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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5 | d | divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400)) | 150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration c of the acid in the mixture ? | "do not need any computation 30 % - - - - - - - - - - - 21 % - - - - - - - - - 12 % if volume of both sol . were equal the concentration c would be 21 % = 1 / 5 , but 12 % is more than 3 times only possibility is 1 / 6 d" | a = 150 * 30
b = a / 100
c = 400 * 12
d = c / 100
e = b + d
f = 150 + 400
g = e / f
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a ) 299 , b ) 267 , c ) 269 , d ) 400 , e ) 276 | d | multiply(divide(2000, add(2000, add(multiply(3000, 2), multiply(4000, 2)))), 2000) | a , b and c invests rs . 2000 , rs . 3000 and rs . 4000 in a business . after one year a removed his money ; b and c continued the business for one more year . if the net profit after 2 years be rs . 3200 , then a ' s share in the profit is ? | "2 * 12 : 3 * 12 : 4 * 24 1 : 3 : 4 1 / 8 * 3200 = 400 answer : d" | a = 3000 * 2
b = 4000 * 2
c = a + b
d = 2000 + c
e = 2000 / d
f = e * 2000
|
a ) 1 second , b ) 2.5 second , c ) 4 second , d ) 3.6 second , e ) 4.5 second | d | divide(100, multiply(100, const_0_2778)) | a train is moving at a speed of 100 km / hour . the length of the train is 100 metre . in how much time will it cross a pole ? | in order to cross the pole , the train will need to cover its own length of 100 meter . to cover 100 x 1000 meter , train takes 60 minutes or 60 x 60 = 3600 second . so , to cover 100 meter , train would take 3600 x 100 / 100 x 1000 = 3.6 second . answer : d . | a = 100 * const_0_2778
b = 100 / a
|
a ) 4 , b ) 10 , c ) 18 , d ) 19 , e ) 25 | b | sqrt(subtract(power(14, const_2), power(subtract(sqrt(subtract(power(14, const_2), power(9, const_2))), 1), const_2))) | a ladder 14 feet long is leaning against a wall that is perpendicular to level ground . the bottom of the ladder is 9 feet from the base of the wall . if the top of the ladder slips down 1 feet , how many feet will the bottom of the ladder slip ? | "14 ^ 2 - 9 ^ 2 = 115 it means that the height is equal to 10.72 . since the top of the ladder slips down 1 feet , then the height of the wall = 10.72 - 1 = 9.72 the bottom = sqrt ( 14 ^ 2 - 9.72 ^ 2 ) = sqrt ( 196 - 94.47 ) = 10.07 ~ = 10 ans is b" | a = 14 ** 2
b = 14 ** 2
c = 9 ** 2
d = b - c
e = math.sqrt(d)
f = e - 1
g = f ** 2
h = a - g
i = math.sqrt(h)
|
a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 1200 , e ) none of these | d | multiply(30, 40) | a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 30 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | "explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 30 = rs . 175 cp on 40 metres = 175 x 40 = rs . 7000 profit earned on 40 metres cloth = rs . 8200 – rs . 7000 = rs . 1200 . answer : option d" | a = 30 * 40
|
a ) 5 , b ) 6 , c ) 7 , d ) 18 , e ) 9 | d | divide(330, multiply(add(60, 6), const_0_2778)) | a train 330 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 330 * 3 / 55 = 18 sec . answer : option d" | a = 60 + 6
b = a * const_0_2778
c = 330 / b
|
a ) 37 1 / 2 days , b ) 22 days , c ) 31 days , d ) 22 days , e ) 25 1 / 2 days | a | add(80, const_100) | a completes 80 % of a work in 20 days . then b also joins and a and b together finish the remaining work in 3 days . how long does it need for b if he alone completes the work ? | work done by a in 20 days = 80 / 100 = 8 / 10 = 4 / 5 work done by a in 1 day = ( 4 / 5 ) / 20 = 4 / 100 = 1 / 25 - - - ( 1 ) work done by a and b in 3 days = 20 / 100 = 1 / 5 ( because remaining 20 % is done in 3 days by a and b ) work done by a and b in 1 day = 1 / 15 - - - ( 2 ) work done by b in 1 day = 1 / 15 – 1 / 25 = 2 / 75 = > b can complete the work in 75 / 2 days = 37 ½ days option a | a = 80 + 100
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a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | a | divide(3, multiply(3, 2)) | if - 3 / ( a - 3 ) = 3 / ( a + 2 ) , then a = ? | multiply all terms of the given equation by ( a - 3 ) ( a + 2 ) , simplify and solve ( a - 3 ) ( a + 2 ) [ - 3 / ( a - 3 ) ] = ( a - 3 ) ( a + 2 ) [ 3 / ( a + 2 ) ] - 3 ( a + 2 ) = 3 ( a - 3 ) a = 1 / 2 correct answer a | a = 3 * 2
b = 3 / a
|
a ) 391 , b ) 400 , c ) 410 , d ) 423 , e ) 445 | b | divide(460, add(const_1, divide(15, const_100))) | company p had 15 percent more employees in december than it had in january . if company p had 460 employees in december , how many employees did it have in january ? | given : january * 1.15 = december - - > january * 1.15 = 460 - - > january = 460 / 1.15 = 400 . answer : b . | a = 15 / 100
b = 1 + a
c = 460 / b
|
a ) a ) 426 , b ) b ) 526 , c ) c ) 207 , d ) d ) 165 , e ) e ) 297 | d | multiply(divide(782, add(divide(3, 4), add(divide(1, 2), divide(1, 3)))), divide(1, 3)) | if rs . 782 be divided into 3 parts , proportional to 1 / 2 : 1 / 3 : 3 / 4 , then the second part is ? | given ratio = 1 / 2 : 2 / 3 : 3 / 4 = 6 : 4 : 9 1 st part = 782 * 4 / 19 = rs . 165 answer : d | a = 3 / 4
b = 1 / 2
c = 1 / 3
d = b + c
e = a + d
f = 782 / e
g = 1 / 3
h = f * g
|
a ) 12 , b ) 29 , c ) 27 , d ) 17 , e ) 99 | d | divide(subtract(117, multiply(const_3, 5)), multiply(const_3, const_2)) | a number is doubled and 5 is added . if the resultant is trebled , it becomes 117 . what is that number ? | "explanation : let the number be x . therefore , 3 ( 2 x + 5 ) = 117 6 x + 15 = 117 6 x = 102 x = 17 answer : d" | a = 3 * 5
b = 117 - a
c = 3 * 2
d = b / c
|
a ) 306.05 , b ) 306.00 , c ) 306.04 , d ) 306.05 , e ) 306.01 | a | subtract(multiply(5000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 5000) | find out the c . i on rs . 5000 at 4 % p . a . compound half - yearly for 1 1 / 2 years . | "a = 5000 ( 51 / 50 ) 3 = 5306.04 5000 - - - - - - - - - - - 306.04 answer : a" | a = 4 / 100
b = 1 + a
c = 3 / 2
d = b ** c
e = 5000 * d
f = e - 5000
|
a ) 3 , b ) 76 , c ) 8 , d ) 9 , e ) 10 | d | add(divide(subtract(24, 3), 3), const_2) | what is the greatest of 3 consecutive integers whose sum is 24 ? | explanation : the sum of three consecutive integers can be written as n + ( n + 1 ) + ( n + 2 ) = 3 n + 3 if the sum is 24 , we need to solve the equation 3 n + 3 = 24 ; = > 3 n = 21 ; = > n = 7 the greatest of the three numbers is therefore 7 + 2 = 9 answer : d ) 9 | a = 24 - 3
b = a / 3
c = b + 2
|
a ) $ 2000 , b ) $ 2500 , c ) $ 2400 , d ) $ 3120 , e ) $ 1540 | c | divide(multiply(divide(multiply(120, 10), subtract(15, 10)), const_100), 10) | i sold a book at a profit of 10 % . had i sold it for $ 120 more , 15 % would have been gained . find the cost price ? | "115 % of cost - 110 % of cost = $ 120 5 % of cost = $ 120 cost = 120 * 100 / 5 = $ 2400 answer is c" | a = 120 * 10
b = 15 - 10
c = a / b
d = c * 100
e = d / 10
|
a ) 10 , b ) 776 , c ) 66 , d ) 12 , e ) 99 | a | subtract(multiply(40, divide(25, const_100)), multiply(divide(4, 5), 25)) | how much is 25 % of 40 is greater than 4 / 5 of 25 ? | "( 25 / 100 ) * 40 – ( 4 / 5 ) * 25 10 - 20 = 10 answer : a" | a = 25 / 100
