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 ) 7 / 16 , b ) 7 / 15 , c ) 10 / 21 , d ) 14 / 17 , e ) 1 / 2 | d | divide(multiply(subtract(const_10, multiply(divide(1, 5), const_10)), multiply(divide(7, 6), multiply(divide(1, 5), const_10))), add(multiply(multiply(divide(1, 5), const_10), multiply(divide(1, 5), const_10)), multiply(subtract(const_10, multiply(divide(1, 5), const_10)), multiply(divide(7, 6), multiply(divide(1, 5), const_10))))) | a lemonade stand sold only small and large cups of lemonade on tuesday . 1 / 5 of the cups sold were small and the rest were large . if the large cups were sold for 7 / 6 as much as the small cups , what fraction of tuesday ' s total revenue was from the sale of large cups ? | let the total # of cups be 10 . # of small cups sold 1 / 5 * 10 = 2 ; # of large cups sold 10 - 2 = 8 ; let the price of small cup be $ 6 , then the price of larges cup would be 7 / 6 * 6 = $ 7 ; revenue from small cups : 2 * $ 6 = $ 12 ; revenue from large cups cups : 8 * $ 7 = $ 56 ; fraction of total revenue from large cups : 56 / ( 56 + 12 ) = 14 / 17 . answer : d . | a = 1 / 5
b = a * 10
c = 10 - b
d = 7 / 6
e = 1 / 5
f = e * 10
g = d * f
h = c * g
i = 1 / 5
j = i * 10
k = 1 / 5
l = k * 10
m = j * l
n = 1 / 5
o = n * 10
p = 10 - o
q = 7 / 6
r = 1 / 5
s = r * 10
t = q * s
u = p * t
v = m + u
w = h / v
|
a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 120,500 , e ) 100,000 | d | divide(multiply(multiply(add(const_2, const_3), const_1000), 12), const_2) | if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 14,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years later , when luke is ready for college ? | answer equals d in 48 years . i thought by 50 th year it would reach 120,500 . options should have been separated more widely for clarity . | a = 2 + 3
b = a * 1000
c = b * 12
d = c / 2
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 9 , e ) 1 / 12 | d | divide(multiply(divide(4, 12), const_100), multiply(divide(12, 4), const_100)) | 12 is 4 % of a , and 4 is 12 % of b . c equals b / a . what is the value of c ? | "4 a / 100 = 12 a = 300 12 b / 100 = 4 b = 100 / 3 c = b / a = 100 / ( 3 * 300 ) = 1 / 9 the answer is d ." | a = 4 / 12
b = a * 100
c = 12 / 4
d = c * 100
e = b / d
|
a ) 48 , b ) 50 , c ) 22 , d ) 27 , e ) 221 | b | divide(subtract(15, add(3, multiply(180, multiply(multiply(divide(divide(divide(3, 10), 180), 8), add(8, 1)), subtract(40, 10))))), multiply(subtract(40, 10), multiply(divide(2, 3), multiply(divide(divide(divide(3, 10), 180), 8), add(8, 1))))) | a contractor undertook to make 15 km of roadway in 40 weeks . in 10 weeks , 3 km was complete by 180 men working 8 hours a day . the men then agreed to work 1 hour a day overtime , and some boys were engaged to assist them , the work was finished in the stipulated time ( 40 weeks ) . how many boys were employed , if the work of 3 boys is equal to that of 2 men ? | explanation : let the capacity of man = 3 units , and boy = 2 units per hour . now total work = 3 Γ 180 Γ 8 Γ 7 Γ 10 = 3 km . - - - - - - - ( 1 ) let k boys were recruited . now total work = ( 3 Γ 180 + 2 Γ k ) Γ 9 Γ 7 Γ 30 = 12 km . - - - - - - ( 2 ) by dividing 2 nd equation by 1 st , β ( 540 + 2 k ) Γ 9 Γ 7 Γ 303 Γ 180 Γ 8 Γ 7 Γ 10 = 4 ( 540 + 2 k ) Γ 9 Γ 7 Γ 303 Γ 180 Γ 8 Γ 7 Γ 10 = 4 β k = 50 answer : b | a = 3 / 10
b = a / 180
c = b / 8
d = 8 + 1
e = c * d
f = 40 - 10
g = e * f
h = 180 * g
i = 3 + h
j = 15 - i
k = 40 - 10
l = 2 / 3
m = 3 / 10
n = m / 180
o = n / 8
p = 8 + 1
q = o * p
r = l * q
s = k * r
t = j / s
|
a ) 3.3 , b ) 3.4 , c ) 3.5 , d ) 3.6 , e ) 3.7 | b | multiply(divide(multiply(add(7, 1.2), subtract(7, 1.2)), add(add(7, 1.2), subtract(7, 1.2))), const_2) | a man can row 7 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . how far is the place ? | "m = 7 s = 1.2 ds = 7 + 1.2 = 8.2 us = 7 - 1.2 = 5.8 x / 8.2 + x / 5.8 = 1 x = 3.40 . answer : b" | a = 7 + 1
b = 7 - 1
c = a * b
d = 7 + 1
e = 7 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | floor(add(divide(log(const_2), log(add(const_1, divide(33.3, const_100)))), const_1)) | find the least number of complete years in which a sum of money put out at 33.3 % compound interest will be more than double of itself ? | "3 years answer : a" | a = math.log(2)
b = 33 / 3
c = 1 + b
d = math.log(c)
e = a / d
f = e + 1
g = math.floor(f)
|
a ) 15 , b ) 20 , c ) 25 , d ) 35 , e ) 45 | b | add(multiply(sqrt(divide(subtract(138, 131), const_2)), const_100), sqrt(subtract(138, divide(subtract(138, 131), const_2)))) | the sum of the squares of three numbers is 138 , while the sum of their products taken two at a time is 131 . their sum is | "explanation : let the numbers be a , b and c . then , a 2 + b 2 + c 2 = 138 and ( ab + bc + ca ) = 131 ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 138 + 2 x 131 = 400 = > ( a + b + c ) = 400 β β β β = 20 . option b" | a = 138 - 131
b = a / 2
c = math.sqrt(b)
d = c * 100
e = 138 - 131
f = e / 2
g = 138 - f
h = math.sqrt(g)
i = d + h
|
a ) 1 / 3 , b ) ΒΌ , c ) 9 / 25 , d ) 5 / 16 , e ) 0 | d | divide(add(3, const_2), multiply(const_4, const_4)) | if a number n is chosen at random from the set of two - digit integers whose digits are both prime numbers , what is the probability w that n is divisible by 3 ? | prime digits are : 2 , 3 , 5 , 7 total number of 2 digit # s with both digits prime are : 4 * 4 = 16 out of these numbers divisible by 3 = 33 , 27 , 57 , 72 and 75 . i had to find the numbers manually using the 4 numbers above . = > prob = 5 / 16 . ans d . took me 3 : 20 mins . | a = 3 + 2
b = 4 * 4
c = a / b
|
a ) a ) 400 , b ) b ) 650 , c ) c ) 500 , d ) d ) 550 , e ) e ) 600 | b | subtract(subtract(multiply(add(add(3, 5), const_2), 170), multiply(5, 150)), multiply(3, 100)) | a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 170 . find the unknown rate of two towels ? | "10 * 170 = 1700 3 * 100 + 5 * 150 = 1050 1700 β 1050 = 650 b" | a = 3 + 5
b = a + 2
c = b * 170
d = 5 * 150
e = c - d
f = 3 * 100
g = e - f
|
a ) 4637 , b ) 4737 , c ) 4937 , d ) 5937 , e ) 5978 | c | subtract(multiply(add(5, const_1), 5600), add(add(add(add(5266, 5768), 5922), 5678), 6029)) | a grocer has a sale of rs . 5266 , rs . 5768 , rs . 5922 , rs . 5678 and rs . 6029 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 5600 ? | "total sale for 5 months = rs . ( 5266 + 5768 + 5922 + 5678 + 6029 ) = rs . 28663 . required sale = rs . [ ( 5600 x 6 ) - 28663 ] = rs . ( 33600 - 28663 ) = rs . 4937 . answer : c" | a = 5 + 1
b = a * 5600
c = 5266 + 5768
d = c + 5922
e = d + 5678
f = e + 6029
g = b - f
|
a ) 400 , b ) 800 , c ) 1250 , d ) 2500 , e ) 10 000 | b | divide(40, divide(2, 40)) | in a certain pond , 40 fish were caught , tagged , and returned to the pond . a few days later , 40 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 ? | "the percent of tagged fish in the second catch is 2 / 40 * 100 = 5 % . we are told that 5 % approximates the percent of tagged fish in the pond . since there are 40 tagged fish , then we have 0.05 x = 40 - - > x = 800 . answer : b ." | a = 2 / 40
b = 40 / a
|
