Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 57 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 57 / 3 = 19 kmph . answer : c" | a ) 15 , b ) 17 , c ) 19 , d ) 18 , e ) 16 | c | subtract(57, divide(multiply(57, const_2), const_3)) | multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)| | physics |
a dishonest dealer professes to sell goods at the cost price but uses a weight of 650 grams per kg , what is his percent ? | "explanation : 650 - - - 350 100 - - - ? = > 53.8 % answer : d" | a ) 15 % , b ) 25 % , c ) 65 % , d ) 53.8 % , e ) 35 % | d | subtract(multiply(divide(const_100, 650), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)| | gain |
the slant height of a cone is 12 cm and radius of the base is 4 cm , find the curved surface of the cone . | "π * 12 * 4 = 48 answer : c" | a ) 33 , b ) 88 , c ) 48 , d ) 77 , e ) 27 | c | multiply(multiply(const_pi, 4), 12) | multiply(n1,const_pi)|multiply(n0,#0)| | geometry |
what is the remainder when ( 55 ) ( 57 ) is divided by 7 ? | "( 55 ) ( 57 ) = ( 56 - 1 ) ( 56 + 1 ) = 56 ^ 2 - 1 which is 1 less than a multiple of 7 . then the remainder will be 6 . the answer is e ." | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | e | reminder(multiply(57, 55), 7) | multiply(n0,n1)|reminder(#0,n2)| | general |
the sum of ages of 4 children born 2 years different each is 40 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 2 ) , ( x + 4 ) , ( x + 6 ) years . then , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) = 40 4 x + 12 = 40 = > 4 x = 28 x = 7 x + 6 = 7 + 6 = 13 answer : a" | a ) 13 , b ) 9 , c ) 16 , d ) 17 , e ) 18 | a | divide(add(add(add(add(2, const_4), add(2, const_4)), add(const_4, const_4)), 40), 4) | add(n1,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)| | general |
the ratio , by volume , of bleach to detergent to water in a certain solution is 2 : 40 : 100 . the solution will be altered so that the ratio of bleach to detergent is tripled while the ratio of detergent to water is halved . if the altered solution will contain 200 liters of water , how many liters of detergent will ... | "b : d : w = 2 : 40 : 100 bnew / dnew = ( 1 / 3 ) * ( 2 / 40 ) = ( 1 / 60 ) dnew / wnew = ( 1 / 2 ) * ( 40 / 100 ) = ( 1 / 5 ) wnew = 200 dnew = wnew / 5 = 200 / 5 = 40 so , answer will be e" | a ) 60 , b ) 55 , c ) 50 , d ) 45 , e ) 40 | e | multiply(divide(200, 100), divide(40, const_2)) | divide(n3,n2)|divide(n1,const_2)|multiply(#0,#1)| | other |
a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 75 paise per sq m is | "explanation : area to be plastered = [ 2 ( l + b ) × h ] + ( l × b ) = [ 2 ( 25 + 12 ) × 6 ] + ( 25 × 12 ) = 744 sq m cost of plastering = 744 × ( 75 / 100 ) = rs . 558 answer : d" | a ) rs . 258 , b ) rs . 358 , c ) rs . 458 , d ) rs . 558 , e ) none of these | d | multiply(divide(75, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | divide(n3,const_100)|multiply(n0,n2)|multiply(n1,n2)|multiply(n0,n1)|multiply(#1,const_2)|multiply(#2,const_2)|add(#4,#5)|add(#6,#3)|multiply(#7,#0)| | physics |
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 140 pages ? | "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 ... | a ) 12 , b ) 18 , c ) 20 , d ) 42 , e ) 30 | d | divide(multiply(140, add(multiply(4, const_2), 4)), 40) | multiply(n0,const_2)|add(n0,#0)|multiply(n4,#1)|divide(#2,n1)| | physics |
there is a total of 100 marbles in a box , each of which is red , green , blue , or white . if one marble is drawn from the box at random , the probability that it will be white is 1 / 4 and the probability that it will be green is 1 / 5 . what is the probability that the marble will be either red or blue ? | "p ( red or blue ) = 1 - p ( white ) - p ( green ) = 20 / 20 - 5 / 20 - 4 / 20 = 11 / 20 the answer is e ." | a ) 2 / 3 , b ) 3 / 5 , c ) 7 / 10 , d ) 9 / 20 , e ) 11 / 20 | e | subtract(1, add(divide(1, 4), divide(1, 5))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|subtract(n1,#2)| | general |
you and your friend spent a total of $ 11 for lunch . your friend spent $ 3 more than you . how much did your friend spend on their lunch ? | "my lunch = l , my friends lunch = l + 1 ( l ) + ( l + 3 ) = 11 l + l + 3 - 3 = 11 - 3 2 l = 8 l = 4 my friends lunch l + 3 = 4 + 3 = $ 7 , the answer is c" | a ) $ 9 , b ) $ 3 , c ) $ 7 , d ) $ 6 , e ) $ 5 | c | add(divide(subtract(11, 3), const_2), 3) | subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)| | general |
what is the square root of 9 ? | "3 x 3 = 9 answer b" | a ) 1 , b ) 3 , c ) 2 , d ) 4 , e ) 9 | b | circle_area(divide(9, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other |
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 |
haresh went to buy fruits . she bought 1 apple , 3 bananas , 5 oranges and paid 21 rupees . another person bought 7 apple , 5 bananas , 3 oranges and paid 51 rupees . if i want to buy 1 apple , 1 banana , 1 orange then how much i have to pay ? | there can be multiple possible rates of apple , banana and orange to satisfy given conditions , but cost of 1 apple , 1 banana , 1 orange will be rs 9 . 1 x + 3 y + 5 z = 21 . . . . . . . . ( 1 ) 7 x + 5 y + 3 z = 51 . . . . . . . . ( 2 ) adding ( 1 ) and ( 2 ) , we get 8 x + 8 y + 8 z = 72 8 * ( x + y + z ) = 72 x + y... | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 7 | b | divide(add(21, 51), add(1, 7)) | add(n3,n7)|add(n0,n4)|divide(#0,#1) | general |
5 cylinders are built successively , the first is 4 meters tall and the following cylinders are each 4 meters taller than the previous . if the perimeter is 10 pi m , what is the total surface area of the 5 cylinders , excluding the bases of each cylinder ? | s . a . = h x p + 2 x a we need to excude the bottom area , so the formula becomes : s . a . = h x p + a p = 2 pi x r , so 10 pi = 2 pi x r r = 5 since we have 5 cylinders with different heights the equation becomes s . a . = ( 4 + 8 + 12 + 16 + 20 ) x 10 pi + 5 x pi x r ^ 2 60 x 10 pi + 5 pi x 5 ^ 2 600 pi + 125 pi = ... | ['a ) 725 pi sq m', 'b ) 1025 pi sq m', 'c ) 600 pi sq m', 'd ) 125 pi sq m', 'e ) 850 pi sq m'] | a | surface_cylinder(divide(10, const_2), add(add(multiply(4, const_4), add(multiply(4, const_3), add(4, multiply(4, const_2)))), add(multiply(4, const_3), add(4, multiply(4, const_2))))) | divide(n3,const_2)|multiply(n1,const_2)|multiply(n1,const_3)|multiply(n1,const_4)|add(n1,#1)|add(#4,#2)|add(#5,#3)|add(#6,#5)|surface_cylinder(#0,#7) | geometry |
