regisss/math_qa huggingface dataset:
Problem: what least number must be subtracted from 9679 so that the remaining number is divisible by 15 ?
Options: a ) 3 , b ) 1 , c ) 16 , d ) 4 , e ) 15
Rationale: "on dividing 9679 by 15 , we get remainder = 4 . required number be subtracted = 4 answer : d"
Answer: d
Problem: the ages of two person differ by 20 years . if 8 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively
Options: a ) 33 , 13 , b ) 25 , 5 , c ) 29 , 9 , d ) 50 , 30 , e ) 20,10
Rationale: let their ages be x and ( x + 20 ) years . then , 5 ( x - 8 ) = ( x + 20 - 8 ) = > 4 x = 52 = > x = 13 their present ages are 33 years and 13 year . answer : a
Answer: a
Problem: a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ?
Options: a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 %
Rationale: "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is = . 25 / 1 * 100 = 25 % d is the answer ."
Answer: d
Problem: the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 17 . find the sum :
Options: a ) 4500 , b ) 7500 , c ) 5000 , d ) 6800 , e ) none of these
Rationale: sol . ( d ) let the sum be 100 . therefore , si = 100 × 5 × 2100 = 10100 × 5 × 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 − 100100 ( 1 + 5100 ) 2 − 100 ∴ = 100 × 21 × 2120 × 20 − 100 = 414 = 100 × 21 × 2120 × 20 − 100 = 414 difference of ci and si = 41 ⁄ 4 - 10 = 1 ⁄ 4 if the difference is 1 ⁄ 4 , the sum = 100 = > if the difference is 17 , the sum = 400 × 17 = 6800 answer d
Answer: d
Problem: a train speeds past a pole in 20 seconds and a platform 100 m long in 25 seconds . its length is ?
Options: a ) 188 m , b ) 876 m , c ) 251 m , d ) 150 m , e ) 400 m
Rationale: "let the length of the train be x meters and its speed be y m / sec . they , x / y = 20 = > y = x / 20 x + 100 / 25 = x / 20 x = 400 m . answer : e"
Answer: e
Problem: a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 850 in 4 years . the sum is ?
Options: a ) s . 738 , b ) s . 750 , c ) s . 650 , d ) s . 730 , e ) s . 735
Rationale: "s . i . for 1 year = ( 850 - 825 ) = rs . 25 s . i . for 3 years = 25 * 3 = rs . 75 principal = ( 825 - 75 ) = rs . 750 . answer : b"
Answer: b
Problem: the difference between a number and its three - fourth is 100 . what is 25 % of that number ?
Options: a ) 120 , b ) 100 , c ) 125 , d ) 150 , e ) none of these
Rationale: "explanation : solution : let the number be x . then , x - 3 / 4 x = 100 = > x / 4 = 100 = > x = 400 . 25 % of x = 25 % of 400 = 25 / 100 * 400 = 100 . answer : b"
Answer: b
Problem: the average of 1 st 3 of 4 numbers is 16 and of the last 3 are 15 . if the sum of the first and the last number is 11 . what is the last numbers ?
Options: a ) 2 , b ) 4 , c ) 6 , d ) 5 , e ) 7
Rationale: "a + b + c = 48 b + c + d = 45 a + d = 11 a – d = 3 a + d = 11 2 d = 8 d = 4 answer b"
Answer: b
Problem: john left home and drove at the rate of 45 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 50 mph to reach his destination . how many miles did john drive ?
Options: a ) 235 miles . , b ) 245 miles . , c ) 240 miles . , d ) 265 miles . , e ) 275 miles .
Rationale: the total distance d traveled by john is given by d = 45 * 2 + 3 * 50 = 240 miles . answer c
Answer: c
Problem: what least number must be subtracted from 3830 so that the remaining number is divisible by 15 ?
Options: a ) 3 , b ) 1 , c ) 5 , d ) 11 , e ) 15
Rationale: "on dividing 3830 by 15 , we get remainder = 5 . required number be subtracted = 5 answer : c"
Answer: c
Problem: in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what isthe probability that theextracted ball would turn on to be white , if all possible hypothesis concerning the color of the balls that initially in the bag were equally possible ?
Options: a ) 2 / 3 , b ) 5 / 8 , c ) 3 / 5 , d ) 4 / 7 , e ) 5 / 8
Rationale: "since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . ∴ required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b"
Answer: b
Problem: find a sum for 1 st 8 prime number ' s ?
Options: a ) 25 , b ) 28 , c ) 30 , d ) 77 , e ) 36
Rationale: "required sum = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 ) = 77 note : 1 is not a prime number option d"
Answer: d
Problem: if 1,000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ?
Options: a ) 100 , b ) 10 , c ) 1,000 , d ) 10,000 , e ) 100,000
Rationale: "given that 1,000 microns = 1 decimeter = 1 , 000,000 angstroms so , 1 micron = 1 , 000,000 / 1,000 = 1,000 answer : c"
Answer: c
Problem: a flagpole 18 meters high casts a shadow of length 45 meters . if a building under similar conditions casts a shadow of length 50 meters , what is the height of the building ( in meters ) ?
Options: a ) 16 , b ) 20 , c ) 24 , d ) 28 , e ) 32
Rationale: "the height : length ratio will be equal in both cases . 18 / 45 = x / 50 x = 20 the answer is b ."
Answer: b
Problem: if 7 ^ ( x - y ) = 343 and 7 ^ ( x + y ) = 16807 , what is the value of x ?
Options: a ) 3 , b ) 4 , c ) 2 , d ) 1 , e ) 7
Rationale: 7 ^ ( x - y ) = 343 = 7 ^ 3 = > x - y = 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - ( equation 1 ) 7 ^ ( x + y ) = 16807 = 7 ^ 5 = > x + y = 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - ( equation 2 ) ( equation 1 ) + ( equation 2 ) = > 2 x = 3 + 5 = 8 = > x = 8 / 2 = 4 answer is b
Answer: b
Problem: 242 students represent x percent of the boys at a school . if the boys at the school make up 50 % of the total school population of x students , what is x ?
Options: a ) 120 , b ) 150 , c ) 180 , d ) 200 , e ) 220
Rationale: "let b be the number of boys in the school . 242 = xb / 100 b = 0.5 x 24200 = 0.5 x ^ 2 x ^ 2 = 48400 x = 220 the answer is e ."
