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a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | d | divide(const_2, 3) | if an integer n is to be selected at random from 1 to 99 , inclusive , what is probability n ( n + 1 ) will be divisible by 3 ? | "for n ( n + 1 ) to be a multiple of 3 , either n or n + 1 has to be a multiple of 3 . thus n must be of the form 3 k or 3 k - 1 , but not 3 k + 1 . the probability is 2 / 3 . the answer is d ." | a = 2 / 3
|
a ) 11 , b ) 10 , c ) 28 , d ) 24 , e ) 35 | e | multiply(divide(subtract(3267, 2420), 2420), const_100) | a sum of money deposited at c . i . amounts to rs . 2420 in 2 years and to rs . 3267 in 3 years . find the rate percent ? | "explanation : 2420 - - - 847 100 - - - ? = > 35 % answer : option e" | a = 3267 - 2420
b = a / 2420
c = b * 100
|
a ) 23 , b ) 11 , c ) 2 , d ) 10 , e ) 12 | e | subtract(212, multiply(50, const_4)) | workers in a factory store products inside two different sized boxes . the bigger of the boxes can store 50 products while the smaller box can store 40 products . if a fresh batch of products numbering 212 in total were to be stored , what is the least number of products that will not be boxed ? | total products to be stored 212 the big box can store 50 small box can store 40 storing 212 products in big box 212 / 50 = 4 r 12 least number of toys that will be left unboxed = 12 answer : e | a = 50 * 4
b = 212 - a
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | min(min(divide(15, add(2, divide(1, 2))), floor(divide(16, add(divide(3, 4), 2)))), divide(8, add(divide(1, 3), 1))) | a recipe requires 2 1 / 2 ( mixed number ) cups of flour 2 3 / 4 ( mixed number ) cups of sugar and 1 1 / 3 ( mixed number ) cups of milk to make one cake . victor has 15 cups if flour , 16 cups of sugar and 8 cups of milk . what is the greatest number of cakes william can make using this recipe ? | "less work up front : go through each item and see what the greatest number of cakes you can make with each . the lowest of these will be the right answer . flour : 15 cups , we need 2.5 cups each . just keep going up the line to see how many cakes we can make : that means i can make 2 cakes with 5 cups , so 6 cakes overall with 15 cups . i ' ve already got the answer narrowed to either a or b . sugar : 16 cups , we need 2.75 cups each . same principle . i can make 2 cups with 5.5 cups , so to make 6 cakes i ' d need 16.5 cups . i do n ' t have that much sugar , so we ' re limited to 5 cakes . no need to even do milk because we ' re already at 5 . sugar will be the limiting factor . answer is a" | a = 1 / 2
b = 2 + a
c = 15 / b
d = 3 / 4
e = d + 2
f = 16 / e
g = math.floor(f)
h = min(c)
i = 1 / 3
j = i + 1
k = 8 / j
l = min(h)
|
a ) 19 % , b ) 12.3 % , c ) 25 % , d ) 40 % , e ) 9.4 % | b | multiply(subtract(const_1, divide(multiply(const_100, const_100), multiply(subtract(const_100, 5), add(const_100, 20)))), const_100) | in a hostel , the number of students decreased by 5 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the previous year ? | cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 5 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.95 n ) * ( 1.2 p ) = > f ( new ) = f / ( 0.95 * 1.2 ) = > f ( new ) = 0.877 f therefore the new cost of food must be 87.7 % of the old cost , or the cost of food must decrease by 12.28 % ( optionb ) | a = 100 * 100
b = 100 - 5
c = 100 + 20
d = b * c
e = a / d
f = 1 - e
g = f * 100
|
a ) 7 , b ) 14 , c ) 25 , d ) 27 , e ) 30 | d | add(subtract(add(multiply(floor(divide(44, 3)), 3), 21), multiply(floor(divide(44, 8)), 8)), 3) | at 15 : 00 there were 21 students in the computer lab . at 15 : 03 and every three minutes after that , 3 students entered the lab . if at 15 : 10 and every ten minutes after that 8 students left the lab , how many students were in the computer lab at 15 : 44 ? | initial no of students + 3 * ( 1 + no of possible 3 minute intervals between 15 : 03 and 15 : 44 ) - 8 * ( 1 + no of possible 10 minute intervals between 15 : 10 and 15 : 44 ) 20 + 3 * 14 - 8 * 4 = 27 d | a = 44 / 3
b = math.floor(a)
c = b * 3
d = c + 21
e = 44 / 8
f = math.floor(e)
g = f * 8
h = d - g
i = h + 3
|
a ) 0.49 , b ) 0.48 , c ) 0.41 , d ) 0.42 , e ) 0.411 | d | multiply(divide(70, multiply(multiply(const_4, const_5), const_5)), divide(60, multiply(multiply(const_4, const_5), const_5))) | if a speaks the truth 70 % of the times , b speaks the truth 60 % of the times . what is the probability that they tell the truth at the same time | "explanation : probability that a speaks truth is 70 / 100 = 0.7 probability that b speaks truth is 60 / 100 = 0.6 since both a and b are independent of each other so probability of a intersection b is p ( a ) Γ p ( b ) = 0.7 Γ 0.6 = 0.42 answer : d" | a = 4 * 5
b = a * 5
c = 70 / b
d = 4 * 5
e = d * 5
f = 60 / e
g = c * f
|
['a ) 660', 'b ) 770', 'c ) 880', 'd ) 900', 'e ) 990'] | e | multiply(const_pi, multiply(21, 15)) | find the curved surface area , if the radius of a cone is 21 m and slant height is 15 m ? | cone curved surface area = Γ― β¬ rl 22 / 7 Γ£ β 21 Γ£ β 15 = 66 Γ£ β 15 = 990 m ( power 2 ) answer is e . | a = 21 * 15
b = math.pi * a
|
a ) 280 , b ) 400 , c ) 540 , d ) 650 , e ) 840 | d | divide(divide(divide(130, subtract(const_1, divide(3, 5))), divide(3, 4)), divide(2, 3)) | of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year . if 130 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ? | "of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month : 2 / 3 * 3 / 4 = 1 / 2 survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year : ( 1 - 3 / 5 ) * 1 / 2 = 1 / 5 survived the first year . 130 geese survived the first year : 1 / 5 * ( total ) = 130 - - > ( total ) = 650 . answer : d ." | a = 3 / 5
b = 1 - a
c = 130 / b
d = 3 / 4
e = c / d
f = 2 / 3
g = e / f
|
a ) 299 , b ) 190 , c ) 276 , d ) 270 , e ) 281 | b | subtract(multiply(multiply(divide(72, const_3600), const_1000), 22), 250) | a goods train runs at the speed of 72 km / hr and crosses a 250 m long platform in 22 sec . what is the length of the goods train ? | "speed = 72 * 5 / 18 = 20 m / sec . time = 22 sec . let the length of the train be x meters . then , ( x + 250 ) / 22 = 20 x = 190 m . answer : b" | a = 72 / 3600
b = a * 1000
c = b * 22
d = c - 250
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | multiply(6, divide(9, 7)) | john was thrice as old as tom 6 years ago . john will be 9 / 7 times as old as tom in 6 years . how old is tom today ? | "j - 6 = 3 ( t - 6 ) , so j = 3 t - 12 j + 6 = 9 / 7 * ( t + 6 ) 7 j + 42 = 9 t + 54 7 ( 3 t - 12 ) + 42 = 9 t + 54 12 t = 96 t = 8 the answer is b ." | a = 9 / 7
b = 6 * a
|
