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 ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | multiply(const_60, divide(50, multiply(const_100, const_10))) | two printing presses begin printing currency at the same time and at constant speeds . press f produces 5 - dollar bills at the rate of 1,000 bills per minute . press t produces 20 - dollar bills at the rate of 200 bills per minute . once the machines have begun printing , how many seconds does it take for press f to produce 50 dollars more currency than press t ? | press f produces $ 5 bills @ 1000 per minute ( 60 sec ) so $ 500 in 6 secs press t produces $ 20 bills 200 per minute ( 60 sec ) so $ 400 in 6 secs . . so we can see f produces $ 100 in 6 secs or will print $ 50 in 3 secs . . ans 3 secs . . b | a = 100 * 10
b = 50 / a
c = const_60 * b
|
a ) 140 % , b ) 152 % , c ) 165 % , d ) 176 % , e ) 190 % | c | multiply(divide(multiply(divide(9, const_100), add(const_100, 10)), divide(multiply(6, const_100), const_100)), const_100) | last year sandy saved 6 % of her annual salary . this year , she made 10 % more money than last year , and she saved 9 % of her salary . the amount saved this year was what percent of the amount she saved last year ? | let last year ' s salary be x . last year , sandy save 0.06 x this year , sandy saved 0.09 * 1.1 x = 0.099 x 0.099 x / 0.06 x = 99 / 60 = 1.65 = 165 % the answer is c . | a = 9 / 100
b = 100 + 10
c = a * b
d = 6 * 100
e = d / 100
f = c / e
g = f * 100
|
a ) $ 68 , b ) $ 40 , c ) $ 60 , d ) $ 100 , e ) $ 90 | a | multiply(divide(2, add(3, 2)), 170) | rahul can do a work in 3 days while rajesh can do the same work in 2 days . both of them finish the work together and get $ 170 . what is the share of rahul ? | "rahul ' s wages : rajesh ' s wages = 1 / 3 : 1 / 2 = 2 : 3 rahul ' s share = 170 * 2 / 5 = $ 68 answer is a" | a = 3 + 2
b = 2 / a
c = b * 170
|
a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 3 : 8 , e ) none | c | divide(add(multiply(6, const_2), 6), add(add(multiply(6, const_2), 6), 6)) | the ratio between sumit ' s and prakash ' s age at present is 2 : 3 . sumit is 6 years younger than prakash . the ratio of sumit ' s age to prakash ' s age after 6 years will be : | explanation : let their ages be 2 x and 3 x years . 3 x - 2 x = 6 or x = 6 sumit ' s age = 12 years , prakash ' s age = 24 years ratio of their ages = 18 : 24 = 3 : 4 . correct option : c | a = 6 * 2
b = a + 6
c = 6 * 2
d = c + 6
e = d + 6
f = b / e
|
a ) 50 km , b ) 100 km , c ) 60 km , d ) 70 km , e ) 80 km | b | multiply(10, divide(20, subtract(12, 10))) | if a person walks at 12 km / hr instead of 10 km / hr , he would have walked 20 km more . the actual distance traveled by him is : | "let the actual distance travelled be x km . x / 10 = ( x + 20 ) / 12 12 x = 10 x + 200 2 x = 200 x = 100 km . answer : b" | a = 12 - 10
b = 20 / a
c = 10 * b
|
a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30 | e | add(multiply(6, const_2), multiply(6, divide(subtract(45, multiply(6, const_2)), add(6, 5)))) | cara leaves her home and walks toward don ' s house . two hours later , don leaves his home and walks toward cara ' s house . the distance between their homes is 45 kilometers , cara ' s walking speed is 6 km / h , and don ' s walking speed is 5 km / h . how many kilometers will cara walk before she meets don ? | cara walks 12 km in the first two hours so there are 33 km remaining . when don starts walking , they complete a total of 11 km per hour . they will meet three hours after don starts walking . since cara walks for 5 hours , she walks 30 km . the answer is e . | a = 6 * 2
b = 6 * 2
c = 45 - b
d = 6 + 5
e = c / d
f = 6 * e
g = a + f
|
a ) 1 / 8 , b ) 1 / 6 , c ) 1 / 4 , d ) 3 / 4 , e ) 7 / 8 | a | multiply(subtract(const_1, divide(3, 4)), divide(1, 2)) | in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 3 / 4 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and curved | good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 3 / 4 of the garden is green , so , since all the flowers must be either green or yellow , we know that 1 / 4 are yellow . we ' re also told there is an equal probability of straight or curved , 1 / 2 . we want to find out the probability of something being yellow and straight , pr ( yellow and straight ) . so if we recall , the probability of two unique events occurring simultaneously is the product of the two probabilities , pr ( a and b ) = p ( a ) * p ( b ) . so we multiply the two probabilities , pr ( yellow ) * pr ( curved ) = 1 / 4 * 1 / 2 = 1 / 8 , or a . | a = 3 / 4
b = 1 - a
c = 1 / 2
d = b * c
|
a ) 105 , b ) 180 , c ) 210 , d ) 240 , e ) 315 | a | divide(divide(36, subtract(multiply(divide(const_4.0, 21), divide(5, 5)), multiply(divide(const_2.0, 3), divide(10, 7)))), 1) | 10 / 21 of 3 / 5 of a number is greater than 4 / 7 of 1 / 5 of the same number by 36 . what is half of that number ? | "let no . be x 10 / 21 * 3 / 5 * x - 4 / 7 * 1 / 5 * x = 8 by further solving 30 x / 105 - 4 x / 35 = 36 18 x / 105 = 36 x = 210 we have to find x / 2 = 210 / 2 = 105 answer : a" | a = 4 / 0
b = 5 / 5
c = a * b
d = 2 / 0
e = 10 / 7
f = d * e
g = c - f
h = 36 / g
i = h / 1
|
a ) 233 , b ) 552 , c ) 376 , d ) 287 , e ) 166 | b | multiply(add(35, add(35, multiply(subtract(12, const_1), 2))), divide(12, 2)) | the speed of a bus increases by 2 km after every one hour . if the distance travelling in the first one hour was 35 km . what was the total distance travelled in 12 hours ? | given that distance travelled in 1 st hour = 35 km and speed of the bus increases by 2 km after every one hour hence distance travelled in 2 nd hour = 37 km hence distance travelled in 3 rd hour = 39 km . . . total distance travelled = [ 35 + 37 + 39 + . . . ( 12 terms ) ] this is an arithmetic progression ( ap ) with first term , a = 35 , number of terms , n = 12 and common difference , d = 2 . the sequence a , ( a + d ) , ( a + 2 d ) , ( a + 3 d ) , ( a + 4 d ) , . . . is called an arithmetic progression ( ap ) where a is the first term and d is the common difference of the ap sum of the first n terms of an arithmetic progression ( ap ) , sn = n 2 [ 2 a + ( n − 1 ) d ] where n = number of terms hence , [ 35 + 37 + 39 + . . . ( 12 terms ) ] = s 12 = 122 [ 2 × 35 + ( 12 − 1 ) 2 ] = 6 [ 70 + 22 ] = 6 × 92 = 552 hence the total distance travelled = 552 km answer : b | a = 12 - 1
b = a * 2
c = 35 + b
d = 35 + c
e = 12 / 2
f = d * e
|
a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30 | c | divide(add(add(multiply(const_2, 11), multiply(const_2, const_3.0)), sqrt(add(multiply(11, subtract(84, multiply(multiply(const_2, 7), multiply(const_2, 11)))), power(add(multiply(const_2, 11), multiply(const_2, 7)), const_2)))), const_2) | a tailor trims 11 feet from opposite edges of a square piece of cloth , and 7 feet from the other two edges . if 84 square feet of cloth remain , what was the length of a side of the original piece of cloth ? | "let the original side of the square be x . ( x - 22 ) * ( x - 14 ) = 84 = 6 * 14 x = 28 the answer is c ." | a = 2 * 11
