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a ) 70 , b ) 35 , c ) 200 , d ) 280 , e ) 140 | e | add(const_3, const_4) | what is the smallest integer that is multiple of 5 , 7,20 | "it is the lcm of 5 , 7 and 20 which is 140 . the answer is e ." | a = 3 + 4
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a ) 0.5 , b ) 0.75 , c ) 1.0 , d ) 1.25 , e ) 1.5 | e | divide(divide(add(3, 5), 5), divide(add(7, 3), 5)) | in the xy - coordinate system , what is the slope of the line that goes through the point ( 7 , 3 ) and is equidistant from the two points p = ( 5 , 5 ) and q = ( 13 , 7 ) ? | "first , get the middle coordinate between ( 5,5 ) and ( 13,7 ) . . . x = 5 + ( 13 - 5 ) / 2 = 9 y = 5 + ( 7 - 5 ) / 2 = 6 second , get the slope of ( 9,6 ) and ( 7,3 ) . m = 6 - 3 / 9 - 7 = 3 / 2 = 1.5 answer : e" | a = 3 + 5
b = a / 5
c = 7 + 3
d = c / 5
e = b / d
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a ) 2 / 5 , b ) 3 / 7 , c ) 5 / 7 , d ) 17 / 35 , e ) 24 / 35 | d | divide(add(multiply(const_2, const_2), multiply(const_3, const_2)), multiply(7, subtract(7, 1))) | if x is to be chosen at random from the integers between 1 to 7 , inclusive , and y is to be chosen at random from the integers between 8 and 12 , inclusive , what is the probability that x + y will be even ? | "x + y will be even if x and y are both even or both odd . p ( x and y are both even ) = 3 / 7 * 3 / 5 = 9 / 35 p ( x and y are both odd ) = 4 / 7 * 2 / 5 = 8 / 35 p ( x + y is even ) = 9 / 35 + 8 / 35 = 17 / 35 the answer is d ." | a = 2 * 2
b = 3 * 2
c = a + b
d = 7 - 1
e = 7 * d
f = c / e
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a ) 1 / 2 , b ) 2 / 3 , c ) 1 , d ) 5 / 4 , e ) 2 | d | divide(subtract(1, multiply(const_3, const_2)), subtract(multiply(const_3, const_2), 8)) | hammers and wrenches are manufactured at a uniform weight per hammer and a uniform weight per wrench . if the total weight of 1 hammers and three wrenches is one - third that of 8 hammers and 5 wrenches , then the total weight of one wrench is how many times that of one hammer ? | "x be the weight of a hammer and y be the weight of a wrench . ( 1 x + 3 y ) = 1 / 3 * ( 8 x + 5 y ) 3 ( 1 x + 3 y ) = ( 8 x + 5 y ) 3 x + 9 y = 8 x + 5 y 4 y = 5 x y = 5 x / 4 ans - d" | a = 3 * 2
b = 1 - a
c = 3 * 2
d = c - 8
e = b / d
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a ) 100 kgs , b ) 88.5 kgs , c ) 86.5 kgs , d ) 67.5 kgs , e ) 88.2 kgs | a | divide(multiply(add(const_1, 3), 175), 7) | 3 friends a , b , c went for week end party to mcdonald β s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 175 kg then find the average weight of all the 7 rounds ? | "average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 175 / 7 = 100 kgs answer : a" | a = 1 + 3
b = a * 175
c = b / 7
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a ) 76 sec , b ) 67 sec , c ) 30 sec , d ) 36 sec , e ) 23 sec | c | divide(add(360, 90), multiply(subtract(63, 9), divide(divide(const_10, const_2), divide(subtract(63, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 360 m ahead of the engine of a 90 m long train running at 63 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 63 - 9 = 54 km / hr . = 54 * 5 / 18 = 15 m / sec . distance to be covered = 360 + 90 = 450 m . time taken = 450 / 15 = 30 sec . answer : c" | a = 360 + 90
b = 63 - 9
c = 10 / 2
d = 63 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
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a ) 41 , b ) 45 , c ) 46 , d ) 47 , e ) 48 | a | multiply(subtract(multiply(divide(subtract(const_100, 18), const_100), divide(add(const_100, 72), const_100)), const_1), const_100) | if price of t . v set is reduced by 18 % , then its sale increases by 72 % , find net effect on sale value | "- a + b + ( ( - a ) ( b ) / 100 ) = - 18 + 72 + ( - 18 * 72 ) / 100 = - 18 + 72 - 13 = 41 answer : a" | a = 100 - 18
b = a / 100
c = 100 + 72
d = c / 100
e = b * d
f = e - 1
g = f * 100
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a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 177.50 , d ) rs . 258.50 , e ) none of these | c | divide(subtract(multiply(155, add(add(1, 1), 2)), add(126, 135)), 2) | teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 155 per kg , the price of the third variety per kg will be ? | "explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \ / mean price ( rs . 155 ) / \ x Γ’ Λ β 155 22.50 = > x Γ’ Λ β ( 155 / 22.50 ) = 1 . = > x Γ’ Λ β 155 = 22.50 . = > x = 177.50 rs . answer : c" | a = 1 + 1
b = a + 2
c = 155 * b
d = 126 + 135
e = c - d
f = e / 2
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a ) 2.91 , b ) 5.911 , c ) 5.485 , d ) 5.986 , e ) 2.999 | c | add(add(4.75, divide(303, const_1000)), divide(432, const_1000)) | solution for 4.75 + . 303 + . 432 | "4.75 + . 303 + . 432 = 0 0 = 0 - 4.75 - 0.303 - 0.432 0 = - 5.485 answer : c" | a = 303 / 1000
b = 4 + 75
c = 432 / 1000
d = b + c
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a ) 93 , b ) 225 , c ) 288 , d ) 324 , e ) 336 | a | divide(70, subtract(const_1, divide(const_1, const_4))) | what number is 70 more than one - fourth of itself ? | 1 / 4 x + 70 = x that means 70 = 3 / 4 x x = ( 70 * 4 ) / 3 = 280 / 3 = 93 a is the answer . | a = 1 / 4
b = 1 - a
c = 70 / b
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a ) 500 , b ) 600 , c ) 800 , d ) 900 , e ) 1200 | c | add(divide(multiply(subtract(550, 150), 3), 2), divide(multiply(150, 4), 3)) | 2 / 3 rd of the boys and 3 / 4 th of the girls of a school participate in a function . if the no . of participating students is 550 , out of which 150 are girls , what is the total no . of students in the school ? | let total number of boys be x and total number of girls be y . y = 400 = > x = 600 and ^ = i 50 = > y = 200 now , x + y = 800 c | a = 550 - 150
b = a * 3
c = b / 2
d = 150 * 4
e = d / 3
f = c + e
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a ) 46 m , b ) 60 m , c ) 65 m , d ) 78 m , e ) 80 m | c | divide(add(divide(5300, 26.50), multiply(const_2, 30)), const_4) | length of a rectangular plot is 30 mtr more than its breadth . if the cost of fencin g the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 30 ) metres . perimeter = 5300 / 26.5 m = 200 m . 2 [ ( x + 30 ) + x ] = 200 2 x + 30 = 100 2 x = 70 x = 35 . hence , length = x + 30 = 65 m c" | a = 5300 / 26
b = 2 * 30
c = a + b
d = c / 4
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a ) 1.5 , b ) 1.6 , c ) 1.7 , d ) 1.8 , e ) 1.9 | a | multiply(divide(30, const_100), 5) | how many litres of pure acid are there in 5 litres of a 30 % solution | explanation : question of this type looks a bit typical , but it is too simple , as below . . . it will be 5 * 30 / 100 = 1.5 answer : option a | a = 30 / 100
