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a ) 12 , b ) 13 , c ) 35 , d ) 40 , e ) 59 | d | add(27, 11) | a number when divided by a divisor leaves a remainder of 27 . when twice the original number is divided by the same divisor , the remainder is 11 . what is the value of the divisor ? | "let the number is n , the divisor = d , i will make the two equations - n = xd + 27 2 n = yd + 11 where x and y are integers solving them : d ( y - 2 x ) = 40 as d is also integer and 40 is a prime number , the d should be 40 to satisfy the above equation . hence answer is ' d '" | a = 27 + 11
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a ) 1 : 7 , b ) 1 : 8 , c ) 1 : 3 , d ) 1 : 1 , e ) 1 : 216 | e | divide(power(1, 6), power(6, 6)) | the triplicate ratio of 1 : 6 is ? | "1 ^ 3 : 6 ^ 3 = 1 : 216 answer : e" | a = 1 ** 6
b = 6 ** 6
c = a / b
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a ) 232 m , b ) 288 m , c ) 324 m , d ) 231 m , e ) 236 m | b | subtract(224, multiply(14, speed(224, 32))) | for a race a distance of 224 meters can be covered by p in 14 seconds and q in 32 seconds . by what distance does p defeat q eventually ? | explanation : this is a simple speed time problem . given conditions : = > speed of p = 224 / 14 = 16 m / s = > speed of q = 224 / 32 = 7 m / s = > difference in time taken = 18 seconds therefore , distance covered by p in that time = 16 m / s x 18 seconds = 288 metres answer : b | a = 14 * speed
b = 224 - a
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a ) 661 , b ) 600 , c ) 620 , d ) 616 , e ) none of these | d | multiply(divide(multiply(2200, 40), const_100), divide(subtract(const_100, 30), const_100)) | in an office , totally there are 2200 employees and 40 % of the total employees are males . 30 % of the males in the office are at - least 50 years old . find the number of males aged below 50 years ? | "number of male employees = 2200 * 40 / 100 = 880 required number of male employees who are less than 50 years old = 880 * ( 100 - 30 ) % = 880 * 75 / 100 = 616 . answer : d" | a = 2200 * 40
b = a / 100
c = 100 - 30
d = c / 100
e = b * d
|
a ) 4 , b ) 5 , c ) 9 , d ) 8 , e ) 10 | a | subtract(subtract(divide(50, 5), 3), 3) | the sum of ages of 5 children born at theintervals of 3 years each is 50 years . what is the age of the youngest child ? | "the age of children x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 50 5 x = 20 , x = 4 . correct answer ( a )" | a = 50 / 5
b = a - 3
c = b - 3
|
a ) 131 , b ) 135 , c ) 139 , d ) 147 , e ) 188 | e | divide(add(multiply(const_2, 275), 35), const_3) | if jake loses 35 pounds , he will weigh thrice as much as his sister . together they now weigh 275 pounds . what is jake ' s present weight , in pounds ? | j = jake β s current weight , in pounds s = sister β s current weight , in pounds we are told that β if jake loses 8 pounds , he will weigh twice as much as his sister . we put this into an equation : j β 35 = 3 s j = 3 s + 35 ( equation 1 ) next , we are told that β together they now weigh 275 pounds . β we can also put this into an equation . j + s = 275 ( equation 2 ) to solve this equation , we can substitute 2 s + 8 from equation 1 for the variable j in equation 2 : 3 s + 35 = 275 - s 4 s = 240 s = 60 j + 60 = 275 j = 215 answer : e | a = 2 * 275
b = a + 35
c = b / 3
|
a ) 42 minutes , b ) 14 minutes , c ) 54 minutes , d ) 40 minutes 20 seconds , e ) none of these | c | multiply(divide(900, subtract(add(40, 30), 20)), const_3) | pipe a fills a tank of capacity 900 liters at the rate of 40 liters a minute . another pipe b fills the same tank at the rate of 30 liters a minute . a pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute . if pipe a is kept open for a minute and then closed and pipe b is open for a minute and then closed and then pipe c is open for a minute and then closed and the cycle is repeated , when will the tank be full ? | "in one cycle they fill 40 + 30 - 20 = 50 liters 900 = 50 * n = > n = 18 here n = number of cycles . total time = 18 * 3 = 54 as in one cycle there are 3 minutes . thus 54 minutes answer : c" | a = 40 + 30
b = a - 20
c = 900 / b
d = c * 3
|
a ) 15 , b ) 30 , c ) 32 , d ) 33 , e ) 46 | c | subtract(divide(subtract(subtract(100, 6), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(100, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | how many even number in the range between 6 to 100 inclusive are not divisible by 3 | "we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 6 , 8,10 , 12,14 . . . 100 using ap formula , we can say 100 = 10 + ( n - 1 ) * 2 or n = 48 . for 6 , 6 , 12,18 , . . . 96 using ap formula , we can say 96 = 12 + ( n - 1 ) * 6 or n = 16 . hence , only divisible by 2 but not 3 = 48 - 16 = 32 . hence , answer c" | a = 100 - 6
b = a - 2
c = b / 2
d = 100 - 2
e = 3 * 4
f = d - e
g = f - 3
h = g - 3
i = h / 3
j = i / 2
k = c - j
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(multiply(6, 8), multiply(3, 6)), add(6, 4)) | 6 workers should finish a job in 8 days . after 3 days came 4 workers join them . how many days z do they need to finish the same job ? | "let rate of one worker be r = > ( 6 * r ) * 8 = 1 ( rate * time = work ) = > r = 1 / 48 = > work remaining after 3 days 1 - ( 3 * 6 ) / 48 = 30 / 48 after 4 ppl joined in ( ( 6 + 4 ) * time ) / 48 = 30 / 48 time z = 3 days to finish the task imo a" | a = 6 * 8
b = 3 * 6
c = a - b
d = 6 + 4
e = c / d
|
a ) 70 , b ) 96 , c ) 108 , d ) 120 , e ) 150 | d | divide(subtract(32, 20), subtract(divide(20, const_100), divide(10, const_100))) | of the diplomats who attended a summit conference : 20 spoke latin , 32 did not speak russian and 20 % of the diplomats spoke neither latin nor russian . if 10 % of the diplomats spoke both latin and russian , then how many diplomats attended the conference ? | 2 x 2 matrix will be the easiest way to calculate this . text in black : given statements text in red : calculated values thus d = 120 is the correct answer | a = 32 - 20
b = 20 / 100
c = 10 / 100
d = b - c
e = a / d
|
a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 3 / 4 , e ) 3 / 5 | d | divide(const_2, 3) | if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 3 children have two children of the same sex and one of the opposite sex ? | "let boy be represented by b and girl by g . the possible outcomes of two children of same sex and one of opposite sex can be : bbg or gbb or bgb or ggb or bgg or gbg ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 / 2 * 1 / 2 ) 1 / 8 + 1 / 8 + 1 / 8 + 1 / 8 + 1 / 8 + 1 / 8 = 6 / 8 = 3 / 4 answer : ( d )" | a = 2 / 3
|
a ) 75.45 , b ) 76.45 , c ) 77.45 , d ) 74.45 , e ) 73.45 | a | divide(add(multiply(12, subtract(12, 20)), multiply(10, subtract(12, 30))), add(12, 10)) | a man buys 12 lts of liquid which contains 20 % of the liquid and the rest is water . he then mixes it with 10 lts of another mixture with 30 % of liquid . what is the % of water in the new mixture ? | "20 % in 12 lts is 2.4 . so water = 12 - 2.4 = 9.6 lts . 30 % of 10 lts = 3 . so water in 2 nd mixture = 10 - 3 = 7 lts . now total quantity = 12 + 10 = 22 lts . total water in it will be 9.6 + 7 = 16.6 lts . % of water = ( 100 * 16.6 ) / 22 = 75.45 answer : a" | a = 12 - 20