b = 2 * 3
c = a + b
d = 2 * 7
e = 2 * 11
f = d * e
g = 84 - f
h = 11 * g
i = 2 * 11
j = 2 * 7
k = i + j
l = k ** 2
m = h + l
n = math.sqrt(m)
o = c + n
p = o / 2
|
a ) rs . 4082.40 , b ) rs . 1024.21 , c ) rs . 2810.6 , d ) rs . 3214 , e ) none of these | a | multiply(multiply(multiply(multiply(6, 1.44), divide(add(const_100, 40), const_100)), multiply(6, divide(add(const_100, 25), const_100))), 45) | an order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth . what be the cost of a carpet whose length and breadth are 40 % more and 25 % more respectively than the first carpet . given that the ratio of carpet is rs . 45 per sq . m ? | explanation : length of the first carpet = ( 1.44 ) ( 6 ) = 8.64 cm area of the second carpet = 8.64 ( 1 + 40 / 100 ) 6 ( 1 + 25 / 100 ) = 51.84 ( 1.4 ) ( 5 / 4 ) sq m = ( 12.96 ) ( 7 ) sq m cost of the second carpet = ( 45 ) ( 12.96 x 7 ) = 315 ( 13 - 0.04 ) = 4095 - 12.6 = rs . 4082.40 answer is a | a = 6 * 1
b = 100 + 40
c = b / 100
d = a * c
e = 100 + 25
f = e / 100
g = 6 * f
h = d * g
i = h * 45
|
a ) 13.67 % , b ) 14.67 % , c ) 15.67 % , d ) 16.67 % , e ) 17.67 % | d | subtract(divide(multiply(const_100, 7), subtract(7, 1)), const_100) | in a office work is distribute between p persons . if 1 / 7 members are absent then work increased for each person is ? | "let total % of work is 100 % total person = p 1 / 7 person are absent of total person . so absent person is 1 / 7 p ie p / 7 . left person is , p - p / 7 = 6 p / 7 . p person do the work 100 % 1 person do the work 100 * p % 6 p / 7 person do the work ( 100 * p * 7 ) / 6 p % = 116.67 % work increased for each person is = ( 116.67 - 100 ) % = 16.67 % answer : d" | a = 100 * 7
b = 7 - 1
c = a / b
d = c - 100
|
a ) 22877 , b ) 27778 , c ) 20000 , d ) 27999 , e ) 17799 | c | divide(multiply(multiply(25, const_100), multiply(16, const_100)), multiply(20, 10)) | a courtyard is 25 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 = ( 2500 × 1600 / 20 × 10 ) = 20000 answer : c" | a = 25 * 100
b = 16 * 100
c = a * b
d = 20 * 10
e = c / d
|
a ) 173 , b ) 126 , c ) 153 , d ) 143 , e ) 133 | b | divide(multiply(700, subtract(const_100, add(add(44, 28), 10))), const_100) | in a school of 700 boys , 44 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ? | "44 + 28 + 10 = 82 % 100 – 82 = 18 % 700 * 18 / 100 = 126 answer : b" | a = 44 + 28
b = a + 10
c = 100 - b
d = 700 * c
e = d / 100
|
a ) 2 % , b ) 2.5 % , c ) 3.7 % , d ) 4.6 % , e ) 5 % | b | subtract(subtract(add(const_100, 30), divide(multiply(add(const_100, 30), 20), const_100)), const_100) | a shop keeper marked 30 % above the cost price and offered 20 % discount then find it ' s net profit ? | "net profit = 30 - 20 + ( 30 * ( - 25 ) / 100 ) = 2.5 % answer is b" | a = 100 + 30
b = 100 + 30
c = b * 20
d = c / 100
e = a - d
f = e - 100
|
a ) 12 , b ) 75 , c ) 88 , d ) 54 , e ) 90 | e | divide(add(100, 80), const_2) | the speed of a car is 100 km in the first hour and 80 km in the second hour . what is the average speed of the car ? | "s = ( 100 + 80 ) / 2 = 90 kmph answer : e" | a = 100 + 80
b = a / 2
|
a ) a ) 78 , b ) b ) 82 , c ) c ) 95 , d ) d ) 91 , e ) e ) 85 | c | divide(add(subtract(multiply(100, 10), 60), 10), 10) | the average marks of 10 students in a class is 100 . but a student mark is wrongly noted as 60 instead of 10 then find the correct average marks ? | "correct avg marks = 100 + ( 10 - 60 ) / 10 avg = 100 - 5 = 95 answer is c" | a = 100 * 10
b = a - 60
c = b + 10
d = c / 10
|
a ) 600 , b ) 900 , c ) 1200 , d ) 1350 , e ) 1500 | b | divide(subtract(multiply(300, 50), multiply(300, 20)), subtract(20, 10)) | there are 300 seniors at morse high school , and 50 % of them have cars . of the remaining grades ( freshmen , sophomores , and juniors ) , only 10 % of them have cars . if 20 % of all the students at morse have cars , how many students are in the 3 lower grades ? | let x be the number of students in the lower three grades . 0.1 x + 150 = 0.2 ( x + 300 ) 0.1 x = 900 x = 900 the answer is b . | a = 300 * 50
b = 300 * 20
c = a - b
d = 20 - 10
e = c / d
|
a ) $ 150 , b ) $ 248.75 , c ) $ 199 , d ) $ 171.6 , e ) $ 190 | c | floor(multiply(20, 9.95)) | carrie likes to buy t - shirts at the local clothing store . they cost $ 9.95 each . one day , she bought 20 t - shirts . how much money did she spend ? | $ 9.95 * 20 = $ 199 . answer is c . | a = 20 * 9
b = math.floor(a)
|
a ) 229 , b ) 240 , c ) 400 , d ) 277 , e ) 221 | c | multiply(20, multiply(90, const_0_2778)) | a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 90 km / hr . what is the length of the platform ? | "speed = 90 * 5 / 18 = 25 m / sec . length of the train = 25 * 20 = 500 m . let the length of the platform be x m . then , ( x + 500 ) / 36 = 25 = > x = 400 m . answer : c" | a = 90 * const_0_2778
b = 20 * a
|
a ) a ) 40 , b ) b ) 60 , c ) c ) 80 , d ) d ) 120 , e ) e ) 140 | b | multiply(divide(160, 8), const_3) | in a mixed college 160 students are there in one class . out of this 160 students 5 / 8 students are girls . how many boys are there ? | "total number of students : 160 total girls : 160 * 5 / 8 = 100 total boys : 160 - 100 = 60 answer is b" | a = 160 / 8
b = a * 3
|
a ) 23 , b ) 41 , c ) 48 , d ) 90 , e ) 86 | d | divide(multiply(multiply(85, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100)) | a ferry can transport 85 tons of vehicles . automobiles range in weight from 1,800 to 3,200 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ? | "to get maximum vehicles we must take into consideration the minimum weight i . e 1800 pounds here since , 1 ton = 2000 pounds 85 tons will be 170,000 pounds from the answer choices : let max number of vehicles be 86 total weight will be = 90 * 1800 = 162000 pounds , which is lesser than the maximum weight allowed . ans : d" | a = 85 * 2
b = a * 1000
c = 1000 + 100
d = c + 100
e = d + 100
f = e + 100
g = f + 100
h = g + 100
i = b / h
|
a ) 14 % , b ) 25 % , c ) 28 % , d ) 34 % , e ) 50 % | b | multiply(divide(add(multiply(divide(10, const_100), 50), 10), add(50, 10)), const_100) | if 10 gallons of grape juice are added to 50 gallons of a mixture , which contains 10 percent grape juice then what percent of the resulting mixture is grape juice ? | "official solution : if we start with 40 gallons of a mixture that is 10 % grape juice , then we have : 50 × 0.10 = 5 gallons of grape juice . 50 × 0.90 = 45 gallons of other components . if we add 10 gallons of grape juice , we will end up with 15 gallons of grape juice and 45 gallons of other components , and we will have a total of 60 gallons of the mixture . so 15 / 60 of the new mixture is grape juice . now we convert this to a percent : percent grape juice = 25 / 100 = 25 % . the correct answer is choice ( b )" | a = 10 / 100
b = a * 50
c = b + 10
d = 50 + 10
e = c / d
f = e * 100
|
