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 ) 3 : 2 , b ) 2 : 1 , c ) 4 : 3 , d ) 3 : 5 , e ) 3 : 1 | b | add(subtract(add(1, 5), add(2, const_2.0)), const_1) | in what ratio p : q should the mixture p of milk and water in the ratio of 2 : 1 be mixed with another mixture q of milk and water in the ratio 1 : 5 so that the resultant mixture contains equal quantities of milk and water ? | "( 2 / 3 ) * p + ( 1 / 6 ) * q = ( 1 / 3 ) * p + ( 5 / 6 ) * q 2 p = 4 q p / q = 2 / 1 the answer is b ." | a = 1 + 5
b = 2 + 2
c = a - b
d = c + 1
|
a ) 1,108 , b ) 2,900 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 350) | at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 350 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 350 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 350 x 8 = 2800 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 2900 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 2900 ( answer b )" | a = 200 / 22
b = a * 350
|
a ) 1 , b ) 10 , c ) 19 , d ) 40 , e ) 21 | d | subtract(multiply(divide(80, 2), add(divide(80, 2), 1)), multiply(divide(add(79, 1), 2), add(divide(subtract(79, 1), 2), 1))) | ifaequals the sum of the even integers from 2 to 80 , inclusive , andbequals the sum of the odd integers from 1 to 79 , inclusive , what is the value of a - b ? | "this is a solution from beatthegmat : even numbers : ( 80 - 2 ) / 2 + 1 = 40 even integers . ( 80 + 2 ) / 2 = 41 is the average of the even set . sum = avg * ( # of elements ) = 41 * 40 = 1640 = a odd numbers : ( 79 - 1 ) / 2 + 1 = 40 odd integers . ( 79 + 1 ) / 2 = 40 is the average of the odd set . sum = avg * ( # of elements ) = 40 * 40 = 1600 = b a - b = 1640 - 1600 = 40 . answer : d" | a = 80 / 2
b = 80 / 2
c = b + 1
d = a * c
e = 79 + 1
f = e / 2
g = 79 - 1
h = g / 2
i = h + 1
j = f * i
k = d - j
|
a ) 150 , b ) 155 , c ) 160 , d ) 165 , e ) none | a | divide(divide(46.5, const_100), divide(31, const_100)) | if 31 % of a number is 46.5 , the number is ? | "answer let 31 % of a = 46.5 ⇒ ( 31 x a ) / 100 = 46.5 ∴ a = ( 46.5 x 100 ) / 31 = 150 correct option : a" | a = 46 / 5
b = 31 / 100
c = a / b
|
a ) 256 , b ) 512 , c ) 768 , d ) 6144 , e ) 1536 | d | multiply(add(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(0.6, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), 0.6), const_10) | if an average hard drive had a capacity of 0.6 tb in 2000 , and average hard drive capacities double every 5 years , what will be the average hard drive capacity in 2050 ? | "0.6 * 2 ^ 10 = 0.6 * 1024 = 614.4 the answer is d ." | a = 0 * 6
b = a * 2
c = b * 2
d = c * 2
e = d * 2
f = e * 2
g = f * 2
h = g * 2
i = h * 2
j = i * 2
k = j + 0
l = k * 10
|
a ) 18 , b ) 16 , c ) 34 , d ) 25 , e ) 12 | a | subtract(multiply(5, const_4), const_2) | two digits in father ' s age are the same as the digit in the son ' s age , but in reverse order . in 5 years father will be twice as old as son will be then . what is the difference in their current age ? | father ' s age = 10 x + y so son ' s age = 10 y + x . . after 5 years , 10 x + y + 5 = 2 * ( 10 y + x + 5 ) . . . . so 5 + 19 y = 8 x . . . check for odd values of y , y = 1 satisfies the eqn with x = 3 . . . so ages are 31 and 13 and ans as found correctly by u is a . . 18 | a = 5 * 4
b = a - 2
|
a ) 187 m , b ) 278 m , c ) 300 m , d ) 150 m , e ) 267 m | c | 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 . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 18 ) m = 300 m . answer : c" | a = 60 * 1000
b = a / 3600
c = b * 18
|
a ) 2000 , b ) 3000 , c ) 4000 , d ) 5000 , e ) 6200 | c | subtract(multiply(const_2, 5050), divide(add(multiply(const_2, 5050), subtract(multiply(const_2, 6250), multiply(const_2, 5200))), const_2)) | the average monthly income of a and b is rs . 5050 . the average monthly income of b and c is rs . 6250 and the average monthly income of a and c is rs . 5200 . what is the monthly income of a ? | "explanation : let monthly income of a = a monthly income of b = b monthly income of c = c a + b = 2 × 5050 . . . . ( equation 1 ) b + c = 2 × 6250 . . . . ( equation 2 ) a + c = 2 × 5200 . . . . ( equation 3 ) ( equation 1 ) + ( equation 3 ) - ( equation 2 ) = > a + b + a + c - ( b + c ) = ( 2 × 5050 ) + ( 2 × 5200 ) - ( 2 × 6250 ) = > 2 a = 2 ( 5050 + 5200 - 6250 ) = > a = 4000 i . e . , monthly income of a = 4000 answer is c" | a = 2 * 5050
b = 2 * 5050
c = 2 * 6250
d = 2 * 5200
e = c - d
f = b + e
g = f / 2
h = a - g
|
a ) 24 , b ) 30 , c ) 36 , d ) 42 , e ) 48 | a | subtract(power(const_2.0, const_4), const_4) | x , y , and z are different prime numbers . the product x ^ 3 * y ^ 2 * z ^ 1 is divisible by how many different positive numbers ? | "the exponents of x ^ 3 * y ^ 2 * z ^ 1 are 3 , 2 , and 1 . the number of factors is ( 3 + 1 ) ( 2 + 1 ) ( 1 + 1 ) = 24 the answer is a ." | a = 2 ** 0
b = a - 4
|
a ) 21 years , b ) 22 years , c ) 23 years , d ) 24 years , e ) 25 years | d | divide(subtract(28, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 28 years older than his son . in four years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 28 . = > 2 ( x + 4 ) = ( x + 28 + 4 ) = > 2 x + 8 = x + 32 = > x = 24 years answer : option d" | a = 2 * 2
b = a - 2
c = 28 - b
d = 2 - 1
e = c / d
|
a ) 12 , b ) 18 , c ) 20 , d ) 22 , e ) 23 | e | multiply(sqrt(divide(divide(529, 3), const_3)), const_3) | the length of a rectangular floor is more than its breadth by 200 % . if rs . 529 is required to paint the floor at the rate of rs . 3 / sq m , what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 529 / 3 = 176.33 sq m l b = 176.33 i . e . , l * l / 3 = 176.33 l ^ 2 = 529 = > l = 23 e" | a = 529 / 3
b = a / 3
c = math.sqrt(b)
d = c * 3
|
a ) 25 m , b ) 45 m , c ) 40 m , d ) 50 m , e ) 62 m | c | multiply(5, 8) | walking at 5 / 6 th of its usual speed a cab is 8 mnts late . find its usual time to cover the journey ? | "new speed = 5 / 6 th of usual speed new time = 6 / 5 th of usual time 6 / 5 ut - ut = 8 m ut / 5 = 8 m ut = 40 m answer is c" | a = 5 * 8
|
a ) 4.05 days , b ) 5.05 days , c ) 14.76 days , d ) 8.05 days , e ) 15.05 days | b | inverse(divide(add(inverse(16), add(inverse(4), inverse(12))), const_2)) | a and b can do a piece of work in 4 days . b and c can do it in 12 days and a and c in 16 days . working together they will complete the work in how many days ? | "a + b = 1 / 4 b + c = 1 / 12 c + a = 1 / 16 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 ( a + b + c ) = 1 / 4 + 1 / 12 + 1 / 16 = 19 / 48 a + b + c = 19 / 96 96 / 19 = 5.05 days answer : b" | a = 1/(16)
b = 1/(4)
c = 1/(12)
d = b + c
e = a + d
f = e / 2
g = 1/(f)
|
a ) 7,000 , b ) 24,000 , c ) 74,880 , d ) 100,000 , e ) 168,000 | d | multiply(const_4, const_10) | a certain machine produces 780 units of product p per hour . working continuously at this constant rate , this machine will produce how many units of product p in 4 days ? | "since 7 days consist of 24 * 4 hours the total is 96 hours . since every hour the machine produces 780 units of product p the total product during 96 hours is 96 * 780 = 74,880 . correct option : d" | a = 4 * 10
