regisss/math_qa huggingface dataset:
Problem: in a college , 50 % of total 400 arts students are locals . 25 % of students from 100 science students are locals and 85 % of total 120 commerce students are locals . what is the total percentage of locals from arts , science and commerce .
Options: a ) 322 , b ) 340 , c ) 333 , d ) 327 , e ) 500
Rationale: locals from arts = 50 % of 400 = 200 locals from science = 25 % of 100 = 25 locals from commerce = 85 % of 120 = 102 total locals = 200 + 25 + 102 = 327 d
Answer: d
Problem: 12 : 6 seconds : : ? : 8 minutes
Options: a ) 10 , b ) 15 , c ) 16 , d ) 25 , e ) 30
Rationale: "12 * 8 = 6 * x x = 16 answer : c"
Answer: c
Problem: a canoe has two oars , left and right . each oar either works or breaks . the failure or non - failure of each oar is independent of the failure or non - failure of the other . you can still row the canoe with one oar . the probability that the left oar works is 3 / 5 . the probability that the right oar works is also 3 / 5 . what is the probability y that you can still row the canoe ?
Options: a ) 9 / 25 , b ) 10 / 25 , c ) 6 / 10 , d ) 2 / 3 , e ) 21 / 25
Rationale: "simply look at the question from the other side . what is the probability that you can ’ t row the canoe ? this would be 2 / 5 x 2 / 5 = 4 / 25 . using the idea that the probability of something happening is 1 – the probability that it doesn ’ t happen , you can use the following equation to reach the right answer y : 1 – 4 / 25 = 21 / 25 . answer choice e ."
Answer: e
Problem: if jake loses 15 pounds , he will weigh twice as much as his sister . together they now weigh 132 pounds . what is jake ’ s present weight , in pounds ?
Options: a ) 63 , b ) 93 , c ) 139 , d ) 147 , e ) 188
Rationale: "lets say j is the weight of jack and s is the wt of his sister . if he loses 15 pounds , he s twice as heavy as his sister . j - 15 = 2 * s also , together they weight 132 pounds j + s = 132 solvong the 2 equation , we get j = 93 pounds ! b"
Answer: b
Problem: a and b complete a work in 7 days . a alone can do it in 14 days . if both together can do the work in how many days ?
Options: a ) 4.6 days , b ) 4.78 days , c ) 5.65 days , d ) 3.77 days , e ) 5.75 days
Rationale: "1 / 7 + 1 / 14 = 3 / 14 14 / 3 = 4.6 days answer : a"
Answer: a
Problem: a car traveled the first quarter of a certain distance at 4 times the speed it traveled the remaining distance . what proportion of the total time traveled , was the time taken to travel the first quarter of the distance ?
Options: a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) 1 / 13
Rationale: these problems can be solved through algebra or sly number picking . being a big fan of solving problems with numbers , let ' s pick a total distance divisible by 4 ( say 40 ) so we can break it up into quarters , and a speed that can easily be one fourth , say 10 . each quarter is thus 10 kilometers ( or miles or feet or angstroms for all it matters ) , and the runner ' s speed is 10 km / h for the first quarter and 5 / 2 km / h for the remaining quarters . he ' ll take 1 hour to do the first quarter and then 4 hours for the second quarter , 4 hours for the third and 4 hours for the fourth . on total he will take 13 hours to complete this race , of which 1 hour was spent on the first quarter . so 1 / 13 . answer e .
Answer: e
Problem: one fourth of one third of two fifth of a number is 17 . what will be 40 % of that number
Options: a ) 140 , b ) 204 , c ) 180 , d ) 200 , e ) 250
Rationale: "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 17 then x = 17 * 30 = 510 40 % of 510 = 204 answer : option b"
Answer: b
Problem: what is the decimal equivalent of ( 1 / 5 ) ^ 2 ?
Options: a ) 0.0016 , b ) 0.04 , c ) 0.16 , d ) 0.25 , e ) 0.5
Rationale: "( 1 / 5 ) ² = ( 1 / 5 ) ( 1 / 5 ) = 1 / 25 approach # 1 : use long division to divide 25 into 1 to get 1 / 25 = 0.04 b"
Answer: b
Problem: at an international conference , “ red ” world countries and “ blue ” world countries are the only participants . the ratio of “ red ” world participants to “ blue ” world participants is 10 : 5 . if one - third of “ red ” world participants are left - handed and two - thirds of “ blue ” world participants are left - handed , then what is the fraction of the participants who are left - handed ?
Options: a ) 4 / 10 , b ) 4 / 9 , c ) 6 / 5 , d ) 9 / 2 , e ) 6 / 4
Rationale: red : blue = 10 : 5 let red = 10 x and blue = 5 x 1 / 3 of red are left handed = > 1 / 3 * 10 x = 10 x / 3 red left handed 2 / 3 of blue are left handed = > 2 / 3 * 5 x = 10 x / 3 blue left handed fraction of participants who are left handed = total left handed / total participants = ( red left handed + blue left handed ) / total participants = ( 10 x / 3 + 10 x / 3 ) / ( 10 x + 5 x ) = 20 / 45 = 4 / 9 answer : b
Answer: b
Problem: the average salary / head of allthe workers in a workshop is rs . 750 , if the average salary / head of 5 technician is rs . 900 and the average salary / head of the rest is rs . 700 , the total no . of workers in the work - shop is ?
Options: a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26
Rationale: let the total number of workers be y . so sum of salary for all workers = sum of salary of 5 technician + sum of salary for other y - 5 workers . 5 x 900 + 700 ( y - 5 ) = 750 y ⇒ 4500 + 700 y - 3500 = 750 y ⇒ 50 y = 1000 ∴ y = 20 so total number of workers = 20 b
Answer: b
Problem: the average age of 10 men increases by 6 years when two women are included in place of two men of ages 18 and 22 years . find the average age of the women ?
Options: a ) 37 , b ) 26 , c ) 30 , d ) 50 , e ) 11
Rationale: "explanation : 18 + 22 + 10 * 6 = 100 / 2 = 50 answer : d"
Answer: d
Problem: a started a business with an investment of rs . 70000 and after 3 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 43000 , then the share of b is ?