a ) 40 % , b ) 44 % , c ) 50 % , d ) 56 % , e ) 60 % | d | multiply(subtract(const_1, divide(add(multiply(divide(40, const_100), 4), divide(60, const_100)), add(4, const_1))), const_100) | an artist who needs to buy only paint and canvas finds that the cost of canvas has decreased by 40 percent and the cost of paint has decreased by 60 percent . if paint previously accounted for 4 times as much of the expense of painting as did canvas , by what percent did the total cost for paint and canvas decrease ? | paint : canvas cost is 4 : 1 . so , paint accounts for 80 % of cost and canvas accounts for 20 % . canvas , after decrease by 40 % will be ( 100 - 40 ) % of 20 % = > 60 % of 20 % = > 12 % of original paint , after decrease by 60 % will be ( 100 - 60 ) % of 80 % = > 40 % of 80 % = > 32 % of original new cost as a % of old cost = 32 + 12 = 44 % of original . so , overall cost saving = 100 - 44 = 56 % therefore , ( d ) | a = 40 / 100
b = a * 4
c = 60 / 100
d = b + c
e = 4 + 1
f = d / e
g = 1 - f
h = g * 100
|
a ) 27 , b ) 70 , c ) 60 , d ) 80 , e ) 24 | c | divide(add(55, 65), const_2) | a man can row upstream at 55 kmph and downstream at 65 kmph , and then find the speed of the man in still water ? | "us = 55 ds = 65 m = ( 55 + 65 ) / 2 = 60 answer : c" | a = 55 + 65
b = a / 2
|
a ) s 510 , b ) s 780 , c ) s 880 , d ) s 480 , e ) s 980 | a | divide(multiply(subtract(multiply(110, 65), multiply(subtract(110, multiply(2.5, const_2)), subtract(65, multiply(2.5, const_2)))), 60), const_100) | a rectangular grassy plot 110 m . by 65 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 60 paise per sq . metre | "area of the plot = 110 m * 65 m = 7150 sq . m area of plot excluding gravel = 105 m * 60 m = 6300 sq . m area of gravel = 7150 sq . m - 6300 sq . m = 850 sq . m cost of building it = 850 sq . m * 60 = 51000 p in rs = 51000 / 100 = rs 510 answer : a" | a = 110 * 65
b = 2 * 5
c = 110 - b
d = 2 * 5
e = 65 - d
f = c * e
g = a - f
h = g * 60
i = h / 100
|
a ) 20 seconds , b ) 30 seconds , c ) 40 seconds , d ) 50 seconds , e ) none of these | c | divide(add(360, 140), divide(multiply(45, const_1000), const_3600)) | a train is 360 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length | "explanation : speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 ∗ 2 / 25 = 40 seconds option c" | a = 360 + 140
b = 45 * 1000
c = b / 3600
d = a / c
|
a ) 10 , b ) 27 , c ) 87 , d ) 23 , e ) 93 | a | divide(multiply(20, 10), subtract(40, 20)) | in some quantity of ghee , 60 % is pure ghee and 40 % is vanaspati . if 10 kg of pure ghee is added , then the strength of vanaspati ghee becomes 20 % . the original quantity was : | "explanation : let the original quantity be x kg . vanaspati ghee in x kg = ( 40 x / 100 ) kg = ( 2 x / 5 ) kg . now , ( 2 x / 5 ) / ( x + 10 ) = 20 / 100 < = > 2 x / ( 5 x + 50 ) = 1 / 5 < = > 5 x = 50 < = > x = 10 . answer : a ) 10 kg" | a = 20 * 10
b = 40 - 20
c = a / b
|
a ) 944.9755 , b ) 944.9655 , c ) 944.9565 , d ) 946.9565 , e ) 944.9546 | e | divide(926, divide(9.026, 926)) | evaluate 926 + 9.026 + 0.926 + 9.0026 | "926 + 9.026 + 0.926 + 9.0026 = 944.9546 option e" | a = 9 / 26
b = 926 / a
|
a ) 1 , b ) 2 , c ) 3 , d ) 7 , e ) 9 | a | add(add(const_4, const_3), const_2) | what is the units digit of the expression 17 ^ 7 - 2 ? | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 1 ( last digit ) 7 ^ 5 = 7 ( last digit ) and the cycle repeats after every 4 powers therefore , last digit of 17 ^ 7 = 3 3 - 2 = 1 answer a" | a = 4 + 3
b = a + 2
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | add(add(divide(30, 3), floor(divide(30, power(3, const_2)))), floor(divide(30, power(3, 3)))) | if v is the product of the integers from 1 to 30 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of v ? | v = 30 ! 8 v = 30 x 29 x 28 x 27 x 26 x 25 x 24 x 24 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 09 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 out of these 30 , 27 , 24 , 21 , 18 , 15 , 12 , 09 , 06 , 3 are factors of 3 3 x 10 , 3 x 3 x 3 , 3 x 8 , 3 x 3 x 2 , 3 x 5 , 3 x 4 , 3 x 3 x 3 , 3 x 2 , 3 so we have a total of 14 three ' s . . . therefore the maximum value of k can be 14 ( c ) | a = 30 / 3
b = 3 ** 2
c = 30 / b
d = math.floor(c)
e = a + d
f = 3 ** 3
g = 30 / f
h = math.floor(g)
i = e + h
|
a ) 25 , 20 , b ) 30 , 15 , c ) 50 , 15 , d ) none of these , e ) 75,35 | c | multiply(add(const_2, const_3), multiply(2, const_3)) | find the side of the largest possible square slabs which can be paved on the floor of a room 2 m 50 cm long and 1 m 50 cm broad . also find the number of such slabs to pave the floor . | explanation : hcf ( 250 , 150 ) = 50 cm the number slabs = 250 * 150 / 50 * 50 = 15 answer : c | a = 2 + 3
b = 2 * 3
c = a * b
|
a ) 1200 m , b ) 2000 m , c ) 1500 m , d ) 1000 m , e ) 4000 m | d | multiply(200, subtract(const_2, const_1)) | a train speeds past a pole in 25 seconds and a platform 200 m long in 30 seconds . its length is : | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 25 = > y = x / 25 x + 200 / 30 = x / 25 x = 1000 m . answer : option d" | a = 2 - 1
b = 200 * a
|
['a ) 1 : 9', 'b ) 2 : 9', 'c ) 3 : 9', 'd ) 5 : 9', 'e ) none of them'] | a | power(power(divide(1, 27), inverse(const_3)), const_2) | two cubes have their volumes in the ratio 1 : 27 . find the ratio of their surface areas . | let their edges be a and b . then , = a ^ 3 / b ^ 3 = 1 / 27 ( or ) ( a / b ) ^ 3 = ( 1 / 3 ) 3 ( or ) ( a / b ) = ( 1 / 3 ) . therefore , ratio of their surface area = 6 a ^ 2 / 6 b ^ 2 = a ^ 2 / b ^ 2 = ( a / b ) ^ 2 = 1 / 9 , i . e . 1 : 9 . answer is a . | a = 1 / 27
b = 1/(3)
c = a ** b
d = c ** 2
|
a ) 24 % , b ) 34 % , c ) 22 % , d ) 18 % , e ) 8.5 % | c | multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(14, const_100), const_4), divide(14, const_100))), const_100) | one fourth of a solution that was 14 % 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 * 0.14 + 25 * x = 100 * 0.16 - - > x = 0.22 . answer : c ." | a = 16 / 100
b = a * 4
c = 14 / 100
d = c * 4
e = 14 / 100
f = d - e
g = b - f
h = g * 100
|
a ) 1 / 5 , b ) 1 / 7 , c ) 1 / 10 , d ) 1 / 12 , e ) 1 / 15 | c | divide(1, 18) | a certain fraction has the same ratio to 1 / 18 , as 2 / 5 does to 2 / 9 . what is this certain fraction ? | "x / ( 1 / 18 ) = ( 2 / 5 ) / ( 2 / 9 ) x = 2 * 9 * 1 / 18 * 5 * 2 = 1 / 10 the answer is c ." | a = 1 / 18
|
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | e | add(10, const_4) | how many odd numbers between 10 and 2000 are the squares of integers ? | "the numbers are the squares of 5 , 7 , 9 , . . . , 43 which includes 20 numbers . the answer is e ." | a = 10 + 4
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a ) 74 , b ) 76 , c ) 78 , d ) 80 , e ) 82 | a | divide(add(multiply(58, 3), multiply(98, 2)), add(2, 3)) | for a certain exam , a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean . what was the mean score e for the exam ? | "a score of 58 was 2 standard deviations below the mean - - > 58 = mean - 2 d a score of 98 was 3 standard deviations above the mean - - > 98 = mean + 3 d solving above for mean e = 74 . answer : a ." | a = 58 * 3
