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a ) 180 cm , b ) 220 cm , c ) 240 cm , d ) 270 cm , e ) 300 cm | e | add(triangle_perimeter(50, 50, 50), triangle_perimeter(50, 50, 50)) | an equilateral triangle t 2 is formed by joining the mid points of the sides of another equilateral triangle t 1 . a third equilateral triangle t 3 is formed by joining the mid - points of t 2 and this process is continued indefinitely . if each side of t 1 is 50 cm , find the sum of the perimeters of all the triangles . | "we have 50 for first triangle , when we join mid - points of first triangle we get the second equilateral triangle then the length of second one is 25 and continues . so we have 50,25 , 12.5 , . . . we have ratio = 1 / 2 , and it is gp type . sum of infinite triangle is a / 1 - r = 50 / 1 - ( 1 / 2 ) = 100 equilateral triangle perimeter is 3 a = 3 * 100 = 300 . so option e ." | a = triangle_perimeter + (
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a ) 5 square meters , b ) 36 square meters , c ) 40 square meters , d ) 83.3 square meters , e ) 120 square meters | c | divide(subtract(subtract(300, 180), 40), const_2) | three walls have wallpaper covering a combined area of 300 square meters . by overlapping the wallpaper to cover a wall with an area of 180 square meters , the area that is covered by exactly two layers of wallpaper is 40 square meters . what is the area that is covered with three layers of wallpaper ? | "300 - 180 = 120 sq m of the wallpaper overlaps ( in either two layers or three layers ) if 36 sq m has two layers , 120 - 40 = 80 sq m of the wallpaper overlaps in three layers . 80 sq m makes two extra layers hence the area over which it makes two extra layers is 40 sq m . answer ( c ) ." | a = 300 - 180
b = a - 40
c = b / 2
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a ) 16 , b ) 18 , c ) 32 , d ) 19 , e ) 12 | c | divide(multiply(40, 4), add(divide(40, 40), divide(multiply(2, 40), 20))) | a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 4 x km . | "total time taken = x / 40 + 2 x / 20 hours = 5 x / 40 = x / 8 hours average speed = 4 x / ( x / 8 ) = 32 kmph answer : c" | a = 40 * 4
b = 40 / 40
c = 2 * 40
d = c / 20
e = b + d
f = a / e
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a ) - 0.0053 , b ) - 0.0003 , c ) 0.0007 , d ) 0.0046 , e ) 0.0153 | d | subtract(subtract(2.0254, divide(divide(add(multiply(const_2, const_10), const_3), const_100), const_100)), 2.0254) | if n = 2.0254 and n * is the decimal obtained by rounding n to the nearest hundredth , what is the value of n * – n ? | "n * = 2.03 n * - n = 2.03 - 2.0254 0.0046 answer : d" | a = 2 * 10
b = a + 3
c = b / 100
d = c / 100
e = 2 - 254
f = e - 2
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a ) 217 , b ) 287 , c ) 400 , d ) 288 , e ) 171 | a | divide(240, power(add(const_1, divide(5, const_100)), 2)) | find the sum lend at c . i . at 5 p . c per annum will amount to rs . 240 in 2 years ? | "explanation : 240 = p ( 21 / 20 ) 2 p = 217 answer : a" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 240 / c
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a ) 4 / 13 , b ) 6 / 13 , c ) 8 / 21 , d ) 11 / 21 , e ) 17 / 42 | b | add(multiply(divide(7, add(7, 7)), divide(subtract(7, const_1), subtract(add(7, 7), const_1))), multiply(divide(subtract(7, const_1), subtract(add(7, 7), const_1)), divide(7, add(7, 7)))) | a bag contains 7 green balls and 7 white balls . if two balls are drawn simultaneously , what is the probability that both balls are the same colour ? | "the total number of ways to draw two balls is 14 c 2 = 91 the number of ways to draw two green balls is 7 c 2 = 21 the number of ways to draw two white balls is 7 c 2 = 21 p ( two balls of the same colour ) = 42 / 91 = 6 / 13 the answer is b ." | a = 7 + 7
b = 7 / a
c = 7 - 1
d = 7 + 7
e = d - 1
f = c / e
g = b * f
h = 7 - 1
i = 7 + 7
j = i - 1
k = h / j
l = 7 + 7
m = 7 / l
n = k * m
o = g + n
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a ) 17 , b ) 23 , c ) 77 , d ) 26 , e ) 30 | b | add(multiply(subtract(const_100, 60), divide(15, const_100)), multiply(divide(50, const_100), subtract(subtract(const_100, 60), multiply(subtract(const_100, 60), divide(15, const_100))))) | a vendor sells 60 percent of apples he had and throws away 15 percent of the remainder . next day he sells 50 percent of the remainder and throws away the rest . what percent of his apples does the vendor throw ? | explanation : let , the number of apples be 100 . on the first day he sells 60 % apples i . e . , 60 apples . remaining apples = 40 . he throws 15 % of the remaining i . e . , 15 % of 40 = 6 . now he has 40 − 6 = 34 apples the next day he throws 50 % of the remaining 34 apples i . e . , 17 . therefore , in total he throws 6 + 17 = 23 apples . answer : b | a = 100 - 60
b = 15 / 100
c = a * b
d = 50 / 100
e = 100 - 60
f = 100 - 60
g = 15 / 100
h = f * g
i = e - h
j = d * i
k = c + j
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a ) 2816 , b ) 5500 , c ) 3300 , d ) 1100 , e ) 4400 | a | multiply(circumface(8), 14) | the radius of a cylinder is 8 m , height 14 m . the volume of the cylinder is : | "cylinder volume = ï € r ( power 2 ) h = 22 / 7 ã — 8 ã — 8 ã — 14 = 2816 m ( power 3 ) answer is a ." | a = circumface * (
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a ) 800 , b ) 900 , c ) 1600 , d ) 1100 , e ) 1200 | c | divide(multiply(subtract(880, 800), 4), divide(20, const_100)) | average of money that group of 4 friends pay for rent each month is $ 800 . after one persons rent is increased by 20 % the new mean is $ 880 . what was original rent of friend whose rent is increased ? | 0.2 x = 4 ( 880 - 800 ) 0.2 x = 320 x = 1600 answer c | a = 880 - 800
b = a * 4
c = 20 / 100
d = b / c
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a ) 5.5 kg , b ) 11 kg , c ) 30 kg , d ) 6.5 kg , e ) 71 kg | d | subtract(43, divide(subtract(add(multiply(30, subtract(43, add(30, 1))), 43), multiply(subtract(subtract(43, add(30, 1)), 1), 30)), const_2)) | when a student joe , weighing 43 kg , joins a group of students whose average weight is 30 kg , the average weight goes up by 1 kg . subsequently , if two students , excluding joe , leave the group the average weight comes back to 30 kg . what is the difference between the average weight of the two students who left and the weight of joe ? | "after two persons leave the group the average remains the same . that means the weight of the two persons = 43 + 30 = 73 so , the average the two persons = 36.5 that gives the answer 43 - 36.5 = 6.5 answer d" | a = 30 + 1
b = 43 - a
c = 30 * b
d = c + 43
e = 30 + 1
f = 43 - e
g = f - 1
h = g * 30
i = d - h
j = i / 2
k = 43 - j
