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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a third of arun â € ™ s marks in mathematics exceed a half of his marks in english by 10 . if he got 230 marks in two subjects together how many marks did he got in english ? | "let arun â € ™ s marks in mathematics and english be x and y then ( 1 / 3 ) x - ( 1 / 2 ) y = 10 2 x - 3 y = 60 â € ¦ â € ¦ > ( 1 ) x + y = 230 â € ¦ â € ¦ . > ( 2 ) solving ( 1 ) and ( 2 ) x = 150 and y = 80 answer is d ." | a ) 140,90 , b ) 180,60 , c ) 160,90 , d ) 150,80 , e ) none of them | d | add(divide(subtract(230, 10), add(const_1, divide(const_1, const_2))), 10) | divide(const_1,const_2)|subtract(n1,n0)|add(#0,const_1)|divide(#1,#2)|add(n0,#3)| | general |
how many times are the hands of a clock at right angle in 5 days ? | in 1 day , they are at right angles 44 times . in 5 days , they are at right angles 220 times . answer : option b | a ) 202 , b ) 220 , c ) 210 , d ) 212 , e ) 222 | b | divide(multiply(multiply(multiply(const_12, const_2), 5), subtract(multiply(const_12, const_4), const_4)), multiply(const_12, const_2)) | multiply(const_12,const_2)|multiply(const_12,const_4)|multiply(n0,#0)|subtract(#1,const_4)|multiply(#2,#3)|divide(#4,#0) | physics |
sebastian bought a meal at a restaurant and left a 15 % tip . with the tip , he paid exactly $ 36.57 . how much did the meal cost without the tip ? | the tip is a percent increase of 15 % , which is 115 % . let x equal the price before the tip . thus , 115 % of this price equals $ 36.57 : 1.15 x = 36.57 divide both sides by 1.15 : = > x = 36.57 / 1.15 = 31.80 correct answer a ) $ 31.80 | a ) $ 31.80 , b ) $ 29.91 , c ) $ 30.15 , d ) $ 30.60 , e ) $ 30.85 | a | subtract(36.57, multiply(36.57, divide(15, const_100))) | divide(n0,const_100)|multiply(n1,#0)|subtract(n1,#1) | general |
what should be the least number to be added to the 4499 number to make it divisible by 9 ? | "answer : 1 option : a" | a ) 1 , b ) 17 , c ) 18 , d ) 77 , e ) 26 | a | subtract(9, reminder(4499, 9)) | reminder(n0,n1)|subtract(n1,#0)| | general |
let c be defined as the sum of all prime numbers between 0 and 28 . what is c / 3 | "prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 sum , c = 100 c / 3 = 11 / 3 answer b" | a ) 155 , b ) 100 / 3 , c ) 61 , d ) 47 , e ) 43 | b | add(divide(3, const_10), power(const_2, add(const_2, const_4))) | add(const_2,const_4)|divide(n2,const_10)|power(const_2,#0)|add(#1,#2)| | general |
we define that k @ j is the product of j number from k in increasing order for positive integers k , j . for example , 6 @ 4 = 6 * 7 * 8 * 9 . if a = 2020 and b = 2120 , what is the value e of the ratio a / b ? | "e - > a / b = 20 * 21 * … … * 39 / 21 * 22 * … . * 39 * 40 = 20 / 40 = 1 / 2 . therefore , the answer is a ." | a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(2020, 2020), add(divide(2020, 2020), divide(2020, 2020))) | divide(n6,n6)|add(#0,#0)|divide(#0,#1)| | general |
how many books each of volume 100 meter cube can be packed into a crate of volume 4000 meter cube ? | gud question with a simple concept . in geo if we want to insert one shape into another we need to know the dimensions of the two shapes . in above with volume given , we can come up with different shapes , so we cant know the answer for ex : 4000 m 3 can be 400 * 10 or 40 * 100 or just 4000 * 1 we do n ' t know , so w... | ['a ) 100', 'b ) 120', 'c ) can not be determined', 'd ) 180', 'e ) 200'] | c | multiply(4000, const_1) | multiply(n1,const_1) | geometry |
jane started baby - sitting when she was 20 years old . whenever she baby - sat for a child , that child was no more than half her age at the time . jane is currently 32 years old , and she stopped baby - sitting 10 years ago . what is the current age of the oldest person for whom jane could have baby - sat ? | check two extreme cases : jane = 20 , child = 10 , years ago = 32 - 20 = 12 - - > child ' s age now = 10 + 12 = 22 ; jane = 22 , child = 11 , years ago = 32 - 22 = 10 - - > child ' s age now = 11 + 10 = 21 . answer : c . | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | add(divide(20, const_2), subtract(32, 20)) | divide(n0,const_2)|subtract(n1,n0)|add(#0,#1) | general |
if integer k is equal to the sum of all even multiples of 25 between 295 and 615 , what is the greatest prime factor of k ? | if we break down what the stem is asking what is the sum of all mult of 50 between 300 and 600 . using arithmetic progression to find n : 600 = 300 + ( n - 1 ) 50 300 + 50 = 50 n 350 = 50 n = > n = 7 the sum would be : 11 * mean mean = [ 600 + 300 ] / 2 = 450 7 * 450 = 4950 b | a ) 5 , b ) 7 , c ) 11 , d ) 13 , e ) 17 | b | divide(add(divide(subtract(multiply(floor(divide(615, 25)), 25), multiply(floor(divide(295, 25)), 25)), 25), const_1), const_2) | divide(n2,n0)|divide(n1,n0)|floor(#0)|floor(#1)|multiply(n0,#2)|multiply(n0,#3)|subtract(#4,#5)|divide(#6,n0)|add(#7,const_1)|divide(#8,const_2) | general |
sonika deposited rs . 8000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ? | "( 8000 * 3 * 2 ) / 100 = 480 9200 - - - - - - - - 9680 answer : a" | a ) 9680 , b ) 4280 , c ) 2789 , d ) 7892 , e ) 2792 | a | add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(9200, 8000), 3), 8000)), 8000), 3), 8000) | divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)| | gain |
a certain city with a population of 60,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district what is the minimum possible population that the least populated district could have ? | "let x = number of people in smallest district x * 1.1 = number of people in largest district x will be minimised when the number of people in largest district is maximised 10 * x * 1.1 = 11 x = total number of people in other districts so we have 11 x + x = 60 k x = 5,000 answer : e" | a ) a ) 10,700 , b ) b ) 10,800 , c ) c ) 10,900 , d ) d ) 11,000 , e ) e ) 5,000 | e | multiply(multiply(const_4, const_2), const_100) | multiply(const_2,const_4)|multiply(#0,const_100)| | general |
the perimeter of a rectangular yard is completely surrounded by a fence that measures 10 meters . what is the length of the yard if the area of the yard is 6 meters squared ? | "perimeter of rectangular yard = 2 ( l + b ) = 10 - - > l + b = 5 area = l * b = 6 b = 5 - l l ( 5 - l ) = 6 5 l - l ^ 2 = 6 l ^ 2 - 5 l + 6 = 0 upon simplifying we get l = 2 or 3 . only 3 is there in the answer choice . answer : e" | a ) 8 , b ) 1 , c ) 5 , d ) 7 , e ) 3 | e | subtract(const_4, const_3) | subtract(const_4,const_3)| | geometry |
