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 |
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a ) 2 % , b ) 8.85 % , c ) 5.85 % , d ) 6.85 % , e ) 1 % | d | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(19, const_100))), const_100)) | the tax on a commodity is diminished by 19 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 81 * 115 = 9315 - - - - - - - - - - - 10000 - - - - - - - - - - - 685 100 - - - - - - - - - - - ? = > 6.85 % decrease answer : d" | a = 15 / 100
b = 1 + a
c = 19 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 75 km , b ) 50 km , c ) 60 km , d ) 85 km , e ) 95 km | a | multiply(const_60, divide(multiply(divide(15, const_60), 50), 50)) | when a train travels at a speed of 60 kmph , it reaches the destination on time . when the same train travels at a speed of 50 kmph , it reaches its destination 15 min late . what is the length of journey ? | "let x be the time reached with the speed 60 km / h 50 km / h - - - - > x + 15 distance is equal so 60 ( km / h ) × xhr = 50 ( km / h ) × ( x + 15 ) hr so 60 x = 50 x + 750 so the would be in km and x = 75 so 75 km answer : a" | a = 15 / const_60
b = a * 50
c = b / 50
d = const_60 * c
|
a ) 33.33 % , b ) 25 % , c ) 75 % , d ) 66.66 % , e ) none of these | d | multiply(subtract(divide(const_100, subtract(const_100, 40)), const_1), const_100) | if x is less than y by 40 % then y exceed x by : | "using formula ( x / ( 100 - x ) * 100 ) where x is percentage decrease ( here it is 40 % ) = > 40 / ( 100 - 40 ) * 100 = 66.66 % answer : d" | a = 100 - 40
b = 100 / a
c = b - 1
d = c * 100
|
a ) 8 , b ) 11 , c ) 12 , d ) 15 , e ) 16 | c | divide(multiply(add(divide(20, const_100), const_1), 22), add(add(divide(20, const_100), const_1), const_1)) | p is 20 % more efficient than q . p can complete a work in 22 days . if p and q work together , how many days will it take to complete the same work ? | "the work done by p in 1 day = 1 / 22 let work done by q in 1 day = q q × ( 120 / 100 ) = 1 / 22 q = 100 / ( 22 × 120 ) = 10 / ( 22 × 12 ) the work done by p and q in 1 day = 1 / 22 + 10 / ( 22 × 12 ) = 22 / ( 22 × 12 ) = 1 / 12 p and q together can do the work in 12 days . the answer is c ." | a = 20 / 100
b = a + 1
c = b * 22
d = 20 / 100
e = d + 1
f = e + 1
g = c / f
|
a ) 50.15 , b ) 52.06 , c ) 51.25 , d ) 53.06 , e ) none of the above | b | divide(add(add(add(multiply(55, 50), multiply(35, 60)), multiply(45, 55)), multiply(42, 45)), add(add(add(55, 35), 45), 42)) | a school has 4 section of chemistry in class x having 55 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student . | "required average marks = 55 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 55 + 35 + 45 + 42 = 2750 + 2100 + 2475 + 1890 / 162 = 9215 â „ 177 = 52.06 answer b" | a = 55 * 50
b = 35 * 60
c = a + b
d = 45 * 55
e = c + d
f = 42 * 45
g = e + f
h = 55 + 35
i = h + 45
j = i + 42
k = g / j
|
a ) s . 1200 , b ) s . 1542 , c ) s . 1528 , d ) s . 1540 , e ) s . 1549 | a | multiply(4, divide(2600, add(add(4, 2), const_3))) | rs . 2600 is divided so that 4 times the first share , thrice the 2 nd share and twice the third share amount to the same . what is the value of the third share ? | "a + b + c = 2600 4 a = 3 b = 2 c = x a : b : c = 1 / 4 : 1 / 3 : 1 / 2 = 3 : 4 : 6 6 / 13 * 2600 = rs . 1200 answer : a" | a = 4 + 2
b = a + 3
c = 2600 / b
d = 4 * c
|
a ) 240 , b ) 270 , c ) 295 , d ) 268 , e ) 252 | d | divide(add(1355, 15), subtract(6, const_1)) | the difference of two numbers is 1355 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1355 ) . x + 1355 = 6 x + 15 5 x = 1340 x = 268 smaller number = 270 . answer d" | a = 1355 + 15
b = 6 - 1
c = a / b
|
a ) 5 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 7 % | a | divide(multiply(subtract(490, 465.5), const_100), 490) | if a radio is purchased for rs 490 and sold for rs 465.50 find the loss % ? | c . p = rs 490 s . p = rs 465.50 loss = c . p - s . p = 490 - 465.50 = rs 24.50 loss % = ( loss * 100 ) / c . p % = ( 24.50 * 100 ) / 490 % = 5 % answer a 5 % | a = 490 - 465
b = a * 100
c = b / 490
|
a ) 65 km , b ) 100 km , c ) 120 km , d ) 130 km , e ) 75 km | d | divide(6.5, subtract(const_1, add(divide(3, 5), divide(7, 20)))) | if a person cover 3 / 5 part of his journey by train , 7 / 20 part of his journey by bus and rest 6.5 km . by walk . what is total distance of his journey ? | let total distance be = x km . then , 3 x / 5 + 7 x / 20 + 6.5 = x ; 12 x + 7 x + 130 = 20 x ; x = 130 hence total distance = 130 km . answer d | a = 3 / 5
b = 7 / 20
c = a + b
d = 1 - c
e = 6 / 5
|
a ) 16 , b ) 72 , c ) 112 , d ) 128 , e ) 142 | d | multiply(2, divide(144, add(2, 16))) | water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 144 grams of water ? | "see h / o = 2 / 16 add 1 to both sides : h / o + 1 = 2 / 16 + 1 take lcm ( h + o ) / o = ( 2 + 16 ) / 16 water / o = 18 / 16 now that we have the water to oxygen ratio , put in the amount of water i . e . 144 gm = > o = 144 * 16 / 18 = 128 answer : d" | a = 2 + 16
b = 144 / a
c = 2 * b
|
a ) 276 , b ) 299 , c ) 224 , d ) 345 , e ) 365 | c | multiply(16, 14) | the h . c . f . of two numbers is 16 and the other two factors of their l . c . m . are 13 and 14 . the larger of the two numbers is : | "clearly , the numbers are ( 16 x 13 ) and ( 16 x 14 ) . larger number = ( 23 x 14 ) = 224 . answer : option c" | a = 16 * 14
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | divide(subtract(subtract(add(add(21, 18), 12), 4), 39), 2) | at a certain resort , each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants . the 3 restaurants are the family buffet , the dining room , and the snack bar . exactly 21 employees are trained to work in the family buffet , 18 are trained to work in the dining room , and 12 are trained to work in the snack bar . if 4 employees are trained to work in exactly 2 restaurants , how many employees are trained to work in all 3 restaurants ? | "39 = 21 + 18 + 12 - 4 - 2 x 2 x = 21 + 18 + 12 - 4 - 39 = 47 - 39 = 8 x = 4 c" | a = 21 + 18
b = a + 12
c = b - 4
d = c - 39
e = d / 2
|
a ) 3168 , b ) 3268 , c ) 3368 , d ) 3468 , e ) 3568 | a | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(4, 4)), multiply(const_4, const_2))), const_10) | how many 4 digit number contain number 2 ? | "total 4 digit no . = 9 * 10 * 10 * 10 = 9000 not containing 2 = 8 * 9 * 9 * 9 = 5832 total 4 digit number contain 2 = 9000 - 5832 = 3168 answer : a" | a = 1000 - 10