b = a * 5
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a ) 800 , b ) 880 , c ) 920 , d ) 950 , e ) 980 | b | subtract(add(500, 600), 220) | in a college students can play cricket or basketball . 500 play cricket . 220 played both and 600 played basketball . what is the total strength of college ? | p ( c ) = 500 p ( b ) = 600 p ( c n b ) = 220 p ( c u b ) = p ( c ) + p ( b ) - p ( c n b ) = 500 + 600 - 220 = 880 answer : b | a = 500 + 600
b = a - 220
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a ) 256 , b ) 5122 , c ) 768 , d ) 2048 , e ) 1536 | d | multiply(add(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(0.2, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), 0.2), const_10) | if an average hard drive had a capacity of 0.2 tb in 2000 , and average hard drive capacities double every 5 years , what will be the average hard drive capacity in 2050 ? | "0.2 * 2 ^ 10 = 0.2 * 1024 = 204.8 the answer is d ." | a = 0 * 2
b = a * 2
c = b * 2
d = c * 2
e = d * 2
f = e * 2
g = f * 2
h = g * 2
i = h * 2
j = i * 2
k = j + 0
l = k * 10
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a ) 21 , b ) 22 , c ) 23 , d ) 25 , e ) 28 | a | divide(subtract(add(add(24, 3), 24), subtract(11, 2)), subtract(11, subtract(11, 2))) | the cricket team of 11 members is 24 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of theremaining players is 1 year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 24 + 27 ) = 9 ( x - 1 ) = > 11 x - 9 x = 42 = > 2 x = 42 = > x = 21 . so , average age of the team is 21 years . a" | a = 24 + 3
b = a + 24
c = 11 - 2
d = b - c
e = 11 - 2
f = 11 - e
g = d / f
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a ) 0.0005 , b ) 0.0004 , c ) 0.0006 , d ) 0.0002 , e ) 0.0001 | b | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 0.032) | what is the least number . which should be added to 0.0320 to make it a perfect square ? | 0.0320 + 0.0004 = 0.0324 ( 0.18 ) ^ 2 answer : b | a = 12 + 4
b = a + 2
c = b / 100
d = 12 + 4
e = d + 2
f = e / 100
g = c * f
h = g - 0
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a ) 12 , b ) 6 , c ) 88 , d ) 77 , e ) 14 | b | subtract(21, divide(21, add(divide(2, 5), const_1))) | a 21 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ? | "1 : 2 / 5 = 5 : 2 2 / 7 * 21 = 6 answer : b" | a = 2 / 5
b = a + 1
c = 21 / b
d = 21 - c
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a ) 50 , b ) 40 , c ) 267 , d ) 29 , e ) 27 | b | subtract(add(200, 340), 500) | a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 340 . how much does c have ? | "a + b + c = 500 a + c = 200 b + c = 340 - - - - - - - - - - - - - - a + b + 2 c = 540 a + b + c = 500 - - - - - - - - - - - - - - - - c = 40 answer : b" | a = 200 + 340
b = a - 500
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a ) 381 , b ) 332 , c ) 383 , d ) 334 , e ) 385 | b | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 15) | light glows for every 15 seconds . how many times did it between 1 : 57 : 58 and 3 : 20 : 47 am | "the diff in sec between 1 : 57 : 58 and 3 : 20 : 47 is 4969 sec , 4969 / 13 = 331 . so total 332 times light ll glow answer : b" | a = 2 + 47
b = 2 + const_60
c = 20 + b
d = c * const_60
e = a + d
f = e / 15
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a ) 31 , b ) 36 , c ) 29 , d ) 53 , e ) 57 | c | add(29, const_1) | the average age of 29 students in a group is 15 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ? | age of the teacher = ( 29 * 16 - 29 * 15 ) = 29 years . answer : c | a = 29 + 1
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a ) 34 , b ) 67 , c ) 85 , d ) 25 , e ) 38 | a | multiply(divide(add(const_100, 20), add(add(const_100, 20), const_100)), const_100) | in may mrs lee ' s earnings were 30 percent of the lee family ' s total income . in june mrs lee earned 20 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "lets say the family income is 100 in may , lee earned 30 family income is 70 in june , lee earned 20 % more than may , so it is ( 30 + 20 * 30 / 100 = 36 ) family income is same 70 in june lee ' s income percent is 36 * 100 / 106 ~ 34 ans is a" | a = 100 + 20
b = 100 + 20
c = b + 100
d = a / c
e = d * 100
|
a ) 32 : 99 , b ) 82 : 31 , c ) 24 : 45 , d ) 34 : 89 , e ) 35 : 21 | c | divide(add(multiply(6000, 4), multiply(divide(9000, const_3), multiply(2, 4))), add(multiply(9000, multiply(2, const_3)), multiply(subtract(9000, divide(9000, const_3)), multiply(2, const_3)))) | a and b invests rs . 6000 and rs . 9000 in a business . after 4 months , a withdraws half of his capital and 2 months later , b withdraws one - third of his capital . in what ratio should they share the profits at the end of the year ? | "a : b ( 6000 * 4 ) + ( 3000 * 8 ) : ( 9000 * 6 ) + ( 6000 * 6 ) 48000 : 90000 24 : 45 answer : c" | a = 6000 * 4
b = 9000 / 3
c = 2 * 4
d = b * c
e = a + d
f = 2 * 3
g = 9000 * f
h = 9000 / 3
i = 9000 - h
j = 2 * 3
k = i * j
l = g + k
m = e / l
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a ) 4 , b ) 0 , c ) 2 , d ) 5 , e ) none of the above | b | divide(subtract(power(8, 1), power(8, 1)), 8) | what is the remainder when 8 ^ 1 + 8 ^ 2 + 8 ^ 3 + . . . + 8 ^ 9 is divided by 8 ? | "notice that in the brackets we have the sum of 9 even multiples of 8 , which yields remainder of 0 upon division by 8 . answer : b" | a = 8 ** 1
b = 8 ** 1
c = a - b
d = c / 8
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a ) 3 , b ) 9 , c ) 12 , d ) 19 , e ) 25 | c | add(add(power(2, 2), multiply(2, 2)), 4) | if [ [ x ] ] = x ^ 2 + 2 x + 4 , what is the value of [ [ 2 ] ] ? | these functions questions might look intimidating , but they just test your knowledge about how well you can substitute values [ [ x ] ] = x ^ 2 + 2 x + 4 [ [ 2 ] ] = 2 ^ 2 + 2 * 2 + 4 = 12 . option c | a = 2 ** 2
b = 2 * 2
c = a + b
d = c + 4
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a ) 1 % , b ) 1.1 % , c ) 9.1 % , d ) 11.81 % , e ) 11.50 % | d | divide(multiply(13, const_100), add(13, const_100)) | the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 13 percent , what was the annual interest rate last year ? | 13 = 1.1 * x x = 11.81 % answer d ) | a = 13 * 100
b = 13 + 100
c = a / b
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a ) 81000 , b ) 78468.75 , c ) 81008 , d ) 81066 , e ) 81022 | b | add(add(62000, multiply(divide(1, 8), 62000)), multiply(divide(1, 8), add(62000, multiply(divide(1, 8), 62000)))) | every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 62000 ? | "62000 * 9 / 8 * 9 / 8 = 78468.75 answer : b" | a = 1 / 8
b = a * 62000
c = 62000 + b
d = 1 / 8
e = 1 / 8
f = e * 62000
g = 62000 + f
h = d * g
i = c + h
