regisss/math_qa huggingface dataset:
Problem: jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 25 / 36 , what is the probability that the next one she draws will be pink ?
Options: a ) 1 / 6 , b ) 4 / 7 , c ) 3 / 7 , d ) 16 / 49 , e ) 40 / 49
Rationale: "the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 25 / 36 = ( 5 / 6 ) * ( 5 / 6 ) therefore probability that the next one she draws is pink = 1 / 6 option ( a )"
Answer: a
Problem: if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 253 games will be played during the tournament . what is the number of participants ?
Options: a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25
Rationale: "let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 253 n * ( n - 1 ) = 506 = 23 * 22 ( trial and error ) the answer is c ."
Answer: c
Problem: it takes 10 days for digging a trench of 100 m long , 50 m broad and 10 m deep . what length of trench , 25 m broad and 15 m deep can be dug in 30 days ?
Options: a ) 400 m , b ) 200 m , c ) 100 m , d ) 89 m , e ) 79 m
Rationale: more days , more length ( direct ) less breadth , more length ( indirect ) more depth , less length ( indirect days 10 : 30 ; breadth 25 : 50 ; : : 100 : x depth 15 : 10 ; : . 10 * 25 * 15 * x = 30 * 50 * 10 * 100 x = ( 30 * 50 * 10 * 100 ) / 10 * 25 * 15 = 400 so the required length = 400 m answer : a
Answer: a
Problem: if 20 men can build a water fountain 56 metres long in 3 days , what length of a similar water fountain can be built by 35 men in 3 days ?
Options: a ) 40 m , b ) 64 m , c ) 77 m , d ) 89 m , e ) 98 m
Rationale: "explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 3 : 3 : : 56 : x therefore ( 20 x 3 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 60 = 98 hence , the required length is 98 m . answer : e"
Answer: e
Problem: p has $ 63 more than what q and r together would have had if both b and c had 1 / 9 of what p has . how much does p have ?
Options: a ) $ 69 , b ) $ 72 , c ) $ 75 , d ) $ 78 , e ) $ 81
Rationale: p = ( 2 / 9 ) * p + 63 ( 7 / 9 ) * p = 63 p = 81 the answer is e .
Answer: e
Problem: a train running at the speed of 110 km / hr crosses a pole in 9 sec . what is the length of the train ?
Options: a ) 298 m , b ) 275 m , c ) 208 m , d ) 988 m , e ) 299 m
Rationale: "speed = 110 * 5 / 18 = 275 / 9 m / sec length of the train = speed * time = 275 / 9 * 9 = 275 m answer : b"
Answer: b
Problem: if 2 ^ y = 4 ^ ( 5 x + 3 ) and 3 ^ ( x - 7 ) = 9 ^ y , what is the value of x + y ?
Options: a ) - 10 , b ) - 5 , c ) - 4 , d ) 3 , e ) 7
Rationale: 2 ^ y = 4 ^ ( 5 x + 3 ) 2 ^ y = 2 ^ 2 ( 5 x + 3 ) y = 10 x + 6 . . . . . . . . . . . 1 3 ^ ( x - 7 ) = 9 ^ y 3 ^ ( x - 7 ) = 3 ^ 2 y x - 7 = 2 y . . . . . . . . . . . . . 2 put value of y = 10 x + 6 in eq 2 x - 7 = 2 ( 10 x + 6 ) x - 7 = 20 x + 12 19 x = - 19 x = - 1 therefore , y = - 10 + 6 y = - 4 x + y = - 1 - 4 = - 5 answer : b
Answer: b
Problem: a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 140 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ?
Options: a ) 28 sec , b ) 16 sec , c ) 38 sec , d ) 18 sec , e ) 17 sec
Rationale: "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 140 = 380 m . time taken = 380 / 10 = 38 sec . answer : c"
Answer: c
Problem: of the 75 house in a development , 50 have a two - car garage , 40 have an in - the - ground swimming pool , and 35 have both a two - car garage and an in - the - ground swimming pool . how many houses in the development have neither a two - car garage nor an in - the - ground swimming pool ?
Options: a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30
Rationale: "neither car nor garage = total - garage - ( swim - common ) = 75 - 50 - ( 40 - 35 ) = 75 - 55 = 20 answer c"
Answer: c
Problem: if the price of a certain computer increased 30 percent from a dollars to 351 dollars , then 2 a =
Options: a ) 540 , b ) 570 , c ) 619 , d ) 649 , e ) 700
Rationale: "before price increase price = a after 30 % price increase price = a + ( 30 / 100 ) * a = 1.3 a = 351 ( given ) i . e . a = 351 / 1.3 = $ 270 i . e . 2 a = 2 * 270 = 540 answer : option a"
Answer: a
Problem: in a certain large company , the ratio of college graduates with a graduate degree to non - college graduates is 1 : 8 , and ratio of college graduates without a graduate degree to non - college graduates is 2 : 3 . if one picks a random college graduate at this large company , what is the probability w this college graduate has a graduate degree ?
Options: a ) 1 / 11 , b ) 1 / 12 , c ) 1 / 13 , d ) 3 / 19 , e ) 3 / 43
Rationale: "in believe the answer is d . please see below for explanation . 0 ) we are told the following ratios cgd - college graduate with degree ncg - non college graduate cgn - college graduate no degree cgd ncg cgn 1 8 3 2 in order to make cgd and cgn comparable we need to find the least common multiple of 8 and 3 and that is 24 multiplying the first ratio by 3 and the second ratio by 8 we get cgd ncg cgn 3 24 16 if one picks a random college graduate at this large company , what is the probability this college graduate has a graduate degree ? nr of cgd = 3 nr of cg = 3 + 16 = 19 probability w of cgd / ( cg ) - > 3 / 19 answer d"
Answer: d
Problem: what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 25 % ?
Options: a ) 20 % , b ) 26 % , c ) 42 % , d ) 27 % , e ) 12.5 %
Rationale: "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 75 2 / 3 sp 1 = 75 sp 1 = 112.50 100 - - - 12.5 = > 12.5 % answer : e"
Answer: e
Problem: how many 5 - digit numbers are there , if the two leftmost digits are even , the other digits are odd and the digit 2 can not appear more than once in the number .