how long does a train 150 m long traveling at 50 kmph takes to cross a bridge of 250 m in length ? | "d = 150 + 250 = 400 m s = 50 * 5 / 18 = 125 / 9 t = 400 * 125 / 9 = 28.8 sec answer : a" | a ) 28.8 sec , b ) 16.8 sec , c ) 15.2 sec , d ) 25.4 sec , e ) 16.2 sec | a | divide(add(150, 250), multiply(50, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
if 7 and 6 are factors of 60 n , what is the minimum value of n ? | "60 n / 7 * 6 should be integer = > 2 * 2 * 3 * 5 * n / 7 * 2 * 3 = 2 * 5 * n / 7 must be an integer for this to be true n must multiple of 7 , thus min of n = 7 hence c" | a ) 2 , b ) 3 , c ) 7 , d ) 14 , e ) 56 | c | lcm(7, 6) | lcm(n0,n1)| | other |
a train 50 m long passes a platform 100 m long in 10 seconds . the speed of the train in m / sec is ? | speed of train = distance covered / time . = ( 50 + 100 ) / 10 = 15 m / sec . answer : d | a ) 150 , b ) 50 , c ) 10 , d ) 15 , e ) 12 | d | divide(add(50, 100), 10) | add(n0,n1)|divide(#0,n2) | physics |
how many liters of water must be evaporated from 50 liters of a 4 - percent sugar solution to get a 10 - percent solution ? | "4 % of a 50 liter solution is 2 l which is 10 % of the solution at the end . the solution at the end must be 20 l . we need to evaporate 30 liters . the answer is a ." | a ) 30 , b ) 32 , c ) 34 , d ) 36 , e ) 38 | a | subtract(50, multiply(divide(50, const_100), 10)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
in a class of 70 students , 10 did not opt for math . 5 did not opt for science and 3 did not opt for either . how many students of the class opted for both math and science ? | "70 students total 10 did not opt for math 5 did not opt for science 3 did not opt for either total of 40 students in math and 2 did not opt for sci but did for math 40 - 2 = 38 38 students of the class opted for both math and science answer : e" | a ) 23 , b ) 25 , c ) 27 , d ) 37 , e ) 38 | e | subtract(subtract(70, 10), subtract(subtract(70, 5), 3)) | subtract(n0,n1)|subtract(n0,n2)|subtract(#1,n3)|subtract(#0,#2)| | other |
562 ã — 99 = ? | "562 ã — ( 100 - 1 ) 56200 - 562 = 55638 answer c" | a ) 55684 , b ) 55432 , c ) 55638 , d ) 44530 , e ) 44538 | c | multiply(divide(562, 99), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the sale price sarees listed for rs . 550 after successive discount is 18 % and 12 % is ? | "550 * ( 88 / 100 ) * ( 82 / 100 ) = 396 answer : d" | a ) 298 , b ) 237 , c ) 342 , d ) 396 , e ) 291 | d | subtract(subtract(550, divide(multiply(550, 18), const_100)), divide(multiply(subtract(550, divide(multiply(550, 18), const_100)), 12), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
an empty wooden vessel weighs 12 % of its total weight when filled with paint . if the weight of a partially filled vessel is one half that of a completely filled vessel , what fraction of the vessel is filled . | "an empty wooden vessel weighs 12 % of its total weight when filled with paint : vessel = 0.12 ( vessel + paint ) ; 12 v = v + p ( so the weight of completely filled vessel is 12 v ) p = 11 v ( so the weight of the paint when the vessels is completely filled is 11 v ) . the weight of a partially filled vessel is one ha... | a ) 3 / 5 , b ) 5 / 11 , c ) 1 / 24 , d ) 4 / 9 , e ) 2 / 5 | b | divide(subtract(divide(12, const_2), const_1), subtract(12, const_1)) | divide(n0,const_2)|subtract(n0,const_1)|subtract(#0,const_1)|divide(#2,#1)| | gain |
a side of beef lost 50 percent of its weight in processing . if the side of beef weighed 750 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 50 / 100 ) * x = 750 = > x = ( 750 * 100 ) / 50 = 1500 answer e" | a ) 191 , b ) 355 , c ) 737 , d ) 840 , e ) 1,500 | e | divide(multiply(750, const_100), subtract(const_100, 50)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
50 % of major airline companies equip their planes with wireless internet access . 70 % of major airlines offer passengers free on - board snacks . what is the greatest possible percentage of major airline companies that offer both wireless internet and free on - board snacks ? | "to maximize the percentage of companies offering both , let ' s assume that all 50 % of companies which offer wireless internet also offer snacks . the answer is d ." | a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 70 % | d | multiply(50, const_1) | multiply(n0,const_1)| | general |
a work which could be finished in 12 days was finished 3 days earlier after 10 more men joined . the number of men employed was ? | x - - - - - - - 12 ( x + 10 ) - - - - 6 x * 12 = ( x + 10 ) 6 x = 10 \ answer : e | a ) 22 , b ) 20 , c ) 88 , d ) 71 , e ) 10 | e | divide(multiply(multiply(3, const_2), 10), subtract(12, multiply(3, const_2))) | multiply(n1,const_2)|multiply(n2,#0)|subtract(n0,#0)|divide(#1,#2) | physics |
a producer of tea blends two varieties of tea from two tea gardens one costing rs 18 per kg and another rs 20 per kg in the ratio 5 : 3 . if he sells the blended variety at rs 23 per kg , then his gain percent is | explanation : suppose he bought 5 kg and 3 kg of tea . cost price = rs . ( 5 x 18 + 3 x 20 ) = rs . 150 . selling price = rs . ( 8 x 23 ) = rs . 184 . profit = 184 - 150 = 34 so , profit % = ( 34 / 150 ) * 100 = 23 % option b | a ) 12 % , b ) 23 % , c ) 14 % , d ) 15 % , e ) 16 % | b | divide(multiply(subtract(multiply(23, add(5, 3)), add(multiply(5, 18), multiply(3, 20))), const_100), add(multiply(5, 18), multiply(3, 20))) | add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|add(#1,#2)|multiply(n4,#0)|subtract(#4,#3)|multiply(#5,const_100)|divide(#6,#3) | gain |
a can run 4 times as fast as b and gives b a start of 69 m . how long should the race course be so that a and b might reach in the same time ? | "speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 69 m he gains 69 * ( 4 / 3 ) i . e 92 m answer : e" | a ) 70 m , b ) 60 m , c ) 80 m , d ) 65 m , e ) 92 m | e | add(multiply(4, divide(divide(69, 4), subtract(4, const_1))), 69) | divide(n1,n0)|subtract(n0,const_1)|divide(#0,#1)|multiply(n0,#2)|add(n1,#3)| | physics |
a car travels uphill at 30 km / hr and downhill at 40 km / hr . it goes 100 km uphill and 50 km downhill . find the average speed of the car ? | "avg speed = total distance / total time . total distance traveled = 100 + 50 = 150 km ; time taken for uphill journey = 100 / 30 = 10 / 3 ; time taken for down hill journey = 50 / 40 = 5 / 4 ; avg speed = 150 / ( 10 / 3 + 5 / 4 ) = 33 kmph answer : b" | a ) 32 kmph , b ) 33 kmph , c ) 34 kmph , d ) 35 kmph , e ) 36 kmph | b | divide(add(100, 50), add(divide(100, 30), divide(50, 40))) | add(n2,n3)|divide(n2,n0)|divide(n3,n1)|add(#1,#2)|divide(#0,#3)| | general |
