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 ) 8.05 , b ) 9 , c ) 10 , d ) 15 , e ) 15.5 | a | divide(sqrt(divide(add(add(power(subtract(divide(add(add(10, 1010), 1010), const_3), 1010), const_2), power(subtract(divide(add(add(10, 1010), 1010), const_3), 1010), const_2)), power(subtract(divide(add(add(10, 1010), 1010), const_3), 10), const_2)), subtract(const_3, const_1))), multiply(add(const_4, const_3), 10)) | calculate the standard deviation of each data set { 10 , 1010 , 1010 } | standard deviation data set a = β [ ( ( 9 - 10 ) 2 + ( 10 - 10 ) 2 + ( 11 - 10 ) 2 + ( 7 - 10 ) 2 + ( 13 - 10 ) 2 ) / 5 ] = 2 standard deviation data set b = β [ ( ( 10 - 10 ) 2 + ( 10 - 10 ) 2 + ( 10 - 10 ) 2 + ( 10 - 10 ) 2 + ( 10 - 10 ) 2 ) / 5 ] = 0 standard deviation data set c = β [ ( ( 1 - 10 ) 2 + ( 1 - 10 ) 2 + ( 10 - 10 ) 2 + ( 19 - 10 ) 2 + ( 19 - 10 ) 2 ) / 5 ] = 8.05 option a | a = 10 + 1010
b = a + 1010
c = b / 3
d = c - 1010
e = d ** 2
f = 10 + 1010
g = f + 1010
h = g / 3
i = h - 1010
j = i ** 2
k = e + j
l = 10 + 1010
m = l + 1010
n = m / 3
o = n - 10
p = o ** 2
q = k + p
r = 3 - 1
s = q / r
t = math.sqrt(s)
u = 4 + 3
v = u * 10
w = t / v
|
a ) 9 , 12,21 , b ) 15 , 20,25 , c ) 18 , 24,42 , d ) 24 , 15,17 , e ) 30 , 12,7 | c | sqrt(divide(18144, add(power(7, const_2), add(power(3, 4), power(4, 4))))) | the ratio of three numbers is 3 : 4 : 7 and their product is 18144 . the numbers are | "let the numbers be 3 x , 4 x and 7 x 3 x x 4 x x 7 x = 18144 x 3 = 216 x = 6 the numbers are 18 , 24,42 answer c 18 , 24,42" | a = 7 ** 2
b = 3 ** 4
c = 4 ** 4
d = b + c
e = a + d
f = 18144 / e
g = math.sqrt(f)
|
a ) 35 % , b ) 28 % , c ) 26 % , d ) 20 % , e ) 25 % | e | multiply(subtract(3, inverse(4)), const_100) | john and david can finish a job together in 3 hours . if john can do the job by himself in 4 hours , what percent of the job does david do ? | "you can also plug in numbers . for example , bob and alice work at a donut factory and make 12 donuts which is the job ( i picked this as a smart number ) . john on his own works 12 / 4 = 3 donuts per hour . john and david work 12 / 3 = 4 donuts per hour so david works 1 donuts / hour to find out the percentage , david works 1 donuts / hr x 3 hours = 3 donuts per hour . therefore 3 donuts / 12 donuts = 1 / 4 = 25 % answer : e" | a = 1/(4)
b = 3 - a
c = b * 100
|
a ) 200 , b ) 100 , c ) 50 , d ) 1200 , e ) 500 | b | multiply(multiply(multiply(const_2, const_2), 5), 5) | the length of the longest tape in cm which can be used to measure exactly , the length 6 m ; 5 m ; and 12 m is : | the three lengths in cm are 600,500 & 1200 . hcf of 600 , 500 & 1200 is 100 hence , the answer is 100 cm . answer : b | a = 2 * 2
b = a * 5
c = b * 5
|
a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) 8 hours | a | divide(84, add(16, 5)) | a boat can travel with a speed of 16 km / hr in still water . if the rate of stream is 5 km / hr , then find the time taken by the boat to cover distance of 84 km downstream . | "explanation : it is very important to check , if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( 16 + 5 ) = 21 kmph time = distance / speed = 84 / 21 = 4 hours answer is a" | a = 16 + 5
b = 84 / a
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a ) 3.05 , b ) 3.50035 , c ) 3.501 , d ) 3.5035 , e ) 3.5 | e | multiply(divide(7.007, 2.002), const_100) | 7.007 / 2.002 = | "7.007 / 2.002 = 7007 / 2002 = 7 ( 1001 ) / 2 ( 1001 ) = 7 / 2 = 3.5 the answer is e ." | a = 7 / 7
b = a * 100
|
a ) 15.41 , b ) 16.41 , c ) 15.61 , d ) 15 , e ) 16.91 | b | add(40, const_1) | the average of first 40 prime numbers is ? | "explanation : average = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 ) / 12 = 197 / 12 = 16.41 ( approx ) answer is b" | a = 40 + 1
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a ) rs . 1386 , b ) rs . 1764 , c ) rs . 1650 , d ) rs . 2268 , e ) none of these | c | add(divide(198, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 198) | the true discount on a bill due 9 months hence at 16 % per annum is rs . 198 . the amount of the bill is | "solution 32.5 let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 198 . β΄ x 16 x 9 / 12 x 1 / 100 } = 198 or x = 1650 . β΄ p . w . = rs . 1650 . answer c" | a = 4 * 3
b = 9 / a
c = b * 16
d = c / 100
e = 198 / d
f = e + 198
|
a ) 140 , b ) 143 , c ) 144 , d ) 145 , e ) 150 | a | multiply(divide(56, const_2), 5) | find the number , difference between number and its 3 / 5 is 56 . | "explanation : let the number = x , then , x - ( 3 / 5 ) x = 56 , = > ( 2 / 5 ) x = 56 = > 2 x = 56 * 5 , = > x = 140 answer : option a" | a = 56 / 2
b = a * 5
|
a ) 22 , b ) 36 , c ) 99 , d ) 77 , e ) 12 | b | divide(600, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 600 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 600 * 3 / 50 = 36 sec . answer : b" | a = 63 - 3
b = a * const_0_2778
c = 600 / b
|
a ) 35 , b ) 30 , c ) 45 , d ) 40 , e ) 50 | b | add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3))) | nicky and cristina are running a 200 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . b | a = 3 * 12
b = a / 3
c = 3 * 12
d = 5 - 3
e = c / d
f = b + e
|
a ) 8 sec , b ) 10 sec , c ) 12 sec , d ) 14 sec , e ) 15 sec | a | multiply(divide(divide(200, const_1000), subtract(54, 36)), const_3600) | a train 200 m long is running with a speed of 54 kmph . in what will it pass a car whose speed at 36 kmph in the direction opposite to that in which the train is going | "explanation : speed of the train relative to man = ( 54 + 36 ) kmph = 90 Γ 5 / 18 m / sec = 25 m / sec . time taken by the train to cross the car = time taken by it to cover 200 m at 25 m / sec = ( 200 Γ 1 / 25 ) sec = 8 sec answer : option a" | a = 200 / 1000
b = 54 - 36
c = a / b
d = c * 3600
|
a ) 2 % , b ) 1 % , c ) 6 % , d ) 5 % , e ) 8 % | c | multiply(divide(divide(const_3, 4), add(multiply(const_3, const_4), add(const_0_25, const_0_25))), const_100) | a sum of 12,500 amounts to 15,500 in 4 years at the rate of simple interest . what is the rate of interest ? | c 6 % s . i . = ( 15500 - 12500 ) = 3000 . rate = ( 100 x 3000 ) / ( 12500 x 4 ) % = 6 % | a = 3 / 4
b = 3 * 4
c = const_0_25 + const_0_25
d = b + c
e = a / d
f = e * 100
|
a ) 11 : 6 , b ) 12 : 7 , c ) 13 : 7 , d ) 65 : 36 , e ) 13 : 6 | d | divide(multiply(65, 8), multiply(72, 4)) | car a runs at the speed of 65 km / hr & reaches its destination in 8 hr . car b runs at the speed of 72 km / h & reaches its destination in 4 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 65 Γ£ β 8 = 520 km distance travelled by car b = 72 Γ£ β 4 = 288 km ratio = 520 / 288 = 65 : 36 d" | a = 65 * 8
