Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5
values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6
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x is a positive integer less than 600 . when x is divided by 7 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ? | "the nubmer which when divided by 7 leaves remainder 1 should be of the form 7 k + 1 this number when divided by 3 leaves remainder 2 . so , ( 7 k + 1 ) - 2 should be divisible by 3 or 7 k - 1 should be divisible by 3 . we now put the values of k starting from 0 to find first number divisible by 3 we find 1 st number a... | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | e | subtract(add(multiply(reminder(7, 600), 3), reminder(3, 600)), reminder(1, 600)) | reminder(n1,n0)|reminder(n3,n0)|reminder(n2,n0)|multiply(n3,#0)|add(#3,#1)|subtract(#4,#2)| | general | E |
a man can row his boat with the stream at 12 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 12 s = ? s = ( 12 - 4 ) / 2 = 4 kmph answer : e" | a ) 1 kmph , b ) 4 kmph , c ) 98 kmph , d ) 6 kmph , e ) 4 kmph | e | divide(subtract(12, 4), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain | E |
find the value of 3 + 2 • ( 8 – 3 ) | "3 + 2 • ( 8 – 3 ) = 3 + 2 ( 5 ) = 3 + 2 × 5 = 3 + 10 = 13 answer : ( d )" | a ) ( a ) 25 , b ) ( b ) 13 , c ) ( c ) 17 , d ) ( d ) 24 , e ) ( e ) 15 | d | multiply(add(divide(2, 8), 3), 8) | divide(n1,n2)|add(n0,#0)|multiply(#1,n2)| | general | D |
running at the same rate , 8 identical machines can produce 560 paperclips a minute . at this rate , how many paperclips could 15 machines produce in 6 minutes ? | "8 machines produce 560 in 1 min 8 machines produce 560 * 6 in 6 min 15 machine produce 560 * 6 * ( 15 / 8 ) in 6 minutes 560 * 6 * 15 / 8 = 6300 answer is b ." | a ) 1344 , b ) 6300 , c ) 8400 , d ) 50400 , e ) 67200 | b | multiply(multiply(560, divide(15, 8)), 6) | divide(n2,n0)|multiply(n1,#0)|multiply(n3,#1)| | gain | B |
if a farmer sells 5 of his goats , his stock of feed will last for 4 more days than planned , but if he buys 10 more goats , he will run out of feed 3 days earlier than planned . if no goats are sold or bought , the farmer will be exactly on schedule . how many goats does the farmer have ? | say farmer has n goat and he is good for d days . : - we have 3 equations given in question : - ( n - 5 ) * d + 4 = ( n + 10 ) * ( d - 3 ) = n * d solving these : ( you can solve 1 st and 3 rd and 2 nd and 3 rd together ) we get : 10 d - 3 n = 30 4 n - 5 d = 20 = > n = 20 ans b it is ! | a ) 12 , b ) 20 , c ) 48 , d ) 55 , e ) 60 | b | multiply(subtract(multiply(3, 4), 10), 10) | multiply(n1,n3)|subtract(#0,n2)|multiply(n2,#1) | general | B |
the difference of two numbers is 1375 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1375 ) . x + 1375 = 6 x + 15 5 x = 1360 x = 272 smaller number = 270 . answer b" | a ) 240 , b ) 272 , c ) 295 , d ) 360 , e ) 252 | b | divide(add(1375, 15), subtract(6, const_1)) | add(n0,n2)|subtract(n1,const_1)|divide(#0,#1)| | general | B |
a grocer has a sale of rs . 6435 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6900 ? | "total fr 5 mnths = ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs 34009 . reqd . sale = rs . [ ( 6900 * 6 ) - 34009 ] = rs . ( 41400 - 34009 ) = rs . 7391 . answer : e" | a ) s . 4991 , b ) s . 5991 , c ) s . 6001 , d ) s . 6991 , e ) s . 7391 | e | subtract(multiply(add(5, const_1), 6900), add(add(add(add(6435, 6927), 6855), 7230), 6562)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general | E |
presently the ratio between the ages of dan and james is 6 : 5 . after 4 years dan will be 28 . what is the present age of james ? | let the present ages dan and james be 6 x years and 5 x years respectively 6 x + 4 = 28 6 x = 24 x = 4 kim ' s age = 5 x = 20 years answer is a | a ) 20 , b ) 19 , c ) 21 , d ) 18 , e ) 22 | a | subtract(28, multiply(const_4, const_2)) | multiply(const_2,const_4)|subtract(n3,#0) | other | A |
the average of 5 quantities is 10 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ? | answer : a ( 5 x 10 - 3 x 4 ) / 2 = 19 | a ) 19 , b ) 14 , c ) 8 , d ) 9.5 , e ) none of these | a | divide(subtract(multiply(5, 10), multiply(3, 4)), 2) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4) | general | A |
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 47 , the how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 47 5 x = 45 x = 9 . hence , b ' s age = 2 x = 19 years . d )" | a ) 10 years , b ) 12 years , c ) 14 years , d ) 18 years , e ) 16 years | d | divide(multiply(subtract(47, const_2), const_2), add(const_4, const_1)) | add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)| | general | D |
what is the number of integers from 1 to 1000 ( inclusive ) that are divisible by neither 11 nor by 34 ? | "normally , i would use the method used by bunuel . it ' s the most accurate . but if you are looking for a speedy solution , you can use another method which will sometimes give you an estimate . looking at the options ( most of them are spread out ) , i wont mind trying it . ( mind you , the method is accurate here s... | a ) 884 , b ) 890 , c ) 892 , d ) 910 , e ) 945 | b | subtract(1000, subtract(add(divide(1000, 11), divide(1000, 34)), divide(1000, multiply(11, 34)))) | divide(n1,n2)|divide(n1,n3)|multiply(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,#4)|subtract(n1,#5)| | other | B |
solve for x and check : - 200 x = 1600 | "solution : dividing each side by - 200 , we obtain ( - 200 x / - 200 ) = ( 1600 / - 200 ) therefore : x = - 8 check : - 200 x = 1600 ( - 200 * - 8 ) = 1600 1600 = 1600 answer : c" | a ) 2000 , b ) 2573 , c ) 1600 , d ) 2950 , e ) none of these | c | multiply(200, divide(1600, 200)) | divide(n1,n0)|multiply(n0,#0)| | general | C |
consider the word rotor . whichever way you read it , from left to right or from right to left , you get the same word . such a word is known as palindrome . find the maximum possible number of 5 - letter palindromes | the first letter from the right can be chosen in 26 ways because there are 26 alphabets . having chosen this , the second letter can be chosen in 26 ways . = > the first two letters can be chosen in 26 ã — 26 = 67626 ã — 26 = 676 ways having chosen the first two letters , the third letter can be chosen in 26 ways . = >... | a ) 1700 , b ) 7570 , c ) 1576 , d ) 17576 , e ) 500 | d | power(add(add(add(const_10, const_10), const_3), const_3), subtract(5, const_2)) | add(const_10,const_10)|subtract(n0,const_2)|add(#0,const_3)|add(#2,const_3)|power(#3,#1) | general | D |