the price of a bushel of corn is currently $ 3.20 , and the price of a peck of wheat is $ 7.80 . the price of corn is increasing at a constant rate of 5 x cents per day while the price of wheat is decreasing at a constant rate of x ( 2 ^ 1 / 2 ) - x cents per day . what is the approximate price when a bushel of corn co... | "i tried using time / rate approach : - initial price difference = 7.80 - 3.20 = 4.60 price of corn increasing by 5 x price of wheat decreasing by x ( 1.4 ) - x = . 4 x since both the quantities are moving towards reducing the price gap hence : - relative increase = 5 x + . 4 x let t be the time by which gap is filled ... | a ) $ 4.50 , b ) $ 5.10 , c ) $ 5.30 , d ) $ 7.50 , e ) $ 5.60 | d | add(3.20, multiply(divide(subtract(7.80, 3.20), add(5, subtract(sqrt(2), 1))), 5)) | sqrt(n3)|subtract(n1,n0)|subtract(#0,n4)|add(n2,#2)|divide(#1,#3)|multiply(n2,#4)|add(n0,#5)| | general |
a person saved $ 10 in buying an item on sale . if he spent $ 400 for the item , approximately how much percent he saved in the transaction ? | "actual price = 400 + 10 = $ 410 saving = 10 / 410 * 100 = 100 / 41 = 2.4 % approximately answer is d" | a ) 8 % , b ) 10 % , c ) 2 % , d ) 2.4 % , e ) 6 % | d | add(floor(multiply(divide(10, add(10, 400)), const_100)), const_1) | add(n0,n1)|divide(n0,#0)|multiply(#1,const_100)|floor(#2)|add(#3,const_1)| | general |
the diagonals of the two squares are in the ratio of 7 : 9 find the ratio of their area | let the diagonals of the square be 7 x and 9 x respectively . ratio of their areas = 1 / 2 ( 7 x ) ^ 2 : 1 / 2 ( 9 x ) = 49 x ^ 2 : 81 x ^ 2 = 49 : 81 answer ( c ) | ['a ) 78 : 89', 'b ) 5 : 48', 'c ) 49 : 81', 'd ) 74 : 36', 'e ) 25 : 49'] | c | divide(divide(power(7, const_2), const_2), divide(power(9, const_2), const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,#3) | geometry |
in a certain parking lot , 10 % of the cars are towed for parking illegally . however 80 % of the cars which are parked illegally are not towed . what percentage of cars in the parking lot are parked illegally ? | "let x be the number of cars and let y be the number of cars parked illegally . 10 % * x = 20 % * y y / x = 1 / 2 = 50 % the answer is c ." | a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | c | divide(subtract(const_1, divide(80, const_100)), divide(10, const_100)) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|divide(#2,#1)| | gain |
for what value of â € œ k â € will the equation ( 2 kx 2 + 7 kx + 2 ) = 0 have equal roots ? | "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 ( 7 k ) ^ 2 - 4 * 2 k * 2 = 0 by solving this equation we get k = 0 , k = 16 / 49 answer : b" | a ) 2 / 7 , b ) 16 / 49 , c ) 16 / 25 , d ) 7 / 1 , e ) 7 / 2 | b | divide(power(2, add(2, 2)), power(7, 2)) | add(n0,n0)|power(n2,n0)|power(n0,#0)|divide(#2,#1)| | general |
in how many ways can 5 different rings be worn in 4 particular fingers ? ( some fingers may get more than one ring and some may get no rings . ) can somebody explain ? | first ring can be worn in 4 ways ( on any of the four fingers ) ; second ring can be worn in 5 ways ( as it can go on any of four fingers - 4 ways ; plus it can go below the first one - 1 ) ; third ring can be worn in 6 ways ( the same logic as for the second ring ) ; fourth ring can be worn in 7 ways ; fifth ring can ... | a ) 6720 , b ) 7720 , c ) 7820 , d ) 7950 , e ) 8120 | a | multiply(multiply(multiply(multiply(4, 5), add(const_4, const_2)), add(add(const_4, const_2), const_1)), add(add(add(const_4, const_2), const_1), const_1)) | add(const_2,const_4)|multiply(n0,n1)|add(#0,const_1)|multiply(#0,#1)|add(#2,const_1)|multiply(#2,#3)|multiply(#4,#5) | general |
bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 50 white marbles . how many red marbles could be in bag a ? | 6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 k w = 50 = 6 k + 4 k = > k = 5 total red ' s in bag a will be 2 k = 10 b | a ) 1 , b ) 5 , c ) 4 , d ) 6 , e ) 8 | b | divide(50, add(multiply(3, 2), 4)) | multiply(n1,n2)|add(n5,#0)|divide(n6,#1) | other |
a box contains 6 pairs of shoes ( 12 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ? | the problem with your solution is that we do n ' t choose 1 shoe from 12 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 11 as after taking one at random there are 11 shoes left and only one is the pair of the first ... | a ) 1 / 11 , b ) 1 / 20 , c ) 1 / 19 , d ) 1 / 10 , e ) 1 / 9 | a | divide(const_1, subtract(12, const_1)) | subtract(n1,const_1)|divide(const_1,#0) | general |
the number of stamps that kaye and alberto had were in the ration of 5 : 3 respectively . after kaye gave alberto 12 of her stamps , the ration of the number of kaye had to the number of alberto had was 8 : 6 . as a result of the gift , kaye had how many more stamps than alberto ? | c k 1 = 5 x a 1 = 3 x k 2 = 5 x - 12 a 2 = 3 x + 12 k 2 / a 2 = 8 / 6 ( 5 x - 12 ) / ( 3 x + 12 ) = 8 / 6 6 * ( 5 x - 12 ) = 8 * ( 3 x + 12 ) 30 x - 72 = 24 x + 96 6 x = 168 x = 28 k 2 = 5 * 28 - 12 = 128 a 2 = 3 * 28 + 12 = 96 k 2 - a 2 = 32 answer : d | a ) 20 , b ) 30 , c ) 40 , d ) 32 , e ) 90 | d | multiply(8, divide(lcm(add(5, 3), add(8, 6)), add(8, 6))) | add(n0,n1)|add(n3,n4)|lcm(#0,#1)|divide(#2,#1)|multiply(n3,#3) | other |
you hold some gold in a vault as an investment . over the past year the price of gold increases by 30 % . in order to keep your gold in the vault , you must pay 2 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year . | "( 100 % + 30 % ) * ( 100 % - 2 % ) = 130 * 0.98 = 127.4 % an increase of 27.4 % your gold holdings have increased in value by 27.4 % . the answer is b" | a ) 27.5 % , b ) 27.4 % , c ) 27.6 % , d ) 27.7 % , e ) 28.4 % | b | subtract(30, divide(30, 2)) | divide(n0,n1)|subtract(n0,#0)| | gain |
a number x is 11 times another number y . the percentage that y is less than x is | "say y = 1 and x = 11 . then y = 1 is less than x = 11 by ( 11 - 1 ) / 11 * 100 = 10 / 11 * 100 = 90.9 % . answer : b ." | a ) 12.5 % , b ) 90.9 , c ) 91.7 , d ) 11 % , e ) 1 % | b | multiply(divide(subtract(11, const_1), 11), const_100) | subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)| | general |
ravi ' s 4 days average income is 1025.68 . but in a form he should write his average income as the greatest positive even integer less than or equal to his 4 days average income . what is the difference between his real average and form filling average ? | since ravi ' s 4 days average income for form filling is defined as the greatest positive even integer less than or equal to his avg income , then avg income for form filling = 1024 ( the greatest positive even integer less than or equal to 1025.68 is 1024 ) . hence the difference = 1025.68 - 1024 = 1.68 answer : c . | a ) 2 , b ) 4.3 , c ) 1.68 , d ) 1.5 , e ) 3.2 | c | subtract(1025.68, subtract(floor(1025.68), const_1)) | floor(n1)|subtract(#0,const_1)|subtract(n1,#1) | general |