Answer: e
Problem: a certain number of workers can do a work in 75 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ?
Options: a ) 65 , b ) 30 , c ) 28 , d ) 24 , e ) 32
Rationale: "number of workers = 10 * ( 75 - 10 ) / 10 = 65 answer is a"
Answer: a
Problem: a man spends 2 / 5 of his salary on house rent , 3 / 10 of his salary on food and 1 / 8 of his salary on conveyance . if he has $ 1400 left with him , find his expenditure on food and conveyance .
Options: a ) $ 8600 , b ) $ 2400 , c ) $ 1000 , d ) $ 3000 , e ) $ 3400
Rationale: remaining part of salary left = 1 - ( 2 / 5 + 3 / 10 + 1 / 8 ) = 1 - ( 33 / 40 ) = 7 / 40 let the monthly salary be $ x then , 7 / 40 of x = $ 1400 x = ( 1400 * 40 / 7 ) = $ 8600 expenditure on food = $ ( 3 / 10 * 800 ) = $ 2400 expenditure on conveyance = $ ( 1 / 8 * 8000 ) = $ 1000 $ 2400 + $ 1000 = $ 3400 so the answer is option e ) $ 3400
Answer: e
Problem: some of 15 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 30 % . what fraction of the original paint was replaced ?
Options: a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 3 / 2 , e ) 3 / 5
Rationale: "let total paint = 1 let amount replaced = x 15 ( 1 - x ) + 25 x = 30 x = 3 / 2 answer : d"
Answer: d
Problem: find the compound interest on $ 1200 for 2 years at 20 % p . a . if ci is component yearly ?
Options: a ) $ 120 , b ) $ 150 , c ) $ 548 , d ) $ 250 , e ) $ 300
Rationale: a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 2 = $ 1728 ci = $ 548 answer is c
Answer: c
Problem: it takes avery 2 hours to build a brick wall while tom can do it in 4 hour . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ?
Options: a ) 70 , b ) 20 , c ) 50 , d ) 60 , e ) 30
Rationale: avery takes 2 hours tom takes 4 hours efficiency of avery is 1 / 2 units / hr efficiency of tom is 1 / 4 units / hr combined efficiency of tom and avery is 1 / 2 + 1 / 4 = 3 / 4 units / hr since they worked for 1 hour they completed 3 / 4 units of work and 1 / 4 units of work is left which is to be completed by tom ( since avery left ) so time taken by tom to complete the remaining work will be 1 / 4 / 1 / 4 hours = > 1 * 60 = 60 minutes . . . answer will be ( d )
Answer: d
Problem: how many of the integers between 25 and 45 are even ?
Options: a ) 21 , b ) 20 , c ) 11 , d ) 10 , e ) 9
Rationale: number start between 25 to 45 is 20 numbers half of them is even . . which is 10 answer : d
Answer: d
Problem: bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 28 more . what was his original selling price ?
Options: a ) $ 200 , b ) $ 260 , c ) $ 320 , d ) $ 380 , e ) $ 440
Rationale: "let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 28 0.07 = 28 p = 400 the original selling price was 400 * 1.1 = 440 . the correct answer is e ."
Answer: e
Problem: if 0.4 of a number is equal to 0.07 of another number , the ratio of the numbers i
Options: a ) 2 : 3 , b ) 7 : 40 , c ) 3 : 20 , d ) 20 : 3 , e ) none
Rationale: "sol . 0.4 a = 0.08 b â ‡ ” a / b = 0.07 / 0.40 = 7 / 40 . â ˆ ´ a : b = 7 : 40 . answer b"
Answer: b
Problem: sixty 5 percent of a number is 21 less than 4 fifth of that number . what is the number
Options: a ) 140 , b ) 165 , c ) 150 , d ) 142 , e ) 174
Rationale: let the number be x . then , 4 * x / 5 – ( 65 % of x ) = 21 4 x / 5 – 65 x / 100 = 21 5 x = 2100 x = 140 . answer a 140
Answer: a
Problem: a wooden cube whose edge length is 7 inches is composed of smaller cubes with edge lengths of one inch . the outside surface of the large cube is painted red and then it is split up into its smaller cubes . if one cube is randomly selected from the small cubes , what is the probability that the cube will have at least one red face ?
Options: a ) 59.8 % , b ) 63.6 % , c ) 67.2 % , d ) 70.5 % , e ) 73.6 %
Rationale: "there are a total of 7 * 7 * 7 = 343 cubes . all the exterior cubes will have at least one face painted red . the interior is formed by 5 * 5 * 5 = 125 cubes . the number of cubes with at least one side painted red is 343 - 125 = 218 cubes the probability that a cube has at least one side painted red is 218 / 343 which is about 63.6 % the answer is b ."
Answer: b
Problem: two numbers a and b are such that the sum of 7 % of a and 4 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b .
Options: a ) 2 : 1 , b ) 1 : 2 , c ) 4 : 9 , d ) 1 : 1 , e ) 3 : 2
Rationale: "explanation : 7 % of a + 4 % of b = 2 / 3 ( 6 % of a + 8 % of b ) 7 a / 100 + 4 b / 100 = 2 / 3 ( 6 a / 100 + 8 b / 100 ) ⇒ 7 a + 4 b = 2 / 3 ( 6 a + 8 b ) ⇒ 21 a + 12 b = 12 a + 16 b ⇒ 9 a = 4 b ⇒ ab = 4 / 9 ⇒ a : b = 4 : 9 answer : option c"
Answer: c
Problem: the average weight of a class is x pounds . when a new student weighing 50 pounds joins the class , the average decreases by 1 pound . in a few months the student ’ s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students ’ weights changed . what is the value of x ?
Options: a ) 85 , b ) 86 , c ) 88 , d ) 90 , e ) 92
Rationale: "when the student weighs 80 pounds the average weight is x - 1 pounds ; when the student weighs 110 pounds the average weight is x + 4 pounds . so , the increase in total weight of 110 - 80 = 30 pounds corresponds to the increase in average weight of ( x + 4 ) - ( x - 1 ) = 5 pounds , which means that there are 30 / 5 = 6 students ( including the new one ) . so , initially there were 5 student . total weight = 5 x + 80 = 6 ( x - 1 ) - - > x = 85 pounds . answer : a ."
