options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) - 2 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | d | add(add(const_4, 2), subtract(9, const_4)) | what is the sum of all possible solutions to the equation β 2 x ^ 2 β x β 9 = x + 1 ? | "first of all notice that since lhs is the square root of a number , it must be non - negative ( the square root function can not give negative result ) , then the rhs must also be non - negative : x + 1 β₯ 0 - - > x β₯ β 1 square the equation : 2 x ^ 2 β x β 9 = x ^ 2 + 2 x + 1 - - > x ^ 2 β 3 x β 10 = 0 - - > x = - 2 o... | a = 4 + 2
b = 9 - 4
c = a + b
|
a ) 18 , b ) 27 , c ) 98 , d ) 27 , e ) 48 | e | multiply(subtract(divide(30, 10), const_1), 24) | pipe a can fill a tank in 10 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ? | "let the pipe b be closed after x minutes . 30 / 10 - x / 24 = 1 = > x / 24 = 30 / 10 - 1 = 2 = > x = 2 * 24 = 48 . answer : e" | a = 30 / 10
b = a - 1
c = b * 24
|
a ) 44 , b ) 54 , c ) 46 , d ) 47 , e ) 48 | b | multiply(subtract(multiply(divide(subtract(const_100, 18), const_100), divide(add(const_100, 88), const_100)), const_1), const_100) | if price of t . v set is reduced by 18 % , then its sale increases by 88 % , find net effect on sale value | "- a + b + ( ( - a ) ( b ) / 100 ) = - 18 + 88 + ( - 18 * 88 ) / 100 = - 18 + 88 - 16 = 54 answer : b" | a = 100 - 18
b = a / 100
c = 100 + 88
d = c / 100
e = b * d
f = e - 1
g = f * 100
|
a ) 14 , b ) 19 , c ) 33 , d ) 38 , e ) 56 | a | subtract(divide(38, const_2), 5) | a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 38 now , the son age 5 years back was : | "explanation : let the son ' s present age be x years . then , ( 38 - x ) = x x = 19 . son ' s age 5 years back = ( 19 - 5 ) = 14 years . answer : a" | a = 38 / 2
b = a - 5
|
a ) 10 , b ) 15 , c ) 8 , d ) 12 , e ) 20 | a | divide(multiply(20, 5), 10) | 20 machines can do a work in 5 days . how many machines are needed to complete the work in 10 days ? | "required number of machines = 20 * 5 / 10 = 10 answer is a" | a = 20 * 5
b = a / 10
|
a ) 12 , b ) 180 , c ) 24 , d ) 28 , e ) 40 | b | multiply(4, divide(225, sqrt(add(power(4, const_2), power(3, const_2))))) | in x - y plane , there is a right triangle abc ( β b = 90 o ) . if the length of ac is 225 and the slope of line segment ac is 4 / 3 , what is the length of ab ? | "slope = change in vertical direction / change in horizontal direction = 4 / 3 change in vertical direction = 4 x = ab change in horizontal direction = 3 x = bc ab ^ 2 + bc ^ 2 = 225 ^ 2 16 x ^ 2 + 9 x ^ 2 = 225 ^ 2 25 x ^ 2 = 225 ^ 2 x ^ 2 = 25 * 9 ^ 2 x = 45 therefore ab = 45 * 4 = 180 answer : b" | a = 4 ** 2
b = 3 ** 2
c = a + b
d = math.sqrt(c)
e = 225 / d
f = 4 * e
|
a ) 10 % , b ) 25 % , c ) 33 % , d ) 50 % , e ) 67 % | d | multiply(add(const_1, const_10), subtract(subtract(18, 12), const_1)) | at company x , senior sales representatives visit the home office once every 18 days , and junior sales representatives visit the home office once every 12 days . the number of visits that a junior sales representative makes in a 2 - year period is approximately what percent greater than the number of visits that a sen... | "each 36 - day period , senior representatives visit the home office 2 times while junior representatives visit 3 times , thus 50 % more . the answer is d ." | a = 1 + 10
b = 18 - 12
c = b - 1
d = a * c
|
a ) 20 , b ) 21 , c ) 23 , d ) 25 , e ) 26 | b | subtract(36, add(add(8, const_2), 8)) | set a of 8 positive integers may have the same element and have 36 . and set b of 8 positive integers must have different elements and have 36 . when m and n are the greatest possible differences between 36 and other elements β sums in set a and set b , respectively , m - n = ? | this is maximum - minimum . hence , 36 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 29 and 36 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 8 . so , 29 - 8 = 21 . the correct answer is b . | a = 8 + 2
b = a + 8
c = 36 - b
|
a ) 648 , b ) 320 , c ) 252 , d ) 225 , e ) 26 | b | add(add(divide(subtract(1100, 100), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(6, const_2)) | how many times digit 6 is used while writing numbers from 100 to 1100 ? | "there are 100 numbers which begin with 600 next , in every 10 numbers such as 100 to 110 , 110 to 120 , 120 to 130 6 comes at least once . number of such intervals = end limit - first no . / interval . our range of numbers is 100 - 1000 1000 - 100 = 900 / 10 = 90 number of 10 s interval in this is 90 . so 90 ' 6 s ' s... | a = 1100 - 100
b = a / 10
c = 10 + 1
d = 10 + 1
e = c * d
f = b + e
g = 6 * 2
h = f + g
|
a ) $ 3.36 , b ) $ 6.85 , c ) $ 8.46 , d ) $ 10.46 , e ) $ 11.85 | d | add(multiply(2, 3.49), multiply(4, 0.87)) | what is the total cost of 2 sandwiches at $ 3.49 each and 4 sodas at $ 0.87 each ? | "answer = d 2 * 3.49 + 4 * 0.87 = 2 ( 3.50 - 0.01 ) + 4 ( 1.00 - 0.13 ) = 7 + 4 - 0.02 - 0.52 = 11 - 0.54 = 10.46" | a = 2 * 3
b = 4 * 0
c = a + b
|
a ) 3 : 2 , b ) 1 : 2 , c ) 3 : 1 , d ) 5 : 2 , e ) 2 : 3 | e | divide(subtract(7, 5), subtract(10, 7)) | gold is 10 times as heavy as water and copper is 5 times as heavy as water . in what ratio should these be mixed to get an alloy 7 times as heavy as water ? | "g = 10 w c = 5 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 10 w + 5 wx = x + 1 * 7 w 10 + 5 x = 7 ( x + 1 ) x = 3 / 2 ratio of gold with copper = 1 : 3 / 2 = 2 : 3 answer is e" | a = 7 - 5
b = 10 - 7
c = a / b
|
a ) 1 : 3 , b ) 2 : 3 , c ) 3 : 4 , d ) 6 : 7 , e ) none of these | d | divide(subtract(6.30, 5.70), subtract(7.00, 6.30)) | find the ratio in which rice at rs . 7.00 a kg be mixed with rice at rs . 5.70 a kg to produce a mixture worth rs . 6.30 a kg ? | "solution required ratio = 60 : 70 = 6 : 7 answer d" | a = 6 - 30
b = 7 - 0
c = a / b
|
a ) 3 , b ) 2 , c ) 1 / 4 , d ) 3 / 5 , e ) 5 / 6 | d | divide(1, divide(add(subtract(9, 1), 6), 6)) | the vertex of a rectangle are ( 1 , 0 ) , ( 9 , 0 ) , ( 1 , 6 ) and ( 9 , 6 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ? | "if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 5 , 3 ) . the slope of a line passing through ( 0,0 ) and ( 5 , 3 ) is 3 / 5 . the answer is d ." | a = 9 - 1