b = 40 * a
c = 4 / 5
d = c * 25
e = b - d
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | subtract(divide(360, 10), divide(360, 12)) | a certain car uses 12 gallons of gasoline in traveling 360 miles . in order for the car to travel the same distance using 10 gallons of gasoline , by how many miles per gallon must the car ’ s gas mileage be increased ? | "360 / 10 = 36 . the difference is 36 - 30 = 6 . answer c" | a = 360 / 10
b = 360 / 12
c = a - b
|
a ) 14 , b ) 16 , c ) 26 , d ) 44 , e ) 65 | d | subtract(70, subtract(subtract(const_100, 60), divide(multiply(subtract(const_100, 60), 35), const_100))) | in an election between two candidates - lange and sobel - 70 % of the voters voted for sobel . of the election ` s voters , 60 % were male . if 35 % of the female voters voted for lange , what percentage of the male voters voted for sobel ? | m f total l 16 14 30 s 44 70 tot 60 40 100 35 % of 40 = 14 - - - - - - female voters that means 16 male voted for l so 60 - 16 = 44 d is correct | a = 100 - 60
b = 100 - 60
c = b * 35
d = c / 100
e = a - d
f = 70 - e
|
a ) 27.5 litres , b ) 26.67 litres , c ) 27.67 litres , d ) 28.57 litres , e ) 26 litres | d | multiply(100, subtract(const_1, sqrt(divide(25, 49)))) | a 100 - litre mixture of milk and water contains 49 litres of milk . ' x ' litres of this mixture is removed and replaced with an equal quantum of water . if the process is repeated once , then the concentration of the milk stands reduced at 25 % . what is the value of x ? | "working formula . . . initial concentration * initial volume = final concentration * final volume . let x is the part removed from 100 lts . 49 % ( 1 - x / 100 ) ^ 2 = 25 % * 100 % ( 1 - x / 100 ) ^ 2 = 25 / 49 - - - - - - > ( 1 - x / 100 ) ^ 2 = ( 5 / 7 ) ^ 2 100 - x = 500 / 7 x = 28.57 . . . ans d" | a = 25 / 49
b = math.sqrt(a)
c = 1 - b
d = 100 * c
|
a ) 46 , b ) 52 , c ) 48 , d ) 49 , e ) 50 | b | add(add(46, multiply(3, 2)), const_1) | the standard deviation of a normal distribution of data is 2 , and 3 standard deviations below the mean is greater than 46 . what is a possible value for the mean of the distribution ? | "the standard deviation ( { sd } ) = 2 ; 3 standard deviations below the mean is greater than 46 : { mean } - 3 * { sd } > 46 ; { mean } - 6 > 46 ; { mean } > 52 . answer : b ." | a = 3 * 2
b = 46 + a
c = b + 1
|
a ) 4096 , b ) 4055 , c ) 4092 , d ) 4085 , e ) 4095 | e | subtract(add(multiply(4, const_1000), multiply(const_10, const_10)), add(const_3, const_2)) | 4 friends a , b , c and d decided to share a lot of apples . each of them had half of the total plus half an apple in order . after each of them took their share 3 time , no apples were left . how many apples were there ? | whenever the rate of reduction is ' half of the total and half of it ' , the answer is always ( 2 ^ n ) - 1 , where ' n ' is the number of times the process is repeated . here , the process is repeated 12 times . so answer is ( 2 ^ 12 ) - 1 = 4095 answer e | a = 4 * 1000
b = 10 * 10
c = a + b
d = 3 + 2
e = c - d
|
a ) 2 : 4 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none of these | a | divide(sqrt(4), sqrt(16)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 16 hours and 4 hours respectively . the ratio of their speeds is : | "let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 4 : â ˆ š 16 = 2 : 4 . answer a" | a = math.sqrt(4)
b = math.sqrt(16)
c = a / b
|
a ) 8 , b ) 12 , c ) 12 , d ) 14 , e ) 16 | b | subtract(add(6, 8), 2) | a certain family has 3 sons : richard is 6 years older than david , and david is 8 years older than scott . if in 8 years , richard will be twice as old as scott , then how old was david 2 years ago ? | let ' s say age of richard is ` ` r ' ' age of david is ` ` d ' ' age of scott is ` ` s ' ' now richard is 6 years older than david , i . e . r = d + 6 david is 8 years older than scott i . e . d = s + 8 if in 8 years , richard will be twice as old as scott i . e . r + 8 = 2 x ( s + 8 ) i . e . r + 8 = 2 s + 16 i . e . r = 2 s + 8 but r = d + 6 = ( s + 8 ) + 6 = s + 14 therefore , 2 s + 8 = s + 14 i . e . s = 6 i . e . r = 20 i . e . d = 14 now , how old was david 2 years ago ? i . e . d - 2 = 14 - 2 = 12 years answer : b | a = 6 + 8
b = a - 2
|
a ) 4500 , b ) 2678 , c ) 2933 , d ) 2761 , e ) 2882 | c | divide(multiply(multiply(4, 22), multiply(2, const_1000)), multiply(const_1, const_60)) | a river 4 m deep and 22 m wide is flowing at the rate of 2 kmph the amount of water that runs into the sea per minute is ? | ( 2000 * 4 * 22 ) / 60 = 2933 m 3 answer : c | a = 4 * 22
b = 2 * 1000
c = a * b
d = 1 * const_60
e = c / d
|
a ) 35 , b ) 37 , c ) 42 , d ) 50 , e ) 69 | e | add(divide(368, gcd(gcd(40, 144), 368)), add(divide(40, gcd(gcd(40, 144), 368)), divide(144, gcd(gcd(40, 144), 368)))) | a drink vendor has 40 liters of maaza , 144 liters of pepsi and 368 liters of sprite . he wants to pack them in cans , so that each can contains the same number of liters of a drink , and does n ' t want to mix any two drinks in a can . what is the least number of cans required ? | the number of liters in each can = hcf of 40 , 144 and 368 = 8 liters . number of cans of maaza = 40 / 8 = 5 number of cans of pepsi = 144 / 8 = 18 number of cans of sprite = 368 / 8 = 46 the total number of cans required = 5 + 18 + 46 = 69 cans . answer : e | a = math.gcd(40, 144)
b = math.gcd(a, 368)
c = 368 / b
d = math.gcd(40, 144)
e = math.gcd(d, 368)
f = 40 / e
g = math.gcd(40, 144)
h = math.gcd(g, 368)
i = 144 / h
j = f + i
k = c + j
|
a ) 1 , b ) 2 , c ) 3 , d ) 6 , e ) 12 | c | multiply(divide(subtract(divide(multiply(25, 20), const_100), 3), 2), 3) | if x is 3 times larger than n , and if 2 n + 3 is 20 % of 25 , what is the value of ? | then the wordings of question are misleading . moreover , it ' d also be a bad way to convey x = 4 n , if it really wanted to convey so . from other questions that i have seen , proper wordings could be : if x is three times as large as n or simply , if x is three times n . c | a = 25 * 20
b = a / 100
c = b - 3
d = c / 2
e = d * 3
|
a ) 10 , b ) 7 , c ) 6 , d ) 5 , e ) 8 | e | add(3, 5) | a one - foot stick is marked in 1 / 3 and 1 / 5 portion . how many total markings will there be , including the end points ? | "lcm of 15 = 15 1 / 3 marking are ( table of 5 ) 0 . . . . . . 5 . . . . . . 10 . . . . . 15 ( total = 4 ) 1 / 5 marking are ( table of 3 ) 0 . . . . . . . 3 . . . . . . 6 . . . . . . 9 . . . . . . . . 12 . . . . . . 15 ( total = 6 ) overlapping markings are 0 . . . . . . . . 15 ( total = 2 ) total markings = 4 + 6 - 2 = 8 answer = e" | a = 3 + 5
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a ) 16 , b ) 18 , c ) 28 , d ) 30 , e ) 34 | e | multiply(divide(6, subtract(9, 7)), 7) | sachin is younger than rahul by 6 years . if their ages are in the respective ratio of 7 : 9 , how old is sachin ? | "let rahul ' s age be x years . then , sachin ' s age = ( x - 6 ) years . ( x - 6 ) / x = 7 / 9 9 x - 54 = 7 x 2 x = 54 x = 27 hence , sachin ' s age = ( x - 7 ) = 34 years . ans - e" | a = 9 - 7
b = 6 / a
c = b * 7
|