mahesh can do a piece of work in 60 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | "work done by mahesh in 60 days = 20 * 1 / 60 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 2 / 3 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 3 / 2 = 45 days answer is a" | a ) 45 , b ) 25 , c ) 37 , d ) 41 , e ) 30 | a | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 60))), 30)) | divide(const_1,n0)|multiply(n1,#0)|subtract(const_1,#1)|divide(#2,n2)|divide(const_1,#3)| | physics |
a land owner needs to fence his semicircular land leaving an opening of length of 3 m for access . the radius of the semicircular plot is 7 m . how long would be the fence in meters ? | perimeter of the semicircle = ï € xd / 2 + d = ï € x 7 + 14 = 22 + 14 = 36 m length of the fence = 36 - 3 = 33 m answer : b | a ) 22 , b ) 33 , c ) 36 , d ) 44 , e ) 51 | b | subtract(add(multiply(7, divide(add(add(const_2, const_10), const_10), 7)), multiply(const_2, 7)), 3) | add(const_10,const_2)|multiply(n1,const_2)|add(#0,const_10)|divide(#2,n1)|multiply(n1,#3)|add(#4,#1)|subtract(#5,n0) | physics |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 21 kmph respectively . when they meet , it is found that one train has travelled 60 km more than the other one . the distance between the two stations is ? | d 444 km 1 h - - - - - 5 ? - - - - - - 60 12 h rs = 16 + 21 = 37 t = 12 d = 37 * 12 = 444 | a ) 222 km , b ) 333 km , c ) 555 km , d ) 444 km , e ) 666 km | d | add(multiply(divide(60, subtract(21, 16)), 16), multiply(divide(60, subtract(21, 16)), 21)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3) | physics |
a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 10000 that a person who randomly guesses the answer to each question will be a winner ? | "a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 1000 that a person who randomly guesses the answer to each question will be a winner ? a . ... | a ) 5 , b ) 10 , c ) 50 , d ) 100 , e ) 1000 | d | multiply(const_1000, divide(1, 10000)) | divide(n0,n1)|multiply(#0,const_1000)| | general |
if a = 105 and a ^ 3 = 21 × 25 × 45 × q , what is the value of q ? | "a = 105 = 3 * 5 * 7 a ^ 3 = 21 × 25 × 45 × q = > a ^ 3 = ( 7 * 3 ) x ( 5 * 5 ) x ( 3 ^ 2 * 5 ) x q = > a ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x q = > ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x q q = 7 ^ 2 = 49 answer d" | a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 54 | d | divide(power(105, 3), multiply(multiply(21, 25), 45)) | multiply(n2,n3)|power(n0,n1)|multiply(n4,#0)|divide(#1,#2)| | general |
what number should replace the question mark ? 5 , 24 , 11 , 20 , 17 , 16 , - - - ? | "answer : b 5 , 24 , 11 , 20 , 17 , 16 , 23 ? there are two alternate sequences : + 6 and - 4 ." | a ) 20 , b ) 23 , c ) 22 , d ) 21 , e ) 26 | b | add(add(add(5, 5), 5), 5) | add(n0,n0)|add(n0,#0)|add(#1,n0)| | general |
the average ( arithmetic mean ) of 4 different integers is 72 . if the largest integer is 90 , what is the least possible value of the smallest integer ? | total of integers = 72 * 4 = 288 lowest of the least possible integer is when the middle 2 intergers are at the maximum or equal to the highest possible integer . but all integers are distinct . so if the largest integer is 90 , then the middle 2 will be 88 and 89 lowest of least possible integer = 288 - ( 90 + 89 + 88... | a ) 1 , b ) 19 , c ) 21 , d ) 30 , e ) 33 | c | subtract(subtract(subtract(multiply(72, 4), 90), subtract(90, const_1)), subtract(90, const_2)) | multiply(n0,n1)|subtract(n2,const_1)|subtract(n2,const_2)|subtract(#0,n2)|subtract(#3,#1)|subtract(#4,#2) | general |
stacy has a 66 page history paper due in 6 days . how many pages per day would she have to write to finish on time ? | 66 / 6 = 11 answer : c | a ) 9 , b ) 8 , c ) 11 , d ) 8.5 , e ) 6 | c | divide(66, 6) | divide(n0,n1) | physics |
jane makes toy bears . when she works with an assistant , she makes 80 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 w ? | c . let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . w = [ ( 2 - 1 ) / 1 ] * 100 % = 100 % | a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(const_10, subtract(subtract(const_100, 80), 10)), const_100) | subtract(const_100,n0)|subtract(#0,n1)|divide(const_10,#1)|multiply(#2,const_100) | physics |
diana took out a charge account at the general store and agreed to pay 7 % simple annual interest . if she charges $ 75 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | 1.07 * $ 75 = $ 80.25 the answer is e . | a ) $ 79.25 , b ) $ 79.50 , c ) $ 79.75 , d ) $ 80.00 , e ) $ 80.25 | e | add(75, multiply(divide(7, const_100), 75)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1) | general |
a rower whose speed is 9 km / hr in still water rows to a certain point upstream and back to the starting point in a river which flows at 3 km / hr . what is the rower ' s average speed ( in km / hr ) for the total journey ? | "time upstream = d / 6 time downstream = d / 12 total time = d / 6 + d / 12 = d / 4 average speed = 2 d / ( d / 4 ) = 8 km / hr the answer is b ." | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | divide(add(subtract(9, const_0.5), add(9, 3)), const_2) | add(n0,const_0.5)|subtract(n0,n1)|add(#0,#1)|divide(#2,const_2)| | general |
a man completes of a job in 10 days . at this rate , how many more days will it takes him to finish the job ? | work done = 5 / 8 balance work = ( 1 - 5 / 8 ) = 3 / 8 let the required number of days be x . then , ( 5 / 8 ) : ( 3 / 8 ) = : : 10 : x ( 5 / 8 ) * x = ( 3 / 8 ) * 10 x = ( 3 / 8 ) * 10 * ( 8 / 5 ) x = 6 . answer is a . | a ) 6 , b ) 12 , c ) 7 , d ) 15 , e ) 5 | a | divide(multiply(10, subtract(const_1, divide(add(const_2, const_3), multiply(const_2, const_4)))), divide(add(const_2, const_3), multiply(const_2, const_4))) | add(const_2,const_3)|multiply(const_2,const_4)|divide(#0,#1)|subtract(const_1,#2)|multiply(n0,#3)|divide(#4,#2) | gain |
