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a ) 10 % , b ) 15 % , c ) 20 % , d ) 40 % , e ) none | c | divide(multiply(const_3, const_100), 15) | the rate at which a sum becomes 4 times of itself in 15 years at s . i . , will be : | sol . let sum = x . then , s . i . = 3 x β΄ rate = [ 100 * s . i . / p * t ] = [ 100 * 3 x / x * 15 ] % = 20 % . answer c | a = 3 * 100
b = a / 15
|
a ) 144 , b ) 131 , c ) 330 , d ) 390 , e ) 445 | c | add(divide(multiply(20, subtract(20, const_1)), const_2), multiply(20, 7)) | 20 business executives and 7 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ? | "there are 20 business exec and in each handshake 2 business execs are involved . hence 20 c 2 = 190 also , each of 20 exec will shake hand with every 7 other chairmen for total of 140 handshake . total = 190 + 140 = 330 ans : c" | a = 20 - 1
b = 20 * a
c = b / 2
d = 20 * 7
e = c + d
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a ) 11 / 20 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 3 / 4 | a | divide(subtract(add(divide(divide(add(add(multiply(const_100, multiply(const_2, add(const_1, const_4))), multiply(const_4, const_100)), multiply(const_4, multiply(const_2, add(const_1, const_4)))), 20), 9), 9), 6), 20) | a total of $ 20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 9 percent , respectively . if the total interest on the two certificates was $ 1,440 at the end of one year , what fractional part of the $ 20.000 was invested at the higher rate ? | x * 6 / 100 * 1 + ( 20000 - x ) * 9 / 100 * 1 = 1440 6 x - 8 x = 144000 - 180000 = > - 2 x = - 18000 = > x = 9000 so 11000 / 20000 = 11 / 20 answer - a | a = 1 + 4
b = 2 * a
c = 100 * b
d = 4 * 100
e = c + d
f = 1 + 4
g = 2 * f
h = 4 * g
i = e + h
j = i / 20
k = j / 9
l = k + 9
m = l - 6
n = m / 20
|
a ) 173 , b ) 163 , c ) 153 , d ) 143 , e ) 133 | c | divide(multiply(850, subtract(const_100, add(add(44, 28), 10))), const_100) | in a school of 850 boys , 44 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ? | "44 + 28 + 10 = 82 % 100 β 82 = 18 % 850 * 18 / 100 = 153 answer : c" | a = 44 + 28
b = a + 10
c = 100 - b
d = 850 * c
e = d / 100
|
a ) 20162 , b ) 18322 , c ) 13032 , d ) 18662 , e ) none of these | a | multiply(109, 109) | 109 Γ 109 + 91 Γ 91 = ? | "explanation : ( a + b ) 2 + ( a β b ) 2 = 2 ( a 2 + b 2 ) ( reference : basic algebraic formulas ) 1092 + 912 = ( 100 + 9 ) 2 + ( 100 β 9 ) 2 = 2 ( 1002 + 92 ) = 2 ( 10000 + 81 ) = 20162 . answer : option a" | a = 109 * 109
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a ) $ 015 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45 | d | divide(multiply(add(const_1, const_1), 0.60), add(const_1, const_2)) | on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.60 per glass on the first day , what was the price q per glass on the second day ? | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.6 = 3 * x - - > x = $ 0.4 . answer : d ." | a = 1 + 1
b = a * 0
c = 1 + 2
d = b / c
|
a ) β 100 , b ) 0.02 , c ) 0.25 , d ) 4 , e ) 8 | b | divide(divide(2, 2), const_1000) | if x = 3 and y = β 2 , what is the value of ( x β 2 y ) ^ y ? | "quickly we can spot that answer is neither integer nor negative . eliminate a , de answer : b" | a = 2 / 2
b = a / 1000
|
a ) 288 , b ) 320 , c ) 2,940 , d ) 3,200 , e ) 28,800 | c | floor(divide(multiply(divide(multiply(multiply(3, 100000000), const_10), 100000000), 98), const_1000)) | last year , for every 100 million vehicles that travelled on a certain highway , 98 vehicles were involved in accidents . if 3 billion vehicles travelled on the highway last year , how many of those vehicles were involved in accidents ? ( 1 billion = 100000000 ) | to solve we will set up a proportion . we know that β 100 million vehicles is to 98 accidents as 3 billion vehicles is to x accidents β . to express everything in terms of β millions β , we can use 3,000 million rather than 3 billion . creating a proportion we have : 100 / 98 = 3,000 / x cross multiplying gives us : 100 x = 3,000 * 98 x = 30 * 98 = 2,940 correct answer is c . | a = 3 * 100000000
b = a * 10
c = b / 100000000
d = c * 98
e = d / 1000
f = math.floor(e)
|
a ) 3.5 , b ) 2.7 , c ) 2.9 , d ) 2.3 , e ) 2.1 | a | divide(125, multiply(130, const_0_2778)) | in what time will a train 125 m long cross an electric pole , it its speed be 130 km / hr ? | "speed = 130 * 5 / 18 = 36 m / sec time taken = 125 / 36 = 3.5 sec . answer : a" | a = 130 * const_0_2778
b = 125 / a
|
a ) 60 % , b ) 65 % , c ) 70 % , d ) 75 % , e ) 80 % | c | subtract(add(add(85, 80), 5), const_100) | if 85 percent of the test takers taking an old paper and pencil gmat exam answered the first question on a given math section correctly , and 80 percent of the test takers answered the second question correctly , and 5 percent of the test takers answered neither question correctly , what percent answered both correctly ? | "{ total } = { first correctly } + { second correctly } - { both correctly } + { neither correctly } 100 = 85 + 80 - { both correctly } + 5 { both correctly } = 70 . answer : c ." | a = 85 + 80
b = a + 5
c = b - 100
|
a ) 53 , b ) 47 , c ) 48 , d ) 59 , e ) 71 | e | multiply(43, 23) | 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , ( . . . ) | "explanation : all are prime numbers in their order , starting from 23 hence , next number is 71 answer : e" | a = 43 * 23
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a ) a ) 6.24 , b ) b ) 8 , c ) c ) 10 , d ) d ) 9.1 , e ) e ) 24 | d | max(multiply(subtract(add(55, 10), const_1), subtract(divide(10, 30), 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 30 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 / 30 = 10 * 2 = 20 time difference = 9.1 ans : d" | a = 55 + 10
b = a - 1
c = 10 / 30
d = 10 / 55
e = c - d
f = b * e
g = max(f)
|
a ) 66 , b ) 77 , c ) 42 , d ) 88 , e ) 12 | c | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 96), const_2) | 96 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | "x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 96 = 42 . answer : c" | a = 4 + 3
b = 4 + 3
c = b + 2
d = a / c
e = d * 96
f = e / 2
|
a ) 40 , b ) 44 , c ) 48 , d ) 50 , e ) 38 | c | multiply(8, multiply(const_3, const_2)) | the length of a side of a hexagon is 8 inches . what is the perimeter ? | "hexagon . it means 6 equal sides . p = 6 ( 8 ) = 48 inches answer c" | a = 3 * 2
b = 8 * a
|
a ) 140 , b ) 150 , c ) 260 , d ) 275 , e ) 280 | d | divide(11, subtract(724.04, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 11 . if n / j = 724.04 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 724.04 here 724 is the quotient . given that remainder = 11 so , 724.04 = 724 + 11 / j so , j = 275" | a = 4 * 10