a ) 86.5 kg , b ) 66.5 kg , c ) 46.5 kg , d ) 65.5 kg , e ) 16.5 kg | d | divide(add(67, add(64, const_1)), const_2) | in arun ' s opinion , his weight is greater than 64 kg but leas than 72 kg . his brother does not agree with arun and he thinks that arun ' s weight is greater than 60 kg but less than 70 kg . his mother ' s view is that his weight can not be greater than 67 kg . if all of them are correct in their estimation , what is the average of diferent probable weights of arun ? | "let arun ' s weight be x kg . according to arun , 64 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 67 . the values satisfying all the above conditions are 65 and 66 . required average = ( 65 + 66 ) / 2 = 65.5 kg answer : d" | a = 64 + 1
b = 67 + a
c = b / 2
|
a ) 26 , b ) 48 , c ) 47 , d ) 36 , e ) 25 | d | multiply(divide(add(1.10, 0.9), add(110, 90)), const_3600) | two trains are moving in opposite directions with speed of 110 km / hr and 90 km / hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds | "explanation : relative speed = 110 + 90 = 200 km / hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 / 200 hr = 1 / 100 hr = 3600 / 100 = 36 seconds answer : option d" | a = 1 + 10
b = 110 + 90
c = a / b
d = c * 3600
|
a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 2 / 5 | e | divide(const_4, add(0, const_10)) | in the xy - plane , a triangle has vertices ( 0,0 ) , ( 4,0 ) and ( 4,10 ) . if a point ( a , b ) is selected at random from the triangular region , what is the probability that a - b > 0 ? | the area of the right triangle is ( 1 / 2 ) * 4 * 10 = 20 . only the points ( a , b ) below the line y = x satisfy a - b > 0 . the part of the triangle which is below the line y = x has an area of ( 1 / 2 ) ( 4 ) ( 4 ) = 8 . p ( a - b > 0 ) = 8 / 20 = 2 / 5 the answer is e . | a = 0 + 10
b = 4 / a
|
a ) 1 / 574 , b ) 2 / 574 , c ) 1 / 287 , d ) 2 / 287 , e ) 3 / 287 | d | divide(multiply(4, subtract(4, const_1)), multiply(add(38, 4), subtract(add(38, 4), const_1))) | into a room you store 38 black bottles and 4 green bottles , and you turn off the light . if you ask to someone look for 2 bottles without turn on the light , what is the probability of both the selected bottles will be green ? | there are 4 green bottles in a group of 38 + 4 that is 42 . thus the probability of picking 2 green bottles is : ( 4 / 42 ) ( 3 / 41 ) = 2 / 287 . the correct answer is d . | a = 4 - 1
b = 4 * a
c = 38 + 4
d = 38 + 4
e = d - 1
f = c * e
g = b / f
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a ) 9000 , b ) 2778 , c ) 6000 , d ) 2889 , e ) 6612 | a | multiply(multiply(const_1, const_12), divide(27000, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 27000 . a ' s share is ? | x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 27000 = 9000 answer : a | a = 1 * 12
b = 1 * 12
c = 12 - 6
d = c * 2
e = b + d
f = 12 - 8
g = f * 3
h = e + g
i = 27000 / h
j = a * i
|
a ) 3 kmph , b ) 7 kmph , c ) 13 kmph , d ) 6.5 kmph , e ) 7 : 3 kmph | b | divide(add(divide(50, 5), divide(120, 5)), const_2) | a man swims downstream 120 km and upstream 50 km taking 5 hours each time ; what is the speed of the current ? | "explanation : 120 - - - 5 ds = 24 ? - - - - 1 50 - - - - 5 us = 10 ? - - - - 1 s = ? s = ( 24 - 10 ) / 2 = 7 answer : option b" | a = 50 / 5
b = 120 / 5
c = a + b
d = c / 2
|
a ) 18 , b ) 36 , c ) 72 , d ) 110 , e ) 108 | d | multiply(divide(198, add(add(1, const_2), multiply(const_2, 3))), subtract(multiply(const_2, 3), 1)) | pat , kate and mark charged a total of 198 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much times as mark , how many more hours did mark charge to the project than kate . | "let kate charge for x hours , then pat charged for 2 x and mat - for 6 x . so , 2 x + 6 x + x = 198 - total hours charged for , x = 22 . mat charged 6 x - x or 5 x for more hours than kate , or for 110 hours . d is correct" | a = 1 + 2
b = 2 * 3
c = a + b
d = 198 / c
e = 2 * 3
f = e - 1
g = d * f
|
a ) 4 . , b ) 6 . , c ) 7 . , d ) 8 . , e ) 10 . | e | add(multiply(18, divide(const_1, 1)), const_1) | in the junior basketball league there are 18 teams , 1 / 3 of them are bad and ½ are rich . what ca n ' t be the number of teams that are rich and bad ? | "total teams = 18 bad teams = ( 1 / 3 ) * 18 = 6 rich teams = 9 so maximum value that the both rich and bad can take will be 9 . so e = 10 can not be that value . ans e ." | a = 1 / 1
b = 18 * a
c = b + 1
|
a ) 3 / 1 , b ) 3 / 2 , c ) 5 / 2 , d ) 5 / 3 , e ) none of these | b | divide(33, 22) | kul is 22 years old and saras is 33 years . find the ratio of saras ' s age to kul ' s age . | explanation : kul is 22 years old . saras is 33 years old . saras ' s age : kul ' s age = 33 : 22 = 3 : 2 answer : b | a = 33 / 22
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a ) 93.9 , b ) 96.3 , c ) 93.2 , d ) 96.7 , e ) 96.21 | c | add(subtract(100, 7), divide(1, 5)) | what is the cp of rs 100 stock at 7 discount , with 1 / 5 % brokerage ? | "explanation : use the formula , cp = 100 â € “ discount + brokerage % cp = 100 - 7 + 1 / 5 93.2 thus the cp is rs 93.2 . answer : c" | a = 100 - 7
b = 1 / 5
c = a + b
|
a ) 12 , b ) 45 , c ) 87 , d ) 56 , e ) 23 | d | multiply(triangle_area(sqrt(14), sqrt(14)), 8) | right triangle pqr is the base of the prism in the figure above . if pq = pr = â ˆ š 14 and the height of the prism is 8 , what is the volume of the prism ? | "volume of prism = area of base * height = 1 / 2 * ( square root of 14 ) * ( square root of 14 ) * 8 = 56 answer : d" | a = math.sqrt(14)
b = math.sqrt(14)
c = triangle_area * (
|
a ) 15,000 , b ) 15,500 , c ) 16,000 , d ) 16,500 , e ) 17,000 | a | divide(divide(subtract(multiply(const_1000, const_100), subtract(subtract(const_3600, const_100), const_1000)), const_1000), add(multiply(add(const_1, divide(const_10, const_100)), subtract(6, const_1)), const_1)) | a city with a population of 97,500 is to be divided into 6 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? | the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 97,500 = p + 5 ( 1.1 ) p 6.5 p = 97,5000 p = 15,000 the answer is a . | a = 1000 * 100
b = 3600 - 100
c = b - 1000
d = a - c
e = d / 1000
f = 10 / 100
g = 1 + f
h = 6 - 1
i = g * h
j = i + 1
k = e / j
|
a ) 792 , b ) 892 , c ) 8920 , d ) 802 , e ) 9920 | c | subtract(multiply(1800, multiply(add(const_1, divide(const_0_25, const_4)), add(const_1, divide(const_0_25, const_4)))), 1800) | find the c . i . on a sum of rs . 1800 for 6 months at 20 % per annum , interest being compounded quarterly ? | "c . i . = 1800 ( 21 / 20 ) ^ 2 - 1800 = 792 answer : c" | a = const_0_25 / 4
b = 1 + a
c = const_0_25 / 4
d = 1 + c
e = b * d
f = 1800 * e
g = f - 1800
|
a ) 1004 , b ) 1000 , c ) 3600 , d ) 1201 , e ) 3601 | e | subtract(add(const_1000, const_1000), multiply(add(const_1000, const_1000), 15)) | in a recent election , james received 15 percent of the 10000 votes cast . to win the election , a candidate needed to receive more than 51 percent of the vote . how many additional votes would james have needed to win the election ? | "james = ( 15 / 100 ) * 10000 = 1500 votes to win = ( 51 / 100 ) * total votes + 1 = ( 51 / 100 ) * 10000 + 1 = 5101 remaining voted needed to win election = 5101 - 1500 = 3601 answer : option e" | a = 1000 + 1000