in a sample of 800 high school students in which all students are either freshmen , sophomores , juniors , or seniors , 22 percent are juniors and 74 percent are not sophomores . if there are 160 seniors , how many more freshmen than sophomores are there among the sample of students ? | "sophomores = 26 % juniors = 22 % seniors = 160 / 800 = 20 % total : 68 % 1 - 68 % = 32 % ( freshmen ) 0,32 * 800 - 0,26 * 800 = 256 - 208 = 48 . answer choice b" | a ) 42 , b ) 48 , c ) 56 , d ) 208 , e ) 256 | b | subtract(divide(multiply(subtract(subtract(subtract(const_100, multiply(divide(160, 800), const_100)), 22), subtract(const_100, 74)), 800), const_100), divide(multiply(subtract(const_100, 74), 800), const_100)) | divide(n3,n0)|subtract(const_100,n2)|multiply(#0,const_100)|multiply(n0,#1)|divide(#3,const_100)|subtract(const_100,#2)|subtract(#5,n1)|subtract(#6,#1)|multiply(n0,#7)|divide(#8,const_100)|subtract(#9,#4)| | gain |
a student needs 60 % of the marks on a test to pass the test . if the student gets 80 marks and fails the test by 100 marks , find the maximum marks set for the test . | "60 % = 180 marks 1 % = 3 marks 100 % = 300 marks the answer is b ." | a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 450 | b | divide(add(80, 100), divide(60, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | gain |
a football player scores 2 goals in his fifth match thus increasing his average goals score by 0.1 . the total number of goals in his 5 matches would be | "while this question can be solved with a rather straight - forward algebra approach ( as the other posters have noted ) , it can also be solved by testing the answers . one of those numbers must be the total number of goals . . . from a tactical standpoint , it ' s best to test either answer b or answer d , so if the ... | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | c | add(subtract(multiply(const_4, 2), multiply(multiply(const_4, 5), 0.1)), 2) | multiply(n0,const_4)|multiply(n2,const_4)|multiply(n1,#1)|subtract(#0,#2)|add(n0,#3)| | general |
if the weight of 12 meters long rod is 14 kg . what is the weight of 6 meters long rod ? | "answer ∵ weight of 12 m long rod = 14 kg ∴ weight of 1 m long rod = 14 / 12 kg ∴ weight of 6 m long rod = 14 x 6 / 12 = 7 kg option : c" | a ) 8 kg . , b ) 10.8 kg . , c ) 7 kg . , d ) 18.0 kg , e ) none | c | divide(multiply(6, 14), 12) | multiply(n1,n2)|divide(#0,n0)| | physics |
a man can row downstream at 18 kmph and upstream at 16 kmph . find the speed of the man in still water and the speed of stream respectively ? | "explanation : let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 18 - - - ( 1 ) and x - y = 16 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 34 = > x = 17 , y = 1 . answer : option e" | a ) 13 , 3 , b ) 12 , 6 , c ) 15 , 3 , d ) 14 , 4 , e ) 17 , 1 | e | divide(divide(add(18, 16), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
after giving a discount of rs . 45 the shopkeeper still gets a profit of 20 % , if the cost price is rs . 180 . find the markup % ? | cost price = 180 s . p = 180 * 120 / 100 = 216 disc = 45 so . . . mark price = 216 + 45 = 261 . . . . . . mark up % = 261 - 180 / 180 = 81 / 180 = . 45 or 45 % answer : c | a ) 0.4 , b ) 0.55 , c ) 0.45 , d ) 0.48 , e ) 0.58 | c | divide(subtract(add(add(180, divide(180, divide(const_100, 20))), 45), 180), 180) | divide(const_100,n1)|divide(n2,#0)|add(n2,#1)|add(n0,#2)|subtract(#3,n2)|divide(#4,n2) | gain |
if x , y , and z are positive integers and 3 x = 2 y = 4 z , then the least possible value of x + y + z is | "given 3 x = 2 y = 4 z x + y + z in terms of x = x + ( 3 x / 2 ) + ( 3 x / 4 ) = 26 x / 8 = 13 x / 4 now checking with each of the answers and see which value gives a minimum integer value . a x = 4 / 13 * 15 , not an integer b , c , e can be ruled out similarly . d is minimum value as x = 13 * 4 / 13 = 4 answer is e" | a ) 15 , b ) 25 , c ) 30 , d ) 42 , e ) 13 | e | add(subtract(divide(multiply(multiply(3, 2), 4), 3), divide(multiply(multiply(3, 2), 4), 2)), divide(multiply(multiply(3, 2), 4), 4)) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n2)|divide(#1,n0)|divide(#1,n1)|subtract(#3,#4)|add(#2,#5)| | general |
a train travelling 80 kmph to pass through a tunnel of 70 km at 5 : 12 am . the train leaves the tunnel at 5 : 18 am . find the length of train ? | distance = speed * time let x is length of train 5 : 18 - 5 : 12 = 6 minutes = 6 / 60 hr = 1 / 10 hr now d = ( x + 7 ) = 80 kmph * 1 / 10 x = 1 km answer : a | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(multiply(divide(subtract(18, 12), const_60), 80), divide(70, const_10)) | divide(n1,const_10)|subtract(n5,n3)|divide(#1,const_60)|multiply(n0,#2)|subtract(#3,#0) | physics |
if x is equal to the sum of the integers from 30 to 50 , inclusive , and y is the number of even integers from 30 to 50 , inclusive , what is the value of x + y ? | since the integers from 30 - 50 are consecutive we can use the ` ` median x # of terms formula ' ' to determine x . there are 21 terms and the median is 40 . this gives a total of 21 x 40 = 840 . there are 11 even integers from 30 - 50 = y . therefore , x + y = 851 . answer : e | a ) 810 , b ) 811 , c ) 830 , d ) 850 , e ) 851 | e | add(multiply(divide(add(30, 50), const_2), add(subtract(50, 30), const_1)), add(divide(subtract(50, 30), const_2), const_1)) | add(n0,n1)|subtract(n1,n0)|add(#1,const_1)|divide(#1,const_2)|divide(#0,const_2)|add(#3,const_1)|multiply(#2,#4)|add(#5,#6)| | general |
of the 100 people in a room , 2 / 5 are women . if 1 / 4 of the people are married , what is the maximum number of women in the room who could be unmarried ? | women = 2 / 5 * 100 = 40 married = 1 / 4 * 100 = 25 unmarried = 75 max ( un - married women ) = 40 answer e | a ) 32 , b ) 34 , c ) 36 , d ) 38 , e ) 40 | e | multiply(100, divide(2, 5)) | divide(n1,n2)|multiply(n0,#0) | general |
the length of the bridge , which a train 160 meters long and travelling at 45 km / hr can cross in 30 seconds , is : | "speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 160 + x ) / 30 = 25 / 2 = = > 2 ( 160 + x ) = 750 = = > x = 215 m . answer : option d" | a ) 230 , b ) 244 , c ) 245 , d ) 215 , e ) 236 | d | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 160) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
