regisss/math_qa huggingface dataset:
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 152 per kg , the price of the third variety per kg will be ?
Options: a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 174.50 , 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 . 152 ) / \ x â ˆ ’ 152 22.50 = > x â ˆ ’ ( 152 / 22.50 ) = 1 . = > x â ˆ ’ 152 = 22.50 . = > x = 174.50 rs . answer : c"
Answer: c
Problem: when positive integer n is divided by 3 , the remainder is 1 . when n is divided by 8 , the remainder is 5 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 15 ?
Options: a ) 1 , b ) 2 , c ) 5 , d ) 19 , e ) 20
Rationale: "when positive integer n is divided by 3 , the remainder is 1 i . e . , n = 3 x + 1 values of n can be one of { 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 . . . . . . . . . . . . . . 49 , 52 , 59 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 8 , the remainder is 5 . . i . e . , n = 7 y + 5 values of n can be one of { 5 , 13 , 21 , 29 , 37 , 45 , 53 , 61 . . . . . . . . } combining both the sets we get n = { 13 , 37 , . . . . . . . . . . . } what is the smallest positive integer p , such that ( n + p ) is a multiple of 15 or 15 x in case of n = 13 p = 2 so for min value of p , we take min value of n . b is the answer ."
Answer: b
Problem: what are the last two digits of ( 301 * 402 * 502 * 604 * 646 * 547 * 448 * 349 ) ^ 2
Options: a ) 96 , b ) 76 , c ) 56 , d ) 36 , e ) 16
Rationale: "( ( 301 * 402 * 502 * 604 * 646 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 2,4 , 6,7 , 8,9 ; 3 & 5 are missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,02 , 04,46 and final three as 47 * 48 * 49 . solving for only last two digits and multiplying them we get : ( ( 04 * 04 * 46 ) ( 56 * 49 ) ) ^ 2 = ( 36 * 44 ) ^ 2 = 84 ^ 2 = 56 hence answer is c"
Answer: c
Problem: p runs 25 % faster than q so p gives q a 60 meter head start . if the race ends in a tie , how far ( in meters ) did p run in the race ?
Options: a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500
Rationale: "let d be the race distance that p ran . let t be the time it took to complete the race . let v be q ' s speed . t = d / 1.25 v = ( d - 60 ) / v d = 1.25 d - 75 0.25 d = 75 d = 300 meters . the answer is c ."
Answer: c
Problem: joe invested a certain sum of money in a simple interest bond whose value grew to $ 560 at the end of 3 years and to $ 660 at the end of another 5 years . what was the rate of interest in which he invested his sum ?
Options: a ) 3 % , b ) 4 % , c ) 5 % , d ) 6 % , e ) 7 %
Rationale: "in 5 years , the value grew $ 100 , so the simple interest was $ 20 per year . in 3 years , the total interest was 3 * $ 20 = $ 60 the principal is $ 560 - $ 60 = 500 . the interest rate is $ 20 / $ 500 = 4 % the answer is b ."
Answer: b
Problem: the average marks of a class of 25 students is 40 and that of another class of 30 students is 60 . find the average marks of all the students ?
Options: a ) 52.6 , b ) 52.4 , c ) 52.1 , d ) 50.9 , e ) 52.9
Rationale: "sum of the marks for the class of 25 students = 25 * 40 = 1000 sum of the marks for the class of 30 students = 30 * 60 = 1800 sum of the marks for the class of 55 students = 1000 + 1800 = 2800 average marks of all the students = 2800 / 55 = 50.9 . answer : d"
Answer: d
Problem: a lady grows cabbage in her garden that is in the shape of a square . each cabbage takes 1 square foot of area in her garden . this year , she has increased her output by 211 cabbages when compared to last year . the shape of the area used for growing the cabbage has remained a square in both these years . how many cabbages did she produce this year ?
Options: a ) 11236
b ) 11025
c ) 14400
d ) 12696
e ) can not be determined
Rationale: explanatory answer the shape of the area used for growing cabbages has remained a square in both the years . let the side of the square area used for growing cabbages this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing cabbages last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of cabbages grown has increased by 211 , the area would have increased by 211 sq ft because each cabbage takes 1 sq ft space . hence , x 2 - y 2 = 211 ( x + y ) ( x - y ) = 211 . 211 is a prime number and hence it will have only two factors . i . e . , 211 and 1 . therefore , 211 can be expressed as product of 2 numbers in only way = 211 * 1 i . e . , ( x + y ) ( x - y ) = 211 * 1 so , ( x + y ) should be 211 and ( x - y ) should be 1 . solving the two equations we get x = 106 and y = 105 . therefore , number of cabbages produced this year = x 2 = 1062 = 11236 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice d . 12696 is not the square of any number . check choice a : if this year ' s produce is 11236 , last year ' s produce would have been 11236 - 211 = 11025 11025 is the square of 105 . so , 11236 is the answer . choice a is the correct answer .
Answer: a
Problem: two equally sized jugs full of water are each emptied into two separate unequally sized empty jugs , x and y . now , jug x is 1 / 3 full , while jug y is 2 / 3 full . if water is poured from jug x into jug y until jug y is filled , what fraction of jug x then contains water ?
Options: a ) 0 , b ) 1 / 15 , c ) 2 / 15 , d ) 1 / 6 , e ) 2 / 10
Rationale: "suppose the water in each jug is l liters cx x ( 1 / 3 ) = l cx = 3 l liters cx is capacity of x cy x ( 2 / 3 ) = l cy = 3 l / 2 liters cy is capacity of y now , y is 3 l / 2 - l empty = l / 2 empty so , we can put only l / 2 water in jug y from jug x jug x ' s remaining water = l - l / 2 = l / 2 fraction of x which contains water = water / cx = ( l / 2 ) / 3 l = 1 / 6 answer will be d"
Answer: d
Problem: find the number of factors of 9321 ?