the arithmetic mean and standard deviation of a certain normal distribution are 17.5 and 2.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | "mean = 17.5 two standard deviations is 2.5 + 2.5 = 5.0 there could be two calues for this . mean + two standard deviations = 22.5 mean - two standard deviations = 12.5 answer choice has 12.5 and so e is the answer ." | a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | e | subtract(17.5, multiply(2, 2.5)) | multiply(n1,n2)|subtract(n0,#0)| | general | E |
the size of a television screen is given as the length of the screen ' s diagonal . if the screens were flat , then the area of a square 18 - inch screen would be how many square inches greater than the area of a square 16 - inch screen ? | pythogoras will help here ! let the sides be x and diagonal be d then d ^ 2 = 2 x ^ 2 and area = x ^ 2 now plug in the given diagonal values to find x values and then subtract the areas ans will be 18 ^ 2 / 2 - 16 ^ 2 / 2 = 68 / 2 = 34 ans c . | ['a ) 2', 'b ) 4', 'c ) 34', 'd ) 38', 'e ) 40'] | c | divide(subtract(power(18, const_2), power(16, const_2)), const_2) | power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|divide(#2,const_2) | geometry | C |
the current birth rate per certain number of people is 52 , whereas corresponding death rate is 16 per same number of people . if the net growth rate in terms of population increase is 1.2 percent , find number of persons . ( initally ) | "sol . net growth on x = ( 52 - 16 ) = 36 . net growth on 100 = ( 36 / x ã — 100 ) % = 1.2 % . then x = 3000 answer : c" | a ) 4000 , b ) 2000 , c ) 3000 , d ) 5000 , e ) 1000 | c | multiply(const_100, divide(subtract(52, 16), 1.2)) | subtract(n0,n1)|divide(#0,n2)|multiply(#1,const_100)| | gain | C |
the probability that a computer company will get a computer hardware contract is 2 / 3 and the probability that it will not get a software contract is 5 / 9 . if the probability of getting at least one contract is 4 / 5 , what is the probability that it will get both the contracts ? | "explanation : let , a ≡ event of getting hardware contract b ≡ event of getting software contract ab ≡ event of getting both hardware and software contract . p ( a ) = 2 / 3 , p ( ~ b ) = 5 / 9 = > p ( b ) = 1 - ( 5 / 9 ) = 4 / 9 . a and b are not mutually exclusive events but independent events . so , p ( at least on... | a ) 14 / 45 , b ) 4 / 9 , c ) 4 / 5 , d ) 4 / 6 , e ) none of these | a | subtract(add(divide(2, 3), subtract(const_1, divide(2, 9))), divide(9, 5)) | divide(n0,n1)|divide(n0,n3)|divide(n3,n5)|subtract(const_1,#1)|add(#0,#3)|subtract(#4,#2)| | other | A |
roses can be purchased individually for $ 4.50 , one dozen for $ 36 , or two dozen for $ 50 . what is the greatest number of roses that can be purchased for $ 680 ? | buy as many $ 50 deals as possible . we can by 650 / 50 = 13 two dozen roses , thus total of 13 * 24 = 312 roses . we are left with 680 - 650 = $ 30 . we can buy 30 / 4.5 = ~ 6 roses for that amount . total = 312 + 6 = 318 . answer : c . | a ) 156 , b ) 162 , c ) 318 , d ) 324 , e ) 325 | c | add(multiply(floor(divide(680, 50)), multiply(const_12, const_2)), floor(divide(subtract(680, multiply(floor(divide(680, 50)), 50)), 4.5))) | divide(n3,n2)|multiply(const_12,const_2)|floor(#0)|multiply(n2,#2)|multiply(#2,#1)|subtract(n3,#3)|divide(#5,n0)|floor(#6)|add(#7,#4) | general | C |
if k ^ 3 is divisible by 180 , what is the least possible value of integer k ? | "180 = 2 ^ 2 * 3 ^ 2 * 5 therefore k must include at least 2 * 3 * 5 = 30 . the answer is b ." | a ) 12 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | b | divide(divide(180, const_2), const_2) | divide(n1,const_2)|divide(#0,const_2)| | general | B |
daal is now being sold at rate rs . 20 a kg . during last month its rate was rs 16 per kg . by how reduce percent should a family its consumption so as to keep the expenditure fixed ? | last month rate = 16 per kg this month rate = 20 per kg if 20 = 100 % then 16 = ? therefore % = ( 16 * 100 ) / 20 = 80 % to reduce consumption 100 - 80 = 20 % answer : b | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | subtract(const_100, multiply(inverse(divide(20, 16)), const_100)) | divide(n0,n1)|inverse(#0)|multiply(#1,const_100)|subtract(const_100,#2) | gain | B |
how many positive integers between 1 and 100 are there such that they are multiples of 15 ? | "multiples of 15 = 15 , 30,45 , - - - - - 90 number of multiples of 15 = > 90 - 15 / 15 + 1 = 6 answer is e" | a ) 5 , b ) 4 , c ) 7 , d ) 1 , e ) 6 | e | divide(subtract(100, 1), 15) | subtract(n1,n0)|divide(#0,n2)| | general | E |
two twins sisters sita and geeta were standing back to back and suddenly they started running in opposite directions for 10 km each . then they turned left and ran for another 7.5 km . what is the distance ( in kilometers ) between the the two twins when they stop ? | the distance between them is the hypotenuse of a right angle triangle with sides 15 km and 20 km . the hypotenuse = sqrt ( 15 ^ 2 + 20 ^ 2 ) = 25 the answer is c . | a ) 21 , b ) 23 , c ) 25 , d ) 27 , e ) 30 | c | sqrt(add(power(multiply(7.5, const_2), const_2), power(multiply(10, const_2), const_2))) | multiply(n1,const_2)|multiply(n0,const_2)|power(#0,const_2)|power(#1,const_2)|add(#2,#3)|sqrt(#4) | physics | C |
the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 972 sq m , then what is the breadth of the rectangular plot ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 972 3 b 2 = 972 b 2 = 324 b = 18 m . answer : option c" | a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 14 | c | sqrt(divide(972, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry | C |
r is the set of positive even integers less than 201 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "r is the set of positive even integers less than 201 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 2,4 , 6,8 , 10,12 . . . s = 4,16 , 36,64 . . . numbers : 4 , 16 , 36 , 64 , 100 , 144 , and 196 are even integers ( less than 201 ) that are in ... | a ) none , b ) two , c ) four , d ) five , e ) seven | e | subtract(const_4, const_1) | subtract(const_4,const_1)| | physics | E |