Options: a ) 33888 , b ) 36000 , c ) 27778 , d ) 27772 , e ) 81122
Rationale: ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 9 ) = 7 : 36 total profit = rs . 43000 share of b = 36 / 43 ( 43000 ) = rs . 36000 answer : b
Answer: b
Problem: working together , printer a and printer b would finish the task in 10 minutes . printer a alone would finish the task in 30 minutes . how many pages does the task contain if printer b prints 5 pages a minute more than printer a ?
Options: a ) 125 , b ) 150 , c ) 175 , d ) 200 , e ) 225
Rationale: "10 * a + 10 * b = x pages in 10 mins printer a will print = 10 / 30 * x pages = 1 / 3 * x pages thus in 10 mins printer printer b will print x - 1 / 3 * x = 2 / 3 * x pages also it is given that printer b prints 5 more pages per min that printer a . in 10 mins printer b will print 50 more pages than printer a thus 2 / 3 * x - 1 / 3 * x = 50 = > x = 150 pages answer : b"
Answer: b
Problem: the ratio of boys to girl in a school is 5 : 4 . if there are 1500 boys in the school , how many girls are there ?
Options: a ) 300 , b ) 1200 , c ) 1500 , d ) 1800 , e ) 2000
Rationale: think of the ratio as 5 parts : 4 parts . divide 1500 by 5 to find 1 ` ` part ' ' of the ratio . 1500 / 5 = 300 multiply this by 4 to get 4 ` ` parts ' ' of the ratio . 300 * 4 = 1200 there are 1200 girls in the school . the answer is b
Answer: b
Problem: a tank can be filled by two pipes a and b in 60 minutes and 40 minutes respectively . how many minutes will it take to fill the tank from empty state if b is used for the first half time and then a and b fill it together for the other half
Options: a ) 15 mins , b ) 20 mins , c ) 25 mins , d ) 30 mins , e ) none of these
Rationale: "explanation : let the total time be x mins . part filled in first half means in x / 2 = 1 / 40 part filled in second half means in x / 2 = 1 / 60 + 1 / 40 = 1 / 24 total = x / 2 ∗ 1 / 40 + x / 2 ∗ 1 / 24 = 1 = > x / 2 ( 1 / 40 + 1 / 24 ) = 1 = > x / 2 ∗ 1 / 15 = 1 = > x = 30 mins option d"
Answer: d
Problem: the diameter of a garden roller is 1.4 m and it is 2 m long . how much area will it cover in 4 revolutions ? ( use ï € = 22 â  „ 7 )
Options: a ) 35.2 m 2 , b ) 44 m 2 , c ) 48 m 2 , d ) 36 m 2 , e ) none of these
Rationale: "required area covered in 5 revolutions = 4 ã — 2 ï € rh = 4 ã — 2 ã — 22 â  „ 7 ã — 0.7 ã — 2 = 35.2 m 2 answer a"
Answer: a
Problem: in a group of ducks and cows , the total number of legs are 48 more than twice the no . of heads . find the total no . of buffaloes .
Options: a ) 28 , b ) 12 , c ) 24 , d ) 16 , e ) 19
Rationale: "let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 48 = > 2 x = 48 = > x = 24 c"
Answer: c
Problem: two - third of a positive number and 64 / 216 of its reciprocal are equal . the number is :
Options: a ) 5 / 12 , b ) 12 / 5 , c ) 25 / 144 , d ) 2 / 3 , e ) 146 / 25
Rationale: "let the number be x . then , 2 / 3 x = 64 / 216 * 1 / x x 2 = 64 / 216 * 3 / 2 = 64 / 144 = 4 / 9 x = 2 / 3 answer : d"
Answer: d
Problem: when positive integer m is divided by positive integer n , the remainder is 12 . if m / n = 24.2 , what is the value of n ?
Options: a ) 120 , b ) 60 , c ) 30 , d ) 24 , e ) 12
Rationale: n = decimal part of 24.2 * remainder i . e . ( 12 ) hence 0.2 * n = 12 n = 60 . b
Answer: b
Problem: if x and y are positive integers and 12 x = 21 y what is the least possible value of xy ?
Options: a ) 9 , b ) 28 , c ) 63 , d ) 84 , e ) 252
Rationale: "12 x = 21 y = > x / y = 7 / 4 = > 4 x = 7 y 4 ( 3 ) = 7 ( 3 ) = > x * y = 9 a"
Answer: a
Problem: a train crosses a platform of 350 m in 15 sec , same train crosses another platform of length 500 m in 20 sec . then find the length of the train ?
Options: a ) 100 m , b ) 180 m , c ) 159 m , d ) 250 m , e ) 152 m
Rationale: "length of the train be ‘ x ’ x + 350 / 15 = x + 500 / 20 4 x + 1400 = 3 x + 1500 x = 100 m answer : a"
Answer: a
Problem: 0.001 ã — 0.001 = ?
Options: a ) 0.1 , b ) 0.001 , c ) 0.01 , d ) 1.0 e - 06 , e ) none of these
Rationale: "0.001 ã — 0.001 = ? or , ? = 0.000001 answer d"
Answer: d
Problem: if doubling a number and adding 20 to the result gives the same answer as multiplying the number by 8 and taking away 4 from the product , the number is
Options: a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 7
Rationale: solution let the number be x . = then , 2 x + 20 = 8 x – 4 = 6 x = 24 ‹ = › x = 4 . answer c
Answer: c
Problem: x + ( 1 / x ) = 3.5 find x ^ 2 + ( 1 / x ^ 2 )
Options: a ) 2.25 , b ) 3.25 , c ) 10.25 , d ) 5.25 , e ) 6.25
Rationale: squaring on both sides ( x + 1 / x ) ^ 2 = 3.5 ^ 2 x ^ 2 + 1 / x ^ 2 = 12.25 - 2 x ^ 2 + 1 / x ^ 2 = 10.25 answer : c
Answer: c
Problem: what is the difference between the largest number and the least number written with the digits 9 , 2 , 1 , 5 ?