Options: a ) 2200 , b ) 2295 , c ) 2375 , d ) 2380 , e ) 2385
Rationale: n = ( 4 * 5 - 1 ) * 5 * 5 * 5 = 2375 where 4 cases of first digit { 2,4 , 6,8 } 5 cases of second digit { 0 , 2,4 , 6,8 } 1 case of 22 for two leftmost digit 5 cases of third digit { 1 , 3,5 , 7,9 } 5 cases of fourth digit { 1 , 3,5 , 7,9 } 5 cases of fifth digit { 1 , 3,5 , 7,9 } c
Answer: c
Problem: a bucket full of nuts was discovered by the crow living in the basement . the crow eats a sixth of the total number of nuts in 4 hours . how many hours i total will it take the crow to finish a quarter of the nuts ?
Options: a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14
Rationale: "in one hour , the crow eats 1 / 24 of the nuts . ( 1 / 4 ) / ( 1 / 24 ) = 6 hours the answer is a ."
Answer: a
Problem: alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 1081 . find the cost of each ice - cream cup ?
Options: a ) 25 , b ) 66 , c ) 77 , d ) 99 , e ) 45
Rationale: "let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1081 96 + 225 + 490 + 6 x = 1081 6 x = 270 = > x = 45 . answer : e"
Answer: e
Problem: selling an kite for rs . 30 , a shop keeper gains 40 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ?
Options: a ) 8 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 26 %
Rationale: "explanation : marked price = rs . 30 c . p . = 100 / 140 * 30 = rs . 21.42 sale price = 90 % of rs . 30 = rs . 27 required gain % = 5.57 / 21.42 * 100 = 26 % . answer : e"
Answer: e
Problem: how many factors of 30 are odd numbers greater than 1 ?
Options: a ) 3 , b ) 3 , c ) 5 , d ) 6 , e ) 7
Rationale: "prime factors of 30 are 2 ^ 1,3 ^ 1,5 ^ 1 total divisors = ( power if a prime factor + 1 ) total no . of odd factors ( 3,5 , ) = ( 1 + 1 ) ( 1 + 1 ) = 4 since we need odd divisors other than 1 = > 4 - 1 = 3 odd divisors b is the answer"
Answer: b
Problem: at a certain restaurant , the average ( arithmetic mean ) number of customers served for the past x days was 75 . if the restaurant serves 120 customers today , raising the average to 80 customers per day , what is the value of x ?
Options: a ) 2 , b ) 5 , c ) 7 , d ) 8 , e ) 20
Rationale: "withoutusing the formula , we can see that today the restaurant served 40 customers above the average . the total amount above the average must equal total amount below the average . this additional 40 customers must offset the “ deficit ” below the average of 80 created on the x days the restaurant served only 75 customers per day . 40 / 5 = 8 days . choice ( a ) . withthe formula , we can set up the following : 80 = ( 75 x + 120 ) / ( x + 1 ) 80 x + 80 = 75 x + 120 5 x = 40 x = 8 answer choice ( d )"
Answer: d
Problem: dividing by 3 ⁄ 6 and then multiplying by 5 ⁄ 6 is the same as dividing by what number ?
Options: a ) 31 ⁄ 5 , b ) 16 ⁄ 5 , c ) 20 ⁄ 9 , d ) 9 ⁄ 20 , e ) 5 ⁄ 3
Rationale: "say x / 3 / 6 * 5 / 6 = x * 6 / 3 * 5 / 6 = x * 5 / 3 e"
Answer: e
Problem: an inspector rejects 15 % of the meters as defective . how many will he examine to reject 15 ?
Options: a ) a ) 100 , b ) b ) 120 , c ) c ) 250 , d ) d ) 200 , e ) e ) 160
Rationale: "then , 15 % of x = 15 ( 15 / 100 ) x = 15 x = ( 15 * 100 * ) / 15 = 100 answer is a"
Answer: a
Problem: pipe a fills a tank of capacity 700 liters at the rate of 40 liters a minute . another pipe b fills the same tank at the rate of 30 liters a minute . a pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute . if pipe a is kept open for a minute and then closed and pipe b is open for a minute and then closed and then pipe c is open for a minute and then closed and the cycle is repeated , when will the tank be full ?
Options: a ) 42 minutes , b ) 14 minutes , c ) 39 minutes , d ) 40 minutes 20 seconds , e ) none of these
Rationale: "in one cycle they fill 40 + 30 - 20 = 50 liters 700 = 50 * n = > n = 14 here n = number of cycles . total time = 14 * 3 = 42 as in one cycle there are 3 minutes . thus 42 minutes answer : a"
Answer: a
Problem: the area of a parallelogram is 72 sq m and its altitude is twice the corresponding base . then the length of the base is ?
Options: a ) 6 , b ) 16 , c ) 8 , d ) 36 , e ) none
Rationale: "2 x * x = 72 = > x = 6 answer : a"
Answer: a
Problem: a man can row upstream at 10 kmph and downstream at 20 kmph , and then find the speed of the man in still water ?
Options: a ) 11 , b ) 77 , c ) 25 , d ) 88 , e ) 34
Rationale: "us = 20 ds = 30 m = ( 30 + 20 ) / 2 = 25 answer : c"
Answer: c
Problem: a and b can finish a work together in 12 days , and b and c together in 16 days . if a alone works for 5 days and then b alone continues for 7 days , then remaining work is done by c in 13 days . in how many days can c alone finish the complete work ?
Options: a ) 22 days , b ) 21 days , c ) 25 days , d ) 24 days , e ) 23 days
Rationale: here lcm of 12 and 16 is taken as total work . ( becomes easy to solve ) assume total work = 48 units then workdone by ( a + b ) in one day = 48 / 12 = 4 units similarly , by ( b + c ) in one day = 48 / 16 = 3 units now according to question , a works 5 days , b for 7 days and c for 13 days to complete total work so , 5 a + 7 b + 13 c = 48 units 5 ( a + b ) + 2 ( b + c ) + 11 c = 48 units 5 * 4 + 2 * 3 + 11 c = 48 units 11 c = 22 units c = 2 units ( c does 2 units of work daily ) therefore , 48 / 2 = 24 days c requires 24 days to complete the work alone . answer d
Answer: d
Problem: the ages of two person , differ by 20 years . if 5 years ag , the elder one be 5 times as old as the younger one their present ages ( in years ) are respectively
Options: a ) 30 , 10 , b ) 2010 , c ) 3515 , d ) 5117 , e ) 20,17
Rationale: let their ages be x and ( x + 20 ) years ( x - 5 ) * 5 = ( x + 20 - 5 ) after solving this we get x = 10 years the age of elder one = 10 + 20 = 30 years so the present ages are 30 and 10 years answer : a
Answer: a
Problem: the price of stock increased by 8 % last year and decreased by 6 % this year . what is the net percentage change in the price of the stock ?
