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a ) 495 , b ) 70 , c ) 245 , d ) 425 , e ) 555 | d | add(1, multiply(multiply(subtract(12, 4), 4), 4)) | in a group of 12 engineers , 4 engineers have a phd degree ; the others have only an ms or bs degree . a team of 4 engineers is to be chosen which must have at least 1 engineer with a phd , how many different teams can be chosen ? | "the problem asks for a combination , since order does n ' t matter . now , selecting r items from a set of n gives the combination formula : ncr = n ! / r ! ( n - r ) ! n = 12 r = 4 so , total teams is 12 c 4 = 12 ! / ( 4 ! ( 12 - 4 ) ! ) = 495 , and n = 12 - 4 = 8 r = 4 for teams without a phd is 8 c 4 = 8 ! / ( 4 ! ( 8 - 4 ) ! ) = 70 , so , teams with at least 1 phd = 495 - 70 = 425 answer : d" | a = 12 - 4
b = a * 4
c = b * 4
d = 1 + c
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a ) rs . 10100 , b ) rs . 10110 , c ) rs . 10123.20 , d ) rs . 10135.50 , e ) rs . 10167.20 | c | subtract(multiply(power(add(divide(12, const_100), const_1), 3), 25000), 25000) | what will be the ci on a sum of rs . 25000 after 3 years at the rate of 12 per year ? | amount = rs . 25000 x 1 + 12 3 100 = rs . 25000 x 28 x 28 x 28 25 25 25 = rs . 35123.20 c . i . = rs . ( 35123.20 - 25000 ) = rs . 10123.20 c | a = 12 / 100
b = a + 1
c = b ** 3
d = c * 25000
e = d - 25000
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a ) 50 km , b ) 30 km , c ) 18 km , d ) 16 km , e ) 97 km | b | multiply(10, divide(10, subtract(15, 10))) | if a person walks at 15 km / hr instead of 10 km / hr , he would have walked 10 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 10 = ( x + 10 ) / 15 5 x - 150 = > x = 30 km . answer : b" | a = 15 - 10
b = 10 / a
c = 10 * b
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a ) 1450 , b ) 1971 , c ) 9676 , d ) 1679 , e ) 2691 | a | multiply(subtract(rectangle_area(add(85, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(85, 55)), 2) | a rectangular grass field is 85 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 85 + 55 + 2.5 * 2 ) 2 * 2.5 = > 725 725 * 2 = rs . 1450 answer : a" | a = 2 * 5
b = 85 + a
c = 2 * 5
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
a ) 16 seconds , b ) 15.4 seconds , c ) 26 seconds , d ) 12 seconds , e ) 18 seconds | b | divide(60, multiply(add(18, 3), const_0_2778)) | the speed at which a man can row a boat in still water is 18 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 60 metres ? | "speed of the boat downstream = 18 + 3 = 21 kmph = 21 * 5 / 18 = 35 / 9 m / s hence time taken to cover 60 m = 60 * 9 / 35 = 15.4 seconds . answer : b" | a = 18 + 3
b = a * const_0_2778
c = 60 / b
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a ) 3 sec , b ) 8 sec , c ) 9 sec , d ) 5 sec , e ) 7 sec | b | divide(add(170, 100), multiply(add(50, 70), const_0_2778)) | two trains are moving at 50 kmph and 70 kmph in opposite directions . their lengths are 170 m and 100 m respectively . the time they will take to pass each other completely is ? | "70 + 50 = 120 * 5 / 18 = 100 / 3 mps d = 170 + 100 = 270 m t = 270 * 3 / 100 = 8 sec answer : b" | a = 170 + 100
b = 50 + 70
c = b * const_0_2778
d = a / c
|
a ) 28 , b ) 27 , c ) 26 , d ) 22 , e ) 24 | e | add(subtract(96, multiply(19, 4)), 4) | having scored 96 runs in the 19 th inning , a cricketer increases his average score by 4 . what will be his average score after 19 innings ? | "explanation : let the average score of the first 18 innings be n 18 n + 96 = 19 ( n + 4 ) = > n = 20 so , average score after 19 th innings = x + 4 = 24 . answer : e" | a = 19 * 4
b = 96 - a
c = b + 4
|
a ) 10 cm , b ) 30 cm , c ) 25 cm , d ) 20 cm , e ) 35 cm | d | multiply(12, const_4) | what is the greatest possible length which can be used to measure exactly the lengths 8 m , 4 m 20 cm and 12 m 20 cm ? | "required length = hcf of 800 cm , 420 cm , 1220 cm = 20 cm answer : option d" | a = 12 * 4
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a ) 4 , b ) 3 , c ) 2 , d ) 0 , e ) 6 | e | subtract(25, reminder(1019, 25)) | what least number should be added to 1019 , so that the sum is completely divisible by 25 ? | 1019 ã · 25 = 40 with remainder = 19 19 + 6 = 25 . hence 6 should be added to 1019 so that the sum will be divisible by 25 answer : option e | a = 25 - reminder
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['a ) 12 : 7', 'b ) 10 : 7', 'c ) 8 : 7', 'd ) 4 : 3', 'e ) none'] | d | divide(multiply(divide(286, add(8, 5)), 8), add(multiply(divide(286, add(8, 5)), 5), 22)) | in a school having roll strength 286 , the ratio of boys and girls is 8 : 5 . if 22 more girls get admitted into the school , the ratio of boys and girls becomes | solution : boys : girls = 8 : 5 ; ( let the boys = 8 x ; girl = 5 x ) total strength = 286 ; 8 x + 5 x = 286 ; 13 x = 286 ; or , x = 286 / 13 = 22 ; boys = 176 and girls = 110 ; 22 more girls get admitted then number of girls become , ( 5 x + 22 ) = 110 + 22 = 132 ; now , new ratio of boys and girls = 176 : 132 = 4 : 3 . answer : option d | a = 8 + 5
b = 286 / a
c = b * 8
d = 8 + 5
e = 286 / d
f = e * 5
g = f + 22
h = c / g
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a ) - 1 , b ) 0 , c ) 6 , d ) 2 , e ) 3 | c | add(multiply(power(add(1, sqrt(2)), 2), power(subtract(add(1, sqrt(2)), 2), 2)), 4) | if x = 1 + √ 2 , then what is the value of x 4 - 4 x 3 + 4 x 2 + 5 ? | "answer x = 1 + √ 2 ∴ x 4 - 4 x 3 + 4 x 2 + 5 = x 2 ( x 2 - 4 x + 4 ) + 5 = x 2 ( x - 2 ) 2 + 5 = ( 1 + √ 2 ) 2 ( 1 + √ 2 - 2 ) 2 + 5 = ( √ 2 + 1 ) 2 ( √ 2 - 1 ) 2 + 5 = [ ( √ 2 ) 2 - ( 1 ) 2 ] 2 + 5 = ( 2 - 1 ) 2 = 1 + 5 = 6 correct option : c" | a = math.sqrt(2)
b = 1 + a
c = b ** 2
d = math.sqrt(2)
e = 1 + d
f = e - 2
g = f ** 2
h = c * g
i = h + 4
|
['a ) 15', 'b ) 25', 'c ) 35', 'd ) 30', 'e ) none of these'] | b | divide(add(325, const_100), 17) | the ratio of sum of squares of first n natural numbers to square of sum of first n natural numbers is 17 : 325 . the value of n is | sum of 1 st n natural no . s is n ( n + 1 ) / 2 , sum of sqaures of 1 st n natural no . s is n ( n + 1 ) ( 2 n + 1 ) / 6 , the ratio of sum of squares of first n natural numbers to square of sum of first n natural numbers = n ( n + 1 ) ( 2 n + 1 ) * 2 * 2 / ( n * n * ( n + 1 ) * ( n + 1 ) ) = 2 ( 2 n + 1 ) / ( 3 n ( n + 1 ) ) now by hit and trial for the given options , if we put 25 then this will give 17 / 325 therefore the answer is 25 answer : b | a = 325 + 100
b = a / 17
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a ) 1000 , b ) 1055 , c ) 1065 , d ) 1075 , e ) 1080 | b | add(multiply(8, 70), multiply(9, 55)) | harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 70 × 8 = 560 . cost of 9 kg of mangoes = 55 × 9 = 490 . total cost he has to pay = 560 + 490 = 1055 . b )" | a = 8 * 70
b = 9 * 55
c = a + b