roy was suffering from severe headaches . he went to see his doctor and the doctor gave him 5 tablets asking him to take one tablet every 15 minutes . how much time will it take roy to consume all the 5 tablets ? | a 60 min roy will be able to consume all the five tablets in an hour . tablet 1 will be taken in 0 min . tablet 2 will be taken in 15 min . tablet 3 will be taken in 30 min . tablet 4 will be taken in 45 min . tablet 5 will be taken in 60 min . | a ) 60 min , b ) 50 min , c ) 70 min , d ) 65 min , e ) 80 min | a | subtract(multiply(5, 15), 15) | multiply(n0,n1)|subtract(#0,n1) | general |
if one root of the equation 2 x ^ 2 + 3 x – k = 0 is 7 , what is the value of k ? | we just enter this root into the equation in order to recieve an equation to find the answer ! 2 * 7 ^ 2 + 3 * 7 - k = 0 k = 98 + 21 = 119 the answer is c | a ) 100 , b ) 110 , c ) 119 , d ) 120 , e ) 112 | c | add(multiply(2, power(7, const_2)), multiply(3, 7)) | multiply(n2,n4)|power(n4,const_2)|multiply(n0,#1)|add(#2,#0) | general |
what will be the compound interest on a sum of rs . 24,500 after 3 years at the rate of 12 % p . a . ? | "explanation : amount = [ 24500 * ( 1 + 12 / 100 ) 3 ] = 24500 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 35123.20 c . i . = ( 34420.74 - 24500 ) = rs . 9920.736 answer : a" | a ) s . 9920.736 , b ) s . 10823.20 , c ) s . 10123.20 , d ) s . 10123.29 , e ) s . 10123.28 | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
at a garage sale , all of the items were sold at different prices . if the price of a radio sold at the garage sale was both the 14 th highest price and the 21 st lowest price among the prices of the items sold , how many items were sold at the garage sale ? | there were 13 items sold at a higher price than the radio and 20 items sold at a lower price than the radio . including the radio , there were 13 + 20 + 1 = 34 items sold . the answer is b . | a ) 33 , b ) 34 , c ) 35 , d ) 36 , e ) 37 | b | subtract(add(21, 14), const_1) | add(n0,n1)|subtract(#0,const_1) | other |
what distance will be covered by a city bus moving at 72 kmph in 30 seconds ? | 72 kmph = 72 * 5 / 18 = 20 mps dist = speed * time = 20 * 30 = 600 m . answer c | a ) 200 m , b ) 300 m , c ) 600 m , d ) 500 m , e ) 400 m | c | multiply(multiply(72, const_0_2778), 30) | multiply(n0,const_0_2778)|multiply(n1,#0) | physics |
the cost of 3 pens and 5 pencils is rs . 260 . also the cost of one pen and one pencil is in the ratio of 5 : 1 respectively . what is the cost of one dozen pens ? | explanation : let the cost of one pen is ‘ 5 x ’ and pencil is ‘ x ’ 3 x 5 x + 5 x = rs . 260 15 x + 5 x = rs . 260 x = 260 / 20 = 13 : . cost of 1 pen = 5 x = 5 x 13 = 65 : . cost of 12 pens , i . e . ( one dozen ) = 65 x 12 = rs . 780 answer : option d | a ) rs . 200 , b ) rs . 250 , c ) rs . 300 , d ) rs . 780 , e ) none of these | d | multiply(multiply(const_3, const_4), divide(260, add(3, 1))) | add(n0,n4)|multiply(const_3,const_4)|divide(n2,#0)|multiply(#2,#1) | other |
in a certain archery competition , points were awarded as follows : the first place winner receives 11 points , the second place winner receives 7 points , the third place winner receives 5 points and the fourth place winner receives 2 points . no other points are awarded . john participated several times in the compet... | "107800 = 2 * 2 * 2 * 5 * 5 * 7 * 7 * 11 john participated 8 times . the answer is e ." | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | floor(sqrt(divide(107800, multiply(multiply(multiply(11, 7), 5), 2)))) | multiply(n0,n1)|multiply(n2,#0)|multiply(n3,#1)|divide(n4,#2)|sqrt(#3)|floor(#4)| | general |
according to a recent student poll , 4 / 5 out of 25 members of the finance club are interested in a career in investment banking . if two students are chosen at random , what is the probability that at least one of them is interested in investment banking ? | "20 students are interested , 5 are not interested prob = 1 - 5 c 2 / 25 c 2 = 1 - ( 5 * 4 / ( 25 * 24 ) ) = 1 - 1 / 30 = 29 / 30 answer : a" | a ) 29 / 30 , b ) 4 / 49 , c ) 2 / 7 , d ) 45 / 49 , e ) 13 / 14 | a | divide(subtract(choose(25, const_2), choose(subtract(25, multiply(25, divide(4, 5))), const_2)), choose(25, const_2)) | choose(n2,const_2)|divide(n0,n1)|multiply(n2,#1)|subtract(n2,#2)|choose(#3,const_2)|subtract(#0,#4)|divide(#5,#0)| | gain |
one side of a rectangular field is 16 m and one of its diagonal is 17 m . find the area of the field . | "solution other side = √ ( 17 ) 2 - ( 16 ) 2 = √ 289 - 256 = √ 33 = 5.7 m . ∴ area = ( 16 x 5.7 ) m 2 = 91.2 m 2 . answer c" | a ) 100 , b ) 120 , c ) 91.2 , d ) 180 , e ) none | c | rectangle_area(16, sqrt(subtract(power(17, const_2), power(16, const_2)))) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3)| | geometry |
a certain clock marks every hour by striking a number of times equal to the hour , and the time required for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stroke and the end of the last stroke is 11 seconds . at 12 : 00 , how many seconds el... | "at 6 : 00 it ' ll chime 6 times . if we assume that the time taken to chime is x , then time between chimes is also x . so you have 6 chimes , which is 6 x and 5 time intervals between chimes . this means that 11 x = 11 seconds . thus x = 1 seconds . by a similar logic , at 12 : 00 , there are 12 chimes and 11 interva... | a ) a . 72 , b ) b . 50 , c ) c . 48 , d ) d . 46 , e ) e . 23 | e | multiply(divide(11, add(6, add(const_1, const_4))), add(12, add(6, add(const_1, const_4)))) | add(const_1,const_4)|add(n0,#0)|add(n3,#1)|divide(n2,#1)|multiply(#2,#3)| | general |
- 24 x 29 + 36 = ? | "given exp . = - 24 x ( 30 - 1 ) + 36 = - ( 24 x 30 ) + 24 + 36 = - 720 + 60 = - 660 answer is d" | a ) 2436 , b ) 2801 , c ) - 2801 , d ) - 660 , e ) none of them | d | multiply(subtract(const_1, const_2), subtract(multiply(24, 29), 36)) | multiply(n0,n1)|subtract(const_1,const_2)|subtract(#0,n2)|multiply(#1,#2)| | general |
an alloy weighing 20 ounces is 70 percent gold . how many ounces of pure gold must be added to create an alloy that is 70 percent gold ? | "in 24 ounces , gold is 20 * ( 70 / 100 ) = 14 ounces . now we add x ounces of pure gold to make it 90 % gold . so 14 + x = ( 24 + x ) * 70 / 100 = > x = 9 . answer is b ." | a ) 6 , b ) 9 , c ) 12 , d ) 24 , e ) 48 | b | divide(multiply(divide(70, const_100), 20), divide(70, const_100)) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|divide(#2,#1)| | gain |