b = 1000 + 1000
c = b * 15
d = a - c
|
a ) 39760 , b ) 49760 , c ) 59760 , d ) 69760 , e ) 70000 | d | subtract(power(const_10, 5), multiply(multiply(multiply(multiply(const_10, 9), add(5, const_3)), add(5, const_2)), add(5, const_1))) | how many 5 digit ( digit = 0 - 9 ) zip codes can exist in which at least one digit is repeated ? | zip codes with 5 digits ( all five digits can be repeated ) : 10 ^ 5 = 100,000 zip codes with no digit being repeated : 10 * 9 * 8 * 7 * 6 = 30,240 zip codes with at least one digit being repeated : 100,000 - 30,240 = 69,760 = 69760 ans : d | a = 10 ** 5
b = 10 * 9
c = 5 + 3
d = b * c
e = 5 + 2
f = d * e
g = 5 + 1
h = f * g
i = a - h
|
a ) 1200 m , b ) 1400 m , c ) 1600 m , d ) 1800 m , e ) 2200 m | b | divide(400, multiply(subtract(54, 2), const_0_2778)) | a train 400 m long is running at a speed of 54 km / hr . if it crosses a tunnel in 2 minutes . then what is the length of tunnel ? | "speed of train in m / sec = ( speed of train in km / hr ) x 5 / 18 ∴ speed of train in m / sec = 54 x 5 / 18 = 15 m / sec time taken to cross the tunnel = 2 minutes = 120 sec therefore , total length covered by train in crossing the tunnel = speed of train x time taken to cross the tunnel = 15 m / sec x 120 sec = 1800 m total length covered by train in crossing the tunnel = length of train + length of tunnel given , length of train = 400 m and calculated above , length covered by train in crossing the tunnel = 1800 m therefore , length of tunnel = 1800 m - 400 m = 1400 m correct answer : option : b" | a = 54 - 2
b = a * const_0_2778
c = 400 / b
|
a ) 2.19 km , b ) 2.22 km , c ) 2.00 km , d ) 1.19 km , e ) 1.99 km | a | multiply(divide(multiply(3, 6), subtract(6, 3)), divide(add(12, 10), const_60)) | if i walk at 3 km / h , i miss the bus by 12 minutes . if i walk at 6 km / h , i reach 10 minutes before the arrival of the bus . how far i walk to reach the bus stand ? | "d = product of speed difference of time / difference of speed d = 3 x 6 / 60 [ 12 â ˆ ’ ( â ˆ ’ 10 ) / 6 - 3 ] [ here , â € “ ve sign indicates before the schedule time ] â ‡ ’ d = 2.19 km answer a" | a = 3 * 6
b = 6 - 3
c = a / b
d = 12 + 10
e = d / const_60
f = c * e
|
a ) 22 , b ) 24 , c ) 28 , d ) 32 , e ) 44 | b | multiply(2, 12) | g ( x ) is defined as the product of all even integers k such that 0 < k ≤ x . for example , g ( 14 ) = 2 × 4 × 6 × 8 × 10 × 12 × 14 . if g ( e ) is divisible by 4 ^ 11 , what is the smallest possible value for e ? | "g ( e ) = 4 ^ 11 = 2 ^ 22 . so we have to find a product with atleast 22 2 ' s in it . in option 1 22 the total no of 2 ' s = [ 22 / 2 ] + [ 22 / 4 ] + [ 22 / 8 ] + [ 22 / 16 ] = 11 + 5 + 2 + 1 = 19 in option 2 24 the total no of 2 ' s = [ 24 / 2 ] + [ 24 / 4 ] + [ 24 / 8 ] + [ 24 / 16 ] = 12 + 6 + 3 + 1 = 22 . hence b" | a = 2 * 12
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a ) 44 , b ) 64 , c ) 70 , d ) 80 , e ) 84 | a | subtract(divide(multiply(divide(220, 5), 12), 2), 220) | the ratio of boarders to day students at a school was originally 5 to 12 . however , after a number of new boarders join the initial 220 boarders , the ratio changed to 1 to 2 . if no boarders became day students and vice versa , and no students left the school , how many new boarders joined the school ? | "let x be the number of new boarders . the ratio changed from 5 : 12 up to 1 : 2 = 6 : 12 . 220 / ( 220 + x ) = 5 / 6 x = 44 the answer is a ." | a = 220 / 5
b = a * 12
c = b / 2
d = c - 220
|
a ) 15 mins , b ) 12 mins , c ) 13 mins , d ) 17 mins , e ) 20 mins | a | add(11, subtract(11, 7)) | recipe requires 15 mins to cook . you have 7 mins hourglass and a 11 mins hourglass with you . how will you measure 15 mins using them ? | correct answer : a put up both hour glasses . time starts when 7 min hourglass turns empty . and 11 min hourglass will run for 4 min . now flip 11 min hour glass upside down as soon as it runs empty . it will be empty in another 11 mins . and time is up . so 4 + 11 = 15 mins | a = 11 - 7
b = 11 + a
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a ) 95 , b ) 64 , c ) 1012 , d ) 9 , e ) 0 | a | subtract(multiply(add(divide(subtract(multiply(7, 8), 6), 2), 9), 3), 7) | what will be the result of 7 ^ 8 - 6 / 2 + 9 ^ 3 + 3 + 12 ? | the multiplication ( e . g . 7 ^ 8 ) and division ( e . g . 6 / 2 ) is done first and then the sum ( e . g . 3 + 12 ) and subtraction ( e . g . 56 - 3 ) , and the answer is 95 , option a . | a = 7 * 8
b = a - 6
c = b / 2
d = c + 9
e = d * 3
f = e - 7
|
a ) $ 24 , b ) $ 27 , c ) $ 59 , d ) $ 75 , e ) $ 35 | c | subtract(86, add(multiply(3, 3), multiply(3, 6))) | erik ' s mother gave him $ 86 to go to the store . erik bought 3 loaves of bread and 3 cartons of orange juice . each loaf of bread cost $ 3 and each carton of orange juice cost $ 6 . how much money does erik have left ? | step 1 : find the cost of the loaves of bread . 3 × $ 3 = $ 9 step 2 : find the cost of the orange juice . 3 × $ 6 = $ 18 step 3 : find the total cost of the groceries . $ 9 + $ 18 = $ 27 step 4 : find the amount of money left . $ 86 – $ 27 = $ 59 erik has $ 59 left . answer is c . | a = 3 * 3
b = 3 * 6
c = a + b
d = 86 - c
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a ) $ 255260 , b ) $ 435260 , c ) $ 125260 , d ) $ 115260 , e ) $ 155260 | d | add(multiply(multiply(subtract(1, divide(22, const_100)), subtract(1, divide(22, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(22, const_2), const_100), sqrt(const_100))) | the value of a machine depreciates at 22 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ? | "the value of the machine after two years = 0.78 * 0.78 * 1 , 50,000 = $ 91,260 sp such that a profit of $ 24,000 is made = 91,260 + 24,000 = $ 1 , 15,260 d" | a = 22 / 100
b = 1 - a
c = 22 / 100
d = 1 - c
e = b * d
f = 100 * 100
g = math.sqrt(100)
h = f * g
i = math.sqrt(100)
j = i / 2
k = j * 100
l = k * 100
m = h + l
n = e * m
o = 22 + 2
p = o * 100
q = math.sqrt(100)
r = p * q
s = n + r
|
a ) 180 , b ) 120 , c ) 90 , d ) 85 , e ) 70 | b | divide(multiply(60, divide(50, const_100)), subtract(divide(80, const_100), divide(65, const_100))) | a team won 50 percent of its first 60 games in a particular season , and 80 percent of its remaining games . if the team won a total of 65 percent of its games that season , what was the total number of games that the team played ? | "65 % is 15 % - points above 50 % and 15 % - points below 80 % . thus the ratio of ` ` the first 60 games ' ' to ` ` remaining games ' ' is 1 : 1 . so the team played a total of 60 + 60 = 120 games . the answer is b ." | a = 50 / 100
b = 60 * a
c = 80 / 100
d = 65 / 100
e = c - d
f = b / e