b = a + 2
c = 100 + b
d = 724 - 4
e = 11 / d
|
a ) 1 / 6 , b ) 3 / 10 , c ) 1 / 2 , d ) 5 / 6 , e ) 8 / 9 | a | divide(subtract(subtract(2, 1), 4), multiply(2, 4)) | a can complete the job in 2 hours and b can complete the same job in 4 hours . a works for 1 hour and then b joins and both complete the job . what fraction of the job did b complete | a = 1 / 6 | a = 2 - 1
b = a - 4
c = 2 * 4
d = b / c
|
a ) 11 , b ) 12 , c ) 33 , d ) 15 , e ) 36 | c | multiply(subtract(divide(const_100, subtract(const_100, 25)), const_1), const_100) | if a ' s income is 25 % less than b ' s . by what % b ' s income is greater than a ? | if b ' s income is rs 100 , a ' s income = rs 75 % b ' s income is greater than a = 100 * 25 / 75 = 33.33 % answer : c | a = 100 - 25
b = 100 / a
c = b - 1
d = c * 100
|
a ) 21 : 11 , b ) 11 : 21 , c ) 22 : 23 , d ) 25 : 23 , e ) none of these | a | divide(multiply(56000, const_12), multiply(44000, add(const_4, const_3))) | x starts a business with rs . 56000 . y joins in the business after 4 months with rs . 44000 . 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 = 56000 * 12 : 44000 * 8 = 56 * 12 : 44 * 8 = 7 * 3 : 11 = 21 : 11 . answer : option a" | a = 56000 * 12
b = 4 + 3
c = 44000 * b
d = a / c
|
a ) a ) 1040 , b ) b ) 1045 , c ) c ) 965 , d ) d ) 1060 , e ) e ) 1075 | c | add(multiply(8, 70), multiply(9, 45)) | tom purchased 8 kg of apples at the rate of 70 per kg and 9 kg of mangoes at the rate of 45 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg apples = 70 Γ 8 = 560 . cost of 9 kg of mangoes = 45 Γ 9 = 405 . total cost he has to pay = 560 + 405 = 965 . c )" | a = 8 * 70
b = 9 * 45
c = a + b
|
a ) 3 / 5 , b ) 5 / 8 , c ) 6 / 7 , d ) 7 / 8 , e ) 6 / 7 | b | divide(add(1, add(divide(1, const_2), add(divide(1, add(1, 3)), divide(3, add(1, 3))))), add(1, 3)) | in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what is theprobability that the extracted ball would turn onto be white , if all possible hypothesisconcerning the color of the balls that initially in the bag were equally possible ? | "since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . β΄ required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b" | a = 1 / 2
b = 1 + 3
c = 1 / b
d = 1 + 3
e = 3 / d
f = c + e
g = a + f
h = 1 + g
i = 1 + 3
j = h / i
|
a ) 15 seconds , b ) 18 seconds , c ) 25 seconds , d ) 30 seconds , e ) 12 seconds | e | divide(36, subtract(6, 3)) | nicky and cristina are running a race . since cristina is faster than nicky , she gives him a 36 meter head start . if cristina runs at a pace of 6 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "used pluging in method say t is the time for cristina to catch up with nicky , the equation will be as under : for nicky = n = 3 * t + 36 for cristina = c = 6 * t @ t = 12 , n = 72 c = 72 right answer ans : e" | a = 6 - 3
b = 36 / a
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a ) 586411 , b ) 586413 , c ) 486411 , d ) 486413 , e ) none of these | b | multiply(divide(587, 999), const_100) | 587 * 999 = ? | "explanation : trick : 587 * ( 1000 - 1 ) = 587000 - 587 = 586413 option b" | a = 587 / 999
b = a * 100
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a ) 14 , b ) 15 , c ) 16 , d ) 18 , e ) 20 | a | divide(multiply(divide(const_1000, const_2), const_1000), multiply(divide(divide(multiply(divide(const_1000, const_2), const_1000), 35), 12), 30)) | it takes 35 identical printing presses 12 hours to print 500,000 papers . how many hours would it take 30 of these printing presses to print 500,000 papers ? | "35 printing presses can do 1 / 12 of the job each hour . 30 printing presses can do 6 / 7 * 1 / 12 = 1 / 14 of the job each hour . the answer is a ." | a = 1000 / 2
b = a * 1000
c = 1000 / 2
d = c * 1000
e = d / 35
f = e / 12
g = f * 30
h = b / g
|
a ) rs 4000 , b ) rs 5000 , c ) rs 6000 , d ) rs 7000 , e ) rs 8000 | b | divide(300, multiply(divide(3, const_100), 2)) | a sum was put at si at certain rate for 2 years . had it been put at 3 % higher rate , it would have fetched rs . 300 more . find the sum | let p be the sum . and x be the interest rate . ( 2 * ( x + 3 ) * p / 100 ) - ( 2 * x * p / 100 ) = 300 therefore 6 p / 100 = 300 p = rs 5000 answer : b | a = 3 / 100
b = a * 2
c = 300 / b
|
a ) $ 314.16 , b ) $ 314.17 , c ) $ 314.18 , d ) $ 314.19 , e ) $ 314.20 | d | add(314.16, divide(const_3, const_100)) | 9 people decided to split the restaurant bill evenly . if the bill was $ 314.16 dollars , how much money did they 1 cent is the smallest unit ? | this is equivalent to finding the first number that is divisible by 9 that occurs after 31416 . in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 31416 = 15 and 18 is divisible by 9 . hence , we need to add 3 to this number for it to be divisible by 9 correct option : d | a = 3 / 100
b = 314 + 16
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a ) a ) 1951609 , b ) b ) 1946025 , c ) c ) 1951602 , d ) d ) 1951603 , e ) e ) 1951604 | b | multiply(divide(1395, 1395), const_100) | 1395 x 1395 = ? | "1395 x 1395 = ( 1395 ) 2 = ( 1400 - 5 ) 2 = ( 1400 ) 2 + ( 5 ) 2 - ( 2 x 1400 x 5 ) = 1960000 + 25 - 14000 = 1960025 - 14000 = 1946025 . answer : b" | a = 1395 / 1395
b = a * 100
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a ) 60 , b ) 40 , c ) 100 , d ) 28 , e ) 50 | b | divide(multiply(add(subtract(48, 10), 12), 4), 5) | x Γ· 4 Γ 5 + 10 β 12 = 48 , then x = | follow reverse process 48 + 12 = 60 β 10 = 50 Γ· 5 = 10 Γ Γ 4 = 40 option b | a = 48 - 10
b = a + 12
c = b * 4
d = c / 5
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a ) 2071 , b ) 2436 , c ) 2801 , d ) - 2801 , e ) - 2071 | e | multiply(subtract(const_1, const_2), subtract(multiply(84, 29), 365)) | - 84 x 29 + 365 = ? | "e given exp . = - 84 x ( 30 - 1 ) + 365 = - ( 84 x 30 ) + 84 + 365 = - 2520 + 449 = - 2071" | a = 1 - 2
b = 84 * 29
c = b - 365
d = a * c
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a ) 288 , b ) 278 , c ) 800 , d ) 267 , e ) 6400 | e | divide(multiply(add(divide(multiply(650, 20), const_100), 190), const_100), 5) | 5 % of a number is more than 20 % of 650 by 190 . find the number ? | "( 5 / 100 ) * x Γ’ β¬ β ( 20 / 100 ) * 650 = 190 1 / 20 x = 320 x = 6400 answer : e" | a = 650 * 20
b = a / 100
c = b + 190
d = c * 100
e = d / 5
|
a ) 12 : 15 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 1 : 2 | a | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(3, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 5 : 3 . the ratio of economy cars to sport utility vehicles is 4 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 5 : 3 - - > e : l = 5 : 3 = 20 : 12 . the ratio of economy cars to sport utility vehicles is 4 : 3 - - > e : s = 4 : 3 = 20 : 15 . thus , l : s = 12 : 15 . answer : a ." | a = 4 * 3