a can do a piece of work in 10 days . he works at it for 4 days and then b finishes it in 9 days . in how many days can a and b together finish the work ? | "4 / 10 + 9 / x = 1 = > x = 15 1 / 10 + 1 / 15 = 1 / 6 = > 6 days answer : a" | a ) 6 , b ) 7 , c ) 5 , d ) 4 , e ) 1 | a | inverse(add(divide(const_1, 10), divide(const_1, multiply(9, const_2)))) | divide(const_1,n0)|multiply(n2,const_2)|divide(const_1,#1)|add(#0,#2)|inverse(#3)| | physics |
if a wholesaler distributes items to several outlets ( a , b , c and d ) in the ratio of 1 / 2 : 1 / 11 : 1 / 12 : 1 / 13 then find the total number of items the wholesaler distributes ? | "here , a : b : c : d = 1 / 2 : 1 / 11 : 1 / 12 : 1 / 13 1 ) l . c . m of 2 : 11 : 12 : 13 is 1716 2 ) find the number of books each friend received - - - - - - - - - ( to find no . of books each friend has , multiply the ratio with the l . c . m . calculated ) a = ( 1 / 2 ) x 1716 = 858 b = ( 1 / 11 ) x 1716 = 156 c =... | a ) 289 , b ) 1289 , c ) 1389 , d ) 1281 , e ) 128 | b | add(add(multiply(const_100, const_4), const_100), add(multiply(12, const_4), add(const_4, const_3))) | add(const_3,const_4)|multiply(const_100,const_4)|multiply(n5,const_4)|add(#1,const_100)|add(#0,#2)|add(#3,#4)| | general |
the sum of ages of 5 children born 3 years different each is 60 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 60 5 x = 30 x = 6 . x + 12 = 6 + 12 = 18 d" | a ) 17 , b ) 19 , c ) 16 , d ) 18 , e ) 21 | d | divide(add(add(add(add(3, const_4), add(3, const_4)), add(const_4, const_4)), 60), 5) | add(n1,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)| | general |
given that p is a positive even integer with a positive units digit , if the units digit of p ^ 3 minus the units digit of p ^ 2 is equal to 0 , what is the units digit of p - 3 ? | "p is a positive even integer with a positive units digit - - > the units digit of p can be 2 , 4 , 6 , or 8 - - > in order the units digit of p ^ 3 - p ^ 2 to be 0 , the units digit of p ^ 3 and p ^ 2 must be the same . thus the units digit of p can be 0 , 1 , 5 or 6 . intersection of values is 6 , thus the units digi... | a ) 10 , b ) 8 , c ) 7 , d ) 3 , e ) 11 | d | add(add(3, 3), 3) | add(n0,n0)|add(n3,#0)| | general |
a 240 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | "s = 240 / 6 * 18 / 5 = 144 kmph answer : b" | a ) 229 , b ) 144 , c ) 278 , d ) 126 , e ) 112 | b | multiply(divide(240, 6), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
a person travels equal distances with speeds of 2 km / hr , 6 km / hr , 6 km / hr . and takes a total time of 11 minutes . find the total distance ? | "let the each distance be x km total distance = 3 x then total time , ( x / 2 ) + ( x / 6 ) + ( x / 6 ) = 11 / 60 x = 0.22 total distance = 3 * 0.22 = 0.66 km = 660 meters correct option is c" | a ) 1 km , b ) 500 mts , c ) 660 mts , d ) 2 km , e ) 250 mts | c | multiply(multiply(divide(divide(11, const_60), add(add(divide(const_1, 2), divide(const_1, 6)), divide(const_1, 6))), const_3), const_1000) | divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3)|multiply(#7,const_1000)| | physics |
natasha climbs up a hill , and descends along the same way she went up . it takes her 4 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 3 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | let the distance to the top be x , so the total distance traveled by natasha is 2 x . the total time is 4 + 2 = 6 hours the average speed = total distance / total time taken = 2 x / 6 = x / 3 the average speed of the complete journey is 3 km / hour x / 3 = 3 x = 9 km the average speed while climbing = distance / time =... | a ) 1.75 , b ) 2.25 , c ) 2.5 , d ) 2.75 , e ) 3.25 | b | divide(divide(multiply(add(4, 2), 3), 2), 4) | add(n0,n1)|multiply(n2,#0)|divide(#1,n1)|divide(#2,n0) | physics |
mike drives his new corvette from san francisco to las vegas , a journey of 640 miles . he drives the first half of the trip at an average rate of 80 miles per hour , but has to slow down for the second half of his journey . if the second half of the trip takes him 200 percent longer than the first half , what is his a... | "veritas prepofficial solution correct answer : c using the formula : time = distance / rate , we find that mike takes 4 hours to cover the first 320 miles of his trip . since the 2 nd 320 miles take 200 % longer than the first , it takes mike 8 hours longer , or 12 hours . ( note : 200 % longer than the first half is ... | a ) p = 26.7 , b ) p = 30.0 , c ) p = 40.0 , d ) p = 53.3 , e ) p = 60.0 | c | divide(640, add(add(multiply(divide(divide(640, const_2), 80), const_2), divide(divide(640, const_2), 80)), divide(divide(640, const_2), 80))) | divide(n0,const_2)|divide(#0,n1)|multiply(#1,const_2)|add(#1,#2)|add(#3,#1)|divide(n0,#4)| | physics |
a group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 1 students left over . what is the sum of the two smallest possible values of n ? | "4 x + 1 = 5 y + 1 . . . . . . . . . . . ie : 4 x - 5 y = 2 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 = 21 if y = 6 thus x = not possible fraction if y = 8 thus x = 10 n = 41 21 + 41 = 62 . . . . . b" | a ) 33 , b ) 62 , c ) 49 , d ) 53 , e ) 86 | b | add(add(multiply(5, const_2), 1), add(multiply(5, multiply(const_2, 1)), 1)) | multiply(n2,const_2)|multiply(const_2,n3)|add(n3,#0)|multiply(n2,#1)|add(n3,#3)|add(#2,#4)| | general |
the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 12 percent , what was the annual interest rate last year ? | "12 = 1.1 * x x = 10.90 % answer e )" | a ) 1 % , b ) 1.1 % , c ) 9.1 % , d ) 10 % , e ) 10.9 % | e | divide(multiply(12, const_100), add(12, const_100)) | add(n1,const_100)|multiply(n1,const_100)|divide(#1,#0)| | gain |
if a - b = 6 and a ^ 2 + b ^ 2 = 100 , find the value of ab | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 100 - 36 = 64 = > ab = 32 answer : b" | a ) 10 , b ) 32 , c ) 15 , d ) 18 , e ) 19 | b | multiply(multiply(add(6, divide(subtract(sqrt(100), 6), 2)), divide(subtract(sqrt(100), 6), 2)), 2) | sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)| | general |
the average amount with a group of 7 numbers is rs . 20 . if the newly joined member has rs . 56 with him , what was the average amount with the group before his joining the group ? | total members in the group = 7 average amount = rs . 20 total amount with them = 7 * 20 = rs . 140 one number has rs . 56 . so , the amount with remaining 6 people = 140 - 56 = rs . 84 the average amount with them = 84 / 6 = rs . 14 . answer : e | a ) s . 13 , b ) s . 12 , c ) s . 15 , d ) s . 22 , e ) s . 14 | e | divide(subtract(multiply(20, 7), 56), subtract(7, const_1)) | multiply(n0,n1)|subtract(n0,const_1)|subtract(#0,n2)|divide(#2,#1) | general |
