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rates for having a manuscript typed at a certain typing service are $ 5 per page for the first time a page is typed and $ 2 per page each time a page is revised . if a certain manuscript has 100 pages , of which 40 were revised only once , 10 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ? | 32 / ( 2 * 2 ) = 8 days answer : e | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | multiply(multiply(32, divide(const_1, const_2)), divide(const_1, const_2)) | divide(const_1,const_2)|multiply(n0,#0)|multiply(#0,#1) | physics |
some persons can do a piece of work in 32 days . two times the number of these people will do half of that work in ? | "the units digit of 33 ^ 2 is the units digit of 3 * 3 = 9 which is 9 . the units digit of 17 ^ 3 is the units digit of 7 * 7 * 7 = 343 which is 3 . the units digit of 39 ^ 2 is the units digit of 9 * 9 = 81 which is 1 . the units digit of 9 * 3 * 1 = 27 is 7 . the answer is d ." | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | divide(add(multiply(factorial(33), factorial(2)), multiply(factorial(33), factorial(3))), 33) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
what is the units digit of 33 ^ 2 * 17 ^ 3 * 39 ^ 2 ? | they can make a team of 3 men and 2 women . the number of ways to do this is 6 c 3 * 4 c 2 = 20 * 6 = 120 they can make a team of 2 men and 3 women . the number of ways to do this is 6 c 2 * 4 c 3 = 15 * 4 = 60 the total number of distinct groups is 180 . the answer is c . | a ) 120 , b ) 150 , c ) 180 , d ) 210 , e ) 240 | c | add(add(multiply(multiply(6, 5), 4), multiply(6, 5)), multiply(6, 5)) | multiply(n1,n3)|multiply(n2,#0)|add(#1,#0)|add(#2,#0) | general |
a department of 10 people - 6 men and 4 women - needs to send a team of 5 to a conference . if they want to make sure that there are no more than 3 members of the team from any one gender , how many distinct groups are possible to send ? | "percentage error in calculated area = ( 3 + 3 + ( 3 Γ£ β 3 ) / 100 ) % = 9.09 % answer : a" | a ) 9.09 % , b ) 4.02 % , c ) 4 % , d ) 3 % , e ) 2 % | a | divide(multiply(subtract(square_area(add(const_100, 3)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | gain |
an error 3 % in excess is made while measuring the side of a square . what is the percentage of error in the calculated area of the square ? | "first , you must find the total weight of the mixture given that 80 % of it will be dough . 75 % * total = 40 = > ( 75 / 100 ) total = 40 = > total = 4000 / 75 = > total = 53.33 oz , from there , you must find 25 % of the total 53.33 oz of the mixture . 25 % * total = > ( 25 / 100 ) ( 53.33 ) = 13.33 oz choclate used , not forgetting that the question asks how much chocolate is left over we must subtract the chocolate used from the initial chocolate . 20 - 13.3 = 6.7 oz chocolate left over . answer : d" | a ) 8.5 , b ) 9.5 , c ) 10.5 , d ) 6.7 , e ) 6.5 | d | multiply(divide(25, const_100), 25) | divide(n2,const_100)|multiply(n2,#0)| | gain |
uncle bruce is baking chocolate chip cookies . he has 40 ounces of dough ( with no chocolate ) and 20 ounces of chocolate . how much chocolate is left over if he uses all the dough but only wants the cookies to consist of 25 % chocolate ? | "1 / 10 + 1 / 15 - 1 / x = 1 / 15 x = 10 10 * 20 = 200 answer : c" | a ) 150 liters , b ) 540 liters , c ) 200 liters , d ) 560 liters , e ) 580 liters | c | multiply(multiply(inverse(subtract(add(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 20)), divide(const_1, 15))), const_3), 15) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n3)|add(#0,#1)|add(#4,#2)|subtract(#5,#3)|inverse(#6)|multiply(#7,const_3)|multiply(n1,#8)| | physics |
two pipes a and b can separately fill a tank in 10 and 15 minutes respectively . a third pipe c can drain off 20 liters of water per minute . if all the pipes are opened , the tank can be filled in 15 minutes . what is the capacity of the tank ? | "a : b = 90000 : 25000 = 90 : 25 = 18 : 5 answer : d" | a ) 9 : 2 , b ) 3 : 2 , c ) 18 : 20 , d ) 18 : 5 , e ) 17 : 4 | d | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | multiply(const_3,const_3)|add(#0,const_1)|multiply(#1,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3)| | gain |
a and b started a business investing rs . 90,000 and rs 25,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "let x be the money invested at 3 % . 1.03 x + 1.04 ( 1000 - x ) = 1036.70 . 0.01 x = 1040 - 1036.70 . 0.01 x = 3.30 . x = 330 . the answer is b ." | a ) $ 300 , b ) $ 330 , c ) $ 360 , d ) $ 390 , e ) $ 420 | b | divide(subtract(multiply(multiply(const_100, multiply(add(const_2, const_3), const_2)), add(divide(4, const_100), const_1)), add(add(multiply(const_100, multiply(add(const_2, const_3), const_2)), multiply(multiply(add(const_2, const_3), const_2), 3)), multiply(const_2, const_3))), subtract(add(divide(4, const_100), const_1), add(divide(3, const_100), const_1))) | add(const_2,const_3)|divide(n2,const_100)|divide(n1,const_100)|multiply(const_2,const_3)|add(#1,const_1)|add(#2,const_1)|multiply(#0,const_2)|multiply(#6,const_100)|multiply(#6,n1)|subtract(#4,#5)|add(#7,#8)|multiply(#4,#7)|add(#10,#3)|subtract(#11,#12)|divide(#13,#9)| | gain |
we invested a total of $ 1,000 . we invested one part of the money at 3 % and the rest of the money at 4 % . the total investment with interest at the end of the year was $ 1 , 036.70 . how much money did we invest at 3 % ? | "notice that we can find the number of 2 and 3 digit numbers by just assuming the first digit can also be zero : 0 1 1 1 2 2 2 3 3 3 5 5 number of possibilities = 4 * 4 * 4 = 64 . then , just add up the number of 1 digits numbers = 4 , so total is 64 + 4 = 68 . answer : e" | a ) 48 , b ) 52 , c ) 66 , d ) 84 , e ) 68 | e | divide(factorial(subtract(add(const_4, 1), const_1)), multiply(factorial(1), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | probability |
how many positive integers less than 400 can be formed using the numbers 1 , 2 , 3 and 5 for the digits ? | explanation : let forks = 4 x , spoons = 4 x & knives = 3 x . now , 4 x = 16 hence x = 4 . number of knives = 3 x = 12 . answer : c | a ) 8 , b ) 4 , c ) 12 , d ) 16 , e ) 14 | c | multiply(divide(16, 4), 3) | divide(n3,n0)|multiply(n2,#0) | other |
forks , spoons , and knives in drawer are in the ratio of 4 : 4 : 3 . if there are 16 forks , the number of knives in the drawer is : | "450 = 2 x 3 ^ 2 x 5 ^ 2 now we need two 2 s , one 3 and one 5 to make it perfect cube . so x = 2 ^ 2 x 3 x 5 = 60 . answer is c ." | a ) 2 , b ) 15 , c ) 30 , d ) 60 , e ) 120 | c | add(const_3, const_4) | add(const_3,const_4)| | geometry |
what is the smallest positive integer x such that 450 x is the cube of a positive integer ? | "c number of valid votes = 80 % of 6000 = 4800 . valid votes polled by other candidate = 45 % of 4800 = ( 45 / 100 x 4800 ) = 2160 ." | a ) 2800 , b ) 2700 , c ) 2160 , d ) 2200 , e ) 2300 | c | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 6000) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)| | gain |
