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 ) 50 , b ) 99 , c ) 77 , d ) 26 , e ) 23 | a | divide(multiply(36.00001, divide(multiply(subtract(46, 36), const_1000), const_3600)), const_2) | two trains of equal are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 36.00001 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 46 - 36 = 10 km / hr . = 10 * 5 / 18 = 25 / 9 m / sec . 2 x / 36 = 25 / 9 = > x = 50 . answer : a" | a = 46 - 36
b = a * 1000
c = b / 3600
d = 36 * 1
e = d / 2
|
a ) 24 , b ) 40 , c ) 44 , d ) 54 , e ) 64 | b | divide(multiply(56, 10), 14) | two numbers n and 14 have lcm = 56 and gcf = 10 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 14 × n = 56 × 10 n = 56 × 10 / 14 = 40 correct answer b" | a = 56 * 10
b = a / 14
|
a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | b | subtract(50, multiply(divide(50, const_100), 4)) | how many liters of water must be evaporated from 50 liters of a 2 percent sugar solution to get a 4 percent sugar solution ? | "let x be the amount that needs to be evaporated . 0.02 ( 50 ) = 0.04 ( 50 - x ) 0.04 x = 2 - 1 x = 1 / 0.04 = 25 liters the answer is b ." | a = 50 / 100
b = a * 4
c = 50 - b
|
a ) 30 % , b ) 33.33 % , c ) 40 % , d ) 60 % , e ) 50 % | b | multiply(divide(subtract(60, add(multiply(2, 2), multiply(2, 2))), 60), const_100) | a batsman scored 60 runs which included 2 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 60 - ( 2 x 4 + 2 x 6 ) = 60 - ( 20 ) = 20 now , we need to calculate 20 is what percent of 60 . = > 20 / 60 * 100 = 33.33 % option b" | a = 2 * 2
b = 2 * 2
c = a + b
d = 60 - c
e = d / 60
f = e * 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 8 | c | multiply(3, divide(5, 5)) | if 3 people can do 3 times of a particular work in 3 days , then how many days would it take 5 people to do 5 times of that particular work ? | 3 people can do the work one time in one day . 1 person can do 1 / 3 of the work in one day . 5 people can do 5 / 3 of the work in one day . 5 people can do 5 times the work in 3 days . the answer is c . | a = 5 / 5
b = 3 * a
|
a ) 80 , b ) 20 , c ) 30 , d ) 41 , e ) 15 | a | multiply(1720, subtract(const_1, divide(multiply(500, multiply(2, multiply(2, const_3))), add(multiply(350, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(500, multiply(2, multiply(2, const_3))))))) | tom opened a shop investing rs . 500 . jose joined him 2 months later , investing rs . 350 . they earned a profit of rs . 1720 after completion of one year . what will be jose ' s share of profit ? | "sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 1720 = 80 a" | a = 2 * 3
b = 2 * a
c = 500 * b
d = 2 * 3
e = 2 * d
f = e - 2
g = 350 * f
h = 2 * 3
i = 2 * h
j = 500 * i
k = g + j
l = c / k
m = 1 - l
n = 1720 * m
|
a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 94.50 , e ) none of these | d | divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(8748, multiply(3, 4)))), multiply(4, sqrt(divide(8748, multiply(3, 4))))), 25), const_100) | the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 8748 sq . m , the cost of fencing the field @ 25 paise per metre is | "solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 8748 ⇔ 12 x 2 = 8748 ⇔ x 2 = 729 ⇔ x = 27 . so , length = 81 m and breadth = 108 m . perimeter = [ 2 ( 81 + 108 ) ] m = 378 m . ∴ cost of fencing = rs . ( 0.25 × 378 ) = rs . 94.50 . answer d" | a = 3 * 4
b = 8748 / a
c = math.sqrt(b)
d = 3 * c
e = 3 * 4
f = 8748 / e
g = math.sqrt(f)
h = 4 * g
i = rectangle_perimeter * (
j = i / 25
|
a ) 10 d , b ) 0 d , c ) 15 d , d ) 5 d , e ) 20 d | a | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | the angle between the minute hand and the hour hand of a clock when the time is 4.20 , is : | "angle traced hr 13 / 3 = ( 360 / 12 * 13 / 3 ) = 130 traced by min hand 20 min = ( 360 / 60 * 20 ) = 120 req = ( 130 - 120 ) = 10 d answer a" | a = 3 * 4
b = 3 * 4
c = b - 1
d = a / c
e = 4 + 1
f = 3 * 4
g = f - 1
h = e * g
i = d * h
j = const_60 / 2
k = i - j
l = 3 * 4
m = l - 1
n = k * m
o = n / 2
|
a ) 15.2 % , b ) 16.0 % , c ) 20.0 % , d ) 23.2 % , e ) 24.8 % | d | multiply(const_100, subtract(multiply(multiply(divide(add(40, const_100), const_100), divide(add(40, const_100), const_100)), divide(subtract(const_100, 40), const_100)), const_1)) | the positive numbers w , x , y , and z are such that x is 40 percent greater than y , y is 20 percent greater than z , and w is 20 percent less than x . what percent greater than z is w ? | "my strategy is same as thedobermanbut instead take z = 100 , which makes life a bit easy . as : z = 100 y = 120 ( 20 % greater than z ) z = 144 ( 20 % greater than y ) now calculate w 20 % less than z = 144 * 80 / 100 = 115.2 now by just looking , relation between w and z : w - z / z * 100 = 23.2 - answer d" | a = 40 + 100
b = a / 100
c = 40 + 100
d = c / 100
e = b * d
f = 100 - 40
g = f / 100
h = e * g
i = h - 1
j = 100 * i
|
a ) 18 days , b ) 38 days , c ) 40 days , d ) 48 days , e ) 50 days | c | divide(multiply(10, 80), 20) | if 10 men do a work in 80 days , in how many days will 20 men do it ? | "10 * 80 = 20 * x x = 40 days answer : c" | a = 10 * 80
b = a / 20
|
a ) 128 , b ) 140 , c ) 123 , d ) 110 , e ) 150 | a | subtract(divide(8400, 60), 12) | a trader sells 60 meters of cloth for rs . 8400 at the profit of rs . 12 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 8400 / 60 = rs . 140 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 140 - rs . 12 = rs . 128 . answer : a" | a = 8400 / 60
b = a - 12
|
a ) 66 , b ) 26 , c ) 42 , d ) 27 , e ) 21 | e | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 48), const_2) | 48 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | "x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 48 = 21 answer : e" | a = 4 + 3
b = 4 + 3
c = b + 2
d = a / c
e = d * 48
f = e / 2
|
a ) 104 , b ) 190 , c ) 109 , d ) 278 , e ) 211 | c | divide(624, 6) | a car covers a distance of 624 km in 6 â ½ hours . find its speed ? | explanation : 624 / 6 = 104 kmph answer : c | a = 624 / 6
|
a ) 17 , b ) 22 , c ) 21 , d ) 24 , e ) 25 | b | subtract(23, const_1) | when average age of 23 members are 0 , how many members greater than 0 ? | "average of 23 numbers = 0 . sum of 23 numbers ( 0 x 23 ) = 0 . it is quite possible that 22 of these numbers may be positive and if their sum is a then 23 rd number is ( - a ) answer is 22 ( b )" | a = 23 - 1
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | divide(add(multiply(9, 2), 9), 18) | if 2 : 9 : : x : 18 , then find the value of x | "explanation : treat 2 : 9 as 2 / 9 and x : 18 as x / 18 , treat : : as = so we get 2 / 9 = x / 18 = > 9 x = 36 = > x = 4 option c" | a = 9 * 2
b = a + 9
c = b / 18
|
a ) 5 inches , b ) 6 inches , c ) 9 inches , d ) 10 inches , e ) 21 inches | e | multiply(35, divide(1.5, add(const_1, 1.5))) | during the first two weeks of january , the total rainfall in springdale was 35 inches . if the rainfall during the second week was 1.5 times the rainfall during the first week , what was the rainfall during the second week of january ? | total rainfall in 2 weeks = 35 inches . assume the rainfall in second week = 1 . x rainfall in first week = x total rainfall = 2.5 x = 35 inches x = 14 and 1.5 x = 21 rainfall during second week = 21 inches option e | a = 1 + 1
b = 1 / 5
c = 35 * b
|