Answer: a
Problem: the speed of a car is 140 km in the first hour and 40 km in the second hour . what is the average speed of the car ?
Options: a ) 72 kmph , b ) 75 kmph , c ) 30 kmph , d ) 90 kmph , e ) 82 kmph
Rationale: "s = ( 140 + 40 ) / 2 = 90 kmph answer : d"
Answer: d
Problem: a train of 50 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ?
Options: a ) 6 , b ) 3 , c ) 4 , d ) 4.5 , e ) 2
Rationale: "d = 50 * 60 + 1500 = 3000 m t = 4500 / 60 * 18 / 5 = 270 sec = 4.5 mins answer : d"
Answer: d
Problem: how much space , in cubic units , is left vacant when maximum number of 5 x 5 x 5 cubes are fitted in a rectangular box measuring 10 x 15 x 18 ?
Options: a ) 50 , b ) 150 , c ) 250 , d ) 350 , e ) 450
Rationale: "no of cubes that can be accommodated in box = ( 10 * 15 * 18 ) / ( 5 * 5 * 5 ) 15 * 10 in numerator can be perfectly divided by 5 * 5 in denominator . side with length 18 ca n ' t be perfectly divided by 5 and hence is the limiting factor . closet multiple of 5 less that 18 is 15 . so vacant area in cube = = 15 * 10 * ( 18 - 15 ) = 15 * 10 * 3 = 450 ans - e"
Answer: e
Problem: 12 buckets of water fill a tank when the capacity of each bucket is 49 litres . how many buckets will be needed to fill the same tank , if the capacity of each bucket is 7 litres ?
Options: a ) 39 , b ) 84 , c ) 80 , d ) 82 , e ) 86
Rationale: capacity of the tank = ( 12 ã — 49 ) litre number of buckets required of capacity of each bucket is 17 litre = 12 ã — 49 / 7 = 12 ã — 7 = 84 answer is b
Answer: b
Problem: if n is an integer and 101 n ^ 2 is less than or equal to 8100 , what is the greatest possible value of n ?
Options: a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
Rationale: "101 * n ^ 2 < = 8100 n ^ 2 < = 8100 / 101 which will be less than 81 since 8100 / 100 = 81 which is the square of 9 next closest value of n where n ^ 2 < = 81 is 8 ans b"
Answer: b
Problem: if the perimeter of a rectangular garden is 950 m , its length when its breadth is 100 m is ?
Options: a ) 338 m , b ) 778 m , c ) 375 m , d ) 276 m , e ) 971 m
Rationale: "2 ( l + 100 ) = 950 = > l = 375 m answer : c"
Answer: c
Problem: a certain league has two divisions . the respective divisions had 11 and 13 teams qualify for the playoffs . each division held its own double - elimination tournament - - where a team is eliminated from the tournament upon losing two games - - in order to determine its champion . the 4 division champions then played in a single - elimination tournament - - where a team is eliminated upon losing one game - - in order to determine the overall league champion . assuming that there were no ties and no forfeits , what is the maximum number of games that could have been played in order to determine the overall league champion ?
Options: a ) 41 , b ) 42 , c ) 45 , d ) 47 , e ) 49
Rationale: overall , there can be 47 minuses . tus , it is d .
Answer: d
Problem: if the ratio of a to b is 5 to 3 and the ratio of b to c is 1 to 5 , what is the ratio of a to c ?
Options: a ) 4 / 15 , b ) 1 / 3 , c ) 2 / 5 , d ) 4 / 5 , e ) 7 / 6
Rationale: "a : b = 5 : 3 - - 1 b : c = 1 : 5 = > b : c = 3 : 15 - - 2 from 1 and 2 , we get a : c = 4 : 15 answer b"
Answer: b
Problem: it takes printer a 4 more minutes more than printer b to print 40 pages . working together , the two printers can print 50 pages in 6 minutes . how long will it take printer a to print 100 pages ?
Options: a ) 12 , b ) 18 , c ) 20 , d ) 24 , e ) 30
Rationale: "if it takes 4 more minutes for a to print 40 pages than it takes b , it takes 5 more minutes for a to print 50 pages than it takes b . thus if b is the number of minutes than b takes to print 50 pages , we can write : 1 / b + 1 / ( b + 5 ) = 1 / 6 ( since in 1 minute , they print 1 / 6 th of the 50 page job ) 6 ( 2 b + 5 ) = b ( b + 5 ) b ^ 2 - 7 b - 30 = 0 ( b - 10 ) ( b + 3 ) = 0 b = 10 thus it takes a 15 minutes to print 50 pages and 15 * 100 / 50 = 30 minutes to print 100 pages ( answer e )"
Answer: e
Problem: what does the hexa number e 78 in radix 7 .
Options: a ) 12455 , b ) 14153 , c ) 14256 , d ) 13541 , e ) 14541
Rationale: conversion of hexa decimal to decimal ( e 78 ) = ( 8 * 16 ^ 0 ) + ( 7 * 16 ^ 1 ) + ( 14 * 16 ^ 2 ) = 8 + 112 + 3584 = 3704 conversion of decimal to radix 7 3704 / 7 : reminder : 1 and dividend : 529 529 / 7 : reminder : 4 and dividend : 75 75 / 7 : reminder : 5 and dividend : 10 10 / 7 : reminder : 3 and dividend : 1 1 / 7 : reminder 1 then finally result is : 13541 answer : d
Answer: d
Problem: rs . 1170 is divided so that 4 times the first share , thrice the 2 nd share and twice the third share amount to the same . what is the value of the third share ?
Options: a ) rs . 720 , b ) rs . 920 , c ) rs . 537 , d ) rs . 540 , e ) rs . 637
Rationale: a + b + c = 1170 4 a = 3 b = 2 c = x a : b : c = 1 / 4 : 1 / 3 : 1 / 2 = 3 : 4 : 6 6 / 13 * 1170 = rs . 540 answer : d
Answer: d
Problem: how much is 55 % of 40 is greater than 4 / 5 of 25 ?