when positive integer n is divided by positive integer j , the remainder is 18 . if n / j = 134.08 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 134.08 here 134 is the quotient . given that remainder = 18 so , 134.08 = 134 + 18 / j so , j = 225 ans e" | a ) 22 , b ) 56 , c ) 78 , d ) 112 , e ) 225 | e | divide(18, subtract(134.08, add(const_100, add(multiply(const_4, const_10), const_2)))) | multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3)| | general | E |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.75 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 250 ? | "to maximize number of hot dogs with 250 $ total number of hot dogs bought in 250 - pack = 22.95 * 10 = 229.5 $ amount remaining = 250 - 229.5 = 20.5 $ total number of hot dogs bought in 20 - pack = 3.05 * 6 = 18.3 $ amount remaining = 20.5 - 18.3 = 2.2 $ total number of hot dogs bought in 8 - pack = 1.55 * 1 = 1.55 $ ... | a ) 1,108 , b ) 2,100 , c ) 2,108 , d ) 2,628 , e ) 2,256 | d | multiply(divide(250, 22.95), 250) | divide(n6,n5)|multiply(n4,#0)| | general | D |
if one root of the equation 2 x ^ 2 + 3 x – k = 0 is 5 , 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 * 5 ^ 2 + 3 * 5 - k = 0 k = 50 + 15 = 65 the answer is a" | a ) 65 , b ) 79 , c ) 68 , d ) 87 , e ) 90 | a | add(multiply(2, power(5, 2)), multiply(3, 5)) | multiply(n2,n4)|power(n4,n0)|multiply(n0,#1)|add(#2,#0)| | general | A |
a boatman goes 3 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes . how long will it take to go 5 km in stationary water ? | "speed ( upstream ) = 3 / 1 = 3 kmhr speed ( downstream ) = 1 / ( 10 / 60 ) = 6 kmhr speed in still water = 1 / 2 ( 3 + 6 ) = 4.5 kmhr time taken in stationary = 5 / 4.5 = 1 hrs 7 min answer : b" | a ) 40 minutes , b ) 1 hour 7 min , c ) 1 hour 15 min , d ) 1 hour 30 min , e ) 1 hour 10 min | b | divide(5, divide(add(multiply(divide(1, 10), const_60), divide(3, 3)), const_2)) | divide(n0,n0)|divide(n2,n3)|multiply(#1,const_60)|add(#0,#2)|divide(#3,const_2)|divide(n4,#4)| | physics | B |
a and b start walking towards each other at 7 am at speed of 12 kmph and 13 kmph . they were initially 25 km apart . at what time do they meet ? | "time of meeting = distance / relative speed = 25 / 13 + 12 = 25 / 25 = 1 hr after 7 pm = 8 am answer is e" | a ) 8 pm , b ) 6 am , c ) 7 am , d ) 10 am , e ) 8 am | e | add(7, divide(25, add(12, 13))) | add(n1,n2)|divide(n3,#0)|add(n0,#1)| | physics | E |
how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 390 m in length ? | "d = 110 + 390 = 500 m s = 60 * 5 / 18 = 50 / 3 t = 500 * 3 / 50 = 30 sec answer : d" | a ) 18.9 sec , b ) 88.9 sec , c ) 22.9 sec , d ) 30.00 sec , e ) 72.0 sec | d | divide(add(110, 390), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics | D |
given two sets a = { 11 , 44 , 55 } and b = { 01 } , if one number is chosen from each set at random , what is the probability that the sum of both numbers is an even number | one way to look at it : the number from set a can be anything . the number selected from set b will determine whether the sum is odd or even . for example , if a 4 is selected from set a , we need a 0 from set b to get an even sum . if a 5 is selected from set a , we need a 1 from set b to get an even sum . and so on .... | a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 7 , d ) 1 / 2 , e ) 3 | d | divide(divide(add(11, 44), const_2), 55) | add(n0,n1)|divide(#0,const_2)|divide(#1,n2) | general | D |
in an office , 30 percent of the workers have at least 5 years of service , and a total of 16 workers have at least 10 years of service . if 90 percent of the workers have fewer than 10 years of service , how many of the workers have at least 5 but fewer than 10 years of service ? | "( 10 / 100 ) workers = 16 = > number of workers = 160 ( 30 / 100 ) * workers = x + 16 = > x = 48 answer b" | a ) 480 , b ) 48 , c ) 50 , d ) 144 , e ) 160 | b | divide(subtract(divide(multiply(divide(16, divide(10, const_100)), 90), const_100), multiply(divide(16, divide(10, const_100)), divide(const_1, const_2))), multiply(const_2, const_4)) | divide(n3,const_100)|divide(const_1,const_2)|multiply(const_2,const_4)|divide(n2,#0)|multiply(n4,#3)|multiply(#3,#1)|divide(#4,const_100)|subtract(#6,#5)|divide(#7,#2)| | gain | B |
the area of sector of a circle whose radius is 12 metro and whose angle at the center is 36 â ° is ? | "36 / 360 * 22 / 7 * 12 * 12 = 45.3 m 2 answer : b" | a ) 52.6 , b ) 45.3 , c ) 52.8 , d ) 52.1 , e ) 52.2 | b | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(36, divide(const_3600, const_10))) | add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|power(n0,const_2)|add(#2,const_2)|divide(n1,#1)|divide(#4,#0)|multiply(#6,#3)|multiply(#5,#7)| | geometry | B |
a metallic sheet is of rectangular shape with dimensions 100 m x 50 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 10 m , the volume of the box ( in m cube ) is : | "explanation : l = ( 100 - 20 ) m = 80 m , [ because 10 + 10 = 20 ] b = ( 50 - 20 ) m = 30 m , h = 10 m . volume of the box = ( 80 x 30 x 8 ) m cube = 24000 m cube . option d" | a ) 42500 m cube , b ) 20000 m cube , c ) 44140 m cube , d ) 24000 m cube , e ) none of these | d | volume_rectangular_prism(subtract(100, multiply(10, const_2)), subtract(50, multiply(10, const_2)), 10) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry | D |
a certain industrial loom weaves 0.132 meters of cloth every second . approximately how many seconds will it take for the loom to weave 15 meters of cloth ? | "let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.132 : 15 : : 1 : x = > 0.132 * x = 15 = > x = 15 / 0.132 = > x = 114 answer : a" | a ) 114 , b ) 115 , c ) 116 , d ) 117 , e ) 118 | a | divide(15, 0.132) | divide(n1,n0)| | physics | A |
a factory produces 1 defective bulb out of 10 bulbs a yr . if it produces 870 bulbs a yr , how many defective bulbs are produced ? | 10 out of 1 is defective 20 out of 2 is defective 100 out of 10 is defective and so on 800 out of 80 is defective 70 out of 7 is defective 80 + 7 = 87 answer : d | a ) 84 , b ) 85 , c ) 86 , d ) 87 , e ) 88 | d | divide(870, 10) | divide(n2,n1) | other | D |