Options: a ) 3 , b ) 6 , c ) 8 , d ) 16 , e ) 32
Rationale: "9321 = 3 x 13 x 239 = 31 x 13 1 x 239 1 ; add one to all the powers i . e . , ( 1 + 1 ) x ( 1 + 1 ) x ( 1 + 1 ) = 8 ; we get eight factors 1 , 3 , 13 , 39 , 239 , 717 , 3107 , 9321 answer : c"
Answer: c
Problem: the average ( arithmetic mean ) of 20 , 40 , and 60 is 5 more than the average of 10 , 60 , and what number ?
Options: a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 55
Rationale: "a 1 = 120 / 3 = 40 a 2 = a 1 - 5 = 35 sum of second list = 35 * 3 = 105 therefore the number = 105 - 70 = 35 b"
Answer: b
Problem: on dividing a number by 357 , we get 39 as remainder . on dividing the same number by 17 , what will be the remainder ?
Options: a ) 4 , b ) 5 , c ) 8 , d ) 7 , e ) 2
Rationale: "let x be the number and y be the quotient . then , x = 357 * y + 39 = ( 17 * 21 * y ) + ( 17 * 2 ) + 5 = 17 * ( 21 y + 2 ) + 5 . required number = 5 . answer is b"
Answer: b
Problem: a man â € ™ s current age is ( 2 / 5 ) of the age of his father . after 12 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ?
Options: a ) 60 , b ) 45 , c ) 38 , d ) 50 , e ) 39
Rationale: "let , father â € ™ s current age is a years . then , man â € ™ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 12 ] = ( 1 / 2 ) ( a + 12 ) 2 ( 2 a + 60 ) = 5 ( a + 12 ) a = 60 a"
Answer: a
Problem: potato chips are on sale , buy 2 get 1 free . there are only 9 bags of chips left ( 4 plain and 5 bbq ) . a shopper randomly grabs 3 bags . what is the probability the shopper will grab exactly 3 bags of bbq chips ?
Options: a ) 2 / 3 , b ) 1 / 3 , c ) 1 / 5 , d ) 5 / 42 , e ) 7 / 42
Rationale: combination probability formula : ncr = n ! / [ r ! ( n - r ) ! ] total possible , select 3 bags from 9 bags = 9 c 3 = 9 ! / [ 3 ! ( 9 - 3 ) ! ] = 84 . to have 3 bbq there must be 0 plain , select 0 plain from 4 = 4 c 0 = 1 . and , select 3 bbq from 5 = 5 c 3 = 10 . 3 bbq and 0 plain = ( 1 ) ( 10 ) = 10 probability = ( number outcomes favorable ) / ( total number outcomes ) = 10 / 84 = 5 / 42 answer : d
Answer: d
Problem: two spherical balls lie on the ground touching . if one of the balls has a radius of 10 cm , and the point of contact is 12 cm above the ground , what is the radius of the other ball ( in centimeters ) ?
Options: a ) 14 , b ) 15 , c ) 16 , d ) 18 , e ) 20
Rationale: "a straight line will join the two centers and the point of contact , thus making similar triangles . 2 / 10 = ( r - 12 ) / r 2 r = 10 r - 120 r = 15 the answer is b ."
Answer: b
Problem: there are two bags a and b . a contains 6 red flowers and 3 pink flowers . where as bag b contains 2 red flowers and 7 pink flowers . one flower is chosen from a bag randomly . what is the probability that the flower chosen is pink ?
Options: a ) 5 / 8 , b ) 5 / 6 , c ) 5 / 1 , d ) 5 / 9 , e ) 5 / 2
Rationale: bag a 6 red + 3 pink bag b 2 red + 7 pink probability of picking 1 pink from bag a is 3 / 9 probability of picking 1 pink from bag b is 7 / 9 total = 3 / 9 + 7 / 9 = 10 / 9 and probability of selecting a bag is 1 / 2 so finally probability of choosing a pink flower = 1 / 2 ( 10 / 9 ) = 5 / 9 answer : d
Answer: d
Problem: the length of the bridge , which a train 180 meters long and travelling at 45 km / hr can cross in 30 seconds , is ?
Options: a ) 195 m , b ) 156 m , c ) 245 m , d ) 156 m , e ) 156 m
Rationale: "speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 180 + x ) / 30 = 25 / 2 = = > 2 ( 180 + x ) = 750 = = > x = 195 m . answer : a"
Answer: a
Problem: a mixture of 50 litres of milk and water contains 10 % water . how many litres of water should be added to the mixture so that the mixture contains 15 % water ?
Options: a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 10
Rationale: quantity of milk in the mixture = 90 / 100 ( 50 ) = 45 litres . after adding water , milk would form 60 % of the mixture . hence , if quantity of mixture after adding x liters of water , ( 60 ) / 100 x = 45 = > x = 75 hence 75 - 50 = 25 litres of water must be added . answer : c
Answer: c
Problem: a train 100 m long crosses a platform 125 m long in 30 sec ; find the speed of the train ?
Options: a ) 36 , b ) 77 , c ) 27 , d ) 33 , e ) 99
Rationale: "d = 100 + 125 = 225 t = 30 s = 225 / 30 * 18 / 5 = 27 kmph answer : c"
Answer: c
Problem: a small company is planning to rent either computer a or computer b to print customer mailing lists . both computer a and computer b must be rented on an hourly basis . the rental fee is based only on the amount of time the computer is turned on . it will cost 40 percent more per hour to rent computer a than to rent computer b . computer b would , however , require 20 hours more than computer a to do the job . if either computer a , or computer b were rented the total cost to rent the computer would be $ 650.00 . what would be the approximate hourly charge to rent computer b ?
Options: a ) $ 9.28 , b ) $ 11.30 , c ) $ 13.20 , d ) $ 17.80 , e ) $ 22.10
Rationale: "pa = price of a pb = price of b ta = time for a to complete the job tb = time for b to complete the job given pa = 1.4 pb ta + 20 = tb pa * ta = pb * tb = 650 1.4 pb * ( tb - 20 ) = pb * tb 1.4 pb tb - pb tb = 1.4 pb * 20 0.4 pbtb = 28 pb tb = 28 / 0.4 = 70 pb = 650 / 70 ~ 9.28 a"
Answer: a
Problem: each of the 20 boxes in a certain shipment weighs either 10 pounds or 20 pounds , and the average ( arithmetic mean ) weight of the boxes in the shipment is 18 pounds . if the average weight of the boxes in the shipment is to be reduced to 16 pounds by removing some of the 20 - pound boxes , how many 20 - pound boxes must be removed ?