a ) 2 / 3 , b ) 1 / 2 , c ) 3 / 4 , d ) 5 / 6 , e ) 1 / 3 | a | add(divide(const_3, 6), divide(const_1, 6)) | if you roll one fair 6 - sided die , what is the probability that the number is even or less than 3 ? | the numbers which satisfy the conditions are 1 , 2 , 4 , and 6 . the probability is 4 / 6 = 2 / 3 the answer is a . | a = 3 / 6
b = 1 / 6
c = a + b
|
a ) 12080 , b ) 18060 , c ) 20060 , d ) 20040 , e ) none of them | c | multiply(multiply(multiply(260, 20), const_2), const_2) | a fifth of arun â € ™ s marks in mathematics exceed a third of his marks in english by 20 . if he got 260 marks in two subjects together how many marks did he got in english ? | let arun â € ™ s marks in mathematics and english be x and y then ( 1 / 5 ) x - ( 1 / 3 ) y = 20 3 x - 5 y = 300 â € ¦ â € ¦ > ( 1 ) x + y = 260 â € ¦ â € ¦ . > ( 2 ) solving ( 1 ) and ( 2 ) x = 200 and y = 60 answer is c . | a = 260 * 20
b = a * 2
c = b * 2
|
a ) 1 : 7 , b ) 1 : 1 , c ) 1 : 2 , d ) 1 : 9 , e ) 1 : 3 | e | divide(subtract(multiply(divide(70, add(8, 27)), 8), divide(multiply(divide(70, add(4, 3)), 3), add(1, 4))), subtract(multiply(divide(70, add(8, 27)), 27), multiply(divide(multiply(divide(70, add(4, 3)), 3), add(1, 4)), 4))) | in a colony of 70 residents , the ratio of the number of men and women is 4 : 3 . among the women , the ratio of the educated to the uneducated is 1 : 4 . if the ratio of the number of education to uneducated persons is 8 : 27 , then find the ratio of the number of educated and uneducated men in the colony ? | "number of men in the colony = 4 / 7 ( 70 ) = 40 number of women in the colony = 3 / 7 ( 70 ) = 30 number of educated women in the colony = 1 / 5 ( 30 ) = 6 number of uneducated women in the colony = 4 / 5 ( 30 ) = 24 number of educated persons in the colony = 8 / 35 ( 70 ) = 16 as 6 females are educated , remaining 10 educated persons must be men . number of uneducated men in the colony = 40 - 10 = 30 number of educated men and uneducated men are in the ratio 10 : 30 = > 1 : 3 answer : e" | a = 8 + 27
b = 70 / a
c = b * 8
d = 4 + 3
e = 70 / d
f = e * 3
g = 1 + 4
h = f / g
i = c - h
j = 8 + 27
k = 70 / j
l = k * 27
m = 4 + 3
n = 70 / m
o = n * 3
p = 1 + 4
q = o / p
r = q * 4
s = l - r
t = i / s
|
a ) 47 , b ) 46 , c ) 54 , d ) 58 , e ) 62 | a | add(add(subtract(40, const_3), 7), subtract(10, 7)) | 40 persons like apple . 7 like orange and mango dislike apple . 10 like mango and apple and dislike orange . 4 like all . how many people like apple ? | "orange + mango - apple = 7 mango + apple - orange = 10 apple = 40 orange + mango + apple = 4 40 + 10 + 4 - 7 = 47 like apple answer : a" | a = 40 - 3
b = a + 7
c = 10 - 7
d = b + c
|
a ) 30000 , b ) 60000 , c ) 50000 , d ) 90000 , e ) 70000 | b | multiply(add(multiply(multiply(const_100, const_10), const_100), subtract(multiply(multiply(const_100, const_10), const_100), multiply(multiply(const_2, const_100), const_100))), divide(1, add(divide(const_3, const_2), 1))) | one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 50,000 as total savings , how much has he saved in public provident fund ? | "let savings in n . s . c and p . p . f . be rs . x and rs . ( 150000 - x ) respectively . then , ( 1 / 3 ) x = ( 1 / 2 ) ( 150000 - x ) ( x / 3 ) + ( x / 2 ) = 75000 5 x / 6 = 75000 x = 75000 x 6 / 5 = 90000 savings in public provident fund = rs . ( 150000 - 90000 ) = rs . 60000 answer is b ." | a = 100 * 10
b = a * 100
c = 100 * 10
d = c * 100
e = 2 * 100
f = e * 100
g = d - f
h = b + g
i = 3 / 2
j = i + 1
k = 1 / j
l = h * k
|
a ) 12 kms , b ) 16 kms , c ) 20 kms , d ) 24 kms , e ) none of above | d | divide(multiply(10, multiply(6, 4)), add(6, 4)) | a girl goes to her school from her house at a speed of 6 km / hr and returns at a speed of 4 km / hr . if she takes 10 hours in going and coming back , the distance between her school and house is | let distance be d 10 = d / 4 + d / 6 answer : d | a = 6 * 4
b = 10 * a
c = 6 + 4
d = b / c
|
a ) 90 , b ) 80 , c ) 70 , d ) 60 , e ) 50 | a | multiply(divide(15, const_2), 12) | if the sum of the 4 th term and the 12 th term of an arithmetic progression is 12 , what is the sum of the first 15 terms of the progression ? | "4 th term + 12 th term = 12 i . e . , ( a + 3 d ) + ( a + 11 d ) = 12 now , sum of first 15 terms = ( 15 / 2 ) * [ 2 a + ( 15 - 1 ) d ] = ( 15 / 2 ) * [ 2 a + 14 d ] = ( 15 / 2 ) * 12 - - - - - - - - - - - - - - - from ( 1 ) = 90 answer : a" | a = 15 / 2
b = a * 12
|
a ) 2525 , b ) 2975 , c ) 3225 , d ) 3775 , e ) 2753 | d | divide(add(51, 52), 53) | ( 51 + 52 + 53 + . . . + 100 ) = ? | "explanation : this is an a . p . in which a = 51 , l = 100 and n = 50 . sum = n ( a + l ) = 50 x ( 51 + 100 ) = ( 25 x 151 ) = 3775 . 2 2 answer : d" | a = 51 + 52
b = a / 53
|
a ) 80 . , b ) 75 . , c ) 70 . , d ) 65 . , e ) 54 . | a | divide(add(multiply(divide(30, add(30, 30)), 50), multiply(divide(30, add(30, 30)), 40)), divide(add(30, 30), const_60)) | a car was driving at 50 km / h for 30 minutes , and then at 40 km / h for another 30 minutes . what was its average speed ? | "driving at 50 km / h for 30 minutes , distance covered = 50 * 1 / 2 = 25 km driving at 40 km / h for 30 minutes , distance covered = 40 * 1 / 2 = 20 km average speed = total distance / total time = 45 / 1 = 45 km / h answer : a" | a = 30 + 30
b = 30 / a
c = b * 50
d = 30 + 30
e = 30 / d
f = e * 40
g = c + f
h = 30 + 30
i = h / const_60
j = g / i
|
a ) 8 , b ) 7 , c ) 6 , d ) 15 , e ) 3 | d | add(add(2, 3), add(2, 3)) | a and b walk around a circular track . they start at 9 a . m . from the same point in the opposite directions . a and b walk at a speed of 2 rounds per hour and 3 rounds per hour respectively . how many times shall they cross each other before 11 : 00 a . m . ? | "sol . relative speed = ( 2 + 3 ) = 5 rounds per hour . so , they cross each other 5 times in an hour . hence , they cross each other 15 times before 11 : 00 a . m . answer d" | a = 2 + 3
b = 2 + 3
c = a + b
|
a ) 10000 , b ) 8000 , c ) 12000 , d ) 6000 , e ) 14000 | b | subtract(24000, multiply(const_60, const_100)) | a started a business with an investment of rs . 10000 and after 7 months b joined him investing rs . 12000 . if the profit at the end of a year is rs . 24000 , then the share of b is ? | "ratio of investments of a and b is ( 10000 * 12 ) : ( 12000 * 5 ) = 2 : 1 total profit = rs . 24000 share of b = 1 / 3 ( 24000 ) = rs . 8000 answer : b" | a = const_60 * 100
b = 24000 - a
|
a ) 400 , b ) 750 , c ) 1250 , d ) 2500 , e ) 10 000 | b | divide(30, divide(2, 30)) | in a certain pond , 30 fish were caught , tagged , and returned to the pond . a few days later , 30 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the pond ? | "if x is total number of fish in the pond : 4 = 30 / x * 100 = > x = 750 so answer is b" | a = 2 / 30
b = 30 / a
|
a ) 389 m , b ) 350 m , c ) 289 m , d ) 799 m , e ) 1000 m | e | subtract(multiply(speed(300, 9), 39), 300) | a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 9 sec . what is the length of the platform ? | speed = 300 / 9 = 100 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 100 / 3 = > x = 1000 m . answer : e | a = speed * (
b = a - 39
|
a ) 2777 , b ) 1500 , c ) 3000 , d ) 2776 , e ) 2881 | c | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | a and b ’ s salaries together amount to rs . 4,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is a ’ s salary ? | ( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 4000 4 b = 4000 = > b = 1000 a = 3000 answer c | a = 2 + 3