in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 6000 , the number of valid votes that the other candidate got , was : | "10 * 6 = 4 * x x = 15 answer : b" | a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | b | multiply(6, divide(10, 4)) | divide(n0,n1)|multiply(n2,#0)| | physics |
10 : 4 seconds : : ? : 6 minutes | "speed of the train = 54 km / hr = ( 54 Γ 10 ) / 36 m / s = 15 m / s length of the train = speed Γ time taken to cross the man = 15 Γ 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) / 15 = > ( 300 + l ) / 15 = 36 = > 300 + l = 15 Γ 36 = 540 = > l = 540 - 300 = 240 meter answer is a ." | a ) 240 , b ) 250 , c ) 260 , d ) 230 , e ) 220 | a | multiply(multiply(const_0_2778, 54), subtract(36, 20)) | multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)| | physics |
a train passes a platform in 36 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km / hr , the length of the platform is | "let the amount lent at 2 % be rs . x amount lent at 10 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 2 / 100 x + 10 / 100 ( 6000 - x ) = 600 - 3 x / 100 = > 600 - 3 / 100 x = 450 = > x = 1875 amount lent at 10 % = 4125 required ratio = 5 : 11 answer : a" | a ) 5 : 11 , b ) 5 : 6 , c ) 5 : 2 , d ) 5 : 8 , e ) 5 : 2 | a | divide(divide(subtract(multiply(450, const_100), multiply(6000, 2)), subtract(10, 2)), divide(subtract(multiply(450, const_100), multiply(6000, 2)), subtract(10, 2))) | multiply(n3,const_100)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|divide(#4,#4)| | gain |
rs . 6000 is lent out in two parts . one part is lent at 2 % 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 ? | "the number of exhaustive outcomes is 36 . let e be the event of getting doublet on the dies is 6 / 36 = 1 / 6 p ( e ) = 1 / 6 . a )" | a ) 1 / 6 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2 | a | divide(const_6, multiply(const_6, const_6)) | multiply(const_6,const_6)|divide(const_6,#0)| | probability |
if two dice are thrown together , the probability of getting a doublet on the dice is | "8 Γ 8 = 4 ^ 2 Γ 2 ^ 2 so total factors = ( 3 + 1 ) ( 3 + 1 ) = 16 answer : d" | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | d | add(power(const_2, const_2), const_2) | power(const_2,const_2)|add(#0,const_2)| | other |
how many different positive integers are factors of 64 ? | "ratio of the sides = Γ’ Β³ Γ’ Λ Ε‘ 343 : Γ’ Β³ Γ’ Λ Ε‘ 512 = 7 : 8 ratio of surface areas = 49 : 64 answer : e" | a ) 12 : 24 , b ) 8 : 16 , c ) 45 : 25 , d ) 18 : 56 , e ) 49 : 64 | e | power(divide(343, 512), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry |
the ratio of the volumes of two cubes is 343 : 512 . what is the ratio of their total surface areas ? | "explanation : for an income of rs . 9 , investment = rs . 100 . for an income of rs 6 , investment = rs . 100 / 9 x 6 = rs 66.66 market value of rs . 100 stock = rs . 66.66 answer is e" | a ) rs 66.55 , b ) rs 68.55 , c ) rs 69.55 , d ) rs 65.55 , e ) rs 66.66 | e | multiply(divide(const_100, 9), 6) | divide(const_100,n1)|multiply(n0,#0)| | gain |
a 6 % stock yields 9 % . the market value of the stock is : | "the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 25 / 36 = ( 5 / 6 ) * ( 5 / 6 ) therefore probability that the next one she draws is pink = 1 / 6 option ( a )" | a ) 1 / 6 , b ) 4 / 7 , c ) 3 / 7 , d ) 16 / 49 , e ) 40 / 49 | a | subtract(const_1, sqrt(divide(25, 36))) | divide(n0,n1)|sqrt(#0)|subtract(const_1,#1)| | general |
jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 25 / 36 , what is the probability that the next one she draws will be pink ? | "let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 253 n * ( n - 1 ) = 506 = 23 * 22 ( trial and error ) the answer is c ." | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | c | divide(add(sqrt(add(multiply(multiply(253, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general |
if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 253 games will be played during the tournament . what is the number of participants ? | more days , more length ( direct ) less breadth , more length ( indirect ) more depth , less length ( indirect days 10 : 30 ; breadth 25 : 50 ; : : 100 : x depth 15 : 10 ; : . 10 * 25 * 15 * x = 30 * 50 * 10 * 100 x = ( 30 * 50 * 10 * 100 ) / 10 * 25 * 15 = 400 so the required length = 400 m answer : a | a ) 400 m , b ) 200 m , c ) 100 m , d ) 89 m , e ) 79 m | a | divide(multiply(multiply(multiply(30, 50), 10), 100), multiply(15, multiply(10, 25))) | multiply(n2,n6)|multiply(n0,n4)|multiply(n0,#0)|multiply(n5,#1)|multiply(n1,#2)|divide(#4,#3) | physics |
it takes 10 days for digging a trench of 100 m long , 50 m broad and 10 m deep . what length of trench , 25 m broad and 15 m deep can be dug in 30 days ? | "explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 3 : 3 : : 56 : x therefore ( 20 x 3 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 60 = 98 hence , the required length is 98 m . answer : e" | a ) 40 m , b ) 64 m , c ) 77 m , d ) 89 m , e ) 98 m | e | multiply(divide(56, multiply(20, 3)), multiply(35, 3)) | multiply(n0,n2)|multiply(n3,n4)|divide(n1,#0)|multiply(#2,#1)| | physics |
if 20 men can build a water fountain 56 metres long in 3 days , what length of a similar water fountain can be built by 35 men in 3 days ? | p = ( 2 / 9 ) * p + 63 ( 7 / 9 ) * p = 63 p = 81 the answer is e . | a ) $ 69 , b ) $ 72 , c ) $ 75 , d ) $ 78 , e ) $ 81 | e | divide(63, subtract(const_1, multiply(divide(1, 9), const_2))) | divide(n1,n2)|multiply(#0,const_2)|subtract(const_1,#1)|divide(n0,#2) | general |
p has $ 63 more than what q and r together would have had if both b and c had 1 / 9 of what p has . how much does p have ? | "speed = 110 * 5 / 18 = 275 / 9 m / sec length of the train = speed * time = 275 / 9 * 9 = 275 m answer : b" | a ) 298 m , b ) 275 m , c ) 208 m , d ) 988 m , e ) 299 m | b | multiply(divide(multiply(110, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
a train running at the speed of 110 km / hr crosses a pole in 9 sec . what is the length of the train ? | 2 ^ y = 4 ^ ( 5 x + 3 ) 2 ^ y = 2 ^ 2 ( 5 x + 3 ) y = 10 x + 6 . . . . . . . . . . . 1 3 ^ ( x - 7 ) = 9 ^ y 3 ^ ( x - 7 ) = 3 ^ 2 y x - 7 = 2 y . . . . . . . . . . . . . 2 put value of y = 10 x + 6 in eq 2 x - 7 = 2 ( 10 x + 6 ) x - 7 = 20 x + 12 19 x = - 19 x = - 1 therefore , y = - 10 + 6 y = - 4 x + y = - 1 - 4 = - 5 answer : b | a ) - 10 , b ) - 5 , c ) - 4 , d ) 3 , e ) 7 | b | add(divide(add(divide(add(negate(7), negate(multiply(multiply(2, 3), 2))), subtract(multiply(2, const_10), const_1)), negate(7)), 2), divide(add(negate(7), negate(multiply(multiply(2, 3), 2))), subtract(multiply(2, const_10), const_1))) | multiply(n0,n3)|multiply(n0,const_10)|negate(n5)|multiply(n0,#0)|subtract(#1,const_1)|negate(#3)|add(#2,#5)|divide(#6,#4)|add(#7,#2)|divide(#8,n0)|add(#9,#7) | general |
if 2 ^ y = 4 ^ ( 5 x + 3 ) and 3 ^ ( x - 7 ) = 9 ^ y , what is the value of x + y ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 140 = 380 m . time taken = 380 / 10 = 38 sec . answer : c" | a ) 28 sec , b ) 16 sec , c ) 38 sec , d ) 18 sec , e ) 17 sec | c | divide(add(240, 140), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)| | general |