find the missing figures : ? % of 50 = 2.125 | ( i ) let x % of 50 = 2.125 . then , ( x / 100 ) * 50 = 2.125 x = ( 2.125 * 2 ) = 4.25 answer is e . | a ) 8.55 , b ) 6.55 , c ) 8.75 , d ) 7.75 , e ) 4.25 | e | divide(multiply(2.125, const_100), 50) | multiply(n1,const_100)|divide(#0,n0) | gain |
a person borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 8 % p . a for 2 years . find his gain in the transaction per year . | explanation : the person borrows rs . 5000 for 2 years at 4 % p . a . simple interest simple interest that he needs to pay = prt / 100 = 5000 × 4 × 2 / 100 = 400 he also lends it at 8 % p . a for 2 years simple interest that he gets = prt / 100 = 5000 × 8 × 2 / 100 = 800 his overall gain in 2 years = rs . 800 - rs . 40... | a ) 167.5 , b ) 150 , c ) 200 , d ) 112.5 , e ) 212.5 | c | divide(subtract(divide(multiply(multiply(5000, 8), 2), const_100), divide(multiply(multiply(5000, 4), 2), const_100)), 2) | multiply(n0,n3)|multiply(n0,n2)|multiply(n1,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|divide(#6,n1) | gain |
the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 114 ounces , what is the volume , in cubic inches , of 63 ounces of this substance ? | "112 ounces of a substance has a volume of 48 cubic inches 63 ounces of a substance has a volume of ( 48 / 114 ) * 63 = 26 cubic inches answer b" | a ) 27 , b ) 26 , c ) 42 , d ) 64 , e ) 147 | b | multiply(divide(48, 114), 63) | divide(n0,n1)|multiply(n2,#0)| | geometry |
the side of a square is increased by 5 % then how much % does its area increases ? | "a = 100 a 2 = 10000 a = 105 a 2 = 11025 - - - - - - - - - - - - - - - - 10000 - - - - - - - - - 1025 100 - - - - - - - ? = > 10.25 % answer : b" | a ) 15.00 % , b ) 10.25 % , c ) 10.00 % , d ) 12.25 % , e ) 12.50 % | b | divide(multiply(subtract(square_area(add(const_100, 5)), 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)| | geometry |
a circular mat with radius 10 inches is placed on a square tabletop , each of whose sides is 24 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? | c . it is a circle inscribed in a square . square side = 24 - - - > square ( table ) area = 24 ^ 2 circle diameter = 20 - - - > circle area = pir ^ 2 = 100 pi ( where pi = ~ 3.14 ) covered fraction = 100 * 3.14 / 24 * 24 = ~ 314 / 24 * 24 = 0.5451 c | ['a ) 5 / 12', 'b ) 2 / 5', 'c ) 0.5451', 'd ) 3 / 4', 'e ) 5 / 6'] | c | divide(circle_area(10), square_area(24)) | circle_area(n0)|square_area(n1)|divide(#0,#1) | geometry |
money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 1500 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ? | "x = 8 % 112 / x years = 112 / 8 = 14 years now , money triples every 14 years therefore , in 14 yrs , if $ 1500 triples to $ 4500 , in 28 years , it will again triple to $ 4500 * 3 = $ 13,500 answer c" | a ) $ 3,750 , b ) $ 5,600 , c ) $ 13,500 , d ) $ 15,000 , e ) $ 22,500 | c | multiply(1500, power(const_3, divide(28, divide(112, 8)))) | divide(n0,n2)|divide(n3,#0)|power(const_3,#1)|multiply(n1,#2)| | gain |
the sum of ages of 5 children born 3 years different each is 75 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 75 5 x = 45 x = 9 . x + 12 = 9 + 12 = 21 e" | a ) 7 , b ) 9 , c ) 16 , d ) 18 , e ) 21 | e | divide(add(add(add(add(3, const_4), add(3, const_4)), add(const_4, const_4)), 75), 5) | add(n1,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)| | general |
in how many ways can a teacher write an answer key for a mini - quiz that contains 3 true - false questions followed by 2 multiples - choice questions with 4 answer choices each , if the correct answers to all true - false questions can not be the same ? | 2 multiples - choice questions can be answered in = 4 x 4 = 16 ways 3 true - false questions can be answered in = 2 x 2 x 2 = 8 ways but out of the 8 ways , 2 ways [ ( true - true - true ) ( false - false - false ) ] will contain same answers thus 3 true - false questions can be answered in = 2 x 2 x 2 = 6 ways total w... | a ) 88 , b ) 90 , c ) 96 , d ) 98 , e ) 102 | c | multiply(subtract(multiply(const_2, multiply(2, 2)), 2), multiply(const_4, const_4)) | multiply(const_4,const_4)|multiply(n1,n1)|multiply(#1,const_2)|subtract(#2,n1)|multiply(#0,#3) | general |
a 50 - liter solution of alcohol and water is 5 percent alcohol . if 1.5 liters of alcohol and 8.5 liters of water are added to this solution , what percent of the solution produced is alcohol ? | "50 l * . 05 = 2.5 l of alc , 50 l - 2.5 l = 47.5 l of water 2.5 + 1.5 = 4 l of alcohol in new solution 47.5 l + 8.5 l = 56 l of water 4 l / 56 l = 1 / 14 this is ~ 7 % answer : e" | a ) 5.5 % , b ) 6 % , c ) 6 1 / 3 % , d ) 6 2 / 3 % , e ) 7 % | e | multiply(divide(add(multiply(50, divide(5, const_100)), 1.5), add(50, add(1.5, 8.5))), const_100) | add(n2,n3)|divide(n1,const_100)|add(n0,#0)|multiply(n0,#1)|add(n2,#3)|divide(#4,#2)|multiply(#5,const_100)| | general |
there are 3 departments having students 72 , 5824 . in an exam they have to be seated in rooms such that each room has equal number of students and each room has students of one type only ( no mixing of departments ) . find the minimum number of rooms required ? | we need to take gcd which is 2 thus all the rooms will have 2 students of the same dept 1 ) 72 / 2 = 36 2 ) 58 / 2 = 29 3 ) 24 / 2 = 12 total no . of min rooms reqd = 36 + 12 + 29 = 77 answer : e | a ) 73 , b ) 74 , c ) 75 , d ) 76 , e ) 77 | e | add(add(divide(72, const_2), const_12), divide(add(add(divide(72, const_2), const_12), const_10), const_2)) | divide(n1,const_2)|add(#0,const_12)|add(#1,const_10)|divide(#2,const_2)|add(#1,#3) | general |
a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 120 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "e 120 * t = 1 km = > t = 1 / 120 km / h v * ( t + 2 / 3600 ) = 1 v ( 1 / 120 + 2 / 3600 ) = 1 = > v = 112.5 km / h" | a ) 121.5 , b ) 122 , c ) 122.5 , d ) 113 , e ) 112.5 | e | divide(1, divide(add(multiply(const_3600, divide(1, 120)), 2), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
an article is bought for rs . 765 and sold for rs . 1000 , find the gain percent ? | "765 - - - - 235 100 - - - - ? = > 30.7 % answer : c" | a ) 30.6 % , b ) 30.5 % , c ) 30.7 % , d ) 30.8 % , e ) 30.9 % | c | subtract(const_100, divide(multiply(1000, const_100), 765)) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)| | gain |
the price of a cycle is reduced by 25 per cent . the new price is reduced by a further 70 per cent . the two reductions together are equal to a single reduction of | "price = p initially price reduced by 25 % which means new price is 3 / 4 p now on this new price further 70 percent is reduced which means the new price is merely 30 percent of 3 / 4 p = = > ( 3 / 4 ) x ( 3 / 10 ) p = 9 / 40 p is the new price after both deduction which is 22.5 percent of the original value p . this i... | a ) 45 % , b ) 40 % , c ) 35 % , d ) 77.5 % , e ) 30 % | d | subtract(const_100, multiply(divide(subtract(const_100, 70), const_100), subtract(const_100, 25))) | subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|multiply(#2,#1)|subtract(const_100,#3)| | general |