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a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | d | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 39 ? | "first , let ' s translate the expression 21 # 39 , using the definition given : 21 # 39 = 21 × 24 × 27 × 30 × 33 × 36 × 39 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 39 = 3 ^ 7 ( 7 × 8 × 9 × 10 × 11 × 12 × 13 ) now let ' s replace each consecutive integer with its prime factorization : 21 # 39 = 3 ^ 7 ( 7 × 2 ^ 3 × 3 ^ 2 × ( 2 × 5 ) × 11 × ( 2 ^ 2 × 3 ) × 13 ) let ' s group the prime bases : 21 # 39 = 2 ^ 6 × 3 ^ 10 × 5 × 7 × 11 × 13 the sum of the exponents is 6 + 10 + 1 + 1 + 1 + 1 = 20 the answer is d ." | a = 2 + 2
b = 3 + a
c = 2 + 2
d = b + c
e = 2 + 2
f = 3 + e
g = d + f
|
a ) rs . 200 , b ) rs . 250 , c ) rs . 300 , d ) rs . 150 , e ) rs . 600 | e | multiply(multiply(3, const_4), divide(200, add(3, 1))) | the cost of 3 pens and 5 pencils is rs . 200 . also the cost of one pen and one pencil is in the ratio of 5 : 1 respectively . what is the cost of one dozen pens ? | "explanation : let the cost of one pen is ‘ 5 x ’ and pencil is ‘ x ’ 3 x 5 x + 5 x = rs . 200 15 x + 5 x = rs . 200 x = 200 / 20 = 10 : . cost of 1 pen = 5 x = 5 x 10 = 50 : . cost of 12 pens , i . e . ( one dozen ) = 50 x 12 = rs . 600 answer : option e" | a = 3 * 4
b = 3 + 1
c = 200 / b
d = a * c
|
a ) 20 , b ) 21 , c ) 23 , d ) 25 , e ) 26 | c | divide(subtract(add(25, add(25, 5)), multiply(const_3.0, const_3.0)), const_2) | the captain of a cricket team of 11 members is 25 years old and the wicket keeper is 5 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 25 + 30 ) = 9 ( x - 1 ) 11 x - 9 x = 46 2 x = 46 x = 23 . the average age of the team is 23 years . the answer is c ." | a = 25 + 5
b = 25 + a
c = 3 * 0
d = b - c
e = d / 2
|
a ) 48 % , b ) 54 % , c ) 60 % , d ) 64 % , e ) 68 % | e | multiply(divide(subtract(divide(72, const_100), divide(55, const_100)), add(subtract(divide(72, const_100), divide(55, const_100)), subtract(divide(80, const_100), divide(72, const_100)))), const_100) | solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 72 % percent carbonated water , what percent of the volume of the mixture is p ? | 72 % is 8 % - points below 80 % and 17 % - points above 55 % . so the ratio of solution p to solution q is 17 : 8 . mixture p is 17 / 25 = 68 % of the volume of mixture pq . the answer is e . | a = 72 / 100
b = 55 / 100
c = a - b
d = 72 / 100
e = 55 / 100
f = d - e
g = 80 / 100
h = 72 / 100
i = g - h
j = f + i
k = c / j
l = k * 100
|
a ) 10 , b ) 19 , c ) 20 , d ) 9 , e ) 12 | d | divide(3, 6) | there are 3 girls and 6 boys . how many kids are there ? | d | a = 3 / 6
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a ) 11 , b ) 32 , c ) 24 , d ) 48 , e ) 50 | c | divide(multiply(48, 8), 16) | two numbers n and 16 have lcm = 48 and gcf = 8 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 16 * n = 48 * 8 n = 48 * 8 / 16 = 24 correct answer c" | a = 48 * 8
b = a / 16
|
a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 80 % | d | multiply(const_100, add(divide(2, 5), divide(3, 10))) | if y > 0 , ( 2 y ) / 5 + ( 3 y ) / 10 is what percent of y ? | can be reduced to 2 y / 5 + 3 y / 10 = 2 y / 5 = 70 % d | a = 2 / 5
b = 3 / 10
c = a + b
d = 100 * c
|
a ) 512 cm , b ) 520 cm , c ) 508 cm , d ) 510 cm , e ) 554 cm | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 200), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 200 inches into centimeter ? | "1 inch = 2.54 cm 200 inches = 200 * 2.54 = 508 cm answer is c" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 200
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 30 , b ) 300 , c ) 720 , d ) 25200 , e ) 18000 | d | multiply(multiply(const_3, const_60), const_60) | if an object travels at seven feet per second , how many feet does it travel in one hour ? | "speed = 7 feet per second . 1 hour = 60 x 60 seconds = 3600 seconds . total no of feet traveled in 1 hour = 3600 x 7 = 25200 answer d" | a = 3 * const_60
b = a * const_60
|
a ) 2337 , b ) 1227 , c ) 3327 , d ) 5457 , e ) 7677 | a | subtract(2344, multiply(multiply(12, 3), 2)) | evaluate : 2344 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 2344 - 12 * 3 * 2 = 2344 - 8 = 2337 correct answer a" | a = 12 * 3
b = a * 2
c = 2344 - b
|
a ) 18 , b ) 21.6 , c ) 22 , d ) 22.5 , e ) 27 | d | multiply(multiply(15, divide(add(const_100, 20), const_100)), divide(add(const_100, 20), const_100)) | a distributor sells a product through an on - line store , which take a commission of 20 % of the price set by the distributor . the distributor obtains the product from a producer at the price of $ 15 per item . what is the price that the buyer observers on - line if the distributor wants to maintain a 20 % profit on the cost of the item ? | "producer price = $ 15 ; the distributor wants to maintain a 20 % profit on the cost of the item , thus he must get $ 15 * 1.2 = $ 18 after the store takes a commission of 20 % of the final price - - > ( final price ) * 0.8 = $ 18 - - > ( final price ) = $ 22.5 . answer : d ." | a = 100 + 20
b = a / 100
c = 15 * b
d = 100 + 20
e = d / 100
f = c * e
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a ) 45 % , b ) 55 % , c ) 35 % , d ) 40 % , e ) 44.44 % | e | multiply(subtract(const_1, divide(const_100, add(const_100, 80))), const_100) | the product of x and y is a constant . if the value of x is increased by 80 % , by what percentage must the value of y be decreased ? | "x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.8 x ( 180 ) i . e . 180 * new ' y ' = 10000 i . e . new ' y ' = 10000 / 180 = 55.55 i . e . y decreases from 100 to 55.55 i . e . decrease of 44.44 % answer : option e" | a = 100 + 80
b = 100 / a
c = 1 - b
d = c * 100
|
a ) 8.5 seconds , b ) 2.8 seconds , c ) 3.5 seconds , d ) 2.5 seconds , e ) 4 seconds | e | divide(100, multiply(90, const_0_2778)) | in what time will a train 100 meters long cross an electric pole , if its speed is 90 km / hr | "first convert speed into m / sec speed = 90 * ( 5 / 18 ) = 25 m / sec time = distance / speed = 100 / 25 = 4 seconds answer : e" | a = 90 * const_0_2778
b = 100 / a
|
a ) 38 . , b ) 40 . , c ) 66 . , d ) 44 . , e ) 46 . | c | subtract(multiply(sqrt(divide(847, 7)), 7), sqrt(divide(847, 7))) | the roof of an apartment building is rectangular and its length is 7 times longer than its width . if the area of the roof is 847 feet squared , what is the difference between the length and the width of the roof ? | "answer is c : 66 let w be the width , so length is 7 w . therefore : w * 7 w = 847 , solving for , w = 11 , so 7 w - w = 6 w = 6 * 11 = 66" | a = 847 / 7
b = math.sqrt(a)
c = b * 7
d = 847 / 7
e = math.sqrt(d)
f = c - e
|