Options: a ) 6084 , b ) 3788 , c ) 2077 , d ) 8262 , e ) 1812
Rationale: "explanation : 1259 9521 - - - - - - - - - - - - 8262 answer : d"
Answer: d
Problem: the ratio of flour to water to sugar in a recipe is 7 : 2 : 1 . the ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar . if the new recipe calls for 2 cups of water , how much sugar is required ?
Options: a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
Rationale: "the ratio of flour to water is 14 : 2 . the ratio of flour to sugar is 3.5 : 1 = 14 : 4 . the new ratio of flour to water to sugar is 14 : 2 : 4 if we need 2 cups of water , then we need 4 cups of sugar . the answer is c ."
Answer: c
Problem: a train , 130 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is
Options: a ) 270 m , b ) 245 m , c ) 235 m , d ) 220 m , e ) 240 m
Rationale: explanation : assume the length of the bridge = x meter total distance covered = 130 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 130 + x ) / 30 m / s = > 45 × ( 10 / 36 ) = ( 130 + x ) / 30 = > 45 × 10 × 30 / 36 = 130 + x = > 45 × 10 × 10 / 12 = 130 + x = > 15 × 10 × 10 / 4 = 130 + x = > 15 × 25 = 130 + x = 375 = > x = 375 - 130 = 245 answer : option b
Answer: b
Problem: the unit digit in the product 1258 * 6754 * 4512 * 9783 is ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
Rationale: "unit digit in the given product = unit digit in 8 * 4 * 2 * 3 = 2 answer is c"
Answer: c
Problem: a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 900 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ?
Options: a ) 671.5 , b ) 600 , c ) 672.5 , d ) 673 , e ) 773.5
Rationale: "many approaches are possible , one of them : let the distance be 1 kilometer . time to cover this distance at 900 kilometers per hour is 1 / 900 hours = 3,600 / 900 seconds = 4 seconds ; time to cover this distance at regular speed is 900 + 2 = 902 seconds = 902 / 3,600 hours = 1 / 600 hours ; so , we get that to cover 1 kilometer 1 / 600 hours is needed - - > regular speed 600 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : b ."
Answer: b
Problem: a school has 4 section of chemistry in class x having 70 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student .
Options: a ) 50.9 , b ) 52.9 , c ) 51.9 , d ) 53.25 , e ) none of the above
Rationale: "required average marks = 70 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 70 + 35 + 45 + 42 = 3500 + 2100 + 2475 + 1890 / 192 = 9965 â  „ 192 = 51.90 answer c"
Answer: c
Problem: how much is 40 % of 60 is greater than 4 / 5 of 25 ?
Options: a ) 2 , b ) 5 , c ) 12 , d ) 8 , e ) 4
Rationale: "( 40 / 100 ) * 60 â € “ ( 4 / 5 ) * 25 24 - 20 = 4 answer : e"
Answer: e
Problem: today joelle opened an interest - bearing savings account and deposited $ 5,000 . if the annual interest rate is 5 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ?
Options: a ) $ 6512.50 , b ) $ 7512.50 , c ) $ 5512.50 , d ) $ 4512.50 , e ) $ 5250.00
Rationale: "interest for 1 st year = 5000 * 5 / 100 = 250 interest for 2 nd year = 5250 * 5 / 100 = 262.50 total = 5000 + 250 + 262.50 = 5512.50 answer : c"
Answer: c
Problem: if the sides of a triangle are 30 cm , 26 cm and 10 cm , what is its area ?
Options: a ) 120 , b ) 110 , c ) 130 , d ) 140 , e ) 150
Rationale: "the triangle with sides 30 cm , 26 cm and 10 cm is right angled , where the hypotenuse is 30 cm . area of the triangle = 1 / 2 * 26 * 10 = 130 cm 2 answer : option c"
Answer: c
Problem: a store owner estimates that the average price of type a products will increase by 15 % next year and that the price of type b products will increase by 18 % next year . this year , the total amount paid for type a products was $ 4600 and the total price paid for type b products was $ 8400 . according to the store owner ' s estimate , and assuming the number of products purchased next year remains the same as that of this year , how much will be spent for both products next year ?
Options: a ) $ 15,202 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155
Rationale: "cost of type a products next year = 1.15 * 4600 = 5290 cost of type b products next year = 1.18 * 8400 = 9912 total 5290 + 9912 = 15202 answer : a"
Answer: a
Problem: when positive integer n is divided by positive integer p , the quotient is 18 , with a remainder of 7 . when n is divided by ( p + 2 ) , the quotient is 10 and the remainder is 3 . what is the value of n ?
Options: a ) 151 , b ) 331 , c ) 511 , d ) 43 , e ) 87
Rationale: "n / p = 18 7 / p = 18 p + 7 n / ( p + 2 ) = 10 1 / ( p + 2 ) = 10 p + 20 + 3 solving these two equations we get p = 2 n = 43 answer is d"
Answer: d
Problem: ashok borrowed some money at the rate of 6 % p . a . for the first two years , at the rate of 9 % p . a . for the next 3 years and at the rate of 14 % p . a . for the period beyond 5 years . if he pays a total interest of 11400 / - at the end of 9 years , how much money did he borrow ?
Options: a ) 16060 , b ) 14000 , c ) 18000 , d ) 12000 , e ) none of these
Rationale: we have , si = p × r × t / 100 ∴ 11400 = p × 6 × 2 / 100 + p × 9 × 3 / 100 + p × 14 × 4 / 100 or , 12 p + 27 p + 56 p = 11400 × 100 or , 95 p = 11400 × 100 ∴ p = 12000 answer d
Answer: d
Problem: ` ` i am 8 times as old as you were when i was as old as you are ` ` , said a man to his son . find out their present ages if the sum of the their ages is 75 years .
Options: a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27
Rationale: present age of boy x and previous age of boy y so present age of dad : 8 y and previous age is x so the gap between duration must be same so 8 y - x = x - y - - > x = 9 y / 2 and x + 8 y = 75 ( given ) - - > 9 y / 2 + y = 75 so y = 6 and and x = 27 therefore fathers age is 8 y = 8 * 6 = 48 and child ' s age = x = 27 answer : e
Answer: e
Problem: ada and paul received their scores on 3 tests . on the first test , ada ' s score was 10 points higher than paul ' s score . on the second test , ada ' s score was 4 points higher than paul ' s score . if paul ' s average ( arithmetic mean ) score on the 3 tests was 4 points higher than ada ' s average score on the 3 tests , then paul ' s score on the third test was how many points higher than ada ' s score ?
