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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product of two co - prime numbers is 117 . their l . c . m should be : | "h . c . f of co - prime numbers is 1 . so , l . c . m = 117 / 1 = 117 . answer : b" | a ) 1 , b ) 117 , c ) 116 , d ) equal to their h . c . f , e ) can not be calculated | b | multiply(117, const_1) | multiply(n0,const_1)| | general | B |
a bowl contains equal numbers of red , orange , green , blue , and yellow candies . kaz eats all of the green candies and half of the orange ones . next , he eats half of the remaining pieces of each color . finally , he eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 32 % of the original number . what percent of the red candies remain ? | "let x be the original number of each color so there are a total of 5 x candies . kaz eats all of the green candies and half of the orange ones . there are 0 green candies and 0.5 x orange candies remaining . he eats half of the remaining pieces of each color . there are 0.25 x orange candies , and 0.5 x each of red , yellow , and blue candies . he eats red and yellow candies in equal proportions . orange + blue + red + yellow = 0.75 x + red + yellow = 1.6 x red + yellow = 0.85 x red = 0.425 x , since red = yellow . the answer is d ." | a ) 27.5 % , b ) 33.3 % , c ) 35.5 % , d ) 42.5 % , e ) 47.5 % | d | multiply(divide(divide(subtract(32, add(divide(divide(const_100, add(const_2, const_3)), const_2), divide(divide(divide(const_100, add(const_2, const_3)), const_2), const_2))), const_2), divide(const_100, add(const_2, const_3))), const_100) | add(const_2,const_3)|divide(const_100,#0)|divide(#1,const_2)|divide(#2,const_2)|add(#2,#3)|subtract(n0,#4)|divide(#5,const_2)|divide(#6,#1)|multiply(#7,const_100)| | general | D |
village x has a population of 68000 , which is decreasing at the rate of 1200 per year . village y has a population of 42000 , which is increasing at the rate of 800 per year . in how many years will the population of the two villages be equal ? | let the population of two villages be equal after p years then , 68000 - 1200 p = 42000 + 800 p 2000 p = 26000 p = 13 answer is e . | a ) 15 , b ) 19 , c ) 11 , d ) 18 , e ) 13 | e | divide(subtract(68000, 42000), add(800, 1200)) | add(n1,n3)|subtract(n0,n2)|divide(#1,#0) | general | E |
running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 5 w widgets . | "i am getting 12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + 2 days . 1 / ( y + 2 ) + 1 / y = 5 / 12 ( 5 / 12 is because ( 5 / 4 ) w widgets are done in 3 days . so , x widgets will be done in 12 / 5 days or 5 / 12 th of a widget in a day ) solving , we have y = 4 = > x takes 6 days to doing x widgets . so , he will take 30 days to doing 5 w widgets . answer : d" | a ) 4 , b ) 6 , c ) 8 , d ) 30 , e ) 12 | d | multiply(multiply(5, 2), 3) | multiply(n0,n4)|multiply(n3,#0)| | general | D |
9.009 / 5.005 = | "9.009 / 5.005 = 9009 / 5005 = 9 ( 1001 ) / 5 ( 1001 ) = 9 / 5 = 1.8 the answer is e ." | a ) 1.08 , b ) 1.4 , c ) 1.8018 , d ) 1.4014 , e ) 1.8 | e | multiply(divide(9.009, 5.005), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | E |
the average monthly income of p and q is rs . 2050 . the average monthly income of q and r is rs . 5250 and the average monthly income of p and r is rs . 6200 . the monthly income ofq is : | "explanation : let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 2050 x 2 ) = 4100 . . . . ( i ) q + r = ( 5250 x 2 ) = 10500 . . . . ( ii ) p + r = ( 6200 x 2 ) = 12400 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 27000 or p + q + r = 13500 . . . . ( iv ) subtracting ( iii ) from ( iv ) , we get p = 1100 . p ' s monthly income = rs . 1100 . answer : a" | a ) 1100 , b ) 2100 , c ) 1200 , d ) 3100 , e ) 1300 | a | subtract(add(2050, 6200), 5250) | add(n0,n2)|subtract(#0,n1)| | general | A |
a store owner estimates that the average price of type a products will increase by 20 % next year and that the price of type b products will increase by 11 % next year . this year , the total amount paid for type a products was $ 3500 and the total price paid for type b products was $ 8600 . according to the store owner ' s estimate , and assuming the number of products purchased next year remains the same as that of this year , how much will be spent for both products next year ? | "cost of type a products next year = 1.20 * 3500 = 4200 cost of type b products next year = 1.11 * 8300 = 9546 total 4200 + 9546 = 13746 answer : a" | a ) $ 13,746 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | a | multiply(divide(const_3, const_4), const_1000) | divide(const_3,const_4)|multiply(#0,const_1000)| | general | A |
what is difference between biggest and smallest fraction among 2 / 3 , 3 / 4 , 4 / 5 and 5 / 4 | "explanation : 2 / 3 = . 66 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 4 = 1.25 so biggest is 5 / 4 and smallest is 2 / 3 their difference is 5 / 4 - 2 / 3 = 7 / 12 option c" | a ) 2 / 5 , b ) 3 / 5 , c ) 7 / 12 , d ) 1 / 7 , e ) none of these | c | subtract(divide(4, 5), divide(2, 3)) | divide(n3,n5)|divide(n0,n1)|subtract(#0,#1)| | general | C |
a man complete a journey in 40 hours . he travels first half of the journey at the rate of 20 km / hr and second half at the rate of 30 km / hr . find the total journey in km . | "0.5 x / 20 + 0.5 x / 30 = 40 - - > x / 20 + x / 30 = 80 - - > 5 x = 60 x 80 - - > x = ( 60 x 80 ) / 5 = 960 km . answer : a" | a ) 960 km , b ) 224 km , c ) 230 km , d ) 232 km , e ) 234 km | a | multiply(const_2, divide(multiply(multiply(20, 30), 40), add(20, 30))) | add(n1,n2)|multiply(n1,n2)|multiply(n0,#1)|divide(#2,#0)|multiply(#3,const_2)| | physics | A |
if x and y are positive integers and x = 5 y + 2 , what is the remainder when x is divided by 5 ? | "this question asks what is . . . ( the answer ) , so we know that the answer will be consistent . as such , we can test values to quickly get the solution . we ' re told that x and y are positive integers and x = 5 y + 2 . we ' re asked for the remainder when x is divided by 5 . if . . . . y = 1 x = 7 7 / 5 = 1 remainder 2 final answer : c" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(add(2, const_1), 5) | add(n1,const_1)|divide(#0,n0)| | general | C |
if a student loses 5 kilograms , he will weigh twice as much as his sister . together they now weigh 110 kilograms . what is the student ' s present weight in kilograms ? | let x be the weight of the sister . then the student ' s weight is 2 x + 5 . x + ( 2 x + 5 ) = 110 3 x = 105 x = 35 kg then the student ' s weight is 75 kg . the answer is e . | a ) 71 , b ) 72 , c ) 73 , d ) 74 , e ) 75 | e | subtract(110, divide(subtract(110, 5), const_3)) | subtract(n1,n0)|divide(#0,const_3)|subtract(n1,#1) | other | E |