yesterday ' s closing prices of 1,760 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a hi... | "lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 1760 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h ... | a ) 960 , b ) 726 , c ) 1,100 , d ) 1,320 , e ) 1,694 | a | multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100))) | divide(n1,const_100)|multiply(n1,const_100)|add(#0,const_1)|subtract(#1,const_10)|add(#2,const_1)|subtract(#3,const_10)|divide(#5,#4)|multiply(#2,#6)| | gain |
if a certain toy store ' s revenue in november was 2 / 5 of its revenue in december and its revenue in january was 1 / 5 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "n = 2 d / 5 j = n / 5 = 2 d / 25 the average of november and january is ( n + j ) / 2 = 12 d / 25 / 2 = 6 d / 25 d is 25 / 6 times the average of november and january . the answer is e ." | a ) 5 / 2 , b ) 10 / 3 , c ) 15 / 4 , d ) 25 / 3 , e ) 25 / 6 | e | divide(1, divide(add(divide(2, 5), multiply(divide(2, 5), divide(1, 5))), const_2)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|add(#0,#2)|divide(#3,const_2)|divide(n2,#4)| | general |
john was thrice as old as tom 6 years ago . john will be 3 / 2 times as old as tom in 6 years . how old is tom today ? | "j - 6 = 3 ( t - 6 ) , so j = 3 t - 12 j + 6 = 3 / 2 * ( t + 6 ) 2 j + 12 = 3 t + 18 2 ( 3 t - 12 ) + 12 = 3 t + 18 3 t = 30 t = 10 the answer is c ." | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | c | multiply(6, divide(3, 2)) | divide(n1,n2)|multiply(n0,#0)| | general |
a rainstorm increased the amount of water stored in state j reservoirs from 130 billion gallons to 160 billion gallons . if the storm increased the amount of water in the reservoirs to 60 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the... | "let total capacity be x we know 160 = 0.60 x x = 160 / 0.60 = 266.67 prior to storm , we had 130 bn gallons 266.67 - 130 = 136.67 answer : c" | a ) 92 , b ) 85 , c ) 136.67 , d ) 140.67 , e ) 150 | c | divide(divide(multiply(160, const_100), 60), const_2) | multiply(n1,const_100)|divide(#0,n2)|divide(#1,const_2)| | general |
what is the 8 th digit to the right of the decimal point in the decimal equivalent of 73 / 11 ? | "73 / 11 = 6.6363 . . . . 63 is non - terminating repeating decimal . the 8 th digit to the right of decimal point will be 3 . answer b" | a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 2 | b | divide(73, 11) | divide(n1,n2)| | general |
if rs . 510 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ? | "( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 510 * 2 / 17 = rs . 60 b ' s share = 510 * 3 / 17 = rs . 90 c ' s share = 510 * 12 / 17 = rs . 360 . answer : b" | a ) rs . 300 , b ) rs . 360 , c ) rs . 389 , d ) rs . 368 , e ) rs . 323 | b | divide(510, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1)) | divide(n3,n4)|divide(n1,n2)|multiply(#1,#0)|add(#0,#2)|add(#3,n3)|divide(n0,#4)| | general |
if p # q denotes the least common multiple of p and q , then p = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ? | "there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , ... | a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72 | c | add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,const_2)|add(n3,#3)| | general |
if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 6 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 6 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 6 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 6 = d" | a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 3 | d | multiply(const_3.0, 1) | multiply(n5,n3)| | general |
there are 8 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 7 teams . . . so total no of matches = 8 x 7 = 56 . now , each match is played twice = > 56 x 2 but 2 teams play a match = > 56 x 2 / 2 = 56 . ans : d" | a ) 15 , b ) 16 , c ) 28 , d ) 56 , e ) 64 | d | divide(multiply(8, subtract(8, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
a shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 3.5 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 12 . if he buys 4 a rupee , his cp = 3.5 he selling at 3.5 a rupee , his sp = 4 profit = sp - cp = 4 - 3.5 = 0.5 profit percent = 0.5 / 3.5 * 100 = 14.28 % answer : c" | a ) 13.2 % , b ) 15.2 % , c ) 14.28 % , d ) 32 1 / 3 % , e ) 33 1 / 2 % | c | divide(multiply(3.5, const_100), 4) | multiply(n1,const_100)|divide(#0,n0)| | gain |
solution a is made up of alcohol and water mixed in the ratio of 21 : 4 by volume ; solution b is made up of alcohol and water mixed in the ratio of 2 : 3 by volume . if solution a and solution b are mixed in the ratio of 5 : 6 by volume , what percent of the resultant mixture is alcohol ? | let mixture after mixing a and b is 110 ml ( number assumed for calculation because of 5 : 6 ) now solution a is 50 ml and solu b is 60 ml further in a 100 ml a contains 84 ml alch so 50 ml contain 42 ml in mix b 100 ml solution contain 40 ml alc , 60 ml of b , it will be 24 ml alcohol . so 42 + 24 = 66 in 110 ml solut... | a ) 32.5 % , b ) 40 % , c ) 60 % , d ) 65 % , e ) can not be determined | c | multiply(add(multiply(divide(21, add(21, 4)), divide(5, add(5, 6))), multiply(divide(2, add(2, 3)), divide(6, add(5, 6)))), const_100) | add(n0,n1)|add(n4,n5)|add(n2,n3)|divide(n0,#0)|divide(n4,#1)|divide(n2,#2)|divide(n5,#1)|multiply(#3,#4)|multiply(#5,#6)|add(#7,#8)|multiply(#9,const_100) | other |
at veridux corporation , there are 180 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 135 male associates , how many female managers are there ? | "well , first let â € ™ s take care of the â € œ totals â € . the numbers in the â € œ totals â € row must add up . if 90 are females , the other 180 â € “ 90 = 90 must be males . similarly , the numbers in the â € œ totals â € column must add up . if 40 are managers , then the other 180 â € “ 40 = 140 must be as... | a ) 95 , b ) 20 , c ) 85 , d ) 75 , e ) 35 | c | multiply(40, const_1) | multiply(n2,const_1)| | general |