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a ) 1 / 2 , b ) 16 / 7 , c ) 3 / 2 , d ) 4 / 5 , e ) 4 / 7 | b | divide(divide(16, 15), divide(2, 15)) | if p ( a ) = 2 / 15 , p ( b ) = 7 / 15 , and p ( a Γ’ Λ Βͺ b ) = 16 / 15 find p ( a | b ) | "p ( a | b ) = p ( a Γ’ Λ Βͺ b ) / p ( b ) p ( a | b ) = ( 16 / 15 ) / ( 7 / 15 ) = 16 / 7 . b" | a = 16 / 15
b = 2 / 15
c = a / b
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a ) 75 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | b | multiply(divide(multiply(multiply(divide(8, const_2), divide(8, const_2)), 10), multiply(multiply(divide(10, const_2), divide(10, const_2)), 8)), const_100) | tanks m and b are each in the shape of a right circular cylinder . the interior of tank m has a height of 10 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank m is what percent of the capacity of tank b ? | b . for m , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 10 = 160 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi m / b = 160 pi / 200 pi = 0.8 | a = 8 / 2
b = 8 / 2
c = a * b
d = c * 10
e = 10 / 2
f = 10 / 2
g = e * f
h = g * 8
i = d / h
j = i * 100
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a ) 36 , b ) 302 , c ) 272 , d ) 292 , e ) 271 | a | divide(add(add(multiply(30, 20), multiply(10, 40)), multiply(65, 40)), const_100) | in a class , 20 % of the students were absent for an exam . 30 % failed by 20 marks and 10 % just passed . find the average score of the class if the remaining students scored an average of 65 marks and the cut off for passing the exam is 40 . | explanation : let total students be 100 . hence , marks of 20 students is 0 as they were absent . 30 students scored 20 marks ( as they failed by 20 marks ) 10 students scored 40 marks ( just passed ) 40 students scored 65 marks ( average of 65 ) therefore , average score of the class is : ( 0 + 600 + 400 + 2600 ) / 100 = 3600 / 100 = 36 answer : a | a = 30 * 20
b = 10 * 40
c = a + b
d = 65 * 40
e = c + d
f = e / 100
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a ) 150 , b ) 225 , c ) 285 , d ) 315 , e ) 374 | c | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 510), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | a library has an average of 510 visitors on sundays and 240 on other days . the avearge number of visitors per day in a month of 30 days beginning with a sunday is ? | "since the month begin with sunday , so there will be five sundays in the month required average = ( 510 * 5 + 240 * 25 ) / 30 = 8550 / 30 = 285 answer is c" | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 510
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 240
l = e + k
m = l / 30
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a ) 75 , b ) 65 , c ) 85 , d ) 95 , e ) 60 | e | add(multiply(8, 2.5), 40) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 40 kg . what is the weight of the new person ? | "total increase in weight = 8 Γ 2.5 = 20 if x is the weight of the new person , total increase in weight = x β 40 = > 20 = x - 40 = > x = 20 + 40 = 60 answer is e ." | a = 8 * 2
b = a + 40
|
a ) 16.5 , b ) 16.0 , c ) 16.4 , d ) 14 , e ) 16.1 | d | divide(add(110, 170), multiply(72, const_0_2778)) | how long does a train 110 m long traveling at 72 kmph takes to cross a bridge of 170 m in length ? | "d = 110 + 170 = 280 m s = 72 * 5 / 18 = 20 t = 280 * 1 / 20 = 14 sec answer : d" | a = 110 + 170
b = 72 * const_0_2778
c = a / b
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a ) 120 , b ) 135 , c ) 140 , d ) 309 , e ) 369 | d | multiply(divide(subtract(250, 14), subtract(5, const_1)), 5) | find large number from below question the difference of two numbers is 250 . on dividing the larger number by the smaller , we get 5 as quotient and the 14 as remainder | "let the smaller number be x . then larger number = ( x + 250 ) . x + 250 = 5 x + 14 4 x = 236 x = 59 large number = 59 + 250 = 309 d" | a = 250 - 14
b = 5 - 1
c = a / b
d = c * 5
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a ) 14 / 52 , b ) 15 / 52 , c ) 16 / 52 , d ) 17 / 52 , e ) 18 / 52 | c | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of drawn an ace or a space or both from a dew of cards . | "there are 13 spades in a standard deck of cards . there are four aces in a standard deck of cards . one of the aces is a spade . so , 13 + 4 - 1 = 16 spades or aces to choose from . since we have a total of 52 cards , the probability of selecting an ace or a spade is 16 / 52 answer : c" | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
|
a ) w β 126 , b ) w β 112 , c ) w β 14 , d ) w + 14 , e ) w + 126 | b | multiply(14, divide(4, 4)) | the water level in a reservoir has been dropping at the rate of 14 inches per day . exactly 4 days ago , the water level was at w inches . what will be the water level exactly 4 days from now if the rate at which the level is dropping remains the same ? | "drop = 14 inches / day 4 days ago = w , means now it ' s equal w - 56 and in 4 days = w - 56 - 56 = w - 112 answer b" | a = 4 / 4
b = 14 * a
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a ) 2 % , b ) 5 % , c ) 28 % , d ) 30 % , e ) 95 % | e | multiply(divide(subtract(subtract(const_100, 80), multiply(divide(multiply(const_10, const_3), const_100), subtract(const_100, 40))), 40), const_100) | thirty percent of the women in a college class are science majors , and the non - science majors make up 80 % of the class . what percentage of the men are non - science majors if 40 % of the class are men ? | "3 / k + 2 / m = 6 / t assuming total # is 100 : [ science - women ] will have - 0.3 * 60 = 18 [ non - science - women ] will have - 42 [ science - men ] will have = 20 - 18 = 2 [ non - science - men ] will have - 38 s 0 38 / 40 * 100 = 95 % answer - e" | a = 100 - 80
b = 10 * 3
c = b / 100
d = 100 - 40
e = c * d
f = a - e
g = f / 40
h = g * 100
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a ) 2378 , b ) 2697 , c ) 250 , d ) 5000 , e ) 6971 | c | subtract(multiply(4, divide(1000, add(add(add(const_2, const_1), const_1), const_4))), multiply(const_2, divide(1000, add(add(add(const_2, const_1), const_1), const_4)))) | an amount of rs . 1000 is to be distributed amongst p , q , r and s such that β p β gets twice as that of β q β and β s β gets 4 times as that of β r β . if β q β and β r β are to receive equal amount , what is the difference between the amounts received by s and p ? | explanation : we have , p = 2 q & s = 4 r further q = r & p + q + r + s = 1,000 thus we get , 2 q + q + q + 4 q = 1,000 8 q = 1,000 or q = rs . 125 thus , r = rs . 125 , p = 250 & s = rs . 500 hence , the required difference = ( s β p ) = ( 500 β 250 ) = rs . 250 answer : c | a = 2 + 1