Options: a ) 29 , b ) 776 , c ) 2 , d ) 12 , e ) 99
Rationale: "( 55 / 100 ) * 40 – ( 4 / 5 ) * 25 22 - 20 = 2 answer : c"
Answer: c
Problem: find large number from below question the difference of two numbers is 1515 . on dividing the larger number by the smaller , we get 16 as quotient and the 15 as remainder
Options: a ) 1209 , b ) 1615 , c ) 1245 , d ) 1300 , e ) 1635
Rationale: "let the smaller number be x . then larger number = ( x + 1515 ) . x + 1515 = 16 x + 15 15 x = 1500 x = 100 large number = 100 + 1515 = 1615 answer : b"
Answer: b
Problem: the price of a t . v . set worth rs . 60000 is to be paid in 20 installments of rs . 1000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ?
Options: a ) 22678 , b ) 59000 , c ) 26788 , d ) 19000 , e ) 26711
Rationale: "money paid in cash = rs . 1000 balance payment = ( 60000 - 1000 ) = rs . 59000 answer : b"
Answer: b
Problem: p and q can complete a work in 15 days and 10 days respectively . they started the work together and then q left after 2 days . p alone completed the remaining work . the work was finished in - - - days .
Options: a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28
Rationale: "explanation : work done by p in 1 day = 1 / 15 work done by q in 1 day = 1 / 10 work done by p and q in 1 day = 1 / 15 + 1 / 10 = 1 / 6 work done by p and q in 2 days = 2 × ( 1 / 6 ) = 1 / 3 remaining work = 1 – 1 / 3 = 2 / 3 time taken by p to complete the remaining work 2 / 3 = ( 2 / 3 ) / ( 1 / 15 ) = 10 days total time taken = 2 + 10 = 12 days answer : option a"
Answer: a
Problem: a rectangular grass field is 65 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ?
Options: a ) 1350 , b ) 1971 , c ) 9676 , d ) 1679 , e ) 1250
Rationale: "area = ( l + b + 2 d ) 2 d = ( 65 + 55 + 2.5 * 2 ) 2 * 2.5 = > 625 625 * 2 = rs . 1250 answer : e"
Answer: e
Problem: how many liters of water must be evaporated from 50 liters of a 3 percent sugar solution to get a 5 percent sugar solution ?
Options: a ) 2 , b ) 4 , c ) 6 , d ) 10 , e ) 20
Rationale: "3 % sugar solution means that currently there are 1.5 liters of sugar and this amount must be 5 % of a reduced final amount ( 50 - x ) 1.5 = 0.05 * ( 50 - x ) 0.05 x = 1 x = 20 answer e"
Answer: e
Problem: the mall charges 40 cents for the first hour of parking and $ 3 for each additional hour until the customer reaches 4 hours , after that the parking fee is $ 2 per hour . if mr . park parked his car in the mall for 7 hours and 30 minutes , how much is he going to pay ?
Options: a ) $ 6.8 , b ) $ 12 . , c ) $ 13 . , d ) $ 14.5 . , e ) $ 15 .
Rationale: "0.40 + 2 * 2 + 1 * 2 + 0.40 = $ 6.8 answer : $ 6.8"
Answer: a
Problem: david spends $ 350 buying his favorite dolls . if he buys only small lemonhead dolls , which are $ 2 cheaper than the large lemonhead dolls , he could buy 20 more dolls than if he were to buy only large lemonhead dolls . how much does a large lemonhead doll cost ?
Options: a ) $ 5 , b ) $ 7 , c ) $ 6 , d ) $ 8 , e ) $ 9
Rationale: if large doll costs $ 5 , then he can buy 350 / 5 = 70 large dolls and 350 / 3 = 116 small dolls . difference is more than 20 . discard . if large doll costs $ 7 , then he can buy 350 / 7 = 50 large dolls and 350 / 5 = 70 small dolls . difference is 20 , which is we wanted . answer b .
Answer: b
Problem: ratio of ages of 3 persons is 4 : 7 : 9 , 8 years ago , the sum of their ages was 56 . find their present ages
Options: a ) 16 , 35,36 , b ) 12 , 28,36 , c ) 16 , 28,27 , d ) 16 , 28,36 , e ) none of these
Rationale: explanation : let the present ages are 4 x , 7 x , 9 x . = > ( 4 x - 8 ) + ( 7 x - 8 ) + ( 9 x - 8 ) = 56 = > 20 x = 80 = > x = 4 so their present ages are : 16 , 28,36 answer : option d
Answer: d
Problem: if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 12,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years later , when luke is ready for college ?
Options: a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 100,000
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: if 12 ! / 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: "this question is asking how many power of 3 exists in 12 ! easiest way is to identify factor of three 12 - > 4 * 3 = > 1 power of 3 9 - > 3 * 3 = > 2 power of 3 6 - > 2 * 3 = > 1 power of 3 3 - > 1 * 3 = > 1 power of 3 total equal 5 answer : c"
Answer: c
Problem: arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 13 from the product of their ages a year from today . what is their age today ?
Options: a ) 2 . , b ) 6 . , c ) 5 . , d ) 7 . , e ) 9 .
Rationale: ad = ( a + 1 ) ( d + 1 ) - 13 0 = a + d - 12 a + d = 12 a = d ( as they are twin brothers ) a = d = 6 b is the answer
Answer: b
Problem: a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 350 bolts ?
Options: a ) 84 , b ) 72 , c ) 120 , d ) 144 , e ) 180
Rationale: old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 350 bolts 350 * 6 / 25 = 84 mins ans a
Answer: a
Problem: what is the minimum possible range in scores of the 3 test - takers ? 3 people each took 5 tests . if the ranges of their scores in the 5 practice tests were 18 , 26 and 32 .
Options: a ) 26 , b ) 32 , c ) 18 , d ) 20 , e ) 30
Rationale: i simply looked at the 3 different possible scores for each individual test : 18 , 32,26 we have to find the minimum range : 32 - 18 = 14 32 - 26 = 6 26 - 18 = 8 the find the minimum range , you have to make the set of the 5 scores as small as possible . which means that 4 of the 5 scores of each individual person is zero . 6 * 5 = 30 answer : e
Answer: e
Problem: a leak in the bottom of a tank can empty the full tank in 8 hours . an inlet pipe fills water at the rate of 6 litres a minute . when the tank is full , the inlet is opened and due to the leak , the tank is empty in 12 hours . how many litres does the cistern hold ?