b = 72 * 4
c = a / b
|
a ) 87 , b ) 89 , c ) 90 , d ) 93 , e ) 95 | d | subtract(multiply(4, add(85, 2)), multiply(85, 3)) | jerry β s average ( arithmetic mean ) score on the first 3 of 4 tests is 85 . if jerry wants to raise his average by 2 points , what score must he earn on the fourth test ? | "total score on 3 tests = 85 * 3 = 255 jerry wants the average to be = 87 hence total score on 4 tests should be = 87 * 4 = 348 score required on the fourth test = 348 - 255 = 93 option d" | a = 85 + 2
b = 4 * a
c = 85 * 3
d = b - c
|
a ) a ) 32 , b ) b ) 35 , c ) c ) 39 , d ) d ) 40 , e ) e ) 45 | a | divide(multiply(20, 8), subtract(8, 3)) | a number exceeds by 20 from its 3 / 8 part . then the number is ? | "x β 3 / 8 x = 20 x = 32 answer : a" | a = 20 * 8
b = 8 - 3
c = a / b
|
a ) 23 % , b ) 62 % , c ) 92 % , d ) 64 % , e ) 22 % | d | subtract(power(5, const_2), power(3, const_2)) | the radius of the two circular fields is in the ratio 3 : 5 the area of the first field is what percent less than the area of the second ? | "r = 3 Ο r 2 = 9 r = 5 Ο r 2 = 25 25 Ο β 16 Ο 100 - - - - ? = > 64 % answer : d" | a = 5 ** 2
b = 3 ** 2
c = a - b
|
a ) 350 , b ) 420 , c ) 510 , d ) 320 , e ) 280 | a | lcm(lcm(25, 35), 50) | find the lowest common multiple of 25 , 35 and 50 . | lcm = 2 * 5 * 5 * 7 = 350 . answer is a | a = math.lcm(25, 35)
b = math.lcm(a, 50)
|
a ) 191 , b ) 355 , c ) 737 , d ) 892 , e ) 1,560 | d | divide(multiply(580, const_100), subtract(const_100, 35)) | a side of beef lost 35 percent of its weight in processing . if the side of beef weighed 580 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 65 / 100 ) * x = 580 = > x = ( 580 * 100 ) / 65 = 892 answer d" | a = 580 * 100
b = 100 - 35
c = a / b
|
a ) $ 24 million , b ) $ 120 million , c ) $ 144 million , d ) $ 240 million , e ) $ 528 million | e | subtract(multiply(288, divide(const_12, const_2)), multiply(1.20, const_1000)) | country x imported approximately $ 1.20 billion of goods in 1996 . if country x imported $ 288 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year , by how much would country xs 1997 imports exceed those of 1996 ? | "convert units to millions as answer is in millions 1996 imports = $ 1.20 bill = $ 1200 mill i . e . 1200 / 12 = $ 100 mill / month 1997 imports = $ 288 mill / 2 month i . e . $ 144 mill / month difference / month = 144 - 100 = 44 difference / year = $ 44 mill * 12 = $ 528 mill answer : e" | a = 12 / 2
b = 288 * a
c = 1 * 20
d = b - c
|
a ) 23.3 kmph , b ) 25.3 kmph , c ) 22.5 kmph , d ) 22.3 kmph , e ) 22.9 kmph | d | divide(add(21, 21), const_2) | a man goes from a to b at a speed of 21 kmph and comes back to a at a speed of 21 kmph . find his average speed for the entire journey ? | "distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 21 ) + ( d / 24 ] = ( 2 d ) / [ 15 d / 168 ) = > 22.3 kmph . answer : d" | a = 21 + 21
b = a / 2
|
a ) 6 kmph , b ) 5 kmph , c ) 2 kmph , d ) 8 kmph , e ) 7 kmph | e | divide(subtract(divide(72, 3), divide(30, 3)), const_2) | a man rows his boat 72 km downstream and 30 km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 72 / ( 3 ) = 24 kmph speed upstream = d / t = 30 / ( 3 ) = 10 kmph the speed of the stream = ( 24 - 10 ) / 2 = 7 kmph answer : e" | a = 72 / 3
b = 30 / 3
c = a - b
d = c / 2
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | add(divide(subtract(5.60, multiply(0.70, const_2)), 0.60), const_2) | a certain fruit stand sold apples for $ 0.70 each and bananas for $ 0.60 each . if a customer purchased both apples and bananas from the stand for a total of $ 5.60 , what total number of apples and bananas did the customer purchase ? | "let ' s start with 1 apple for $ 0.70 . let ' s subtract $ 0.70 from $ 5.60 until we get a multiple of $ 0.60 . $ 5.60 , $ 4.90 , $ 4.20 = 7 * $ 0.60 the customer purchased 7 bananas and 2 apples . the answer is b ." | a = 0 * 70
b = 5 - 60
c = b / 0
d = c + 2
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | subtract(divide(multiply(divide(multiply(add(5, 4), divide(multiply(25, 3), add(3, 2))), 5), 4), add(5, 4)), divide(multiply(25, 2), add(3, 2))) | 25 liters of a mixture is created by mixing liquid p and liquid q in the ratio 3 : 2 . how many liters of liquid q must be added to make the ratio 5 : 4 ? | "let x be the amount of liquid q to be added . ( 2 / 5 ) * 25 + x = ( 4 / 9 ) * ( 25 + x ) 450 + 45 x = 500 + 20 x 25 x = 50 x = 2 the answer is b ." | a = 5 + 4
b = 25 * 3
c = 3 + 2
d = b / c
e = a * d
f = e / 5
g = f * 4
h = 5 + 4
i = g / h
j = 25 * 2
k = 3 + 2
l = j / k
m = i - l
|
a ) rs . 278 , b ) rs . 361 , c ) rs . 369 , d ) rs . 360 , e ) rs . 363 | d | multiply(divide(1560, add(add(2, 3), 4)), 4) | an amount of rs . 1560 was divided among a , b and c in the ratio 1 / 2 : 1 / 3 : 1 / 4 . find the share of c ? | "let the shares of a , b and c be a , b and c respectively . a : b : c = 1 / 2 : 1 / 3 : 1 / 4 a : b : c = 6 / 12 : 4 / 12 : 3 : 12 = 6 : 4 : 3 share of c = 3 / 13 * 1560 = rs . 360 . answer : d" | a = 2 + 3
b = a + 4
c = 1560 / b
d = c * 4
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | subtract(multiply(4, 12), 42) | elena purchased brand x pens for $ 4.00 apiece and brand y for $ 2.20 apiece . if elena purchased a total of 12 of these pens for $ 42.00 , how many brand x pens did she purchase ? | 4 x + 2.8 y = 42 - - > multiply by 2.5 ( to get the integers ) - - > 10 x + 7 y = 105 - - > only one positive integers solutions x = 6 and y = 5 ( how to solve : 7 y must have the last digit of 5 in order the last digit of the sum to be 5 ) . answer : c . | a = 4 * 12
b = a - 42
|
a ) 25 m , b ) 30 m , c ) 32 m , d ) 50 m , e ) 62 m | b | multiply(5, 6) | walking at 5 / 6 th of its usual speed a cab is 6 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 = 6 m ut / 5 = 6 m ut = 30 m answer is b" | a = 5 * 6
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a ) - 2.5 , b ) - 1.5 , c ) - 0.5 , d ) 0.5 , e ) 1.5 | d | add(negate(6), 4) | on the number line , if x is halfway between - 6 and 4 , and if y is halfway between - 2 and 6 , what number is halfway between x and y ? | "x = - 1 and y = 2 . the answer is d ." | a = negate + (