b = a + 6
c = b / 6
d = 1 / c
|
a ) 61 , b ) 60 , c ) 79 , d ) 70 , e ) 78 | b | multiply(divide(10, const_2), multiply(5, 6)) | calculate how many seconds it will take for 4 bells to toll together again , given that they begin to toll together respectively at the intervals of 5 , 6 , 10 and 12 seconds . ? | "lcm of 5 , 6 , 10 and 12 is 60 lcm = 60 answer : b" | a = 10 / 2
b = 5 * 6
c = a * b
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | multiply(divide(const_1, add(const_1, const_4)), divide(const_1, subtract(divide(const_1, 6), divide(const_1, 15)))) | a water tank is one - fifth full . pipe a can fill a tank in 15 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how many minutes will it take to empty or fill the tank completely ? | "the combined rate of filling / emptying the tank = 1 / 15 - 1 / 6 = - 1 / 10 since the rate is negative , the tank will be emptied . a full tank would take 10 minutes to empty . since the tank is only one - fifth full , the time is ( 1 / 5 ) * 10 = 2 minutes the answer is a ." | a = 1 + 4
b = 1 / a
c = 1 / 6
d = 1 / 15
e = c - d
f = 1 / e
g = b * f
|
a ) 2 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 1 / 2 | c | multiply(divide(4, add(4, 3)), divide(3, subtract(add(4, 3), const_1))) | a jar contains 4 black and 3 white balls . if you pick two balls at the same time , what ' s the probability that one ball is black and one is white ? | "p ( 1 st black , 2 nd white ) = 4 / 7 * 3 / 6 = 4 / 14 ; p ( 1 st white , 2 nd black ) = 3 / 7 * 4 / 6 = 4 / 14 . p = 4 / 14 + 4 / 14 = 4 / 7 . answer : c ." | a = 4 + 3
b = 4 / a
c = 4 + 3
d = c - 1
e = 3 / d
f = b * e
|
a ) 3 , b ) 5 , c ) 4 , d ) 2 , e ) 6 | d | divide(multiply(3, 6), add(3, 6)) | a can do a work in 3 days . b can do the same work in 6 days . if both a & b are working together in how many days they will finish the work ? | "a rate = 1 / 3 b rate = 1 / 6 ( a + b ) rate = ( 1 / 3 ) + ( 1 / 6 ) = 1 / 2 a & b finish the work in 2 days correct option is d" | a = 3 * 6
b = 3 + 6
c = a / b
|
a ) 10 , b ) 50 , c ) 55 , d ) 35 , e ) 65 | c | add(subtract(80, 100), subtract(50, 35)) | in an it company , there are a total of 100 employees including 50 programmers . the number of male employees is 80 , including 35 male programmers . how many employees must be selected to guaranty that we have 3 programmers of the same sex ? | "you could pick 50 non - programmers , 2 male programmers , and 2 female programmers , and still not have 3 programmers of the same sex . but if you pick one more person , you must either pick a male or a female programmer , so the answer is 55 . c" | a = 80 - 100
b = 50 - 35
c = a + b
|
a ) 84 , b ) 96 , c ) 104 , d ) 112 , e ) 120 | c | multiply(divide(390, add(add(multiply(3, 2), 4), 5)), 4) | in a certain town , the ratio of ny yankees fans to ny mets fans is 3 : 2 , and the ratio of ny mets fans to boston red sox fans is 4 : 5 . if there are 390 baseball fans in the town , each of whom is a fan of exactly one of those three teams , how many ny mets fans are there in this town ? | "the ratio of yankees : mets : red sox = 6 : 4 : 5 the mets fans are 4 / 15 of the population . ( 4 / 15 ) * 390 = 104 the answer is c ." | a = 3 * 2
b = a + 4
c = b + 5
d = 390 / c
e = d * 4
|
a ) 12 sec , b ) 15 sec , c ) 16 sec , d ) 28 sec , e ) none | d | multiply(divide(280, multiply(36, const_1000)), const_3600) | a train 280 m long , running with a speed of 36 km / hr will pass a tree in | "sol . speed = ( 36 x 5 / 18 ) m / sec . = 10 m / sec . time taken = ( 280 x 1 / 10 ) sec = 28 sec answer d" | a = 36 * 1000
b = 280 / a
c = b * 3600
|
a ) 6 , b ) 9 , c ) 12 , d ) 15 , e ) 18 | b | divide(divide(4, subtract(const_1, divide(const_1, 3))), subtract(const_1, divide(const_1, 3))) | for all real numbers v , an operation is defined by the equation v * = v - v / 3 . if ( v * ) * = 4 , then v = | "( v * ) * = ( v - v / 3 ) - ( v - v / 3 ) / 3 4 = 2 v / 3 - 2 v / 9 = 4 v / 9 v = 9 the answer is b ." | a = 1 / 3
b = 1 - a
c = 4 / b
d = 1 / 3
e = 1 - d
f = c / e
|
a ) 600 , b ) 882 , c ) 374 , d ) 356 , e ) 521 | c | add(340, multiply(340, divide(10, const_100))) | a person buys an article at rs . 340 . at what price should he sell the article so as to make a profit of 10 % ? | "cost price = rs . 340 profit = 10 % of 340 = rs . 34 selling price = cost price + profit = 340 + 34 = 374 answer : c" | a = 10 / 100
b = 340 * a
c = 340 + b
|
a ) 163 , b ) 785 , c ) 852 , d ) 456 , e ) 852 | a | divide(subtract(12401, 13), 76) | on dividing 12401 by a certain number , we get 76 as quotient and 13 as remainder . what is the divisor ? | "divisor * quotient + remainder = dividend divisor = ( dividend ) - ( remainder ) / quotient ( 12401 - 13 ) / 76 = 163 answer ( a )" | a = 12401 - 13
b = a / 76
|
a ) 78 , b ) 56 , c ) 80 , d ) 27 , e ) 40 | c | multiply(40, const_2) | the average of the marks of 10 students in a class is 40 . if the marks of each student are doubled , find the new average ? | "sum of the marks for the 10 students = 10 * 40 = 400 . the marks of each student are doubled , the sum also will be doubled . the new sum = 400 * 2 = 800 . so , the new average = 800 / 10 = 80 . answer : c" | a = 40 * 2
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a ) $ 250 , b ) $ 350 , c ) $ 450 , d ) $ 550 , e ) $ 650 | c | multiply(multiply(power(const_100, const_2), divide(const_4.0, const_100)), divide(const_3, 6)) | how much interest will $ 10,000 earn in 9 months at an annual rate of 6 % ? | "soln : - 9 months = 3 / 4 of year ; 6 % = 6 / 100 = 3 / 50 ; $ 10,000 ( principal ) * 3 / 50 ( interest rate ) * 3 / 4 ( time ) = $ 450 . answer : c" | a = 100 ** 2
b = 4 / 0
c = a * b
d = 3 / 6
e = c * d
|
a ) 120 , b ) 150 , c ) 145 , d ) 108 , e ) 112 | d | divide(multiply(72, 36), const_4) | what is the sum of the greatest common factor and the lowest common multiple of 72 and 36 ? | "prime factorization of 36 = 2 x 2 x 3 x 3 prime factorization of 72 = 2 x 2 x 2 x 3 x 3 gcf = 36 lcm = 72 sum = 108 ans : d" | a = 72 * 36
b = a / 4
|