Options: a ) 7580 , b ) 7960 , c ) 8290 , d ) 8640 , e ) none
Rationale: "solution work done by the inlet in 1 hour = ( 1 / 8 - 1 / 12 ) = 1 / 24 . work done by the inlet in 1 min . = ( 1 / 24 × 1 / 60 ) = 1 / 1440 . volume of 1 / 1440 part = 6 litres . therefore , volume of whole = ( 1440 × 6 ) ‹ = › 8640 litres . answer d"
Answer: d
Problem: how many hours in one year ?
Options: a ) 365 , b ) 52 , c ) 1024 , d ) 8760 , e ) no one knows
Rationale: the day has 24 hours , and the year contains 365 days . so the hours exist in one year is the multiplication of 24 by 365 . so the answer = 365 * 24 = 8760 hours so the correct answer is d
Answer: d
Problem: if k is a non - negative integer and 15 ^ k is a divisor of 759,325 then 3 ^ k - k ^ 3 =
Options: a ) 0 , b ) 1 , c ) 37 , d ) 118 , e ) 513
Rationale: "at first we should understand whether 759325 divisible by 15 . 15 is equal to 3 * 5 so 759325 should be divisible on these integers 759325 is divisible by 5 because of last digit 5 but not divisible by 3 because sum of its digits does n ' t divisible by 3 : 7 + 5 + 9 + 3 + 2 + 5 = 31 and 31 does n ' t divisible by 3 so 759325 can be divisible by 15 ^ k only if k = 0 and 15 ^ k = 1 so 3 ^ k - k ^ 3 = 1 - 0 = 1 answer is b"
Answer: b
Problem: when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 3 times , what is the probability that it will land once on heads and twice tails ?
Options: a ) 1 / 3 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 5
Rationale: "must be once on heads and twice on tails 1 / 2 * 1 / 2 * 1 / 2 = 1 / 8 answer : d"
Answer: d
Problem: what is the smallest integer that is multiple of 7 , 9 and 20
Options: a ) 141 , b ) 180 , c ) 130 , d ) 1260 , e ) 1420
Rationale: "correct answer : d it is the lcm of 7 , 9 and 20 which is 1260"
Answer: d
Problem: if a man lost 4 % by selling oranges at the rate of 36 a rupee at how many a rupee must he sell them to gain 44 % ?
Options: a ) 1 , b ) 8 , c ) 9 , d ) 24 , e ) 3
Rationale: "96 % - - - - 36 144 % - - - - ? 96 / 144 * 36 = 24 answer : d"
Answer: d
Problem: the fisherman sale , all of the prices of the fishes sold were different . if the price of a radio sold at the fisherman sale was both the 4 th highest price and the 13 th lowest price among the prices of the fishes sold , how many fishes were sold at the fisherman sale ?
Options: a ) 13 , b ) 14 , c ) 15 , d ) 41 , e ) 16
Rationale: 3 + 12 + 1 = 16 answer : e
Answer: e
Problem: in a school of 600 students , 45 % wear blue shirts , 23 % wear red shirts , 15 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ?
Options: a ) 102 , b ) 112 , c ) 122 , d ) 132 , e ) 142
Rationale: "45 + 23 + 15 = 83 % 100 – 83 = 17 % 600 * 17 / 100 = 102 the answer is a ."
Answer: a
Problem: the average of the marks of 10 students in a class is 80 . if the marks of each student are doubled , find the new average ?
Options: a ) 80 , b ) 120 , c ) 160 , d ) 270 , e ) 110
Rationale: "sum of the marks for the 10 students = 10 * 80 = 800 . the marks of each student are doubled , the sum also will be doubled . the new sum = 800 * 2 = 1600 . so , the new average = 1600 / 10 = 160 . answer : c"
Answer: c
Problem: in a forest 250 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ?
Options: a ) 150 , b ) 750 , c ) 1,250 , d ) 1,500 , e ) 2,500
Rationale: "the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 250 tagged deers comprise 10 % of total # of deers - - > total # of deers = 250 * 10 = 2,500 . answer : e"
Answer: e
Problem: the area of a sector of a circle of radius 5 cm formed by an arc of length 5.5 cm is ?
Options: a ) 13.78 , b ) 13.67 , c ) 13.75 , d ) 13.98 , e ) 13.28
Rationale: "( 5 * 5.5 ) / 2 = 13.75 answer : c"
Answer: c
Problem: jane makes toy bears . when she works with an assistant , she makes 60 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ?
Options: a ) 70 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 %
Rationale: "we can use fractional equivalents here to solve the problem 80 % = 4 / 5 ; this means that in 1 st case if she prepares 5 bears , in 2 nd case she prepares 9 bears 10 % = 1 / 10 ; this means that in 1 st case if she needs 10 hours , in 2 nd case she needs 9 hours now we come to productivity based on above fractional values the productivity in 1 st case is 0.5 bears / hour and in the 2 nd case it is 1 bear / hour hence the productivity is double with the assistant i . e . the increase in productivity is 70 % a"
Answer: a
Problem: there were totally 100 men . 83 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ?
Options: a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15
Rationale: 100 - ( 100 - 83 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 17 - 25 - 15 - 30 = 100 - 87 = 13 answer : c
Answer: c
Problem: a 450 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec . what is the length of the platform ?
Options: a ) 600 , b ) 525 , c ) 360 , d ) 370 , e ) 380
Rationale: "speed = 450 / 18 = 25 m / sec . let the length of the platform be x meters . then , ( x + 450 ) / 39 = 25 = > x = 975 m . l = 975 - 450 = 525 answer : option b"
Answer: b
Problem: the distance between west - town to east - town is 15 kilometers . two birds start flying simultaneously towards one another , the first leaving from west - town at a speed of 4 kilometers per minute and the second bird , leaving from east - town , at a speed of 1 kilometers per minute . what will be the distance , in kilometers , between the meeting point and west - town ?
Options: a ) 3 . , b ) 7 . , c ) 10 . , d ) 12 , e ) 15 .
