Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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a batsman makes a score of 74 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 74 = total runs scored in 17 innings = > 16 ( x - 3 ) + 74 = 17 x = > 16 x - 48 + 74 = 17 x = > x = 26 answer is c" | a ) 25 , b ) 31 , c ) 26 , d ) 29 , e ) 39 | c | add(subtract(74, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general | C |
two numbers are less than a third number by 30 % and 37 % . how much percent is the second number is less than the first ? | let the third number be x first number = 70 % of x = 7 x / 10 ; second number = 63 % of x = 63 x / 100 ; difference = 7 x / 10 - 63 x / 100 = 7 x / 100 ; required percentage = 7 x / 100 * 10 / 7 x * 100 = 10 % answer is a | a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | a | multiply(subtract(add(30, const_1), 30), const_10) | add(n0,const_1)|subtract(#0,n0)|multiply(#1,const_10) | gain | A |
what annual payment will discharge a debt of rs . 1090 due in 2 years at the rate of 5 % compound interest ? | explanation : let each installment be rs . x . then , x / ( 1 + 5 / 100 ) + x / ( 1 + 5 / 100 ) 2 = 1090 820 x + 1090 * 441 x = 586.21 so , value of each installment = rs . 586.21 answer : option b | a ) 993.2 , b ) 586.21 , c ) 534.33 , d ) 543.33 , e ) 646.33 | b | divide(multiply(power(add(divide(5, const_100), const_1), 2), 1090), 2) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|divide(#3,n1) | gain | B |
a bag contains 7 green balls and 7 white balls . if two balls are drawn simultaneously , what is the probability that both balls are the same colour ? | "the total number of ways to draw two balls is 14 c 2 = 91 the number of ways to draw two green balls is 7 c 2 = 21 the number of ways to draw two white balls is 7 c 2 = 21 p ( two balls of the same colour ) = 42 / 91 = 6 / 13 the answer is b ." | a ) 4 / 13 , b ) 6 / 13 , c ) 8 / 21 , d ) 11 / 21 , e ) 17 / 42 | b | add(multiply(divide(7, add(7, 7)), divide(subtract(7, const_1), subtract(add(7, 7), const_1))), multiply(divide(subtract(7, const_1), subtract(add(7, 7), const_1)), divide(7, add(7, 7)))) | add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#7,#4)|add(#8,#9)| | other | B |
a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 30 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he save ? | "1 st year income = i 1 st year savings = s 1 st year expense = e 1 2 nd year income = 1.3 i 2 nd year savings = 2 s ( 100 % increase ) 2 nd year expense = e 2 e 1 + e 2 = 2 e 1 e 2 = e 1 that means expenses are same during both years . with increase of 30 % income the savings increased by 100 % . or s = . 3 i or s = 30 % of income d is the answer" | a ) 21 % , b ) 22.5 % , c ) 25 % , d ) 30 % , e ) 40 % | d | multiply(divide(subtract(add(add(100, 30), 100), multiply(2, 100)), 100), 100) | add(n0,n1)|multiply(n1,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,n1)|multiply(#4,n1)| | general | D |
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 t this college graduate has a graduate degree ? | "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 t of cgd / ( cg ) - > 3 / 19 answer d" | a ) 1 / 11 , b ) 1 / 12 , c ) 1 / 13 , d ) 3 / 19 , e ) 3 / 43 | d | divide(divide(divide(1, 8), divide(2, 3)), add(divide(divide(1, 8), divide(2, 3)), 1)) | divide(n0,n1)|divide(n2,n3)|divide(#0,#1)|add(#2,n0)|divide(#2,#3)| | other | D |
a company pays 15.5 % dividend to its investors . if an investor buys rs . 50 shares and gets 25 % on investment , at what price did the investor buy the shares ? | "explanation : dividend on 1 share = ( 15.5 * 50 ) / 100 = rs . 7.75 rs . 25 is income on an investment of rs . 100 rs . 7.75 is income on an investment of rs . ( 7.75 * 100 ) / 25 = rs . 31 answer : b" | a ) 25 , b ) 31 , c ) 18 , d ) 19 , e ) 01 | b | divide(multiply(divide(multiply(15.5, 50), const_100), const_100), 25) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_100)|divide(#2,n2)| | gain | B |
in how many seconds will a train 180 meters long pass an oak tree , if the speed of the train is 36 km / hr ? | "speed = 36 * 5 / 18 = 10 m / s time = 180 / 10 = 18 seconds the answer is c ." | a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | c | divide(180, multiply(const_0_2778, 36)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | C |
the width of a rectangular hall is ½ of its length . if the area of the hall is 450 sq . m , what is the difference between its length and breadth ? | "let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 450 = x * 1 x / 2 x ² = 900 x = 30 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 30 / 2 = 15 m answer : d" | a ) 8 m , b ) 10 m , c ) 12 m , d ) 15 m , e ) 17 m | d | divide(sqrt(divide(450, divide(const_1, const_2))), const_2) | divide(const_1,const_2)|divide(n0,#0)|sqrt(#1)|divide(#2,const_2)| | geometry | D |
the visitors of a modern art museum who watched a certain picasso painting were asked to fill in a short questionnaire indicating whether they had enjoyed looking at the picture and whether they felt they had understood it . according to the results of the survey , all 140 visitors who did not enjoy the painting also did not feel they had understood the painting , and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting . if 3 / 4 of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting , then how many visitors answered the questionnaire ? | "if we exclude those cases and take the question at face value , then it seems straightforward . group # 1 = ( did n ' t like , did n ' t understand ) = 120 group # 2 = ( likeunderstood ) = 3 / 4 ( 1 / 4 ) n = 560 n = 480 answer = ( e )" | a ) 90 , b ) 120 , c ) 160 , d ) 360 , e ) 560 | e | divide(140, subtract(const_1, divide(3, 4))) | divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)| | general | E |
evaluate 75 % of 450 + 45 % of 750 | "explanation : = ( 75 / 100 ) * 450 + ( 45 / 100 ) * 750 = 675 option e" | a ) 632 , b ) 642 , c ) 652 , d ) 675 , e ) 572 | e | divide(75, divide(450, 75)) | divide(n1,n0)|divide(n0,#0)| | gain | E |
in august , a cricket team that played 120 matches won 30 % of the games it played . after a continuous winning streak , this team raised its average to 52 % . how many matches did the team win to attain this average ? | "let the no of matches played more = x so , ( 120 + x ) * 52 / 100 = 36 + x by solving we get x = 55 answer : b" | a ) 40 , b ) 55 , c ) 68 , d ) 80 , e ) 98 | b | divide(subtract(multiply(divide(52, const_100), 120), multiply(divide(30, const_100), 120)), subtract(const_1, divide(52, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#0)|subtract(#2,#3)|divide(#5,#4)| | general | B |