Options: a ) 0.2 % , b ) 0.8 % , c ) 1.2 % , d ) 1.52 % , e ) 2 %
Rationale: ( 100 % + 8 % ) * ( 100 % - 6 % ) = 1.08 * 0.94 = 1.0152 = 101.52 % . the net percentage change in the price of the stock is ( + ) 1.52 % the answer is d
Answer: d
Problem: the mass of 1 cubic meter of a substance is 400 kg under certain conditions . what is the volume in cubic centimeters of 1 gram of this substance under these conditions ? ( 1 kg = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters )
Options: a ) 1.5 , b ) 2.5 , c ) 3.5 , d ) 4.5 , e ) 5.5
Rationale: "400 kg - 1 cubic meter ; 400,000 g - 1 cubic meter ; 400,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 / 400,000 = 10 / 4 = 2.5 cubic centimeters . answer : b ."
Answer: b
Problem: for what value of â € œ k â €  will the equation ( 2 kx 2 + 5 kx + 1 ) = 0 have equal roots ?
Options: a ) 2 / 7 , b ) 9 / 4 , c ) 8 / 25 , d ) 7 / 1 , e ) 7 / 2
Rationale: "for a 2 nd degree equation ax 2 + bx _ c = 0 has equal roots the condition is b 2 - 4 ac = 0 in the given equation ( 5 k ) ^ 2 - 4 * 2 k * 1 = 0 by solving this equation we get k = 0 , k = 8 / 25 answer : c"
Answer: c
Problem: in a class of 25 students , 2 students did not borrow any books from the library , 10 students each borrowed 1 book , 8 students each borrowed 2 books , and the rest of the students each borrowed at least 3 books . if the average ( arithmetic mean ) number of books borrowed per student was 2 , what is the maximum number of books that any single student could have borrowed ?
Options: a ) 8 , b ) 10 , c ) 12 , d ) 13 , e ) 15
Rationale: "the average number of books per student was 2 means that total of 2 * 25 = 50 books were borrowed ; 2 + 10 + 8 = 20 students borrowed total of 2 * 0 + 10 * 1 + 8 * 2 = 26 books ; so 50 - 26 = 24 books are left to distribute among 25 - 20 = 5 students , these 5 arethe rest who borrowed at least 3 books ; tomaximizethe number of books one student from above 5 could have borrowed we shouldminimizethe number of books other 4 students from 5 could have borrowed . minimum these 4 students could have borrowed is 4 books per student , so total number of books they could have borrowed is 4 * 3 = 12 books . so the 5 th student could have borrowed is 24 - 12 = 12 books . answer : c ."
Answer: c
Problem: a metallic sphere of radius 12 cm is melted and drawn into a wire , whose radius of cross section is 24 cm . what is the length of the wire ?
Options: a ) 6 cm , b ) 4 cm , c ) 8 cm , d ) 3 cm , e ) 9 cm
Rationale: "volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . π ( 24 ) ^ 2 * h = ( 4 / 3 ) π ( 12 ) ^ 3 = > h = 4 cm answer : b"
Answer: b
Problem: boys and girls in a class are writing letters . there are twice as many girls as boys in the class , and each girl writes 3 more letters than each boy . if boys write 24 of the 90 total letters written by the class , how many letters does each boy write ?
Options: a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12
Rationale: there are twice as many girls as boys in the class - - > g = 2 b . each girl writes 3 more letters than each boy - - > boys write x letters , girls write x + 3 letters . boys write 24 letters - - > bx = 24 . girls write 90 - 24 = 66 letters - - > ( 2 b ) ( x + 3 ) = 66 - - > 2 bx + 6 b = 66 - - > 2 * 24 + 6 b = 66 - - > b = 3 . bx = 24 - - > 3 x = 24 - - > x = 8 . answer : d .
Answer: d
Problem: a man walking at the rate of 5 km / hr crosses a bridge in 15 minutes . the length of the bridge ( in meters ) is :
Options: a ) 600 , b ) 750 , c ) 1000 , d ) 1250 , e ) none of these
Rationale: "speed = ( 5 * 5 / 18 ) m / sec = 25 / 18 m / sec . distance covered in 15 minutes = ( 25 / 18 * 15 * 60 ) m = 1250 m . correct option : d"
Answer: d
Problem: a honey bee flies for 10 seconds , from a daisy to a rose . it immediately continues to a poppy , flying for 6 additional seconds . the distance the bee passed , flying from the daisy to the rose is 14 meters longer than the distance it passed flying from the rose to the poppy . the bee flies to the poppy at 3 meters per second faster than her speed flying to the rose . the bee flies how many meters per second from the daisy to the rose ?
Options: a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10
Rationale: "let the speed be ' s ' and let the distance between rose and poppy be ' x ' the problem boils down to : rose to poppy : s + 3 = x / 6 - - - - - - - 1 daisy to rose : s = ( x + 14 ) / 10 - - - - - - 2 so from 1 we can re write x as x = 6 s + 18 substitute the value of x in 2 gives us s = 2 m / s a"
Answer: a
Problem: mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8010 / - . find out the amount that he borrowed .
Options: a ) s . 5266 , b ) s . 5269 , c ) s . 5228 , d ) s . 5201 , e ) s . 52192
Rationale: "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8010 = p + ( 54 p / 100 ) 8010 = ( 100 p + 54 p ) / 100 8010 = 154 p / 100 p = ( 8010 * 100 ) / 154 = rs . 5201.298 answer : d"
Answer: d
Problem: in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 6500 , the number of valid votes that the other candidate got , was :
Options: a ) 2800 , b ) 2700 , c ) 2900 , d ) 2340 , e ) 2300
Rationale: "d number of valid votes = 80 % of 6500 = 5200 . valid votes polled by other candidate = 45 % of 5200 = ( 45 / 100 x 5200 ) = 2340 ."
Answer: d
Problem: if 9 ! / 3 ^ x is an integer , what is the greatest possible value of x ?
Options: a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
Rationale: 9 - 3 * 3 6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 4 is allowed . imo b .