Rationale: "by end of 1 st min : distance covered by 1 st bird is 4 km and 2 nd is 1 km . total distance = 5 km . by end of 2 nd min : distance covered by 1 st bird is 8 km and 2 nd is 2 km . total distance = 10 km . by end of 3 rd min : distance covered by 1 st bird is 12 km and 2 nd is 3 km . total distance = 15 km . so , meeting point is 12 kms from west - town ( from where 1 st bird started ) hence , answer will be d ."
Answer: d
Problem: if an integer n is to be chosen at random from the integers 1 to 94 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ?
Options: a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 59 / 94
Rationale: "n ( n + 1 ) ( n + 2 ) will be divisible by 8 when n is a multiple of 2 or when ( n + 1 ) is a multiple of 8 . thus when n is even , this whole expression will be divisible by 8 . from 1 to 96 , there are 47 even integers . now when ( n + 1 ) is multiple by 8 , we have 12 such values for ( n + 1 ) probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 = ( 47 + 12 ) / 94 = 59 / 94 = 5 / 8 ans is e"
Answer: e
Problem: what is x if x + 2 y = 10 and y = 1 ?
Options: a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2
Rationale: x = 10 - 2 y x = 10 - 2 . x = 8 answer : b
Answer: b
Problem: what is the median of a set of consecutive integers if the sum of nth number from the beginning and nth number from the end is 50 ?
Options: a ) 10 , b ) 25 , c ) 50 , d ) 75 , e ) 100
Rationale: "surprisingly no one answered this easy one . property of a set of consecutive integerz . mean = median = ( first element + last element ) / 2 = ( second element + last but one element ) / 2 = ( third element + third last element ) / 2 etc . etc . so mean = median = 50 / 2 = 25 answer is b"
Answer: b
Problem: set j consists of 15 consecutive even numbers . if the smallest term in the set is - 10 , what is the range of the positive integers in set j ?
Options: a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 20
Rationale: "since there are only 15 integers , another approach is the just list all 15 . we get : - 10 , - 8 , - 6 , - 4 , - 2 , 0,2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 range of positive integers = 18 - 2 = 16 answer : d"
Answer: d
Problem: a number is selected at random from the first 40 natural numbers . what is the probability that the number is a multiple of either 2 or 21 ?
Options: a ) 21 / 40 , b ) 2 / 5 , c ) 7 / 15 , d ) 4 / 15 , e ) 11 / 30
Rationale: "number of multiples of 2 from 1 through 40 = 40 / 2 = 20 number of multiples of 21 from 1 through 40 = 40 / 21 = 1 number of multiples of 2 and 21 both from 1 through 40 = number of multiples of 21 * 2 ( = 42 ) = 0 total favourable cases = 20 + 1 - 0 = 21 probability = 21 / 40 answer : option a"
Answer: a
Problem: if a , b , c and d are positive integers less than 4 , and 4 ^ a + 3 ^ b + 2 ^ c + 1 ^ d = 78 then what is the value of b / c ?
Options: a ) 3 , b ) 2 , c ) 1 , d ) 1 / 2 , e ) 1 / 3
Rationale: since a , b and c are positive integers less than 4 we have the following possibilities : 4 ^ a = 416 , or 64 3 ^ b = 39 , or 27 2 ^ c = 24 , or 8 trial and error gives us quite quick the solution of 4 ^ a = 64 3 ^ b = 9 2 ^ c = 4 64 + 9 + 4 = 77 i . e . c = 2 and b = 2 - - - - > b / c = 1 / 1 = 1 the correct answer is c
Answer: c
Problem: on a school â € ™ s annual day sweets were to be equally distributed amongst 112 children . but on that particular day , 32 children were absent . thus the remaining children got 6 extra sweets . how many sweets was each child originally supposed to get ?
Options: a ) 28 , b ) 27 , c ) 11 , d ) 15 , e ) 19
Rationale: explanation : let ' k ' be the total number of sweets . given total number of students = 112 if sweets are distributed among 112 children , let number of sweets each student gets = ' l ' = > k / 112 = l . . . . ( 1 ) but on that day students absent = 32 = > remaining = 112 - 32 = 80 then , each student gets ' 6 ' sweets extra . = > k / 80 = l + 6 . . . . ( 2 ) from ( 1 ) k = 112 l substitute in ( 2 ) , we get 112 l = 80 l + 480 32 l = 480 l = 15 therefore , 15 sweets were each student originally supposed to get . answer : d
Answer: d
Problem: the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 16 . find the sum :
Options: a ) 4500 , b ) 7500 , c ) 5000 , d ) 6400 , e ) none of these
Rationale: "sol . ( d ) let the sum be 100 . therefore , si = 100 × 5 × 2 / 100 = 10 and ci = 100 ( 1 + 5 / 100 ) 2 − 100 ∴ = 100 × 21 × 21 / 20 × 20 − 100 = 41 / 4 difference of ci and si = 41 ⁄ 4 - 10 = 1 ⁄ 4 if the difference is 1 ⁄ 4 , the sum = 100 = > if the difference is 16 , the sum = 400 × 16 = 6400 answer d"
Answer: d
Problem: there are two numbers . if 50 % of the first number is added to the second number , then the second number increases to its 5 - fourth . find the ratio of the first number to the second number ?
Options: a ) a ) 1 / 2 , b ) b ) 5 / 8 , c ) c ) 6 , d ) d ) 6 / 7 , e ) e ) 7
Rationale: let the two numbers be x and y . 50 / 100 * x + y = 5 / 4 y = > 1 / 2 x = 1 / 4 y = > x / y = 1 / 2 a )
Answer: a
Problem: if you wish to live to be 100 years old ( hypothetically ) , you must consume 500 calories less than your average daily allowance for your age . if you are in your 60 ' s , and your average daily allowance is 2000 calories per day , how many calories are you allowed in a week ?
Options: a ) 7000 , b ) 10500 , c ) 14000 , d ) 12000 , e ) 3500
Rationale: to determine the amount of calories allowed per day , you need to subtract 500 from 2000 , for a total of 1500 to determine the amount of calories allowed per week , you need to multiply the daily amount allowed ( 1500 ) by the number of days in a week ( 7 ) for a total of 10500 the correct answer is b
Answer: b
Problem: 250 metres long yard , 51 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees
Options: a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 5
Rationale: "51 trees have 50 gaps between them , required distance ( 250 / 50 ) = 15 e"
Answer: e
Problem: a sum of money deposited at c . i . amounts to rs . 2420 in 2 years and to rs . 2783 in 3 years . find the rate percent ?