a ) 13.7 , b ) 14.0 , c ) 14.3 , d ) 14.6 , e ) 14.9 | b | multiply(sqrt(divide(19.6, 10.0)), 10.0) | at 1 : 00 pm , there were 10.0 grams of bacteria . the bacteria increased to x grams at 4 : 00 pm , and 19.6 grams at 7 : 00 pm . if the amount of bacteria present increased by the same fraction during each of the 3 - hour periods , how many grams of bacteria were present at 4 : 00 pm ? | "let x be the factor by which the bacteria increases every three hours . at 4 : 00 pm , the amount of bacteria was 10 x and at 7 : 00 pm it was 10 x ^ 2 . 10 x ^ 2 = 19.6 x ^ 2 = 1.96 x = 1.4 at 4 : 00 pm , the amount of bacteria was 10 ( 1.4 ) = 14 grams . the answer is b ." | a = 19 / 6
b = math.sqrt(a)
c = b * 10
|
a ) 20029 , b ) 20000 , c ) 20289 , d ) 17500 , e ) 20026 | d | divide(350, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 350 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 350 = > x = 17500 answer : d" | a = 5 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 10 / 100
f = 20 / 100
g = e * f
h = d - g
i = 350 / h
|
a ) 15 % , b ) 25 % , c ) 0.125 % , d ) 0.2083 % , e ) none | e | divide(multiply(multiply(divide(480, 3840), const_100), const_100), 60) | farm tax is levied on the 60 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : | "this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 / 60 ) = 20.833 % and... | a = 480 / 3840
b = a * 100
c = b * 100
d = c / 60
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | e | multiply(1, 3) | if 3 spiders make 3 webs in 3 days , then 1 spider will make 1 web in how many days ? | "let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 3 webs 3 : 1 1 x 3 x x = 3 x 1 x 3 = > x = 3 answer is e" | a = 1 * 3
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a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 8 | a | subtract(divide(multiply(18, add(const_4, const_1)), 9), add(const_4, const_1)) | three machines , each working at the same constant rate , together can complete a certain job in 18 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 9 days ? | "rate of one machine = 1 job / ( 18 * 3 ) days let x = number of machines needed to complete the job in 9 days 1 / ( 3 * 18 ) * 9 * x = 1 job x = 6 6 - 3 = 3 answer : a" | a = 4 + 1
b = 18 * a
c = b / 9
d = 4 + 1
e = c - d
|
a ) 380 times , b ) 142 times , c ) 382 times , d ) 392 times , e ) 165 times | e | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 30) | light glows for every 30 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am . | "time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 / 30 ) = 165 times . answer : e" | a = 2 + 47
b = 2 + const_60
c = 20 + b
d = c * const_60
e = a + d
f = e / 30
|
a ) a ) 33 , b ) b ) 98 , c ) c ) 34 , d ) d ) 35 , e ) e ) 62 | a | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 2 : 5 . the first number is : | "( 3 x - 9 ) : ( 5 x - 9 ) = 2 : 5 x = 11 = > 3 x = 33 answer : a" | a = 3 * 5
b = 9 / a
c = 3 * b
d = 3 * 5
e = 9 / d
f = 5 * e
g = c + f
|
a ) 14 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) 33 years | e | divide(subtract(35, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 35 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 35 ) + 2 = 2 ( x + 2 ) x + 37 = 2 x + 4 x = 33 . answer : e" | a = 2 * 2
b = a - 2
c = 35 - b
d = 2 - 1
e = c / d
|
a ) 30000 , b ) 50000 , c ) 90000 , d ) 20000 , e ) 60000 | c | multiply(divide(60000, 4), 6) | p and q invested in a business . the profit earned was divided in the ratio 4 : 6 . if p invested rs 60000 , the amount invested by q is | "let the amount invested by q = q 60000 : q = 4 : 6 β 60000 Γ 6 = 4 q β q = ( 60000 Γ 6 ) / 4 = 90000 answer is c ." | a = 60000 / 4
b = a * 6
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a ) $ 120 , b ) $ 40 , c ) $ 80 , d ) $ 100 , e ) $ 140 | a | multiply(subtract(add(add(add(2, 3), 5), 10), 8), 10) | on charlie ' s cell phone plan , he signed up for 8 gb of data per 4 week period . if he went over the 8 gb per month , it would cost him an extra $ 10.00 per gb . in the first week , he used 2 gb ; in the 2 nd week , he used 3 gb ; in the 3 rd week , he used 5 gb ; and in the 4 th week , he used a whopping 10 gb . how... | to determine how much extra he would need to pay , you have to add all 4 weeks of gb usage ( 2 + 3 + 5 + 10 = 20 ) . to determine the extra amount of gb usage , you have to subtract the the original amount of gb ( 8 ) allowed on his plan . 20 - 8 = 12 to determine the extra cost on his cell phone plan , you have to mul... | a = 2 + 3
b = a + 5
c = b + 10
d = c - 8
e = d * 10
|
a ) 4 min , b ) 2 min , c ) 3 min , d ) 1 min , e ) 10 min | a | multiply(44, divide(1, 11)) | a pipe can fill a cistern in 44 minutes . find the time in which 1 / 11 part of the cistern will be filled ? | "full cistern filled in = 44 minutes 1 / 11 part filled in = 44 * 1 / 11 = 4 minutes answer is a" | a = 1 / 11
b = 44 * a
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a ) 0.0009 , b ) 9.0 e - 05 , c ) 0.09 , d ) 0.009 , e ) 9.0 | b | multiply(divide(0.0003, 0.3), const_100) | 0.0003 x 0.3 = ? | "3 x 3 = 9 . sum of decimal places = 5 0.0003 x 0.3 = 0.00009 answer : option b" | a = 0 / 3
b = a * 100
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a ) e = 16 , b ) e = 32 , c ) e = 64 , d ) 128 , e ) 512 | b | multiply(8, 4) | the weight of a hollow sphere is directly dependent on its surface area . the surface area of a sphere is 4 Ο Β· r ^ 2 , where r is the radius of the sphere . if a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams , a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams e... | "weight directly proportional to 4 pi r ^ 2 now , 4 pi is constant , so , weight is directly proportional to r ^ 2 . when radius = 0.15 , weight = 8 , so ( 0.15 ) ^ 2 proportional to 8 ; ( 0.15 ) ^ 2 * 4 proportional to 8 * 4 , solving further ( 0.15 ) ^ 2 * 2 ^ 2 = ( 0.15 * 2 ) ^ 2 = 0.3 ^ 2 ; so answer = 32 ( b )" | a = 8 * 4
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a ) 150 meter , b ) 105 meter , c ) 140 meter , d ) 135 meter , e ) none of these | b | multiply(divide(multiply(54, const_1000), const_3600), 7) | a train running at the speed of 54 km / hr crosses a pole in 7 seconds . find the length of the train . | "explanation : speed = 54 * ( 5 / 18 ) m / sec = 15 m / sec length of train ( distance ) = speed * time = 15 * 7 = 105 meter option b" | a = 54 * 1000
