regisss/math_qa huggingface dataset:
Problem: ram - leela has $ 100 in her piggy bank . how much will she have in her bank 52 weeks from now if she puts $ 1 in the bank next week , $ 2 two weeks from now , $ 3 3 weeks from now , and continues to increase the amount that she puts in by $ 1 each week ?
Options: a ) 1478 , b ) 1578 , c ) 1678 , d ) 1778 , e ) 1798
Rationale: the dollar deposits are in an a . p . 1,2 , 3,4 . . . 52 with common difference 1 sum of the terms is n ( n + 1 ) / 2 i . e 52 * ( 52 + 1 ) / 2 = 52 * 53 / 2 = 1378 total deposit therefore with chiu - lihas is 100 + 1378 = 1478 $
Answer: a
Problem: a certain mixture of nuts consists of 5 parts almonds to 2 parts walnuts , by weight . what is the number of pounds of almonds in 350 pounds of the mixture ?
Options: a ) 250 , b ) 84 , c ) 40 , d ) 28 , e ) 20
Rationale: almonds : walnuts = 5 : 2 total mixture has 7 parts in a 350 pound mixture , almonds are 5 / 7 ( total mixture ) = 5 / 7 * 350 = 250 pounds answer ( a )
Answer: a
Problem: if x = - 5 / 4 and y = - 3 / 2 , what is the value of the expression - 2 x – y ^ 2 ?
Options: a ) 1 / 4 , b ) - 1 , c ) 5 / 4 , d ) 3 / 2 , e ) 3 / 4
Rationale: x = - 5 / 4 and y = - 3 / 2 = = > - 2 ( - 5 / 4 ) - ( 3 / 2 ) ^ 2 = 10 / 4 - 9 / 4 = 1 / 4 ans : a
Answer: a
Problem: which expression is the greatest
Options: a ) 3252 / 3257 , b ) 3456 / 3461 , c ) 3591 / 3596 , d ) 3641 / 3656 , e ) 3453 / 3458
Rationale: "options can be re - written as ( x - 5 ) x = > 1 - ( 5 / x ) a ) 1 - ( 5 / 3257 ) b ) 1 - ( 5 / 3461 ) c ) 1 - ( 5 / 3596 ) d ) 1 - ( 5 / 3656 ) e ) 1 - ( 5 / 3458 ) to get the largest among these second half should be the least and so denominator to be largest . hence ' d ' ."
Answer: d
Problem: jerry ’ s average ( arithmetic mean ) score on the first 3 of 4 tests is 78 . if jerry wants to raise his average by 2 points , what score must he earn on the fourth test ?
Options: a ) 87 , b ) 89 , c ) 86 , d ) 93 , e ) 95
Rationale: "total score on 3 tests = 78 * 3 = 234 jerry wants the average to be = 80 hence total score on 4 tests should be = 80 * 4 = 320 score required on the fourth test = 320 - 234 = 86 option c"
Answer: c
Problem: guy drives 60 miles to attend a meeting . halfway through , he increases his speed so that his average speed on the second half is 16 miles per hour faster than the average speed on the first half . his average speed for the entire trip is 30 miles per hour . guy drives on average how many miles per hour during the first half of the way ?
Options: a ) 12 , b ) 14 , c ) 16 , d ) 24 , e ) 40
Rationale: let x be the average speed for 1 st half of the distance . then the average speed for 2 nd half of the distance will be x + 16 avg speed = total distance / total time 30 = 60 / { ( 30 / x ) + ( 30 / ( x + 16 ) ) } solving we get x ^ 2 - 14 x - 240 = 0 x = - 10 or 24 x cant be negative hence x = 24 answer : d
Answer: d
Problem: 8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 18 men , 4 women and 10 children together complete the piece of work ?
Options: a ) 5 days , b ) 15 days , c ) 28 days , d ) 4 days , e ) 7 days
Rationale: "explanation : 1 man ’ s 1 day ’ s work = 1 / 8 × 12 = 1 / 96 18 men ’ s 1 day ’ s work = 1 × 18 / 96 = 3 / 16 1 woman ’ s 1 day ’ s work = 1 / 192 4 women ’ s 1 day ’ s work = 1 / 192 × 4 = 1 / 48 1 child ’ s 1 day ’ s work = 1 / 240 10 children ’ s 1 day ’ s work = 1 / 24 therefore , ( 18 men + 4 women + 10 children ) ’ s 1 day ’ s work = 3 / 16 + 1 / 48 + 1 / 24 = 1 / 4 the required no . of days = 4 days answer : option d"
Answer: d
Problem: square a is inscribed in circle b . if the perimeter of a is 32 , what is the circumference of b ?
Options: a ) 6 √ 2 π
b ) 4 √ 2 π
c ) 8 √ 2 π
d ) 7 √ 2 π
e ) 5
Rationale: square forms two right angled triangles . any time we have a right angle triangle inside a circle , the hypotenuse is the diameter . hypotenuse here = diagonal of the square = 8 sqrt ( 2 ) = diameter = > radius = 4 sqrt ( 2 ) circumference of the circle = 2 pi r = 8 pi sqrt ( 2 ) answer is c .
Answer: c
Problem: 15 years hence , rohan will be just 4 times as old as he was 15 years ago . how old is rohan at present ?
Options: a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40
Rationale: let the present age of rohan be x years then , given : x + 15 = 4 ( x - 15 ) x = 25 answer : b
Answer: b
Problem: dan ’ s car gets 32 miles per gallon . if gas costs $ 4 / gallon , then how many miles can dan ’ s car go on $ 54 of gas ?
Options: a ) 236 , b ) 354 , c ) 432 , d ) 512 , e ) 670
Rationale: 54 / 4 = 13.5 gallons 13.5 * 32 = 432 miles the answer is c .
Answer: c
Problem: if the operation ø is defined for all positive integers x and w by x ø w = ( 2 ^ x ) / ( 2 ^ w ) then ( 3 ø 1 ) ø 1 = ?
Options: a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32
Rationale: "3 ø 1 = 2 ^ 3 / 2 ^ 1 = 4 4 ø 1 = 2 ^ 4 / 2 = 8 the answer is c ."
Answer: c
Problem: what will be the remainder when ( 67 ^ 67 ) + 67 is divided by 68 ?
Options: a ) 52 , b ) 62 , c ) 66 , d ) 68 , e ) 72
Rationale: "x ^ n + 1 will be divisible by x + 1 only when n is odd 67 ^ 67 + 1 will be divisible by 67 + 1 ( 67 ^ 67 + 1 ) + 66 , when divided by 68 will give 66 as remainder answer is c"
Answer: c
Problem: walking at the rate of 5 kmph a man cover certain distance in 5 hr . running at a speed of 15 kmph the man will cover the same distance in .