a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 164 | e | add(floor(divide(327, 2)), const_1) | the guests at a football banquet consumed a total of 327 pounds of food . if no individual guest consumed more than 2 pounds of food , what is the minimum number of guests that could have attended the banquet ? | to minimize one quantity maximize other . 163 * 2 ( max possible amount of food a guest could consume ) = 326 pounds , so there must be more than 163 guests , next integer is 164 . answer : e . | a = 327 / 2
b = math.floor(a)
c = b + 1
|
a ) 20 , b ) 5 , c ) 30 , d ) 35 , e ) 67 | a | divide(multiply(floor(subtract(20, divide(divide(multiply(multiply(20, 55), 65), 70), 65))), const_100), 20) | a report consists of 20 sheets each of 55 lines and each such line consist of 65 characters . this report is retyped into sheets each of 65 lines such that each line consists of 70 characters . the percentage reduction in number of sheets is closest to | explanation : total no of characters = 20 * 55 * 65 . from the problem it is clear that 20 * 55 * 65 = x * 65 * 70 . we get x = 15.6 βΌ 16 because these are sheets . therefore 4 less sheets . hencde , there is 20 % decrease in number of sheets . answer : a | a = 20 * 55
b = a * 65
c = b / 70
d = c / 65
e = 20 - d
f = math.floor(e)
g = f * 100
h = g / 20
|
a ) 5 : 2 , b ) 5 : 1 , c ) 4 : 3 , d ) 4 : 1 , e ) 3 : 1 | a | divide(subtract(180, 170), subtract(184, 180)) | students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 184 cms . what was the ratio q of men to women ? | ( a ) q = 184 x 5 + 170 x 2 = 1260 . a | a = 180 - 170
b = 184 - 180
c = a / b
|
a ) 0 , b ) 2 , c ) 5 , d ) 4 , e ) 1 | e | divide(8, 4) | 8 / 4 / 2 = ? | "8 / 4 / 2 = ( 8 / 4 ) / 2 = 2 / 2 = 1 answer is e ." | a = 8 / 4
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | c | divide(multiply(subtract(4.00, 2.00), 100), 2.00) | a wholesaler wishes to sell 100 pounds of mixed nuts at $ 2.00 a pound . she mixes peanuts worth $ 1.50 a pound with cashews worth $ 4.00 a pound . how many pounds of cashews must she use ? | "from the question stem we know that we need a mixture of 100 pounds of peanuts and cashews . if we represent peanuts as x and cashews as y , we get x + y = 100 . since the wholesaler wants to sell the mixture of 100 pounds @ $ 2.50 , we can write this as : $ 2.5 * ( x + y ) = $ 1.5 x + $ 4 y from the equation x + y = 100 , we can rewrite y as y = 100 - x and substitute this into our equation to get : $ 2.5 * ( x + 100 - x ) = $ 1.5 x + $ 4 ( 100 - x ) if you solve for x , you will get x = 60 , and therefore y = 40 . so the wholesaler must use 40 pounds of cashews . you can substitute into the original equation to see that : $ 250 = $ 1.5 ( 60 ) + $ 4 ( 40 ) answer is c ." | a = 4 - 0
b = a * 100
c = b / 2
|
a ) 1.2 % , b ) 1.1 % , c ) 1.0 % , d ) 0.9 % , e ) 0.8 % | b | multiply(divide(add(const_1, divide(multiply(1.2, 800), const_100)), const_1000), const_100) | by weight , liquid x makes up 0.7 percent of solution p and 1.2 percent of solution q . if 200 grams of solution p are mixed with 800 grams of solution q , then liquid x accounts for what percent of the weight of the resulting solution ? | "the number of grams of liquid x is 0.7 ( 200 ) / 100 + 1.2 ( 800 ) / 100 = 1.4 + 9.6 = 11 grams . 11 / 1000 = 1.1 % the answer is b ." | a = 1 * 2
b = a / 100
c = 1 + b
d = c / 1000
e = d * 100
|
a ) 16000 , b ) 27778 , c ) 20000 , d ) 27999 , e ) 17799 | a | divide(multiply(multiply(20, const_100), multiply(16, const_100)), multiply(20, 10)) | a courtyard is 20 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm . the total number of bricks required is ? | "number of bricks = courtyard area / 1 brick area = ( 2000 Γ 1600 / 20 Γ 10 ) = 16000 answer : a" | a = 20 * 100
b = 16 * 100
c = a * b
d = 20 * 10
e = c / d
|
a ) 92 kmph , b ) 98 kmph , c ) 90 kmph , d ) 80 kmph , e ) 82 kmph | c | divide(540, divide(multiply(4, 3), 2)) | a car takes 4 hours to cover a distance of 540 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 4 distence = 540 3 / 2 of 4 hours = 4 * 3 / 2 = 6 hours required speed = 540 / 6 = 90 kmph c )" | a = 4 * 3
b = a / 2
c = 540 / b
|
a ) 107 , b ) 147 , c ) 10 , d ) 296 , e ) none of these | b | multiply(multiply(multiply(divide(3, 8), divide(168, 5)), divide(const_2.0, 549)), 9) | 3 / 8 of 168 * 15 / 5 + x = 549 / 9 + 275 | "explanation : let 3 / 8 of 168 * 15 / 5 + x = 549 / 9 + 275 then , 63 * 15 / 5 + x = 61 + 275 63 * 3 + x = 336 189 + x = 336 x = 147 answer b" | a = 3 / 8
b = 168 / 5
c = a * b
d = 2 / 0
e = c * d
f = e * 9
|
a ) a ) 44 , b ) b ) 77 , c ) c ) 66 , d ) d ) 55 , e ) e ) 37 | a | subtract(divide(multiply(divide(multiply(36, 8), 30), 50), 6), 36) | 36 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 36 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 80 80 β 36 = 44 answer : a" | a = 36 * 8
b = a / 30
c = b * 50
d = c / 6
e = d - 36
|
a ) 9 / 25 , b ) 1 / 5 , c ) 16 / 121 , d ) 105 / 121 , e ) 6 / 25 | d | divide(subtract(121, power(divide(16, const_4), const_2)), 121) | square a has an area of 121 square centimeters . square b has a perimeter of 16 centimeters . if square b is placed within square a and a random point is chosen within square a , what is the probability the point is not within square b ? | "i guess it ' s mean that square b is placed within square aentirely . since , the perimeter of b is 16 , then its side is 16 / 4 = 4 and the area is 4 ^ 2 = 16 ; empty space between the squares is 121 - 16 = 105 square centimeters , so if a random point is in this area then it wo n ' t be within square b : p = favorable / total = 105 / 121 . answer : d ." | a = 16 / 4
b = a ** 2
c = 121 - b
d = c / 121
|
a ) 200 Ο , b ) 240 Ο , c ) 300 Ο , d ) 450 Ο , e ) 1,200 Ο | d | multiply(multiply(multiply(multiply(divide(15, add(multiply(const_2, const_100), multiply(add(const_2, const_3), const_1000))), const_2), divide(add(const_2, multiply(const_2, const_10)), add(const_3, const_4))), 1,320), const_60) | the end of a blade on an airplane propeller is 15 feet from the center . if the propeller spins at the rate of 1,320 revolutions per second , how many miles will the tip of the blade travel in one minute ? ( 1 mile = 5,280 feet ) | "distance traveled in 1 revolution = 2 Ο r = 2 Ο 15 / 5280 revolutions in one second = 1320 revolutions in 60 seconds ( one minute ) = 1320 * 60 total distance traveled = total revolutions * distance traveled in one revolution 1320 * 60 * 2 Ο 15 / 5280 = 450 Ο d is the answer" | a = 2 * 100
b = 2 + 3
c = b * 1000
d = a + c
e = 15 / d
f = e * 2
g = 2 * 10
h = 2 + g
i = 3 + 4
j = h / i
k = f * j
l = k * 1
m = l * const_60
|
a ) 5 , b ) 4 , c ) 4.5 , d ) 3.75 , e ) 3 | d | divide(add(4, subtract(4, 1)), const_2) | a man whose speed is 4 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1 kmph , find his average speed for the total journey ? | "m = 4 s = 1 ds = 5 us = 3 as = ( 2 * 5 * 3 ) / 8 = 3.75 answer : d" | a = 4 - 1
b = 4 + a
c = b / 2
|
a ) 680 , b ) 620 , c ) 652 , d ) 520 , e ) 458 | a | divide(multiply(add(divide(multiply(140, 40), const_100), 80), const_100), 20) | 20 % of a number is more than 40 % of 140 by 80 . find the number ? | "( 20 / 100 ) * x β ( 40 / 100 ) * 140 = 80 1 / 5 x = 136 x = 680 answer : a" | a = 140 * 40