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a ) 2 hr , b ) 2 hr 20 min , c ) 2 hr 40 min , d ) 5 hr 40 min , e ) 6 hr | c | divide(480, const_2) | while working alone at their respective constant rates , server p uploads 480 files in 4 hours and server y uploads 480 files in 8 hours . if all files uploaded by these servers are the same size , how long would it take the two servers , working at the same time and at their respective constant rates , to process a total of 480 files ? | server p processes 480 / 4 files per hour = 120 per hour server y processes 180 / 8 files per hour = 60 per hour total files processed per hour when p and y work together = 120 + 60 per hour = 180 files per hour 480 / 180 = 2 2 / 3 hours = c | a = 480 / 2
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a ) 0.7 , b ) 0 , c ) 0.8 , d ) 8 , e ) 7 | c | divide(subtract(multiply(0.0203, 2.92), multiply(0.0073, 14.5)), 0.7) | 0.0203 x 2.92 / 0.0073 x 14.5 x 0.7 = ? | "= 0.0203 x 2.92 / 0.0073 x 14.5 x 0.7 = 203 x 292 / 73 x 145 x 7 = 4 / 5 = 0.8 answer is c ." | a = 0 * 203
b = 0 * 73
c = a - b
d = c / 0
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a ) 15 , b ) 66 , c ) 77 , d ) 52 , e ) 45 | e | multiply(divide(30, const_60), 90) | the speed of a train is 90 kmph . what is the distance covered by it in 30 minutes ? | "90 * 30 / 60 = 45 kmph answer : e" | a = 30 / const_60
b = a * 90
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a ) 5 , b ) 64 , c ) 62 , d ) 60 , e ) 12 | b | add(const_3, const_4) | what is the smallest positive integer x such that 280 - x is the cube of a positive integer | "given 280 - x is a perfect cube so we will take 216 = 6 * 6 * 6 280 - x = 216 x = 280 - 216 = 64 correct option is b" | a = 3 + 4
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a ) 221 , b ) 235 , c ) 245 , d ) 289 , e ) 260 | b | subtract(divide(multiply(const_1, const_1000), divide(60, 15)), 15) | in a kilometer race , a beats b by 60 meters or 15 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 15 ) / 60 = 250 sec . time taken by a = 250 - 15 = 235 sec . answer : b" | a = 1 * 1000
b = 60 / 15
c = a / b
d = c - 15
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a ) 78.08 , b ) 73.6 , c ) 75.2 , d ) 76.8 , e ) 81.6 | a | multiply(add(add(7.5, 8.3), 8.6), 3.2) | in a certain diving competition , 5 judges score each dive on a scale from 1 to 10 . the point value of the dive is obtained by dropping the highest score and the lowest score and multiplying the sum of the remaining scores by the degree of difficulty . if a dive with a degree of difficulty of 3.2 received scores of 7.5 , 8.3 , 9.0 , 6.0 , and 8.6 , what was the point value of the dive ? | "degree of difficulty of dive = 3.2 scores are 6.0 , 7.5 , 8.0 , 8.5 and 9.0 we can drop 6.0 and 9.0 sum of the remaining scores = ( 7.5 + 8.3 + 8.6 ) = 24 point of value of the dive = 24.4 * 3.2 = 78.08 answer a" | a = 7 + 5
b = a + 8
c = b * 3
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a ) 5 th , b ) 4 th , c ) 6 th , d ) 7 th , e ) 8 th | b | add(3, const_1) | train starts from amritsar to bombay at 9 am . it reaches destination after 3 days at 9 : 30 am . every day a train starts . how many trains does it come across on the way ? | because one train taking 3 days so 1 train first day another at second day and 3 rd one at third day but as given train reaches at 9.30 am insted of 9 . oo am so d same day train already left that train will be 4 th train . answer : b | a = 3 + 1
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a ) − 48 , b ) − 2 , c ) 2 , d ) 9 , e ) 48 | d | subtract(subtract(subtract(subtract(add(add(9, 10), subtract(9, 10)), const_1), const_1), const_1), const_1) | if a ( a - 9 ) = 10 and b ( b - 9 ) = 10 , where a ≠ b , then a + b = | "i . e . if a = - 1 then b = 10 or if a = 10 then b = - 1 but in each case a + b = - 1 + 10 = 9 answer : option d" | a = 9 + 10
b = 9 - 10
c = a + b
d = c - 1
e = d - 1
f = e - 1
g = f - 1
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a ) 15 , b ) 16 , c ) 28 , d ) 56 , e ) 132 | e | divide(multiply(12, subtract(12, const_1)), const_2) | there are 12 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 11 teams . . . so total no of matches = 12 x 11 = 132 . now , each match is played twice = > 132 x 2 but 2 teams play a match = > 132 x 2 / 2 = 132 . answer : e" | a = 12 - 1
b = 12 * a
c = b / 2
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a ) 2 % , b ) 7 % , c ) 9 % , d ) 3 % , e ) 1 % | a | divide(multiply(divide(1, 5), const_100), 10) | at what rate percent per annum will the simple interest on a sum of money be 1 / 5 of the amount in 10 years ? | "let sum = x . then , s . i . = x / 5 , time = 10 years . rate = ( 100 * x ) / ( x * 5 * 10 ) = 2 % answer : a" | a = 1 / 5
b = a * 100
c = b / 10
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a ) 6.3 , b ) 6.9 , c ) 7.1 , d ) 6.1 , e ) 6.2 | c | divide(const_1, add(divide(const_1, 25), add(divide(const_1, 15), divide(const_1, 10)))) | a man can do a job in 15 days . his father takes 10 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ? | "1 day work of the three persons = ( 1 / 15 + 1 / 10 + 1 / 25 ) = 21 / 150 so , all three together will complete the work in 150 / 21 = 7.1 days . answer : c" | a = 1 / 25
b = 1 / 15
c = 1 / 10
d = b + c
e = a + d
f = 1 / e
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a ) 30 , b ) 35 , c ) 20 , d ) 18 , e ) 10 | b | subtract(subtract(add(subtract(90, 11), 22), 44), 22) | in a neighborhood having 90 households , 11 did not have either a car or a bike . if 22 households had a both a car and a bike and 44 had a car , how many had bike only ? | { total } = { car } + { bike } - { both } + { neither } - - > 90 = 44 + { bike } - 22 + 11 - - > { bike } = 57 - - > # those who have bike only is { bike } - { both } = 57 - 22 = 35 . answer : b . | a = 90 - 11
b = a + 22
c = b - 44
d = c - 22
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a ) 38 , b ) 40 , c ) 78 , d ) 88 , e ) 125 | b | multiply(multiply(multiply(divide(subtract(11, 5), 3), const_2), divide(subtract(11, 5), 3)), 5) | the 3 digits of a number add up to 11 . the number is divisible by 5 . the leftmost digit is double the middle digit . what is the product of the 3 digits ? | xyz is a 3 digit integer divisible by 5 . so z can take only 0 or 5 . we are also given that x = 2 y this means x + y ( 2 y + y ) must be a multiple of 3 xy 5 or xy 0 case 1 x + y + 0 = 11 x + y = 11 this ca n ' t be the case because 11 is not a multiple of 3 case 2 x + y + 5 = 11 x + y = 6 solving we get , x = 4 y = 2 so the 3 digit number is 425 thus the product of the digits will be = 4 x 2 x 5 = 40 answer b | a = 11 - 5
b = a / 3
c = b * 2
d = 11 - 5
e = d / 3
f = c * e
g = f * 5