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a ) - 16 , b ) - 8 , c ) 0 , d ) 14 , e ) 16 | b | add(sqrt(9), sqrt(64)) | if x and y are integers such that ( x + 1 ) ^ 2 is less than or equal to 9 and ( y - 1 ) ^ 2 is less than 64 , what is the sum of the maximum possible value of xy and the minimum possible value of xy ? | "( x + 1 ) ^ 2 < = 9 x < = 2 x > = - 4 ( y - 1 ) ^ 2 < 64 y < 9 y > - 7 max possible value of xy is - 4 × - 6 = 24 minimum possible value of xy is - 4 × 8 = - 32 - 32 + 24 = - 8 answer : b" | a = math.sqrt(9)
b = math.sqrt(64)
c = a + b
|
a ) 29 , b ) 70 , c ) 92 , d ) can not be determined , e ) none of these | d | divide(63, subtract(const_10, const_1)) | the difference between a two - digit number and the number obtained by interchanging the two digits is 63 . which is the smaller of the two numbers ? | "explanation : let the ten ' s digit be x and units digit by y . then , ( 10 x + y ) - ( 10 y + x ) = 63 9 ( x - y ) = 63 x - y = 7 thus , none of the numbers can be determined . answer is d" | a = 10 - 1
b = 63 / a
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a ) 3120 , b ) 3330 , c ) 3540 , d ) 3750 , e ) 3960 | e | multiply(330, 12) | a man started driving at a constant speed , from the site of a blast , the moment he heard the blast . he heard a second blast after a time of 30 mins and 12 seconds . if the second blast occurred exactly 30 mins after the first , how many meters was he from the site when he heard the second blast ? ( speed of sound = 330 m / s ) | "the distance the sound traveled to the man is 12 * 330 = 3960 meters the answer is e ." | a = 330 * 12
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a ) 4,514 , b ) 4,475 , c ) 4,521 , d ) 4,428 , e ) 4,349 | e | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | how many positive integers less than 5,000 are evenly divisible by neither 13 nor 21 ? | "integers less than 5000 divisible by 13 5000 / 13 = 333 . something , so 333 integers less than 5000 divisible by 21 5000 / 21 = 238 . # # , so 238 we have double counted some , so take lcm of 13 and 21 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 13 and 21 = 333 + 238 - 47 = 524 now subtract that from 4999 . 4999 - 524 = 4349 answer e ." | a = 4 + 13
b = a - 1
c = math.factorial(b)
d = math.factorial(13)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 900 km , b ) 800 km , c ) 600 km , d ) 690 km , e ) 609 km | c | multiply(divide(multiply(30, add(const_3, const_2)), subtract(40, 30)), 40) | a train leaves delhi at 9 a . m . at a speed of 30 kmph . another train leaves at 2 p . m . at a speed of 40 kmph on the same day and in the same direction . how far from delhi , will the two trains meet ? | explanation : d = 30 * 5 = 150 rs = 40 – 30 = 10 t = 150 / 10 = 15 d = 40 * 15 = 600 km answer : option c | a = 3 + 2
b = 30 * a
c = 40 - 30
d = b / c
e = d * 40
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a ) 2 / 5 , b ) 3 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) none of these | e | subtract(divide(4, 5), divide(2, 5)) | what is difference between biggest and smallest fraction among 2 / 5 , 3 / 4 , 4 / 5 and 5 / 6 | "explanation : 2 / 5 = . 4 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 6 = . 833 so biggest is 5 / 6 and smallest is 2 / 5 their difference is 5 / 6 - 2 / 5 = 13 / 30 option e" | a = 4 / 5
b = 2 / 5
c = a - b
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a ) rs . 56 , b ) rs . 55 , c ) rs . 50 , d ) rs . 90 , e ) rs . 75 | c | multiply(divide(const_100, 10), 5) | a 5 % stock yields 10 % . the market value of the stock is : | "explanation : for an income of rs . 10 , investment = rs . 100 . for an income of rs 5 , investment = rs . 100 / 10 x 5 = rs 50 market value of rs . 100 stock = rs . 50 answer is c" | a = 100 / 10
b = a * 5
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a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 82 kmph | c | divide(divide(subtract(120, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | a train 120 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is | "speed of train relative to man : 120 / 6 * 18 / 5 km / hr = 72 km / hr let speed of train = x therefore x + 6 = 72 x = 72 - 6 x = 66 km / hr answer : c" | a = 6 * const_0_2778
b = a * 6
c = 120 - b
d = c / 6
e = d / const_0_2778
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a ) 15 % , b ) 25 % , c ) 35 % , d ) 45 % , e ) 55 % | b | multiply(divide(subtract(add(divide(32, 2), multiply(subtract(64, 32), 2)), 64), 64), const_100) | every day daniel drives 64 miles back from work . on sunday , daniel drove all the way back from work at a constant speed of x miles per hour . on monday , daniel drove the first 32 miles back from work at ( 2 x ) miles per hour , and the rest of the way at ( x / 2 ) miles per hour . the time it took daniel to drive back from work on monday is longer than the time it took him to drive back from work on sunday by what percent ? | "let ' s test x = 4 . . . . on sunday , daniel drove 64 miles at 4 miles / hour . d = ( r ) ( t ) 64 = ( 4 ) ( t ) 64 / 4 = 16 = t it takes 16 hours to drive home on monday , daniel drove the first 32 miles at ( 2 ) ( 4 ) = 8 miles / hour and the rest of the way ( 32 miles ) at 4 / 2 = 2 miles / hour d = ( r ) ( t ) 32 = ( 8 ) ( t ) 32 / 8 = 4 = t it takes 4 hours for the first part d = ( r ) ( t ) 32 = ( 2 ) ( t ) 32 / 2 = 16 = t it takes 16 hours for the second part total time to drive home on monday = 4 + 16 = 20 hours we ' re asked by what percent 20 hours is greater than 16 hours . 20 / 16 = 1.25 , so it is 25 % greater . b" | a = 32 / 2
b = 64 - 32
c = b * 2
d = a + c
e = d - 64
f = e / 64
g = f * 100
|
a ) 90 deg , b ) 75 deg , c ) 45 deg , d ) 15 deg , e ) 30 deg | a | subtract(multiply(20, multiply(const_3, const_2)), 10) | what is the angle between the hands of a clock when time is 10 : 20 ? | "angle between two hands = 20 h - 11 / 2 m = 20 * 10 - 20 * 11 / 2 = 200 - 110 = 90 deg answer : a" | a = 3 * 2
b = 20 * a
c = b - 10
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a ) 230 , b ) 141 , c ) 66 , d ) 245 , e ) 21 | d | divide(volume_cylinder(divide(14, const_2), 5), const_pi) | the diameter of a cylindrical tin is 14 cm and height is 5 cm . find the volume of the cylinder ? | "r = 7 h = 5 π * 7 * 7 * 5 = 45 π cc answer : d" | a = 14 / 2
b = volume_cylinder / (
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a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | a | divide(multiply(40, 15), 25) | if 40 men take 15 days to to complete a job , in how many days can 25 men finish that work ? | ans . 24 days | a = 40 * 15
b = a / 25