b = 12 * a
c = 12 - 30
d = 10 * c
e = b + d
f = 12 + 10
g = e / f
|
a ) 25 days , b ) 30 days , c ) 20 days , d ) 15 days , e ) 10 days | b | inverse(add(subtract(divide(const_1, 30), subtract(divide(const_1, 20), divide(const_1, 40))), subtract(divide(const_1, 20), divide(const_1, 40)))) | a and b can do a work in 20 days , b and c can do it in 30 days ; a , b and c together can finish it in 40 days . a and c together will do it in ? | a + b + c 1 day work = 1 / 40 a + b 1 day work = 1 / 20 b + c 1 day work = 1 / 30 a + c 1 day work = 2 * 1 / 40 - 1 / 20 + 1 / 30 = 1 / 30 a and c together will do the work in 30 days . answer is b | a = 1 / 30
b = 1 / 20
c = 1 / 40
d = b - c
e = a - d
f = 1 / 20
g = 1 / 40
h = f - g
i = e + h
j = 1/(i)
|
a ) 213444 , b ) 214344 , c ) 214434 , d ) 231444 , e ) 233444 | a | multiply(multiply(multiply(power(add(const_3, const_4), const_2), power(divide(divide(36, const_3), const_2), const_2)), power(const_2, const_2)), power(const_3, const_2)) | the least perfect square , which is divisible by each of 21 , 36 and 66 is : | "explanation : l . c . m . of 21 , 36 , 66 = 2772 . now , 2772 = 2 x 2 x 3 x 3 x 7 x 11 to make it a perfect square , it must be multiplied by 7 x 11 . so , required number = 22 x 32 x 72 x 112 = 213444 answer is a" | a = 3 + 4
b = a ** 2
c = 36 / 3
d = c / 2
e = d ** 2
f = b * e
g = 2 ** 2
h = f * g
i = 3 ** 2
j = h * i
|
a ) 13 , b ) 3 , c ) 15 , d ) 1 , e ) 2 | d | gcd(gcd(subtract(subtract(172, 87), 54), subtract(87, 54)), subtract(172, 87)) | find the greatest number that will divide 54 , 87 and 172 so as to leave the same remainder in each case | explanation : required number = ( 87 - 54 ) , ( 172 - 87 ) , ( 172 - 54 ) = h . c . f of 33,85 and 118 is 1 answer : option d | a = 172 - 87
b = a - 54
c = 87 - 54
d = math.gcd(b, c)
e = 172 - 87
f = math.gcd(d, e)
|
a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 30 , d ) 13 / 18 , e ) 11 / 15 | d | subtract(1, add(multiply(inverse(3), inverse(const_2)), inverse(9))) | sally has a gold credit card with a certain spending limit , and a platinum card with twice the spending limit of the gold card . currently , she has a balance on her gold card that is 1 / 3 of the spending limit on that card , and she has a balance on her platinum card that is 1 / 9 of the spending limit on that card . if sally transfers the entire balance on her gold card to her platinum card , what portion of her limit on the platinum card will remain unspent ? | "let s assume the platinum card spending limit = x gold card spending limit will be = x / 2 balance on gold card is = x / 2 * 1 / 3 = x / 6 platinum card unspent limit is = x - 1 / 9 x = 8 / 9 x so if gold card balance is transferred then the rest unspent will be 8 / 9 x - x / 6 = 13 / 18 x so the ans is d" | a = 1/(3)
b = 1/(2)
c = a * b
d = 1/(9)
e = c + d
f = 1 - e
|
a ) 60 , b ) 77 , c ) 269 , d ) 26 , e ) 91 | a | divide(multiply(52.8, const_100), subtract(const_100, 12)) | the number which exceeds 12 % of it by 52.8 is : | "explanation : let the number be x . then , x β 12 % of x = 52.8 x β ( 12 / 100 ) x = 52.8 x ( 1 β 12 / 100 ) = 52.8 ( 88 / 100 ) x = 52.8 x = ( 100 x 52.8 ) / 88 = 60 answer : a" | a = 52 * 8
b = 100 - 12
c = a / b
|
a ) a ) 45 , b ) b ) 34 , c ) c ) 50 , d ) d ) 67 , e ) e ) 100 | a | divide(divide(multiply(360, 6), 12), const_4) | according to the directions on the can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 3 cans of water to make orange juice . how many 12 ounces cans of the concentrate are required to prepare 360 6 ounces servings of orange juice ? | "its a . total juice rquired = 360 * 6 = 2160 ounce 12 ounce concentate makes = 12 * 4 = 48 ounce juice total cans required = 2160 / 48 = 45 . answer a" | a = 360 * 6
b = a / 12
c = b / 4
|
a ) 2.91 , b ) 2.65 , c ) 2.938 , d ) 2.986 , e ) 2.999 | b | add(add(2.01, divide(3, const_1000)), divide(34, const_1000)) | solution for 2.01 + . 3 + . 34 | "2.01 + . 3 + . 34 = 0 0 = 0 - 2.01 - 0.3 - 0.34 0 = - 2.65 answer : b" | a = 3 / 1000
b = 2 + 1
c = 34 / 1000
d = b + c
|
a ) 1 , b ) 1.5 , c ) 2 , d ) 2.5 , e ) 3 | b | multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 15)), 6)), 15) | two pipes p and q can fill a cistern in 12 and 15 minutes respectively . both are opened together , but at the end of 6 minutes the first is turned off . how many more minutes will it take for the cistern to fill after the first pipe is turned off ? | "let x be the total time it takes for the cistern to fill . 6 / 12 + x / 15 = 1 x / 15 = 1 / 2 x = 7.5 after the first pipe is turned off , it takes 1.5 more minutes to fill the cistern . the answer is b ." | a = 1 / 12
b = 1 / 15
c = a + b
d = c * 6
e = 1 - d
f = e * 15
|
a ) 0.5 , b ) 0.625 , c ) 0.75 , d ) 0.875 , e ) 1.0 | b | divide(divide(add(1, const_2.0), const_2), divide(add(1, 1), 7)) | in the xy - coordinate system , what is the slope of the line that goes through the point ( 1 , 1 ) and is equidistant from the two points p = ( 3 , 7 ) and q = ( 9 , 11 ) ? | "first , get the middle coordinate between ( 3,7 ) and ( 9,11 ) . . . x = 3 + ( 9 - 3 ) / 2 = 6 y = 7 + ( 11 - 7 ) / 2 = 9 second , get the slope of ( 9,6 ) and ( 1,1 ) . m = 6 - 1 / 9 - 1 = 5 / 8 = 0.625 answer : b" | a = 1 + 2
b = a / 2
c = 1 + 1
d = c / 7
e = b / d
|
a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 14 % | e | divide(divide(multiply(40, 60), multiply(80, 60)), add(divide(multiply(40, 60), multiply(80, 60)), 12)) | a certain car can travel 40 minutes on a gallon of gasoline at 60 miles per hour . if the car had started with a full tank and had 12 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 80 miles at 60 mph ? | "total time for travelling 80 miles @ 60 mph = 80 / 60 = 4 / 3 hour = 80 minutes . given , the car uses 1 gallon for every 40 minutes of driving @ 60 mph . thus in 80 minutes it will use = 2 gallons . thus , full tank = 2 + 12 = 14 gallons - - - > 2 / 14 = 14 % of the fuel used . e is the correct answer ." | a = 40 * 60
b = 80 * 60
c = a / b
d = 40 * 60
e = 80 * 60
f = d / e
g = f + 12
h = c / g
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a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | multiply(divide(7.5, 5), multiply(const_4, 2)) | the ratio of flour to water to sugar in a recipe is 11 : 5 : 2 . the ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar . if the new recipe calls for 7.5 cups of water , how much sugar is required ? | the ratio of flour to water is 22 : 5 . the ratio of flour to sugar is 5.5 : 2 = 22 : 8 . the new ratio of flour to water to sugar is 22 : 5 : 8 if we need 7.5 cups of water , then we need 12 cups of sugar . the answer is e . | a = 7 / 5
b = 4 * 2
c = a * b
|
a ) 187 km , b ) 480 km , c ) 278 km , d ) 297 km , e ) 600 km | e | divide(1, 2) | with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 5 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ? | "let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 5 2 x / 15 - x / 8 = 5 = > x = 600 km . answer : e" | a = 1 / 2
