Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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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 at k = 1 thus smallest number will be 7 ( 1 ) + 1 = 8 now , next number will be = 8 + lcm of 37 i . e 29 now we will find number of all such values less than 500 by using the formula for last term of an a . p 8 + ( n - 1 ) 21 = 600 n = 25.42 or n = 25 answer : - e" | 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 since the numbers start from 1 . ) in 1000 consecutive numbers , number of multiples of 11 = 1000 / 11 = 90 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 35 = 1000 / 35 = 28 number of multiples of 11 * 35 i . e . 385 = 1000 / 385 = 2 number of integers from 1 to 1000 that are divisible by neither 11 nor by 35 = 1000 - ( 90 + 28 - 2 ) { using the concept of sets here ) = 890 think : why did i say the method is approximate in some cases ? think what happens if the given range is 11 to 1010 both inclusive ( again 1000 numbers ) what is the number of multiples in this case ? b" | 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 . = > all the three letters can be chosen in 676 ã — 26 = 17576676 ã — 26 = 17576 ways . it implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter . answer d | 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 one of a and b ) = p ( a ) + p ( b ) - p ( ab ) . = > 4 / 5 = ( 2 / 3 ) + ( 4 / 9 ) - p ( ab ) . = > p ( ab ) = 14 / 45 . hence , the required probability is 14 / 45 . answer : a" | 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 both sets . solution : seven answer : e" | 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 $ amount remaining = 2.2 - 1.75 = 0.45 $ this amount is too less to buy any 8 - pack . greatest number of hot dogs one can buy with 250 $ = 250 * 10 + 20 * 6 + 8 * 1 = 2628 answer d" | 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 . so , p ( sum is even ) = p ( select any number from set aandselect the number from set b that makes the sum even ) = p ( select any number from set a ) xp ( select the number from set b that makes the sum even ) = 1 x 1 / 2 = 1 / 2 = d | 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 printing of one lakh books be completed ? | 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 , they all operated for 2 hours and work completed = 2 × ( 37 / 120 ) = 37 / 60 pending work = 1 - 37 / 60 = 23 / 60 hours taken by q an r to complete the pending work = ( 23 / 60 ) / ( 11 / 60 ) = 23 / 11 which is approximately equal to 2 hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm answer : option c | 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 nearest to hundreth and no tenth , so can be eliminated . . 4.9 is our answer . . . c | 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 interest on rs . 2160 for unexpired time = 360 / 2300 × 2660 = 0.16 × 2660 = rs . 416 answer : option c" | 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 after 2 years . so he need not to pay interest for for remaining 1 years . i . e . rs . ( x * 12 * 1 ) / 100 . so the total interest he has to pay is rs . ( ( 1092 * 12 * 3 ) / 100 ) - [ ( ( x * 12 * 2 ) / 100 ) + ( ( x * 12 * 1 ) / 100 ) ] . so , the total amount he has to pay is rs . 1092 + rs . ( ( 1092 * 12 * 3 ) / 100 ) - [ ( ( x * 12 * 2 ) / 100 ) + ( ( x * 12 * 1 ) / 100 ) ] . the total amount he paid = 3 x . ( because he paid rs . x for three times ) 1092 + ( ( 1092 * 12 * 3 ) / 100 ) - [ ( ( x * 12 * 2 ) / 100 ) + ( ( x * 12 * 1 ) / 100 ) ] = 3 x by solving , we will get x = 442 . answer : a" | 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 = > 336 / c = 480 / ( c + 6 ) = > 336 c + 336 * 6 = 480 c = > c = 336 * 6 / 144 = 14 answer a ." | 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 bold one into it = > 30 d - 10 ( 100 / d ) = 100 the we get the final equation 30 d ^ 2 - 1000 = 100 d ( divide 16 ) = > d ^ 2 - 5 d - 150 = 0 ( d - 15 ) ( d + 10 ) = 0 so d = 7.68 days . ans : ( b ) | 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 of those accepted to those rejected would be 99 to 1 . how many t eggs does the plant process per day ? | "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 the no . of eggs inspected remain and 12 more eggs get accepted that means there t = 384 + 12 = 396 eggs accepted and 4 rejected . . . and the ratio will be 99 : 1 bingo . . . this is what the questions says . . . . its always a good idea to start with c ." | 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 for the increased population ? | 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 , 5,5 ) , ( 6,6 , 6,6 ) } x = { ( 1,1 , 1,1 , ) , ( 2,2 , 2,2 ) , ( 3,3 , 3,3 ) , ( 4,4 , 4,4 ) , ( 5,5 , 5,5 ) , ( 6,6 , 6,6 ) } n ( x ) = 6 n ( x ) = 6 hence required probability , = n ( x ) n ( s ) = 664 = n ( x ) n ( s ) = 664 = 1 / 216 c" | 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 have received back ? | "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 total # of chips lost is 16 : x + x - 2 = 16 - - > x = 9 : 9 $ 20 chips were lost and 9 - 2 = 7 $ 100 chips were lost . total worth of chips lost is 9 * 20 + 7 * 100 = $ 880 , so the gambler kept $ 4,000 - $ 880 = $ 3,120 . answer : d" | 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 would be approximately what percent of these combined assets ? | 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 % or 79 % b | 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 he cover while the sun was shining on him ? | 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 12 t . therefore , total distance covered while the sun was shining over him is r = 2 / 14 = 1 / 7 . answer : e | 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 special collection at that time , how many books of the special collection were loaned out during that month ? | "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 : e ." | 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 these conditions ? | "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 is already occupied by a ) ; choose the y coordinate for vertex b ( as x coordinate is fixed by a ) : 3 c 1 , ( 4 - 1 = 3 as 1 vertical dot is already occupied by a ) . 9 c 1 * 4 c 1 * 8 c 1 * 3 c 1 = 864 . answer : c ." | 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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