keats library purchases a number of new books , all in the category of biography , and the library does not acquire any other books . with the addition of the new biographies , the biography collection of the library amounts to 55 % of the total number of books in the library . if prior to the purchase , only 20 % of t... | "let x be the number of new biographies added to the library . let b be the original number of biographies , so the original number of books was 5 b . 0.55 ( 5 b + x ) = b + x 1.75 b = 0.45 x x = 3.9 b the answer is d ." | a ) 270 % , b ) 310 % , c ) 350 % , d ) 390 % , e ) 430 % | d | multiply(divide(divide(subtract(multiply(55, const_100), multiply(20, const_100)), subtract(const_100, 55)), 20), const_100) | multiply(n0,const_100)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#1)|divide(#3,#2)|divide(#4,n1)|multiply(#5,const_100)| | general |
after decreasing 24 % in the price of an article costs rs . 836 . find the actual cost of an article ? | cp * ( 76 / 100 ) = 836 cp = 11 * 100 = > cp = 1100 answer : d | a ) 1667 , b ) 6789 , c ) 1200 , d ) 1100 , e ) 1421 | d | divide(836, subtract(const_1, divide(24, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1) | gain |
two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 36 . find the ratio of their corresponding heights . | "we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 16 / 36 - - > height / height = 4 / 6 . answer : ... | a ) 4 / 6 , b ) 5 / 4 , c ) 3 / 2 , d ) 5 / 7 , e ) 2 / 3 | a | divide(sqrt(16), sqrt(36)) | sqrt(n0)|sqrt(n1)|divide(#0,#1)| | geometry |
in a weight - lifting competition , the total weight of joe ' s two lifts was 1500 pounds . if twice the weight of his first lift was 300 pounds more than the weight of his second lift , what was the weight , in pounds , of his first lift ? | "this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was 1500 pounds . we sum the two variables to obtain : f ... | a ) 225 , b ) 275 , c ) 325 , d ) 350 , e ) 600 | e | divide(add(1500, 300), const_3) | add(n0,n1)|divide(#0,const_3)| | general |
the cash realised on selling a 14 % stock is rs . 109.25 , brokerage being 1 / 4 % is | "explanation : cash realised = rs . ( 109.25 - 0.25 ) = rs . 109 . answer : b" | a ) 123 , b ) 109 , c ) 100 , d ) 156 , e ) 240 | b | subtract(109.25, divide(1, 4)) | divide(n2,n3)|subtract(n1,#0)| | general |
if 47.5 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | "students enrolled in biology are 47.5 % and therefore not enrolled are 52.5 % . so of 880 is 880 * . 525 = 462 answer is d 462" | a ) 110 , b ) 330 , c ) 550 , d ) 462 , e ) 880 | d | multiply(divide(880, const_100), subtract(const_100, 47.5)) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)| | other |
two ships are sailing in the sea on the two sides of a lighthouse . the angle of elevation of the top of the lighthouse is observed from the ships are 30 ° and 45 ° respectively . if the lighthouse is 100 m high , the distance between the two ships is : | let ab be the lighthouse and c and d be the positions of the ships . then , ab = 100 m , acb = 30 ° and adb = 45 ° . ab = tan 30 ° = 1 ac = ab x 3 = 1003 m . ac 3 ab = tan 45 ° = 1 ad = ab = 100 m . ad cd = ( ac + ad ) = ( 1003 + 100 ) m = 100 ( 3 + 1 ) = ( 100 x 2.73 ) m = 273 m . answer = c | a ) 173 m , b ) 200 m , c ) 273 m , d ) 300 m , e ) 373 m | c | multiply(add(sqrt(const_3), const_1), 100) | sqrt(const_3)|add(#0,const_1)|multiply(n2,#1) | physics |
a tank can supply water to a village for 80 days . if a leak at the bottom of the tank drains out 10 liters per day , the supply lasts for 60 days only . for how many days will the supply last if the leak drains out 20 liters per day ? | "losing 10 liters per day results in a loss of 600 liters in 60 days . so , those 600 liters were for 20 days , making daily consumption of the village 30 liters per day . thus the capacity of the tank is 30 * 80 = 2400 liters . losing 20 liters plus 30 liters gives 50 liters per day . at this rate the supply will last... | a ) 42 , b ) 45 , c ) 48 , d ) 50 , e ) 55 | c | divide(multiply(divide(multiply(10, 60), subtract(80, 60)), 80), add(20, divide(multiply(10, 60), subtract(80, 60)))) | multiply(n1,n2)|subtract(n0,n2)|divide(#0,#1)|add(n3,#2)|multiply(n0,#2)|divide(#4,#3)| | physics |
a can do a work in 25 days and b can do it in 10 days . in how many days a and b can do the work ? | "explanation : a ' s 1 day ' s work = 1 / 25 b ' s 1 day ' s work = 1 / 10 they work together = 1 / 25 + 1 / 10 = 7 / 50 = 50 / 7 = 6 8 / 7 days answer : option c" | a ) 20 days , b ) 10 days , c ) 6 8 / 7 days , d ) 8 days , e ) 7 days | c | divide(const_1, subtract(divide(const_1, 10), divide(const_1, 25))) | divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|divide(const_1,#2)| | physics |
a ' and ' b ' are positive integers such that their lcm is 20 and their hcf is 1 . what is the difference between the maximum and minimum possible values of ' a - b ' ? | possible values of a and b can be 54 ; 45 and 120 ; 201 maximum possible value for a - b is 20 - 1 = 19 minimum possible value for a - b is 1 - 20 = - 19 19 - ( - 19 ) = 38 ans is d | a ) 28 , b ) 29 , c ) 34 , d ) 38 , e ) 36 | d | subtract(add(20, 20), add(1, 1)) | add(n0,n0)|add(n1,n1)|subtract(#0,#1) | general |
the size of a television screen is given as the length of the screen ' s diagonal . if the screens were flat , then the area of a square 18 - inch screen would be how many square inches greater than the area of a square 16 - inch screen ? | pythogoras will help here ! let the sides be x and diagonal be d then d ^ 2 = 2 x ^ 2 and area = x ^ 2 now plug in the given diagonal values to find x values and then subtract the areas ans will be 18 ^ 2 / 2 - 16 ^ 2 / 2 = 68 / 2 = 34 ans c . | ['a ) 2', 'b ) 4', 'c ) 34', 'd ) 38', 'e ) 40'] | c | divide(subtract(power(18, const_2), power(16, const_2)), const_2) | power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|divide(#2,const_2) | geometry |
an alloy weighing 24 ounces is 70 percent gold . how many ounces of pure gold must be added to create an alloy that is 80 percent gold ? | "in 24 ounces , gold is 24 * ( 70 / 100 ) = 16.8 ounces . now we add x ounces of pure gold to make it 90 % gold . so 16.8 + x = ( 24 + x ) * 80 / 100 = > x = 12 . answer is c ." | a ) 6 , b ) 9 , c ) 12 , d ) 24 , e ) 48 | c | divide(multiply(divide(80, const_100), 24), divide(70, const_100)) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|divide(#2,#1)| | gain |