b = a / 100
c = b + 80
d = c * 100
e = d / 20
|
a ) 7,000 , b ) 24,000 , c ) 66,000 , d ) 100,000 , e ) 168,000 | c | multiply(const_4, const_10) | a certain machine produces 550 units of product p per hour . working continuously at this constant rate , this machine will produce how many units of product p in 5 days ? | "since 5 days consist of 24 * 5 hours the total is 120 hours . since every hour the machine produces 550 units of product p the total product during 120 hours is 120 * 550 = 66,000 . correct option : c" | a = 4 * 10
|
a ) 18 , b ) 20 , c ) 22 , d ) 23 , e ) 25 | c | divide(subtract(multiply(7, 1000), multiply(7, 780)), subtract(850, 780)) | the average salary / head of all the workers ina workshop is rs . 850 , if the average salary / head of 7 technician is rs . 1000 and the average salary / head of the rest is rs . 780 , the total no . of workers in the work - shop is ? | let the total number of workers be y . so sum of salary for all workers = sum of salary of 7 technician + sum of salary for other y - 7 workers . 7 x 1000 + 780 ( y - 7 ) = 850 y β 7000 + 780 y - 5460 = 850 y β 70 y = 1540 β΄ y = 22 so total number of workers = 22 c | a = 7 * 1000
b = 7 * 780
c = a - b
d = 850 - 780
e = c / d
|
a ) 62 kg , b ) 60 kg , c ) 70 kg , d ) 72 kg , e ) none of these | a | subtract(80, multiply(6, 3)) | the average weight of 6 students decreases by 3 kg when one of them weighing 80 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 = 3 kg = > ( 80 - x ) / 6 = 3 = > x = 62 answer : a" | a = 6 * 3
b = 80 - a
|
a ) 1 , b ) 3.5 , c ) 20 , d ) 49 , e ) 30 | c | divide(1, divide(1, 20)) | if 20 honey bees make 20 grams of honey in 20 days , then 1 honey bee will make 1 gram of honey in how many days ? | explanation : let the required number days be x . less honey bees , more days ( indirect proportion ) less honey , less days ( direct proportion ) honey bees 1 : 20 : : 20 : x honey 20 : 1 = > 1 x 20 x x = 20 x 1 x 20 = > x = 20 . answer : c | a = 1 / 20
b = 1 / a
|
a ) 7 : 15 , b ) 7 : 10 , c ) 7 : 8 , d ) 7 : 4 , e ) 7 : 2 | a | divide(subtract(33, divide(36, divide(add(const_100, 60), const_100))), divide(36, divide(add(const_100, 60), const_100))) | a dishonest person wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing 33 $ per litre so as to make a profit of 60 % on cost when he sells the resulting milk and water mixture for 36 $ . in what ratio should he mix the water and milk ? | "first of all , let ' s consider 1 liter of the stuff he is going to sell - - - naive customers think it ' s pure milk , but we know it ' s some milk - water mixture . he is going to sell this liter of milk - water for $ 36 . this $ 36 should be a 60 % increase over cost . here , we need to think about percentage increases as multipliers . using multipliers ( cost ) * 1.60 = $ 36 cost = 36 / 1.6 = $ 22.5 if he wants a 60 % increase over cost on the sale of one liter of his milk - water , the cost has to be $ 22.5 well , a liter of milk costs $ 33 , so if he is going to use just $ 30 of milk in his mixture , that ' s 22.5 / 33 = 15 / 22 of a liter . if milk is 15 / 22 of the liter , then water is 7 / 22 of the liter , and the ratio of water to milk is 7 : 15 . answer choice ( a )" | a = 100 + 60
b = a / 100
c = 36 / b
d = 33 - c
e = 100 + 60
f = e / 100
g = 36 / f
h = d / g
|
a ) 16 % , b ) 27 % , c ) 32 % , d ) 40 % , e ) 52 % | b | multiply(divide(subtract(65, 48), 65), const_100) | in town x , 65 percent of the population are employed , and 48 percent of the population are employed males . what percent of the employed people in town x are females ? | "we are asked to find the percentage of females in employed people . total employed people 65 % , out of which 48 are employed males , hence 17 % are employed females . ( employed females ) / ( total employed people ) = 17 / 64 = 27 % answer : b ." | a = 65 - 48
b = a / 65
c = b * 100
|
a ) 20 / 21 , b ) 23 / 25 , c ) 24 / 23 , d ) 22 / 5 , e ) 21 / 5 | e | divide(subtract(multiply(9, 3), multiply(const_2, const_3)), add(3, const_2)) | out of 3 consecutive odd numbers 9 times the first number is equal to addition of twice the third number and adding 9 to twice the second . what is the first number ? | description : = > 9 x = 2 ( x + 2 ) + 9 + 2 ( x + 4 ) = > 9 x = 4 x + 21 = > 5 x = 21 x = 21 / 5 answer e | a = 9 * 3
b = 2 * 3
c = a - b
d = 3 + 2
e = c / d
|
a ) 227 , b ) 83 , c ) 23 , d ) 827 , e ) none of these | b | subtract(subtract(subtract(multiply(23, const_4), const_4), const_4), const_1) | find the lcm of 23 , 46,827 | explanation : whenever we have to solve this sort of question , remember the formula . lcm = \ \ begin { aligned } \ \ frac { hcf of denominators } { lcm of numerators } \ \ end { aligned } so answers will be option 2 , please also give attention to the difference in formula of hcf and lcm answer : option b | a = 23 * 4
b = a - 4
c = b - 4
d = c - 1
|
a ) 14 , b ) 12 , c ) 15 , d ) 11 , e ) 10 | a | subtract(multiply(divide(4, 3), 4), 3) | at a certain paint store forest green is made by mixing 4 parts blue paint with 3 parts yellow paint . verdant green is made by mixing 4 parts yellow paint with 3 parts blue paint . how many liters of yellow paint must be added to 42 liters of forest green to change it to verdant green ? | "42 liter of forset green have 24 liter of blue and 18 liter of yellow suppose we add x liter of yellow to make it a verdant green so the ratio of blue to yellow in verdant green is ΒΎ so the equation is blue / yellow = 24 / ( 18 + x ) = ΒΎ 54 + 3 x = 96 = > x = 14 answer : a" | a = 4 / 3
b = a * 4
c = b - 3
|
a ) $ 1,250 , 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 $ 300,000 , 80 % of which was in the first mortgage , what is her approximate monthly interest payment ? | "0.04 x + 0.09 y = 300000 [ 1 ] 0.04 x = 0.80 * 300000 = 240000 [ 2 ] 240000 + 0.09 y = 300000 - - > 0.09 y = 60000 [ 3 ] 240000 / 12 = 20000 [ 4 ] 60000 / 12 = 5000 [ 5 ] adding [ 4,5 ] we get : 25000 [ 6 ] dividing [ 6 ] / 2 to get an average we get 1.25 , ans a" | a = 4 * 2
b = 9 - a
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | subtract(negate(4,8), multiply(subtract(3,2, 4,6), divide(subtract(3,2, 4,6), subtract(1,2, 3,2)))) | 1,2 , 3,2 , 4,6 , 4,8 , _____ | "divide the series : 1 , 2,3 2 , 4,6 4,8 , ? so it is 12 answer : c" | a = negate - (
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a ) 806 , b ) 744 , c ) 912 , d ) 1200 , e ) 1400 | a | divide(multiply(divide(312, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | there were two candidates in an election . winner candidate received 62 % of votes and won the election by 312 votes . find the number of votes casted to the winning candidate ? | "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 312 62 % - - - - - - - - ? = > 806 answer : a" | a = 100 - 62
b = 62 - a
c = b / 100
d = 312 / c
e = d * 62
f = e / 100
|
a ) 272258 , b ) 272358 , c ) 278616 , d ) 274258 , e ) 274358 | c | multiply(divide(5358, 52), const_100) | 5358 x 52 = ? | "5358 x 51 = 5358 x ( 50 + 2 ) = 5358 x 50 + 5358 x 2 = 267900 + 10716 = 278616 . c )" | a = 5358 / 52