b = a + 1
c = b + 4
d = 1000 / c
e = 4 * d
f = 2 + 1
g = f + 1
h = g + 4
i = 1000 / h
j = 2 * i
k = e - j
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a ) 800 , b ) 1,250 , c ) 8,000 , d ) 12,000 , e ) 80,000 | a | multiply(divide(multiply(divide(multiply(4, 10), 50), power(10, const_4)), const_1000), 4) | a certain galaxy is known to comprise approximately 4 x 10 ^ 10 stars . of every 50 million of these stars , one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ? | "total no . of stars on galaxy = 4 * 10 ^ 10 of every 50 million stars , 1 is larger than sun . 1 million = 10 ^ 6 therofore , 50 million = 50 * 10 ^ 6 total no . of stars larger than sun = 4 * 10 ^ 10 / 50 * 10 ^ 6 = 40 * 10 ^ 2 / 5 = 800 therefore answer is a" | a = 4 * 10
b = a / 50
c = 10 ** 4
d = b * c
e = d / 1000
f = e * 4
|
a ) 5 hrs , b ) 10 hrs , c ) 7.5 hrs , d ) 20 hrs , e ) 30 hrs | c | divide(45, 6) | ajay can walk 6 km in 1 hour . in how many hours he can walk 45 km ? | "1 hour he walk 6 km he walk 45 km in = 45 / 6 * 1 = 7.5 hours answer is c" | a = 45 / 6
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a ) rs . 21724.14 , b ) rs . 31724.14 , c ) rs . 51724.14 , d ) rs . 61724.14 , e ) none of these | c | divide(multiply(divide(multiply(54000, const_100), add(const_100, 20)), const_100), subtract(const_100, 13)) | a man sells a car to his friend at 13 % loss . if the friend sells it for rs . 54000 and gains 20 % , the original c . p . of the car was : | "explanation : s . p = rs . 54,000 . gain earned = 20 % c . p = rs . [ 100 / 120 Γ£ β 54000 ] = rs . 45000 this is the price the first person sold to the second at at loss of 13 % . now s . p = rs . 45000 and loss = 13 % c . p . rs . [ 100 / 87 Γ£ β 45000 ] = rs . 51724.14 . correct option : c" | a = 54000 * 100
b = 100 + 20
c = a / b
d = c * 100
e = 100 - 13
f = d / e
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a ) 22 , b ) 897 , c ) 268 , d ) 20 , e ) 48 | e | add(9, add(add(multiply(const_2, 10), 10), 9)) | n 10 years , a will be twice as old as b was 10 years ago . if a is now 9 years older than b the present age of b is | a + 10 = 2 ( b - 10 ) . . . . . . . . ( 1 ) a = b + 9 . . . . . . . . . ( 2 ) from equations . 1 & 2 we get b = 39 a will be 39 + 9 = 48 years old . answer : e | a = 2 * 10
b = a + 10
c = b + 9
d = 9 + c
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a ) rs 429 , b ) rs 412 , c ) rs 400 , d ) rs 129 , e ) rs 122 | c | multiply(divide(const_4, add(const_4, add(const_1, const_2))), 700) | rs . 700 is divided among a , b , c so that a receives half as much as b and b half as much as c . then c ' s share is | "let c = x . then b = x / 2 and a = x / 4 a : b : c = 1 : 2 : 4 . c ' s share rs . [ ( 4 / 7 ) * 700 ) = 400 answer : c" | a = 1 + 2
b = 4 + a
c = 4 / b
d = c * 700
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a ) 10 ^ 6 , b ) 10 ^ 7 , c ) 10 ^ 8 , d ) 10 ^ 9 , e ) 10 ^ 10 | d | divide(0.1, power(0.01, 5)) | the decimal 0.1 is how many times greater than the decimal ( 0.01 ) ^ 5 ? | "0.1 = 10 ^ - 1 ( 0.01 ) ^ 5 = ( 10 ^ - 2 ) ^ 5 = 10 ^ - 10 10 ^ 9 * 10 ^ - 10 = 10 ^ - 1 the answer is d ." | a = 0 ** 1
b = 0 / 1
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a ) a ) 32 , b ) b ) 24 , c ) c ) 39 , d ) d ) 40 , e ) e ) 45 | b | divide(multiply(15, 8), subtract(8, 3)) | a number exceeds by 15 from its 3 / 8 part . then the number is ? | x β 3 / 8 x = 15 x = 24 answer : b | a = 15 * 8
b = 8 - 3
c = a / b
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a ) 654 , b ) 752 , c ) 656 , d ) 657 , e ) 658 | b | multiply(divide(subtract(const_100, 20), const_100), 940.00) | yearly subscription to professional magazines cost a company $ 940.00 . to make a 20 % cut in the magazine budget , how much less must be spent ? | "total cost 940 940 * 20 / 100 = 188 so the cut in amount is 188 the less amount to be spend is 940 - 188 = 752 answer : b" | a = 100 - 20
b = a / 100
c = b * 940
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a ) 600 , b ) 500 , c ) 400 , d ) 300 , e ) none of these | b | divide(add(125, 40), divide(33, const_100)) | a student has to obtain 33 % of the total marks to pass . he got 125 marks and failed by 40 marks . the maximum marks are | explanation : given that the student got 125 marks and still he failed by 40 marks = > the minimum pass mark = 125 + 40 = 165 given that minimum pass mark = 33 % of the total mark = > total mark Γ ( 33 / 100 ) = 165 = > total mark = 16500 / 33 = 500 answer : option b | a = 125 + 40
b = 33 / 100
c = a / b
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a ) 429.3 , b ) 270 , c ) 177 , d ) 166 , e ) 111 | a | floor(divide(8800, add(20, divide(2.5, const_100)))) | find the number of shares that can be bought for rs . 8800 if the market value is rs . 20 each with brokerage being 2.5 % . | explanation : cost of each share = ( 20 + 2.5 % of 20 ) = rs . 20.5 therefore , number of shares = 8800 / 20.5 = 429.3 answer : a | a = 2 / 5
b = 20 + a
c = 8800 / b
d = math.floor(c)
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a ) 25.5 , b ) 2.5 , c ) 12.5 , d ) . 25 , e ) none of these | c | divide(1, 0.08) | 1 / 0.08 is equal to | "explanation : 1 / 0.08 = ( 1 * 100 ) / 8 = 100 / 8 = 12.5 option c" | a = 1 / 0
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a ) 5 : 8 , b ) 5 : 5 , c ) 5 : 9 , d ) 5 : 3 , e ) 5 : 1 | a | divide(3, 2) | the simple form of the ratio 5 / 8 : 3 / 2 is ? | "5 / 8 : 3 / 2 = 5 : 8 answer : a" | a = 3 / 2
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a ) 1.34 , b ) 2.84 , c ) 3.84 , d ) 4.34 , e ) 5 | c | multiply(const_12, divide(multiply(40, divide(40, const_100)), 50)) | a reduction of 40 % in the price of bananas would enable a man to obtain 50 more for rs . 40 , what is reduced price per dozen ? | "40 * ( 40 / 100 ) = 16 - - - 50 ? - - - 12 = > rs . 3.84 answer : c" | a = 40 / 100
b = 40 * a
c = b / 50
d = 12 * c
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a ) 240 , b ) 200 , c ) 160 , d ) 100 , e ) 50 | d | divide(20, subtract(subtract(const_1, inverse(30)), divide(const_1, const_2))) | a student traveled 30 percent of the distance of the trip alone , continued another 20 miles with a friend , and then finished the last half of the trip alone . how many miles long was the trip ? | "let x be the total length of the trip . 0.3 x + 20 miles + 0.5 x = x 20 miles = 0.2 x x = 100 miles the answer is d ." | a = 1/(30)
b = 1 - a
c = 1 / 2
d = b - c
e = 20 / d