b = 98 * 2
c = a + b
d = 2 + 3
e = c / d
|
a ) 45 min , b ) 50 min , c ) 40 min , d ) 48 min , e ) 54 min | e | multiply(divide(multiply(3, add(3, divide(20, const_60))), 9), const_60) | walking at the rate of 3 kmph a man cover certain distance in 3 hr 20 min . running at a speed of 9 kmph the man will cover the same distance in . | "distance = speed * time 3 * 10 / 3 = 10 km new speed = 9 kmph therefore time = d / s = 10 / 9 = 54 min answer : e ." | a = 20 / const_60
b = 3 + a
c = 3 * b
d = c / 9
e = d * const_60
|
a ) 0.5 % , b ) 1 % , c ) 1.5 % , d ) 2.5 % , e ) 3 % | d | multiply(divide(25, 1), const_100) | what percent is 25 gm of 1 kg ? | "1 kg = 1000 gm 25 / 1000 × 100 = 2500 / 1000 = 5 / 2 = 2.5 % d )" | a = 25 / 1
b = a * 100
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a ) 800 , b ) 600 , c ) 1250 , d ) 950 , e ) 1000 | c | multiply(multiply(50, 5), 5) | an airplane covers 50 miles in 1 / 5 hours . how many miles can the airplane cover in 5 hours ? | divide 5 by 1 / 5 = 5 ÷ 1 / 5 = 5 / 1 ÷ 1 / 5 = 5 / 1 * 5 / 1 = 25 . then , multiply 50 by 25 to get 1250 in 5 hours , the airplane will cover 1250 miles answer is c . | a = 50 * 5
b = a * 5
|
a ) 400 , b ) 500 , c ) 505 , d ) none of these , e ) 506 | b | divide(const_100.0, divide(07, 35)) | evaluate 35 / . 07 | "explanation : 35 / . 07 = 3500 / 7 = 500 option b" | a = 7 / 35
b = 100 / 0
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a ) 60 % , b ) 26 % , c ) 30 % , d ) 56 % , e ) 73 % | b | subtract(divide(multiply(add(const_100, 19.7), const_100), subtract(const_100, 5)), const_100) | a shopkeeper sold an article offering a discount of 5 % and earned a profit of 19.7 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 19.70 let marked price be rs . x . then , 95 / 100 x = 119.70 x = 11970 / 95 = rs . 126 now , s . p . = rs . 126 , c . p . = rs . 100 profit % = 26 % . answer : b" | a = 100 + 19
b = a * 100
c = 100 - 5
d = b / c
e = d - 100
|
a ) 0.0006 , b ) 0.006 , c ) 0.06 , d ) 0.6 , e ) 6.0 | a | multiply(divide(0.003, 0.2), const_100) | 0.003 x 0.2 = ? | "3 x 2 = 6 . sum of decimal places = 4 0.003 x 0.2 = 0.0006 answer : option a" | a = 0 / 3
b = a * 100
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a ) 100 marks , b ) 240 marks , c ) 280 marks , d ) 371 marks , e ) 827 marks | b | add(multiply(divide(add(60, 30), subtract(divide(45, const_100), divide(30, const_100))), divide(30, const_100)), 60) | a candidate who gets 30 % of the marks fails by 60 marks . but another candidate who gets 45 % marks gets 30 marks more than necessary for passing . find the number of marks for passing ? | "30 % - - - - - - - - - - - - 60 45 % - - - - - - - - - - - - 30 - - - - - - - - - - - - - - - - - - - - - - 15 % - - - - - - - - - - - - - 90 30 % - - - - - - - - - - - - - - 180 180 + 60 = 240 marks answer : b" | a = 60 + 30
b = 45 / 100
c = 30 / 100
d = b - c
e = a / d
f = 30 / 100
g = e * f
h = g + 60
|
a ) 160 π , b ) 320 π , c ) 400 π , d ) 540 π , e ) 640 π | b | multiply(power(multiply(sqrt(80), const_2), const_2), const_pi) | in may , the groundskeeper at spring lake golf club built a circular green with an area of 80 π square feet . in august , the groundskeeper doubled the distance from the center of the green to the edge of the green . what is the total area of the renovated green ? | "area = π r ^ 2 , so doubling the radius results in an area that is 4 times the original area . 4 ( 80 π ) = 320 π the answer is b ." | a = math.sqrt(80)
b = a * 2
c = b ** 2
d = c * math.pi
|
['a ) 2 pie', 'b ) 11 pie', 'c ) 178 pie', 'd ) 711 pie', 'e ) 1280 pie'] | c | floor(divide(volume_cylinder(divide(160, const_2), multiply(subtract(divide(divide(80, const_2), divide(80, 16)), divide(80, 16)), 16)), power(add(const_10, const_2), const_3))) | 80 pie square inches of material of negligible thickness are required to construct a 1 / 16 scale model of a cylindrical barrel . if the diameter of the base of the barrel is 160 inches , then what is the volume of the barrel , to the nearest cubic foot ? | radius of barrel , r = 80 radius of model , r = 80 / 16 = 5 surface area of the model = 2 ∗ pi ∗ r ( r + h ) = 80 ∗ pi which gives h = 3 and h = 3 ∗ 16 = 48 so , volume of the barrel = pi ∗ r 2 ∗ h = pi ∗ ( 80 ) 2 ∗ ( 48 ) in cubic inches converting in cubic ft results : volume = pi ∗ ( 80 ) 2 ∗ ( 48 ) / ( 12 ) 3 = 178 pi cubic ft approximately answer [ c ] | a = 160 / 2
b = 80 / 2
c = 80 / 16
d = b / c
e = 80 / 16
f = d - e
g = f * 16
h = volume_cylinder / (
i = 10 + 2
j = i ** 3
k = math.floor(h, j)
|
a ) 85 kg , b ) 90 kg , c ) 80 kg , d ) 75 kg , e ) 70 kg | c | add(multiply(20, 2), 40) | the average weight of 20 girls increases by 2 kg when a new girl comes in place of one of them weighing 40 kg . what might be the weight of the new girl ? | "total weight increased = 20 x 2 kg = 40 kg . weight of new person = 40 + 40 kg = 80 kg answer : c" | a = 20 * 2
b = a + 40
|
a ) 3 . , b ) 4 . , c ) 5 . , d ) 6 . , e ) 9 | a | power(subtract(power(5, 2), 2), divide(2, 5)) | a ( 5 , w ^ 3 ) is the ( x , y ) coordinate of point located on the parabola y = x ^ 2 + 2 . what is the value of w ? | "y = x ^ 2 + 2 x = 5 y = 5 ^ 2 + 2 = 27 y is in the form w ^ 3 w ^ 3 = 27 w = 3 answer : a" | a = 5 ** 2
b = a - 2
c = 2 / 5
d = b ** c
|
a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | d | subtract(119, subtract(add(144, 119), 232)) | in a graduating class of 232 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number t and the smallest possible number of students that could have taken both geometry and biology ? | "official solution : first of all , notice that since 144 took geometry and 119 took biology , then the number of students who took both geometry and biology can not be greater than 119 . { total } = { geometry } + { biology } - { both } + { neither } ; 232 = 144 + 119 - { both } + { neither } ; { both } = 31 + { neither } . { both } is minimized when { neither } is 0 . in this case { both } = 31 . the greatest possible number t of students that could have taken both geometry and biology , is 119 . thus , the answer is 119 - 31 = 88 . answer : d ." | a = 144 + 119
b = a - 232
c = 119 - b
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | add(divide(subtract(6.5, multiply(0.8, const_2)), 0.7), const_2) | a certain fruit stand sold apples for $ 0.80 each and bananas for $ 0.70 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.50 , what total number of apples and bananas did the customer purchase ? | let ' s start with 1 apple for $ 0.80 . let ' s subtract $ 0.80 from $ 6.50 until we get a multiple of $ 0.70 . $ 6.50 , $ 5.70 , $ 4.90 = 7 * $ 0.70 the customer purchased 7 bananas and 2 apples . the answer is b . | a = 0 * 8
b = 6 - 5
c = b / 0
d = c + 2
|