b = 3 * 3
c = a / b
d = 3 * 4
e = 3 * 4
f = d / e
g = c / f
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a ) 3 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | d | divide(168, add(24, 4)) | a boat can travel with a speed of 24 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 168 km downstream | "explanation : speed of the boat in still water = 24 km / hr speed of the stream = 4 km / hr speed downstream = ( 22 + 5 ) = 28 km / hr distance travelled downstream = 168 km time taken = distance / speed = 168 / 28 = 6 hours answer : option d" | a = 24 + 4
b = 168 / a
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a ) 6 , b ) 42 , c ) 24 , d ) 18 , e ) 95 | d | subtract(divide(power(negate(10), const_2), 5), const_2) | find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 10 x + 5 = 0 ? | a / b + b / a = ( a 2 + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab ] / ab a + b = - 10 / 1 = - 10 ab = 5 / 1 = 5 hence a / b + b / a = [ ( - 10 ) 2 - 2 ( 5 ) ] / 5 = 90 / 5 = 18 . d ) | a = negate ** (
b = a / 2
c = b - 5
|
a ) $ 496.30 , b ) $ 512.40 , c ) $ 555.56 , d ) $ 574.90 , e ) $ 588.20 | c | divide(750, add(divide(35, const_100), const_1)) | jim is able to sell a hand - carved statue for $ 750 which was a 35 % profit over his cost . how much did the statue originally cost him ? | "750 = 1.35 * x x = 750 / 1.35 = 555.5555556 . . . which rounds to $ 555.56 , which is ( c ) ." | a = 35 / 100
b = a + 1
c = 750 / b
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a ) 122.6 $ , b ) 128.9 $ , c ) 243.7 $ , d ) 298.85 $ , e ) 309.85 $ | e | add(multiply(multiply(add(65, divide(multiply(65, 120), const_100)), 2), 0.6), multiply(multiply(65, 3), 0.6)) | in a fuel station the service costs $ 4.25 per car , every liter of fuel costs 0.6 $ . assuming that you fill up 3 mini - vans and 2 trucks , how much money will the fuel cost to all the cars owners total , if a mini - van ' s tank is 65 liters and a truck ' s tank is 120 % bigger and they are all empty - ? | "service cost of 3 van and 2 truck = 4.25 * ( 3 + 2 ) = 21.25 fuel in 3 van = 3 * 65 = 195 litre fuel in 2 trucks = 2 * 65 ( 1 + 120 / 100 ) = 286 total fuel ( van + truck ) = 481 litre total fuel cost = 481 * 0.6 = 288.6 total cost = fuel + service = 288.6 + 21.25 = 309.85 answer is e" | a = 65 * 120
b = a / 100
c = 65 + b
d = c * 2
e = d * 0
f = 65 * 3
g = f * 0
h = e + g
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a ) 10 hours , b ) 20 hours , c ) 60 hours , d ) 40 hours , e ) 50 hours | c | inverse(subtract(divide(const_1, 10), divide(const_1, add(10, 2)))) | a cistern is normally filled in 10 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ? | "if leakage / hour = 1 / x , then 1 / 10 - 1 / x = 1 / 12 , solving 1 / x = 1 / 60 so in 60 hours full cistern will be empty . answer : c" | a = 1 / 10
b = 10 + 2
c = 1 / b
d = a - c
e = 1/(d)
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a ) 191 , b ) 199 , c ) 193 , d ) 212 , e ) 213 | b | divide(795, const_4) | to be considered for β movie of the year , β a film must appear in at least 1 / 4 of the top - 10 - movies lists submitted by the cinematic academy β s 795 members . what is the smallest number of top - 10 lists a film can appear on and still be considered for β movie of the year β ? | total movies submitted are 795 . as per question we need to take 1 / 4 of 795 to be considered for top 10 movies = 198.75 approximate the value we 199 . imo option b is the correct answer . . . | a = 795 / 4
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a ) 1.7 , b ) 1.9 , c ) 3 , d ) 1.5 , e ) 1.2 | c | divide(9, const_3) | a man can swim in still water at 9 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ? | "m = 9 s = x ds = 9 + x us = 9 - x 9 + x = ( 9 - x ) 2 9 + x = 18 - 2 x 3 x = 9 x = 3 answer : c" | a = 9 / 3
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a ) 3 , b ) 6 , c ) 9 , d ) 10.5 , e ) 12 | e | multiply(6, const_2) | the distance from the x - axis to point p is half the distance from the y - axis to point p . if the coordinates of p are ( x , - 6 ) , how many units is p from the y - axis ? | "the x - axis is 6 units from the point p . thus the y - axis is 12 units from the point p . the answer is e ." | a = 6 * 2
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a ) 9 , b ) 77 , c ) 66 , d ) 55 , e ) 12 | a | subtract(inverse(subtract(multiply(divide(const_1, 8), subtract(const_1, multiply(2, divide(const_1, 24)))), divide(const_1, 24))), add(24, 8)) | a can do a piece of work in 24 days . when he had worked for 2 days b joins him . if the complete work was finished in 8 days . in how many days b alone can finish the work ? | "8 / 24 + 6 / x = 1 x = 9 days answer : a" | a = 1 / 8
b = 1 / 24
c = 2 * b
d = 1 - c
e = a * d
f = 1 / 24
g = e - f
h = 1/(g)
i = 24 + 8
j = h - i
|
a ) 5 litres , b ) 7 litres , c ) 15 litres , d ) can not be determined , e ) none of these | c | subtract(divide(multiply(divide(5, const_100), 10), divide(2, const_100)), 10) | milk contains 5 % water . what quantity of pure milk should be added to 10 litres of milk to reduce this to 2 % ? | "here required quantity of pure milk = 10 Γ ( 5 β 2 ) / 2 = 10 Γ 3 / 2 = 15 litres . answer c" | a = 5 / 100
b = a * 10
c = 2 / 100
d = b / c
e = d - 10
|
a ) 0.2 , b ) 0.08 , c ) 0.16 , d ) 0.14 , e ) 0.15 | b | multiply(divide(40, multiply(multiply(const_5, const_5), const_4)), divide(20, multiply(multiply(const_5, const_5), const_4))) | 40 percent of the members of a study group are women , and 20 percent of those women are lawyers . if one member of the study group is to be selected at random , what is the probability that the member selected is a woman lawyer ? | "say there are 100 people in that group , then there would be 0.4 * 0.20 * 100 = 8 women lawyers , which means that the probability that the member selected is a woman lawyer is favorable / total = 8 / 100 . answer : b" | a = 5 * 5
b = a * 4
c = 40 / b
d = 5 * 5
e = d * 4
f = 20 / e
g = c * f
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a ) 72 , b ) 45 , c ) 48 , d ) 51 , e ) 44 | a | add(45, 30) | two goods trains each 750 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 750 + 750 = 1500 m . required time = 1500 * 6 / 125 = 72 sec . answer : option a" | a = 45 + 30
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a ) rs . 608 , b ) rs . 6076 , c ) rs . 600 , d ) rs . 602 , e ) rs . 601 | c | divide(multiply(const_100, divide(multiply(24, const_100), multiply(2, 10))), multiply(2, 10)) | the banker ' s gain of a certain sum due 2 years hence at 10 % per annum is rs . 24 . the present worth is : | "t . d . = b . g . x 100 = rs . 24 x 100 = rs . 120 . rate x time 10 x 2 p . w . = 100 x t . d . = rs . 100 x 120 = rs . 600 . rate x time 10 x 2 answer : c" | a = 24 * 100