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a ) 123 , b ) 125 , c ) 127 , d ) 137 , e ) 148 | c | subtract(subtract(210, 70), subtract(subtract(210, 55), 42)) | in a class of 210 students , 70 did not opt for math . 55 did not opt for science and 42 did not opt for either . how many students of the class opted for both math and science ? | "210 students total 70 did not opt for math 55 did not opt for science 42 did not opt for either total of 140 students in math and 13 did not opt for sci but did for math 140 - 13 = 127 127 students of the class opted for both math and science answer : c" | a = 210 - 70
b = 210 - 55
c = b - 42
d = a - c
|
a ) 22 % , b ) 23 % , c ) 24.5 % , d ) 24.75 % , e ) 30 % | d | multiply(const_100, divide(multiply(divide(33, const_100), 3), add(1, 3))) | 1 ltr of water is added with 3 ltr of a solution containing 33 % of alcohol in the water . the % of alcohol in the new mixture is ? | "we have a 3 litre solution containing 33 % of alcohol in the water . = > quantity of alcohol in the solution = 3 ã — 33 / 100 now 1 litre of water is added to the solution . = > total quantity of the new solution = 3 + 1 = 4 percentage of alcohol in the new solution = 3 ã — 33 / 100 4 ã — 100 = 3 ã — 331004 ã — 100 = 3 ã — 8.25 / 100 = 24.75 % d" | a = 33 / 100
b = a * 3
c = 1 + 3
d = b / c
e = 100 * d
|
a ) 5 , b ) 12 , c ) 18 , d ) 24 , e ) 28 | e | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(12, 8)), 12)), subtract(12, 9)), divide(const_1, multiply(12, 8))) | 12 welders work at a constant rate they complete an order in 8 days . if after the first day , 9 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | "1 . we need to find out the time taken by 3 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 12 * 8 = 96 days 4 . but on day 1 twelve workers had already worked finishing 1 / 8 of the job . so 3 workers have to finish only 7 / 8 of the job . 5 . total time taken by 3 workers can be got from formula used at ( 2 ) . i . e . , 3 * total time taken = 96 . total time taken by 3 workers to finish the complete job is 96 / 3 = 32 days . 6 . time taken by 6 workers to finish 7 / 8 of the job is 7 / 8 * 32 = 28 days . the answer is choice e" | a = 12 * 8
b = 1 / a
c = b * 12
d = 1 - c
e = 12 - 9
f = d / e
g = 12 * 8
h = 1 / g
i = f / h
|
a ) 1 / 17 , b ) 3 / 17 , c ) 1 / 10 , d ) 5 / 20 , e ) 3 / 10 | d | divide(subtract(20, 15), 20) | a ’ s speed is 20 / 15 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 20 meters , b covers 15 meters in that time . so if the race is 20 meters long , when a reaches the finish line , b would be 5 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 5 meters so that he doesn ’ t need to cover that . in that case , the time required by a ( to cover 20 meters ) would be the same as the time required by b ( to cover 15 meters ) to reach the finish line . so b should get a head start of 5 / 20 th of the race . answer ( d )" | a = 20 - 15
b = a / 20
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a ) 163 , b ) 171 , c ) 177 , d ) 185 , e ) 193 | a | add(multiply(81, const_2), 1) | the sum of the even numbers between 1 and n is 81 * 82 , where n is an odd number , then n = ? | "let n - 1 = 2 a . 2 + 4 + . . . + 2 a = 2 * ( 1 + 2 + . . . + a ) = 2 * ( a ) ( a + 1 ) / 2 = ( a ) ( a + 1 ) = 81 * 82 then a = 81 and n = 163 . the answer is a ." | a = 81 * 2
b = a + 1
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a ) $ 4.50 , b ) $ 6.00 , c ) $ 4.00 , d ) $ 5.50 , e ) $ 4.80 | e | divide(add(multiply(4, 4), 8), 5) | saleem bought 4 baskets with the average cost of $ 4 . if saleem also buys the fifth basket with the price of $ 8 , what is the average ( arithmetic mean ) price of those 5 baskets ? | 4 baskets - > total cost = 4 * avg cost = 4 * 4 = 16 fifth basket cost = 8 total cost for 5 baskets = 16 + 8 = 24 am = 24 / 5 = 4.80 answer : e | a = 4 * 4
b = a + 8
c = b / 5
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a ) 5 km / hr , b ) 1 km / hr , c ) 6 km / hr , d ) 3 km / hr , e ) 7 km / hr | a | divide(30, multiply(2, const_3)) | in a covering a distance of 30 km , abhay takes 2 hours more than sameer . if abhay double his speed , then he would take 1 hour less than sammer . abhay ' s speed is ? | a 5 km / hr let abhay ' s speed be x km / / hr . then , 30 / x - 30 / 2 x = 3 x = 5 km / hr . | a = 2 * 3
b = 30 / a
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a ) 1 / 4 , b ) 1 / 12 , c ) 2 / 13 , d ) 3 / 13 , e ) 3 / 26 | d | divide(multiply(const_4, const_3), 52) | from a pack of 52 cards , 1 card is drawn at random . what is the probability that a jack , queen , or king is drawn ? | the total number of cards is 52 . the number of jacks , queens , or kings is 12 . p ( jack , queen , king ) = 12 / 52 = 3 / 13 the answer is d . | a = 4 * 3
b = a / 52
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a ) 3000 , b ) 3100 , c ) 3200 , d ) 3300 , e ) 3400 | a | divide(1350, multiply(divide(5, const_100), 9)) | a sum was put at simple interest at a certain rate for 9 years had it been put at 5 % higher rate , it would have fetched 1350 more . find the sum . | "difference in s . i . = p × t / 100 ( r 1 − r 2 ) ⇒ 1350 = p × 9 x 5 / 100 ( ∵ r 1 - r 2 = 2 ) ⇒ p = 1350 × 100 / 9 x 5 = 3000 answer a" | a = 5 / 100
b = a * 9
c = 1350 / b
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a ) 18 square inches , b ) 20 square inches , c ) 24 square inches , d ) 28 square inches , e ) 30 square inches | e | divide(subtract(subtract(224, 24), multiply(175, divide(80, const_100))), const_2) | three table runners have a combined area of 224 square inches . by overlapping the runners to cover 80 % of a table of area 175 square inches , the area that is covered by exactly two layers of runner is 24 square inches . what is the area of the table that is covered with three layers of runner ? | "total = a + b + c - ( sum of exactly 2 - group overlaps ) - 2 * ( all three ) + neither 80 % * 175 = 224 - 24 - 2 * ( all three ) + 0 2 * ( all three ) = 224 - 24 - 140 all three = 30 answer : e" | a = 224 - 24
b = 80 / 100
c = 175 * b
d = a - c
e = d / 2
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a ) 98.5 % , b ) 100.1 % , c ) 102.8 % , d ) 104.5 % , e ) 105.0 % | b | divide(multiply(add(const_100, multiply(const_100, const_10)), add(subtract(const_100, 9), const_4)), multiply(const_100, const_10)) | this year , mbb consulting fired 9 % of its employees and left remaining employee salaries unchanged . sally , a first - year post - mba consultant , noticed that that the average ( arithmetic mean ) of employee salaries at mbb was 10 % more after the employee headcount reduction than before . the total salary pool allocated to employees after headcount reduction is what percent of that before the headcount reduction ? | 100 employees getting 1000 $ avg , so total salary for 100 ppl = 100000 9 % reduction in employees lead to 91 employees and a salary increase of 10 % of previous avg salary thus the new avg salary is = 10 % ( 1000 ) + 1000 = 1100 so total salary of 91 employees is 91 * 1100 = 100100 now the new salary is more than previous salary by x % . x = ( 100100 / 100000 ) * 100 = 100.1 % so the answer is b | a = 100 * 10
b = 100 + a
c = 100 - 9
d = c + 4
e = b * d
f = 100 * 10
g = e / f