what is the smallest number which when diminished by 20 , is divisible 15 , 30 , 45 and 60 ? | "required number = ( lcm of 15 , 30 , 45 and 60 ) + 20 = 180 + 20 = 200 option a" | a ) 200 , b ) 220 , c ) 240 , d ) 322 , e ) 342 | a | add(lcm(lcm(15, 30), lcm(45, 60)), 20) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|add(n0,#2)| | general | A |
reeya obtained 65 , 67 , 76 , 82 and 85 out of 100 in different subjects , what will be the average | "explanation : ( 65 + 67 + 76 + 82 + 855 ) = 75 answer : option b" | a ) 70 , b ) 75 , c ) 80 , d ) 85 , e ) 60 | b | divide(add(add(add(add(65, 67), 76), 82), 85), add(const_4, const_1)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general | B |
if an article is sold at 18 % profit instead of 9 % profit , then the profit would be $ 72 more . what is the cost price ? | "9 % * cost price = $ 72 1 % * cost price = $ 72 / 9 = $ 8 the cost price is $ 800 . the answer is d ." | a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 900 | d | multiply(divide(72, 9), const_100) | divide(n2,n1)|multiply(#0,const_100)| | gain | D |
if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer that must divide n is | "possible values of n = 12 , 24 , 36 and 48 . but it is 12 that can divide all the possible values of n . if we consider 48 , it will not divide 12 , 24 and 36 . hence , 12 is the value that must divide n . answer : b" | a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | b | multiply(sqrt(divide(72, 2)), 2) | divide(n1,n0)|sqrt(#0)|multiply(n0,#1)| | general | B |
of all the homes on gotham street , 1 / 3 are termite - ridden , and 2 / 5 of these are collapsing . what fraction of the homes are termite - ridden , but not collapsing ? | let total homes be 15 termite ridden = 1 / 3 ( 15 ) = 5 termite ridden and collapsing = 2 / 5 ( 5 ) = 2 thus homes that are termite ridden , but not collapsing = 5 - 2 = 3 thus required ratio = 3 / 15 = 1 / 5 answer e | a ) a ) 2 / 15 , b ) b ) 3 / 15 , c ) c ) 4 / 5 , d ) d ) 2 / 5 , e ) e ) 1 / 5 | e | divide(add(1, 2), multiply(3, 5)) | add(n0,n2)|multiply(n1,n3)|divide(#0,#1) | general | E |
what percent of 12.4 kg is 20 gms ? | "explanation : required percentage = ( 20 / 12400 * 100 ) % = 4 / 25 % = 0.16 % answer : b ) . 16 %" | a ) 25 , b ) 16 , c ) 18 , d ) 19 , e ) 17 | b | multiply(divide(12.4, 20), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain | B |
a train running at the speed of 180 km / hr crosses a pole in 18 seconds . what is the length of the train ? | "speed = ( 180 x ( 5 / 18 ) m / sec = ( 50 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( 50 ) x 18 ) m = 900 m b" | a ) 800 , b ) 900 , c ) 950 , d ) 1000 , e ) 1050 | b | multiply(divide(multiply(180, const_1000), const_3600), 18) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | B |
in order to obtain an income of rs . 580 from 15 % stock at rs . 100 , one must make an investment of | "to obtain rs . 10 , investment = rs . 100 . to obtain rs . 580 , investment = = rs . 8700 . answer : c" | a ) 5363 , b ) 6240 , c ) 8700 , d ) 5600 , e ) 2732 | c | multiply(divide(100, 15), 580) | divide(n2,n1)|multiply(n0,#0)| | gain | C |
if 2 x + y = 7 and x + 2 y = 5 , then xy / 3 = | "2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 xy / 3 = 3 * 1 / 3 = 1 a" | a ) 1 , b ) 4 / 3 , c ) 17 / 5 , d ) 18 / 5 , e ) 4 | a | divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 3) | multiply(n0,n1)|multiply(n0,n0)|subtract(#0,n3)|subtract(#1,const_1)|divide(#2,#3)|multiply(n0,#4)|subtract(n1,#5)|add(#4,#6)|divide(#7,n4)| | general | A |
9 . the least number which should be added to 28523 so that the sum is exactly divisible by 3 , 5 , 7 and 8 is | lcm of 3 , 5 , 7 and 8 = 840 28523 ÷ 840 = 33 remainder = 803 hence the least number which should be added = 840 - 803 = 37 answer : option d | a ) 41 , b ) 42 , c ) 32 , d ) 37 , e ) 39 | d | subtract(lcm(lcm(lcm(3, 5), 7), 8), reminder(28523, lcm(lcm(lcm(3, 5), 7), 8))) | lcm(n2,n3)|lcm(n4,#0)|lcm(n5,#1)|reminder(n1,#2)|subtract(#2,#3) | general | D |
two trains each 260 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively . in what time will they cross each other completely ? | "explanation : d = 260 m + 260 m = 520 m rs = 80 + 70 = 150 * 5 / 18 = 125 / 3 t = 520 * 3 / 125 = 12.48 sec answer : option c" | a ) 15 sec , b ) 19 sec , c ) 12.48 sec , d ) 10 sec , e ) 11 sec | c | divide(260, multiply(80, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | C |
machine p can print one lakh books in 8 hours . machine q can print the same number of books in 10 hours while machine r can print the same in 12 hours . all the machines started printing at 9 a . m . machine p is stopped at 11 a . m . and the remaining two machines complete work . approximately at what time will the p... | explanation : work done by p in 1 hour = 1 / 8 work done by q in 1 hour = 1 / 10 work done by r in 1 hour = 1 / 12 work done by p , q and r in 1 hour = 1 / 8 + 1 / 10 + 1 / 12 = 37 / 120 work done by q and r in 1 hour = 1 / 10 + 1 / 12 = 22 / 120 = 11 / 60 from 9 am to 11 am , all the machines were operating . ie , the... | a ) 3 pm , b ) 2 pm , c ) 1 : 00 pm , d ) 11 am , e ) 12 am | c | floor(add(11, divide(subtract(const_1, multiply(subtract(11, 9), add(divide(const_1, 12), add(divide(const_1, 8), divide(const_1, 10))))), add(divide(const_1, 10), divide(const_1, 12))))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|subtract(n4,n3)|add(#0,#1)|add(#1,#2)|add(#4,#2)|multiply(#6,#3)|subtract(const_1,#7)|divide(#8,#5)|add(n4,#9)|floor(#10) | physics | C |
the unit digit in the product ( 635 * 767 * 984 * 489 ) is : | "explanation : unit digit in the given product = unit digit in ( 5 * 7 * 4 * 9 ) = 0 answer : a" | a ) 0 , b ) 8 , c ) 3 , d ) 2 , e ) 1 | a | subtract(multiply(multiply(multiply(635, 767), 984), 489), subtract(multiply(multiply(multiply(635, 767), 984), 489), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general | A |
how many seconds will a 700 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 700 * 3 / 50 = 42 sec . answer : e" | a ) 23 sec , b ) 30 sec , c ) 27 sec , d ) 36 sec , e ) 42 sec | e | divide(700, multiply(subtract(63, 3), const_0_2778)) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics | E |