a ) 621 , b ) 276 , c ) 236 , d ) 600 , e ) 1200 | e | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 3000) | a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 3000 . the difference between the profit shares of a and c is ? | "explanation : ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 3000 = > 5 parts out of 15 parts is rs . 3000 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 3000 ) = rs . 1200 answer : e" | a = 12000 / 10000
b = 8000 / 10000
c = a - b
d = c * 3000
|
a ) 11 , b ) 12 , c ) 9 , d ) 14 , e ) 15 | c | subtract(multiply(15, 15), add(multiply(4, 14), multiply(10, 16))) | the average age of 15 students of a class is 15 years . out of these , the average age of 4 students is 14 years and that of the other 10 students is 16 years , the age of the 15 th student is | "explanation : age of the 15 th student = [ 15 * 15 - ( 14 * 5 + 16 * 9 ) ] = 9 years . answer : c" | a = 15 * 15
b = 4 * 14
c = 10 * 16
d = b + c
e = a - d
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a ) 2996 sq m , b ) 4774 sq m , c ) 4312 sq m , d ) 2768 sq m , e ) 2786 sq m | b | subtract(circle_area(divide(704, multiply(const_2, const_pi))), circle_area(divide(660, multiply(const_2, const_pi)))) | the circumferences of two circles are 660 meters and 704 meters . find the difference between the areas of the larger and the smaller circles ? | "let the radii of the smaller and the larger circles be s m and l m respectively . 2 ∏ s = 660 and 2 ∏ l = 704 s = 660 / 2 ∏ and l = 704 / 2 ∏ difference between the areas = ∏ l ^ 2 - ∏ s ^ 2 = ∏ { 330 ^ 2 / ∏ ^ 2 - 352 ^ 2 / ∏ ^ 2 } = 330 ^ 2 / ∏ - 352 ^ 2 / ∏ = ( 330 - 352 ) ( 330 + 352 ) / ∏ = ( 22 ) ( 682 ) / ( 22 / 7 ) = 4774 sq m answer : b" | a = 2 * math.pi
b = 704 / a
c = circle_area - (
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a ) 9 , b ) 10 , c ) 11 , d ) 13 , e ) 15 | d | divide(multiply(add(divide(30, const_100), const_1), 23), add(add(divide(30, const_100), const_1), const_1)) | p is 30 % more efficient than q . p can complete a work in 23 days . if p and q work together , how much time will it take to complete the same work ? | "work done by p in 1 day = 1 / 23 let work done by q in 1 day = q q × ( 130 / 100 ) = 1 / 23 = > q = 100 / ( 23 × 130 ) = 10 / ( 23 × 13 ) work done by p and q in 1 day = 1 / 23 + 10 / ( 23 × 13 ) = 23 / ( 23 × 13 ) = 1 / 13 = > p and q together can do the work in 13 days option d" | a = 30 / 100
b = a + 1
c = b * 23
d = 30 / 100
e = d + 1
f = e + 1
g = c / f
|
['a ) 4', 'b ) 6', 'c ) 9', 'd ) 10', 'e ) 5'] | e | add(add(const_4, const_1), const_1) | suppose you have three identical prisms with congruent equilateral triangles as the end - polygons . suppose you attach them by the rectangular faces so they are perfectly aligned . there will be some large faces created by two or more co - planar faces of the individual prisms : count each such large face as one . given that , how many faces does the resultant solid have | to arrange them as described i did as follows on my scratch paper ( see the file attached ) , where i counted 4 sides and bottom , hence 4 + 1 = 5 . e | a = 4 + 1
b = a + 1
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a ) 24 , b ) 25 , c ) 26 , d ) 48 , e ) 50 | b | subtract(divide(add(50, 0), const_2), divide(add(100, 0), const_2)) | the average ( arithmetic mean ) of the even integers from 0 to 100 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 50 inclusive ? | "so , according to a mean of a set of even numbers from 0 to 100 = ( 0 + 100 ) / 2 = 50 and mean of a set of even numbers from 0 to 50 = ( 0 + 50 ) / 2 = 25 difference = 50 - 25 = 25 answer : b ." | a = 50 + 0
b = a / 2
c = 100 + 0
d = c / 2
e = b - d
|
a ) 12 , b ) 24 , c ) 36 , d ) 38 , e ) 50 | e | divide(factorial(subtract(add(const_4, 30), const_1)), multiply(factorial(30), factorial(subtract(const_4, const_1)))) | how many positive integers less than 10,000 are such that the product of their digits is 30 ? | "30 = 1 * 30 , 2 * 15 , 3 * 10 , 5 * 6 we have the option of using single digits 1,2 , 3,5 and 6 no : of four digit numbers such that the product is 30 the possible combinations are using digits ( 1,2 , 3,5 ) & ( 1,1 , 5 , 6 ) such that the product 30 . no : of different combinations possible using ( 1,2 , 3,5 ) = 4 ! = 24 no : of different combinations possible using ( 5,6 , 1,1 ) = 4 ! / 2 ! = 12 ( ' 1 ' is repeated twice ) total 4 digit numbers so that the product is 30 is 24 + 12 = 36 no : of three digit numbers such that the product is 30 the possible combinations are using digits ( 2,3 , 5 ) & ( 1 , 5,6 ) no : of different combinations possible using ( 2 , 3,5 ) = 3 ! = 6 no : of different combinations possible using ( 1 , 5,6 ) = 3 ! = 6 total 3 digit numbers so that the product is 30 is 6 + 6 = 12 no : of 2 digit numbers such that the product is 30 is 2 ( 56 and 65 ) total numbers under 10000 such that the product is 30 is 36 + 12 + 2 = 50 e" | a = 4 + 30
b = a - 1
c = math.factorial(b)
d = math.factorial(30)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 23 , b ) 27 , c ) 28 , d ) 12 , e ) 18 | e | divide(100, multiply(add(16, 4), const_0_2778)) | the speed at which a man can row a boat in still water is 16 kmph . if he rows downstream , where the speed of current is 4 kmph , what time will he take to cover 100 metres ? | speed of the boat downstream = 16 + 4 = 20 kmph = 20 * 5 / 18 = 5.55 m / s hence time taken to cover 100 m = 100 / 5.55 = 18 seconds . answer : e | a = 16 + 4
b = a * const_0_2778
c = 100 / b
|
a ) 2.04 % , b ) 6.12 % , c ) 8 % , d ) 8.25 % , e ) 10 % | d | multiply(2, const_4) | on the first of the year , james invested x dollars at proudstar bank in an account that yields 2 % in interest every quarter year . at the end of the year , during which he made no additional deposits or withdrawals , he had y dollars in the account . if james had invested the same amount in an account which pays interest on a yearly basis , what must the interest rate be for james to have y dollars at the end of the year ? | "per quarter , interest = 2 % so for a year , interest = 8 % due to quarter cumulation , effective yield ( ytm ) would be slight higher than 8 % answer = 8.25 % = answer : d" | a = 2 * 4
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a ) 285 , b ) 277 , c ) 290 , d ) 662 , e ) 262 | a | subtract(subtract(390, divide(multiply(390, 14), const_100)), divide(multiply(subtract(390, divide(multiply(390, 14), const_100)), 15), const_100)) | the sale price sarees listed for rs . 390 after successive discount is 14 % and 15 % is ? | "390 * ( 86 / 100 ) * ( 85 / 100 ) = 285 answer : a" | a = 390 * 14
b = a / 100
c = 390 - b
d = 390 * 14
e = d / 100
f = 390 - e
g = f * 15
h = g / 100
i = c - h
|
a ) 7,000 , b ) 10,500 , c ) 11,750 , d ) 12,000 , e ) 13,000 | a | divide(divide(50, subtract(divide(subtract(divide(const_60, 42), inverse(2)), const_60), inverse(multiply(const_60, 2)))), const_1000) | with both inlets open , a water tank will be filled with water in 42 minutes . the first inlet alone would fill the tank in 2 hours . if in every minutes the second inlet admits 50 cubic meters of water than the first , what is the capacity of the tank ? | "the work done by inlet a and b together in 1 min = 1 / 42 the work done by inlet a ( first inlet ) in 1 min = 1 / 120 the work done by inlet b ( second inlet ) in 1 min = ( 1 / 42 ) - ( 1 / 120 ) = 1 / 65 difference of work done by b and a = b - a = 50 cubic meter i . e . ( 1 / 65 ) - ( 1 / 120 ) = 50 cubic meter = 7000 cubic meter answer : option a" | a = const_60 / 42