Options: a ) 9 , b ) 14 , c ) 17 , d ) 23 , e ) 26
Rationale: my take is option d ( 23 ) i followed a simple approach ( explained below ) : test 1 : ada ' s score = paul ' s score + 10 test 2 : ada ' s score = paul ' s score + 4 avg . of paul ' s score = 4 points higher than avg . of ada ' s score this implies that : sum of paul ' s score [ 3 tests ] = 12 points higher than sum of ada ' s score [ 3 tests ] ( 12 points higher since 3 points were given in terms of average of 3 scores ) so , paul needs to score 23 points higher than ada in test 3 , since paul needs to compensate for the lower score in test 1 and test 2 ( 26 = 10 + 4 + 12 ) e
Answer: e
Problem: the simple interest on a certain sum of money for 3 years at 8 % per annum is half the compound interest on $ 4000 for 2 years at 10 % per annum . the sum placed on simple interest is ?
Options: a ) $ 1250 , b ) $ 1300 , c ) $ 1750 , d ) $ 2015 , e ) $ 1452
Rationale: c . i . = 4000 * ( 1 + 10 / 100 ) ^ 2 - 4000 = 4000 * 11 / 10 * 11 / 10 - 4000 = $ 840 sum = 420 * 100 / 3 * 8 = $ 1750 answer is c
Answer: c
Problem: teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 156 per kg , the price of the third variety per kg will be ?
Options: a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 178.50 , d ) rs . 258.50 , e ) none of these
Rationale: "explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \ / mean price ( rs . 156 ) / \ x â ˆ ’ 156 22.50 = > x â ˆ ’ ( 156 / 22.50 ) = 1 . = > x â ˆ ’ 156 = 22.50 . = > x = 178.50 rs . answer : c"
Answer: c
Problem: in a company with 48 employees , some part - time and some full - time , exactly ( 1 / 3 ) of the part - time employees and ( 1 / 4 ) of the full - time employees take the subway to work . what is the greatest possible number of employees who take the subway to work ?
Options: a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16
Rationale: "let part time emp = x let full time emp = y then , 48 = x + y . . . . . . . . . ( 1 ) ( 1 / 3 ) x + ( 1 / 4 ) y = no . of ppl taking the subway 4 x + 3 y / 12 = no . of ppl taking the subway using 1 x / 12 + 3 * 48 / 12 = no . of ppl taking the subway so , the minimum value of x has to be 12 . hence to maximize the no . put 36 for x 36 / 12 + 12 / 4 = 15 answer : d"
Answer: d
Problem: what is the maximum number of pieces of birthday cake of size 3 ” by 3 ” that can be cut from a cake 21 ” by 21 ” ?
Options: a ) 5 , b ) 7 , c ) 49 , d ) 20 , e ) 25
Rationale: "the prompt is essentially asking for the maximum number of 3 x 3 squares that can be cut from a larger 21 by 21 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 7 ' pieces ' each , we have ( 7 ) ( 7 ) = 49 total smaller squares ( at maximum ) . c"
Answer: c
Problem: in a certain company , the ratio of the number of managers to the number of non - managers in any department must always be greater than 7 : 37 . in the company , what is the maximum number of non - managers in a department that has 9 managers ?
Options: a ) 44 , b ) 45 , c ) 46 , d ) 47 , e ) 48
Rationale: "9 / 7 * 37 = 47.6 the answer is d ."
Answer: d
Problem: a train covers a distance of 12 km in 10 min . if it takes 7 sec to pass a telegraph post , then the length of the train is ?
Options: a ) 177 m , b ) 189 m , c ) 140 m , d ) 178 m , e ) 188 m
Rationale: "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 7 = 140 m . answer : c"
Answer: c
Problem: in a class of 78 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ?
Options: a ) 26 , b ) 27 , c ) 24 , d ) 18 , e ) 11
Rationale: "explanation : you could solve this by drawing a venn diagram . a simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking french to find the number taking only french . likewise find those taking only german . then we have : total = only french + only german + both + neither 78 = ( 41 - 9 ) + ( 22 - 9 ) + 9 + neither . not enrolled students = 24 answer : c"
Answer: c
Problem: if ( n + 2 ) ! / n ! = 182 , n = ?
Options: a ) 2 / 131 , b ) 9 , c ) 10 , d ) 11 , e ) 12
Rationale: "( n + 2 ) ! / n ! = 182 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 132 from here , we might just test the answer choices . since ( 14 ) ( 13 ) = 182 , we can see that n = 12 e"
Answer: e
Problem: in what ratio must tea at rs . 64 per kg be mixed with tea at rs . 74 per kg so that the mixture must be worth rs . 69 per kg ?
Options: a ) 1 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 5 : 3 , e ) none
Rationale: "required ratio = 500 : 500 = 1 : 1 answer a"
Answer: a
Problem: 36 . a 240 m long train crosses a platform of equal length in 27 s . what is the speed of the train in km / h ?
Options: a ) 48 , b ) 60 , c ) 56 , d ) 64 , e ) 68
Rationale: by the formula , speed = distance / time = ( 240 + 240 ) / 27 x ( 18 / 5 ) = 480 / 27 x 18 / 5 = 64 km / h answer : d
Answer: d
Problem: the base of a triangular field is three times its altitude . if the cost of cultivating the field at rs . 24.68 per hectare be rs . 333.18 , find its base and height .
Options: a ) 100 m
b ) 125 m
c ) 150 m
d ) 278 m
e ) 300 m
Rationale: sol . area of the field = total cost / rate = ( 333.18 / 25.6 ) hectares = 13.5 hectares ( 13.5 x 10000 ) m ^ 2 = 135000 m ^ 2 . let altitude = x metres and base = 3 x metres . then , ( 1 / 2 ) * 3 x * x = 135000 < = > x ^ 2 = 90000 < = > x = 300 . base = 900 m and altitude = 300 m . ans : e
Answer: e
Problem: a , b and c enter into a partnership by investing $ 11000 , $ 15000 and $ 23000 respectively . at the end of 8 months , b receives $ 1540 as his share . find the share of a .