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a ) 12 , b ) 18 , c ) 48 , d ) 20 , e ) 30 | c | divide(multiply(160, add(multiply(4, const_2), 4)), 40) | it takes printer a 4 more minutes more than printer b to print 40 pages . working together , the two printers can print 50 pages in 6 minutes . how long will it take printer a to print 160 pages ? | if it takes 4 more minutes for a to print 40 pages than it takes b , it takes 5 more minutes for a to print 50 pages than it takes b . thus if b is the number of minutes than b takes to print 50 pages , we can write : 1 / b + 1 / ( b + 5 ) = 1 / 6 ( since in 1 minute , they print 1 / 6 th of the 50 page job ) 6 ( 2 b + 5 ) = b ( b + 5 ) b ^ 2 - 7 b - 30 = 0 ( b - 10 ) ( b + 3 ) = 0 b = 10 thus it takes a 15 minutes to print 50 pages and 15 * 160 / 50 = 48 minutes to print 160 pages ( answer c ) | a = 4 * 2
b = a + 4
c = 160 * b
d = c / 40
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a ) 288 , b ) 132 , c ) 772 , d ) 207.3 , e ) 261 | d | multiply(circumface(divide(22, const_2)), 3) | find the cost of fencing around a circular field of diameter 22 m at the rate of rs . 3 a meter ? | "2 * 22 / 7 * 11 = 69.1 69.1 * 3 = rs . 207.3 answer : d" | a = 22 / 2
b = circumface * (
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a ) 121 , b ) 168 , c ) 189 , d ) 190 , e ) 200 | a | subtract(multiply(151, 7), multiply(6, 156)) | there are 6 people in the elevator . their average weight is 156 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person . | solution average of 7 people after the last one enters = 151 . â ˆ ´ required weight = ( 7 x 151 ) - ( 6 x 156 ) = 1057 - 936 = 121 . answer a | a = 151 * 7
b = 6 * 156
c = a - b
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a ) 30 , b ) 12 , c ) 18 , d ) 22 , e ) 08 | a | multiply(2, 15) | each child has 2 pencils and 13 skittles . if there are 15 children , how many pencils are there in total ? | 2 * 15 = 30 . answer is a . | a = 2 * 15
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a ) 3 , b ) 5 , c ) 18 , d ) 6 , e ) 8 | c | add(7, 11) | if a certain number x is divided by 62 , the reminder is 7 . what is the reminder when x + 11 is divided by 31 ? | x can be written as 62 k + 7 or x = 7 , 69,131 , etc . x + 11 = 62 k + 7 + 11 = 62 k + 18 or x + 11 = 18 , 80,142 etc . when divided by 31 , we will get the remainder 18 . c | a = 7 + 11
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a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | e | floor(add(divide(100, 7), divide(100, power(7, const_2)))) | if 7 ^ w is a factor of the product of the integers from 1 to 100 , inclusive , what is the largest value of w ? | so the question just means that we have to find all the multiples w of 7 between 1 to 100 so there are 14 multiples of 7 ( 7 - 98 ) but 49 and 98 contain two 7 ' s as factors so we have to add 14 + 2 = 16 e | a = 100 / 7
b = 7 ** 2
c = 100 / b
d = a + c
e = math.floor(d)
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a ) a ) 400 , b ) b ) 450 , c ) c ) 500 , d ) d ) 550 , e ) e ) 600 | a | subtract(subtract(multiply(add(add(3, 5), const_2), 145), multiply(5, 150)), multiply(3, 100)) | a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 145 . find the unknown rate of two towels ? | "10 * 145 = 1450 3 * 100 + 5 * 150 = 1050 1450 – 1050 = 400 a" | a = 3 + 5
b = a + 2
c = b * 145
d = 5 * 150
e = c - d
f = 3 * 100
g = e - f
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a ) 20 min , b ) 10 min , c ) 15 min , d ) 25 min , e ) 30 min | c | divide(multiply(2, const_60), 8) | if 2 pipes take an hour to fill a tank , then how long should 8 pipes take to fill the same tank ? | if 2 pipes take an hour to fill a tank , then 8 pipes will take 2 * 60 / 8 = 15 mins to dig a ditch of the same type . answer : c | a = 2 * const_60
b = a / 8
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a ) 10 , b ) 20 , c ) 40 , d ) 60 , e ) 12 | e | divide(multiply(6, 4), const_2) | there are 6 chess amateurs playing in villa ' s chess club tournament . if each chess amateur plays with exactly 4 other amateurs , what is the total number of chess games possible to be played in the tournament ? | method 2 : each person is one participant of 4 games . so there are in all 4 * 6 = 24 instances of one participant games . but each game has 2 participants so total number of games = 24 / 2 = 12 e | a = 6 * 4
b = a / 2
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a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 4 / 5 | a | divide(subtract(30, 26), subtract(30, 19)) | a jar full of whisky contains 30 % alcohol . a part of this whisky is replaced by another containg 19 % alcohol and now the percentage of alcohol was found to be 26 % . what quantity of whisky is replaced ? | "let us assume the total original amount of whiskey = 10 ml - - - > 4 ml alcohol and 6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = 4 - 0.4 x new quantity of whiskey added = x ml out of which 0.19 is the alcohol . thus , the final quantity of alcohol = 4 - 0.4 x + 0.19 x - - - - > ( 4 - 0.21 x ) / 10 = 0.26 - - - > x = 20 / 3 ml . per the question , you need to find the x ml removed as a ratio of the initial volume - - - > ( 20 / 3 ) / 10 = 1 / 3 . hence , a is the correct answer ." | a = 30 - 26
b = 30 - 19
c = a / b
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a ) 20 / 9 , b ) 40 / 9 , c ) 40 / 14 , d ) 60 / 9 , e ) 80 / 9 | c | divide(const_1, add(divide(const_1, 4), divide(const_1, 10))) | worker a takes 4 hours to do a job . worker b takes 10 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | one day work of a = 1 / 4 one day work of b = 1 / 10 so one day work of a and b together = 1 / 4 + 1 / 10 = 14 / 40 so total days required = 40 / 14 answer : c | a = 1 / 4
b = 1 / 10
c = a + b
d = 1 / c
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a ) 10 mps , b ) 8 mps , c ) 9 mps , d ) 7 mps , e ) 12 mps | b | multiply(const_0_2778, 30) | express a speed of 30 kmph in meters per second ? | "30 * 5 / 18 = 8 mps answer : b" | a = const_0_2778 * 30
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a ) 192 , b ) 280 , c ) 384 , d ) 400 , e ) 500 | a | divide(multiply(multiply(100, 16), 6), subtract(multiply(5, 16), multiply(5, 6))) | in a maths test , students were asked to find 5 / 16 of a certain number . one of the students by mistake found 5 / 6 th of that number and his answer was 100 more than the correct answer . find the number . | "explanation : let the number be x . 5 * x / 6 = 5 * x / 16 + 100 25 * x / 48 = 100 x = 192 answer a" | a = 100 * 16
b = a * 6
c = 5 * 16
d = 5 * 6
e = c - d
f = b / e
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a ) 2 : 3 , b ) 5 : 1 , c ) 4 : 5 , d ) 7 : 1 , e ) 8 : 1 | b | subtract(9, 6) | a boat running up stram takes 6 hours to cover a certain distance , while it takes 9 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ? | "explanation : let speed of boat is x km / h and speed stream is y km / hr 6 ( x + y ) = 9 ( x - y ) 6 x + 6 y = 9 x - 9 y 15 y = 3 x 5 y = x x / y = 5 / 1 5 : 1 answer : option b" | a = 9 - 6