how many pieces can a square pizza be cut into by making 4 linear cuts ? | 1 cut makes two pieces a second cut can make 4 pieces by cutting through 2 of the pieces a third cut can make 7 pieces by cutting through 3 of the pieces a fourth cut can make 11 pieces by cutting through 4 of the pieces b | ['a ) 10', 'b ) 11', 'c ) 12', 'd ) 13', 'e ) 14'] | b | subtract(multiply(4, const_3), const_1) | multiply(n0,const_3)|subtract(#0,const_1) | geometry | B |
if the price of a certain bond on may 1 st was 2 / 3 the price of the bond on june 1 st and the price of the bond on july 1 st was 50 % greater than the price of the bond on may 1 st . then the price of the bond on june 1 st st was what percent of the average ( arithmetic mean ) price of the bond on may 1 st and july 1 st ? | "the price on june 1 st = 12 ( assume ) ; the price on may 1 st = 2 / 3 * 12 = 8 ; the price on july 1 st = 8 * 1.50 = 12 . the average price of the bond on may 1 st and july 1 st = ( 8 + 12 ) / 2 = 10 . the price of the bond on june 1 st ( 12 ) is 6 / 5 times ( 120 % ) the average price of the bond on may 1 st and july 1 st . answer : c ." | a ) 50 % , b ) 75 % , c ) 120 % , d ) 133 1 / 3 % , e ) 150 % | c | add(multiply(50, const_0_33), add(const_100, const_0_33)) | add(const_0_33,const_100)|multiply(n5,const_0_33)|add(#0,#1)| | general | C |
find the area of trapezium whose parallel sides are 22 cm and 18 cm long , and the distance between them is 15 cm . | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 22 + 18 ) * ( 15 ) = 300 cm 2 answer : c" | a ) 227 , b ) 299 , c ) 300 , d ) 161 , e ) 212 | c | quadrilateral_area(15, 18, 22) | quadrilateral_area(n2,n1,n0)| | physics | C |
a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 25 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | "explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 25 = rs . 180 cp on 40 metres = 180 x 40 = rs . 7200 profit earned on 40 metres cloth = rs . 8200 – rs . 7200 = rs . 1000 . answer : option c" | a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 1200 , e ) none of these | c | multiply(25, 40) | multiply(n0,n2)| | gain | C |
the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.40 per metre . | "explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.40 ) = rs . 5808 . answer : d ) 5808" | a ) 2399 , b ) 3888 , c ) 2999 , d ) 5808 , e ) 2888 | d | multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.40) | divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)| | geometry | D |
the average weight of 5 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ? | "total weight increases = 5 × 1.5 = 7.5 kg so the weight of new person = 65 + 7.5 = 72.5 kg answer c" | a ) 76 kg , b ) 77 kg , c ) 72.5 kg , d ) data inadequate , e ) none of these | c | add(65, multiply(5, 1.5)) | multiply(n0,n1)|add(n2,#0)| | general | C |
a 40 gallon solution of salt and water is 10 % salt . how many gallons of water must be added to the solution in order to decrease the salt to 8 % of the volume ? | "amount of salt = 4.0 assume x gallons of water are added . 4.0 / 40 + x = 8 / 100 400 = 8 x + 320 8 x = 80 x = 10 correct option : a" | a ) 810 , b ) 12 , c ) 13 , d ) 14 , e ) 16 | a | divide(multiply(40, subtract(10, 8)), 8) | subtract(n1,n2)|multiply(n0,#0)|divide(#1,n2)| | general | A |
the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 50 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | "take it as original output = 100 . to meet demand increase by 10 % , then output = 110 . to meet holiday demand , new output increase by 50 % then output equals 165 to restore new holidy demand output to original 100 . final - initial / final * 100 = 65 / 165 * 100 = 39 % approxiamately . option e is correct ." | a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 39 % | e | multiply(divide(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 50), const_100)), const_1), multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 50), const_100))), const_100) | add(n0,const_100)|add(n1,const_100)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|subtract(#4,const_1)|divide(#5,#4)|multiply(#6,const_100)| | general | E |
5301 x 13 = ? | "append 0 before and after : 053010 calculation : 1 x 3 + 0 = 3 ( take 3 as ones digit of the product ) 0 x 3 + 1 = 1 ( take 1 tens digit of the product ) 3 x 3 + 0 = 9 ( take 9 hundreds digit of the product ) 5 x 3 + 3 = 18 ( take 8 as thousands digit of the product , carry over 1 ) 0 x 3 + 5 = 5 ; 5 + 1 = 6 ( take 6 as ten thousands digits of the product ) so , 5301 x 13 = 68913 answer is e ." | a ) 89136 , b ) 31986 , c ) 68910 , d ) 53113 , e ) 68913 | e | multiply(divide(5301, 13), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | E |
compound x contains elements a and b at an approximate ratio , by weight , of 2 : 10 . approximately how many grams of element b are there in 222 grams of compound x ? | "total number of fractions = 2 + 10 = 12 element b constitutes = 10 out of 12 parts of x so in 222 gms of x have 222 * 10 / 12 = 185 gms of b and 222 - 185 = 37 gms of a . cross check : - a / b = 37 / 185 = 2 / 10 ( as given ) ans b" | a ) 54 , b ) 185 , c ) 250 , d ) 270 , e ) 322 | b | divide(multiply(222, 10), add(2, 10)) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)| | other | B |
seven 6 faced dice are thrown together . the probability that all the three show the same number on them is ? | "the three dice can fall in 6 * 6 * 6 * 6 * 6 * 6 * 6 = 279936 ways . hence the probability is 6 / 279936 = 1 / 46656 answer : b" | a ) 1 / 32 , b ) 1 / 46656 , c ) 1 / 33 , d ) 1 / 38 , e ) 1 / 34 | b | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | divide(const_1,n0)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2)| | probability | B |
p is able to do a piece of work in 15 days and q can do the same work in 20 days . if they can work together for 4 days , what is the fraction of work left ? | explanation : p ' s 1 - day work = 1 / 15 q ' s 1 - day work = 1 / 20 work done by ( p + q ) in 1 day = 1 / 15 + 1 / 20 = 7 / 60 . work done by them in 4 days = ( 7 / 60 ) * 4 = 7 / 15 . work left = 1 - ( 7 / 15 ) = 8 / 15 . answer is a | a ) 8 / 15 , b ) 7 / 15 , c ) 1 / 15 , d ) 3 / 15 , e ) none of these | a | subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 15)), 4)) | divide(const_1,n1)|divide(const_1,n0)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3) | physics | A |
what is the remainder if 7 ^ 3 is divided by 100 ? | "7 * 7 * 7 / 100 = 343 / 100 remainder 43 answer : e" | a ) 52 , b ) 35 , c ) 42 , d ) 41 , e ) 43 | e | subtract(divide(100, const_2), multiply(7, 7)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general | E |