b = a / 3600
c = b * 7
|
a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 10 , d ) 1 / 11 , e ) none of above | a | divide(circle_area(divide(25, const_2)), const_2) | what will be the fraction of 25 % | "explanation : it will 25 * 1 / 100 = 1 / 4 option a" | a = 25 / 2
b = circle_area / (
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a ) 2 / 5 , b ) 3 / 5 , c ) 7 / 12 , d ) 1 / 7 , e ) none of these | c | subtract(divide(4, 5), divide(2, 3)) | what is difference between biggest and smallest fraction among 2 / 3 , 3 / 4 , 4 / 5 and 5 / 4 | "explanation : 2 / 3 = . 66 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 4 = 1.25 so biggest is 5 / 4 and smallest is 2 / 3 their difference is 5 / 4 - 2 / 3 = 7 / 12 option c" | a = 4 / 5
b = 2 / 3
c = a - b
|
['a ) 5 : 8', 'b ) 3 : 2', 'c ) 6 : 1', 'd ) 8 : 7', 'e ) 5 : 6'] | b | divide(rectangle_area(3, 2), rectangle_area(divide(divide(rectangle_perimeter(3, 2), const_2), add(4, 1)), multiply(divide(divide(rectangle_perimeter(3, 2), const_2), add(4, 1)), 4))) | an order was placed for a carpet whose length and width were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and width were in the ratio 4 : 1 but were was no change in its perimeter . what is the ratio of the areas of the carpets ? | let the length and width of one carpet be 3 x and 2 x . let the length and width of the other carpet be 4 y and y . 2 ( 3 x + 2 x ) = 2 ( 4 y + y ) 5 x = 5 y x = y the ratio of the areas of the carpet in both cases : = 3 x * 2 x : 4 y * y = 6 x ^ 2 : 4 y ^ 2 = 6 x ^ 2 : 4 x ^ 2 = 6 : 4 = 3 : 2 the answer is b . | a = rectangle_area / (
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a ) 120 , b ) 772 , c ) 288 , d ) 266 , e ) 240 | e | divide(multiply(24, 20), const_2) | if the sides of a triangle are 26 cm , 24 cm and 20 cm , what is its area ? | "the triangle with sides 26 cm , 24 cm and 20 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 / 2 * 24 * 20 = 240 cm 2 answer : e" | a = 24 * 20
b = a / 2
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | divide(multiply(divide(const_1000, const_2), const_1000), multiply(divide(divide(multiply(divide(const_1000, const_2), const_1000), 40), 9), 30)) | it takes 40 identical printing presses 9 hours to print 500,000 papers . how many hours would it take 30 of these printing presses to print 500,000 papers ? | "40 printing presses can do 1 / 9 of the job each hour . 30 printing presses can do 3 / 4 * 1 / 9 = 1 / 12 of the job each hour . the answer is d ." | a = 1000 / 2
b = a * 1000
c = 1000 / 2
d = c * 1000
e = d / 40
f = e / 9
g = f * 30
h = b / g
|
a ) 10 years , b ) 11 years , c ) 12 years , d ) 13 years , e ) 14 years | c | subtract(multiply(15, 15), add(multiply(6, 14), multiply(9, 16))) | the average age of 15 students of a class is 16 years . out of these , the average age of 6 students is 14 years and that of the other 9 students is 16 years . tee age of the 15 th student is ? | "age of the 15 th student = [ 15 * 16 - ( 14 * 6 + 16 * 9 ) ] = ( 240 - 228 ) = 12 years . answer : c" | a = 15 * 15
b = 6 * 14
c = 9 * 16
d = b + c
e = a - d
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a ) 8 , b ) 12 , c ) 16 , d ) 20 , e ) 24 | b | divide(subtract(200, subtract(power(multiply(6, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 6 inches wide on each side , the total area would have been ( m + 200 ) square inches . what is the perimeter of the photograph , in inches ? | "let x and y be the width and length of the photograph . ( x + 2 ) ( y + 2 ) = m and so ( 1 ) xy + 2 x + 2 y + 4 = m ( x + 12 ) ( y + 12 ) = m and so ( 2 ) xy + 12 x + 12 y + 144 = m + 200 let ' s subtract equation ( 1 ) from equation ( 2 ) . 10 x + 10 y + 140 = 200 2 x + 2 y = 12 , which is the perimeter of the photog... | a = 6 * 2
b = a ** 2
c = 1 * 2
d = c ** 2
e = b - d
f = 200 - e
g = f / 2
|
a ) 11 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | subtract(divide(600, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7))) | how many numbers between 100 and 600 are divisible by 2 , 3 , and 7 together ? | "explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 600 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 600 : 600 / 42 = 14.28 hence , 42 * 14 = 588 is the last number divisible by 42 within 60... | a = 2 * 3
b = a * 7
c = 600 / b
d = 2 * 3
e = d * 7
f = 100 / e
g = c - f
|
a ) 17 hr , b ) 19 hr , c ) 22 hr , d ) 24 hr , e ) 26 hr | c | inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 5), 1), 5)))) | a pump can fill a tank with water in 2 hours . because of a leak , it took 2 1 / 5 hours to fill the tank . the leak can drain all the water of the tank in ? | work done by the tank in 1 hour = ( 1 / 2 - 2 1 / 5 ) = 1 / 22 leak will empty the tank in 22 hrs . answer : c | a = 1 / 2
b = 2 * 5
c = b + 1
d = c / 5
e = 1/(d)
f = a - e
g = 1/(f)
|
a ) 8232 , b ) 3282 , c ) 4282 , d ) 8242 , e ) 2382 | a | multiply(3000, power(add(const_1, divide(40, const_100)), 3)) | the salary of a labourer increase consistently by 40 % per year . if his present salary is rs . 3000 then what will his salary be after 3 years ? | explanation : salary = 3000 x 1.4 x 1.4 x 1.4 = 8232 answer : option a | a = 40 / 100
b = 1 + a
c = b ** 3
d = 3000 * c
|
a ) 35 % , b ) 48 % , c ) 55 % , d ) 65 % , e ) 75 % | b | add(divide(multiply(30, subtract(const_100, 40)), const_100), subtract(40, 10)) | in country z , 10 % of the people do not have a university diploma but have the job of their choice , and 30 % of the people who do not have the job of their choice have a university diploma . if 40 % of the people have the job of their choice , what percent of the people have a university diploma ? | setting up a matrix is how i solve this one . diploma no diploma totals job of choice w / diploma job of choice w / o diploma = 10 % job of choice total = 40 % not job of choice with diploma = . 3 x not job of choice w / o diploma = . 7 x total not job of choice = x total with diploma total without diploma total citize... | a = 100 - 40
b = 30 * a
c = b / 100