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a ) 18 % , b ) 31 % , c ) 30 % , d ) 45 % , e ) 12 % | c | multiply(divide(subtract(1900, 1330), 1900), const_100) | the cost price of a radio is rs . 1900 and it was sold for rs . 1330 , find the loss % ? | "1900 - - - - 570 100 - - - - ? = > 30 % answer : c" | a = 1900 - 1330
b = a / 1900
c = b * 100
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a ) 476.75 ft , b ) 450 ft , c ) 480 ft , d ) 490 ft , e ) 500 ft | a | multiply(25, divide(multiply(13, 5280), const_3600)) | someone on a skateboard is traveling 13 miles per hour . how many feet does she travel in 25 seconds ? ( 1 mile = 5280 feet ) | "per second = > 13 * 5280 ft / 60 * 60 = 19.07 ft 25 seconds = > 19.07 * 25 = 476.75 ft answer : a" | a = 13 * 5280
b = a / 3600
c = 25 * b
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a ) 10 / 16 , b ) 6 / 16 , c ) 4 / 16 , d ) 6 / 10 , e ) 5 / 9 | e | divide(divide(subtract(14, 4), add(const_1, const_1)), add(divide(subtract(14, 4), add(const_1, const_1)), 4)) | there are 4 more women than there are men on a local co - ed softball team . if there are a total of 14 players on the team , what is the ratio of men to women ? | "w = m + 4 w + m = 14 m + 4 + m = 14 2 m = 10 m = 5 w = 9 ratio : 5 : 9 ans : e" | a = 14 - 4
b = 1 + 1
c = a / b
d = 14 - 4
e = 1 + 1
f = d / e
g = f + 4
h = c / g
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a ) 349 , b ) 541.75 , c ) 224.37 , d ) 370.833 , e ) none of these | d | divide(divide(8900, 6), 4) | 8900 ÷ 6 ÷ 4 = ? | explanation : given exp . 8900 * 1 / 6 * 1 / 4 = 370.833 answer : d | a = 8900 / 6
b = a / 4
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a ) 9.09 % , b ) 10 % , c ) 12 1 / 2 % , d ) 15 % , e ) 16 2 / 3 % | a | divide(10, add(const_1, divide(10, const_100))) | the markup on a box of apples is 10 percent of the cost . the markup is what percent of the selling price ? ( markup = selling price - cost ) | mp = 0.1 cp sp = cp + 0.1 cp = 1.1 cp hence mp = 0.1 / 1.1 sp = 1 / 1 sp . hence mp is 9.09 % of sp answer a | a = 10 / 100
b = 1 + a
c = 10 / b
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a ) 188 m , b ) 876 m , c ) 251 m , d ) 100 m , e ) 145 m | d | multiply(100, subtract(const_2, const_1)) | a train speeds past a pole in 15 seconds and a platform 100 m long in 40 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 40 = x / 15 x = 100 m . answer : d" | a = 2 - 1
b = 100 * a
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a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 110 | d | multiply(divide(multiply(30, const_1000), const_3600), 12) | a train running at the speed of 30 km / hr crosses a pole in 12 seconds . what is the length of the train ? | "speed = ( 30 x ( 5 / 18 ) m / sec = ( 25 / 3 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( 25 / 3 ) x 12 ) m = 100 m d" | a = 30 * 1000
b = a / 3600
c = b * 12
|
a ) 40 , b ) 80 , c ) 120 , d ) 200 , e ) 220 | a | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | three numbers are in the ratio 3 : 4 : 5 and their l . c . m is 2400 . their h . c . f is : | "let the numbers be 3 x , 4 x and 5 x . then , their l . c . m = 60 x . so , 60 x = 2400 or x = 40 . the numbers are ( 3 * 40 ) , ( 4 * 40 ) and ( 5 * 40 ) . hence , required h . c . f = 40 . answer : a" | a = 3 * 5
b = a * 100
c = 4 * 5
d = b + c
|
a ) 24 , b ) 58 , c ) 26 , d ) 34 , e ) 25 | a | divide(factorial(subtract(add(const_4, 210), const_1)), multiply(factorial(210), factorial(subtract(const_4, const_1)))) | how many positive integers less than 9999 are such that the product of their digits is 210 . | "the prime factorization of 210 is 2 * 3 * 5 * 7 . so one way to make the right kind of number is to use those four digits , in any of the 4 ! = 24 orders you can put them in . notice though that we can also get 210 as a product by multiplying 5 , 6 and 7 . so we can make some 3 - digit numbers with the right product : 3 ! = 6 of them to be exact . but we can also get the right product using the digit 1 along with the digits 5 , 6 , and 7 . again we can arrange those digits in 4 ! = 24 orders . a" | a = 4 + 210
b = a - 1
c = math.factorial(b)
d = math.factorial(210)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 9 am , b ) 12 am , c ) 10.30 am , d ) 12.30 am , e ) 11 am | a | add(divide(add(65, 25), add(20, 25)), 7) | two stations p and q are 65 km apart on a straight track . one train starts from p at 7 a . m . and travels towards q at 20 kmph . another train starts from q at 8 a . m . and travels towards p at a speed of 25 kmph . at what time will they meet ? | assume both trains meet after x hours after 7 am distance covered by train starting from p in x hours = 20 x km distance covered by train starting from q in ( x - 1 ) hours = 25 ( x - 1 ) total distance = 65 = > 20 x + 25 ( x - 1 ) = 65 = > 45 x = 90 = > x = 2 means , they meet after 2 hours after 7 am , ie , they meet at 9 am answer is a . | a = 65 + 25
b = 20 + 25
c = a / b
d = c + 7
|
a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(385, 320), 320), const_100) | a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase w in sales ? | "last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase w = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c ." | a = 385 - 320
b = a / 320
c = b * 100
|
a ) rs . 6500 , b ) rs . 7500 , c ) rs . 7200 , d ) rs . 7273 , e ) rs . 6300 | d | divide(multiply(400, const_100), subtract(const_100, add(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(5, const_100))))) | a man saves 10 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 5 % , he is only able to save rs . 400 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 90 savings = rs . 10 present expenditure 90 + 90 * ( 5 / 100 ) = rs . 94.5 present savings = 100 – 94.5 = rs . 5.50 if savings is rs . 5.50 , salary = rs . 100 if savings is rs . 400 , salary = 100 / 5.5 * 400 = 7273 answer : d" | a = 400 * 100
b = 100 - 10
c = 100 - 10
d = 5 / 100
e = c * d
f = b + e
g = 100 - f
h = a / g
|
a ) 30 / 12 , b ) 5 / 12 , c ) 3 / 4 , d ) 2 / 12 , e ) 3 / 5 | a | divide(subtract(const_1, add(multiply(2, inverse(6)), multiply(2, inverse(8)))), inverse(6)) | working individually , emma can wrap presents for 6 hours and troy can wrap presents in 8 hours . if emma and troy work together but independently at the task for 2 hours , at which point troy leaves , how many remaining hours will it take emma to complete the task alone ? | in first 2 hrs troy will finish 2 / 8 = 1 / 4 of work and emma will finish 2 / 6 work so total 1 / 4 + 1 / 3 = 7 / 12 work is finished and 1 - 7 / 12 = 5 / 12 work remaining . now emma will take ( 5 / 12 ) * 6 = 30 / 12 hrs to finish it . so answer is a . | a = 1/(6)