it takes 2 workers a total of 12 hours to build a giant machine with each worker working at the same rate . if 3 workers start to build the machine at 11 : 00 am , and one extra worker per hour is added beginning at 2 : 00 pm , at what time will the machine be complete ? | 2 workers build 1 / 12 of a machine in one hour . 1 worker builds 1 / 24 of a machine in one hour . in the first 3 hours , 3 workers build 3 * ( 1 / 24 ) * 3 = 9 / 24 of a machine from 2 : 00 to 3 : 00 , 4 workers build another 4 / 24 . the total is 13 / 24 . from 3 : 00 to 4 : 00 , 5 workers build another 5 / 24 . the total is 18 / 24 . from 4 : 00 to 5 : 00 , 6 workers build another 6 / 24 . the total is 24 / 24 . the machine is complete at 5 : 00 . the answer is c . | a ) 4 : 30 , b ) 4 : 45 , c ) 5 : 00 , d ) 5 : 15 , e ) 5 : 30 | c | subtract(divide(multiply(2, 12), 12), const_0_33) | multiply(n0,n1)|divide(#0,n1)|subtract(#1,const_0_33) | general | C |
3 business people wish to invest in a new company . each person is willing to pay one third of the total investment . after careful calculations , they realize that each of them would pay $ 2200 less if they could find two more equal investors . how much is the total investment in the new business ? | total investment between 5 : ( x / 5 ) total investment including 2200 less between 3 people ( x - ( 2200 * 3 ) ) / 3 set both eq . equal to each other . 16,500 answer c ) | a ) $ 11,000 , b ) $ 50,000 , c ) $ 16,500 , d ) $ 6,600 , e ) $ 3,600 | c | divide(reminder(multiply(add(3, const_2), divide(multiply(2200, 3), const_2)), const_1000), const_10) | add(n0,const_2)|multiply(n0,n1)|divide(#1,const_2)|multiply(#0,#2)|reminder(#3,const_1000)|divide(#4,const_10) | general | C |
a and b together can do a piece of work in 6 days and a alone can do it in 14 days . in how many days can b alone can do it ? | "explanation : a and b can do work 1 / 6 in 1 day a alone can do 1 / 14 work in 1 day b alone can do ( 1 / 6 - 1 / 14 ) = 2 / 21 work in 1 day = > complete work can be done in 21 / 2 days by b answer : option c" | a ) 12 days , b ) 15 days , c ) 21 / 2 days , d ) 21 days , e ) 22 days | c | inverse(subtract(inverse(6), inverse(14))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics | C |
a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and b quits 15 days before the project is completed , in how many days total will the project be completed ? | "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 15 days , a can do 3 / 4 of the project . thus a and b must complete 1 / 4 of the project , which takes 3 days . the total number of days is 3 + 15 = 18 . the answer is d ." | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | add(divide(subtract(const_1, multiply(divide(const_1, 30), 15)), add(divide(const_1, 20), divide(const_1, 30))), 15) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)| | physics | D |
in a certain company 20 % of the men and 40 % of the women attended the annual company picnic . if 45 % of all the employees are men . what % of all the employee went to the picnic ? | "total men in company 45 % means total women in company 55 % ( assume total people in company 100 % ) no of men employees attended picnic = 45 x ( 20 / 100 ) = 9 no of women employees attend picnic = 55 x ( 40 / 100 ) = 22 total percentage of employees attend the picnic = 9 + 22 = 31 % answer : a" | a ) 31 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 37 % | a | multiply(add(multiply(divide(45, const_100), divide(20, const_100)), multiply(divide(subtract(const_100, 45), const_100), divide(40, const_100))), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n2)|divide(#3,const_100)|multiply(#0,#1)|multiply(#4,#2)|add(#5,#6)|multiply(#7,const_100)| | gain | A |
the sum of 3 consecutive even numbers is 246 . what are the numbers ? | first x make the first x second x + 2 even numbers , sowe add 2 to get the next third x + 4 add 2 more ( 4 total ) to get the third f + s + t = 246 summeansaddfirst ( f ) plussecond ( s ) plusthird ( t ) ( x ) + ( x + 2 ) + ( x + 4 ) = 246 replace each f , s , and t withwhatwe labeled them x + x + 2 + x + 4 = 246 here the parenthesis are not needed 3 x + 6 = 246 combine like terms x + x + x and 2 + 4 − 6 − 6 subtract 6 fromboth sides 3 x = 240 the variable ismultiplied by 3 3 3 divide both sides by 3 x = 80 our solution for x first 80 replace x in the origional listwith 80 . second ( 80 ) + 2 = 82 the numbers are 80 , 82 , and 84 . third ( 80 ) + 4 = 84 correct answer a | a ) 80 , 8284 , b ) 65 , 6871 , c ) 45 , 4742 , d ) 100 , 10249 , e ) 87 , 8889 | a | divide(subtract(246, add(const_2, const_4)), 3) | add(const_2,const_4)|subtract(n1,#0)|divide(#1,n0) | physics | A |
paul has to secure 50 % marks to clear his exam of class 7 th . he got 50 marks and failed by 10 marks . what is the maximum marks ? | c 120 to pass the exam ravish needs 50 + 10 = 60 marks . = > ( 60 / 50 ) * 100 = 120 | a ) 100 , b ) 110 , c ) 120 , d ) 210 , e ) 200 | c | divide(multiply(add(50, 10), const_100), 50) | add(n0,n3)|multiply(#0,const_100)|divide(#1,n0) | gain | C |
a boat can travel with a speed of 16 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 60 km downstream . | "speed downstream = ( 16 + 4 ) km / hr = 20 km / hr . time taken to travel 60 km downstream = 60 / 20 = 3 hours . answer : b" | a ) 2 hours . , b ) 3 hours . , c ) 4 hours . , d ) 5 hours . , e ) 6 hours . | b | divide(60, add(16, 4)) | add(n0,n1)|divide(n2,#0)| | physics | B |
a train 110 m long is running with a speed of 24 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 24 + 6 = 30 km / hr . = 30 * 5 / 18 = 8.33 m / sec . time taken to pass the men = 110 / 8.33 = 13.2 sec . answer : e" | a ) 7 sec , b ) 6 sec , c ) 8 sec , d ) 14 sec , e ) 13.2 sec | e | divide(110, multiply(add(24, 6), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics | E |
4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 6 mat - weavers in 6 days ? | "let the required number of bottles be x . more weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) wavers 4 : 6 : : 4 : x days 4 : 6 4 * 4 * x = 6 * 6 * 4 x = ( 6 * 6 * 4 ) / ( 4 x 4 ) x = 9 . answer is b ." | a ) 25 , b ) 9 , c ) 39 , d ) 61 , e ) 16 | b | multiply(multiply(4, divide(6, 4)), divide(6, 4)) | divide(n3,n0)|multiply(n0,#0)|multiply(#0,#1)| | gain | B |
population of a city in 20004 was 1000000 . if in 2005 there isan increment of 25 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city atthe end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 25 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1178125 d" | a ) 976374 , b ) 979923 , c ) 980241 , d ) 1178125 , e ) 1083875 | d | multiply(1000000, multiply(multiply(add(const_1, divide(25, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6)| | gain | D |