b = 4 * a
c = 2 * b
d = c * 100
e = 85 / 100
f = 1 - e
g = d * f
h = 95 / 100
i = 1 - h
j = 85 / 100
k = 1 - j
l = i + k
m = g / l
|
a ) 246 , b ) 248 , c ) 1292 , d ) 15,128 , e ) 30,256 | c | multiply(multiply(322, subtract(324, 322)), subtract(324, 322)) | the telephone company wants to add an area code composed of 2 letters to every phone number . in order to do so , the company chose a special sign language containing 324 different signs . if the company used 322 of the signs fully and two remained unused , how many additional area codes can be created if the company uses all 324 signs ? | "# of 2 - letter codes possible from 324 different signs = 324 * 324 . # of 2 - letter codes possible from 322 different signs = 322 * 322 . the difference = 324 ^ 2 - 322 ^ 2 = ( 324 - 322 ) ( 324 + 322 ) = 1292 . answer : c ." | a = 324 - 322
b = 322 * a
c = 324 - 322
d = b * c
|
a ) $ 2.50 , b ) $ 3.00 , c ) $ 3.50 , d ) $ 4.00 , e ) $ 5.00 | e | add(3.5, multiply(0.5, subtract(add(divide(subtract(18, 2), 4), const_1), 2))) | the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 3.50 + 0.50 ( x − 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles ? | number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 / 4 = 4 total number of axles = 5 t = 3.50 + 0.50 ( x − 2 ) = 3.50 + . 5 * 3 = 3.5 + 1.5 = 5 $ answer e | a = 18 - 2
b = a / 4
c = b + 1
d = c - 2
e = 0 * 5
f = 3 + 5
|
a ) 0.5 l , b ) l , c ) 1.5 l , d ) 2 l , e ) 3 l | e | subtract(4, const_1) | a train of length l is traveling at a constant velocity and passes a pole in t seconds . if the same train travelling at the same velocity passes a platform in 4 t seconds , then what is the length of the platform ? | "the train passes a pole in t seconds , so velocity v = l / t ( l + p ) / v = 4 t ( l + p ) / ( l / t ) = 4 t p = 3 l the answer is e ." | a = 4 - 1
|
a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | b | divide(divide(divide(675, const_3), const_3), const_4) | the length of a rectangular plot is thrice its width . if the area of the rectangular plot is 675 sq meters , then what is the width ( in meters ) of the rectangular plot ? | "area = l * w = 3 w ^ 2 = 675 w ^ 2 = 225 w = 15 the answer is b ." | a = 675 / 3
b = a / 3
c = b / 4
|
a ) 72 , b ) 34 , c ) 78 , d ) 37 , e ) 27 | a | multiply(divide(add(70, divide(70, 10)), add(const_100, divide(70, 10))), const_100) | in feb mrs bil ' s earnings were 70 percent of the bil family ' s total income . in june mrs bil earned 10 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs bil ' s earnings were approximately what percent of the bil family ' s total income ? | lets say the family income is 100 in may , bil earned 70 family income is 30 in june , bil earned 10 % more than may , so it is ( 70 + 10 * 70 / 100 = 77 ) family income is same 30 in june bil ' s income percent is 77 * 100 / 107 ~ 72 ans is a | a = 70 / 10
b = 70 + a
c = 70 / 10
d = 100 + c
e = b / d
f = e * 100
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | add(const_2, const_2) | find the number of zeroes at the end of 50 ! | "calculate division by 5 and 25 50 / 5 = 10 50 / 25 = 2 total = 12 answer : c" | a = 2 + 2
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a ) 8 / 19 , b ) 7 / 19 , c ) 10 / 19 , d ) 20 / 19 , e ) 22 / 19 | b | subtract(const_1, divide(choose(subtract(20, 4), const_2), choose(20, const_2))) | a box contains 20 electric bulbs , out of which 4 are defective . two bulbs are chosen at random form this box . the probability that at least one of these is defective is : | "p ( none is defective ) = 16 c 2 / 20 c 2 = 12 / 19 p ( at least one is defective ) = 1 - 12 / 19 = 7 / 19 answer b" | a = 20 - 4
b = math.comb(a, 2)
c = math.comb(20, 2)
d = b / c
e = 1 - d
|
a ) 1 / 5 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 20 , e ) 7 / 20 | d | divide(const_3, multiply(4, 5)) | john and amanda stand at opposite ends of a straight road and start running towards each other at the same moment . their rates are randomly selected in advance so that john runs at a constant rate of 2 , 3 , 4 , or 5 miles per hour and amanda runs at a constant rate of 3 , 4 , 5 , 6 , or 7 miles per hour . what is the probability that john has traveled farther than amanda by the time they meet ? | john will run farther if he runs at 5 mph and amanda runs at 4 mph or 3 mph . in this case , p ( john runs farther ) = 1 / 4 * 2 / 5 = 2 / 20 john will run farther if he runs at 4 mph and amanda runs at 3 mph . in this case , p ( john runs farther ) = 1 / 4 * 1 / 5 = 1 / 20 p ( john runs farther ) = 2 / 20 + 1 / 20 = 3 / 20 the answer is d . | a = 4 * 5
b = 3 / a
|
a ) 50000 , b ) 55000 , c ) 60000 , d ) 70000 , e ) 80804 | e | divide(subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(divide(4, const_100), const_4)), const_4)), multiply(multiply(const_4, const_100), const_100)), subtract(power(add(const_1, divide(divide(2, const_100), const_2)), const_2), const_1)) | john invests $ x at the semi - annual constant compounded rate of 2 percent and also does $ 40,000 at the quarterly constant compounded rate of 4 percent . if the interests are the same after 1 year , what is the value of x ? ? | a = p ( 1 + r / n ) ^ nt a = total amount accrued p = principal deposited r = rate of interest in decimal form n = number of times per year , interest compounded t = time in number of years . . x ( 1 + 0.02 / 2 ) ^ 2 - x = 40,000 ( 1 + 0.04 / 4 ) ^ 4 - 40,000 [ when the principal is subtracted from the total amount accrued , the resulting difference is the interest portion and question states interests are equal ) = > x [ ( 1.01 ) ^ 2 - 1 ] = 40,000 [ ( 1.01 ) ^ 4 - 1 ] = > x [ ( 1.01 ) ^ 2 - 1 ] = 40,000 [ ( 1.01 ) ^ 2 + 1 ] [ ( 1.01 ) ^ 2 - 1 ] - - > using a ^ 2 - b ^ 2 = a + b x a - b formula and cancel common expression on both sides = > x = 40,000 ( 1.0201 + 1 ) = 80804 . hence answer is e . | a = 4 * 100
b = a * 100
c = 4 / 100
d = c / 4
e = 1 + d
f = e ** 4
g = b * f
h = 4 * 100
i = h * 100
j = g - i
k = 2 / 100
l = k / 2
m = 1 + l
n = m ** 2
o = n - 1
p = j / o
|
a ) 10 and 3 , b ) 7 and 10 , c ) 10 and 7 , d ) 3 and 10 , e ) 8 and 5 | e | subtract(add(divide(14, 2), 1), 3) | one side of a rectangle is 3 cm shorter than the other side . if we increase the length of each side by 1 cm , then the area of the rectangle will increase by 14 cm 2 . find the lengths of all sides . | "let x be the length of the longer side x > 3 , then the other side ' s length is x − 3 cm . then the area is s 1 = x ( x - 3 ) cm 2 . after we increase the lengths of the sides they will become ( x + 1 ) and ( x − 3 + 1 ) = ( x − 2 ) cm long . hence the area of the new rectangle will be a 2 = ( x + 1 ) ⋅ ( x − 2 ) cm 2 , which is 14 cm 2 more than the first area . therefore a 1 + 14 = a 2 x ( x − 3 ) + 14 = ( x + 1 ) ( x − 2 ) x 2 − 3 x + 14 = x 2 + x − 2 x − 2 2 x = 16 x = 8 . so , the sides of the rectangle are 8 cm and ( 8 − 3 ) = 5 cm long . so answer is e ." | a = 14 / 2
b = a + 1
c = b - 3