find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1456 ? | "1456 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1300 answer : e" | a ) 1000 , b ) 2217 , c ) 2889 , d ) 2777 , e ) 1300 | e | divide(1456, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)| | general |
what is the 4 digit no . in which the 1 st digit is 1 / 3 of the second , the 3 rd is the sum of the 1 st and 2 nd , and the last is 3 times the second ? | first digit is 1 / 3 second digit = > the numbers can be 1 & 3 , 2 & 6 , 3 & 9 . first + second = third = > we can eliminate 3 & 9 since 3 + 9 = 12 . last is 3 times the second = > we can eliminate option 2 & 6 since 3 * 6 = 18 . hence the number is 1349 c | a ) 1123 , b ) 1234 , c ) 1349 , d ) 1455 , e ) 1567 | c | add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(add(4, 3), add(4, 3))) | add(n0,n3)|multiply(const_10,const_100)|multiply(const_100,const_3)|add(#1,#2)|multiply(#0,#0)|add(#3,#4) | general |
what number is 76 more than one - fourth of itself ? | "1 / 4 x + 76 = x that means 76 = 3 / 4 x x = ( 76 * 4 ) / 3 = 304 / 3 = 101 a is the answer" | a ) 101 , b ) 225 , c ) 304 , d ) 324 , e ) 336 | a | divide(76, subtract(const_1, divide(const_1, const_4))) | divide(const_1,const_4)|subtract(const_1,#0)|divide(n0,#1)| | general |
a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint w , in gallons , that must be purchased in order to measure... | "given w : b = 3 : 5 that means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture . but we want least amount of whiteblack paints for minimum of 2 gallons of mixture , so lets reduce keeping same ratio , 1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture , but we want only 2 gallons , l... | a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | multiply(const_0_25,n2)|multiply(n1,n2)|add(#0,#1)| | general |
someone on a skateboard is traveling 15 miles per hour . how many feet does she travel in 30 seconds ? ( 1 mile = 5280 feet ) | "per second = > 15 * 5280 ft / 60 * 60 = 22 ft 30 seconds = > 22 * 30 = 660 ft answer : b" | a ) 600 ft , b ) 660 ft , c ) 670 ft , d ) 680 ft , e ) 700 ft | b | multiply(30, divide(multiply(15, 5280), const_3600)) | multiply(n0,n3)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
a company produces 72000 bottles of water everyday . if a case can hold 10 bottles of water . how many cases are required by the company to hold its one day production | "number of bottles that can be held in a case = 10 number of cases required to hold 72000 bottles = 72000 / 10 = 7200 cases . so the answer is d = 7200" | a ) 2000 , b ) 4500 , c ) 5000 , d ) 7200 , e ) 9000 | d | divide(72000, 10) | divide(n0,n1)| | physics |
the sum of 99 consecutive integers is 9999 . what is the greatest integer in the set ? | "sum of n numbers = ( n ( 2 a + ( n - 1 ) d ) ) / 2 where n = no of terms a = 1 st term , d - difference between two terms ( ie d = a 2 - a 1 ) here sum = 9999 , n = 99 d = 1 ( since consecutive numbers ) 9999 = ( 99 / 2 ) * ( 2 a + ( 99 - 1 ) 1 ) from this ' a ' ( ie the 1 st term ) = 52 nth term in a sequence : nth t... | a ) 100 , b ) 120 , c ) 149 , d ) 150 , e ) 151 | d | add(add(power(add(add(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), power(add... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
a train of 24 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 3.5 km long ? | "d = 25 * 60 + 3500 = 5000 m t = 5000 / 60 * 18 / 5 = 300 sec = 5 mins answer : c" | a ) 4 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | c | add(divide(multiply(add(24, const_1), 60), const_1000), 3.5) | add(n0,const_1)|multiply(n1,#0)|divide(#1,const_1000)|add(n4,#2)| | physics |
the tax on a commodity is diminished by 20 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 80 * 115 = 9200 - - - - - - - - - - - 10000 - - - - - - - - - - - 800 100 - - - - - - - - - - - ? = > 8 % decrease answer : b" | a ) 9 % decrease , b ) 8 % decrease , c ) 6 % decrease , d ) 1 % decrease , e ) 2 % decrease | b | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(20, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general |
the length of a rectangular floor is more than its breadth by 200 % . if rs . 484 is required to paint the floor at the rate of rs . 3 / sq m , what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 484 / 3 = 161.33 sq m l b = 161.33 i . e . , l * l / 3 = 161.33 l ^ 2 = 484 = > l = 22 . d" | a ) 12 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | d | multiply(sqrt(divide(divide(484, 3), const_3)), const_3) | divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)| | gain |
a train running at the speed of 50 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 50 * ( 5 / 18 ) m / sec = 125 / 9 m / sec length of train ( distance ) = speed * time ( 125 / 9 ) * 9 = 125 meter answer : a" | a ) 125 , b ) 272 , c ) 278 , d ) 277 , e ) 112 | a | multiply(divide(multiply(50, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 33 ? | "first , let ' s translate the expression 21 # 33 , using the definition given : 21 # 33 = 21 × 24 × 27 × 30 × 33 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 33 = 3 ^ 6 ( 7 × 8 × 9 × 10 × 11 ) now let ' s replace each consecutive integer with its prime factorization ... | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | c | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | add(const_2,const_2)|add(n0,#0)|add(#1,#0)|add(#2,#1)| | general |
after decreasing 20 % in the price of an article costs rs . 620 . find the actual cost of an article ? | "cp * ( 80 / 100 ) = 620 cp = 7.75 * 100 = > cp = 775 answer : a" | a ) 775 , b ) 620 , c ) 750 , d ) 800 , e ) 720 | a | divide(620, subtract(const_1, divide(20, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
90 % of the population of a village is 23040 . the total population of the village is ? | "answer ∵ 90 % of p = 23040 ∴ p = ( 23040 x 100 ) / 90 = 25600 correct option : d" | a ) 32256 , b ) 24000 , c ) 24936 , d ) 25600 , e ) none | d | multiply(divide(const_100, 90), 23040) | divide(const_100,n0)|multiply(n1,#0)| | general |
