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 ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | a | divide(add(multiply(const_2, const_2), multiply(const_3, const_2)), multiply(6, subtract(6, 1))) | if x is to be chosen at random from the integers between 1 to 6 , inclusive , and y is to be chosen at random from the integers between 7 and 10 , inclusive , what is the probability that x + y will be even ? | "x + y will be even if x and y are both even or both odd . p ( x and y are both even ) = 3 / 6 * 2 / 4 = 1 / 4 p ( x and y are both odd ) = 3 / 6 * 2 / 4 = 1 / 4 p ( x + y is even ) = 1 / 4 + 1 / 4 = 1 / 2 the answer is a ." | a = 2 * 2
b = 3 * 2
c = a + b
d = 6 - 1
e = 6 * d
f = c / e
|
a ) $ 960 , b ) $ 1,350 , c ) $ 1,725 , d ) $ 2,050 , e ) $ 2,250 | e | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, 10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 7.50), const_1000) | a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 7.50 . if the company ’ s latest order required a truck with at least 288,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | "total volume is 288000 given lbh = 8 * 10 * 12 . the number of hats inside it = 288000 / 10 * 8 * 12 = 300 . price of each hat is 7.5 $ then total value is 300 * 7.5 = 2250 . answer : option e is correct answer . ." | a = 3 * 100
b = 8 * 10
c = a + b
d = c + 4
e = d * 1000
f = 8 * 10
g = f * 12
h = e / g
i = h * 7
j = i / 1000
|
a ) 0.004 % , b ) 0.04 % , c ) 40 % , d ) 4 % , e ) 40 % | c | multiply(divide(multiply(50, 0.08), 10), const_100) | a bowl was filled with 10 ounces of water , and 0.08 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | "total amount of water evaporated each day during a 50 - day period = . 08 * 50 = . 08 * 100 / 2 = 8 / 2 = 4 percent of the original amount of water evaporated during this period = ( 4 / 10 ) * 100 % = 40 % answer c" | a = 50 * 0
b = a / 10
c = b * 100
|
a ) 5 , b ) 6 , c ) 4 , d ) 3 , e ) 2 | c | divide(440, divide(multiply(multiply(10, 440), divide(add(const_100, 10), const_100)), subtract(multiply(440, divide(add(const_100, 10), const_100)), 440))) | machine a and machine b are each used to manufacture 440 sprockets . it takes machine a 10 hours longer to produce 440 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? | "machine b : takes x hours to produce 440 sprockets machine a : takes ( x + 10 ) hours to produce 440 sprockets machine b : in 1 hour , b makes 440 / x sprockets machine a : in 1 hour , a makes 440 / ( x + 10 ) sprockets equating : 1.1 ( 440 / ( x + 10 ) ) = 440 / x 484 / ( x + 10 ) = 440 / x 484 x = 440 x + 4400 44 x = 4400 x = 100 a makes 440 / ( 110 ) = 4 sprockets per hour answer : c" | a = 10 * 440
b = 100 + 10
c = b / 100
d = a * c
e = 100 + 10
f = e / 100
g = 440 * f
h = g - 440
i = d / h
j = 440 / i
|
a ) 72 sec , b ) 132 sec , c ) 192 sec , d ) 252 sec , e ) none | e | multiply(const_3600, divide(divide(add(200, 150), const_1000), subtract(42, 40))) | two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 42 kmph respectively . in how much time will they cross each other , if they are running in the same direction ? | "solution relative speed = ( 42 - 40 ) kmph = 2 kmph = ( 2 x 5 / 18 ) m / sec = ( 10 / 18 ) m / sec time taken = ( 350 x 18 / 10 ) sec = 630 sec . answer e" | a = 200 + 150
b = a / 1000
c = 42 - 40
d = b / c
e = 3600 * d
|
a ) 31 , b ) 21 , c ) 29 , d ) 22 , e ) 12 | 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 three consecutive even numbers is 87 . find the middle number of the three ? | "middle number = 87 / 3 = 29 ans 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 ) 126 , b ) 156 , c ) 180 , d ) 321 , e ) 260 | c | multiply(add(divide(subtract(divide(divide(1170, 6.5), const_2), 10), const_2), add(divide(subtract(divide(divide(1170, 6.5), const_2), 10), const_2), 10)), const_2) | the length of a rectangular plot is 10 mtr more than its width . the cost of fencing the plot along its perimeter at the rate of rs . 6.5 mtr is rs . 1170 . the perimeter of the plot is ? | "sol . let width = x , length = ( 10 + x ) perimeter = 2 ( x + ( 10 + x ) ) = 2 ( 2 x = 10 ) & 2 ( 2 x + 10 ) * 6.5 = 1170 x = 40 required perimeter = 2 ( 40 + 50 ) = 180 c" | a = 1170 / 6
b = a / 2
c = b - 10
d = c / 2
e = 1170 / 6
f = e / 2
g = f - 10
h = g / 2
i = h + 10
j = d + i
k = j * 2
|
a ) 165 ∏ cm , b ) 65 ∏ cm , c ) 12 ∏ cm , d ) 14 ∏ cm , e ) 16 ∏ cm | c | multiply(const_2, multiply(divide(add(multiply(const_3, const_100), add(const_10, const_4)), const_100), divide(divide(sqrt(784), const_2), divide(7, const_3)))) | there are two circles of different radii . the are of a square is 784 sq cm and its side is twice the radius of the larger circle . the radius of the larger circle is 7 - third that of the smaller circle . find the circumference of the smaller circle . | let the radii of the larger and the smaller circles be l cm and s cm respectively . let the side of the square be a cm . a 2 = 784 = ( 4 ) ( 196 ) = ( 22 ) . ( 142 ) a = ( 2 ) ( 14 ) = 28 a = 2 l , l = a / 2 = 14 l = ( 7 / 3 ) s therefore s = ( 3 / 7 ) ( l ) = 6 circumference of the smaller circle = 2 ∏ s = 12 ∏ cm . answer : c | a = 3 * 100
b = 10 + 4
c = a + b
d = c / 100
e = math.sqrt(784)
f = e / 2
g = 7 / 3
h = f / g
i = d * h
j = 2 * i
|
a ) 38 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | a | add(30, const_1) | the average of first four prime numbers greater than 30 is ? | "31 + 37 + 41 + 43 = 152 / 4 = 38 answer : a" | a = 30 + 1
|
a ) 39 : 40 , b ) 39 : 49 , c ) 39 : 42 , d ) 39 : 47 , e ) 17 : 20 | e | divide(add(200, multiply(divide(200, add(subtract(400, const_100), 200)), add(210, multiply(const_2, add(const_3, const_2))))), subtract(400, const_100)) | a and b put in rs . 200 and rs . 400 respectively into a business . a reinvests into the business his share of the first year ' s profit of rs . 210 where as b does not . in what ratio should they divide the second year ' s profit ? | "explanation : 1 : 2 a = 2 / 3 * 210 = 140 340 : 400 17 : 20 answer : e" | a = 400 - 100
b = a + 200
c = 200 / b
d = 3 + 2
e = 2 * d
f = 210 + e
g = c * f
h = 200 + g
i = 400 - 100
j = h / i
|
a ) 150 , b ) 388 , c ) 266 , d ) 299 , e ) 261 | a | multiply(divide(multiply(60, const_1000), const_3600), 9) | a train running at the speed of 60 km / hr crosses a pole in 9 seconds . find the length of the train . | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 9 = 150 meter . answer : a" | a = 60 * 1000
b = a / 3600
c = b * 9
|