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a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | b | multiply(divide(subtract(subtract(add(const_1, divide(60, const_100)), const_1), divide(20, const_100)), add(const_1, divide(60, const_100))), const_100) | a merchant marks goods up by 60 % and then offers a discount on the marked price . the profit that the merchant makes after offering the discount is 20 % . what % discount did the merchant offer ? | "let p be the original price of the goods and let x be the rate after the markup . ( 1.6 p ) * x = 1.2 p x = 1.2 / 1.6 = 0.75 which is a discount of 25 % . the answer is b ." | a = 60 / 100
b = 1 + a
c = b - 1
d = 20 / 100
e = c - d
f = 60 / 100
g = 1 + f
h = e / g
i = h * 100
|
a ) 86 , b ) 165 , c ) 76 , d ) 125 , e ) 26 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(65, 3)) | a student scored an average of 65 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | "given m + p + c = 65 * 3 = 195 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 195 = 125 answer : d" | a = 90 * 2
b = 70 * 2
c = a + b
d = 65 * 3
e = c - d
|
a ) 150 , b ) 200 , c ) 250 , d ) 245 , e ) 225 | c | add(multiply(15, 15), 25) | find the number of trailing zeros in the product of ( 1 ^ 1 ) * ( 5 ^ 5 ) * ( 10 ^ 10 ) * ( 15 ^ 15 ) * ( 20 ^ 20 ) * ( 25 ^ 25 ) β¦ β¦ β¦ . * ( 50 ^ 50 ) . | looking at the numbers it looks like ( 1 x 5 ) ^ 5 ( 2 x 5 ) ^ 10 . . . ( 10 x 5 ) ^ 50 1 . determine the limiting factor . is it 2 or is it 5 ? we know that all the numbers are multiple of 5 but not of 2 . thus , the limiting factor in this case is 2 . let ' s drop all the 5 . then , we count factors of 2 of even multiples . 2 ^ 10 = 10 4 ^ 20 = 20 + 20 6 ^ 30 = 30 8 ^ 40 = 40 + 40 + 40 10 ^ 50 = 50 250 answer : c | a = 15 * 15
b = a + 25
|
a ) a ) 2 , b ) b ) 4 , c ) c ) 5 , d ) d ) 7 , e ) e ) 8 | b | add(divide(subtract(multiply(floor(divide(8, 2)), 2), multiply(add(floor(divide(2, 2)), const_1), 2)), 2), const_1) | how many numbers from 2 to 8 are exactly divisible by 2 ? | "2 / 2 = 1 and 8 / 2 = 4 4 - 1 = 3 3 + 1 = 4 numbers . answer : b" | a = 8 / 2
b = math.floor(a)
c = b * 2
d = 2 / 2
e = math.floor(d)
f = e + 1
g = f * 2
h = c - g
i = h / 2
j = i + 1
|
a ) 215 , b ) 235 , c ) 255 , d ) 275 , e ) 295 | e | subtract(multiply(divide(multiply(45, const_1000), const_3600), 30), 80) | the length of a bridge in meters , which a train 80 - meters long and traveling at 45 km / hr can cross in 30 seconds is ? | "45 km / h = 45000 m / 3600 s = 12.5 m / s in 30 seconds , the train can go 30 ( 12.5 ) = 375 meters let x be the length of the bridge . x + 80 = 375 meters x = 295 meters the answer is e ." | a = 45 * 1000
b = a / 3600
c = b * 30
d = c - 80
|
a ) 8 , b ) 6 , c ) 2 , d ) 4 , e ) 1 | c | divide(divide(divide(divide(divide(3872, const_3), const_3), const_4), const_4), const_4) | find the smallest number which should be divided with 3872 to make it a perfect square . | "3872 = 11 * 11 * 2 * 2 * 2 * 2 * 2 required smallest number = 2 2 is the smallest number which should be divided with 3872 to make it a perfect square . answer : c" | a = 3872 / 3
b = a / 3
c = b / 4
d = c / 4
e = d / 4
|
a ) 560 , b ) 616 , c ) 672 , d ) 200 , e ) 1024 | d | add(multiply(divide(const_3, const_2), const_100), add(multiply(add(const_2, const_3), 59.80), const_3)) | x and y are both integers . if x / y = 59.80 , then what is the sum of all the possible two digit remainders of x / y ? | "remainder = 0.80 = - - > 80 / 100 - - > can be written as ( 80 / 4 ) / ( 100 / 4 ) = 20 / 25 so remainders can be 20 , 40 , 60 , 80 we need the sum of only 2 digit remainders - - > 20 + 40 + 60 + 80 = 200 answer : d" | a = 3 / 2
b = a * 100
c = 2 + 3
d = c * 59
e = d + 3
f = b + e
|
a ) 21 , b ) 26 , c ) 30 , d ) 18 , e ) 16 | d | add(add(11, divide(subtract(11, 5), 2)), 4) | a , b , c , d and e are 5 consecutive points on a straight line . if bc = 2 cd , de = 4 , ab = 5 and ac = 11 , what is the length of ae ? | "a - - - - - b - - - - c - - - - - - - - d - - - - e given : de = 4 - ( i ) ab = 5 - ( ii ) ac = 11 - ( iii ) bc = 2 cd - ( iv ) from ( ii ) and ( iii ) , bc = 6 from ( iv ) , cd = 3 length of ae = ac + cd + de = 11 + 3 + 4 = 18 correct option : d" | a = 11 - 5
b = a / 2
c = 11 + b
d = c + 4
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a ) 1 / 4 , b ) 4 / 5 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | e | subtract(divide(lcm(const_2, const_3), 2.8), const_2) | on a partly cloudy day , derek decides to walk back from work . when it is sunny , he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction t of the total distance did he cover while the sun was shining on him ? | if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 / 4 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or 12 t . therefore , total distance covered while the sun was shining over him is t = 2 / 14 = 1 / 7 . answer : e | a = math.lcm(2, 3)
b = a / 2
c = b - 2
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a ) 229 , b ) 288 , c ) 600 , d ) 888 , e ) 1400 | e | multiply(divide(multiply(35, add(const_3, 2)), subtract(40, 35)), 40) | a train leaves delhi at 9 a . m . at a speed of 35 kmph . another train leaves at 2 p . m . at a speed of 40 kmph on the same day and in the same direction . how far from delhi , will the two trains meet ? | "d = 35 * 5 = 175 rs = 40 β 35 = 5 t = 175 / 5 = 35 d = 40 * 35 = 1400 km answer : e" | a = 3 + 2
b = 35 * a
c = 40 - 35
d = b / c
e = d * 40
|
a ) rs . 10000 , b ) rs . 6000 , c ) rs . 8000 , d ) rs . 4000 , e ) rs . 2000 | a | multiply(2000, add(const_4, const_1)) | rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 2000 . then his monthly salary is | sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 2000 Γ’ β‘ β 20 / 100 x = 2000 Γ’ β‘ β x = 2000 Γ£ β 5 = 10000 . answer a | a = 4 + 1
b = 2000 * a
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a ) 16.5 % , b ) 10 % , c ) 35 % , d ) 55 % , e ) 65 % | b | multiply(divide(subtract(3.5, 3), subtract(7.5, 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 7.5 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 + 0.075 * fringe benefit costs ; so , 35 ( s + f ) = 30 s + 75 f - - > s = 9 f - - > f / s = 1 / 9 - - > f / ( s + f ) = 1 / ( 1 + 9 ) = 1 / 10 = 0.1 . answer : b ." | a = 3 - 5
b = 7 - 5
c = a / b
d = c * 100
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a ) 16 , b ) 4 , c ) 15 , d ) 18 , e ) 12 | b | subtract(const_60, multiply(const_60, divide(42, 45))) | excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 42 kmph . of how many minutes does the train stop per hour ? | "t = 3 / 45 * 60 = 4 answer : b" | a = 42 / 45
b = const_60 * a
c = const_60 - b
|
a ) . 8 , b ) . 09 , c ) . 009 , d ) . 0009 , e ) none of them | a | divide(divide(008, const_1000), divide(01, const_100)) | . 008 / ? = . 01 | "let . 008 / x = . 01 ; then x = . 008 / . 01 = . 8 / 1 = . 8 answer is a" | a = 8 / 1000
b = 1 / 100
c = a / b
|