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a ) 6 minutes , b ) 8 minutes , c ) 5 minutes , d ) 2 minutes , e ) 7 minutes | b | add(multiply(12, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 24)), const_2)), 12), const_60)) | two pipes a and b can fill a tank in 12 and 24 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "required time = 12 * 24 / 12 + 12 = 12 * 24 / 36 = 24 / 3 = 8 minutes answer : b" | a = 12 * 100
b = 1 / 12
c = 1 / 24
d = b + c
e = d * 2
f = 1 - e
g = f * 12
h = g * const_60
i = a + h
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a ) 32.5 , b ) 35 , c ) 48 , d ) 65 , e ) 67.5 | c | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(10, const_100)))), subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(10, const_100))))) | the price of a jacket is reduced by 25 % . during a special sale the price of the jacket is reduced another 10 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 25 % , therefore bringing down the price to $ 75 . 3 ) again it is further discounted by 10 % , therefore bringing down the price to $ 67.5 . 4 ) now 67.5 has to be added by x % in order to equal the original price . 67.5 + ( x % ) 67.5 = 100 . solving this eq for x , we get x = 48.1 answer is c ." | a = 100 - 25
b = 100 - 25
c = 10 / 100
d = b * c
e = a - d
f = 100 - e
g = 100 - 25
h = 100 - 25
i = 10 / 100
j = h * i
k = g - j
l = f / k
m = 100 * l
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a ) 2.0 , b ) 2.5 , c ) 3.0 , d ) 2.1 , e ) 4.0 | e | divide(add(3.5, subtract(3.5, 1.5)), const_2) | a man whose speed is 3.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph , find his average speed for the total journey ? | "m = 3.5 s = 1.5 ds = 5 us = 2 as = ( 2 * 5 * 2 ) / 5 = 4 answer : e" | a = 3 - 5
b = 3 + 5
c = b / 2
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a ) β 1 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | c | subtract(divide(divide(divide(660, add(add(const_3, const_4), const_4)), const_10), const_2), const_2) | if a , b , c , and d are integers ; w , x , y , and z are prime numbers ; w < x < y < z ; and ( wa ) ( xb ) ( yc ) ( zd ) = 660 ( wa ) ( xb ) ( yc ) ( zd ) = 660 , what is the value of ( a + b ) β ( c + d ) ? | 660660 = 2 ^ 2 x 3 ^ 1 x 5 ^ 1 x 11 ^ 1 w < x < y < z = 2 < 3 < 5 < 11 so , can can say - w = 2 x = 3 y = 5 z = 11 hence a = 2 , b = c = d = 1 ( a + b ) β ( c + d ) will be ( 2 + 1 ) β ( 1 + 1 ) = 1 answer will be ( c ) | a = 3 + 4
b = a + 4
c = 660 / b
d = c / 10
e = d / 2
f = e - 2
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a ) 130 % , b ) 300 % , c ) 150 % , d ) 160 % , e ) 170 % | b | add(multiply(subtract(multiply(add(const_1, divide(50, const_100)), const_2), const_1), const_100), const_100) | a man gains 50 % by selling an article for a certain price . if he sells it at double the price , the percentage of profit will be . | explanation : let the c . p . = x , then s . p . = ( 150 / 100 ) x = 3 x / 2 new s . p . = 2 ( 3 x / 2 ) = 3 x / 1 profit = 3 x / 1 - x = 3 x / 1 profit % = ( profit / c . p . ) * 100 = > ( 3 x / 1 ) * ( 1 / x ) * 100 = 300 % option b | a = 50 / 100
b = 1 + a
c = b * 2
d = c - 1
e = d * 100
f = e + 100
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a ) 3 , b ) 2 , c ) 1 , d ) 5 , e ) 4 | d | subtract(subtract(14, 5), subtract(12, 8)) | | 14 - 5 | - | 8 - 12 | = ? | | 14 - 5 | - | 8 - 12 | = | 9 | - | - 4 | = 9 - 4 = 5 correct answer d | a = 14 - 5
b = 12 - 8
c = a - b
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a ) 11 , b ) 15 , c ) 77 , d ) 33 , e ) 88 | b | add(add(3, 7), 5) | in kaya ' s teacher ' s desk there are 3 pink highlighters , 7 yellow highlighters , and 5 blue highlighters . how many highlighters are there in all ? | "add the numbers of highlighters . 3 + 7 + 5 = 15 . answer is b ." | a = 3 + 7
b = a + 5
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a ) 450 , b ) 810 , c ) 900 , d ) 1000 , e ) 1100 | c | multiply(divide(multiply(9, add(9, 1)), 2), multiply(2, const_10)) | the sum of all the digits of the integers from 18 to 21 inclusive is 24 ( 1 + 8 + 1 + 9 + 2 + 0 + 2 + 1 = 24 ) . what is the sum w of all the digits of the integers from 0 to 99 inclusive ? | "we want the sum of the digits from 0 to 99 , so i approximated : 0 - 9 - > 45 - > ( 9 + 0 ) * 10 / 2 40 - 49 - > 85 ( 13 + 4 ) * 10 / 2 90 - 99 - > 135 ( 18 + 9 ) * 10 / 2 we can see at a glance that theweightgoes up as the numbers go up ( meaning the difference between 85 and 45 is 40 , while 135 - 85 is 50 , this means that the second part of this sequence carries more weight for our result ) , so we know that the final answer has to be more than 850 ( 85 * 10 ) but close to it , and that ' s just w = 900 : the answer is c ." | a = 9 + 1
b = 9 * a
c = b / 2
d = 2 * 10
e = c * d
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a ) rs 600 , b ) rs 625 , c ) rs 650 , d ) rs 675 , e ) none of these | b | inverse(multiply(power(divide(4, const_100), const_2), 1)) | the difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4 % per annum is rs 1 . find the sum | explanation : let the sum be p s . i . = p β 4 β 2 / 100 = 2 p / 25 c . i . = p ( 1 + 4 / 100 ) 2 β p = 676 p / 625 β p = 51 p / 625 as , c . i . - s . i = 1 = > 51 p / 625 β 2 p / 25 = 1 = > 51 p β 50 p / 625 = 1 p = 625 option b | a = 4 / 100
b = a ** 2
c = b * 1
d = 1/(c)
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a ) rs . 70 , b ) rs . 36 , c ) rs . 54 , d ) rs . 50 , e ) none | a | divide(8.4, divide(12, const_100)) | the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 8.4 . the true discount is | "solution t . d = [ b . g x 100 / r x t ] = rs . ( 8.4 x 100 / 12 x 1 ) = rs . 70 . answer a" | a = 12 / 100
b = 8 / 4
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a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 45 | d | divide(multiply(20, 3), const_2) | in a basketball game , tim scored 20 points more than joe , but only half as many points as ken . if the 3 players scored a combined total of 100 points , how many points did tim score ? | let joe scored point = x then tim scored = x + 20 ken scored = 2 * ( x + 20 ) = 2 x + 40 as given , x + x + 20 + 2 x + 40 = 100 points 4 x + 60 = 100 x = 100 - 60 / 4 = 10 so tim scored = x + 20 i . e ) 10 + 20 = 30 answer : d | a = 20 * 3
b = a / 2
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a ) $ 2.40 , b ) $ 25.20 , c ) $ 25.40 , d ) $ 25.60 , e ) $ 25.50 | c | add(2.00, multiply(subtract(divide(2.00, divide(1, 5)), 1), 0.60)) | if taxi fares were $ 2.00 for the first 1 / 5 mile and $ 0.60 for each 1 / 5 mile there after , then the taxi fare for a 8 - mile ride was | "in 8 miles , initial 1 / 5 mile charge is $ 2 rest of the distance = 8 - ( 1 / 5 ) = 39 / 5 rest of the distance charge = 39 ( 0.6 ) = $ 23.4 ( as the charge is 0.6 for every 1 / 5 mile ) = > total charge for 4 miles = 2 + 23.4 = 25.4 answer is c" | a = 1 / 5
b = 2 / 0