d = 40 - 10
e = c + d
|
a ) s . 43000 , b ) s . 50000 , c ) s . 57000 , d ) s . 60000 , e ) s . 47500 | e | add(add(multiply(multiply(multiply(20, 15), const_2), 50), multiply(multiply(multiply(20, 5), const_2), 50)), multiply(multiply(multiply(15, 5), const_2), 50)) | in a house a hall is 20 m long , 15 m wide and 5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . 50 per square m ? | "length ( l ) = 20 m , breadth ( b ) = 15 m and height ( h ) = 5 m total area of the hall to be covered with mat = 2 ( lb + bh + hl ) = 2 ( 20 * 15 + 15 * 5 + 5 * 20 ) = 2 ( 300 + 75 + 100 ) = 2 * 475 = 950 sq . m total expenditure = 50 * 950 = rs . 47500 e" | a = 20 * 15
b = a * 2
c = b * 50
d = 20 * 5
e = d * 2
f = e * 50
g = c + f
h = 15 * 5
i = h * 2
j = i * 50
k = g + j
|
a ) 0 , b ) 3 , c ) 6 , d ) 9 , e ) 12 | e | add(10, 2) | there is a 55 cm line marked at each centimeter and an insect is placed at every centimeter . 9 frogs are trained to jump a constant distance . the first one jumps 2 cm in every leap , the second one jumps 3 cm and so on until the 9 th one jumps 10 cm in every leap and they eat any insect that is available at that spot... | only the prime numbers greater than 10 and less than 55 were left . that is 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , and 53 . the total is 12 . the answer is e . | a = 10 + 2
|
a ) rs . 27,000 , b ) rs . 24,000 , c ) rs . 30,000 , d ) rs . 29,000 , e ) none of these | d | multiply(add(multiply(multiply(multiply(const_4, 2), multiply(add(const_2, const_3), const_2)), const_100), multiply(multiply(add(const_2, const_3), const_100), const_100)), divide(divide(multiply(add(const_2, const_3), const_2), const_2), multiply(const_4, const_3))) | jayant opened a shop investing rs . 30,000 . madhu joined him 2 months later , investing rs . 45,000 . they earned a profit of rs . 58,000 after completion of one year . what will be madhu ' s share of profit ? | 30,000 * 12 = 45,000 * 8 1 : 1 madhu ' s share = 1 / 2 * 58,000 i . e . rs . 29,000 answer : d | a = 4 * 2
b = 2 + 3
c = b * 2
d = a * c
e = d * 100
f = 2 + 3
g = f * 100
h = g * 100
i = e + h
j = 2 + 3
k = j * 2
l = k / 2
m = 4 * 3
n = l / m
o = i * n
|
a ) 40 days , b ) 50 days , c ) 60 days , d ) 70 days , e ) 80 days | d | divide(multiply(600, 35), 300) | there is enough provisions for 600 men in an army camp for 35 days . if there were 300 men less , how long will the provision last ? | "exp : we have , m 1 d 1 = m 2 d 2 600 * 35 = 300 * d 2 d 2 = 600 * 35 / 300 = 70 days . answer : d" | a = 600 * 35
b = a / 300
|
a ) 784 , b ) 763 , c ) 120 , d ) 160 , e ) 1260 | a | multiply(divide(multiply(8, subtract(8, const_1)), const_2), divide(multiply(8, subtract(8, const_1)), const_2)) | if 8 parallel lines in a plane is intersected by a family of another 8 parallel lines , how many parallelograms are there in the network thus formed ? | "parallelogram can formed by 2 horizontal and 2 vertical lines for horizontal 8 c 2 for vertical 8 c 2 total parallelogram is 10 c 2 * 8 c 2 = 28 * 28 = 784 answer : a" | a = 8 - 1
b = 8 * a
c = b / 2
d = 8 - 1
e = 8 * d
f = e / 2
g = c * f
|
a ) 13 , b ) 41 , c ) 47 , d ) 48 , e ) 53 | b | subtract(add(floor(divide(100, const_3)), floor(divide(100, add(const_1, const_4)))), multiply(floor(divide(100, multiply(const_3, add(const_1, const_4)))), const_2)) | there are 100 lights which are functional and each is controlled by a separate on / off switch . two children a and b start playing with the switches . a starts by pressing every third switch till he reaches the end . b , thereafter , presses every fifth switch till he too reaches the end . if all switches were in off ... | "number of switches = 100 number of switches turned on by a : 3 , 6 , . . . 99 = 33 number of switches turned on by b : 5 , 10 , . . . . 100 = 20 few switches are turned on by a and later turned off by b : lcm ( 3,5 ) = 15 x = 15 , 30 , . . . . 90 = 6 . subtract the above 6 switches from both a and b as they are turned... | a = 100 / 3
b = math.floor(a)
c = 1 + 4
d = 100 / c
e = math.floor(d)
f = b + e
g = 1 + 4
h = 3 * g
i = 100 / h
j = math.floor(i)
k = j * 2
l = f - k
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | c | divide(add(14, 10), add(2, 1)) | at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 10 pencils and 14 charts have been brought to the meeting , how man... | say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 10 total charts are 11 so a + 2 b = 14 add both : 3 a + 3 b = 24 so a + b = 8 total number of people = 8 c | a = 14 + 10
b = 2 + 1
c = a / b
|
a ) 7000 , b ) 6700 , c ) 6695 , d ) 7645 , e ) 6575 | c | floor(divide(6695, multiply(divide(subtract(const_100, 30), const_100), divide(subtract(const_100, 25), const_100)))) | 30 % people of a village in sri lanka died by bombardment , 25 % of the remainder left the village on account of fear . if now the population is reduced to 6695 , how much was it in the beginning ? | "x * ( 70 / 100 ) * ( 75 / 100 ) = 3515 x = 6695 answer : c" | a = 100 - 30
b = a / 100
c = 100 - 25
d = c / 100
e = b * d
f = 6695 / e
g = math.floor(f)
|
a ) 0.1 % , b ) 1 % , c ) 7 % , d ) 10 % , e ) none | b | divide(multiply(divide(70, const_100), const_100), 70) | i gain 70 paise on rs . 70 . my gain percent is | "solution gain % = ( 0.70 / 70 Γ 100 ) % = 1 % . answer b" | a = 70 / 100
b = a * 100
c = b / 70
|
a ) 22 , b ) 25 , c ) 32 , d ) 99 , e ) 88 | b | multiply(divide(multiply(50, 4), multiply(40, 20)), const_100) | 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day . if the average displacement of water by a man is 4 , then the rise in the water level in the tank will be : | explanation : total volume of water displaced = ( 4 x 50 ) = 200 rise in water level = = 0.25 m = 25 cm answer : b ) 25 cm | a = 50 * 4
b = 40 * 20
c = a / b
d = c * 100
|
a ) 9000 , b ) 26699 , c ) 96000 , d ) 19000 , e ) 26711 | a | subtract(10000, 1000) | the price of a t . v . set worth rs . 10000 is to be paid in 20 installments of rs . 1000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "money paid in cash = rs . 1000 balance payment = ( 10000 - 1000 ) = rs . 9000 answer : a" | a = 10000 - 1000
|