a ) $ 10700 , b ) $ 17000 , c ) $ 180000 , d ) $ 1700 , e ) $ 170000 | e | divide(17000, subtract(1, add(add(divide(1, 5), divide(1, 10)), divide(3, 5)))) | a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 17000 left with him . find salary . . | "[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 17000 , = [ 1 - 9 / 10 ] * total salary = $ 17000 , total salary = $ 17000 * 10 = $ 180000 , correct answer ( e )" | a = 1 / 5
b = 1 / 10
c = a + b
d = 3 / 5
e = c + d
f = 1 - e
g = 17000 / f
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a ) 999 , b ) 7811 , c ) 6555 , d ) 9200 , e ) 7920 | e | lcm(8, 11) | what is the lowest positive integer that is divisible by 8 through 11 , inclusive ? | "the integer should be divisible by : 8 , 9 , 10 and 11 . the least common multiple of these integers is lcm = 8 * 9 * 10 * 11 = 7920 answer : e" | a = math.lcm(8, 11)
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a ) 91.5 , b ) 87.1 , c ) 91.7 , d ) 91.3 , e ) 91.1 | b | multiply(multiply(const_2, divide(multiply(subtract(20, const_3), const_2), add(const_4, const_3))), 20) | the sector of a circle has radius of 20 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 20 ) + 2 ( 20 ) = 47.1 + 40 = 87.1 cm answer : b" | a = 20 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 20
|
a ) $ 144 , b ) $ 130 , c ) $ 80 , d ) $ 110 , e ) $ 129 | e | subtract(multiply(19, 7), multiply(4, 1)) | tara bought 19 cartons of ice cream and 4 cartons of yoghurt . each carton of ice cream cost $ 7 and each carton of yoghurt cost $ 1 . how much more did tara spend on ice cream than on yoghurt ? | step 1 : find the cost of the ice cream . 19 Γ $ 7 = $ 133 step 2 : find the cost of the yoghurt . 4 Γ $ 1 = $ 4 step 3 : find how much more the ice cream cost than the yoghurt . $ 133 β $ 4 = $ 129 tara spent $ 129 more on ice cream . answer is e . | a = 19 * 7
b = 4 * 1
c = a - b
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a ) 41 , b ) 42 , c ) 36 , d ) 44 , e ) 45 | c | divide(factorial(9), multiply(factorial(subtract(9, const_2)), factorial(const_2))) | if 9 boys meet at a reunion and each boy shakes hands exactly once with each of the others , then what is the total number of handshakes | "n ( n - 1 ) / 2 = 9 * 8 / 2 = 36 answer : c" | a = math.factorial(9)
b = 9 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
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a ) 3 : 7 , b ) 4 : 9 , c ) 20 : 13 , d ) 5 : 7 , e ) 6 : 11 | c | divide(multiply(50, 8), multiply(65, 4)) | car a runs at the speed of 50 km / hr and reaches its destination in 8 hours . car b runs at the speed of 65 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ? | "car a travels 50 Γ 8 = 400 km car b travels 65 Γ 4 = 260 km the ratio is 400 : 260 = 40 : 26 = 20 : 13 the answer is c ." | a = 50 * 8
b = 65 * 4
c = a / b
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a ) 210 , b ) 252 , c ) 280 , d ) 320 , e ) 300 | e | divide(multiply(multiply(25, 48), 60), multiply(multiply(8, 6), 5)) | a grocer is storing soap boxes in cartons that measure 25 inches by 48 inches by 60 inches . if the measurement of each soap box is 8 inches by 6 inches by 5 inches , then what is the maximum number of soap boxes that can be placed in each carton ? | "however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = 25 * 48 * 60 dimensions of the soap = 5 * 6 * 8 we get = 5 x 6 x 10 = 300 so the question is why this particular arrangement , in order to maximize number of soaps we need to minimize the space wasted and this is the only config where we dont waste any space so we can expect the maximum number the answer is ( e )" | a = 25 * 48
b = a * 60
c = 8 * 6
d = c * 5
e = b / d
|
a ) 56780 , b ) 78910 , c ) 97479 , d ) 97918 , e ) 97920 | c | divide(multiply(const_1000, const_1000), const_10) | what is the greatest 5 - digit number when divided by 6 , 7 , 8 , 9 , and 10 leaves a remainder of 4 , 5 , 6 , 7 , and 9 respectively ? | when you divide a positive integer by 10 , the remainder will just be the units digit . we know the remainder is 9 when we divide by 10 , so c is the only possible answer . | a = 1000 * 1000
b = a / 10
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a ) $ 550 , b ) $ 600 , c ) $ 500 , d ) $ 400 , e ) $ 450 | c | multiply(const_2.0, divide(multiply(50, divide(2, 3)), divide(const_1, 3))) | a collection of books went on sale , and 2 / 3 of them were sold for $ 5 each . if none of the 50 remaining books were sold , what was the total amount received for the books that were sold ? | "if 50 books constitute 1 / 3 rd of the total , then 2 / 3 rd of the total = 100 books amount received for sold books = 100 * 5 = $ 500 answer : c" | a = 2 / 3
b = 50 * a
c = 1 / 3
d = b / c
e = 2 * 0
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a ) 2.21 , b ) 2.45 , c ) 2.67 , d ) 2.83 , e ) 2.95 | c | multiply(divide(subtract(6, 2), add(add(6, 2), subtract(6, 2))), add(6, 2)) | a rower can row 6 km / h in still water . when the river is running at 2 km / h , it takes the rower 1 hour to row to big rock and back . how many kilometers is it to big rock ? | "let x be the distance to big rock . time = x / 4 + x / 8 = 1 x = 32 / 12 = 2.67 km the answer is c ." | a = 6 - 2
b = 6 + 2
c = 6 - 2
d = b + c
e = a / d
f = 6 + 2
g = e * f
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a ) 6 rs , b ) 7 rs , c ) 8 rs , d ) 5 rs , e ) 1 rs | d | divide(75, multiply(const_3, 5)) | 5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 75 only . men Γ’ β¬ β’ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 75 rs . 5 m + 5 m + 5 m - - - - - 75 rs . 15 m - - - - - - 75 rs . = > 1 m = 5 rs . answer : d" | a = 3 * 5
b = 75 / a
|
a ) 2 , b ) 1.15 , c ) 2.01 , d ) 2.06 , e ) 2.35 | d | divide(divide(multiply(multiply(34.31, 0.473), 1.5), multiply(multiply(7.57, 23.25), 0.0673)), const_10) | the value of ( 34.31 * 0.473 * 1.5 ) / ( 0.0673 * 23.25 * 7.57 ) is close to | "( 34.31 * 0.473 * 1.5 ) / ( 0.0673 * 23.25 * 7.57 ) = 24.343 / 11.845 = 2.06 answer : d" | a = 34 * 31
b = a * 1
c = 7 * 57
d = c * 0
e = b / d
f = e / 10
|
a ) $ 5 , b ) $ 10 , c ) $ 14 , d ) $ 4 , e ) $ 28 | d | subtract(multiply(40, const_2), 76) | elvin ' s monthly telephone bill is the sum of the charge for the calls he made during the month and a fixed monthly charge for internet service . elvin ' s total telephone bill for january was $ 40 and elvin ' s total telephone bill for february was 76 $ . if elvin ' s charge for the calls he made in february was twice the charge for the calls he made in january , what is elvin ' s fixed monthly charge for internet service ? | bill = fixed charge + charge of calls made in jan , bill = fixed charge ( let , y ) + charge of calls made in jan ( let , x ) = $ 40 in feb , bill = fixed charge ( let , y ) + charge of calls made in feb ( then , 2 x ) = $ 76 i . e . x + y = 40 and 2 x + y = 76 take the difference if two equations i . e . ( 2 x + y ) - ( x + y ) = 76 - 40 i . e . x = 36 i . e . fixed monthly charge , y = 4 answer : option d | a = 40 * 2
b = a - 76
|