a ) 0 , b ) 1 , c ) 2 , d ) 5 , e ) 6 | c | floor(multiply(const_100, divide(62, 5000))) | what is the thousandths digit in the decimal equivalent of 62 / 5000 ? | "62 / 5000 = 62 / ( 5 * 10 ^ 3 ) = ( 62 / 5 ) * 10 ^ - 3 = 12.4 * 10 ^ - 3 = . 0124 thousandths digit = 2 answer c" | a = 62 / 5000
b = 100 * a
c = math.floor(b)
|
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | c | divide(180, multiply(const_0_2778, 36)) | in how many seconds will a train 180 meters long pass an oak tree , if the speed of the train is 36 km / hr ? | "speed = 36 * 5 / 18 = 10 m / s time = 180 / 10 = 18 seconds the answer is c ." | a = const_0_2778 * 36
b = 180 / a
|
a ) 19 , b ) 21 , c ) 25 , d ) 22 , e ) 20 | d | add(add(add(const_2.0, 10), 10), 10) | what number should replace the question mark ? 10 , 30 , 14 , 25 , 18 , 20 , - - - ? | "answer : d 10 , 30 , 14 , 25 , 18 , 20 , 22 ? there are two alternate sequences : + 4 and - 5 ." | a = 2 + 0
b = a + 10
c = b + 10
|
a ) 33 , b ) 77.4 , c ) 25 , d ) 22 , e ) 72 | b | multiply(const_3_6, divide(add(250, 180), 20)) | a train of length 250 m crosses a bridge of length 180 m in 20 seconds . what is the speed of train ? | "sol : ( length of train + length of bridge ) = speed of train x time ( 250 + 180 ) = 20 x speed speed = 430 / 20 = 21.5 m / s = 77.4 km / h answer = b" | a = 250 + 180
b = a / 20
c = const_3_6 * b
|
a ) 2 days , b ) 4 days , c ) 5 days , d ) 6 days , e ) 9 days | d | divide(12, subtract(const_3, const_1)) | a can do a work in 12 days . when he had worked for 3 days , b joinedhim . if they complete the work in 3 more days , in how manydays can balone finish the work ? | sax work done by afar 3 days : i j . . remzming war — 1 . work done by ( a + b ) for 1 day : . work done by a for 1 day 6 days d | a = 3 - 1
b = 12 / a
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | divide(log(divide(100, 0.0010101)), log(10)) | if k is an integer and 0.0010101 x 10 ^ k is greater than 100 , what is the least possible value of k ? | "0.0010101 * 10 ^ k > 100 we need to move the decimal point to the right 5 places to get 101.01 this is equivalent to multiplying by 10 ^ 5 . the answer is d ." | a = 100 / 0
b = math.log(a)
c = math.log(10)
d = b / c
|
a ) 1 : 2 , b ) 3 : 2 , c ) 9 : 1 , d ) 5 : 2 , e ) 6 : 5 | c | divide(subtract(18, 9), subtract(19, 18)) | gold is 19 times as heavy as water and copper is 9 times as heavy as water . in what ratio should these be mixed to get an alloy 18 times as heavy as water ? | "g = 19 w c = 9 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 19 w + 9 wx = x + 1 * 18 w 19 + 9 x = 18 ( x + 1 ) x = 1 / 9 ratio of gold with copper = 1 : 1 / 9 = 9 : 1 answer is c" | a = 18 - 9
b = 19 - 18
c = a / b
|
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | e | subtract(power(subtract(73, multiply(add(const_3, const_4), const_10)), subtract(350, multiply(floor(divide(350, const_4)), const_4))), multiply(const_2, const_10)) | find the ones digit of 73 ^ 350 | "3,9 , 7,1 repeats after every four times therefore 350 / 4 = 87 + 2 remainder 9 is the ones digit so e is correct" | a = 3 + 4
b = a * 10
c = 73 - b
d = 350 / 4
e = math.floor(d)
f = e * 4
g = 350 - f
h = c ** g
i = 2 * 10
j = h - i
|
a ) rs . 90 , b ) rs . 100 , c ) rs . 150 , d ) rs . 190 , e ) rs . 200 | b | multiply(divide(subtract(multiply(1, 900), multiply(multiply(const_3, const_4), 9)), multiply(multiply(const_3, const_4), const_1)), const_4) | a man engaged a servant on the condition that he would pay him rs . 900 and a uniform after 1 year service . he served onlyfor 9 months and received uniform & rs . 650 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b" | a = 1 * 900
b = 3 * 4
c = b * 9
d = a - c
e = 3 * 4
f = e * 1
g = d / f
h = g * 4
|
a ) 9 : 8 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 10 : 9 | e | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(2, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 2 . the ratio of economy cars to sport utility vehicles is 5 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 2 - - > e : l = 3 : 2 = 15 : 10 . the ratio of economy cars to sport utility vehicles is 5 : 3 - - > e : s = 5 : 3 = 15 : 9 . thus , l : s = 10 : 9 . answer : e ." | a = 4 * 3
b = 3 * 3
c = a / b
d = 3 * 4
e = 2 * 4
f = d / e
g = c / f
|
a ) - 4 , b ) 4 , c ) - 2 , d ) 1 / 2 , e ) - 2 | e | divide(const_1, 0.5) | in the coordinate plane a slope of the line k is 0.5 times the y - intercept of the line k . what is the x - intercept of the line k ? | as y = 0.5 mx + m , from 0 = 1 / 2 mx + m we get x = - 2 . hence , the correct answer choice is e . | a = 1 / 0
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a ) 65 km , b ) 85 km , c ) 55 km , d ) 72 km , e ) 40 km | d | divide(7, add(divide(const_1, add(21, 3)), divide(const_1, subtract(21, 3)))) | a motorboat can go at 21 kmph and still water . it takes 7 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 3 kmph ? | let the distance between a and b be x km . total time = x / ( 21 + 3 ) + x / ( 21 - 3 ) = 7 = > x / 24 + x / 18 = 7 = > ( 4 x + 3 x ) / 72 = 7 = > x = 72 km . answer : d | a = 21 + 3
b = 1 / a
c = 21 - 3
d = 1 / c
e = b + d
f = 7 / e
|
a ) 450 , b ) 810 , c ) 900 , d ) 1000 , e ) 1100 | c | multiply(divide(multiply(9, add(9, 1)), 2), multiply(2, const_10)) | the sum of all the digits of the integers from 18 to 21 inclusive is 24 ( 1 + 8 + 1 + 9 + 2 + 0 + 2 + 1 = 24 ) . what is the sum r of all the digits of the integers from 0 to 99 inclusive ? | "we want the sum of the digits from 0 to 99 , so i approximated : 0 - 9 - > 45 - > ( 9 + 0 ) * 10 / 2 40 - 49 - > 85 ( 13 + 4 ) * 10 / 2 90 - 99 - > 135 ( 18 + 9 ) * 10 / 2 we can see at a glance that theweightgoes up as the numbers go up ( meaning the difference between 85 and 45 is 40 , while 135 - 85 is 50 , this means that the second part of this sequence carries more weight for our result ) , so we know that the final answer has to be more than 850 ( 85 * 10 ) but close to it , and that ' s just r = 900 : the answer is c ." | a = 9 + 1
b = 9 * a
c = b / 2
d = 2 * 10
e = c * d
|
a ) 3 / 8 , b ) 4 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 3 / 4 | b | divide(subtract(add(divide(divide(add(add(multiply(const_100, multiply(const_2, add(const_1, const_4))), multiply(const_4, const_100)), multiply(const_4, multiply(const_2, add(const_1, const_4)))), 20.000), 8), 8), 4), 20.000) | a total of $ 20,000 was invested in two certificates of deposit at simple annual interest rates of 4 percent and 8 percent , respectively . if the total interest on the two certificates was $ 1,440 at the end of one year , what fractional part of the $ 20.000 was invested at the higher rate ? | "x * 4 / 100 * 1 + ( 20000 - x ) * 8 / 100 * 1 = 1440 4 x - 8 x = 144000 - 160000 = > - 4 x = - 16000 = > x = 4000 / 4 so 16000 / 20000 = 4 / 5 answer - b" | a = 1 + 4
b = 2 * a
c = 100 * b
d = 4 * 100
e = c + d
f = 1 + 4
g = 2 * f
h = 4 * g
i = e + h
j = i / 20
k = j / 8
l = k + 8
m = l - 4
n = m / 20
|