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a ) 1 / 32 , b ) 1 / 28 , c ) 1 / 24 , d ) 1 / 16 , e ) 1 / 14 | c | divide(const_1, add(20, const_4)) | if a randomly selected positive single digit multiple of 3 is multiplied by a randomly selected prime number less than 20 , what is the probability t that this product will be a multiple of 45 ? | "there are 3 single digit multiple of 3 , that is , 3 , 6,9 . there are 8 prime nos less than 20 - 2,3 , 5,7 , 11,13 , 17,19 total outcome - 8 * 3 = 24 favourable outcome = 1 ( 9 * 5 ) hence required probability t = 1 / 24 . answer c ." | a = 20 + 4
b = 1 / a
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a ) 66 % , b ) 64 % , c ) 68 % , d ) 69 % , e ) 72 % | c | subtract(const_100, add(multiply(8, 2), multiply(2, 8))) | uba capital recently bought brand new vehicles for office use . uba capital only went for toyota and honda and bought more of toyota than honda at the ratio of 8 : 2 . if 80 % of the toyota bought and 20 % of the honda bought were suv â € ™ s . how many suv â € ™ s did uba capital buy in the aforementioned purchase ? | "let total no of vehicles bought be 100 , toyota 80 and honda 20 , so total number of suv ' s bought for toyota and honda respectively 80 * 80 / 100 = 64 and 20 * 20 / 100 = 4 so total 68 suv ' s were bought out of 100 vehicles bought . . so required % is 68 % answer : c" | a = 8 * 2
b = 2 * 8
c = a + b
d = 100 - c
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a ) 100 , b ) 10 , c ) 10000 , d ) 100000 , e ) 1000000 | d | power(10, subtract(655, 650)) | 10 ^ ( 655 ) ã · 10 ^ ( 650 ) = ? | "10 ^ ( 655 ) ã · 10 ^ ( 651 ) = 10 ^ ( 655 - 651 ) = 10 ^ 5 = 100000 answer : d" | a = 655 - 650
b = 10 ** a
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a ) 50 % , b ) 20 % , c ) 5 % , d ) 2 % , e ) 0.22 % | e | multiply(divide(multiply(250.10, power(10, add(const_3, const_3))), multiply(115.19, power(10, 9))), const_100) | a corporation that had $ 115.19 billion in profits for the year paid out $ 250.10 million in employee benefits . approximately what percent of the profits were the employee benefits ? ( note : 1 billion = 10 ^ 9 ) | "required answer = [ employee benefit / profit ] * 100 = [ ( 250.10 million ) / ( 115.19 billion ) ] * 100 = [ ( 250.10 * 10 ^ 6 ) / ( 115.19 * 10 ^ 9 ) ] * 100 = ( 2.2 / 1000 ) * 100 = 0.22 % so answer is ( e )" | a = 3 + 3
b = 10 ** a
c = 250 * 10
d = 10 ** 9
e = 115 * 19
f = c / e
g = f * 100
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a ) 1.1 , b ) 2.2 , c ) 2.5 , d ) 2.1 , e ) 2.9 | a | multiply(divide(divide(multiply(divide(10, const_100), 100), 10), multiply(divide(10, const_100), 100)), const_100) | a reduction of 10 % in the price of salt enables a lady to obtain 10 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 10 / 100 ) = 10 - - - 10 ? - - - 1 = > rs . 1 100 - - - 90 ? - - - 1 = > rs . 1.1 answer : a" | a = 10 / 100
b = a * 100
c = b / 10
d = 10 / 100
e = d * 100
f = c / e
g = f * 100
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a ) a ) 10 , b ) b ) 25 , c ) c ) 40 , d ) d ) 55 , e ) e ) 70 | e | subtract(multiply(16, 62.5), multiply(subtract(16, const_1), 62.0)) | the average ( arithmetic mean ) of 16 students first quiz scores in a difficult english class is 62.5 . when one student dropped the class , the average of the remaining scores increased to 62.0 . what is the quiz score of the student who dropped the class ? | "total score of 16 students is 16 * 62.50 = 1000 total score of 15 students is 15 * 62 = 930 so , the score of the person who left is 70 ( 1000 - 930 ) answer will be ( e )" | a = 16 * 62
b = 16 - 1
c = b * 62
d = a - c
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a ) 1 : 64 , b ) 1 : 62 , c ) 1 : 34 , d ) 8 : 27 , e ) 5 : 64 | d | divide(power(2, const_3), power(3, const_3)) | the triplicate ratio of 2 : 3 is ? | 2 ^ 3 : 3 ^ 3 = 8 : 27 answer : d | a = 2 ** 3
b = 3 ** 3
c = a / b
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a ) 1 minutes , b ) 15 minute , c ) 100 minutes , d ) 10000 minutes , e ) 1000 minutes | b | multiply(divide(15, 15), 15) | if 15 lions can kill 15 deers in 15 minutes how long will it take 100 lions to kill 100 deers ? | we can try the logic of time and work , our work is to kill the deers so 15 ( lions ) * 15 ( min ) / 15 ( deers ) = 100 ( lions ) * x ( min ) / 100 ( deers ) hence answer is x = 15 answer : b | a = 15 / 15
b = a * 15
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | multiply(subtract(add(const_1, floor(divide(1076, 23))), divide(1076, 23)), 23) | what is the least number that should be added to 1076 , so the sum of the number is divisible by 23 ? | ( 1076 / 23 ) gives a remainder 18 so we need to add 5 . the answer is e . | a = 1076 / 23
b = math.floor(a)
c = 1 + b
d = 1076 / 23
e = c - d
f = e * 23
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a ) 1 / 40 , b ) 8 / 75 , c ) 75 / 4 , d ) 75 / 8 , e ) 75 / 6 | b | divide(40, 375) | the sum of two numbers is 40 and their product is 375 . what will be the sum of their reciprocals ? | ( 1 / a ) + ( 1 / b ) = ( a + b ) / ab = 40 / 375 = 8 / 75 answer : b | a = 40 / 375
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 2 / 5 , e ) 2 / 7 | a | divide(const_1, const_2) | in a throw of a coin find the probability of getting a tail ? | "s = { h , t } e = { t } p ( e ) = 1 / 2 answer is a" | a = 1 / 2
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a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 550 | b | multiply(100, const_2) | on her annual road trip to visit her family in seal beach , california , traci stopped to rest after she traveled 1 ⁄ 3 of the total distance and again after she traveled 1 ⁄ 2 of the distance remaining between her first stop and her destination . she then drove the remaining 100 miles and arrived safely at her destination . what was the total distance , in miles , from traci ’ s starting point to seal beach ? | "let d = total distance traci traveled 1 / 3 = d / 3 i . e . remaining distance = 2 d / 3 she traveled 1 / 2 th of 2 d / 3 = d / 3 thus : d = ( d / 3 ) + ( d / 3 ) + 100 d = 300 answer : b" | a = 100 * 2
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a ) 408 km , b ) 409 km , c ) 410 km , d ) 412 km , e ) 419 km | b | multiply(inverse(add(inverse(multiply(const_2, 30)), inverse(multiply(25, const_2)))), 15) | pavan travelled for 15 hours . he covered the first half of the distance at 30 kmph and remaining half of the distance at 25 kmph . find the distance travelled by pavan . | let the distance travelled be x km . total time = ( x / 2 ) / 30 + ( x / 2 ) / 25 = 15 = > x / 60 + x / 50 = 15 = > ( 5 x + 6 x ) / 300 = 15 = > x = 409 km answer : b | a = 2 * 30
b = 1/(a)
c = 25 * 2
d = 1/(c)
e = b + d
f = 1/(e)
g = f * 15
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a ) 1 , b ) 8 , c ) 12 , d ) 20 , e ) 7 | d | multiply(5, const_4) | how many quarters are equal to 5 dollars ? | "5 * 4 = 20 quarters answer : d" | a = 5 * 4