b = a * 100
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a ) 54 , b ) 33 , c ) 44 , d ) 66 , e ) 99 | a | multiply(multiply(const_3, 9), const_2) | if x / ( 9 p ) is an odd prime number , where x is a positive integer and p is a prime number , what is the least value of x ? | "x / ( 9 p ) = odd prime number x = odd prime number * 9 p least value of x = lowest odd prime number * 9 * lowest value of p = 3 * 9 * 2 = 54 answer a" | a = 3 * 9
b = a * 2
|
a ) rs . 200 , b ) rs . 300 , c ) rs . 400 , d ) rs . 600 , e ) rs . 500 | c | multiply(900, divide(const_4, divide(900, const_100))) | rs . 900 is divided among maya , annie , saiji so that maya receives half as much as annie and annie half as much as saiji . then saiji β s share is : | let saiji = x . then , annie = x / 2 and maya = x / 4 . therefore , maya : annie : saiji = x / 4 : x / 2 : x = 1 : 2 : 4 . therefore , saiji β s share = rs . 900 * 4 / 9 = rs . 400 . answer : c | a = 900 / 100
b = 4 / a
c = 900 * b
|
a ) 3 , b ) 4 , c ) 5 , d ) 8 , e ) 6 | d | inverse(add(inverse(12), inverse(24))) | a company has two models of computers , model m and model n . operating at a constant rate , a model m computer can complete a certain task in 24 minutes and a model n computer can complete the same task in 12 minutes . if the company used the same number of each model of computer to complete the task in 1 minute , how many model m computers were used ? | let ' s say 1 work is processing 24 gb of data . model m : 1 gb per min model n : 2 gb per min working together , 1 m and 1 n = 3 gb per min so , 8 times as many computers would work at 18 gb per min . so no . of m = 8 answer is d | a = 1/(12)
b = 1/(24)
c = a + b
d = 1/(c)
|
a ) 6 days , b ) 8 days , c ) 12 days , d ) 10 days , e ) 5 days | e | add(inverse(subtract(divide(const_1, 4), divide(const_1, 20))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 4 days . a alone can do it in 20 days . what time b will take to do the work alone ? | "explanation : a and b 1 day ' s work = 1 / 4 a alone can do 1 day ' s work = 1 / 20 what time b will take to do the work alone ? b = ( a + b ) - a = ( 1 / 4 ) - ( 1 / 20 ) = 5 days answer : option e" | a = 1 / 4
b = 1 / 20
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
|
a ) 180 sec , b ) 190 sec , c ) 290 sec , d ) 490 sec , e ) 40 sec | e | subtract(divide(multiply(const_1, const_1000), divide(200, 10)), 10) | in a kilometer race , a beats b by 200 meters or 10 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 10 ) / 200 = 50 sec . time taken by a = 50 - 10 = 40 sec . answer : e" | a = 1 * 1000
b = 200 / 10
c = a / b
d = c - 10
|
a ) 30 kmh , b ) 40 kmh , c ) 42 kmh , d ) 44 kmh , e ) 50 kmh | a | divide(const_3, add(add(divide(const_1, 80), divide(const_1, 15)), divide(const_1, 48))) | if a car went the first third of the distance at 80 kmh , the second third at 15 kmh , and the last third at 48 kmh , what was the average speed of the car for the entire trip ? | "assume d / 3 = 240 ( this number is convenient because it is divisible by 80 , 15 and 48 ) so : 240 = 80 * t 1 = 3 hrs 240 = 15 * t 2 = 16 hrs 240 = 48 * t 3 = 5 hrs t = t 1 + t 2 + t 3 = 24 hrs d = rt ( 240 * 3 ) = r * 24 r = 30 answer : a" | a = 1 / 80
b = 1 / 15
c = a + b
d = 1 / 48
e = c + d
f = 3 / e
|
a ) rs . 45,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 15,000 , e ) none | d | divide(multiply(multiply(add(const_1, const_4), const_1000), 2), 6) | x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 6 . if x invested rs . 5,000 . the amount invested by y is | solution suppose y invested rs . y then , 5000 / y = 2 / 6 Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ y = ( 5000 Γ£ β 6 / 2 ) . Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ y = 15000 . answer d | a = 1 + 4
b = a * 1000
c = b * 2
d = c / 6
|
a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | a | add(divide(subtract(100, add(add(add(2, add(2, 2)), add(add(2, 2), 2)), add(add(add(2, 2), 2), 2))), 5), add(add(add(2, 2), 2), 2)) | in a school with 5 classes , each class has 2 students less than the previous class . how many students are there in the largest class if the total number of students at school is 100 ? | let x be the number of students in the largest class . then x + ( x - 2 ) + ( x - 4 ) + ( x - 6 ) + ( x - 8 ) = 100 5 x - 20 = 100 5 x = 120 x = 24 the answer is a . | a = 2 + 2
b = 2 + a
c = 2 + 2
d = c + 2
e = b + d
f = 2 + 2
g = f + 2
h = g + 2
i = e + h
j = 100 - i
k = j / 5
l = 2 + 2
m = l + 2
n = m + 2
o = k + n
|
a ) 33 1 / 7 % , b ) 33 1 / 6 % , c ) 50 % , d ) 38 1 / 3 % , e ) 33 2 / 3 % | c | subtract(const_100, divide(multiply(900, const_100), 600)) | an article is bought for rs . 600 and sold for rs . 900 , find the gain percent ? | "600 - - - - 300 100 - - - - ? = > 50 % answer : c" | a = 900 * 100
b = a / 600
c = 100 - b
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | add(7, const_1) | mark and ann together were allocated n boxes of cookies to sell for a club project . mark sold 7 boxes less than n and ann sold 2 boxes less than n . if mark and ann have each sold at least one box of cookies , but together they have sold less than n boxes , what is the value of n ? | "if n = 8 mark sold 1 box and ann sold 6 boxes total 7 < 8 answer : e" | a = 7 + 1
|
a ) 10 , b ) 5 , c ) 15 , d ) 7.5 , e ) 12.5 | b | add(add(divide(3, 4), multiply(divide(3, 4), 5)), multiply(const_0_25, 5)) | total 15 cows 5 cow gives each 2 liter milk 5 cow gives each 3 / 4 liter milk 5 cow gives each 1 / 4 liter milk this is split into 3 son per each 5 cows & 5 liter milk how ? | "5 cow 2 liter each = 10 liter 5 cow 3 / 4 liter each = 3 / 4 = 0.75 * 5 = 3.75 5 cow 1 / 4 liter each = 1 / 4 = 0.25 * 5 = 1.25 add 10 + 3.75 + 1.25 = 15 milk split into 3 son each 5 liter then 15 / 3 = 5 answer : b" | a = 3 / 4
b = 3 / 4
c = b * 5
d = a + c
e = const_0_25 * 5
f = d + e
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | divide(subtract(add(multiply(2, 5), 19), add(multiply(const_3, 4), 5)), subtract(multiply(2, 3), multiply(3, const_1))) | given f ( x ) = 3 x β 5 , for what value of x does 2 * [ f ( x ) ] β 19 = f ( x β 4 ) ? | 2 ( 3 x - 5 ) - 19 = 3 ( x - 4 ) - 5 3 x = 12 x = 4 the answer is b . | a = 2 * 5
b = a + 19
c = 3 * 4
d = c + 5
e = b - d
f = 2 * 3
g = 3 * 1
h = f - g
i = e / h
|
a ) 40 / 41 , b ) 20 / 41 , c ) 30 / 41 , d ) 60 / 41 , e ) 80 / 41 | a | divide(8, add(divide(20, const_100), 8)) | a committee is reviewing a total of 20 x black - and - white films and 8 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction of the selected films are in color ? | "say x = y = 10 . in this case we would have : 20 x = 200 black - and - white films ; 8 y = 80 color films . y / x % = 10 / 10 % = 1 % of the black - and - white films , so 2 black - and - white films and all 80 color films , thus total of 82 films were selected . color films thus compose 80 / 82 = 40 / 41 of the selected films . answer : a" | a = 20 / 100
b = a + 8
c = 8 / b
|