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a ) 17 , 18 , b ) 7 , 8 , c ) 10 , 11 , d ) 1 , 2 , e ) 8 , 9 | c | add(add(power(add(add(divide(subtract(subtract(21, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(21, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(21, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(21, const_10), const_2), const_4), const_2), const_2))) | the sum of two consecutive integers is 21 . find the numbers . | "n + ( n + 1 ) = 21 2 n + 1 = 21 2 n = 20 n = 10 answer : c" | a = 21 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 21 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 21 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 21 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 144 mins , b ) 140 mins , c ) 136 mins , d ) 132 minw , e ) none of these | b | multiply(add(const_1, const_4), 35) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 35 minutes , then the slower pipe alone will be able to fill the tank in | "explanation : let the slower pipe alone fill the tank in x minutes then faster will fill in x / 3 minutes . part filled by slower pipe in 1 minute = 1 / x part filled by faster pipe in 1 minute = 3 / x part filled by both in 1 minute = 1 / x + 3 / x = 1 / 35 = > 4 / x = 1 / 35 x = 35 ∗ 4 = 140 mins option b" | a = 1 + 4
b = a * 35
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a ) 0 , b ) 1 , c ) 35 , d ) 120 , e ) 245 | b | subtract(power(5, subtract(const_1, const_1)), power(subtract(const_1, const_1), 5)) | if k is a non - negative integer and 15 ^ k is a divisor of 823,435 then 5 ^ k - k ^ 5 = | "8 + 2 + 3 + 4 + 3 + 5 = 25 , so this number is not divisible by 3 and thus not divisible by 15 . therefore , k = 0 5 ^ k - k ^ 5 = 1 - 0 = 1 the answer is b ." | a = 1 - 1
b = 5 ** a
c = 1 - 1
d = c ** 5
e = b - d
|
a ) 2.5 , b ) 3.0 , c ) 3.5 , d ) 4.0 , e ) 4.5 | d | divide(const_1, add(divide(const_1, 5), divide(divide(const_1, 5), const_3))) | p alone can complete a job in 5 days . the work done by q alone in one day is equal to one - fourth of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? | "p ' s rate is 1 / 5 q ' s rate is 1 / 20 the combined rate is 1 / 5 + 1 / 20 = 1 / 4 if they work together , the job will take 4 days . the answer is d ." | a = 1 / 5
b = 1 / 5
c = b / 3
d = a + c
e = 1 / d
|
a ) 3 , b ) 5 , c ) 4 , d ) - 3 , e ) - 2 | d | power(11, const_2) | what must be added to each of the numbers 7 , 11 and 19 , so that the resulting numbers may be in continued proportion ? | let x be the required number , then ( 7 + x ) : ( 11 + x ) : : ( 11 + x ) : ( 19 + x ) ( 7 + x ) ( 19 + x ) = ( 11 + x ) 2 x 2 + 26 x + 133 = x 2 + 22 x + 121 4 x = - 12 or x = - 3 answer : d | a = 11 ** 2
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a ) 160 , b ) 163 , c ) 164 , d ) 165 , e ) 166 | a | multiply(divide(64, const_2), 5) | find the number , difference between number and its 3 / 5 is 64 . | explanation : let the number = x , then , x - ( 3 / 5 ) x = 64 , = > ( 2 / 5 ) x = 64 = > 2 x = 64 * 5 , = > x = 160 answer : option a | a = 64 / 2
b = a * 5
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a ) 0.8 kmph , b ) 9.9 kmph , c ) 1.3 kmph , d ) 1.2 kmph , e ) 5.7 kmph | a | multiply(const_3_6, divide(8, 36)) | convert the 8 / 36 m / s into kilometers per hour ? | "8 / 36 m / s = 8 / 36 * 18 / 5 = 8 / 10 = 0.8 kmph . answer : a" | a = 8 / 36
b = const_3_6 * a
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a ) 12 , b ) 39 , c ) 54 , d ) 64 , e ) 156 | e | sqrt(multiply(234, 104)) | the mean proportional between 234 and 104 is : | required mean proportion is = ( root of 234 ) ( root of 104 ) = root of ( 13 * 9 * 2 * 13 * 18 ) = ( 13 * 3 * 4 ) = 156 answer : e | a = 234 * 104
b = math.sqrt(a)
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a ) 34 , b ) 43 , c ) 40 , d ) 38.6 , e ) 29 | d | divide(add(add(subtract(43, const_10), const_100), add(subtract(43, const_10), const_100)), add(divide(add(subtract(43, const_10), const_100), 43), divide(add(subtract(43, const_10), const_100), 35))) | x and y are two towns . ganesh covers the distance from x to y at an average speed of 43 km / hr . however , he covers the distance from y to x at an average speed of 35 km / hr . his average speed during the whole journey in km / hr . is : | "solution : average speed = 2 xy / x + y = 2 * 43 * 35 / 43 + 35 = 38.6 answer : d" | a = 43 - 10
b = a + 100
c = 43 - 10
d = c + 100
e = b + d
f = 43 - 10
g = f + 100
h = g / 43
i = 43 - 10
j = i + 100
k = j / 35
l = h + k
m = e / l
|
a ) 420 , b ) 560 , c ) 840 , d ) 960 , e ) 1080 | c | lcm(1, 8) | what is the lowest positive integer that is divisible by each of the integers 1 through 8 , inclusive ? | "the integers are 1 , 2 , 3 , 4 , 5 , 6 , 7 and 8 lowest positive integer that is is divisible by each of the integers 1 through 8 = 2 ^ 3 * 3 * 5 * 7 = 8 * 3 * 5 * 7 = 24 * 35 = 840 answer c" | a = math.lcm(1, 8)
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a ) 11.11 , b ) 10 , c ) 14.14 , d ) 16.15 , e ) 1 | c | sqrt(add(power(10, const_2), power(10, const_2))) | a man walks west 10 miles , then turns right towards north and walks another 10 miles . how far is he from the place he started walking . | as we can see the man moves in configuration of a right angle triangle . the man has walked on the sides and now we need to calculate the length of the hypotenuse which is the distance of the man from the starting point to his current location . since we know the length of two sides , we can find the length of hypotenuse . as per pythagoras : a ^ 2 + b ^ 2 = c ^ 2 a = 10 b = 10 10 ^ 2 + 10 ^ 2 = c ^ 2 100 + 100 = c ^ 2 200 = c ^ 2 c = squareroot 200 c = 14.14 miles ans : c | a = 10 ** 2
b = 10 ** 2
c = a + b
d = math.sqrt(c)
|
a ) 1 / 35 , b ) 3 / 35 , c ) 1 / 21 , d ) 2 / 35 , e ) none | c | divide(60, multiply(6, 210)) | the hcf and lcm of two numbers m and n are respectively 6 and 210 . if m + n = 60 , then 1 / m + 1 / n is equal to | "answer we have , m x n = 6 x 210 = 1260 â ˆ ´ 1 / m + 1 / n = ( m + n ) / mn = 60 / 1260 = 1 / 21 correct option : c" | a = 6 * 210
b = 60 / a
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a ) 2 , b ) 4 , c ) 5 , d ) 8 , e ) 10 | c | subtract(divide(300, 10), divide(300, 12)) | a certain car uses 12 gallons of gasoline in traveling 300 miles . in order for the car to travel the same distance using 10 gallons of gasoline , by how many miles per gallon must the car ’ s gas mileage be increased ? | "300 / 10 = 30 . the difference is 30 - 25 = 5 . answer c" | a = 300 / 10
b = 300 / 12
c = a - b
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a ) 286 m , b ) 350 m , c ) 200 m , d ) 278 m , e ) 267 m | c | subtract(multiply(speed(300, 18), 30), 300) | a 300 m long train crosses a platform in 30 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 30 = 50 / 3 x = 200 m . answer : c" | a = speed * (
b = a - 30
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a ) 5 , b ) 6 , c ) 4 , d ) 8.2 , e ) 5.5 | e | divide(110, multiply(add(65, 7), const_0_2778)) | a train 110 m long is running with a speed of 65 km / hr . in what time will it pass a man who is running at 7 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 65 + 7 = 72 km / hr . = 72 * 5 / 18 = 20 m / sec . time taken to pass the men = 110 * 1 / 20 = 5.5 sec . answer : option e" | a = 65 + 7