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a ) 100 , b ) 500 , c ) 600 , d ) 800 , e ) 200 | e | divide(12, subtract(204.06, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 12 . if n / j = 204.06 , 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 = 204.06 here 204 is the quotient . given that remainder = 12 so , 204.06 = 204 + 12 / j so , j = 200 answer : e" | a = 4 * 10
b = a + 2
c = 100 + b
d = 204 - 6
e = 12 / d
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a ) $ 1.56 , b ) $ 2.40 , c ) $ 3.80 , d ) $ 4.50 , e ) $ 2.80 | d | add(1.00, multiply(subtract(divide(1.00, divide(1, 5)), 1), 0.25)) | if taxi fares were $ 1.00 for the first 1 / 5 mile and $ 0.25 for each 1 / 5 mile there after , then the taxi fare for a 3 - mile ride was | "in 3 miles , initial 1 / 5 mile charge is $ 1 rest of the distance = 3 - ( 1 / 5 ) = 14 / 5 rest of the distance charge = 14 ( 0.25 ) = $ 3.5 ( as the charge is 0.25 for every 1 / 5 mile ) = > total charge for 3 miles = 1 + 3.5 = 4.5 answer is d ." | a = 1 / 5
b = 1 / 0
c = b - 1
d = c * 0
e = 1 + 0
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a ) 3 , b ) 2 , c ) 1 , d ) 4 , e ) 5 | b | multiply(multiply(divide(50, const_100), divide(2, 5)), const_10) | during a particular baseball game , the probability of a team ' s pitcher throwing a strike on each throw is 2 / 5 . what is the least number of times that the pitcher should pitch the ball that will increase the probability of getting a strike at least once to more than 50 % . | rule of subtraction : p ( a ) = 1 - p ( a ' ) rule of multiplication : p ( a ∩ b ) = p ( a ) p ( b ) the probability that the pitcher throws a strike at least once in 2 guesses is 1 - ( 3 / 5 ) ^ 2 = 1 - 9 / 25 = 16 / 25 > 50 % . answer : b | a = 50 / 100
b = 2 / 5
c = a * b
d = c * 10
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a ) - 65 , b ) - 64 , c ) - 32 , d ) 32 , e ) 64 | b | subtract(negate(64), const_1) | y = x ^ 2 + bx + 64 cuts the x axis at ( h , 0 ) and ( k , 0 ) . if h and k are integers , what is the least value of b ? | "as the curve cuts the x - axis at ( h , 0 ) and ( k , 0 ) . therefore h , k are the roots of the quadratic equation . for the quadratic equation is in the form ofax ^ 2 + bx + c = 0 , the product of the roots = c / a = 64 / 1 = 256 and the sum of the roots = - b / a = - b 64 can be expressed as product of two numbers in the following ways : 1 * 64 2 * 32 4 * 16 the sum of the roots is maximum when the roots are 1 and 64 and the maximum sum is 1 + 64 = 65 . the least value possible for b is therefore - 65 . b" | a = negate - (
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a ) 8 % , b ) 6 % , c ) 21 % , d ) 14 % , e ) 17 % | d | subtract(subtract(add(20, const_100), divide(multiply(add(20, const_100), 5), const_100)), const_100) | a merchant marks his goods up by 20 % and then offers a discount of 5 % on the marked price . what % profit does the merchant make after the discount ? | "let the price be 100 . the price becomes 120 after a 20 % markup . now a discount of 5 % on 120 . profit = 114 - 100 14 % answer d" | a = 20 + 100
b = 20 + 100
c = b * 5
d = c / 100
e = a - d
f = e - 100
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a ) rs . 21.81 , b ) rs . 12 , c ) rs . 12.25 , d ) rs . 12.31 , e ) none | a | divide(multiply(16, add(const_100, 20)), subtract(const_100, 12)) | a fruit seller sells mangoes at the rate of rs . 16 per kg and thereby loses 12 % . at what price per kg , he should have sold them to make a profit of 20 % ? | "solution 88 : 16 = 120 : x x = ( 16 ã — 120 / 88 ) = rs . 21.81 hence , s . p per kg = rs . 21.81 answer a" | a = 100 + 20
b = 16 * a
c = 100 - 12
d = b / c
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a ) 45 , b ) 58.33 % , c ) 600 / 11 , d ) 55 , e ) 35 | b | multiply(divide(subtract(120, add(multiply(5, 5), multiply(5, 5))), 120), const_100) | a batsman scored 120 runs which included 5 boundaries and 5 sixes . what percent of his total score did he make by running between the wickets ? | "explanation : number of runs made by running , = > 120 − ( 5 × 4 + 5 × 6 ) . = > 120 − ( 50 ) . = > 70 . hence , the required percentage is : - = > 70 / 120 * 100 = > 58.33 % answer : b" | a = 5 * 5
b = 5 * 5
c = a + b
d = 120 - c
e = d / 120
f = e * 100
|
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | multiply(divide(subtract(30, 6), 4), 3) | right now , the ratio between the ages of sandy and molly is 4 : 3 . after 6 years , sandy ’ s age will be 30 years . what is molly ' s age right now ? | "now , sandy is 30 - 6 = 24 molly ' s age is ( 3 / 4 ) * 24 = 18 the answer is d ." | a = 30 - 6
b = a / 4
c = b * 3
|
a ) 230 m , b ) 240 m , c ) 260 m , d ) 270 m , e ) 250 m | a | subtract(multiply(multiply(72, const_0_2778), 26), 290) | a goods train runs at a speed of 72 kmph and crosses a 290 m long platform in 26 seconds . what is the length of the goods train ? | "s = 290 + x / t 72 * 5 / 18 = 290 + x / 26 x = 230 answer : a" | a = 72 * const_0_2778
b = a * 26
c = b - 290
|
a ) 40 % , b ) 25 % , c ) 35 % , d ) 30 % , e ) 32 % | e | multiply(divide(subtract(multiply(1100, add(const_1, divide(20, const_100))), 1000), 1000), const_100) | two years ago , ram put $ 1000 into a savings account . at the end of the first year , his account had accrued $ 100 in interest bringing his total balance to $ 1100 . the next year , his account balance increased by 20 % . at the end of the two years , by what percent has sam ' s account balance increased from his initial deposit of $ 1000 ? | investment 1000 dollars 1 st year total gained = 100 total amount end of first year = 1100 second year account increased by 20 % = 1100 * 0.2 = 220 therefore total amount by second year end = 1320 so total percentage increase in money = ( 1320 - 1000 ) * 100 / 1000 = 32 % correct answer e = 32 % | a = 20 / 100
b = 1 + a
c = 1100 * b
d = c - 1000
e = d / 1000
f = e * 100
|
a ) 100 , b ) 300 , c ) 400 , d ) 3,000 , e ) 4,000 | c | multiply(divide(12, subtract(99, 96)), const_100) | in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio of those accepted to those rejected would be 99 to 1 . how many r eggs does the plant process per day ? | "straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if the no . of eggs inspected remain and 12 more eggs get accepted that means there r = 384 + 12 = 396 eggs accepted and 4 rejected . . . and the ratio will be 99 : 1 bingo . . . this is what the questions says . . . . its always a good idea to start with c ." | a = 99 - 96
b = 12 / a
c = b * 100
|
a ) 160 , b ) 150 , c ) 100 , d ) 80 , e ) 50 | e | divide(subtract(multiply(200, divide(16, const_100)), 30), subtract(divide(16, const_100), divide(12, const_100))) | an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 30 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 30 gallons then 0.12 a + 0.16 ( 200 - a ) = 30 - - > a = 50 . answer : e ." | a = 16 / 100