a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 40 % | b | divide(divide(multiply(40, 50), multiply(80, 60)), add(divide(multiply(40, 50), multiply(80, 60)), 8)) | a certain car can travel 40 minutes on a gallon of gasoline at 50 miles per hour . if the car had started with a full tank and had 8 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 80 miles at 60 mph ? | "let , tank capacity = t gallon used fuel = ( t - 8 ) gallons distance travelled ( @ 50 miles / hr ) = 80 miles distance travelled in 1 gallon = distance travelled in 40 mins ( @ 50 miles / hr ) = ( 50 / 50 ) * 40 = 40 miles fuel used to travel 80 miles = ( 80 / 40 ) = 2 gallon i . e . used fuel = ( t - 8 ) = 2 gallon i . e . t = 10 gallons i . e . used fuel = ( 2 / 10 ) * 100 = 20 % b" | a = 40 * 50
b = 80 * 60
c = a / b
d = 40 * 50
e = 80 * 60
f = d / e
g = f + 8
h = c / g
|
a ) 46 kg , b ) 48 kg , c ) 70 kg , d ) 72 kg , e ) none of these | a | subtract(86, multiply(8, 5)) | the average weight of 8 students decreases by 5 kg when one of them weighing 86 kg is replaced by a new student . the weight of the student is | "explanation : let the weight of student be x kg . given , difference in average weight = 5 kg = > ( 86 - x ) / 8 = 5 = > x = 46 answer : a" | a = 8 * 5
b = 86 - a
|
a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | b | subtract(8, 5) | a football team lost 5 yards and then gained 8 . what is the team ' s progress ? | "for lost , use negative . for gain , use positive . progress = - 5 + 8 = 3 yards b" | a = 8 - 5
|
a ) 120 sec , b ) 176 sec , c ) 178 sec , d ) 187 sec , e ) 180 sec | e | divide(600, subtract(multiply(60, const_0_2778), multiply(36, const_0_2778))) | a and b go around a circular track of length 600 m on a cycle at speeds of 36 kmph and 60 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 600 / ( 36 * 5 / 18 ) , 600 / ( 60 * 5 / 18 ) } = 180 sec . answer : e" | a = 60 * const_0_2778
b = 36 * const_0_2778
c = a - b
d = 600 / c
|
a ) 337 , b ) 299 , c ) 375 , d ) 662 , e ) 201 | c | divide(1500, const_3) | divide rs . 1500 among a , b and c so that a receives 1 / 3 as much as b and c together and b receives 2 / 3 as a and c together . a ' s share is ? | "a + b + c = 1500 a = 1 / 3 ( b + c ) ; b = 2 / 3 ( a + c ) a / ( b + c ) = 1 / 3 a = 1 / 4 * 1500 = > 375 answer : c" | a = 1500 / 3
|
a ) 28800 , b ) 24000 , c ) 24936 , d ) 25640 , e ) none | a | multiply(divide(const_100, 80), 23040) | 80 % of the population of a village is 23040 . the total population of the village is ? | "answer ∵ 80 % of p = 23040 ∴ p = ( 23040 x 100 ) / 80 = 28800 correct option : a" | a = 100 / 80
b = a * 23040
|
a ) s . 109 , b ) s . 115 , c ) s . 100 , d ) s . 103 , e ) s . 102 | b | multiply(5, divide(1702, add(add(multiply(6, 3), multiply(9, 4)), multiply(4, 5)))) | a , b and c completed a piece of work , a worked for 6 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was rs . 1702 ? | "3 x 4 x 5 x 6 9 4 18 x + 36 x + 20 x = 1702 74 x = 1702 = > x = 23 5 x = 115 rs . answer : b" | a = 6 * 3
b = 9 * 4
c = a + b
d = 4 * 5
e = c + d
f = 1702 / e
g = 5 * f
|
a ) 7.5 , b ) 12.5 , c ) 17.5 , d ) 22.5 , e ) 27.5 | c | multiply(divide(subtract(multiply(18, divide(5, const_60)), multiply(4, divide(5, const_60))), 4), const_60) | a hiker walking at a constant rate of 4 kilometers per hour is passed by a cyclist travelling in the same direction along the same path at a constant rate of 18 kilometers per hour . the cyclist stops and waits for the hiker 5 minutes after passing her while the hiker continues to walk at her constant rate . how many minutes must the cyclist wait until the hiker catches up ? | in 5 minutes , the cyclist travels a distance of ( 5 / 60 ) * 18 = 1.5 km . the time it takes the hiker to complete this distance is 1.5 / 4 hours = 22.5 minutes the cyclist needs to wait 22.5 - 5 = 17.5 minutes the answer is c . | a = 5 / const_60
b = 18 * a
c = 5 / const_60
d = 4 * c
e = b - d
f = e / 4
g = f * const_60
|
a ) 4 : 30 , b ) 5 : 00 , c ) 5 : 30 , d ) 6 : 00 , e ) 6 : 30 | b | divide(subtract(divide(300, 45), 1), add(1, divide(40, 45))) | at 1 : 00 pm , a truck left city p and headed toward city q at a constant speed of 45 km / h . one hour later , a car left city q and headed toward city p along the same road at a constant speed of 40 km / h . if the distance between city p and city q is 300 km , at what time will the truck and the car meet each other ? | "at 2 : 00 pm , the truck and the car are 255 km apart . the truck and the car complete a distance of 85 km each hour . the time it takes to meet is 255 / 85 = 3 hours . they will meet at 5 : 00 pm . the answer is b ." | a = 300 / 45
b = a - 1
c = 40 / 45
d = 1 + c
e = b / d
|
a ) 4 , b ) 4.5 , c ) 5 , d ) 5.5 , e ) 6 | e | divide(add(multiply(75, const_4), multiply(divide(multiply(const_4, 75), add(const_1, const_4)), const_4)), add(75, divide(multiply(const_4, 75), add(const_1, const_4)))) | two cars are driving toward each other . the first car is traveling at a speed of 75 km / h , which is 25 % slower than the second car ' s speed . if the distance between the cars is 1050 km , how many hours will it take until the two cars meet ? | "the speed of the first car is 75 km / h . the speed of the second car is 75 / 0.75 = 100 km / h . the two cars complete a total of 175 km each hour . the time it takes the cars to meet is 1050 / 175 = 6 hours . the answer is e ." | a = 75 * 4
b = 4 * 75
c = 1 + 4
d = b / c
e = d * 4
f = a + e
g = 4 * 75
h = 1 + 4
i = g / h
j = 75 + i
k = f / j
|
a ) 175 seconds , b ) 195 seconds , c ) 155 seconds , d ) 115 seconds , e ) 203 seconds | e | divide(26, 0.128) | an industrial loom weaves 0.128 metres of cloth every second . approximately , how many seconds will it take for the loom to weave 26 metre of cloth ? | "explanation : let the time required by x seconds . then , more cloth means more time ( direct proportion ) so , 0.128 : 1 : : 26 : x = > x = { \ color { blue } \ frac { 26 \ times 1 } { 0.128 } } = > x = 203.13 so time will be approx 203 seconds answer : e" | a = 26 / 0
|
a ) 6 , b ) 8 , c ) 9 , d ) 12 , e ) 14 | c | subtract(38, subtract(add(26, 20), 17)) | in a class of 38 students 26 play football and play 20 long tennis , if 17 play above , many play neither ? | "26 + 20 - 17 = 29 38 - 29 = 9 play neither answer is c" | a = 26 + 20
b = a - 17
c = 38 - b
|
a ) 74 % , b ) 64 % , c ) 84 % , d ) 94 % , e ) 34 % | b | divide(multiply(circle_area(80), const_100), circle_area(const_100)) | if the diameter of circle r is 80 % of the diameter of circle s , the area of circle r is what percent of the area of circle s ? | "let diameter of circle r , dr = 80 and diameter of circle s , ds = 100 radius of circle r , rr = 40 radius of circle s , rs = 50 area of circle r / area of circle s = ( pi * rr ^ 2 ) / ( pi * rs ^ 2 ) = ( 40 / 50 ) ^ 2 = ( 8 / 10 ) ^ 2 = 64 % answer : b" | a = circle_area * (
b = a / 100
|
a ) 1 : 12 , b ) 85 : 31 , c ) 51 : 45 , d ) 58 : 67 , e ) 53 : 58 | e | power(divide(9261, 12167), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 9261 : 12167 . what is the ratio of their total surface areas ? | "ratio of the sides = ³ √ 9261 : ³ √ 12167 = 21 : 23 ratio of surface areas = 212 : 232 = 53 : 58 answer : e" | a = 9261 / 12167