b = 2 * 10
c = a / b
d = 100 * c
e = 2 * 10
f = d / e
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a ) 55 , b ) 56 , c ) 60 , d ) 58 , e ) none of these | c | divide(rectangle_perimeter(90, 60), 5) | a rectangular plot measuring 90 metres by 60 metres is to be enclosed by wire fencing . if the poles of the fence are kept 5 metres apart , how many poles will be needed ? | "solution perimeter of the plot = 2 ( 90 + 60 ) = 300 m . β΄ number of poles = [ 300 / 5 ] = 60 m answer c" | a = rectangle_perimeter / (
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a ) 288 , b ) 146.66 , c ) 188 , d ) 180 , e ) 12 | b | divide(add(1200, 1000), divide(1200, 80)) | a 1200 m long train crosses a tree in 80 sec , how much time will i take to pass a platform 1000 m long ? | "l = s * t s = 1200 / 80 s = 15 m / sec . total length ( d ) = 2200 m t = d / s t = 2200 / 15 t = 146.66 sec answer : b" | a = 1200 + 1000
b = 1200 / 80
c = a / b
|
a ) - 1 , b ) β 2 , c ) 1 , d ) 2 , e ) 0 | c | add(add(7, const_3.0), subtract(subtract(5, 2), 1)) | simplify : ( 7 + 2 ) β ( 5 + 3 + 1 ) - 1 . | "solution : ( 7 + 2 ) β ( 5 + 3 + 1 ) - 1 = 9 - 5 - 3 + 1 - 1 = 9 - 8 + 1 - 1 = 2 - 1 = 1 answer : ( c )" | a = 7 + 3
b = 5 - 2
c = b - 1
d = a + c
|
a ) 343 , b ) 388 , c ) 327 , d ) 88 , e ) 99 | a | multiply(power(7, const_2), 7) | how many shots of 1 cm radius can be prepared from a sphere of 7 cm radius ? | "4 / 3 Γ― β¬ * 7 * 7 * 7 = 4 / 3 Γ― β¬ * 1 * 1 * 1 * x x = 343 answer : a" | a = 7 ** 2
b = a * 7
|
a ) 10.5 % , b ) 18 % , c ) 15 % , d ) 22 % , e ) 11.11 % | e | divide(const_100, multiply(multiply(divide(10, const_100), divide(10, const_100)), const_100)) | on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 10 % 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 ? | "0.1 m = 0.90 e = > e / m = 1 / 9 * 100 = 11.11 % = e . m - # of motorists e - # of motorists exceeding speed" | a = 10 / 100
b = 10 / 100
c = a * b
d = c * 100
e = 100 / d
|
a ) 4.0 , b ) 3.8 , c ) 7.2 , d ) 7.5 , e ) 8.0 | a | divide(multiply(multiply(6, 3), const_2), add(6, 3)) | a river boat leaves silver town and travels upstream to gold town at an average speed of 6 kilometers per hour . it returns by the same route at an average speed of 3 kilometers per hour . what is the average speed for the round - trip in kilometers per hour ? | "pick a number which is lcm of 3 and 6 = 12 . upstream time = 12 / 6 = 2 hrs downstream time = 12 / 3 = 4 hrs total time = 6 hrs total distance = 24 average speed = 24 / 6 = 4.0 km / hr" | a = 6 * 3
b = a * 2
c = 6 + 3
d = b / c
|
a ) 3008 , b ) 2004 , c ) 1008 , d ) 2016 , e ) 3000 | b | divide(4008, const_2) | abcd is a square where ab = Γ’ Λ Ε‘ 4008 . let x be a point on ab and y be a point on cd such that ax = cy . compute the area of trapezoid axyd . | "note that trapezoids axy d and bxy c are congruent , so the area of axy d is always 4008 / 2 = 2004 . correct answer b" | a = 4008 / 2
|
a ) 1.7 , b ) 2.8 , c ) 3.6 , d ) 2.7 , e ) 1.1 | d | divide(add(divide(45, 5), divide(72, 5)), const_2) | a man swims downstream 72 km and upstream 45 km taking 5 hours each time ; what is the speed of the current ? | "72 - - - 5 ds = 14.4 ? - - - - 1 45 - - - - 5 us = 9 ? - - - - 1 s = ? s = ( 14.4 - 9 ) / 2 = 2.7 answer : d" | a = 45 / 5
b = 72 / 5
c = a + b
d = c / 2
|
a ) 24 , b ) 26 , c ) 28 , d ) 32 , e ) 44 | c | add(multiply(multiply(1, 7), const_100), multiply(2, 7)) | two numbers are in the ratio 1 : 2 , if 7 is added to both , their ratio changes to 3 : 5 . the greatest number is | "let the numbers be x and 2 x then , ( x + 7 ) / ( 2 x + 7 ) = 3 / 5 x = 14 there fore answer is 28 answer c 28" | a = 1 * 7
b = a * 100
c = 2 * 7
d = b + c
|
a ) 6.9 , b ) 69.0 , c ) 0.6845 , d ) 0.6859 , e ) 0.615 | e | divide(multiply(82, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 82 % of 3 / 4 ? | "82 % * ( 3 / 4 ) = 0.82 * 0.75 = 0.615 answer : e" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 82 * n
p = o / 100
|
a ) 200 , b ) 220 , c ) 240 , d ) 180 , e ) 260 | d | multiply(divide(const_3, add(divide(const_2, divide(300, 100)), const_1)), 100) | points x , y , and z lie , in that order , on a straight railroad track . the distance from point x to point y is twice the distance from point y to point z . a train traveled from point x to point z without stopping . the train ' s average speed when traveling from point x to point y was 300 miles per hour and the train ' s average speed when traveling from point y to point z was 100 miles per hour . what was the train ' s average speed , in miles per hour , when traveling from point x to point z ? | average speed = distance / time because we are looking for average speed we can pick a distance for the variable d . speed x - y = 300 speed y - z = 100 average speed = total distance / total rate rate = distance / time x = = = = = = = = = = = = = = = = = = = = y = = = = = = = = = = z if x - y is twice the length of y - z then let x - y = 2 d and let y - z = d average speed = 3 d / ( 2 d / 300 ) + ( d / 100 ) 3 d / ( 2 d / 300 ) + ( 3 d / 300 ) 3 d / ( 5 d / 300 ) 900 d / 5 d average speed = 180 answer : d | a = 300 / 100
b = 2 / a
c = b + 1
d = 3 / c
e = d * 100
|
a ) 220 meter , b ) 225 meter , c ) 230 meter , d ) 235 meter , e ) none of these | c | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 270) | a 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "explanation : as trains are running in opposite directions so their relative speed will get added so , relative speed = 120 + 80 = 200 kmph = 200 * ( 5 / 18 ) = 500 / 9 m / sec let the length of other train is x meter then x + 270 / 9 = 500 / 9 = > x + 270 = 500 = > x = 230 so the length of the train is 230 meters option c" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 270
|
a ) $ 1250 , b ) $ 1060 , c ) $ 1350 , d ) $ 900 , e ) $ 1000 | d | multiply(subtract(10000, 2000), multiply(subtract(const_1, divide(const_1, 10)), divide(const_1, 10))) | if xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 2000 papers and 20 % discount after first 10000 papers , how much will it cost to buy 20000 sheets of xerox paper ? | "30 sec approach - solve it using approximation 20000 sheet at full price , 5 cent = 1000 20000 sheet at max discount price , 4 cent = 800 your ans got to be between these two . ans d it is ." | a = 10000 - 2000
b = 1 / 10
c = 1 - b
d = 1 / 10
e = c * d
f = a * e
|