a ) 60 sec , b ) 82 sec , c ) 95 sec , d ) 100 sec , e ) 120 sec | a | divide(500, multiply(30, const_0_2778)) | how many seconds does puja take to cover a distance of 500 m , if she runs at a speed of 30 km / hr ? | hint : time = distance / speed we see that the distance is given in metres while the speed is given in km / hr and the answer is asked in seconds . so , convert km / hr into m / s by multiplying 5 / 18 m / s to the given value of speed . 30 km / hr = 30 x 5 / 18 = 75 / 9 m / sec i . e . place these values in the formula : time = 500 x 9 / 75 = 60 sec answer is a | a = 30 * const_0_2778
b = 500 / a
|
a ) 18 , b ) 40 , c ) 48 , d ) 32 , e ) 56 | a | add(multiply(divide(2, multiply(40, 4)), 4), multiply(divide(2, multiply(40, 4)), 40)) | the l . c . m . of 2 numbers is 40 . the numbers are in the ratio 4 : 5 . find their sum ? | "let the numbers be 4 x and 5 x l . c . m . = 20 x 20 x = 40 x = 2 the numbers are = 8 and 10 required sum = 8 + 10 = 18 answer is a" | a = 40 * 4
b = 2 / a
c = b * 4
d = 40 * 4
e = 2 / d
f = e * 40
g = c + f
|
a ) 22 , b ) 11 , c ) 25 , d ) 14 , e ) 16 | d | subtract(add(multiply(5, 4), 94), 100) | the sum of the ages 4 members of a family 5 years ago was 94 . today , when the daughter has been married off and replaced by a daughter - in - law , the sum of their ages is 100 . assuming that there has been no other change in the family structure and all members are alive , what is the difference in the ages of daughter and daughter - in - law ? | solution : sum of ages of 4 members 5 years ago = 94 = > sum of present ages of 4 members = 94 + 4 * 5 = 114 difference in the sum of the ages = difference in the ages of daughter and daughter - in - law difference in the sum of the ages = 114 - 100 = 14 = > difference in the ages of daughter and daughter - in - law = 14 answer d | a = 5 * 4
b = a + 94
c = b - 100
|
a ) 87 , b ) 18 , c ) 42 , d ) 16 , e ) 10 | c | divide(divide(1600, 20), divide(1400, multiply(35, 21))) | 35 binders can bind 1400 books in 21 days . how many binders will be required to bind 1600 books in 20 days ? | "binders books days 35 1400 21 x 1600 20 x / 35 = ( 1600 / 1400 ) * ( 21 / 20 ) = > x = 42 answer : c" | a = 1600 / 20
b = 35 * 21
c = 1400 / b
d = a / c
|
a ) 8 / 5 , b ) 22 , c ) 3 , d ) 11 / 3 , e ) 4 | b | divide(multiply(subtract(const_12, const_1), subtract(inverse(subtract(const_12, const_3)), inverse(const_10))), subtract(inverse(add(const_1, const_4)), inverse(subtract(const_12, const_3)))) | for each month of a given year except december , a worker earned the same monthly salary and donated one - tenth of that salary to charity . in december , the worker earned n times his usual monthly salary and donated one - fifth of his earnings to charity . if the worker ' s charitable contributions totaled one - sixth of his earnings for the entire year , what is the value of n ? | "let monthly salary for each of the 11 months except december was x , then 11 x * 1 / 10 + nx * 1 / 5 = 1 / 6 ( 11 x + nx ) ; 11 / 10 + n / 5 = 1 / 6 ( 11 + n ) 11 + 2 n / 10 = 11 + n / 6 = > 66 + 12 n = 110 + 10 n = > 2 n = 44 n = 22 answer : b" | a = 12 - 1
b = 12 - 3
c = 1/(b)
d = 1/(10)
e = c - d
f = a * e
g = 1 + 4
h = 1/(g)
i = 12 - 3
j = 1/(i)
k = h - j
l = f / k
|
a ) 1 / 4 , b ) 3 / 8 , c ) 3 / 16 , d ) 5 / 32 , e ) 7 / 32 | d | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting exactly 4 heads in a single throw of five fair coins ? | "one possible case is hhhht . p ( hhhht ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 there are 5 c 4 = 5 possible cases . p ( 4 heads ) = 5 * 1 / 32 = 5 / 32 the answer is d ." | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
|
a ) a ) 7 , b ) b ) 5 , c ) c ) 4 , d ) d ) 18 , e ) e ) 30 | e | divide(subtract(1212, multiply(add(12, 8), divide(3, 5))), 40) | 40 Γ ? + ( 12 + 8 ) Γ 3 / 5 = 1212 | "explanation : = > 40 Γ ? + ( 12 + 8 ) Γ 3 / 5 = 1212 = > 40 Γ ? = 1212 - ( 12 + 8 ) Γ 3 / 5 = > 40 Γ ? = 1212 - 20 Γ 3 / 5 = > 40 Γ ? = 1212 - 20 Γ 3 / 5 = 1200 = > ? = 1200 / 40 = 30 answer : option e" | a = 12 + 8
b = 3 / 5
c = a * b
d = 1212 - c
e = d / 40
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a ) 144 , b ) 288 , c ) 12 , d ) 256 , e ) 2880 | e | multiply(factorial(4), factorial(5)) | in how many ways 4 boys and 5 girls can be seated in a row so that they are alternate . | "solution : let the arrangement be , g b g b g b g b g 4 boys can be seated in 4 ! ways . girl can be seated in 3 ! ways . required number of ways , = 4 ! * 5 ! = 2880 . answer : option e" | a = math.factorial(4)
b = math.factorial(5)
c = a * b
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a ) 3 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | a | divide(divide(multiply(multiply(48, 12), 4), 72), 8) | in a manufacturing plant , it takes 48 machines 4 hours of continuous work to fill 8 standard orders . at this rate , how many hours of continuous work by 72 machines are required to fill 12 standard orders ? | "the choices give away the answer . . 48 machines take 4 hours to fill 8 standard orders . . in next eq we are doubling the machines from 48 to 72 , but the work is not doubling ( only 1 1 / 2 times ) , = 4 * 48 / 72 * 12 / 8 = 4 ans a" | a = 48 * 12
b = a * 4
c = b / 72
d = c / 8
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(subtract(2, multiply(multiply(3, 3), 2)), subtract(const_1, multiply(3, 3))) | f ( x ) is a function such that f ( x ) + 3 f ( 8 - x ) = x for all real numbers x . find the value of f ( 2 ) | "f ( x ) + 3 f ( 8 - x ) = f ( 2 ) + 3 f ( 6 ) = 2 : x = 2 above f ( 6 ) + 3 f ( 2 ) = 6 : x = 6 above f ( 6 ) = 6 - 3 f ( 2 ) : solve equation c for f ( 6 ) f ( 2 ) + 3 ( 6 - 3 f ( 2 ) ) = 2 : substitute f ( 2 ) = 2 : solve above equation correct answer c" | a = 3 * 3
b = a * 2
c = 2 - b
d = 3 * 3
e = 1 - d
f = c / e
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a ) 1.5 : 5 , b ) 2 : 5 , c ) 3 : 5 , d ) 4 : 11 , e ) 4 : 5 | d | divide(power(64, const_0_33), power(1331, const_0_33)) | two cubes of their volumes in the ratio 64 : 1331 . the ratio of their surface area is : | "the ratio of their surface area is 64 : 1331 4 : 11 answer is d ." | a = 64 ** const_0_33
b = 1331 ** const_0_33
c = a / b
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a ) 4 / 25 , b ) 9 / 37 , c ) 2 / 5 , d ) 8 / 15 , e ) 2 / 3 | b | divide(divide(multiply(30, 30), const_100), add(divide(multiply(30, 30), const_100), divide(multiply(40, subtract(const_100, 30)), const_100))) | at joel β s bookstore , the current inventory is 30 % historical fiction . of the historical fiction books , 30 % are new releases , while 40 % of the other books are new releases . what fraction of all new releases are the historical fiction new releases ? | "let there be 100 books in all historic fiction books = 30 % of total = 30 other books = 70 new historic fiction = 30 % of 30 = 9 other new books = 40 % of 70 = 28 total new books = 37 fraction = 9 / 37 ans : b" | a = 30 * 30
b = a / 100
c = 30 * 30
d = c / 100
e = 100 - 30
f = 40 * e
g = f / 100
h = d + g
i = b / h