Options: a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 15
Rationale: "if the average of 10 - pound and 20 - pound boxes is 18 , the ratio of 10 - pound boxes : 20 - pound boxes is 1 : 4 . so out of 20 boxes , 4 are 10 - pound boxes and 16 are 20 - pound boxes . if the average of 10 and 20 - pound boxes is to be 16 , the ratio of 10 - pound boxes : 20 - pound boxes should be 2 : 3 . the number of 10 pound boxes remains the same so we still have 4 of them . to get a ratio of 2 : 3 , the number of 20 - pound boxes must be 6 . we need to remove 10 of the 20 - pound boxes . the answer is b ."
Answer: b
Problem: a television have side 6 * 5 . its screen size is 5 * 4 . calculate the percentage of unscreened part t . v .
Options: a ) 30 , b ) 25 , c ) 33.33 , d ) 44.44 , e ) 28
Rationale: unscreened part = 30 - 20 = 10 . % unscreened part = 10 / 30 * 100 = 33.33 % answer c
Answer: c
Problem: how many 5 digit number contain number 3 ?
Options: a ) 31512 , b ) 32512 , c ) 33512 , d ) 34512 , e ) 37512
Rationale: "total 5 digit no . = 9 * 10 * 10 * 10 * 10 = 90000 not containing 3 = 8 * 9 * 9 * 9 * 9 = 52488 total 5 digit number contain 3 = 90000 - 52488 = 37512 answer : e"
Answer: e
Problem: a train crosses a post in 15 seconds and a platform 100 m long in 25 seconds . its length is
Options: a ) 150 m , b ) 300 m , c ) 400 m , d ) 180 m , e ) 160 m
Rationale: explanation : assume x is the length of the train and v is the speed x / v = 15 = > v = x / 15 ( x + 100 ) / v = 25 = > v = ( x + 100 ) / 25 ie , x / 15 = ( x + 100 ) / 25 = > 5 x = 3 x + 300 = > x = 300 / 2 = 150 answer : option a
Answer: a
Problem: how many factors does 49 ^ 2 have ?
Options: a ) 2 , b ) 8 , c ) 24 , d ) 25 , e ) 36
Rationale: "36 ^ 2 = 6 * 6 * 6 * 6 = 2 ^ 4 * 3 ^ 4 total factors = ( 4 + 1 ) * ( 4 + 1 ) = 6 * 6 = 36 answer e ."
Answer: e
Problem: a cube of side one meter length is cut into small cubes of side 25 cm each . how many such small cubes can be obtained ?
Options: a ) 56 , b ) 60 , c ) 62 , d ) 64 , e ) 65
Rationale: "along one edge , the number of small cubes that can be cut = 100 / 25 = 4 along each edge 25 cubes can be cut . ( along length , breadth and height ) . total number of small cubes that can be cut = 4 * 4 * 4 = 64 answer : d"
Answer: d
Problem: last week john spent 10 percent of his wages on recreation . this week , his wages are 10 percent less than last week ʼ s wages and he spent 40 percent of his wages on recreation . the amount he spends on recreation this week is what percent of the amount he spent on recreation last week
Options: a ) 180 % , b ) 360 % , c ) 200 % , d ) 220 % , e ) 250 %
Rationale: "say john ' s wages last week were $ 100 , so he spent 0.10 * 100 = $ 10 on recreation ; this week ' s wages is 0.90 * 100 = $ 90 , so he spends 0.4 * 90 = $ 36 on recreation ; 36 / 10 = 3.60 , hence the amount he spends on recreation this week is 360 % of the amount he spent on recreation last week : 10 * 3.6 = 36 answer : b"
Answer: b
Problem: company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 8 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ?
Options: a ) 7 / 17 , b ) 14 / 31 , c ) 4 / 9 , d ) 17 / 35 , e ) 1 / 2
Rationale: "# of basic stereos was 2 / 3 of total and # of deluxe stereos was 1 / 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 8 / 5 units . total time for basic would be 10 * 1 = 10 and total time for deluxe would be 5 * 8 / 5 = 8 - - > total time for both of them would be 10 + 8 = 18 - - > deluxe / total = 8 / 18 = 4 / 9 . answer : c ."
Answer: c
Problem: the age of mother one decade age was 3 times the age of her daughter . 10 years hence , mother ’ s age will be two times that of her daughter . the proposition of their current ages is :
Options: a ) 7 : 3 , b ) 6 : 3 , c ) 8 : 3 , d ) 2 : 3 , e ) 6 : 3
Rationale: a 7 : 3 let , ages of mother and daughter 1 decade ago was 3 a and a years correspondingly then , ( 3 a + 10 ) + 10 = 2 [ ( a + 10 ) + 10 ] 3 a + 20 = 2 a + 40 x = 20 needed ratio = ( 3 a + 10 ) : ( a + 10 ) = 70 : 30 = 7 : 3
Answer: a
Problem: a line with slope 2 intersects a line with slope 6 at the point ( 40 , 30 ) . what is the distance between the x - intercepts of these lines ?
Options: a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12
Rationale: the line with slope 2 passing through ( 40 , 30 ) has x - intercept 40 − ( 30 / 2 ) . the line with slope 6 passing through ( 40 , 30 ) has x - intercept 40 − ( 30 / 6 ) . so the difference in the x - intercepts is ( 30 / 2 ) − ( 30 / 6 ) = 10 . correct answer d
Answer: d
Problem: is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 47 , then how old is b ?
Options: a ) 7 , b ) 8 , c ) 9 , d ) 18 , e ) 11
Rationale: let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 47 5 x = 45 = > x = 9 hence , b ' s age = 2 x = 18 years . answer : d
Answer: d
Problem: right now , al and eliot have bank accounts , and al has more money than eliot . the difference between their two accounts is 1 / 12 of the sum of their two accounts . if al ’ s account were to increase by 10 % and eliot ’ s account were to increase by 20 % , then al would have exactly $ 20 more than eliot in his account . how much money does eliot have in his account right now ?