Options: a ) $ 811 , b ) $ 922 , c ) $ 1033 , d ) $ 1144 , e ) $ 1255
Rationale: "the ratio of capital of a , b and c = 11000 : 15000 : 23000 = 11 : 15 : 23 a ' s share = ( 11 / 15 ) * 1560 = $ 1144 the answer is d ."
Answer: d
Problem: if each year the population of the country grows by 30 % , how many years will elapse before the population of the country doubles ?
Options: a ) 2 , b ) 4 , c ) 5 , d ) 3 , e ) 7
Rationale: "till year 2000 , population is 100 . year 2001 : population becomes 130 . . . . . . . . . . . . . 1 year elapsed year 2002 : population becomes 169 . . . . . . . . . . . . . 2 year elapsed year 2004 : population > 200 . . . . . . . . . . . . . . . . . . 3 year elapsed answer : d"
Answer: d
Problem: if the median of a list of numbers is m , the first quartile of the list is the median of the numbers in the list that are less than m . what is the first quartile of the list of numbers 42 , 24 , 30 , 34 , 26 , 36 , 33 and 35 ?
Options: a ) 33 , b ) 28 , c ) 27 , d ) 24 , e ) 23
Rationale: "it is given that a quartile is the middle number of all numbers less than median . . so lets arrange the number in ascending order - 42 , 24 , 30 , 34 , 26 , 36 , 33 and 35 24 , 26 , 30 , 33 , 34 , 35 , 36 , 42 . . . numbers less than median are 24 , 26 , 30 , 33 . . the median of these numbers = center of 26 and 30 = 28 b"
Answer: b
Problem: how many integers are divisible by 3 between 20 ! and 20 ! + 20 inclusive ?
Options: a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
Rationale: "b - 7 20 ! is divisible by 3 there are 6 numbers between 10 ! and 10 ! + 20 that are divisible by 3 . hence 7"
Answer: b
Problem: the speed of a boat in still water is 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 27 minutes is :
Options: a ) 9.25 , b ) 5.25 , c ) 7.25 , d ) 6.25 , e ) 11.25
Rationale: "explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 27 / 60 ) ) km = 11.25 km . answer : e"
Answer: e
Problem: if 4 people are selected from a group of 10 married couples , what is the probability that none of them would be married to each other ?
Options: a ) 1 / 33 , b ) 2 / 63 , c ) 1 / 3 , d ) 43 / 62 , e ) 11 / 12
Rationale: "if we are to select 4 people from 10 couples without any restriction , how many ways can we make the selection ? 20 ! / 4 ! 16 ! = 4845 if we are to select 4 people from 10 couples with restriction that no married couple can both make it to the group , only a representative ? 10 ! / 4 ! 6 ! = 210 but we know that to select a person from each couple , take 2 possibilities 210 * 2 * 2 * 2 * 2 = 3360 probability = desired / all possibilities = 3360 / 4845 = 43 / 62 answer : d"
Answer: d
Problem: what is the least number of square tiles required to pave the floor of a room 50 m 00 cm long and 11 m 25 cm broad ?
Options: a ) 724 , b ) 804 , c ) 814 , d ) 9000 , e ) none
Rationale: solution length of largest tile = h . c . f . of 5000 cm & 1125 cm = 25 cm . area of each tile = ( 25 x 25 ) cm 2 ∴ required number of tiles = [ 5000 x 1125 / 25 x 25 ] = 9000 . answer d
Answer: d
Problem: in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 65 liters and they are all empty , how much money total will it cost to fuel all cars ?
Options: a ) 320 $ , b ) 380 $ , c ) 420 $ , d ) 450 $ , e ) 528 $
Rationale: "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 65 ) = 21 + 507 = > 528 hence answer will be ( e ) 528"
Answer: e
Problem: if the average of 37 , 69 , 47 and x is 46.5 , what is x ?
Options: a ) 37 , b ) 41 , c ) 50 , d ) 59 , e ) 33
Rationale: "x = 33 because : ( 37 + 69 + 47 + x ) / 4 = 46.5 ( 37 + 69 + 47 + x ) / 4 * 4 = 46.5 * 4 ( 37 + 69 + 47 + x ) = 186 ( 37 + 69 + 47 + x ) - 153 = 186 - 153 x = 33 therefore , the answer is e , 33 ."
Answer: e
Problem: joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 60 . he was allowed to drop his lowest score . after doing so , his average test score improved to 65 . what is the lowest test score that was dropped ?
Options: a ) 20 , b ) 45 , c ) 55 , d ) 65 , e ) 80
Rationale: "the arithmetic mean of 4 equally weighted tests was 60 . so what we can assume is that we have 4 test scores , each 60 . he dropped his lowest score and the avg went to 65 . this means that the lowest score was not 60 and other three scores had given the lowest score 5 each to make it up to 60 too . when the lowest score was removed , the other 3 scores got their 5 back . so the lowest score was 3 * 5 = 15 less than 60 . so the lowest score = 60 - 15 = 45 answer ( b )"
Answer: b
Problem: if the number 992 , 132,24 x is divisible by 11 , what must be the value of x ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 0
Rationale: multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 992 , 132,24 x sum of digits at odd places = 9 + 2 + 3 + 2 + x = 16 + x ( i ) sum of digits at even places = 9 + 1 + 2 + 4 = 16 ( ii ) ( i ) - ( ii ) = 16 + x - 16 = x - 0 hence x should be = 0 to make this a multiple of 11 ( 0 ) option e
Answer: e
Problem: in kaya ' s teacher ' s desk there are 9 pink highlighters , 8 yellow highlighters , and 5 blue highlighters . how many highlighters are there in all ?
Options: a ) 11 , b ) 22 , c ) 77 , d ) 33 , e ) 88
Rationale: add the numbers of highlighters . 9 + 8 + 5 = 22 . answer is b .