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | b | divide(48, const_10) | how many integers from 0 to 48 , inclusive , have a remainder of 1 when divided by 3 ? | "my ans is also c . 17 . explanation : 1 also gives 1 remainder when divided by 3 , another number is 4 , then 7 and so on . hence we have an arithmetic progression : 1 , 4 , 7 , 10 , . . . . . 46 , which are in the form 3 n + 1 . now we have to find out number of terms . tn = a + ( n - 1 ) d , where tn is the nth term of an ap , a is the first term and d is the common difference . so , 46 = 1 + ( n - 1 ) 3 or , ( n - 1 ) 3 = 45 or , n - 1 = 15 or , n = 16 b" | a = 48 / 10
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | add(add(add(const_1, add(const_1, const_1)), const_1), const_1) | what is the prime factors β number of 36 ? | "prime factors β number , as i assume , for a number x = a ^ n * b ^ m * c ^ o * d ^ p . . . is = n + m + o + p . . . so , 24 = 2 ^ 2 * 3 ^ 2 prime factors β number will be 2 + 2 = 4 . hence , answer is c ." | a = 1 + 1
b = 1 + a
c = b + 1
d = c + 1
|
a ) 55 , b ) 99 , c ) 88 , d ) 76 , e ) 12 | a | divide(add(250, 300), multiply(subtract(72, 36), const_0_2778)) | how much time will a train of length 250 m moving at a speed of 72 kmph take to cross another train of length 300 m , moving at 36 kmph in the same direction ? | "the distance to be covered = sum of their lengths = 250 + 300 = 550 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 / 18 = 10 mps . time required = d / s = 550 / 10 = 55 sec . answer : a" | a = 250 + 300
b = 72 - 36
c = b * const_0_2778
d = a / c
|
a ) 36 , b ) 39 , c ) 42 , d ) 45 , e ) 48 | d | divide(divide(20, subtract(const_1, divide(const_1, 3))), subtract(const_1, divide(const_1, 3))) | for all real numbers v , an operation is defined by the equation v * = v - v / 3 . if ( v * ) * = 20 , then v = | "( v * ) * = ( v - v / 3 ) - ( v - v / 3 ) / 3 20 = 2 v / 3 - 2 v / 9 = 4 v / 9 v = 45 the answer is d ." | a = 1 / 3
b = 1 - a
c = 20 / b
d = 1 / 3
e = 1 - d
f = c / e
|
a ) 1865113 , b ) 1775123 , c ) 1764613 , d ) 1675123 , e ) none of them | c | multiply(4300231, power(add(const_4, const_1), const_4)) | ( 4300231 ) - ? = 2535618 | "let 4300231 - x = 2535618 then x = 4300231 - 2535618 = 1764613 answer is c" | a = 4 + 1
b = a ** 4
c = 4300231 * b
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 13 | b | subtract(add(divide(36, subtract(subtract(multiply(const_10, 2), const_10), const_1)), multiply(divide(36, subtract(subtract(multiply(const_10, 2), const_10), const_1)), 2)), subtract(multiply(divide(36, subtract(subtract(multiply(const_10, 2), const_10), const_1)), 2), divide(36, subtract(subtract(multiply(const_10, 2), const_10), const_1)))) | the difference of 2 digit no . & the no . obtained by interchanging the digits is 36 . what is the difference the sum and the number if the ratio between the digits of the number is 1 : 2 ? | let the number be xy . given xy β yx = 36 . this means the number is greater is than the number got on reversing the digits . this shows that the ten β s digit x > unit digit y . also given ratio between digits is 1 : 2 = > x = 2 y ( 10 x + y ) β ( 10 y + x ) = 36 = > x β y = 4 = > 2 y β y = 4 . hence , ( x + y ) β ( x β y ) = 3 y β y = 2 y = 8 b | a = 10 * 2
b = a - 10
c = b - 1
d = 36 / c
e = 10 * 2
f = e - 10
g = f - 1
h = 36 / g
i = h * 2
j = d + i
k = 10 * 2
l = k - 10
m = l - 1
n = 36 / m
o = n * 2
p = 10 * 2
q = p - 10
r = q - 1
s = 36 / r
t = o - s
u = j - t
|
a ) 12.5 , b ) 26.2 , c ) 6.5 , d ) 7.5 , e ) 12 | d | add(15, divide(subtract(36, 26), 5)) | the average of 5 numbers is calculated as 15 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "5 * 15 + 36 β 26 = 75 / 10 = 7.5 answer : d" | a = 36 - 26
b = a / 5
c = 15 + b
|
a ) 10 % , b ) 11.1 % , c ) 25 % , d ) 30 % , e ) 35 % | b | multiply(divide(subtract(const_1, divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 10), const_100)), const_100) | the length of a rectangle is reduced by 10 % . by what % would the width have to be increased to maintain the original area ? | "sol . required change = ( 10 * 100 ) / ( 100 - 10 ) = 11.1 % b" | a = 100 - 10
b = a / 100
c = 1 - b
d = 100 - 10
e = d / 100
f = c / e
g = f * 100
|
a ) s 4000 , b ) s 2800 , c ) s 1800 , d ) s 1500 , e ) s 1550 | e | divide(subtract(multiply(3500, divide(5, const_100)), 144), subtract(divide(5, const_100), divide(3, const_100))) | rs 3500 is divided into two parts such that one part is put out at 3 % and the other at 5 % . if the annual interest earned from both the investments be rs 144 , find the first part . | "explanation : average rate = ( 144 / 3500 ) * 100 = 4.11 ratio = 8.85 : 11.14 so , first part = ( 8.85 / 20 ) * 3500 = rs 1550 . answer : e" | a = 5 / 100
b = 3500 * a
c = b - 144
d = 5 / 100
e = 3 / 100
f = d - e
g = c / f
|
a ) 7 sec , b ) 6 sec , c ) 4 sec , d ) 2 sec , e ) 9 sec | b | divide(110, multiply(add(60, 6), const_0_2778)) | a train 110 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 = 110 * 3 / 55 = 6 sec . answer : b" | a = 60 + 6
b = a * const_0_2778
c = 110 / b
|
a ) 444440 , b ) 610000 , c ) 666640 , d ) 995456 , e ) 880000 | d | multiply(multiply(power(4, const_3), add(add(add(5, 4), 3), 2)), add(add(add(multiply(const_100, const_10), const_100), const_10), const_1)) | what is the sum of all 4 digit integers formed using the digits 2 , 3 , 4 and 5 ( repetition is allowed ) | n = 4 * 4 * 4 * 4 = 256 < x > = ( 5555 + 2222 ) / 2 = 3888.5 sum = number of integers x average value n * < x > = 256 * 3888.5 = 711040 answer = d | a = 4 ** 3
b = 5 + 4
c = b + 3
d = c + 2
e = a * d
f = 100 * 10
g = f + 100
h = g + 10
i = h + 1
j = e * i
|
a ) 55 m , b ) 60 m , c ) 80 m , d ) 82 m , e ) 84 m | a | divide(add(divide(5300, 26.50), multiply(const_2, 10)), const_4) | length of a rectangular plot is 10 mtr more than its breadth . if the cost of fencing 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 + 10 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 10 ) + x ] = 200 2 x + 10 = 100 2 x = 90 x = 45 . hence , length = x + 10 = 55 m a" | a = 5300 / 26
b = 2 * 10
c = a + b
d = c / 4
|
['a ) 8 m', 'b ) 2 m', 'c ) 7 m', 'd ) 10 m', 'e ) 6 β 2 m'] | e | multiply(sqrt(const_2), divide(24, const_4)) | what is the diagonal of a square field whose perimeter is 24 m ? | sol : let the side of the square be x meters . therefore , 4 x = 24 x = 6 diagonal of the square is = β 2 * ( 6 ) ^ 2 = 6 β 2 answer : e | a = math.sqrt(2)
b = 24 / 4
c = a * b
|
a ) 3.12 , b ) 8 , c ) 10 , d ) 15 , e ) 24 | a | max(multiply(subtract(add(55, 5), const_1), subtract(divide(5, 35), divide(5, 55))), const_4) | due to construction , the speed limit along an 5 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 5 miles stretch = 5 * 60 / 55 = 5 * 12 / 11 = 5.45 new time in minutes to cross 5 miles stretch = 5 * 60 / 35 = 5 * 12 / 7 = 8.57 time difference = 3.12 ans : a" | a = 55 + 5