a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 140 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "neither car nor garage = total - garage - ( swim - common ) = 75 - 50 - ( 40 - 35 ) = 75 - 55 = 20 answer c" | a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | c | subtract(75, add(add(subtract(50, 35), subtract(40, 35)), 35)) | subtract(n1,n3)|subtract(n2,n3)|add(#0,#1)|add(n3,#2)|subtract(n0,#3)| | other |
of the 75 house in a development , 50 have a two - car garage , 40 have an in - the - ground swimming pool , and 35 have both a two - car garage and an in - the - ground swimming pool . how many houses in the development have neither a two - car garage nor an in - the - ground swimming pool ? | "before price increase price = a after 30 % price increase price = a + ( 30 / 100 ) * a = 1.3 a = 351 ( given ) i . e . a = 351 / 1.3 = $ 270 i . e . 2 a = 2 * 270 = 540 answer : option a" | a ) 540 , b ) 570 , c ) 619 , d ) 649 , e ) 700 | a | multiply(divide(351, divide(add(const_100, 30), const_100)), 2) | add(n0,const_100)|divide(#0,const_100)|divide(n1,#1)|multiply(n2,#2)| | general |
if the price of a certain computer increased 30 percent from a dollars to 351 dollars , then 2 a = | "in believe the answer is d . please see below for explanation . 0 ) we are told the following ratios cgd - college graduate with degree ncg - non college graduate cgn - college graduate no degree cgd ncg cgn 1 8 3 2 in order to make cgd and cgn comparable we need to find the least common multiple of 8 and 3 and that is 24 multiplying the first ratio by 3 and the second ratio by 8 we get cgd ncg cgn 3 24 16 if one picks a random college graduate at this large company , what is the probability this college graduate has a graduate degree ? nr of cgd = 3 nr of cg = 3 + 16 = 19 probability w of cgd / ( cg ) - > 3 / 19 answer d" | 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 |
in a certain large company , the ratio of college graduates with a graduate degree to non - college graduates is 1 : 8 , and ratio of college graduates without a graduate degree to non - college graduates is 2 : 3 . if one picks a random college graduate at this large company , what is the probability w this college graduate has a graduate degree ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 75 2 / 3 sp 1 = 75 sp 1 = 112.50 100 - - - 12.5 = > 12.5 % answer : e" | a ) 20 % , b ) 26 % , c ) 42 % , d ) 27 % , e ) 12.5 % | e | subtract(divide(subtract(const_100, 25), divide(2, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain |
what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 25 % ? | n = ( 4 * 5 - 1 ) * 5 * 5 * 5 = 2375 where 4 cases of first digit { 2,4 , 6,8 } 5 cases of second digit { 0 , 2,4 , 6,8 } 1 case of 22 for two leftmost digit 5 cases of third digit { 1 , 3,5 , 7,9 } 5 cases of fourth digit { 1 , 3,5 , 7,9 } 5 cases of fifth digit { 1 , 3,5 , 7,9 } c | a ) 2200 , b ) 2295 , c ) 2375 , d ) 2380 , e ) 2385 | c | multiply(add(multiply(const_3, const_4), add(const_3, const_4)), power(5, const_3)) | add(const_3,const_4)|multiply(const_3,const_4)|power(n0,const_3)|add(#0,#1)|multiply(#3,#2) | general |
how many 5 - digit numbers are there , if the two leftmost digits are even , the other digits are odd and the digit 2 can not appear more than once in the number . | "in one hour , the crow eats 1 / 24 of the nuts . ( 1 / 4 ) / ( 1 / 24 ) = 6 hours the answer is a ." | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | a | divide(divide(const_1, const_4), divide(divide(const_1, add(const_2, const_3)), 4)) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,#0)|divide(#2,n0)|divide(#1,#3)| | general |
a bucket full of nuts was discovered by the crow living in the basement . the crow eats a sixth of the total number of nuts in 4 hours . how many hours i total will it take the crow to finish a quarter of the nuts ? | "let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1081 96 + 225 + 490 + 6 x = 1081 6 x = 270 = > x = 45 . answer : e" | a ) 25 , b ) 66 , c ) 77 , d ) 99 , e ) 45 | e | divide(subtract(subtract(subtract(1081, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6) | multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)| | general |
alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 1081 . find the cost of each ice - cream cup ? | "explanation : marked price = rs . 30 c . p . = 100 / 140 * 30 = rs . 21.42 sale price = 90 % of rs . 30 = rs . 27 required gain % = 5.57 / 21.42 * 100 = 26 % . answer : e" | a ) 8 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 26 % | e | multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(40, const_100))), divide(multiply(30, const_100), add(40, const_100))), const_100) | add(n1,const_100)|divide(n0,const_100)|multiply(n0,const_100)|subtract(const_100,n2)|divide(#2,#0)|multiply(#1,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)| | gain |
selling an kite for rs . 30 , a shop keeper gains 40 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ? | "prime factors of 30 are 2 ^ 1,3 ^ 1,5 ^ 1 total divisors = ( power if a prime factor + 1 ) total no . of odd factors ( 3,5 , ) = ( 1 + 1 ) ( 1 + 1 ) = 4 since we need odd divisors other than 1 = > 4 - 1 = 3 odd divisors b is the answer" | a ) 3 , b ) 3 , c ) 5 , d ) 6 , e ) 7 | b | divide(30, multiply(const_10, const_2)) | multiply(const_10,const_2)|divide(n0,#0)| | other |
how many factors of 30 are odd numbers greater than 1 ? | "withoutusing the formula , we can see that today the restaurant served 40 customers above the average . the total amount above the average must equal total amount below the average . this additional 40 customers must offset the β deficit β below the average of 80 created on the x days the restaurant served only 75 customers per day . 40 / 5 = 8 days . choice ( a ) . withthe formula , we can set up the following : 80 = ( 75 x + 120 ) / ( x + 1 ) 80 x + 80 = 75 x + 120 5 x = 40 x = 8 answer choice ( d )" | a ) 2 , b ) 5 , c ) 7 , d ) 8 , e ) 20 | d | subtract(divide(subtract(120, 80), subtract(80, 75)), divide(subtract(120, const_100), const_100)) | subtract(n1,n2)|subtract(n2,n0)|subtract(n1,const_100)|divide(#0,#1)|divide(#2,const_100)|subtract(#3,#4)| | general |
at a certain restaurant , the average ( arithmetic mean ) number of customers served for the past x days was 75 . if the restaurant serves 120 customers today , raising the average to 80 customers per day , what is the value of x ? | "say x / 3 / 6 * 5 / 6 = x * 6 / 3 * 5 / 6 = x * 5 / 3 e" | a ) 31 β 5 , b ) 16 β 5 , c ) 20 β 9 , d ) 9 β 20 , e ) 5 β 3 | e | multiply(divide(6, 3), divide(5, 6)) | divide(n1,n0)|divide(n2,n3)|multiply(#0,#1)| | general |
dividing by 3 β 6 and then multiplying by 5 β 6 is the same as dividing by what number ? | "then , 15 % of x = 15 ( 15 / 100 ) x = 15 x = ( 15 * 100 * ) / 15 = 100 answer is a" | a ) a ) 100 , b ) b ) 120 , c ) c ) 250 , d ) d ) 200 , e ) e ) 160 | a | divide(multiply(15, const_100), 15) | multiply(n1,const_100)|divide(#0,n0)| | gain |