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a ) 337 , b ) 437 , c ) 457 , d ) 537 , e ) 412 | b | add(multiply(subtract(10, const_1), subtract(50, const_2)), 10) | what is the sum of the digits of integer k , if k = ( 10 ^ 50 - 50 ) | "there are 51 digits in 10 ^ 50 when we subtract 50 from it , there will be 50 digits left . 10 ^ 50 can be written as 9999999 . . . . ( 50 times ) + 1 so , 10 ^ 50 - 50 = 9999999 . . . . ( 50 times ) + 1 - 50 = 9999999 . . . . ( 50 times ) - 49 consider the last 2 digits , 99 - 49 = 50 the last 2 digits will be 50 . and our number would be 99999 . . . . . . 99950 with 2 less 9 s after subtraction . number of 9 s left are 48 and the last two digits are 50 the sum of the digits will be ( 48 * 9 ) + 5 + 0 = 437 answer : - b" | a = 10 - 1
b = 50 - 2
c = a * b
d = c + 10
|
a ) 30 , b ) 40 , c ) 36 , d ) 56 , e ) 66 | b | add(lcm(lcm(6, 9), lcm(12, 18)), 4) | what is the least number which when divided by 6 , 9 , 12 and 18 leaves remainder 4 in each care ? | "explanation : lcm of 6 , 9 , 12 and 18 is 36 required number = 36 + 4 = 40 answer : option b" | a = math.lcm(6, 9)
b = math.lcm(12, 18)
c = math.lcm(a, b)
d = c + 4
|
a ) 4.37 % , b ) 5 % , c ) 6 % , d ) 8.75 % , e ) none | c | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | the population of a town increased from 1 , 75,000 to 2 , 80,000 in a decade . the average percent increase of population per year is | "solution increase in 10 years = ( 280000 - 175000 ) = 105000 . increase % = ( 105000 / 175000 Γ£ β 100 ) % = 60 % . required average = ( 60 / 10 ) % = 6 % . answer c" | a = 10 * 1000
b = a * 10
c = b - 1000
d = c - 1000
e = 2 + 3
f = e * 100
g = d - f
h = 3 + 4
i = h * 10
j = 2 + 3
k = i + j
l = k * 1000
m = g / l
n = m - 1
o = n / 10
p = o * 100
q = p + 4
|
a ) 10 sec , b ) 15 sec , c ) 12 sec , d ) 11 sec , e ) 16 sec | a | divide(200, divide(multiply(add(69, 3), const_1000), const_3600)) | a bullet train 200 m long is running with a speed of 69 kmph . in what time will it pass a man who is running at 3 kmph in the direction opposite to that in which the bullet train is going ? | "a 10 sec speed of the bullet train relative to man = ( 69 + 3 ) kmph = 72 * 5 / 18 m / sec = 60 / 3 m / sec . time taken by the bullet train to cross the man = time taken by it to cover 200 m at ( 60 / 3 ) m / sec = ( 200 * 3 / 60 ) sec = 10 sec" | a = 69 + 3
b = a * 1000
c = b / 3600
d = 200 / c
|
a ) 8 sec , b ) 1 sec , c ) 9 sec , d ) 6 sec , e ) 2 sec | c | divide(90, multiply(36, const_0_2778)) | in what time will a railway train 90 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 90 / 36 * 18 / 5 = 9 sec answer : c" | a = 36 * const_0_2778
b = 90 / a
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a ) s . 6000 , b ) s . 9000 , c ) s . 10800 , d ) s . 9942 , e ) s . 9980 | d | divide(multiply(6800, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), 5), const_100))) | after successive discounts of 20 % , 10 % and 5 % a certain good is sold for rs . 6800 . find the actual price of the good . | "let actual price was 100 . after three successive discount this will become , 100 = = 20 % discount = > 80 = = 10 % discount = > 72 = = 5 % discount = 68.4 now compare , 68.4 = 6800 1 = 6800 / 68.4 100 = ( 6800 * 100 ) / 68.4 = rs . 9942 . answer : option d" | a = 6800 * 100
b = 100 - 20
c = 100 - 20
d = c * 10
e = d / 100
f = b - e
g = 100 - 20
h = 100 - 20
i = h * 10
j = i / 100
k = g - j
l = k * 5
m = l / 100
n = f - m
o = a / n
|
a ) 38 metre , b ) 28 metre , c ) 20 metre , d ) 15 metre , e ) 28 metre | c | subtract(160, multiply(divide(160, 32), 28)) | a can run 160 metre in 28 seconds and b in 32 seconds . by what distance a beat b ? | "clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 160 / 32 = 5 m / s distance covered by b in 4 seconds = speed Γ£ β time = 5 Γ£ β 4 = 20 metre i . e . , a beat b by 20 metre answer is c" | a = 160 / 32
b = a * 28
c = 160 - b
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | c | subtract(multiply(20, 8), multiply(6, 20)) | a club wants to mix 20 pounds of candy worth $ 8.00 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 * 8 + 5 * w ) / ( 20 + w ) = > 120 + 6 w = 160 + 5 w = > w = 40 answer c | a = 20 * 8
b = 6 * 20
c = a - b
|
a ) 400 , b ) 267 , c ) 287 , d ) 480 , e ) 811 | a | divide(multiply(600, const_100), add(const_100, 30)) | by selling an article at rs . 600 , a profit of 30 % is made . find its cost price ? | "sp = 600 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 600 * [ 100 / ( 100 + 50 ) ] = 600 * [ 100 / 150 ] = rs . 400 answer : a" | a = 600 * 100
b = 100 + 30
c = a / b
|
a ) 75 % , b ) 80 % , c ) 45 % , d ) 55 % , e ) 65 % | b | subtract(multiply(70, const_3), add(60, 70)) | a student gets 60 % in one subject , 70 % in the other . to get an overall of 70 % how much should get in third subject . | "let the 3 rd subject % = x 60 + 70 + x = 3 * 70 130 + x = 210 x = 210 - 130 = 80 answer : b" | a = 70 * 3
b = 60 + 70
c = a - b
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a ) 42.85 % , b ) 66.68 % , c ) 66.766 % , d ) 86.66 % , e ) 66.65 % | a | multiply(divide(30, subtract(const_100, 30)), const_100) | if a ' s height is 30 % less than that of b , how much percent b ' s height is more than that of a ? | excess of b ' s height over a ' s = [ ( 30 / ( 100 - 30 ) ] x 100 % = 42.85 % answer : a ) | a = 100 - 30
b = 30 / a
c = b * 100
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a ) 1 / 4 , b ) 1 / 2 , c ) 1 , d ) 2 , e ) 3 | d | multiply(divide(13, add(13, 13)), 2) | if 13 = 13 w / ( 1 - w ) , then ( 2 w ) 2 = | "13 - 13 w = 13 w 26 w = 13 w = 1 / 2 2 w = 1 2 w * 2 = 1 * 2 = 2 answer : d" | a = 13 + 13
b = 13 / a
c = b * 2
|
a ) 16 m , b ) 32 m , c ) 27 m , d ) 26 m , e ) 76 m | a | divide(add(add(sqrt(subtract(power(4, const_2), power(2, const_2))), 2), add(sqrt(subtract(power(4, const_2), power(2, const_2))), 2)), 2) | what is the perimeter of a square field whose diagonal is 4 β 2 ? | "4 a = 16 m answer : a" | a = 4 ** 2
b = 2 ** 2
c = a - b
d = math.sqrt(c)
e = d + 2
f = 4 ** 2
g = 2 ** 2
h = f - g
i = math.sqrt(h)
j = i + 2
k = e + j
l = k / 2
|
a ) 18 , b ) 20 , c ) 22 , d ) 26 , e ) 28 | a | divide(multiply(9, 13), const_2) | if the sides of a triangle are 4 cm , 9 cm and 13 cm , what is its area ? | "the triangle with sides 4 cm , 9 cm and 13 cm is right angled , where the hypotenuse is 13 cm . area of the triangle = 1 / 2 * 4 * 9 = 18 cm 2 answer : option a" | a = 9 * 13
b = a / 2
|
a ) 8 , b ) 2 , c ) 10 , d ) 15 , e ) 18 | b | subtract(65, subtract(add(45, 30), 12)) | of the 65 cars on a car lot , 45 have air - bag , 30 have power windows , and 12 have both air - bag and power windows . how many of the cars on the lot have neither air - bag nor power windows ? | total - neither = all air bag + all power windows - both or 65 - neither = 45 + 30 - 12 = 63 . = > neither = 2 , hence b . answer : b | a = 45 + 30
b = a - 12
c = 65 - b