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a ) 67 % . , b ) 42.8 % . , c ) 60 % . , d ) 68 % . , e ) 80 % . | b | multiply(divide(1, 7), const_100) | if two positive numbers are in the ratio 1 / 10 : 1 / 7 , then by what percent is the second number more than the first ? | "given ratio = 1 / 10 : 1 / 7 = 7 : 10 let first number be 7 x and the second number be 10 x . the second number is more than first number by 3 x . required percentage = 3 x / 7 x * 100 = 42.8 % . answer : b" | a = 1 / 7
b = a * 100
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a ) 5.7 , b ) 57 , c ) 570 , d ) 0.57 , e ) 0.057 | b | divide(712.5, 12.5) | if 7125 ¸ 1.25 = 5700 < the value of 712.5 ÷ 12.5 is : | given 7125 / 1.25 = 5700 712.5 / 12.5 = 71.25 / 1.25 = 7125 * 1 / 1.25 * 100 = 5700 / 100 = 57 answer : b | a = 712 / 5
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(45, 33) | how many of the positive factors of 33 are not factors of 45 ? | "factors of 33 - 1 , 3 , 11 , 33 , factors of 45 - 1 , 3 , 5 , 9 , 15 , 45 , comparing both , we have three factors of 33 which are not factors of 45 - 11 , 33 , the answer is c" | a = 45 / 33
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a ) 862 , b ) 1067 , c ) 1977 , d ) 1056 , e ) 1097 | a | divide(1120, add(const_1, divide(multiply(5, 6), const_100))) | find the principle on a certain sum of money at 5 % per annum for 6 years if the amount being rs . 1120 ? | 1120 = p [ 1 + ( 5 * 6 ) / 100 ] p = 862 answer : a | a = 5 * 6
b = a / 100
c = 1 + b
d = 1120 / c
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a ) 430 , b ) 440 , c ) 450 , d ) 460 , e ) 470 | c | subtract(500, subtract(add(40, 20), 10)) | a certain high school has 500 students . of these students , 40 are taking music , 20 are taking art , and 10 are taking both music and art . how many students are taking neither music nor art ? | we ' re given a series of facts to work with : 1 ) a certain high school has 500 students . 2 ) of these students : x are taking music , y are taking art , and z are taking both music and art . we ' re asked how many students are taking neither music nor art ? let ' s test x = 40 y = 20 z = 10 so , we have 40 students taking music , 20 taking art and 10 taking both music and art . 30 student taking just music 10 student taking just art 10 student taking both music and art total = 50 students we ' re asked for the total number of students who are taking neither course . that is 500 - 50 = 450 . c | a = 40 + 20
b = a - 10
c = 500 - b
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a ) 0 , b ) 50 , c ) 450 , d ) 693 , e ) 500 | d | subtract(multiply(multiply(add(const_3, const_4), const_1000), divide(1, 10)), multiply(divide(divide(1, 10), const_100), multiply(add(const_3, const_4), const_1000))) | when 1 / 10 percent of 7,000 is subtracted from 1 / 10 of 7,000 , the difference is | we can break this problem into two parts : 1 ) what is 1 / 10 percent of 7,000 ? 2 ) what is 1 / 10 of 7,000 ? to calculate 1 / 10 percent of 7,000 we must first remember to divide 1 / 10 by 100 . so we have : ( 1 / 10 ) / ( 100 ) to divide a number by 100 means to multiply it by 1 / 100 , so we have : 1 / 10 x 1 / 100 = 1 / 1,000 thus , 1 / 10 percent of 7,000 = 1 / 1,000 x 7,000 = 7 . now let ' s concentrate on part 2 . we need to calculate 1 / 10 of 7,000 . to do this we simply multiply 1 / 10 by 7,000 . 1 / 10 x 7,000 = 700 the answer to part 1 is 7 , and the answer to part 2 is 700 . their difference is 700 – 7 = 693 . answer d . | a = 3 + 4
b = a * 1000
c = 1 / 10
d = b * c
e = 1 / 10
f = e / 100
g = 3 + 4
h = g * 1000
i = f * h
j = d - i
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a ) rs . 2500 , b ) rs . 1500 , c ) rs . 1400 , d ) rs . 1600 , e ) rs . 1300 | b | divide(210, divide(multiply(4, add(3, divide(1, 2))), const_100)) | what sum of money will produce rs . 210 as simple interest in 4 years at 3 1 / 2 percent ? | 210 = ( p * 4 * 7 / 2 ) / 100 p = 1500 answer : b | a = 1 / 2
b = 3 + a
c = 4 * b
d = c / 100
e = 210 / d
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a ) 138 , b ) 148 , c ) 150 , d ) 162 , e ) 192 | e | multiply(subtract(5, divide(multiply(90, 5), add(90, 160))), const_60) | while driving from a - ville to b - town , harriet drove at a constant speed of 90 kilometers per hour . upon arriving in b - town , harriet immediately turned and drove back to a - ville at a constant speed of 160 kilometers per hour . if the entire trip took 5 hours , how many minutes did it take harriet to drive from a - ville to b - town ? | 5 hr = 300 min . if harriet spend equal hrs on each leg she will spend 150 min on each . since speed a - b is less than speed b - a and distance on each leg is the same , time spent on a - b is more than 150 min , which mean we can eliminate ans . a , b and c . now let plug in ans . d or e and verify which one give same distance on each leg . d . t = 162 min * leg a - b - - - > d = 90.192 / 60 = 17280 / 60 * leg b - a - - - - > d = 160 * 108 / 60 = 17280 / 60 so the correct ans . ise | a = 90 * 5
b = 90 + 160
c = a / b
d = 5 - c
e = d * const_60
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a ) 100 , b ) 250 , c ) 750 , d ) 5625 , e ) 6000 | e | multiply(multiply(multiply(64, 25), divide(1, const_2)), 7.5) | the milk level in a rectangular box measuring 64 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons ) | 6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 64 * 25 * 1 / 2 = 800 feet ^ 3 of milk must be removed , which equals to 800 * 7.5 = 6000 gallons . answer : e . | a = 64 * 25
b = 1 / 2
c = a * b
d = c * 7
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a ) $ 1,000 , b ) $ 960 , c ) $ 240 , d ) $ 200 , e ) $ 150 | c | divide(add(divide(multiply(1000, 20), const_100), 1000), multiply(divide(1000, const_100), const_2)) | a computer store offers employees a 20 % discount off the retail price . if the store purchased a computer from the manufacturer for $ 1000 dollars and marked up the price 20 % to the final retail price , how much would an employee save if he purchased the computer at the employee discount ( 20 % off retail price ) as opposed to the final retail price . | "cost price = 1000 profit = 20 % = 20 % of 1000 = 200 selling price = cp + profit sp = 1200 a discount of 20 % to employees means 20 % off on 1200 so 20 % of 1200 = 240 so a employee thinks he saved 240 . . answer : c" | a = 1000 * 20
b = a / 100
c = b + 1000
d = 1000 / 100
e = d * 2
f = c / e
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a ) $ 1800 , b ) $ 2500 , c ) $ 3000 , d ) $ 3120 , e ) $ 1540 | a | divide(multiply(divide(multiply(90, 10), subtract(15, 10)), const_100), 10) | i sold a book at a profit of 10 % . had i sold it for $ 90 more , 15 % would have been gained . find the cost price ? | "115 % of cost - 110 % of cost = $ 90 5 % of cost = $ 90 cost = 90 * 100 / 5 = $ 1800 answer is a" | a = 90 * 10
b = 15 - 10
c = a / b
d = c * 100
e = d / 10
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a ) 3.6 , b ) 6 , c ) 18 , d ) can not be determined , e ) none of these | e | sqrt(divide(multiply(972, const_100), 1200)) | reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 972 as interest at the end of the loan period , what was the rate of interest ? | "let rate = r % and time = r years . then , 1200 x r x r / 100 = 972 12 r 2 = 972 r 2 = 81 r = 9 . answer : e" | a = 972 * 100
b = a / 1200
c = math.sqrt(b)