Options: a ) 12 min , b ) 36 min , c ) 40 min , d ) 48 min , e ) 60 min
Rationale: distance = speed * time 5 * 5 = 25 km new speed = 15 kmph therefore time = 25 / 15 = 5 / 3 = 36 min answer : b
Answer: b
Problem: subash can copy 50 pages in 10 hrs . subash and prakash together can copy 300 pages in 40 hours . in how much time prakash can copy 18 pages .
Options: a ) 8 , b ) 10 , c ) 2 , d ) 14 , e ) 16
Rationale: subhas ' s 1 hr copy page = 50 / 10 = 5 page ( subhas + prakash ) ' s 1 hr copy page = 300 / 40 = 7.5 page from above prakash ' s 1 hr copy page = 2.5 page so time taken in 30 page ' s copy = ( 5 / 2.5 ) = 2 hrs answer : c
Answer: c
Problem: a pair of articles was bought for $ 720 at a discount of 10 % . what must be the marked price of each of the article ?
Options: a ) $ 300 , b ) $ 500 , c ) $ 350 , d ) $ 400 , e ) $ 600
Rationale: "s . p . of each of the article = 720 / 2 = $ 360 let m . p = $ x 90 % of x = 360 x = 360 * 100 / 90 = $ 400 answer is d"
Answer: d
Problem: average expenditure of a person for the first 3 days of a week is rs . 330 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is :
Options: a ) 350 , b ) 370 , c ) 390 , d ) 430 , e ) none
Rationale: "explanation : assumed mean = rs . 330 total excess than assumed mean = 4 × ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 / 7 = rs . 40 therefore , average expenditure for 7 days = rs . 330 + rs . 40 = rs . 370 correct option : b"
Answer: b
Problem: set x consists of 10 integers and has median of 30 and a range of 30 . what is the value of the greatest possible integer that can be present in the set ?
Options: a ) 32 , b ) 37 , c ) c . 40 , d ) 43 , e ) 50
Rationale: "note that both median and range do not restrict too many numbers in the set . range is only concerned with the smallest and greatest . median only cares about the middle . quick check of each option starting from the largest : ( e ) 50 range of 20 means the smallest integer will be 30 . so 20 can not lie in between and hence can not be the median . ( d ) 43 range of 20 means the smallest integer will be 23 . so 20 can not lie in between and hence can not be the median . ( c ) 40 range of 20 means the smallest integer will be 20 . 20 can lie in between such as : 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 40 , 50 this is possible . hence it is the greatest such number . answer ( e )"
Answer: e
Problem: in a class there are 55 pupil , out of them 10 are in debate only and 18 in singing only . then how many in both ?
Options: a ) 132 , b ) 26 , c ) 17 , d ) 11 , e ) 12
Rationale: explanation : total pupil = 55 debate + singing = 10 + 18 = 28 the intersection for two = 55 â € “ 10 â € “ 28 = 17 play both games . answer : c
Answer: c
Problem: in a hostel , the number of students decreased by 8 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the previous year ?
Options: a ) 19 % , b ) 15 % , c ) 25 % , d ) 40 % , e ) 9.4 %
Rationale: "cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 8 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.92 n ) * ( 1.2 p ) = > f ( new ) = f / ( 0.92 * 1.2 ) = > f ( new ) = 0.906 f therefore the new cost of food must be 90.6 % of the old cost , or the cost of food must decrease by 9.4 % answer : e"
Answer: e
Problem: the total circumference of two circles is 88 . if the first circle has a circumference that is exactly twice the circumference of the second circle , then what is the approximate sum of their two radii ?
Options: a ) 5.7 , b ) 6.0 , c ) 6.7 , d ) 9.7 , e ) 14.0
Rationale: "let r = radius of smaller circle . let r = radius of larger circle therefore : 2 π r + 2 π r = 88 where 2 r = r thus : 2 π r + 4 π r = 88 6 π r = 88 r = approx 4.7 π r + 2 r π = 88 3 π r = 88 r = approx 9.3 r + r = approx 14.0 answer : e"
Answer: e
Problem: how many multiples of 2 are there between 101 and 999 ?
Options: a ) 250 , b ) 440 , c ) 510 , d ) 575 , e ) 449
Rationale: "2 multiples are 102 , 104,106 , - - - - - - - - - , 994 , 996,998 it should be mentioned whether 1 and 89 are inclusive . the answer is ( 998 - 102 ) / 2 + 1 = 449 answer is e"
Answer: e
Problem: find the smallest number which when divided by 13 and 16 leaves respective remainders of 2 and 5 .
Options: a ) 128 , b ) 197 , c ) 127 , d ) 182 , e ) 091
Rationale: "explanation : let ' n ' is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5 . required number = ( lcm of 13 and 16 ) - ( common difference of divisors and remainders ) = ( 208 ) - ( 11 ) = 197 . answer : b"
Answer: b
Problem: how much is 80 % of 40 is greater than 4 / 5 of 20 ?
Options: a ) 12 , b ) 27 , c ) 18 , d ) 16 , e ) 81
Rationale: "( 80 / 100 ) * 40 â € “ ( 4 / 5 ) * 20 32 - 16 = 16 answer : d"
Answer: d
Problem: if the price of sugar rises from rs . 10 per kg to rs . 13 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by
Options: a ) 15 % , b ) 20 % , c ) 23 % , d ) 30 % , e ) none
Rationale: "sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 10 = x × 13 = x = 77 kg . ∴ reduction in consumption = 23 % . answer c"
Answer: c
Problem: a train crosses a bridge of length 800 m in 45 seconds and a lamp post on the bridge in 15 seconds . what is the length of the train in metres ?
Options: a ) 375 m , b ) 750 m , c ) 400 m , d ) 800 m , e ) 300 m
Rationale: "let length of train = l case - 1 : distance = 800 + l ( while crossing the bridge ) time = 45 seconds i . e . speed = distance / time = ( 800 + l ) / 45 case - 2 : distance = l ( while passing the lamp post ) time = 15 seconds i . e . speed = distance / time = ( l ) / 15 but since speed has to be same in both cases so ( 800 + l ) / 45 = ( l ) / 15 i . e . 800 + l = 3 l i . e . 2 l = 800 i . e . l = 400 answer : option c"
Answer: c
Problem: the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 20 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ?