b = 200 * a
c = b - 30
d = 16 / 100
e = 12 / 100
f = d - e
g = c / f
|
a ) 5 liters , b ) 10 liters , c ) 15 liters , d ) 8 liters , e ) 6 liters | d | subtract(24, divide(multiply(24, 40), 60)) | what quantity of water should taken out to concentrate 24 liters of 40 % acidic liquid to 60 % acidic liquid ? | "required answer is = 24 ( 60 - 40 ) / 60 = 8 liters answer is d" | a = 24 * 40
b = a / 60
c = 24 - b
|
a ) 1 , b ) 2 , c ) 17 / 5 , d ) 18 / 5 , e ) 4 | b | divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 2) | if 2 x + y = 7 and x + 2 y = 5 , then 2 xy / 3 = ? | "2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 ( 2 * 3 * 1 ) / 3 = 6 / 3 = 2 b" | a = 7 * 2
b = a - 5
c = 2 * 2
d = c - 1
e = b / d
f = 7 * 2
g = f - 5
h = 2 * 2
i = h - 1
j = g / i
k = 2 * j
l = 7 - k
m = e + l
n = m / 2
|
a ) 1 : 1 , b ) 3 : 2 , c ) 1 : 3 , d ) 2 : 5 , e ) 5 : 7 | a | divide(subtract(multiply(divide(50, add(4, 6)), 4), divide(multiply(divide(50, add(3, 2)), 2), add(1, 3))), subtract(multiply(divide(50, add(4, 6)), 6), multiply(divide(multiply(divide(50, add(3, 2)), 2), add(1, 3)), 3))) | in a colony of 50 residents , the ratio of the number of men and women is 3 : 2 . among the women , the ratio of the educated to the uneducated is 1 : 3 . if the ratio of the number of education to uneducated persons is 4 : 6 , then find the ratio of the number of educated and uneducated men in the colony ? | "number of men in the colony = 3 / 5 ( 50 ) = 30 number of women in the colony = 2 / 5 ( 50 ) = 20 number of educated women in the colony = 1 / 4 ( 20 ) = 5 number of uneducated women in the colony = 3 / 4 ( 20 ) = 15 number of educated persons in the colony = 4 / 10 ( 50 ) = 20 as 5 females are educated , remaining 15 educated persons must be men . number of uneducated men in the colony = 30 - 15 = 15 number of educated men and uneducated men are in the ratio 15 : 15 = > 1 : 1 answer : a" | a = 4 + 6
b = 50 / a
c = b * 4
d = 3 + 2
e = 50 / d
f = e * 2
g = 1 + 3
h = f / g
i = c - h
j = 4 + 6
k = 50 / j
l = k * 6
m = 3 + 2
n = 50 / m
o = n * 2
p = 1 + 3
q = o / p
r = q * 3
s = l - r
t = i / s
|
a ) 8 % , b ) 9 % , c ) 11 % , d ) 12.5 % , e ) 14.8 % | c | multiply(divide(subtract(multiply(divide(2, 15), 135), 18), multiply(divide(2, 15), 135)), const_100) | a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 135 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage is dose : weight : : 2 : 15 . now if weight is 135 ( multiplying factor is 9 : ( 135 / 15 ) ) then typical dosage would be 2 * 9 = 18 cc . dosage = 18 cc . dosage is greater by 2 cc . % dosage is greater : ( 2 / 18 ) * 100 = 11.11 % c is the answer ." | a = 2 / 15
b = a * 135
c = b - 18
d = 2 / 15
e = d * 135
f = c / e
g = f * 100
|
a ) 150 meter , b ) 145 meter , c ) 160 meter , d ) 135 meter , e ) none of these | c | multiply(divide(multiply(72, const_1000), const_3600), 8) | a train running at the speed of 72 km / hr crosses a pole in 8 seconds . find the length of the train . | "explanation : speed = 72 * ( 5 / 18 ) m / sec = 20 m / sec length of train ( distance ) = speed * time = 20 * 8 = 160 meter option c" | a = 72 * 1000
b = a / 3600
c = b * 8
|
a ) 6 , 030,053 , b ) 6 , 002,001 , c ) 6 , 030,055 , d ) 6 , 030,056 , e ) 6 , 030,057 | b | divide(power(1,000, 2), power(1,000, 2)) | 1,000 ^ 2 + 1,001 ^ 2 = | "interesting problem . i think key is to notice that all the given answer choices differs in last two digits . therefore , our entire focus should be to figure out how the given terms contribute to last two digits of total . 1000 ^ 2 - > 00 1001 ^ 1 - > 01 total - > * 1 answer b ." | a = 1 ** 0
b = 1 ** 0
c = a / b
|
a ) 26.34 litres , b ) 27.36 litres , c ) 28 litres , d ) 29.16 litres , e ) 30 litres | d | multiply(power(subtract(const_1, divide(4, 40)), const_3), 40) | a container contains 40 litres of milk . from this container 4 litres of milk was taken out and replaced by water . this process was repeated further two times . how much milk is now contained by the container ? | "amount of milk left after 3 operations = [ 40 ( 1 - 4 / 40 ) ^ 3 ] litres = 40 x 9 / 10 x 9 / 10 x 9 / 10 = 29.16 litres answer : d" | a = 4 / 40
b = 1 - a
c = b ** 3
d = c * 40
|
a ) a ) 8239 , b ) b ) 2900 , c ) c ) 6000 , d ) d ) 2393 , e ) e ) 2009 | c | multiply(multiply(subtract(4, 3), 3000), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 3000 more than d , what is b ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x rs . respectively . then , 4 x - 3 x = 3000 = > x = 3000 . b ' s share = rs . 2 x = 2 * 3000 = rs . 6000 . answer : c" | a = 4 - 3
b = a * 3000
c = b * 3
|
a ) 4063 , b ) 3846 , c ) 5351 , d ) 6000 , e ) 6154 | b | divide(multiply(5000, divide(5, const_100)), divide(6.5, const_100)) | last year a certain bond price with a face value of 5000 yielded 5 % of its face value in interest . if that interest was approx 6.5 of the bond ' s selling price approx what was the bond ' s selling price ? | "interest = 0.05 * 5000 = 0.065 * selling price - - > selling price = 0.05 * 5000 / 0.065 - - > selling price = ~ 3,846 answer : b ." | a = 5 / 100
b = 5000 * a
c = 6 / 5
d = b / c
|
a ) 2 ! , b ) 5 ! , c ) 4 ! , d ) 3 ! , e ) 6 ! | c | subtract(divide(divide(5, const_1), const_3), const_3) | in how many ways can 5 people be arranged in a circle ? | "( 5 − 1 ) ! = 4 ! = 24 ways ans - c" | a = 5 / 1
b = a / 3
c = b - 3
|
a ) 3 , b ) 9 , c ) 10 , d ) 12 , e ) 14 | c | subtract(subtract(add(subtract(subtract(subtract(subtract(subtract(subtract(subtract(30, 4), 4), 2), 2), const_1), const_3.0), const_1), 2), 3), 4) | if 3 a – 2 b – 2 c = 30 and √ 3 a - √ ( 2 b + 2 c ) = 4 , what is the value of a + b + c ? | "when we look at the two equations , we can relize some similarity , so lets work on it . . 3 a – 2 b – 2 c = 32 can be written as √ 3 a ^ 2 - √ ( 2 b + 2 c ) ^ 2 = 32 { √ 3 a - √ ( 2 b + 2 c ) } { √ 3 a + √ ( 2 b + 2 c ) } = 32 . . or 4 * √ 3 a + √ ( 2 b + 2 c ) = 32 . . or √ 3 a + √ ( 2 b + 2 c ) = 8 . . now lets work on these two equations 1 ) √ 3 a - √ ( 2 b + 2 c ) = 4 . . 2 ) √ 3 a + √ ( 2 b + 2 c ) = 8 . . a ) add the two eq . . √ 3 a + √ ( 2 b + 2 c ) + √ 3 a - √ ( 2 b + 2 c ) = 12 . . 2 √ 3 a = 12 . . or √ 3 a = 6 . . 3 a = 36 . . a = 12 . b ) subtract 1 from 2 . . √ 3 a + √ ( 2 b + 2 c ) - √ 3 a + √ ( 2 b + 2 c ) = 4 . . 2 √ ( 2 b + 2 c ) = 4 . . √ ( 2 b + 2 c ) = 2 . . 2 b + 2 c = 4 . . or b + c = 2 . . from a and b a + b + c = 12 + 2 = 14 . . c" | a = 30 - 4
b = a - 4
c = b - 2
d = c - 2
e = d - 1
f = e - 3
g = f - 1
h = g + 2
i = h - 3
j = i - 4
|
a ) 1 kmph , b ) 2 kmph , c ) 8 kmph , d ) 7 kmph , e ) 5 kmph | b | divide(subtract(8, 4), const_2) | a man can row his boat with the stream at 8 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 8 us = 4 s = ? s = ( 8 - 4 ) / 2 = 2 kmph answer : b" | a = 8 - 4
b = a / 2
|