Options: a ) $ 110 , b ) $ 120 , c ) $ 180 , d ) $ 200 , e ) $ 260
Rationale: "lets assume al have amount a in his bank account and eliot ' s bank account got e amount . we can form an equation from the first condition . a - e = 1 / 12 * ( a + e ) = = > 11 a = 13 e - - - - - - - - - - - - ( 1 ) second condition gives two different amounts , al ' s amount = 1.1 a and eliot ' s amount = 1.2 e 1.1 a = 20 + 1.2 e = = > 11 a = 200 + 12 e - - - - - - - ( 2 ) substituting ( 1 ) in ( 2 ) : 13 e = 200 + 12 e = = > e = 200 d"
Answer: d
Problem: a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 6.5 hours , then the distance covered at 40 kmph is
Options: a ) 130 km , b ) 120 km , c ) 100 km , d ) 280 km , e ) none of these
Rationale: "let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 6.5 or , x / 40 + ( 250 - x ) / 60 = 6.5 or , ( 3 x + 500 - 2 x ) / / 120 = 6.5 or 500 + x = 780 x = 280 answer : d"
Answer: d
Problem: at an election 2 candidates are participated a candidate got 45 % of votes and defeated by 9000 votes . 83 votes are invalid . find the total polled votes ?
Options: a ) 100000 , b ) 85236 , c ) 89562 , d ) 90083 , e ) 95842
Rationale: "winners votes = 100 - 45 = 55 % invalid votes = 83 polled votes = [ ( 100 * 9000 ) / ( 2 * 55 - 100 ) ] + 83 = 90083 answer is d"
Answer: d
Problem: a searchlight on top of the watchtower makes 2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 6 seconds ?
Options: a ) 5 / 6 , b ) 4 / 5 , c ) 3 / 4 , d ) 2 / 3 , e ) 1 / 2
Rationale: "the searchlight completes one revolution every 30 seconds . the probability that the man ' s area will be lit up is 6 / 30 = 1 / 5 . the probability that he will stay in the dark is 1 - 1 / 5 = 4 / 5 the answer is b ."
Answer: b
Problem: a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 12 % while buying and by 20 % while selling . what is his percentage profit ?
Options: a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 40 % , e ) ca n ' t be calculated
Rationale: "the owner buys 100 kg but actually gets 112 kg ; the owner sells 100 kg but actually gives 80 kg ; profit : ( 112 - 80 ) / 80 * 100 = ~ 40 % answer : d ."
Answer: d
Problem: tough and tricky questions : number properties . if ( z + 3 ) / 9 is an integer , what is remainder when z is divided by 9 ?
Options: a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
Rationale: assume the answer choices as the value of z 9 / 9 = 1 ans e
Answer: e
Problem: if x dollars is invested at 9 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 9 percent investment will exceed the annual income from the 8 percent investment by $ 48 . if $ 2,000 is the total amount invested , how much is invested at 9 percent ?
Options: a ) 300 , b ) 250 , c ) 1223.53 , d ) 1133.24 , e ) 776.47
Rationale: "2 equations with 2 unknowns 9 x / 100 - 8 y / 100 = 48 and x + y = 2000 solving these 2 equations , x = 1223.53 and y = 776.47 answer - c"
Answer: c
Problem: if the wheel is 8 cm then the number of revolutions to cover a distance of 948 cm is ?
Options: a ) 11 , b ) 14 , c ) 14 , d ) 12 , e ) 19
Rationale: "2 * 22 / 7 * 8 * x = 948 = > x = 18.9 answer : e"
Answer: e
Problem: the cost price of 11 articles is equal to the selling price of 10 articles . what is the profit percent ?
Options: a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 %
Rationale: "10 * sp = 11 * cp sp = 1.1 * cp the profit percent is 10 % . the answer is b ."
Answer: b
Problem: if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value d of z - x ?
Options: a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
Rationale: "x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number , then what could be minimum possible odd z . if x is some even number y - x > 5 ; y > x + 5 ; minimum value for y = x + 5 + 2 = x + 7 [ note : x + 5 is as even + odd = odd and nearest odd greater than x + 5 is x + 5 + 2 ] minimum value for z = y + 2 = x + 7 + 2 = x + 9 [ note : z = y + 2 because both z and y are odd . difference between two odd numbers is 2 ] d = z - x = x + 9 - x = 9 ans : d"
Answer: d
Problem: if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 70 / r years . if pat ' s parents invested $ 5,000 in a long - term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of the investment 18 years later , when pat is ready for college ?
Options: a ) $ 20000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000
Rationale: since investment doubles in 70 / r years , then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 5,000 * 2 ) * 2 = $ 20,000 ( after 9 years investment will become $ 5,000 * 2 = $ 10,000 and in another 9 years it ' ll become $ 10,000 * 2 = $ 20,000 ) . answer : a .
Answer: a
Problem: how many multiples of 5 are there between 5 to 90 ?
Options: a ) 9 , b ) 16 , c ) 17 , d ) none of these , e ) can not be determined
Rationale: "explanation : as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the units place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 . answer : b"
Answer: b
Problem: instead of multiplying a number by 9 , the number is divided by 9 . what is the percentage of error obtained ?
Options: a ) 97.765 % , b ) 99.765 % , c ) 96.765 % , d ) 95.765 % , e ) 98.765 %
Rationale: "let the number be x the right number is 9 x the wrong number is x / 9 error is ( 9 x - x / 9 ) = 80 x / 9 percentage of error is ( ( 80 x / 9 ) / 9 x ) * 100 = 98.765 % answer : e"
Answer: e
Problem: 1500 men have provisions for 17 days . if 320 more men join them , for how many days will the provisions last now ?
Options: a ) 12.9 , b ) 12.0 , c ) 14 , d ) 15 , e ) 16
Rationale: "1500 * 17 = 1820 * x x = 14 answer : c"
Answer: c
Problem: the cost of carpeting a room 15 meters long with a carpet 75 cm wide at 30 paisa per meter is rs . 36 . the breadth of the room is ?