b = a - 1
c = 5 / 35
d = 5 / 55
e = c - d
f = b * e
g = max(f)
|
a ) 10 , b ) 12 , c ) 16 , d ) 2 , e ) 22 | d | divide(subtract(multiply(2, subtract(40, 4)), multiply(2, 32)), 4) | the average age of an adult class is 40 years . 2 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 year . find what was the original strength of class ? | "let original strength = y then , 40 y + 2 x 32 = ( y + 2 ) x 36 Γ’ β‘ β 40 y + 64 = 36 y + 72 Γ’ β‘ β 4 y = 8 Γ’ Λ Β΄ y = 2 d" | a = 40 - 4
b = 2 * a
c = 2 * 32
d = b - c
e = d / 4
|
a ) 4 : 5 , b ) 4 : 3 , c ) 4 : 4 , d ) 4 : 9 , e ) 3 : 4 | e | divide(sqrt(9), 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 9 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = β b : β a = β 9 : β 16 = 3 : 4 answer : e" | a = math.sqrt(9)
b = math.sqrt(16)
c = a / b
|
a ) 3 , b ) 2 , c ) 5 , d ) 4 , e ) 6 | a | multiply(multiply(divide(divide(47, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5))), const_3), const_1000) | a person travels equal distances with speeds of 3 km / hr , 4 km / hr , 5 km / hr . and takes a total time of 47 minutes . find the total distance ? | "let the each distance be x km total distance = 3 x then total time , ( x / 3 ) + ( x / 4 ) + ( x / 5 ) = 47 / 60 x = 1 total distance = 3 * 1 = 3 km correct option is a" | a = 47 / const_60
b = 1 / 3
c = 1 / 4
d = b + c
e = 1 / 5
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) - 2 , b ) 5 , c ) 7 , d ) 24 , e ) 35 | a | subtract(5, divide(add(multiply(5, const_2), 11), subtract(5, const_2))) | the product of a and b is equal to 11 more than twice the sum of a and b . if b = 5 , what is the value of b - a ? | "ab = 11 + 2 ( a + b ) 5 a = 11 + 2 a + 10 3 a = 21 a = 7 b - a = 5 - 7 = - 2 a is the answer" | a = 5 * 2
b = a + 11
c = 5 - 2
d = b / c
e = 5 - d
|
a ) 16 % , b ) 25 % , c ) 32 % , d ) 40 % , e ) 52 % | b | multiply(divide(subtract(64, 48), 64), const_100) | in town x , 64 percent of the population are employed , and 48 percent of the population are employed males . what percent of the employed people in town x are females ? | "let population be 100 total employed people = 64 % of the population = 64 employed males = 48 % of the population = 48 employed females = 16 % of the population = 16 employed females % in terms of employed people = ( 16 / 64 ) * 100 = 25 % answer is b ." | a = 64 - 48
b = a / 64
c = b * 100
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | a | add(4, multiply(power(2, 2), multiply(const_3, const_3))) | if s is a positive integer , and if the units ' digit of s ^ 2 is 4 and the units ' digit of ( s + 1 ) ^ 2 is 1 , what is the units ' digit of ( s + 2 ) ^ 2 ? | a for me . _ 1 Β² = _ 1 _ 2 Β² = _ 4 _ 3 Β² = _ 9 _ 4 Β² = _ 6 _ 5 Β² = _ 5 _ 6 Β² = _ 6 _ 7 Β² = _ 9 _ 8 Β² = _ 4 _ 9 Β² = _ 1 _ 0 Β² = _ 0 on the list , the only number that squared has the units digit = 4 preceding a number that squared has the units digit = 1 is the _ 8 . so , s = _ 8 , that has square 4 . the next square is 1 andnext nextsquare is 0 . | a = 2 ** 2
b = 3 * 3
c = a * b
d = 4 + c
|
a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | b | add(add(10, subtract(40, 10)), 10) | in a group of 65 people , 40 like cricket , 10 like both cricket and tennis . how many like tennis ? | "make a venn diagram , and enter your data . let the number of people who like only tennis be x x + 10 + 30 = 65 x = 25 so number who like tennis = 10 + 25 = 35 answer b" | a = 40 - 10
b = 10 + a
c = b + 10
|
a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | b | subtract(14.0, multiply(2, 1.5)) | the arithmetic mean and standard deviation of a certain normal distribution are 14.0 and 1.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | "the value which isexactlytwo sd less than the mean is : mean - 2 * sd = 14.0 - 2 * 1.5 = 11 . answer : b ." | a = 2 * 1
b = 14 - 0
|
a ) 24 , b ) 12 , c ) 6 , d ) 4 , e ) 2 | e | divide(divide(18, const_3), const_3) | if a * b denotes the greatest common divisor of a and b , then ( ( 16 * 20 ) * ( 18 * 24 ) ) = ? | the greatest common divisor of 16 and 20 is 4 . hence 16 * 20 = 4 ( note that * here denotes the function not multiplication ) . the greatest common divisor of 18 and 24 is 6 . hence 18 * 24 = 6 . hence ( ( 16 * 20 ) * ( 18 * 24 ) ) = 4 * 6 . the greatest common divisor of 4 and 6 is 2 . answer ; e . | a = 18 / 3
b = a / 3
|
a ) a ) 4 , b ) b ) 8 , c ) c ) 6 , d ) d ) 2 , e ) e ) 1 | b | add(subtract(39, add(30, 1)), 1) | the average weight of a group of boys is 30 kg . after a boy of weight 39 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 39 kg joins the group , total weight of boys = 30 x + 39 so 30 x + 39 = 31 ( x + 1 ) = > x = 8 . answer : b" | a = 30 + 1
b = 39 - a
c = b + 1
|
a ) 52 kmph , b ) 56 kmph , c ) 58 kmph , d ) 30 kmph , e ) 34 kmph | d | subtract(multiply(divide(300, 12), const_3_6), 60) | a man sitting in a train which is traveling at 60 kmph observes that a goods train , traveling in opposite direction , takes 12 seconds to pass him . if the goods train is 300 m long , find its speed | "explanation : relative speed = 300 / 12 m / sec = ( ( 300 / 12 ) Γ ( 18 / 5 ) ) kmph = 90 kmph . speed of goods train = ( 90 - 60 ) kmph = 30 kmph answer : option d" | a = 300 / 12
b = a * const_3_6
c = b - 60
|
a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 80 % | b | multiply(divide(subtract(divide(65, const_100), divide(55, const_100)), add(subtract(divide(65, const_100), divide(55, const_100)), subtract(divide(80, const_100), divide(65, const_100)))), const_100) | solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 65 percent carbonated water , what percent of the volume of the mixture is p ? | "65 % is 15 % - points below 80 % and 10 % - points above 55 % . so the ratio of solution p to solution q is 2 : 3 . mixture p is 2 / 5 = 40 % of the volume of mixture pq . the answer is b ." | a = 65 / 100
b = 55 / 100
c = a - b
d = 65 / 100
e = 55 / 100
f = d - e
g = 80 / 100
h = 65 / 100
i = g - h
j = f + i
k = c / j
l = k * 100
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | subtract(multiply(40, 2), add(multiply(subtract(subtract(40, add(add(multiply(12, 1), 12), 2)), 1), 3), add(multiply(12, 1), multiply(12, 2)))) | in a class of 40 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 12 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ? | "the class borrowed a total of 40 * 2 = 80 books . the 26 students who borrowed 0 , 1 , or 2 books borrowed a total of 12 + 12 * 2 = 36 . to maximize the number of books borrowed by 1 student , let ' s assume that 13 students borrowed 3 books and 1 student borrowed the rest . 80 - 36 - 3 * 13 = 5 the maximum number of books borrowed by any student is 5 . the answer is c ." | a = 40 * 2