b = 4 * 4
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
|
a ) 24 % , b ) 25 % , c ) 30 % , d ) 37.5 % , e ) 40 % | d | multiply(subtract(const_1, divide(multiply(20, const_2), add(add(20, 43), const_1))), const_100) | a bookstore has a shelf that contains biographies which normally sell for $ 20 each and mysteries that normally sell for $ 12 each . during a sale , the biographies and mysteries are discounted at different rates so that a customer saves a total of $ 19 from the normal price by buying 5 discounted biographies and 3 discounted mysteries . if the sum of the discount rates for the two types of books is 43 percent , what is the discount rate on mysteries ? | let b be the discount on biographies and m be the discount on mysteries so . , b + m = 0.435 - - - - - ( 1 ) and ( 20 * 5 + 12 * 3 ) - ( 20 * 5 * ( 1 - b ) + 12 * 3 * ( 1 - m ) ) = 19 - - > 100 ( 1 - ( 1 - b ) ) + 36 ( 1 - ( 1 - m ) = 19 100 b + 36 m = 19 - - - - - - ( 2 ) solving 12 . , we get m = 0.375 = 37.5 % d | a = 20 * 2
b = 20 + 43
c = b + 1
d = a / c
e = 1 - d
f = e * 100
|
a ) 39 , b ) 33 , c ) 34 , d ) 30 , e ) 36 | c | multiply(divide(add(add(floor(divide(15, 4)), const_1), floor(divide(55, 4))), const_2), 4) | calculate the average of all the numbers between 15 and 55 which are divisible by 4 . | explanation : numbers divisible by 4 are 16,20 , 24,28 , 32,36 , 40,44 , 48,52 , , , average = ( 16 + 20 + 24 + 28 + 32 + 36 + 40 + 44 + 48 + 52 ) / 10 = 340 / 10 = 34 answer : c | a = 15 / 4
b = math.floor(a)
c = b + 1
d = 55 / 4
e = math.floor(d)
f = c + e
g = f / 2
h = g * 4
|
a ) 2 , b ) 3 , c ) 0 , d ) 5 , e ) none of the above | c | divide(subtract(power(9, 1), power(9, 1)), 9) | what is the remainder when 9 ^ 1 + 9 ^ 2 + 9 ^ 3 + . . . + 9 ^ 9 is divided by 9 ? | "notice that in the brackets we have the sum of 9 odd multiples of 9 , which yields remainder of 0 upon division by 9 . answer : c" | a = 9 ** 1
b = 9 ** 1
c = a - b
d = c / 9
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a ) rs . 488 , b ) rs . 480 , c ) rs . 482 , d ) rs . 487 , e ) rs . 481 | b | subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 25), const_100)) | by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 25 % ? | "sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 25 % = 25 % of 640 = rs . 160 sp = cp - loss = 640 - 160 = rs . 480 answer : b" | a = 800 * 100
b = 25 + 100
c = a / b
d = 800 * 100
e = 25 + 100
f = d / e
g = f * 25
h = g / 100
i = c - h
|
a ) 10 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 . | a | subtract(add(16, 3), 9) | if ( a + b ) = 16 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 16 = > a = 16 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 16 - b + 3 - c = > 19 - ( b + c ) = > 19 - 9 = 10 . option a . . ." | a = 16 + 3
b = a - 9
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a ) 177 , b ) 150 , c ) 817 , d ) 160 , e ) 616 | d | divide(880, add(add(divide(10, 4), divide(10, 5)), const_1)) | rs . 880 is divided amongst a , b , c so that 4 times a ' s share , 5 times b ' s share and 10 times c ' s share are all equal . find c ' s share ? | a + b + c = 880 4 a = 5 b = 10 c = x a : b : c = 1 / 4 : 1 / 5 : 1 / 10 = 5 : 4 : 2 2 / 11 * 880 = rs . 160 answer : d | a = 10 / 4
b = 10 / 5
c = a + b
d = c + 1
e = 880 / d
|
a ) $ 50 , b ) $ 300 , c ) $ 150 , d ) $ 250 , e ) $ 600 | b | divide(900, 1) | divide $ 900 among a , b in the ratio 1 : 2 . how many $ that a get ? | "sum of ratio terms = 1 + 2 = 3 a = 900 * 1 / 3 = $ 300 answer is b" | a = 900 / 1
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a ) 378 m , b ) 350 m , c ) 380 m , d ) 200 m , e ) 250 m | a | multiply(divide(96, const_3_6), 14) | a car is running at a speed of 96 kmph . what distance will it cover in 14 sec ? | "speed = 96 kmph = 96 * 5 / 18 = 27 m / s distance covered in 14 sec = 27 * 14 = 378 m answer is a" | a = 96 / const_3_6
b = a * 14
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a ) 3 : 2 , b ) 1 : 1 , c ) 1 : 2 , d ) 3 : 4 , e ) 3 : 8 | e | multiply(inverse(multiply(const_2, const_2)), inverse(multiply(const_2, divide(const_1, 3)))) | in a certain aquarium , the number of red fish is 3 times the number of the green fish , and the number of blue fish is half the number of the red fish . if the ratio of green fish to red fish were to be doubled , and the ratio of red fish to blue fish were to be doubled , then the ratio of blue fish to green fish in that aquarium would be | lots of little places to make a mistake or panic here . = ) r = 3 g b = r / 2 g / r = 1 / 3 - - - - > doubled g / r = 2 / 3 r / b = 2 - - - - > doubled r / b = 4 b / g = ( b / r ) * ( r / g ) = ( 1 / 4 ) * ( 3 / 2 ) = 3 / 8 answer : e | a = 2 * 2
b = 1/(a)
c = 1 / 3
d = 2 * c
e = 1/(d)
f = b * e
|
a ) 1 : 3 , b ) 1 : 2 , c ) 1 : 1 , d ) 2 : 1 , e ) 5 : 4 | e | divide(0.845, add(0.8, 0.9)) | in what proportion must flour at $ 0.8 per pound be mixed with flour at $ 0.9 per pound so that the mixture costs $ 0.845 per pound ? | "using weighted average method : let x be the proportion in which the $ 0.8 per pound flour is mixed with $ 0.9 per pound flour . thus 0.8 * x + 0.9 * ( 1 - x ) = 0.845 0.9 – 0.1 x = 0.845 x = 0.55 thus ratio of both flours is 5 : 4 e" | a = 0 + 8
b = 0 / 845
|
a ) 36 mph , b ) 40 mph , c ) 39 mph , d ) 42 mph , e ) 38 mph | c | divide(add(90, 75), add(divide(90, 30), divide(75, 60))) | a car drives 90 miles on local roads at 30 mph , and 75 miles on the highway at 60 mph , what is the average speed of the entire trip ? | so the answer is plainly c . . . . we have a general relation for speed , time and distance : v ( velocity ) * t ( time ) = d ( distance ) for first part we have d = 90 miles , and v = 30 mph so we can obtain time : 30 * t = 90 or t = 90 / 30 = 3 hours the needed time to cover 90 miles in the same way we should divide 75 to 60 to find the needed time to cover 75 miles , so t = 1.25 hours so the total time for covering total distance would be 3 + 1.25 = 4.25 hours and total distance is 90 + 75 = 165 miles final stage : average speed is total distance divide to total time : 165 / 4.25 = 38.8 miles per hour rounded up to 39 mph . . . . | a = 90 + 75
b = 90 / 30
c = 75 / 60
d = b + c
e = a / d
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a ) 2 / 1 , b ) 4 / 3 , c ) 2 / 3 , d ) 3 / 4 , e ) 3 / 2 | b | divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 18)) | in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 18 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 18 ) = 36 / 27 = 4 / 3 . answer : b . | a = 4 + 1