['a ) 8 hours', 'b ) 4 hours 25 minutes', 'c ) 3 hours 15 minutes', 'd ) 6 hours', 'e ) 2 hours'] | e | divide(divide(multiply(250, 6), 12), multiply(6, const_10)) | there is rain at a rate of 12 centimeters per hour all over new jersey . somewhere downtown in new jersey a group of students are waiting for the rain to stop . if the rain filled a tank the with a base area of 250 square centimeters and a depth of 6 centimeters , how long did the students wait for the rain to stop ? | answer is : e , 4 hours and 30 mins the volume of the tank is irrelevant and only height matters since rain fell all over the city . thus , it takes only . 12 / 6 = 2 hours of rain to fill the tank . | a = 250 * 6
b = a / 12
c = 6 * 10
d = b / c
|
a ) 1 , b ) 3 , c ) 2 , d ) 5 , e ) 4 | d | add(divide(12, 3), const_2) | if a and b are positive integers , and a = 3 b + 12 , the greatest common divisor of a and b can not be | "if b is 1 , 2 , 3 , or 4 , then gcd of a and b is 1 , 2 , 3 , and 4 respectively . so , by poe the answer must be d . still : if b is a multiple of 5 , then a is 12 greater than a multiple of 5 , so not a multiple of 5 , so both of them can not be divisive by 5 . answer : d ." | a = 12 / 3
b = a + 2
|
a ) 9 % , b ) 10 % , c ) 11 % , d ) 16 % , e ) 90 % | d | divide(multiply(14, const_100), subtract(const_100, 14)) | during a sale , the price of a pair of shoes is marked down 14 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | "assume the price = 100 price during sale = 86 price after sale = 100 percent increase = 14 / 86 * 100 = 16 % approx . correct option : d" | a = 14 * 100
b = 100 - 14
c = a / b
|
a ) 10 sec , b ) 32 sec , c ) 17 sec , d ) 20 sec , e ) 89 sec | c | divide(add(60, 280), multiply(add(42, 30), const_0_2778)) | two trains of length 60 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 60 + 280 = 400 m . the time required = d / s = 340 / 20 = 20 sec . answer : c" | a = 60 + 280
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 150 , b ) 175 , c ) 200 , d ) 225 , e ) 250 | b | divide(multiply(0.7, 100), 0.4) | one 100 gallons of fuel are purchased at $ 0.70 per gallon . the fuel is consumed at a rate of $ 0.40 worth of fuel per hour . at this rate , how many hours are required to consume the 100 gallons of fuel ? | the fuel is consumed at this rate : $ 0.40 / $ 0.70 = 4 / 7 gallons per hour . the time it takes to consume 100 gallons is : 100 / ( 4 / 7 ) = 175 hours . the answer is b . | a = 0 * 7
b = a / 0
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 3 | a | divide(15, subtract(75, 15)) | a chemist mixes one liter of pure water with x liters of a 75 % salt solution , and the resulting mixture is a 15 % salt solution . what is the value of x ? | "concentration of salt in pure solution = 0 concentration of salt in salt solution = 75 % concentration of salt in the mixed solution = 15 % the pure solution and the salt solution is mixed in the ratio of - - > ( 75 - 15 ) / ( 15 - 0 ) = 4 / 1 1 / x = 4 / 1 x = 1 / 4 answer : a" | a = 75 - 15
b = 15 / a
|
a ) 127 , b ) 688 , c ) 400 , d ) 121 , e ) 123 | c | divide(subtract(600, 580), divide(5, const_100)) | if 5 % more is gained by selling an article for rs . 600 than by selling it for rs . 580 , the cost of the article is ? | "let c . p . be rs . x . then , 5 % of x = 600 - 580 = 20 x / 20 = 20 = > x = 400 answer : c" | a = 600 - 580
b = 5 / 100
c = a / b
|
a ) 40 minutes , b ) 1 hour , c ) 1 hour 15 min , d ) 1 hour 30 min , e ) 1 hour 10 min | c | divide(5, divide(add(multiply(divide(1, 10), const_60), divide(2, 2)), const_2)) | a boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes . how long will it take to go 5 km in stationary water ? | "speed ( upstream ) = 2 / 1 = 2 kmhr speed ( downstream ) = 1 / ( 10 / 60 ) = 6 kmhr speed in still water = 1 / 2 ( 2 + 6 ) = 4 kmhr time taken in stationary = 5 / 4 = 1 hrs 15 min answer : c" | a = 1 / 10
b = a * const_60
c = 2 / 2
d = b + c
e = d / 2
f = 5 / e
|
a ) 15120 , b ) 14500 , c ) 14900 , d ) 14300 , e ) 14000 | a | multiply(divide(add(divide(subtract(50000, add(add(5000, 4000), 5000)), const_3), add(5000, 4000)), 50000), 36000) | a , b , c subscribe rs . 50000 for a business . if a subscribes rs . 4000 more than b and b rs . 5000 more than c , out of a total profit of rs . 36000 , what will be the amount a receives ? | total amount invested = 50000 assume that investment of c = x . then investment of b = 5000 + x , investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000 β 3 x + 14000 = 50000 β 3 x = 50000 β 14000 = 36000 β x = 36000 / 3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a , b and c = 21000 : 17000 : 12000 = 21 : 17 : 12 share of a = total profit Γ 21 / 50 = 36000 Γ 21 / 50 = 15,120 answer is a . | a = 5000 + 4000
b = a + 5000
c = 50000 - b
d = c / 3
e = 5000 + 4000
f = d + e
g = f / 50000
h = g * 36000
|
a ) 68 mph , b ) 56.67 mph , c ) 60 mph , d ) 64 mph , e ) 66.67 mph | c | add(divide(add(multiply(80, 3), multiply(30, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | steve traveled the first 2 hours of his journey at 30 mph and the last 3 hours of his journey at 80 mph . what is his average speed of travel for the entire journey ? | "answer average speed of travel = total distance travelled / total time taken total distance traveled by steve = distance covered in the first 2 hours + distance covered in the next 3 hours . distance covered in the first 2 hours = speed * time = 30 * 2 = 60 miles . distance covered in the next 3 hours = speed * time = 80 * 3 = 240 miles . therefore , total distance covered = 60 + 240 = 300 miles . total time taken = 2 + 3 = 5 hours . hence , average speed = total distance travelled / total time taken = 300 / 5 = 60 miles per hour . choice c" | a = 80 * 3
b = 30 * 2
c = a + b
d = 3 + 2
e = c / d
f = 100 / 3
g = f - const_0_33
h = e + g
|
a ) 79 , b ) 30 , c ) 36 , d ) 66 , e ) 12 | c | multiply(divide(add(600, 100), 70), const_3_6) | a train 600 m long can cross an over brdge of 100 m long in 70 sec and then find the speed of the train ? | length = speed * time speed = l / t s = 700 / 70 s = 10 m / sec speed = 10 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 36 kmph . answer : c | a = 600 + 100
b = a / 70
c = b * const_3_6
|
a ) 20 % , b ) 40 % , c ) 50 % , d ) 70.38 % , e ) 100 % | d | subtract(const_100, divide(subtract(const_100, 60), add(const_1, divide(35, const_100)))) | when sold at a 60 % discount , a sweater nets the merchant a 35 % profit on the wholesale cost at which he initially purchased the item . by what % is the sweater marked up from wholesale at its normal retail price ? | "we should be careful about what are we measuring % on / what is the base . . let the marked up price = 100 . . selling price = 100 - 60 % of 100 = 40 . . profit = 35 % . . therefore the wholesale purchase cost = x . . . . 1.35 x = 40 or x = 29.62 . . . marked price was 100 so . . . so answer is 70.38 % . . d" | a = 100 - 60