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 35 % profit ? | 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 . = 130 % of rs . 1600 = 130 / 100 * 1600 = rs . 2080 . answer : a | a ) 2080 , b ) 2778 , c ) 2299 , d ) 2778 , e ) 2771 | a | multiply(divide(add(const_100, 35), 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 |
running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 10 such machines produce in 4 minutes ? | "let the number of required bottles be x machine 6 : 10 time 1 : 4 = : : 270 : x therefore 6 x 1 x x = 10 x 4 x 270 = > x = ( 10 x 4 x 270 ) / ( 6 ) = > x = 1800 answer : option b" | a ) 648 , b ) 1800 , c ) 2700 , d ) 10800 , e ) none | b | multiply(multiply(divide(270, 6), 4), 10) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
thirty percent of the members of a swim club have passed the lifesaving test . among the members who havenotpassed the test , 40 have taken the preparatory course and 30 have not taken the course . how many members are there in the swim club ? | "30 % of the members have passed the test , thus 70 % have not passed the test . we also know that 30 + 40 = 70 members have not passed the test , thus 0.7 * total = 70 - - > total = 100 . answer : c ." | a ) 60 , b ) 80 , c ) 100 , d ) 120 , e ) 140 | c | divide(add(40, 30), divide(subtract(const_100, 30), const_100)) | add(n0,n1)|subtract(const_100,n1)|divide(#1,const_100)|divide(#0,#2)| | gain |
what is the smallest number which when diminished by 8 , is divisible by 9 , 6 , 12 and 18 ? | "explanation : required number = lcm of ( 9 , 6 , 12 and 18 ) + 8 = 36 + 8 = 44 answer : option c" | a ) 50 , b ) 45 , c ) 44 , d ) 60 , e ) 65 | c | add(lcm(lcm(9, 6), lcm(12, 18)), 8) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|add(n0,#2)| | general |
the ratio of two quantities is 1 to 2 . if each of the quantities is increased by 15 , what is the ratio of these 2 new quantities ? | "the ratio can not be straight way added any quantity . . . 1 : 2 means 1 x : 2 x . . . so when you add a quantity , it becomes 1 x + 15 : 2 x + 15 . . so value of x is must . . ans e" | a ) 5 : 6 , b ) 25 : 27 , c ) 15 : 16 , d ) 20 : 21 , e ) it can not be determined from the information given | e | divide(add(1, 15), add(2, 15)) | add(n0,n2)|add(n1,n2)|divide(#0,#1)| | other |
on a partly cloudy day , derek decides to walk back from work . when it is sunny , he walks at a speed of s miles / hr ( s is an integer ) and when it gets rain cloud , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction of the total distance d... | if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 / 4 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or... | a ) 1 / 8 , b ) 1 / 6 , c ) 1 / 7 , d ) 1 / 5 , e ) 1 / 4 | c | subtract(divide(lcm(const_2, const_3), 2.8), const_2) | lcm(const_2,const_3)|divide(#0,n1)|subtract(#1,const_2) | general |
two tracks are parallel . the first track has 6 checkpoints and the second one has 10 checkpoints . in how many ways can the 6 checkpoints of first track be joined with the 10 checkpoints of the second to form a triangle ? | to make a triangle , you need 2 checkpoints from one track and 1 from the other . you can not have all 3 from the same track since then the points will be in a line ( assuming straight line of track ) you select 2 checkpoints from the first track and one from the second or two from the second track and one from the fir... | ['a ) 120', 'b ) 150', 'c ) 200', 'd ) 270', 'e ) 420'] | e | add(multiply(divide(divide(factorial(10), factorial(subtract(10, const_2))), factorial(const_2)), divide(factorial(6), factorial(subtract(6, const_1)))), multiply(divide(factorial(10), factorial(subtract(const_10, const_1))), divide(divide(factorial(6), const_2), factorial(subtract(6, const_2))))) | factorial(n1)|factorial(const_2)|factorial(n0)|subtract(n1,const_2)|subtract(n0,const_1)|subtract(const_10,const_1)|subtract(n0,const_2)|divide(#2,const_2)|factorial(#3)|factorial(#4)|factorial(#5)|factorial(#6)|divide(#0,#8)|divide(#2,#9)|divide(#0,#10)|divide(#7,#11)|divide(#12,#1)|multiply(#14,#15)|multiply(#16,#13)... | geometry |
an engineer undertakes a project to build a road 15 km long in 300 days and employs 55 men for the purpose . after 100 days , he finds only 2.5 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "55 workers working already let x be the total men required to finish the task in next 200 days 2.5 km done hence remaining is 12.5 km also , work has to be completed in next 200 days ( 300 - 100 = 200 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proport... | a ) a . 43 , b ) b . 45 , c ) c . 55 , d ) d . 68 , e ) e . 83 | e | subtract(divide(multiply(multiply(55, subtract(15, 2.5)), 100), multiply(2.5, subtract(300, 100))), 55) | subtract(n0,n4)|subtract(n1,n3)|multiply(n2,#0)|multiply(n4,#1)|multiply(n3,#2)|divide(#4,#3)|subtract(#5,n2)| | physics |
bob invested $ 2000 in fund a and $ 1000 in fund b . over the next two years , the money in fund a earned a total interest of 12 percent for the two years combined and the money in fund b earned 30 percent annual interest compounded annually . two years after bob made these investments . bob ' s investment in fund a wa... | bob earned 2000 * ( 1 + 0.12 ) in total by fund a and earned 1000 * ( 1 + 0.3 ) ^ 2 in total by fund b . so 2000 * ( 1.12 ) - 1000 * ( 1.69 ) = 550 . the answer is , therefore , ( b ) . | a ) $ 500 , b ) $ 550 , c ) $ 600 , d ) $ 650 , e ) $ 700 | b | subtract(multiply(2000, add(const_1, divide(12, const_100))), multiply(1000, power(add(const_1, divide(30, const_100)), const_2))) | divide(n2,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(n0,#2)|power(#3,const_2)|multiply(n1,#5)|subtract(#4,#6) | general |
the ratio between the sale price and the cost price of an article is 6 : 5 . what is the ratio between the profit and the cost price of that article ? | let c . p . = rs . 5 x and s . p . = rs . 6 x . then , gain = rs . x required ratio = x : 5 x = 1 : 5 answer : b | a ) 2 : 9 , b ) 1 : 5 , c ) 3 : 6 , d ) 2 : 0 , e ) 2 : 1 | b | divide(subtract(6, 5), 5) | subtract(n0,n1)|divide(#0,n1) | other |
how many digits 2 ^ 1000 has ? | "2 ^ 10 = 1.024 * 10 ^ 3 = > 2 ^ 1000 = ( 1.024 ) ^ 100 * 10 ^ 300 therefore 310 digits would be my best guess e" | a ) 31 , b ) 35 , c ) 50 , d ) 99 , e ) 310 | e | floor(add(const_1, multiply(divide(log(2), log(const_10)), 1000))) | log(n0)|log(const_10)|divide(#0,#1)|multiply(n1,#2)|add(#3,const_1)|floor(#4)| | general |