if y is the smallest positive integer such that 8,820 multiplied by y is the square of an integer , then y must be | "i just tried plugging in the numbers and found out that 5 * 8820 = 44 , 100 , which is a square of 210 b" | a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | b | multiply(add(const_2, const_3), const_2) | add(const_2,const_3)|multiply(#0,const_2)| | geometry |
a train 100 m long crosses a platform 125 m long in 40 sec ; find the speed of the train ? | "d = 100 + 125 = 225 t = 40 s = 225 / 40 * 18 / 5 = 20.25 kmph answer : e" | a ) 36.21 , b ) 77.11 , c ) 54.12 , d ) 33.0 , e ) 20.25 | e | subtract(multiply(40, multiply(const_60.0, const_0_2778)), 100) | multiply(const_60.0,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
a group of men decided to do a work in 15 days , but 8 of them became absent . if the rest of the group did the work in 18 days , find the original number of men ? | "original number of men = 8 * 18 / ( 18 - 15 ) = 48 answer is b" | a ) 15 , b ) 48 , c ) 54 , d ) 33 , e ) 45 | b | divide(multiply(8, 18), subtract(18, 15)) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | physics |
x , y , and z are all unique numbers . if x is chosen randomly from the set { 5 , 6 , 7 , 8 , 9 , 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ? | "p ( x is prime ) = 3 / 7 p ( y is prime ) = 1 / 4 if y is prime , then z is not prime since y and z are unique . then the probability is 3 / 7 * 1 / 4 = 3 / 28 the answer is e ." | a ) 1 / 5 , b ) 3 / 20 , c ) 1 / 7 , d ) 2 / 15 , e ) 3 / 28 | e | multiply(divide(const_2, 5), divide(const_1, const_4)) | divide(const_2,n0)|divide(const_1,const_4)|multiply(#0,#1)| | probability |
a and b can do a piece of work in 2 days , b and c in 3 days , c and a in 4 days . how long will c take to do it ? | "2 c = 1 / 3 + 1 / 4 – 1 / 2 = 1 / 12 c = 1 / 24 = > 24 days answer : d" | a ) 14 days , b ) 20 days , c ) 22 days , d ) 24 days , e ) 26 days | d | divide(multiply(2, const_3), subtract(divide(add(divide(multiply(2, const_3), 4), add(divide(multiply(2, const_3), 2), divide(multiply(2, const_3), 3))), const_2), divide(multiply(2, const_3), 2))) | multiply(n0,const_3)|divide(#0,n0)|divide(#0,n1)|divide(#0,n2)|add(#1,#2)|add(#4,#3)|divide(#5,const_2)|subtract(#6,#1)|divide(#0,#7)| | physics |
a tourist does not have to pay tax on the first $ 600 of goods he purchases in country b , but does have to pay a 12 percent tax on the portion of the total value that is in excess of $ 600 . what tax must be paid by a tourist if he buys goods with a total value of $ 1720 ? | "correct answer : e the tourist must pay tax on $ 1720 - $ 600 = $ 1120 . thus , the amount of tax he has to pay is 0.12 ( $ 1120 ) = $ 134.40 . the correct answer is e ." | a ) $ 54.00 , b ) $ 64.80 , c ) $ 90.00 , d ) $ 100.80 , e ) $ 134.40 | e | divide(multiply(subtract(1720, 600), 12), const_100) | subtract(n3,n0)|multiply(n1,#0)|divide(#1,const_100)| | general |
the area of one square is x ^ 2 + 10 x + 25 and the area of another square is 4 x ^ 2 − 20 x + 25 . if the sum of the perimeters of both squares is 36 , what is the value of x ? | "the areas are ( x + 5 ) ^ 2 and ( 2 x - 5 ) ^ 2 . the lengths of the sides are x + 5 and 2 x - 5 . if we add the two perimeters : 4 ( x + 5 ) + 4 ( 2 x - 5 ) = 36 12 x = 36 x = 3 the answer is a ." | a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | a | divide(subtract(36, subtract(multiply(4, divide(10, 2)), 10)), 10) | divide(n1,n0)|multiply(#0,n3)|subtract(#1,n1)|subtract(n7,#2)|divide(#3,n1)| | general |
when magnified 1000 times by an electron microscope , the image of a certain circular piece of tissue has a diameter of 0.3 centimeter . the actual diameter of the tissue , in centimeters , is | it is very easy if x is the diameter , then the magnified length is 1000 x . ince 1000 x = 0.3 then x = 0.3 / 1000 = 0.0003 . the answer is d | a ) 0.005 , b ) 0.002 , c ) 0.001 , d ) 0.0003 , e ) 0.0002 | d | divide(0.3, 1000) | divide(n1,n0) | geometry |
a group of n students can be divided into equal groups of 4 with 3 student left over or equal groups of 5 with 3 students left over . what is the sum of the two smallest possible values of n ? | "4 x + 3 = 5 y + 3 . . . . . . . . . . . ie : 4 x - 5 y = 0 x , y must be > 1 and y is even ie ( 2 , 4,6 , . . etc ) if y = 2 thus x = fraction ( not possible ) if y = 4 thus x = 5 n = 23 if y = 6 thus x = not possible fraction if y = 8 thus x = 10 n = 43 23 + 43 = 66 . . . . . a" | a ) 66 , b ) 46 , c ) 49 , d ) 53 , e ) 86 | a | add(add(multiply(5, const_2), 3), add(multiply(5, multiply(const_2, 3)), 3)) | multiply(n2,const_2)|multiply(const_2,n3)|add(n3,#0)|multiply(n2,#1)|add(n3,#3)|add(#2,#4)| | general |
a 600 meter long train crosses a signal post in 40 seconds . how long will it take to cross a 9 kilometer long bridge , at the same speed ? | "s = 600 / 40 = 15 mps s = 9000 / 15 = 600 sec = 10 min . answer : c" | a ) 4 min , b ) 2 min , c ) 10 min , d ) 9 min , e ) 3 min | c | divide(divide(multiply(9, const_1000), speed(600, 40)), const_60) | multiply(n2,const_1000)|speed(n0,n1)|divide(#0,#1)|divide(#2,const_60)| | physics |
q - 1 : 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 5 ? | "6 two digit numbers divisible by 5 must end in either 0 or 5 . since there is no 0 , all correct answers are a combination of the other digits and 5 . there are six other numbers plus the 5 so there are 6 correct answers . correct answer is d ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | add(add(add(add(add(1, 1), 1), 3), 4), 1) | add(n0,n1)|add(n0,#0)|add(n3,#1)|add(n4,#2)|add(n0,#3)| | general |
a car travels at a speed of 65 miles per hour . how far will it travel in 9 hours ? | "during each hour , the car travels 65 miles . for 9 hours it will travel 65 + 65 + 65 + 65 + 65 + 65 + 65 + 65 + 65 = 9 × 65 = 585 miles correct answer is c ) 585 miles" | a ) 125 miles , b ) 225 miles , c ) 585 miles , d ) 425 miles , e ) 525 miles | c | multiply(65, 9) | multiply(n0,n1)| | physics |
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 |