b = 12 * 1
c = b + 12
d = c + 2
e = 40 - d
f = e - 1
g = f * 3
h = 12 * 1
i = 12 * 2
j = h + i
k = g + j
l = a - k
|
a ) 500 , b ) 550 , c ) 600 , d ) 933 , e ) 750 | d | divide(560, subtract(const_1, divide(40, const_100))) | shop offered 40 % offer for every shirt , smith bought a shirt at rs . 560 . and what was the shop ' s original selling price ? | "sp * ( 60 / 100 ) = 560 sp = 9.33 * 100 = > cp = 933 answer : d" | a = 40 / 100
b = 1 - a
c = 560 / b
|
a ) 33.25 , b ) 89 , c ) 64 , d ) 68 , e ) 90 | c | subtract(add(add(48, 12), divide(40, 8)), subtract(4, const_3)) | the average score of a cricketer for 12 matches is 48 runs . if the average for first 8 matches is 40 , then average for last 4 matches is | explanation : = ( 48 Γ 12 ) β ( 40 Γ 8 ) / 4 = ( 576 β 320 ) / 4 = 64 answer : option c | a = 48 + 12
b = 40 / 8
c = a + b
d = 4 - 3
e = c - d
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a ) 34 , b ) 25 , c ) 48 , d ) 50 , e ) none | a | divide(add(multiply(40, 30), multiply(30, 40)), add(40, 30)) | the average of 40 results is 30 and the average of other 30 results is 40 . what is the average of all the results ? | answer sum of 70 result = sum of 40 result + sum of 30 result . = 40 x 30 + 30 x 40 = 2400 / 70 correct option : a | a = 40 * 30
b = 30 * 40
c = a + b
d = 40 + 30
e = c / d
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a ) 7500 , b ) 3388 , c ) 2665 , d ) 7800 , e ) 2661 | d | divide(2340, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100))) | a candidate got 35 % of the votes polled and he lost to his rival by 2340 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2340 100 % - - - - - - - - - ? = > 7800 answer : d" | a = 35 / 100
b = 1 - a
c = 35 / 100
d = b - c
e = 2340 / d
|
a ) rs . 5,000 , b ) rs . 5,500 , c ) rs . 5,700 , d ) rs . 6,500 , e ) rs . 7,500 | e | multiply(multiply(add(const_4, const_1), const_4), multiply(2, multiply(const_3, const_4))) | a and b started a business in partnership investing rs . 20,000 and rs . 15,000 respectively . after 6 months , c joined them with rs . 20,000 . whatwill be b ' s share in total profit of rs . 25,000 earned at the end of 2 years from the startingof the business ? | "a : b : c = ( 20,000 x 24 ) : ( 15,000 x 24 ) : ( 20,000 x 18 ) = 4 : 3 : 3 . b ' s share = rs . 25000 x 3 / 10 = rs . 7,500 . e" | a = 4 + 1
b = a * 4
c = 3 * 4
d = 2 * c
e = b * d
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a ) 1 / 6 , b ) 2 / 33 , c ) 14 / 33 , d ) 9 / 16 , e ) 3 / 44 | c | multiply(divide(subtract(12, 4), 12), divide(subtract(subtract(12, 4), const_1), subtract(12, const_1))) | in a box of 12 pens , a total of 4 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ? | "method - 1 there are 9 fine pieces of pen and 4 defective in a lot of 12 pens i . e . probability of first pen not being defective = ( 8 / 12 ) i . e . probability of second pen not being defective = ( 7 / 11 ) [ 11 pen remaining with 8 defective remaining considering that first was defective ] probability of both pen being non - defective = ( 8 / 12 ) * ( 7 / 11 ) = 14 / 33 answer : option c" | a = 12 - 4
b = a / 12
c = 12 - 4
d = c - 1
e = 12 - 1
f = d / e
g = b * f
|
a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 4 % , e ) 3.5 % | e | multiply(divide(multiply(50, 0.007), 10), const_100) | a bowl was filled with 10 ounces of water , and 0.007 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | "total amount of water evaporated each day during a 50 - day period = . 007 * 50 = . 007 * 100 / 2 = . 7 / 2 = . 35 percent of the original amount of water evaporated during this period = ( . 35 / 10 ) * 100 % = 3.5 % answer e" | a = 50 * 0
b = a / 10
c = b * 100
|
a ) 102 % , b ) 105 % , c ) 120 % , d ) 135 % , e ) 104 % | e | add(subtract(subtract(30, 20), divide(multiply(20, 30), const_100)), const_100) | the price of a certain painting increased by 30 % during the first year and decreased by 20 % during the second year . the price of the painting at the end of the 2 - year period was what percent of the original price ? | "easiest thing to do : assume that price is 100 price at the end of yr 1 : 100 + 30 = 130 price at the end of year 2 = 130 - 130 * 0.20 = 130 * 0.80 = 104 hence required answer = ( 104 / 100 ) * 100 % = 104 % answer is e ." | a = 30 - 20
b = 20 * 30
c = b / 100
d = a - c
e = d + 100
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | add(floor(divide(subtract(multiply(21, 6), multiply(7, 16)), subtract(21, 16))), const_1) | the least whole number which when subtracted from both the terms of the ratio 6 : 7 to give a ra Ι΅ o less than 16 : 21 , is | explanation : let x is subtracted . then , ( 6 β x ) ( 7 β x ) < 162121 ( 6 οΏ½ x ) < 16 ( 7 οΏ½ x ) = > 5 x > 14 = x > 2.8 least such number is 3 answer : a | a = 21 * 6
b = 7 * 16
c = a - b
d = 21 - 16
e = c / d
f = math.floor(e)
g = f + 1
|
a ) 2.5 , b ) 2.7 , c ) 2.9 , d ) 2.3 , e ) 2.1 | c | divide(134, multiply(169, const_0_2778)) | in what time will a train 134 m long cross an electric pole , it its speed be 169 km / hr ? | "speed = 169 * 5 / 18 = 46.9 m / sec time taken = 134 / 46.9 = 2.9 sec . answer : c" | a = 169 * const_0_2778
b = 134 / a
|
a ) 28 , b ) 50 , c ) 88 , d ) 22 , e ) 32 | e | divide(divide(subtract(75, multiply(multiply(10, const_0_2778), 10)), 5), const_0_2778) | a train 75 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 75 / 10 ) m / sec = ( 15 / 2 ) m / sec . [ ( 15 / 2 ) * ( 18 / 5 ) ] km / hr = 27 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 27 = = > x = 32 km / hr . answer : e" | a = 10 * const_0_2778
b = a * 10
c = 75 - b
d = c / 5
e = d / const_0_2778
|
a ) 73 , b ) 63 , c ) 65 , d ) 70 , e ) 83 | b | subtract(multiply(20, multiply(150, const_0_2778)), 200) | a train 200 m long crosses a platform 150 m long in 20 sec ; find the speed of the train ? | "d = 200 + 150 = 350 t = 20 s = 350 / 20 * 18 / 5 = 63 kmph . answer : b" | a = 150 * const_0_2778
b = 20 * a
c = b - 200
|
a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | d | divide(log(divide(multiply(power(5, 13), power(9, 7)), 3)), log(15)) | if ( 5 ^ 13 ) ( 9 ^ 7 ) = 3 ( 15 ^ x ) , what is the value of x ? | ( 5 ^ 13 ) ( 9 ^ 7 ) = 3 ( 15 ^ x ) = > 5 ^ 13 * 3 ^ 14 = 3 * 3 ^ x * 5 ^ x = > 5 ^ 13 * 3 ^ 14 = 3 ^ ( x + 1 ) * 5 ^ x value of x = 13 answer d | a = 5 ** 13
b = 9 ** 7
c = a * b
d = c / 3
e = math.log(d)
f = math.log(15)
g = e / f
|