b = 35 / 100
c = 1 + b
d = a / c
e = 100 - d
|
a ) 3 / 5 , b ) 1 / 5 , c ) 2 / 5 , d ) 4 / 5 , e ) 1 | a | divide(multiply(choose(3, const_2), choose(3, const_2)), choose(add(3, 3), 4)) | from a group of 3 women and 3 girls , 4 are to be randomly selected . what is the probability that equal numbers of women and girls will be selected ? | using the first example , here is the probability of that exact sequence occurring : wwgg = ( 3 / 6 ) ( 2 / 5 ) ( 3 / 4 ) ( 2 / 3 ) = 36 / 360 = 1 / 10 each of the other 5 options will yield the exact same probability . . . . wgwg = ( 3 / 6 ) ( 3 / 5 ) ( 2 / 4 ) ( 2 / 3 ) = 36 / 360 = 1 / 10 so we have 6 different options that each produce a 1 / 10 chance of occurring . 6 ( 1 / 10 ) = 6 / 10 = 3 / 5 final answer : a | a = math.comb(3, 2)
b = math.comb(3, 2)
c = a * b
d = 3 + 3
e = math.comb(d, 4)
f = c / e
|
a ) 4800 , b ) 4860 , c ) 3825 , d ) 5500 , e ) 5550 | c | add(100, const_1) | what is the sum of natural numbers between 50 and 100 | "a = first number l = last number sn = n / 2 [ a + l ] between 50 and 100 numbers = 51 = > 100 - 50 = 50 + 1 = 51 sn = 51 / 2 Γ£ β 150 = 51 Γ£ β 75 = 3825 answer : c" | a = 100 + 1
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a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | d | multiply(divide(divide(99, const_3), 99), const_2.0) | an integer n between 1 and 99 , inclusive , is to be chosen at random . what is the probability that n ( n + 1 ) will be divisible by 2 ? | "n ( n + 1 ) to be divisible by 2 either n or n + 1 must be a multiples of 2 . in each following group of numbers : { 1 , 2 , 3 } , { 4 , 5 , 6 } , { 7 , 8 , 9 } , . . . , { 97 , 98 , 99 } there are exactly 2 numbers out of 3 satisfying the above condition . for example in { 1 , 2 , 3 } n can be : 2 , or 3 . thus , the overall probability is 2 / 3 . answer : d ." | a = 99 / 3
b = a / 99
c = b * 2
|
a ) $ 1200 , b ) $ 2000 , c ) $ 2150 , d ) $ 2500 , e ) $ 12000 | e | subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(10, const_100)), const_4)), multiply(const_100, const_100)) | carl is facing very difficult financial times and can only pay the interest on a $ 30,000 loan he has taken . the bank charges him a quarterly compound rate of 10 % . what is the approximate interest he pays annually ? | "usually , you are given the annual rate of interest and it is mentioned that it is annual rate . the bank charges him a quarterly compounded annual rate of 40 % . here you find per quarter rate as ( 40 / 4 ) % = 10 % i have actually never seen a question with quarter rate given but since this question did not mentionannual rate of interestand since the options did not make sense with 5 % annual rate of interest , it is apparent that the intent was a 10 % quarterly rate . so the bank charges 10 % every quarter and compounds it in the next quarter . had it been a simple quarterly rate , we would have just found 4 * 10 % of 30,000 = $ 12000 as our answer . but since , the interest is compounded , it will be a bit more than $ 12000 . option ( e ) looks correct ." | a = 100 * 100
b = 10 / 100
c = 1 + b
d = c ** 4
e = a * d
f = 100 * 100
g = e - f
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a ) 2019.625 , b ) 2017.625 , c ) 2013.625 , d ) 2015.625 , e ) 2011.625 | d | divide(divide(subtract(multiply(6000, power(add(const_1, divide(15, const_100)), 2)), 6000), 2), multiply(4, divide(12, const_100))) | the s . i . on a certain sum of money for 4 years at 12 % per annum is half the c . i . on rs . 6000 for 2 years at 15 % per annum . the sum placed on s . i . is ? | c . i . = [ 6000 * ( 1 + 15 / 100 ) 2 - 6000 ] = ( 6000 * 23 / 20 * 23 / 20 - 6000 ) = rs . 1935 . sum = ( 967.5 * 100 ) / ( 4 * 12 ) = rs . 2015.625 answer : d | a = 15 / 100
b = 1 + a
c = b ** 2
d = 6000 * c
e = d - 6000
f = e / 2
g = 12 / 100
h = 4 * g
i = f / h
|
a ) 5 , b ) 9 , c ) 10 , d ) 20 , e ) 30 | e | multiply(subtract(8, 10), 10) | what is the greatest positive integer x such that 2 ^ x is a factor of 8 ^ 10 ? | "8 ^ 10 = ( 2 ^ 3 ) ^ 10 = 2 ^ 30 answer : e" | a = 8 - 10
b = a * 10
|
a ) 20 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | a | multiply(80, divide(1, 4)) | of the 80 people in a room , 1 / 4 are women . if 3 / 4 of the people are married , what is the maximum number of women in the room who could be unmarried ? | "women = 1 / 4 * 80 = 20 married = 3 / 4 * 80 = 60 unmarried = 20 max ( un - married women ) = 20 answer a" | a = 1 / 4
b = 80 * a
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a ) 81 : 121 , b ) 9 : 11 , c ) 729 : 1331 , d ) 1 : 4 , e ) none of these | d | power(divide(125, 1000), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 125 : 1000 . what is the ratio of their total surface areas ? | "ratio of the sides = Γ’ Β³ Γ’ Λ Ε‘ 125 : Γ’ Β³ Γ’ Λ Ε‘ 1000 = 5 : 10 ratio of surface areas = 5 ^ 2 : 10 ^ 2 = 25 : 100 = 1 : 4 answer : d" | a = 125 / 1000
b = 1 / 3
c = a ** b
|
a ) 395 , b ) 275 , c ) 355 , d ) 415 , e ) 398.2 | e | divide(add(add(add(add(subtract(multiply(62, add(const_4, const_1)), add(add(add(12, 42), 78), 104)), 128), 255), 511), 1023), add(const_4, const_1)) | if the mean of numbers 12 , x , 42 , 78 and 104 is 62 , then what is the mean of 128 , 255 , 511 , 1023 and x ? | "the mean of numbers 12 , x , 42 , 78 and 104 is 62 : 12 + x + 42 + 78 + 104 = 62 * 5 - - > x = 74 ; so , the mean of 128 , 255 , 511 , 1023 and x is ( 128 + 255 + 511 + 1023 + 74 ) / 5 = 398.2 . answer : e ." | a = 4 + 1
b = 62 * a
c = 12 + 42
d = c + 78
e = d + 104
f = b - e
g = f + 128
h = g + 255
i = h + 511
j = i + 1023
k = 4 + 1
l = j / k
|
['a ) 25 : 16', 'b ) 24 : 25', 'c ) 5 : 6', 'd ) 4 : 5', 'e ) 4 : 9'] | b | divide(rectangle_area(2, 3), square_area(divide(multiply(add(2, 3), const_2), const_4))) | if the perimeter of square region q and the perimeter of rectangular region r are equal and the sides of r are in the ratio 2 : 3 then the ratio of the area of r to the area of q | we know perimeter of a square ( pq ) = 4 * side perimeter of a rectangle ( pr ) = 2 ( length + breath ) let us assume 40 to be the perimeter of the square ( since we know each side of a square is equal and the perimeter is divisible by 4 , also take in to account the length and breadth of the rectangle is in the ration 2 k : 3 k = 5 k ; we can assume such a number ) therefore , pq = pr = 40 area of the square = 100 sq . units we know 2 ( length + breadth ) = 40 i . e . length + breadth = 20 ( or 5 k = 20 given that l : b ( or b : l ) = 2 : 3 ) therefore length = 8 , breath = 12 area of the rectangle = 8 * 12 = 96 sq . units question asked = area of the rectangle : area of the square = 96 : 100 = = > 24 : 25 = b | a = rectangle_area / (