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a ) 315 , b ) 339 , c ) 288 , d ) 299 , e ) 111 | a | divide(subtract(700, multiply(divide(700, const_2), divide(const_1, add(const_1, const_4)))), const_2) | a large field of 700 hectares is divided into two parts . the difference of the areas of the two parts is one - fifth of the average of the two areas . what is the area of the smaller part in hectares ? | explanation : let the areas of the two parts be x and ( 700 - x ) hectares therefore , so , the two parts are 385 and 315 . hence , area of the smaller = 315 hectares answer : a ) 315 | a = 700 / 2
b = 1 + 4
c = 1 / b
d = a * c
e = 700 - d
f = e / 2
|
a ) 17.1 , b ) 17.3 , c ) 17.5 , d ) 17.7 , e ) 17.2 | d | divide(177, divide(14.5, 1.45)) | if 2994 Γ£ Β· 14.5 = 177 , then 29.94 Γ£ Β· 1.45 = ? | "29.94 / 1.45 = 299.4 / 14.5 = ( 2994 / 14.5 ) x 1 / 10 ) [ here , substitute 177 in the place of 2994 / 14.5 ] = 177 / 10 = 17.7 answer is d ." | a = 14 / 5
b = 177 / a
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a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent t should the number of hours worked be increased ? | correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the increase in hours worked : wh = . 8 wx ( divide both sides by . 8 w ) 1.25 h = x we know that the new number of hours worked t will be 25 % greater than the original number . the answer is c . | a = 10 * 4
b = 100 - 20
c = b / 100
d = c * 10
e = a / d
f = e - 4
g = f / 4
h = g * 100
|
a ) 9 / 10 , b ) 6 / 10 , c ) 2 / 10 , d ) 3 / 10 , e ) 5 / 10 | d | divide(const_3.0, const_10) | tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn has a number which is a multiple of 4 or 15 ? | "explanation : here , s = { 1 , 2 , 3 , 4 , . . . . , 19 , 20 } = > n ( s ) = 20 let e = event of getting a multiple of 4 or 15 = multiples od 4 are { 4 , 8 , 12 , 16 , 20 } and multiples of 15 means multiples of 3 and 5 = { 3 , 6 , 9 , 12 , 15 , 18 , 5 , 10 , 15 , 20 } . = the common multiple is only ( 15 ) . = > e = n ( e ) = 6 required probability = p ( e ) = n ( e ) / n ( s ) = 6 / 20 = 3 / 10 . answer is d" | a = 3 / 0
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a ) 66 , b ) 70 , c ) 95 , d ) 112 , e ) 140 | e | multiply(divide(multiply(105, 4), add(75, 105)), const_60) | cole drove from home to work at an average speed of 75 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 4 hours , how many minutes did it take cole to drive to work ? | first round distance travelled ( say ) = d speed = 75 k / h time taken , t 2 = d / 75 hr second round distance traveled = d ( same distance ) speed = 105 k / h time taken , t 2 = d / 105 hr total time taken = 4 hrs therefore , 4 = d / 75 + d / 105 lcm of 75 and 105 = 525 4 = d / 75 + d / 105 = > 4 = 7 d / 525 + 5 d / 525 = > d = 525 / 3 km therefore , t 1 = d / 75 = > t 1 = 525 / ( 3 x 75 ) = > t 1 = ( 7 x 60 ) / 3 - - in minutes = > t 1 = 140 minutes . e | a = 105 * 4
b = 75 + 105
c = a / b
d = c * const_60
|
a ) 70 , b ) 24 , c ) 35 , d ) 62 , e ) 21 | a | multiply(6, const_4) | what could be the range of a set consisting of odd multiples of ( 6 + 1 ) ? | "range = the difference between the greatest and the smallest numbers in the sequence . our sequence is odd and is a multiple of 7 . every number in that sequence can be represented like this : 7 * ( 2 n + 1 ) where n is any positive integer . range = 7 * ( 2 m + 1 ) - 7 * ( 2 n + 1 ) = 7 * 2 * ( m - n ) = 14 * ( m - n ) . m , n - any positive integers the answer must be divisible by 14 , which is only 70 . the correct answer is a ." | a = 6 * 4
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a ) $ 38,796 . , b ) $ 40,000 . , c ) $ 38,976 . , d ) $ 39,679 . , e ) $ 53,592 . | e | subtract(multiply(log(multiply(55000, multiply(divide(subtract(const_100, 16), const_100), divide(add(const_100, 16), const_100)))), const_3), multiply(const_3, const_3)) | the apartment on king - williams street is an asset that its value is tramping about . from the year 1973 to 1983 it ' s value decreased by 16 % and from 1983 to 1993 it ' s value increased by 16 % . what is the value of the asset in 1993 if in 1973 it was worth $ 55000 ? | 55,000 * 1.16 * . 84 = 53,592 answer e | a = 100 - 16
b = a / 100
c = 100 + 16
d = c / 100
e = b * d
f = 55000 * e
g = math.log(f)
h = g * 3
i = 3 * 3
j = h - i
|
a ) 1 / 2 , b ) 1 , c ) 2 , d ) 5 / 2 , e ) 4 | b | divide(subtract(add(5, 2), 5), 2) | in the xy - coordinate system , if ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 , then k = | "the equation of the line can be re written as y = ( 1 / 2 ) x - 5 / 2 . slope = 1 / 2 , which means for every 1 unit increase in the y co - ordinate , x will increase by 2 units . hence k = 1 . answer : b" | a = 5 + 2
b = a - 5
c = b / 2
|
a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | b | sqrt(54) | the difference between c . i . and s . i . on an amount of $ 15,000 for 2 years is $ 54 . what is the rate of interest per annum ? | "$ 54 is the interest on the first year of interest . let x be the interest rate . the interest after the first year is 15000 * x . the interest on the first year ' s interest is 15000 * x * x 15000 * x ^ 2 = 54 x = 0.06 the answer is b ." | a = math.sqrt(54)
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a ) $ 100 , b ) $ 150 , c ) $ 125 , d ) $ 200 , e ) $ 240 | e | multiply(divide(600, add(divide(2, 3), const_1)), divide(2, 3)) | $ 600 is divided amongst a , b and c so that a may get 2 / 3 as much as b and c together , b may get 6 / 9 as much as a and c together , then the share of a is | "a : ( b + c ) = 2 : 3 a ' s share = 600 * 2 / 5 = $ 240 answer is e" | a = 2 / 3
b = a + 1
c = 600 / b
d = 2 / 3
e = c * d
|
a ) 9 , b ) 2 , c ) 1 , d ) 3 , e ) 6 | d | divide(multiply(532869, 9), 532869) | what is the smallest no . which must be added to 532869 so as to obtain a sum which is divisible by 9 ? | "for 532869 , 5 + 3 + 2 + 8 + 6 + 9 = 33 3 must be added to 532869 to make it divisible by 9 . now , 5 + 3 + 2 + 8 + 7 + 2 = 27 = > 27 is a multiple of 9 and hence 532869 is also divisible by 9 d" | a = 532869 * 9
b = a / 532869
|
a ) 9 , b ) 5 , c ) 7 , d ) 6 , e ) 51 | b | multiply(divide(12, const_60), add(20, 5)) | the speed of a boat in still water is 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 12 minutes is : | "explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 12 / 60 ) ) km = 5 km . answer : b" | a = 12 / const_60
b = 20 + 5
c = a * b
|