a ) 1 / 5 , b ) 3 / 28 , c ) 1 / 7 , d ) 1 / 14 , e ) 5 / 28 | b | divide(multiply(factorial(add(subtract(add(const_4, const_4), const_3), const_1)), factorial(const_3)), factorial(add(const_4, const_4))) | a , b , c , d , e , f , g , h sitting in a row what is the probability that a , b , d are sitting together ? | "total number of arrangement is = 8 ! = 40320 favorable event i . e a , b and d can be arranged in 3 ! and the remaining can be arranged in 6 ! since abd can be in any of the six positions . so 3 ! * 6 ! / 8 ! = 3 / 28 answer : b" | a = 4 + 4
b = a - 3
c = b + 1
d = math.factorial(c)
e = math.factorial(3)
f = d * e
g = 4 + 4
h = math.factorial(g)
i = f / h
|
a ) 201 , b ) 394 , c ) 302 , d ) 301 , e ) 294 | d | add(multiply(21, 14), 7) | the divisor is 21 , the quotient is 14 and the remainder is 7 . what is the dividend ? | d = d * q + r d = 21 * 14 + 7 d = 294 + 7 d = 301 | a = 21 * 14
b = a + 7
|
a ) 8 mph , b ) 5.25 mph , c ) 3.5 mph , d ) 4 mph , e ) 0.5 mph | c | divide(add(1, 6), const_2) | tabby is training for a triathlon . she swims at a speed of 1 mile per hour . she runs at a speed of 6 miles per hour . she wants to figure out her average speed for these two events . what is the correct answer for her ? | "( 1 mph + 6 mph ) / 2 = 3.5 mph correct option is : c" | a = 1 + 6
b = a / 2
|
a ) 993.2 , b ) 551.25 , c ) 534.33 , d ) 543.33 , e ) 564.69 | e | divide(multiply(power(add(divide(5, const_100), const_1), 2), 1050), 2) | what annual payment will discharge a debt of rs . 1050 due in 2 years at the rate of 5 % compound interest ? | "explanation : let each installment be rs . x . then , x / ( 1 + 5 / 100 ) + x / ( 1 + 5 / 100 ) 2 = 1050 820 x + 1050 * 441 x = 564.69 so , value of each installment = rs . 564.69 answer : option e" | a = 5 / 100
b = a + 1
c = b ** 2
d = c * 1050
e = d / 2
|
a ) 15 % , b ) 13.6 % , c ) 65 % , d ) 45 % , e ) 35 % | b | subtract(multiply(divide(const_100, 880), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 880 grams per kg , what is his percent ? | "explanation : 880 - - - 120 100 - - - ? = > 13.6 % answer : b" | a = 100 / 880
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
|
a ) 1140 , b ) 1145 , c ) 1240 , d ) 1340 , e ) 15640 | a | divide(factorial(subtract(21, const_1)), multiply(factorial(subtract(const_4, const_1)), factorial(subtract(subtract(21, const_1), subtract(const_4, const_1))))) | the number of positive integer solutions for the equation x + y + z + t = 21 is | "the number of positive integer solutions for the equatio fx 1 + x 2 + β― + xn = k ( k - 1 ) c ( n - 1 ) - where k is the number and n is number of variable in the equation . 21 - 1 c 4 - 1 = 20 c 3 = 1140 answer : a" | a = 21 - 1
b = math.factorial(a)
c = 4 - 1
d = math.factorial(c)
e = 21 - 1
f = 4 - 1
g = e - f
h = math.factorial(g)
i = d * h
j = b / i
|
a ) 34 % , b ) 33 % , c ) 20 % , d ) 35 % , e ) 30 % | c | divide(multiply(5, const_100), 6) | a shopkeeper buys mangoes at the rate of 6 a rupee and sells them at 5 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 30 . if he buys 6 a rupee , his cp = 5 he selling at 5 a rupee , his sp = 6 profit = sp - cp = 6 - 5 = 1 profit percent = 1 / 5 * 100 = 20 % answer : c" | a = 5 * 100
b = a / 6
|
a ) 5 , b ) 15 , c ) 3 , d ) 95 , e ) 125 | c | divide(subtract(330, divide(330, add(divide(10, const_100), const_1))), 10) | machine p and machine q are each used to manufacture 330 sprockets . it takes machine p 10 hours longer to produce 330 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | "p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 330 / x = 330 / 1.1 x + 10 1.1 ( 330 ) = 330 + 11 x 11 x = 33 x = 3 the answer is c ." | a = 10 / 100
b = a + 1
c = 330 / b
d = 330 - c
e = d / 10
|
a ) 40 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | a | multiply(40, const_1) | at veridux corporation , there are 250 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 160 male associates , how many female managers are there ? | 250 employees : 90 male , 160 female 40 managers , 210 associates 160 male associates implies 50 female associates which means the remaining 40 females must be managers a . 40 | a = 40 * 1
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a ) 24 , b ) 77 , c ) 297 , d ) 265 , e ) 29 | d | multiply(subtract(61, 8), 5) | think of a number , divide it by 5 and add 8 to it . the result is 61 . what is the number thought of ? | "explanation : 61 - 6 = 53 53 x 5 = 265 answer : d" | a = 61 - 8
b = a * 5
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | d | multiply(3, divide(negate(multiply(add(add(7, 4), 3), 8)), subtract(multiply(add(add(7, 4), 3), const_2), multiply(add(add(7, 4), 3), 4)))) | the ratio by weight , measured in pounds , of books to clothes to electronics in a suitcase initially stands at 7 : 4 : 3 . someone removes 8 pounds of clothing from the suitcase , thereby doubling the ratio of books to clothes . how many pounds do the electronics in the suitcase weigh ? | the weights of the items in the suitcase are 7 k , 4 k , and 3 k . if removing 8 pounds of clothes doubles the ratio of books to clothes , then 8 pounds represents half the weight of the clothes . 2 k = 8 pounds and then k = 4 pounds . the electronics weigh 3 ( 4 ) = 12 pounds . the answer is d . | a = 7 + 4
b = a + 3
c = b * 8
d = negate / (
e = 7 + 4
f = e + 3
g = f * 2
h = 7 + 4
i = h + 3
j = i * 4
k = g - j
l = 3 * d
|
a ) 32.5 , b ) 11.12 , c ) 32.1 , d ) 10.8 , e ) 32.3 | d | add(divide(circumface(2.1), const_2), multiply(2.1, const_2)) | the radius of a semi circle is 2.1 cm then its perimeter is ? | "36 / 7 r = 2.1 = 10.80 answer : d" | a = circumface / (
b = a + 2
|
a ) 10 : 17 , b ) 2 : 5 , c ) 10 : 19 , d ) 25 : 7 , e ) 32 : 25 | c | divide(const_10, add(add(multiply(10, const_2), const_2), const_1)) | the dimensions of a rectangular solid are 4 inches , 5 inches , and 10 inches . if a cube , a side of which is equal to one of the dimensions of the rectangular solid , is placed entirely within thespherejust large enough to hold the cube , what the ratio of the volume of the cube to the volume within thespherethat is not occupied by the cube ? | answer : c . | a = 10 * 2
b = a + 2
c = b + 1
d = 10 / c
|
a ) 21 , b ) 63 , c ) 3969 , d ) 147 , e ) 167 | b | sqrt(multiply(189, 21)) | which number should replace both the asterisks in ( * / 21 ) x ( * / 189 ) = 1 ? | "answer let ( y / 21 ) x ( y / 189 ) = 1 then , y 2 = 21 x 189 = 21 x 21 x 9 β΄ y = ( 21 x 3 ) = 63 . option : b" | a = 189 * 21
b = math.sqrt(a)
|