a father purchased dress for his 3 daughters . the dresses are of same color but diff size and they are kept in dark room . what is probability that all the 3 will not choose their own dress ? | answer : b | a ) 22 , b ) 87 , c ) 29 , d ) 26 , e ) 191 | b | divide(choose(const_2, const_1), factorial(3)) | choose(const_2,const_1)|factorial(n0)|divide(#0,#1)| | probability |
if the population of a certain country increases at the rate of two person every 60 seconds , by how many persons does the population increase in 100 minutes ? | "answer = 2 * 100 = 200 answer = e" | a ) 120 , b ) 150 , c ) 240 , d ) 220 , e ) 200 | e | multiply(divide(const_60, 60), 100) | divide(const_60,n0)|multiply(n1,#0)| | physics |
a salt manufacturing company produced a total of 2500 tonnes of salt in january of a particular year . starting from february its production increased by 100 tonnes every month over the previous months until the end of the year . find its ave 66 rage daily production for that year ? | "total production of salt by the company in that year = 2500 + 2600 + 2700 + . . . . + 3600 = 36600 . average monthly production of salt for that year = 36600 / 365 â ‰ ˆ 100 answer : a" | a ) 100 , b ) 105 , c ) 109 , d ) 120 , e ) 90 | a | divide(divide(multiply(add(multiply(2500, const_2), multiply(subtract(const_12, const_1), 100)), const_12), const_2), add(add(multiply(const_3, 100), multiply(multiply(const_2, const_3), const_10)), add(const_2, const_3))) | add(const_2,const_3)|multiply(n0,const_2)|multiply(n1,const_3)|multiply(const_2,const_3)|subtract(const_12,const_1)|multiply(n1,#4)|multiply(#3,const_10)|add(#1,#5)|add(#2,#6)|add(#8,#0)|multiply(#7,const_12)|divide(#10,const_2)|divide(#11,#9)| | general |
1200 men have provisions for 18 days . if 450 more men join them , for how many days will the provisions last now ? | "1200 * 18 = 1650 * x x = 13.1 answer : e" | a ) 12.9 , b ) 12.0 , c ) 12.5 , d ) 12.2 , e ) 13.1 | e | divide(multiply(18, 1200), add(1200, 450)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
when a train travels at a speed of 90 kmph , it reaches the destination on time . when the same train travels at a speed of 50 kmph , it reaches its destination 15 min late . what is the length of journey ? | "let x be the time reached with the speed 90 km / h 50 km / h - - - - > x + 15 distance is equal so 90 ( km / h ) × xhr = 50 ( km / h ) × ( x + 15 ) hr so 90 x = 50 x + 750 so the would be in km and x = 18.75 answer : a" | a ) 18.75 km , b ) 50 km , c ) 60 km , d ) 85 km , e ) 95 km | a | multiply(const_60, divide(multiply(divide(15, const_60), 50), 50)) | divide(n2,const_60)|multiply(n1,#0)|divide(#1,n1)|multiply(#2,const_60)| | physics |
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 40 marks , find the maximum marks set for the test . | "60 % = 120 marks 1 % = 2 marks 100 % = 200 marks the answer is b ." | a ) 180 , b ) 200 , c ) 220 , d ) 240 , e ) 260 | b | divide(add(80, 40), divide(60, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | gain |
what is the sum of all even numbers from 1 to 401 ? | "explanation : 400 / 2 = 200 200 * 201 = 40200 answer : e" | a ) 122821 , b ) 281228 , c ) 281199 , d ) 122850 , e ) 40200 | e | divide(multiply(1, 401), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
the size of a flat - screen television is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square 21 - inch flat - screen television than a square 19 - inch flat - screen television ? | "if we take a square with side length x and draw a diagonal , we get two isosceles right triangles . if we focus on one such right triangle , we see that the legs have length x . square 21 - inch flat - screen television the diagonal ( hypotenuse ) = 21 so , we can apply the pythagorean theorem to get x ² + x ² = 21 ² ... | a ) 42 , b ) 40 , c ) 38 , d ) 36 , e ) 48 | b | subtract(divide(power(21, const_2), const_2), divide(power(19, const_2), const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)| | geometry |
the proportion of copper and zinc in the brass is 13 : 7 . how much zinc will there be in 100 kg of brass ? | 7 / 20 * 100 = 35 answer : b | a ) 37 kg , b ) 35 kg , c ) 85 kg , d ) 45 kg , e ) 25 kg | b | multiply(divide(7, add(7, 13)), 100) | add(n0,n1)|divide(n1,#0)|multiply(n2,#1) | general |
if a man lost 8 % by selling oranges at the rate of 18 a rupee at how many a rupee must he sell them to gain 45 % ? | "92 % - - - - 18 145 % - - - - ? 92 / 145 * 18 = 11.42 answer : c" | a ) 33.56 , b ) 23.68 , c ) 11.42 , d ) 9.56 , e ) 23.55 | c | divide(multiply(subtract(const_100, 8), 18), add(const_100, 45)) | add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)| | gain |
the time it took car p to travel 150 miles was 2 hours less than the time it took car r to travel the same distance . if car p ’ s average speed was 10 miles per hour greater than that of car r , what was car r ’ s average speed , in miles per hour ? | "let speed of car r be = x then speed of car p = x + 10 a / q , ( 150 / x ) - ( 150 / ( x + 10 ) ) = 2 solving for x = 23 miles \ hr . a" | a ) 23 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | a | divide(add(negate(10), sqrt(subtract(power(negate(10), 2), multiply(const_4, negate(divide(multiply(150, 10), 2)))))), 2) | multiply(n0,n2)|negate(n2)|divide(#0,n1)|power(#1,n1)|negate(#2)|multiply(#4,const_4)|subtract(#3,#5)|sqrt(#6)|add(#1,#7)|divide(#8,n1)| | physics |
subtracting 2 % of a from a is equivalent to multiplying a by how much ? | answer let a - 2 % of a = ab . ⇒ ( 98 x a ) / 100 = ab ∴ b = 0.98 correct option : a | a ) 0.98 , b ) 9.4 , c ) 0.094 , d ) 94 , e ) none | a | divide(subtract(const_100, 2), const_100) | subtract(const_100,n0)|divide(#0,const_100) | general |
a cistern 5 m long and 2 m wide contains water up to a breadth of 1 m 10 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 2 * 1.1 + 5 * 1.1 ) ] + 5 * 2 = 25 m square option c" | a ) 42 m sqaure , b ) 49 m sqaure , c ) 25 m sqaure , d ) 28 m sqaure , e ) none of these | c | add(multiply(const_2, add(multiply(add(divide(10, const_100), 1), 2), multiply(add(divide(10, const_100), 1), 5))), multiply(2, 5)) | divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)| | physics |
right now , the ratio between the ages of sandy and molly is 4 : 3 . after 6 years , sandy ’ s age will be 42 years . what is molly ' s age right now ? | "now , sandy is 42 - 6 = 36 molly ' s age is ( 3 / 4 ) * 36 = 27 the answer is b ." | a ) 24 , b ) 27 , c ) 30 , d ) 33 , e ) 36 | b | multiply(divide(subtract(42, 6), 4), 3) | subtract(n3,n2)|divide(#0,n0)|multiply(n1,#1)| | other |
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