b = 2 * a
c = 1/(8)
d = 2 * c
e = b + d
f = 1 - e
g = 1/(6)
h = f / g
|
a ) 20 , b ) 200 , c ) 300 , d ) 90 , e ) 140 | d | subtract(divide(divide(8480, 26.50), const_2), multiply(const_2, 20)) | the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 8480 , what is the length of the plot in metres ? | "let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 8480 [ 4 l - 40 ] = 8480 / 26.50 = 320 4 l = 360 l = 360 / 4 = 90 meters . answer : d" | a = 8480 / 26
b = a / 2
c = 2 * 20
d = b - c
|
a ) 1 / 2 , b ) 5 / 8 , c ) 9 / 14 , d ) 16 / 21 , e ) 9 / 10 | c | divide(add(multiply(divide(24, const_2), add(const_2, const_3)), multiply(24, const_2)), multiply(24, add(const_3, const_4))) | a certain electric - company plan offers customers reduced rates for electricity used between 8 p . m . and 8 a . m . weekdays and 24 hours a day saturdays and sundays . under this plan , the reduced rates z apply to what fraction of a week ? | number of hours between 8 pm to 8 am = 12 number of hours with reduced rates = ( 12 * 5 ) + ( 24 * 2 ) hours with reduced rates z / total number of hours in a week = ( 12 * 5 ) + ( 24 * 2 ) / ( 24 * 7 ) = 108 / ( 24 * 7 ) = 9 / 14 answer : c | a = 24 / 2
b = 2 + 3
c = a * b
d = 24 * 2
e = c + d
f = 3 + 4
g = 24 * f
h = e / g
|
a ) 13 : 18 , b ) 8 : 12 , c ) 1 : 12 , d ) 8 : 17 , e ) 1 : 11 | a | power(divide(125, 216), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 125 : 216 . what is the ratio of their total surface areas ? | "ratio of the sides = ³ √ 125 : ³ √ 216 = 5 : 6 ratio of surface areas = 52 : 62 = 13 : 18 answer : a" | a = 125 / 216
b = 1 / 3
c = a ** b
|
a ) 23 kg , b ) 24 kg , c ) 25 kg , d ) 26 kg , e ) 27 kg | d | subtract(add(multiply(35, const_2), multiply(41, const_2)), multiply(42, const_3)) | the average weight of d , e and f is 42 kg . if the average weight of d and e be 35 kg and that of e and f be 41 kg , what is the weight of e ? | let the weight of d , e and f are a , b and c respectively . average weight of d , e and f = 42 a + b + c = 42 × 3 = 126 - - - equation ( 1 ) average weight of d and e = 35 a + b = 35 × 2 = 70 - - - equation ( 2 ) average weight of e and f = 41 b + c = 41 × 2 = 82 - - - equation ( 3 ) equation ( 2 ) + equation ( 3 ) - equation ( 1 ) = > a + b + b + c - ( a + b + c ) = 70 + 82 - 126 = > b = 152 - 126 = 26 weight of b = 26 kg answer : d | a = 35 * 2
b = 41 * 2
c = a + b
d = 42 * 3
e = c - d
|
a ) 7.19 , b ) 7.17 , c ) 7.2 , d ) 7.35 , e ) 7.11 | d | divide(add(121, 165), multiply(add(75, 65), const_0_2778)) | two trains 121 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 75 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 121 + 165 ) / ( 75 + 65 ) * 18 / 5 t = 7.35 answer : d" | a = 121 + 165
b = 75 + 65
c = b * const_0_2778
d = a / c
|
a ) 87 , b ) 97 , c ) 27 , d ) 21 , e ) 56 | e | subtract(multiply(12, 15), add(const_10, const_1)) | find the smallest number which when divided by 12 and 15 leaves respective remainders of 8 and 11 . | "let ' n ' is the smallest number which divided by 12 and 15 leaves respective remainders of 8 and 11 . required number = ( lcm of 12 and 15 ) - ( common difference of divisors and remainders ) = ( 60 ) - ( 4 ) = 56 . answer : e" | a = 12 * 15
b = 10 + 1
c = a - b
|
a ) 8 % , b ) 10 % , c ) 21 % , d ) 15 % , e ) 17 % | a | subtract(subtract(add(20, const_100), divide(multiply(add(20, const_100), 10), const_100)), const_100) | a merchant marks his goods up by 20 % and then offers a discount of 10 % on the marked price . what % profit does the merchant make after the discount ? | "let the price be 100 . the price becomes 120 after a 20 % markup . now a discount of 10 % on 120 . profit = 108 - 100 8 % answer a" | a = 20 + 100
b = 20 + 100
c = b * 10
d = c / 100
e = a - d
f = e - 100
|
a ) 52 , b ) 37 , c ) 39 , d ) 43 , e ) 42 | a | add(subtract(90, multiply(2, 20)), 2) | a batsman in his 20 th innings makes a score of 90 , and thereby increases his average by 2 . what is his average after the 20 th innings ? he had never been ’ not out ’ . | "average score before 20 th innings = 90 - 2 × 20 = 50 average score after 20 th innings = > 50 + 2 = 52 answer : a" | a = 2 * 20
b = 90 - a
c = b + 2
|
a ) 350 , b ) 353 , c ) 354 , d ) 356 , e ) 357 | b | add(add(add(add(const_60, 4), 4), multiply(add(add(const_60, 4), 4), 4)), add(multiply(3, 1), multiply(multiply(3, 2), 2))) | s ( n ) is a n - digit number formed by attaching the first n perfect squares , in order , into one integer . for example , s ( 1 ) = 1 , s ( 2 ) = 14 , s ( 3 ) = 149 , s ( 4 ) = 14916 , s ( 5 ) = 1491625 , etc . how many digits are in s ( 99 ) ? | "focus on the points where the number of digits in squares change : 1 , 2 , 3 - single digit squares . first 2 digit number is 10 . 4 , 5 , . . . 9 - two digit squares . to get 9 , the last number with two digit square , think that first 3 digit number is 100 which is 10 ^ 2 . so 9 ^ 2 must be the last 2 digit square . 10 , 11 , 12 , . . . 31 - three digit squares . to get 31 , think of 1000 - the first 4 digit number . it is not a perfect square but 900 is 30 ^ 2 . 32 ^ 2 = 2 ^ 10 = 1024 , the first 4 digit square . 32 - 99 - four digit squares . to get 99 , think of 10,000 - the first 5 digit number which is 100 ^ 2 . so number of digits in s ( 99 ) = 3 * 1 + 6 * 2 + 22 * 3 + 68 * 4 = 3 + 12 + 66 + 272 = 353 ; answer : b" | a = const_60 + 4
b = a + 4
c = const_60 + 4
d = c + 4
e = d * 4
f = b + e
g = 3 * 1
h = 3 * 2
i = h * 2
j = g + i
k = f + j
|
a ) 276 , b ) 295 , c ) 322 , d ) 345 , e ) 354 | a | multiply(23, 12) | the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 10 and 12 . the larger of the two numbers is : | "clearly , the numbers are ( 23 x 10 ) and ( 23 x 12 ) . larger number = ( 23 x 12 ) = 276 . answer : option a" | a = 23 * 12
|
a ) 3533 , b ) 3833 , c ) 3333 , d ) 3633 , e ) 3733 | e | add(add(subtract(divide(divide(1210, const_10), const_10), const_1), subtract(subtract(divide(1210, const_10), const_1), subtract(divide(divide(1210, const_10), const_10), const_1))), multiply(subtract(subtract(1210, const_1), subtract(divide(1210, const_10), const_1)), const_3)) | how many figures are required to number the pages the pages of a book containing 1210 pages ? | "1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 999 = 900 * 3 = 2700 1000 to 1210 = 211 * 4 = 844 - - - - - - - - - - - 3733 answer : e" | a = 1210 / 10
b = a / 10
c = b - 1
d = 1210 / 10
e = d - 1
f = 1210 / 10
g = f / 10
h = g - 1
i = e - h
j = c + i
k = 1210 - 1
l = 1210 / 10
m = l - 1
n = k - m
o = n * 3
p = j + o
|