a ) 1.25 % , b ) 3.75 % , c ) 6.25 % , d ) 6.67 % , e ) 15 % | e | multiply(divide(multiply(multiply(const_100, const_100), divide(6, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 10,000 gallons of a solution that is 6 percent sodium chloride by volume . if 6,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "the remaining solution will be approximately what percent sodium chloride ? means : what percent of the remaining solution is sodium chloride . now , since the remaining solution is 10,000 - 6,000 = 4,000 gallons and sodium chloride is 600 gallons ( 6 % of initial solution of 10,000 gallons ) then sodium chloride is 600 / 4,000 * 100 = 15 % of the remaining solution of 4,000 gallons . answer : e ." | a = 100 * 100
b = 6 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
|
a ) 15 min , b ) 13 min , c ) 12 min , d ) 10 min , e ) 17 min | d | add(multiply(12, subtract(const_1, multiply(add(inverse(12), inverse(20)), 30))), 30) | two pipes a and b can fill a cistern in 12 and 20 minutes respectively , and a third pipe c can empty it in 30 minutes . how long will it take to fill the cistern if all the three are opened at the same time ? | "1 / 12 + 1 / 20 - 1 / 30 = 1 / 10 10 / 1 = 10 answer : d" | a = 1/(12)
b = 1/(20)
c = a + b
d = c * 30
e = 1 - d
f = 12 * e
g = f + 30
|
a ) 768 , b ) 243 , c ) 246 , d ) 625 , e ) 248 | e | add(231, 48) | two tests had the same maximum mark . the pass percentages in the first and the second test were 40 % and 45 % respectively . a candidate scored 231 marks in the second test and failed by 48 marks in that test . find the pass mark in the first test ? | "let the maximum mark in each test be m . the candidate failed by 48 marks in the second test . pass mark in the second test = 231 + 48 = 279 45 / 100 m = 279 pass mark in the first test = 40 / 100 m = 40 / 45 * 279 = 248 . answer : e" | a = 231 + 48
|
a ) 100 , b ) 133 , c ) 150 , d ) 167 , e ) 188 | a | multiply(20, divide(200, add(20, 20))) | trains a and b start simultaneously from stations 200 miles apart , and travel the same route toward each other on adjacent parallel tracks . if train a and train b travel at a constant rate of 20 miles per hour and 20 miles per hour , respectively , how many miles will train a have traveled when the trains pass each other , to the nearest mile ? | "since we know the distance ( 200 ) and the combined rate ( 40 ) , we plug it into the formula : distance = rate * time 200 = 40 * time we can solve for the time they will meet cause we added the rate of train a and train b together . so the time will be 200 / 40 from dividing 40 on both sides to isolate time in the equation above . time will be 5 hours so now you can plug that in for train a β s distance . distance = rate * time distance = 20 * 5 distance = 100 according to answer choice a ." | a = 20 + 20
b = 200 / a
c = 20 * b
|
a ) 0.01 , b ) 0.02 , c ) 0.03 , d ) 0.1 , e ) 0.2 | e | divide(subtract(add(floor(multiply(add(add(13.165, 7.686), 11.545), const_100)), const_1), add(floor(multiply(11.545, const_100)), add(floor(multiply(13.165, const_100)), floor(multiply(7.686, const_100))))), const_10) | x = 13.165 y = 7.686 z = 11.545 the number a is obtained by first rounding the values of x , y , and z to the hundredths place and then adding the resulting values . the number b is obtained by first adding the values of x , y , and z and then rounding the sum to the hundredths place . what is the value of a β b ? | rounding off the values : x = 13.17 y = 7.69 z = 11.55 a = 13.17 + 7.69 + 11.55 = 32.41 x + y + z = 13.165 + 7.686 + 11.545 = 32.206 b = 32.21 a - b = 0.2 option e | a = 13 + 165
b = a + 11
c = b * 100
d = math.floor(c)
e = d + 1
f = 11 * 545
g = math.floor(f)
h = 13 * 165
i = math.floor(h)
j = 7 * 686
k = math.floor(j)
l = i + k
m = g + l
n = e - m
o = n / 10
|
a ) 0.1 , b ) 0.001 , c ) 0.01 , d ) 0.0001 , e ) 1 | c | divide(1, const_100) | how is 1 % expressed as a decimal fraction ? | "1 / 100 = 0.01 answer : c" | a = 1 / 100
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a ) $ 21,000 , b ) $ 18,000 , c ) $ 15,000 , d ) $ 14,500 , e ) $ 4,000 | d | divide(add(divide(subtract(350, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(350, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000) | salesperson a ' s compensation for any week is $ 350 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of a ' s total sales for that week . for what amount of total weekly sales would both salepeople earn the same compensation ? | "sometime , setting up an equation is an easy way to go with : 350 + 0.06 ( x - 1000 ) = 0.08 x x = 14,500 ans : d" | a = 6 / 100
b = a * 1
c = 350 - b
d = 8 / 100
e = 6 / 100
f = d - e
g = c / f
h = 6 / 100
i = h * 1
j = 350 - i
k = 8 / 100
l = 6 / 100
m = k - l
n = j / m
o = g + n
p = o / 1
|
a ) s . 800 , b ) s . 810 , c ) s . 820 , d ) s . 900 , e ) s . 920 | b | multiply(2430, divide(1, 3)) | a , b , c and d enter into partnership . a subscribes 1 / 3 of the capital b 1 / 4 , c 1 / 5 and d the rest . how much share did a get in a profit of rs . 2430 ? | "2430 * 1 / 3 = 810 option b" | a = 1 / 3
b = 2430 * a
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a ) 0.5 , b ) 1.5 , c ) 2.5 , d ) 3.5 , e ) 4.5 | b | add(divide(divide(8, const_2), 4), divide(divide(8, const_2), 8)) | a person walks at a speed of 4 km / hr and runs at a speed of 8 km / hr . how many hours will the person require to cover a distance of 8 km , if the person completes half of the distance by walking and the other half by running ? | "time = 4 / 4 + 4 / 8 = 12 / 8 = 1.5 hours the answer is b ." | a = 8 / 2
b = a / 4
c = 8 / 2
d = c / 8
e = b + d
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a ) 79 , b ) 99 , c ) 47 , d ) 89 , e ) 45 | c | divide(add(add(add(add(45, 35), 52), 47), 55), divide(const_10, const_2)) | dacid obtained 45 , 35 , 52 , 47 and 55 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | average = ( 45 + 35 + 52 + 47 + 55 ) / 5 = 234 / 5 = 47 . answer : c | a = 45 + 35
b = a + 52
c = b + 47
d = c + 55
e = 10 / 2
f = d / e
|
a ) 0.36 , b ) 3.6 , c ) 9000 , d ) 3,600 , e ) 36,000 | c | multiply(divide(1, add(const_2, const_4)), multiply(divide(const_3600, const_10), const_100)) | if anangletis defined as 1 percent of 1 degree , then how many anglets are there in a quarter circle ? | "1 degree * 1 / 100 = 1 anglet so 1 degree = 100 anglets = > 90 degrees = 9000 anglets answer - c" | a = 2 + 4
b = 1 / a
c = 3600 / 10
d = c * 100
e = b * d
|
a ) 2 , b ) 5 , c ) 7 , d ) 24 , e ) 35 | b | subtract(9, divide(add(multiply(9, const_2), 10), subtract(9, const_2))) | the product of a and b is equal to 10 more than twice the sum of a and b . if b = 9 , what is the value of b - a ? | "ab = 10 + 2 ( a + b ) 9 a = 10 + 2 a + 18 7 a = 28 a = 4 b - a = 9 - 4 = 5 b is the answer" | a = 9 * 2
b = a + 10
c = 9 - 2
d = b / c
e = 9 - d
|