Answer: b
Problem: if 21 ! / 3 ^ x is an integer , what is the greatest possible value of x ?
Options: a ) 3 , b ) 4 , c ) 5 , d ) d , e ) 7
Rationale: "21 - 7 * 3 18 - 3 * 3 * 2 15 - 5 * 3 12 - 4 * 3 9 - 3 * 3 6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 9 is allowed . imo d ."
Answer: d
Problem: the average of 13 result is 60 . average of the first 7 of them is 57 and that of the last 7 is 61 . find the 8 th result ?
Options: a ) 35 , b ) 37 , c ) 46 , d ) 48 , e ) 50
Rationale: sum of all the 13 results = 13 * 60 = 780 sum of the first 7 of them = 7 * 57 = 399 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 399 - 780 = 46 . c
Answer: c
Problem: 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , ( . . . )
Options: a ) 53 , b ) 67 , c ) 48 , d ) 59 , e ) 45
Rationale: "explanation : all are prime numbers in their order , starting from 23 hence , next number is 67 answer : b"
Answer: b
Problem: if x / ( 9 p ) is an even prime number , where x is a positive integer and p is a prime number , what is the least value of x ?
Options: a ) 22 , b ) 33 , c ) 44 , d ) 66 , e ) 99
Rationale: "x / ( 11 p ) = even prime number x = even prime number * 11 p least value of x = lowest even prime number * 11 * lowest value of p = 2 * 11 * 2 = 44 answer c"
Answer: c
Problem: after decreasing 25 % in the price of an article costs rs . 1500 . find the actual cost of an article ?
Options: a ) 1400 , b ) 1300 , c ) 1200 , d ) 2000 , e ) 1500
Rationale: "cp * ( 75 / 100 ) = 1500 cp = 20 * 100 = > cp = 2000 answer : d"
Answer: d
Problem: if the sides of a triangle are 52 cm , 48 cm and 20 cm , what is its area ?
Options: a ) 480 cm 2 , b ) 765 cm 2 , c ) 216 cm 2 , d ) 197 cm 2 , e ) 275 cm 2
Rationale: "the triangle with sides 52 cm , 48 cm and 20 cm is right angled , where the hypotenuse is 52 cm . area of the triangle = 1 / 2 * 48 * 20 = 480 cm 2 answer : a"
Answer: a
Problem: 5 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 3443 , how much was it in the beginning ?
Options: a ) 4264 , b ) 2776 , c ) 4400 , d ) 2871 , e ) 881
Rationale: "x * ( 95 / 100 ) * ( 85 / 100 ) = 3443 x = 4264 answer : a"
Answer: a
Problem: adam borrowed some money at the rate of 6 % p . a . for the first two years , at the rate of 9 % p . a . for the next 3 years , and at the rate of 14 % p . a . for the period beyond 4 years . if he pays a total interest of 11900 at the end of 9 years , how much money did he borrow ?
Options: a ) 10,526 , b ) 12,526 , c ) 14,000 , d ) 16,000 , e ) 16,536
Rationale: let the sum borrowed be x . then , ( x × 6 × 21 / 00 ) + ( x × 9 × 3 / 100 ) + ( x × 14 × 4 / 100 ) = 11900 ⇒ ( 3 ⁄ 25 x + 27 ⁄ 100 x + 14 ⁄ 25 x ) = 11400 ⇒ 95 ⁄ 100 x = 11900 ⇒ x = ( 11900 × 100 / 95 ) = 12526 hence , sum borrowed = 12,526 answer b
Answer: b
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 6 men , 4 women and 10 children together complete the piece of work ?
Options: a ) 5 days , b ) 15 days , c ) 28 days , d ) 8 days , e ) 7 days
Rationale: "explanation : 1 man ’ s 1 day ’ s work = 1 / 8 × 12 = 1 / 96 6 men ’ s 1 day ’ s work = 1 × 6 / 96 = 1 / 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 , ( 6 men + 4 women + 10 children ) ’ s 1 day ’ s work = 1 / 16 + 1 / 48 + 1 / 24 = 1 / 8 the required no . of days = 8 days answer : option d"
Answer: d
Problem: if henry were to add 6 gallons of water to a tank that is already 3 / 4 full of water , the tank would be 7 / 8 full . how many gallons of water would the tank hold if it were full ?
Options: a ) 25 , b ) 48 , c ) 64 , d ) 80 , e ) 96
Rationale: "7 / 8 x - 3 / 4 x = 6 galls 1 / 8 * x = 6 gallons x = 48 gallons answer b"
Answer: b
Problem: in what time will a railway train 120 m long moving at the rate of 70 kmph pass a telegraph post on its way ?
Options: a ) 3 sec , b ) 4 sec , c ) 5 sec , d ) 6 sec , e ) 7 sec
Rationale: "t = 120 / 70 * 18 / 5 = 6 sec answer : d"
Answer: d
Problem: there are 400 employees in a room . 99 % are managers . how many managers must leave the room to bring down the percentage of managers to 98 % ?
Options: a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250
Rationale: "there are 396 managers and 4 others . the 4 others would compose 2 % of the total number of people if there were 200 people in the room . thus 200 managers must leave . the answer is d ."
Answer: d
Problem: each of the cucumbers in 100 pounds of cucumbers is composed of 99 % water , by weight . after some of the water evaporates , the cucumbers are now 97 % water by weight . what is the new weight of the cucumbers , in pounds ?
Options: a ) 2 , b ) 33 , c ) 92 , d ) 96 , e ) 98
Rationale: "out of 100 pounds 99 % or 99 pounds is water and 1 pound is non - water . after somewaterevaporates the cucumbers become 97 % water and 3 % of non - water , so now 1 pound of non - water composes 3 % of cucucmbers , which means that the new weight of cucumbers is 1 / 0.03 = 34 pounds . answer : b ."
Answer: b
Problem: the two trains of lengths 400 m , 600 m respectively , running at same directions . the faster train can cross the slower train in 180 sec , the speed of the slower train is 48 km . then find the speed of the faster train ?
Options: a ) 76 kmph , b ) 68 kmph , c ) 87 kmph , d ) 56 kmph , e ) 10 kmph
Rationale: length of the two trains = 600 m + 400 m speed of the first train = x speed of the second train = 48 kmph 1000 / x - 48 = 180 1000 / x - 48 * 5 / 18 = 180 50 = 9 x - 120 x = 68 kmph answer : b
Answer: b
Problem: a light flashes every 5 seconds , how many times will it flash in ? of an hour ?