b = 1 / 3
c = a ** b
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 25 | a | add(inverse(subtract(divide(const_1, 3), divide(const_1, 12))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 3 days . if a alone can do it in 12 days . in how many days can b alone do it ? | "a 4 1 / 3 â € “ 1 / 12 = 1 / 4 = > 4" | a = 1 / 3
b = 1 / 12
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
|
a ) 100 , b ) 150 , c ) 200 , d ) 250 , e ) 300 | c | multiply(divide(subtract(18, 10), subtract(30, 18)), 300) | solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 300 milliliters of solution x to create a solution that is 18 percent alcohol by volume ? | "18 % is 8 % - points higher than 10 % but 12 % - points lower than 30 % . thus there should be 3 parts of solution x for 2 parts of solution y . we should add 200 ml of solution y . the answer is c ." | a = 18 - 10
b = 30 - 18
c = a / b
d = c * 300
|
a ) 0.35 , b ) 3.5 , c ) 35 , d ) 350 , e ) 700 | e | multiply(divide(multiply(divide(multiply(multiply(125, 150), 175), multiply(multiply(25, 50), 75)), 20), 100), const_100) | if 125 % of j is equal to 25 % of k , 150 % of k is equal to 50 % of l , and 175 % of l is equal to 75 % of m , then 20 % of m is equal to what percent of 100 % of j ? | "imo answer should be 350 . . . consider j = 10 , then k = 50 , l = 150 and m = 350 . . . . 20 % of 350 , comes out to be 70 . . . . 100 % of 10 is 10 . . . . ( 70 * 100 ) / 10 = 700 . . . . ans : e" | a = 125 * 150
b = a * 175
c = 25 * 50
d = c * 75
e = b / d
f = e * 20
g = f / 100
h = g * 100
|
a ) 342 , b ) 352 , c ) 322 , d ) 332 , e ) 362 | b | subtract(subtract(multiply(divide(2200, const_10), const_2), const_12), const_12) | a sum of rs . 2200 has been divided among a , b and c such that a gets 1 / 4 of what b gets and b gets 1 / 5 of what c gets . what is b ' s share ? | "let c ' s share = rs . x then b ' s share = x / 5 a ' s share = x / 5 × 1 / 4 = x / 20 hence , x + x / 5 + x / 20 = 2200 ⇒ x ( 1 + 1 / 5 + 1 / 20 ) = 2200 ⇒ x ( 25 / 20 ) = 2200 ⇒ x ( 5 / 4 ) = 2200 ⇒ x = 2200 × 4 / 5 = 440 × 4 = rs . 1760 b ' s share = x / 5 = 1760 / 5 = rs . 352 answer is b ." | a = 2200 / 10
b = a * 2
c = b - 12
d = c - 12
|
a ) $ 0.75 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | a | add(multiply(const_0.25, subtract(5, 1)), 0.25) | at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 9.50 + 0.25 ( s – 1 ) . an employee with a salary grade of 4 receives how many more dollars per hour than an employee with a salary grade of 1 ? | "salary grade of 4 is p ( 4 ) = 9.50 + 0.25 ( 4 – 1 ) = 9.50 + 0.25 * 3 ; salary grade of 1 is p ( 1 ) = 9.50 + 0.25 ( 1 – 1 ) = 9.50 ; p ( 4 ) - p ( 1 ) = 9.50 + 0.25 * 4 - 9.50 = 0.75 . answer : a ." | a = 5 - 1
b = 0 * 25
c = b + 0
|
a ) 7250 , b ) 7200 , c ) 7820 , d ) 6000 , e ) 2799 | a | subtract(8000, 750) | the price of a t . v . set worth rs . 8000 is to be paid in 20 installments of rs . 750 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "money paid in cash = rs . 750 balance payment = ( 8000 - 1000 ) = rs . 7250 answer : a" | a = 8000 - 750
|
a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | c | subtract(const_100, subtract(add(40, 40), 15)) | in an examination , 40 % of total students failed in hindi , 40 % failed in english and 15 % in both . the percentage of these who passed in both the subjects is : | "pass percentage = 100 - ( 40 + 45 - 15 ) = 100 - 70 = 30 answer : c" | a = 40 + 40
b = a - 15
c = 100 - b
|
['a ) 10', 'b ) 15', 'c ) 20', 'd ) 25', 'e ) 30'] | c | multiply(subtract(16, 6), const_2) | some students are standing in a circle in which 6 th and the 16 th student are standing opposite to each other . find how many students were present there . | ( n / 2 ) - x = 6 n - x = 16 solving these two = > n = number of students = 20 answer : c | a = 16 - 6
b = a * 2
|
a ) 15730 , b ) 15240 , c ) 12456 , d ) 11452 , e ) 10002 | a | multiply(multiply(divide(add(10, const_100), const_100), 13000), divide(add(10, const_100), const_100)) | if the annual increase in the population of a town is 10 % and the present number of people is 13000 , what will the population be in 2 years ? | the required population is = 13000 ( 1 + 10 / 100 ) ^ 2 = 13000 * 11 / 10 * 11 / 10 = 15730 answer is a | a = 10 + 100
b = a / 100
c = b * 13000
d = 10 + 100
e = d / 100
f = c * e
|
a ) 2532.05 , b ) 2552.32 , c ) 2524.34 , d ) 7250.75 , e ) 9012.55 | d | subtract(multiply(46,000, power(add(const_1, divide(9, const_100)), 20)), 46,000) | find the compound interest on $ 46,000 at 20 % per annum for 9 months , compounded quarterly | "principal = $ 46000 ; time = 9 months = 3 quarters ; rate = 20 % per annum = 5 % per quarter . amount = $ [ 46000 x ( 1 + ( 5 / 100 ) ) ^ 3 ] = $ 53250.75 ci . = $ ( 53250.75 - 46000 ) = $ 7250.75 answer d ." | a = 9 / 100
b = 1 + a
c = b ** 20
d = 46 * 0
e = d - 46
|
a ) 0.23255 , b ) 0.14544 , c ) 0.25632 , d ) 0.35466 , e ) 0.63435 | a | multiply(divide(divide(480, divide(60, const_100)), subtract(subtract(subtract(add(add(add(add(add(44000, 44000), 44000), add(44000, 44000)), add(44000, 44000)), 44000), const_3600), const_3600), divide(480, divide(60, const_100)))), const_100) | lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 44000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is : | total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 344000 x \ therefore required percentage = \ inline \ frac { 800 x } { 344000 } \ times 100 = 0.23255 a | a = 60 / 100
b = 480 / a
c = 44000 + 44000
d = c + 44000
e = 44000 + 44000
f = d + e
g = 44000 + 44000
h = f + g
i = h + 44000
j = i - 3600
k = j - 3600
l = 60 / 100
m = 480 / l
n = k - m
o = b / n
p = o * 100
|
a ) 2 , b ) 33 , c ) 38 , d ) 35 , e ) 14 | a | divide(10, 5) | how many positive integer solutions does the equation 5 x + 10 y = 100 have ? | formula : ( constant ) / ( lcm of two nos ) = 100 / ( 5 * 10 ) = 2 answer : a | a = 10 / 5
|
a ) 140 , b ) 150 , c ) 160 , d ) 170 , e ) 220 | e | multiply(10, multiply(const_2, divide(sqrt(3025), divide(10, const_2)))) | the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 3025 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 3025 sq . units = > side of square = √ 3025 = 55 units the radius of the circle = side of the square = 55 units length of the rectangle = 2 / 5 * 55 = 22 units given that breadth = 10 units area of the rectangle = lb = 22 * 10 = 220 sq . units answer : option e" | a = math.sqrt(3025)
b = 10 / 2
c = a / b
d = 2 * c
e = 10 * d
|