set s consists of integers { 4 , 7 , 10 , 14 , 15 } . if integer n is included in the set , the average ( arithmetic mean ) of set s will increase by 50 % . what is the value of integer n ? | "the average of the numbers in set s is 10 . if we increase the mean by 50 % , the new mean is 15 . thus , on average , 5 numbers increase by 5 . therefore n = 15 + 25 = 40 the answer is d ." | a ) 28 , b ) 32 , c ) 36 , d ) 40 , e ) 44 | d | add(add(10, 14), 14) | add(n2,n3)|add(n3,#0)| | general | D |
if log 2 = 0.3010 and log 3 = 0.4771 , the value of log 5 ( 512 ) | "log 5 ( 512 ) = log ( 512 ) / log 5 = log 2 ^ 9 / log ( 10 / 2 ) = 9 log 2 / ( log 10 - log 2 ) = ( 9 x 0.3010 ) / ( 1 - 0.3010 ) = 2.709 / 0.699 = 2709 / 699 = 3.876 answer is a ." | a ) 3.876 , b ) 2.967 , c ) 2.87 , d ) 3.912 , e ) 1.9 | a | multiply(0.3010, divide(0.3010, 0.4771)) | divide(n1,n3)|multiply(n1,#0)| | other | A |
on an order of 4 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | "clearly , the retailer gets 1 dozen out of 5 dozens free . equivalent discount = 1 / 5 * 100 = 20 % . answer b ) 20 %" | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | subtract(const_100, multiply(divide(const_3.0, const_4), const_100)) | divide(const_3.0,const_4)|multiply(#0,const_100)|subtract(const_100,#1)| | general | B |
how many pieces of 0.85 meteres can be cut from a rod 42.5 meteres long | explanation : we need so simple divide 42.5 / 0.85 , = ( 4250 / 85 ) = 50 option c | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | divide(42.5, 0.85) | divide(n1,n0) | physics | C |
the value for d = ( 0.889 × 55 ) / 9.97 to the nearest tenth is | if we read the q , we can easily home on to the answer , , the numerator is clearly between 40 and 50 . . denomiator is close to 10 . . so d = ( 0.889 × 55 ) / 9.97 is between 4 and 5 . . it may tempt us to solve it since ther are two values between 4 and 5 . . but the catch is innearest tenth 4.63 can be called neares... | a ) 0.5 , b ) 4.63 , c ) 4.9 , d ) 7.7 , e ) 49.1 | c | divide(floor(multiply(divide(multiply(0.889, 55), 9.97), const_10)), const_10) | multiply(n0,n1)|divide(#0,n2)|multiply(#1,const_10)|floor(#2)|divide(#3,const_10) | general | C |
a and b started a business investing rs . 21,000 and rs . 28,000 respectively . out of a total profit of rs . 14,000 , b ’ s share is : | "ratio of their shares = 21000 : 28000 = 3 : 4 . b ’ s share = rs . 14000 * 4 / 7 = rs . 8000 answer : e" | a ) 7600 , b ) 7700 , c ) 7800 , d ) 7900 , e ) 8000 | e | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 14,000), multiply(14,000, 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 | E |
a shopkeeper has 280 kg of apples . he sells 50 % of these at 20 % profit and remaining 60 % at 30 % profit . find his % profit on total . | "if the total quantity was 100 then 50 x 20 % + 60 x 30 % = 28 this profit will remain same for any total quantity unless the % of products remains the same . hence ' d ' is the answer" | a ) 24 % , b ) 25 % , c ) 26 % , d ) 28 % , e ) 35 % | d | divide(multiply(subtract(add(multiply(divide(multiply(280, 50), const_100), divide(add(const_100, 20), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280) | add(n2,const_100)|add(n4,const_100)|multiply(n0,n1)|multiply(n0,n3)|divide(#2,const_100)|divide(#0,const_100)|divide(#3,const_100)|divide(#1,const_100)|multiply(#4,#5)|multiply(#6,#7)|add(#8,#9)|subtract(#10,n0)|multiply(#11,const_100)|divide(#12,n0)| | gain | D |
the true discount on a bill of rs . 2660 is rs . 360 . what is the banker ' s discount ? | "explanation : f = rs . 2660 td = rs . 360 pw = f - td = 2660 - 360 = rs . 2300 true discount is the simple interest on the present value for unexpired time = > simple interest on rs . 2300 for unexpired time = rs . 360 banker ' s discount is the simple interest on the face value of the bill for unexpired time = simple... | a ) rs . 432 , b ) rs . 422 , c ) rs . 416 , d ) rs . 442 , e ) none of these | c | multiply(divide(360, subtract(2660, 360)), 2660) | subtract(n0,n1)|divide(n1,#0)|multiply(n0,#1)| | gain | C |
if a ( a - 4 ) = 21 and b ( b - 4 ) = 21 , where a ≠ b , then a + b = | "a ( a - 4 ) = 21 and b ( b - 4 ) = 21 = > a , b must be integers and if a is - 3 or 7 , b will be 7 and - 3 respectively = > a + b = 4 ans : c" | a ) − 48 , b ) − 2 , c ) 4 , d ) 46 , e ) 48 | c | subtract(subtract(subtract(subtract(add(add(4, 21), subtract(4, 21)), const_1), const_1), const_1), const_1) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1)| | general | C |
working together , wayne and his son can shovel the entire driveway in three hours . if wayne can shovel two times as fast as his son can , how many hours would it take for his son to shovel the entire driveway on his own ? | "w : the time for wyane to do the job s : the time for his son to do the job we have 1 / w + 1 / s = 1 / 2 and w = 2 s then we have 1 / ( 2 * s ) + 1 / s = 1 / 2 < = > 3 / ( 2 * s ) = 1 / 2 < = > s = 3 ans : b" | a ) 4 , b ) 3 , c ) 8 , d ) 9 , e ) 12 | b | multiply(multiply(add(inverse(multiply(const_2, const_4)), const_1), const_3), multiply(const_2, const_4)) | multiply(const_2,const_4)|inverse(#0)|add(#1,const_1)|multiply(#2,const_3)|multiply(#3,#0)| | physics | B |
a man drives at a speed of 40 miles / hr . his wife left 30 mins . late with 50 miles / hr speed . when will they meet ? | in 30 minutes the man would have droven 20 miles , when his wife starts . now both are driving in the same direction . so speed of wife related to man is 10 ( 50 - 40 ) miles / hr . so wife will cover 20 miles in 2 hours to meet the man . answer : b | a ) 1 hours , b ) 2 hours , c ) 3 hours , d ) 4 hours , e ) 5 hours | b | divide(multiply(40, divide(30, const_60)), subtract(50, 40)) | divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1) | physics | B |
what annual installment will discharge a debt of rs . 1092 due in 3 years at 12 % simple interest ? | "let the installment be rs . x if he keeps the total amount for 3 years , he have to pay the interest of rs . ( 1092 * 12 * 3 ) / 100 . but he paid rs . x after 1 year . so he need not to pay interest for rs . x for remaining 2 years . i . e he need not to pay rs . ( x * 12 * 2 ) / 100 . and he paid another rs . x afte... | a ) 442 , b ) 441 , c ) 440 , d ) 320 , e ) 532 | a | multiply(multiply(const_100.0, divide(12, 1092)), 3) | divide(n2,const_100)|multiply(n0,#0)|multiply(n1,#1)| | gain | A |