Options: a ) 720 , b ) 600 , c ) 650 , d ) 700 , e ) 750
Rationale: "1 flash = 5 sec for 1 min = 12 flashes so for 1 hour = 12 * 60 = 720 flashes . answer : a"
Answer: a
Problem: what is the units digit of 3333 ^ ( 333 ) * 3333 ^ ( 222 ) ?
Options: a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8
Rationale: "each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits . here ' s another way to organize the information . we ' re given [ ( 2222 ) ^ 333 ] [ ( 3333 ) ^ 222 ] we can ' combine ' some of the pieces and rewrite this product as . . . . ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) [ ( 2222 ) ^ 111 ] ( 2222 ) ( 3333 ) = a big number that ends in a 6 taking a number that ends in a 6 and raising it to a power creates a nice pattern : 6 ^ 1 = 6 6 ^ 2 = 36 6 ^ 3 = 216 etc . thus , we know that ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) will be a gigantic number that ends in a 6 . 2 ^ 111 requires us to figure out thecycleof the units digit . . . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 so , every 4 powers , the pattern of the units digits repeats ( 2 , 4 , 8 , 6 . . . . . 2 , 4 , 8 , 6 . . . . ) . 111 = 27 sets of 4 with a remainder of 3 . . . . this means that 2 ^ 111 = a big number that ends in an 8 so we have to multiply a big number that ends in a 6 and a big number that ends in an 8 . ( 6 ) ( 8 ) = 48 , so the final product will be a gigantic number that ends in an 6 . final answer : d"
Answer: d
Problem: a 6 litre sol is 40 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ?
Options: a ) a . 0.6 , b ) b . 1.2 , c ) c . 2.1 , d ) d . 3 , e ) e . 5.4
Rationale: "40 % of 6 = 2.4 50 % of 6 = 3 shortage is 0.6 so we need to have 0.6 / 50 % to get 50 % alcohol content . = 1.2 b"
Answer: b
Problem: in how many ways can the integer 54 be expressed as a product of two different positive integers ?
Options: a ) 10 , b ) 8 , c ) 5 , d ) 4 , e ) 2
Rationale: "54 = ( 3 ^ 3 ) * 2 since 54 is not a perfect square , no of ways = 4 answer d"
Answer: d
Problem: find the number of different prime factors of 2400
Options: a ) 4 , b ) 2 , c ) 3 , d ) 5 , e ) 6
Rationale: "explanation : l . c . m of 2400 = 3 x 2 x 2 x 2 x 2 x 2 x 5 x 5 3 , 2 , 5 number of different prime factors is 3 . answer : option c"
Answer: c
Problem: in arun ' s opinion , his weight is greater than 63 kg but leas than 72 kg . his brother does not agree with arun and he thinks that arun ' s weight is greater than 60 kg but less than 70 kg . his mother ' s view is that his weight can not be greater than 66 kg . if all of them are correct in their estimation , what is the average of diferent probable weights of arun ?
Options: a ) 86.5 kg , b ) 64.5 kg , c ) 46.5 kg , d ) 26.5 kg , e ) 16.5 kg
Rationale: "let arun ' s weight be x kg . according to arun , 63 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 66 . the values satisfying all the above conditions are 64 and 65 . required average = ( 64 + 65 ) / 2 = 64.5 kg answer : b"
Answer: b
Problem: a cat leaps 5 leaps for every 4 leaps of a dog , but 2 leaps of the dog are equal to 3 leaps of the cat . what is the ratio of the speed of the cat to that of the dog ?
Options: a ) 5 : 6 , b ) 3 : 2 , c ) 4 : 8 , d ) 1 : 2 , e ) 7 : 8
Rationale: given ; 2 dog = 3 cat ; or , dog / cat = 3 / 2 ; let cat ' s 1 leap = 2 meter and dogs 1 leap = 3 meter . then , ratio of speed of cat and dog = 2 * 5 / 3 * 4 = 5 : 6 . ' ' answer : 5 : 6 ;
Answer: a
Problem: the percentage profit earned by selling an article for rs . 1920 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 15 % profit ?
Options: a ) 2000 , b ) 2778 , c ) 1840 , d ) 2778 , e ) 2771
Rationale: "let c . p . be rs . x . then , ( 1920 - x ) / x * 100 = ( x - 1280 ) / x * 100 1920 - x = x - 1280 2 x = 3200 = > x = 1600 required s . p . = 115 % of rs . 1600 = 115 / 100 * 1600 = rs . 1840 . answer : c"
Answer: c
Problem: a is an integer greater than 19 but less than 32 , b is an integer greater than 29 but less than 42 , what is the range of a / b ?
Options: a ) 1 / 4 , b ) 1 / 2 , c ) 3 / 4 , d ) 1 , e ) 5 / 4
Rationale: "min value of a / b will be when b is highest and a is lowest - - - > a = 20 and b = 40 so , a / b = 1 / 2 max value of a / b will be when b is lowest and a is highest - - - > a = 30 and b = 30 so , a / b = 1 range is 1 - ( 1 / 2 ) = 1 / 2 . answer should be b"
Answer: b
Problem: rs . 2500 is divided into two parts such that if one part be put out at 5 % simple interest and the other at 6 % , the yearly annual income may be rs . 125 . how much was lent at 5 % ?
Options: a ) 2333 , b ) 2777 , c ) 2688 , d ) 1000 , e ) 2500
Rationale: "( x * 5 * 1 ) / 100 + [ ( 2500 - x ) * 6 * 1 ] / 100 = 125 x = 2500 answer : e"
Answer: e
Problem: jerry travels 8 miles at an average speed of 40 miles per hour , stops for 13 minutes , and then travels another 20 miles at an average speed of 60 miles per hour . what is jerry ’ s average speed , in miles per hour , for this trip ?
Options: a ) 48 , b ) 42.5 , c ) 44 , d ) 50 , e ) 52.5
Rationale: "total time taken by jerry = ( 8 / 40 ) * 60 minutes + 13 minutes + ( 20 / 60 ) * 60 minutes = 35 minutes average speed = total distance / total time = ( 8 + 20 ) miles / ( 35 / 60 ) hours = 28 * 60 / 35 = 48 miles per hour answer : option a"
Answer: a
Problem: the sum of the first n positive perfect squares , where n is a positive integer , is given by the formula n ^ 3 / 3 + c * n ^ 2 + n / 6 , where c is a constant . what is the sum of the first 18 positive perfect squares ?