Options: a ) 28 % , b ) 40 % , c ) 64.8 % , d ) 70 % , e ) 72 %
Rationale: "let initial price be 1000 price in day 1 after 10 % discount = 900 price in day 2 after 10 % discount = 810 price in day 3 after 20 % discount = 648 so , price in day 3 as percentage of the sale price on day 1 will be = 648 / 900 * 100 = > 72 % answer will definitely be ( e )"
Answer: e
Problem: the sum of three consecutive numbers is 87 . the greatest among these three number is :
Options: a ) 26 , b ) 28 , c ) 29 , d ) 30 , e ) 31
Rationale: "let the numbers be x , x + 1 and x + 2 then , x + ( x + 1 ) + ( x + 2 ) = 87 3 x = 84 x = 28 greatest number , ( x + 2 ) = 30 . answer : d"
Answer: d
Problem: a train travels 290 km in 4.5 hours and 400 km in 5.5 hours . find the average speed of train .
Options: a ) 80 kmph , b ) 69 kmph , c ) 70 kmph , d ) 90 kmph , e ) none of these
Rationale: "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 290 + 400 = 690 km thus , total speed = 10 hrs or , average speed = 690 / 10 or , 69 kmph . answer : b"
Answer: b
Problem: what is the average of 12 , 13 , 14 , 510 , 520 , 530 , 1,115 , 1,120 , and 1 , 125,2140 , 2345 ?
Options: a ) 419 , b ) 551 , c ) 601 , d ) 620 , e ) 858.54
Rationale: "add 12 , 13 , 14 , 510 , 520 , 530 , 1,115 , 1,120 , and 1,125 , 2140 , 2345 grouping numbers together may quicken the addition sum = 9444 4959 / 11 = 858.54 . e"
Answer: e
Problem: when a certain tree was first planted , it was 4 feet tall , and the height of the tree increased by a constant amount each year for the next 6 years . at the end of the 6 th year , the tree was 1 / 3 taller than it was at the end of the 4 th year . by how many feet did the height of the tree increase each year ?
Options: a ) 3 / 10 , b ) 2 , c ) 1 / 2 , d ) 2 / 3 , e ) 6 / 5
Rationale: "say , the tree grows by x feet every year . then , 4 + 6 x = ( 1 + 1 / 3 ) ( 4 + 4 x ) or , x = 2 answer b"
Answer: b
Problem: how many integerskgreater than 100 and less than 800 are there such that if the hundreds and the units digits ofkare reversed , the resulting integer is k + 99 ?
Options: a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90
Rationale: "numbers will be like 102 = > 201 = 102 + 99 203 = > 302 = 103 + 99 so the hundereth digit and units digit are consecutive where unit digit is bigger than hundred digit . there will be six pairs of such numbers for every pair there will 10 numbers like for 12 = > 102 , 112,132 , 142,152 , 162,172 , 182,192 . total = 6 * 10 = 60 hence b ."
Answer: b
Problem: local kennel has cats and dogs in the ratio of 5 : 10 . if there are 40 fewer cats than dogs , how many dogs are in the kennel ?
Options: a ) 80 , b ) 70 , c ) 60 , d ) 50 , e ) 45
Rationale: "lets work with the data given to us . we know that there ratio of cats to dogs is 5 : 10 or cats 5 dogs 10 we can write number of cats as 5 x and number of dogs as 10 x and we know that 10 x - 5 x = 40 ( therefore 5 x = 40 = > x = 8 ) then # of dogs = 10 x 8 = 80 answer is a"
Answer: a
Problem: for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 45 true ?
Options: a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9
Rationale: "answer b ( a + b ) ( a - b ) = 45 3 cases for ( a + b ) , ( a - b ) 45 , 1 15 , 3 9 , 5 answer b"
Answer: b
Problem: if the number 892 , 142,24 x is divisible by 11 , what must be the value of x ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 0 , e ) 5
Rationale: "multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 892 , 142,24 x sum of digits at odd places = 8 + 2 + 4 + 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 d"
Answer: d
Problem: a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio w of milk and water in the resultant mixture ?
Options: a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2
Rationale: "he 20 litre mixture contains milk and water in the ratio of 3 : 2 . therefore , there will be 12 litres of milk in the mixture and 8 litres of water in the mixture . step 1 . when 10 litres of the mixture is removed , 6 litres of milk is removed and 4 litres of water is removed . therefore , there will be 6 litres of milk and 4 litres of water left in the container . it is then replaced with pure milk of 10 litres . now the container will have 16 litres of milk and 4 litres of water . step 2 . when 10 litres of the new mixture is removed , 8 litres of milk and 2 litres of water is removed . the container will have 8 litres of milk and 2 litres of water in it . now 10 litres of pure milk is added . therefore , the container will have 18 litres of milk and 2 litres of water in it at the end of the second step . therefore , the ratio of milk and water is 18 : 2 or 9 : 1 . shortcut . we are essentially replacing water in the mixture with pure milk . let w _ o be the amount of water in the mixture originally = 8 litres . let w _ r be the amount of water in the mixture after the replacements have taken place . then , { w _ r } / { w _ o } = ( 1 - r / m ) ^ n where r is the amount of the mixture replaced by milk in each of the steps , m is the total volume of the mixture and n is the number of times the cycle is repeated . hence , { w _ r } / { w _ o }   = ( 1 / 2 ) ^ 2   = 1 / 4 therefore w , w _ r   = { w _ o } / 4 = 8 / 4   = 2 litres . b"
Answer: b
Problem: of the families in city x in 1992 , 60 percent owned a personal computer . the number of families in city x owning a computer in 1993 was 50 percent greater than it was in 1992 , and the total number of families in city x was 3 percent greater in 1993 than it was in 1992 . what percent of the families in city x owned a personal computer in 1993 ?
Options: a ) 68.99 % , b ) 66.55 % , c ) 91.23 % , d ) 77.77 % , e ) 87.37 %
Rationale: "say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 60 number of families owning computer in 1998 = 60 * 150 / 100 = 90 number of families in 1998 = 103 the percentage = 90 / 103 * 100 = 87.37 % . answer : e"
Answer: e
Problem: find the value of ( √ 1.21 ) / ( √ 0.81 ) + ( √ 1.00 ) / ( √ 0.49 ) is
Options: a ) 2.65 , b ) 145 / 63 , c ) 155 / 63 , d ) 125 / 63 , e ) 185 / 63
Rationale: ( √ 1.21 ) / ( √ 0.81 ) + ( √ 1.00 ) / ( √ 0.49 ) 11 / 9 + 10 / 7 = > 2.65 answer is a
Answer: a
Problem: louie takes out a 3 - month loan of $ 1000 . the lender charges him 10 % interest per month compounded monthly . the terms of the loan state that louie must repay the loan in 3 equal monthly payments . to the nearest dollar , how much does louis have to pay each month ?