b = 1/(2)
c = a - b
d = c / const_60
e = const_60 * 2
f = 1/(e)
g = d - f
h = 50 / g
i = h / 1000
|
a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | b | divide(subtract(divide(40, const_2), sqrt(subtract(power(divide(40, const_2), const_2), multiply(const_4, 91)))), const_2) | if a rectangular billboard has an area of 91 square feet and a perimeter of 40 feet , what is the length of each of the shorter sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 91 and a perimeter of 40 . we ' re asked for the length of one of the shorter sides of the rectangle . since the answers are all integers , and the area is 91 , the shorter side will almost certainly be less than 10 ( since 10 x 10 = 100 , but we ' re not dealing with a square ) . let ' s test answer b : 7 if . . . the shorter side = 7 . . . the area = 91 . . . . 91 / 7 = 13 = the longer side perimeter = 7 + 7 + 13 + 13 = 40 b" | a = 40 / 2
b = 40 / 2
c = b ** 2
d = 4 * 91
e = c - d
f = math.sqrt(e)
g = a - f
h = g / 2
|
a ) 9 : 25 , b ) 9 : 35 , c ) 81 : 124 , d ) 81 : 126 , e ) 81 : 129 | a | power(divide(27, 125), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 27 : 125 . what is the ratio of their total surface areas ? | "ratio of the sides = 3 √ 27 : 3 √ 125 = 3 : 5 ratio of surface areas = 3 ^ 2 : 5 ^ 2 = 9 : 25 answer : option a" | a = 27 / 125
b = 1 / 3
c = a ** b
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a ) 1,250 , b ) 1,000 , c ) 850 , d ) 750 , e ) 350 | a | multiply(1, 1) | the ratio of the first and second class fares between two stations is 3 : 1 and that of the number of passengers travelling between the two stations by the first and the second class is 1 : 50 . if in a day , rs . 1 , 325 are collected from the passengers travelling between the two stations , then the amount collected from the second class from the second class passengers is | explanation : let the fares be 3 x and x respectively for the first and second class . suppose in a day passengers travelling between the two stations be 1 and 50 . then by given condition , 3 x × 1 + x × 50 = 1,325 = > x = 25 the amount collected from second class = 25 × 50 = rs . 1250 answer : a | a = 1 * 1
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a ) 189 cm 2 , b ) 150 cm 2 , c ) 127 cm 2 , d ) 177 cm 2 , e ) 90 cm 2 | e | multiply(multiply(divide(const_1, const_2), add(4, 5)), 20) | find the area of the quadrilateral of one of its diagonals is 20 cm and its off sets 5 cm and 4 cm ? | "1 / 2 * 20 ( 5 + 4 ) = 90 cm 2 answer : e" | a = 1 / 2
b = 4 + 5
c = a * b
d = c * 20
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a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | e | subtract(multiply(20, 10.00), multiply(6.00, 20)) | a club wants to mix 20 pounds of candy worth $ 10.00 per pound with candy worth $ 5.00 per pound to reduce the cost of the mixture to $ 6.00 per pound . how many pounds of the $ 5.00 per pound candy should be used ? | "let number of pounds of 5 $ candy to be used be w 6 = ( 20 * 10 + 5 * w ) / ( 20 + w ) = > 120 + 6 w = 200 + 5 w = > w = 80 answer e" | a = 20 * 10
b = 6 * 0
c = a - b
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a ) 126 , b ) 136 , c ) 146 , d ) 156 , e ) 166 | d | divide(multiply(12, 312), 24) | the reciprocal of the hcf and lcm of two are 1 / 12 and 1 / 312 . if one of the number is 24 then other no . is | "reciprocal of the hcf and lcm of two are 1 / 12 and 1 / 312 so , hcf = 12 , lcm = 312 lcm * hcf = product of two numbers = a * b = > b = lcm * hcf / a so , other = 12 * 312 / 24 = 156 answer : d" | a = 12 * 312
b = a / 24
|
a ) 15 days , b ) 16.32 days , c ) 12.72 days , d ) 17 days , e ) 18 days | c | divide(const_1, add(divide(const_1, 20), divide(const_1, 35))) | a can do a piece of work 20 days . b can do work in 35 days . in how many days they will complete the work together ? | "lcm = 140 , ratio = 20 : 35 = 4 : 7 no of days = 140 / ( 4 + 7 ) = 90 / 5 = 12.72 days answer : c" | a = 1 / 20
b = 1 / 35
c = a + b
d = 1 / c
|
a ) 338 , b ) 823 , c ) 833 , d ) 337 , e ) 733 | c | subtract(1000, add(divide(multiply(1000, 15), const_100), divide(multiply(2, 1000), const_100))) | the sale price shirts listed for rs . 1000 after successive discount is 15 % and 2 % is ? | "1000 * ( 85 / 100 ) * ( 98 / 100 ) = 833 answer : c" | a = 1000 * 15
b = a / 100
c = 2 * 1000
d = c / 100
e = b + d
f = 1000 - e
|
a ) 4 : 49 , b ) 1 : 3 , c ) 3 : 21 , d ) 1 : 7 , e ) 3 : 49 | a | divide(4, const_60) | what is the ratio of 4 / 7 to the product 3 * ( 7 / 3 ) ? | "4 / 7 / 21 / 3 = 4 / 49 . . . imo option a ." | a = 4 / const_60
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a ) 20 % , b ) 42 % , c ) 44 % , d ) 92 % , e ) 90 % | d | multiply(subtract(const_1, divide(5, subtract(const_100, 30))), const_100) | exactly 5 % of the reporters for a certain wire service cover local politics in country x . if 30 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ? | "you are correct . people who cover local politics are a subset of people who cover politics . 30 % of reporters who cover politics do not cover local politics so 70 % do cover local politics . reporters covering local politics = 70 % of reporters covering politics = 5 % of all reporters reporters covering politics / all reporters = 5 / 70 = 1 / 14 reporters not covering politics / all reporters = 13 / 14 = 92 % = d" | a = 100 - 30
b = 5 / a
c = 1 - b
d = c * 100
|
a ) $ 87952 , b ) $ 93750 , c ) $ 89542 , d ) $ 95820 , e ) $ 84256 | b | divide(add(35000, 40000), subtract(const_1, divide(20, const_100))) | kanul spent $ 35000 in buying raw materials , $ 40000 in buying machinery and 20 % of the total amount he had as cash with him . what was the total amount ? | "let the total amount be x then , ( 100 - 20 ) % of x = 35000 + 40000 80 % of x = 75000 80 x / 100 = 75000 x = $ 93750 answer is b" | a = 35000 + 40000
b = 20 / 100
c = 1 - b
d = a / c
|
a ) 16 , b ) 12 , c ) 15 , d ) 18 , e ) 10 | d | multiply(18, const_1) | the total age of a and b is 18 years more than the total age of b and c . c is how many years younger than a . ? | "( a + b ) - ( b - c ) = 18 a - c = 18 answer is d" | a = 18 * 1
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a ) 91.5 , b ) 91.4 , c ) 97 , d ) 91.3 , e ) 91.1 | c | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | the sector of a circle has radius of 21 cm and central angle 150 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 150 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 55 + 42 = 97 cm answer : c" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
|