a ) 500 , b ) 600 , c ) 700 , d ) 800 , e ) 900 | b | divide(144, multiply(divide(4, const_100), 6)) | a sum was put at simple interest at a certain rate for 6 years had it been put at 4 % higher rate , it would have fetched 144 more . find the sum . | "difference in s . i . = p Γ t / 100 ( r 1 β r 2 ) β 144 = p Γ 6 x 4 / 100 ( β΅ r 1 - r 2 = 2 ) β p = 144 Γ 100 / 6 Γ 4 = 600 answer b" | a = 4 / 100
b = a * 6
c = 144 / b
|
a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 3 | a | subtract(427398, multiply(floor(divide(427398, 10)), 10)) | what least no . must be subtracted from 427398 so that remaining no . is divisible by 10 ? | "explanation : on dividing 427398 by 10 we get the remainder 8 , so 8 should be subtracted option a" | a = 427398 / 10
b = math.floor(a)
c = b * 10
d = 427398 - c
|
a ) 480 , b ) 1600 , c ) 1920 , d ) 2080 , e ) 2400 | e | divide(subtract(multiply(7, 4000), 4000), add(7, 3)) | the number of students enrolled at school xx this year is 7 percent more than it was last year . the number of students enrolled at school yy this year is 3 percent more than it was last year . if school xx grew by 40 more students than school yy did , and if there were 4000 total enrolled students last year at both schools , how many students were enrolled at school yy last year ? | given info : no . of students enrolled in school x is = 7 % more than previous year ' s strength = 7 x / 100 no . of students enrolled in school y is = 3 % more than previous year ' s strength = 3 y / 100 given that ` ` this year ' ' school x have 40 students more than school y . so the increase in strength can be written as 7 x / 100 = 40 + 3 y / 100 7 x = 4000 + 3 y . . . . . . . . . . . . . . . . . . . . ( 1 ) and total students on previous year are x + y = 4000 . . . . . . . . . . . . . . . . . . . . . . ( 2 ) equating & solving we get , y = 2400 ans is e | a = 7 * 4000
b = a - 4000
c = 7 + 3
d = b / c
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a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2 | b | divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10))) | a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio t of milk and water in the resultant mixture ? | "he 20 litre mixture contains milk and water in the ratio of 3 : 2 . therefore , there will be 12 litres of milk in the mixture and 8 litres of water in the mixture . step 1 . when 10 litres of the mixture is removed , 6 litres of milk is removed and 4 litres of water is removed . therefore , there will be 6 litres of milk and 4 litres of water left in the container . it is then replaced with pure milk of 10 litres . now the container will have 16 litres of milk and 4 litres of water . step 2 . when 10 litres of the new mixture is removed , 8 litres of milk and 2 litres of water is removed . the container will have 8 litres of milk and 2 litres of water in it . now 10 litres of pure milk is added . therefore , the container will have 18 litres of milk and 2 litres of water in it at the end of the second step . therefore , the ratio of milk and water is 18 : 2 or 9 : 1 . shortcut . we are essentially replacing water in the mixture with pure milk . let w _ o be the amount of water in the mixture originally = 8 litres . let w _ r be the amount of water in the mixture after the replacements have taken place . then , { w _ r } / { w _ o } = ( 1 - r / m ) ^ n where r is the amount of the mixture replaced by milk in each of the steps , m is the total volume of the mixture and n is the number of times the cycle is repeated . hence , { w _ r } / { w _ o } Β = ( 1 / 2 ) ^ 2 Β = 1 / 4 therefore , tw _ r Β = { w _ o } / 4 = 8 / 4 Β = 2 litres . b" | a = 3 + 2
b = 3 / a
c = 20 - 10
d = b * c
e = d + 10
f = e / 20
g = 20 - 10
h = f * g
i = h + 10
j = 3 + 2
k = 2 / j
l = 20 - 10
m = k * l
n = m / 20
o = 20 - 10
p = n * o
q = i / p
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a ) $ 252 , b ) $ 135 , c ) $ 90 , d ) $ 60 , e ) $ 54 | a | multiply(const_2.0, divide(multiply(36, divide(2, 3)), divide(const_1, 3))) | a collection of books went on sale , and 2 / 3 of them were sold for $ 3.50 each . if none of the 36 remaining books were sold , what was the total amount received for the books that were sold ? | "since 2 / 3 of the books in the collection were sold , 1 / 3 were not sold . the 36 unsold books represent 1 / 3 of the total number of books in the collection , and 2 / 3 of the total number of books equals 2 ( 36 ) or 72 . the total proceeds of the sale was 72 ( $ 3.50 ) or $ 252 . the best answer is therefore a . answer : a ." | a = 2 / 3
b = 36 * a
c = 1 / 3
d = b / c
e = 2 * 0
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a ) 31 , b ) 35 , c ) 29 , d ) 25 , e ) 30 | e | divide(add(multiply(36, 35), multiply(15, 14)), add(35, 14)) | the average runs scored by a batsman in 35 matches is 36 . in the next 14 matches the batsman scored an average of 15 runs . find his average in all the 30 matches ? | total score of the batsman in 35 matches = 1260 . total score of the batsman in the next 14 matches = 210 . total score of the batsman in the 30 matches = 1470 . average score of the batsman = 1470 / 49 = 30 . answer : e | a = 36 * 35
b = 15 * 14
c = a + b
d = 35 + 14
e = c / d
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a ) 16 , b ) 23 , c ) 19 , d ) 20 , e ) 22 | b | subtract(divide(subtract(multiply(18, 5), 40), const_2), const_2) | the average ( arithmetic mean ) of the 5 positive integers k , m , r , s , and t is 18 , and k < m < r < s < t . if t is 40 , what is the greatest possible value of the median of the 5 integers ? | "we need to find the median which is the third value when the numbers are in increasing order . since k < m < r < s < t , the median would be r . the average of the positive integers is 18 which means that in effect , all numbers are equal to 18 . if the largest number is 40 , it is 22 more than 18 . we need r to be maximum so k and m should be as small as possible to get the average of 16 . since all the numbers are positive integers , k and m can not be less than 1 and 2 respectively . 1 is 17 less than 18 and 2 is 16 less than 18 which means k and m combined are 33 less than the average . 40 is already 22 more than 18 and hence we only have 33 - 22 = 11 extra to distribute between r and s . since s must be greater than r , r can be 18 + 5 = 23 and s can be 18 + 6 = 24 . so r is 23 answer ( b )" | a = 18 * 5
b = a - 40
c = b / 2
d = c - 2
|