a ) 974 , b ) 1067 , c ) 1977 , d ) 1056 , e ) 1097 | a | divide(1120, add(const_1, divide(multiply(5, 3), const_100))) | find the principle on a certain sum of money at 5 % per annum for 3 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 5 * 3 ) / 100 ] p = 974 answer : a" | a = 5 * 3
b = a / 100
c = 1 + b
d = 1120 / c
|
a ) 6 min , b ) 4 min , c ) 2 min , d ) 5 min , e ) 8 min | a | multiply(divide(const_1, add(const_1, const_4)), divide(const_1, subtract(divide(const_1, 6), divide(const_1, 10)))) | a water tank is two - fifth full . pipe a can fill a tank in 10 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how long will it take to empty or fill the tank completely ? | "clearly , pipe b is faster than pipe a and so , the tank will be emptied . part to be emptied = 2 / 5 part emptied by ( a + b ) in 1 minute = ( 1 / 6 - 1 / 10 ) = 1 / 15 1 / 15 : 2 / 5 : : 1 : x x = ( 2 / 5 * 1 * 15 ) = 6 min . so , the tank will be emptied in 6 min . answer : a" | a = 1 + 4
b = 1 / a
c = 1 / 6
d = 1 / 10
e = c - d
f = 1 / e
g = b * f
|
a ) 3 , b ) 4 , c ) 5 , d ) 8 , e ) 15 | b | divide(reminder(31, 23), const_2) | how many two digit prime numbers are there in which both digits are prime numbers ? ( for example , 23 is one of these numbers but 31 is not , since 1 is not a prime number . ) | the second digit can only be 3 or 7 , so the choice quickly narrows down to 23 , 27 , 33 , 37 , 53 , 57 , 73 , and 77 . of these , 27 , 33 , and 57 are divisible by 3 , and 77 by 7 , leaving 23 , 37 , 53 , and 73 . it is easy to see that none of these is divisible by 2 , 3 , 5 , or 7 , and there is no need to look at g... | a = reminder / (
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a ) 7500 , b ) 3388 , c ) 2665 , d ) 8200 , e ) 2661 | d | divide(2460, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100))) | a candidate got 35 % of the votes polled and he lost to his rival by 2460 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2460 100 % - - - - - - - - - ? = > 8200 answer : d" | a = 35 / 100
b = 1 - a
c = 35 / 100
d = b - c
e = 2460 / d
|
a ) 20 , b ) 14 , c ) 16 , d ) 5 , e ) 8 | b | divide(127, 9) | { 8 , 46 , 53 , 127 } what number should be inserted in the set of 7 values to get a median of 9 ? | median is the middle number of a sorted distribution . in case of even number of items , median will be the average of middle two values . the question asks fora new number that will cause the median of new set at 9 . mentionable that given set is already sorted and number of items of new set will be 7 + 1 = 8 . to get... | a = 127 / 9
|
a ) 12.5 % , b ) 40 % , c ) 67 % , d ) 125 % , e ) none | c | multiply(divide(2, 3), const_100) | the ratio 2 : 3 expressed as a percent equals | "solution 2 : 3 = 2 / 3 = ( 2 / 3 x 100 ) % . = 66.6 % . answer c" | a = 2 / 3
b = a * 100
|
a ) 2 / 69 , b ) 2 / 60 , c ) 2 / 63 , d ) 2 / 35 , e ) 2 / 38 | d | multiply(divide(1, 7), divide(2, 5)) | two brother x and y appeared for an exam . the probability of selection of x is 1 / 7 and that of b is 2 / 5 . find the probability that both of them are selected . | "explanation : let a be the event that x is selected and b is the event that y is selected . p ( a ) = 1 / 7 , p ( b ) = 2 / 5 . let c be the event that both are selected . p ( c ) = p ( a ) Γ£ β p ( b ) as a and b are independent events : = ( 1 / 7 ) Γ£ β ( 2 / 5 ) = 2 / 35 answer : d ) 2 / 35" | a = 1 / 7
b = 2 / 5
c = a * b
|
a ) 11 , 1315 , b ) 7 , 2 , 10 , c ) 7 , 9 , 10 , d ) 7 , 9 , 11 , e ) 7 , 9 , 29 | a | subtract(multiply(divide(subtract(subtract(17, const_1), add(3, const_1)), const_2), const_2), const_1) | 3 consecutive odd integers are in increasing order such that the sum of the last two integers is 17 more than the first integer . find the 3 integers ? | explanation : let the three consecutive odd integers be x , x + 2 and x + 4 respectively . x + 4 + x + 2 = x + 17 = > x = 11 hence three consecutive odd integers are 11 , 13 and 15 answer : a | a = 17 - 1
b = 3 + 1
c = a - b
d = c / 2
e = d * 2
f = e - 1
|
a ) 13 / 10 , b ) 4 / 5 , c ) 11 / 10 , d ) 7 / 10 , e ) 9 / 10 | e | subtract(const_1, multiply(add(divide(const_1, 10), divide(const_1, 5)), 3)) | p is able to do a piece of work in 5 days and q can do the same work in 10 days . if they can work together for 3 days , what is the fraction of work completed ? | "explanation : amount of work p can do in 1 day = 1 / 5 amount of work q can do in 1 day = 1 / 10 amount of work p and q can do in 1 day = 1 / 5 + 1 / 10 = 3 / 10 amount of work p and q can together do in 3 days = 3 Γ ( 3 / 10 ) = 9 / 10 answer : option e" | a = 1 / 10
b = 1 / 5
c = a + b
d = c * 3
e = 1 - d
|
a ) 24 , b ) 28 , c ) 30 , d ) 32 , e ) 35 | c | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(di... | the sum of three consecutive multiples of 3 is 81 . what is the largest number ? | "let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 81 9 x = 72 x = 8 largest number = 3 x + 6 = 30 . answer : c" | a = 3 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 3 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 3 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 3 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 15 , b ) 20 , c ) 25 , d ) 28 , e ) 30 | e | subtract(50, multiply(divide(50, const_100), 10)) | how many liters of water must be evaporated from 50 liters of a 4 percent sugar solution to get a 10 percent sugar solution ? | "let x be the amount that needs to be evaporated . 0.04 ( 50 ) = 0.1 ( 50 - x ) 0.1 x = 5 - 2 x = 3 / 0.1 = 30 liters the answer is e ." | a = 50 / 100
b = a * 10
c = 50 - b
|
a ) 10.6 % , b ) 3.65 % , c ) 6.25 % , d ) 8.32 % , e ) 11 % | c | divide(const_100, 16) | at what rate percent per annum will a sum of money double in 16 years ? | "let principle = p s . i . = p t = 16 yrs rate = 100 * p / p * 16 = 25 / 4 % = 6.25 % answer is c" | a = 100 / 16
|
a ) $ 960 , b ) $ 1,875 , c ) $ 1,725 , d ) $ 2,050 , e ) $ 2,250 | b | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, 10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 7.50), const_1000) | a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 7.50 . if the company β s latest order required a truck with at least 240,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | "number of boxes = total volume / volume of one box = 240,000 / ( 8 * 10 * 12 ) = 250 one box costs 7.50 , so 250 box will cost = 250 * 7.5 = 1875 b is the answer" | a = 3 * 100