a ) 12 , b ) 20 , c ) 24 , d ) 30 , e ) 36 | b | inverse(subtract(divide(const_3, 36), divide(const_2, 60))) | tom drives from town w to town b , driving at a constant speed of 60 miles per hour . from town b tom immediately continues to town c . the distance between w and b is twice the distance between b and c . if the average speed of the whole journey was 36 mph , then what is tom ' s speed driving from b to c in miles per hour ? | let ' s assume that it takes 4 hours to go from point w to b . then the distance between them becomes 240 which makes distance between b and c 120 . ( 240 + 120 ) / ( 4 + x ) gives us the average speed which is 36 . you find x = 6 . so the question simplifies itself to 120 / 6 = 20 hence the answer is b . | a = 3 / 36
b = 2 / 60
c = a - b
d = 1/(c)
|
a ) 5.2 , b ) 6.5 , c ) 5.3 , d ) 5.9 , e ) 5.1 | b | add(12, const_1) | the average of first 12 natural numbers is ? | "sum of 12 natural no . = 156 / 2 = 78 average = 78 / 12 = 6.5 answer : b" | a = 12 + 1
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a ) 60 % , b ) 55 % , c ) 70 % , d ) 75 % , e ) 80 % | b | subtract(add(add(85, 65), 5), const_100) | if 85 percent of the test takers taking an old paper and pencil gmat exam answered the first question on a given math section correctly , and 65 percent of the test takers answered the second question correctly , and 5 percent of the test takers answered neither question correctly , what percent answered both correctly ? | "{ total } = { first correctly } + { second correctly } - { both correctly } + { neither correctly } 100 = 85 + 65 - { both correctly } + 5 { both correctly } = 55 . answer : b ." | a = 85 + 65
b = a + 5
c = b - 100
|
a ) 21.06 , b ) 23 , c ) 26 , d ) 23.69 , e ) 22.56 | a | divide(add(add(add(add(add(add(12, 13), 14), 51), 52), 53), 12), add(const_3, const_4)) | what is the average of 12 , 13 , 14 , 51 , 52 , 53 , 1 , 11.5 , 1 , 1.20 , and 11.25 ? | "add 12 , 13 , 14 , 51 , 52 , 53 , 1 , 11.5 , 1,12 , and 11.25 grouping numbers together may quicken the addition sum = 231.75 231.75 / 11 = 21.06 a" | a = 12 + 13
b = a + 14
c = b + 51
d = c + 52
e = d + 53
f = e + 12
g = 3 + 4
h = f / g
|
a ) 20 % , b ) 29 % , c ) 10 % , d ) 80 % , e ) 42.5 % | e | subtract(divide(subtract(const_100, 5), divide(2, 3)), const_100) | what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 5 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 95 2 / 3 sp 1 = 95 sp 1 = 142.5 100 - - - 42.5 = > 42.5 % answer : e" | a = 100 - 5
b = 2 / 3
c = a / b
d = c - 100
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a ) $ 14 , b ) $ 2 , c ) $ 10 , d ) $ 16 , e ) $ 18 | d | subtract(multiply(add(240, divide(multiply(divide(25, const_100), 240), subtract(const_1, divide(25, const_100)))), subtract(const_1, divide(20, const_100))), 240) | a jewelry store purchased a pair of earrings for $ 240 and was selling it at a price that equaled the purchase price of the earrings plus a markup that was 25 percent of the selling price . after some time a jewelry store owner decided to decrease the selling price by 20 percent . what was the jewelry store ' s gross profit on this sale ? | sale price ( sp ) = 240 + markup ( mp ) - - > mp = sp - 240 and given mp = sp / 4 ( 25 % is 1 / 4 th ) so sp / 4 = sp - 240 3 sp / 4 = 240 sp = 320 now a discount of 20 % is given so new sp is . 8 * 320 = 256 profit = 256 - 240 = 16 answer is d | a = 25 / 100
b = a * 240
c = 25 / 100
d = 1 - c
e = b / d
f = 240 + e
g = 20 / 100
h = 1 - g
i = f * h
j = i - 240
|
a ) 92 , b ) 96 , c ) 100 , d ) 104 , e ) 108 | b | divide(36, subtract(1, add(add(add(divide(1, 4), divide(1, 8)), divide(1, 12)), divide(1, 6)))) | if 1 / 4 of the passengers on a ship are from north america , 1 / 8 are europeans , 1 / 12 are from africa , 1 / 6 are from asia and the remaining 36 people are citizens of other continents , then how many passengers are on board the ship ? | "1 / 4 + 1 / 8 + 1 / 12 + 1 / 6 = ( 6 + 3 + 2 + 4 ) / 24 = 15 / 24 let x be the number of passengers on the ship . 36 = ( 9 / 24 ) x = 3 x / 8 x = 96 the answer is b ." | a = 1 / 4
b = 1 / 8
c = a + b
d = 1 / 12
e = c + d
f = 1 / 6
g = e + f
h = 1 - g
i = 36 / h
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a ) 37.9 , b ) 37.2 , c ) 37.1 , d ) 37.82 , e ) 37.8 | e | multiply(subtract(8, 1), divide(subtract(subtract(100, divide(100, 5)), divide(100, 8)), divide(100, 8))) | a hare and a tortoise have a race along a circle of 100 yards diameter . the tortoise goes in one direction and the hare in the other . the hare starts after the tortoise has covered 1 / 5 of its distance and that too leisurely . the hare and tortoise meet when the hare has covered only 1 / 8 of the distance . by what factor should the hare increase its speed so as to tie the race ? | assume the circumference of the circle is 200 meters . hare and tortoise started at the same point but moves in the opposite direction . it is given that by that time tortoise covered 40 m ( 1 / 5 th of the distance ) , hare started and both met after hare has covered 25 . this implies , in the time hare has covered 25 m , hare has covered 200 - 40 - 25 = 135 meters . so hare : tortoise speeds = 25 : 135 = 5 : 27 now hare and tortoise has to reach the starting point means , hare has to cover 175 meters and tortoise has to cover only 25 meters in the same time . as time = distancespeed = 2527 = 1755 × kdistancespeed = 2527 = 1755 × k ie . , hare has to increase its speed by a factor k . solving we get k = 37.8 answer : e | a = 8 - 1
b = 100 / 5
c = 100 - b
d = 100 / 8
e = c - d
f = 100 / 8
g = e / f
h = a * g
|
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | e | sqrt(divide(volume_cylinder(divide(20, const_2), 20), multiply(const_pi, 20))) | a certain rectangular crate measures 20 feet by 20 feet by 20 feet . a cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces . what should the radius of the tank be if it is to be of the largest possible volume ? | "for max volume of cylinder ( pi * r ^ 2 * h ) we need to max out r ^ 2 * h . we do n ' t know what the dimensions of the crate refer to . . therefore for max vol base should be 20 x 20 i . e . of radius 20 / 2 = 10 e" | a = 20 / 2
b = volume_cylinder / (
c = math.pi * 20
d = math.sqrt(b)
|
a ) $ 125 , b ) $ 135 , c ) $ 145 , d ) $ 155 , e ) $ 165 | a | divide(divide(20, divide(subtract(const_100, 80), const_100)), divide(80, const_100)) | jack got paid today for his part - time job as a juggler , and noticed that 20 % of his paycheck went to taxes . jack wants to spend 80 % of what ' s left on a new bike , and put the rest into savings . if jack puts $ 20 into savings , how much was his original paycheck before taxes ? | let p be jack ' s original paycheck value . the value after taxes t is given by t = p - 20 % p let s be the amount put into savings , given by s = t - 80 % t for s = $ 20 , solve for p $ 20 = ( p - 20 % p ) - 80 % ( p - 20 % p ) p = $ 125 correct answer a | a = 100 - 80
b = a / 100
c = 20 / b
d = 80 / 100
e = c / d