a ) 52 , b ) 54 , c ) 56 , d ) 58 , e ) 60 | c | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | how many positive integers less than 7,000 are there in which the sum of the digits equals 5 ? | "basically , the question asks how many 4 digit numbers ( including those in the form 0 xxx , 00 xx , and 000 x ) have digits which add up to 5 . think about the question this way : we know that there is a total of 5 to be spread among the 4 digits , we just have to determine the number of ways it can be spread . let x represent a sum of 1 , and | represent a seperator between two digits . as a result , we will have 5 x ' s ( digits add up to the 5 ) , and 3 | ' s ( 3 digit seperators ) . so , for example : xx | x | x | x = 2111 | | xxx | xx = 0032 etc . there are 8 c 3 ways to determine where to place the separators . hence , the answer is 8 c 3 = 56 . c" | a = 4 + 5
b = a - 1
c = math.factorial(b)
d = math.factorial(5)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 0.004 , b ) 0.11111 , c ) 2.775 , d ) 3.6036 , e ) 36.036 | b | inverse(add(divide(4, 0.03), divide(4, 0.37))) | 4 / [ ( 1 / 0.03 ) + ( 1 / 0.37 ) ] = ? | "approximate . 1 / . 03 = 100 / 3 = 33 1 / . 37 = 100 / 37 = 3 denominator becomes 33 + 3 = 36 4 / 36 = . 1111 answer ( b )" | a = 4 / 0
b = 4 / 0
c = a + b
d = 1/(c)
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a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 4 | b | multiply(divide(const_1, multiply(add(const_100, 28), divide(const_1, subtract(const_100, 28)))), 16) | by selling 16 pencils for a rupee a man loses 28 % . how many for a rupee should he sell in order to gain 28 % ? | "72 % - - - 16 128 % - - - ? 72 / 128 * 16 = 9 answer : b" | a = 100 + 28
b = 100 - 28
c = 1 / b
d = a * c
e = 1 / d
f = e * 16
|
a ) 200 % , b ) 210 % , c ) 300 % , d ) 310 % , e ) none of these | c | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(10, const_100)), const_2)), const_100) | if the radius of a circle is diminished by 10 % , then the area is diminished by : | "explanation : let the original radius be r cm . new radius = 2 r area = Ο r 2 new area = Ο 2 r 2 = 4 Ο r 2 increase in area = ( 4 Ο r 2 β Ο r 2 ) = 3 Ο r 2 increase percent = 3 Ο r 2 / Ο r 2 β 100 = 300 % option c" | a = 100 / 100
b = 100 / 100
c = 10 / 100
d = b - c
e = d ** 2
f = a - e
g = f * 100
|
a ) $ 140 , b ) $ 160 , c ) $ 220 , d ) $ 232 , e ) $ 260 | d | subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(24, const_12)) | a parking garage rents parking spaces for $ 10 per week or $ 24 per month . how much does a person save in a year by renting by the month rather than by the week ? | "10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 24 = 288 $ 520 - 288 = 232 ans d" | a = 3 + 2
b = 10 * a
c = b + 2
d = c * 10
e = 24 * 12
f = d - e
|
a ) 8 kmph , b ) 5 kmph , c ) 4 kmph , d ) 6 kmph , e ) 7 kmph | b | divide(add(divide(12, 2), divide(12, 3)), 2) | a boat running downstream covers a distance of 12 km in 2 hours while for covering the same distance upstream , it takes 3 hours . what is the speed of the boat in still water ? | "speed downstream = 12 / 2 = 6 kmph speed upstream = 12 / 3 = 4 kmph speed of the boat in still water = 6 + 4 / 2 = 5 kmph answer : b" | a = 12 / 2
b = 12 / 3
c = a + b
d = c / 2
|
['a ) 288', 'b ) 289', 'c ) 200', 'd ) 112', 'e ) 178'] | c | divide(square_area(20), const_2) | what is the area of a square field whose diagonal of length 20 m ? | d 2 / 2 = ( 20 * 20 ) / 2 = 200 answer : c | a = square_area / (
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a ) 3 % , b ) 4 % , c ) 5 % , d ) 6 % , e ) 7 % | d | multiply(divide(divide(subtract(add(multiply(multiply(add(const_2, const_3), const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)), add(multiply(multiply(const_3, const_4), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100))), add(multiply(multiply(const_3, const_4), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100))), 4), const_100) | a sum of rs . 12,500 amounts to rs . 15,500 in 4 years at the rate of simple interest . what is the rate of interest ? | s . i . = rs . ( 15500 - 12500 ) = rs . 3000 . rate = ( 100 x 3000 ) / ( 12500 x 4 ) % = 6 % answer : d | a = 2 + 3
b = a * 3
c = 2 + 3
d = c * 2
e = d * 100
f = b * e
g = 2 + 3
h = g * 100
i = f + h
j = 3 * 4
k = 2 + 3
l = k * 2
m = l * 100
n = j * m
o = 2 + 3
p = o * 100
q = n + p
r = i - q
s = 3 * 4
t = 2 + 3
u = t * 2
v = u * 100
w = s * v
x = 2 + 3
y = x * 100
z = w + y
A = r / z
B = A / 4
C = B * 100
|
a ) loss of rs . 200 , b ) loss of rs . 100 , c ) profit of rs . 100 , d ) profit of rs . 200 , e ) none of these | d | subtract(add(multiply(15000, subtract(const_1, divide(4, const_100))), multiply(8000, add(const_1, divide(10, const_100)))), add(15000, 8000)) | ravi purchased a refrigerator and a mobile phone for rs . 15000 and rs . 8000 respectively . he sold the refrigerator at a loss of 4 percent and the mobile phone at a profit of 10 percent . overall he make a . | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 10 / 100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 800 = rs . 200 as this is positive , an overall profit of rs . 200 was made . answer : d" | a = 4 / 100
b = 1 - a
c = 15000 * b
d = 10 / 100
e = 1 + d
f = 8000 * e
g = c + f
h = 15000 + 8000
i = g - h
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a ) 80 , b ) 90 , c ) 30 , d ) 72 , e ) 60 | d | multiply(add(8, const_1), 8) | tim came second in math . when his mother asked him how much he had scored , he answered that he got the sum of the first 8 even numbers . his mother immediately worked out the answer . how much had he scored in math ? | d 72 sum = ( n x n ) + n hence , 8 x 8 = 64 + 8 = 72 | a = 8 + 1
b = a * 8
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a ) 26 / 18 , b ) 42 / 15 , c ) 48 / 63 , d ) 12 / 36 , e ) 150 / 23 | e | divide(const_1, divide(add(add(inverse(5), inverse(15)), inverse(25)), 5)) | a and b can do a work in 5 days , b and c in 15 days and c and a in 25 days . in how many days will the work be completed , if all three of them work together ? | "one day work of a and b = 1 / 5 one day work of b and c = 1 / 15 one day work of c and a = 1 / 25 2 ( a + b + c ) = 1 / 5 + 1 / 15 + 1 / 25 2 ( a + b + c ) = 23 / 75 ( a + b + c ) = 23 / 150 number of days required = 150 / 23 days . answer : e" | a = 1/(5)
b = 1/(15)
c = a + b
d = 1/(25)
e = c + d
f = e / 5
g = 1 / f
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a ) 50 , b ) 55 , c ) 62 , d ) 77 , e ) 66 | e | divide(multiply(22, 300), const_100) | 22 percent of 300 | 1 % of 300 = 3 22 % of 300 = 22 * 3 = 66 answer : e | a = 22 * 300
b = a / 100