joe invested a certain sum of money in a simple interest bond whose value grew to $ 460 at the end of 3 years and to $ 560 at the end of another 5 years . what was the rate of interest in which he invested his sum ? | "in 5 years , the value grew $ 100 , so the simple interest was $ 20 per year . in 3 years , the total interest was 3 * $ 20 = $ 60 the principal is $ 460 - $ 60 = 400 . the interest rate is $ 20 / $ 400 = 5 % the answer is c ." | a ) 3 % , b ) 4 % , c ) 5 % , d ) 6 % , e ) 7 % | c | multiply(divide(divide(subtract(560, 460), 5), subtract(460, multiply(divide(subtract(560, 460), 5), 3))), const_100) | subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|divide(#1,#3)|multiply(#4,const_100)| | gain | C |
how many integerskgreater than 100 and less than 900 are there such that if the hundreds and the units digits ofkare reversed , the resulting integer is k + 99 ? | "numbers will be like 102 = > 201 = 102 + 99 203 = > 302 = 103 + 99 so the hundereth digit and units digit are consecutive where unit digit is bigger than hundred digit . there will be seven pairs of such numbers for every pair there will 10 numbers like for 12 = > 102 , 112,132 , 142,152 , 162,172 , 182,192 . total = 7 * 10 = 70 hence c ." | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | add(multiply(divide(100, const_10), multiply(const_2, const_4)), divide(100, const_10)) | divide(n0,const_10)|multiply(const_2,const_4)|multiply(#0,#1)|add(#0,#2)| | general | C |
if x is 20 percent more than y and y is 70 percent less than z , then x is what percent of z ? | z = 100 ; y = 30 so x = 36 x as % of z = 36 / 100 * 100 = > 36 % answer will be ( d ) | a ) 500 % , b ) 250 % , c ) 500 / 3 % , d ) 36 % , e ) 60 % | d | multiply(add(const_1, divide(20, const_100)), subtract(const_100, 70)) | divide(n0,const_100)|subtract(const_100,n1)|add(#0,const_1)|multiply(#2,#1) | general | D |
if x < y < z and y - x > 3 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ? | "we have : 1 ) x < y < z 2 ) y - x > 3 3 ) x = 2 k ( x is an even number ) 4 ) y = 2 n + 1 ( y is an odd number ) 5 ) z = 2 p + 1 ( z is an odd number ) 6 ) z - x = ? least value z - x = 2 p + 1 - 2 k = 2 p - 2 k + 1 = 2 ( p - k ) + 1 - that means that z - x must be an odd number . we are asked to find the least value , so we have to pick the least numbers since y is odd and x is even , y - x must be odd . since y - x > 3 , the least value for y - x must be 5 , the least value for x must be 2 , and , thus , the least possible value for y must be 7 ( y - 2 = 5 , y = 7 ) 2 < 7 < z , since z is odd , the least possible value for z is 9 z - x = 9 - 2 = 7 answer a" | a ) 7 , b ) 6 , c ) 8 , d ) 9 , e ) 10 | a | add(add(3, const_2), const_2) | add(n0,const_2)|add(#0,const_2)| | general | A |
a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 5 % dividend at the end of the year , then how much does he get ? | "solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 5 / 100 x 12000 ) = rs . 600 . answer b" | a ) rs . 500 , b ) rs . 600 , c ) rs . 650 , d ) rs . 720 , e ) none | b | multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, 100)), add(100, multiply(100, divide(20, 100))))), divide(5, 100)) | divide(n3,n1)|divide(n2,n1)|multiply(const_10,const_1000)|multiply(const_1000,const_4)|multiply(n1,const_4)|add(#2,#3)|multiply(n1,#1)|add(#5,#4)|add(n1,#6)|divide(#7,#8)|multiply(n1,#9)|multiply(#0,#10)| | gain | B |
the average ( arithmetic mean ) of the even integers from 0 to 40 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 20 inclusive ? | "the sum of even numbers from 0 to n is 2 + 4 + . . . + n = 2 ( 1 + 2 + . . . + n / 2 ) = 2 ( n / 2 ) ( n / 2 + 1 ) / 2 = ( n / 2 ) ( n / 2 + 1 ) the average is ( n / 2 ) ( n / 2 + 1 ) / ( n / 2 + 1 ) = n / 2 the average of the even numbers from 0 to 40 is 40 / 2 = 20 the average of the even numbers from 0 to 20 is 20 / 2 = 10 the answer is c ." | a ) 5 , b ) 8 , c ) 10 , d ) 15 , e ) 20 | c | subtract(divide(add(20, 0), const_2), divide(add(40, 0), const_2)) | add(n2,n3)|add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)| | general | C |
machine a can finish a job in 4 hours , machine Π² can finish the job in 12 hours , and machine Ρ can finish the job in 8 hours . how many hours will it take for a , b , and Ρ together to finish the job ? | "the combined rate is 1 / 4 + 1 / 12 + 1 / 8 = 11 / 24 of the job per hour . the time to complete the job is 24 / 11 hours . the answer is d ." | a ) 15 / 4 , b ) 18 / 7 , c ) 21 / 8 , d ) 24 / 11 , e ) 28 / 13 | d | inverse(add(add(inverse(4), inverse(12)), inverse(8))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|inverse(#4)| | physics | D |
one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 50,000 as total savings , how much has he saved in public provident fund ? | "let savings in n . s . c and p . p . f . be rs . x and rs . ( 150000 - x ) respectively . then , ( 1 / 3 ) x = ( 1 / 2 ) ( 150000 - x ) ( x / 3 ) + ( x / 2 ) = 75000 5 x / 6 = 75000 x = 75000 x 6 / 5 = 90000 savings in public provident fund = rs . ( 150000 - 90000 ) = rs . 60000 answer is b ." | a ) 30000 , b ) 60000 , c ) 50000 , d ) 90000 , e ) 70000 | b | multiply(add(multiply(multiply(const_100, const_10), const_100), subtract(multiply(multiply(const_100, const_10), const_100), multiply(multiply(const_2, const_100), const_100))), divide(1, add(divide(const_3, const_2), 1))) | divide(const_3,const_2)|multiply(const_10,const_100)|multiply(const_100,const_2)|add(n0,#0)|multiply(#1,const_100)|multiply(#2,const_100)|divide(n0,#3)|subtract(#4,#5)|add(#4,#7)|multiply(#8,#6)| | general | B |
a = 4 ^ 15 - 625 ^ 3 and a / x is an integer , where x is a positive integer greater than 1 , such that it does not have a factor p such that 1 < p < x , then how many different values for x are possible ? | "this is a tricky worded question and i think the answer is should be d not c . . . here is my reason : the stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . the stem wants to say that x is a prime number . because any prime number has no factor grater than 1 and itself . on the other hand the stem says that x could get how many different number not must get different number ( this is very important issue ) as our friends say , if we simplify numerator more we can obtain : 5 ^ 12 ( 5 ^ 3 - 1 ) = 5 ^ 12 ( 124 ) = 5 ^ 12 ( 31 * 2 * 2 ) divided by x and we are told that this fraction is an integer . so , x could be ( not must be ) 5 , 31 , or 2 ! ! ! so , x could get 2 different values and answer is c . . . ." | a ) none , b ) one , c ) two , d ) three , e ) four | c | subtract(15, multiply(3, const_4)) | multiply(n3,const_4)|subtract(n1,#0)| | general | C |
a reduction of 18 % in the price of oil enables a house wife to obtain 8 kgs more for rs . 1080 , what is the reduced price for kg ? | "1080 * ( 18 / 100 ) = 194 - - - - 8 ? - - - - 1 = > rs . 24.25 answer : c" | a ) 72.15 , b ) 27.24 , c ) 24.25 , d ) 28.46 , e ) 20.9 | c | divide(divide(multiply(1080, 18), const_100), 8) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain | C |