a ) rs . 2.04 , b ) rs . 2.12 , c ) rs . 2.23 , d ) rs . 3 , e ) rs . 3.04 | a | subtract(subtract(multiply(5000, power(add(1, divide(divide(4, const_2), const_100)), const_2)), 5000), divide(multiply(5000, 4), const_100)) | what is the difference between the ci on rs . 5000 for 1 years at 4 % per annum ci yearly and half yearly ? | c . i . when interest compounded yearly = rs . 5000 x 1 + 4 x 1 + x 4 100 100 = rs . 5000 x 26 x 51 25 50 = rs . 5304 . c . i . when interest is compounded half - yearly = rs . 5000 x 1 + 2 3 100 = rs . 5000 x 51 x 51 x 51 50 50 50 = rs . 5306.04 difference = rs . ( 5306.04 - 5304 ) = rs . 2.04 a | a = 4 / 2
b = a / 100
c = 1 + b
d = c ** 2
e = 5000 * d
f = e - 5000
g = 5000 * 4
h = g / 100
i = f - h
|
a ) 5 % , b ) 20 % , c ) 50 % , d ) 350 % , e ) 500 % | d | multiply(divide(140, 40), const_100) | 140 is what percent of 40 ? | "40 * x = 140 - - > x = 3.5 - - > 3.5 expressed as percent is 350 % . answer : d ." | a = 140 / 40
b = a * 100
|
a ) $ 600 , b ) $ 740 , c ) $ 850 , d ) $ 980 , e ) $ 1,140 | d | multiply(add(add(subtract(subtract(56, 20), 10), multiply(subtract(const_1, divide(40, const_100)), 10)), multiply(subtract(const_1, divide(15, const_100)), 20)), 20) | tickets to a certain concert sell for $ 20 each . the first 10 people to show up at the ticket booth received a 40 % discount , and the next 20 received a 15 % discount . if 56 people bought tickets to the concert , what was the total revenue from ticket sales ? | "price of 1 ticket = 20 $ revenue generated from sales of first 10 tickets = 10 * ( 60 / 100 * 20 ) = 10 * 12 = 120 revenue generated from sales of next 20 tickets = 20 * ( 85 / 100 * 20 ) = 20 * 17 = 340 revenue generated from sales of last 26 tickets = 20 * 26 = 520 revenue generated from sales of 56 tickets = 120 + 340 + 520 = 980 $ answer d" | a = 56 - 20
b = a - 10
c = 40 / 100
d = 1 - c
e = d * 10
f = b + e
g = 15 / 100
h = 1 - g
i = h * 20
j = f + i
k = j * 20
|
a ) 200 , b ) 230 , c ) 140 , d ) 170 , e ) none of these | d | multiply(subtract(divide(subtract(29, 3), 2), 3), add(divide(subtract(29, 3), 2), 4)) | in the annual cultural programme of indra prastha there was this math quiz going on . yudhisthir faced this last question that will bring the champion ' s trophy . what will be thesolution for the given problem ? the function f is defined is defined asf ( 2 x + 3 ) = ( x - 3 ) ( x + 4 ) what is f ( 29 ) ? | explanation : f ( 2 x + 3 ) = ( x - 3 ) ( x + 4 ) put x = 13 f ( 2 * 13 + 3 ) = ( 13 - 3 ) * ( 13 + 4 ) = 10 * 17 f ( 29 ) = 170 answer : d | a = 29 - 3
b = a / 2
c = b - 3
d = 29 - 3
e = d / 2
f = e + 4
g = c * f
|
a ) 30 , b ) 32 , c ) 35 , d ) 36 , e ) 40 | b | divide(70, add(divide(35, 48), divide(subtract(70, 35), 24))) | a driver goes on a trip of 70 kilometers , the first 35 kilometers at 48 kilometers per hour and the remaining distance at 24 kilometers per hour . what is the average speed of the entire trip in kilometers per hour ? | "the time for the first part of the trip was 35 / 48 hours . the time for the second part of the trip was 35 / 24 hours . the total time fro the trip was 35 / 48 + 35 / 24 = 105 / 48 = 35 / 16 hours . the average speed for the trip was 70 / ( 35 / 16 ) = 32 kph the answer is b ." | a = 35 / 48
b = 70 - 35
c = b / 24
d = a + c
e = 70 / d
|
a ) 10 % , b ) 35 % , c ) 45 % , d ) 66.7 % , e ) 90 % | e | multiply(divide(subtract(divide(multiply(3, multiply(multiply(3, 5), 4)), 4), divide(multiply(divide(multiply(3, multiply(multiply(3, 5), 4)), 4), 3), 5)), subtract(multiply(multiply(3, 5), 4), divide(multiply(const_2, multiply(multiply(3, 5), 4)), 3))), const_100) | one day a car rental agency rented 2 / 3 of its cars , including 3 / 5 of its cars with cd players . if 3 / 4 of its cars have cd players , what percent of the cars that were not rented had cd players ? | let us try to solve this by getting some number for total cars . let there be 3,4 and 5 = 60 cars . . . ( we have taken denominators of 2 / 3,3 / 4 and 3 / 5 least common number ) . 2 / 3 ( 60 ) = 40 cars . . . total rented cars , then 1 / 3 ( 60 ) = 20 cars were not rented . 3 / 4 ( 60 ) = 45 cars with cd players . . . total cdp cars now 3 / 5 ( 3 / 4 ) = 3 / 5 ( 45 ) = 27 cdp cars and these are cars that we rented . non rented cdp = total cdp - rented cdp = 45 - 27 = 18 . 18 / 20 = 90 % . answer : e | a = 3 * 5
b = a * 4
c = 3 * b
d = c / 4
e = 3 * 5
f = e * 4
g = 3 * f
h = g / 4
i = h * 3
j = i / 5
k = d - j
l = 3 * 5
m = l * 4
n = 3 * 5
o = n * 4
p = 2 * o
q = p / 3
r = m - q
s = k / r
t = s * 100
|
a ) 5 , b ) 7 , c ) 9 , d ) 10 , e ) 12 | b | subtract(11, subtract(subtract(multiply(const_2, const_4), const_3), const_1)) | if 183 is divisible by 11 , find the value of the smallest natural no . n ? | in aptitude tests , we get questions on divisibility by 11 . a number is divisible by 11 , when the difference between the sum of digits at even places and at odd places is 0 or multiple of 11 the given number is n 183 . ( sum of digits at even places ) – ( sum of digits at odd places ) = 0 ( 8 + n ) - ( 3 + 1 ) = 0 ( 8 + n ) - 4 = 0 here the value of n must be 7 b | a = 2 * 4
b = a - 3
c = b - 1
d = 11 - c
|
a ) 1 / 15 , b ) 1 / 10 , c ) 1 / 9 , d ) 1 / 6 , e ) 1 / 3 | b | divide(divide(factorial(6), multiply(factorial(2), factorial(3))), factorial(6)) | joshua and jose work at an auto repair center with 3 other workers . for a survey on health care insurance , 2 of the 6 workers will be randomly chosen to be interviewed . what is the probability that joshua and jose will both be chosen ? | two methods 1 ) probability of chosing josh first = 1 / 5 probability of chosing jose second = 1 / 4 total = 1 / 20 probability of chosing jose first = 1 / 5 probability of chosing josh second = 1 / 4 total = 1 / 20 final = 1 / 20 + 1 / 20 = 1 / 10 b | a = math.factorial(6)
b = math.factorial(2)
c = math.factorial(3)
d = b * c
e = a / d
f = math.factorial(6)
g = e / f
|
a ) 20 days , b ) 65 days , c ) 86 days , d ) 45 days , e ) 17 days | a | divide(10, subtract(const_1, divide(add(10, 10), 40))) | a can do a piece of work in 40 days ; b can do the same in 40 days . a started alone but left the work after 10 days , then b worked at it for 10 days . c finished the remaining work in 10 days . c alone can do the whole work in ? | "10 / 40 + 10 / 40 + 10 / x = 1 x = 20 days answer : a" | a = 10 + 10
b = a / 40
c = 1 - b
d = 10 / c
|
a ) 22 seconds , b ) 12 seconds , c ) 10 seconds , d ) 18 seconds , e ) 28 seconds | e | subtract(multiply(divide(7, 56), 280), 7) | in a 280 meters race a beats b by 56 m or 7 seconds . a ' s time over the course is : | "b runs 56 m in 7 sec . = > b runs 280 m in 7 / 56 * 280 = 35 seconds since a beats b by 7 seconds , a runs 280 m in ( 35 - 7 ) = 28 seconds hence , a ' s time over the course = 28 seconds answer : e" | a = 7 / 56
b = a * 280
c = b - 7
|