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a ) 45 / 12 , b ) 23 / 12 , c ) 40 / 33 , d ) 34 / 12 , e ) 43 / 5 | a | divide(subtract(const_1, multiply(add(divide(const_1, 9), divide(const_1, 12)), 3)), divide(const_1, 9)) | working individually , allison can sew dresses for 9 hours and al can sew dresses in 12 hours . if allison and al work together but independently at the task for 3 hours , at which point al leaves , how many remaining hours will it take allison to complete the task alone ? | in first 3 hrs al will finish 3 / 12 = 1 / 4 of work and allison will finish 3 / 9 = 1 / 3 work so total 1 / 4 + 1 / 3 = 7 / 12 work is finished and 1 - 7 / 12 = 5 / 12 work remaining . now allison will take ( 5 / 12 ) * 9 = 45 / 12 hrs to finish it . so answer is a . | a = 1 / 9
b = 1 / 12
c = a + b
d = c * 3
e = 1 - d
f = 1 / 9
g = e / f
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a ) 5 % , b ) 10 % , c ) 20 % , d ) 25 % , e ) 33 % | d | multiply(subtract(const_1, divide(const_3, add(3, const_1))), const_100) | 3 friends are planning to attend a concert . if another friend also goes with them , they will get a group discount on their tickets . if the new total price of the tickets with the group discount is the same as the original total price of the tickets without the group discount , how much is the discount ? | let x be the original price of one ticket . the total original cost is 3 x . the new cost is 4 y , where y is the discounted price of one ticket . 3 x = 4 y y = 3 x / 4 = 0.75 x which is a discount of 25 % . the answer is d . | a = 3 + 1
b = 3 / a
c = 1 - b
d = c * 100
|
a ) rs . 60 , b ) rs . 108 , c ) rs . 110 , d ) rs . 112 , e ) rs . 122 | b | multiply(divide(90, subtract(540, 90)), 540) | the true discount on a bill of rs . 540 is rs . 90 . the banker ' s discount is : | "p . w . = rs . ( 540 - 90 ) = rs . 450 . s . i . on rs . 450 = rs . 90 . s . i . on rs . 540 = rs . ( 90 / 450 ) x 540 = rs . 108 . b . d . = rs . 108 . answer : b" | a = 540 - 90
b = 90 / a
c = b * 540
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a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 20 | e | divide(3, subtract(96.15, floor(96.15))) | when positive integer x is divided by positive integer y , the remainder is 3 . if x / y = 96.15 , what is the value of y ? | by the definition of a remainder , the remainder here is equal to 3 / y . the remainder in decimal form is given as . 15 therefore , 3 / y = . 15 solve for y and get 20 . e | a = math.floor(96, 15)
b = 96 - 15
c = 3 / b
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a ) rs . 8928 , b ) rs . 89228 , c ) rs . 8911 , d ) rs . 8925 , e ) rs . 8922 | d | divide(multiply(4216.25, const_100), multiply(9, 5)) | fetched a total simple interest of rs . 4216.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ? | explanation : principal = rs . = rs . = rs . 8925 answer : d | a = 4216 * 25
b = 9 * 5
c = a / b
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a ) 16 , b ) 18 , c ) 19 , d ) 22 , e ) 24 | a | add(15, divide(subtract(36, 26), 10)) | the average of 10 numbers is calculated as 15 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "explanation : 10 * 15 + 36 β 26 = 160 / 10 = 16 a )" | a = 36 - 26
b = a / 10
c = 15 + b
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a ) 68 , b ) 70.4 , c ) 86 , d ) 105.6 , e ) 117.6 | e | add(98, multiply(divide(20, const_100), 98)) | if x is 20 percent greater than 98 , then x = | "x = 98 * 1.2 = 117.6 so the answer is e ." | a = 20 / 100
b = a * 98
c = 98 + b
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | a | divide(subtract(450, multiply(75, const_2)), 30) | 30 pens and 75 pencils were purchased for 450 . if the average price of a pencil was 2.00 , find the average price of a pen . | since average price of a pencil = 2 β΄ price of 75 pencils = 150 β΄ price of 30 pens = ( 450 β 150 ) = 300 β΄ average price of a pen = 360 β 60 = 10 answer a | a = 75 * 2
b = 450 - a
c = b / 30
|
a ) 250 km , b ) 300 km , c ) 500 km , d ) 450 km , e ) 350 km | c | multiply(100, 5) | a person is traveling at 100 km / hr and reached his destiny in 5 then find the distance ? | "t = 5 hr s = 100 km / hr d = t * s = 100 * 5 = 500 km answer is c" | a = 100 * 5
|
a ) 69 , b ) 135 , c ) 139 , d ) 147 , e ) 188 | a | add(multiply(divide(subtract(99, 29), const_3), const_2), 29) | if jake loses 29 pounds , he will weigh twice as much as his sister . together they now weigh 99 pounds . what is jake ' s present weight , in pounds ? | "j = jake β s current weight , in pounds s = sister β s current weight , in pounds we are told that β if jake loses 8 pounds , he will weigh twice as much as his sister . we put this into an equation : j β 29 = 2 s j = 2 s + 29 ( equation 1 ) next , we are told that β together they now weigh 89 pounds . β we can also put this into an equation . j + s = 89 ( equation 2 ) to solve this equation , we can substitute 2 s + 8 from equation 1 for the variable j in equation 2 : 2 s + 29 = 89 - s 3 s = 60 s = 20 j + 20 = 89 j = 69 answer : a" | a = 99 - 29
b = a / 3
c = b * 2
d = c + 29
|
a ) 8.5 , b ) 6.0 , c ) 9.5 , d ) 9.0 , e ) 8.25 | b | divide(subtract(18, 6), const_2) | a man can row downstream at the rate of 18 kmph and upstream at 6 kmph . find the man β s rate in still water and rate of current ? | "rate of still water = 1 / 2 ( down stream + upstream ) = 1 / 2 ( 18 + 6 ) = 12 kmph rate of current = 1 / 2 ( down stream - upstream ) = 1 / 2 ( 18 - 6 ) = 1 / 2 ( 12 ) = 6 kmph answer is b ." | a = 18 - 6
b = a / 2
|
a ) 5 , b ) 8 , c ) 10 , d ) 15 , e ) 20 | c | subtract(divide(add(20, 0), const_2), divide(add(40, 0), const_2)) | the average ( arithmetic mean ) of the even integers from 0 to 40 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 20 inclusive ? | "the sum of even numbers from 0 to n is 2 + 4 + . . . + n = 2 ( 1 + 2 + . . . + n / 2 ) = 2 ( n / 2 ) ( n / 2 + 1 ) / 2 = ( n / 2 ) ( n / 2 + 1 ) the average is ( n / 2 ) ( n / 2 + 1 ) / ( n / 2 + 1 ) = n / 2 the average of the even numbers from 0 to 40 is 40 / 2 = 20 the average of the even numbers from 0 to 20 is 20 / 2 = 10 the answer is c ." | a = 20 + 0
b = a / 2
c = 40 + 0
d = c / 2
e = b - d
|
a ) 264 , b ) 259 , c ) 269 , d ) 270 , e ) 282 | b | subtract(lcm(lcm(8, 11), 24), 5) | what is the smallest number h which when increased by 5 is completely divisible by 8 , 11 and 24 ? | this question includes a number of great number property shortcuts that you can take advantage of : 1 ) the question asks for the smallest number that , when increased by 5 is divisible by 8 , 11 and 24 . since the answers are numbers , we can test the answers . 2 ) any number that is divisible by 24 is also divisible by 8 , so we really just need to consider the 11 and the 24 . 3 ) to be divisible by 24 , the end number must be even . since we ' re adding 5 to get to that end number , the starting number must be odd . the last shortcut allows us to eliminate answers a , d and e . between answers b and c , testing b first will prove that , when increased by 5 , the end sum is divisible by both 11 and 24 . final answer : b | a = math.lcm(8, 11)