c = b - 1
d = c * 0
e = 2 + 0
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a ) 603 , b ) 703 , c ) 800 , d ) 903 , e ) 1000 | a | divide(multiply(9, 670), 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 670 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 670 , loss of pyarelal = 670 * 9 x / 10 x = 603 answer : a" | a = 9 * 670
b = 1 + 9
c = a / b
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a ) 57600 , b ) 44000 , c ) 34936 , d ) 25640 , e ) none | a | multiply(divide(const_100, 40), 23040) | 40 % of the population of a village is 23040 . the total population of the village is ? | "answer β΅ 40 % of p = 23040 β΄ p = ( 23040 x 100 ) / 40 = 57600 correct option : a" | a = 100 / 40
b = a * 23040
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a ) 287 , b ) 900 , c ) 289 , d ) 276 , e ) 207 | b | multiply(multiply(54, const_0_2778), 30) | what distance will be covered by a bus moving at 54 kmph in 30 seconds ? | "54 kmph = 54 * 5 / 18 = 30 mps d = speed * time = 30 * 30 = 900 m . answer : b" | a = 54 * const_0_2778
b = a * 30
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a ) 8 , b ) 12 , c ) 16 , d ) 24 , e ) 11 | e | divide(multiply(22, 2), const_4) | if ( 1 / 2 ) ^ 22 ( 1 / 81 ) ^ k = 1 / 18 ^ 22 , then k = | i ' m going to focus on denominator only . . ( 2 ^ 22 ) . ( ( 3 ^ 4 ) ^ k = 18 ^ 22 ( 2 ^ 22 ) . ( ( 3 ^ 4 k ) = ( 2 . 3 ^ 2 ) ^ 22 ( 2 ^ 22 ) . ( ( 3 ^ 4 k ) = ( 2 ^ 24 ) . ( 3 ^ 2 ) ^ 22 hence 4 k = 44 k = 11 answer e i hope it ' s quite clear | a = 22 * 2
b = a / 4
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a ) 450 , b ) 540 , c ) 542 , d ) 829 , e ) 279 | a | multiply(multiply(inverse(subtract(add(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 45)), divide(const_1, 15))), const_3), 15) | two pipes a and b can separately fill a tank in 10 and 15 minutes respectively . a third pipe c can drain off 45 liters of water per minute . if all the pipes are opened , the tank can be filled in 15 minutes . what is the capacity of the tank ? | "1 / 10 + 1 / 15 - 1 / x = 1 / 15 x = 10 10 * 45 = 450 answer : a" | a = 1 / 10
b = 1 / 15
c = a + b
d = 1 / 45
e = c + d
f = 1 / 15
g = e - f
h = 1/(g)
i = h * 3
j = i * 15
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a ) 6 cm , b ) 8.25 cm , c ) 11.25 cm , d ) 16.37 cm , e ) 20.62 cm | d | divide(volume_cube(17), multiply(20, 15)) | a cube of edge 17 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 15 cm , find the rise in water level ? | "increase in volume = volume of the cube = 17 * 17 * 17 cm ^ 3 rise in water level = volume / area = 17 * 17 * 17 / 20 * 15 = 16.37 cm answer is d" | a = volume_cube / (
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a ) 277 , b ) 270 , c ) 291 , d ) 266 , e ) 121 | c | multiply(520, divide(7, const_100)) | find the simple interest on rs . 520 for 7 months at 8 paisa per month ? | "i = ( 520 * 7 * 8 ) / 100 = 291 answer : c" | a = 7 / 100
b = 520 * a
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a ) 1200 , b ) 1000 , c ) 1100 , d ) 1050 , e ) 1250 | c | divide(900, subtract(const_1, divide(10, const_100))) | a person incurs 10 % loss by selling a laptop for $ 900 . at what price should the watch be sold to earn 10 % profit ? | "let the new s . p . = x then ( 100 - loss % ) : ( 1 st s . p . ) = ( 100 + gain % ) : ( 2 nd s . p . ) ( 100 - 10 ) / 900 = ( 100 + 10 ) / x x = 110 * 900 / 90 = 1100 new price = $ 1100 correct option is c" | a = 10 / 100
b = 1 - a
c = 900 / b
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a ) 290 / 289 , b ) 122 / 121 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these | b | add(power(divide(const_1, const_1), const_2), power(divide(const_1, 11), const_2)) | product of two natural numbers is 11 . then , the sum of reciprocals of their squares is | "explanation : if the numbers are a , b , then ab = 17 , as 17 is a prime number , so a = 1 , b = 17 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 11 ( 2 ) = 122 / 121 option b" | a = 1 / 1
b = a ** 2
c = 1 / 11
d = c ** 2
e = b + d
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a ) 16.5 % , b ) 20 % , c ) 33 % , d ) 55 % , e ) 65 % | c | multiply(divide(subtract(3.5, 3), subtract(4.5, 3)), const_100) | a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe - benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 4.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs at the beginning of 1993 ? | "the amount by which employment costs rose is equal to 0.035 ( salary costs + fringe benefit costs ) ; on the other hand the amount by which employment costs rose is equal to 0.03 * salary costs + 0.045 * fringe benefit costs ; so , 35 ( s + f ) = 30 s + 45 f - - > s = 2 f - - > f / s = 1 / 2 - - > f / ( s + f ) = 1 / ( 1 + 2 ) = 1 / 3 = 0.33 . answer : c ." | a = 3 - 5
b = 4 - 5
c = a / b
d = c * 100
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a ) 18 , b ) 21 , c ) 24 , d ) 27 , e ) 28 | b | divide(add(multiply(2, 30), 45), add(2, 3)) | sandy gets 3 marks for each correct sum and loses 2 marks for each incorrect sum . sandy attempts 30 sums and obtains 45 marks . how many sums did sandy get correct ? | let x be the correct sums and ( 30 - x ) be the incorrect sums . 3 x - 2 ( 30 - x ) = 45 5 x = 105 x = 21 the answer is b . | a = 2 * 30
b = a + 45
c = 2 + 3
d = b / c
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a ) 6 , b ) 7 , c ) 8 , d ) 12 , e ) 14 | d | inverse(subtract(divide(const_1, 3), divide(const_1, 4))) | bruce and anne can clean their house in 4 hours working together at their respective constant rates . if anne β s speed were doubled , they could clean their house in 3 hours working at their respective rates . how many q hours does it currently take anne to clean the house on her own ? | "lets suppose anne and bruce take a and b hrs working separately so in 1 hour they can together finish 1 / a + 1 / b portion of the work which equals 1 / 4 ( as the work is completed in 4 hours ) after anne doubles her rate of work the portion completed by the both is 1 / a + 2 / b which is equal to q = 1 / 3 ( as the work is completed in q = 3 hours ) solving these 2 equations we can find b as 12 so , d" | a = 1 / 3
b = 1 / 4
c = a - b
d = 1/(c)
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a ) 260 , b ) 250 , c ) 150 , d ) 225 , e ) 325 | a | subtract(multiply(200, const_2), multiply(160, const_2)) | if the average ( arithmetic mean ) of a and b is 160 , and c β a = 200 , what is the average of b and c ? | "a + b / 2 = 160 = > a + b = 320 a = c - 200 . . . sub this value c - 200 + b = 320 = > c + b = 520 = > c + b / 2 = 260 answer : a" | a = 200 * 2
b = 160 * 2
c = a - b