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a ) 7.5 , b ) 6 , c ) 5.5 , d ) 6.5 , e ) 5 | a | divide(15, 2) | the ratio of the present ages of ramesh and mahesh is 2 : 5 . 10 years hence , the ratio of their ages will be 10 : 15 . find the difference in their present ages ? | let the present ages of ramesh and mahesh be 2 x and 5 x years respectively . ( 2 x + 10 ) / ( 5 x + 10 ) = 10 / 15 6 x + 30 = 10 x + 20 = > x = 2.5 difference in their ages will be the same at all times . this difference = difference of their present ages = > 5 x - 2 x = 3 x = > 7.5 years answer : a | a = 15 / 2
|
['a ) 1 or 7', 'b ) 3 or 21', 'c ) 4 or 28', 'd ) 2 or 14', 'e ) 5 or 14'] | c | max(add(sqrt(subtract(power(20, const_2), power(divide(32, const_2), const_2))), sqrt(subtract(power(20, const_2), power(divide(24, const_2), const_2)))), subtract(sqrt(subtract(power(20, const_2), power(divide(24, const_2), const_2))), sqrt(subtract(power(20, const_2), power(divide(32, const_2), const_2))))) | what is the distance between two parallel chords of length 32 cm and 24 cm in a circle of radius 20 cm ? | since the chord length = 2 sqrrt ( r ^ 2 - d ^ 2 ) chord length = 32 , r = 20 then we get d = 12 chord length = 24 , r = 20 then we get d = 16 then we add / subtract we get option c ) 4 or 28 answer : c | a = 20 ** 2
b = 32 / 2
c = b ** 2
d = a - c
e = math.sqrt(d)
f = 20 ** 2
g = 24 / 2
h = g ** 2
i = f - h
j = math.sqrt(i)
k = e + j
l = 20 ** 2
m = 24 / 2
n = m ** 2
o = l - n
p = math.sqrt(o)
q = 20 ** 2
r = 32 / 2
s = r ** 2
t = q - s
u = math.sqrt(t)
v = p - u
w = max(k)
|
a ) 2200 , b ) 2000 , c ) 2100 , d ) 2300 , e ) none | c | add(2000, multiply(2000, divide(5, const_100))) | one boss decided to increase the salary of an employee by 5 % . how much will he get if his salary was rs . 2000 ? | solution : actual salary is 2000 rupees 5 % of 2000 = 2000 * 5 / 100 = 100 so 100 rupees increased = 2000 + 100 = 2100 he will get salary of 2100 answer c | a = 5 / 100
b = 2000 * a
c = 2000 + b
|
a ) 35 , b ) 25 , c ) 20 , d ) 27 , e ) none | d | multiply(divide(189, multiply(7, 9)), 7) | a ratio between two numbers is 7 : 9 and their l . c . m . is 189 . the second number is | "sol . let the required numbers be 7 x and 9 x . then , their l . c . m . is 63 x . β΄ 63 x = 189 β x = 3 . hence , the second number is 27 . answer d" | a = 7 * 9
b = 189 / a
c = b * 7
|
a ) 64 % , b ) 66 % , c ) 68 % , d ) 70 % , e ) 72 % | b | divide(add(multiply(70, 60), multiply(80, subtract(100, 70))), const_100) | in a certain accounting class of 100 students , 70 % of the students took the final exam on the assigned day while the rest of the students took the exam on a make - up date . if the students on the assigned day had an average score of 60 % , and the students on the make - up date had an average score of 80 % , what was the average score for the entire class ? | 70 % of the class scored 60 % and 30 % of the class scored 80 % . the difference between 60 % and 80 % is 20 % . the average will be 60 % + 0.3 ( 20 % ) = 66 % . the answer is b . | a = 70 * 60
b = 100 - 70
c = 80 * b
d = a + c
e = d / 100
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a ) 6 , b ) 32 , c ) 120 , d ) 720 , e ) 820 | d | divide(divide(factorial(32), factorial(subtract(32, 6))), divide(factorial(32), multiply(factorial(6), factorial(subtract(32, 6))))) | if p ( 32 , 6 ) = kc ( 32 , 6 ) , then what is the value of k ? | solution since 32 p 6 = k 32 c 6 β 32 ! / ( 32 - 6 ) ! = k 32 ! / ( 32 - 6 ) ! β k = 6 ! = 720 . answer d | a = math.factorial(32)
b = 32 - 6
c = math.factorial(b)
d = a / c
e = math.factorial(32)
f = math.factorial(6)
g = 32 - 6
h = math.factorial(g)
i = f * h
j = e / i
k = d / j
|
a ) 0 , b ) 2 , c ) 4 , d ) 8 , e ) more than 8 | d | subtract(add(subtract(10, 5), 5), 2) | for how many values of q , is | | | q - 5 | - 10 | - 5 | = 2 ? ( those ls are mods ) | i think its 8 | | | q - 5 | - 10 | - 5 | = 2 let | q - 5 | = a which makes above | | a - 10 | - 5 | = 2 let | a - 10 | = b which makes | b - 5 | = 2 now for the above b can take 3 , 7 for every b = 3 a can have 13 , 7 and for b = 7 a can have 17 and 3 so ' a ' has four solutions 13 , 7 , 17 and 3 for a = 13 ; q has 18 or - 8 thus has 2 for every combination hence 4 x 2 = 8 answer d | a = 10 - 5
b = a + 5
c = b - 2
|
a ) 3 , b ) 4.5 , c ) 6 , d ) d ) 5.5 , e ) e ) 5 | c | subtract(40, 30) | a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 60 km / hr respectively . b starts three hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "the table you made does n ' t make sense to me . all three meet at the same point means the distance they cover is the same . we know their rates are 30 , 40 and 60 . say the time taken by b is t hrs . then a takes 3 + t hrs . and we need to find the time taken by k . distance covered by a = distance covered by b 30 * ( 3 + t ) = 40 * t t = 9 hrs distance covered by b = distance covered by k 40 * t = 60 * time taken by k time taken by k = 40 * 9 / 60 = 6 hrs time taken by a = 3 + t = 3 + 9 = 12 hrs time taken by k = 6 hrs so k starts 12 - 6 = 6 hrs after a . ( answer c )" | a = 40 - 30
|
a ) $ 1.40 , b ) $ 2.40 , c ) $ 3.40 , d ) $ 3.19 , e ) $ 5.40 | d | multiply(subtract(const_1, multiply(divide(add(const_100, 25), const_100), divide(const_1, const_2))), 8.50) | at joes steakhouse the hourly wage for a chef is 25 % greater than that of a dishwasher , and the hourly wage of a dishwasher is half as much as the hourly wage of a manager . if a managers wage is $ 8.50 per hour , how much less than a manager does a chef earn each hour ? | "manager wages per hour = $ 8.50 dishwasher wages per hour = half of manager ' s wages . = 1 / 2 ( $ 8.50 ) = = > $ 4.25 chef wages per hour = 25 % greater than dishwasher wages - - > 25 % of $ 4.25 = ( 25 * ( $ 4.25 ) ) / 100 - - > ( $ 106.25 ) / 100 - - > $ 1.06 therefore , chef wages per hour = $ 4.25 + $ 1.06 = = > $ 5.31 difference of wages between manager and chef = $ 8.50 - $ 5.31 = = > $ 3.19 answer : d" | a = 100 + 25
b = a / 100
c = 1 / 2
d = b * c
e = 1 - d
f = e * 8
|
a ) 33 , b ) 29 , c ) 17 , d ) 10 , e ) 16 | c | divide(divide(divide(add(multiply(add(const_1, const_2), divide(20, const_2)), divide(multiply(const_3, const_3), const_100)), divide(add(const_100, 25), const_100)), divide(add(20, const_100), const_100)), divide(add(const_100, 18), const_100)) | a manufacturer sells a pair of glasses to a wholesale dealer at a profit of 18 % . the wholesaler sells the same to retailer at a profit of 20 % . the retailer in turn sells them to a customer for rs . 30.09 , there by earning a profit of 25 % . the cost price for the manufacturer is ? | "explanation : let the cost price for the manufacturer be rs . x . then , 125 % of 120 % of 118 % of x = 30.09 . 125 / 100 * 120 / 100 * 118 / 100 * x = 3009 / 100 177 / 100 x = 3009 / 100 = > x = 17 answer : c" | a = 1 + 2