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a ) 55 , b ) 60 , c ) 73 , d ) 82 , e ) 91 | e | subtract(100, divide(subtract(100, 73), const_3)) | a teacher grades students β tests by subtracting twice the number of incorrect responses from the number of correct responses . if student b answers each of the 100 questions on her test and receives a score of 73 , how many questions did student b answer correctly ? | "a score of 73 brings to mind that it is an odd number even though you are subtracting an even number ( twice the incorrect responses ) out of the correct responses score . so the correct responses score must be an odd number too ( odd - even = odd ) . since the overall score is 73 , the number of correct responses must be more than 73 since the student answered all questions . only 91 satisfies both these conditions . answer ( e )" | a = 100 - 73
b = a / 3
c = 100 - b
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a ) 78 , b ) 80 , c ) 85 , d ) 88 , e ) 90 | b | divide(multiply(multiply(30, 48), 12), volume_cube(divide(12, const_2))) | a box measuring 30 inches long by 48 inches wide by 12 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ? | "least number of cubes will be required when the cubes that could fit in are biggest . 6 is the biggest number that could divide all three , 30 , 48 and 12 . thus side of cube must be 6 , and total number of cubes = 30 / 6 * 48 / 6 * 12 / 6 = 80 ans b ." | a = 30 * 48
b = a * 12
c = 12 / 2
d = b / volume_cube
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a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 22 % , e ) 30 % | b | divide(const_100, multiply(multiply(divide(10, const_100), divide(20, const_100)), const_100)) | on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 20 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? | "say there are 100 motorists . { # of motorists who exceed speed & receive tickets } + { # of motorists who exceed speed & do n ' t receive tickets } = { total # of motorist who exceed speed } ; given : { # of motorists who exceed speed & receive tickets } = 10 ; also , if { total # of motorist who exceed speed } = x , then 0.2 x = { # of motorists who exceed speed & do n ' t receive tickets } ; 10 + 0.2 x = x - - > x = 12.5 answer : b ." | a = 10 / 100
b = 20 / 100
c = a * b
d = c * 100
e = 100 / d
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a ) 8.5 , b ) 8.0 , c ) 9.5 , d ) 9.0 , e ) 8.25 | a | divide(subtract(24, 7), const_2) | a man can row downstream at the rate of 24 kmph and upstream at 7 kmph . find the man β s rate in still water and rate of current ? | "rate of still water = 1 / 2 ( down stream + upstream ) = 1 / 2 ( 24 + 7 ) = 15.5 kmph rate of current = 1 / 2 ( down stream - upstream ) = 1 / 2 ( 24 - 7 ) = 1 / 2 ( 17 ) = 8.5 kmph answer is a ." | a = 24 - 7
b = a / 2
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a ) 76 kg , b ) 77 kg , c ) 72.5 kg , d ) data inadequate , e ) none of these | c | add(65, multiply(5, 1.5)) | the average weight of 5 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ? | "total weight increases = 5 Γ 1.5 = 7.5 kg so the weight of new person = 65 + 7.5 = 72.5 kg answer c" | a = 5 * 1
b = 65 + a
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a ) 4 : 5 , b ) 4 : 2 , c ) 4 : 4 , d ) 4 : 8 , e ) 4 : 1 | b | divide(sqrt(16), sqrt(4)) | 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 4 hours and 16 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 16 : β 4 = 4 : 2 answer : b" | a = math.sqrt(16)
b = math.sqrt(4)
c = a / b
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a ) 90 % , b ) 99 % , c ) 100 % , d ) 101 % , e ) 110 % | b | multiply(10, 10) | on july 1 of last year , total employees at company e was decreased by 10 percent . without any change in the salaries of the remaining employees , the average ( arithmetic mean ) employee salary was 10 percent more after the decrease in the number of employees than before the decrease . the total of the combined salaries of all the employees at company e after july 1 last year was what percent e of thatbeforejuly 1 last year ? | "the total number of employees = n the average salary = x total salary to all emplyoees = xn after the total number of employees = n - 0.1 n = 0.9 n the average salary = x + 10 % of x = 1.1 x total salary to all emplyoees = 0.9 n ( 1.1 x ) total salary after as a % of total salary before e = [ 0.9 n ( 1.1 x ) ] / xn = 0.99 or 99 % = b" | a = 10 * 10
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a ) 3 , b ) 4 , c ) 12 , d ) 32 , e ) 35 | d | subtract(70, reminder(3, 7)) | when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 70 . | "first , let us say i have a number n which is divisible by 5 and by 7 . we all agree that it will be divisible by 35 , the lcm of 5 and 7 . now , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 1 , we can say the number is of the form n = 5 a + 1 e . g . 5 + 1 , 10 + 1 , 15 + 1 , 20 + 1 , 25 + 1 , 30 + 1 , 35 + 1 etc and n = 7 b + 1 e . g . 7 + 1 , 14 + 1 , 21 + 1 , 28 + 1 , 35 + 1 etc so when it is divided by the lcm , 35 , it will give 1 as remainder ( as is apparent above ) next , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 3 , we can say the number is of the form n = 5 a + 1 and n = 7 b + 3 now , the only thing you should try to understand here is that when n is divided by 5 and if i say the remainder is 1 , it is the same as saying the remainder is - 4 . e . g . when 6 is divided by 5 , remainder is 1 because it is 1 more than a multiple of 5 . i can also say it is 4 less than the next multiple of 5 , ca n ' t i ? 6 is one more than 5 , but 4 less than 10 . therefore , we can say n = 5 x - 4 and n = 7 y - 4 ( a remainder of 3 when divided by 7 is the same as getting a remainder of - 4 ) now this question is exactly like the question above . so when you divide n by 70 , remainder will be - 4 i . e . n will be 4 less than a multiple of 70 . so you must add 4 to n to make it a multiple of 70 d" | a = 70 - reminder
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a ) 5500 , b ) 4500 , c ) 2500 , d ) 6970 , e ) none | a | subtract(multiply(const_10, 6), 6) | the difference between the place values of 6 and 5 in the number 826533 is | sol . = ( place value of 6 ) β ( place value of 5 ) = ( 6000 - 500 ) = 5500 answer a | a = 10 * 6
b = a - 6
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a ) 4 days , b ) 6 days , c ) 8 days , d ) 18 days , e ) 28 days | a | inverse(add(divide(const_1, 12), multiply(const_2, divide(const_1, 12)))) | a works twice as fast as b . if b can complete a work in 12 days independently , the number of days in which a and b can together finish the work in : | "explanation : ratio of rates of working of a and b = 2 : 1 . so , ratio of times taken = 1 : 2 . b ' s 1 day ' s work = 1 / 12 a ' s 1 day ' s work = 1 / 6 ; ( 2 times of b ' s work ) ( a + b ) ' s 1 day ' s work = ( 1 / 6 + 1 / 12 ) = 3 / 12 = 1 / 4 so , a and b together can finish the work in 4 days . answer is a" | a = 1 / 12
b = 1 / 12
c = 2 * b
d = a + c
e = 1/(d)