the number of timeshare condos available at sunset beach is 3 / 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 400 , what is the difference between the number of condos available at sunset beach and the number of condos available at playa del mar ? | "let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 / 5 x we know , x + 3 / 5 x = 400 hence , x = 250 . so , number of timeshare condos available at playa del mar = 250 the difference between the number of condos available at sunset beach and the number of condos available at playa del mar = x - 3 / 5 x = 2 / 5 x = 2 / 5 ( 250 ) = 100 the correct answer is d" | a ) 60 , b ) 90 , c ) 120 , d ) 100 , e ) 240 | d | add(divide(multiply(400, 3), 5), multiply(3, 5)) | multiply(n0,n2)|multiply(n0,n1)|divide(#0,n1)|add(#2,#1)| | general | D |
rs . 6000 is lent out in two parts . one part is lent at 5 % p . a simple interest and the other is lent at 10 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | let the amount lent at 5 % be rs . x amount lent at 10 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 5 / 100 x + 10 / 100 ( 6000 - x ) = 600 - 1 x / 20 = > 600 - 1 / 20 x = 450 = > x = 3000 amount lent at 10 % = 3000 required ratio = 3000 : 3000 = 5 : 5 answer : b | a ) 5 : 1 , b ) 5 : 5 , c ) 5 : 8 , d ) 5 : 4 , e ) 5 : 2 | b | divide(divide(subtract(multiply(450, const_100), multiply(6000, 5)), subtract(10, 5)), divide(subtract(multiply(450, const_100), multiply(6000, 5)), subtract(10, 5))) | multiply(n3,const_100)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|divide(#4,#4) | gain | B |
a fruit seller had some oranges . he sells 40 % oranges and still has 480 oranges . how many oranges he had originally ? | "60 % of oranges = 480 100 % of oranges = ( 480 × 100 ) / 6 = 800 total oranges = 700 answer : a" | a ) 800 , b ) 710 , c ) 720 , d ) 730 , e ) 740 | a | add(480, multiply(480, divide(40, const_100))) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | gain | A |
if a and b get profits of rs . 4,000 and rs . 2,000 respectively at the end of year then ratio of their investments are | "ratio = 4000 / 2000 = 2 : 1 answer : e" | a ) 4 : 1 , b ) 1 : 4 , c ) 3 : 2 , d ) 2 : 3 , e ) 2 : 1 | e | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | add(const_2,const_3)|multiply(const_2,const_3)|multiply(#1,const_100)|multiply(#0,const_2)|multiply(#2,const_100)|divide(#4,#3)|divide(#4,#5)| | gain | E |
x alone can do a piece of work in 15 days and y alone can do it in 10 days . x and y undertook to do it for rs . 720 . with the help of z they finished it in 5 days . how much is paid to z ? | solution : in one day x can finish 1 / 15 th of the work . in one day y can finish 1 / 10 th of the work . let us say that in one day z can finish 1 / zth of the work . when all the three work together in one day they can finish 1 / 15 + 1 / 10 + 1 / z = 1 / 5 th of the work . therefore , 1 / z = 1 / 30 . ratio of their efficiencies = 1 / 15 : 1 / 10 : 1 / 30 = 2 : 3 : 1 . therefore z receives 1 / 6 th of the total money . according to their efficiencies money is divided as 240 : 360 : 120 . hence , the share of z = rs . 120 . answer b | a ) rs . 360 , b ) rs . 120 , c ) rs . 240 , d ) rs . 300 , e ) none | b | multiply(divide(subtract(subtract(divide(const_1, 5), divide(const_1, 10)), divide(const_1, 15)), divide(const_1, 5)), 720) | divide(const_1,n3)|divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|subtract(#3,#2)|divide(#4,#0)|multiply(n2,#5) | physics | B |
find 95 × × 97 | "here both numbers are less than 100 . so they are deficient of - 5 and - 3 compared with 100 . so answer : e" | a ) 92 / 198 , b ) 92 / 12 , c ) 92 / 13 , d ) 92 / 10 , e ) 92 / 15 | e | divide(95, 97) | divide(n0,n1)| | general | E |
if a is a positive integer , and if the units digit of a ^ 2 is 1 and the units digit of ( a + 1 ) ^ 2 is 0 , what is the units digit of ( a + 2 ) ^ 2 ? | "if the units digit of a ^ 2 is 1 , then the units digit of a is either 1 or 9 . if the units digit of ( a + 1 ) ^ 2 is 0 , then the units digit of a + 1 is 0 . to satisfy both conditions , the units digit of a must be 9 . then a + 2 has the units digit of 1 , thus the units digit of ( a + 2 ) ^ 2 will be 1 . the answer is a ." | a ) 1 . , b ) 3 . , c ) 5 . , d ) 7 . , e ) 9 . | a | power(add(multiply(1, 2), 2), 2) | multiply(n0,n1)|add(n0,#0)|power(#1,n0)| | general | A |
an article with cost price of 280 is sold at 30 % profit . what is the selling price ? | "sp = 1.30 * 280 = 364 answer : b" | a ) 198 , b ) 364 , c ) 369 , d ) 207 , e ) 210 | b | add(280, multiply(280, divide(30, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain | B |
i bought two books ; for rs . 360 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | "x * ( 85 / 100 ) = ( 360 - x ) 119 / 100 x = 210 answer : b" | a ) 197 , b ) 210 , c ) 189 , d ) 278 , e ) 268 | b | divide(multiply(360, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | add(n2,const_100)|subtract(const_100,n1)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)| | gain | B |
a circular rim a having a diameter of 45 inches is rotating at the rate of x inches / min . another circular rim b with a diameter of 30 inches is rotating at the rate of y inches / min . what is the value of y in terms of x , if both the rims reach their starting positions at the same time after every rotation . | t = s 1 / v 1 = s 2 / v 2 or , 45 / x = 30 / y or , y = 45 x / 30 = 3 x / 2 ( answer a ) | ['a ) 3 x / 2', 'b ) 4 x / 5', 'c ) 7 x / 5', 'd ) 5 x / 7', 'e ) 3 x / 4'] | a | divide(multiply(const_3, 45), const_2) | multiply(n0,const_3)|divide(#0,const_2) | geometry | A |
two vessels contains equal number of mixtures milk and water in the ratio 4 : 2 and 5 : 1 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ? | "the ratio of milk and water in the new vessel is = ( 4 / 6 + 5 / 6 ) : ( 2 / 6 + 1 / 6 ) = 9 / 6 : 3 / 6 = 3 : 1 answer is a" | a ) 3 : 1 , b ) 9 : 13 , c ) 5 : 11 , d ) 11 : 3 , e ) 15 : 4 | a | divide(add(multiply(4, divide(add(5, 1), add(4, 2))), 5), add(multiply(2, divide(add(5, 1), add(4, 2))), 1)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n2,#3)|add(n3,#4)|divide(#5,#6)| | other | A |
a new home buyer pays 4 % annual interest on her first mortgage and 9 % annual interest on her second mortgage . if she borrowed a total of $ 310,000 , 80 % of which was in the first mortgage , what is her approximate monthly interest payment ? | 0.04 x + 0.09 y = 310000 [ 1 ] 0.04 x = 0.80 * 310000 = 248000 [ 2 ] 248000 + 0.09 y = 310000 - - > 0.09 y = 62000 [ 3 ] 248000 / 12 = 258333.3333 [ 4 ] 62000 / 12 = 5166.67 [ 5 ] adding [ 4,5 ] we get : 25833 [ 6 ] dividing [ 6 ] / 2 to get an average we get 1.292 , ans a | a ) $ 1,292 , b ) $ 1,733 , c ) $ 3,466 , d ) $ 12,917 , e ) $ 20,796 | a | subtract(9, multiply(4, const_2)) | multiply(n0,const_2)|subtract(n1,#0) | general | A |