an inspector rejects 15 % of the meters as defective . how many will he examine to reject 15 ? | "in one cycle they fill 40 + 30 - 20 = 50 liters 700 = 50 * n = > n = 14 here n = number of cycles . total time = 14 * 3 = 42 as in one cycle there are 3 minutes . thus 42 minutes answer : a" | a ) 42 minutes , b ) 14 minutes , c ) 39 minutes , d ) 40 minutes 20 seconds , e ) none of these | a | multiply(divide(700, subtract(add(40, 30), 20)), const_3) | add(n1,n2)|subtract(#0,n3)|divide(n0,#1)|multiply(#2,const_3)| | physics |
pipe a fills a tank of capacity 700 liters at the rate of 40 liters a minute . another pipe b fills the same tank at the rate of 30 liters a minute . a pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute . if pipe a is kept open for a minute and then closed and pipe b is open for a minute and then closed and then pipe c is open for a minute and then closed and the cycle is repeated , when will the tank be full ? | "2 x * x = 72 = > x = 6 answer : a" | a ) 6 , b ) 16 , c ) 8 , d ) 36 , e ) none | a | sqrt(divide(72, const_2)) | divide(n0,const_2)|sqrt(#0)| | geometry |
the area of a parallelogram is 72 sq m and its altitude is twice the corresponding base . then the length of the base is ? | "us = 20 ds = 30 m = ( 30 + 20 ) / 2 = 25 answer : c" | a ) 11 , b ) 77 , c ) 25 , d ) 88 , e ) 34 | c | divide(add(10, 20), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a man can row upstream at 10 kmph and downstream at 20 kmph , and then find the speed of the man in still water ? | here lcm of 12 and 16 is taken as total work . ( becomes easy to solve ) assume total work = 48 units then workdone by ( a + b ) in one day = 48 / 12 = 4 units similarly , by ( b + c ) in one day = 48 / 16 = 3 units now according to question , a works 5 days , b for 7 days and c for 13 days to complete total work so , 5 a + 7 b + 13 c = 48 units 5 ( a + b ) + 2 ( b + c ) + 11 c = 48 units 5 * 4 + 2 * 3 + 11 c = 48 units 11 c = 22 units c = 2 units ( c does 2 units of work daily ) therefore , 48 / 2 = 24 days c requires 24 days to complete the work alone . answer d | a ) 22 days , b ) 21 days , c ) 25 days , d ) 24 days , e ) 23 days | d | divide(const_1, divide(subtract(const_1, add(divide(5, 12), divide(const_2, 16))), subtract(add(5, 13), 7))) | add(n2,n4)|divide(n2,n0)|divide(const_2,n1)|add(#1,#2)|subtract(#0,n3)|subtract(const_1,#3)|divide(#5,#4)|divide(const_1,#6) | physics |
a and b can finish a work together in 12 days , and b and c together in 16 days . if a alone works for 5 days and then b alone continues for 7 days , then remaining work is done by c in 13 days . in how many days can c alone finish the complete work ? | let their ages be x and ( x + 20 ) years ( x - 5 ) * 5 = ( x + 20 - 5 ) after solving this we get x = 10 years the age of elder one = 10 + 20 = 30 years so the present ages are 30 and 10 years answer : a | a ) 30 , 10 , b ) 2010 , c ) 3515 , d ) 5117 , e ) 20,17 | a | add(20, const_10) | add(n0,const_10) | general |
the ages of two person , differ by 20 years . if 5 years ag , the elder one be 5 times as old as the younger one their present ages ( in years ) are respectively | ( 100 % + 8 % ) * ( 100 % - 6 % ) = 1.08 * 0.94 = 1.0152 = 101.52 % . the net percentage change in the price of the stock is ( + ) 1.52 % the answer is d | a ) 0.2 % , b ) 0.8 % , c ) 1.2 % , d ) 1.52 % , e ) 2 % | d | subtract(multiply(multiply(divide(add(const_100, 8), const_100), divide(subtract(const_100, 6), const_100)), const_100), const_100) | add(n0,const_100)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|multiply(#4,const_100)|subtract(#5,const_100) | general |
the price of stock increased by 8 % last year and decreased by 6 % this year . what is the net percentage change in the price of the stock ? | "400 kg - 1 cubic meter ; 400,000 g - 1 cubic meter ; 400,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 / 400,000 = 10 / 4 = 2.5 cubic centimeters . answer : b ." | a ) 1.5 , b ) 2.5 , c ) 3.5 , d ) 4.5 , e ) 5.5 | b | divide(multiply(1,000, 1,000), multiply(400, 1,000)) | multiply(n4,n4)|multiply(n1,n4)|divide(#0,#1)| | geometry |
the mass of 1 cubic meter of a substance is 400 kg under certain conditions . what is the volume in cubic centimeters of 1 gram of this substance under these conditions ? ( 1 kg = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | "for a 2 nd degree equation ax 2 + bx _ c = 0 has equal roots the condition is b 2 - 4 ac = 0 in the given equation ( 5 k ) ^ 2 - 4 * 2 k * 1 = 0 by solving this equation we get k = 0 , k = 8 / 25 answer : c" | a ) 2 / 7 , b ) 9 / 4 , c ) 8 / 25 , d ) 7 / 1 , e ) 7 / 2 | c | divide(power(2, add(2, 2)), power(5, 2)) | add(n0,n0)|power(n2,n0)|power(n0,#0)|divide(#2,#1)| | general |
for what value of Γ’ β¬ Ε k Γ’ β¬ Β will the equation ( 2 kx 2 + 5 kx + 1 ) = 0 have equal roots ? | "the average number of books per student was 2 means that total of 2 * 25 = 50 books were borrowed ; 2 + 10 + 8 = 20 students borrowed total of 2 * 0 + 10 * 1 + 8 * 2 = 26 books ; so 50 - 26 = 24 books are left to distribute among 25 - 20 = 5 students , these 5 arethe rest who borrowed at least 3 books ; tomaximizethe number of books one student from above 5 could have borrowed we shouldminimizethe number of books other 4 students from 5 could have borrowed . minimum these 4 students could have borrowed is 4 books per student , so total number of books they could have borrowed is 4 * 3 = 12 books . so the 5 th student could have borrowed is 24 - 12 = 12 books . answer : c ." | a ) 8 , b ) 10 , c ) 12 , d ) 13 , e ) 15 | c | add(subtract(multiply(25, 2), add(add(10, multiply(8, 2)), multiply(2, 3))), 3) | multiply(n0,n5)|multiply(n4,n5)|multiply(n1,n6)|add(n2,#1)|add(#3,#2)|subtract(#0,#4)|add(n6,#5)| | general |
in a class of 25 students , 2 students did not borrow any books from the library , 10 students each borrowed 1 book , 8 students each borrowed 2 books , and the rest of the students each borrowed at least 3 books . if the average ( arithmetic mean ) number of books borrowed per student was 2 , what is the maximum number of books that any single student could have borrowed ? | "volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . Ο ( 24 ) ^ 2 * h = ( 4 / 3 ) Ο ( 12 ) ^ 3 = > h = 4 cm answer : b" | a ) 6 cm , b ) 4 cm , c ) 8 cm , d ) 3 cm , e ) 9 cm | b | divide(multiply(const_4, divide(power(12, const_3), power(24, const_2))), const_3) | power(n0,const_3)|power(n1,const_2)|divide(#0,#1)|multiply(#2,const_4)|divide(#3,const_3)| | physics |
a metallic sphere of radius 12 cm is melted and drawn into a wire , whose radius of cross section is 24 cm . what is the length of the wire ? | there are twice as many girls as boys in the class - - > g = 2 b . each girl writes 3 more letters than each boy - - > boys write x letters , girls write x + 3 letters . boys write 24 letters - - > bx = 24 . girls write 90 - 24 = 66 letters - - > ( 2 b ) ( x + 3 ) = 66 - - > 2 bx + 6 b = 66 - - > 2 * 24 + 6 b = 66 - - > b = 3 . bx = 24 - - > 3 x = 24 - - > x = 8 . answer : d . | a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | divide(24, divide(subtract(90, multiply(3, 24)), multiply(3, const_2))) | multiply(n0,n1)|multiply(n0,const_2)|subtract(n2,#0)|divide(#2,#1)|divide(n1,#3) | general |