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a ) 70 , b ) 71 , c ) 72 , d ) 73 , e ) 74 | b | subtract(104, divide(subtract(104, 5), const_3)) | if a student loses 5 kilograms , he will weigh twice as much as his sister . together they now weigh 104 kilograms . what is the student ' s present weight in kilograms ? | "let x be the weight of the sister . then the student ' s weight is 2 x + 5 . x + ( 2 x + 5 ) = 104 3 x = 99 x = 33 kg then the student ' s weight is 71 kg . the answer is b ." | a = 104 - 5
b = a / 3
c = 104 - b
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a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | subtract(13, 5) | a football team lost 5 yards and then gained 13 . what is the team ' s progress ? | "for lost , use negative . for gain , use positive . progress = - 5 + 13 = 8 yards e" | a = 13 - 5
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a ) 10 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | multiply(subtract(2, const_4), 100) | find the value of x from logx 100 = 2 | "solution : logb 1000 = 3 we can write it as , b 3 = 1000 b 3 = 103 so from the above equation b = 10 answer is a" | a = 2 - 4
b = a * 100
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a ) 1 : 2 , b ) 2 : 3 , c ) 1 : 3 , d ) 4 : 1 , e ) none of these | d | divide(multiply(52000, const_12), multiply(26000, add(const_4, const_3))) | x starts a business with rs . 52000 . y joins in the business after 6 months with rs . 26000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 52000 * 12 : 26000 * 6 = 52 * 12 : 26 * 6 = 24 : 6 = 4 : 1 . answer : option d" | a = 52000 * 12
b = 4 + 3
c = 26000 * b
d = a / c
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a ) 550 , b ) 551 , c ) 560 , d ) 561 , e ) 572 | d | add(divide(multiply(add(divide(subtract(60, 40), const_2), const_1), add(multiply(const_2, 40), multiply(divide(subtract(60, 40), const_2), const_2))), const_2), add(divide(subtract(60, 40), const_2), const_1)) | if x is equal to the sum of the even integers from 40 to 60 inclusive , and y is the number of even integers from 40 to 60 inclusive , what is the value of x + y ? | this is a perfect example of why you should not use formulas without understanding them properly . if you understand them , you will not make a mistake and will save time . the formula quoted by the original poster : n ( n + 1 ) is absolutely fine . but one needs to understand that n is the number of even terms starting from the first even term . ( i discuss why this is so here : sum of even numbers from 40 to 60 using this formula will be : 30 * 31 - 19 * 20 = 10 ( 3 * 31 - 19 * 2 ) = 550 since number of terms is 11 , required sum is 561 but , i would not use this formula for this question and would do it the way many of you have done : average = 50 ( it is the middle number ) , number of terms = 11 ( no formula again . any 10 consecutive integers have 5 even and 5 odd numbers . 41 to 60 will have 10 even integers and 40 is the 11 th one ) sum = 50 * 11 + 11 = 561 answer is d | a = 60 - 40
b = a / 2
c = b + 1
d = 2 * 40
e = 60 - 40
f = e / 2
g = f * 2
h = d + g
i = c * h
j = i / 2
k = 60 - 40
l = k / 2
m = l + 1
n = j + m
|
a ) 15 hrs , b ) 18 hrs , c ) 19 hrs , d ) 17 hrs , e ) 16 hrs | a | subtract(divide(17, subtract(3, 2)), 2) | a monkey start climbing up a tree 17 ft tall . each hour it hops 3 ft and slips back 2 ft . how much time would it take the monkey to reach the top . | "if monkey hops 3 ft and slips back 2 ft in a hour , it means the monkey hops ( 3 ft - 2 ft ) = 1 ft / hr . similarly in 14 hrs it wil be 14 ft . bt since the height of the tree is 17 ft , so if the monkey hops up the tree in the next hr i . e 15 th hr then it reaches at the top of the tree . hence it takes 15 hrs for monkey to reach at the top answer : a" | a = 3 - 2
b = 17 / a
c = b - 2
|
a ) 7000 , b ) 7900 , c ) 6000 , d ) 5000 , e ) 4000 | b | divide(79, divide(subtract(7, 6), const_100)) | in a competitive examination in state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 79 more candidates got selected than a . what was the number of candidates appeared from each state ? | "state a and state b had an equal number of candidates appeared . in state a , 6 % candidates got selected from the total appeared candidates in state b , 7 % candidates got selected from the total appeared candidates but in state b , 79 more candidates got selected than state a from these , it is clear that 1 % of the total appeared candidates in state b = 79 = > total appeared candidates in state b = 79 x 100 = 7900 = > total appeared candidates in state a = total appeared candidates in state b = 7900" | a = 7 - 6
b = a / 100
c = 79 / b
|
a ) 18 : 20 , b ) 10 : 9 , c ) 2 : 5 , d ) 5 : 2 , e ) 3 : 2 | a | divide(multiply(divide(3, 5), 3), multiply(divide(const_2, const_3), const_3)) | the price ratio of oranges to lemons is 2 : 3 . the price ratio of grapefruits to lemons is 3 : 5 . what is the ratio , by price , of grapefruit to oranges ? | let price of lemons be 30 ( since that is common ) so , price of oranges = 20 & price of lemons is = 30 further price of grapefruit will be 6 * 3 = 18 thus price , of grapefruit to oranges = 18 : 20 answer : a | a = 3 / 5
b = a * 3
c = 2 / 3
d = c * 3
e = b / d
|
a ) 160 , b ) 170 , c ) 180 , d ) 190 , e ) 200 | d | multiply(divide(subtract(3.64, divide(multiply(16, 180), const_1000)), 4), const_1000) | 20 carrots on a scale weigh 3.64 kg . when 4 carrots are removed from the scale , the average weight of the 16 carrots is 180 grams . what is the average weight ( in grams ) of the 4 carrots which were removed ? | "16 * 180 = 2880 . the other 4 carrots weigh a total of 760 grams . the average weight is 760 / 4 = 190 grams . the answer is d ." | a = 16 * 180
b = a / 1000
c = 3 - 64
d = c / 4
e = d * 1000
|
a ) rs . 380 , b ) rs . 600 , c ) rs . 420 , d ) rs . 400 , e ) rs . 480 | d | multiply(multiply(3, subtract(divide(const_1, 3), add(divide(const_1, 6), divide(const_1, 8)))), 3200) | a can do a particular work in 6 days . b can do the same work in 8 days . a and b signed to do it for rs . 3200 . they completed the work in 3 days with the help of c . how much is to be paid to c ? | "explanation : amount of work a can do in 1 day = 1 / 6 amount of work b can do in 1 day = 1 / 8 amount of work a + b can do in 1 day = 1 / 6 + 1 / 8 = 7 / 24 amount of work a + b + c can do = 1 / 3 amount of work c can do in 1 day = 1 / 3 - 7 / 24 = 1 / 24 work a can do in 1 day : work b can do in 1 day : work c can do in 1 day = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 amount to be paid to c = 3200 Γ ( 1 / 8 ) = 400 answer : option d" | a = 1 / 3
b = 1 / 6
c = 1 / 8
d = b + c
e = a - d
f = 3 * e
g = f * 3200
|
a ) 36 kmph , b ) 42 kmph , c ) 48 kmph , d ) 52 kmph , e ) 64 kmph | a | multiply(divide(300, subtract(40, 10)), const_3_6) | a train requires 10 seconds to pass a pole while it requires 40 seconds to cross a stationary train which is 300 mtrs long . find the speed of the train . | "in 10 s the train crosses the pole and in 40 sec the train crosses one more stationary train in 30 sec the train travels a distance of 300 mtrs speed = 300 / 30 = 10 m / s = 10 * 18 / 5 = 36 kmph answer : a" | a = 40 - 10
b = 300 / a
c = b * const_3_6
|