a ) 2 , b ) - 1 , c ) 4 , d ) - 5 , e ) 6 | b | divide(subtract(46, 2), 4) | if | 4 x + 2 | = 46 , what is the sum of all the possible values of x ? | "there will be two cases 4 x + 2 = 46 or 4 x + 2 = - 46 = > x = 11 or x = - 12 sum of both the values will be - 12 + 11 = - 1 answer is b" | a = 46 - 2
b = a / 4
|
a ) 130 , b ) 100 , c ) 125 , d ) 175 , e ) 225 | a | divide(subtract(multiply(divide(780, const_3), const_4), 780), const_2) | there are 780 male and female participants in a meeting . half the female participants and one - quarterof the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "let m be the number of male participants and f be the number of female articipants in the meeting . thetotal number of participants is given as 780 . hence , we have m + f = 780 now , we have that half the female participants and one - quarter of the male participants are democrats . let d equal the number of the democrats . then we have the equation f / 2 + m / 4 = d now , we have that one - third of the total participants are democrats . hence , we have the equation d = 780 / 3 = 260 solving the three equations yields the solution f = 260 , m = 520 , and d = 260 . the number of female democratic participants equals half the female participants equals 260 / 2 = 130 . answer : a" | a = 780 / 3
b = a * 4
c = b - 780
d = c / 2
|
a ) 425 , b ) 345 , c ) 375 , d ) 380 , e ) 400 | a | multiply(divide(680, 5), 3) | there are 680 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | "in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 680 . hence 3 x + 5 x = 680 solve for x 8 x = 680 x = 85 number of boys 3 x = 3 Γ 85 = 255 number of girls 5 x = 5 Γ 85 = 425 a" | a = 680 / 5
b = a * 3
|
a ) 719 / 720 , b ) 1 / 120 , c ) 2 / 233 , d ) 3 / 543 , e ) 1 / 720 | b | divide(const_1, multiply(multiply(6, 5), const_4)) | in a clothing store , there are 6 different colored neckties ( orange , yellow , green , blue , and indigo ) and 5 different colored shirts ( orange , yellow , green , blue , and indigo ) that must be packed into boxes for gifts . if each box can only fit one necktie and one shirt , what is the probability that all of the boxes will contain a necktie and a shirt of the same color ? | 5 ties and 5 shirts . . . red tie can take any of 5 shirts . . orange can take any of the remaining 4 shirts yellow any of remaining 3 . . and so on till last indigo chooses the 1 remaining . . total ways = 5 * 4 * 3 * 2 * 1 = 120 out of this 120 , only 1 way will have same colour tie and shirt . . prob = 1 / 120 b | a = 6 * 5
b = a * 4
c = 1 / b
|
a ) 13 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | divide(325, add(subtract(26, 2), const_1)) | 325 metres long yard , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "26 trees have 25 gaps between them , required distance ( 325 / 25 ) = 13 a" | a = 26 - 2
b = a + 1
c = 325 / b
|
a ) 1 , b ) 2 , c ) 4 , d ) 5 , e ) 7 | d | multiply(multiply(4, 4), divide(1, 4)) | in the coordinate plane , points ( x , 1 ) and ( 4 , y ) are on line k . if line k passes through the origin and has slope 1 / 4 , then x + y = | "line k passes through the origin and has slope 1 / 4 means that its equation is y = 1 / 4 * x . thus : ( x , 1 ) = ( 4 , 1 ) and ( 4 , y ) = ( 4,1 ) - - > x + y = 4 + 1 = 5 . answer : d" | a = 4 * 4
b = 1 / 4
c = a * b
|
a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) 60 | e | subtract(multiply(add(20, const_1), 5), 45) | the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 45 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ? | "20 x + 45 = 21 ( x β 5 ) x = 60 answer : e" | a = 20 + 1
b = a * 5
c = b - 45
|
a ) a ) 9 , b ) b ) 3 , c ) c ) 12 , d ) d ) 6 , e ) e ) 10 | b | subtract(reminder(89, const_10), subtract(add(add(reminder(457, const_10), const_1), reminder(457, const_10)), reminder(89, const_10))) | q and f represent two distinct digits . if the number 457 q 89 f is divisible by 36 , what is the value of ( q + f ) ? | a no . divisible by 36 means it is div by 49 . to be div by 4 last 2 digits to be multiple of 4 so , f could be either 2 or 6 only similarly for a number to be div by 9 its sum must be multiple of 9 first . filtering we find q could be either 1 , 47 only . to be divisible by 9 only 1 works . ( if we choose f = 6 then q have to be 6 but as per condition qf are distinct ) so q = 1 f = 2 ans : - 3 . b | a = reminder - (
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a ) 84.4 , b ) 70.2 , c ) 80.1 , d ) 60.8 , e ) 62.5 | e | multiply(subtract(const_1, divide(20, 50)), const_100) | the total population of dogs in a community is estimated to be 50 % male and 50 % female . the total number of black dogs is 20 % greater than the total number of male black dogs . if the total number of female dogs is 8 times more than the number of female black dogs what percentage of male dogs is black ? | since we are dealing in percentage let us pick 100 as the number of black male dogs . that means that the total number of black dogs is = 120 ( 20 % more ) , therefore the number of black female dogs is 20 . the total number of female dogs is 8 x the number of black female = 20 * 8 = 160 female dogs . male dogs are 50 % of the dogs , and because there are 160 female dogs , then there must also be 160 male dogs therefore percentage of male dogs that are black = 100 / 160 * 100 = 62.5 correct option is e | a = 20 / 50
b = 1 - a
c = b * 100
|
a ) 0.005 , b ) 0.002 , c ) 0.001 , d ) 0.0003 , e ) 0.0002 | d | divide(0.3, 1,000) | when magnified 1,000 times by an electron microscope , the image of a certain circular piece of tissue has a diameter of 0.3 centimeter . the actual diameter of the tissue , in centimeters , is | "it is very easy if x is the diameter , then the magnified length is 1000 x . ince 1000 x = 0.3 then x = 0.3 / 1000 = 0.0003 . the answer is d" | a = 0 / 3
|
a ) 50 % , b ) 85 % , c ) 25 % , d ) 75 % , e ) none of above | a | multiply(divide(25, 50), const_100) | the ratio 25 : 50 expressed as percent equals to | "explanation : actually it means 25 is what percent of 50 , which can be calculated as , ( 25 / 50 ) * 100 = 25 * 2 = 50 answer : option a" | a = 25 / 50
b = a * 100
|
a ) 23 years , b ) 22 years , c ) 21 years , d ) 20 years , e ) 19 years | d | divide(subtract(22, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 22 years older than his son . in two years , his age will be twice the age of his son . what is the present age of his son ? | "let present age of the son = x years then , present age the man = ( x + 22 ) years given that , in 2 years , man ' s age will be twice the age of his son Γ’ β‘ β ( x + 22 ) + 2 = 2 ( x + 2 ) Γ’ β‘ β x = 20 answer : d" | a = 2 * 2
b = a - 2
c = 22 - b
d = 2 - 1
e = c / d
|
a ) 2 . , b ) 4 . , c ) 5 . , d ) 6 , e ) 8 . | b | divide(multiply(8, 18), 36) | 18 beavers , working together in a constant pace , can build a dam in 8 hours . how many hours will it take 36 beavers that work at the same pace , to build the same dam ? | "total work = 18 * 8 = 144 beaver hours 36 beaver * x = 144 beaver hours x = 144 / 36 = 4 answer : b" | a = 8 * 18
b = a / 36
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a ) 8 , b ) 13 , c ) 28 , d ) 6 , e ) 2 | d | multiply(divide(2, 4), const_100) | 2 + 4 | d | a = 2 / 4