during a certain two - week period , 75 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 2 / 3 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedies , in terms of a , were rented during that two - week period ? | "movies : 75 % comedies . 25 % remaining genre . now in this 25 % , there are only 2 categories . action movies and drama movies . if action = x ; drama movies = 2 x / 3 . total 5 x / 3 . 5 x / 3 = 25 ; x = 15 action movies : 15 % drama movies : 10 % we can say that out of 100 z , : comedies : 75 z action : 15 z drama : 10 z now action movies were a this means : a = 15 z . z = a / 15 comedies : 75 z = 75 * ( a / 15 ) 5 a a is the answer ." | a ) 5 a , b ) 10 a , c ) 15 a , d ) 18 a , e ) 25 a | a | floor(divide(75, add(2, const_1))) | add(n1,const_1)|divide(n0,#0)|floor(#1)| | general | A |
there are 192 items that are members of set u . of these items , 49 are members of set b , 59 are not members of either of set a or set b , and 23 are members of both sets a and b . how many of the members of set u are members of set a ? | "you had the answer almost right . the x = 84 refers to only set a . however what ' s being asked is how many members are part of set a . this will include : 1 . only set a 2 . set a and set b so the answer is set a = 84 + set ab = 84 + 23 = 17 c" | a ) 72 , b ) 85 , c ) 107 , d ) 98 , e ) 108 | c | subtract(add(subtract(192, 59), 23), 49) | subtract(n0,n2)|add(n3,#0)|subtract(#1,n1)| | other | C |
if f ( f ( n ) ) + f ( n ) = 2 n + 3 , f ( 0 ) = 1 then f ( 2012 ) = ? | "f ( f ( 0 ) ) + f ( 0 ) = 2 ( 0 ) + 3 β β f ( 1 ) = 3 - 1 = 2 , f ( 1 ) = 2 f ( f ( 1 ) ) + f ( 1 ) = 2 ( 1 ) + 3 β β f ( 2 ) = 5 - 2 = 3 , f ( 2 ) = 3 f ( f ( 2 ) ) + f ( 2 ) = 2 ( 2 ) + 3 β β f ( 3 ) = 7 - 3 = 4 , f ( 3 ) = 4 . . . . . . . . . . . . . . f ( 2012 ) = 2013 ans : a" | a ) 2013 , b ) 2088 , c ) 270 , d ) 1881 , e ) 1781 | a | add(1, 2012) | add(n3,n4)| | general | A |
find out the square of a number which when doubled exceeds its one nineth by 17 ? | "let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 9 ) + 17 = > 18 p = p + 153 = > p = 9 p ^ 2 = 9 ^ 2 = 81 answer : c" | a ) 16 , b ) 25 , c ) 81 , d ) 26 , e ) 17 | c | power(divide(17, subtract(const_2, divide(const_1, add(const_4, const_1)))), const_2) | add(const_1,const_4)|divide(const_1,#0)|subtract(const_2,#1)|divide(n0,#2)|power(#3,const_2)| | general | C |
if you add all the numbers on your mobile phone except 9 , what is the answer ? | "we have to add 0 to 8 to find the answer . therefore 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 answer is d" | a ) 45 , b ) 28 , c ) 65 , d ) 36 , e ) 42 | d | multiply(divide(add(const_1, subtract(const_10, const_1)), const_2), subtract(const_10, const_1)) | subtract(const_10,const_1)|add(#0,const_1)|divide(#1,const_2)|multiply(#2,#0)| | general | D |
dacid obtained 45 , 35 , 52 , 47 and 55 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | average = ( 45 + 35 + 52 + 47 + 55 ) / 5 = 234 / 5 = 47 . answer : c | a ) 79 , b ) 99 , c ) 47 , d ) 89 , e ) 45 | c | divide(add(add(add(add(45, 35), 52), 47), 55), divide(const_10, const_2)) | add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1) | general | C |
a bag marked at $ 200 is sold for $ 120 . the rate of discount is ? | "rate of discount = 80 / 200 * 100 = 40 % answer is e" | a ) 10 % , b ) 25 % , c ) 20 % , d ) 50 % , e ) 40 % | e | multiply(divide(subtract(200, 120), 200), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain | E |
the sum of four consecutive numbers is 102 . the greatest among these three number is : | "let the numbers be x , x + 1 , x + 2 and x + 3 then , x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) = 102 4 x = 96 x = 24 greatest number , ( x + 3 ) = 27 answer : e" | a ) 26 , b ) 28 , c ) 29 , d ) 30 , e ) 27 | e | divide(add(102, const_1), const_2) | add(n0,const_1)|divide(#0,const_2)| | physics | E |
one fourth of one third of two fifth of a number is 16 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 16 then x = 16 * 30 = 480 40 % of 480 = 192 answer : option d" | a ) 140 , b ) 150 , c ) 180 , d ) 192 , e ) 250 | d | divide(multiply(divide(16, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100)| | gain | D |
if the farmer sells 75 of his chickens , his stock of feed will last for 20 more days than planned , but if he buys 100 more chickens , he will run out of feed 15 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ? | let v denote the volume of feed one chicken consumes per day . then the total volume of feed in stock will be v β d β c where d is the number of days the feed will last if the number of chickens does not change and cc is the current number of chickens . from the question it follows that v ( d + 20 ) ( c β 75 ) = vdc v ( d β 15 ) ( c + 100 ) = vdc the first equation simplifies to 20 c β 75 d = 1500 . the second equation simplifies to ( β 15 ) c + 100 d = 1500 after dividing everything by 5 we get the linear system : 4 c β 15 d = 300 ( β 3 ) c + 20 d = 300 solving it we get c = 300 d = 60 answer : e | a ) 60 , b ) 120 , c ) 240 , d ) 275 , e ) 300 | e | multiply(divide(multiply(20, divide(75, divide(subtract(100, 75), subtract(20, 15)))), subtract(20, divide(75, divide(subtract(100, 75), subtract(20, 15))))), divide(subtract(100, 75), subtract(20, 15))) | subtract(n2,n0)|subtract(n1,n3)|divide(#0,#1)|divide(n0,#2)|multiply(n1,#3)|subtract(n1,#3)|divide(#4,#5)|multiply(#6,#2) | general | E |
a garden center sells a certain grass seed in 5 - pound bags at $ 13.85 per bag , 10 - pound bags at $ 20.40 per bag , and 25 - pound bags $ 32.25 per bag . if a customer is to buy at least 65 pounds of the grass seed , but no more than 80 pounds , what is the least possible cost of the grass seed that the customer will buy ? | "there can be 2 cases 1 ) 25 + 25 + 10 + 5 = $ 98.75 or 2 ) 25 + 25 + 25 = $ 96.75 c" | a ) $ 94.03 , b ) $ 96.75 , c ) $ 98.75 , d ) $ 102.07 , e ) $ 105.3 | c | add(add(multiply(const_2, 32.25), multiply(const_1, 20.40)), multiply(const_1, 13.85)) | multiply(n5,const_2)|multiply(n3,const_1)|multiply(n1,const_1)|add(#0,#1)|add(#3,#2)| | general | C |
what is the average of 120 , 130 , 140 , 510 , 520 , 530 , 1115 , 1120 , and 1125 ? | "add 120 , 130 , 140 , 510 , 520 , 530 , 1115 , 1120 , and 1125 grouping numbers together may quicken the addition sum = 5310 5310 / 9 = 590 . d" | a ) 419 , b ) 551 , c ) 601 , d ) 590 , e ) 721 | d | divide(add(add(add(add(add(add(1115, 130), 140), 510), 520), 530), 1115), add(const_3, const_4)) | add(n6,n1)|add(const_3,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|add(n6,#5)|divide(#6,#1)| | general | D |
the workforce of company x is 60 % female . the company hired 30 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ? | "let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 30 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 30 ) 0.05 x = 16.5 x = 330 - this is quantity of employees before adding 30 men so after adding it will be 360 answer is d" | a ) 160 , b ) 220 , c ) 240 , d ) 360 , e ) 420 | d | add(divide(multiply(divide(55, const_100), 30), subtract(divide(60, const_100), divide(55, const_100))), 30) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)| | gain | D |