an aeroplane covers a certain distance at a speed of 240 kmph in 5 hours . to cover the same distance in 123 hours , it must travel at a speed of : | speed and time are inversely proportional ⇒ speed ∝ 1 time ( when distance is constant ) here distance is constant and speed and time are inversely proportionalspeed ∝ 1 time ⇒ speed 1 speed 2 = time 2 time 1 ⇒ 240 speed 2 = ( 123 ) 5 ⇒ 240 speed 2 = ( 53 ) 5 ⇒ 240 speed 2 = 13 ⇒ speed 2 = 240 × 3 = 720 km / hr answer : d | a ) 234 , b ) 377 , c ) 720 , d ) 378 , e ) 268 | d | divide(divide(multiply(240, 5), add(const_1, divide(const_2, const_3))), const_2) | divide(const_2,const_3)|multiply(n0,n1)|add(#0,const_1)|divide(#1,#2)|divide(#3,const_2) | physics | D |
if two positive numbers are in the ratio 1 / 7 : 1 / 5 , then by what percent is the second number more than the first ? | "given ratio = 1 / 7 : 1 / 5 = 5 : 7 let first number be 5 x and the second number be 7 x . the second number is more than first number by 2 x . required percentage = 2 x / 5 x * 100 = 40 % . answer : d" | a ) 70 % , b ) 90 % , c ) 60 % , d ) 40 % , e ) 65 % | d | multiply(divide(1, 5), const_100) | divide(n0,n3)|multiply(#0,const_100)| | general | D |
if 40 % of a number is equal to two - third of another number , what is the ratio of first number to the second number ? | "let 40 % of a = 2 / 3 b then 40 a / 100 = 2 b / 3 2 a / 5 = 2 b / 3 a / b = 5 / 3 a : b = 5 : 3 answer is d" | a ) 2 : 5 , b ) 1 : 4 , c ) 3 : 7 , d ) 5 : 3 , e ) 2 : 3 | d | divide(divide(const_1, const_4), divide(40, const_100)) | divide(const_1,const_4)|divide(n0,const_100)|divide(#0,#1)| | general | D |
a contractor undertakes to do a job within 100 days and hires 10 people to do it . after 20 days , he realizes that one fourth of the work is done so he fires 2 people . in how many more days w will the work get over ? | "we can also use the concept of man - days here 100 days - - > 10 men so the job includes 100 * 10 = 1000 man - days after 20 days 1 / 4 of job is completed so 1 / 4 x 1000 man - days = 250 man - days job is done now the balance job = 1000 - 250 = 750 man - days worth of job since 2 men are fired so b / l men = 8 therefore total no . of days of job = 750 man - day / 8 days = 375 / 4 = 94 days ( approx . ) now since this is total and ques . is asking for additional no . of days , so 94 - 20 = 74 days the nearest approx . to answer is 75 ans : c ( 75 days )" | a ) 60 , b ) w = 70 , c ) w = 75 , d ) w = 80 , e ) w = 100 | c | divide(multiply(divide(multiply(10, 20), const_0_25), subtract(const_1, const_0_25)), subtract(10, 2)) | multiply(n1,n2)|subtract(const_1,const_0_25)|subtract(n1,n3)|divide(#0,const_0_25)|multiply(#3,#1)|divide(#4,#2)| | physics | C |
what is the total surface area in square meters of a rectangular solid whose length is 5 meters , width is 4 meters , and depth is 1 meters ? | "surface area of a cuboid = 2 ( lb + bh + lh ) = 2 ( 5 * 4 + 4 * 1 + 5 * 1 ) = 2 ( 20 + 4 + 5 ) = 2 * 29 = 58 m 2 answer : d" | a ) 48 m 2 , b ) 40 m 2 , c ) 50 m 2 , d ) 58 m 2 , e ) 62 m 2 | d | surface_rectangular_prism(4, 5, 1) | surface_rectangular_prism(n0,n1,n2)| | geometry | D |
the cash difference between the selling prices of an book at a profit of 4 % and 6 % is $ 3 . the ratio of the two selling prices is : | "c 52 : 53 let c . p . of the book be $ x . then , required ratio = 104 % of x / 106 % of x = 104 / 106 = 52 / 53 = 52 : 53" | a ) 50 : 53 , b ) 58 : 53 , c ) 52 : 53 , d ) 42 : 36 , e ) 38 : 53 | c | divide(add(const_100, 4), add(const_100, 6)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)| | gain | C |
if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 4 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 4 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 4 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 4 = e" | a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 4 | e | multiply(3, 1) | multiply(n2,n3)| | general | E |
a train speeds past a pole in 15 seconds and a platfrom 100 m long in 25 seconds . its length is : | "sol . let the length of the train be x metres and its speed be y m / sec . they , x / y = 15 ⇒ y = x / 15 ∴ x + 100 / 25 = x / 15 ⇔ x = 150 m . answer d" | a ) 100 m , b ) 125 m , c ) 130 m , d ) 150 m , e ) none | d | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics | D |
the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 55 kg . what might be the weight of the new person ? | "c 75 kg total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 55 + 20 ) kg = 75 kg ." | a ) 56 kg , b ) 90 kg , c ) 75 kg , d ) data inadequate , e ) none of these | c | add(multiply(8, 2.5), 55) | multiply(n0,n1)|add(n2,#0)| | general | C |
a certain company reported that the revenue on sales increased 20 % from 2000 to 2003 , and increased 40 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ? | "assume the revenue in 2000 to be 100 . then in 2003 it would be 120 and and in 2005 140 , so from 2003 to 2005 it increased by ( 140 - 120 ) / 120 = 20 / 120 = 17 % answer : e ." | a ) 50 % , b ) 40 % , c ) 35 % , d ) 32 % , e ) 17 % | e | multiply(divide(subtract(add(const_1, divide(40, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_100) | divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(#2,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain | E |
a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 900 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "many approaches are possible , one of them : let the distance be 1 kilometer . time to cover this distance at 900 kilometers per hour is 1 / 900 hours = 3,600 / 900 seconds = 4 seconds ; time to cover this distance at regular speed is 900 + 2 = 902 seconds = 902 / 3,600 hours = 1 / 600 hours ; so , we get that to cover 1 kilometer 1 / 600 hours is needed - - > regular speed 600 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : b ." | a ) 671.5 , b ) 600 , c ) 672.5 , d ) 673 , e ) 773.5 | b | divide(1, divide(add(multiply(const_3600, divide(1, 900)), 2), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics | B |
if a rectangular billboard has an area of 91 square feet and a perimeter of 40 feet , what is the length of each of the shorter sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 91 and a perimeter of 40 . we ' re asked for the length of one of the shorter sides of the rectangle . since the answers are all integers , and the area is 91 , the shorter side will almost certainly be less than 10 ( since 10 x 10 = 100 , but we ' re not dealing with a square ) . let ' s test answer b : 7 if . . . the shorter side = 7 . . . the area = 91 . . . . 91 / 7 = 13 = the longer side perimeter = 7 + 7 + 13 + 13 = 40 b" | a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | b | divide(subtract(divide(40, const_2), sqrt(subtract(power(divide(40, const_2), const_2), multiply(const_4, 91)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry | B |