Options: a ) 333 , b ) 383 , c ) 402 , d ) 433 , e ) 483
Rationale: after 1 st month : ( 1000 ) ( 1.1 ) - x = 1100 - x after 2 nd month : ( 1100 - x ) ( 1.1 ) - x = 1210 - 2.21 x after 3 rd month : ( 1210 - 2.21 x ) ( 1.1 ) - x = 1331 - 3.31 x now , the amount after the last payment in 3 rd month must bring the total to 0 . hence : 1331 - 3.31 x = 0 x = 1331 / 3.31 = 402.11 the answer is c
Answer: c
Problem: if the perimeter of a rectangular garden is 600 m , its length when its breadth is 200 m is ?
Options: a ) 299 m , b ) 777 m , c ) 200 m , d ) 100 m , e ) 128 m
Rationale: "2 ( l + 200 ) = 600 = > l = 100 m answer : d"
Answer: d
Problem: rs 3000 is divided into two parts such that one part is put out at 3 % and the other at 5 % . if the annual interest earned from both the investments be rs 144 , find the first part .
Options: a ) s 400 , b ) s 280 , c ) s 300 , d ) s 350 , e ) s 310
Rationale: explanation : average rate = ( 144 / 3000 ) * 100 = 4.8 ratio = 2 : 18 so , first part = ( 2 / 20 ) * 3000 = rs 300 . answer : c
Answer: c
Problem: the area of a square is equal to three times the area of a rectangle of dimensions 3 cm * 9 cm . what is the perimeter of the square ?
Options: a ) 30 , b ) 36 , c ) 46 , d ) 66 , e ) 76
Rationale: "area of the square = s * s = 3 ( 3 * 9 ) = > s = 9 cm perimeter of the square = 4 * 9 = 36 cm . answer : option b"
Answer: b
Problem: a machine , working at a constant rate , manufactures 18 dies in 25 minutes . how many dies does it make in 1 hr 15 min ?
Options: a ) 55 , b ) 53 , c ) 54 , d ) 52 , e ) 50
Rationale: "change 1 hr 15 min to 75 min . for this , we need to set up a simple proportion of dies per time 18 / 25 = s / 75 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . we can cancel the common factor of 25 in the two denominators . 18 / 1 = s / 3 s = 3 * 18 s = 54 the machine would be 54 dies in 1 hr 15 min . answer : c"
Answer: c
Problem: a room 5.44 m long and 3.74 m broad is to be paved with square tiles . the least number of square tiles required to cover the floor is :
Options: a ) 176
b ) 192
c ) 184
d ) 162
e ) 172
Rationale: area of the room = 544 * 374 sq . cm size of largest square tile = hcf of 544 cm & 374 cm area of 1 tile = 34 * 34 sq . cm therefore , number of tiles = ( 544 * 374 / 34 * 34 ) = 176 answer : a
Answer: a
Problem: find the maximum value of n such that 50 ! is perfectly divisible by 2520 ^ n
Options: a ) 7 , b ) 8 , c ) 9 , d ) 6 , e ) 5
Rationale: 2520 = 2 ^ 3 * 3 ^ 2 * 5 * 7 here , 7 is the highest prime , so find the no . of 7 ' s in 50 ! only no . of 7 ' s in 50 ! = [ 50 / 7 ] + [ 50 / 7 ^ 2 ] = 7 + 1 = 8 answer : b
Answer: b
Problem: in two triangle , the ratio of the areas is 4 : 3 and that of their heights is 3 : 4 the ratio of their bases is
Options: a ) 16 : 9
b ) 9 : 16
c ) 9 : 12
d ) 16 : 12
e ) 4 : 12
Rationale: explanation : given a 1 / a 2 = 4 / 3 , h 1 / h 2 = 3 / 4 also , we know that a 1 / a 2 = ( 1 / 2 × b 1 × h 1 ) / ( 1 / 2 × b 2 × h 2 ) on substituting in above equation , we get b 1 / b 2 = 16 / 9 therefore b 1 : b 2 = 16 : 9 answer : option a
Answer: a
Problem: a number exceeds by 30 from its 3 / 8 part . then the number is ?
Options: a ) a ) 32 , b ) b ) 35 , c ) c ) 39 , d ) d ) 40 , e ) e ) 48
Rationale: "x – 3 / 8 x = 30 x = 48 answer : e"
Answer: e
Problem: the h . c . f . of two numbers is 12 and their l . c . m . is 520 . if one of the numbers is 480 , then the other is :
Options: a ) 13 , b ) 18 , c ) 21 , d ) 24 , e ) 38
Rationale: "other number = ( 12 x 520 ) / 480 = 13 . answer : a"
Answer: a
Problem: the radius of a cylinder is 12 m , height 21 m . the lateral surface area of the cylinder is :
Options: a ) 1584
b ) 1854
c ) 1458
d ) 1485
e ) none of them
Rationale: lateral surface area = 2 π rh = 2 × 22 / 7 × 12 × 21 = 44 × 36 = 1584 m ( power 2 ) answer is a .
Answer: a
Problem: in an election between two candidates , the first candidate got 80 % of the votes and the second candidate got 240 votes . what was the total number of votes ?
Options: a ) 600 , b ) 900 , c ) 1200 , d ) 1500 , e ) 1800
Rationale: "let v be the total number of votes . 0.2 v = 240 v = 1200 the answer is c ."
Answer: c
Problem: how many of the positive factors of 19 are not factors of 29 ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
Rationale: "factors of 19 - 1 , 19 factors of 29 - 1 , 29 comparing both , we have three factors of 19 which are not factors of 29 - 19 , the answer is b"
Answer: b
Problem: a retailer buys a radio for rs 225 . his overhead expenses are rs 30 . he sellis the radio for rs 300 . the profit percent of the retailer is
Options: a ) 10 % , b ) 50 % , c ) 25 % , d ) 17.6 % , e ) none of these
Rationale: "explanation : cost price = ( 225 + 30 ) = 255 sell price = 300 gain = ( 45 / 255 ) * 100 = 17.6 % . answer : d"
Answer: d
Problem: the number of boxes in a warehouse can be divided evenly into 6 equal shipments by boat or 27 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ?