a ) 50 % , b ) 40 % , c ) 25 % , d ) 20 % , e ) 10 % | c | multiply(divide(subtract(multiply(2, divide(add(3, 1), add(3, 2))), subtract(3, multiply(3, divide(add(3, 1), add(3, 2))))), add(3, 1)), const_100) | in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what z percent of all the people in smithtown are left - handed women ? | "looking at the ratio we can take total number of people = 20 . . ans 5 / 20 or 25 % c" | a = 3 + 1
b = 3 + 2
c = a / b
d = 2 * c
e = 3 + 1
f = 3 + 2
g = e / f
h = 3 * g
i = 3 - h
j = d - i
k = 3 + 1
l = j / k
m = l * 100
|
a ) 167 sec , b ) 190 sec , c ) 176 sec , d ) 115 sec , e ) 123 sec | d | subtract(divide(multiply(const_1, const_1000), divide(80, 10)), 10) | in a kilometer race , a beats b by 80 meters or 10 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 10 ) / 80 = 125 sec . time taken by a = 125 - 10 = 115 sec . answer : d" | a = 1 * 1000
b = 80 / 10
c = a / b
d = c - 10
|
a ) 2 , b ) 3 , c ) 6 , d ) 12 , e ) 24 | a | divide(divide(divide(divide(divide(divide(144, const_2), const_2), const_2), const_2), const_3), const_3) | if n is the smallest integer such that 144 times n is the square of an integer , what is the value of n ? | "144 can written as = 2 * 2 * 2 * 2 * 3 * 3 - - > 2 ^ 4 * 3 ^ 2 - - - ( 1 ) so for 144 * n to be a square of an integer , the integer should have even powers to the prime numbers it composed of . here 2 already has even power - > so n has to be 3 to make the power of 2 in ( 1 ) even option a is correct" | a = 144 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 3
f = e / 3
|
a ) 39 , b ) 50 , c ) 60 , d ) 54 , e ) 60 | d | add(subtract(86, 25), const_1) | claire has a total of 86 pets consisting of gerbils and hamsters only . one - quarter of the gerbils are male , and one - third of the hamsters are male . if there are 25 males altogether , how many gerbils does claire have ? | "g + h = 86 . . . 1 ; g / 4 + h / 3 = 25 . . . . 2 or 3 g + 4 h = 25 * 12 = 300 g = 86 - h or 3 ( 86 - h ) + 4 h = 300 h = 300 - 258 = 42 then g = 96 - 42 = 54 d" | a = 86 - 25
b = a + 1
|
a ) 18 , b ) 20 , c ) 24 , d ) 33.75 , e ) 42 | d | multiply(const_3, divide(60, const_10)) | jackie has two solutions that are 4 percent sulfuric acid and 12 percent sulfuric acid by volume , respectively . if these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid , approximately how many liters of the 12 percent solution will be required ? | "let a = amount of 4 % acid and b = amount of 12 % acid . now , the equation translates to , 0.04 a + . 12 b = . 05 ( a + b ) but a + b = 60 therefore . 04 a + . 12 b = . 05 ( 60 ) = > 4 a + 12 b = 300 but b = 60 - a therefore 4 a + 12 ( 60 - a ) = 300 = > 16 a = 420 hence a = 26.25 . b = 60 - 26.25 = 33.75 answer : d" | a = 60 / 10
b = 3 * a
|
['a ) 0', 'b ) 36', 'c ) 180', 'd ) 216', 'e ) 384'] | e | multiply(subtract(10, multiply(4, const_2)), multiply(12, 16)) | how much space , in cubic units , is left vacant when maximum number of 4 x 4 x 4 cubes are fitted in a rectangular box measuring 10 x 12 x 16 ? | no of cubes that can be accommodated in box = ( 10 * 12 * 16 ) / ( 4 * 4 * 4 ) 12 * 16 in numerator can be perfectly divided by 4 * 4 in denominator . side with length 10 ca n ' t be perfectly divided by 4 and hence is the limiting factor . closet multiple of 4 less that 10 is 8 . so vacant area in cube = = 12 * 16 * ( 10 - 8 ) = 12 * 16 * 2 = 384 answer : e | a = 4 * 2
b = 10 - a
c = 12 * 16
d = b * c
|
a ) 7 : 22 , b ) 7 : 24 , c ) 7 : 26 , d ) 7 : 28 , e ) 7 : 30 | c | divide(8, subtract(34, const_3)) | it is currently 8 : 34 pm . at what time in the morning was it exactly 39,668 minutes ago ? | divide 39,668 by 60 to convert to minutes : 39,668 / 60 = 661 r 8 . that is 661 hours , 8 minutes . all of the answer choices are during the same hour before : 34 , so we can assume the 661 hours takes the time back into the 7 am hour to 7 : 34 am . going back 8 more minutes yields the time 7 : 26 am . c | a = 34 - 3
b = 8 / a
|
a ) 12 , b ) 15 , c ) 18 , d ) 24 , e ) 27 | e | multiply(multiply(const_3, 3), 3) | a palindrome is a number that reads the same forward and backward . for example . 2442 and 111 are palindromes . if 5 - digit palindromes are formed using one or more of the digits , 1 , 2 , 3 , how many such palindromes are possible ? | xyzyx x can be 1 , 2 , or 3 , thus 3 options . y can be 1 , 2 , or 3 , thus 3 options . z can be 1 , 2 , or 3 , thus 3 options . total 3 ^ 3 = 27 . answer : e . | a = 3 * 3
b = a * 3
|
a ) 36 , b ) 50 , c ) 28 , d ) 26 , e ) 65 | e | divide(divide(subtract(125, multiply(multiply(20, const_0_2778), 20)), 20), const_0_2778) | a train 125 m long passes a man , running at 20 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 20 ) km / hr . x - 20 = 45 = = > x = 65 km / hr answer : e" | a = 20 * const_0_2778
b = a * 20
c = 125 - b
d = c / 20
e = d / const_0_2778
|
a ) 50.5 % , b ) 44.4 % , c ) 22.2 % , d ) 17.6 % , e ) 25 % | d | multiply(divide(subtract(20, 14), add(20, 14)), const_100) | if 20 % of ( x - y ) = 14 % of ( x + y ) , then what percent of x is y ? | "20 % of ( x - y ) = 14 % of ( x + y ) 20 / 100 ( x - y ) = 14 / 100 ( x + y ) 6 x = 34 y required percentage = y / x * 100 = 6 y / 34 y * 100 = 17.64 % answer is d" | a = 20 - 14
b = 20 + 14
c = a / b
d = c * 100
|
a ) rs . 260 , b ) rs . 360 , c ) rs . 350 , d ) rs . 364 , e ) rs . 362 | b | divide(510, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1)) | if rs . 510 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively : | "( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 510 * 2 / 17 = rs . 60 b ' s share = 510 * 3 / 17 = rs . 90 c ' s share = 510 * 12 / 17 = rs . 360 . answer : b" | a = 2 / 3
b = 1 / 4
c = a * b
d = 1 / 4
e = c + d
f = e + 1
g = 510 / f
|
a ) a ) 140 , b ) b ) 134 , c ) c ) 148 , d ) d ) 158 , e ) e ) 160 | a | add(multiply(15, 9), 5) | what is the dividend . divisor 15 , the quotient is 9 and the remainder is 5 ? | "d = d * q + r d = 15 * 9 + 5 d = 135 + 5 d = 140 answer a" | a = 15 * 9
b = a + 5
|
a ) 4 : 3 , b ) 8 : 7 , c ) 4 : 1 , d ) 6 : 5 , e ) none of these | b | divide(multiply(multiply(3, 2), const_4), multiply(7, 3)) | an order was placed for the supply of a carper whose length and breadth were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and breadth were in the ratio 7 : 3 but were was no change in its parameter . find the ratio of the areas of the carpets in both the cases . | "explanation : let the length and breadth of the carpet in the first case be 3 x units and 2 x units respectively . let the dimensions of the carpet in the second case be 7 y , 3 y units respectively . from the data , . 2 ( 3 x + 2 x ) = 2 ( 7 y + 3 y ) = > 5 x = 10 y = > x = 2 y required ratio of the areas of the carpet in both the cases = 3 x * 2 x : 7 y : 3 y = 6 x 2 : 21 y 2 = 6 * ( 2 y ) 2 : 21 y 2 = 6 * 4 y 2 : 21 y 2 = 8 : 7 answer is b" | a = 3 * 2
b = a * 4
c = 7 * 3
d = b / c
|