boys and girls in a class are writing letters . there are twice as many girls as boys in the class , and each girl writes 3 more letters than each boy . if boys write 24 of the 90 total letters written by the class , how many letters does each boy write ? | "speed = ( 5 * 5 / 18 ) m / sec = 25 / 18 m / sec . distance covered in 15 minutes = ( 25 / 18 * 15 * 60 ) m = 1250 m . correct option : d" | a ) 600 , b ) 750 , c ) 1000 , d ) 1250 , e ) none of these | d | multiply(divide(multiply(5, const_1000), const_60), 15) | multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)| | physics |
a man walking at the rate of 5 km / hr crosses a bridge in 15 minutes . the length of the bridge ( in meters ) is : | "let the speed be ' s ' and let the distance between rose and poppy be ' x ' the problem boils down to : rose to poppy : s + 3 = x / 6 - - - - - - - 1 daisy to rose : s = ( x + 14 ) / 10 - - - - - - 2 so from 1 we can re write x as x = 6 s + 18 substitute the value of x in 2 gives us s = 2 m / s a" | a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | a | divide(add(multiply(6, 3), 14), 10) | multiply(n1,n3)|add(n2,#0)|divide(#1,n0)| | physics |
a honey bee flies for 10 seconds , from a daisy to a rose . it immediately continues to a poppy , flying for 6 additional seconds . the distance the bee passed , flying from the daisy to the rose is 14 meters longer than the distance it passed flying from the rose to the poppy . the bee flies to the poppy at 3 meters per second faster than her speed flying to the rose . the bee flies how many meters per second from the daisy to the rose ? | "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8010 = p + ( 54 p / 100 ) 8010 = ( 100 p + 54 p ) / 100 8010 = 154 p / 100 p = ( 8010 * 100 ) / 154 = rs . 5201.298 answer : d" | a ) s . 5266 , b ) s . 5269 , c ) s . 5228 , d ) s . 5201 , e ) s . 52192 | d | divide(8010, add(const_1, divide(multiply(6, 9), const_100))) | multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n3,#2)| | gain |
mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8010 / - . find out the amount that he borrowed . | "d number of valid votes = 80 % of 6500 = 5200 . valid votes polled by other candidate = 45 % of 5200 = ( 45 / 100 x 5200 ) = 2340 ." | a ) 2800 , b ) 2700 , c ) 2900 , d ) 2340 , e ) 2300 | d | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 6500) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)| | gain |
in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 6500 , the number of valid votes that the other candidate got , was : | 9 - 3 * 3 6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 4 is allowed . imo b . | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | add(divide(9, 3), divide(3, divide(9, 3))) | divide(n0,n1)|divide(n1,#0)|add(#0,#1) | general |
if 9 ! / 3 ^ x is an integer , what is the greatest possible value of x ? | "21 - 7 * 3 18 - 3 * 3 * 2 15 - 5 * 3 12 - 4 * 3 9 - 3 * 3 6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 9 is allowed . imo d ." | a ) 3 , b ) 4 , c ) 5 , d ) d , e ) 7 | d | add(divide(21, 3), divide(3, divide(21, 3))) | divide(n0,n1)|divide(n1,#0)|add(#0,#1)| | general |
if 21 ! / 3 ^ x is an integer , what is the greatest possible value of x ? | sum of all the 13 results = 13 * 60 = 780 sum of the first 7 of them = 7 * 57 = 399 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 399 - 780 = 46 . c | a ) 35 , b ) 37 , c ) 46 , d ) 48 , e ) 50 | c | subtract(add(multiply(7, 57), multiply(7, 61)), multiply(13, 60)) | multiply(n2,n3)|multiply(n2,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2) | general |
the average of 13 result is 60 . average of the first 7 of them is 57 and that of the last 7 is 61 . find the 8 th result ? | "explanation : all are prime numbers in their order , starting from 23 hence , next number is 67 answer : b" | a ) 53 , b ) 67 , c ) 48 , d ) 59 , e ) 45 | b | multiply(43, 23) | multiply(const_3.0,n0)| | general |
23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , ( . . . ) | "x / ( 11 p ) = even prime number x = even prime number * 11 p least value of x = lowest even prime number * 11 * lowest value of p = 2 * 11 * 2 = 44 answer c" | a ) 22 , b ) 33 , c ) 44 , d ) 66 , e ) 99 | c | multiply(multiply(9, const_2), const_2) | multiply(n0,const_2)|multiply(#0,const_2)| | general |
if x / ( 9 p ) is an even prime number , where x is a positive integer and p is a prime number , what is the least value of x ? | "cp * ( 75 / 100 ) = 1500 cp = 20 * 100 = > cp = 2000 answer : d" | a ) 1400 , b ) 1300 , c ) 1200 , d ) 2000 , e ) 1500 | d | divide(1500, subtract(const_1, divide(25, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
after decreasing 25 % in the price of an article costs rs . 1500 . find the actual cost of an article ? | "the triangle with sides 52 cm , 48 cm and 20 cm is right angled , where the hypotenuse is 52 cm . area of the triangle = 1 / 2 * 48 * 20 = 480 cm 2 answer : a" | a ) 480 cm 2 , b ) 765 cm 2 , c ) 216 cm 2 , d ) 197 cm 2 , e ) 275 cm 2 | a | divide(multiply(48, 20), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
if the sides of a triangle are 52 cm , 48 cm and 20 cm , what is its area ? | "x * ( 95 / 100 ) * ( 85 / 100 ) = 3443 x = 4264 answer : a" | a ) 4264 , b ) 2776 , c ) 4400 , d ) 2871 , e ) 881 | a | floor(divide(3443, multiply(divide(subtract(const_100, 5), const_100), divide(subtract(const_100, 15), const_100)))) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)| | general |
5 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 3443 , how much was it in the beginning ? | let the sum borrowed be x . then , ( x Γ 6 Γ 21 / 00 ) + ( x Γ 9 Γ 3 / 100 ) + ( x Γ 14 Γ 4 / 100 ) = 11900 β ( 3 β 25 x + 27 β 100 x + 14 β 25 x ) = 11400 β 95 β 100 x = 11900 β x = ( 11900 Γ 100 / 95 ) = 12526 hence , sum borrowed = 12,526 answer b | a ) 10,526 , b ) 12,526 , c ) 14,000 , d ) 16,000 , e ) 16,536 | b | subtract(divide(11900, add(add(divide(multiply(6, const_2), const_100), divide(multiply(9, 3), const_100)), divide(multiply(14, 4), const_100))), multiply(const_12, const_1000)) | multiply(n0,const_2)|multiply(n1,n2)|multiply(n3,n4)|multiply(const_1000,const_12)|divide(#0,const_100)|divide(#1,const_100)|divide(#2,const_100)|add(#4,#5)|add(#7,#6)|divide(n5,#8)|subtract(#9,#3) | gain |