find the least number which when divided by 5 , 67 , and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder | explanation : l . c . m . of 5,6 , 7,8 = 840 . required number is of the form 840 k + 3 least value of k for which ( 840 k + 3 ) is divisible by 9 is k = 2 . required number = ( 840 x 2 + 3 ) = 1683 answer is a | a ) 1683 , b ) 1684 , c ) 1685 , d ) 1582 , e ) 1584 | a | add(multiply(lcm(lcm(lcm(5, multiply(const_2, const_3)), add(const_3, const_4)), 8), const_2), 3) | add(const_3,const_4)|multiply(const_2,const_3)|lcm(n0,#1)|lcm(#0,#2)|lcm(n2,#3)|multiply(#4,const_2)|add(n3,#5) | general | A |
an alloy is to contain copper and zinc in the ratio 5 : 3 . the zinc required to be melted with 40 kg of copper is ? | let the required quantity of copper be x kg 5 : 3 : : 40 : x 5 x = 40 * 3 x = 24 kg answer is d | a ) 15 kg , b ) 20 kg , c ) 8 kg , d ) 24 kg , e ) 32 kg | d | divide(multiply(40, 3), 5) | multiply(n1,n2)|divide(#0,n0) | other | D |
on a scale of map , 0.4 cm represents 5.3 km . if the distance between the points on the map is 64 cm , the actual distance between these points is : | explanation : let the actual distance be x km . then , more distance on the map , more is the actual distance ( direct proportion ) = > 0.4 : 64 : : 5.3 : x = > 0.4 x = 64 x 5.3 = > x = 64 x 5.3 / 0.4 = > x = 848 answer : d | a ) 9 km , b ) 72.5 km , c ) 190.75 km , d ) 848 km , e ) none of these | d | multiply(divide(5.3, 0.4), 64) | divide(n1,n0)|multiply(n2,#0) | physics | D |
a scale 7 ft . 6 inches long is divided into 5 equal parts . find the length of each part . | "explanation : total length of scale in inches = ( 7 * 12 ) + 6 = 90 inches length of each of the 5 parts = 90 / 5 = 18 inches answer : d" | a ) 20 inches , b ) 77 inches , c ) 66 inches , d ) 18 inches , e ) 66 inches | d | divide(add(multiply(7, const_12), 6), 5) | multiply(n0,const_12)|add(n1,#0)|divide(#1,n2)| | general | D |
6.2 is what percent of 1000 ? | "10 % of 1000 is 100 . 6.2 is way less than 100 ( 10 % of 1000 ) . so it can not be 62 % of 1000 . eliminate option a 1 % of 1000 is 10 . 6.2 is less than 10 . so it cant be 6.2 % . eliminate option b ( 0.5 % ) i . e . half of 1 % of 1000 is 5 . but 6.2 is greater than 5 . . so it has to be greater than 0.5 % . . . hence answer is 0.62 % , answer : c" | a ) 62 % , b ) 6.2 % , c ) 0.62 % , d ) 0.062 % , e ) 0.0062 % | c | multiply(divide(6.2, 1000), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain | C |
a cricketer scored 152 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 152 - ( 12 x 4 + 2 x 6 ) = 152 - ( 60 ) = 92 now , we need to calculate 92 is what percent of 152 . = > 92 / 152 * 100 = 60.52 % answer : c" | a ) 46.2 % , b ) 54.54 % , c ) 60.52 % , d ) 70 % , e ) none of these | c | multiply(divide(subtract(152, add(multiply(12, const_4), multiply(2, multiply(2, const_3)))), 152), const_100) | multiply(n1,const_4)|multiply(n2,const_3)|multiply(n2,#1)|add(#0,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)| | general | C |
after decreasing 20 % in the price of an article costs rs . 600 . find the actual cost of an article ? | "cp * ( 80 / 100 ) = 600 cp = 7.75 * 100 = > cp = 775 answer : a" | a ) 775 , b ) 620 , c ) 53 , d ) 530 , e ) 220 | a | divide(600, subtract(const_1, divide(20, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain | A |
in a class of students , 1 / 2 of the number of girls is equal to 1 / 6 of the total number of students . what is the ratio of boys to girls in the class ? | "( 1 / 2 ) g = ( 1 / 6 ) ( b + g ) 6 g = 2 b + 2 g 4 g = 2 b b / g = 2 / 1 . the answer is c ." | a ) 1 / 2 , b ) 2 / 3 , c ) 2 / 1 , d ) 3 / 2 , e ) 4 / 3 | c | divide(subtract(divide(1, 2), divide(1, 6)), divide(1, 6)) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,#1)| | general | C |
how many positive integers less than 120 are there such that they are multiples of 13 or multiples of 12 but not both ? | "for 13 : 13 . . . 117 = 13 * 9 = 117 for 12 : 12 . . . 120 = 12 * 10 = 120 but there is one integer 13 * 12 . so n = ( 9 ) + ( 10 ) = 19 b" | a ) 13 , b ) 19 , c ) 15 , d ) 16 , e ) 17 | b | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general | B |
a profit of rs . 20000 is to be distributed among a , b , c in the proportion 2 : 3 : 5 . what will be the difference between b β s and c β s shares | explanation : b ' s share = rs 20000 * ( 3 / 10 ) = rs 6000 c ' s share = rs 20000 * ( 5 / 10 ) = rs 10000 difference in b ' s and c ' s shares = rs 4000 answer : a | a ) 4000 , b ) 3000 , c ) 5500 , d ) 4500 , e ) none of these | a | subtract(multiply(5, divide(20000, add(add(2, 3), 5))), multiply(3, divide(20000, add(add(2, 3), 5)))) | add(n1,n2)|add(n3,#0)|divide(n0,#1)|multiply(n3,#2)|multiply(n2,#2)|subtract(#3,#4) | gain | A |
sandy is younger than molly by 18 years . if the ratio of their ages is 7 : 9 , how old is sandy ? | "let sandy ' s age be 7 x and let molly ' s age be 9 x . 9 x - 7 x = 18 x = 9 sandy is 63 years old . the answer is d ." | a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70 | d | multiply(divide(7, subtract(9, 7)), 18) | subtract(n2,n1)|divide(n1,#0)|multiply(n0,#1)| | other | D |
in may , the groundskeeper at spring lake golf club built a circular green with an area of 70 Ο square feet . in august , the groundskeeper doubled the distance from the center of the green to the edge of the green . what is the total area of the renovated green ? | "area = Ο r ^ 2 , so doubling the radius results in an area that is 4 times the original area . 4 ( 70 Ο ) = 280 Ο the answer is c ." | a ) 1000 Ο , b ) 400 Ο , c ) 280 Ο , d ) 200 Ο , e ) 20 Ο | c | multiply(power(multiply(sqrt(70), const_2), const_2), const_pi) | sqrt(n0)|multiply(#0,const_2)|power(#1,const_2)|multiply(#2,const_pi)| | geometry | C |
a man buys an item at rs . 1300 and sells it at the loss of 20 percent . then what is the selling price of that item | explanation : here always remember , when ever x % loss , it means s . p . = ( 100 - x ) % of c . p when ever x % profit , it means s . p . = ( 100 + x ) % of c . p so here will be ( 100 - x ) % of c . p . = 80 % of 1300 = 80 / 100 * 1300 = 1040 option d | a ) rs . 660 , b ) rs . 760 , c ) rs . 860 , d ) rs . 1040 , e ) none of these | d | multiply(1300, subtract(const_1, divide(20, const_100))) | divide(n1,const_100)|subtract(const_1,#0)|multiply(n0,#1) | gain | D |
a certain company reported that the revenue on sales increased 15 % from 2000 to 2003 , and increased 30 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ? | "assume the revenue in 2000 to be 100 . then in 2003 it would be 115 and and in 2005 130 , so from 2003 to 2005 it increased by ( 130 - 115 ) / 115 = 15 / 115 = 13 % . answer : e ." | a ) 50 % , b ) 40 % , c ) 35 % , d ) 32 % , e ) 13 % | e | multiply(divide(subtract(add(const_1, divide(30, const_100)), add(const_1, divide(15, const_100))), add(const_1, divide(15, const_100))), const_100) | divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(#2,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain | E |