Answer: b
Problem: a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 500 bolts ?
Options: a ) 36 , b ) 72 , c ) 120 , d ) 144 , e ) 180
Rationale: "old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 500 bolts 500 * 6 / 25 = 120 mins ans c"
Answer: c
Problem: the average mark of a class of thirty two students is 75 . if 3 students whose marks are 28 and 34 are removed , then find the approximate average mark of the remaining students of the class .
Options: a ) 71 , b ) 74 , c ) 78 , d ) 70 , e ) 80
Rationale: exp . total mark of 32 students = 75 * 32 = 2400 , total mark after the removal of 2 students = 2400 – ( 28 + 34 ) = 2400 – 62 = 2338 approximate average mark = 2338 / ( 32 - 2 ) = 2338 / 30 = 78 answer : c
Answer: c
Problem: a can do a piece of work in 6 days , and b can do it in 5 days . if c , who can do the work in 12 days , joins them , how long will they take to complete the work ?
Options: a ) 2 days , b ) 3.5 days , c ) 2.2 days , d ) 4 days , e ) 5.7 days
Rationale: a , b , and c do the work in = 6 * 5 * 12 / 6 * 5 + 5 * 12 + 6 * 12 = 360 / 162 = 2.2 days answer is c
Answer: c
Problem: the perimeter of a triangle is 39 cm and the inradius of the triangle is 1.5 cm . what is the area of the triangle ?
Options: a ) 22 , b ) 35 , c ) 77 , d ) 54 , e ) 29
Rationale: "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 1.5 * 39 / 2 = 29 cm 2 answer : e"
Answer: e
Problem: the average age of 20 students of a class is 20 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years , the age of the 20 th student is
Options: a ) 186 , b ) 120 , c ) 152 , d ) 220 , e ) 220
Rationale: "explanation : age of the 20 th student = [ 20 * 20 - ( 14 * 5 + 16 * 9 ) ] = ( 400 - 214 ) = 186 years . answer : a"
Answer: a
Problem: tom , working alone , can paint a room in 6 hours . peter and john , working independently , can paint the same room in 3 hours and 5 hours , respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ?
Options: a ) 19 / 35 , b ) 23 / 42 , c ) 27 / 55 , d ) 31 / 63 , e ) 35 / 74
Rationale: "tom paints 1 / 6 of the room in the first hour . tom and peter paint 1 / 6 + 1 / 3 = 1 / 2 of the room in the next hour for a total of 4 / 6 . the three people then paint the remaining 2 / 6 in a time of ( 2 / 6 ) / ( 21 / 30 ) = 10 / 21 hours peter worked for 31 / 21 hours so he painted 31 / 21 * 1 / 3 = 31 / 63 of the room . the answer is d ."
Answer: d
Problem: little roshini had $ 10.50 . he spent some money on sweets and gave to his two friends $ 3.40 each . how much money was spent for sweets ?
Options: a ) $ 4.70 , b ) $ 4.80 , c ) $ 3.85 , d ) $ 3.80 , e ) $ 3.70
Rationale: roshini spent and gave to his two friends a total of x + 3.40 + 3.40 = $ 3.65 money spent for sweet = 10.50 - 6.80 = $ 3.70 correct answer is e ) $ 3.70
Answer: e
Problem: a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 9 hours . if the company used the same number of each type of machine to do job in 12 hours , how many machine r were used ?
Options: a ) 3 , b ) 4 , c ) 18 / 5 , d ) 9 , e ) 12
Rationale: "yes there is a typo in the question , i got the same ques on my gmat prep last week , and the questions goes as : a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 9 hours . if the company used the same number of each type of machine to do job in 2 hours , how many machine r were used ? so for a job to be done in 2 hours r = 1 / 2 r _ a ( rate of machine r ) = 1 / 36 r _ s ( rate of machine s ) = 1 / 9 lets say x machines are used to attain the desired rate , thus x / 36 + x / 9 = 1 / 2 ( desired r = 1 / 2 i . e . to complete the job in 2 hours ) ( x + 4 x ) / 36 = 1 / 2 5 x / 36 = 1 / 2 x = 18 / 5 . qa = 18 / 5 ( answer c )"
Answer: c
Problem: a batsman makes a score of 110 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning .
Options: a ) 40 , b ) 50 , c ) 60 , d ) 45 , e ) 56
Rationale: "let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 110 = 11 x x = 110 - 50 = 60 answer is c"
Answer: c
Problem: if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 1 hour ?
Options: a ) 100 , b ) 120 , c ) 150 , d ) 180 , e ) 240
Rationale: "answer = 4 * 60 = 240 answer is e"
Answer: e
Problem: the difference of two numbers is 8 and one - fourth of their sum is 6 . find the numbers .
Options: a ) 16,8 , b ) 10,2 , c ) 17,6 , d ) 14,10 , e ) 12,4
Rationale: "let numbers be x and y . equation ( i ) : x - y = 8 equation ( ii ) : ( x + y ) / 4 = 6 solve system of equations : x = 16 , y = 8 a is the correct answer ."
Answer: a
Problem: a retailer bought a shirt at wholesale and marked it up 80 % to its initial price of $ 45 . by how many more dollars does he need to increase the price to achieve a 100 % markup ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: "x is the buy price . x + 80 % x = 1,8 x is the sale price = 45 so x , the buy price becomes 45 / 1,8 = 25 % 100 mark up means x + 100 % x = 2 x so 50 . 50 - 45 = 5 answer : e"
Answer: e
Problem: indu gave bindu rs . 1250 on compound interest for 2 years at 8 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ?
Options: a ) 1 , b ) 2 , c ) 8 , d ) 4 , e ) 5
Rationale: "1250 = d ( 100 / 4 ) 8 d = 8 answer : c"
Answer: c
Problem: the ratio between the length and the breadth of a rectangular park is 1 : 2 . if a man cycling along the boundary of the park at the speed of 6 km / hr completes one round in 6 min , then the area of the park ( in sq . m ) is ?