Options: a ) 1,010 , b ) 1,164 , c ) 1,240 , d ) 1,316 , e ) 2,109
Rationale: first we need to find the constant ' c ' . the easiest way to find this is for the sum of the first two perfect squares for 1 and 2 = 1 and 4 respectively . hence lhs = 1 + 4 and plug n = 2 for rhs and simplify to get c = 1 / 2 . plug values of n = 18 and c = 1 / 2 into the equation and simplify to get the answer 2109 . option e .
Answer: e
Problem: surface area of two spheres are in the ratio 1 : 4 what is the ratio of their volumes ?
Options: a ) 1 : 9
b ) 1 : 8
c ) 1 : 3
d ) 1 : 4
e ) 1 : 5
Rationale: 1 : 8 answer : b
Answer: b
Problem: if the area of a circle decreases by 30 % , then the radius of a circle decreases by
Options: a ) 20 % , b ) 16 % , c ) 36 % , d ) 64 % , e ) none of these
Rationale: "if area of a circle decreased by x % then the radius of a circle decreases by ( 100 − 10 √ 100 − x ) % = ( 100 − 10 √ 100 − 30 ) % = ( 100 − 10 √ 70 ) % = 100 - 84 = 16 % answer b"
Answer: b
Problem: two unbiased coins are tossed . what is the probability of getting at most one head ?
Options: a ) 2 / 3 , b ) 1 , c ) 3 / 4 , d ) 2 , e ) 1 / 2
Rationale: "s = { hh , tt , ht , th } e = event of getting at most one head . e = { tt , ht , th } . p ( e ) = n ( e ) / n ( s ) = 3 / 4 answer is option c"
Answer: c
Problem: the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 25 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ?
Options: a ) 347 , b ) 987 , c ) 1070 , d ) 1371 , e ) 1667
Rationale: "let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 25 = 2 r a = 12 / 5 r and r = 75 a = 180 required total cost = 4 * 180 + 3 * 75 + 5 * 25 = 720 + 225 + 125 = $ 1070 c"
Answer: c
Problem: in the graduating class of a certain college , 48 percent of the students are male and 52 percent are female . in this class 40 percent of the male and 20 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ?
Options: a ) 0.9 , b ) 0.7 , c ) 0.45 , d ) 0.3 , e ) 0.25
Rationale: "percent of students who are 25 years old or older is 0.4 * 48 + 0.2 * 52 = ~ 30 , so percent of people who are less than 25 years old is 100 - 30 = 70 . answer : b ."
Answer: b
Problem: a train 300 m long , running with a speed of 54 km / hr will pass a tree in ?
Options: a ) 17 sec , b ) 16 sec , c ) 20 sec , d ) 14 sec , e ) 12 sec
Rationale: "speed = 54 * 5 / 18 = 15 m / sec time taken = 300 * 1 / 15 = 20 sec answer : c"
Answer: c
Problem: a , b and c rents a pasture for rs . 899 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ?
Options: a ) 270 , b ) 279 , c ) 267 , d ) 255 , e ) 552
Rationale: "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 899 = 279 answer : b"
Answer: b
Problem: a student got twice as many sums wrong as he got right . if he attempted 27 sums in all , how many did he solve correctly ?
Options: a ) 12 , b ) 16 , c ) 18 , d ) 9 , e ) 12
Rationale: "explanation : suppose the boy got x sums right and 2 x sums wrong . then , x + 2 x = 27 3 x = 27 x = 9 . answer : d"
Answer: d
Problem: the compound interest on a sum for 2 years is rs . 832 and the simple interest on the same sum for the same period is rs . 800 . the difference between the compound and simple interest for 3 years will be
Options: a ) rs . 48 , b ) rs . 98.56 , c ) rs . 66.56 , d ) rs . 66.58 , e ) none of these
Rationale: explanation : given that simple interest for 2 years is rs . 800 i . e . , simple interest for 1 st year is rs . 400 and simple interest for 2 nd year is also rs . 400 compound interest for 1 st year will be 400 and compound interest for 2 nd year will be 832 - 400 = 432 you can see that compound interest for 2 nd year is more than simple interest for 2 nd year by 432 - 400 = rs . 32 i . e , rs . 32 is the interest obtained for rs . 400 for 1 year rate , r = 100 × si / pt = ( 100 × 32 ) / ( 400 × 1 ) = 8 % difference between compound and simple interest for the 3 rd year = simple interest obtained for rs . 832 = prt / 100 = ( 832 × 8 × 1 ) / 100 = rs . 66.56 total difference between the compound and simple interest for 3 years = 32 + 66.56 = rs . 98.56 answer : option b
Answer: b
Problem: when a merchant imported a certain item , he paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1000 . if the amount of the import tax that the merchant paid was $ 94.50 , what was the total value of the item ?
Options: a ) $ 2,350 , b ) $ 2,850 , c ) $ 3,250 , d ) $ 3,400 , e ) $ 3,750
Rationale: let x be the value in excess of $ 1,000 . 0.07 x = 94.5 x = $ 1,350 the total value was $ 1,350 + $ 1,000 = $ 2,350 . the answer is a .
Answer: a
Problem: the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 8463 for a computer table , then what was the cost price of the computer table ?
Options: a ) 6727 , b ) 6825 , c ) 6728 , d ) 6725 , e ) 2871
Rationale: ": cp = sp * ( 100 / ( 100 + profit % ) ) = 8463 ( 100 / 124 ) = rs . 6825 . answer : b"
Answer: b
Problem: 15 beavers , working together in a constant pace , can build a dam in 4 hours . how many hours will it take 20 beavers that work at the same pace , to build the same dam ?
Options: a ) 2 . , b ) 4 . , c ) 5 . , d ) 6 , e ) 3 .