b = math.lcm(a, 24)
c = b - 5
|
a ) 14 % , b ) 16 % , c ) 17 % , d ) 78 % , e ) 28 % | a | multiply(divide(subtract(1500, 1290), 1500), const_100) | the cost price of a radio is rs . 1500 and it was sold for rs . 1290 , find the loss % ? | "1500 - - - - 210 100 - - - - ? = > 14 % answer : a" | a = 1500 - 1290
b = a / 1500
c = b * 100
|
a ) a ) 4500 , b ) b ) 5200 , c ) c ) 6900 , d ) d ) 7520 , e ) e ) 6000 | a | divide(900, divide(subtract(60, subtract(const_100, 60)), const_100)) | in an election a candidate who gets 60 % of the votes is elected by a majority of 900 votes . what is the total number of votes polled ? | "let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 60 ) % of x = 40 % of x 60 % of x - 40 % of x = 900 20 x / 100 = 900 x = 900 * 100 / 20 = 4500 answer is a" | a = 100 - 60
b = 60 - a
c = b / 100
d = 900 / c
|
a ) 2.29 , b ) 2.75 , c ) 4.25 , d ) 4.5 , e ) none of these | c | multiply(17, 17) | ( 17 ) 3.75 x ( 17 ) ? = 178 | "solution let ( 17 ) 3.75 * ( 17 ) x = 178 . then , ( 17 ) 3.5 + x = ( 17 ) 8 . β΄ 3.75 + x = 8 β x = ( 8 - 3.75 ) β x = 4.25 answer c" | a = 17 * 17
|
a ) 1.2 km , b ) 2.4 km , c ) 3.2 km , d ) 3.6 km , e ) 4.6 km | d | multiply(add(15, 3), divide(12, const_60)) | the speed of a boat in still water in 15 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 12 minutes is : | "speed downstream = ( 15 + 3 ) kmph = 18 kmph . distance travelled = 18 x 12 / 60 km = 3.6 km . answer : d" | a = 15 + 3
b = 12 / const_60
c = a * b
|
a ) 10 , b ) 1000 , c ) 100 , d ) 0.01 , e ) 0.0001 | b | multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100)) | if 1000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ? | "1000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter 1000 microns = 1 , 000,000 angstroms 1 micron = 1 , 000,000 / 1,000 = 1,000 answer : b" | a = 100 * 100
b = 1 / a
c = 100 * 100
d = b * c
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | divide(multiply(multiply(10, 7), 18), multiply(15, 12)) | 10 men , working 7 hours a day can complete a work in 18 days . how many hours a day must 15 men work to complete the same work in 12 days ? | the number of hours required to complete the work is 10 * 7 * 18 = 1260 15 Γ 12 Γ ( x ) = 1260 x = 7 the answer is b . | a = 10 * 7
b = a * 18
c = 15 * 12
d = b / c
|
a ) 5 , b ) 3 , c ) 6 , d ) 2 , e ) 1 | b | subtract(4, divide(subtract(22, 1), add(20, 1))) | calculate the value of n from the below equation : y ^ 4 Γ’ Λ β 20 y + 1 = 22 | use elimination method to find the correct option . you find that of all the options 3 is the correct value for y answer : b | a = 22 - 1
b = 20 + 1
c = a / b
d = 4 - c
|
a ) 32.2 , b ) 32.98 , c ) 42.3 , d ) 48.6 , e ) 32.4 | d | add(37, const_1) | the average of first five prime numbers greater than 37 is ? | "41 + 43 + 47 + 53 + 59 = 243 / 5 = 48.6 answer : d" | a = 37 + 1
|
a ) 3 / 4 , b ) 1 / 2 , c ) 1 / 7 , d ) 1 / 8 , e ) 4 / 3 | b | divide(add(divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2)), divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2))), divide(divide(factorial(add(6, 6)), factorial(subtract(add(6, 6), const_2))), factorial(const_2))) | a bag contains 6 black and 6 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ? | "let number of balls = ( 6 + 6 ) = 12 . number of white balls = 6 . p ( drawing a white ball ) = 6 / 12 = 1 / 2 . option b ." | a = math.factorial(6)
b = 6 - 2
c = math.factorial(b)
d = a / c
e = math.factorial(2)
f = d / e
g = math.factorial(6)
h = 6 - 2
i = math.factorial(h)
j = g / i
k = math.factorial(2)
l = j / k
m = f + l
n = 6 + 6
o = math.factorial(n)
p = 6 + 6
q = p - 2
r = math.factorial(q)
s = o / r
t = math.factorial(2)
u = s / t
v = m / u
|
a ) 3.86 % , b ) 4.86 % , c ) 5.86 % , d ) 6.86 % , e ) 7.86 % | d | multiply(divide(subtract(subtract(67, multiply(67, divide(10, const_100))), 56.16), subtract(67, multiply(67, divide(10, const_100)))), const_100) | the list price of an article is rs . 67 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | 67 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 56.16 x = 6.86 % answer : d | a = 10 / 100
b = 67 * a
c = 67 - b
d = c - 56
e = 10 / 100
f = 67 * e
g = 67 - f
h = d / g
i = h * 100
|
a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | d | subtract(divide(add(multiply(divide(45, 60), 4), 4), subtract(divide(60, 60), divide(45, 60))), 4) | karen places a bet with tom that she will beat tom in a car race by 4 miles even if karen starts 4 minutes late . assuming that karen drives at an average speed of 60 mph and tom drives at an average speed of 45 mph , how many w miles will tom drive before karen wins the bet ? | "let k and t be the speeds of karen and tom respectively . t be the time that karen will travel - - - - > t + 4 / 60 will be the total time tom will travel by the time the distance between karen and tom is 4 miles . thus , per the question , k ( t ) - t ( t + 4 / 60 ) = 4 - - - > t = 7 / 15 hours thus the distance traveled by tom when karen is 4 miles ahead of him w : t * ( t + 4 / 60 ) = 45 ( 7 / 15 + 4 / 60 ) = 24 miles . d is the correct answer ." | a = 45 / 60
b = a * 4
c = b + 4
d = 60 / 60
e = 45 / 60
f = d - e
g = c / f
h = g - 4
|
a ) 4 / 9 , b ) 2 / 3 , c ) 3 / 2 , d ) 9 / 2 , e ) 9 / 4 | a | divide(power(2, const_2), power(3, const_2)) | rectangle a has sides a and b , and rectangle b has sides c and d . if a / c = b / d = 2 / 3 , what is the ratio of rectangle a β s area to rectangle b β s area ? | "the area of rectangle a is ab . c = 3 a / 2 and d = 3 b / 2 . the area of rectangle b is cd = 9 ab / 4 . the ratio of rectangle a ' s area to rectangle b ' s area is ab / ( 9 ab / 4 ) = 4 / 9 . the answer is a ." | a = 2 ** 2
b = 3 ** 2
c = a / b
|
a ) 0.05 , b ) 0.15 , c ) 0.45 , d ) 0.5 , e ) 0.65 | e | subtract(add(0.8, 0.25), 0.4) | the probability that event a occurs is 0.4 , and the probability that events a and b both occur is 0.25 . if the probability that either event a or event b occurs is 0.8 , what is the probability that event b will occur ? | "p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.8 = 0.4 + p ( b ) - 0.25 p ( b ) = 0.65 ans : e" | a = 0 + 8
b = a - 0
|
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