b = 45 / a
c = 45 - b
d = 4 + 1
e = 45 / d
f = e + 18
g = c / f
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a ) 45 , b ) 36 , c ) 40 , d ) 50 , e ) 48 | a | divide(original_price_before_loss(20, 90), divide(original_price_before_gain(20, 60), 20)) | a man sold 20 articles for $ 60 and gained 20 % . how many articles should he sell for $ 90 to incur a loss 20 % ? | "production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 required production costs for a loss of 20 % : $ 90 * ( 100 % + 20 % ) = $ 108 number of articles to be sold for $ 108 to incur a 20 % loss : $ 108 / $ 2.40 = 45 thus , solution a is correct ." | a = original_price_before_loss / (
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a ) 1 / 6 , b ) 1 / 3 , c ) 10 / 21 , d ) 4 / 10 , e ) 3 / 2 | d | multiply(subtract(const_2, 2), const_2) | if x > 3000 , then the value of ( 3 x ) / ( 2 x ^ 1.21 - 1111 ^ 1.3 ) is closest to ? | "assume x = 3002 ( 3 x ) / ( 2 x ^ 1.21 - 1111 ^ 1.3 ) = 9006 / ( 2 * ( 3002 ) ^ 1.21 - 1111 ^ 1.3 ) = 9006 / 23154 = = 4 / 10 d" | a = 2 - 2
b = a * 2
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a ) 235 miles . , b ) 245 miles . , c ) 255 miles . , d ) 265 miles . , e ) 275 miles . | e | add(multiply(55, 2), multiply(3, 55)) | john left home and drove at the rate of 55 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination . how many miles did john drive ? | "the total distance d traveled by john is given by d = 55 * 2 + 3 * 55 = 275 miles . answer e" | a = 55 * 2
b = 3 * 55
c = a + b
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a ) 4 , b ) 10 , c ) 15 , d ) 16 , e ) 17 | d | divide(30, divide(multiply(multiply(2, multiply(multiply(add(2, const_3), 2), const_3)), 2), 30)) | six bells commence tolling together and toll at intervals of 2 , 4,6 , 8,10 and 12 seconds respectively . in 30 m 9 nutes , how many times do they toll together ? | "solution l . c . m . of 2,4 , 6,8 , 10,12 is 120 . so , the bells will toll together after every 120 seconds , i . e . , 2 minutes in 30 minutes , they will toll together left [ ( 30 / 2 ) + 1 ] = 16 times . answer d" | a = 2 + 3
b = a * 2
c = b * 3
d = 2 * c
e = d * 2
f = e / 30
g = 30 / f
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a ) 3 minutes , b ) 6 minutes , c ) 8 minutes , d ) 9 minutes , e ) 10 minutes | a | divide(subtract(6, divide(6, divide(add(2, 6), subtract(6, 2)))), const_1) | a man cycling along the road noticed that every 6 minutes a bus overtakes him and every 2 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed , what is the time interval between consecutive buses ? | "let ' s say the distance between the buses is d . we want to determine interval = \ frac { d } { b } , where b is the speed of bus . let the speed of cyclist be c . every 6 minutes a bus overtakes cyclist : \ frac { d } { b - c } = 6 , d = 6 b - 6 c ; every 2 minutes cyclist meets an oncoming bus : \ frac { d } { b + c } = 2 , d = 2 b + 2 c ; d = 6 b - 6 c = 2 b + 2 c , - - > b = 2 c , - - > d = 6 b - 3 b = 3 b . interval = \ frac { d } { b } = \ frac { 3 b } { b } = 3 answer : a ( 3 minutes ) ." | a = 2 + 6
b = 6 - 2
c = a / b
d = 6 / c
e = 6 - d
f = e / 1
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a ) 287 , b ) 269 , c ) 450 , d ) 200 , e ) 230 | c | divide(square_area(30), const_2) | what is the area of a square field whose diagonal of length 30 m ? | "d 2 / 2 = ( 30 * 30 ) / 2 = 450 answer : c" | a = square_area / (
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a ) 19 , b ) 20 , c ) 32 , d ) 41 , e ) 53 | a | subtract(multiply(add(floor(divide(1056, 25)), const_1), 25), 1056) | what is the least number should be added to 1056 , so the sum of the number is completely divisible by 25 ? | "( 1056 / 25 ) gives remainder 6 19 + 6 = 23 , so we need to add 19 a" | a = 1056 / 25
b = math.floor(a)
c = b + 1
d = c * 25
e = d - 1056
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a ) s 425 , 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)))), 50), 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 50 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 * 50 = 42500 p in rs = 42500 / 100 = rs 425 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 * 50
i = h / 100
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a ) 13.64 % , b ) 15.5 % , c ) 14 % , d ) 14.15 % , e ) 14.95 % | d | divide(multiply(multiply(40, 60), divide(1212, const_100)), multiply(40, subtract(60, 7))) | a man bought 40 shares of rs . 60 at 7 discount , the rate of dividend being 1212 % the rate of interest obtained is | "explanation : face value of a share = rs . 60 he bought each share at rs . 60 - rs . 7 = rs . 53 number of shares = 40 dividend = 12 1 / 2 % = 25 / 2 % dividend per share = 60 × 25 / 2 × 100 = rs . 7.5 total dividend = ( 40 × 7.5 ) ie , he got a dividend of ( 40 × 7.5 ) for an investment of rs . ( 40 × 53 ) interest obtained = 40 × 7.5 × 100 / 40 × 53 = 14.15 % answer : option d" | a = 40 * 60
b = 1212 / 100
c = a * b
d = 60 - 7
e = 40 * d
f = c / e
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a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 7 | d | subtract(853845, multiply(floor(divide(853845, 9)), 9)) | if 853845 bars of chocolate need to be placed in little boxes and each box can contain at most 9 bars of chocolate . how many bars of chocolate will be in the last unfilled box ? | the number of bars of chocolate that can be placed in the little boxes of 9 bars is the first number that is divisible by 9 that occurs before 853845 . in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 853845 = 33 and 27 is divisible by 9 . hence , we need to remove 6 to this number for it to be divisible by 9 correct option : d | a = 853845 / 9
b = math.floor(a)
c = b * 9
d = 853845 - c
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a ) 245 , b ) 265 , c ) 295 , d ) 325 , e ) 375 | b | subtract(multiply(divide(multiply(45, const_1000), const_3600), 30), 110) | the length of a bridge in meters , which a train 110 - meters long and traveling at 45 km / hr can cross in 30 seconds is ? | "45 km / h = 45000 m / 3600 s = 12.5 m / s in 30 seconds , the train can go 30 ( 12.5 ) = 375 meters let x be the length of the bridge . x + 110 = 375 meters x = 265 meters the answer is b ." | a = 45 * 1000
b = a / 3600
c = b * 30