b = a * const_0_2778
c = 110 / b
|
a ) 2200 , b ) 2614.9 , c ) 3300 , d ) 1100 , e ) 4400 | b | multiply(circumface(8), 13) | the radius of a cylinder is 8 m , height 13 m . the volume of the cylinder is : | "cylinder volume = ï € r ( power 2 ) h = 22 / 7 ã — 8 ã — 8 ã — 13 = 2614.9 m ( power 3 ) answer is b ." | a = circumface * (
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a ) 53 % , b ) 58 % , c ) 60 % , d ) 64 % , e ) 75 % | c | multiply(divide(subtract(const_100, 25), add(25, const_100)), const_100) | marketing executives for a certain chewing gum company projected a 25 percent increase in revenue this year over that of last year , but revenue this year actually decreased by 25 % . what percent of the projected revenue was the actual revenue ? | "last year revenue = 100 ( assume ) ; this year revenue = 75 ; projected revenue = 125 . actual / projected * 100 = 75 / 125 * 100 = 60 % . answer : c ." | a = 100 - 25
b = 25 + 100
c = a / b
d = c * 100
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a ) 610 , b ) 710 , c ) 810 , d ) 900 , e ) 1000 | c | divide(multiply(9, 900), add(1, 9)) | ashok and pyarelal invested money together in a business and share a capital of ashok is 1 / 9 of that of pyarelal . if the incur a loss of rs 900 then loss of pyarelal ? | "let the capital of pyarelal be x , then capital of ashok = x / 9 so ratio of investment of pyarelal and ashok = x : x / 9 = 9 x : x hence out of the total loss of 900 , loss of pyarelal = 900 * 9 x / 10 x = 810 answer : c" | a = 9 * 900
b = 1 + 9
c = a / b
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | multiply(divide(divide(subtract(975, 750), 750), 5), const_100) | at what rate percent on simple interest will rs . 750 amount to rs . 975 in 5 years ? | "225 = ( 750 * 5 * r ) / 100 r = 6 % . answer : e" | a = 975 - 750
b = a / 750
c = b / 5
d = c * 100
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a ) 4 / 3 , b ) 5 / 3 , c ) 7 / 3 , d ) 8 / 3 , e ) 10 / 3 | d | divide(subtract(4, multiply(divide(3, 5), 4)), divide(3, 5)) | a container holds 4 quarts of diesel and 4 quarts of petrol . how many quarts of water must be added to the container to create a mixture that is 3 parts dieselto 5 parts water by petrol ? | an alternate to the alligation method is the more direct / algebraic method : let x be the amount of petrol to be added . new total amount of water = 4 + x total amount of diesel = 4 new total = 4 + 4 + x = 8 + x final ratio required ( for petrol ) = 5 / ( 5 + 3 ) = 5 / 8 thus , ( 4 + x ) / ( 8 + x ) = 5 / 8 - - > solving for x you get x = 8 / 3 . d is thus the correct answer . | a = 3 / 5
b = a * 4
c = 4 - b
d = 3 / 5
e = c / d
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | divide(subtract(12.80, 0.50), 1.5) | at a certain bowling alley , it costs $ 0.50 to rent bowling shoes for the day and $ 1.5 to bowl 1 game . if a person has $ 12.80 and must rent shoes , what is the greatest number of complete games that person can bowl in one day ? | "after renting bowling shoes the person is left with $ 12.80 - $ 0.5 = $ 12.30 , which is enough for 12.3 / 1.5 < 9 = ~ 8 . answer : b ." | a = 12 - 80
b = a / 1
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a ) 7 , b ) 14 , c ) 28 , d ) 35 , e ) 49 | e | power(divide(divide(divide(divide(divide(divide(1512, const_2), const_2), const_2), const_3), const_3), const_3), const_2) | tough and tricky questions : number properties . what is the smallest positive integer x such that 1512 x is the cube of a positive integer ? | we want to know the smallest x that will make 1512 x a cube of some number . let ' s call that number y . let ' s first figure out what we ' re working with . the prime factorization of 450 can be visualized : . . . . . . . . . 1512 . . . . . . . . / . . . . . . . \ . . . . . . 216 . . . . . . 7 . . . . . / . . \ . . . . . . . 36 . . . . 6 . . . . . . / . \ . . . . / . \ . . . . 6 … . 6 … 2 … 3 . / . \ . . / \ 2 . . 3 2 … 3 so , we have 2 * 2 * 2 * 3 * 3 * 3 * 7 that can be multiplied together to get 1512 . now we need to figure out what we need to make 1512 * x into a cube of y ( y ^ 3 = 1512 * x ) . we have two 5 s , two 3 s , and one 2 . to arrange these numbers in identical triples ( 2 , 3,7 ) , we need at least one more 5 , one 3 , and two 2 ' s . each of these triples will give us the value of y ( 2 * 3 * 7 = 42 ) , which , multiplied by itself three times , gives us 1512 * x . looking at the factors we need to complete the triples , we get 7 * 7 = 49 . we know this is the smallest number possible because prime factors by definition can not be broken down any further . therefore , we can go with answer choice d . if time permits , we can do a sanity check . we calculated that y should be 2 * 3 * 7 , or 42 . 42 * 42 * 42 = 74088 . also , 1512 * 49 = 74088 . answer : e | a = 1512 / 2
b = a / 2
c = b / 2
d = c / 3
e = d / 3
f = e / 3
g = f ** 2
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a ) 4 , b ) 16 , c ) 8 , d ) 1 , e ) 2 | b | multiply(multiply(4, divide(8, 4)), divide(8, 4)) | 4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 8 mat - weavers in 8 days ? | explanation : let the required number of mats be x more mat - weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) hence we can write as ( mat - weavers ) 4 : 8 ( days ) 4 : 8 } : : 4 : x ⇒ 4 × 4 × x = 8 × 8 × 4 ⇒ x = 2 × 2 × 4 = 16 answer : option b | a = 8 / 4
b = 4 * a
c = 8 / 4
d = b * c
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a ) 10 , b ) 15 , c ) 6 , d ) 4 , e ) 5 | d | subtract(power(4, const_2), multiply(2, 6)) | if x + y = - 4 , and x = 6 / y , what is the value of x ^ 2 + y ^ 2 ? | x ^ 2 + y ^ 2 should make you think of these formulas : ( x + y ) ( x + y ) = x ^ 2 + y ^ 2 + 2 xy we already know ( x + y ) = - 4 and x * y = 6 ( x + y ) ( x + y ) = ( - 4 ) ( - 4 ) = x ^ 2 + y ^ 2 + 2 * ( 6 ) x ^ 2 + y ^ 2 = 16 - 12 = 4 answer : d | a = 4 ** 2
b = 2 * 6
c = a - b
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a ) 1 / 4 , b ) 1 / 10 , c ) 7 / 15 , d ) 8 / 15 , e ) 9 / 13 | d | subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 4)) | a can do a work in 15 days and b in 20 days . if they work on it together for 4 days then the fraction of the work that is left is ? | "explanation : a ’ s one day work = 1 / 15 b ’ s one day work = 1 / 20 ( a + b ) ’ s one day work is = 1 / 15 + 1 / 20 = 7 / 60 their 4 days work = 4 x 7 / 60 = 7 / 15 remaining work = 8 / 15 answer : option d" | a = 1 / 15
b = 1 / 20
c = a + b
d = c * 4
e = 1 - d
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a ) rs . 40 , b ) rs . 50 , c ) rs . 49 , d ) rs . 59 , e ) rs . 44 | e | divide(add(66, 22), const_2) | the profit obtained by selling an article for rs . 66 is the same as the loss obtained by selling it for rs . 22 . what is the cost price of the article ? | s . p 1 - c . p = c . p – s . p 2 66 - c . p = c . p - 22 2 c . p = 66 + 22 ; c . p = 88 / 2 = 44 answer : e | a = 66 + 22
b = a / 2