Options: a ) 27 , b ) 33 , c ) 54 , d ) 81 , e ) 162
Rationale: "explanations 1 ) this tells us that the number of boxes is evenly divisible by both 6 and 27 ; in other words , it ’ s a common multiple of 6 and 27 . the question says : what ’ s the smallest value it could have ? in other words , what ’ s the lcm of 6 and 27 ? ( this question is one example of a real - world set - up where the question is actually asking for the lcm . ) step ( a ) : 6 = 2 * 3 27 = 3 * 3 * 3 step ( b ) : 6 = 2 * 3 27 = 3 * 3 * 3 gcf = 3 step ( c ) : 6 = 3 * 2 27 = 3 * 9 step ( d ) lcm = 3 * 2 * 9 = 54 thus , 54 is the lcm of 6 and 27 . answer : c ."
Answer: c
Problem: if it is assumed that 60 percent of those who receive a questionnaire by mail will respond and 210 responses are needed , what is the minimum number of questionnaires that should be mailed ?
Options: a ) 290 , b ) 320 , c ) 350 , d ) 380 , e ) 410
Rationale: let x be the minimum number of questionnaires to be mailed . 0.6 x = 210 x = 350 the answer is c .
Answer: c
Problem: a lemonade stand sold only small and large cups of lemonade on tuesday . 4 / 5 of the cups sold were small and the rest were large . if the large cups were sold for 7 / 6 as much as the small cups , what fraction of tuesday ' s total revenue was from the sale of large cups ?
Options: a ) ( a ) 7 / 16 , b ) ( b ) 7 / 15 , c ) ( c ) 10 / 21 , d ) ( d ) 7 / 31 , e ) ( e ) 1 / 2
Rationale: "a simpler way i guess would be to think that in total 5 cups were sold . out of which 4 are small and 1 is large . now let the small ones cost $ 6 . so the large ones would cost $ 7 . so , 4 * 6 = 24 and 1 * 7 = 7 . total revenue was 24 + 7 = 31 and large cup sales as found above is 7 therefore answer is 7 / 31 d"
Answer: d
Problem: 96 % of the population of a city is 23040 . find the total population of the city ?
Options: a ) 30000 , b ) 28000 , c ) 24000 , d ) 25000 , e ) 25500
Rationale: "population consider as x x * ( 96 / 100 ) = 23040 x = 240 * 100 = = > 24000 answer c"
Answer: c
Problem: the average weight of 10 men is increased by 2 ½ kg when one of the men who weighs 58 kg is replaced by a new man . what is the weight of the new man ?
Options: a ) 80 kg , b ) 83 kg , c ) 70 kg , d ) 75 kg , e ) 85 kg
Rationale: since the average has increased by 2.5 kg , the weight of the man who stepped in must be equal to 58 + 10 x 2.5 58 + 25 = 83 kg ans : ' b '
Answer: b
Problem: if 6 log ( 4 * 5 ^ 2 ) = x , find x
Options: a ) 10 , b ) 11 , c ) 7 , d ) 12 , e ) 9
Rationale: 6 ( log 2 ^ 2 * 5 ^ 2 ) = x 6 log ( 5 * 2 ) ^ 2 = x 6 * 2 log ( 5 * 2 ) = x 12 log 10 = x log 10 base 10 = 1 so 12 * 1 = x x = 12 answer : d
Answer: d
Problem: in the rectangular coordinate system , what is the x - intercept of a line passing through ( 10 , 3 ) and ( − 10 , − 7 ) ?
Options: a ) 4 , b ) 2 , c ) 0 , d ) − 2 , e ) − 4
Rationale: "slope = rise / run = 10 / 20 = 1 / 2 the equation of the line is y = ( 1 / 2 ) x + b 3 = ( 1 / 2 ) ( 10 ) + b b = - 2 the equation of the line is y = ( 1 / 2 ) x - 2 to find the x - intercept , let y = 0 : 0 = ( 1 / 2 ) x - 2 x = 4 the answer is a ."
Answer: a
Problem: the price of a t . v . set worth rs . 40000 is to be paid in 20 installments of rs . 2500 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 ?
Options: a ) 29997 , b ) 28088 , c ) 27098 , d ) 37500 , e ) 2799
Rationale: "money paid in cash = rs . 2500 balance payment = ( 40000 - 2500 ) = rs . 37500 . answer : d"
Answer: d
Problem: when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 12 - inch boxes . if the university pays $ 0.50 for every box , and if the university needs 2.4 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ?
Options: a ) $ 250 , b ) $ 275 , c ) $ 510 , d ) $ 1,250 , e ) $ 2,550
Rationale: "the volume of each box is 20 * 20 * 12 = 4800 cubic inches . number of boxes = 2 , 400,000 / 4800 = 500 boxes total cost = 500 × $ 0.5 = $ 250 the answer is a ."
Answer: a
Problem: boy sells a book for rs . 450 he gets a loss of 10 % , to gain 10 % , what should be the sp ?
Options: a ) rs . 320 , b ) rs . 450 , c ) rs . 550 , d ) rs . 640 , e ) rs . 680
Rationale: "find selling price to gain 10 % . now , we are asked to find selling price to gain 10 % profit . hint : selling price = ( 100 + gain % ) × c . p . 100 selling price = ( 100 + 10 ) × 500 100 selling price = ( 110 ) × 500 100 therefore , selling price = rs . 550 c"
Answer: c
Problem: in country z , 10 % of the people do not have a university diploma but have the job of their choice , and 25 % of the people who do not have the job of their choice have a university diploma . if 20 % of the people have the job of their choice , what percent of the people have a university diploma ?
Options: a ) 30 % , b ) 45 % , c ) 55 % , d ) 65 % , e ) 75 %
Rationale: "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 = 20 % not job of choice with diploma = . 25 x not job of choice w / o diploma = . 75 x total not job of choice = x total with diploma total without diploma total citizen = 100 if 20 % of people have their job of choice , then 80 % of people do not have their job of choice . 25 % of 80 % = 20 % . we can also see that 10 % of the people have their job of choice and a diploma ( 20 % - 10 % = 10 % ) . 10 % + 20 % = 30 % . therefore 30 % of the people in country z have a diploma . ans a"
Answer: a
Problem: a batsman makes a score of 84 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ?
Options: a ) 19 , b ) 29 , c ) 36 , d ) 49 , e ) 59
Rationale: "let the average after 17 th inning = x then average after 16 th inning = ( x - 3 ) therefore 16 ( x - 3 ) + 84 = 17 x therefore x = 36 answer : c"
Answer: c
Problem: of the 120 passengers on flight 750 , 40 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 / 3 of the passengers in first class are male , how many females are there in coach class ?