adam borrowed some money at the rate of 6 % p . a . for the first two years , at the rate of 9 % p . a . for the next 3 years , and at the rate of 14 % p . a . for the period beyond 4 years . if he pays a total interest of 11900 at the end of 9 years , how much money did he borrow ? | "explanation : 1 man β s 1 day β s work = 1 / 8 Γ 12 = 1 / 96 6 men β s 1 day β s work = 1 Γ 6 / 96 = 1 / 16 1 woman β s 1 day β s work = 1 / 192 4 women β s 1 day β s work = 1 / 192 Γ 4 = 1 / 48 1 child β s 1 day β s work = 1 / 240 10 children β s 1 day β s work = 1 / 24 therefore , ( 6 men + 4 women + 10 children ) β s 1 day β s work = 1 / 16 + 1 / 48 + 1 / 24 = 1 / 8 the required no . of days = 8 days answer : option d" | a ) 5 days , b ) 15 days , c ) 28 days , d ) 8 days , e ) 7 days | d | inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 6), multiply(inverse(multiply(48, 4)), 4)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|inverse(#0)|inverse(#1)|inverse(#2)|multiply(n6,#3)|multiply(n2,#4)|multiply(n4,#5)|add(#6,#7)|add(#9,#8)|inverse(#10)| | physics |
8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 6 men , 4 women and 10 children together complete the piece of work ? | "7 / 8 x - 3 / 4 x = 6 galls 1 / 8 * x = 6 gallons x = 48 gallons answer b" | a ) 25 , b ) 48 , c ) 64 , d ) 80 , e ) 96 | b | multiply(6, divide(const_1, subtract(divide(7, 8), divide(3, 4)))) | divide(n3,n4)|divide(n1,n2)|subtract(#0,#1)|divide(const_1,#2)|multiply(n0,#3)| | general |
if henry were to add 6 gallons of water to a tank that is already 3 / 4 full of water , the tank would be 7 / 8 full . how many gallons of water would the tank hold if it were full ? | "t = 120 / 70 * 18 / 5 = 6 sec answer : d" | a ) 3 sec , b ) 4 sec , c ) 5 sec , d ) 6 sec , e ) 7 sec | d | divide(120, multiply(70, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
in what time will a railway train 120 m long moving at the rate of 70 kmph pass a telegraph post on its way ? | "there are 396 managers and 4 others . the 4 others would compose 2 % of the total number of people if there were 200 people in the room . thus 200 managers must leave . the answer is d ." | a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250 | d | divide(subtract(multiply(400, divide(99, const_100)), multiply(400, divide(98, const_100))), subtract(const_1, divide(98, const_100))) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#1)|subtract(#2,#3)|divide(#5,#4)| | gain |
there are 400 employees in a room . 99 % are managers . how many managers must leave the room to bring down the percentage of managers to 98 % ? | "out of 100 pounds 99 % or 99 pounds is water and 1 pound is non - water . after somewaterevaporates the cucumbers become 97 % water and 3 % of non - water , so now 1 pound of non - water composes 3 % of cucucmbers , which means that the new weight of cucumbers is 1 / 0.03 = 34 pounds . answer : b ." | a ) 2 , b ) 33 , c ) 92 , d ) 96 , e ) 98 | b | multiply(divide(subtract(100, 99), subtract(100, 97)), 100) | subtract(n0,n1)|subtract(n0,n2)|divide(#0,#1)|multiply(#2,n0)| | gain |
each of the cucumbers in 100 pounds of cucumbers is composed of 99 % water , by weight . after some of the water evaporates , the cucumbers are now 97 % water by weight . what is the new weight of the cucumbers , in pounds ? | length of the two trains = 600 m + 400 m speed of the first train = x speed of the second train = 48 kmph 1000 / x - 48 = 180 1000 / x - 48 * 5 / 18 = 180 50 = 9 x - 120 x = 68 kmph answer : b | a ) 76 kmph , b ) 68 kmph , c ) 87 kmph , d ) 56 kmph , e ) 10 kmph | b | add(48, multiply(divide(add(400, 600), 180), const_3_6)) | add(n0,n1)|divide(#0,n2)|multiply(#1,const_3_6)|add(n3,#2) | physics |
the two trains of lengths 400 m , 600 m respectively , running at same directions . the faster train can cross the slower train in 180 sec , the speed of the slower train is 48 km . then find the speed of the faster train ? | "1 flash = 5 sec for 1 min = 12 flashes so for 1 hour = 12 * 60 = 720 flashes . answer : a" | a ) 720 , b ) 600 , c ) 650 , d ) 700 , e ) 750 | a | divide(const_3600, 5) | divide(const_3600,n0)| | physics |
a light flashes every 5 seconds , how many times will it flash in ? of an hour ? | "each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits . here ' s another way to organize the information . we ' re given [ ( 2222 ) ^ 333 ] [ ( 3333 ) ^ 222 ] we can ' combine ' some of the pieces and rewrite this product as . . . . ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) [ ( 2222 ) ^ 111 ] ( 2222 ) ( 3333 ) = a big number that ends in a 6 taking a number that ends in a 6 and raising it to a power creates a nice pattern : 6 ^ 1 = 6 6 ^ 2 = 36 6 ^ 3 = 216 etc . thus , we know that ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) will be a gigantic number that ends in a 6 . 2 ^ 111 requires us to figure out thecycleof the units digit . . . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 so , every 4 powers , the pattern of the units digits repeats ( 2 , 4 , 8 , 6 . . . . . 2 , 4 , 8 , 6 . . . . ) . 111 = 27 sets of 4 with a remainder of 3 . . . . this means that 2 ^ 111 = a big number that ends in an 8 so we have to multiply a big number that ends in a 6 and a big number that ends in an 8 . ( 6 ) ( 8 ) = 48 , so the final product will be a gigantic number that ends in an 6 . final answer : d" | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
what is the units digit of 3333 ^ ( 333 ) * 3333 ^ ( 222 ) ? | "40 % of 6 = 2.4 50 % of 6 = 3 shortage is 0.6 so we need to have 0.6 / 50 % to get 50 % alcohol content . = 1.2 b" | a ) a . 0.6 , b ) b . 1.2 , c ) c . 2.1 , d ) d . 3 , e ) e . 5.4 | b | subtract(6, multiply(const_2, multiply(divide(40, const_100), 6))) | divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_2)|subtract(n0,#2)| | gain |
a 6 litre sol is 40 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ? | "54 = ( 3 ^ 3 ) * 2 since 54 is not a perfect square , no of ways = 4 answer d" | a ) 10 , b ) 8 , c ) 5 , d ) 4 , e ) 2 | d | subtract(divide(divide(54, const_1), const_3), const_3) | divide(n0,const_1)|divide(#0,const_3)|subtract(#1,const_3)| | general |
in how many ways can the integer 54 be expressed as a product of two different positive integers ? | "explanation : l . c . m of 2400 = 3 x 2 x 2 x 2 x 2 x 2 x 5 x 5 3 , 2 , 5 number of different prime factors is 3 . answer : option c" | a ) 4 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | c | add(const_2, const_2) | add(const_2,const_2)| | other |
find the number of different prime factors of 2400 | "let arun ' s weight be x kg . according to arun , 63 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 66 . the values satisfying all the above conditions are 64 and 65 . required average = ( 64 + 65 ) / 2 = 64.5 kg answer : b" | a ) 86.5 kg , b ) 64.5 kg , c ) 46.5 kg , d ) 26.5 kg , e ) 16.5 kg | b | divide(add(66, add(63, const_1)), const_2) | add(n0,const_1)|add(n4,#0)|divide(#1,const_2)| | general |
in arun ' s opinion , his weight is greater than 63 kg but leas than 72 kg . his brother does not agree with arun and he thinks that arun ' s weight is greater than 60 kg but less than 70 kg . his mother ' s view is that his weight can not be greater than 66 kg . if all of them are correct in their estimation , what is the average of diferent probable weights of arun ? | given ; 2 dog = 3 cat ; or , dog / cat = 3 / 2 ; let cat ' s 1 leap = 2 meter and dogs 1 leap = 3 meter . then , ratio of speed of cat and dog = 2 * 5 / 3 * 4 = 5 : 6 . ' ' answer : 5 : 6 ; | a ) 5 : 6 , b ) 3 : 2 , c ) 4 : 8 , d ) 1 : 2 , e ) 7 : 8 | a | divide(multiply(divide(2, 3), 5), 4) | divide(n2,n3)|multiply(n0,#0)|divide(#1,n1) | other |