d = c - 110
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a ) 2000 , b ) 3000 , c ) 4000 , d ) 5000 , e ) 6000 | c | subtract(36000, multiply(divide(8, 9), 36000)) | income and expenditure of a person are in the ratio 9 : 8 . if the income of the person is rs . 36000 , then find his savings ? | "let the income and the expenditure of the person be rs . 9 x and rs . 7 x respectively . income , 9 x = 36000 = > x = 4000 savings = income - expenditure = 9 x - 8 x = x so , savings = rs . 4000 . answer : c" | a = 8 / 9
b = a * 36000
c = 36000 - b
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a ) 311 , b ) 322 , c ) 313 , d ) 314 , e ) 385 | a | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 16) | light glows for every 16 seconds . how many times did it between 1 : 57 : 58 and 3 : 20 : 47 am | "the diff in sec between 1 : 57 : 58 and 3 : 20 : 47 is 4969 sec , 4969 / 16 = 310 . so total 311 times light ll glow answer : a" | a = 2 + 47
b = 2 + const_60
c = 20 + b
d = c * const_60
e = a + d
f = e / 16
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a ) 18 , b ) 12 , c ) 9 , d ) 14 , e ) 16 | d | sqrt(294) | the difference between c . i . and s . i . on an amount of rs . 15,000 for 2 years is rs . 294 . what is the rate of interest per annum ? | "explanation : [ 15000 * ( 1 + r / 100 ) 2 - 15000 ] - ( 15000 * r * 2 ) / 100 = 294 15000 [ ( 1 + r / 100 ) 2 - 1 - 2 r / 100 ] = 294 15000 [ ( 100 + r ) 2 - 10000 - 200 r ] / 10000 = 294 r 2 = ( 294 * 2 ) / 3 = 196 = > r = 14 rate = 14 % answer : option d" | a = math.sqrt(294)
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a ) 1 hours , b ) 2 hours , c ) 2 / 3 hours , d ) 1 / 3 hours , e ) 2 / 5 hours | a | subtract(divide(multiply(divide(const_1, const_2), 30), subtract(60, 30)), divide(const_1, const_2)) | a thief goes away with a santro car at a speed of 30 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 60 kmph when will the owner over take the thief from the start ? | "- - - - - - - - - - - 30 - - - - - - - - - - - - - - - - - - - - | 60 30 d = 30 rs = 60 â € “ 30 = 30 t = 30 / 30 = 1 hours answer : a" | a = 1 / 2
b = a * 30
c = 60 - 30
d = b / c
e = 1 / 2
f = d - e
|
a ) a ) 82390 , b ) b ) 29000 , c ) c ) 20005 , d ) d ) 23930 , e ) e ) 20000 | e | multiply(multiply(subtract(4, 3), 10000), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 10000 more than d , what is b ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x rs . respectively . then , 4 x - 3 x = 1000 = > x = 10000 . b ' s share = rs . 2 x = 2 * 10000 = rs . 20000 . answer : e" | a = 4 - 3
b = a * 10000
c = b * 3
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a ) 25 % , b ) 30 % , c ) 35 % , d ) 40 % , e ) 50 % | d | multiply(divide(multiply(subtract(const_1, divide(const_2.0, 3)), multiply(divide(1, 2), const_100)), subtract(const_100, multiply(divide(1, 2), const_100))), const_100) | one night a certain hotel rented 1 / 2 of its rooms , including 2 / 3 of their air conditioned rooms . if 3 / 5 of its rooms were air conditioned , what percent of the rooms that were not rented were air conditioned ? | "the rooms which were not rented is 1 / 2 the ac rooms which were not rented is ( 1 / 3 ) * ( 3 / 5 ) = 1 / 5 the percentage of unrented rooms which were ac rooms is ( 1 / 5 ) / ( 1 / 2 ) = 2 / 5 = 40 % the answer is d ." | a = 2 / 0
b = 1 - a
c = 1 / 2
d = c * 100
e = b * d
f = 1 / 2
g = f * 100
h = 100 - g
i = e / h
j = i * 100
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a ) 115 , b ) 117 , c ) 116 , d ) 114 , e ) 112 | b | divide(multiply(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)), multiply(add(24, 24), add(24, 24))) | a class is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the class room ? | explanation : length = 6 m 24 cm = 624 cm width = 4 m 32 cm = 432 cm hcf of 624 and 432 = 48 number of square tiles required = ( 624 x 432 ) / ( 48 x 48 ) = 13 x 9 = 117 . answer is b | a = 6 * 100
b = a + 24
c = 4 * 100
d = c + 32
e = b * d
f = 24 + 24
g = 24 + 24
h = f * g
i = e / h
|
a ) 0.2 , b ) 0.5 , c ) 0.6 , d ) 0.8 , e ) 1.0 | d | divide(multiply(divide(multiply(8, 5), const_100), 128), const_100) | 128 % of 5 / 8 = | "should be simple . 1.28 * 5 / 8 = 6.4 / 8 = 0.8 correct option : d" | a = 8 * 5
b = a / 100
c = b * 128
d = c / 100
|
a ) 20 , b ) 99 , c ) 27 , d ) 22 , e ) 29 | a | multiply(divide(multiply(4, 5), subtract(multiply(4, 4), multiply(3, 4))), 4) | mixture contains alcohol and water in the ratio 4 : 3 . if 4 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture . | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 16 x = 4 ( 3 x + 5 ) 4 x = 20 x = 5 quantity of alcohol = ( 4 x 5 ) litres = 20 litres . answer : a" | a = 4 * 5
b = 4 * 4
c = 3 * 4
d = b - c
e = a / d
f = e * 4
|
a ) 300 , b ) 120 , c ) 150 , d ) 170 , e ) 270 | b | divide(multiply(20, 12), const_2) | the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 12 terms of the arithmetic progression ? | "n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n / 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 12 . . . we get 12 / 2 [ 2 a + 14 d ] = 15 [ a + 7 d ] = 12 * 10 = 120 . . . answer is b . . ." | a = 20 * 12
b = a / 2
|
a ) 20,196 , b ) 19,780 , c ) 19,680 , d ) 19,380 , e ) none of these | a | floor(divide(divide(multiply(add(multiply(multiply(add(multiply(add(const_1, const_4), const_2), 8), const_100), multiply(add(const_1, const_4), const_2)), multiply(add(const_1, const_4), const_100)), add(const_100, 8)), const_100), multiply(multiply(add(const_1, const_4), const_100), const_2))) | sonika bought a v . c . r . at the list price of 18,700 . if the rate of sales tax was 8 % , find the amount she had to pay for purchasing the v . c . r . | "sol . list price of v . c . r . = 18,700 rate of sales tax = 8 % ∴ sales tax = 8 % of 18,700 = 8 ⁄ 100 × 18700 = 1496 so , total amount which sonika had to pay for purchasing the v . c . r . = 18,700 + 1496 = 20,196 . answer a" | a = 1 + 4
b = a * 2
c = b + 8
d = c * 100
e = 1 + 4
f = e * 2
g = d * f
h = 1 + 4
i = h * 100
j = g + i
k = 100 + 8
l = j * k
m = l / 100
n = 1 + 4
o = n * 100
p = o * 2
q = m / p
r = math.floor(q)
|
a ) 275 m , b ) 360 m , c ) 375 m , d ) 420 m , e ) 440 m | c | multiply(multiply(54, const_0_2778), 25) | if the speed of a man is 54 km per hour , then what is the distance traveled by him in 25 seconds ? | the distance traveled in 25 sec = 54 * ( 5 / 18 ) * 25 = 375 m answer : c | a = 54 * const_0_2778