the average age of an adult class is 40 years . 12 new students with an avg age of 32 years join the class . thereforedecreasing the average by 4 year . find what was theoriginal strength of class ? | "let original strength = y then , 40 y + 12 x 32 = ( y + 12 ) x 36 ⇒ 40 y + 384 = 36 y + 432 ⇒ 4 y = 48 ∴ y = 12 b" | a ) 10 , b ) 12 , c ) 16 , d ) 20 , e ) 22 | b | divide(subtract(multiply(12, subtract(40, 4)), multiply(12, 32)), 4) | multiply(n1,n2)|subtract(n0,n3)|multiply(n1,#1)|subtract(#2,#0)|divide(#3,n3)| | general | B |
two airplanes take off from one airfield at noon . one flies due east at 202 miles per hour while the other flies directly northeast at 283 miles per hour . approximately how many miles apart are the airplanes at 2 p . m . ? | "e in two hours : the plane flying east will be 404 miles away from airport . the other plane will be 566 miles away from airport . 566 / 404 = ~ 1.4 = ~ sqrt ( 2 ) this means that planes formed a right isocheles triangle = > sides of such triangles relate as 1 : 1 : sqrt ( 2 ) = > the planes are 404 miles apart . e" | a ) 166 , b ) 332 , c ) 400 , d ) 483 , e ) 404 | e | sqrt(subtract(power(multiply(283, 2), 2), power(multiply(202, 2), 2))) | multiply(n1,n2)|multiply(n0,n2)|power(#0,n2)|power(#1,n2)|subtract(#2,#3)|sqrt(#4)| | physics | E |
the average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day . during the first 7 days , his average wages was $ 87 per day and the average wages during the last 7 days was $ 90 per day . what was his wage on the 8 th day ? | "average daily wage of a worker for 15 consecutive working days = 90 $ during the first 7 days , the daily average daily wage = 87 $ during the last 7 days , the daily average daily wage = 90 $ wage on 8 th day = 90 * 15 - ( 87 * 7 + 90 * 7 ) = 1350 - ( 609 + 630 ) = 1350 - 1239 = 111 answer e" | a ) $ 83 , b ) $ 90 , c ) $ 92 , d ) $ 97 , e ) $ 111 | e | subtract(multiply(90, 15), add(multiply(87, 7), multiply(90, 7))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)| | physics | E |
a and b invests rs . 3000 and rs . 6000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ? | "( 3 * 6 + 6 * 6 ) : ( 6 * 12 ) 18 : 24 = > 3 : 4 . answer : e" | a ) 9 : 6 , b ) 3 : 8 , c ) 3 : 1 , d ) 9 : 9 , e ) 3 : 4 | e | divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(6000, add(6, 6))) | add(n2,n2)|multiply(n0,n2)|multiply(n0,const_2)|multiply(n2,#2)|multiply(n1,#0)|add(#1,#3)|divide(#5,#4)| | gain | E |
how many integers from 101 to 700 , inclusive , remains the value unchanged when the digits were reversed ? | "question is asking for palindrome first digit possibilities - 1 through 6 = 6 7 is not possible here because it would result in a number greater than 7 ( i . e 707 , 717 . . ) second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 6 * 10 = 60 answer is b ." | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | b | divide(700, const_10) | divide(n1,const_10)| | general | B |
x can do a piece of work in 40 days . he works at it for 8 days and then y finished it in 16 days . how long will y take to complete the work ? | "work done by x in 8 days = 8 * 1 / 40 = 1 / 5 remaining work = 1 - 1 / 5 = 4 / 5 4 / 5 work is done by y in 16 days whole work will be done by y in 16 * 5 / 4 = 20 days answer is e" | a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | e | multiply(16, inverse(subtract(const_1, divide(8, 40)))) | divide(n1,n0)|subtract(const_1,#0)|inverse(#1)|multiply(n2,#2)| | physics | E |
the average monthly salary of 20 employees in an organisation is rs . 1300 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 1400 * 21 - 1300 * 20 ) = rs . 3400 . answer : d" | a ) 3600 , b ) 3890 , c ) 88798 , d ) 3400 , e ) 2891 | d | subtract(multiply(add(1300, 100), add(20, const_1)), multiply(1300, 20)) | add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general | D |
at what price must an book costing $ 47.50 be marked in order that after deducting 20 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "c $ 62.50 cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 80 / 100 ) = 59.375 mp = 74.2" | a ) 72.5 , b ) 55.5 , c ) 64.2 , d ) 82.5 , e ) 60.5 | c | multiply(divide(divide(multiply(47.50, add(const_100, 25)), const_100), subtract(const_100, 20)), const_100) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | gain | C |
nicky and cristina are running a 200 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . b | a ) 35 , b ) 30 , c ) 45 , d ) 40 , e ) 50 | b | add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3))) | multiply(n1,n3)|subtract(n2,n3)|divide(#0,n3)|divide(#0,#1)|add(#2,#3) | physics | B |
3 - fourth of two - third of 3 - seventh of a number is 27 . what is 10 % of that number ? | explanation : solution : assume the number be x . then , 3 / 4 of 2 / 3 of 3 / 7 of x = 27 . x = 27 * 7 / 3 * 3 / 2 * 4 / 3 . x = 126 . ' . 10 % of 126 = 10 / 100 * 126 = 12.6 answer : a | a ) 12.6 % , b ) 6.3 % , c ) 27 % , d ) 25.2 % , e ) none of these | a | divide(multiply(27, 10), multiply(multiply(divide(3, add(const_3, const_4)), multiply(divide(3, const_4), divide(const_2, const_3))), const_100)) | add(const_3,const_4)|divide(n0,const_4)|divide(const_2,const_3)|multiply(n2,n3)|divide(n0,#0)|multiply(#1,#2)|multiply(#4,#5)|multiply(#6,const_100)|divide(#3,#7) | gain | A |
the present ratio of students to teachers at a certain school is 30 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ? | "lets say t = 10 - - - > s = 300 - - - > 350 / 15 = a bit less than 25 ( a and b will only make this ratio lesser , eliminate a and b ) go for t = 12 - - > s = 30 * 12 = 360 - - - > 360 + 50 / 12 + 5 = less than 25 . either you can now go further with t = 15 or mark e as the answer . check : t = 15 - - - > s = 30 * 15 = 450 - - - > 450 + 50 / 20 = 25 . answer : e" | a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | e | divide(add(30, 25), 25) | add(n0,n4)|divide(#0,n4)| | other | E |
the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 90 metres ? | "speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 90 m = 90 / 5 = 18 seconds . answer : e" | a ) 16 seconds , b ) 76 seconds , c ) 26 seconds , d ) 12 seconds , e ) 18 seconds | e | divide(90, multiply(add(15, 3), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics | E |
john want to buy a $ 100 trouser at the store , but he think it ’ s too expensive . finally , it goes on sale for $ 30 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it ’ s 100 – 30 = 70 . the “ original ” is our starting point ; in this case , it ’ s 100 . ( 70 / 100 ) * 100 = ( 0.7 ) * 100 = 70 % . d" | a ) 20 % , b ) 30 % , c ) 40 % , d ) 70 % , e ) 80 % | d | subtract(100, 30) | subtract(n0,n1)| | general | D |