jill has 42 gallons of water stored in quart , half - gallon , and one gallon jars . she has equal numbers of each size jar holding the liquid . what is the total number of water filled jars ? | let the number of each size of jar = wthen 1 / 4 w + 1 / 2 w + w = 42 1 3 / 4 w = 42 w = 24 the total number of jars = 3 w = 72 answer : d | a ) 3 , b ) 6 , c ) 9 , d ) 72 , e ) 14 | d | multiply(divide(42, add(const_1, add(const_0_25, divide(const_1, const_2)))), const_3) | divide(const_1,const_2)|add(#0,const_0_25)|add(#1,const_1)|divide(n0,#2)|multiply(#3,const_3)| | general | D |
the value of a machine depreciates at 20 % per annum . if its present value is rs . 1 , 50,000 , at what price should it be sold after two years such that a profit of rs . 26,000 is made ? | "the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = rs . 96,000 sp such that a profit of rs . 24,000 is made = 96,000 + 26,000 = rs . 1 , 22,000 answer : c" | a ) rs . 1 , 10,000 , b ) rs . 1 , 20,000 , c ) rs . 1 , 22,000 , d ) rs . 1 , 21,000 , e ) none of these | c | add(multiply(multiply(subtract(1, divide(20, const_100)), subtract(1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100))) | add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(n1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10)| | gain | C |
a car traveled 480 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 6 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "let the speed in highway be h mpg and in city be c mpg . h = c + 6 h miles are covered in 1 gallon 462 miles will be covered in 462 / h . similarly c miles are covered in 1 gallon 336 miles will be covered in 336 / c . both should be same ( as car ' s fuel capacity does not change with speed ) = > 336 / c = 480 / h = ... | a ) 14 , b ) 16 , c ) 21 , d ) 22 , e ) 27 | a | divide(336, divide(subtract(480, 336), 6)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics | A |
tom read a book containing 100 pages by reading the same number of pages each day . if he would have finished the book 10 days earlier by reading 30 pages a day more , how many days did tom spend reading the book ? | actually u can set up 2 equation p - - stands for the pages d - - stands for the days 1 ) p * d = 100 ( we want to find the days , sop = 100 / d ) 2 ) ( p + 30 ) ( d - 10 ) = 100 = > pd - 10 p + 30 d - 300 = 100 as the 1 ) stated u can put 1 ) into 2 ) = > 100 - 10 p + 30 d - 300 = 100 = > 30 d - 10 p = 100 put the bol... | a ) 7 , b ) 7.68 , c ) 8 , d ) 9 , e ) 10 | b | divide(subtract(sqrt(add(multiply(multiply(30, 10), const_4), power(30, const_2))), 30), const_2) | multiply(n1,n2)|power(n2,const_2)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n2)|divide(#5,const_2) | general | B |
what is the cp of rs 100 stock at 5 discount , with 1 / 5 % brokerage ? | "explanation : use the formula , cp = 100 â € “ discount + brokerage % cp = 100 - 5 + 1 / 5 95.2 thus the cp is rs 95.2 . answer : c" | a ) 96.9 , b ) 96.3 , c ) 95.2 , d ) 96.7 , e ) 96.21 | c | add(subtract(100, 5), divide(1, 5)) | divide(n2,n3)|subtract(n0,n1)|add(#0,#1)| | gain | C |
two numbers n and 14 have lcm = 56 and gcf = 10 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 14 × n = 56 × 10 n = 56 × 10 / 14 = 40 correct answer b" | a ) 24 , b ) 40 , c ) 44 , d ) 54 , e ) 64 | b | divide(multiply(56, 10), 14) | multiply(n1,n2)|divide(#0,n0)| | physics | B |
the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 2.0 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 – 6 = 666 * 2.0 = 1332 answer : b" | a ) s . 1014 , b ) s . 1332 , c ) s . 999 , d ) s . 1085 , e ) s . 1020 | b | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 2.0), 3) | multiply(n3,const_2)|sqrt(n0)|multiply(#1,const_4)|subtract(#2,#0)|multiply(n2,#3)|multiply(#4,n1)| | physics | B |
the greatest number that divides 2928 and 3240 leaving remainders 5 and 1 respectively is : | "explanation : 2928 - 5 = 2923 , 3240 - 1 = 3239 highest number that can divide 2923 and 3239 is hcf of numbers . hcf of 2923 and 3239 = 79 answer : c" | a ) 74 , b ) 78 , c ) 79 , d ) 81 , e ) 85 | c | divide(subtract(3240, 1), gcd(2928, 3240)) | gcd(n0,n1)|subtract(n1,n3)|divide(#1,#0)| | general | C |
a student scored an average of 65 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | "given m + p + c = 65 * 3 = 195 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 195 = 125 answer : d" | a ) 86 , b ) 165 , c ) 76 , d ) 125 , e ) 26 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(65, 3)) | multiply(n2,const_2)|multiply(n3,const_2)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general | D |
the price of a car and ac are in the ratio 3 : 2 . if the scooter costs $ 500 more than the ac . then the price of the ac ? | let the price of the car and ac is 3 x and 2 x 3 x - 2 x = 500 x = 500 price of ac = 3 * 500 = $ 1500 answer is a | a ) 1500 , b ) 200 , c ) 1700 , d ) 2000 , e ) 2500 | a | multiply(500, const_3) | multiply(n2,const_3) | other | A |
what is the measure of the angle e made by the diagonals of the any adjacent sides of a cube . | c . . 60 degrees all the diagonals are equal . if we take 3 touching sides and connect their diagonals , we form an equilateral triangle . therefore , each angle would be 60 . . c | ['a ) 30', 'b ) 45', 'c ) 60', 'd ) 75', 'e ) 90'] | c | divide(const_180, const_3) | divide(const_180,const_3) | geometry | C |
how many hours are there in 1200 minutes ? | we know that there are 60 minutes in 1 hour . divide the number of minutes by the number of minutes in 1 hour . we get , divide 1200 by 601200 ÷ 60 = 20 so there are 20 hours in 1200 minutes . answer is b | a ) 10 , b ) 20 , c ) 85 , d ) 86 , e ) 88 | b | divide(1200, const_60) | divide(n0,const_60) | physics | B |
if albert ’ s monthly earnings rise by 14 % , he would earn $ 678 . if , instead , his earnings rise by only 15 % , how much ( in $ ) would he earn this month ? | "= 678 / 1.14 ∗ 1.15 = 683 = 683 answer is c" | a ) 643 , b ) 689 , c ) 683 , d ) 690 , e ) 693 | c | multiply(divide(678, add(const_1, divide(14, const_100))), add(const_1, divide(15, const_100))) | divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n1,#3)|multiply(#2,#4)| | gain | C |
in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio o... | "straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if th... | a ) 100 , b ) 300 , c ) 400 , d ) 3,000 , e ) 4,000 | c | multiply(divide(12, subtract(99, 96)), const_100) | subtract(n3,n0)|divide(n2,#0)|multiply(#1,const_100)| | general | C |