a ) 42 , b ) 36 , c ) 108 , d ) 120 , e ) 124 | c | multiply(const_3_6, divide(divide(add(120, 120), 4), const_2)) | two trains are running in opposite directions in the same speed . the length of each train is 120 meter . if they cross each other in 4 seconds , the speed of each train ( in km / hr ) is | "explanation : distance covered = 120 + 120 = 240 m time = 4 s let the speed of each train = v . then relative speed = v + v = 2 v 2 v = distance / time = 240 / 4 = 60 m / s speed of each train = v = 60 / 2 = 30 m / s = 30 Γ 36 / 10 km / hr = 108 km / hr answer : option c" | a = 120 + 120
b = a / 4
c = b / 2
d = const_3_6 * c
|
a ) 43 , b ) 42 , c ) 22 , d ) 21 , e ) 20 | a | add(divide(subtract(1000, 700), 7), const_1) | how many multiples of 7 are there between 700 and 1000 , inclusive ? | "there are exactly 301 numbers between 700 to 1000 . . 701 to 800 = 100 801 to 1000 = 200 . . . total 300 numbers , count 700 too . . 301 / 7 = 43 . . . ans option a ." | a = 1000 - 700
b = a / 7
c = b + 1
|
a ) 7 , b ) 8 , c ) 6 , d ) 10 , e ) 11 | c | divide(subtract(12.8, 0.5), 2) | at a certain bowling alley , it costs $ 0.50 to rent bowling shoes for the day and $ 2 to bowl 1 game . if a person has $ 12.80 and must rent shoes , what is the greatest number of complete games that person can bowl in one day ? | after renting bowling shoes the person is left with $ 12.80 - $ 0.5 = $ 12.30 , which is enough for 12.3 / 2 < 7 = ~ 6 . answer : c . | a = 12 - 8
b = a / 2
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 3 , e ) none of these | b | divide(multiply(130, 250), multiply(750, 65)) | if a * b * c = 130 , b * c * d = 65 , c * d * e = 750 and d * e * f = 250 the ( a * f ) / ( c * d ) = ? | "explanation : a Γ’ Λ β b Γ’ Λ β c / b Γ’ Λ β c Γ’ Λ β d = 130 / 65 = > a / d = 2 d Γ’ Λ β e Γ’ Λ β f / c Γ’ Λ β d Γ’ Λ β e = 250 / 750 = > f / c = 1 / 3 a / d * f / c = 2 * 1 / 3 = 2 / 3 answer : b" | a = 130 * 250
b = 750 * 65
c = a / b
|
a ) 7 hours , b ) 7.1 hours , c ) 7.2 hours , d ) 7.3 hours , e ) none of these | c | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 9))) | a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time cistern will get filled ? | "explanation : when we have question like one is filling the tank and other is empting it , then we subtraction as , filled in 1 hour = 1 / 4 empties in 1 hour = 1 / 9 net filled in 1 hour = 1 / 4 - 1 / 9 = 5 / 36 so cistern will be filled in 36 / 5 hours i . e . 7.2 hours option c" | a = 1 / 4
b = 1 / 9
c = a - b
d = 1 / c
|
a ) 197600 , b ) 168000 , c ) 278000 , d ) 192700 , e ) none of them | d | subtract(986, multiply(multiply(237, 986), 37)) | evaluate : 986 x 237 - 986 x 37 | "986 x 237 - 986 x 37 = 986 x ( 237 - 37 ) = 986 x 200 = 197200 . answer is d ." | a = 237 * 986
b = a * 37
c = 986 - b
|
a ) 495 , b ) 550 , c ) 555 , d ) 600 , e ) 640 | e | divide(multiply(25, 55), const_4) | what is the sum of the odd integers from 25 to 55 , inclusive ? | "the mean is 40 . sum = mean ( # of elements ) there are 16 odd numbers between 25 - 55 inclusive . 16 * 40 = 640 e" | a = 25 * 55
b = a / 4
|
a ) 1 / 3 , b ) 2 / 3 , c ) 4 / 3 , d ) 5 / 3 , e ) 7 / 3 | d | divide(subtract(divide(2, 3), divide(1, 4)), divide(1, 4)) | in a class of students , 2 / 3 of the number of girls is equal to 1 / 4 of the total number of students . what is the ratio of boys to girls in the class ? | "( 2 / 3 ) g = ( 1 / 4 ) ( b + g ) 8 g = 3 b + 3 g 5 g = 3 b b / g = 5 / 3 . the answer is d ." | a = 2 / 3
b = 1 / 4
c = a - b
d = 1 / 4
e = c / d
|
a ) 5 / 13 , b ) 6 / 13 , c ) 9 / 13 , d ) 13 / 9 , e ) 13 / 6 | c | divide(subtract(85, 40), subtract(85, 20)) | a portion of the 85 % solution of chemicals was replaced with an equal amount of 20 % solution of chemicals . as a result , 40 % solution of chemicals resulted . what part of the original solution was replaced ? | "let the original amount of 85 percent solution = x and amount replaced be y 0.85 ( x - y ) + 0.2 y = 0.4 x 0.85 x - 0.85 y + 0.2 y = 0.4 x 0.45 x = 0.65 y 9 x = 13 y y = 9 x / 13 so , amount replaced = 9 / 13 of original amount answer : c" | a = 85 - 40
b = 85 - 20
c = a / b
|
['a ) 4 hours', 'b ) 11 hours', 'c ) 7 hours 15 minutes', 'd ) 3 hours', 'e ) 14 hours'] | d | divide(30, 10) | rain is falling at a rate of 10 centimeters per hour all over north carolina . somewhere downtown in north carolina a group of people are waiting for the rain to stop . if the rain filled a round puddle the with a base area of 300 square centimeters and a depth of 30 centimeters , how long did the people wait for the rain to stop ? | answer is : d , 3 hours the volume of the puddle is irrelevant and only height matters since rain fell all over the city . thus , it takes only 30 / 10 = 3 hours of rain to fill the puddle | a = 30 / 10
|
a ) 123 , b ) 127 , c ) 235 , d ) 305 , e ) 10 | e | gcd(subtract(1816, 6), subtract(1442, 12)) | the greatest number which on dividing 1442 and 1816 leaves remainders 12 and 6 respectively , is : | explanation : required number = h . c . f . of ( 1442 - 12 ) and ( 1816 - 6 ) = h . c . f . of 1430 and 1810 = 10 . answer : e | a = 1816 - 6
b = 1442 - 12
c = math.gcd(a, b)
|
a ) β 4 , b ) β 3 , c ) β 2 , d ) β 1 , e ) 0 | b | subtract(multiply(3, const_2), multiply(1, const_2)) | if the average ( arithmetic mean ) of x + 1 , x + 3 , and x + 5 is 0 , then x = | "( x + 1 + x + 3 + x + 5 ) / 3 = 0 = > 3 x + 9 = 0 = > x = - 3 answer b" | a = 3 * 2
b = 1 * 2
c = a - b
|
a ) 9800000 , b ) 2000000 , c ) 7500000 , d ) 1200000 , e ) none of these | b | divide(multiply(multiply(multiply(4, const_100), multiply(2, const_100)), multiply(4, const_100)), multiply(multiply(4, 2), 2)) | a wooden box of dimensions 4 m x 2 m x 4 m is to carry rectangularboxes of dimensions 4 cm x 2 cm x 2 cm . the maximum number ofboxes that can be carried in the wooden box , is | explanation : number = ( 400 * 200 * 400 ) / 4 * 2 * 2 = 2000000 answer : b | a = 4 * 100
b = 2 * 100
c = a * b
d = 4 * 100
e = c * d
f = 4 * 2
g = f * 2
h = e / g
|
a ) rs . 415 , b ) rs . 405 , c ) rs . 450 , d ) rs . 345 , e ) rs . 564 | b | multiply(750, divide(9, const_100)) | find the simple interest on rs . 750 for 9 months at 6 paisa per month ? | "explanation : i = ( 750 * 9 * 6 ) / 100 = 405 answer : option b" | a = 9 / 100
b = 750 * a
|
a ) 68 , b ) 72 , c ) 76 , d ) 80 , e ) 84 | b | multiply(divide(1182, add(460, 525)), const_60) | if two projectiles are launched at the same moment from 1182 km apart and travel directly towards each other at 460 km per hour and 525 km per hour respectively , how many minutes will it take for them to meet ? | the projectiles travel a total of 985 km per hour . the time to meet is 1182 / 985 = 1.2 hours = 72 minutes the answer is b . | a = 460 + 525
b = 1182 / a
c = b * const_60
|
a ) 1.0875 days , b ) 0.1875 days , c ) 0.0675 days , d ) 0.075 days , e ) 0.0175 days | d | inverse(add(inverse(20), inverse(40))) | a and b complete a work in 20 days . a alone can do it in 40 days . if both together can do the work in how many days ? | "1 / 20 + 1 / 40 = 0.075 days answer : d" | a = 1/(20)
b = 1/(40)
c = a + b
d = 1/(c)
|
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