b = a * 100
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a ) 16 , b ) 18 , c ) 20 , d ) 24 , e ) 30 | b | multiply(4, divide(30, add(4, 6))) | maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 30 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h , what is the distance traveled by brad ? | "time taken = total distance / relative speed total distance = 30 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken = 30 / 10 = 3 hrs distance traveled by brad = brad ' s speed * time taken = 6 * 3 = 18 kms . . . answer - b" | a = 4 + 6
b = 30 / a
c = 4 * b
|
a ) 17 hr , b ) 19 hr , c ) 10 hr , d ) 24 hr , e ) 26 hr | e | inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 6), 1), 6)))) | a pump can fill a tank with water in 2 hours . because of a leak , it took 2 1 / 6 hours to fill the tank . the leak can drain all the water of the tank in ? | "work done by the tank in 1 hour = ( 1 / 2 - 2 1 / 6 ) = 1 / 26 leak will empty the tank in 26 hrs . answer : e" | a = 1 / 2
b = 2 * 6
c = b + 1
d = c / 6
e = 1/(d)
f = a - e
g = 1/(f)
|
a ) 1628.4 , b ) 1534 , c ) 1492 , d ) 240 , e ) none of these | d | multiply(divide(add(multiply(5, 20), multiply(subtract(10, 5), 4)), subtract(10, subtract(10, 5))), 10) | 10 people went to a hotel for combine dinner party 5 of them spent rs . 20 each on their dinner and rest spent 4 more than the average expenditure of all the 10 . what was the total money spent by them . | "solution : let average expenditure of 10 people be x . then , 10 x = 20 * 5 + 5 * ( x + 4 ) ; or , 10 x = 20 * 5 + 5 x + 20 ; or , x = 24 ; so , total money spent = 24 * 10 = rs . 240 . answer : option d" | a = 5 * 20
b = 10 - 5
c = b * 4
d = a + c
e = 10 - 5
f = 10 - e
g = d / f
h = g * 10
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | divide(30, divide(20, 10)) | a sum is divided among w , x and y in such a way that for each rupee w gets , x gets 30 paisa and y gets 20 paisa . if the share of w is rs . 10 , what is the total amount ? | w : x : y = 100 : 30 : 20 20 : 6 : 4 20 - - - 10 30 - - - ? = > 15 answer : d | a = 20 / 10
b = 30 / a
|
a ) 1485 , b ) 2700 , c ) 3300 , d ) 4860 , e ) 5400 | a | multiply(divide(factorial(divide(12, 4)), const_2), divide(factorial(12), multiply(factorial(subtract(12, 4)), factorial(4)))) | a plant manager must assign 12 new workers to one of five shifts . she needs a first , second , and third shift , and two alternate shifts . each of the shifts will receive 4 new workers . how many different ways can she assign the new workers ? | whatever : my take selecting team of 4 out of 12 to assign to the shifts = 12 c 4 = 495 ways . now 4 out of 12 means total of 3 group possible . so putting them in shifts = counting methode : first , second , third , = 3 * 2 * 1 = 6 here alt and alt are the same : so 6 / 2 = 3 ways . total ways of selecting = ( selecting 4 out of 12 ) * arranging those teams in shifts = 495 * 3 = 1485 ans : a | a = 12 / 4
b = math.factorial(a)
c = b / 2
d = math.factorial(12)
e = 12 - 4
f = math.factorial(e)
g = math.factorial(4)
h = f * g
i = d / h
j = c * i
|
a ) 84 . , b ) 75 . , c ) 70 . , d ) 65 . , e ) 54 . | a | divide(add(multiply(divide(30, add(30, 45)), 66), multiply(divide(45, add(30, 45)), 100)), divide(add(30, 45), const_60)) | a car was driving at 66 km / h for 30 minutes , and then at 100 km / h for another 45 minutes . what was its average speed ? | "driving at 66 km / h for 30 minutes , distance covered = 60 * 1 / 2 = 30 km driving at 100 km / h for 45 minutes , distance covered = 100 * 3 / 4 = 75 km average speed = total distance / total time = 105 / 1.25 = 84 km / h answer : a" | a = 30 + 45
b = 30 / a
c = b * 66
d = 30 + 45
e = 45 / d
f = e * 100
g = c + f
h = 30 + 45
i = h / const_60
j = g / i
|
a ) 10 , b ) 16 , c ) 37 , d ) 29 , e ) 22 | b | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 30), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 20), const_100)), const_100)), const_100)) | if the price of a book is first decreased by 30 % and then increased by 20 % , then the net change in the price will be : | "explanation : let the original price be rs . 100 . decreased by 30 % = 70 then increased 20 % on rs 70 = 70 + 14 = 84 net change in price = 100 - 84 = 16 answer : b" | a = 100 * 30
b = a / 100
c = 100 - b
d = c / 100
e = 100 * 20
f = e / 100
g = 100 + f
h = g / 100
i = d * h
j = i * 100
k = 100 - j
|
a ) 13 / 2 , b ) 17 / 3 , c ) 15 / 2 , d ) 20 / 3 , e ) 9 | d | divide(multiply(10, 4), subtract(10, 4)) | a man can do a piece of work in 10 days , but with the help of his son , he can do it in 4 days . in what time can the son do it alone ? | "son ' s 1 day ' s work = ( 1 / 4 ) - ( 1 / 10 ) = 3 / 20 the son alone can do the work in 20 / 3 days answer is d" | a = 10 * 4
b = 10 - 4
c = a / b
|
a ) - 2 , b ) - 4 , c ) - 6 , d ) - 8 , e ) - 9 | b | subtract(88, 62) | find the next term 88 , 62 , . . ? | 8 , 8 , 6 , 2 , x i guess each differ in d range of 2 8 - 8 = 0 8 - 6 = 2 6 - 2 = 4 2 - x should be ` ` 6 ' ' 2 - x = 6 x = - 4 answer : b | a = 88 - 62
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a ) 55 : 23 , b ) 52 : 33 , c ) 51 : 52 , d ) 45 : 53 , e ) 51 : 34 | c | divide(add(const_100, 2), add(const_100, 4)) | the cash difference between the selling prices of an book at a profit of 2 % and 4 % is $ 3 . the ratio of the two selling prices is : | "let c . p . of the book be $ x . then , required ratio = 102 % of x / 104 % of x = 102 / 104 = 51 / 52 = 51 : 52 c" | a = 100 + 2
b = 100 + 4
c = a / b
|
a ) 11.3 sec , b ) 12.1 sec , c ) 13.1 sec , d ) 16.5 sec , e ) 12.7 sec | a | divide(add(100, 135), multiply(75, const_0_2778)) | how long does a train 100 m long running at the speed of 75 km / hr takes to cross a bridge 135 m length ? | "speed = 75 * 5 / 18 = 20.8 m / sec total distance covered = 100 + 135 = 235 m . required time = 235 / 20.8 ' = 11.3 sec . answer : a" | a = 100 + 135
b = 75 * const_0_2778
c = a / b
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a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 18 | d | add(const_3, const_4) | what is the smallest positive integer x , such that 7000 x is a perfect cube ? | "take out the factors of 7000 x that will come 10 ^ 3 * 7 . for perfect cube you need every no . raise to the power 3 . for 7000 x to be a perfect cube , need two 7 that means 49 . e is the answer ." | a = 3 + 4
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a ) 5 / 2 , b ) 3 / 2 , c ) 2 / 3 , d ) 2 / 7 , e ) 0 | d | divide(add(1, 2), add(2, 5)) | if x = 1 - 5 t and y = 2 t - 1 , then for what value of t does x = y ? | "we are given x = 1 β 5 t and y = 2 t β 1 , and we need to determine the value for t when x = y . we should notice that both x and y are already in terms of t . thus , we can substitute 1 β 5 t for x and 2 t β 1 for y in the equation x = y . this gives us : 1 β 5 t = 2 t β 1 2 = 7 t 2 / 7 = t the answer is d ." | a = 1 + 2
b = 2 + 5
c = a / b