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a ) 23 , b ) 28 , c ) 30 , d ) 32 , e ) 34 | b | add(add(add(add(add(add(add(add(const_2, const_3), add(const_2, const_3)), add(add(const_2, const_3), const_2)), add(5, const_2)), add(add(5, const_2), const_2)), add(add(add(5, const_2), const_2), const_4)), add(add(add(add(5, const_2), const_2), const_4), const_2)), add(add(add(add(add(5, const_2), const_2), const_4), const_2), const_4)) | find a sum for 1 st 5 prime no . ' s ? | "required sum = ( 2 + 3 + 5 + 7 + 11 ) = 28 note : 1 is not a prime number option b" | a = 2 + 3
b = 2 + 3
c = a + b
d = 2 + 3
e = d + 2
f = c + e
g = 5 + 2
h = f + g
i = 5 + 2
j = i + 2
k = h + j
l = 5 + 2
m = l + 2
n = m + 4
o = k + n
p = 5 + 2
q = p + 2
r = q + 4
s = r + 2
t = o + s
u = 5 + 2
v = u + 2
w = v + 4
x = w + 2
y = x + 4
z = t + y
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a ) 9 / 25 , b ) 10 / 25 , c ) 6 / 10 , d ) 2 / 3 , e ) 21 / 25 | e | add(multiply(divide(3, 5), divide(3, 5)), add(multiply(divide(3, 5), divide(const_2, 5)), multiply(divide(3, 5), divide(const_2, 5)))) | a canoe has two oars , left and right . each oar either works or breaks . the failure or non - failure of each oar is independent of the failure or non - failure of the other . you can still row the canoe with one oar . the probability that the left oar works is 3 / 5 . the probability that the right oar works is also 3 / 5 . what is the probability w that you can still row the canoe ? | "simply look at the question from the other side . what is the probability that you can ’ t row the canoe ? this would be 2 / 5 x 2 / 5 = 4 / 25 . using the idea that the probability of something happening is 1 – the probability that it doesn ’ t happen , you can use the following equation to reach the right answer w : 1 – 4 / 25 = 21 / 25 . answer choice e ." | a = 3 / 5
b = 3 / 5
c = a * b
d = 3 / 5
e = 2 / 5
f = d * e
g = 3 / 5
h = 2 / 5
i = g * h
j = f + i
k = c + j
|
a ) 4 , b ) 3 , c ) 8 , d ) 9 , e ) 12 | b | multiply(multiply(add(inverse(multiply(const_2, const_4)), const_1), const_3), multiply(const_2, const_4)) | working together , wayne and his son can shovel the entire driveway in three hours . if wayne can shovel two times as fast as his son can , how many hours would it take for his son to shovel the entire driveway on his own ? | "w : the time for wyane to do the job s : the time for his son to do the job we have 1 / w + 1 / s = 1 / 2 and w = 2 s then we have 1 / ( 2 * s ) + 1 / s = 1 / 2 < = > 3 / ( 2 * s ) = 1 / 2 < = > s = 3 ans : b" | a = 2 * 4
b = 1/(a)
c = b + 1
d = c * 3
e = 2 * 4
f = d * e
|
a ) 135 m , b ) 125 m , c ) 129 m , d ) 120 m , e ) 165 m | b | multiply(multiply(90, const_0_2778), 5) | if a train , travelling at a speed of 90 kmph , crosses a pole in 5 sec , then the length of train is ? | "explanation : d = 90 * 5 / 18 * 5 = 125 m answer : option b" | a = 90 * const_0_2778
b = a * 5
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a ) 518 , b ) 452 , c ) 468 , d ) 502 , e ) 396 | c | divide(multiply(divide(96, subtract(1, divide(5, 9))), 13), subtract(7, 1)) | karthik read 7 / 13 th of a book in 1 st week and 5 / 9 th of the remaining book in 2 nd week . if there were 96 pages unread after 2 nd week , how many pages were there in the book ? | solution 96 pages were unread so on second week have 5 / 9 so unread pages = 1 - 5 / 9 = 4 / 9 96 * 9 / 4 = 216 then , on first week unread pages were 216 total pages = 216 * 13 / 6 = 468 so answer is c | a = 5 / 9
b = 1 - a
c = 96 / b
d = c * 13
e = 7 - 1
f = d / e
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a ) 0.012 , b ) 0.12 , c ) 0.0012 , d ) 0.00012 , e ) none of them | c | divide(0.0204, 17) | find the quotient : 0.0204 / 17 | 204 / 17 = 12 . dividend contains 4 places of decimal . 0.2040 / 17 = 0.0012 . answer is c . | a = 0 / 204
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a ) 300 , b ) 400 , c ) 250 , d ) 350 , e ) 450 | b | add(multiply(divide(400, 50), 30), multiply(divide(400, 50), 20)) | the consumption of diesel per hour of a bus varies directly as square of its speed . when the bus is travelling at 50 kmph its consumption is 1 litre per hour . if each litre costs $ 30 and other expenses per hous is $ 20 , then what would be the minimum expenditure required to cover a distance of 400 km ? | "50 kmph consumption is 1 lt / hr so 400 km will take 8 hrs and the consumption is 8 lt for entire distance . 1 lt costs $ 30 so 8 lt costs $ 240 extra expenses for 1 hr - $ 20 8 hrs - $ 160 total expense - $ 240 + $ 160 = $ 400 answer : b" | a = 400 / 50
b = a * 30
c = 400 / 50
d = c * 20
e = b + d
|
a ) 57 , b ) 67 , c ) 77 , d ) 87 , e ) 97 | c | sqrt(multiply(59.29, const_100)) | a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 59.29 , the number of the member is the group is : | "money collected = ( 59.29 x 100 ) paise = 5929 paise . number of members = 5929 = 77 . answer : c" | a = 59 * 29
b = math.sqrt(a)
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a ) 10 , b ) 20 , c ) 30 , d ) 60 , e ) 70 | d | add(multiply(sqrt(30), const_10), const_3) | if a ^ 2 + b ^ 2 = 30 and ab = 10 , what is the value of the expression ( a - b ) ^ 2 + ( a + b ) ^ 2 ? | "( a - b ) ^ 2 = a ^ 2 + b ^ 2 - 2 ab = 30 - 20 = 10 ( a + b ) ^ 2 = a ^ 2 + b ^ 2 + 2 ab = 30 + 20 = 50 so ( a + b ) ^ 2 + ( a - b ) ^ 2 = 50 + 10 = 60 d" | a = math.sqrt(30)
b = a * 10
c = b + 3
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a ) rs . 500 , b ) rs . 400 , c ) rs . 800 , d ) rs . 600 , e ) rs . 200 | d | multiply(30, 20) | a can do a work in 30 days , b can do it in 20 days . they together under took to do a piece of work for rs . 1000 , what is the share of b ? | explanation : share of their work days = 30 : 20 share of their work = 20 : 30 share of b ’ s money = 3 / 5 * 1000 = 600 answer : option d | a = 30 * 20
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a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 6 | b | divide(divide(6, const_3), const_2) | 3 x ^ 2 - 6 x + 3 = 0 find the above equation find the value of x | a = 3 , b = - 6 , c = 3 x 1,2 = ( 6 ± √ ( ( - 6 ) 2 - 4 × 3 × 3 ) ) / ( 2 × 3 ) = ( 6 ± √ ( 36 - 36 ) ) / 6 = ( 6 ± 0 ) / 6 x 1 = x 2 = 1 b | a = 6 / 3
b = a / 2
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a ) 8 % , b ) 1 % , c ) 20 % , d ) 80 % , e ) none | c | multiply(subtract(const_1, divide(divide(4, const_100), divide(5, const_100))), const_100) | the income of a broker remains unchanged though the rate of commission is increased from 4 % to 5 % . the percentage of slump in business is : | "explanation : let the business value changes from x to y . then , 4 % of x = 5 % of y or 4 / 100 × x = 5 / 100 × y or y = 4 / 5 x change in business = ( x − 4 / 5 x ) = 1 / 5 x percentage slump in business = ( 1 / 5 x × 1 / x × 100 ) % correct option : c" | a = 4 / 100
b = 5 / 100
c = a / b
d = 1 - c
e = d * 100