p and q started a business investing rs . 96,000 and rs . 28,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 96000 : 28000 = 24 : 7 answer : b" | a ) 17 : 6 , b ) 24 : 7 , c ) 25 : 6 , d ) 21 : 3 , e ) 17 : 4 | b | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | add(n2,const_3)|add(#0,const_3)|multiply(n2,#0)|multiply(#1,#2)|multiply(#2,const_3)|add(#0,#3)|add(#0,#4)|divide(#5,#6)| | gain | B |
a wildlife preserve is being planned for 8000 rhinoceroses . the preserve is to contain a total of 10000 acres of watering area , plus 100 acres of grazing area for each rhinoceros . if the number of rhinoceroses is expected to increase by 10 percent , how many 1000 acres should the preserve have in order to provide fo... | number of rhinos = 8000 watering area = 10,000 acres number of rhino to increase by 10 percent , then number of rhino = 8800 grazing area for a rhino = 100 total grazing area for 8800 rhinos = 8800 * 100 = 8 , 80,000 total area required for the wildlife preserve = 8 , 80,000 + 10,000 = 8 , 90,000 = 890 * 1000 answer e | a ) 340 , b ) 330 , c ) 320 , d ) 310 , e ) 890 | e | divide(add(add(multiply(100, 8000), divide(multiply(100, 8000), 10)), 10000), 1000) | multiply(n0,n2)|divide(#0,n3)|add(#1,#0)|add(n1,#2)|divide(#3,n4) | general | E |
how many positive integers , from 2 to 100 , inclusive , are not divisible by even integers greater than 1 ? | "the no will be of the form 2 ^ n to achieve this . since any other form will have odd no in the prime factorization . hence we need to find solution of n for equation - 1 < 2 ^ n < 101 . 2 ^ 6 = 64 , 2 ^ 7 - 128 hence n can take values from 1 to 6 . hence ans - ( b ) 6" | a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 50 | b | divide(subtract(multiply(multiply(2, multiply(2, multiply(2, multiply(2, 2)))), 2), const_4), const_10) | multiply(n0,n0)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|multiply(n0,#3)|subtract(#4,const_4)|divide(#5,const_10)| | general | B |
a lady builds 9 cm length , 12 cm width , and 3 cm height box using 3 cubic cm cubes . what is the minimum number of cubes required to build the box ? | "number of cubes required = volume of box / volume of cube = 9 * 12 * 3 / 3 = 108 cubes answer : b" | a ) 107 , b ) 108 , c ) 109 , d ) 110 , e ) 111 | b | divide(multiply(multiply(9, 12), 3), 3) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)| | geometry | B |
the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 45 , find the other ? | "other number = 12 * 600 / 45 = 160 answer is b" | a ) 100 , b ) 160 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 45) | multiply(n0,n2)| | physics | B |
a number when divided by 20 leaves 6 as a remainder . what will be the remainder if the number is divided by 19 ? | let the minimum whole number be 20 + 6 = 26 where 6 is the remainder . 26 when divided by 19 then leaves 7 . correct answer a . | a ) 7 , b ) 6 , c ) 5 , d ) 11 , e ) 13 | a | reminder(add(20, 6), 19) | add(n0,n1)|reminder(#0,n2) | general | A |
4 dice are thrown simultaneously on the board . find the probability show the same face . | "the total number of elementary events associated to the random experiments of throwing four dice simultaneously is : = 6 × 6 × 6 × 6 = 64 = 6 × 6 × 6 × 6 = 64 n ( s ) = 64 n ( s ) = 64 let xx be the event that all dice show the same face . x = { ( 1,1 , 1,1 , ) , ( 2,2 , 2,2 ) , ( 3,3 , 3,3 ) , ( 4,4 , 4,4 ) , ( 5,5 ,... | a ) 1 / 213 , b ) 1 / 215 , c ) 1 / 216 , d ) 2 / 113 , e ) 3 / 114 | c | multiply(divide(const_3, add(const_3, const_3)), divide(const_3, add(const_3, const_3))) | add(const_3,const_3)|divide(const_3,#0)|multiply(#1,#1)| | probability | C |
a boat goes 100 km downstream in 10 hours , and 75 m upstream in 15 hours . the speed of the stream is ? | 100 - - - 10 ds = 10 ? - - - - 1 75 - - - - 15 us = 5 ? - - - - - 1 s = ( 10 - 5 ) / 2 = 2 21 / 2 kmph answer : d | a ) 22 1 / 9 kmph , b ) 22 8 / 2 kmph , c ) 22 1 / 8 kmph , d ) 22 1 / 2 kmph , e ) 22 1 / 4 kmph | d | divide(subtract(divide(100, 10), divide(75, 15)), const_2) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,const_2) | physics | D |
( 3 * 10 ^ 2 ) * ( 4 * 10 ^ - 2 ) = ? | 3 * 10 ^ 2 = 300 4 * 10 ^ - 2 = 0.04 ( 3 * 10 ^ 2 ) * ( 4 * 10 ^ - 2 ) = 300 * 0.04 = 12.00 the answer is option b | a ) 14 , b ) 12 , c ) 1200 , d ) 1.2 , e ) 14.11 | b | multiply(multiply(3, power(10, 2)), multiply(4, power(10, negate(2)))) | negate(n2)|power(n1,n2)|multiply(n0,#1)|power(n1,#0)|multiply(n3,#3)|multiply(#2,#4) | general | B |
a gambler bought $ 4,000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 16 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could ha... | "in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that... | a ) $ 2,040 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 3,120 , e ) $ 1,400 | d | subtract(multiply(const_3, const_1000), add(multiply(divide(add(16, 2), 2), 20), multiply(subtract(divide(add(16, 2), 2), 2), 100))) | add(n3,n5)|multiply(const_1000,const_3)|divide(#0,n5)|multiply(n1,#2)|subtract(#2,n5)|multiply(n2,#4)|add(#3,#5)|subtract(#1,#6)| | general | D |
16 balls are numbered 1 to 16 . a ball is drawn and then another ball is drawn without replacement . what is the probability that both balls have even numbers ? | "p ( 1 st ball is even ) = 8 / 16 p ( 2 nd ball is also even ) = 7 / 15 p ( both balls are even ) = 8 / 16 * 7 / 15 = 7 / 30 the answer is d ." | a ) 4 / 11 , b ) 5 / 17 , c ) 6 / 25 , d ) 7 / 30 , e ) 8 / 37 | d | multiply(divide(add(const_4, const_4), 16), divide(subtract(add(const_4, const_4), 1), subtract(16, 1))) | add(const_4,const_4)|subtract(n0,n1)|divide(#0,n0)|subtract(#0,n1)|divide(#3,#1)|multiply(#2,#4)| | other | D |
company kw is being sold , and both company a and company b were considering the purchase . the price of company kw is 30 % more than company a has in assets , and this same price is also 100 % more than company b has in assets . if companies a and b were to merge and combine their assets , the price of company kw woul... | let the price of company a ' s assets be 100 price of assets of kw is 30 % more than company a ' s assets which is 130 price of assets of kw is 100 % more than company b ' s assets which means price of company b ' s assets is half the price of kw = 65 a + b = 165 kw = 130 kw / ( a + b ) * 100 = 130 / 165 * 100 = 78.78 ... | a ) 66 % , b ) 79 % , c ) 86 % , d ) 116 % , e ) 150 % | b | multiply(divide(add(100, 30), add(100, divide(add(100, 30), const_2))), const_100) | add(n0,n1)|divide(#0,const_2)|add(n1,#1)|divide(#0,#2)|multiply(#3,const_100) | gain | B |