a ) 30 , b ) 58 , c ) 60 , d ) 90 , e ) 120 | b | multiply(6, inverse(subtract(1, add(divide(1, 5), divide(2, 3))))) | in traveling from a dormitory to a certain city , a student went 1 / 5 of the way by foot , 2 / 3 of the way by bus , and the remaining 6 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "i believe there is a better way to do it . basically one of the options should satisfy the given criteria . 60 did 1 / 5 * 60 = 12 2 / 3 * 60 = 40 so total distance 52 + remaining 6 = 58 answer b" | a = 1 / 5
b = 2 / 3
c = a + b
d = 1 - c
e = 1/(d)
f = 6 * e
|
a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 12 | b | divide(9, subtract(96.12, floor(96.12))) | when positive integer x is divided by positive integer y , the remainder is 9 . if x / y = 96.12 , what is the value of y ? | "x = qy + 9 , where q is the quotient x / y = ( q y + 9 ) / y = q + ( 9 / y ) = 96.12 = 96 + 0.12 = > q = 96 9 / y = 0.12 = > y = 75 answer is b ." | a = math.floor(96, 12)
b = 96 - 12
c = 9 / b
|
a ) 20 , b ) 18 , c ) 16 , d ) 15 , e ) 15.39 | e | divide(subtract(sqrt(add(multiply(multiply(15, 5), const_4), power(15, const_2))), 15), const_2) | tom read a book containing 1000 pages by reading the same number of pages each day . if he would have finished the book 5 days earlier by reading 15 pages a day more , how many days did tom spend reading the book ? | "actually u can set up 2 equation p - - stands for the pages d - - stands for the days 1 ) p * d = 480 ( we want to find the days , sop = 480 / d ) 2 ) ( p + 15 ) ( d - 5 ) = 1000 = > pd - 5 p + 15 d - 75 = 1000 as the 1 ) stated u can put 1 ) into 2 ) = > 1000 - 5 p + 15 d - 75 = 1000 = > 15 d - 5 p = 75 put the bold one into it = > 15 d - 5 ( 480 / d ) = 75 the we get the final equation 16 d ^ 2 - 2400 = 80 d ( divide 16 ) = > d ^ 2 - 5 d - 160 = 0 so d = 15.39 days . ans : ( e )" | a = 15 * 5
b = a * 4
c = 15 ** 2
d = b + c
e = math.sqrt(d)
f = e - 15
g = f / 2
|
a ) 2532 , b ) 2552 , c ) 2524 , d ) 2522 , e ) 2512 | d | subtract(multiply(16,000, power(add(const_1, divide(9, const_100)), 20)), 16,000) | find the compound interest on $ 16,000 at 20 % per annum for 9 months , compounded quarterly | "principal = $ 16000 ; time = 9 months = 3 quarters ; rate = 20 % per annum = 5 % per quarter . amount = $ [ 16000 x ( 1 + ( 5 / 100 ) ) ^ 3 ] = $ 18522 . ci . = $ ( 18522 - 16000 ) = $ 2522 . answer d ." | a = 9 / 100
b = 1 + a
c = b ** 20
d = 16 * 0
e = d - 16
|
['a ) 12', 'b ) 42', 'c ) 24', 'd ) 28', 'e ) 26'] | c | subtract(multiply(16, 21), multiply(24, subtract(16, const_3))) | the number of students in each section of a school is 24 . after admitting new students , three new sections were started . now , the total number of sections is 16 and there are 21 students in each section . the number of new students admitted is : | original number of sections = 16 - 3 = 13 original number of students = 24 x 13 = 312 present number of students = 21 x 16 = 336 number of new students admitted = 336 - 312 = 24 so the answer is option c ) 24 . | a = 16 * 21
b = 16 - 3
c = 24 * b
d = a - c
|
a ) 7 km / hr , b ) 7.2 km / hr , c ) 5.2 km / hr , d ) 10 km / hr , e ) 5 km / hr | b | divide(divide(600, const_1000), divide(5, const_60)) | a man crosses a 600 m long street in 5 minutes . what is his speed in km per hour ? | speed = 600 / 5 * 60 = 2 m / sec converting m / sec to km / hr . = 2 * 18 / 5 = 7.2 km / hr answer is b | a = 600 / 1000
b = 5 / const_60
c = a / b
|
a ) 300 , b ) 240 , c ) 220 , d ) 200 , e ) 180 | e | multiply(divide(15, const_2), 24) | if the sum of the 4 th term and the 12 th term of an arithmetic progression is 24 , what is the sum of the first 15 terms of the progression ? | "4 th term + 12 th term = 24 i . e . , ( a + 3 d ) + ( a + 11 d ) = 24 now , sum of first 15 terms = ( 15 / 2 ) * [ 2 a + ( 15 - 1 ) d ] = ( 15 / 2 ) * [ 2 a + 14 d ] = ( 15 / 2 ) * 24 - - - - - - - - - - - - - - - from ( 1 ) = 180 answer : e" | a = 15 / 2
b = a * 24
|
a ) 1236 , b ) 3528 , c ) 4096 , d ) 4608 , e ) 6561 | b | multiply(multiply(add(const_4, const_3), add(const_4, const_3)), multiply(add(4, 4), multiply(3, 3))) | how many 4 - digit positive integers are there , where each digit is positive , and no 3 adjacent digits are same ? | first digit . . 9 posibilities second digit , 8 possibilities third digit , 7 possibilities fourth digit , 7 possibilities . 9 * 8 * 7 * 7 = 3528 . b | a = 4 + 3
b = 4 + 3
c = a * b
d = 4 + 4
e = 3 * 3
f = d * e
g = c * f
|
a ) a : 45 , b ) b : 25 , c ) c : 37.5 , d ) d : 36 , e ) e : 42 | c | divide(15, subtract(const_1, sqrt(divide(9, add(9, 16))))) | 15 lts are taken of from a container full of liquid a and replaced with liquid b . again 15 more lts of the mixture is taken and replaced with liquid b . after this process , if the container contains liquid a and b in the ratio 9 : 16 , what is the capacity of the container w ? | "if you have a 37.5 liter capacity , you start with 37.5 l of a and 0 l of b . 1 st replacement after the first replacement you have 37.5 - 15 = 22.5 l of a and 15 l of b . the key is figuring out how many liters of a and b , respectively , are contained in the next 15 liters of mixture to be removed . the current ratio of a to total mixture is 22.5 / 37.5 ; expressed as a fraction this becomes ( 45 / 2 ) / ( 75 / 2 ) , or 45 / 2 * 2 / 75 . canceling the 2 s and factoring out a 5 leaves the ratio as 9 / 15 . note , no need to reduce further as we ' re trying to figure out the amount of a and b in 15 l of solution . 9 / 15 of a means there must be 6 / 15 of b . multiply each respective ratio by 15 to get 9 l of a and 6 l of b in the next 15 l removal . final replacement the next 15 l removal means 9 liters of a and 6 liters of b are removed and replaced with 15 liters of b . 22.5 - 9 = 13.5 liters of a . 15 liters of b - 6 liters + 15 more liters = 24 liters of b . test to the see if the final ratio = 9 / 16 ; 13.5 / 24 = ( 27 / 2 ) * ( 1 / 24 ) = 9 / 16 . choice c is correct ." | a = 9 + 16
b = 9 / a
c = math.sqrt(b)
d = 1 - c
e = 15 / d
|
a ) a ) 0 , b ) b ) 3 , c ) c ) 4 , d ) d ) 6 , e ) e ) 7 | a | divide(multiply(45, 30), 15) | a number when divided by 45 , gives 30 as quotient and 0 as remainder . what will be the remainder when dividing the same number by 15 | explanation : p ÷ 45 = 30 = > p = 30 * 45 = 1350 p / 15 = 1350 / 15 = 90 , remainder = 0 answer : option a | a = 45 * 30
b = a / 15
|