a ) a ) 47 , b ) b ) 45.6 , c ) c ) 44 , d ) d ) 48.5 , e ) e ) 49 | d | divide(add(multiply(45, subtract(30, add(3, 3))), multiply(95, 3)), 30) | in a class of 30 students in an examination in maths 3 students scored 95 marks each , 3 get zero each and the average of the rest was 45 . what is the average of the whole class ? | "explanation : total marks obtained by a class of 25 students = 3 * 95 + 3 * 0 + 26 * 45 = 1455 average marks of whole class = 1455 / 30 = 48.5 answer : option d" | a = 3 + 3
b = 30 - a
c = 45 * b
d = 95 * 3
e = c + d
f = e / 30
|
a ) a ) 200 , b ) b ) 190 , c ) c ) 180 , d ) d ) 170 , e ) e ) 160 | a | subtract(add(200, 350), 350) | a , b and c have rs . 350 between them , a and c together have rs . 200 and b and c rs . 350 . how much does c have ? | "a + b + c = 350 a + c = 200 b + c = 350 - - - - - - - - - - - - - - a + b + 2 c = 550 a + b + c = 350 - - - - - - - - - - - - - - - - c = 200 answer : a" | a = 200 + 350
b = a - 350
|
a ) 0.0016 , b ) 0.0625 , c ) 0.16 , d ) 0.278 , e ) 0.5 | d | power(divide(1, const_4.0), 2) | what is the decimal equivalent of ( 1 / 6 ) ^ 2 ? | "( 1 / 6 ) ² = ( 1 / 6 ) ( 1 / 6 ) = 1 / 36 approach # 1 : use long division to divide 36 into 1 to get 1 / 36 = 0.0278 d" | a = 1 / 4
b = a ** 2
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | subtract(15, const_1) | there are 32 balls which are red , blue or green . if 15 balls are green and the sum of red balls and green balls is less than 23 , at most how many red balls are there ? | "r + g + b = 32 g = 15 r + g < 23 = > r + 15 < 23 = > r < 8 = > at most 7 red balls answer : d" | a = 15 - 1
|
a ) 400 , b ) 525 , c ) 750 , d ) 850 , e ) none | c | divide(divide(multiply(90, const_1000), divide(const_60, const_1)), const_2) | the length of a train and that of a platform are equal . if with a speed of 90 k / hr , the train crosses the platform in one minute , then the length of the train ( in metres ) is : | "sol . speed = [ 90 * 5 / 18 ] m / sec = 25 m / sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x metres . then , 2 x / 60 = 25 ⇔ x = 25 * 60 / 2 = 750 answer c" | a = 90 * 1000
b = const_60 / 1
c = a / b
d = c / 2
|
a ) 14 , b ) 13 , c ) 8 , d ) 7 , e ) 5 | c | add(subtract(add(11, 18), subtract(30, 3)), subtract(18, 11)) | of 30 applicants for a job , 11 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "c . 8 30 - 3 = 27 27 - 11 - 18 = - 8 then 8 are in the intersection between 4 years experience and degree . answer c" | a = 11 + 18
b = 30 - 3
c = a - b
d = 18 - 11
e = c + d
|
a ) $ 900 , b ) $ 300 , c ) $ 600 , d ) $ 700 , e ) $ 800 | c | divide(300, subtract(const_1, divide(2, 4))) | linda spent 2 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 300 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 2 / 4 = 1 / 2 on a tv but the tv cost her $ 300 . so 1 / 2 of her savings is $ 300 . so her original savings are 2 times $ 300 = $ 600 correct answer c" | a = 2 / 4
b = 1 - a
c = 300 / b
|
a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 4.5 kmph | e | divide(subtract(12, 3), const_2) | what is the speed of the stream if a canoe rows upstream at 3 km / hr and downstream at 12 km / hr | "sol . speed of stream = 1 / 2 ( 12 - 3 ) kmph = 4.5 kmph . answer e" | a = 12 - 3
b = a / 2
|
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | c | add(divide(subtract(multiply(multiply(3, 2), 3), multiply(3, 2)), 2), subtract(multiply(multiply(3, 2), 3), multiply(3, 2))) | the ratio of the number of females to males at a party was 1 : 2 but when 3 females and 3 males left , the ratio became 1 : 3 . how many people were at the party originally ? | "the total number of people are x females + 2 x males . 3 * ( x - 3 ) = 2 x - 3 x = 6 there were 3 x = 18 people at the party originally . the answer is c ." | a = 3 * 2
b = a * 3
c = 3 * 2
d = b - c
e = d / 2
f = 3 * 2
g = f * 3
h = 3 * 2
i = g - h
j = e + i
|
a ) a ) 360 , b ) b ) 150 , c ) c ) 180 , d ) d ) 200 , e ) e ) 220 | a | divide(multiply(divide(30, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | one fourth of one third of two fifth of a number is 30 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 30 then x = 30 * 30 = 900 40 % of 900 = 360 answer : option a" | a = 1 / 4
b = 1 / 3
c = a * b
d = 2 + 3
e = 2 / d
f = c * e
g = 30 / f
h = g * 40
i = h / 100
|
a ) 5 , b ) 12.5 , c ) 14.5 , d ) 12.75 , e ) 25 | d | divide(multiply(5, 255), const_100) | 5 percent of 255 = | 255 / 100 = 2.55 1 % = 2.55 5 % = 2.55 * 5 = 12.75 answer : d | a = 5 * 255
b = a / 100
|
a ) 10 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | b | divide(multiply(add(20, 5), 4), 5) | a certain number of men can do a work in 20 days . if there were 4 men less it could be finished in 5 days more . how many men are there ? | the original number of men = 4 ( 20 + 5 ) / 5 = 20 men answer is b | a = 20 + 5
b = a * 4
c = b / 5
|
a ) 150 meter , b ) 299 meter , c ) 666 meter , d ) 90 meter , e ) 144 meter | d | multiply(divide(multiply(36, const_1000), const_3600), 9) | a train running at the speed of 36 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 36 * ( 5 / 18 ) m / sec = 10 m / sec length of train ( distance ) = speed * time ( 10 ) * 9 = 90 meter answer : d" | a = 36 * 1000
b = a / 3600
c = b * 9
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | d | subtract(power(add(3, 2), 2), multiply(11, const_4)) | if n is a prime number greater than 3 , what is the remainder when n ^ 2 is divided by 11 ? | "there are several algebraic ways to solve this question including the one under the spoiler . but the easiest way is as follows : since we can not have two correct answersjust pick a prime greater than 3 , square it and see what would be the remainder upon division of it by 11 . n = 5 - - > n ^ 2 = 25 - - > remainder upon division 25 by 11 is 3 . answer : d ." | a = 3 + 2
b = a ** 2
c = 11 * 4
d = b - c
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | e | multiply(power(2, 2), 2) | point ( m , n ) is on the circle represented by m ^ 2 + n ^ 2 = 10 , and m , n are integers . how many such points are possible ? | m ^ 2 + n ^ 2 = 10 and m , n are integers means that 10 is the sum of two perfect squares . 10 is the sum of only one pair of perfect squares 1 and 9 . so , there can be 8 such points , 4 in each quadrant : ( 1 , 3 ) ; ( 1 , - 3 ) ; ( - 1 , 3 ) ; ( - 1 , - 3 ) ; ( 3 , 1 ) ; ( 3 , - 1 ) ; ( - 3 , 1 ) ; ( - 3 , - 1 ) . answer : e . | a = 2 ** 2
b = a * 2
|
a ) 188 cm 2 , b ) 150 cm 2 , c ) 168 cm 2 , d ) 195 cm 2 , e ) 987 cm 2 | d | multiply(multiply(divide(const_1, const_2), add(6, 9)), 26) | find the area of the quadrilateral of one of its diagonals is 26 cm and its off sets 9 cm and 6 cm ? | "1 / 2 * 26 ( 9 + 6 ) = 195 cm 2 answer : d" | a = 1 / 2
b = 6 + 9
c = a * b
d = c * 26
|