it takes joey the postman 1 hours to run a 2 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 6 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 3 miles an hour let the other half be x miles an hour now , avg speed = 5 mile an hour 2 * 2 * x / 3 + x = 6 4 x = 6 x + 18 = > 2 x = 18 x = 9 a" | a ) 9 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | divide(2, subtract(divide(multiply(const_2, 2), 6), 1)) | multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2)| | physics | A |
a set consists of 20 numbers , all are even or multiple of 5 . if 9 numbers are even and 12 numbers are multiple of 5 , how many numbers is multiple of 10 ? | "{ total } = { even } + { multiple of 5 } - { both } + { nether } . since { neither } = 0 ( allare even or multiple of 5 ) then : 20 = 9 + 12 - { both } + 0 ; { both } = 1 ( so 1 number is both even and multiple of 5 , so it must be a multiple of 10 ) . answer : b ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | b | subtract(20, 12) | subtract(n0,n3)| | general | B |
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 cloudy , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction r of the total distance did... | 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 / 4 , b ) 4 / 5 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | e | subtract(divide(lcm(const_2, const_3), 2.8), const_2) | lcm(const_2,const_3)|divide(#0,n1)|subtract(#1,const_2)| | general | E |
linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 210 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 210 . so 1 / 4 of her savings is $ 210 . so her original savings are 4 times $ 210 = $ 840 correct answer a" | a ) $ 840 , b ) $ 800 , c ) $ 1000 , d ) $ 700 , e ) $ 1500 | a | divide(210, subtract(const_1, divide(3, 4))) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)| | general | A |
for what value of â € œ k â € will the equation ( 2 kx 2 + 4 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 ( 4 k ) ^ 2 - 4 * 2 k * 2 = 0 by solving this equation we get k = 0 , k = 1 answer : a" | a ) 1 , b ) 9 / 4 , c ) 16 / 25 , d ) 7 / 1 , e ) 7 / 2 | a | divide(power(2, add(2, 2)), power(4, 2)) | add(n0,n0)|power(n2,n0)|power(n0,#0)|divide(#2,#1)| | general | A |
a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 70 percent of books that were loaned out are returned and there are 60 books in the spec... | "i did n ' t understand how did we get 100 ? total = 75 books . 65 % of books that were loaned out are returned - - > 100 % - 70 % = 30 % of books that were loaned out are not returned . now , there are 60 books , thus 76 - 60 = 16 books are not returned . { loaned out } * 0.30 = 7 - - > { loaned out } = 50 . answer : ... | a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 50 | e | divide(subtract(75, 60), subtract(const_1, divide(70, const_100))) | divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)| | gain | E |
a 12 % stock yields 8 % . the market value of the stock is : | "solution to obtain rs . 8 , investment = rs . 100 . to obtain rs . 12 , investment = rs . ( 100 / 8 x 12 ) = rs . 150 ∴ market value of rs . 100 stock = rs . 150 answer b" | a ) rs . 72 , b ) rs . 150 , c ) rs . 112.50 , d ) rs . 116.50 , e ) none of these | b | multiply(divide(const_100, 8), 12) | divide(const_100,n1)|multiply(n0,#0)| | gain | B |
if the price of petrol increases by 44 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ? | "explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 44 ) i . e 144 . now , his consumption should be reduced to : - = ( 144 − 100 ) / 144 ∗ 100 . hence , the consumption should be reduced to 30.6 % . answer : e" | a ) 25 % , b ) 20 % , c ) 16.67 % , d ) 33.33 % , e ) none of these | e | multiply(subtract(const_1, divide(const_100, add(const_100, 44))), const_100) | add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | general | E |
right triangle abc is to be drawn in the xy - plane so that the right angle is at a and ab is parallel to the y - axis . if the x - and y - coordinates of a , b , and c are to be integers that are consistent with the inequalities - 4 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet ... | "we have the rectangle with dimensions 9 * 4 ( 9 horizontal dots and 4 vertical ) . ab is parallel to y - axis and ac is parallel to x - axis . choose the ( x , y ) coordinates for vertex a : 9 c 1 * 4 c 1 ; choose the x coordinate for vertex c ( as y coordinate is fixed by a ) : 8 c 1 , ( 9 - 1 = 8 as 1 horizontal dot... | a ) 54 , b ) 432 , c ) 864 , d ) 2,916 , e ) 148,824 | c | multiply(multiply(4, subtract(4, const_1)), multiply(9, 4)) | multiply(n0,n3)|subtract(n0,const_1)|multiply(n0,#1)|multiply(#2,#0)| | geometry | C |
what is the probability of getting only 1 head in a single throw of four fair coins ? | "one possible case is httt . p ( httt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 there are 4 c 1 = 4 possible cases . p ( 1 head ) = 4 * 1 / 16 = 1 / 4 the answer is c ." | a ) 1 / 2 , b ) 3 / 8 , c ) 1 / 4 , d ) 1 / 5 , e ) 5 / 8 | c | divide(const_2, choose(add(const_3, const_3), const_3)) | add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)| | probability | C |
a started a business with an investment of rs . 10000 and after 7 months b joined him investing rs . 12000 . if the profit at the end of a year is rs . 24000 , then the share of b is ? | "ratio of investments of a and b is ( 10000 * 12 ) : ( 12000 * 5 ) = 2 : 1 total profit = rs . 24000 share of b = 1 / 3 ( 24000 ) = rs . 8000 answer : b" | a ) 10000 , b ) 8000 , c ) 12000 , d ) 6000 , e ) 14000 | b | subtract(24000, multiply(const_60, const_100)) | multiply(const_100,const_60)|subtract(n3,#0)| | gain | B |
1907 x 1907 = ? | "1907 x 1907 = ( 1907 ) 2 = ( 1900 + 7 ) 2 = ( 1900 ) 2 + ( 7 ) 2 + ( 2 x 1900 x 7 ) = 3610000 + 49 + 26600 . = 3636649 . d )" | a ) a ) 3623216 , b ) b ) 3624216 , c ) c ) 3624316 , d ) d ) 3636649 , e ) e ) 3625216 | d | multiply(divide(1907, 1907), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | D |
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