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a ) 1266 , b ) 3000 , c ) 2866 , d ) 2999 , e ) 2678 | b | multiply(subtract(12, 7), divide(2400, subtract(7, 3))) | an amount of money is to be divided between p , q and r in the ratio of 3 : 7 : 12 . if the difference between the shares of p and q is rs . 2400 , what will be the difference between q and r ' s share ? | "4 - - - 2000 5 - - - ? = > 3000 answer : b :" | a = 12 - 7
b = 7 - 3
c = 2400 / b
d = a * c
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a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28 | d | add(divide(60, add(0.75, 4)), divide(60, add(0.75, 4))) | a car ferry can hold up to 60 tons of cargo . what is the greatest number of vehicles that the ferry can carry if half the vehicles are cars with an average ( arithmetic mean ) weight of 0.75 tons and half of the vehicles are trucks with an average ( arithmetic mean ) weight of 4 tons ? | "the weight of one car and one truck is 4.75 tons . 60 / 4.75 = 12 plus a remainder the ferry could carry 12 cars and 12 trucks for a total of 24 vehicles . the answer is d ." | a = 0 + 75
b = 60 / a
c = 0 + 75
d = 60 / c
e = b + d
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a ) s . 5000 , b ) s . 6500 , c ) s . 5800 , d ) s . 6000 , e ) s . 6200 | b | divide(65, multiply(divide(10, const_100), divide(10, const_100))) | if difference between compound interest and simple interest on a sum at 10 % p . a . for 2 years is rs . 65 then sum is | "p ( r / 100 ) ^ 2 = c . i - s . i p ( 10 / 100 ) ^ 2 = 65 6500 answer : b" | a = 10 / 100
b = 10 / 100
c = a * b
d = 65 / c
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 5 , d ) 5 / 6 , e ) none of these | e | divide(subtract(const_6, 1), const_6) | in a throw of dice what is the probability of ge æ « ng number greater than 1 | "explanation : number greater than 1 is 2 , 3 , 4 , 5 & 6 , so only 5 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 5 / 6 answer : e" | a = 6 - 1
b = a / 6
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a ) 83.7 % , b ) 85.2 % , c ) 86.8 % , d ) 78.8 % , e ) 69.4 % | a | add(multiply(divide(divide(10, const_100), subtract(1, divide(1, 7))), const_100), 2) | the price of an item is discounted 7 percent on day 1 of a sale . on day 2 , the item is discounted another 7 percent , and on day 3 , it is discounted an additional 10 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "let initial price be 100 price in day 1 after 7 % discount = 93 price in day 2 after 7 % discount = 86.49 price in day 3 after 10 % discount = 77.84 so , price in day 3 as percentage of the sale price on day 1 will be = 77.84 / 93 * 100 = > 83.7 % answer will definitely be ( a )" | a = 10 / 100
b = 1 / 7
c = 1 - b
d = a / c
e = d * 100
f = e + 2
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a ) 16.16 % , b ) 15.76 % , c ) 14.14 % , d ) 13.13 % , e ) 12.12 % | b | subtract(const_100, multiply(multiply(add(const_1, divide(4, const_100)), subtract(const_1, divide(19, const_100))), const_100)) | a volunteer organization is recruiting new members . in the fall they manage to increase their number by 4 % . by the spring however membership falls by 19 % . what is the total change in percentage from fall to spring ? | "( 100 % + 4 % ) * ( 100 % - 19 % ) = 1.04 * . 81 = 0.8424 . 1 - 0.8424 = 15.76 % lost = - 15.76 % the answer is b the organization has lost 15.76 % of its total volunteers from fall to spring ." | a = 4 / 100
b = 1 + a
c = 19 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
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a ) 195 / 63 , b ) 2.5 , c ) 155 / 63 , d ) 125 / 63 , e ) 185 / 63 | b | add(divide(sqrt(1.21), sqrt(0.81)), divide(sqrt(0.81), sqrt(0.49))) | find the value of ( √ 1.21 ) / ( √ 0.81 ) + ( √ 0.81 ) / ( √ 0.49 ) is | "( √ 1.21 ) / ( √ 0.81 ) + ( √ 0.81 ) / ( √ 0.49 ) 11 / 9 + 9 / 7 = > 2.50 answer is b" | a = math.sqrt(1)
b = math.sqrt(0)
c = a / b
d = math.sqrt(0)
e = math.sqrt(0)
f = d / e
g = c + f
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a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | c | divide(3, divide(48, const_10)) | a cubical tank is filled with water to a level of 3 feet . if the water in the tank occupies 48 cubic feet , to what fraction of its capacity is the tank filled with water ? | "the volume of water in the tank is h * l * b = 48 cubic feet . since h = 3 , then l * b = 16 and l = b = 4 . since the tank is cubical , the capacity of the tank is 4 * 4 * 4 = 64 . the ratio of the water in the tank to the capacity is 48 / 64 = 3 / 4 the answer is c ." | a = 48 / 10
b = 3 / a
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a ) 240 , b ) 288 , c ) 273 , d ) 877 , e ) 361 | c | multiply(divide(630, add(add(multiply(6000, 8), multiply(subtract(6000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(6000, 8), multiply(subtract(6000, 1000), subtract(const_12, 8)))) | a and b began business with rs . 6000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of b . | "( 6 * 8 + 5 * 4 ) : ( 4 * 8 + 5 * 4 ) 17 : 13 13 / 30 * 630 = 273 answer : c" | a = 6000 * 8
b = 6000 - 1000
c = 12 - 8
d = b * c
e = a + d
f = 4000 * 8
g = 4000 + 1000
h = 12 - 8
i = g * h
j = f + i
k = e + j
l = 630 / k
m = 6000 * 8
n = 6000 - 1000
o = 12 - 8
p = n * o
q = m + p
r = l * q
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a ) 12 , b ) 9 , c ) 8 , d ) 7 , e ) 6 | e | add(divide(subtract(18, 3), 5), 3) | there are 18 balls in a jar . you take out 3 blue balls without putting them back inside , and now the probability of pulling out a blue ball is 1 / 5 . how many blue balls were there in the beginning ? | there are 18 balls in a jar . you take out 3 blue balls without putting 3 blue balls + 15 / 5 = 6 answer : e | a = 18 - 3
b = a / 5
c = b + 3
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a ) 25 % , b ) 30 % , c ) 35 % , d ) 40 % , e ) 45 % | c | multiply(divide(70, multiply(const_100, const_2)), const_100) | sue ' s trail mix is 30 % nuts and 70 % dried fruit . jane ' s trail mix is 60 % nuts and 40 % chocolate chips . if the combined mixture of sue and jane ' s trails mix contains 45 % nuts , what percent of the combined mixture is dried fruit ? | 45 % is 15 % - points above 30 % and 15 % - points below 60 % . so the ratio of sue ' s mix to jane ' s mix is 1 : 1 . 1 / 2 * 70 % = 35 % the answer is c . | a = 100 * 2
b = 70 / a
c = b * 100
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a ) 40 , b ) 52 , c ) 68 , d ) 80 , e ) 98 | d | divide(subtract(multiply(divide(52, const_100), 120), multiply(divide(20, const_100), 120)), subtract(const_1, divide(52, const_100))) | in august , a cricket team that played 120 matches won 20 % of the games it played . after a continuous winning streak , this team raised its average to 52 % . how many matches did the team win to attain this average ? | "let the no of matches played more = x so , ( 120 + x ) * 52 / 100 = 24 + x by solving we get x = 80 answer : d" | a = 52 / 100
b = a * 120
c = 20 / 100
d = c * 120
e = b - d
f = 52 / 100
g = 1 - f
h = e / g
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