a sum was put at simple interest at a certain rate for 6 years had it been put at 4 % higher rate , it would have fetched 144 more . find the sum . | "difference in s . i . = p × t / 100 ( r 1 − r 2 ) ⇒ 144 = p × 6 x 4 / 100 ( ∵ r 1 - r 2 = 2 ) ⇒ p = 144 × 100 / 6 × 4 = 600 answer b" | a ) 500 , b ) 600 , c ) 700 , d ) 800 , e ) 900 | b | divide(144, multiply(divide(4, const_100), 6)) | divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)| | gain | B |
in a certain game , each player scores either 2 points or 5 points . if n players score 2 points and m players score 5 points , and the total number of points scored is 50 , what is the least possible positive difference q between n and m ? | "we have equation 2 n + 5 m = 50 we have factor 2 in first number and we have factor 5 in second number . lcm ( 2 , 5 ) = 10 so we can try some numbers and we should start from 5 because it will be less list than for 2 2 * 5 = 10 and n should be equal 20 4 * 5 = 20 and n should be equal 15 6 * 5 = 30 and n should be equal 10 8 * 5 = 40 and n should be equal 5 10 * 5 = 50 and n should be equal 0 third variant give us the mininal difference n - m = 10 - 6 = 4 and there is some mistake in my way of thinking because we do n ' t have such answer ) if we change the task and will seek for difference between m and n than minimal result q will be 8 - 5 = 3 and answer b" | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(5, 2) | subtract(n1,n0)| | general | B |
linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 200 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 200 . so 1 / 4 of her savings is $ 200 . so her original savings are 4 times $ 200 = $ 800 correct answer d" | a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 900 | d | divide(200, subtract(const_1, divide(3, 4))) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)| | general | D |
which number need to add to 859622 to get a no . exactly divisible by 456 ? | "dividend = quotient * divisor + reminder 859622 / 456 gives quotient = 1885 and reminder = 62 . so , the next number divisible by 456 is 456 places infront of 456 * 1885 which means 456 – 62 = 394 should be added to 859622 . e" | a ) 345466 , b ) 465767 , c ) 565676 , d ) 645469 , e ) 859622 | e | multiply(456, subtract(add(floor(divide(859622, 456)), const_1), divide(859622, 456))) | divide(n0,n1)|floor(#0)|add(#1,const_1)|subtract(#2,#0)|multiply(n1,#3)| | general | E |
there is enough provisions for 100 girls in a hostel for 50 days . if there were 20 men less , how long will the provision last ? | we have , m 1 d 1 = m 2 d 2 100 * 50 = 20 * d 2 d 2 = 100 * 50 / 20 = 250 days . answer : a | a ) 250 , b ) 255 , c ) 260 , d ) 265 , e ) 270 | a | add(divide(multiply(100, 20), const_10), 50) | multiply(n0,n2)|divide(#0,const_10)|add(n1,#1) | physics | A |
the workforce of company x is 60 % female . the company hired 26 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ? | "let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 26 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 26 ) 0.05 x = 14.3 x = 286 - this is quantity of employees before adding 26 men so after adding it will be 312 answer is c" | a ) 160 , b ) 220 , c ) 312 , d ) 360 , e ) 420 | c | add(divide(multiply(divide(55, const_100), 26), subtract(divide(60, const_100), divide(55, const_100))), 26) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)| | gain | C |
what will be the cost of house to paint which area equal to 484 sq ft , if the price per foot of building is rs . 20 | let the side of the square plot be a ft . a 2 = 484 = > a = 22 length of the fence = perimeter of the plot = 4 a = 88 ft . cost of building the fence = 88 * 20 = rs . 1760 . answer : b | a ) 1800 , b ) 1760 , c ) 1400 , d ) 2600 , e ) 3600 | b | multiply(multiply(sqrt(484), const_4), 20) | sqrt(n0)|multiply(#0,const_4)|multiply(n1,#1) | geometry | B |
an article with cost price of 192 is sold at 25 % profit . what is the selling price ? | "sp = 1.25 * 192 = 240 answer : d" | a ) 198 , b ) 200 , c ) 204 , d ) 240 , e ) 210 | d | add(192, multiply(192, divide(25, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain | D |
a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 35 % ? | "cost price = rs . 500 profit = 35 % of 500 = rs . 175 selling price = cost price + profit = 500 + 175 = 675 answer : e" | a ) 600 , b ) 887 , c ) 256 , d ) 654 , e ) 675 | e | add(500, multiply(500, divide(35, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain | E |
the cost of an article is decreased by 20 % . if the original cost is $ 40 , find the decrease cost . | "original cost = $ 40 decrease in it = 20 % of $ 40 = 20 / 100 ã — 40 = 800 / 100 = $ 8 therefore , decrease cost = $ 40 - $ 8 = $ 32 answer : b" | a ) 33 , b ) 32 , c ) 68 , d ) 36 , e ) 38 | b | divide(multiply(40, 20), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain | B |
if s = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } , how much less is the mean of the numbers in s than the median of the numbers in s ? | "mean = ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) / 7 = 4 median = 4 difference = 4 - 4 = 0 option a" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | multiply(divide(add(add(add(add(add(2, 1), 3), 4), 5), 6), add(const_2, const_4)), divide(add(3, 4), const_2)) | add(n0,n1)|add(const_2,const_4)|add(n2,n3)|add(n2,#0)|divide(#2,const_2)|add(n3,#3)|add(n4,#5)|add(n5,#6)|divide(#7,#1)|multiply(#8,#4)| | general | A |
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 86 minutes , then the slower pipe alone will be able to fill the tank in | "solution let the slower pipe alone fill the tank in x minutes . then , faster pipes will fill it in x / 3 minutes . therefore , 1 / x + 3 / x = 1 / 36 ‹ = › 4 / x = 1 / 36 ‹ = › x = 144 min . answer c" | a ) 81 min , b ) 108 min , c ) 144 min , d ) 192 min , e ) none | c | multiply(add(const_1, const_4), 86) | add(const_1,const_4)|multiply(n0,#0)| | physics | C |
the average weight of a group of boys is 30 kg . after a boy of weight 40 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 40 kg joins the group , total weight of boys = 30 x + 40 so 30 x + 40 = 31 ( x + 1 ) = > x = 9 . answer : e" | a ) 14 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | e | add(subtract(40, add(30, 1)), 1) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general | E |
mary can do a piece of work in 28 days . rosy is 40 % more efficient than mary . the number of days taken by rosy to do the same piece of work is ? | "ratio of times taken by mary and rosy = 140 : 100 = 14 : 10 suppose rosy takes x days to do the work . 14 : 10 : : 28 : x = > x = 20 days . hence , rosy takes 20 days to complete the work . answer : c" | a ) 22 , b ) 24 , c ) 20 , d ) 25 , e ) 27 | c | divide(28, add(const_1, divide(40, const_100))) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | physics | C |