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | a | subtract(negate(11), multiply(subtract(10, 8), divide(subtract(10, 8), subtract(7, 10)))) | 7 , 10 , 8 , 11 , 9 , 12 , ___ | "10 = 7 + 3 8 = 10 - 2 11 = 8 + 3 9 = 11 - 2 12 = 9 + 3 similarly 12 - 2 = 10 answer : a" | a = negate - (
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a ) a ) rs . 6 , 602.50 , b ) b ) rs . 6 , 601.50 , c ) c ) rs . 6 , 603.50 , d ) d ) rs . 6 , 604.50 , e ) e ) rs . 6 , 605.50 | a | subtract(multiply(power(add(const_1, divide(divide(20, const_100), const_2)), const_4), 20000), multiply(power(add(divide(20, const_100), const_1), const_2), 20000)) | how much more would rs . 20000 fetch , after two years , if it is put at 20 % p . a . compound interest payable half yearly than if is put at 20 % p . a . compound interest payable yearly ? | "20000 ( 11 / 10 ) 4 - 20000 ( 6 / 5 ) 2 = 6 , 602.50 answer : a" | a = 20 / 100
b = a / 2
c = 1 + b
d = c ** 4
e = d * 20000
f = 20 / 100
g = f + 1
h = g ** 2
i = h * 20000
j = e - i
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a ) $ 250 , b ) $ 350 , c ) $ 450 , d ) $ 60 , e ) $ 650 | d | multiply(multiply(power(const_100, const_2), divide(10, const_100)), divide(const_3, 10)) | how much interest will $ 10,000 earn in 9 months at an annual rate of 10 % ? | "soln : - 9 months = 3 / 4 of year ; 10 % = 10 / 100 = 1 / 10 ; $ 10,000 ( principal ) * 3 / 50 ( interest rate ) * 1 / 10 ( time ) = $ 60 . answer : d" | a = 100 ** 2
b = 10 / 100
c = a * b
d = 3 / 10
e = c * d
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a ) $ 1,333 , b ) $ 1,733 , c ) $ 3,466 , d ) $ 13,333 , e ) $ 20,796 | a | subtract(9, multiply(4, const_2)) | a new home buyer pays 4 % annual interest on her first mortgage and 9 % annual interest on her second mortgage . if she borrowed a total of $ 320,000 , 80 % of which was in the first mortgage , what is her approximate monthly interest payment ? | "making the numbers a little easy looking : monthly payment = 320,000 * ( 80 / 100 * 4 / 100 + 20 / 100 * 9 / 100 ) / 12 = 3200 / 12 * ( 3.2 + 1.8 ) = 16000 / 12 = 1333 answer : a" | a = 4 * 2
b = 9 - a
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a ) 5 : 8 , b ) 5 : 3 , c ) 5 : 2 , d ) 5 : 9 , e ) 2 : 5 | c | divide(multiply(5, 4), 8) | a dog takes 5 leaps for every 8 leaps of a hare . if one leap of the dog is equal to 4 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 5 * 4 ) leaps of hare : 8 leaps of hare = 20 : 8 - - > 10 : 4 - - > 5 : 2 answer : c" | a = 5 * 4
b = a / 8
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a ) 33.33 % , b ) 35 % , c ) 27.75 % , d ) 35.75 % , e ) 32.25 % | a | multiply(divide(50, subtract(200, 50)), const_100) | by selling 200 apples , a fruit - seller gains the selling price of 50 apples . find the gain percent ? | sp = cp + g 200 sp = 200 cp + 50 sp 150 sp = 200 cp 150 - - - 50 cp 100 - - - ? = > 33.33 % answer : a | a = 200 - 50
b = 50 / a
c = b * 100
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a ) 12 days , b ) 16 days , c ) 18 days , d ) 11 days , e ) 38 days | a | add(divide(subtract(const_1, multiply(add(divide(const_1, 24), divide(const_1, 30)), 8)), add(divide(const_1, multiply(add(const_2, const_3), multiply(const_2, 8))), add(divide(const_1, 24), divide(const_1, 30)))), 8) | a , b and c can do a piece of work in 24 days , 30 days and 40 days respectively . they began the work together but c left 8 days before the completion of the work . in how many days was the work completed ? | "one day work of a , b and c = 1 / 24 + 1 / 30 + 1 / 40 = 1 / 10 work done by a and b together in the last 8 days = 8 * ( 1 / 24 + 1 / 30 ) = 3 / 5 remaining work = 2 / 5 the number of days required for this initial work = 4 days . the total number of days required = 8 + 4 = 12 days . answer : a" | a = 1 / 24
b = 1 / 30
c = a + b
d = c * 8
e = 1 - d
f = 2 + 3
g = 2 * 8
h = f * g
i = 1 / h
j = 1 / 24
k = 1 / 30
l = j + k
m = i + l
n = e / m
o = n + 8
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a ) 53 , b ) 54 , c ) 55 , d ) 56 , e ) 57 | c | divide(subtract(multiply(floor(12.4), 4), 26), subtract(12.4, floor(12.4))) | a man whose bowling average is 12.4 , takes 4 wickets for 26 runs and there by decreases his average by 0.4 . the number of wickets taken by him before his last match is ? | "12.4 * x + 26 = ( 4 + x ) 12 solve equation x = 55 answer : c" | a = math.floor(12, 4)
b = a * 4
c = b - 26
d = math.floor(12, 4)
e = 12 - 4
f = c / e
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a ) 35 , b ) 28 , c ) 40 , d ) 44 , e ) 42 | c | subtract(60, multiply(sqrt(25), divide(subtract(60, 48), sqrt(9)))) | an engine moves at the speed of 60 kmph without any coaches attached to it . speed of the train reduces at the rate that varies directly as the square root of the number of coaches attached . when 9 coaches are attached speed decreases to 48 kmph . what will be the speed of train when 25 coaches are attached . | 1 . no . of coaches = 9 sqr root = 3 speed decreases by 12 12 = k * 3 k = 4 no . of coaches = 25 swr root = 5 decrease = 5 * 4 = 20 new speed = 60 - 20 = 40 c | a = math.sqrt(25)
b = 60 - 48
c = math.sqrt(9)
d = b / c
e = a * d
f = 60 - e
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a ) 2 % , b ) 5 % , c ) 14 % , d ) 28 % , e ) 63 % | b | floor(multiply(subtract(divide(10, 63), divide(12, 80)), const_100)) | a survey was sent to 80 customers , 12 of whom responded . then the survey was redesigned and sent to another 63 customers , 10 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "rate of first survey = 12 / 80 rate of second survey = 10 / 63 % response rate increase ( 10 / 63 - 12 / 80 ) / ( 12 / 80 ) = 5 % answer is b = 5 %" | a = 10 / 63
b = 12 / 80
c = a - b
d = c * 100
e = math.floor(d)
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a ) 298 , b ) 288 , c ) 120 , d ) 776 , e ) 180 | e | divide(const_100.0, subtract(divide(const_100.0, 10), 9)) | a train covers a distance of 12 km in 10 min . if it takes 9 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 9 = 180 m . answer : e" | a = 100 / 0
b = a - 9
c = 100 / 0
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a ) 2 : 3 , b ) 5 : 3 , c ) 1 : 4 , d ) 3 : 2 , e ) 3 : 1 | b | divide(subtract(60, divide(60, add(5, 1))), add(divide(60, add(5, 1)), 20)) | in a mixture of 60 litres the ratio of milk to water is 5 : 1 . additional 20 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | "given that milk / water = 5 x / x and 5 x + x = 60 - - > x = 10 . thus milk = 5 x = 50 liters and water = x = 10 liters . new ratio = 50 / ( 20 + 10 ) = 50 / 30 = 5 / 3 . answer is b" | a = 5 + 1
b = 60 / a
c = 60 - b
d = 5 + 1
e = 60 / d
f = e + 20
g = c / f
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a ) 240 meters , b ) 320 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined | b | subtract(multiply(divide(multiply(72, const_1000), const_3600), 34), multiply(divide(multiply(72, const_1000), const_3600), 18)) | a train traveling at 72 kmph crosses a platform in 34 seconds and a man standing on the platform in 18 seconds . what is the length of the platform in meters ? | "speed of train = 72 * ( 5 / 18 ) = 20 m / s lets consider the man as a stationery point on the platform . crossing the point gives us the length of the train . lt = 20 * 18 = 360 m . crossing the platform gives us the length of trainlength of platform . l ( t + p ) = 20 * 34 = 680 m . so , length of platform = 680 - 360 = 320 m imo , answer b" | a = 72 * 1000