if 20 % of a is the same as 30 % of b , then a : b is : | "expl : 20 % of a i = 30 % of b = 20 a / 100 = 30 b / 100 = 3 / 2 = 3 : 2 answer : d" | a ) 5 : 4 , b ) 5 : 3 , c ) 4 : 3 , d ) 3 : 2 , e ) 1 : 3 | d | divide(divide(30, const_100), divide(20, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)| | gain | D |
a merchant has 1000 kg of sugar part of which he sells at 8 % profit and the rest at 18 % profit . he gains 14 % on the whole . the quantity sold at 18 % profit is | explanation : by the rule of alligation , we have profit % by selling 1 st part profit % by selling 2 nd part 8 18 net % profit 14 18 - 14 = 4 14 - 8 = 6 = > quantity of part 1 : quantity of part 2 = 4 : 6 = 2 : 3 total quantity is given as 1000 kg so quantity of part 2 ( quantity sold at 18 % profit ) = 1000 Γ 35 = 600 kg answer : option c | a ) 300 , b ) 400 , c ) 600 , d ) 500 , e ) none of these | c | multiply(multiply(add(const_2, const_3), const_2), multiply(subtract(add(divide(14, const_100), const_1), add(const_1, divide(8, const_100))), 1000)) | add(const_2,const_3)|divide(n3,const_100)|divide(n1,const_100)|add(#1,const_1)|add(#2,const_1)|multiply(#0,const_2)|subtract(#3,#4)|multiply(n0,#6)|multiply(#5,#7) | gain | C |
two cars started at the same time , from the same point , driving along the same road . the rate of the first car is 50 mph and the rate of the second car is 60 mph . how long will it take for the distance between the two cars to be 30 miles ? | car 1 50 50 50 total = 150 car 2 60 60 60 total = 180 differece = 30 miles after 3 hrs . answer : c | a ) 1 hrs , b ) 2 hrs , c ) 3 hrs , d ) 4 hrs , e ) 5 hrs | c | divide(30, subtract(60, 50)) | subtract(n1,n0)|divide(n2,#0) | physics | C |
if money is invested at r percent interest , compounded annually , the amount of investment will double in approximately 70 / r years . if pat ' s parents invested $ 5000 in a long term bond that pays 4 percent interest , compounded annually , what will be the approximate total amount of investment 36 years later , when pat is ready for college ? | "since investment doubles in 70 / r years then for r = 4 it ' ll double in 70 / 4 = ~ 18 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 36 years investment will double twice and become ( $ 5,000 * 2 ) * 2 = $ 20,000 ( after 18 years investment will become $ 5,000 * 2 = $ 10,000 and in another 18 years it ' ll become $ 10,000 * 2 = $ 20,000 ) . answer : a ." | a ) $ 20000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | a | multiply(multiply(5000, const_2), const_2) | multiply(n1,const_2)|multiply(#0,const_2)| | general | A |
if a = { 1 , 3 , 5 } , b = { 3 , 5 , 6 } . find a βͺ b | "a = { 1 , 3,5 } b = { 3 , 5,6 } therefore , correct answer : a βͺ b = { 1 , 3 , 5 , 6 } b" | a ) { 1,5 } , b ) { 1,3 , 5,6 } , c ) { 2,6 } , d ) { 8,9 } , e ) { 4,12 } | b | add(divide(add(add(add(add(1, 3), 5), 3), 5), add(const_4, const_1)), 5) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)|add(n2,#5)| | general | B |
the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 80 metres ? | "speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 80 m = 80 / 5 = 16 seconds . answer : a" | a ) 16 seconds , b ) 76 seconds , c ) 26 seconds , d ) 12 seconds , e ) 18 seconds | a | divide(80, multiply(add(15, 3), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics | A |
there are 7 red shoes & 3 green shoes . if two of red shoes are drawn what is the probability of getting red shoes | "taking 2 red shoe the probablity is 7 c 2 from 10 shoes probablity of taking 2 red shoe is 7 c 2 / 10 c 2 = 7 / 15 answer : d" | a ) 1 / 13 , b ) 1 / 14 , c ) 1 / 15 , d ) 7 / 15 , e ) 7 / 16 | d | divide(choose(7, const_2), choose(add(7, 3), const_2)) | add(n0,n1)|choose(n0,const_2)|choose(#0,const_2)|divide(#1,#2)| | probability | D |
on a certain farm the ratio of horses to cows is 5 : 1 . if the farm were to sell 15 horses and buy 15 cows , the ratio of horses to cows would then be 17 : 7 . after the transaction , how many more horses than cows would the farm own ? | originally , there were 5 k horses and k cows . 7 ( 5 k - 15 ) = 17 ( k + 15 ) 35 k - 17 k = 255 + 105 18 k = 360 k = 20 the difference between horses and cows is ( 5 k - 15 ) - ( k + 15 ) = 4 k - 30 = 50 the answer is c . | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | subtract(subtract(multiply(5, divide(add(multiply(15, 17), multiply(15, 7)), subtract(multiply(5, 7), 17))), 15), add(divide(add(multiply(15, 17), multiply(15, 7)), subtract(multiply(5, 7), 17)), 15)) | multiply(n2,n4)|multiply(n2,n5)|multiply(n0,n5)|add(#0,#1)|subtract(#2,n4)|divide(#3,#4)|add(n2,#5)|multiply(n0,#5)|subtract(#7,n2)|subtract(#8,#6) | other | C |
bottle r contains 250 capsules and costs $ 6.25 . bottle t contains 100 capsules and costs $ 3.0 . what is the difference between the cost per capsule for bottle r and the cost per capsule for bottle t ? | "cost per capsule in r is 6.25 / 250 = 0.625 / 25 = 0.025 cost per capsule in t is 3.00 / 100 = 0.03 the difference is 0.005 the answer is e" | a ) $ 0.25 , b ) $ 0.12 , c ) $ 0.05 , d ) $ 0.003 , e ) $ 0.005 | e | subtract(divide(3.0, 100), divide(6.25, 250)) | divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)| | general | E |
x does a work in 10 days . y does the same work in 15 days . in how many days they together will do the same work ? | "x ' s 1 day ' s work = 1 / 10 y ' s 1 day ' s work = 1 / 15 ( x + y ) ' s 1 day ' s work = ( 1 / 10 + 1 / 15 ) = 1 / 6 both together will finish the work in 6 days . correct option is d" | a ) 10 , b ) 15 , c ) 3 , d ) 6 , e ) 12 | d | inverse(add(divide(const_1, 10), divide(const_1, 15))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)| | physics | D |
the distance from city a to city b is 120 miles . while driving from city a to city b , bob drives at a constant speed of 40 miles per hour . alice leaves city a 30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ? | the time it takes bob to drive to city b is 120 / 40 = 3 hours . alice needs to take less than 2.5 hours for the trip . alice needs to exceed a constant speed of 120 / 2.5 = 48 miles per hour . the answer is b . | a ) 45 , b ) 48 , c ) 50 , d ) 52 , e ) 54 | b | divide(120, subtract(divide(120, 40), divide(30, const_60))) | divide(n0,n1)|divide(n2,const_60)|subtract(#0,#1)|divide(n0,#2) | physics | B |