b = a * 25
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a ) a ) 300 , b ) b ) 150 , c ) c ) 130 , d ) d ) 90 , e ) e ) 210 | a | multiply(100, 3) | soja finished 2 / 3 of a book . she calculated that she finished 100 more pages than she has yet to read . how long is her book ? | let x be the total number of pages in the book , then she finished 2 / 3 * x pages . then she has x − 2 / 3 * x = 1 / 3 * x pages left . 2 / 3 * x − 1 / 3 * x = 100 1 / 3 * x = 100 x = 300 so the book is 300 pages long . answer is a . | a = 100 * 3
|
a ) 21 st , b ) 22 nd , c ) 33 rd , d ) 24 th , e ) none of these | c | subtract(multiply(2, 18), 1) | a monkey ascends a greased pole 18 metres high . he ascends 2 metres in first minute and slips down 1 metre in the alternate minute . in which minute , he reaches the top ? | "in 2 minutes , he ascends = 1 metre â ˆ ´ 16 metres , he ascends in 32 minutes . â ˆ ´ he reaches the top in 33 rd minute . answer c" | a = 2 * 18
b = a - 1
|
a ) 50 , b ) 48 , c ) 65 , d ) 66 , e ) 72 | d | divide(add(multiply(10, 60), multiply(15, 70)), 25) | the average score of a cricketer in 10 matches is 60 and in other 15 matches is 70 . then find the average score in all the 25 matches ? | "average in 25 matches = ( 10 * 60 + 15 * 70 ) / 10 + 15 = 66 answer is d" | a = 10 * 60
b = 15 * 70
c = a + b
d = c / 25
|
a ) 15 , b ) 30 , c ) 36 , d ) 40 , e ) 45 | c | subtract(multiply(divide(30, const_100), 180), multiply(divide(const_1, const_3), multiply(divide(30, const_100), 180))) | one - third less than 30 % of 180 is equal to : | "lots of ways to tackle this . 30 % of 180 = 54 1 / 3 of 45 = 18 so , 1 / 3 less than 54 is equal to 54 - 18 = 36 answer : c" | a = 30 / 100
b = a * 180
c = 1 / 3
d = 30 / 100
e = d * 180
f = c * e
g = b - f
|
a ) 100 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | a | divide(subtract(multiply(35, 120), multiply(120, 15)), subtract(39, 15)) | the average of marks obtained by 120 boys was 35 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | "let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 35 24 x = 4200 - 1800 = > x = 2400 / 24 x = 100 . hence , the number of boys passed = 100 . answer : a" | a = 35 * 120
b = 120 * 15
c = a - b
d = 39 - 15
e = c / d
|
a ) 38 , b ) 47 , c ) 50 , d ) 53 , e ) 93 | e | add(divide(450, 5), 3) | a whale goes on a feeding frenzy that lasts for 5 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 450 kilos of plankton , how many kilos did he consume on the third hour ? | "if you list the amount eaten each hour , you ' ll get an equally spaced list , increasing by 3 each hour . in any equally spaced list , the median equals the mean . here , the mean is 450 / 5 = 90 , so the median is also 90 , and that is the amount eaten in the 2 nd hour . we need to add 3 to find the total eaten in the next hour , so the answer is 93 option e" | a = 450 / 5
b = a + 3
|
a ) $ 500 , b ) $ 1000 , c ) $ 1500 , d ) $ 2000 , e ) $ 2500 | b | multiply(multiply(subtract(4, 3), 500), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets $ 500 more than d , what is b ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x respectively . then , 4 x - 3 x = 500 x = $ 500 b ' s share = 2 x = 2 * $ 500 = $ 1000 the answer is b ." | a = 4 - 3
b = a * 500
c = b * 3
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | c | multiply(1, 5) | if 5 spiders make 5 webs in 5 days , then 1 spider will make 1 web in how many days ? | "let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 5 webs 5 : 1 1 x 5 x x = 5 x 1 x 5 = > x = 5 answer is c" | a = 1 * 5
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a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 6000 , e ) s . 6880 | d | divide(multiply(multiply(multiply(24, 4), 4), 150000), multiply(multiply(multiply(10, 16), 6), 10)) | 10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 150000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 150000 p = rs . [ ( 24 * 4 * 4 * 150000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 57600000 / 9600 ) = > p = rs . 6000 answer : d" | a = 24 * 4
b = a * 4
c = b * 150000
d = 10 * 16
e = d * 6
f = e * 10
g = c / f
|
a ) 6150 , b ) 6250 , c ) 6350 , d ) 6450 , e ) 6550 | d | divide(2451, subtract(subtract(const_1, divide(31, const_100)), divide(31, const_100))) | a candidate got 31 % of the votes polled and he lost to his rival by 2451 votes . how many votes were cast ? | "let x be the total number of votes . 0.31 x + 2451 = 0.69 x 0.38 x = 2451 x = 2451 / 0.38 = 6450 the answer is d ." | a = 31 / 100
b = 1 - a
c = 31 / 100
d = b - c
e = 2451 / d
|
a ) $ 250 , b ) $ 300 , c ) $ 500 , d ) $ 600 , e ) $ 1000 | d | multiply(2000, divide(3, const_100)) | find the simple interest on $ 2000 for 3 years at 10 % per annum ? | "si = ptr / 100 = 2000 * 3 * 10 / 100 = $ 600 answer is d" | a = 3 / 100
b = 2000 * a
|
a ) 100 , b ) 120 , c ) 250 , d ) 200 , e ) 160 | a | divide(multiply(10, const_100), 10) | an inspector rejects 10 % of the meters as defective . how many will he examine to reject 10 ? | "then , 10 % of x = 10 ( 10 / 100 ) x = 10 x = ( 10 * 100 * ) / 10 = 100 answer is a" | a = 10 * 100
b = a / 10
|
a ) 973 , b ) 6973 , c ) 3996 , d ) 6084 , e ) none of these | c | subtract(multiply(const_10, 4), 4) | the difference between the place value and the face value of 4 in the numeral 854973 is | "( place value of 4 ) - ( face value of 4 ) = ( 4000 - 4 ) = 3996 answer : option c" | a = 10 * 4
b = a - 4
|
a ) 36.1 , b ) 36.5 , c ) 36.22 , d ) 36.12 , e ) 36.18 | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 36 . it was found later that an observation 29 was wrongly taken as 23 . the corrected new mean is : | "explanation : correct sum = ( 36 * 50 + 29 - 23 ) = 1806 . correct mean = = 1806 / 50 = 36.12 answer : d ) 36.12" | a = 36 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) $ 2,750 , b ) $ 5,500 , c ) $ 11,000 , d ) $ 22,000 , e ) $ 44,000 | a | multiply(divide(multiply(divide(550, 2), divide(550, 2)), subtract(605, 550)), 2) | shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $ 550 as interest . he invested the remaining in a bond that paid compound interest , interest being compounded annually , for the same 2 years at the same rate of interest and received $ 605 as interest . what was the value of his total savings before investing in these two bonds ? | "si for 1 st year = 550 / 2 = 275 275 + 275 + si on ( 275 ) = 605 = = > si on 275 = 55 = = > si on 275 = 275 * r * 1 / 100 = 55 r = 20 % now using si for 2 years on t / 2 of the money ( t / 2 ) * 20 * 2 / 100 = 550 t = 550 * 5 = 2750 answer : a" | a = 550 / 2