Options: a ) 15 , b ) 10 , c ) 28 , d ) 24 , e ) 82
Rationale: "explanation : 2420 - - - 363 100 - - - ? = > 15 % answer : option a"
Answer: a
Problem: if an integer n is to be selected at random from 1 to 100 , inclusive , what is probability n ( n + 1 ) will be divisible by 16 ?
Options: a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4
Rationale: "because n ( n + 1 ) is always an even product of even * odd or odd * even factors , there is a probability of 1 that that it will be divisible by 2 , and , thus , a probability of 1 / 2 that it will be divisible by 4 and , thus , a probability of 1 / 4 that it will be divisible by 8 and , thus , a probability of 1 / 8 that it will be divisible by 16 1 * 1 / 8 = 1 / 8 answer : b"
Answer: b
Problem: how many zeros are there in 50 !
Options: a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
Rationale: "to find the number of trailing zeroes in 50 ! factorial we must find the number of integral values in : 50 / 5 + 50 / 5 ^ 2 + 50 / 5 ^ 3 + . . . . . . . . . = 10 + 2 + 0 + 0 . . . . = 12 answer : c"
Answer: c
Problem: eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 450 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy : freddy )
Options: a ) 8 / 3 , b ) 2 / 1 , c ) 8 / 5 , d ) 5 / 8 , e ) 5 / 3
Rationale: "distance traveled by eddy = 600 km time taken by eddy = 3 hours average speed of eddy = 450 / 3 = 150 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 150 / 75 = 2 / 1 answer b"
Answer: b
Problem: pipe x that can fill a tank in an hour and pipe y that can fill the tank in half an hour are opened simultaneously when the tank is empty . pipe y is shut 15 minutes before the tank overflows . when will the tank overflow ?
Options: a ) 35 mins , b ) 30 mins , c ) 40 mins , d ) 32 mins , e ) 36 mins
Rationale: the last 15 minutes only pipe x was open . since it needs 1 hour to fill the tank , then in 15 minutes it fills 1 / 4 th of the tank , thus 3 / 4 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 3 / 4 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 3 / 4 ) / 3 = 1 / 4 hours = 15 minutes . total time = 15 + 15 = 30 minutes . answer : b
Answer: b
Problem: a and b can complete a work in 15 days and 10 day . they started doing the work together but after 2 days b had to leave and a alone completed the remaining work . the whole work was completed in ?
Options: a ) 10 days , b ) 12 days , c ) 15 days , d ) 8 days , e ) 16 days
Rationale: "a + b 1 day work = 1 / 15 + 1 / 10 = 1 / 6 work done by a and b in 2 days = 1 / 6 * 2 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 now 1 / 15 work is done by a in 1 day 2 / 3 work will be done by a in 15 * 2 / 3 = 10 days total time taken = 10 + 2 = 12 days answer is b"
Answer: b
Problem: the speeds of three asteroids were compared . asteroids x - 13 and y - 14 were observed for identical durations , while asteroid z - 15 was observed for 2 seconds longer . during its period of observation , asteroid y - 14 traveled three times the distance x - 13 traveled , and therefore y - 14 was found to be faster than x - 13 by 6000 kilometers per second . asteroid z - 15 had an identical speed as that of x - 13 , but because z - 15 was observed for a longer period , it traveled five times the distance x - 13 traveled during x - 13 ' s inspection . asteroid x - 13 traveled how many kilometers during its observation ?
Options: a ) 500 , b ) 1,600 / 3 , c ) 1,000 , d ) 1,500 , e ) 2,500
Rationale: "x 13 : ( t , d , s ) y 14 : ( t , 3 d , s + 6000 mi / hour ) z 15 : ( t + 2 seconds , s , 5 d ) d = ? distance = speed * time x 13 : d = s * t x 14 : 3 d = ( s + 6000 ) * t = = = > 3 d = ts + 6000 t z 15 : 5 d = s * ( t + 2 t ) = = = > 5 d = st + 2 st = = = > 5 d - 2 st = st 3 d = 5 d - 2 st + 6000 t - 2 d = - 2 st + 6000 t 2 d = 2 st - 6000 t d = st - 3000 t x 13 : d = s * t st - 3000 t = s * t s - 3000 = s - 1500 = s i got to this point and could n ' t go any further . this seems like a problem where i can set up individual d = r * t formulas and solve but it appears that ' s not the case . for future reference how would i know not to waste my time setting up this problem in the aforementioned way ? thanks ! ! ! the distance of z 15 is equal to five times the distance of x 13 ( we established that x 13 is the baseline and thus , it ' s measurements are d , s , t ) s ( t + 2 ) = 5 ( s * t ) what clues would i have to know to set up the equation in this fashion ? is it because i am better off setting two identical distances together ? st + 2 s = 5 st t + 2 = 5 t 2 = 4 t t = 1 / 2 we are looking for distance ( d = s * t ) so we need to solve for speed now that we have time . speed y 14 - speed x 13 speed = d / t 3 d / t - d / t = 6000 ( remember , t is the same because both asteroids were observed for the same amount of time ) 2 d = 6000 2 = 3000 d = s * t d = 3000 * ( 1 / 2 ) d = 1500 answer : d"
Answer: d
Problem: working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 12 , 30 , and 40 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ?
Options: a ) 4 / 11 , b ) 1 / 2 , c ) 15 / 22 , d ) 120 / 7 , e ) 11 / 4
Rationale: "p 1 takes 12 hrs rate for p 2 p 3 together = 1 / 30 + 1 / 40 = 7 / 120 therefore they take 120 / 7 ratio = 120 / 7 = d"
Answer: d
Problem: the roof of an apartment building is rectangular and its length is 3 times longer than its width . if the area of the roof is 768 feet squared , what is the difference between the length and the width of the roof ?
Options: a ) 32 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 46 .