Options: a ) 40 , b ) 48 , c ) 50 , d ) 52 , e ) 56
Rationale: "number of passengers on flight = 120 number of female passengers = . 4 * 120 = 48 number of passengers in first class = ( 10 / 100 ) * 120 = 12 number of passengers in coach class = ( 90 / 100 ) * 120 = 108 number of male passengers in first class = 1 / 3 * 12 = 4 number of female passengers in first class = 12 - 4 = 8 number of female passengers in coach class = 48 - 8 = 40 answer a"
Answer: a
Problem: the maximum numbers of students among them 451 pens and 410 toys can be distributed in such a way that each student gets the same number of pens and same number of toys is
Options: a ) 41 , b ) 910 , c ) 1001 , d ) 1911 , e ) none
Rationale: "olution required number of students . = h . c . f of 451 and 410 . â € ¹ = â € º 41 . answer a"
Answer: a
Problem: find the average of all the numbers between 4 and 32 which are divisible by 5 .
Options: a ) 17.5 , b ) 20 , c ) 24 , d ) 30 , e ) 32
Rationale: "solution average = ( 5 + 10 + 15 + 20 + 25 + 30 ) / 6 ) = 105 / 6 = 17.5 . answer a"
Answer: a
Problem: a cylinder with base radius of 8 cm and height of 2 cm is melted to form a cone of height 6 cm . the radius of the cone will be :
Options: a ) 8 cm
b ) 9 cm
c ) 10 cm
d ) 11 cm
e ) none
Rationale: sol . let the radius of the cone be r cm . then , 1 / 3 ∏ * r ² * 6 = ∏ * 8 * 8 * 2 ⇔ r ² = [ 8 * 8 * 2 * 3 / 6 ] = 64 ⇔ r = 8 cm . answer a
Answer: a
Problem: if @ is a binary operation defined as the difference between an integer n and the product of n and 5 , then what is the largest positive integer n such that the outcome of the binary operation of n is less than 18 ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: "@ ( n ) = 5 n - n we need to find the largest positive integer such that 5 n - n < 18 . then 4 n < 18 and n < 4.5 the largest possible integer is n = 4 . the answer is d ."
Answer: d
Problem: a runs twice as fast as b and gives b a start of 60 m . how long should the racecourse be so that a and b might reach in the same time ?
Options: a ) 120 m . , b ) 80 m . , c ) 150 m . , d ) 100 m . , e ) none of the above
Rationale: ratio of speeds of a and b is 2 : 1 b is 60 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 60 m is 60 / 1 = 60 m so the total time taken by a and b to reach = 2 * 60 = 120 m answer : a
Answer: a
Problem: how long does a train 100 m long running at the speed of 80 km / hr takes to cross a bridge 142 m length ?
Options: a ) 10.7 sec , b ) 10.9 sec , c ) 10.1 sec , d ) 15.1 sec , e ) 12.7 sec
Rationale: "speed = 80 * 5 / 18 = 22.2 m / sec total distance covered = 100 + 142 = 242 m . required time = 242 / 22.2 ' = 10.9 sec . answer : b"
Answer: b
Problem: a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 5 days before the project is completed , in how many days total will the project be completed ?
Options: a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24
Rationale: "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 5 days , b can do 1 / 6 of the project . thus a and b must complete 5 / 6 of the project , which takes 10 days . the total number of days is 10 + 5 = 15 . the answer is b ."
Answer: b
Problem: angelo and isabella are both salespersons . in any given week , angelo makes $ 580 in base salary plus 8 percent of the portion of his sales above $ 1,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of money ?
Options: a ) 23,500 , b ) 24,500 , c ) 25,000 , d ) 26,500 , e ) 27,500
Rationale: "let the weekly sales of both = x 580 + ( x − 1000 ) 8 / 100 = 10 / 100 x x = 25000 answer : c"
Answer: c
Problem: a dessert recipe calls for 50 % melted chocolate and 50 % raspberry puree to make a particular sauce . a chef accidentally makes 15 cups of the sauce with 30 % melted chocolate and 70 % raspberry puree instead . how many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50 % of each ?
Options: a ) 1.5 , b ) 2.14 , c ) 3 , d ) 4.5 , e ) 5
Rationale: "yes , we assume that the mix is homogeneous . otherwise , we will not be able to solve the question . you have 15 cups of sauce with 30 % chocolate . you also have unlimited amount of pure chocolate sauce . now you need to mix these two in such a way that you get total 15 cups of sauce with 50 % chocolate . using scale method : w 1 / w 2 = ( 100 - 50 ) / ( 50 - 30 ) = 5 / 2 w 1 - amount of 30 % chocolate sauce w 2 - amount of pure chocolate sauce so for every 5 / 2 cups of 30 % chocolate sauce , we need 2 cup of pure chocolate sauce . this will give us 7 cups of 50 % chocolate sauce . but we need 15 cups of 50 % chocolate sauce . so we need to mix 5 * 15 / 7 = 12.5 cups of 40 % chocolate sauce with 1 * 15 / 7 = 2.14 cups of pure chocolate sauce . answer ( b )"
Answer: b
Problem: the price of commodity x increases by 45 cents every year , while the price of commodity y increases by 20 cents every year . in 2001 , the price of commodity x was $ 5.20 and the price of commodity y was $ 7.30 . in which year will the price of commodity x be 35 cents less than the price of commodity y ?
Options: a ) 2006 , b ) 2007 , c ) 2008 , d ) 2009 , e ) 2010
Rationale: "the price of commodity x increases 25 cents each year relative to commodity y . the price difference is $ 2.10 and commodity x needs to be 35 cents less than commodity y . $ 1.75 / 25 cents = 7 years the answer is 2001 + 7 years = 2008 . the answer is c ."
Answer: c
Problem: if log 1087.5 = 1.9421 , then the number of digits in ( 875 ) 10 is ?
Options: a ) 30 , b ) 28 , c ) 27 , d ) 26 , e ) 25
Rationale: "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 1087.5 + 1 ) = 10 ( 1.9421 + 1 ) = 10 ( 2.9421 ) = 29.421 x = antilog ( 29.421 ) therefore , number of digits in x = 30 . answer : a"
Answer: a
Problem: a river 4 m deep and 65 m wide is flowing at the rate of 6 kmph the amount of water that runs into the sea per minute is ?