Options: a ) 6 meters , b ) 8 meters , c ) 9 meters , d ) 10 meters , e ) 12 meters
Rationale: length of carpet = total cost / rate = 3600 / 30 = 120 m area of carpet = ( 120 x 75 ) / 100 m 2 = 90 m 2 ∴ area of the room = 90 m 2 breadth of the room = area / length = 90 / 15 m = 6 m answer : a
Answer: a
Problem: the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 7 minutes , then the area of the park ( in sq . m ) is :
Options: a ) 153601 , b ) 153600 , c ) 153602 , d ) 117600 , e ) 153604
Rationale: "perimeter = distance covered in 7 min . = ( 12000 / 60 ) x 7 m = 1400 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1400 or x = 140 . length = 420 m and breadth = 280 m . area = ( 420 x 280 ) m 2 = 117600 m 2 . answer : d"
Answer: d
Problem: james took a 3 - hour bike ride . in the second hour he traveled 6 miles , which was 20 percent farther than he traveled the first hour . if he traveled 25 percent farther in the third hour than he did in the second hour , how many miles did jose travel during the entire ride ?
Options: a ) 54.0 , b ) 54.9 , c ) 18.5 , d ) 57.0 , e ) 63.0
Rationale: "let the distance travelled in the first hour be x . thus , 1.2 x = 6 , x = 5 . now , the distance travelled in the 3 rd hour = 6 + 1 / 4 â ˆ — 6 = 7.5 the only option ending with a 0.5 in the decimal place is c . answer : c"
Answer: c
Problem: what least number must be added to 9599 , so that the sum is completely divisible by 100 ?
Options: a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8
Rationale: "if we divide 9599 by 100 remainder is 99 100 - 99 = 1 answer : a"
Answer: a
Problem: in august , a cricket team that played 120 matches won 28 % of the games it played . after a continuous winning streak , this team raised its average to 52 % . how many matches did the team win to attain this average ?
Options: a ) 40 , b ) 52 , c ) 60 , d ) 80 , e ) 98
Rationale: "let the no of matches played more = x so , ( 120 + x ) * 52 / 100 = 33.6 + x by solving we get x = 60 answer : c"
Answer: c
Problem: a fish can go upstream at 35 kmph and downstream at 55 kmph , and then find the speed of the fish in still water ?
Options: a ) 45 , b ) 55 , c ) 35 , d ) 65 , e ) 70
Rationale: us = 35 ds = 55 m = ( 35 + 55 ) / 2 = 45 answer : a
Answer: a
Problem: two numbers are respectively 20 % and 25 % more than a third number . so find the percentage that is first of the second is ?
Options: a ) 96 , b ) 20 , c ) 35 , d ) 78 , e ) 89
Rationale: "first - 120 second - 125 third - 100 125 - - - - 120 100 - - - - - 96 answer a"
Answer: a
Problem: a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 450 . find the total worth of the stock ?
Options: a ) 20029 , b ) 22500 , c ) 20289 , d ) 20027 , e ) 20026
Rationale: "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 450 = > x = 22500 answer : b"
Answer: b
Problem: the number which exceeds 16 % of it by 126 is :
Options: a ) 150 , b ) 52 , c ) 58 , d ) 60 , e ) 62
Rationale: "solution solution let the number be x . x - 16 % of x = 126 x - 16 / 100 x = 126 x - 4 / 25 x = 126 21 / 25 x = 126 x = ( 126 x 25 / 21 ) = 150 answer a"
Answer: a
Problem: a person ' s present age is two - fifth of the age of his mother . after 8 years , he will be one - half of the age of his mother . how old id the mother at present ?
Options: a ) 46 , b ) 49 , c ) 40 , d ) 21 , e ) 19
Rationale: "let the mother ' s present age be x years . then , the person ' s present age = 2 / 5 x years . ( 2 / 5 x + 8 ) = 1 / 2 ( x + 8 ) 2 ( 2 x + 40 ) = 5 ( x + 8 ) = > x = 40 answer : c"
Answer: c
Problem: 4.509 / 10.02 =
Options: a ) 0.45 , b ) 0.3505 , c ) 0.3509 , d ) 0.351 , e ) 0.3527
Rationale: "4.509 / 10.02 4509 / 1002 = 4.5 move the comma two places to the left as we have 2 decimal places from the divisor : 0.45 . answer : a"
Answer: a
Problem: the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 5 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ?
Options: a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25
Rationale: "let the average age of the whole team be x years . 11 x - ( 26 + 31 ) = 9 ( x - 1 ) 11 x - 9 x = 48 2 x = 48 x = 24 . the average age of the team is 24 years . the answer is d ."
Answer: d
Problem: the ratio of type of pen and type of notebook is 5 : 4 . and in 80 ways one can start writing . tell me the no . of pen and no . of notebook .
Options: a ) 6,12 , b ) 10,6 , c ) 10,8 , d ) 8,4 , e ) 6,8
Rationale: let the ratio constant be x . no . of pen = 5 x , no . of notebook = 4 x . total no . of ways of start writing = 5 x * 4 x = 20 ( square of x ) = 80 x = â ˆ š 4 = 2 pen = 10 , notebook = 8 answer c
Answer: c
Problem: joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 50 . he was allowed to drop his lowest score . after doing so , his average test score improved to 55 . what is the lowest test score that was dropped ?
Options: a ) 20 , b ) 35 , c ) 55 , d ) 65 , e ) 80
Rationale: "the arithmetic mean of 4 equally weighted tests was 50 . so what we can assume is that we have 4 test scores , each 50 . he dropped his lowest score and the avg went to 55 . this means that the lowest score was not 50 and other three scores had given the lowest score 5 each to make it up to 50 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 50 . so the lowest score = 50 - 15 = 35 answer ( b )"
Answer: b
Problem: jamshid and irwin decided to make up their own version of basketball . in addition to being able to score 2 - point baskets , they created ways to score 5 - , 11 - , and 13 - point baskets . if at halftime the product of the point values of irwin ’ s baskets was 2420 , how many 11 - point baskets e did irwin score ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
Rationale: first thing i noticed was that all these numbers are prime . . . prime factorization gives the following step 1 : 2 * 1210 step 2 : 2 * 11 * 110 step 4 : 2 * 11 * 11 * 10 step 5 : 2 * 11 * 11 * 2 * 5 so he scored e = 2 11 - point baskets in my opinion . answer c
Answer: c
Problem: a train is running at a speed of 40 km / hr and it crosses a post in 18 seconds . what is the length of the train ?