Rationale: "answer is a : 42 let w be the width , so length is 3 w . therefore : w * 4 w = 768 , solving for , w = 16 , so 3 w - w = 2 w = 2 * 16 = 32"
Answer: a
Problem: the sum of two numbers is 528 and their h . c . f . is 33 . the number of pairs of numbers satisfying the above conditions is :
Options: a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 15
Rationale: "explanation : let the required numbers be 33 a and 33 b . then , 33 a + 33 b = 528 . = > a + b = 16 . now , co - primes with sum 16 are ( 1 , 15 ) , ( 3 , 13 ) , ( 5 , 11 ) and ( 7 , 9 ) . â ˆ ´ required numbers are ( 33 * 1 , 33 * 15 ) , ( 33 * 3 , 33 * 13 ) , ( 33 * 5 , 33 * 11 ) , ( 33 x 7 , 33 x 9 ) . the number of such pairs is 4 . answer is a"
Answer: a
Problem: a certain fraction is equivalent to 2 / 5 . if the numerator of the fraction is increased by 3 and the denominator is doubled , the new fraction is equivalent to 1 / 3 . what is the sum of the numerator and denominator of the original fraction ?
Options: a ) 21 / 4 , b ) 35 , c ) 28 , d ) 26 , e ) 21
Rationale: x / y = 2 / 5 - > 1 ( x + 3 ) / 2 y = 1 / 3 - > 2 divide 1 by 2 : = > 2 x / ( x + 3 ) = 6 / 5 = > 10 x = 6 x + 18 = > x = 3 / 2 = > y = 5 / 2 * 3 / 2 = 15 / 4 so x + y = 21 / 4 the answer is a
Answer: a
Problem: it takes 30 days to fill a laboratory dish with bacteria . if the size of the bacteria doubles each day , how long did it take for the bacteria to fill 1 / 64 of the dish ?
Options: a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28
Rationale: the bacteria doubles each day , so after 29 days , the dish was half full . after 28 days , the dish was one quarter full . after 27 days , the dish was one eighth full . after 26 days , the dish was one sixteenth full . after 25 days , the dish was 1 / 32 full . after 24 days , the dish was 1 / 64 full . the answer is c .
Answer: c
Problem: in what time will a train 95 m long cross an electric pole , it its speed be 214 km / hr ?
Options: a ) 1.6 sec , b ) 2.9 sec , c ) 2.7 sec , d ) 8.7 sec , e ) 8.5 sec
Rationale: "speed = 214 * 5 / 18 = 59 m / sec time taken = 95 / 59 = 1.6 sec . answer : a"
Answer: a
Problem: q - 2 : how many two digit numbers of distinct digits can be formed by using digits 1 , 2 , 3 , 4 , 5 , 6 and 7 such that the numbers are divisible by 3 ?
Options: a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 14
Rationale: concept : a number will be divisible by 3 if sum of all the digits of the number is divisible by 3 here the possible sum of the two distinct digits can be 3 , 6 , 9 and 12 only in order to satisfy the given condition if sum of digits = 3 , no . = 12 , 21 - - - 2 cases if sum of digits = 6 , no . = 15 , 24 , 42 , 51 - - - 4 cases if sum of digits = 9 , no . = 27 , 36 , 45 , 54 , 63 , 72 - - - 6 cases if sum of digits = 12 , no . = 57 , 75 - - - 2 cases total cases = 2 + 4 + 6 + 2 = 14 cases answer : option e
Answer: e
Problem: car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 65 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ?
Options: a ) 15 , b ) 20 , c ) 40 , d ) 47 , e ) 49
Rationale: "car y began travelling after 72 minutes or 1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so , 35 ( t + 1.2 ) = 65 t t = 1.4 distance traveled by car x from the time car y began traveling until both cars stopped is 35 x 1.4 = 49 miles answer : - e"
Answer: e
Problem: the l . c . m . of two numbers is 42 . the numbers are in the ratio 2 : 3 . then sum of the number is :
Options: a ) 28 , b ) 32 , c ) 35 , d ) 53 , e ) 54
Rationale: let the numbers be 2 x and 3 x . then , their l . c . m . = 6 x . so , 6 x = 42 or x = 7 . the numbers are 14 and 21 . hence , required sum = ( 14 + 21 ) = 35 . answer : option c
Answer: c
Problem: if 20 % of a = b , then b % of 20 is the same as :
Options: a ) 4 % of a , b ) 8 % of a , c ) 12 % of a , d ) 9 % of a , e ) 1 % of a
Rationale: "20 % of a = b b % of 20 = = = = 4 % of a . answer : a"
Answer: a
Problem: a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 150 . find the unknown rate of two towels ?
Options: a ) 400 , b ) 450 , c ) 500 , d ) 550 , e ) 600
Rationale: "10 * 150 = 1500 3 * 100 + 5 * 150 = 1050 1500 – 1050 = 450 b"
Answer: b
Problem: a present value of a machine is $ 400 . its value depletiation rate is 25 % per annum then find the machine value after 2 years ?
Options: a ) $ 125 , b ) $ 100 , c ) $ 200 , d ) $ 215 , e ) $ 225
Rationale: "p = $ 400 r = 25 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 400 * 3 / 4 * 3 / 4 = $ 225 approximately answer is e"
Answer: e
Problem: a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 3 / 4 % his yearly income diminishes by rs . 60.50 , his capital is ?
Options: a ) 24,200 , b ) 28,888 , c ) 24,600 , d ) 24,628 , e ) 24,6012
Rationale: "let the capital be rs . x . then , ( x * 8 * 1 ) / 100 - ( x * 31 / 4 * 1 / 100 ) = 60.50 32 x - 31 x = 6050 * 4 x = 24,200 . answer : a"
Answer: a
Problem: a merchant purchased a jacket for $ 60 and then determined a selling price that equalled the purchase price of the jacket plus a markup that was 20 percent of the selling price . during a sale , the merchant discounted the selling price by 20 percent and sold the jacket . what was the merchant ’ s gross profit on this sale ?
Options: a ) $ 0 , b ) $ 3 , c ) $ 4 , d ) $ 12 , e ) $ 15
Rationale: "actual cost = $ 60 sp = actual cost + mark up = actual cost + 20 % sp = 60 * 100 / 80 on sale sp = 80 / 100 ( 60 * 100 / 80 ) = 60 gross profit = $ 0 answer is a"
Answer: a