b = 550 / 2
c = a * b
d = 605 - 550
e = c / d
f = e * 2
|
a ) 25 % , b ) 40 % , c ) 35 % , d ) 8 % , e ) 12 % | c | multiply(divide(350, 1), const_100) | what percent is 350 gm of 1 kg ? | "1 kg = 1000 gm 350 / 1000 ã — 100 = 35000 / 1000 = 35 % answer is c" | a = 350 / 1
b = a * 100
|
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | e | divide(subtract(multiply(30, 7), multiply(3, 7)), 7) | 7 people average age is 30 . youngest person age is 3 . find average of the people when youngest was born . | "average age of people = 30 so have total age = 210 before 7 years we have to deduct each person age by seven years 210 - 21 = 189 so average age would be 189 / 7 = 27 answer : e" | a = 30 * 7
b = 3 * 7
c = a - b
d = c / 7
|
a ) 3 , b ) 9 , c ) 12 , d ) 15 , e ) 24 | e | add(8, const_1) | the average of first five multiples of 8 is : | "solution average = 8 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 120 / 5 = 24 answer e" | a = 8 + 1
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a ) 1 / 6 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 4 | a | divide(multiply(power(2, 5), power(9, 2)), multiply(power(8, 2), power(3, 5))) | what is value of ( ( 2 ^ 5 ) * ( 9 ^ 2 ) ) / ( ( 8 ^ 2 ) * ( 3 ^ 5 ) ) = ? | = > ( ( 2 ^ 5 ) * ( ( 3 ^ 2 ) ^ 2 ) / ( ( 2 ^ 3 ) ^ 2 ) * ( 3 ^ 5 ) ) = > ( ( 2 ^ 5 ) * ( 3 ^ 4 ) ) / ( ( 2 ^ 6 ) * ( 3 ^ 5 ) ) = > 1 / ( 2 * 3 ) = > 1 / 6 option a is answer | a = 2 ** 5
b = 9 ** 2
c = a * b
d = 8 ** 2
e = 3 ** 5
f = d * e
g = c / f
|
['a ) 3 / 5', 'b ) 7 / 16', 'c ) 12 / 16', 'd ) 24 / 12', 'e ) 2 / 3'] | b | divide(subtract(circle_area(divide(16, const_2)), circle_area(divide(12, const_2))), circle_area(divide(16, const_2))) | two circular frames are kept one above the other . frame x has a diameter of 16 cm and frame y has a diameter of 12 cm . what fraction of the surface of frame x is not covered by frame y ? | the questions asks us to find the surface which is not covered by the frame y i . e . , area of the surface not covered . where as circumference is the length along the edge of the circle , 2 * pi * r implies the length of the curve pi * r ^ 2 implies area enclosed by that curve . . hence area of the circle is considered for this problem area of the frame y = pi * r ^ 2 , where r = 6 = > pi * 36 area of the frame x = pi * 64 surface not covered by the frame y = pi * 64 - pi * 36 = pi * 28 fraction of frame x ' s surface is not covered by a frame y = pi * 28 / total area of the frame x = 28 / 64 = 7 / 16 answer is b | a = 16 / 2
b = circle_area - (
c = 12 / 2
d = b / circle_area
|
a ) 4.5 , b ) 5.5 , c ) 6.5 , d ) 7.5 , e ) 3.5 | e | divide(140, 40) | a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 140 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 140 kilometers ? 140 ÷ 40 = 3.5 × 1 gallon = 3.5 gallons correct answer e" | a = 140 / 40
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a ) 5 kg , b ) 10 kg , c ) 15 kg , d ) 20 kg , e ) 25 kg | b | divide(multiply(divide(20, const_100), 10), subtract(divide(40, const_100), divide(20, const_100))) | in some quantity of ghee , 60 % of pure ghee and 40 % of is vanaspati . if 10 kg of pure ghee is added , then the strength of vanaspati ghee becomes 20 % . the original quantity was ? | let the original quantity be x kg vanaspati ghee in x kg = 40 / 100 * x = 2 x / 5 kg ( 2 x / 5 ) / ( x + 10 ) = 20 / 100 2 x / ( 5 x + 50 ) = 1 / 5 5 x = 50 x = 10 kg answer is b | a = 20 / 100
b = a * 10
c = 40 / 100
d = 20 / 100
e = c - d
f = b / e
|
a ) 7 / 10 , b ) 15 / 14 , c ) 4 / 5 , d ) 5 / 4 , e ) can not be determined | a | divide(divide(42, 6), divide(42, 4.2)) | jack and jill are marathon runners . jack can finish a marathon ( 42 km ) in 6 hours and jill can run a marathon in 4.2 hours . what is the ratio of their average running speed ? ( jack : jill ) | "average speed of jack = distance / time = 42 / 6 = 7 average speed of jill = 42 / ( 4.2 ) = 10 ratio of average speed of jack to jill = 7 / 10 answer a" | a = 42 / 6
b = 42 / 4
c = a / b
|
a ) 524.48 , b ) 556.02 , c ) 58.3 , d ) 585.64 , e ) 595.46 | c | multiply(128.5, power(add(const_4, const_1), const_4)) | ( 128.5 x 4 ) + ( 13.8 x 5 ) = ? x 10 | "explanation : ? = ( 128.5 x 4 ) + ( 13.8 x 5 ) / 10 = 514 + 69 / 10 = 58.3 answer : option c" | a = 4 + 1
b = a ** 4
c = 128 * 5
|
a ) 11 , b ) 12 , c ) 10 , d ) 2 , e ) 7 | c | subtract(56, add(12, 34)) | when a natural number n is successively divided by 12 , 34 the remainders are 56 , 78 . what will be the sum of the remainders if the order of the division is reversed ? | 1 2 3 4 5 6 7 8 leave the top right - most number 8 start with bottom right - most number 5 8 * 3 + 7 = 31 31 * 2 + 6 = 68 68 * 1 + 5 = 73 this is the number required now , do the successive division in the reverse order the sum of the remainders is 10 hence , the correct option is c | a = 12 + 34
b = 56 - a
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a ) 160 hr , b ) 200 hr , c ) 120 hr , d ) 140 hr , e ) 180 hr | b | multiply(divide(add(divide(252, add(5, 2)), divide(252, subtract(5, 2))), const_60), const_100) | speed of a boat in still water is 5 km / hr and speed of the stream is 2 kmph . a man rows to a place at a distance of 252 km and comes back to the starting point the total time taken by him is . | explanation : speed up stream = 5 - 2 = 3 km / hr speed down stream = 5 + 2 = 7 km / hr total time taken = 252 / 3 + 252 / 7 = 84 + 36 = 120 hr answer : option b | a = 5 + 2
b = 252 / a
c = 5 - 2
d = 252 / c
e = b + d
f = e / const_60
g = f * 100
|
a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | e | add(divide(subtract(add(40, 2), 30), 1.5), 30) | each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 42 hours if harry and james were paid the same amount last week , how many hours did harry work last week ? | "james worked for 41 hours hence he earned 40 * x + 2 * 2 x = 44 x dollars ; we know that harry also earned the same 44 x dollars , out of which he earned 30 x dollars for thefirst 30 hoursplus 14 x additional dollars . since for each additional hour he gets 1.5 x dollars then he worked for 14 x / 1.5 x = 9 additional hours , so harry worked for total of 30 + 9 = 39 hours . answer : e ." | a = 40 + 2
b = a - 30
c = b / 1
d = c + 30