Rationale: "total work = 15 * 4 = 60 beaver hours 20 beaver * x = 60 beaver hours x = 60 / 20 = 3 answer : e"
Answer: e
Problem: find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 14 cm
Options: a ) 178 cm 2 , b ) 179 cm 2 , c ) 266 cm 2 , d ) 167 cm 2 , e ) 197 cm 2
Rationale: "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 14 ) = 266 cm 2 answer : c"
Answer: c
Problem: the sum q of prime numbers that are greater than 60 but less than 70 is
Options: a ) 67 , b ) 128 , c ) 191 , d ) 197 , e ) 260
Rationale: a prime number is a number that has only two factors : 1 and itself . therefore , a prime number is divisible by two numbers only . let ' s list the numbers from 61 to 69 . 61 , 62 , 63 , 64 , 65 , 66 , 67 , 68 , 69 immediately we can eliminate the even numbers because they are divisible by 2 and thus are not prime . we are now left with : 61 , 63 , 65 , 67 , 69 we can next eliminate 65 because 65 is a multiple of 5 . we are now left with 61 , 63 , 67 , 69 . to eliminate any remaining values , we would look at those that are multiples of 3 . if you don ’ t know an easy way to do this , just start with a number that is an obvious multiple of 3 , such as 60 , and then keep adding 3 . we see that 60 , 63 , 66 , 69 are all multiples of 3 and therefore are not prime . thus , we can eliminate 63 and 69 from the list because they are not prime . finally , we are left with 61 and 67 , and we must determine whether they are divisible by 7 . they are not , and therefore they must be both prime . thus , the sum q of 61 and 67 is 128 . answer b .
Answer: b
Problem: at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 400 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ?
Options: a ) 1,108 , b ) 3,300 , c ) 2,108 , d ) 2,124 , e ) 2,256
Rationale: "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 400 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 400 x 8 = 3200 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 3300 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 3300 ( answer b )"
Answer: b
Problem: if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 990 , what is the least possible value of n ?
Options: a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
Rationale: "you need to primarily find the primes of 990 : 2 , 3,3 , 5,11 the integer must contain all these primes by knowing 11 is an prime , we know the answer can be b , c , d or e as these all contain 11 . 11 = 1 , 2,3 , 4,5 , 6,7 , 8,9 , 10,11 11 contains the 2 , 3,5 and 11 explicitly and the second 3 comes from the 6 ( 2 * 3 ) , therefore the answer is 11 . this is because we know a number that contains all the primes of 990 will be a multiple of 990 . answer : b"
Answer: b
Problem: the probability that a man will be alive for 10 more yrs is 1 / 2 & the probability that his wife will alive for 10 more yrs is 1 / 3 . the probability that none of them will be alive for 10 more yrs , is
Options: a ) 1 / 2 , b ) 1 , c ) 1 / 3 , d ) 3 / 4 , e ) 2
Rationale: "sol . required probability = pg . ) x p ( b ) = ( 1 — d x ( 1 — i ) = : x 1 = 1 / 3 ans . ( c )"
Answer: c
Problem: if a * b = 2 a - 3 b + ab , then 3 * 5 + 5 * 3 is equal to
Options: a ) 22 , b ) 25 , c ) 26 , d ) 28 , e ) 23
Rationale: 3 * 5 + 5 * 3 = ( 2 x 3 - 3 x 5 + 3 x 5 ) + ( 2 x 5 - 3 x 3 + 5 x 3 ) = 22 answer a 22
Answer: a
Problem: a special municipal payroll tax charges not tax on a payroll less than $ 200,000 and only 0.2 % on a company ’ s payroll above $ 200,000 . if belfried industries paid $ 200 in this special municipal payroll tax , then they must have had a payroll of ?
Options: a ) $ 180,000 , b ) $ 202,000 , c ) $ 220,000 , d ) $ 300,000 , e ) $ 2 , 200,000
Rationale: answer : d , ( with different approach ) : the 200 paid is 0.2 % of the additional amount above 200,000 . let it be x now 0.2 % of x = 200 therefore x = 100,000 total = 200,000 + x = 300,000
Answer: d
Problem: if 25 ^ 5 × 5 ^ ( - 1 ) = ( 125 ) ^ x , then what is the value of x ?
Options: a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
Rationale: 25 ^ 5 × 5 ^ ( - 1 ) = ( 125 ) ^ x ( 5 ^ 2 ) ^ 5 × 5 ^ ( - 1 ) = 5 ^ 3 x 5 ^ 10 x 5 ^ ( - 1 ) = 5 ^ 3 x ; since all of the bases are the same now , we can equate the exponents in the next step 10 - 1 = 3 x 9 = 3 x x = 3 ans . b ) 3
Answer: b
Problem: a lady starts from p towards q and realizes that at a point r , if he walks 50 km further he will be at a point s , which is as far away from r as it is from q . what is the distance between p and q if the distance between p and r is half the distance from r to q ? ( assume that p , q , r and s are all on the same straight line )
Options: a ) 150 km , b ) 200 km , c ) 250 km , d ) 125 km , e ) 155 km
Rationale: p ___ 50 _____ r ___ 50 _____ s ____ 50 ___ q the above figure gives the locations of p , r , s & q in relation to each other . answer : a
Answer: a
Problem: how many odd factors does 250 have ?
Options: a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8
Rationale: "start with the prime factorization : 250 = 2 * 5 * 7 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 , 1 } plus 1 to each = { 2 , 2 } product = 2 * 2 = 4 therefore , there are 4 odd factors of 250 . answer : b ."
Answer: b
Problem: in kaya ' s teacher ' s desk there are 24 pink highlighters , 28 yellow highlighters , and 25 blue highlighters . how many highlighters are there in all ?
Options: a ) 11 , b ) 22 , c ) 77 , d ) 33 , e ) 88
Rationale: "add the numbers of highlighters . 24 + 28 + 25 = 77 . answer is c ."
Answer: c
Problem: a train speeds past a pole in 10 seconds and a platform 50 m long in 20 seconds . its length is :
Options: a ) 30 m . , b ) 40 m . , c ) 60 m . , d ) 50 m . , e ) 70 m .
Rationale: "let the length of the train be x meters and its speed be y m / sec . they , x / y = 10 = > y = x / 10 x + 50 / 20 = x / 10 x = 50 m . answer : option d"
Answer: d
Problem: the sum of two numbers is 528 and their h . c . f is 33 . the number of pairs of numbers satisfying the above condition is
Options: a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8
Rationale: "let the required numbers be 33 a and 33 b . then 33 a + 33 b = 528 \ inline \ fn _ jvn \ rightarrow a + b = 16 . now , co - primes with sum 16 are ( 1,15 ) , ( 3,13 ) , ( 5,11 ) and ( 7,9 ) . \ inline \ fn _ jvn \ therefore required numbers are ( 33 x 1 , 33 x 15 ) , ( 33 x 3 , 33 x 13 ) , ( 33 x 5 , 33 x 11 ) , ( 33 x 7 , 33 x 9 ) the number of such pairs is 4 answer : a"
Answer: a
Problem: the average age of 50 students in a class is 10 years . if teacher ' s age is also included then average increases 1 year then find the teacher ' s age ?