a special municipal payroll tax charges not tax on a payroll less than $ 200,000 and only 0.1 % on a company β s payroll above $ 200,000 . if belfried industries paid $ 400 in this special municipal payroll tax , then they must have had a payroll of ? | "answer : e , ( with different approach ) : the 400 paid is 0.1 % of the additional amount above 200,000 . let it be x now 0.1 % of x = 400 therefore x = 400,000 total = 200,000 + x = 600,000" | a ) $ 180,000 , b ) $ 202,000 , c ) $ 220,000 , d ) $ 400,000 , e ) $ 600,000 | e | multiply(multiply(400, const_100), const_10) | multiply(n3,const_100)|multiply(#0,const_10)| | general | E |
john makes $ 60 a week from his job . he earns a raise and now makes $ 80 a week . what is the % increase ? | "increase = ( 20 / 60 ) * 100 = ( 2 / 6 ) * 100 = 33.33 % . b" | a ) 16 % , b ) 33.33 % , c ) 17 % , d ) 17.61 % , e ) 17.56 % | b | multiply(divide(subtract(80, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain | B |
a student multiplied a number by 3 / 5 instead of 5 / 3 , what is the percentage error in the calculation ? | "let the number be x . then , ideally he should have multiplied by x by 5 / 3 . hence correct result was . by mistake he multiplied x by 3 / 5 . hence the result with error = then , error = error % = = 64 % answer : b" | a ) 22 , b ) 64 , c ) 365 , d ) 29 , e ) 36 | b | multiply(divide(subtract(divide(5, 3), divide(3, 5)), divide(5, 3)), const_100) | divide(n1,n0)|divide(n0,n1)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)| | general | B |
if x is a positive number and 1 / 2 the square root of x is the cube root of x , then x = | 1 / 2 the square root of x is cube root of x . if x = 64 . . then 1 / 2 the square root of x = 4 and cube of x is 64 . option a . | a ) 64 , b ) 32 , c ) 16 , d ) 4 , e ) 1 | a | power(2, multiply(const_2, const_3)) | multiply(const_2,const_3)|power(n1,#0) | general | A |
the average speed of a car decreased by 3 miles per hour every successive 8 - minutes interval . if the car traveled 3.6 miles in the fifth 8 - minute interval , what was the average speed of the car , in miles per hour , in the first 8 minute interval ? | ( 3.6 miles / 8 minutes ) * 60 minutes / hour = 27 mph let x be the original speed . x - 4 ( 3 ) = 27 x = 39 mph the answer is c . | a ) 31 , b ) 35 , c ) 39 , d ) 43 , e ) 47 | c | add(add(add(add(divide(3.6, divide(8, const_60)), 3), 3), 3), 3) | divide(n1,const_60)|divide(n2,#0)|add(n0,#1)|add(n0,#2)|add(n0,#3)|add(n0,#4) | physics | C |
the mean of 30 values was 140 . it was detected on rechecking that one value 145 was wrongly copied as 135 for the computation of the mean . find the correct mean . | "corrected mean = 140 Γ 30 β 135 + 145 / 30 = 4200 β 135 + 145 / 30 = 4210 / 30 = 140.33 answer b" | a ) 151 , b ) 140.33 , c ) 152 , d ) 148 , e ) none of the above | b | divide(add(multiply(30, 140), subtract(145, 135)), 30) | multiply(n0,n1)|subtract(n2,n3)|add(#0,#1)|divide(#2,n0)| | general | B |
the total number of students in grades 1 and 2 is 30 more than the total number of students in grades 2 and 5 . how much lesser is the number of students in grade 5 as compared to grade 1 ? | ( grade 1 + grade 2 ) - ( grade 2 + grade 5 ) = 30 grade 1 - grade 5 = 30 answer : b | a ) 20 , b ) 30 , c ) 10 , d ) 40 , e ) 15 | b | divide(add(add(add(add(add(add(add(30, 5), 2), 2), 2), const_3), const_4), const_12), 2) | add(n2,n4)|add(n1,#0)|add(n1,#1)|add(n1,#2)|add(#3,const_3)|add(#4,const_4)|add(#5,const_12)|divide(#6,n1) | general | B |
the area of a circle is increased by 1500 % . by what percent has the diameter of the circle increased ? | "the area of the circle is increased by 1500 % , thus the area is increased 16 times . the area of a circle it proportional to the square of the diameter ( area = Ο d ^ 2 / 4 ) , therefore the diameter must increase 4 times ( diameter increase 4 times = area increase 16 times ) , which is increase by 500 % . answer : b ." | a ) 100 % , b ) 500 % , c ) 300 % , d ) 600 % , e ) 800 % | b | multiply(const_100, divide(const_2, const_2)) | divide(const_2,const_2)|multiply(#0,const_100)| | geometry | B |
cole drove from home to work at an average speed of 70 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ? | "let the distance one way be x time from home to work = x / 70 time from work to home = x / 105 total time = 2 hrs ( x / 70 ) + ( x / 105 ) = 2 solving for x , we get x = 84 time from home to work in minutes = ( 84 ) * 60 / 70 = 72 minutes ans = c" | a ) 66 , b ) 70 , c ) 72 , d ) 75 , e ) 78 | c | multiply(divide(multiply(105, 2), add(70, 105)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)| | physics | C |
john bought a shirt on sale for 25 % off the original price and another 25 % off the discounted price . if the final price was $ 15 , what was the price before the first discount ? | "let x be the price before the first discount . the price after the first discount is x - 25 % x ( price after first discount ) a second discount of 25 % of the discounted price after which the final price is 15 ( x - 25 % x ) - 25 % ( x - 25 % x ) = 15 solve for x x = $ 26.66 correct answer d" | a ) $ 45.10 , b ) $ 34.31 , c ) $ 28.44 , d ) $ 26.66 , e ) $ 65.23 | d | divide(multiply(multiply(const_100, const_100), 15), subtract(multiply(subtract(const_100, 25), const_100), multiply(subtract(const_100, 25), 25))) | multiply(const_100,const_100)|subtract(const_100,n0)|multiply(n2,#0)|multiply(#1,const_100)|multiply(n0,#1)|subtract(#3,#4)|divide(#2,#5)| | gain | D |
simplify : 500 x 500 - 200 x 200 | "( 500 ) ^ 2 - ( 200 ) ^ 2 = ( 500 + 200 ) ( 500 - 200 ) = 700 x 300 = 210000 . answer is d ." | a ) 761200 , b ) 761400 , c ) 761800 , d ) 210000 , e ) none of them | d | add(multiply(500, 500), multiply(200, 200)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | general | D |
the product of two positive integers is 1023 and their difference is 2 . what is the bigger number ? | "let ' s use trial and error to find the two numbers . 32 * 30 = 960 ( too low ) 33 * 31 = 1023 the answer is c ." | a ) 29 , b ) 31 , c ) 33 , d ) 35 , e ) 37 | c | subtract(add(multiply(divide(1023, divide(1023, 2)), const_4), const_10), const_1) | divide(n0,n1)|divide(n0,#0)|multiply(#1,const_4)|add(#2,const_10)|subtract(#3,const_1)| | general | C |
a number whose fifth part increased by 3 is equal to its fourth part diminished by 3 is ? | "answer let the number be n . then , ( n / 5 ) + 3 = ( n / 4 ) - 3 Γ’ β‘ β ( n / 4 ) - ( n / 5 ) = 6 Γ’ β‘ β ( 5 n - 4 n ) / 20 = 6 Γ’ Λ Β΄ n = 120 option : a" | a ) 120 , b ) 180 , c ) 200 , d ) 220 , e ) none | a | divide(add(3, 3), subtract(divide(const_1, 3), divide(const_1, add(const_1, 3)))) | add(n0,n1)|add(const_1,n0)|divide(const_1,n0)|divide(const_1,#1)|subtract(#2,#3)|divide(#0,#4)| | general | A |
a cube with its sides numbered 1 through 6 is rolled twice , first landing on a and then landing on b . if any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6 , what is the probability t that a + b is prime ? | "total # of outcomes is 6 * 6 = 36 ; favorable outcomes : a - b - - > prime 1 - 1 - - > 2 ; 1 - 2 - - > 3 ; 2 - 1 - - > 3 ; 1 - 4 - - > 5 ; 4 - 1 - - > 5 ; 2 - 3 - - > 5 ; 3 - 2 - - > 5 ; 1 - 6 - - > 7 ; 6 - 1 - - > 7 ; 2 - 5 - - > 7 ; 5 - 2 - - > 7 ; 3 - 4 - - > 7 ; 4 - 3 - - > 7 ; 6 - 5 - - > 11 ; 5 - 6 - - > 11 . total of 15 favorable outcomes t = 15 / 36 . answer : c ." | a ) 0 , b ) 1 / 12 , c ) 5 / 12 , d ) 7 / 18 , e ) 4 / 9 | c | multiply(add(const_12, const_3), power(divide(1, 6), const_2)) | add(const_12,const_3)|divide(n0,n1)|power(#1,const_2)|multiply(#0,#2)| | general | C |