a ) 60 kg , b ) 59 kg , c ) 62 kg , d ) 63 kg , e ) 61 kg | d | subtract(add(multiply(88, const_2), multiply(72, const_2)), multiply(75, const_3)) | the average weight of a , b and c is 75 kg . if the average weight of a and b be 88 kg and that of b and c be 72 kg , then the weight of b is : | "explanation let a , b , c represent their respective weights . then , we have : a + b + c = ( 75 x 3 ) = 225 â € ¦ . ( i ) a + b = ( 88 x 2 ) = 176 â € ¦ . ( ii ) b + c = ( 72 x 2 ) = 144 â € ¦ . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 288 â € ¦ . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 63 . b â € ™ s weight = 63 kg . answer d" | a = 88 * 2
b = 72 * 2
c = a + b
d = 75 * 3
e = c - d
|
a ) 56 , b ) 45 , c ) 58.2 , d ) 58 , e ) 46 | d | add(multiply(11, divide(26.1, 9)), 26.1) | a and b are mixed together in the ratio 9 : 11 . what is the weight of mixture , if 26.1 kg of a has been consumed in it ? | explanation : for 9 kg a , mixture = ( 9 + 11 ) kg . for 26.1 kg a , mixture = ( ( 20 / 9 ) x 26.1 ) kg = 58 kg . answer : d | a = 26 / 1
b = 11 * a
c = b + 26
|
a ) 11 , b ) 12 , c ) 18 , d ) 14 , e ) 15 | c | add(12, add(4, 2)) | find the constant k so that : - x 2 - ( k + 12 ) x - 8 = - ( x - 2 ) ( x - 4 ) | "- x 2 - ( k + 12 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 12 ) x - 8 = - x 2 + 6 x - 8 - ( k + 12 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 18 : solve the above for k correct answer c" | a = 4 + 2
b = 12 + a
|
a ) $ 10,000 , b ) $ $ 28 , 378.37 , c ) $ 42,000 , d ) $ 50,000 , e ) $ 60,000 | b | add(divide(divide(divide(multiply(add(divide(20, const_100), 1), multiply(subtract(add(const_1000, const_60), const_10), const_1000)), add(multiply(subtract(31, 1), add(divide(20, const_100), 1)), 1)), add(divide(20, const_100), 1)), const_100), add(multiply(const_100, const_2), const_3)) | last year , company x paid out a total of $ 1 , 050,000 in salaries to its 31 employees . if no employee earned a salary that is more than 20 % greater than any other employee , what is the lowest possible salary that any one employee earned ? | "employee 1 earned $ x ( say ) employee 2 will not earn more than $ 1.2 x therfore , to minimize the salary of any one employee , we need to maximize the salaries of the other 30 employees ( 1.2 x * 30 ) + x = 1 , 050,000 solving for x = $ 28 , 378.37 answer b" | a = 20 / 100
b = a + 1
c = 1000 + const_60
d = c - 10
e = d * 1000
f = b * e
g = 31 - 1
h = 20 / 100
i = h + 1
j = g * i
k = j + 1
l = f / k
m = 20 / 100
n = m + 1
o = l / n
p = o / 100
q = 100 * 2
r = q + 3
s = p + r
|
a ) 5 , b ) 7 , c ) 12 , d ) 13 , e ) 14 | b | divide(divide(divide(divide(350, const_2), const_3), const_4), divide(const_10, const_2)) | if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 350 , what is the least possible value of n ? | "350 = 2 * 5 * 5 * 7 so the least value forncan be 7 . b" | a = 350 / 2
b = a / 3
c = b / 4
d = 10 / 2
e = c / d
|
a ) 4 : 9 , b ) 1 : 4 , c ) 1 : 8 , d ) 1 : 18 , e ) 1 : 13 | a | divide(power(2, const_2), power(3, const_2)) | the duplicate ratio of 2 : 3 is ? | "2 ^ 2 : 3 ^ 2 = 4 : 9 answer : a" | a = 2 ** 2
b = 3 ** 2
c = a / b
|
a ) 100 , b ) 250 , c ) 750 , d ) 3750 , e ) 5625 | d | multiply(volume_rectangular_prism(40, 25, divide(6, add(const_10, const_2))), 7.5) | the water level in a rectangular swimming pool measuring 40 feet by 25 feet is to be lowered by 6 inches . how many gallons of water must be removed ? ( 1 cu ft = 7.5 gallons ) | "6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 40 * 25 * 1 / 2 = 500 feet ^ 3 of water must be removed , which equals to 500 * 7.5 = 3750 gallons . answer : d ." | a = 10 + 2
b = 6 / a
c = volume_rectangular_prism * (
|
a ) 81.8 , b ) 81.1 , c ) 81.2 , d ) 81.5 , e ) 81.9 | d | add(81, divide(const_1, const_2)) | 3 maths classes : a , b and c take an algebra test . the average score of class a is 83 . the average score of class b is 76 . the average score of class c is 85 . the average score of class a and b is 79 and average score of class b and c is 81 . what is the average score of classes a , b , c ? | let the number of students in classes a , b and c be p , q and r respectively . then , total score of a = 83 p , total score of b = 76 q , total score of c = 85 r . also given that , ( 83 p + 76 q ) / ( p + q ) = 79 = > 4 p = 3 q . ( 76 q + 85 r ) / ( q + r ) = 81 = > 4 r = 5 q , = > q = 4 p / 3 and r = 5 p / 3 therefore , average score of a , b , c = ( 83 p + 76 q + 85 r ) / ( p + q + r ) = 978 / 12 = 81.5 answer : d | a = 1 / 2
b = 81 + a
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(subtract(subtract(add(add(19, 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 19 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 = 19 + 18 + 12 - 4 - 2 x 2 x = 19 + 18 + 12 - 4 - 39 = 45 - 39 = 6 x = 3 answer : b" | a = 19 + 18
b = a + 12
c = b - 4
d = c - 39
e = d / 2
|
a ) 432 , b ) 212 , c ) 75 , d ) 992 , e ) 212 | c | multiply(multiply(divide(40, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(40, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | the length of rectangle is thrice its breadth and its perimeter is 40 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 40 l = 15 b = 5 lb = 15 * 5 = 75 answer : c" | a = 3 * 2
b = 1 * 2
c = a + b
d = 40 / c
e = d * 3
f = 3 * 2
g = 1 * 2
h = f + g
i = 40 / h
j = e * i
|
a ) 3.34 sq m , b ) 98.8 sq m , c ) 67.8 sq m , d ) 27.9 sq m , e ) 19.9 sq m | a | subtract(circle_area(add(divide(4, const_2), divide(25, const_100))), circle_area(divide(4, const_2))) | a 25 cm wide path is to be made around a circular garden having a diameter of 4 meters . approximate area of the path is square meters is ? | "area of the path = area of the outer circle - area of the inner circle = ∏ { 4 / 2 + 25 / 100 } 2 - ∏ [ 4 / 2 ] 2 = ∏ [ 2.252 - 22 ] = ∏ ( 0.25 ) ( 4.25 ) { ( a 2 - b 2 = ( a - b ) ( a + b ) } = ( 3.14 ) ( 1 / 4 ) ( 17 / 4 ) = 53.38 / 16 = 3.34 sq m answer : a" | a = 4 / 2
b = 25 / 100
c = a + b
d = circle_area - (
|
a ) 142.286 , b ) 251.228 , c ) 34 , d ) 200 , e ) 150.627 | a | divide(add(700, 130), divide(multiply(21, const_1000), const_3600)) | a train is 700 meter long is running at a speed of 21 km / hour . in what time will it pass a bridge of 130 meter length ? | "speed = 21 km / hr = 21 * ( 5 / 18 ) m / sec = 35 / 6 m / sec total distance = 700 + 130 = 830 meter time = distance / speed = 830 * ( 6 / 35 ) = 142.286 seconds . answer : a" | a = 700 + 130
b = 21 * 1000
c = b / 3600
d = a / c
|
a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 8 , d ) 1 / 16 , e ) 3 / 16 | b | multiply(power(divide(const_1, const_2), 3), multiply(choose(const_4, const_3), divide(const_1, const_2))) | a coin is tossed 4 times . what is the probability of getting exactly 3 heads ? | the number of possible outcomes is 2 ^ 4 = 16 there are 4 ways to get exactly 3 heads . p ( exactly 3 heads ) = 4 / 16 = 1 / 4 the answer is b . | a = 1 / 2
b = a ** 3
c = math.comb(4, 3)
d = 1 / 2
e = c * d
f = b * e
|