Options: a ) 190 metres , b ) 160 metres , c ) 200 metres , d ) 120 metres , e ) 250 metres
Rationale: speed of the train , v = 40 km / hr = 40000 / 3600 m / s = 400 / 36 m / s time taken to cross , t = 18 s distance covered , d = vt = ( 400 / 36 ) × 18 = 200 m distance covered is equal to the length of the train = 200 m correct answer is 200 metres
Answer: c
Problem: 70 % of 300 + 25 % of 400 - ? = 256
Options: a ) 48 , b ) 68 , c ) 54 , d ) 60 , e ) none of these
Rationale: "explanation : solution : let 70 % of 300 + 25 % of 400 - x = 256 . then , x = ( 70 / 100 * 300 ) + ( 25 / 100 * 400 ) - 256 = 210 + 100 - 256 = 54 . answer : c"
Answer: c
Problem: riya and priya set on a journey . riya moves eastward at a speed of 30 kmph and priya moves westward at a speed of 50 kmph . how far will be priya from riya after 45 minutes
Options: a ) 25 kms , b ) 60 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms
Rationale: "total eastward distance = 30 kmph * 3 / 4 hr = 22.50 km total westward distance = 50 kmph * 3 / 4 hr = 37.5 km total distn betn them = 22.5 + 37.5 = 60 km ans 60 km answer : b"
Answer: b
Problem: let c be defined as the sum of all prime numbers between 0 and 23 . what is c / 3
Options: a ) 155 , b ) 129 , c ) 61 , d ) 25.3 , e ) 43
Rationale: "prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 sum , c = 77 c / 3 = 25.3 answer d"
Answer: d
Problem: a cistern which could be filled in 10 hours takes 2 hour more to be filled owing to a leak in its bottom . if the cistern is full , in what time will the leak empty it ?
Options: a ) 50 hours , b ) 80 hours , c ) 100 hours , d ) 150 hours , e ) 110 hours
Rationale: "let the leak empty the full cistern in x hours 10 x / x - 10 = 10 + 1 x = 110 hours answer is e"
Answer: e
Problem: when the positive integer x is divided by 9 , the remainder is 5 . what is the remainder when 5 x is divided by 9 ?
Options: a ) 0 , b ) 1 , c ) 3 , d ) 4 , e ) 7
Rationale: "i tried plugging in numbers x = 9 q + 5 x = 14 5 x = 70 5 x / 9 = 63 + 7 remainder is 7 . answer is e ."
Answer: e
Problem: when positive integer x is divided by positive integer y , the remainder is 7 . if x / y = 86.1 , what is the value of y ?
Options: a ) 72 , b ) 71 , c ) 70 , d ) 30 , e ) 45
Rationale: "when positive integer x is divided by positive integer y , the remainder is 7 - - > x = qy + 7 ; x / y = 86.1 - - > x = 86 y + 0.1 y ( so q above equals to 86 ) ; 0.1 y = 7 - - > y = 70 . answer : c ."
Answer: c
Problem: find the simple interest on $ 10000 at 8 % per annum for 12 months ?
Options: a ) $ 410 , b ) $ 500 , c ) $ 650 , d ) $ 800 , e ) $ 1000
Rationale: "p = $ 10000 r = 8 % t = 12 / 12 years = 1 year s . i . = p * r * t / 100 = 10000 * 8 * 1 / 100 = $ 800 answer is d"
Answer: d
Problem: a fruit - salad mixture consists of apples , peaches , and grapes in the ratio 6 : 5 : 2 , respectively , by weight . if 78 pounds of the mixture is prepared , the mixture includes how many more pounds of apples than grapes ?
Options: a ) 24 , b ) 12 , c ) 9 , d ) 6 , e ) 4
Rationale: "we can first set up our ratio using variable multipliers . we are given that a fruit - salad mixture consists of apples , peaches , and grapes , in the ratio of 6 : 5 : 2 , respectively , by weight . thus , we can say : apples : peaches : grapes = 6 x : 5 x : 2 x we are given that 39 pounds of the mixture is prepared so we can set up the following question and determine a value for x : 6 x + 5 x + 2 x = 78 13 x = 78 x = 6 now we can determine the number of pounds of apples and of grapes . pounds of grapes = ( 2 ) ( 6 ) = 12 pounds of apples = ( 6 ) ( 6 ) = 36 thus we know that there are 36 - 12 = 24 more pounds of apples than grapes . answer is a ."
Answer: a
Problem: if ( a + b ) = 11 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ?
Options: a ) 16 . , b ) 8 . , c ) 5 . , d ) 2 . , e ) - 2 .
Rationale: "given a + b = 11 = > a = 11 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 11 - b + 3 - c = > 14 - ( b + c ) = > 14 - 9 = 5 . option c . . ."
Answer: c
Problem: from a vessel on the first day , 1 / 3 rd of the liquid evaporates . on the second day 3 / 4 th of the remaining liquid evaporates . what fraction of the volume is present at the end of the 2 day
Options: a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7
Rationale: let x be the volume . . . after 1 st day volume remaining = ( x - x / 3 ) = 2 x / 3 after 2 st day volume remaining = ( 2 x / 3 ) - ( ( 2 x / 3 ) * ( 3 / 4 ) ) = ( 2 x / 3 ) ( 1 - 3 / 4 ) = ( 2 x / 3 ) * ( 1 / 4 ) = x / 6 answer : d
Answer: d
Problem: the ratio of the incomes of uma and bala is 8 : 7 and the ratio of their expenditure is 7 : 6 . if at the end of the year , each saves $ 2000 then the income of uma is ?