a ) 6.0 mile , b ) 3.0 mile , c ) 4.0 mile , d ) 3.45 mile , e ) 12 mile | d | multiply(divide(subtract(10, multiply(24, divide(add(1, 5), const_60))), add(5, add(1, 5))), 5) | stacy and heather are 10 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour fast than heather ' s constant rate of 5 miles / hour . if heather starts her journey 24 minutes after stacy , how far from the original destination has heather walked when the two meet ? . | "ss - stacy ' s speed = 6 m / hr sh - heather ' s speed = 5 m / hr in 24 minutes stacy will cover = ( 24 / 60 ) * 6 = 2.4 miles now since both are walking in opposite directions , add their speeds - 6 + 5 = 11 m / hr and distance to cover is 10 - 2.4 = 7.6 time taken = distance / speed = 7.6 / 11 = 0.69 hrs heather will cover = 5 * 0.69 = 3.45 miles . answer d" | a = 1 + 5
b = a / const_60
c = 24 * b
d = 10 - c
e = 1 + 5
f = 5 + e
g = d / f
h = g * 5
|
a ) 22 , b ) 37 , c ) 30 , d ) 33 , e ) 18 | d | divide(add(26, 40), const_2) | a man can row upstream at 26 kmph and downstream at 40 kmph , and then find the speed of the man in still water ? | "us = 26 ds = 40 m = ( 40 + 26 ) / 2 = 33 answer : d" | a = 26 + 40
b = a / 2
|
a ) 25 % , b ) 50 % , c ) 62.5 % , d ) 72.5 % , e ) 75 % | a | divide(multiply(add(add(96, 2), add(1, 1)), add(1, 1)), 8) | if n is an integer from 1 to 96 ( inclusive ) , what is the probability for n * ( n + 1 ) * ( n + 2 ) being divisible by 8 ? | ( a ) 25 % , to be divisible by 8 the number needs to have 3 2 s in it . only a multiple of 4 can provide that . number of numbers divisible by 4 = 96 / 4 = 24 . so p ( 8 ) = 24 / 96 = 25 % . | a = 96 + 2
b = 1 + 1
c = a + b
d = 1 + 1
e = c * d
f = e / 8
|
a ) 25 % , b ) 30 % , c ) 50 % , d ) 20 % , e ) 10 % | d | divide(multiply(15, const_100), subtract(90, 15)) | by selling an article for $ 90 , a person gains $ 15 . what is the gain % ? | "s . p . = $ 90 gain = $ 15 c . p . = 90 - 15 = 75 gain % = 15 / 75 * 100 = 20 % answer is d" | a = 15 * 100
b = 90 - 15
c = a / b
|
a ) 104345 , b ) 107375 , c ) 108385 , d ) 109395 , e ) 105355 | d | add(102325, multiply(multiply(153, 2), const_10)) | on multiplying a number b by 153 , the result obtained was 102325 . however , it is found that both the 2 ' s are wrong . find the correct result . | the only thing you actually know about the correct number b is that it is divisible by 153 and has 5 as a factor . you should immediately try to find the factors of 153 and look for them in the options . 153 = 9 * 17 divisibility by 9 is easy to check . only ( d ) satisfies . | a = 153 * 2
b = a * 10
c = 102325 + b
|
a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | a | divide(add(9, const_10), multiply(multiply(const_2, const_3), const_10)) | if a computer program generates 3 - digit odd numbers using the numbers 9 , 7 , 0 , 8 , and 1 , and the digits can be repeated , what is the probability that the number formed from the program is a multiple of 3 ? | a | a = 9 + 10
b = 2 * 3
c = b * 10
d = a / c
|
a ) 5 , b ) 9 , c ) 40 , d ) 50 , e ) 25 | b | divide(subtract(multiply(48, 45), multiply(divide(add(const_100, 20), const_100), multiply(40, 45))), subtract(multiply(70, divide(add(const_100, 20), const_100)), 48)) | how many pounds of salt at 70 cents / lb must be mixed with 45 lbs of salt that costs 40 cents / lb so that a merchant will get 20 % profit by selling the mixture at 48 cents / lb ? | "selling price is 48 cents / lb for a 20 % profit , cost price should be 40 cents / lb ( cp * 6 / 5 = 48 ) basically , you need to mix 40 cents / lb ( salt 1 ) with 70 cents / lb ( salt 2 ) to get a mixture costing 45 cents / lb ( salt avg ) weight of salt 1 / weight of salt 2 = ( salt 2 - saltavg ) / ( saltavg - salt 1 ) = ( 70 - 45 ) / ( 45 - 40 ) = 5 / 1 we know that weight of salt 1 is 45 lbs . weight of salt 2 must be 9 lbs . answer ( b )" | a = 48 * 45
b = 100 + 20
c = b / 100
d = 40 * 45
e = c * d
f = a - e
g = 100 + 20
h = g / 100
i = 70 * h
j = i - 48
k = f / j
|
a ) 116 , b ) 25 , c ) 30 , d ) 24 , e ) 15 | c | multiply(5, multiply(const_3, const_2)) | the length of a side of a hexagon is 5 inches . what is the perimeter ? | "hexagon . it means 6 equal sides . p = 5 + 5 + 5 + 5 + 5 + 5 = 10 + 10 + 10 = 30 inches answer c" | a = 3 * 2
b = 5 * a
|
a ) 28 % , b ) 34 % , c ) 22 % , d ) 18 % , e ) 8.5 % | a | multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(12, const_100), const_4), divide(12, const_100))), const_100) | one fourth of a solution that was 12 % salt by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent salt by weight ? | consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.12 + 25 * x = 100 * 0.16 - - > x = 0.28 . answer : a . | a = 16 / 100
b = a * 4
c = 12 / 100
d = c * 4
e = 12 / 100
f = d - e
g = b - f
h = g * 100
|
a ) 8 % , b ) 13 % , c ) 10 % , d ) 11 % , e ) 12 % | b | multiply(divide(4, add(28, 4)), const_100) | a house wife saved $ 4 in buying an item on sale . if she spent $ 28 for the item , approximately how much percent she saved in the transaction ? | "actual price = 28 + 4 = $ 32 saving = 4 / 32 * 100 = 13 % approximately answer is b" | a = 28 + 4
b = 4 / a
c = b * 100
|
a ) 600,400 , b ) 500,500 , c ) 300,700 , d ) 800,200 , e ) 200,400 | e | divide(multiply(3, 6), add(3, 6)) | a can do a work in 3 days . b can do the same work in 6 days . both a & b together will finish the work and they got $ 1000 from that work . find their shares ? | "ratio of their works a : b = 3 : 6 ratio of their wages a : b = 1 : 2 a ' s share = ( 1 / 5 ) 1000 = 200 b ' s share = ( 2 / 5 ) 1000 = 400 correct option is e" | a = 3 * 6
b = 3 + 6
c = a / b
|
a ) 3 / 4 , b ) 11 / 16 , c ) 7 / 8 , d ) 3 / 16 , e ) 5 / 16 | d | multiply(divide(45, const_60), divide(15, const_60)) | two friends deepak and rajeev have agreed to meet at a definite spot on a particular day between 9 pm and 10 pm . the first person to come waits for some time and leaves . if the other one does not turn up by that time . if deepak comes first , he waits for 45 minutes and if rajeev comes first , he waits for 15 minutes . what is the probability of meeting between deepak and rajeev if their arrival times are independent of eachother and each person arrives during the indicated period at random ? | deepak : 45 / 60 = 3 / 4 rajeev : 15 / 60 = 1 / 4 as these are independent events total probability = ( 3 / 4 ) * ( 1 / 4 ) = 3 / 16 answer : d | a = 45 / const_60
b = 15 / const_60
c = a * b
|
a ) 39 , b ) 50 , c ) 66 , d ) 57 , e ) 60 | c | add(subtract(90, 25), const_1) | claire has a total of 90 pets consisting of gerbils and hamsters only . one - quarter of the gerbils are male , and one - third of the hamsters are male . if there are 25 males altogether , how many gerbils does claire have ? | g + h = 90 . . . 1 ; g / 4 + h / 3 = 25 . . . . 2 or 3 g + 4 h = 25 * 12 = 300 g = 90 - h or 3 ( 90 - h ) + 4 h = 300 h = 300 - 270 = 30 then g = 96 - 30 = 66 c | a = 90 - 25