a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28 | d | add(multiply(4, 5), 4) | the ratio of pens to pencils is 5 to 6 . there are 4 more pencils than pens . how many pencils are there ? | "let the number of pens be 5 x and the number of pencils be 6 x . 6 x - 5 x = 4 x = 4 the number of pencils is 24 . the answer is d ." | a = 4 * 5
b = a + 4
|
a ) 227 , b ) 450 , c ) 460 , d ) 480 , e ) 171 | b | multiply(12, multiply(54, const_0_2778)) | a train passes a station platform in 42 sec and a man standing on the platform in 12 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 12 = 180 m . let the length of the platform be x m . then , ( x + 180 ) / 42 = 15 = > x = 450 m . answer : b" | a = 54 * const_0_2778
b = 12 * a
|
a ) 30 , b ) 40 , c ) 60 , d ) none , e ) can not be determined | c | divide(45, 75) | how many pieces of 75 cm can be cut from a rope 45 meters long ? | "explanation : total pieces of 75 cm that can be cut from a rope of 45 meters long is = ( 45 meters ) / ( 75 cm ) = ( 45 meters ) / ( 0.75 meters ) = 60 answer : c" | a = 45 / 75
|
a ) 5 : 8 , b ) 3 : 1 , c ) 3 : 7 , d ) 9 : 5 , e ) 9 : 8 | b | divide(divide(subtract(96, 75), subtract(96, 68)), subtract(const_1, divide(subtract(96, 75), subtract(96, 68)))) | in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 75 per kg ? | "( 96 - 75 ) / ( 75 - 68 ) = 21 / 7 = 3 / 1 answer : b" | a = 96 - 75
b = 96 - 68
c = a / b
d = 96 - 75
e = 96 - 68
f = d / e
g = 1 - f
h = c / g
|
a ) 18.45 % , b ) 18.46 % , c ) 18.47 % , d ) 18.48 % , e ) none of these | b | subtract(const_100, multiply(divide(add(6, const_100), add(30, const_100)), const_100)) | a shopkeeper fixes the marked price of an item 30 % above its cost price . the percentage of discount allowed to gain 6 % is | "explanation : let the cost price = rs 100 then , marked price = rs 130 required gain = 6 % , so selling price = rs 106 discount = 130 - 106 = 24 discount % = ( 24 / 130 ) * 100 = 18.46 % option b" | a = 6 + 100
b = 30 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 23 , b ) 20 , c ) 15 , d ) 26 , e ) 28 | a | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120))) | the product of two numbers is 120 and the sum of their squares is 289 . find the sum ? | "nos a and b ab = 120 and a 2 + b 2 = 289 ( a + b ) 2 = 529 a + b = root ( 529 ) = 23 answer a" | a = 2 * 120
b = 289 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 120
f = d + e
g = math.sqrt(f)
|
a ) 470 , b ) 468 , c ) 465 , d ) 351 , e ) 485 | d | add(divide(divide(26, divide(divide(divide(divide(divide(26, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(26, divide(divide(divide(divide(divide(26, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of first 26 natural numbers | "explanation : sum of n natural numbers = n ( n + 1 ) / 2 = 26 ( 26 + 1 ) / 2 = 26 ( 27 ) / 2 = 351 answer : option d" | a = 26 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 26 / e
g = f / 2
h = 26 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 26 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
|
['a ) ( n - m )', 'b ) mn / ( m + n )', 'c ) ( nm )', 'd ) ( n - m ) / ( m + n )', 'e ) none of these'] | b | divide(multiply(const_1, const_3), add(const_1, const_3)) | in an electric circuit , two resistors with resistances m and n are connected in parallel . in this case , if p is the combined resistance of these two resistors , then the reciprocal of p is equal to the sum of the reciprocals of m and n . what is p in terms of m and n ? | the wording is a bit confusing , though basically we are told that 1 / p = 1 / m + 1 / n , from which it follows that p = mn / ( m + n ) . answer : b | a = 1 * 3
b = 1 + 3
c = a / b
|
a ) 2.25 , b ) 7 , c ) 4.25 , d ) 5.25 , e ) 6.25 | b | subtract(power(3, 2), 2) | x + ( 1 / x ) = 3 find x ^ 2 + ( 1 / x ^ 2 ) | "squaring on both sides ( x + 1 / x ) ^ 2 = 3 ^ 2 x ^ 2 + 1 / x ^ 2 = 9 - 2 x ^ 2 + 1 / x ^ 2 = 7 answer : b" | a = 3 ** 2
b = a - 2
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a ) 443 , b ) 450 , c ) 465 , d ) 468 , e ) 475 | c | divide(add(multiply(8, 410), multiply(add(410, 65), subtract(52, 8))), 52) | a certain debt will be paid in 52 installments from january 1 to december 31 of a certain year . each of the first 8 payments is to be $ 410 ; each of the remaining payments is to be $ 65 more than each of the first 8 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the year ? | "total number of installments = 52 payment per installment for the first 8 installments = 410 payment per installment for the remaining 32 installments = 410 + 65 = 475 average = ( 8 * 410 + 44 * 475 ) / 52 = 465 answer c" | a = 8 * 410
b = 410 + 65
c = 52 - 8
d = b * c
e = a + d
f = e / 52
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a ) 300 , b ) 120 , c ) 150 , d ) 170 , e ) 200 | e | divide(multiply(20, 20), const_2) | the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 20 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 = 20 . . . we get 20 / 2 [ 2 a + 14 d ] = 20 [ a + 7 d ] = 20 * 10 = 200 . . . answer is e . . ." | a = 20 * 20
b = a / 2
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a ) rs . 1200 , b ) rs . 1500 , c ) rs . 1860 , d ) rs . 2000 , e ) none of these | c | multiply(divide(divide(multiply(15, const_2), 10), add(divide(multiply(15, const_2), 10), divide(multiply(15, const_2), 15))), 3100) | a alone can finish a work in 10 days which b alone can finish in 15 days . if they work together and finish it , then out of a total wages of rs . 3100 , a will get : | explanation : ratio of working days of a : b = 10 : 15 therefore , their wages ratio = reverse ratio = 15 : 10 therefore , a will get 15 units of ratio total ratio = 25 1 unit of ratio = 3100 / 25 = 124 so , a ’ s amount = 124 × 15 = rs . 1860 . answer : option c | a = 15 * 2
b = a / 10
c = 15 * 2
d = c / 10
e = 15 * 2
f = e / 15
g = d + f
h = b / g
i = h * 3100
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a ) $ 20 , b ) $ 40 , c ) $ 50 , d ) $ 80 , e ) $ 60 | e | subtract(divide(50, divide(50, const_100)), multiply(divide(50, divide(50, const_100)), divide(40, const_100))) | crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price , and thus increased the profit from the sale of each key chain from 40 % of the selling price to 50 % of the selling price . if the manufacturing cost is now $ 50 , what was it before the decrease ? | "deargoodyear 2013 , i ' m happy to help . this is a relatively straightforward problem , not very challenging . btw , crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the 1970 s . manufacturing now is $ 50 . they now are making a 50 % profit , so the selling price must be $ 100 . they had this same selling price , $ 100 , before they made the change , and had a profit of 40 % , so the manufacturing must have been $ 60 . answer = ( e ) ." | a = 50 / 100
b = 50 / a
c = 50 / 100
d = 50 / c
e = 40 / 100