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a ) 1 / 35 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 5 , e ) 4 / 5 | d | divide(subtract(divide(35, const_100), divide(50, const_100)), subtract(divide(25, const_100), divide(50, const_100))) | some of 50 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 35 % . what fraction of the original paint was replaced ? | "35 % is 10 % - points above 25 % and 15 % - points below 50 % . thus the ratio of 25 % - solution to 50 % - solution is 3 : 2 . 3 / 5 of the original paint was replaced . the answer is d ." | a = 35 / 100
b = 50 / 100
c = a - b
d = 25 / 100
e = 50 / 100
f = d - e
g = c / f
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a ) 12.5 % , b ) 13 % , c ) 15 % , d ) 17 % , e ) 20 % | e | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 64)), const_100), multiply(const_100, const_100)) | the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 64 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.64 x 1.25 = 0.8 decrease in price = ( 0.2 / 1 ) × 100 = 20 % e )" | a = 100 * 100
b = 100 + 25
c = b * 64
d = a - c
e = d * 100
f = 100 * 100
g = e / f
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | a | subtract(divide(multiply(multiply(30, subtract(10, 2)), 10), multiply(2, subtract(30, 10))), 30) | an engineer undertakes a project to build a road 10 km long in 30 days and employs 30 men for the purpose . after 10 days , he finds only 2 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "30 workers working already let x be the total men required to finish the task in next 20 days 2 km done hence remaining is 8 km also , work has to be completed in next 20 days ( 30 - 10 = 20 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion hence , x = ( 30 * 8 * 10 ) / ( 2 * 20 ) thus , x = 60 thus , more men needed to finish the task = 60 - 30 = 30 answer : a" | a = 10 - 2
b = 30 * a
c = b * 10
d = 30 - 10
e = 2 * d
f = c / e
g = f - 30
|
a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 21 | e | add(add(add(const_4, 3), add(3, const_2)), 3) | the number 155 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "7 ^ 2 + 5 ^ 2 + 9 ^ 2 = 49 + 25 + 81 = 155 7 + 5 + 9 = 21 hence answer is e" | a = 4 + 3
b = 3 + 2
c = a + b
d = c + 3
|
a ) 25 , b ) 43 , c ) 14 , d ) 36 , e ) 12 | d | subtract(multiply(add(21, const_3), const_2), multiply(const_3, const_4)) | two digits in john ' s age are the same as the digit in the wilson ' s age , but in reverse order . in 21 years john will be twice as old as wilson will be then . what is the difference in their current age ? | john ' s age = 10 x + y so wilson ' s age = 10 y + x . . after 21 years , 10 x + y + 21 = 2 * ( 10 y + x + 21 ) . . . . so 21 + 19 y = 8 x . . . check for odd values of y , y = 1 satisfies the eqn with x = 5 . . . so ages are 51 and 15 and ans as found correctly by u is d . . 36 | a = 21 + 3
b = a * 2
c = 3 * 4
d = b - c
|
a ) $ 58 , b ) $ 42 , c ) $ 65 , d ) $ 48 , e ) $ 62 | a | divide(add(add(add(add(45, 50), 60), 65), 70), 5) | a cab driver 5 days income was $ 45 , $ 50 , $ 60 , $ 65 , $ 70 . then his average income is ? | avg = sum of observations / number of observations avg income = ( 45 + 50 + 60 + 65 + 70 ) / 5 = 58 answer is a | a = 45 + 50
b = a + 60
c = b + 65
d = c + 70
e = d / 5
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a ) 1 : 3 , b ) 3 : 5 , c ) 3 : 4 , d ) 3 : 7 , e ) 1 : 5 | b | multiply(divide(4, 5), divide(3, 4)) | if a : b is 3 : 4 and b : c is 4 : 5 then a : c is equal to | the two ratios given are having the same number 4 for b in both the ratios . hence - a : b = 3 : 4 b : c = 4 : 5 = > a : c = 3 : 5 answer b | a = 4 / 5
b = 3 / 4
c = a * b
|
a ) 5 / 12 , b ) 1 / 15 , c ) 2 / 13 , d ) 2 / 15 , e ) 1 / 17 | a | divide(divide(factorial(subtract(9, 2)), multiply(factorial(subtract(subtract(9, 2), 3)), factorial(3))), divide(factorial(9), multiply(factorial(subtract(9, 3)), factorial(3)))) | in a box of 9 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | first , there are 7 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 9 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 7 c 3 / 9 c 3 = 35 / 84 = 5 / 12 answer is a | a = 9 - 2
b = math.factorial(a)
c = 9 - 2
d = c - 3
e = math.factorial(d)
f = math.factorial(3)
g = e * f
h = b / g
i = math.factorial(9)
j = 9 - 3
k = math.factorial(j)
l = math.factorial(3)
m = k * l
n = i / m
o = h / n
|
a ) 20 % , b ) 26 % , c ) 42 % , d ) 27 % , e ) 23 % | e | subtract(divide(subtract(const_100, 18), 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 18 % ? | sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 82 2 / 3 sp 1 = 82 sp 1 = 123 100 - - - 23 = > 23 % answer : e | a = 100 - 18
b = 2 / 3
c = a / b
d = c - 100
|
a ) 2.25 hours , b ) 8 hours , c ) 12 hours , d ) 14.25 hours , e ) 16 hours | b | divide(multiply(6, 8), 6) | 8 identical machines , working alone and at their constant rates , take 6 hours to complete a job lot . how long would it take for 6 such machines to perform the same job ? | "let each machine do 1 unit of work for 1 hour 8 machines - - > 8 units of work in 1 hour for 6 hours = 8 * 6 = 48 units of total work is done . now this 48 units of total work must be done by 6 machines 6 units of work ( 6 machines ) - - - > 1 hour for 48 units of work 6 * 8 - - - > 1 * 8 hours b 8 hours" | a = 6 * 8
b = a / 6
|
a ) s . 429 , b ) s . 480 , c ) s . 429 , d ) s . 128 , e ) s . 384 | e | subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 40), const_100)) | by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 40 % ? | "sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 40 % = 40 % of 640 = rs . 256 sp = cp - loss = 640 - 256 = rs . 384 answer : e" | a = 800 * 100
b = 25 + 100
c = a / b
d = 800 * 100
e = 25 + 100
f = d / e
g = f * 40
h = g / 100
i = c - h
|
a ) 20 liters , b ) 25 liters , c ) 30 liters , d ) 15 liters , e ) 18 liters | a | subtract(divide(multiply(divide(10, const_100), 20), divide(5, const_100)), 20) | milk contains 10 % water . what content of pure milk should be added to 20 liters of milk to reduce this to 5 % ? | quantity of water in 20 liters = 10 % of 20 liters = 2 liters let x liters of pure milk be added . then , 2 / ( 20 + x ) = 5 / 100 5 x = 100 x = 20 liters answer is a | a = 10 / 100
b = a * 20
c = 5 / 100
d = b / c
e = d - 20
