7)
|
a ) 2277 , b ) 5000 , c ) 1000 , d ) 2651 , e ) 1971 | b | divide(5225, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 5225 . what was his original salary ? | "x * ( 110 / 100 ) * ( 95 / 100 ) = 5225 x * ( 11 / 10 ) * ( 1 / 100 ) = 55 x = 5000 answer : b" | a = 10 / 100
b = 1 + a
c = 5 / 100
d = 1 - c
e = b * d
f = 5225 / e
|
a ) $ 16.32 , b ) $ 18.00 , c ) $ 10.125 , d ) $ 24.48 , e ) $ 28.80 | c | multiply(divide(subtract(const_100, 10), const_100), multiply(0.15, 75)) | the regular price per can of a certain brand of soda is $ 0.15 . if the regular price per can is discounted 10 percent when the soda is purchased in 24 - can cases , what is the price of 75 cans of this brand of soda purchased in 24 - can cases ? | the discounted price of one can of soda is ( 0.9 ) ( $ 0.15 ) , or $ 0.135 therefore , the price of 75 cans of soda at the discounted price would be ( 75 ) ( $ 0.135 ) = 10.125 answer : c . | a = 100 - 10
b = a / 100
c = 0 * 15
d = b * c
|
a ) 888 , b ) 333 , c ) 555 , d ) 1221 , e ) 889 | d | divide(add(multiply(multiply(666, 666), 666), multiply(multiply(555, 555), 555)), subtract(add(multiply(666, 666), multiply(555, 555)), multiply(666, 555))) | solve : - 666 x 666 x 666 + 555 x 555 x 555 = ? ( 666 x 666 - 666 x 555 + 555 x 555 ) | given exp . = ( a 3 + b 3 ) = ( a + b ) = ( 666 + 555 ) = 1221 ( a 2 - ab + b 2 ) answer d | a = 666 * 666
b = a * 666
c = 555 * 555
d = c * 555
e = b + d
f = 666 * 666
g = 555 * 555
h = f + g
i = 666 * 555
j = h - i
k = e / j
|
a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 12 | a | divide(11.52, subtract(96.12, floor(96.12))) | when positive integer x is divided by positive integer y , the remainder is 11.52 . if x / y = 96.12 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 11.52 - - > x = qy + 11.52 ; x / y = 96.12 - - > x = 96 y + 0.12 y ( so q above equals to 96 ) ; 0.12 y = 11.52 - - > y = 96 . answer : a ." | a = math.floor(96, 12)
b = 96 - 12
c = 11 / 52
|
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