b = a / 3600
c = b * 34
d = 72 * 1000
e = d / 3600
f = e * 18
g = c - f
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['a ) 1 / sqrt 2', 'b ) 1', 'c ) sqrt 2', 'd ) sqrt 3', 'e ) 2 sqrt 3'] | d | sqrt(add(const_1, 2)) | square abcd is the base of the cube while square efgh is the cube ' s top face such that point e is above point a , point f is above point b etc . what is the distance between the midpoint of edge ab and the midpoint of edge eh if the area of square abcd is 2 ? | distance from mid point of ab to ad = sqrt [ ( 1 / sqrt 2 ) ^ 2 + ( 1 / sqrt 2 ) ^ 2 ] = 1 the distance between the midpoint of edge ab and the midpoint of edge eh = sqrt [ 1 ^ 2 + ( sqrt 2 ) ^ 2 ] = sqrt 3 . answer : d | a = 1 + 2
b = math.sqrt(a)
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a ) 12 , b ) 11 , c ) 10 , d ) 27 , e ) 22 | c | subtract(20, 10) | in an ngo the daily average wages of 20 illiterate employees is decreased from rs . 25 to rs . 10 , thus the average salary of all the literate and illiterate employees is decreased rs . 10 per day . the no . of educated employees working in the ngo is : | explanation : total employees = \ inline \ frac { ( 25 - 10 ) \ times 20 } { 10 } = 30 hence number of educated employees = 30 - 20 = 10 answer : c ) 10 | a = 20 - 10
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a ) 180 , b ) 181 , c ) 182 , d ) 178 , e ) 168 | a | add(add(add(50, 48), 44), 46) | you have been given a physical balance and 7 weights of 52 , 50 , 48 , 44 , 45 , 46 and 78 kgs . keeping weights on one pan and object on the other , what is the maximum you can weigh less than 183 kgs . | "add the max weights 52 + 50 + 78 = 180 answer : a" | a = 50 + 48
b = a + 44
c = b + 46
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a ) e 982 , b ) 27 , c ) 12 , d ) 14.6 , e ) 28 | d | subtract(const_60, multiply(const_60, divide(34, 45))) | excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 34 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 11 / 45 * 60 = 14.6 answer : option d" | a = 34 / 45
b = const_60 * a
c = const_60 - b
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a ) 20,000 , b ) 21,000 , c ) 28,200 , d ) 20,500 , e ) none of these | c | divide(add(multiply(subtract(const_12, 5), 4000), multiply(multiply(const_2, multiply(const_100, const_100)), 5)), multiply(const_100, const_10)) | a , b and c start a business each investing 20,000 . after 5 months a withdrew 5000 , b withdrew 4000 and c invests 6000 more . at the end of the year , a total profit of 69,900 was recorded . find the share of b | "ratio of the capitals of a , b and c = 20000 × 5 + 15000 × 7 : 20000 × 5 + 16000 × 7 : 20000 × 5 + 26000 × 7 = 205000 : 212000 : 282000 = 205 : 212 : 282 . b ’ s share = ( 69900 × 212 ⁄ 699 ) = 21200 ; answer c" | a = 12 - 5
b = a * 4000
c = 100 * 100
d = 2 * c
e = d * 5
f = b + e
g = 100 * 10
h = f / g
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a ) 10 , b ) 7 , c ) 6 , d ) 12 , e ) 13 | b | divide(multiply(add(multiply(divide(8, 5), 3), 5), 10), 14) | 5 men or 8 women do equal amount of work in a day . a job requires 3 men and 5 women to finish the job in 10 days how many woman are required to finish the job in 14 days . | 1 man can do the work of 8 / 5 = 1.6 women equivalent of 3 * 1.6 + 5 = 9.8 women take 10 days so women needed to do the work in 14 days = [ 10 / 14 ] * 9.8 = 7 answer : b | a = 8 / 5
b = a * 3
c = b + 5
d = c * 10
e = d / 14
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a ) 5 hrs , b ) 10 hrs , c ) 15 hrs , d ) 20 hrs , e ) 30 hrs | a | divide(50, 10) | ajay can walk 10 km in 1 hour . in how many hours he can walk 50 km ? | 1 hour he walk 10 km he walk 50 km in = 50 / 10 * 1 = 5 hours answer is a | a = 50 / 10
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a ) 76.4 % , b ) 78 % , c ) 77.4 % , d ) 72 % , e ) 73.9 % | d | divide(add(75, 65), const_2) | factory x ' s bulbs work for over 5000 hours in 75 % of cases , whereas factory y ' s bulbs work for over 5000 hours in 65 % of cases . it is known that factory x supplies 70 % of the total bulbs available . what is the chance that a purchased bulb will work for longer than 5000 hours ? | "for x , 70 % of 75 % will work . for y , 30 % of 65 % will work . * 30 % is the rest of the bulb supply in the market . so , the probability that a purchased bulb will work is : 0.70 ( 0.75 ) = . 525 0.30 ( 0.65 ) = 0.195 the combined probability then is 52.5 + 19.5 = 72 % ans d" | a = 75 + 65
b = a / 2
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a ) 2112 m , b ) 2704 m , c ) 2278 m , d ) 2288 m , e ) 2378 m | a | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 1500), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 1500 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 1500 resolutions . = 1500 * 2 * 22 / 7 * 22.4 = 211200 cm = 2112 m answer : a" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 1500
i = h / 100
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a ) 210 m , b ) 220 m , c ) 230 m , d ) 240 m , e ) 200 m | e | subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 300) | a 300 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds . what is the lenght of other train . | relative speeds = ( 120 + 80 ) km / hr = 200 km / hr = ( 200 * 5 / 18 ) m / s = ( 500 / 9 ) m / s let length of train be xm x + 300 / 9 = 500 / 9 x = 200 ans is 200 m answer : e | a = 120 + 80
b = a * const_0_2778
c = 9 * b
d = c - 300
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a ) 12 , b ) 14 , c ) 45 , d ) 8 , e ) 65 | d | inverse(add(divide(divide(const_100, subtract(const_100, 20)), 12), divide(multiply(divide(divide(const_100, subtract(const_100, 20)), 12), 20), const_100))) | by selling 12 pens for a rupee a woman loses 20 % . how many for a rupee should he sell in order to gain 20 % ? | "d 80 % - - - 12 120 % - - - ? 80 / 120 * 12 = 8" | a = 100 - 20
b = 100 / a
c = b / 12
d = 100 - 20
e = 100 / d
f = e / 12
g = f * 20
h = g / 100
i = c + h
j = 1/(i)
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(16, 26) | how many of the positive factors of 26 , 16 and how many common factors are there in numbers ? | "factors of 26 - 1 , 2 , 13 and 26 factors of 16 - 1 , 2 , 4 , 8 and 16 comparing both , we have three common factors of 45,16 - 2 answer : b" | a = 16 / 26
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | d | subtract(subtract(add(add(63, 18), 37), const_100), const_10) | find unit digit of n when n = 63 ^ 1 ! + 2 ! + . . . + 63 ! + 18 ^ 1 ! + 2 ! + . . . + 18 ! + 37 ^ 1 ! + 2 ! + . . . 37 ! | 3,8 and 7 have a power cycle of 4 , i . e . the units digit in each case will repeat after every 4 th power . for eg : 3 ^ 1 = 3 , 3 ^ 2 = 9 , 3 ^ 3 = 27 , 3 ^ 4 = 81 , 3 ^ 5 = 243 all the powers given ( 1 ! + 2 ! + . . . . ) , are multiples of 4 . it is so because the last 2 digits of the total sum will be 00 , for each one of them , which make them divisible by 4 . thus , the given problem boils down to 3 ^ 4 + 8 ^ 4 + 7 ^ 4 = 1 + 6 + 1 = 8 thus , the units digit is 8 . answer : d | a = 63 + 18
b = a + 37
c = b - 100
d = c - 10