b = a + 1
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | c | add(add(const_4, const_3), const_2) | what is the units digit of 2222 ^ ( 333 ) * 3333 ^ ( 444 ) ? | "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 ] [ ( 3333 ) ^ 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 4 . final answer : c" | a = 4 + 3
b = a + 2
|
a ) 18 % , b ) 20 % , c ) 17 % , d ) 18 % , e ) none of these | c | subtract(subtract(add(const_100, 30), multiply(add(const_100, 30), divide(10, const_100))), const_100) | a shopkeeper labeled the price of his articles so as to earn a profit of 30 % on the cost price . he then sold the articles by offering a discount of 10 % on the labeled price . what is the actual percent profit earned in the deal ? | explanation : let the cp of the article = rs . 100 . then labeled price = rs . 130 . sp = rs . 130 - 10 % of 130 = rs . 130 - 13 = rs . 117 . gain = rs . 117 – rs . 100 = rs . 17 therefore , gain / profit percent = 17 % . answer : option c | a = 100 + 30
b = 100 + 30
c = 10 / 100
d = b * c
e = a - d
f = e - 100
|
a ) a ) 1 , b ) b ) 2 , c ) c ) 3 , d ) d ) 6 , e ) e ) 12 | d | multiply(subtract(subtract(3, 10), 10), multiply(multiply(10, const_0_25), subtract(subtract(3, 10), 10))) | what is the area inscribed by the lines y = 10 , x = 1 , y = 3 x + 1 on an xy - coordinate plane ? | "first , let ' s graph the lines y = 10 and x = 1 at this point , we need to find the points where the line y = 3 x + 1 intersects the other two lines . for the vertical line , we know that x = 1 , so we ' ll plug x = 1 into the equation y = 3 x + 1 to get y = 3 ( 1 ) + 1 = 4 perfect , when x = 1 , y = 5 , so one point of intersection is ( 1,4 ) for the horizontal line , we know that y = 10 , so we ' ll plug y = 10 into the equation y = 3 x + 1 to get 10 = 3 x + 1 . solve to get : x = 3 so , when y = 10 , x = 3 , so one point of intersection is ( 3,10 ) now add these points to our graph and sketch the line y = 3 x + 1 at this point , we can see that we have the following triangle . the base has length 2 and the height is 6 area = ( 1 / 2 ) ( base ) ( height ) = ( 1 / 2 ) ( 2 ) ( 6 ) = 6 answer : d" | a = 3 - 10
b = a - 10
c = 10 * const_0_25
d = 3 - 10
e = d - 10
f = c * e
g = b * f
|
a ) 10 , b ) 12.5 , c ) 25 , d ) 12 , e ) 14.5 | b | multiply(divide(100, multiply(400, 2)), const_100) | what is rate of interest if principal . amount be 400 , simple interest 100 and time 2 year . | "s . i = ( p * r * t ) / 100 100 = 800 r / 100 r = 100 / 8 = 12.5 % answer b" | a = 400 * 2
b = 100 / a
c = b * 100
|
a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 7 , d ) 1 / 2 , e ) 4 / 7 | b | divide(multiply(4, const_1), add(multiply(4, const_1), multiply(4, const_1))) | harold works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 4 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? | "the time from beginning of march to the end of september is 7 months . if x is the average monthly tip for all months other than august then his august month tip will be 4 * x his total tip for the 7 months = 6 * ( average tip for the months other than august ) + 4 x = 10 x august tips as a fraction of total tips = 4 x / 10 x = 2 / 5 . so b" | a = 4 * 1
b = 4 * 1
c = 4 * 1
d = b + c
e = a / d
|
a ) 100 m , b ) 120 m , c ) 140 m , d ) 160 m , e ) 170 cm | c | divide(12, subtract(divide(12, 10), 7)) | a train covers a distance of 12 km in 10 minutes . if it takes 7 seconds to pass a telegraph post , then the length of the train is | "explanation : speed = 12 / 10 x 60 km / hr = 72 x 5 / 18 m / sec = 20 m / sec . length of the train = ( speed x time ) = ( 20 x 7 ) m = 140 m answer : option c" | a = 12 / 10
b = a - 7
c = 12 / b
|
a ) 1 : 21 , b ) 12 : 7 , c ) 1 : 15 , d ) 21 : 17 , e ) 13 : 13 | b | power(divide(6859, 1331), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 6859 : 1331 . what is the ratio of their total surface areas ? | "ratio of the sides = ³ √ 6859 : ³ √ 1331 = 19 : 11 ratio of surface areas = 192 : 112 = 12 : 7 answer : b" | a = 6859 / 1331
b = 1 / 3
c = a ** b
|
a ) 400 , b ) 410 , c ) 460 , d ) 420 , e ) 430 | d | subtract(divide(multiply(multiply(3500, 14), 3), const_100), divide(multiply(multiply(3500, 10), 3), const_100)) | if a lends rs . 3500 to b at 10 % per annum and b lends the same sum to c at 14 % per annum then the gain of b in a period of 3 years is ? | "( 3500 * 4 * 3 ) / 100 = > 420 answer : d" | a = 3500 * 14
b = a * 3
c = b / 100
d = 3500 * 10
e = d * 3
f = e / 100
g = c - f
|
a ) 19 , b ) 36 , c ) 11 , d ) 19 , e ) 114 | b | multiply(add(add(const_1, const_2), const_3), 6) | a take twice as much time as b or thrice as much time to finish a piece of work . working together , they can finish the work in 6 days . b can do the work alone in ? | "suppose a , b and c take x , x / 2 and x / 3 respectively to finish the work . then , ( 1 / x + 2 / x + 3 / x ) = 1 / 6 6 / x = 1 / 6 = > x = 36 so , b takes 18 hours to finish the work . answer : b" | a = 1 + 2
b = a + 3
c = b * 6
|
a ) 43 , b ) 45 , c ) 47 , d ) 49 , e ) 51 | d | divide(add(multiply(multiply(36, const_0_2778), 14), 350), multiply(36, const_0_2778)) | a train running at a speed of 36 km / h passes an electric pole in 14 seconds . in how many seconds will the whole train pass a 350 - meter long platform ? | "let the length of the train be x meters . when a train crosses an electric pole , the distance covered is its own length x . speed = 36 km / h = 36000 m / 3600 s = 10 m / s x = 14 * 10 = 140 m . the time taken to pass the platform = ( 140 + 350 ) / 10 = 49 seconds the answer is d ." | a = 36 * const_0_2778
b = a * 14
c = b + 350
d = 36 * const_0_2778
e = c / d
|
a ) 367 , b ) 380 , c ) 400 , d ) 406 , e ) 412 | a | multiply(divide(subtract(divide(360, subtract(const_1, divide(const_1, 10))), 360), 12), 110) | mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 110 mangoes . | "mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 120 mangoes . method 1 : let price per mango = x . let number of mangoes be n . then , nx = 360 . now price = 0.9 x ; number of mangoes = n + 12 . total amount = 0.9 x * ( n + 12 ) = 360 . nx = 0.9 nx + 10.8 x = > 0.1 nx = 10.8 x = > n = 108 = > x = 360 / 108 = 3.33 original price of 110 mangoes = 110 * 3.33 = 367 . answer a" | a = 1 / 10
b = 1 - a
c = 360 / b
d = c - 360
e = d / 12
f = e * 110
|
a ) 2 : 9 , b ) 3 : 8 , c ) 8 : 5 , d ) 5 : 4 , e ) 7 : 3 | d | divide(5, divide(20, 5)) | there are two numbers . if 20 % of the first number is added to the second number , then the second number increases to its 5 - fourth . what is the ratio of the first number to the second number ? | let the two numbers be x and y . ( 2 / 10 ) * x + y = ( 5 / 4 ) * y ( 2 / 10 ) * x = ( 1 / 4 ) * y x / y = 10 / 8 = 5 / 4 the answer is d . | a = 20 / 5