a man sitting in a train which is travelling at 50 kmph observes that a goods train , travelling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . | "relative speed = ( 280 β 9 ) m / sec = ( 280 β 9 Γ 18 β 5 ) kmph = 112 kmph . β΄ speed of goods train = ( 112 β 50 ) kmph = 62 kmph . answer a" | a ) 62 kmph , b ) 58 kmph , c ) 52 kmph , d ) 50 kmph , e ) none of these | a | subtract(multiply(divide(280, 9), const_3_6), 50) | divide(n2,n1)|multiply(#0,const_3_6)|subtract(#1,n0)| | physics | A |
if paint costs $ 3.20 per quart , and a quart covers 120 square feet , how much will it cost to paint the outside of a cube 10 feet on each edge ? | "total surface area = 6 a ^ 2 = 6 * 10 * 10 = 600 each quart covers 20 sqr ft thus total number of quarts = 600 / 120 = 5 cost will be 5 * 3.2 = $ 16 ans : b" | a ) $ 1.60 , b ) $ 16.00 , c ) $ 96.00 , d ) $ 108.00 , e ) $ 196.00 | b | multiply(divide(3.20, 120), surface_cube(10)) | divide(n0,n1)|surface_cube(n2)|multiply(#0,#1)| | geometry | B |
by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 600 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 600 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 600 ) / 100 + ( 1.8 * 700 ) / 100 = 17.4 gms % of liquid x in resultant mixture = ( 17.4 / 1000 ) * 100 = 1.74 % a" | a ) 1.74 % , b ) 1.94 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(600, 0.8), multiply(700, 1.8)), const_1000) | multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,const_1000)| | gain | A |
when the positive integer x is divided by 9 , the remainder is 5 . what is the remainder when 7 x is divided by 9 ? | i tried plugging in numbers x = 9 q + 5 x = 14 7 x = 98 7 x / 9 = 9 * 10 + 8 remainder is 8 answer is e . | a ) 0 , b ) 1 , c ) 3 , d ) 4 , e ) 8 | e | reminder(multiply(5, 7), 9) | multiply(n1,n2)|reminder(#0,n0) | general | E |
two trains of equal length are running on parallel lines in the same directions at 46 km / hr . and 36 km / hr . the faster trains pass the slower train in 36 seconds . the length of each train is : | "explanation : the relative speed of train is 46 - 36 = 10 km / hr = ( 10 x 5 ) / 18 = 25 / 9 m / s 10 Γ 518 = 259 m / s in 36 secs the total distance travelled is 36 x 25 / 9 = 100 m . therefore the length of each train is = 100 / 2 = 50 m . answer b" | a ) 82 m , b ) 50 m , c ) 72 m , d ) 80 m , e ) none of these | b | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 36), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general | B |
if the cost price of 50 articles is equal to the selling price of 15 articles , then the gain or loss percent is ? | "percentage of profit = 35 / 15 * 100 = 233 % answer : e" | a ) 16 , b ) 127 , c ) 12 , d ) 18 , e ) 233 | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 15), 50)), divide(multiply(const_100, 15), 50))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain | E |
we define that k @ j is the product of j number from k in increasing order for positive integers k , j . for example , 6 @ 4 = 6 * 7 * 8 * 9 . if a = 2020 and b = 2120 , what is the value w of the ratio a / b ? | "w - > a / b = 20 * 21 * β¦ β¦ * 39 / 21 * 22 * β¦ . * 39 * 40 = 20 / 40 = 1 / 2 . therefore , the answer is a ." | a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(2020, 2020), add(divide(2020, 2020), divide(2020, 2020))) | divide(n6,n6)|add(#0,#0)|divide(#0,#1)| | general | A |
a train 300 m long can cross an electric pole in 10 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 300 / 10 s = 30 m / sec speed = 30 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 108 kmph answer : e" | a ) 88 kmph , b ) 89 kmph , c ) 72 kmph , d ) 16 kmph , e ) 108 kmph | e | divide(divide(300, const_1000), divide(10, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics | E |
800 is increased by 110 % . find the final number . | explanation final number = initial number + 110 % ( original number ) = 800 + 110 % ( 800 ) = 800 + 880 = 1680 . answer e | a ) 1200 , b ) 1210 , c ) 1180 , d ) 1190 , e ) 1680 | e | add(800, multiply(800, divide(110, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain | E |
a rectangular courtyard , the sides of which are in the ratio of 4 : 3 , cost rs . 600 for paving at 50 p per m 2 ; find the length of the diagonal of the courtyard ? | 1 m 2 - - - - 1 / 2 ? - - - - - 600 = > 1200 m 2 4 x * 3 x = 1200 = > x = 10 answer : b | ['a ) 17 m', 'b ) 45 m', 'c ) 54 m', 'd ) 77 m', 'e ) 34 m'] | b | subtract(sqrt(add(power(multiply(4, sqrt(divide(divide(600, divide(50, const_100)), multiply(4, 3)))), 2), power(multiply(3, sqrt(divide(divide(600, divide(50, const_100)), multiply(4, 3)))), 2))), add(const_2, const_3)) | add(const_2,const_3)|divide(n3,const_100)|multiply(n0,n1)|divide(n2,#1)|divide(#3,#2)|sqrt(#4)|multiply(n0,#5)|multiply(n1,#5)|power(#6,n4)|power(#7,n4)|add(#8,#9)|sqrt(#10)|subtract(#11,#0) | geometry | B |
if n = 8 ^ 8 β 5 , what is the units digit of n ? | "8 ^ 8 - 8 = 8 ( 8 ^ 7 - 1 ) = = > 8 ( 2 ^ 21 - 1 ) last digit of 2 ^ 21 is 2 based on what explanation livestronger is saying . 2 ^ 24 - 1 yields 2 - 1 = 1 as the unit digit . now on multiply this with 5 , we get unit digit as 5 answer : e" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | e | divide(log(5), log(power(8, 8))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general | E |
in a weight - lifting competition , the total weight of joe ' s two lifts was 600 pounds . if twice the weight of his first lift was 300 pounds more than the weight of his second lift , what was the weight , in pounds , of his first lift ? | "this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was 600 pounds . we sum the two variables to obtain : f + s = 600 we are also given that twice the weight of his first lift was 300 pounds more than the weight of his second lift . we express this as : 2 f = 300 + s 2 f β 300 = s we can now plug in ( 2 f β 300 ) for s into the first equation , so we have : f + 2 f β 300 = 600 3 f = 900 f = 300 answer is c" | a ) 225 , b ) 275 , c ) 300 , d ) 350 , e ) 400 | c | divide(add(600, 300), const_3) | add(n0,n1)|divide(#0,const_3)| | general | C |
a boat moves down stream at the rate of 1 km in 6 minutes and upstream at the rate of 1 km in 10 minutes . the speed of current is | if speed ( in kmph ) of the boat = b and current = c , then in downstream time taken ( in hrs . ) = 1 / ( b + c ) = 6 / 60 = 1 / 10 or b + c = 10 - - - ( i ) & in upstream time taken ( in hrs . ) = 1 / ( b - c ) = 10 / 60 = 1 / 6 or b - c = 6 - - - ( ii ) from ( i ) & ( ii ) , b = 8 , c = 2 answer : a | a ) 2 kmph , b ) 3 kmph , c ) 4 kmph , d ) 5 kmph , e ) 6 kmph | a | divide(subtract(divide(1, divide(6, const_60)), divide(1, divide(10, const_60))), const_2) | divide(n1,const_60)|divide(n3,const_60)|divide(n0,#0)|divide(n0,#1)|subtract(#2,#3)|divide(#4,const_2) | physics | A |