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a ) 1670 , b ) 1600 , c ) 1680 , d ) 1900 , e ) 1230 | c | add(1200, multiply(1200, divide(40, const_100))) | a person buys an article at $ 1200 . at what price should he sell the article so as to make a profit of 40 % ? | c 1680 cost price = $ 1200 profit = 40 % of 1200 = $ 480 selling price = cost price + profit = 1200 + 480 = 1680 | a = 40 / 100
b = 1200 * a
c = 1200 + b
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a ) 55 % , b ) 60 % , c ) 70 % , d ) 75 % , e ) 80 % | d | multiply(subtract(const_1, divide(25, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))), subtract(const_1, divide(75, multiply(multiply(const_2, const_5), multiply(const_2, const_5))))) | there is a 75 % chance that tigers will not win at all during the whole season . there is a 25 % chance that germany will not play at all in the whole season . what is the greatest possible probability that the tigers will win and germany will play during the season ? | there is a 75 % chance that tigers will not win at all during the whole season we can infer that there is 25 % chance tigers will win . similarly there is a 25 % chance that germany will not play at all in the whole season we can also infer that there is 75 % chance that germany will play . answer d | a = 2 * 5
b = 2 * 5
c = a * b
d = 25 / c
e = 1 - d
f = 2 * 5
g = 2 * 5
h = f * g
i = 75 / h
j = 1 - i
k = e * j
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a ) 120 , b ) 100 , c ) 135 , d ) 140 , e ) 160 | e | multiply(multiply(20, 4), divide(8, 4)) | in the coordinate plane , points ( x , 8 ) and ( 20 , y ) are on line k . if line k passes through the origin and has slope 1 / 4 , then x * y = | line k passes through the origin and has slope 1 / 4 means that its equation is y = 1 / 4 * x . thus : ( x , 8 ) = ( 32 , 8 ) and ( 20 , y ) = ( 20,5 ) - - > x * y = 32 * 5 = 160 . answer : e | a = 20 * 4
b = 8 / 4
c = a * b
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a ) 4900 , b ) 3000 , c ) 1250 , d ) 1200 , e ) 1225 | e | add(divide(divide(35, divide(divide(divide(divide(divide(35, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(35, divide(divide(divide(divide(divide(35, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of first 35 odd numbers | "explanation : n 2 = 352 = 1225 answer : option e" | a = 35 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 35 / e
g = f / 2
h = 35 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 35 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
|
a ) 16 , b ) 10 , c ) 28 , d ) 4 , e ) 2 | c | divide(subtract(multiply(add(3, const_1), 20), multiply(2, const_12)), const_2) | bill is golfing with 3 friends and can either buy generic golf tees that are packaged by the dozen or the higher quality aero flight tees that come by the pair . what is the minimum number of packages of aero flight tees bill must purchase to ensure that he has at least 20 golf tees for each member of his foursome , if he will buy no more than 2 packages of the generic golf tees ? | at least 20 golf tees for each member of his foursome = total of at least 4 * 20 = 80 tees . 2 packages of the generic golf tees , that are packaged by the dozen = 2 * 12 = 24 tees . so , bill must by at least 16 aero tees . they come by the pair , hence he must by at least 56 / 2 = 28 packages of aero flight tees . answer : c . | a = 3 + 1
b = a * 20
c = 2 * 12
d = b - c
e = d / 2
|
a ) 11 kmph , b ) 12 kmph , c ) 14 kmph , d ) 17 kmph , e ) none of these | c | subtract(subtract(20, 3), 3) | a man ' s speed with the current is 20 kmph and speed of the current is 3 kmph . the man ' s speed against the current will be | "explanation : if you solved this question yourself , then trust me you have a all very clear with the basics of this chapter . if not then lets solve this together . speed with current is 20 , speed of the man + it is speed of the current speed in still water = 20 - 3 = 17 now speed against the current will be speed of the man - speed of the current = 17 - 3 = 14 kmph option c" | a = 20 - 3
b = a - 3
|
a ) $ 24 million , b ) $ 120 million , c ) $ 144 million , d ) $ 240 million , e ) $ 888 million | e | subtract(multiply(388, divide(const_12, const_2)), multiply(1.44, const_1000)) | country x imported approximately $ 1.44 billion of goods in 1996 . if country x imported $ 388 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year , by how much would country xs 1997 imports exceed those of 1996 ? | "convert units to millions as answer is in millions 1996 imports = $ 1.44 bill = $ 1440 mill i . e . 1440 / 12 = $ 120 mill / month 1997 imports = $ 388 mill / 2 month i . e . $ 194 mill / month difference / month = 194 - 120 = 74 difference / year = $ 74 mill * 12 = $ 988 mill answer : e" | a = 12 / 2
b = 388 * a
c = 1 * 44
d = b - c
|
a ) 113 , b ) 150 , c ) 225 , d ) 250 , e ) 500 | a | sqrt(divide(multiply(90, const_100), divide(70, const_100))) | 90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 70 % of the total school population of x students , what is x ? | "90 = x / 100 * 70 / 100 * x = > x ^ 2 = 9 * 10000 / 7 = > x = 113 a" | a = 90 * 100
b = 70 / 100
c = a / b
d = math.sqrt(c)
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(divide(5, const_2), multiply(43, 43)) | what is the remainder when 43 ^ 88 is divided by 5 ? | "the units digit of the exponents of 3 cycle in a group of 4 : { 3 , 9 , 7 , 1 } 88 has the form 4 k so the units digit of 43 ^ 88 is 1 . the remainder when dividing by 5 is 1 . the answer is b ." | a = 5 / 2
b = 43 * 43
c = a - b
|
a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | c | divide(990, multiply(const_10, const_2)) | how many factors of 990 are odd numbers greater than 1 ? | when factorized , 990 has 5 prime factors . of these prime factors 4 are odd and 1 is even . hence total number of odd factors is 2 * 2 * 2 * 2 ( 16 ) , which includes 1 . the total number of odd factors greater than 1 are 15 . ( option c ) | a = 10 * 2
b = 990 / a
|
a ) 9 / 25 , b ) 1 / 5 , c ) 16 / 25 , d ) 3 / 5 , e ) 6 / 25 | c | divide(subtract(25, power(divide(12, const_4), const_2)), 25) | square a has an area of 25 square centimeters . square b has a perimeter of 12 centimeters . if square b is placed within square a and a random point is chosen within square a , what is the probability the point is not within square b ? | "i guess it ' s mean that square b is placed within square aentirely . since , the perimeter of b is 12 , then its side is 12 / 4 = 3 and the area is 3 ^ 2 = 9 empty space between the squares is 25 - 9 = 16 square centimeters , so if a random point is in this area then it wo n ' t be within square b : p = favorable / total = 16 / 25 . answer : c" | a = 12 / 4
b = a ** 2
c = 25 - b
d = c / 25
|
['a ) 40', 'b ) 50', 'c ) 60', 'd ) 70', 'e ) 80'] | c | subtract(divide(divide(5300, 26.5), const_2), multiply(const_2, 20)) | the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ 26.50 per metre is rs . 5300 , what is the length of the plot in metres ? | let breadth = x metres . then , length = ( x + 20 ) metres . perimeter = 5300 / 26.50 m = 200 m . 2 [ ( x + 20 ) + x ] = 200 2 x + 20 = 100 2 x = 80 x = 40 . hence , length = x + 20 = 60 m . answer : c | a = 5300 / 26
b = a / 2
c = 2 * 20
d = b - c
|