Options: a ) 59 , b ) 55 , c ) 61 , d ) 45 , e ) 36
Rationale: "total age of 50 students = 50 * 10 = 500 total age of 51 persons = 51 * 11 = 561 age of teacher = 561 - 500 = 61 years answer is c"
Answer: c
Problem: the denominator of a fraction is 6 greater than the numerator . if the numerator and the denominator are increased by 1 , the resulting fraction is equal to 4 â  „ 5 . what is the value of the original fraction ?
Options: a ) 17 / 23 , b ) 23 / 29 , c ) 29 / 35 , d ) 31 / 37 , e ) 33 / 39
Rationale: "let the numerator be x . then the denominator is x + 6 . x + 1 / x + 7 = 4 / 5 . 5 x + 5 = 4 x + 28 . x = 23 . the original fraction is 23 / 29 . the answer is b ."
Answer: b
Problem: consider the sets tn = { n , n + 1 , n + 2 , n + 3 , n + 4 ) , where n = 1 , 2 , 3 , … , 96 . how many of these sets contain 6 or any integral multiple thereof ( i . e . , any one of the numbers 6 , 12 , 18 , … ) ?
Options: a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 84
Rationale: explanation : if n = 1 , then the set t 1 = { 1 , 23 , 45 } , and it does not have 6 or any multiples . n = 2 to n = 6 has 6 in the set . n = 7 , has the set t 7 = { 7 , 89 , 1011 } , and no 6 or multiples . so 1 in every 6 members do not have 6 or multiples of 6 . so , till n = 96 , there are 16 sets of “ 6 members ” ( 16 * 6 = 96 ) and 16 sets do not have 6 or its multiples , while the remaining 80 sets have . answer : a
Answer: a
Problem: car z travels 55 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour , but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour . how many miles does car z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per hour ?
Options: a ) 320 , b ) 375.2 , c ) 400 , d ) 408.3 , e ) 440
Rationale: the question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour . we therefore first calculate the fuel efficiency at that speed . the stem tells us that at 45 miles / hour , the car will run 55 miles / gallon and at 60 miles / hour , that distance decreases by 20 % . we can therefore conclude that the car will travel 44 miles / gallon at a constant speed of 60 miles / gallon . with 10 gallons of fuel , the car can therefore travel 44 miles / gallon * 10 gallons = 440 miles . answer e .
Answer: e
Problem: a certain farmer pays $ 70 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is 360 feet by 605 feet ? ( 43,560 square feet = 1 acre )
Options: a ) $ 5,330 , b ) $ 3,360 , c ) $ 1,350 , d ) $ 360 , e ) $ 1050
Rationale: basically the question an error . 1 acre = 43,560 square feet and if it is then the answer is 1050 ( e )
Answer: e
Problem: if a square mirror has a 20 - inch diagonal , what is the approximate perimeter t of the mirror , in inches ?
Options: a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120
Rationale: "if you draw the square and diagonal inside the square . u can see square becomes part of two triangles opposite to each other . and we know the property of the triangle , addition of two sides of triangle must be greater than its diagonal in order to complete the triangle . and each side must be less than 20 and perimeter t must be less than 80 , so we can eliminate answer choice c , d and e . so side 1 + side 2 > 20 , that means side 1 or side 2 must be > 10 . so we can eliminate the answer choice a . now we are left with is b"
Answer: b
Problem: a clock shows the time as 9 a . m . if the minute hand gains 6 minutes every hour , how many minutes will the clock gain by 6 p . m . ?
Options: a ) 30 min , b ) 35 min , c ) 45 min , d ) 50 min , e ) 54 min
Rationale: "there are 9 hours in between 9 a . m . to 6 p . m . 9 * 6 = 54 minutes . answer : e"
Answer: e
Problem: 20 people went to a hotel for combine dinner party 12 of them spent rs . 70 each on their dinner and rest spent 4 more than the average expenditure of all the 20 . what was the total money spent by them .
Options: a ) 1628.4 , b ) 1534 , c ) 1453 , d ) 1496 , e ) none of these
Rationale: "solution : let average expenditure of 20 people be x . then , 20 x = 12 * 70 + 8 * ( x + 4 ) ; or , 20 x = 12 * 70 + 8 x + 32 ; or , x = 72.667 ; so , total money spent = 72.67 * 20 = rs . 1453.4 . answer : option c"
Answer: c
Problem: how many integers k greater than 100 and less than 600 are there such that if the hundreds and the unit digits of k are reversed , the resulting integer is k + 99 ?
Options: a ) 40 , b ) 60 , c ) 70 , d ) 80 , e ) 90
Rationale: "not sure if this is the shortest . . but this is how i did this there are 8 sets of integers with hundreds and units digits exchanged that satisfies k + 99 . 1 . 102 | 201 ( satisfies k + 99 , where k = 102 ) 2 . 203 | 302 ( satisfies k + 99 , where k = 203 ) 3 . . . . 4 . 405 | 504 each set has 10 such numbers . 1 . 102 | 201 ( still k + 99 holds good ) 2 . 112 | 211 3 . 122 | 221 4 . 132 | 231 5 . . . . 6 . . . . 7 . . . . 8 . . . . 9 . 182 | 281 10 . 192 | 291 therefore , 4 sets with 10 such number in each set will give 4 x 10 = 40 integers . a"
Answer: a
Problem: the ratio of the two natural numbers is 5 : 6 . if a certain number is added to both the numbers , the ratio becomes 7 : 8 . if the larger number exceeds the smaller number by 10 , find the number added ?
Options: a ) 17 , b ) 14 , c ) 10 , d ) 16 , e ) 20
Rationale: "let the two numbers be 5 x and 6 x . let the numbers added to both so that their ratio becomes 7 : 8 be k . ( 5 x + k ) / ( 6 x + k ) = 7 / 8 = > 40 x + 8 k = 42 x + 7 k = > k = 2 x . 6 x - 5 x = 10 = > x = 10 k = 2 x = 20 . answer : c"
Answer: c