sandy is younger than molly by 14 years . if the ratio of their ages is 7 : 9 , how old is sandy ? | "let sandy ' s age be 7 x and let molly ' s age be 9 x . 9 x - 7 x = 14 x = 7 sandy is 49 years old . the answer is b ." | a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70 | b | multiply(divide(7, subtract(9, 7)), 14) | subtract(n2,n1)|divide(n1,#0)|multiply(n0,#1)| | other | B |
the sector of a circle has radius of 21 cm and central angle 144 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 144 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 52.8 + 42 = 94.8 cm answer : d" | a ) 80 cm , b ) 82 cm , c ) 90 cm , d ) 94.8 cm , e ) 95 cm | d | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics | D |
the probability that a man will be alive for 10 more yrs is 3 / 4 & the probability that his wife will alive for 10 more yrs is 1 / 5 . the probability that none of them will be alive for 10 more yrs , is | "sol . required probability = pg . ) x p ( b ) = ( 1 β d x ( 1 β i ) = : x 1 = 1 / 5 ans . ( d )" | a ) 1 / 2 , b ) 1 , c ) 2 / 3 , d ) 1 / 5 , e ) 2 | d | multiply(subtract(1, divide(3, 4)), subtract(3, divide(3, 5))) | divide(n4,n2)|divide(n1,n5)|subtract(n1,#0)|subtract(n1,#1)|multiply(#2,#3)| | general | D |
if a certain number x is divided by 82 , the reminder is 5 . what is the reminder when x + 17 is divided by 41 ? | "x can be written as 82 k + 5 or x = 5 , 87,169 , etc . x + 17 = 82 k + 5 + 17 = 82 k + 22 or x + 17 = 22,104 , 186 etc . when divided by 41 , we will get the remainder 22 . d" | a ) 3 , b ) 5 , c ) 6 , d ) 22 , e ) 18 | d | add(5, 17) | add(n1,n2)| | general | D |
a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6000 ? | "explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6000 * 6 ) β 30,130 ] = rs . ( 36000 β 30,130 ) = rs . 5870 answer a" | a ) rs . 5870 , b ) rs . 5991 , c ) rs . 6020 , d ) rs . 6850 , e ) none of these | a | subtract(multiply(add(5, const_1), 6000), add(add(add(add(5420, 5660), 6200), 6350), 6500)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general | A |
tough and tricky questions : word problems . a salesman ' s income consists of commission and base salary . his weekly income totals over the past 5 weeks have been $ 406 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) income over the next 5 weeks be to increase his average weekly income to $ 500 over the 10 - week period ? | official solution : ( b ) first , we need to add up the wages over the past 5 weeks : $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070 . to average $ 500 over 10 weeks , the salesman would need to earn : $ 500 Γ 10 = $ 5000 . subtract $ 2070 from $ 5000 to determine how much he would need to earn , in total , over the next 5 weeks to average $ 500 for the 10 weeks : $ 5000 β $ 2070 = $ 2930 . dividing $ 2930 by 5 will give us the amount he needs to earn on average over the next 5 weeks : $ 2930 / 5 = $ 586 . the correct answer is choice ( b ) . | a ) $ 570 , b ) $ 586 , c ) $ 630 , d ) $ 686 , e ) $ 715 | b | add(500, subtract(500, divide(add(add(add(add(406, 413), 420), 436), 395), 5))) | add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|divide(#3,n0)|subtract(n7,#4)|add(n7,#5) | general | B |
the average of 6 observations is 16 . a new observation is included and the new average is decreased by 1 . the seventh observation is ? | "let seventh observation = x . then , according to the question we have = > ( 96 + x ) / 7 = 15 = > x = 9 hence , the seventh observation is 9 . answer : c" | a ) 1 , b ) 3 , c ) 9 , d ) 6 , e ) 7 | c | subtract(multiply(subtract(16, 1), add(6, 1)), multiply(16, 6)) | add(n0,n2)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#2)|subtract(#3,#1)| | general | C |
solve the given equation 9 ^ y = 3 ^ 12 , what is y ? | 3 ^ 2 y = 3 ^ 12 2 y = 12 , therefore y = 6 e | a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(12, divide(log(9), log(3))) | log(n0)|log(n1)|divide(#0,#1)|divide(n2,#2) | general | E |
the triplicate ratio of 1 : 4 is ? | "1 ^ 3 : 4 ^ 3 = 1 : 64 answer : d" | a ) 1 : 7 , b ) 1 : 8 , c ) 1 : 3 , d ) 1 : 64 , e ) 1 : 2 | d | divide(power(1, 4), power(4, 4)) | power(n0,n1)|power(n1,n1)|divide(#0,#1)| | other | D |
a and b can do a work in 8 days , b and c can do the same work in 12 days . a , b and c together can finish it in 6 days . a and c together will do it in ? | ( a + b + c ) ' s 1 day work = 1 / 6 ; ( a + b ) ' s 1 day work = 1 / 8 ( b + c ) ' s 1 day work = 1 / 12 ( a + c ) ' s 1 day work = ( 2 * 1 / 6 ) - ( 1 / 8 + 1 / 12 ) = ( 1 / 3 - 5 / 24 ) = 1 / 8 so , a and c together will do the work in 8 days . answer a | a ) 8 , b ) 4 , c ) 6 , d ) 12 , e ) 10 | a | divide(multiply(8, add(divide(8, const_2), const_2)), 6) | divide(n0,const_2)|add(#0,const_2)|multiply(n0,#1)|divide(#2,n2) | physics | A |
if the price of a certain computer increased 30 percent from d dollars to 377 dollars , then 2 d = | "before price increase price = d after 30 % price increase price = d + ( 30 / 100 ) * d = 1.3 d = 377 ( given ) i . e . d = 377 / 1.3 = $ 290 i . e . 2 d = 2 * 290 = 580 answer : option e" | a ) 540 , b ) 570 , c ) 619 , d ) 649 , e ) 580 | e | multiply(divide(377, divide(add(const_100, 30), const_100)), 2) | add(n0,const_100)|divide(#0,const_100)|divide(n1,#1)|multiply(n2,#2)| | general | E |
10 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 2 women and 10 children take to complete the work ? | "1 women ' s 1 day work = 1 / 70 1 child ' s 1 day work = 1 / 140 ( 2 women + 10 children ) ' s 1 day work = ( 2 / 70 + 10 / 140 ) = 1 / 10 2 women and 10 children will complete the work in 10 days . e" | a ) 4 , b ) 5 , c ) 7 , d ) 8 , e ) 10 | e | inverse(add(divide(2, multiply(10, 7)), divide(10, multiply(10, 14)))) | multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)| | physics | E |
how many positive integers less than 50 have a reminder 01 when divided by 5 ? | "1 also gives the remainder of 1 when divided by 5 . so , there are total of 10 numbers . answer : a ." | a ) 10 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | a | divide(factorial(subtract(add(const_4, 01), const_1)), multiply(factorial(01), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general | A |
two numbers are 30 % and 37 % are less than a third number . how much percent is the second number less than the first ? | i ii iii 70 63 100 70 - - - - - - - - 7 100 - - - - - - ? = > 10 % answer : c | a ) 19 % , b ) 80 % , c ) 10 % , d ) 90 % , e ) 15 % | c | multiply(divide(subtract(subtract(const_100, 30), subtract(const_100, 37)), subtract(const_100, 30)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100) | gain | C |
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