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a ) 6 , b ) 12 , c ) 14 , d ) 16 , e ) 20 | e | divide(multiply(subtract(45, 20), 8), 10) | how many gallons of milk that is 10 percent butter - fat must be added to 8 gallons of milk that is 45 percent butterfat to obtain milk that is 20 percent butterfat ? | equate the fat : 0.1 x + 0.45 * 8 = 0.2 ( x + 8 ) - - > x = 20 . answer : e . | a = 45 - 20
b = a * 8
c = b / 10
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['a ) 40 cm', 'b ) 45 cm', 'c ) 50 cm', 'd ) 52 cm', 'e ) 57 cm'] | c | multiply(add(subtract(multiply(add(3, 4), 3), multiply(4, 3)), add(subtract(multiply(add(3, 4), 3), multiply(4, 3)), add(3, 4))), const_2) | if the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm , a square with the same area as the original rectangle would result . find the perimeter of the original rectangle . | explanation : let x and y be the length and breadth of the rectangle respectively . then , x - 4 = y + 3 or x - y = 7 - - - - ( 1 ) area of the rectangle = xy ; area of the square = ( x - 4 ) ( y + 3 ) ( x - 4 ) ( y + 3 ) = xy < = > 3 x - 4 y = 12 - - - - ( 2 ) solving 1 & 2 , we get x = 16 and y = 9 . perimeter of the rectangle = 2 ( x + y ) = [ 2 ( 16 + 9 ) ] cm = 50 cm . answer : option c | a = 3 + 4
b = a * 3
c = 4 * 3
d = b - c
e = 3 + 4
f = e * 3
g = 4 * 3
h = f - g
i = 3 + 4
j = h + i
k = d + j
l = k * 2
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a ) 16 , b ) 18 , c ) 20 , d ) 70 , e ) 30 | d | multiply(4, divide(110, add(4, 7))) | 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 110 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 7 km / h , what is the distance traveled by brad ? | "time taken = total distance / relative speed total distance = 110 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 7 + 4 = 11 kms / hr time taken = 110 / 11 = 10 hrs distance traveled by brad = brad ' s speed * time taken = 7 * 10 = 70 kms . . . answer - d" | a = 4 + 7
b = 110 / a
c = 4 * b
|
a ) 0.75 d , b ) 0.76 d , c ) 0.765 d , d ) 0.775 d , e ) 0.805 d | c | subtract(divide(subtract(const_100, 15), const_100), multiply(divide(subtract(const_100, 15), const_100), divide(10, const_100))) | a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 15 % . staff are allowed a further 10 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? | "successive doscounts = - a - b + ab / 100 putting the values , total discount = - 15 - 10 + 150 / 100 = - 25 + 1.5 = 23.5 % discount hence the final price of the dress = d - 0.235 d = 0.765 d correct option : c" | a = 100 - 15
b = a / 100
c = 100 - 15
d = c / 100
e = 10 / 100
f = d * e
g = b - f
|
a ) 11 / 25 , b ) 7 / 12 , c ) 5 / 13 , d ) 5 / 12 , e ) 6 / 17 | a | divide(subtract(11, multiply(const_2, const_3)), 11) | in a simultaneous throw of a pair of dice , find the probability of getting a total more than 11 | total number of cases = 10 * 10 = 100 favourable cases = [ ( 2,10 ) , ( 3,9 ) , ( 3,10 ) , ( 4,8 ) , ( 4,9 ) , ( 4,10 ) , ( 5,7 ) , ( 5,8 ) , ( 5,9 ) , ( 5,10 ) , ( 6,6 ) , ( 6,7 ) , ( 6,8 ) , ( 6,9 ) , ( 6,10 ) , ( 7,5 ) , ( 7,6 ) , ( 7,7 ) , ( 7,8 ) , ( 7,9 ) , ( 7,10 ) , ( 8,4 ) , ( 8,5 ) , ( 8,6 ) , ( 8,7 ) , ( 8,8 ) , ( 8,9 ) , ( 8,10 ) , ( 9,3 ) , ( 9,4 ) , ( 9,5 ) , ( 9,6 ) , ( 9,7 ) , ( 9,8 ) , ( 9,10 ) , ( 10,2 ) , ( 10,3 ) , ( 10,4 ) , ( 10,5 ) , ( 10,6 ) , ( 10,7 ) , ( 10,8 ) , ( 10,9 ) , ( 10,10 ) ] = 44 so probability = 44 / 100 = 11 / 25 answer is a | a = 2 * 3
b = 11 - a
c = b / 11
|
a ) $ 240 , b ) $ 300 , c ) $ 360 , d ) $ 420 , e ) $ 480 | b | divide(210, divide(subtract(const_100, 30), const_100)) | sandy had $ 210 left after spending 30 % of the money she took for shopping . how much money did sandy take along with her ? | "let the money sandy took for shopping be x . 0.7 x = 210 x = 300 the answer is b ." | a = 100 - 30
b = a / 100
c = 210 / b
|
a ) 18 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | a | add(add(subtract(multiply(2, 2), multiply(3, 2)), multiply(2, 2)), add(subtract(multiply(2, 2), multiply(3, 2)), multiply(2, 2))) | if a ã — b = 2 a - 3 b + ab , then 2 ã — 5 + 5 ã — 2 is equal to : | "explanation : 2 ã — 5 + 5 ã — 2 = ( 2 ã — 2 - 3 ã — 5 + 2 ã — 5 ) + ( 2 ã — 5 - 3 ã — 2 + 5 ã — 3 ) = ( 4 - 15 + 10 + 10 - 6 + 15 ) = 18 . answer : a" | a = 2 * 2
b = 3 * 2
c = a - b
d = 2 * 2
e = c + d
f = 2 * 2
g = 3 * 2
h = f - g
i = 2 * 2
j = h + i
k = e + j
|
a ) 2 / 131 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | d | subtract(add(const_4, const_4), const_1) | if ( n + 2 ) ! / n ! = 156 , n = ? | "( n + 2 ) ! / n ! = 156 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 156 from here , we might just test the answer choices . since ( 13 ) ( 12 ) = 156 , we can see that n = 11 d" | a = 4 + 4
b = a - 1
|
a ) 230 , b ) 274 , c ) 306 , d ) 354 , e ) 424 | c | multiply(18, subtract(18, const_1)) | 18 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | "2 * 18 c 2 = 2 * 153 = 306 the answer is c ." | a = 18 - 1
b = 18 * a
|
a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | d | add(add(multiply(subtract(12, const_1), 2), divide(10, 2)), divide(10, 2)) | in a garden , there are 10 rows and 12 columns of mango trees . the distance between the two trees is 2 metres and a distance of three metres is left from all sides of the boundary of the garden . what is the length of the garden ? | "between the 12 mango trees , there are 11 gaps and each gap has 2 meter length also , 3 meter is left from all sides of the boundary of the garden . hence , length of the garden = ( 11 ã — 2 ) + 3 + 3 = 28 meter answer is d ." | a = 12 - 1
b = a * 2
c = 10 / 2
d = b + c
e = 10 / 2
f = d + e
|
a ) 1.2 , b ) 1.5 , c ) 1.8 , d ) 2.1 , e ) 2.4 | a | subtract(6, multiply(divide(multiply(6, 40), const_100), const_2)) | a 6 - liter solution is 40 % alcohol . how many liters of pure alcohol must be added to produce a solution that is 50 % alcohol ? | let x be the amount of pure alcohol required . 0.4 ( 6 ) + x = 0.5 ( x + 6 ) 0.5 x = 3 - 2.4 x = 1.2 liters the answer is a . | a = 6 * 40
b = a / 100
c = b * 2
d = 6 - c
|
a ) 1500 , b ) 1600 , c ) 2000 , d ) 2100 , e ) 2500 | c | add(2100, divide(multiply(2100, 5), const_100)) | the present population of a town is 2100 . population increase rate is 5 % p . a . find the population of town before 1 years ? | "p = 2100 r = 5 % required population of town = p / ( 1 + r / 100 ) ^ t = 2100 / ( 1 + 5 / 100 ) = 2100 / ( 21 / 20 ) = 2000 ( approximately ) answer is c" | a = 2100 * 5
b = a / 100
c = 2100 + b
|
a ) 8 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | subtract(subtract(divide(divide(multiply(25, 8), const_10), const_2), const_0_25), const_0_25) | a math teacher has 25 cards , each of which is in the shape of a geometric figure . half of the cards are rectangles , and a third of the cards are rhombuses . if 8 cards are squares , what is the maximum possible number of cards that re circles . | "a square is a special kind of rhombus ( sides are perpendicular ) a square is a special kind of rectangles ( sides with same length ) among the 25 cards with have : 15 rectangles 10 rhombus 8 squares among the 15 rectangles , there could be 8 special ones ( with sides of same length ) that are squares . that lets at least 7 rectangles that are not square . among the 10 rectangles , there could be 8 special ones ( with sides perpendicular ) that are squares . that lets at least 2 rhombus that are not square . we have 8 squares . so the minimum different cards that represent a square , a rhombus or a rectangle is 2 + 7 + 8 = 17 which means that the maximum number of circles that you could have is 25 - 17 = 8 answer ( a )" | a = 25 * 8