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | a | multiply(divide(subtract(const_1, add(divide(const_1, 3), divide(const_1, 2))), divide(const_1, 2)), const_60) | it takes avery 3 hours to build a brick wall while tom can do it in 2 hours . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ? | "avery takes 3 hours tom takes 2 hours efficiency of avery is 1 / 3 units / hr efficiency of tom is 1 / 2 units / hr combined efficiency of tom and avery is 1 / 3 + 1 / 2 = 5 / 6 units / hr since they worked for 1 hour they completed 5 / 6 units of work and 1 / 6 units of work is left which is to be completed by tom ( since avery left ) so time taken by tom to complete the remaining work will be 1 / 6 / 1 / 2 hours = > 1 / 3 * 60 = 20 minutes . . . answer will be ( a )" | a = 1 / 3
b = 1 / 2
c = a + b
d = 1 - c
e = 1 / 2
f = d / e
g = f * const_60
|
a ) $ 12,000 , b ) $ 18,000 , c ) $ 15,000 , d ) $ 4,500 , e ) $ 4,000 | a | divide(add(divide(subtract(360, multiply(divide(6, const_100), 2,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(360, multiply(divide(6, const_100), 2,000)), subtract(divide(8, const_100), divide(6, const_100)))), 2,000) | salesperson a ' s compensation for any week is $ 360 plus 6 percent of the portion of a ' s total sales above $ 2,000 for that week . salesperson b ' s compensation for any week is 8 percent of a ' s total sales for that week . for what amount of total weekly sales would both salepeople earn the same compensation ? | "sometime , setting up an equation is an easy way to go with : 360 + 0.06 ( x - 2000 ) = 0.08 x x = 12000 ans : a" | a = 6 / 100
b = a * 2
c = 360 - b
d = 8 / 100
e = 6 / 100
f = d - e
g = c / f
h = 6 / 100
i = h * 2
j = 360 - i
k = 8 / 100
l = 6 / 100
m = k - l
n = j / m
o = g + n
p = o / 2
|
a ) 4.5 , b ) 5 , c ) 6.5 , d ) 7.5 , e ) 8.5 | b | divide(200, 40) | a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 200 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 180 kilometers ? 200 ÷ 40 = 5 × 1 gallon = 5 gallons correct answer b" | a = 200 / 40
|
a ) 23 / 200 , b ) 11 / 100 , c ) 7 / 50 , d ) 3 / 20 , e ) 4 / 25 | a | add(multiply(subtract(const_1, divide(9, 10)), subtract(const_1, divide(3, 5))), multiply(subtract(const_1, divide(7, 8)), divide(3, 5))) | at a monthly meeting , 3 / 5 of the attendees were males and 7 / 8 of the male attendees arrived on time . if 9 / 10 of the female attendees arrived on time , what fraction of the attendees at the monthly meeting did not arrive on time ? | males who did not arrive on time are 1 / 8 * 3 / 5 = 3 / 40 of the attendees . females who did not arrive on time are 1 / 10 * 2 / 5 = 2 / 50 of the attendees . the fraction of all attendees who did not arrive on time is 3 / 40 + 2 / 50 = 23 / 200 the answer is a . | a = 9 / 10
b = 1 - a
c = 3 / 5
d = 1 - c
e = b * d
f = 7 / 8
g = 1 - f
h = 3 / 5
i = g * h
j = e + i
|
a ) 950 , b ) 1050 , c ) 2150 , d ) 2050 , e ) 1000 | b | multiply(choose(15, 2), choose(10, 1)) | there are 15 boys and 10 girls in a class . if three students are selected at random , in how many ways that 1 girl and 2 boys are selected ? | "n ( s ) = sample space = 25 c 3 = 2300 e = event that 1 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 15 and 1 girl from 10 = 15 c 2 * 10 c 1 = 1050 ans - b" | a = math.comb(15, 2)
b = math.comb(10, 1)
c = a * b
|
a ) 55 days , b ) 77 days , c ) 20 days , d ) 88 days , e ) 15 days | e | divide(multiply(25, 12), 20) | 12 men can complete a piece of work in 25 days . in how many days can 20 men complete that piece of work ? | "12 * 25 = 20 * x = > x = 15 days answer : e" | a = 25 * 12
b = a / 20
|
a ) rs . 2000 , b ) rs . 10000 , c ) rs . 15000 , d ) rs . 20000 , e ) none of these | d | divide(7200, divide(multiply(3, 12), const_100)) | consider a lady took a loan from a bank at the rate of 12 % p . a . simple interest . after 3 years she had to pay rs . 7200 interest only for the period . the principal amount borrowed by her was | explanation : principal = rs . ( 100 × 7200 / 12 × 3 ) = > rs . 20,000 . answer : d | a = 3 * 12
b = a / 100
c = 7200 / b
|
a ) a ) 7 , b ) b ) 5 , c ) c ) 9 , d ) d ) 8 , e ) e ) none of the above | b | divide(150, 30) | how many pages do you need if you have to write 150 lines of information into pages that each hold 30 lines ? | sol . 30 lines per page total 150 lines = 150 / 30 = 5 answer : b | a = 150 / 30
|
a ) 5 : 1 , b ) 10 : 1 , c ) 15 : 1 , d ) 20 : 1 , e ) 25 : 1 | b | divide(multiply(10, 2), 2) | a dog takes 10 leaps for every 2 leaps of a hare . if one leap of the dog is equal to 2 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 10 * 2 ) leaps of hare : 2 leaps of hare = 20 : 2 . answer : b ) 10 : 1" | a = 10 * 2
b = a / 2
|
a ) 100 , b ) 120 , c ) 130 , d ) 150 , e ) 160 | d | multiply(divide(multiply(60, const_1000), const_3600), 9) | a train running at the speed of 60 km / hr crosses a pole in 9 seconds . 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 * 9 ) m = 150 m . answer : option d" | a = 60 * 1000
b = a / 3600
c = b * 9
|
a ) 4 , b ) 16 , c ) 8 , d ) 8 √ 2 , e ) can not be determined from the information provided | b | sqrt(add(power(multiply(sqrt(64), sqrt(const_2)), const_2), power(multiply(sqrt(64), sqrt(const_2)), const_2))) | triangle xyz is an isosceles right triangle . if side xy is longer than side yz , and the area of the triangle is 64 , what is the measure of side xy ? | "ans b . . 16 . xy being larger means it is the hyp . . area = ( 1 / 2 ) * ( yz ) ^ 2 = 64 or yz = 8 * \ sqrt { 2 } . . therefore hyp = xy = 16" | a = math.sqrt(64)
b = math.sqrt(2)
c = a * b
d = c ** 2
e = math.sqrt(64)
f = math.sqrt(2)
g = e * f
h = g ** 2
i = d + h
j = math.sqrt(i)
|
a ) s . 684 , b ) s . 500 , c ) s . 540 , d ) s . 740 , e ) s . 840 | a | subtract(780, multiply(divide(subtract(1020, 780), 5), 2)) | a sum of money lent out at s . i . amounts to rs . 780 after 2 years and to rs . 1020 after a further period of 5 years . the sum is ? | "s . i for 5 years = ( 1020 - 780 ) = rs . 240 . s . i . for 2 years = 240 / 5 * 2 = rs . 96 . principal = ( 780 - 96 ) = rs . 684 answer : a" | a = 1020 - 780
b = a / 5
c = b * 2
d = 780 - c
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | e | subtract(15, const_1) | there are 40 balls which are red , blue or green . if 15 balls are green and the sum of red balls and green balls is less than 25 , at most how many red balls are there ? | "r + g + b = 40 g = 15 r + g < 25 = > r + 15 < 25 = > r < 10 = > at most 6 red balls answer : e" | a = 15 - 1
|
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