a ) $ 12.20 , b ) $ 12.50 , c ) $ 12.55 , d ) $ 12.70 , e ) $ 13.00 | a | add(divide(add(10, 14), const_2), add(const_0_25, const_0_25)) | a vendor buys 10 t - shirts at an average price of $ 14 per t - shirt . he then buys 15 more t - shirts at an average price of $ 11 per t - shirt . what is the average price s per t - shirt that the vendor paid for these purchases ? | "correct answer : a explanation : the relevant formula for this problem is average s = ( sum ) / ( number of terms ) . another way to look at the formula is sum = average x number of terms . for the first purchase , the vendor ' s sum ( total cost ) was $ 140 , since 14 x 10 = 140 . for the second purchase , the vendor ' s cost was $ 165 , since 11 x 15 = 165 . the grand sum is then $ 140 + $ 165 , which equals $ 305 . the total number of shirts purchased was 25 , so to get the average price per shirt , we divide 305 by 25 , which equals $ 12.20 . as a result , the correct answer is a . note : a relative understanding of weighted average offers a shortcut to this problem . because the true average of 11 and 14 is 12.5 , but the vendor sells more shirts at the lower price than at the higher price , the weighted average must be less than $ 12.50 ; only answer choice a is a possibility ." | a = 10 + 14
b = a / 2
c = const_0_25 + const_0_25
d = b + c
|
a ) 61.90 % , b ) 52.20 % , c ) 56.20 % , d ) 70.45 % , e ) 74.41 % | a | multiply(const_100, divide(divide(multiply(add(30, const_100), 50), const_100), add(const_100, 5))) | of the families in city x in 1904 , 50 percent owned a personal computer . the number of families in city x owning a computer in 1908 was 30 percent greater than it was in 1904 , and the total number of families in city x was 5 percent greater in 1908 than it was in 1904 . what percent of the families in city x owned a personal computer in 1908 ? | "say a 100 families existed in 1904 then the number of families owning a computer in 1904 - 50 number of families owning computer in 1908 = 50 * 130 / 100 = 65 number of families in 1908 = 105 the percentage = 65 / 105 * 100 = 61.90 % . option : a" | a = 30 + 100
b = a * 50
c = b / 100
d = 100 + 5
e = c / d
f = 100 * e
|
a ) 21 % , b ) 25 % , c ) 69 % , d ) 31 % , e ) 12 % | e | subtract(subtract(add(40, const_100), divide(multiply(add(40, const_100), 20), const_100)), const_100) | a merchant marks his goods up by 40 % and then offers a discount of 20 % on the marked price . what % profit does the merchant make after the discount ? | "let the price be 100 . the price becomes 140 after a 40 % markup . now a discount of 20 % on 140 . profit = 112 - 100 12 % answer e" | a = 40 + 100
b = 40 + 100
c = b * 20
d = c / 100
e = a - d
f = e - 100
|
a ) 16 a , b ) 13 a , c ) 14 a , d ) 11 2 / 3 a , e ) 12 a | d | floor(divide(70, add(4, const_1))) | during a certain two - week period , 70 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 4 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedies , in terms of a , were rented during that two - week period ? | "total movies = 100 . comedies = 70 . action + drama = 30 . since there were 5 times as many dramas as action movies , then action + 4 * action = 30 - - > action = a = 6 . comedies = 70 = 14 a . answer : d ." | a = 4 + 1
b = 70 / a
c = math.floor(b)
|
a ) 776 , b ) 1044 , c ) 299 , d ) 257 , e ) 125 | b | divide(multiply(divide(288, divide(subtract(58, subtract(const_100, 58)), const_100)), 58), const_100) | there were two candidates in an election . winner candidate received 58 % of votes and won the election by 288 votes . find the number of votes casted to the winning candidate ? | "w = 58 % l = 42 % 58 % - 42 % = 16 % 16 % - - - - - - - - 288 58 % - - - - - - - - ? = > 1044 answer : b" | a = 100 - 58
b = 58 - a
c = b / 100
d = 288 / c
e = d * 58
f = e / 100
|
a ) 305 , b ) 315 , c ) 345 , d ) 325 , e ) 335 | b | divide(divide(8, subtract(multiply(divide(4, 15), divide(5, 7)), multiply(divide(2, 5), divide(4, 9)))), 2) | 4 / 15 of 5 / 7 of a number is greater than 4 / 9 of 2 / 5 of the same number by 8 . what is half of that number ? | let the number be x . then 4 / 15 of 5 / 7 of x - 4 / 9 of 2 / 5 of x = 8 4 / 21 x - 8 / 45 x = 8 ( 4 / 21 - 8 / 45 ) x = 8 ( 60 - 56 ) / 315 x = 8 4 / 315 x = 8 x = ( 8 * 315 ) / 4 = 630 1 / 2 x = 315 hence required number = 315 answer is b . | a = 4 / 15
b = 5 / 7
c = a * b
d = 2 / 5
e = 4 / 9
f = d * e
g = c - f
h = 8 / g
i = h / 2
|
a ) 7 % , b ) 13 % , c ) 14 % , d ) 15 % , e ) 16 % | a | divide(multiply(subtract(multiply(20, add(5, 3)), add(multiply(5, 18), multiply(3, 20))), const_100), add(multiply(5, 18), multiply(3, 20))) | a producer of tea blends two varieties of tea from two tea gardens one costing rs 18 per kg and another rs 20 per kg in the ratio 5 : 3 . if he sells the blended variety at rs 20 per kg , then his gain percent is | "explanation : suppose he bought 5 kg and 3 kg of tea . cost price = rs . ( 5 x 18 + 3 x 20 ) = rs . 150 . selling price = rs . ( 8 x 20 ) = rs . 160 . profit = 160 - 150 = 10 so , profit % = ( 10 / 150 ) * 100 = 7 % option a" | a = 5 + 3
b = 20 * a
c = 5 * 18
d = 3 * 20
e = c + d
f = b - e
g = f * 100
h = 5 * 18
i = 3 * 20
j = h + i
k = g / j
|
a ) 33 , b ) 32 , c ) 31 , d ) 30 , e ) 29 | a | add(divide(subtract(subtract(200, const_2), add(100, const_2)), 3), const_1) | what is the total number of y integers between 100 and 200 that are divisible by 3 ? | yes there is a different way of arriving at that answer . . . . u can also use airthmetic progression to get the answer since the first term to be divisble by 3 is 102 . . take that as a . . the starting no and since 198 is the last digit to be divisible by 3 take that as n . . . since the difference is 3 take that as d no u have to find what term is 198 take that as nth term the formula for that is n = a + ( n - 1 ) * d 198 = 102 + ( n - 1 ) * 3 from this u get n = 33 | a = 200 - 2
b = 100 + 2
c = a - b
d = c / 3
e = d + 1
|
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | c | divide(add(sqrt(add(multiply(multiply(136, const_2), const_4), const_1)), const_1), const_2) | if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 136 games will be played during the tournament . find the number of participants . | let p be the number of participants . pc 2 = 136 ( p ) ( p - 1 ) = 272 = 17 * 16 p = 17 the answer is c . | a = 136 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
|
a ) 2 , b ) 4 , c ) 5 , d ) 67 , e ) 2 6 / 17 | e | inverse(add(inverse(20), add(inverse(4), inverse(8)))) | a , b and c can do a piece of work in 4 days , 8 days and 20 days respectively . how long will they taken , if all the three work together ? | "1 / 4 + 1 / 8 + 1 / 20 = 17 / 40 = > 40 / 17 = > 2 6 / 17 days answer : e" | a = 1/(20)
b = 1/(4)
c = 1/(8)
d = b + c
e = a + d
f = 1/(e)
|
a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 7 | d | subtract(divide(subtract(90, 40), subtract(45, 40)), const_1) | for the past n days , the average ( arithmetic mean ) daily production at a company was 40 units . if today ' s production of 90 units raises the average to 45 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 40 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 40 n + 90 = 45 * ( n + 1 ) - - > n = 9 answer : d ." | a = 90 - 40
b = 45 - 40
c = a / b
d = c - 1
|