b = 8 * 10
c = a + b
d = c + 4
e = d * 1000
f = 8 * 10
g = f * 12
h = e / g
i = h * 7
j = i / 1000
|
a ) 144 , b ) 131 , c ) 115 , d ) 90 , e ) 45 | c | add(divide(multiply(10, subtract(10, const_1)), const_2), multiply(10, 7)) | 10 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 10 business exec and in each handshake 2 business execs are involved . hence 10 c 2 = 45 also , each of 10 exec will shake hand with every 7 other chairmen for total of 70 handshake . total = 45 + 70 = 115 ans : c" | a = 10 - 1
b = 10 * a
c = b / 2
d = 10 * 7
e = c + d
|
a ) 1 : 6 , b ) 6 : 1 , c ) 2 : 3 , d ) 4 : 3 , e ) 3 : 2 | a | divide(add(16, divide(2, 3)), const_100) | in what ratio must water be mixed with milk to gain 16 ( 2 / 3 ) % on selling the mixture at cost price ? | "explanation : let c . p . of 1 litre milk be re . 1 . s . p . of 1 litre of mixture = re . 1 , gain = 50 / 3 % c . p . of 1 litre of mixture = ( 100 x 3 / 350 x 1 ) = 6 / 7 by the rule of alligation , we have : ratio of water and milk = 1 / 7 : 6 / 7 = 1 : 6 . answer is a" | a = 2 / 3
b = 16 + a
c = b / 100
|
a ) 2.5 min , b ) 1.8 min , c ) 1.5 min , d ) 3.5 min , e ) 2 min | c | add(multiply(2, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 2), divide(const_1, 6)), const_2)), 2), const_60)) | two pipes a and b can fill a tank in 2 and 6 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by the first pipe in 1 minute = 1 / 2 part filled by the second pipe in 1 minute = 1 / 6 net part filled by pipe a and pipe b in 1 minute = ( 1 / 2 ) + ( 1 / 6 ) = 2 / 3 i . e , pipe a and b together can fill the tank in 3 / 2 minutes = 1.5 minutes answer is c" | a = 2 * 100
b = 1 / 2
c = 1 / 6
d = b + c
e = d * 2
f = 1 - e
g = f * 2
h = g * const_60
i = a + h
|
a ) 144 kmph , b ) 124 kmph , c ) 134 kmph , d ) 150 kmph , e ) 160 kmph | a | multiply(divide(400, subtract(18, 8)), const_3_6) | a train requires 8 seconds to pass a pole while it requires 18 seconds to cross a stationary train which is 400 mtrs long . find the speed of the train . | in 8 s the train crosses the pole and in 18 sec the train crosses one more stationary train in 10 sec the train travels a distance of 400 mtrs speed = 400 / 10 = 40 m / s = 40 ( 3600 / 1000 ) = 40 * 18 / 5 = 144 kmph answer : a | a = 18 - 8
b = 400 / a
c = b * const_3_6
|
a ) 133 , b ) 168 , c ) 189 , d ) 190 , e ) 200 | a | subtract(multiply(151, 7), multiply(6, 154)) | there are 6 people in the elevator . their average weight is 154 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person . | "solution average of 7 people after the last one enters = 151 . Γ’ Λ Β΄ required weight = ( 7 x 151 ) - ( 6 x 154 ) = 1057 - 924 = 133 . answer a" | a = 151 * 7
b = 6 * 154
c = a - b
|
a ) 1.5 , b ) 0.5 , c ) 6.25 , d ) 0.25 , e ) 6.0 | d | subtract(6.25, floor(6.25)) | for any number z , z * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.25 β 6.25 * ? | "since z * is defined as the greatest positive even integer less than or equal to z , then 6.25 * = 6 ( the greatest positive even integer less than or equal to 6.25 is 6 ) . hence , 6.25 β 6.25 * = 6.25 - 6 = 0.25 answer : d ." | a = math.floor(6, 25)
b = 6 - 25
|
a ) 4.2 , b ) 7.1 , c ) 7.2 , d ) 7.5 , e ) 8.0 | a | divide(multiply(multiply(3, 7), const_2), add(3, 7)) | a river boat leaves silver town and travels upstream to gold town at an average speed of 3 kilometers per hour . it returns by the same route at an average speed of 7 kilometers per hour . what is the average speed for the round - trip in kilometers per hour ? | pick a number which is lcm of 7 and 3 = 21 . upstream time = 21 / 3 = 7 hrs downstream time = 21 / 7 = 3 hrs total time = 10 hrs total distance = 42 average speed = 42 / 10 = 4.2 km / hr | a = 3 * 7
b = a * 2
c = 3 + 7
d = b / c
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a ) 62.12 , b ) 63 , c ) 63.33 , d ) 64 , e ) 65 | c | subtract(subtract(multiply(divide(400, add(add(6, 6), 5)), const_3), const_4), const_3) | 6 friends visited a book stall and brought 6 books , and everyone decided to contribute equally to the total bill of $ 400 . if one of the friends had a coupon for 5 % off the total bill , and if each friend still contributed equally after the coupon was applied to the bill , how much did each friend pay ? | at the non - discounted price , each friend would pay $ 66.66 , as $ 400 divided by 6 friends is $ 66.66 per friend . but if the bill is 5 % off , then each friend would pay 5 % less . 5 % of $ 66.66 is $ 3.33 , so each friend saves $ 3.33 and pays the remaining $ 63.33 correct option : option c | a = 6 + 6
b = a + 5
c = 400 / b
d = c * 3
e = d - 4
f = e - 3
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a ) 20,20 , b ) 46,14 , c ) 25,15 , d ) 30,10 , e ) none of these | b | subtract(add(divide(multiply(32, 5), subtract(5, const_1)), 5), 32) | the ages of two persons differ by 32 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "explanation : let their ages be x and ( x + 32 ) years . 5 ( x - 5 ) = ( x + 32 - 5 ) or 4 x = 52 or x = 14 . their present ages are 46 years and 14 years option b" | a = 32 * 5
b = 5 - 1
c = a / b
d = c + 5
e = d - 32
|
a ) 80 kg , b ) 85 kg , c ) 75 kg , d ) 100 kg , e ) 110 kg | c | add(multiply(8, 5), 35) | the average weight of 8 person ' s increases by 5 kg when a new person comes in place of one of them weighing 35 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 5 ) kg = 40 kg . weight of new person = ( 35 + 40 ) kg = 75 kg . answer : c" | a = 8 * 5
b = a + 35
|
a ) 51.2 , b ) 59.5 , c ) 52.8 , d ) 52.5 , e ) 52.1 | a | divide(add(multiply(22, 40), multiply(28, 60)), add(22, 28)) | the average marks of a class of 22 students is 40 and that of another class of 28 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 22 students = 22 * 40 = 880 sum of the marks for the class of 28 students = 28 * 60 = 1680 sum of the marks for the class of 50 students = 880 + 1680 = 2560 average marks of all the students = 2560 / 50 = 51.2 answer : a" | a = 22 * 40