Options: a ) 25000 , b ) 26000 , c ) 27000 , d ) 28000 , e ) 29000
Rationale: "rate of water flow - 6 kmph - - 6000 / 60 - - 100 m / min depth of river - - 4 m width of river - - 65 m vol of water per min - - 100 * 4 * 65 - - - 26000 answer b"
Answer: b
Problem: in a garden , there are 10 rows and 17 columns of mango trees . the distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden . the length of the garden is
Options: a ) 20 m , b ) 22 m , c ) 34 m , d ) 26 m , e ) 28 m
Rationale: "explanation : each row contains 17 plants . there are 15 gapes between the two corner trees ( 16 x 2 ) metres and 1 metre on each side is left . therefore length = ( 32 + 2 ) m = 34 m . answer : c"
Answer: c
Problem: a person spent rs . 6,040 from his salary on food and 8,000 on house rent . after that he was left with 70 % of his monthly salary . what is his monthly salary ?
Options: a ) 46,800 , b ) 66,800 , c ) 56,800 , d ) 26,800 , e ) 76,800
Rationale: "total money spent on food and house rent = 6,040 + 8,000 = 14,040 which is 100 - 70 = 30 % of his monthly salary ∴ his salary = 14040 x 100 / 30 = 46800 answer : a"
Answer: a
Problem: a basketball is dropped from a height of 40 feet . if it bounces back up to a height that is exactly half of its previous height , and it stops bouncing after hitting the ground for the fourth time , then how many total feet will the ball have traveled after 3 full bounces .
Options: a ) 50 , b ) 55 , c ) 110 , d ) 75 , e ) 80
Rationale: "initial distance = 40 feet first bounce = 20 feet up + 20 feet down = 40 feet second bouche = 10 feet up + 10 feet down = 20 feet third bounce = 5 feet up and 5 feet down = 10 feet total distance covered = 40 + 40 + 20 + 10 = 110 answer is c"
Answer: c
Problem: 80 370 860 1550 ? 3530
Options: a ) 900 , b ) 2440 , c ) 750 , d ) 244 , e ) 960
Rationale: "10 ^ 2 - 20 = 80 20 ^ 2 - 30 = 370 30 ^ 2 - 40 = 860 40 ^ 2 - 50 = 1550 50 ^ 2 - 60 = 2440 60 ^ 2 - 70 = 3530 . answer : b"
Answer: b
Problem: a sum fetched a total simple interest of 4052.25 at the rate of 9 % . p . a . in 5 years . what is the sum ?
Options: a ) 9005 , b ) 8925 , c ) 2345 , d ) 6474 , e ) 8723
Rationale: principal = ( 100 x 4052.25 ) / ( 9 x 5 ) = 405225 / 45 = 9005 . answer a
Answer: a
Problem: a rotameter is a device that measures flow of liquid and gases . when measuring liquid phase flows , 2.5 inches represent 60 liters per minute of liquid . with gas measurements the rotameter moves 50 % of the movement he moves with the liquid phase . how many liters of gas passed through the rotameter if it measured 4 inches ?
Options: a ) 176 , b ) 192 , c ) 202 , d ) 218 , e ) 284
Rationale: in case of liquid - 2.5 inches represents 60 lit / min . in case of gas - 50 % of 2.5 inches represents 60 lit / min 1.25 inches represents 60 lit / min 4 inches will represent 60 * 4 / 1.25 = 192 b is the answer
Answer: b
Problem: if a trader sold two cars each at rs . 325475 and gains 10 % on the first and loses 10 % on the second , then his profit or loss percent on the whole is ?
Options: a ) 1.44 % , b ) 1.74 % , c ) 1.84 % , d ) 1.47 % , e ) 1.00 %
Rationale: "sp of each car is rs . 325475 , he gains 10 % on first car and losses 10 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / 100 = ( 10 ) ( 10 ) / 100 % = 1.00 % answer : e"
Answer: e
Problem: how many positive integers less than 130 are there such that they are multiples of 13 or multiples of 12 but not both ?
Options: a ) 16 , b ) 18 , c ) 20 , d ) 24 , e ) 28
Rationale: "for 13 : 13 . . . 130 = 13 * 10 for 12 : 12 . . . 130 = 12 * 10 but there is one integer 13 * 12 . so n = ( 10 ) + ( 10 ) = 20 c"
Answer: c
Problem: the marks obtained by polly and sandy are in the ratio 5 : 6 and those obtained by sandy and willy are in the ratio of 3 : 2 . the marks obtained by polly and willy are in the ratio of . . . ?
Options: a ) 3 : 2 , b ) 5 : 4 , c ) 7 : 6 , d ) 9 : 8 , e ) 11 : 10
Rationale: "polly : sandy = 5 : 6 sandy : willy = 3 : 2 = 6 : 4 polly : sandy : willy = 5 : 6 : 4 polly : willy = 5 : 4 the answer is b ."
Answer: b
Problem: what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 7 ?
Options: a ) 5 , b ) 14 , c ) 10 , d ) 20 , e ) 30
Rationale: what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 7 ? 9 ^ 7 = ( 3 ^ 2 ) ^ 7 = 3 ^ 14 b . 14
Answer: b
Problem: if 70 percent of a class answered the first question on a certain test correctly , 55 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ?
Options: a ) 45 % , b ) 40 % , c ) 35 % , d ) 30 % , e ) 25 %
Rationale: "70 % answered the first question correctly and 20 % answered neither correctly . then 10 % missed the first question but answered the second question correctly . then the percent who answered both correctly is 55 % - 10 % = 45 % . the answer is a ."
Answer: a
Problem: each machine of type a has 3 steel parts and 2 chrome parts . each machine of type b has 2 steel parts and 4 chrome parts . if a certain group of type a and type b machines has a total of 50 steel parts and 66 chrome parts , how many machines are in the group
Options: a ) 11 , b ) 21 , c ) 31 , d ) 61 , e ) 51
Rationale: "look at the below representation of the problem : steel chrome total a 3 2 50 > > no . of type a machines = 50 / 5 = 10 b 2 4 66 > > no . of type b machines = 66 / 6 = 11 so the answer is 21 i . e b . hope its clear ."
Answer: b
Problem: a batsman in his 12 th innings makes a score of 75 and thereby increases his average by 1 runs . what is his average after the 12 th innings if he had never been ‘ not out ’ ?
Options: a ) 42 , b ) 43 , c ) 44 , d ) 64 , e ) 46
Rationale: "let ‘ x ’ be the average score after 12 th innings ⇒ 12 x = 11 × ( x – 1 ) + 75 ∴ x = 64 answer d"
Answer: d
Problem: students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 188 cms . what was the ratio of men to women ?