b = a / 10
c = b / 2
d = c - const_0_25
e = d - const_0_25
|
a ) a ) 0.9 , b ) b ) 0.6 , c ) c ) 0.45 , d ) d ) 0.3 , e ) e ) 0.25 | b | subtract(const_1, multiply(divide(40, const_100), divide(subtract(const_100, 30), const_100))) | in the graduating class of a certain college , 48 percent of the students are male and 52 percent are female . in this class 40 percent of the male and 30 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | "percent of students who are 25 years old or older is 0.4 * 48 + 0.3 * 52 = ~ 34 , so percent of people who are less than 25 years old is 100 - 34 = 66 . answer : b ." | a = 40 / 100
b = 100 - 30
c = b / 100
d = a * c
e = 1 - d
|
a ) 6 , b ) 9 , c ) 12 , d ) 34 , e ) 48 | d | divide(multiply(divide(90, const_100), 26), divide(70, const_100)) | an alloy weighing 26 ounces is 70 percent gold . how many ounces of pure gold must be added to create an alloy that is 90 percent gold ? | in 24 ounces , gold is 26 * ( 70 / 100 ) = 18.2 ounces . now we add x ounces of pure gold to make it 90 % gold . so 18.2 + x = ( 24 + x ) * 90 / 100 = > x = 34 . answer is d . | a = 90 / 100
b = a * 26
c = 70 / 100
d = b / c
|
a ) a ) 100 , b ) b ) 150 , c ) c ) 250 , d ) d ) 200 , e ) e ) 160 | b | divide(multiply(15, const_100), 10) | an inspector rejects 10 % of the meters as defective . how many will he examine to reject 15 ? | "then , 10 % of x = 15 ( 10 / 100 ) x = 15 x = ( 15 * 100 * ) / 10 = 150 answer is b" | a = 15 * 100
b = a / 10
|
a ) 297 , b ) 879 , c ) 342 , d ) 240 , e ) 269 | d | subtract(subtract(400, divide(multiply(400, 25), const_100)), divide(multiply(subtract(400, divide(multiply(400, 25), const_100)), 20), const_100)) | the sale price sarees listed for rs . 400 after successive discount is 25 % and 20 % is ? | "400 * ( 75 / 100 ) * ( 80 / 100 ) = 240 answer : d" | a = 400 * 25
b = a / 100
c = 400 - b
d = 400 * 25
e = d / 100
f = 400 - e
g = f * 20
h = g / 100
i = c - h
|
a ) 15 / 10 , b ) 19 / 10 , c ) 20 / 19 , d ) 16 / 10 , e ) 27 / 10 | b | divide(80, multiply(add(69, 82), const_0_2778)) | two trains of length 120 m and 210 m are 80 m apart . they start moving towards each other on parallel tracks , at speeds 69 kmph and 82 kmph . after how much time will the trains meet ? | "they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 69 + 82 ) * 5 / 18 = 41.9 mps . the time required = d / s = 80 / 41.9 = 19 / 10 sec . answer : b" | a = 69 + 82
b = a * const_0_2778
c = 80 / b
|
a ) 10 % , b ) 15 % , c ) 25 % , d ) 30 % , e ) 20 % | e | multiply(divide(subtract(8, 10), 10), const_100) | a shopkeeper sold 10 articles at the cost price of 8 articles . then find the profit % or lost % ? | "here 10 articles selling price = 8 articles cost price so the difference = 10 - 8 = 2 % of profit = 2 * 100 / 10 = 20 % correct option is e" | a = 8 - 10
b = a / 10
c = b * 100
|
a ) 76 days , b ) 48 days , c ) 98 days , d ) 31 days , e ) 22 days | b | inverse(subtract(inverse(16), inverse(24))) | a and b can finish a work in 16 days while a alone can do the same work in 24 days . in how many days b alone will complete the work ? | "b = 1 / 16 – 1 / 24 = 1 / 48 = > 48 days answer : b" | a = 1/(16)
b = 1/(24)
c = a - b
d = 1/(c)
|
a ) . 9 , b ) . 09 , c ) . 009 , d ) . 0009 , e ) none of them | a | divide(divide(009, const_1000), divide(01, const_100)) | . 009 / ? = . 01 | "let . 009 / x = . 01 ; then x = . 009 / . 01 = . 9 / 1 = . 9 answer is a" | a = 9 / 1000
b = 1 / 100
c = a / b
|
a ) 200 m , b ) 700 m , c ) 800 m , d ) 100 metre , e ) 1004 m | d | multiply(subtract(const_1, divide(add(4, divide(1, 2)), 5)), const_1000) | a can run a kilometer race in 4 1 / 2 min while b can run same race in 5 min . how many meters start can a give b in a kilometer race , so that the race mat end in a dead heat ? | explanation : a can give b ( 5 min - 4 1 / 2 min ) = 30 sec start . the distance covered by b in 5 min = 1000 m . distance covered in 30 sec = ( 1000 * 30 ) / 300 = 100 m . a can give b 100 m start . answer : d | a = 1 / 2
b = 4 + a
c = b / 5
d = 1 - c
e = d * 1000
|
a ) $ 1.00 , b ) $ 1.25 , c ) $ 12.00 , d ) $ 6.00 , e ) $ 20.00 | c | subtract(30.00, multiply(subtract(10, 2.00), 2.00)) | if 10 people contributed a total of $ 30.00 toward a gift and each of them contributed at least $ 2.00 , then the maximum possible amount any one person could have contributed is | "c for me 9 people with 2 $ each - > maximum = 12" | a = 10 - 2
b = a * 2
c = 30 - 0
|
a ) 62.5 , b ) 150 , c ) 225 , d ) 112.5 , e ) 212.5 | a | divide(subtract(divide(multiply(multiply(5000, 61), 2), const_100), divide(multiply(multiply(5000, 5), 2), const_100)), 2) | a person borrows rs . 5000 for 2 years at 5 % p . a . simple interest . he immediately lends it to another person at 61 ⁄ 4 % p . a for 2 years . find his gain in the transaction per year . | "explanation : the person borrows rs . 5000 for 2 years at 5 % p . a . simple interest simple interest that he needs to pay = prt / 100 = 5000 × 5 × 2 / 100 = 500 he also lends it at 6 1 ⁄ 4 % p . a for 2 years simple interest that he gets = prt / 100 = 5000 × 25 / 4 × 2 / 100 = 625 his overall gain in 2 years = rs . 625 - rs . 500 = rs . 125 his overall gain in 1 year = 125 / 2 = rs . 62.5 answer : option a" | a = 5000 * 61
b = a * 2
c = b / 100
d = 5000 * 5
e = d * 2
f = e / 100
g = c - f
h = g / 2
|
a ) 40 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | a | divide(add(102, 138), 6) | a student chose a number , multiplied it by 6 , then subtracted 138 from the result and got 102 . what was the number he chose ? | "solution : let xx be the number he chose , then 6 ⋅ x − 138 = 102 6 x = 240 x = 40 answer a" | a = 102 + 138
b = a / 6
|
a ) 1 / 18 , b ) 1 / 6 , c ) 1 / 3 , d ) 1 / 2 , e ) 5 / 6 | e | subtract(add(12, divide(12, 6)), divide(1, 6)) | in a certain lottery , the probability that a number between 12 and 20 , inclusive , is drawn is 1 / 6 . if the probability that a number 12 or larger is drawn is 1 / 3 , what is the probability that a number less than or equal to 20 is drawn ? | "you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set 1 : number 12 or larger set 2 : number 20 or smaller 1 = p ( set 1 ) + p ( set 2 ) - p ( set 1 and set 2 ) ( combined probability is 1 because every number will be either 12 or moreor 20 or lessor both ) 1 / 3 + p ( set 2 ) - 1 / 6 = 1 p ( set 2 ) = 5 / 6 answer ( e )" | a = 12 / 6
b = 12 + a
c = 1 / 6
d = b - c
|
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