a cat leaps 5 leaps for every 4 leaps of a dog , but 2 leaps of the dog are equal to 3 leaps of the cat . what is the ratio of the speed of the cat to that of the dog ? | "let c . p . be rs . x . then , ( 1920 - x ) / x * 100 = ( x - 1280 ) / x * 100 1920 - x = x - 1280 2 x = 3200 = > x = 1600 required s . p . = 115 % of rs . 1600 = 115 / 100 * 1600 = rs . 1840 . answer : c" | a ) 2000 , b ) 2778 , c ) 1840 , d ) 2778 , e ) 2771 | c | multiply(divide(add(const_100, 15), const_100), divide(add(1920, 1280), const_2)) | add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(#1,const_2)|multiply(#2,#3)| | gain |
the percentage profit earned by selling an article for rs . 1920 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 15 % profit ? | "min value of a / b will be when b is highest and a is lowest - - - > a = 20 and b = 40 so , a / b = 1 / 2 max value of a / b will be when b is lowest and a is highest - - - > a = 30 and b = 30 so , a / b = 1 range is 1 - ( 1 / 2 ) = 1 / 2 . answer should be b" | a ) 1 / 4 , b ) 1 / 2 , c ) 3 / 4 , d ) 1 , e ) 5 / 4 | b | subtract(divide(subtract(32, const_1), add(29, const_1)), divide(add(19, const_1), subtract(42, const_1))) | add(n2,const_1)|add(n0,const_1)|subtract(n1,const_1)|subtract(n3,const_1)|divide(#2,#0)|divide(#1,#3)|subtract(#4,#5)| | general |
a is an integer greater than 19 but less than 32 , b is an integer greater than 29 but less than 42 , what is the range of a / b ? | "( x * 5 * 1 ) / 100 + [ ( 2500 - x ) * 6 * 1 ] / 100 = 125 x = 2500 answer : e" | a ) 2333 , b ) 2777 , c ) 2688 , d ) 1000 , e ) 2500 | e | divide(subtract(125, divide(multiply(6, 2500), const_100)), subtract(divide(5, const_100), divide(6, const_100))) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,n2)|divide(#2,const_100)|subtract(#0,#1)|subtract(n3,#3)|divide(#5,#4)| | gain |
rs . 2500 is divided into two parts such that if one part be put out at 5 % simple interest and the other at 6 % , the yearly annual income may be rs . 125 . how much was lent at 5 % ? | "total time taken by jerry = ( 8 / 40 ) * 60 minutes + 13 minutes + ( 20 / 60 ) * 60 minutes = 35 minutes average speed = total distance / total time = ( 8 + 20 ) miles / ( 35 / 60 ) hours = 28 * 60 / 35 = 48 miles per hour answer : option a" | a ) 48 , b ) 42.5 , c ) 44 , d ) 50 , e ) 52.5 | a | divide(add(8, 20), add(add(divide(8, 40), divide(13, 60)), divide(20, 60))) | add(n0,n3)|divide(n0,n1)|divide(n2,n4)|divide(n3,n4)|add(#1,#2)|add(#4,#3)|divide(#0,#5)| | physics |
jerry travels 8 miles at an average speed of 40 miles per hour , stops for 13 minutes , and then travels another 20 miles at an average speed of 60 miles per hour . what is jerry β s average speed , in miles per hour , for this trip ? | first we need to find the constant ' c ' . the easiest way to find this is for the sum of the first two perfect squares for 1 and 2 = 1 and 4 respectively . hence lhs = 1 + 4 and plug n = 2 for rhs and simplify to get c = 1 / 2 . plug values of n = 18 and c = 1 / 2 into the equation and simplify to get the answer 2109 . option e . | a ) 1,010 , b ) 1,164 , c ) 1,240 , d ) 1,316 , e ) 2,109 | e | divide(divide(multiply(multiply(18, add(18, const_1)), add(multiply(18, const_2), const_1)), 6), divide(multiply(multiply(18, add(18, const_1)), add(multiply(18, const_2), const_1)), 6)) | add(n4,const_1)|multiply(n4,const_2)|add(#1,const_1)|multiply(n4,#0)|multiply(#2,#3)|divide(#4,n3)|divide(#5,#5) | general |
the sum of the first n positive perfect squares , where n is a positive integer , is given by the formula n ^ 3 / 3 + c * n ^ 2 + n / 6 , where c is a constant . what is the sum of the first 18 positive perfect squares ? | 1 : 8 answer : b | ['a ) 1 : 9', 'b ) 1 : 8', 'c ) 1 : 3', 'd ) 1 : 4', 'e ) 1 : 5'] | b | power(sqrt(divide(1, 4)), const_3) | divide(n0,n1)|sqrt(#0)|power(#1,const_3) | geometry |
surface area of two spheres are in the ratio 1 : 4 what is the ratio of their volumes ? | "if area of a circle decreased by x % then the radius of a circle decreases by ( 100 β 10 β 100 β x ) % = ( 100 β 10 β 100 β 30 ) % = ( 100 β 10 β 70 ) % = 100 - 84 = 16 % answer b" | a ) 20 % , b ) 16 % , c ) 36 % , d ) 64 % , e ) none of these | b | multiply(subtract(const_1, sqrt(divide(subtract(const_100, 30), const_100))), const_100) | subtract(const_100,n0)|divide(#0,const_100)|sqrt(#1)|subtract(const_1,#2)|multiply(#3,const_100)| | geometry |
if the area of a circle decreases by 30 % , then the radius of a circle decreases by | "s = { hh , tt , ht , th } e = event of getting at most one head . e = { tt , ht , th } . p ( e ) = n ( e ) / n ( s ) = 3 / 4 answer is option c" | a ) 2 / 3 , b ) 1 , c ) 3 / 4 , d ) 2 , e ) 1 / 2 | c | negate_prob(divide(const_1, power(const_2, const_3))) | power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)| | probability |
two unbiased coins are tossed . what is the probability of getting at most one head ? | "let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 25 = 2 r a = 12 / 5 r and r = 75 a = 180 required total cost = 4 * 180 + 3 * 75 + 5 * 25 = 720 + 225 + 125 = $ 1070 c" | a ) 347 , b ) 987 , c ) 1070 , d ) 1371 , e ) 1667 | c | add(add(multiply(4, multiply(divide(24, 10), divide(multiply(25, 6), 2))), multiply(3, divide(multiply(25, 6), 2))), multiply(5, 25)) | divide(n1,n0)|multiply(n2,n4)|multiply(n4,n7)|divide(#1,n3)|multiply(#0,#3)|multiply(n6,#3)|multiply(n5,#4)|add(#6,#5)|add(#7,#2)| | general |
the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 25 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ? | "percent of students who are 25 years old or older is 0.4 * 48 + 0.2 * 52 = ~ 30 , so percent of people who are less than 25 years old is 100 - 30 = 70 . answer : b ." | a ) 0.9 , b ) 0.7 , c ) 0.45 , d ) 0.3 , e ) 0.25 | b | subtract(const_1, multiply(divide(40, const_100), divide(subtract(const_100, 20), const_100))) | divide(n2,const_100)|subtract(const_100,n3)|divide(#1,const_100)|multiply(#0,#2)|subtract(const_1,#3)| | general |
in the graduating class of a certain college , 48 percent of the students are male and 52 percent are female . in this class 40 percent of the male and 20 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | "speed = 54 * 5 / 18 = 15 m / sec time taken = 300 * 1 / 15 = 20 sec answer : c" | a ) 17 sec , b ) 16 sec , c ) 20 sec , d ) 14 sec , e ) 12 sec | c | multiply(divide(300, multiply(54, const_1000)), const_3600) | multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)| | physics |
a train 300 m long , running with a speed of 54 km / hr will pass a tree in ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 899 = 279 answer : b" | a ) 270 , b ) 279 , c ) 267 , d ) 255 , e ) 552 | b | multiply(divide(899, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|multiply(#5,#1)| | general |