a ) 7.18 , b ) 7.12 , c ) 7.16 , d ) 7.55 , e ) 7.82 | e | divide(add(150, 165), multiply(add(80, 65), const_0_2778)) | two trains 150 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 150 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.82 answer : e" | a = 150 + 165
b = 80 + 65
c = b * const_0_2778
d = a / c
|
a ) 125 m , b ) 150 m , c ) 187 m , d ) 167 m , e ) 197 m | a | multiply(divide(multiply(50, const_1000), const_3600), 9) | a train running at the speed of 50 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 50 * 5 / 18 = 125 / 9 m / sec length of the train = speed * time = 125 / 9 * 9 = 125 m answer : a" | a = 50 * 1000
b = a / 3600
c = b * 9
|
a ) 5 / 12 , b ) 12 / 5 , c ) 25 / 144 , d ) 144 / 25 , e ) 146 / 25 | a | sqrt(divide(multiply(25, const_3), multiply(216, const_2))) | two - third of a positive number and 25 / 216 of its reciprocal are equal . the number is : | "let the number be x . then , 2 / 3 x = 25 / 216 * 1 / x x 2 = 25 / 216 * 3 / 2 = 25 / 144 x = 5 / 12 answer : a" | a = 25 * 3
b = 216 * 2
c = a / b
d = math.sqrt(c)
|
a ) 2.5 % , b ) 11 % , c ) 5 % , d ) 15 % , e ) 25 % | b | multiply(divide(subtract(50, 40), add(50, 40)), const_100) | if 50 % of ( x - y ) = 40 % of ( x + y ) then what percent of x is y ? | "50 % of ( x - y ) = 40 % of ( x + y ) ( 50 / 100 ) ( x - y ) = ( 40 / 100 ) ( x + y ) 5 ( x - y ) = 4 ( x + y ) x = 9 y x = 9 y therefore required percentage = ( ( y / x ) x 100 ) % = ( ( y / 9 y ) x 100 ) = 11 % answer is b ." | a = 50 - 40
b = 50 + 40
c = a / b
d = c * 100
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 7 | e | inverse(add(divide(subtract(35, 25), subtract(60, 35)), const_1)) | a certain quantity of 60 % solution is replaced with 25 % solution such that the new concentration is 35 % . what is the fraction of the solution that was replaced ? | "let ' s say that the total original mixture a is 100 ml the original mixture a thus has 60 ml of alcohol out of 100 ml of solution you want to replace some of that original mixture a with another mixture b that contains 25 ml of alcohol per 100 ml . thus , the difference between 60 ml and 25 ml is 35 ml per 100 ml of mixture . this means that every time you replace 100 ml of the original mixture a by 100 ml of mixture b , the original alcohol concentration will decrease by 35 % . the question says that the new mixture , let ' s call it c , must be 35 % alcohol , a decrease of only 25 % . therefore , 25 out of 35 is 5 / 7 and e is the answer ." | a = 35 - 25
b = 60 - 35
c = a / b
d = c + 1
e = 1/(d)
|
a ) 12 , b ) 15 , c ) 18 , d ) 20 , e ) 25 | d | divide(divide(300, 5), 3) | two dogsled teams raced across a 300 mile course in wyoming . team a finished the course in 3 fewer hours than team t . if team a ' s average speed was 5 mph greater than team t ' s , what was team t ' s average mph ? | "this is a very specific format that has appeared in a handful of real gmat questions , and you may wish to learn to recognize it : here we have a * fixed * distance , and we are given the difference between the times and speeds of two things that have traveled that distance . this is one of the very small number of question formats where backsolving is typically easier than solving directly , since the direct approach normally produces a quadratic equation . say team t ' s speed was s . then team t ' s time is 300 / s . team a ' s speed was then s + 5 , and team a ' s time was then 300 / ( s + 5 ) . we need to find an answer choice for s so that the time of team a is 3 less than the time of team t . that is , we need an answer choice so that 300 / ( s + 5 ) = ( 300 / s ) - 3 . you can now immediately use number properties to zero in on promising answer choices : the times in these questions will always work out to be integers , and we need to divide 300 by s , and by s + 5 . so we want an answer choice s which is a factor of 300 , and for which s + 5 is also a factor of 300 . so you can rule out answers a and c immediately , since s + 5 wo n ' t be a divisor of 300 in those cases ( sometimes using number properties you get to the correct answer without doing any other work , but unfortunately that ' s not the case here ) . testing the other answer choices , if you try answer d , you find the time for team t is 15 hours , and for team a is 12 hours , and since these differ by 3 , as desired , d is correct ." | a = 300 / 5
b = a / 3
|
a ) 4 mph , b ) 2.5 mph , c ) 3 mph , d ) 2 mph , e ) none of these | d | divide(subtract(sqrt(add(multiply(power(10, const_2), const_4), power(multiply(divide(36, divide(90, const_60)), const_2), const_2))), multiply(divide(36, divide(90, const_60)), const_2)), const_2) | a boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream . if the speed of the boat in still water is 10 mph , the speed of the stream is : | explanation : speed of the boat in still water = 10 mph let speed of the stream be x mph then , speed downstream = ( 10 + x ) mph speed upstream = ( 10 - x ) mph time taken to travel 36 miles upstream - time taken to travel 36 miles downstream = 90 / 60 hours = > 36 / ( 10 β x ) β 36 / ( 10 + x ) = 3 / 2 = > 12 / ( 10 β x ) β 12 / ( 10 + x ) = 1 / 2 = > 24 ( 10 + x ) β 24 ( 10 β x ) = ( 10 + x ) ( 10 β x ) = > 240 + 24 x β 240 + 24 x = ( 100 β x 2 ) = > 48 x = 100 β x 2 = > x 2 + 48 x β 100 = 0 = > ( x + 50 ) ( x β 2 ) = 0 = > x = - 50 or 2 . answer : option d | a = 10 ** 2
b = a * 4
c = 90 / const_60
d = 36 / c
e = d * 2
f = e ** 2
g = b + f
h = math.sqrt(g)
i = 90 / const_60
j = 36 / i
k = j * 2
l = h - k
m = l / 2
|
a ) 75.4 feet , b ) 98.4 feet , c ) 95.4 feet , d ) 85.4 feet , e ) 92.4 feet | c | subtract(rectangle_perimeter(25, divide(880, 25)), 25) | a rectangular field has to be fenced on three sides leaving a side of 25 feet uncovered . if the area of the field is 880 sq . feet , how many feet of fencing will be required ? | "area of the field = 880 sq . feet . length of the adjacent sides are 25 feet and 880 / 25 = 35.2 feet . required length of the fencing = 25 + 35.2 + 35.2 = 95.4 feet answer : c" | a = 880 / 25
b = rectangle_perimeter - (
|
a ) 2.25 , b ) 3.25 , c ) 4.25 , d ) 5.25 , e ) 6.25 | c | subtract(power(2.5, 2), 2) | x + ( 1 / x ) = 2.5 find x ^ 2 + ( 1 / x ^ 2 ) | "squaring on both sides ( x + 1 / x ) ^ 2 = 2.5 ^ 2 x ^ 2 + 1 / x ^ 2 = 6.25 - 2 x ^ 2 + 1 / x ^ 2 = 4.25 answer : c" | a = 2 ** 5
b = a - 2
|
a ) 1 / 4 , b ) 1 / 3 , c ) 5 / 13 , d ) 5 / 12 , e ) 1 / 2 | d | add(multiply(divide(3, 6), divide(3, 6)), multiply(divide(2, 6), divide(const_3, 6))) | a = { 0 , 1 , - 3 , 6 , - 8 , - 10 } b = { - 1 , 2 , - 4 , 7 , 6 , - 9 } if a is a number that is randomly selected from set a , and b is a number that is randomly selected from set b , what is the probability that ab < 0 ? | for the product of 2 numbers to be negative either of them must be positive or negative : p ( positive , negative ) = 2 / 6 * 3 / 6 = 6 / 36 ; p ( negative , positive ) = 3 / 6 * 3 / 6 = 9 / 36 p = 6 / 36 + 9 / 36 p = 15 / 36 p = 5 / 12 answer : d | a = 3 / 6
b = 3 / 6
c = a * b
d = 2 / 6
e = 3 / 6
f = d * e
g = c + f
|
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