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a ) 0 , b ) - 2 , c ) - 45 , d ) - 49 , e ) - 51 | c | add(add(negate(24), const_1), add(add(negate(24), const_1), const_1)) | the sum of all the integers k such that β 24 < k < 24 is | "- 23 - - - - - - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - - - 23 values upto + 23 cancels outwe are left with only - 23 - 22 sum of which is - 45 . hence option d . c" | a = negate + (
b = a + 1
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a ) 18 , b ) 19 , c ) 20 , d ) 31 , e ) 22 | d | subtract(add(21, 11), const_1) | a student is ranked 21 th from right and 11 th from left . how many students are there in totality ? | "from right 21 , from left 11 total = 21 + 11 - 1 = 31 answer : d" | a = 21 + 11
b = a - 1
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a ) 437 , b ) 299 , c ) 322 , d ) 345 , e ) 355 | a | multiply(23, 19) | the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 13 and 19 . the larger of the two numbers is : | "clearly , the numbers are ( 23 x 13 ) and ( 23 x 19 ) . larger number = ( 23 x 19 ) = 437 . answer : option a" | a = 23 * 19
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a ) $ 900 , b ) $ 800 , c ) $ 960 , d ) $ 700 , e ) $ 1500 | c | divide(240, subtract(const_1, divide(3, 4))) | linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 240 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 240 . so 1 / 4 of her savings is $ 240 . so her original savings are 4 times $ 240 = $ 960 correct answer c" | a = 3 / 4
b = 1 - a
c = 240 / b
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a ) 11 , b ) 30 , c ) 99 , d ) 48 , e ) 61 | d | divide(800, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 800 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 800 * 3 / 50 = 48 sec . answer : d | a = 63 - 3
b = a * const_0_2778
c = 800 / b
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a ) 19.63 % , b ) 20 % , c ) 37.5 % , d ) 25 % , e ) 37.5 % | a | multiply(subtract(add(const_100, 10), add(13, subtract(const_100, 20))), const_2) | a man cheats while buying as well as while selling . while buying he takes 10 % more than what he pays for and while selling he gives 20 % less than what he claims to . find the profit percent , if he sells at 13 % below the cost price of the claimed weight . | "there is a one step calculation method too . it requires more thought but is faster . the man takes 10 % more than what he pays for . so if he claims to take 100 pounds , he pays $ 100 but he actually takes 110 pounds for which he will take from the customer $ 110 . hence , in effect , there is a 10 % mark up . while selling , he sells 20 % less . this means , he claims to sell 100 pounds and gets $ 100 but actually sells only 80 pounds and should have got only $ 80 for it . so this is again a mark up of $ 20 on $ 80 which is 25 % . but he also sells at 13 % less ( 1 + m 1 % ) ( 1 + m 2 % ) ( 1 - d % ) = ( 1 + p % ) 11 / 10 * 5 / 4 * 87 / 100 = ( 1 + p % ) profit % = 19.63 % a" | a = 100 + 10
b = 100 - 20
c = 13 + b
d = a - c
e = d * 2
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a ) 1.204 , b ) 1.9874 , c ) 1.3564 , d ) 1.2547 , e ) 1.6547 | a | add(multiply(const_4, 1.8061), divide(log(const_100), log(const_10))) | if log 64 = 1.8061 , then the value of log 16 will be ( approx ) ? | "log 64 = 1.8061 log 4 ^ 3 = 1.8061 log 4 = 0.6020 2 log 4 = 1.2040 log 4 ^ 2 = 1.2040 log 16 = 1.2040 ( approx ) answer a" | a = 4 * 1
b = math.log(100)
c = math.log(10)
d = b / c
e = a + d
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a ) $ 1.40 , b ) $ 1.58 , c ) $ 2.00 , d ) $ 2.64 , e ) $ 2.79 | e | divide(add(add(multiply(3, 2.25), multiply(4, 1.5)), multiply(4, 1)), 6) | the student manager of the university theater estimated that for every 6 movie tickets sold on valentine ' s weekend , the theater will sell 3 packs of grain crackers at $ 2.25 each , 4 bottles of a beverage at $ 1.50 each , and 4 chocolate bars at $ 1.00 each . find the approximate average ( arithmetic mean ) amount of estimated snack sales per movie ticket sold . | 3 * 2.25 + 4 * 1.50 + 4 * 1 = 6.75 + 6 + 4 = 16.75 . this is the sale for 6 tickets avg sale per ticket : 16.75 / 6 = $ 2 . 79 answer : e | a = 3 * 2
b = 4 * 1
c = a + b
d = 4 * 1
e = c + d
f = e / 6
|
a ) 15 , b ) 8 , c ) 18 , d ) none of these , e ) can not be determined | c | divide(subtract(divide(50, divide(5, 4)), multiply(subtract(5, const_1), 4)), 4) | the sum of the ages of 5 children born at the intervals of 4 years each is 50 years . what is the age of the eldest child ? | "explanation : let x = the youngest child . each of the other four children will then be x + 4 , x + 8 , x + 12 , x + 16 . we know that the sum of their ages is 50 . so , x + ( x + 4 ) + ( x + 8 ) + ( x + 12 ) + ( x + 16 ) = 50 x = 2 the youngest child is 2 years old . age of eldest child is = 2 + 16 = 18 answer : c" | a = 5 / 4
b = 50 / a
c = 5 - 1
d = c * 4
e = b - d
f = e / 4
|
a ) 12 , b ) 8 , c ) 88 , d ) 77 , e ) 14 | b | subtract(28, divide(28, add(divide(2.00001, 5), const_1))) | a 28 cm long wire is to be cut into two pieces so that one piece will be 2.00001 / 5 th of the other , how many centimeters will the shorter piece be ? | "1 : 2 / 5 = 5 : 2 2 / 7 * 28 = 8 answer : b" | a = 2 / 1
b = a + 1
c = 28 / b
d = 28 - c
|
a ) 120 , b ) 76 , c ) 108 , d ) 105 , e ) 86 | c | add(subtract(multiply(multiply(multiply(const_2, const_3), const_3), 3), multiply(multiply(const_2, const_3), const_3)), multiply(subtract(multiply(multiply(multiply(const_2, const_3), const_3), 3), multiply(multiply(const_2, const_3), const_3)), const_2)) | eighteen years ago , a father was 3 times as old as his son . now the father is only twice as old as his son . then the sum of the present ages of the son and the father is : | let the present ages of the father and son be 2 x and x years respectively . then , ( 2 x - 18 ) = 3 ( x - 18 ) x = 36 . required sum = ( 2 x + x ) = 3 x = 108 years . answer is c | a = 2 * 3
b = a * 3
c = b * 3
d = 2 * 3
e = d * 3
f = c - e
g = 2 * 3
h = g * 3
i = h * 3
j = 2 * 3
k = j * 3
l = i - k
m = l * 2
n = f + m
|
a ) 1240 , b ) 1120 , c ) 1190 , d ) 1978 , e ) none of these | d | add(828, divide(multiply(828, const_100), multiply(12, 6))) | the banker ' s gain on a sum due 6 years hence at 12 % per annum is rs . 828 . what is the banker ' s discount ? | "explanation : td = ( bg Γ 100 ) / tr = ( 828 Γ 100 ) / ( 6 Γ 12 ) = rs . 1150 bg = bd β td = > 828 = bd - 1150 = > bd = 1978 answer : option d" | a = 828 * 100
b = 12 * 6
c = a / b
d = 828 + c
|