Options: a ) $ 16800 , b ) $ 16500 , c ) $ 16000 , d ) $ 16300 , e ) $ 16200
Rationale: let the income of uma and bala be $ 8 x and $ 7 x let their expenditures be $ 7 y and $ 6 y 8 x - 7 y = 2000 - - - - - - - 1 ) 7 x - 6 y = 2000 - - - - - - - 2 ) from 1 ) and 2 ) x = 1000 uma ' s income = 8 x = 8 * 2000 = $ 16000 answer is c
Answer: c
Problem: the average of 6 numbers is 30 . if the average of first 4 is 25 and that of last 3 is 35 , the fourth number is :
Options: a ) 25 , b ) 26 , c ) 18 , d ) 19 , e ) 10
Rationale: let the six numbers be , a , b , c , d , e , f . a + b + c + d + e + f = 30 × 6 = 180 - - - - ( 1 ) a + b + c + d = 25 × 4 = 100 - - - - ( 2 ) d + e + f = 35 × 3 = 105 - - - - ( 3 ) add 2 nd and 3 rd equations and subtract 1 st equation from this . d = 25 answer : a
Answer: a
Problem: how many of the positive factors of 15 , 45 and how many common factors are there in numbers ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: factors of 15 - 1 , 3 , 5 , and 15 factors of 45 - 1 , 3 , 9 , 15 and 45 comparing both , we have three common factors of 45,16 - 3 answer c
Answer: c
Problem: divide rs . 32000 in the ratio 1 : 9 ?
Options: a ) 12000 , 20000 , b ) 12000 , 200098 , c ) 12000 , 20007 , d ) 3200 , 28800 , e ) 12000 , 20001
Rationale: "1 / 10 * 32000 = 3200 9 / 10 * 32000 = 28800 answer : d"
Answer: d
Problem: if a car had traveled 25 kmh faster than it actually did , the trip would have lasted 30 minutes less . if the car went exactly 75 km , at what speed did it travel ?
Options: a ) 30 kmh , b ) 40 kmh , c ) 45 kmh , d ) 50 kmh , e ) 55 kmh
Rationale: time = distance / speed difference in time = 1 / 2 hrs 75 / x - 75 / ( x + 25 ) = 1 / 2 substitute the value of x from the options . - - > x = 50 - - > 75 / 50 - 75 / 75 = 3 / 2 - 1 = 1 / 2 answer : d
Answer: d
Problem: in the faculty of reverse - engineering , 240 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 80 % of the total ?
Options: a ) 515 . , b ) 545 . , c ) 661 . , d ) 644 . , e ) 666 .
Rationale: "answer is c : 661 solution : total number of students studying both are 423 + 240 - 134 = 529 ( subtracting the 134 since they were included in the both the other numbers already ) . so 80 % of total is 529 , so 100 % is approx . 661 ."
Answer: c
Problem: machine a and machine b are each used to manufacture 770 sprockets . it takes machine a 10 hours longer to produce 770 sprockets than machine b . machine b produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machineaproduce ?
Options: a ) 6 , b ) 6.6 , c ) 60 , d ) 100 , e ) 7.7
Rationale: "time taken by b = t time taken by a = t + 10 qty produced by a = q qty produced by b = 1.1 q for b : t ( 1.1 q ) = 770 qt = 700 for a : ( t + 10 ) ( q ) = 770 qt + 10 q = 770 700 + 10 q = 770 q = 7 so a can produce 7 / hour . then b can produce = 7 ( 1.1 ) = 7.7 / hour . e"
Answer: e
Problem: for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 5 + 3 and * 5 + 5 , inclusive ?
Options: a ) none , b ) one , c ) two , d ) three , e ) four
Rationale: generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 5 would be divisible by all numbers from 1 to 5 . = > * 5 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 5 is divisible by 3 ) in fact adding anyprimenumber between 1 to 5 to * 5 will definitely be divisible . so the answer is none ( a ) ! supposing if the question had asked for prime numbers between * 5 + 3 and * 5 + 11 then the answer would be 1 . for * 5 + 3 and * 5 + 13 , it is 2 and so on . . . a
Answer: a
Problem: the total marks obtained by a student in mathematics and physics is 60 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together .
Options: a ) 40 , b ) 30 , c ) 25 , d ) data inadequate , e ) none of these .
Rationale: "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 60 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 60 + 20 ) / 2 = 40 . answer : a"
Answer: a
Problem: two numbers are respectively 40 % and 30 % more than a third number . the second number expressed in terms of percentage of the first is ?
Options: a ) 95 % , b ) 93 % , c ) 90 % , d ) 80 % , e ) none of these
Rationale: "here , x = 40 and y = 30 ; therefore second number = [ [ ( 100 + y ) / ( 100 + x ) ] x 100 ] % of first number = [ [ ( 100 + 30 ) / ( 100 + 40 ) ] x 100 ] % of first number = 92.8 % of the first answer : b"
Answer: b
Problem: a man can row downstream at 22 kmph and upstream at 14 kmph . find the speed of the man in still water and the speed of stream respectively ?
Options: a ) 13 , 3 , b ) 18 , 4 , c ) 15 , 3 , d ) 14 , 4 , e ) none of these
Rationale: "explanation : let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 22 - - - ( 1 ) and x - y = 14 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 36 = > x = 18 , y = 4 . answer : option b"
Answer: b
Problem: 8.5 × 6.4 + 4.5 × 11.6 = ? ÷ 4
Options: a ) 426.2 , b ) 426.4 , c ) 106.6 , d ) 422.4 , e ) 522.5
Rationale: "explanation : ( 54.4 + 52.2 ) × 4 = 426.4 answer : option b"
Answer: b
Problem: a cubical tank is filled with water to a level of 2 feet . if the water in the tank occupies 50 cubic feet , to what fraction of its capacity is the tank filled with water ?
Options: a ) 1 / 3
b ) 2 / 3
c ) 3 / 4
d ) 2 / 5
e ) 5 / 6
Rationale: the volume of water in the tank is h * l * b = 50 cubic feet . since h = 2 , then l * b = 25 and l = b = 5 . since the tank is cubical , the capacity of the tank is 5 * 5 * 5 = 125 . the ratio of the water in the tank to the capacity is 50 / 125 = 2 / 5 the answer is d .