f = d * e
g = b - f
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a ) 138 , b ) 148 , c ) 150 , d ) 162 , e ) 168 | e | multiply(subtract(5, divide(multiply(110, 5), add(110, 140))), const_60) | while driving from a - ville to b - town , harriet drove at a constant speed of 110 kilometers per hour . upon arriving in b - town , harriet immediately turned and drove back to a - ville at a constant speed of 140 kilometers per hour . if the entire trip took 5 hours , how many minutes did it take harriet to drive from a - ville to b - town ? | "5 hr = 300 min . if harriet spend equal hrs on each leg she will spend 150 min on each . since speed a - b is less than speed b - a and distance on each leg is the same , time spent on a - b is more than 150 min , which mean we can eliminate ans . a , b and c . now let plug in ans . d or e and verify which one give same distance on each leg . e . t = 168 min * leg a - b - - - > d = 110.168 / 60 = 18480 / 60 * leg b - a - - - - > d = 140 * 132 / 60 = 18480 / 60 so the correct ans . ise" | a = 110 * 5
b = 110 + 140
c = a / b
d = 5 - c
e = d * const_60
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a ) 7 , b ) 9 , c ) 11 , d ) 12 , e ) 13 | d | subtract(multiply(7, const_10), add(add(add(add(add(add(add(add(1, 2), 4), 5), 6), 9), 9), 10), 12)) | if the average of 1 , 2 , 4 , 5 , 6 , 9 , 9 , 10 , 12 and x is 7 , what is the value of x ? | "sum of the deviations of the numbers in the set from the mean is always zero 1 , 2 , 4 , 5 , 6 , 9 , 9 , 10 , 12 mean is 7 so the list is - 6 - 5 - 3 - 2 - 1 + 2 + 2 + 3 + 5 . . . this shud total to zero but this is - 5 , hence we need a number that is 5 more than the mean to get a + 5 and make it zero hence the answer is 7 + 5 = 12 d" | a = 7 * 10
b = 1 + 2
c = b + 4
d = c + 5
e = d + 6
f = e + 9
g = f + 9
h = g + 10
i = h + 12
j = a - i
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a ) 89 , b ) 20 , c ) 40 , d ) 88 , e ) 34 | c | divide(add(200, 200), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 200 m ahead of the engine of a 200 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 200 + 200 = 360 m . time taken = 400 / 10 = 40 sec . answer : option c" | a = 200 + 200
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
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a ) 484 , b ) 726 , c ) 1,200 , d ) 1,320 , e ) 1,694 | c | multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100))) | yesterday ' s closing prices of 2,200 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a higher price today than yesterday ? | "lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 2200 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h = 1200 . c is my answer ." | a = 20 * 100
b = a - 10
c = b - 10
d = 20 / 100
e = 1 + d
f = e + 1
g = c / f
h = 20 / 100
i = 1 + h
j = g * i
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['a ) 8.04 kg', 'b ) 8.14 kg', 'c ) 8.24 kg', 'd ) 9.04 kg', 'e ) none of these'] | a | divide(multiply(subtract(multiply(multiply(50, 40), 23), multiply(multiply(subtract(50, multiply(2, const_3)), subtract(40, multiply(2, const_3))), subtract(23, const_3))), 0.5), const_1000) | the dimensions of an open box are 50 cm , 40 cm and 23 cm . its thickness is 2 cm . if 1 cubic cm of metal used in the box weighs 0.5 gms , find the weight of the box . | explanation : volume of the metal used in the box = external volume - internal volume = [ ( 50 x 40 x 23 ) - ( 44 x 34 x 20 ) ] cm 3 = 16080 cm 3 weight of the metal = [ ( 16080 x 0.5 ) / 1000 ] kg = 8.04 kg . answer : a | a = 50 * 40
b = a * 23
c = 2 * 3
d = 50 - c
e = 2 * 3
f = 40 - e
g = d * f
h = 23 - 3
i = g * h
j = b - i
k = j * 0
l = k / 1000
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a ) 228 , b ) 65.28 , c ) 48.32 , d ) 27 , e ) 21.1 | b | add(45, 30) | two goods trains each 680 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 680 + 680 = 1360 m . required time = 1360 * 6 / 125 = 65.28 sec . answer : b" | a = 45 + 30
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a ) 343 , b ) 677 , c ) 790 , d ) 867 , e ) 932 | c | divide(add(212, 25), divide(30, const_100)) | mike needs 30 % to pass . if he scored 212 marks and falls short by 25 marks , what was the maximum marks he could have got ? | "if mike had scored 25 marks more , he could have scored 30 % therefore , mike required 212 + 25 = 237 marks let the maximum marks be m . then 30 % of m = 237 ( 30 / 100 ) × m = 237 m = ( 237 × 100 ) / 30 m = 23700 / 30 m = 790 answer : c" | a = 212 + 25
b = 30 / 100
c = a / b
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a ) 17 / 30 , b ) 2 / 5 , c ) 7 / 15 , d ) 4 / 15 , e ) 11 / 30 | d | divide(add(floor(divide(30, 5)), divide(30, 12)), 30) | a number is selected at random from the first 30 natural numbers . what is the probability that the number is a multiple of either 5 or 12 ? | "number of multiples of 5 from 1 through 30 = 30 / 5 = 6 number of multiples of 12 from 1 through 30 = 30 / 12 = 2 number of multiples of 5 and 12 both from 1 through 30 = number of multiples of 12 * 5 ( = 60 ) = 0 total favourable cases = 6 + 2 - 0 = 8 probability = 8 / 30 = 4 / 15 answer : option d" | a = 30 / 5
b = math.floor(a)
c = 30 / 12
d = b + c
e = d / 30
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a ) 450 km , b ) 480 km , c ) 500 km , d ) 540 km , e ) 590 km | d | divide(multiply(420, 6), divide(add(multiply(3, 4), 2), 3)) | an aeroplane covers a certain distance of 420 kmph in 6 hours . to cover the same distance in 4 2 / 3 hours , it must travel at a speed of ? | "speed of aeroplane = 420 kmph distance travelled in 6 hours = 420 * 6 = 2520 km speed of aeroplane to acver 2520 km in 14 / 3 = 2520 * 3 / 14 = 540 km answer : option ' d '" | a = 420 * 6
b = 3 * 4
c = b + 2
d = c / 3
e = a / d
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a ) − 5 , b ) − 2 , c ) 0 , d ) 2 , e ) 4 | a | subtract(add(power(negate(5), 2), 16), multiply(10, negate(5))) | for what value of x between − 5 and 5 , inclusive , is the value of x ^ 2 − 10 x + 16 the greatest ? | we can see from the statement that two terms containing x , x ^ 2 will always be positive and - 10 x will be positive if x is - ive . . so the equation will have greatest value if x is - ive , and lower the value of x , greater is the equation . so - 5 will give the greatest value . . ans a | a = negate ** (
b = a + 2
c = b - 16
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a ) 20 % , b ) 27.5 % , c ) 10 % , d ) 80 % , e ) 90 % | b | subtract(divide(subtract(const_100, 15), 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 15 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 85 2 / 3 sp 1 = 85 sp 1 = 127.5 100 - - - 27.5 = > 27.5 % answer : b" | a = 100 - 15
b = 2 / 3
c = a / b
d = c - 100
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