b = 28 * 60
c = a + b
d = 22 + 28
e = c / d
|
a ) - 512 , b ) - 257 , c ) - 256 , d ) - 513 , e ) 128 | d | subtract(negate(512), const_1) | y = x ^ 2 + bx + 512 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 of ax ^ 2 + bx + c = 0 , the product of the roots = c / a = 512 / 1 = 256 and the sum of the roots = - b / a = - b 512 can be expressed as product of two number... | a = negate - (
|
a ) 250 , b ) 170 , c ) 148 , d ) 157 , e ) 150 | d | multiply(multiply(const_pi, 5), 10) | the slant height of a cone is 10 cm and radius of the base is 5 cm , find the curved surface of the cone . | "Ο * 5 * 10 = 157 answer : d" | a = math.pi * 5
b = a * 10
|
a ) 290 cm 2 , b ) 360 cm 2 , c ) 270 cm 2 , d ) 280 cm 2 , e ) 260 cm 2 | b | multiply(30, 12) | find the area of a parallelogram with base 30 cm and height 12 cm ? | "area of a parallelogram = base * height = 30 * 12 = 360 cm 2 answer : b" | a = 30 * 12
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a ) - 29 , b ) - 19 , c ) 18 , d ) 29 , e ) 39 | c | subtract(57, subtract(45, add(subtract(28, 37), 15))) | if 45 - [ 28 - { 37 - ( 15 - * ) } ] = 57 , then * is equal to : | "45 - [ 28 - { 37 - ( 15 - * ) } ] = 57 = > 45 - [ 28 - { 37 - 15 + * } ] = 57 45 - [ 28 - 37 + 15 - * ] = 57 = > 45 [ 43 - 37 - * ] = 57 45 - [ 6 - * ] = 57 = > 45 - 6 + * = 57 39 + * = 57 = > * = 57 - 39 = 18 answer : c" | a = 28 - 37
b = a + 15
c = 45 - b
d = 57 - c
|
a ) 22 , b ) 190 , c ) 277 , d ) 204 , e ) 208 | d | subtract(divide(multiply(const_1, const_1000), divide(51, 11)), 11) | in a kilometer race , a beats b by 51 meters or 11 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 11 ) / 51 = 215 sec . time taken by a = 215 - 11 = 204 sec . answer : d" | a = 1 * 1000
b = 51 / 11
c = a / b
d = c - 11
|
a ) 488 , b ) 996 , c ) 228 , d ) 456 , e ) 1986 | b | multiply(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 5) | how many three - digit numbers that do not contain the digits 2 or 5 are there ? | "we can have 7 digits ( 1 , 3,4 , 6,7 , 8,9 ) for the first place ( hundred ' s place ) . and similarly 8 digits for tenth ' s and unit digit . ( 0,1 , 3,4 , 6,7 , 8,9 ) so in total 7 * 8 * 8 = 488 hence b" | a = 10 * 10
b = 10 - 1
c = a * b
d = c * 5
|
a ) 100 , b ) 150 , c ) 225 , d ) 112.5 , e ) 212.5 | a | divide(subtract(divide(multiply(multiply(5000, 6), 2), const_100), divide(multiply(multiply(5000, 4), 2), const_100)), 2) | a person borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 % p . a for 2 years . find his gain in the transaction per year . | "explanation : the person borrows rs . 5000 for 2 years at 4 % p . a . simple interest simple interest that he needs to pay = prt / 100 = 5000 Γ 4 Γ 2 / 100 = 400 he also lends it at 6 % p . a for 2 years simple interest that he gets = prt / 100 = 5000 Γ 6 Γ 2 / 100 = 600 his overall gain in 2 years = rs . 600 - rs . 4... | a = 5000 * 6
b = a * 2
c = b / 100
d = 5000 * 4
e = d * 2
f = e / 100
g = c - f
h = g / 2
|
a ) 500 , b ) 550 , c ) 746 , d ) 700 , e ) 750 | c | divide(560, subtract(const_1, divide(25, const_100))) | shop offered 25 % offer for every shirt , smith bought a shirt at rs . 560 . and what was the shop ' s original selling price ? | sp * ( 75 / 100 ) = 560 sp = 7.46 * 100 = > cp = 746 answer : c | a = 25 / 100
b = 1 - a
c = 560 / b
|
a ) 724 , b ) 804 , c ) 11413 , d ) 844 , e ) none | c | divide(multiply(add(multiply(9, const_100), 9), multiply(10, const_100)), power(9, const_2)) | what is the least number of square tiles required to pave the floor of a room 10 m 17 cm long and 9 m 9 cm broad ? | "solution length of largest tile = h . c . f . of 1017 cm & 909 cm = 9 cm . area of each tile = ( 9 x 9 ) cm 2 β΄ required number of tiles = [ 1017 x 909 / 9 x 9 ] = 11413 . answer c" | a = 9 * 100
b = a + 9
c = 10 * 100
d = b * c
e = 9 ** 2
f = d / e
|
a ) 8 hours , b ) 9 / 5 hours , c ) 10 / 3 hours , d ) 6 hours , e ) 11 / 3 hours | c | multiply(9, divide(20, subtract(15, 9))) | if a person walks at 15 km / hr instead of 9 km / hr , he would have walked 20 km more . the time traveled by him is ? | "let the actual distance traveled be x km . then , x / 9 = ( x + 20 ) / 15 5 x - 3 x = 60 = > x = 30 km . travel time is = 30 / 9 hours = 10 / 3 hours answer : c" | a = 15 - 9
b = 20 / a
c = 9 * b
|
a ) 21 , b ) 24 , c ) 27 , d ) 36 , e ) 45 | c | add(divide(subtract(72, add(add(const_3, const_3), 3)), 3), add(const_3, const_3)) | the sum of 3 consecutive odd natural numbers , each divisible by 3 is 72 . what is the largest among them ? | let the 3 consecutive odd numbers each divisible by 3 be x , x + 3 , x + 6 given , x + x + 3 + x + 6 = 72 3 x = 63 x = 21 the numbers are 21 , 24 , 27 therefore the largest number is 27 answer c . | a = 3 + 3
b = a + 3
c = 72 - b
d = c / 3
e = 3 + 3
f = d + e
|
a ) 1 , b ) 2 , c ) 0.5 , d ) 1.5 , e ) 0 | e | multiply(divide(subtract(12, 4), const_2), const_1) | if sum of two no . is 12 and subtraction to that is 4 . so how much min . should added to smaller no . to make it even ? | let smaller no . be x . bigger no . = 12 - x . 12 - x - x = 4 . 12 - 2 x = 4 2 x = 8 , x = 4 . so 4 is itself even no . so , zero should be added . answer e | a = 12 - 4
b = a / 2
c = b * 1
|
a ) rs . 2000 , b ) rs . 10000 , c ) rs . 15000 , d ) rs . 20000 , e ) none of these | c | divide(5400, divide(multiply(3, 12), const_100)) | consider a lady took a loan from a bank at the rate of 12 % p . a . simple interest . after 3 years she had to pay rs . 5400 interest only for the period . the principal amount borrowed by her was | "explanation : principal = rs . ( 100 Γ 5400 / 12 Γ 3 ) = > rs . 15,000 . answer : c" | a = 3 * 12
b = a / 100
c = 5400 / b
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a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(385, 320), 320), const_100) | a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase q in sales ? | "last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase q = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c ." | a = 385 - 320
b = a / 320
c = b * 100
|
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