a ) 0.125 % , b ) 1.25 % , c ) 11.7 % , d ) 125 % , e ) 0.152 % | c | multiply(divide(divide(multiply(170, 5.17), add(add(9.18, 5.17), 2.05)), 170), const_100) | irin , ingrid and nell bake chocolate chip cookies in the ratio of 9.18 : 5.17 : 2.05 . if altogether they baked a batch of 170 cookies , what percent of the cookies did nell bake ? | "9.18 x + 5.17 x + 2.05 x = 16.4 x = 170 cookies x = 170 / 16.4 = 10 ( approx ) so , nell baked 10 * 2.05 cookies or 20 cookies ( approx ) % share = 20 / 170 = 11.7 approx hence , answer is c ." | a = 170 * 5
b = 9 + 18
c = b + 2
d = a / c
e = d / 170
f = e * 100
|
a ) 275 , b ) 287.5 , c ) 288 , d ) 233.5 , e ) 245.6 | b | multiply(125, subtract(const_2, const_1)) | a train speeds past a pole in 23 seconds and a platform 125 m long in 33 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 23 = > y = x / 23 x + 125 / 33 = x / 23 x = 287.5 m . answer : b" | a = 2 - 1
b = 125 * a
|
a ) $ 14.40 , b ) $ 14.00 , c ) $ 11.00 , d ) $ 9.60 , e ) $ 10.00 | e | subtract(12.5, multiply(divide(25, const_100), 12.5)) | a bookseller sells his books at a 25 % markup in price . if he sells a book for $ 12.5 , how much did he pay for it ? | "let the cost price of book = x selling price of book = 12.5 $ markup % = 25 ( 125 / 100 ) x = 12.5 = > x = 10 answer e" | a = 25 / 100
b = a * 12
c = 12 - 5
|
a ) rs . 9000.30 , b ) rs . 9720 , c ) rs . 10123.20 , d ) rs . 10483.20 , e ) none of these | c | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 25,000 after 3 years at the rate of 12 p . c . p . a . ? | "explanation : amount = rs . [ 25000 x ( 1 + 12 / 100 ) 3 ] = rs . 35123.20 c . i . = rs . ( 35123.20 - 25000 ) = rs . 10123.20 answer is c" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) 14090 , b ) 16010 , c ) 15060 , d ) 14930 , e ) 16075 | a | subtract(add(5000, 12000), add(multiply(5000, divide(15, const_100)), multiply(divide(18, const_100), 12000))) | a soft drink company had 5000 small and 12000 big bottles in storage . if 15 % of small 18 % of big bottles have been sold , then the total bottles remaining in storage is | 5000 + 12000 - ( 0.15 * 5000 + 0.18 * 12000 ) = 14090 . answer : a . | a = 5000 + 12000
b = 15 / 100
c = 5000 * b
d = 18 / 100
e = d * 12000
f = c + e
g = a - f
|
a ) 100 , b ) 160 , c ) 120 , d ) 200 , e ) none of these | d | multiply(divide(24, multiply(divide(4, add(const_1, const_4)), divide(3, add(const_4, const_4)))), divide(250, const_100)) | 4 - fifths of 3 - eighths of a number is 24 . what is 250 per cent of that number ? | let the number = x . 4 ⁄ 5 × 3 ⁄ 8 x = 24 or x = 24 × 2 × 5 / 3 = 80 ∴ 250 per cent of the number = 250 ⁄ 100 × 80 = 200 answer d | a = 1 + 4
b = 4 / a
c = 4 + 4
d = 3 / c
e = b * d
f = 24 / e
g = 250 / 100
h = f * g
|
a ) $ 430 , b ) $ 1620 , c ) $ 1650 , d ) $ 1110 , e ) $ 1770 | d | add(add(multiply(100, 10), multiply(40, const_4.0)), multiply(multiply(10, 6), const_2)) | rates for having a manuscript typed at a certain typing service are $ 10 per page for the first time a page is typed and $ 6 per page each time a page is revised . if a certain manuscript has 100 pages , of which 40 were revised only once , 10 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ? | "for 100 - 40 - 10 = 50 pages only cost is 10 $ per page for the first time page is typed - 50 * 10 = 500 $ ; for 40 pages the cost is : first time 5 $ + 6 $ of the first revision - 40 * ( 5 + 6 ) = 440 $ ; for 10 pages the cost is : first time 5 $ + 6 $ of the first revision + 6 $ of the second revision - 10 ( 5 + 6 + 6 ) = 170 $ ; total : 500 + 440 + 170 = 1110 $ . answer : d ." | a = 100 * 10
b = 40 * 4
c = a + b
d = 10 * 6
e = d * 2
f = c + e
|
a ) 8163 , b ) 8633 , c ) 8663 , d ) 8636 , e ) none of these | b | divide(add(7921, 9409), const_2) | find the mean proportional between 7921 & 9409 ? | "explanation : formula = √ a × b a = 7921 and b = 9409 √ 7921 × 9409 = 89 × 97 = 8633 answer : option b" | a = 7921 + 9409
b = a / 2
|
a ) 330 , b ) 315 , c ) 312.5 , d ) 370 , e ) 350 | c | multiply(divide(25, 1250), const_100) | 25 % of 1250 | 1 % of 1250 is = 12.50 25 % of 1250 is = 12.50 * 25 = 312.5 answer : c | a = 25 / 1250
b = a * 100
|
a ) 59 : 31 , b ) 3 : 2 , c ) 118 : 126 , d ) 193 : 122 , e ) 201 : 132 | a | divide(add(add(multiply(7, multiply(add(2, 1), add(3, 2))), multiply(2, multiply(add(7, 3), add(3, 2)))), multiply(3, multiply(add(7, 3), add(2, 1)))), add(add(multiply(3, multiply(add(2, 1), add(3, 2))), multiply(1, multiply(add(7, 3), add(3, 2)))), multiply(2, multiply(add(7, 3), add(2, 1))))) | there are 3 vessels of equal capacity . vessel a contains milk and water in the ratio 7 : 3 ; vessel b contains milk and water in the ratio 2 : 1 and vessel c contains milk and water in the ratio 3 : 2 . if the mixture in all the 3 vessels is mixed up . what will be the ratio of milk and water ? | 7 : 3 = > 7 x + 3 x = 10 x 2 : 1 = > 2 y + 1 y = 3 y 3 : 2 = > 3 z + 2 z = 5 z 10 x = 3 y = 5 z take lcm of 10 , 3,5 = 30 or simply ; x = 3 y = 10 z = 6 so , ratio of milk : water = ( 7 x + 2 y + 3 z ) / ( 3 x + y + 2 z ) = 59 / 31 ans : a | a = 2 + 1
b = 3 + 2
c = a * b
d = 7 * c
e = 7 + 3
f = 3 + 2
g = e * f
h = 2 * g
i = d + h
j = 7 + 3
k = 2 + 1
l = j * k
m = 3 * l
n = i + m
o = 2 + 1
p = 3 + 2
q = o * p
r = 3 * q
s = 7 + 3
t = 3 + 2
u = s * t
v = 1 * u
w = r + v
x = 7 + 3
y = 2 + 1
z = x * y
A = 2 * z
B = w + A
C = n / B
|
a ) 16 , b ) 24 , c ) 60 , d ) 100 , e ) 240 | d | multiply(multiply(const_3, 5), multiply(5, 2)) | from a total of 5 boys and 5 girls , how many 4 - person committees can be selected if the committee must have exactly 2 boys and 2 girls ? | "answer = d = 100 no of 4 person committees that can be formed = 5 c 2 * 5 c 2 = 100 answer d" | a = 3 * 5
b = 5 * 2
c = a * b
|
a ) 10.8 , b ) 18 , c ) 25 , d ) 30 , e ) none | d | multiply(108, const_0_2778) | a train moves with a speed of 108 kmph . its speed in metres per second is | "sol . 108 kmph = [ 108 x 5 / 18 ] m / sec = 30 m / sec . answer d" | a = 108 * const_0_2778
|
a ) 4.5 , b ) 3 , c ) 4 , d ) 3.5 , e ) 8.5 | a | divide(multiply(power(132, 7), power(132, 132)), 11.5) | ( 132 ) ^ 7 × ( 132 ) ^ ? = ( 132 ) ^ 11.5 . | "( 132 ) ^ 7 × ( 132 ) ^ x = ( 132 ) ^ 11.5 = > 7 + x = 11.5 = > x = 11.5 - 7 = 4.5 answer is a" | a = 132 ** 7
b = 132 ** 132
c = a * b
d = c / 11
|
a ) 47 , b ) 53 , c ) 39 , d ) 43 , e ) 42 | b | add(subtract(85, multiply(3, 10)), 3) | a batsman in his 10 th innings makes a score of 85 , and thereby increases his average by 3 . what is his average after the 10 th innings ? he had never been ’ not out ’ . | "average score before 10 th innings = 85 - 3 × 10 = 50 average score after 10 th innings = > 50 + 3 = 53 answer : b" | a = 3 * 10
b = 85 - a
c = b + 3
|