yearly subscription to professional magazines cost a company $ 940.00 . to make a 50 % cut in the magazine budget , how much less must be spent ? | "total cost 940 940 * 50 / 100 = 470 so the cut in amount is 470 the less amount to be spend is 940 - 470 = 470 answer : e" | a ) 654 , b ) 655 , c ) 656 , d ) 657 , e ) 470 | e | multiply(divide(subtract(const_100, 50), const_100), 940.00) | subtract(const_100,n1)|divide(#0,const_100)|multiply(n0,#1)| | general |
price of a book increases 15 % successively ( 2 times ) what is the new price of the book more compared to that of the old price : | new price is 1.15 * 1.15 * old price = 1.3225 * old price increase in price = 0.3225 * old price % increase = 100 * 0.3225 * old price / old price = 32.25 % answer : a | a ) 32.25 % , b ) 23.34 % , c ) 36 % , d ) 39 % , e ) 29 % | a | multiply(subtract(multiply(divide(add(const_100, 15), const_100), divide(add(const_100, 15), const_100)), const_1), const_100) | add(n0,const_100)|divide(#0,const_100)|multiply(#1,#1)|subtract(#2,const_1)|multiply(#3,const_100) | general |
there are 7 fictions and 6 non - fictions . how many cases are there such that 2 fictions and 2 non - fictions are selected from them ? | "number of ways of selecting 2 fiction books = 7 c 2 number of ways of selecting 2 non fiction books = 6 c 2 7 c 2 * 6 c 2 = 21 * 15 = 315 answer : c" | a ) 90 , b ) 120 , c ) 315 , d ) 180 , e ) 200 | c | divide(multiply(multiply(7, const_4), multiply(6, 7)), power(factorial(2), 2)) | factorial(n2)|multiply(n0,const_4)|multiply(n0,n1)|multiply(#1,#2)|power(#0,n2)|divide(#3,#4)| | general |
andrew purchased 6 kg of grapes at the rate of 74 per kg and 9 kg of mangoes at the rate of 59 per kg . how much amount did he pay to the shopkeeper ? | cost of 6 kg grapes = 74 × 6 = 444 . cost of 9 kg of mangoes = 59 × 9 = 531 . total cost he has to pay = 444 + 531 = 975 b | a ) 1000 , b ) 975 , c ) 980 , d ) 1020 , e ) 1080 | b | add(multiply(6, 74), multiply(9, 59)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1) | gain |
the ratio of the incomes of uma and bala is 4 : 3 and the ratio of their expenditure is 3 : 2 . if at the end of the year , each saves $ 5000 then the income of uma is ? | "let the income of uma and bala be $ 4 x and $ 3 x let their expenditures be $ 3 y and $ 2 y 4 x - 3 y = 5000 - - - - - - - 1 ) 3 x - 2 y = 5000 - - - - - - - 2 ) from 1 ) and 2 ) x = 5000 uma ' s income = 4 x = 4 * 5000 = $ 20000 answer is e" | a ) $ 24000 , b ) $ 23000 , c ) $ 22000 , d ) $ 21000 , e ) $ 20000 | e | multiply(4, 5000) | multiply(n0,n4)| | other |
3 / 4 of 1 / 2 of 2 / 5 of 5040 = ? | c 756 ? = 5040 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 756 | a ) 392 , b ) 229 , c ) 756 , d ) 493 , e ) 540 | c | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5040) | divide(n3,n5)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3)|multiply(n6,#4) | general |
there are 17 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | "the total number of stations = 19 from 19 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in 19 p ₂ ways . ² ⁰ p ₂ = 19 * 18 = 342 . answer : d" | a ) 156 , b ) 167 , c ) 157 , d ) 342 , e ) 380 | d | multiply(add(17, const_1), add(add(17, const_1), const_1)) | add(n0,const_1)|add(#0,const_1)|multiply(#0,#1)| | physics |
there are 18 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ? | "the total number of stations = 20 from 20 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 20 p 2 ways . 20 p 2 = 20 * 19 = 380 answer : b" | a ) 800 , b ) 380 , c ) 360 , d ) 478 , e ) 566 | b | multiply(add(18, const_2), subtract(add(18, const_2), const_1)) | add(n0,const_2)|subtract(#0,const_1)|multiply(#0,#1)| | physics |
tickets numbered 1 to 30 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn has a number which is a multiple of 3 or 5 ? | "here , the sample space s = ( 1 , 2 , 3 , 4 , 5 , . . . , 29 , 30 ) . let e = the event of getting a multiple of 3 or 5 . e = ( 3 , 6 , 9 , 12 , 15 , 18 , 5 , 10 , 20 , 21 , 24 , 25 , 27 , 30 ) p ( e ) = n ( e ) / n ( s ) = 14 / 30 = 7 / 15 answer : b" | a ) 2 / 15 , b ) 7 / 15 , c ) 10 / 30 , d ) 7 / 30 , e ) 8 / 30 | b | divide(3, const_10) | divide(n2,const_10)| | general |
a 160 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | s = 160 / 6 * 18 / 5 = 96 kmph answer : a | a ) 96 kmph , b ) 94 kmph , c ) 92 kmph , d ) 86 kmph , e ) 76 kmph | a | multiply(divide(160, 6), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6) | physics |
an alloy of copper and zinc contains copper and zinc in the ratio 6 : 4 . another alloy of copper and zinc contains copper and zinc in the ratio 3 : 7 . in what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc ? | "let alloy _ 1 be x units , and let alloy _ 2 be y units . so , fraction of copper in alloy _ 1 = 6 x / 10 , and fraction of zinc in alloy _ 1 = 4 x / 10 similarly , fraction of copper in alloy _ 2 = 3 y / 10 , and fraction of zinc in alloy _ 2 = 7 y / 10 . mixing them , we get copper = 6 x / 10 + 3 y / 10 ; zinc = 4 x... | a ) 12 : 1 , b ) 1 : 2 , c ) 4 : 1 , d ) 2 : 5 , e ) 2 : 6 | b | divide(subtract(divide(6, add(3, 7)), divide(3, 7)), subtract(divide(3, 7), divide(3, add(3, 7)))) | add(n2,n3)|divide(n2,n3)|divide(n0,#0)|divide(n2,#0)|subtract(#2,#1)|subtract(#1,#3)|divide(#4,#5)| | general |
a tank contains 7,500 gallons of a solution that is 6 percent sodium chloride by volume . if 2,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "we start with 7,500 gallons of a solution that is 6 % sodium chloride by volume . this means that there are 0.06 x 7,500 = 450 gallons of sodium chloride . when 2,000 gallons of water evaporate we are left with 5,500 gallons of solution . from here we can determine what percent of the 5,500 gallon solution is sodium c... | a ) 5.18 % , b ) 6.18 % , c ) 7.18 % , d ) 8.18 % , e ) . 7 % | d | multiply(divide(multiply(multiply(const_100, const_100), divide(6, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | add(const_2,const_3)|divide(n1,const_100)|multiply(const_100,const_100)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#4,const_100)|multiply(#0,#6)|add(#7,#5)|subtract(#2,#8)|divide(#3,#9)|multiply(#10,const_100)| | gain |
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