Options: a ) 600 m , b ) 6000 m , c ) 20000 m , d ) 100 m , e ) 2000 m
Rationale: "perimeter = distance covered in 6 min . = ( 6000 x 6 ) / 60 m = 600 m . let length = 1 x metres and breadth = 2 x metres . then , 2 ( 1 x + 2 x ) = 600 or x = 100 . length = 100 m and breadth = 200 m . area = ( 100 x 200 ) m 2 = 20000 m c"
Answer: c
Problem: at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 7.50 + 0.25 ( s – 1 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1 ?
Options: a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75
Rationale: oa is definitely wrong . the answer should be c .
Answer: c
Problem: the lateral surface area of cuboid length 12 m , breadth 8 m and height 6 m .
Options: a ) 540 , b ) 240 , c ) 440 , d ) 140 , e ) 340
Rationale: "cuboid lateral surface = 2 h ( l + b ) = 2 × 6 ( 20 ) = 240 m ( power 2 ) answer is b"
Answer: b
Problem: running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 12 such machines produce in 4 minutes ?
Options: a ) 648 , b ) 1800 , c ) 2700 , d ) 2160 , e ) none of these
Rationale: "solution let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) â ˆ ´ 6 ã — 1 ã — x = 12 ã — 4 ã — 270 â ‡ ” x = 12 x 4 x 270 / 6 = 2160 . answer d"
Answer: d
Problem: of the 400 employees in a certain company , 25 percent will be relocated to city x and the remaining 75 percent will be relocated to city y . however , 40 percent of the employees prefer city y and 60 percent prefer city x . what is the highest possible number of employees who will be relocated to the city they prefer ?
Options: a ) 65 , b ) 100 , c ) 115 , d ) 130 , e ) 260
Rationale: "240 prefer x ( group 1 ) ; 160 prefer y ( group 2 ) . city y needs 300 people : letall 160 who prefer y ( entire group 2 ) be relocated there , the rest 140 will be those who prefer x from group 1 ; city x needs 100 people : 240 - 140 = 100 from group 1 will be relocated to x , which they prefer . so , the highest possible number of employees who will be relocated to the city they prefer is 160 + 100 = 260 . answer : e ."
Answer: e
Problem: two friends plan to walk along a 22 - km trail , starting at opposite ends of the trail at the same time . if friend p ' s rate is 20 % faster than friend q ' s , how many kilometers will friend p have walked when they pass each other ?
Options: a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
Rationale: "if q complete x kilometers , then p completes 1.2 x kilometers . x + 1.2 x = 22 2.2 x = 22 x = 10 then p will have have walked 1.2 * 10 = 12 km . the answer is c ."
Answer: c
Problem: a car traveled 40 % of the way from town x to town y at an average speed of 40 mph . the car traveled at an average speed of v mph for the remaining part of the trip . the average speed for the entire trip was 50 mph . what is v in mph ?
Options: a ) 50 , b ) 45 , c ) 55 , d ) 60 , e ) 65
Rationale: "assume total distance = 100 miles time taken for 40 miles = 40 / 40 = 1 hour time taken for the rest of the 60 miles = 60 / v hours . average speed = 50 therefore the total time needed = 2 hours . 2 = 1 + 60 / v hence v = 60 mph answer : d"
Answer: d
Problem: the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , what is the weight of b ?
Options: a ) 28 , b ) 30 , c ) 32 , d ) 34 , e ) 31
Rationale: "let the weight of a , b and c are a , b and c respectively . average weight of a , b and c = 45 a + b + c = 45 × 3 = 135 - - - equation ( 1 ) average weight of a and b = 40 a + b = 40 × 2 = 80 - - - equation ( 2 ) average weight of b and c = 43 b + c = 43 × 2 = 86 - - - equation ( 3 ) equation ( 2 ) + equation ( 3 ) - equation ( 1 ) = > a + b + b + c - ( a + b + c ) = 80 + 86 - 135 = > b = 80 + 86 - 135 = 166 - 135 = 31 weight of b = 31 kg answer is e ."
Answer: e
Problem: bert and rebecca were looking at the price of a condominium . the price of the condominium was 80 % more than bert had in savings , and separately , the same price was also 20 % more than rebecca had in savings . what is the ratio of what bert has in savings to what rebecca has in savings .
Options: a ) 1 : 4 , b ) 4 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) 3 : 4
Rationale: "suppose bert had 100 so price becomes 180 , this 180 = 1.2 times r ' s saving . . so r ' s saving becomes 150 so required ratio is 100 : 150 = 2 : 3 answer : c"
Answer: c
Problem: of the 1000 inhabitants of a town , 60 % are males of whom 120 % are literate . if , of all the inhabitants , 25 % are literate , then what percent of the females of the town are literate ?
Options: a ) 32.5 % , b ) 32.9 % , c ) 92.5 % , d ) 38.5 % , e ) 37.5 %
Rationale: "explanation : number of males = 60 % of 1000 = 600 . number of females = ( 1000 - 600 ) = 400 . number of literates = 25 % of 1000 = 250 . number of literate males = 20 % of 600 = 120 . number of literate females = ( 250 - 120 ) = 130 . required pecentage = ( 130 / 400 * 100 ) % = 32.5 % . answer : a ) 32.5 %"
Answer: a
Problem: if 4 < x < 6 < y < 10 , then what is the greatest possible positive integer difference of x and y ?
Options: a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
Rationale: "4 < x < 6 < y < 10 ; 4 < x y < 10 4 + y < x + 10 y - x < 6 . positive integer difference is 5 ( for example y = 9.5 and x = 4.5 ) answer : c ."
Answer: c
Problem: a car traveling at a certain constant speed takes 15 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 48 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ?
Options: a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43
Rationale: "48 * t = 1 km = > t = 1 / 48 km / h v * ( t + 15 / 3600 ) = 1 v ( 1 / 48 + 15 / 3600 ) = 1 v ( 90 / 3600 ) = 1 v = 40 km / h the answer is b ."
Answer: b
Problem: in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.75 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ?
Options: a ) 320 $ , b ) 420 $ , c ) 490 $ , d ) 516 $ , e ) 680 $
Rationale: 12 * 1.75 + 0.75 * 12 * 55 = 516 hence - d
Answer: d
Problem: the sum of the first n positive perfect squares , where n is a positive integer , is given by the formula n 3 / 3 + cn 2 + n / 6 , where c is a constant . what is the sum of the first 15 positive perfect squares ?