a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | c | divide(40, divide(divide(40, const_2), 12)) | a runner runs the 40 miles from marathon to athens at a constant speed . halfway through the run she injures her foot , and continues to run at half her previous speed . if the second half takes her 12 hours longer than the first half , how many hours did it take the runner to run the second half ? | "the runner runs the first 20 miles at speed v and the second 20 miles at speed v / 2 . the time t 2 to run the second half must be twice the time t 1 to run the first half . t 2 = 2 * t 1 = t 1 + 12 t 1 = 12 and so t 2 = 24 . the answer is c ." | a = 40 / 2
b = a / 12
c = 40 / b
|
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | d | multiply(divide(multiply(multiply(40, divide(40, 40)), const_2), 40), const_2) | the racing magic takes 40 seconds to circle the racing track once . the charging bull makes 40 rounds of the track in an hour . if they left the starting point together , how many minutes will it take for them to meet at the starting point for the second time ? | "time taken by racing magic to make one circle = 40 seconds time taken bycharging bullto make one circle = 60 mins / 40 = 1.5 mins = 90 seconds lcm of 90 and 40 seconds = 360 seconds = 6 mins time taken for them to meet at the starting point for the second time = 6 mins * 2 = 12 mins answer d" | a = 40 / 40
b = 40 * a
c = b * 2
d = c / 40
e = d * 2
|
a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | e | multiply(divide(subtract(add(divide(32, 2), multiply(subtract(96, 32), 2)), 96), 96), const_100) | every day daniel drives 96 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 96 miles at 4 miles / hour . d = ( r ) ( t ) 96 = ( 4 ) ( t ) 96 / 4 = 24 = t it takes 24 hours to drive home on monday , daniel drove the first 32 miles at ( 2 ) ( 4 ) = 8 miles / hour and the rest of the way ( 64 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 ) 64 = ( 2 ) ( t ) 64 / 2 = 32 = t it takes 32 hours for the second part total time to drive home on monday = 4 + 32 = 36 hours we ' re asked by what percent 36 hours is greater than 32 hours . 36 / 32 = 1.5 , so it is 50 % greater . e" | a = 32 / 2
b = 96 - 32
c = b * 2
d = a + c
e = d - 96
f = e / 96
g = f * 100
|
a ) 150 meter , b ) 145 meter , c ) 140 meter , d ) 135 meter , e ) 210 meter | e | multiply(divide(multiply(108, const_1000), const_3600), 7) | a train running at the speed of 108 km / hr crosses a pole in 7 seconds . find the length of the train . | "explanation : speed = 108 * ( 5 / 18 ) m / sec = 30 m / sec length of train ( distance ) = speed * time = 30 * 7 = 210 meter option e" | a = 108 * 1000
b = a / 3600
c = b * 7
|
a ) 375 m , b ) 750 / 3 m , c ) 2500 / 3 m , d ) 800 m , e ) 300 m | c | multiply(divide(2500, subtract(120, 30)), 30) | a train crosses a bridge of length 2500 m in 120 seconds and a lamp post on the bridge in 30 seconds . what is the length of the train in metres ? | "let length of train = l case - 1 : distance = 2500 + l ( while crossing the bridge ) time = 120 seconds i . e . speed = distance / time = ( 2500 + l ) / 120 case - 2 : distance = l ( while passing the lamp post ) time = 30 seconds i . e . speed = distance / time = ( l ) / 30 but since speed has to be same in both cases so ( 2500 + l ) / 120 = ( l ) / 30 i . e . 4 l = 2500 + l i . e . 3 l = 2500 i . e . l = 2500 / 3 answer : option c" | a = 120 - 30
b = 2500 / a
c = b * 30
|
a ) 15345 , b ) 15645 , c ) 15625 , d ) 15342 , e ) 15683 | c | add(divide(divide(250, divide(divide(divide(divide(divide(250, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(250, divide(divide(divide(divide(divide(250, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of all odd number upto 250 . | "explanation : number of odd numbers upto 250 = n / 2 = 250 / 2 = 125 sum of first 125 odd numbers = ( 125 ) 2 = 15625 answer : option c" | a = 250 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 250 / e
g = f / 2
h = 250 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 250 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
|
a ) 8000000 , b ) 1000000 , c ) 7500000 , d ) 1200000 , e ) none of these | a | divide(multiply(multiply(multiply(6, const_100), multiply(7, const_100)), multiply(8, const_100)), multiply(multiply(8, 7), 7)) | a wooden box of dimensions 8 m x 7 m x 6 m is to carry rectangularboxes of dimensions 2 cm x 7 cm x 3 cm . the maximum number ofboxes that can be carried in the wooden box , is | explanation : number = ( 800 * 700 * 600 ) / 2 * 7 * 3 = 8000000 answer : a | a = 6 * 100
b = 7 * 100
c = a * b
d = 8 * 100
e = c * d
f = 8 * 7
g = f * 7
h = e / g
|
a ) 12 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | a | divide(subtract(const_1, multiply(5, divide(const_1, 15))), divide(const_1, 18)) | x can finish a work in 18 days . y can finish the same work in 15 days . yworked for 5 days and left the job . how many days does x alone need to finish the remaining work ? | work done by x in 1 day = 1 / 18 work done by y in 1 day = 1 / 15 work done by y in 5 days = 5 / 15 = 1 / 3 remaining work = 1 β 1 / 3 = 2 / 3 number of days in which x can finish the remaining work = ( 2 / 3 ) / ( 1 / 18 ) = 12 a | a = 1 / 15
b = 5 * a
c = 1 - b
d = 1 / 18
e = c / d
|
a ) 140 , b ) 160 , c ) 220 , d ) 250 , e ) 265 | e | divide(divide(multiply(60, const_1000), 60), multiply(divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3)), divide(120, const_100))) | the diameter of the wheel of a car is 120 m . how many revolution / min mustthe wheel makeing order to keep a speed of 60 km / hour approximately ? | "distance to be covered in 1 min . = ( 60 x 1000 ) / ( 60 ) m = 1000 m . circumference of the wheel = ( 2 x ( 22 / 7 ) x 0.60 ) m = 3.77 m . number of revolutions per min . = ( 1000 / 3.77 ) = 265 e" | a = 60 * 1000
b = a / 60
c = 2 * 10
d = c + 2
e = 4 + 3
f = d / e
g = 120 / 100
h = f * g
i = b / h
|
a ) 10 , b ) 99 , c ) 7.5 , d ) 55 , e ) 21 | c | add(inverse(subtract(divide(const_1, 5), divide(const_1, 15))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 5 days . if a alone can do it in 15 days . in how many days can b alone do it ? | "1 / 5 β 1 / 15 = 2 / 15 = > 7.5 answer : c" | a = 1 / 5
b = 1 / 15
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
|
a ) 0.125 , b ) 0.25 , c ) 0.375 , d ) 0.5 , e ) 0.666 | c | multiply(power(divide(const_1, const_2), 3), 3) | if a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exectly twice in 3 consecutive flips ? | "p = ncm * p ^ m * q ^ ( n - m ) n - the total number of trials . m - the number of trials with heads . n - m - the number of trials with tails . p - probability of heads . q - probability of tails . p = 3 c 2 * ( 1 / 2 ) ^ 2 * ( 1 / 2 ) ^ 1 = 3 / 8 = 0.375 answer : c" | a = 1 / 2
b = a ** 3
c = b * 3
|
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