b = 20 / 2
c = a * b
d = 3 * 3
e = d / 100
f = c + e
g = 100 + 25
h = g / 100
i = f / h
j = 20 + 100
k = j / 100
l = i / k
m = 100 + 18
n = m / 100
o = l / n
|
a ) 477 , b ) 384 , c ) 279 , d ) 276 , e ) 291 | b | add(multiply(divide(60, subtract(21, 11)), 11), multiply(divide(60, subtract(21, 11)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 11 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 11 + 21 = 32 t = 12 d = 32 * 12 = 384 . answer : b" | a = 21 - 11
b = 60 / a
c = b * 11
d = 21 - 11
e = 60 / d
f = e * 21
g = c + f
|
a ) 39 yr , b ) 42 yr , c ) 45 yr , d ) 32 yr , e ) 50 yr | a | add(add(multiply(const_2, 10), 10), 9) | in 10 years , a will be twice as old as b was 10 years ago . if a is now 9 years older than b , the present age of b is ? | "let b ' s present age = x years then a ' s present age = x + 9 years x + 9 + 10 = 2 ( x - 10 ) x + 19 = 2 x - 20 x = 39 years answer is a" | a = 2 * 10
b = a + 10
c = b + 9
|
a ) 5 , b ) 2 , c ) 8 , d ) 7 , e ) 9 | c | add(const_2, const_3) | if 89 / 99 = 0.898 , what is the 83 rd digit to the right of the decimal point of the fraction ? | "we are not concerned what 89 / 99 means . . we have to look at the decimal . . 0.898 means 0.898989 . . . . so right of decimal , all odd numbered are 8 and all even numbered are 9 . . here 83 is odd , so ans is 8 answer is c" | a = 2 + 3
|
a ) $ 220.85 , b ) $ 214.85 , c ) $ 114.85 , d ) $ 314.85 , e ) $ 104.85 | c | subtract(subtract(200.5, 35.25), multiply(25.2, const_2)) | little krish had $ 200.50 . he spent $ 35.25 on sweets and gave to his two friends $ 25.20 each . how much money was left ? | krish spent and gave to his two friends a total of 35.25 + 25.20 + 25.20 = $ 85.65 money left 200.50 - 85.65 = $ 114.85 correct answer is c ) $ 114.85 | a = 200 - 5
b = 25 * 2
c = a - b
|
a ) 18 , b ) 36 , c ) 72 , d ) 90 , e ) 65 | e | subtract(divide(117, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), divide(divide(117, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), multiply(3, const_2))) | pat , kate , and mark charged a total of 117 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much time as mark , how many more hours did mark charge to the project than kate ? | "65 all u do is do 2 : 1 : 6 = > 2 x + x + 6 x = 117 = > x = 13 26 : 13 : 78 78 - 13 = 65 answer e" | a = 1 / 3
b = 1 + a
c = 3 * 2
d = 1 / c
e = b + d
f = 117 / e
g = 1 / 3
h = 1 + g
i = 3 * 2
j = 1 / i
k = h + j
l = 117 / k
m = 3 * 2
n = l / m
o = f - n
|
a ) 1 / 6 , b ) 125 / 216 , c ) 1 / 36 , d ) 81 / 216 , e ) 91 / 216 | e | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting at least one six in a single throw of three unbiased dice ? | "find the number of cases in which none of the digits show a ' 6 ' . i . e . all three dice show a number other than ' 6 ' , 5 * 5 * 5 = 125 cases . total possible outcomes when three dice are thrown = 216 . the number of outcomes in which at least one die shows a ' 6 ' = total possible outcomes when three dice are thrown - number of outcomes in which none of them show ' 6 ' . = 216 - 125 = 91 . the required probability = 91 / 216 . answer : e" | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
|
a ) - 4 , b ) 4 , c ) - 1 / 4 , d ) 1 / 4 , e ) 2 / 3 | e | divide(const_1, 3) | in the coordinate plane a slope of the line k is 3 / 2 times the x - intercept of the line k . what is the y - intercept of the line k ? | as 3 / 2 y = mx + c , from 3 / 2 y = 0 + m we get y = 2 / 3 . hence , the correct answer choice is e . | a = 1 / 3
|
a ) 9800000 , b ) 1000000 , c ) 7500000 , d ) 4000000 , e ) none of these | d | divide(multiply(multiply(multiply(8, const_100), multiply(5, const_100)), multiply(const_4.0, const_100)), multiply(multiply(8, 10), 10)) | a wooden box of dimensions 8 m x 10 m x 6 m is to carry rectangularboxes of dimensions 4 cm x 5 cm x 6 cm . the maximum number ofboxes that can be carried in the wooden box , is | explanation : number = ( 800 * 1000 * 600 ) / 4 * 5 * 6 = 4000000 answer : d | a = 8 * 100
b = 5 * 100
c = a * b
d = 4 * 0
e = c * d
f = 8 * 10
g = f * 10
h = e / g
|
a ) 1 / 15 , b ) 25 / 57 , c ) 35 / 256 , d ) 1 / 221 , e ) 2 / 15 | d | divide(multiply(divide(52, const_4), divide(52, const_4)), choose(52, const_2)) | from a pack of 52 cards , two cards are drawn together at random . what is the probability of both the cards being kings ? | "let s be the sample space . then , n ( s ) = 52 c 2 = ( 52 x 51 ) / ( 2 x 1 ) = 1326 . let e = event of getting 2 kings out of 4 . n ( e ) = 4 c 2 = ( 4 x 3 ) / ( 2 x 1 ) = 6 . p ( e ) = n ( e ) / n ( s ) = 6 / 1326 = 1 / 221 . option d" | a = 52 / 4
b = 52 / 4
c = a * b
d = math.comb(52, 2)
e = c / d
|
a ) 50.25 , b ) 52.25 , c ) 51.95 , d ) 53.25 , e ) none of the above | c | divide(add(add(add(multiply(65, 50), multiply(35, 60)), multiply(45, 55)), multiply(42, 45)), add(add(add(65, 35), 45), 42)) | a school has 4 section of chemistry in class x having 65 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student . | required average marks = 65 Γ£ β 50 + 35 Γ£ β 60 + 45 Γ£ β 55 + 42 Γ£ β 45 / 65 + 35 + 45 + 42 = 3250 + 2100 + 2475 + 1890 / 187 = 9715 Γ’ Β β 187 = 51.95 answer c | a = 65 * 50
b = 35 * 60
c = a + b
d = 45 * 55
e = c + d
f = 42 * 45
g = e + f
h = 65 + 35
i = h + 45
j = i + 42
k = g / j
|
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | a | divide(multiply(300, const_3_6), add(75, 15)) | the speed at which a girl can row a boat in still water is 75 kmph . if he rows downstream , where the speed of current is 15 kmph , what time will he take to cover 300 metres ? | "speed of the boat downstream = 75 + 15 = 90 kmph = 90 * 5 / 18 = 25 m / s hence time taken to cover 400 m = 400 / 25 = 16 seconds . answer : a" | a = 300 * const_3_6
b = 75 + 15
c = a / b
|
a ) 1 / 16 , b ) 3 / 11 , c ) 13 / 20 , d ) 3 / 10 , e ) 1 / 10 | a | multiply(divide(const_1, const_2), divide(const_1, const_4)) | x , y , and z are all unique numbers . if x is chosen randomly from the set { 8 , 9 , 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ? | "p ( x is prime ) = 1 / 4 p ( y is prime ) = 1 / 4 if y is prime , then z is not prime since y and z are unique . then the probability is 1 / 4 * 1 / 4 = 1 / 16 the answer is a ." | a = 1 / 2
b = 1 / 4
c = a * b
|
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