Options: a ) 1,010 , b ) 1,164 , c ) 1,240 , d ) 1,316 , e ) 1,476
Rationale: "for n = 1 , 1 = 1 / 3 + c + 1 / 6 1 = 1 / 2 + c = > c = 1 / 2 15 * 15 * 15 / 3 + 1 / 2 * 15 * 15 + 15 / 6 = 1240 answer : c"
Answer: c
Problem: a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 15 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ?
Options: a ) 1276
b ) 1200
c ) 2832
d ) 1800
e ) 1236
Rationale: explanation : l * 30 = 1200 è l = 40 40 + 30 + 50 = 120 120 * 15 = 1800 answer : option d
Answer: d
Problem: there are 7 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ?
Options: a ) 21 , b ) 30 , c ) 45 , d ) 60 , e ) 90
Rationale: "7 players are there . two players play one game with one another . so 7 c 2 = 7 * 6 / 2 = 21 so option a is correct"
Answer: a
Problem: in a graduating class of 238 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ?
Options: a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31
Rationale: "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 238 = 25 greatest - smallest is 144 - 25 = 119 so answer must be b . 119"
Answer: b
Problem: the diagonals of a rhombus are 12 cm and 15 cm . find its area ?
Options: a ) 30 , b ) 45 , c ) 90 , d ) 120 , e ) 180
Rationale: "area = 0.5 x 12 x 15 = 90 cm â ² answer : c"
Answer: c
Problem: there are 50 students in a class . if 10 % are absent on a particular day , find the number of students present in the class .
Options: a ) 43 , b ) 36 , c ) 28 , d ) 45 , e ) 11
Rationale: "number of students absent on a particular day = 10 % of 50 i . e . , 10 / 100 × 50 = 5 therefore , the number of students present = 50 - 5 = 45 students . answer : d"
Answer: d
Problem: the speed at which a woman can row a boat in still water is 300 kmph . if he rows downstream , where the speed of current is 42 kmph , what time will he take to cover 190 metres ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: "speed of the boat downstream = 300 + 42 = 342 kmph = 342 * 5 / 18 = 95 m / s hence time taken to cover 190 m = 190 / 95 = 2 seconds . answer : b"
Answer: b
Problem: in a circle with a radius of 6 , what is the area of the biggest rectangle that can be cut out of this circle ?
Options: a ) 12
b ) 96
c ) 50
d ) 72
e ) 144
Rationale: the biggest rectangle would be a square , whose diagonal is 12 ( 6 + 6 ) if side of square is a , 144 = a ^ 2 + a ^ 2 = 2 a ^ 2 so area of square = a ^ 2 = 144 / 2 = 72 answer : d
Answer: d
Problem: the speed of a boat in still water in 42 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 44 minutes is :
Options: a ) 86.6 km , b ) 46.6 km , c ) 33 km , d ) 35.6 km , e ) 26.6 km
Rationale: "speed downstream = ( 42 + 3 ) = 45 kmph time = 44 minutes = 44 / 60 hour = 11 / 15 hour distance travelled = time × speed = 11 / 15 × 45 = 33 km answer : c"
Answer: c
Problem: find the total average marks of all the students in 2 separate classes , if the average marks of students in the first class of 39 students is 45 and that of another class of 35 students is 70 .
Options: a ) 55.8 , b ) 51.8 , c ) 53.8 , d ) 56.8 , e ) 52.8
Rationale: sum of the marks for the class of 39 students = 39 * 45 = 1755 sum of the marks for the class of 35 students = 35 * 70 = 2450 sum of the marks for the class of 74 students = 1755 + 2450 = 4205 average marks of all the students = 4205 / 74 = 56.8 answer : d
Answer: d
Problem: on an order of 4 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of :
Options: a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 %
Rationale: "clearly , the retailer gets 1 dozen out of 5 dozens free . equivalent discount = 1 / 5 * 100 = 20 % . answer b ) 20 %"
Answer: b
Problem: in an examination , 300 students appeared . out of these students ; 27 % got first division , 54 % got second division and the remaining just passed . assuming that no student failed ; find the number of students who just passed .
Options: a ) 33 , b ) 57 , c ) 54 , d ) 99 , e ) 01
Rationale: "the number of students with first division = 27 % of 300 = 27 / 100 × 300 = 8100 / 100 = 81 and , the number of students with second division = 54 % of 300 = 54 / 100 × 300 = 16200 / 100 = 162 therefore , the number of students who just passed = 300 – ( 81 + 162 ) = 57 answer : b"
Answer: b
Problem: x starts a business with rs . 77000 . y joins in the business after 5 months with rs . 42000 . what will be the ratio in which they should share the profit at the end of the year ?
Options: a ) 21 : 23 , b ) 23 : 45 , c ) 22 : 7 , d ) 25 : 29 , e ) none of these
Rationale: "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 77000 * 12 : 42000 * 7 = 77 * 12 : 42 * 7 = 11 * 2 : 7 = 22 : 7 . answer : option c"
Answer: c
Problem: what is the % change in the area of a rectangle when its length increases by 10 % and its width decreases by 10 % ?
Options: a ) 0 % , b ) 20 % increase , c ) 20 % decrease , d ) 1 % decrease , e ) insufficient data
Rationale: "explanatory answer whenever you encounter problems like this , use a numerical example and then proceed . for ease of computation , it is safe in most cases , to assume the length to be 100 units and the width to be 100 units . ( remember , a square is a rectangle too and the problem works the same way when you assume different values for length and width . computation becomes a bit tedious with different values for length and width ) area of a rectangle = length * width = 100 * 100 = 10,000 sq units . when the length increases by 10 % , the new length becomes 110 units . and as the width decreases by 10 % , new width becomes 90 units . therefore , new area = 110 * 90 = 9900 sq units . new area is 100 sq units lesser than the original area . % change in area = ( ( change in area ) / ( original area ) ) * 100 = ( 100 / 10,000 ) * 100 = 1 % decrease in area the correct choice is ( d )"
Answer: d