Answer: d
Problem: a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 210 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ?
Options: a ) 88 , b ) 45 , c ) 36 , d ) 80 , e ) 12
Rationale: "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 210 = 450 m . time taken = 450 / 10 = 45 sec . answer : b"
Answer: b
Problem: a cistern is normally filled in 12 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ?
Options: a ) 10 hours , b ) 84 hours , c ) 30 hours , d ) 40 hours , e ) 50 hours
Rationale: "if leakage / hour = 1 / x , then 1 / 12 - 1 / x = 1 / 14 , solving 1 / x = 1 / 84 so in 84 hours full cistern will be empty . answer : b"
Answer: b
Problem: what is the units digit of the product of the first 100 odd numbers ?
Options: a ) 1 , b ) 3 , c ) 0 , d ) 5 , e ) 7
Rationale: 1 * 3 * 5 * 7 . . . . . . . . . . . . . . . . . . . will end up in 5 in the units place answer : d
Answer: d
Problem: how many positive integers between 20 and 2000 ( both are inclusive ) are there such that they are multiples of 10 ?
Options: a ) 201 , b ) 193 , c ) 200 , d ) 199 , e ) 195
Rationale: "multiples of 10 = 20 , 30,40 - - - - - , 1990,2000 number of multiples of 10 = > 2000 - 20 / 10 + 1 = 199 answer is d"
Answer: d
Problem: what is the cost of leveling the field in the form of parallelogram at the rate of rs . 50 / 10 sq . metre , whose base & perpendicular distance from the other side being 54 m & 24 m respectively ?
Options: a ) rs . 5280 , b ) rs . 5500 , c ) rs . 6000 , d ) rs . 6480 , e ) rs . 7680
Rationale: area of the parallelogram = length of the base * perpendicular height = 54 * 24 = 1296 m . total cost of levelling = rs . 6480 d
Answer: d
Problem: a and b together can complete a piece of work in 4 days . if a alone can complete the same work in 12 days , in how many days , in how many days can b alone complete that work ?
Options: a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 5
Rationale: "( a + b ) ' s 1 day ' s work = 1 / 4 a ' s 1 day ' s work = 1 / 12 b ' s 1 days ' s work = ( 1 / 4 ) - ( 1 / 12 ) = 1 / 6 hence , b alone can complete the work in 6 days . answer is a"
Answer: a
Problem: the area of a square is added to one of its sides , and the perimeter is then subtracted from this total , the result is 4 . what is the length of one side ?
Options: a ) 5
b ) 4
c ) 3
d ) 2
e ) 1
Rationale: the equation is ; side + area - perimeter = s + a - p = s + s ^ 2 - 4 s = s ( 1 + s - 4 ) . by plugging in the answers we can test the answers quickly ; then , 4 is the only possible answer . answer : b
Answer: b
Problem: a , b and c rents a pasture for rs . 870 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should a pay ?
Options: a ) 240 , b ) 270 , c ) 276 , d ) 271 , e ) 272
Rationale: "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 8 / 29 * 870 = 240 answer : a"
Answer: a
Problem: caleb spends $ 68.50 on 50 hamburgers for the marching band . if single burgers cost $ 1.00 each and double burgers cost $ 1.50 each , how many double burgers did he buy ?
Options: a ) 5 , b ) 10 , c ) 37 , d ) 40 , e ) 45
Rationale: solution - lets say , single hamburgersxand double hamburgersy given that , x + y = 50 and 1 x + 1.5 y = 68.50 . by solving the equations y = 37 . ans c .
Answer: c
Problem: in a graduating class of 234 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 ) 115 , 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 - 234 = 29 greatest - smallest is 144 - 29 = 115 so answer must be b . 115"
Answer: b
Problem: two trains are moving at 60 kmph and 70 kmph in opposite directions . their lengths are 150 m and 100 m respectively . the time they will take to pass each other completely is ?
Options: a ) 5 sec , b ) 6 sec , c ) 7 sec , d ) 8 sec , e ) 9 sec
Rationale: "70 + 60 = 130 * 5 / 18 = 325 / 9 mps d = 150 + 100 = 250 m t = 250 * 9 / 325 = 7 sec answer : c"
Answer: c
Problem: one fourth of a solution that was 11 % salt by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent salt by weight ?
Options: a ) 24 % , b ) 34 % , c ) 22 % , d ) 31 % , e ) 8.5 %
Rationale: "consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.11 + 25 * x = 100 * 0.16 - - > x = 0.31 . answer : d ."
Answer: d
Problem: 1297 x 1297 = ?
Options: a ) a ) 1682209 , b ) b ) 1951601 , c ) c ) 1951602 , d ) d ) 1951603 , e ) e ) 1951604
Rationale: "1297 x 1297 = ( 1297 ) 2 = ( 1300 - 3 ) 2 = ( 1300 ) 2 + ( 3 ) 2 - ( 2 x 1300 x 3 ) = 1690000 + 9 - 7800 = 1690009 - 7800 = 1682209 . answer : a"
Answer: a
Problem: 80 percent of the members of a study group are women , and 40 percent of those women are lawyers . if one member of the study group is to be selected at random , what is the probability that the member selected is a woman lawyer ?
Options: a ) 0.45 , b ) 0.55 , c ) 0.65 , d ) 0.32 , e ) 0.35
Rationale: say there are 100 people in that group , then there would be 0.8 * 0.40 * 100 = 32 women lawyers , which means that the probability that the member selected is a woman lawyer is favorable / total = 32 / 100 . answer : d
Answer: d
Problem: if a and b are positive integers and ( 2 ^ a ) ^ b = 2 ^ 5 , what is the value of 2 ^ a * 2 ^ b ?
Options: a ) 8 , b ) 64 , c ) 16 , d ) 32 , e ) 4
Rationale: "2 ^ ab = 2 ^ 5 therefore ab = 5 either a = 1 or 5 or b = 5 or 1 therefore 2 ^ a * 2 ^ b = 2 ^ ( a + b ) = 2 ^ 6 = 64 b"
Answer: b