a ) 150 , b ) 190 , c ) 250 , d ) 80 , e ) 130 | a | add(add(divide(subtract(700, 400), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(2, const_2)) | how many times digit 2 is used while writing numbers from 400 to 700 ? | "in 400 to 500 there are 50 two ' s in 500 to 600 there are 50 two ' s in 600 to 700 there are 50 two ' s so total is 150 two ' s correct option : a" | a = 700 - 400
b = a / 10
c = 10 + 1
d = 10 + 1
e = c * d
f = b + e
g = 2 * 2
h = f + g
|
a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 28 | b | multiply(2, add(divide(subtract(48, multiply(2, const_2)), add(6, 2)), const_2)) | lionel left his house and walked towards walt ' s house , 48 miles away . two hours later , walt left his house and ran towards lionel ' s house . if lionel ' s speed was 2 miles per hour and walt ' s 6 miles per hour , how many miles had lionel walked when he met walt ? | in the first 2 hours lionel at the rate of 2 miles per hour covered distance = rate * time = 2 * 2 = 4 miles . so , the distance between him and walt was 48 - 4 = 44 miles when walt left his house . now , their combined rate to cover this distance was 2 + 6 = 8 miles per hour , hence they will meet ( they will cover that distance ) in time = distance / rate = 44 / 8 = 5.5 hours . total time that lionel was walking is 2 + 5.5 = 7.5 hours , which means that he covered in that time interval distance = rate * time = 2 * 7.5 = 15 miles . answer : b . | a = 2 * 2
b = 48 - a
c = 6 + 2
d = b / c
e = d + 2
f = 2 * e
|
a ) a ) 6.24 , b ) b ) 8 , c ) c ) 29.1 , d ) d ) 15 , e ) e ) 24 | c | max(multiply(subtract(add(55, 10), const_1), subtract(divide(10, 15), divide(10, 55))), const_4) | due to construction , the speed limit along an 10 - mile section of highway is reduced from 55 miles per hour to 15 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 10 miles stretch = 10 * 60 / 55 = 10 * 12 / 11 = 10.9 new time in minutes to cross 10 miles stretch = 10 * 60 / 15 = 10 * 4 = 40 time difference = 29.1 ans : c" | a = 55 + 10
b = a - 1
c = 10 / 15
d = 10 / 55
e = c - d
f = b * e
g = max(f)
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a ) 71 : 76 , b ) 71 : 131 , c ) 71 : 145 , d ) 71 : 167 , e ) 71 : 113 | a | power(divide(2744, 3375), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 2744 : 3375 . what is the ratio of their total surface areas ? | "ratio of the sides = Β³ β 2744 : Β³ β 3375 = 14 : 15 ratio of surface areas = 142 : 152 = 71 : 76 answer : a" | a = 2744 / 3375
b = 1 / 3
c = a ** b
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a ) 3 : 2 , b ) 9 : 2 , c ) 18 : 20 , d ) 1 : 4 , e ) 18 : 4 | b | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | a and b started a business investing rs . 90,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "a : b = 90000 : 20000 = 90 : 20 = 18 : 4 = 9 : 2 answer : b" | a = 3 * 3
b = 3 * 3
c = b + 1
d = a * c
e = d + 2
f = 3 * 3
g = f + 1
h = 2 * g
i = e / h
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(2000, 662) | the compound interest on $ 2000 at 10 % per annum is $ 662 . the period ( in years ) is ? | a = p ( 1 + r / 100 ) ^ t 2662 = 2000 ( 1 + 10 / 100 ) ^ t ( 11 / 10 ) ^ t = 2662 / 2000 ( 11 / 10 ) ^ t = ( 11 / 10 ) ^ 3 t = 3 years answer is b | a = 2000 / 662
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a ) $ 880 , b ) $ 990 , c ) $ 1,350 , d ) $ 1,100 , e ) $ 1,210 | c | subtract(multiply(150, divide(const_100, 10)), 150) | if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 150 , how much more remains to be paid ? | "10 / 100 p = 150 > > p = 150 * 100 / 10 = 1500 1500 - 150 = 1350 answer : c" | a = 100 / 10
b = 150 * a
c = b - 150
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a ) 35 , b ) 50 , c ) 88 , d ) 22 , e ) 12 | a | divide(divide(subtract(125, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 125 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 15 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 15 ) m / sec = ( 25 / 3 ) m / sec . [ ( 25 / 3 ) * ( 18 / 5 ) ] km / hr = 30 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 30 = = > x = 35 km / hr . answer : a" | a = 5 * const_0_2778
b = a * 5
c = 125 - b
d = c / 5
e = d / const_0_2778
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a ) 90 , b ) 98 , c ) 106 , d ) 120 , e ) 136 | a | add(multiply(negate(15), power(subtract(add(3, 2), 3), const_2)), 150) | an object thrown directly upward is at a height of h feet after t seconds , where h = - 15 ( t - 3 ) ^ 2 + 150 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ? | we see that h will be a maximum h = 150 when t - 3 = 0 , that is when t = 3 . at t = 5 , h = - 15 ( 5 - 3 ) ^ 2 + 150 = - 15 ( 4 ) + 150 = 90 the answer is a . | a = negate * (
b = 3 + 2
c = b - 3
d = c ** 2
e = a + d
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a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 7 | d | subtract(multiply(subtract(13, 1), add(6, 1)), multiply(13, 6)) | the average of 6 observations is 13 . a new observation is included and the new average is decreased by 1 . the seventh observation is ? | let seventh observation = x . then , according to the question we have = > ( 78 + x ) / 7 = 12 = > x = 6 . hence , the seventh observation is 6 . answer : d | a = 13 - 1
b = 6 + 1
c = a * b
d = 13 * 6
e = c - d
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a ) 4 , b ) 2 , c ) 1 / 4 , d ) 2 / 5 , e ) 5 / 6 | d | divide(1, divide(add(subtract(9, 1), 4), 4)) | the vertex of a rectangle are ( 1 , 0 ) , ( 9 , 0 ) , ( 1 , 4 ) and ( 9 , 4 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ? | "if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 5 , 2 ) . the slope of a line passing through ( 0,0 ) and ( 5 , 2 ) is 2 / 5 . the answer is d ." | a = 9 - 1
b = a + 4
c = b / 4
d = 1 / c
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a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | e | add(add(divide(subtract(100, 40), 5), 1), 1) | a test has 100 questions . each question has 5 options , but only 1 option is correct . if test - takers mark the correct option , they are awarded 1 point . however , if an answer is incorrectly marked , the test - taker loses 0.25 points . no points are awarded or deducted if a question is not attempted . a certain group of test - takers attempted different numbers of questions , but each test - taker still received the same net score of 40 . what is the maximum possible number of such test - takers ? | "a correct answers get you 1 point , an incorrect answer gets you minus 1 / 4 point and a skipped question gets you 0 points . since there are 200 total questions , there are a variety of ways to get a total of 40 points . let c be the number of correct answers and let i be the number of incorrect answers . to get 40 points , a test taker must have at least 40 correct answers . then c = > 40 . for every correct question above 40 , the test taker has 4 incorrect answers . then , the i = 4 * ( c - 40 ) . also , i + c < = 100 . thus 5 c < = 260 and so c < = 52 . then 40 < = c < = 52 and c can have 13 possible values . the answer is e ." | a = 100 - 40