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a ) 30 , b ) 60 , c ) 90 , d ) 180 , e ) 540 | c | multiply(divide(divide(factorial(6), factorial(subtract(6, const_2))), const_2), factorial(3)) | 6 students are equally divided into 3 groups , then , the 3 groups were assigned to 3 different topics . how many different arrangements w are possible ? | 90 is the number of ways you can assign 3 teams formed out of 12 people to 3 different tasks . but now you can order the 3 tasks in 3 ! ways . t 1 t 2 t 3 or t 2 t 1 t 3 . . . . etc etc . i was confused between 90 and 540 but since question used the wordarrangementsdecided to go with complete arrangements w including the order of tasks . could you explain the highlighted step . . . i ' m getting 90 = 15 * 3 ! suppose the students are numbered 12 , 34 , 56 and tasks are x , y and z one of the 15 possible ways of forming teams is 12 , 34 , 56 . these teams can be assigned to 3 tasks in 3 ! = 6 ways x - - y - - z 12 - - 34 - - 56 12 - - 56 - - 34 34 - - 12 - - 56 34 - - 56 - - 12 56 - - 12 - - 34 56 - - 34 - - 12 so the answer should be 15 * 6 = 90 but now you can fruther decide which task you want to perform first x y or z . . = c | a = math.factorial(6)
b = 6 - 2
c = math.factorial(b)
d = a / c
e = d / 2
f = math.factorial(3)
g = e * f
|
a ) 195 / 63 , b ) 145 / 63 , c ) 155 / 63 , d ) 125 / 63 , e ) 3.075 | e | add(divide(sqrt(1.5), sqrt(0.81)), divide(sqrt(1.44), sqrt(0.49))) | find the value of ( √ 1.5 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) is | "( √ 1.5 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) = > 3.075 answer is e ." | a = math.sqrt(1)
b = math.sqrt(0)
c = a / b
d = math.sqrt(1)
e = math.sqrt(0)
f = d / e
g = c + f
|
a ) 12765 , b ) 121227 , c ) 26887 , d ) 19977 , e ) 26992 | a | add(add(add(const_1000, const_1000), divide(40000, 4)), add(add(multiply(const_60, const_10), const_100), add(const_60, 5))) | a completes a piece of work in 3 / 4 of the time in b does , b takes 4 / 5 of the time in c does . they got a profit of rs . 40000 how much b gets ? | assume c takes 20 days . now b takes 4 / 5 ( 20 ) = 16 days . a takes 3 / 4 ( 16 ) = 12 now their efficiencies ratio = 1 / 20 : 1 / 16 : 1 / 12 = 12 : 15 : 20 b ' s share in the profit of rs . 40000 = 15 / 47 ( 40000 ) = rs . 12765 answer : a | a = 1000 + 1000
b = 40000 / 4
c = a + b
d = const_60 * 10
e = d + 100
f = const_60 + 5
g = e + f
h = c + g
|
a ) 42 , b ) 43 , c ) 44 , d ) 45 , e ) 46 | c | add(add(power(add(add(divide(subtract(subtract(87, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(87, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(87, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(87, const_10), const_2), const_4), const_2), const_2))) | the sum of two consecutive number is 87 . which is the larger number ? | "let consecutive number be x , x + 1 therefore sum of the consecutive number is x + x + 1 = 87 2 x + 1 = 87 2 x = 86 x = 43 therefore larger number is x + 1 = 44 answer : c" | a = 87 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 87 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 87 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 87 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 110 , b ) 1100 , c ) 9900 , d ) 10000 , e ) 12100 | c | multiply(multiply(subtract(const_10, const_1), multiply(add(6, 4), subtract(16, 5))), const_10) | right triangle pqr is to be constructed in the xy - plane so that the right angle is at p and pr is parallel to the x - axis . the x and y coordinates of p , q and r are to be integers that satisfy the inequalitites - 4 ≤ x ≤ 5 and 6 ≤ y ≤ 16 . how many different triangles with these properties could be constructed ? | total values for x = 10 ; y = 11 x 1 , y 1 = 10 * 11 . . . . . . . . . . . . . . . . . . . . . . . coordinates of 1 st pnt x 2 , y 2 = 9 * 1 ( y 2 = y 1 ) . . . . . . . . . . . . . . . coordinates of 2 nd pnt y coordinates will be same as that of 1 st pnt bcoz it is parallel to x axis x 3 , y 3 = 1 * 10 ( x 2 = x 3 ) . . . . . . . . . . coordinates of 3 rd pt . x coordinates will be same as that of 2 nd point bcoz to make a right angle it has to be parallel to y axis tot ways = 10 * 11 * 9 * 1 * 1 * 10 = 9900 ans : c | a = 10 - 1
b = 6 + 4
c = 16 - 5
d = b * c
e = a * d
f = e * 10
|
a ) 5 kg , b ) 11 kg , c ) 30 kg , d ) 36.5 kg , e ) 71 kg | a | subtract(40, divide(subtract(add(multiply(30, subtract(40, add(30, 1))), 40), multiply(subtract(subtract(40, add(30, 1)), 1), 30)), const_2)) | when a student joe , weighing 40 kg , joins a group of students whose average weight is 30 kg , the average weight goes up by 1 kg . subsequently , if two students , excluding joe , leave the group the average weight comes back to 30 kg . what is the difference between the average weight of the two students who left and the weight of joe ? | "after two persons leave the group the average remains the same . that means the weight of the two persons = 40 + 30 = 70 so , the average the two persons = 35 that gives the answer 40 - 35 = 5 answer a" | a = 30 + 1
b = 40 - a
c = 30 * b
d = c + 40
e = 30 + 1
f = 40 - e
g = f - 1
h = g * 30
i = d - h
j = i / 2
k = 40 - j
|
a ) a . 72 , b ) b . 50 , c ) c . 48 , d ) d . 69 , e ) e . 44 | d | multiply(divide(33, add(6, add(const_1, const_4))), add(12, add(6, add(const_1, const_4)))) | a certain clock marks every hour by striking a number of times equal to the hour , and the time required for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stroke and the end of the last stroke is 33 seconds . at 12 : 00 , how many seconds elapse between the beginning of the first stroke and the end of the last stroke ? | "at 6 : 00 it ' ll chime 6 times . if we assume that the time taken to chime is x , then time between chimes is also x . so you have 6 chimes , which is 6 x and 5 time intervals between chimes . this means that 11 x = 33 seconds . thus x = 3 seconds . by a similar logic , at 12 : 00 , there are 12 chimes and 11 intervals so the total time is ( 12 + 11 ) x = 23 x = 69 seconds . answer d" | a = 1 + 4
b = 6 + a
c = 33 / b
d = 1 + 4
e = 6 + d
f = 12 + e
g = c * f
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | multiply(divide(69, const_100), 20) | john bought a total of 20 mangoes and oranges . each mango costs 80 cents and each orange costs 60 cents . if the average price of the 20 mangoes and oranges that john originally purchased was 69 cents , then how many oranges needs to return to raise the average price of his purchase to 72 cents ? | "let number of mangoes be x , number of oranges be 12 - x 0.80 x + ( 20 - x ) 0.60 / 20 = 0.69 solving for x , we get x = 9 - - > mangoes 9 , oranges 11 now , number of oranges to be returned be y 0.80 * 9 + ( 11 - y ) * 0.60 / 20 - y = 0.72 solving for y , y = 5 ans : b" | a = 69 / 100
b = a * 20
|
a ) 230 m , b ) 240 m , c ) 260 m , d ) 270 m , e ) 280 m | e | subtract(multiply(multiply(72, const_0_2778), 26), 240) | a goods train runs at a speed of 72 kmph and crosses a 240 m long platform in 26 seconds . what is the length of the goods train ? | "s = 240 + x / t 72 * 5 / 18 = 240 + x / 26 x = 280 answer : e" | a = 72 * const_0_2778
b = a * 26
c = b - 240
|