a , b and c rents a pasture for rs . 899 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "explanation : suppose the boy got x sums right and 2 x sums wrong . then , x + 2 x = 27 3 x = 27 x = 9 . answer : d" | a ) 12 , b ) 16 , c ) 18 , d ) 9 , e ) 12 | d | divide(27, add(const_1, const_2)) | add(const_1,const_2)|divide(n0,#0)| | general |
a student got twice as many sums wrong as he got right . if he attempted 27 sums in all , how many did he solve correctly ? | explanation : given that simple interest for 2 years is rs . 800 i . e . , simple interest for 1 st year is rs . 400 and simple interest for 2 nd year is also rs . 400 compound interest for 1 st year will be 400 and compound interest for 2 nd year will be 832 - 400 = 432 you can see that compound interest for 2 nd year is more than simple interest for 2 nd year by 432 - 400 = rs . 32 i . e , rs . 32 is the interest obtained for rs . 400 for 1 year rate , r = 100 Γ si / pt = ( 100 Γ 32 ) / ( 400 Γ 1 ) = 8 % difference between compound and simple interest for the 3 rd year = simple interest obtained for rs . 832 = prt / 100 = ( 832 Γ 8 Γ 1 ) / 100 = rs . 66.56 total difference between the compound and simple interest for 3 years = 32 + 66.56 = rs . 98.56 answer : option b | a ) rs . 48 , b ) rs . 98.56 , c ) rs . 66.56 , d ) rs . 66.58 , e ) none of these | b | add(subtract(832, 800), multiply(832, divide(subtract(832, 800), divide(800, 2)))) | divide(n2,n0)|subtract(n1,n2)|divide(#1,#0)|multiply(n1,#2)|add(#3,#1) | general |
the compound interest on a sum for 2 years is rs . 832 and the simple interest on the same sum for the same period is rs . 800 . the difference between the compound and simple interest for 3 years will be | let x be the value in excess of $ 1,000 . 0.07 x = 94.5 x = $ 1,350 the total value was $ 1,350 + $ 1,000 = $ 2,350 . the answer is a . | a ) $ 2,350 , b ) $ 2,850 , c ) $ 3,250 , d ) $ 3,400 , e ) $ 3,750 | a | floor(divide(add(divide(94.5, divide(7, const_100)), 1000), const_1000)) | divide(n0,const_100)|divide(n2,#0)|add(n1,#1)|divide(#2,const_1000)|floor(#3) | general |
when a merchant imported a certain item , he paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1000 . if the amount of the import tax that the merchant paid was $ 94.50 , what was the total value of the item ? | ": cp = sp * ( 100 / ( 100 + profit % ) ) = 8463 ( 100 / 124 ) = rs . 6825 . answer : b" | a ) 6727 , b ) 6825 , c ) 6728 , d ) 6725 , e ) 2871 | b | divide(8463, add(const_1, divide(24, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 8463 for a computer table , then what was the cost price of the computer table ? | "total work = 15 * 4 = 60 beaver hours 20 beaver * x = 60 beaver hours x = 60 / 20 = 3 answer : e" | a ) 2 . , b ) 4 . , c ) 5 . , d ) 6 , e ) 3 . | e | divide(multiply(4, 15), 20) | multiply(n0,n1)|divide(#0,n2)| | physics |
15 beavers , working together in a constant pace , can build a dam in 4 hours . how many hours will it take 20 beavers that work at the same pace , to build the same dam ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 14 ) = 266 cm 2 answer : c" | a ) 178 cm 2 , b ) 179 cm 2 , c ) 266 cm 2 , d ) 167 cm 2 , e ) 197 cm 2 | c | quadrilateral_area(14, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 14 cm | a prime number is a number that has only two factors : 1 and itself . therefore , a prime number is divisible by two numbers only . let ' s list the numbers from 61 to 69 . 61 , 62 , 63 , 64 , 65 , 66 , 67 , 68 , 69 immediately we can eliminate the even numbers because they are divisible by 2 and thus are not prime . we are now left with : 61 , 63 , 65 , 67 , 69 we can next eliminate 65 because 65 is a multiple of 5 . we are now left with 61 , 63 , 67 , 69 . to eliminate any remaining values , we would look at those that are multiples of 3 . if you don β t know an easy way to do this , just start with a number that is an obvious multiple of 3 , such as 60 , and then keep adding 3 . we see that 60 , 63 , 66 , 69 are all multiples of 3 and therefore are not prime . thus , we can eliminate 63 and 69 from the list because they are not prime . finally , we are left with 61 and 67 , and we must determine whether they are divisible by 7 . they are not , and therefore they must be both prime . thus , the sum q of 61 and 67 is 128 . answer b . | a ) 67 , b ) 128 , c ) 191 , d ) 197 , e ) 260 | b | add(add(60, const_1), subtract(70, const_3)) | add(n0,const_1)|subtract(n1,const_3)|add(#0,#1) | general |
the sum q of prime numbers that are greater than 60 but less than 70 is | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 400 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 400 x 8 = 3200 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 3300 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 3300 ( answer b )" | a ) 1,108 , b ) 3,300 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 400) | divide(n6,n5)|multiply(n4,#0)| | general |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 400 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "you need to primarily find the primes of 990 : 2 , 3,3 , 5,11 the integer must contain all these primes by knowing 11 is an prime , we know the answer can be b , c , d or e as these all contain 11 . 11 = 1 , 2,3 , 4,5 , 6,7 , 8,9 , 10,11 11 contains the 2 , 3,5 and 11 explicitly and the second 3 comes from the 6 ( 2 * 3 ) , therefore the answer is 11 . this is because we know a number that contains all the primes of 990 will be a multiple of 990 . answer : b" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | b | divide(divide(divide(divide(990, const_2), const_3), const_4), divide(const_10, const_2)) | divide(n1,const_2)|divide(const_10,const_2)|divide(#0,const_3)|divide(#2,const_4)|divide(#3,#1)| | general |
if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 990 , what is the least possible value of n ? | "sol . required probability = pg . ) x p ( b ) = ( 1 β d x ( 1 β i ) = : x 1 = 1 / 3 ans . ( c )" | a ) 1 / 2 , b ) 1 , c ) 1 / 3 , d ) 3 / 4 , e ) 2 | c | multiply(subtract(1, divide(1, 2)), subtract(1, divide(1, 3))) | divide(n1,n2)|divide(n1,n5)|subtract(n1,#0)|subtract(n1,#1)|multiply(#2,#3)| | general |
the probability that a man will be alive for 10 more yrs is 1 / 2 & the probability that his wife will alive for 10 more yrs is 1 / 3 . the probability that none of them will be alive for 10 more yrs , is | 3 * 5 + 5 * 3 = ( 2 x 3 - 3 x 5 + 3 x 5 ) + ( 2 x 5 - 3 x 3 + 5 x 3 ) = 22 answer a 22 | a ) 22 , b ) 25 , c ) 26 , d ) 28 , e ) 23 | a | add(multiply(2, 3), multiply(3, 5)) | multiply(n0,n1)|multiply(n1,n3)|add(#0,#1) | general |
if a * b = 2 a - 3 b + ab , then 3 * 5 + 5 * 3 is equal to | answer : d , ( with different approach ) : the 200 paid is 0.2 % of the additional amount above 200,000 . let it be x now 0.2 % of x = 200 therefore x = 100,000 total = 200,000 + x = 300,000 | a ) $ 180,000 , b ) $ 202,000 , c ) $ 220,000 , d ) $ 300,000 , e ) $ 2 , 200,000 | d | multiply(multiply(200, const_100), const_10) | multiply(n3,const_100)|multiply(#0,const_10) | general |