a ) $ 900 , b ) $ 720 , c ) $ 675 , d ) $ 450 , e ) $ 225 | b | multiply(divide(1800, add(add(5, 4), 1)), 4) | in the johnsons ' monthly budget , the dollar amounts allocated to household expenses , food , and miscellaneous items are in the ratio 5 : 4 : 1 , respectively . if the total amount allocated to these 3 categories is $ 1800 , what is the amount allocated to food ? | to solve this question , it will be best to first express the given ratio using variable multipliers . thus , we have : household expenses : food : miscellaneous = 5 x : 4 x : x since we are told that the total amount allocated to these categories is $ 1,800 we can set up the equation : 5 x + 4 x + x = 1,800 10 x = 1,800 x = 180 thus , the total amount allocated to food is 4 x 180 = $ 720 . answer b . | a = 5 + 4
b = a + 1
c = 1800 / b
d = c * 4
|
a ) 2080 , b ) 1700 , c ) 2350 , d ) 2500 , e ) 8000 | a | divide(multiply(multiply(13, 2600), 16), add(multiply(16, 16), 4)) | one ton has 2600 pounds , and one pound has 16 ounces . how many packets containing wheat weighing 16 pounds and 4 ounces each would totally fill a gunny bag of capacity 13 tons ? | "16 pounds and 4 ounces = 16 * 16 + 4 = 260 ounces . 13 tons = 13 * 2600 pound = 13 * 2600 * 16 ounces . hence the answer is ( 13 * 2600 * 16 ) / 260 = 2080 . answer : a ." | a = 13 * 2600
b = a * 16
c = 16 * 16
d = c + 4
e = b / d
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(divide(5, const_2), multiply(8, 8)) | what is the remainder when 8 ^ 381 is divided by 5 ? | i also agree that the remainder is ' 3 ' ( using the last digit of the powers of 7 ) . could we have the official answer please ? d | a = 5 / 2
b = 8 * 8
c = a - b
|
a ) 7 : 3 , b ) 2 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 2 : 5 | b | divide(subtract(15.8, 15.4), subtract(16.4, 15.8)) | the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.4 years , the ratio of the number of boys to the number of girls in the class is | "explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.4 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.4 ) < = > k = 0.4 / 0.6 = 2 / 3 . required ratio = 2 / 3 : 1 = 2 : 3 . answer : b" | a = 15 - 8
b = 16 - 4
c = a / b
|
a ) 70 , b ) 245 , c ) 150 , d ) 35 , e ) 250 | e | divide(35, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100))) | in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 35 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ? | "since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 35 , so 0.14 * { total } = 35 - - > { total } = 250 . answer : e ." | a = 100 - 30
b = a / 100
c = 20 / 100
d = b * c
e = 35 / d
|
a ) 5 / 9 , b ) 3 / 15 , c ) 23 / 30 , d ) 22 / 30 , e ) 53 / 90 | d | divide(add(multiply(multiply(5, 2), 2), 2), multiply(2, multiply(3, 5))) | of the female students at barkely university , 5 / 6 are on the honor roll . of the male students , 2 / 3 are on the honor roll . if 2 / 5 of the students are female , what fraction of all the students are on the honor roll ? | let the total students be 30 given 2 / 5 of the students are females = 12 then males = 3 / 5 = 18 5 / 6 of the females are on honor roll = 10 males on the honor roll = 2 / 3 = 12 total students on honor roll = 12 + 10 = 22 fraction = 22 / 30 d | a = 5 * 2
b = a * 2
c = b + 2
d = 3 * 5
e = 2 * d
f = c / e
|
a ) 150 , b ) 278 , c ) 179 , d ) 300 , e ) 191 | d | multiply(divide(multiply(60, const_1000), const_3600), 18) | a train running at the speed of 60 km / hr crosses a pole in 18 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 ) * 18 = 300 meter answer : d" | a = 60 * 1000
b = a / 3600
c = b * 18
|
['a ) 432 sq m', 'b ) 363 sq m', 'c ) 452 sq m', 'd ) 428 sq m', 'e ) 525 sq m'] | b | multiply(multiply(divide(88, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(88, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | the length of rectangle is thrice its breadth and its perimeter is 88 m , find the area of the rectangle ? | 2 ( 3 x + x ) = 88 l = 33 b = 11 lb = 33 * 11 = 363 answer : b | a = 3 * 2
b = 1 * 2
c = a + b
d = 88 / c
e = d * 3
f = 3 * 2
g = 1 * 2
h = f + g
i = 88 / h
j = e * i
|
a ) 81 : 169 , b ) 81 : 122 , c ) 81 : 124 , d ) 81 : 126 , e ) 81 : 129 | a | power(divide(729, 2197), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 729 : 2197 . what is the ratio of their total surface areas ? | "ratio of the sides = 3 √ 729 : 3 √ 2197 = 9 : 13 ratio of surface areas = 9 ^ 2 : 13 ^ 2 = 81 : 169 answer : option a" | a = 729 / 2197
b = 1 / 3
c = a ** b
|
a ) 40 th minute , b ) 41 st minute , c ) 43 rd minute , d ) 42 nd minute , e ) 45 th minute | b | add(multiply(multiply(const_4, 2), 2), 1) | a monkey ascends a greased pole 22 meters high . he ascends 2 meters in the first minute and then slips down 1 meter in the alternate minute . if this pattern continues until he climbs the pole , in how many minutes would he reach at the top of the pole ? | "the money is climbing 1 meter in 2 min . this pattern will go on till he reaches 20 meters . i mean this will continue for first 20 * 2 = 40 mins . he would have reached 20 meters . after that he will climb 2 meters and he will reach the pole . so total time taken = 40 + 1 = 41 mins . so , asnwer will be b" | a = 4 * 2
b = a * 2
c = b + 1
|
a ) 32 metre , b ) 28 metre , c ) 23 metre , d ) 15 metre , e ) 28 metre | a | subtract(256, multiply(divide(256, 32), 28)) | a can run 256 metre in 28 seconds and b in 32 seconds . by what distance a beat b ? | "clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 256 / 32 = 8 m / s distance covered by b in 4 seconds = speed ã — time = 8 ã — 4 = 32 metre i . e . , a beat b by 32 metre answer is a" | a = 256 / 32
b = a * 28
c = 256 - b
|
a ) 1 , b ) 2 , c ) 1 and 1 / 2 , d ) - 1 / 2 , e ) - 1 | b | divide(2, subtract(4, const_1)) | if | x | = 4 x - 2 , then x = ? | "answer : approach : substituted option a i . e x = 2 . inequality satisfied . b" | a = 4 - 1
b = 2 / a
|
a ) 7 , b ) 14 , c ) 42 , d ) 49 , e ) 98 | d | power(7, const_2) | two circular signs are to be painted . if the diameter of the larger sign is 7 times that of the smaller sign , how many times more paint is needed to paint the larger sign ? ( we can assume that a given amount of paint covers the same area on both signs . ) | let r be the radius of the smaller sign . then the diameter of the smaller sign is 2 r , the diameter of the larger sign is 14 r , and the radius of the larger sign is 7 r . the area a of the smaller sign is a = pir ^ 2 . the area of the larger sign is pi ( 7 r ) ^ 2 = 49 pir ^ 2 = 49 a . since the area is 49 times larger , we need 49 times more paint for the larger sign . the answer is d . | a = 7 ** 2
|
a ) $ 1,250 , b ) $ 1,450 , c ) $ 1,650 , d ) $ 1,850 , e ) $ 2,050 | d | floor(divide(add(divide(76.50, divide(9, const_100)), 1,000), 1,000)) | when a merchant imported a certain item , he paid a 9 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 76.50 , what was the total value of the item ? | "let x be the value in excess of $ 1,000 . 0.09 x = 76.5 x = $ 850 the total value was $ 850 + $ 1,000 = $ 1,850 . the answer is d ." | a = 9 / 100
b = 76 / 50
c = b + 1
d = c / 1
e = math.floor(d)
|
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