a ) 1 / 3 , b ) 2 / 3 , c ) 5 / 7 , d ) 7 / 5 , e ) 3 / 2 | c | divide(subtract(divide(const_1, const_2), subtract(subtract(const_1, divide(const_2, const_3)), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))), subtract(divide(const_1, const_2), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))) | a certain ball team has an equal number of right - and left - handed players . on a certain day , two - thirds of the players were absent from practice . of the players at practice that day , two - third were right handed . what is the ratio of the number of right - handed players who were not at practice that day to the number of left handed players who were not at practice ? | "say the total number of players is 18 , 9 right - handed and 9 left - handed . on a certain day , two - thirds of the players were absent from practice - - > 12 absent and 6 present . of the players at practice that day , one - third were right - handed - - > 6 * 2 / 3 = 4 were right - handed and 2 left - handed . the number of right - handed players who were not at practice that day is 9 - 4 = 5 . the number of left - handed players who were not at practice that days is 9 - 2 = 7 . the ratio = 5 / 7 . answer : c" | a = 1 / 2
b = 2 / 3
c = 1 - b
d = 2 / 3
e = 1 - d
f = 2 / 3
g = e * f
h = c - g
i = a - h
j = 1 / 2
k = 2 / 3
l = 1 - k
m = 2 / 3
n = l * m
o = j - n
p = i / o
|
a ) 30 , b ) 31 , c ) 32 , d ) 33 , e ) 34 | c | divide(subtract(multiply(5, 20), multiply(12, subtract(5, const_2))), const_2) | martin bought 5 packet milk at an average price ( arithmetic mean ) of 20 ¢ . if martin returned two packet to the retailer , and the average price of the remaining millk packet was 12 ¢ , then what is the average price , in cents , of the two returned milk packets ? | total price of 5 packet milk = 5 * 20 = 100 total price of 3 packet milk = 3 * 12 = 36 total price of 2 packet returned milk = 100 - 36 = 64 average price of 2 packet milk = 64 / 2 = 32 correct option answer : c | a = 5 * 20
b = 5 - 2
c = 12 * b
d = a - c
e = d / 2
|
a ) 22 . , b ) 24 . , c ) 30 . , d ) 36 , e ) 38 . | c | divide(multiply(18, 20), 12) | 20 beavers , working together in a constant pace , can build a dam in 18 hours . how many hours will it take 12 beavers that work at the same pace , to build the same dam ? | total work = 20 * 18 = 360 beaver hours 12 beaver * x = 360 beaver hours x = 360 / 12 = 30 answer : c | a = 18 * 20
b = a / 12
|
a ) 75 , b ) 81 , c ) 71 , d ) 79 , e ) none of these | d | add(divide(subtract(146, multiply(const_2, const_2)), const_2), multiply(const_4, const_2)) | a , b , c , d and e are ve consecutive odd numbers the sum of a and c is 146 . what is the value of e ? | a + c = 146 ⇒ a + a + 4 = 146 ⇒ a = 146 - 4 / 2 = 71 ∴ e = a + 8 = 71 + 8 = 79 answer d | a = 2 * 2
b = 146 - a
c = b / 2
d = 4 * 2
e = c + d
|
a ) 4 , b ) 6 , c ) 5 , d ) 8 , e ) 7 | a | multiply(multiply(9, 9), divide(5, 9)) | in the coordinate plane , points ( x , 5 ) and ( 9 , y ) are on line k . if line k passes through the origin and has slope 5 / 9 , then x - y = | "line k passes through the origin and has slope 1 / 4 means that its equation is y = 5 / 9 * x . thus : ( x , 5 ) = ( 9 , 5 ) and ( 9 , y ) = ( 9,5 ) - - > x - y = 9 - 5 = 4 . answer : a" | a = 9 * 9
b = 5 / 9
c = a * b
|
a ) 26 , b ) 22 , c ) 25 , d ) 30 , e ) 29 | d | add(add(multiply(3, 11), 5), 3) | find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 3 | "( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 3 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 3 = 2 ^ 11 x 2 ^ 11 x 7 ^ 5 x 11 ^ 3 = 2 ^ 22 x 7 ^ 5 x 11 ^ 3 total number of prime factors = ( 22 + 5 + 3 ) = 30 . answer is d ." | a = 3 * 11
b = a + 5
c = b + 3
|
a ) 1 / 25 , b ) 1 / 15 , c ) 1 / 3 , d ) 1 / 2 , e ) 1 | c | add(add(add(divide(multiply(20, subtract(20, const_1)), multiply(add(add(add(20, 20), 6), 4), subtract(add(add(add(20, 20), 6), 4), const_1))), divide(multiply(20, subtract(20, const_1)), multiply(add(add(add(20, 20), 6), 4), subtract(add(add(add(20, 20), 6), 4), const_1)))), divide(multiply(6, add(4, const_1)), multiply(add(add(add(20, 20), 6), 4), subtract(add(add(add(20, 20), 6), 4), const_1)))), divide(multiply(4, const_3), multiply(add(add(add(20, 20), 6), 4), subtract(add(add(add(20, 20), 6), 4), const_1)))) | a jar contains 20 red , 20 blue , 6 yellow and 4 pink balls . if two balls are to be selected at random and without replacement , approximately what is the probability both of them to be the same colour ? | ok so jar contains 20 red 20 blue 6 yellow and 4 pink . . adding the number of balls we get 50 total . . now its a counting + probability exercise . . 2 stages exist : pick first ball then second ball . . denominator for first ball = 50 since there are 50 balls denominator for second ball = 49 since only 49 would remain for each type ball i constructed probability of getting 2 of same . . i used a counting method . . first ball 20 chances blue and 2 nd ball 19 chances blue and so on and so forth blue = ( 20 * 19 ) / ( 50 * 49 ) rounded to 400 / 2500 red = ( 20 * 19 ) / ( 50 * 49 ) rounded to 400 / 2500 yellow = ( 6 * 5 ) / ( 50 * 49 ) rounded to 30 / 2500 pink = ( 4 * 3 ) / ( 49 * 50 ) rounded to 12 / 2500 then i summed these individual probabilities to get 842 / 2500 this is ~ 1 / 3 as 800 * 3 = 2400 ans : c | a = 20 - 1
b = 20 * a
c = 20 + 20
d = c + 6
e = d + 4
f = 20 + 20
g = f + 6
h = g + 4
i = h - 1
j = e * i
k = b / j
l = 20 - 1
m = 20 * l
n = 20 + 20
o = n + 6
p = o + 4
q = 20 + 20
r = q + 6
s = r + 4
t = s - 1
u = p * t
v = m / u
w = k + v
x = 4 + 1
y = 6 * x
z = 20 + 20
A = z + 6
B = A + 4
C = 20 + 20
D = C + 6
E = D + 4
F = E - 1
G = B * F
H = y / G
I = w + H
J = 4 * 3
K = 20 + 20
L = K + 6
M = L + 4
N = 20 + 20
O = N + 6
P = O + 4
Q = P - 1
R = M * Q
S = J / R
T = I + S
|
a ) 7 , b ) 20 , c ) 28 , d ) 14 , e ) 19 | a | divide(divide(7, 8), divide(1, 8)) | diana is painting statues . she has 7 / 8 of a gallon of paint remaining . each statue requires 1 / 8 gallon of paint . how many statues can she paint ? | "number of statues = all the paint ÷ amount used per statue = 7 / 8 ÷ 1 / 8 = 7 / 8 * 8 / 1 = 7 / 1 = 7 answer is a ." | a = 7 / 8
b = 1 / 8
c = a / b
|
a ) 5 minutes , b ) 10 minutes , c ) 15 minutes , d ) 18 minutes , e ) 20 minutes | b | divide(10, add(divide(10, 24), add(divide(10, 40), divide(10, 30)))) | jonathan can type a 10 page document in 40 minutes , susan can type it in 30 minutes , and jack can type it in 24 minutes . working together , how much time will it take them to type the same document ? | "you may set up common equation like this : job / a + job / b + job / c = job / x memorize this universal formula , you will need it definitely for gmat . and find x from this equation in this specific case , the equation will look like this : 10 / 40 + 10 / 30 + 10 / 24 = 10 / x if you solve this equation , you get the same answer b ( 10 )" | a = 10 / 24
b = 10 / 40
c = 10 / 30
d = b + c
e = a + d
f = 10 / e
|
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