Options: a ) 5 : 2 , b ) 5 : 1 , c ) 4 : 3 , d ) 5 : 4 , e ) 3 : 1
Rationale: "we ' re given a few facts to work with : 1 ) the average height of the females is 170 cm 2 ) the average height of the males is 188 cm 3 ) the average of the group is 180 cm we ' re asked for the ratio of men to women . w = number of women m = number of men ( 170 w + 188 m ) / ( w + m ) = 180 170 w + 188 m = 180 w + 180 m 8 m = 10 w 4 m = 5 w m / w = 5 / 4 the ratio of men to women is 5 t ổ 4 . d"
Answer: d
Problem: the average ( arithmetic mean ) of the integers from 100 to 200 , inclusive , is how much lesser than the average of the integers from 150 to 950 , inclusive ?
Options: a ) 350 , b ) 475 , c ) 300 , d ) 425 , e ) 400
Rationale: "for an ap the mean or average of series is average of first and last term . so , average of numbers between 150 to 950 , inclusive = ( 150 + 950 ) / 2 = 550 average of numbers between 100 to 200 , inclusive = ( 100 + 200 ) / 2 = 150 difference = 550 - 150 = 400 answer is e"
Answer: e
Problem: the present ages of 3 persons are in the proportion of 4 : 7 : 9 . 8 years ago , the sum of their ages was 116 . find their present ages .
Options: a ) 20 , 35,45 , b ) 28 , 49,63 , c ) 16 , 28,36 , d ) 16 , 28,46 , e ) none of these
Rationale: let the present ages of three persons be 4 k , 7 k and 9 k respectively . ( 4 k - 8 ) + ( 7 k - 8 ) + ( 9 k - 8 ) = 116 20 k = 140 k = 7 therefore , then present ages are 28 , 49,63 . answer : b
Answer: b
Problem: a car traveling at a certain constant speed takes 5 seconds longer to travel 1 km than it would take to travel 1 km at 90 km / hour . at what speed , in km / hr , is the car traveling ?
Options: a ) 70 , b ) 80 , c ) 74 , d ) 75 , e ) 78
Rationale: "time to cover 1 kilometer at 80 kilometers per hour is 1 / 90 hours = 3,600 / 90 seconds = 40 seconds ; time to cover 1 kilometer at regular speed is 40 + 5 = 45 seconds = 45 / 3,600 hours = 1 / 80 hours ; so , we get that to cover 1 kilometer 1 / 80 hours is needed - - > regular speed 80 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : b ."
Answer: b
Problem: if the price of gasoline increases by 25 % and ron intends to spend only 15 % more on gasoline , by how much percent should he reduce the quantity of gasoline that he buys ?
Options: a ) 6.66 % , b ) 8 % , c ) 10 % , d ) 12 % , e ) 12.5 %
Rationale: "let price of original price of gasoline = 10 $ and let original consumption of gasoline = 10 amount spend on gasoline originally = 10 * 10 = 100 $ the price of gasoline increases by 25 % and ron intends to spend only 15 % more on gasoline increased price of gasoline = 12.5 $ now ron intends to spend on gasoline = 1.15 * 100 = 115 $ consumption of gasoline needs to be = 115 / 12.5 = 1150 / 125 = 46 / 5 = 9.2 reduction in consumption of gasoline = 10 - 9.2 = . 8 % reduction in consumption of gasoline = . 8 / 10 * 100 % = 8 % answer b"
Answer: b
Problem: a rower can row 7 km / h in still water . when the river is running at 1 km / h , it takes the rower 1 hour to row to big rock and back . how many kilometers is it to big rock ?
Options: a ) 3.19 , b ) 3.28 , c ) 3.35 , d ) 3.43 , e ) 3.56
Rationale: "let x be the distance to big rock . time = x / 6 + x / 8 = 1 x = 48 / 14 = 3.43 km the answer is d ."
Answer: d
Problem: a cistern is normally filled in 4 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full , the leak will empty it in ?
Options: a ) 33 , b ) 88 , c ) 40 , d ) 99 , e ) 12
Rationale: "1 / 4 - 1 / x = 1 / 6 x = 12 answer : e"
Answer: e
Problem: 25.25 / 1500 is equal to :
Options: a ) 1.012526333 , b ) 0.012625333 , c ) 0.125263333 , d ) 0.126253333 , e ) 0.016833333
Rationale: "25.25 / 2000 = 2525 / 200000 = 0.016833333 answer : e"
Answer: e
Problem: if jake loses 20 pounds , he will weigh twice as much as his sister . together they now weigh 224 pounds . what is jake ’ s present weight , in pounds ?
Options: a ) 131 , b ) 135 , c ) 156 , d ) 177 , e ) 188
Rationale: "lets say j is the weight of jack and s is the wt of his sister . if he loses 20 pounds , he s twice as heavy as his sister . j - 20 = 2 * s also , together they weight 224 pounds j + s = 224 solvong the 2 equation , we get j = 156 pounds ! c"
Answer: c
Problem: in a certain lottery , the probability that a number between 12 and 20 , inclusive , is drawn is 1 / 6 . if the probability that a number 12 or larger is drawn is 1 / 2 , what is the probability that a number less than or equal to 20 is drawn ?
Options: a ) 1 / 18 , b ) 1 / 6 , c ) 2 / 3 , d ) 1 / 2 , e ) 5 / 6
Rationale: "you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set 1 : number 12 or larger set 2 : number 20 or smaller 1 = p ( set 1 ) + p ( set 2 ) - p ( set 1 and set 2 ) ( combined probability is 1 because every number will be either 12 or moreor 20 or lessor both ) 1 / 2 + p ( set 2 ) - 1 / 6 = 2 / 3 p ( set 2 ) = 2 / 3 answer ( c )"
Answer: c
Problem: if the sides of a rectangle are increased by 15 % , what is the percentage increase in the area ?
Options: a ) 32.25 % , b ) 40 % , c ) 37.50 % , d ) 36.75 % , e ) 32 %
Rationale: "say both sides of the rectangle are equal to 100 ( so consider that we have a square ) . in this case the area is 100 * 100 = 10,000 . the area of new rectangle would be 115 * 115 = 13,225 , which is 32.25 % greater than the old area . answer : a ."
Answer: a
Problem: a fashion designer sold a pair of jeans to a retail store for 40 percent more than it cost to manufacture the pair of jeans . a customer bought the pair of jeans for 40 percent more than the retailer paid for them . the price the customer paid was what percent greater than the cost of manufacturing the jeans ?
Options: a ) 65 % , b ) 70 % , c ) 75 % , d ) 85 % , e ) 96 %
Rationale: "find the product of the two increases : ( 14 / 10 ) * ( 14 / 10 ) which is 1.96 and a 96 % increase . e"
Answer: e