alex takes a loan of $ 9,000 to buy a used truck at the rate of 9 % simple interest . calculate the annual interest to be paid for the loan amount . | from the details given in the problem principle = p = $ 9,000 and r = 9 % or 0.09 expressed as a decimal . as the annual interest is to be calculated , the time period t = 1 . plugging these values in the simple interest formula , i = p x t x r = 9,000 x 1 x 0.09 = 810.00 annual interest to be paid = $ 810 answer : b | a ) 680 , b ) 810 , c ) 800 , d ) 730 , e ) 750 | b | divide(multiply(multiply(multiply(9, const_100), sqrt(const_100)), 9), const_100) | multiply(n1,const_100)|sqrt(const_100)|multiply(#0,#1)|multiply(n1,#2)|divide(#3,const_100) | gain | B |
how much greater is the combined area in square inches of the front and back of a rectangular sheet of paper measuring 11 inches by 17 inches than that of a rectangular sheet of paper measuring 8.5 inches by 11 inches ? | "area of sheet a = 11 * 17 area of sheet b = 11 * 8.5 difference in area = 11 * 8.5 required % = ( 11 * 8.5 ) * 100 / ( 11 * 8.5 ) % = 100 % answer : c" | a ) 50 % , b ) 87 % , c ) 100 % , d ) 187 % , e ) 200 % | c | multiply(divide(subtract(multiply(rectangle_area(11, 17), const_2), multiply(rectangle_area(8.5, 11), const_2)), rectangle_area(11, 17)), const_100) | rectangle_area(n0,n1)|rectangle_area(n0,n2)|multiply(#0,const_2)|multiply(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#5,const_100)| | geometry | C |
36 people { a 1 , a 2 … a 36 } meet and shake hands in a circular fashion . in other words , there are totally 36 handshakes involving the pairs , { a 1 , a 2 } , { a 2 , a 3 } , … , { a 35 , a 36 } , { a 36 , a 1 } . then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is | ans : { a 1 , a 2 } , { a 2 , a 3 } , { a 3 , a 4 } , { a 4 , a 5 } , { a 5 , a 6 } , { a 6 , a 7 } … , { a 35 , a 36 } , { a 36 , a 1 } from the above arrangement , if we separate a 3 , a 6 , a 9 , . . . . . a 36 . total 12 persons the reamining persons must have shaked hand with atleast one person . so answer is 12 . answer : d | ['a ) 76', 'b ) 55', 'c ) 44', 'd ) 12', 'e ) 91'] | d | divide(36, const_3) | divide(n0,const_3) | geometry | D |
how many different positive integers are factors of 40 ? | 2 * 20 4 * 10 8 * 5 answer : d | a ) 3 , b ) 2 , c ) 1 , d ) 6 , e ) 4 | d | add(power(const_2, const_2), const_2) | power(const_2,const_2)|add(#0,const_2) | other | D |
if n = 7 ^ 11 – 7 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 7 ^ 3 - 7 . now 7 ^ 3 has last digit 3 . since 7 ^ 11 ( or for that matter 7 ^ 3 ) is greater than 7 , we subtract 7 from 13 ( the 10 + 3 - - > 10 has come from carry over from the tenth place ) . thus 13 - 7 = 6 is the answer . option d | a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | d | divide(log(7), log(power(7, 11))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general | D |
how many of the integers between 45 and 105 are even ? | "number start between 45 to 105 is 60 numbers half of them is even . . which is 30 answer : b" | a ) 21 , b ) 30 , c ) 11 , d ) 10 , e ) 9 | b | divide(subtract(105, 45), const_2) | subtract(n1,n0)|divide(#0,const_2)| | general | B |
two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 18 mph between the 2 friends until the friends meet . how many miles does the pigeon travel ? | "a 54 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 18 mph = 54 miles" | a ) 54 , b ) 66 , c ) 80 , d ) 36 , e ) 96 | a | multiply(18, divide(divide(36, 2), 6)) | divide(n1,n3)|divide(#0,n0)|multiply(n2,#1)| | physics | A |
8 people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire rest of the car , then the share of each of the remaining persons increased by ? | when there are eight people , the share of each person is 1 / 8 th the total cost answer : b | a ) 1 / 9 , b ) 1 / 8 , c ) 1 / 7 , d ) 7 / 8 , e ) 8 / 7 | b | divide(const_1, 8) | divide(const_1,n0) | general | B |
in a tree , 3 / 7 of the birds are robins while the rest are bluejays . if 1 / 3 of the robins are female and 3 / 5 of the bluejays are female , what fraction of the birds in the tree are male ? | "the fraction of birds that are male robins is ( 2 / 3 ) ( 3 / 7 ) = 2 / 7 . the fraction of birds that are male bluejays is ( 2 / 5 ) ( 4 / 7 ) = 8 / 35 . the total fraction of male birds is 2 / 7 + 8 / 35 = 18 / 35 . the answer is c ." | a ) 11 / 35 , b ) 14 / 35 , c ) 18 / 35 , d ) 31 / 70 , e ) 37 / 70 | c | add(multiply(divide(3, 7), divide(const_2.0, 5)), multiply(divide(3, 7), divide(1, 3))) | divide(n0,n1)|divide(n0,n5)|divide(n5,n1)|divide(n2,n3)|multiply(#0,#1)|multiply(#2,#3)|add(#4,#5)| | general | C |
what is the least number which when divided by 5 , 6 , 7 and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder ? | "lcm of 5 , 6 , 7 and 8 = 840 hence the number can be written in the form ( 840 k + 3 ) which is divisible by 9 . if k = 1 , number = ( 840 × 1 ) + 3 = 843 which is not divisible by 9 . if k = 2 , number = ( 840 × 2 ) + 3 = 1683 which is divisible by 9 . hence 1683 is the least number which when divided by 5 , 6 , 7 and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder . answer : b" | a ) 1108 , b ) 1683 , c ) 2007 , d ) 3363 , e ) 1436 | b | add(lcm(lcm(5, 6), lcm(7, 8)), 3) | lcm(n0,n1)|lcm(n2,n3)|lcm(#0,#1)|add(n4,#2)| | general | B |
a number when divided by a divisor leaves a remainder of 21 . when twice the original number is divided by the same divisor , the remainder is 11 . what is the value of the divisor ? | "let the number is n , the divisor = d , i will make the two equations - n = xd + 21 2 n = yd + 11 where x and y are integers solving them : d ( y - 2 x ) = 34 as d is also integer and 34 is a prime number , the d should be 34 to satisfy the above equation . hence answer is ' d '" | a ) 12 , b ) 13 , c ) 34 , d ) 37 , e ) 59 | d | add(21, 11) | add(n0,n1)| | general | D |
a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 105 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 × 7 : 12 × 5 : 15 × 3 = 2 × 7 : 12 × 1 : 3 × 3 = 14 : 12 : 9 amount that c should pay = 105 × 9 / 35 = 3 × 9 = 27 answer is b ." | a ) 35 , b ) 27 , c ) 25 , d ) 15 , e ) 55 | b | multiply(105, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | multiply(n4,n5)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(#3,#0)|divide(#0,#4)|multiply(n6,#5)| | general | B |
what is 5 + 7 | d | a ) 2 , b ) 4 , c ) 10 , d ) 12 , e ) 11 | d | circle_area(divide(5, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | general | D |
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