a ) 1 / 10 , b ) 3 / 10 , c ) 1 / 4 , d ) 7 / 10 , e ) 9 / 10 | c | divide(subtract(100, subtract(add(50, 40), 15)), 100) | a certain manufacturer of cake , muffin , and bread mixes has 100 buyers , of whom 50 purchases cake mix , 40 purchase muffin mix , and 15 purchase both cake mix and muffin mix . if a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix nor muffin mix ? | "c + m + b - cm - mb - cb - 2 cmb = 100 c - cake buyers , m - muffin and b - bread buyers . cm , mb , cb and cmb are intersecting regions . the question asks for people who have bought only bread mixes = b - cb - mb - 2 cmb has to be found out . 50 + 40 + b - cb - mb - 15 - 2 cmb = 100 b - cb - mb - 2 cmb = 25 hence the probability = 25 / 100 = 1 / 4 . c" | a = 50 + 40
b = a - 15
c = 100 - b
d = c / 100
|
a ) 8 % increase , b ) 4 % decrease , c ) 6 % increase , d ) 66 % increase , e ) 66 % decrease | a | subtract(divide(multiply(add(80, const_100), subtract(const_100, 40)), const_100), const_100) | when the price of an article was reduced by 40 % its sale increased by 80 % . what was the net effect on the sale ? | "if n items are sold for $ p each , revenue is $ np . if we reduce the price by 40 % , the new price is 0.6 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.6 p ) ( 1.8 n ) = 1.08 np , which is 1.08 times the old revenue , so is 8 % greater . answer : a" | a = 80 + 100
b = 100 - 40
c = a * b
d = c / 100
e = d - 100
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | d | divide(900, const_10) | how many integers from 101 to 900 , inclusive , remains the value unchanged when the digits were reversed ? | "question is asking for palindrome first digit possibilities - 1 through 8 = 8 9 is not possible here because it would result in a number greater than 9 ( i . e 909 , 919 . . ) second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 8 * 10 = 80 answer is d ." | a = 900 / 10
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 10 | e | add(add(const_4, const_3), const_2) | what is the units digit of 2222 ^ ( 333 ) * 4444 ^ ( 222 ) ? | "each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits . here ' s another way to organize the information . we ' re given [ ( 2222 ) ^ 333 ] [ ( 4444 ) ^ 222 ] we can ' combine ' some of the pieces and rewrite this product as . . . . ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) [ ( 2222 ) ^ 111 ] ( 2222 ) ( 3333 ) = a big number that ends in a 6 taking a number that ends in a 6 and raising it to a power creates a nice pattern : 6 ^ 1 = 6 6 ^ 2 = 36 6 ^ 3 = 216 etc . thus , we know that ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) will be a gigantic number that ends in a 6 . 2 ^ 111 requires us to figure out thecycleof the units digit . . . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 so , every 4 powers , the pattern of the units digits repeats ( 2 , 4 , 8 , 6 . . . . . 2 , 4 , 8 , 6 . . . . ) . 111 = 27 sets of 4 with a remainder of 3 . . . . this means that 2 ^ 111 = a big number that ends in an 8 so we have to multiply a big number that ends in a 6 and a big number that ends in an 8 . ( 6 ) ( 8 ) = 48 , so the final product will be a gigantic number that ends in an 8 . final answer : e" | a = 4 + 3
b = a + 2
|
a ) 33 , b ) 30 , c ) 38 , d ) 37 , e ) 27 | b | divide(10, subtract(const_1, divide(2, 3))) | the ratio of petrol and kerosene in the container is 3 : 2 when 10 liters of the mixture is taken out and is replaced by the kerosene , the ratio become 2 : 3 . then total quantity of the mixture in the container is : | explanation : pool : kerosene 3 : 2 ( initially ) 2 : 3 ( after replacement ) \ frac { remaining \ : ( or \ : left ) \ : quantity } { initial \ : quantity } = \ left ( 1 - \ frac { replaced \ : quantity } { total \ : quantity } \ right ) ( for petrol ) \ frac { 2 } { 3 } = \ left ( 1 - \ frac { 10 } { k } \ right ) \ rightarrow \ frac { 1 } { 3 } = \ frac { 10 } { k } \ rightarrow k = 30 l therefore the total quantity of the mixture in the container is 30 liters . answer : b ) 30 | a = 2 / 3
b = 1 - a
c = 10 / b
|
a ) 456780 , b ) 678910 , c ) 997479 , d ) 997917 , e ) 997920 | d | divide(multiply(const_1000, const_1000), 10) | what is the greatest 6 - digit number when divided by 6 , 7 , 8 , 9 , and 10 leaves a remainder of 3 , 4 , 5 , 6 and 7 respectively ? | when you divide a positive integer by 10 , the remainder will just be the units digit . we know the remainder is 7 when we divide by 10 , so d is the only possible answer . | a = 1000 * 1000
b = a / 10
|
a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent q should the number of hours worked be increased ? | "correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the increase in hours worked : wh = . 8 wx ( divide both sides by . 8 w ) 1.25 h = x we know that the new number of hours worked will be q = 25 % greater than the original number . the answer is c ." | a = 10 * 4
b = 100 - 20
c = b / 100
d = c * 10
e = a / d
f = e - 4
g = f / 4
h = g * 100
|
a ) 4830 , b ) 5040 , c ) 5250 , d ) 5460 , e ) 5680 | b | factorial(6) | if each digit in the set a = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } is used exactly once , in how many ways can the digits be arranged ? | "7 ! = 5040 the answer is b ." | a = math.factorial(6)
|
a ) 15 , b ) 26 , c ) 28 , d ) 4 , e ) none of these | d | floor(sqrt(20)) | if the sum of a number and its square is 20 , what is the number ? | "let the number be x . then , x + x 2 = 20 ( x + 5 ) ( x - 4 ) = 0 x = 4 answer : d" | a = math.sqrt(20)
b = math.floor(a)
|
a ) 172 , b ) 180 , c ) 192 , d ) 200 , e ) 216 | c | multiply(add(add(8, subtract(8, 4)), 4), add(subtract(add(8, subtract(8, 4)), divide(8, const_2)), 4)) | roy is now 8 years older than julia and half of that amount older than kelly . if in 4 years , roy will be twice as old as julia , then in 4 years what would be roy ’ s age multiplied by kelly ’ s age ? | "r = j + 8 = k + 4 r + 4 = 2 ( j + 4 ) ( j + 8 ) + 4 = 2 j + 8 j = 4 r = 12 k = 8 in 4 years ( r + 4 ) ( k + 4 ) = 16 * 12 = 192 the answer is c ." | a = 8 - 4
b = 8 + a
c = b + 4
d = 8 - 4
e = 8 + d
f = 8 / 2
g = e - f
h = g + 4
i = c * h
|
a ) $ 5050.50 , b ) $ 5100.50 , c ) $ 5200.50 , d ) $ 5300.50 , e ) $ 5400.50 | b | multiply(multiply(multiply(add(const_2, const_3), const_100), multiply(add(const_2, const_3), const_2)), power(add(divide(divide(4, const_4), const_100), const_1), const_2)) | an investor deposited $ 5,000 to open a new savings account that earned 4 percent annual interest , compounded quarterly . if there were no other transactions in the account , what was the amount of money in the account 6 months after the account was opened ? | the amount in the account after 6 months is 1.01 * 1.01 ( $ 5,000 ) = $ 5100.50 the answer is b . | a = 2 + 3
b = a * 100
c = 2 + 3
d = c * 2
e = b * d
f = 4 / 4
g = f / 100
h = g + 1
i = h ** 2
j = e * i
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | divide(30, divide(48, 16)) | in a group of people , if 30 people were made to stand in each column , 16 columns could be formed . if 48 people were made to stand in a column , how many columns could be formed ? | 16 * 30 = 48 * n n = 10 the answer is a . | a = 48 / 16
b = 30 / a
|
a ) 216 , b ) 200 , c ) 212 , d ) 220 , e ) 210 | a | divide(multiply(60, 18), 5) | 18 buckets of water fill a tank when the capacity of each bucket is 60 litres . how many buckets will be needed to fill the same tank , if the capacity of each bucket is 5 litres ? | "capacity of the tank = ( 18 ã — 60 ) litre number of buckets required of capacity of each bucket is 17 litre = 18 ã — 60 / 5 = 18 ã — 12 = 216 answer is a" | a = 60 * 18
b = a / 5
|
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