b = 5 / a
|
a ) 56 seconds , b ) 18 seconds , c ) 40 seconds , d ) 19 seconds , e ) 45 seconds | a | divide(add(560, 140), divide(multiply(45, const_1000), const_3600)) | a train is 560 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 560 + 140 = 700 meter time = distance / speed = 700 * ( 2 / 25 ) = 56 seconds answer : a" | a = 560 + 140
b = 45 * 1000
c = b / 3600
d = a / c
|
a ) 140 , b ) 99 , c ) 88 , d ) 160 , e ) 12 | d | multiply(10, multiply(const_2, divide(sqrt(1600), divide(10, const_2)))) | the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 1600 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 1600 sq . units = > side of square = √ 1600 = 40 units the radius of the circle = side of the square = 40 units length of the rectangle = 2 / 5 * 40 = 16 units given that breadth = 10 units area of the rectangle = lb = 16 * 10 = 160 sq . units answer : d" | a = math.sqrt(1600)
b = 10 / 2
c = a / b
d = 2 * c
e = 10 * d
|
a ) 36 , b ) 38 , c ) 40 , d ) can not be determined , e ) none of these | c | add(28, const_1) | the average age of 28 students in a group is 11 years . when teacher ’ s age is included to it , the average increases by one . what is the teacher ’ s age in years ? | "age of the teacher = ( 29 × 12 – 28 × 11 ) years = 40 years . answer c" | a = 28 + 1
|
a ) 2 kmph , b ) 7 kmph , c ) 98 kmph , d ) 6 kmph , e ) 4 kmph | a | divide(subtract(10, 6), const_2) | a man can row his boat with the stream at 10 km / h and against the stream in 6 km / h . the man ' s rate is ? | "ds = 10 us = 6 s = ? s = ( 10 - 6 ) / 2 = 2 kmph answer : a" | a = 10 - 6
b = a / 2
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 11 | e | subtract(12, const_1) | if the product of 12 integers is negative , at most how many of the integers can be negative ? | "the product of 12 integers is negative thus an odd number of integers need to be negative to have a negative product we are asked at most how many are required . so , the highest odd integer before 6 , i . e . 11 correct option : e" | a = 12 - 1
|
a ) 99 , b ) 98 , c ) 97 , d ) 96 , e ) 95 | d | add(multiply(const_2, const_3), subtract(const_100, const_10)) | a number is said to be prime saturated if the product of all the different positive prime factors of d is less than the square root of d . what is the greatest two digit prime saturated integer ? | "clearly d a number is said to be prime saturated if the product of all the different positive prime factors of d is less than the square root of d . 96 has more number of smaller prime factor thats the clue ! ! = d" | a = 2 * 3
b = 100 - 10
c = a + b
|
a ) 5 . , b ) 10 . , c ) 14 . , d ) 15 . , e ) 20 . | c | multiply(divide(add(28, divide(28, const_2)), 6), const_2) | the distance from steve ' s house to work is 28 km . on the way back steve drives twice as fast as he did on the way to work . altogether , steve is spending 6 hours a day on the roads . what is steve ' s speed on the way back from work ? | "time is in the ratio 2 : 1 : : to : fro office therefore , 2 x + 1 x = 6 hrs time take to come back - 2 hrs , distance travelled - 28 km = > speed = 14 kmph c" | a = 28 / 2
b = 28 + a
c = b / 6
d = c * 2
|
a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | e | add(multiply(0.25, subtract(5, 1)), 0.25) | at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 9.50 + 0.25 ( s – 2 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1 ? | oa is definitely wrong . the answer should be e . | a = 5 - 1
b = 0 * 25
c = b + 0
|
a ) 10 , b ) 30 , c ) 190 , d ) 60 , e ) 90 | c | divide(multiply(20, subtract(20, const_1)), const_2) | there are 20 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "20 players are there . two players play one game with one another . so 20 c 2 = 20 * 19 / 2 = 190 so option c is correct" | a = 20 - 1
b = 20 * a
c = b / 2
|
['a ) 296', 'b ) 252', 'c ) 344', 'd ) 388', 'e ) none of these'] | c | add(add(power(add(add(divide(subtract(subtract(36, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(36, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(36, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(36, const_10), const_2), const_4), const_2), const_2))) | the sum of four consecutive even numbers is 36 . find the sum of the squares of these numbers ? | let the four numbers be x , x + 2 , x + 4 and x + 6 . = > x + x + 2 + x + 4 + x + 6 = 36 = > 4 x + 12 = 36 = > x = 6 the numbers are 6 , 8 , 10 and 12 . sum of their squares = 62 + 82 + 102 + 122 = 36 + 64 + 100 + 144 = 344 . answer : c | a = 36 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 36 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 36 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 36 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 8 , b ) 9 , c ) 11 , d ) 14 , e ) 15 | d | divide(multiply(add(divide(40, const_100), const_1), 24), add(add(divide(40, const_100), const_1), const_1)) | p is 40 % more efficient than q . p can complete a work in 24 days . if p and q work together , how many days will it take to complete the same work ? | "the work done by p in 1 day = 1 / 24 let work done by q in 1 day = q q × ( 140 / 100 ) = 1 / 24 q = 100 / ( 24 × 140 ) = 10 / ( 24 × 14 ) the work done by p and q in 1 day = 1 / 24 + 10 / ( 24 × 14 ) = 24 / ( 24 × 14 ) = 1 / 14 p and q together can do the work in 14 days . the answer is d ." | a = 40 / 100
b = a + 1
c = b * 24
d = 40 / 100
e = d + 1
f = e + 1
g = c / f
|
a ) 50 km , b ) 56 km , c ) 70 km , d ) 80 km , e ) 46 km | a | multiply(10, divide(20, subtract(14, 10))) | if a person walks at 14 km / hr instead of 10 km / hr , he would have walked 20 km more . the actual distance travelled by him is : | "let the actual distance travelled be x km . then , x / 10 = ( x + 20 ) / 14 14 x = 10 x + 200 4 x = 200 x = 50 km . answer : a" | a = 14 - 10
b = 20 / a
c = 10 * b
|
a ) none of these , b ) 342 , c ) 324 , d ) 312 , e ) can not be determined | c | divide(add(422, 404), 2) | ( 422 + 404 ) 2 − ( 4 × 422 × 404 ) = ? | "explanation : given equation is in the form ( a + b ) 2 − 4 ab where a = 422 and b = 404 hence answer = ( a + b ) 2 − 4 ab = ( a − b ) 2 = ( 422 − 404 ) 2 = 182 = 324 , answer : option c" | a = 422 + 404
b = a / 2
|
a ) 73 , b ) 32 , c ) 34 , d ) 43 , e ) 112 | e | sqrt(divide(multiply(square_area(8), 98), inverse(const_2))) | the length of the rectangular field is double its width . inside the field there is square shaped pond 8 m long . if the area of the pond is 1 / 98 of the area of the field . what is the length of the field ? | "explanation : a / 98 = 8 * 8 = > a = 8 * 8 * 98 x * 2 x = 8 * 8 * 98 x = 56 = > 2 x = 112 answer : option e" | a = square_area * (
b = a / 98
c = 1/(2)
d = math.sqrt(b)
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.