b = a / 5
c = b + 1
d = c + 1
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a ) 4554 , b ) 4657 , c ) 8457 , d ) 3458 , e ) 5254 | a | add(add(multiply(multiply(const_1000, const_1), const_4), divide(multiply(const_1000, const_1), const_2)), multiply(add(const_4, const_1), const_10)) | a women in her conversation said ` ` if u reverse my own age , in figures represent my husbands age . he is of course senior to me and difference between our age is one 0 ne - eleventh of their sum . what is the woman ' s and her husbands age ? | let man , s age be xy the woman ' s age is yx then 10 x + y - 10 y - x = 1 / 11 ( 10 x + y + 10 y + x ) 9 x - 9 y = x + y 8 x = 10 y x / y = 10 / 8 = 5 / 4 so xy = 54 and yx = 45 answer : a | a = 1000 * 1
b = a * 4
c = 1000 * 1
d = c / 2
e = b + d
f = 4 + 1
g = f * 10
h = e + g
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a ) rs 840 , b ) rs 320 , c ) rs 810 , d ) rs 680 , e ) none of these | c | multiply(divide(multiply(9, 6), multiply(4, 10)), 600) | if 4 men working 10 hours a day earn rs . 600 per week , then 9 men working 6 hours a day will earn how much per week ? | "explanation : ( men 4 : 9 ) : ( hrs / day 10 : 6 ) : : 600 : x hence 4 * 10 * x = 9 * 6 * 600 or x = 9 * 6 * 600 / 4 * 10 = 810 answer : c" | a = 9 * 6
b = 4 * 10
c = a / b
d = c * 600
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a ) 1 / 6 , b ) 2 / 9 , c ) 5 / 6 , d ) 7 / 9 , e ) 8 / 9 | c | divide(const_5, 6) | a dog breeder currently has 9 breeding dogs . 6 of the dogs have exactly 1 littermate , and 3 of the dogs have exactly 2 littermates . if 2 dogs are selected at random , what is the probability q that both selected dogs are not littermates ? | "we have three pairs of dogs for the 6 with exactly one littermate , and one triplet , with each having exactly two littermates . so , in fact there are two types of dogs : those with one littermate - say a , and the others with two littermates - b . work with probabilities : choosing two dogs , we can have either one dog of type b or none ( we can not have two dogs both of type b ) . the probability of choosing one dog of type b and one of type a is 3 / 9 * 6 / 8 * 2 = 1 / 2 ( the factor of 2 for the two possibilities ba and ab ) . the probability of choosing two dogs of type a which are not littermates is 6 / 9 * 4 / 8 = 1 / 3 ( choose one a , then another a which is n ' t the previous one ' s littermate ) . the required probability is 1 / 2 + 1 / 3 = 5 / 6 . find the probability for the complementary event : choose aa or bb . probability of choosing two dogs of type a who are littermates is 6 / 9 * 1 / 8 = 1 / 12 . probability of choosing two dogs of type b ( who necessarily are littermates ) is 3 / 9 * 2 / 8 q = 1 / 12 . again , we obtain 1 - ( 1 / 12 + 1 / 12 ) = 5 / 6 . answer : c" | a = 5 / 6
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a ) 57 , b ) 67 , c ) 45 , d ) 47 , e ) 97 | c | sqrt(multiply(20.25, const_100)) | a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 20.25 , the number of the member is the group is : | "money collected = ( 20.25 x 100 ) paise = 2025 paise numbers of members = 2025 squareroot = 45 answer c" | a = 20 * 25
b = math.sqrt(a)
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a ) 187 m , b ) 350 m , c ) 267 m , d ) 287 m , e ) 675 m | e | subtract(multiply(speed(300, 12), 39), 300) | a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 12 sec . what is the length of the platform ? | "speed = 300 / 12 = 25 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 25 = > x = 675 m . answer : e" | a = speed * (
b = a - 39
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a ) s . 5266 , b ) s . 5269 , c ) s . 5228 , d ) s . 5218 , e ) s . 5526 | e | divide(8510, add(const_1, divide(multiply(6, 9), const_100))) | mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8510 / - . find out the amount that he borrowed . | "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8510 = p + ( 54 p / 100 ) 8510 = ( 100 p + 54 p ) / 100 8510 = 154 p / 100 p = ( 8510 * 100 ) / 154 = rs . 5525.974 answer : e" | a = 6 * 9
b = a / 100
c = 1 + b
d = 8510 / c
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a ) 40 , b ) 44 , c ) 80 , d ) 88 , e ) 48 | d | divide(subtract(power(24, const_2), 400), const_2) | if the sum of two numbers is 24 and the sum of their squares is 400 , then the product of the numbers is | "according to the given conditions x + y = 24 and x ^ 2 + y ^ 2 = 400 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 24 ^ 2 = 400 + 2 xy so xy = 176 / 2 = 88 answer : d" | a = 24 ** 2
b = a - 400
c = b / 2
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a ) 4 , b ) 8 , c ) 12 , d ) 9 , e ) 18 | d | divide(power(6, const_2), 4) | if the area of a square with sides of length 6 centimeters is equal to the area of a rectangle with a width of 4 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 6 ^ 2 = l * 4 = > l = 36 / 4 = 9 answer d" | a = 6 ** 2
b = a / 4
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a ) two , b ) three , c ) four , d ) five , e ) six | d | add(floor(subtract(divide(subtract(17, 5), 5), divide(subtract(1, 5), 5))), const_2) | how many integer values t are there for x such that 1 < 3 x + 5 < 17 ? | 1 < 3 x + 5 < 17 = > - 4 < 3 x < 12 = > - 4 / 3 < x < 4 x can take integer values t - 10 , 1 , 2 , 3 answer d | a = 17 - 5
b = a / 5
c = 1 - 5
d = c / 5
e = b - d
f = math.floor(e)
g = f + 2
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a ) 45 , b ) 48 , c ) 51 , d ) 42 , e ) 54 | d | add(divide(117, 3), 3) | the sum of 3 consecutive multiples of 3 is 117 . what is the largest number ? | let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 117 9 x = 108 x = 12 largest number = 3 x + 6 = 42 answer : d | a = 117 / 3
b = a + 3
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a ) 80 % , b ) 85 % , c ) 82 % , d ) 70 % , e ) 75 % | a | divide(const_100, divide(45, subtract(81, 45))) | if the cost price of 81 chocolates is equal to the selling price of 45 chocolates , the gain percent is : | "explanation : solution : let c . p . of each chocolate be re . 1 . then , c . p . of 45 chocolates = rs . 45 ; s . p . of 45 chocolates = rs . 81 . . ' . gain % = 36 * 100 / 45 = 80 % answer : a" | a = 81 - 45
b = 45 / a
c = 100 / b
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