Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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if a car went the first third of the distance at 60 kmh , the second third at 24 kmh , and the last third at 48 kmh , what was the average speed of the car for the entire trip ? | "assume d / 3 = 240 ( this number is convenient because it is divisible by 60 , 24 and 48 ) so : 240 = 60 * t 1 = 4 hrs 240 = 24 * t 2 = 10 hrs 240 = 48 * t 3 = 5 hrs t = t 1 + t 2 + t 3 = 19 hrs d = rt ( 240 * 3 ) = r * 19 r = 38 answer : a" | a ) 38 kmh , b ) 40 kmh , c ) 42 kmh , d ) 44 kmh , e ) 50 kmh | a | divide(const_3, add(add(divide(const_1, 60), divide(const_1, 24)), divide(const_1, 48))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_3,#4)| | physics |
a reduction of 10 % in the price of oil enables a house wife to obtain 6 kgs more for rs . 900 , what is the reduced price for kg ? | "900 * ( 10 / 100 ) = 90 - - - - 6 ? - - - - 1 = > rs . 15 answer : b" | a ) 13 , b ) 15 , c ) 25 , d ) 17 , e ) 19 | b | divide(divide(multiply(900, 10), const_100), 6) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain |
the triplicate ratio of 3 : 5 is ? | "3 ^ 3 : 5 ^ 3 = 27 : 125 answer : a" | a ) 27 : 125 , b ) 1 : 62 , c ) 1 : 34 , d ) 3 : 64 , e ) 5 : 64 | a | divide(power(const_2.0, 5), power(5, 5)) | power(const_2.0,n1)|power(n1,n1)|divide(#0,#1)| | other |
a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 200 when the rate to profit rises from 5 to 7 percent . find the capital of a ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 200 a capital = 15000 * 2 / 3 = 10000 answer : e" | a ) 3377 , b ) 2899 , c ) 2500 , d ) 2778 , e ) 10000 | e | divide(multiply(200, const_100), 2) | multiply(n2,const_100)|divide(#0,n0)| | general |
find out the square of a number which when doubled exceeds its quarter by 7 ? | "a 16 explanation : let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 4 ) + 7 = > 8 p = p + 28 = > p = 4 p ^ 2 = 4 ^ 2 = 16 ." | a ) 16 , b ) 11 , c ) 31 , d ) 29 , e ) 19 | a | power(divide(7, 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 |
a sports equipment store sold ping pong rackets for a total of $ 490 . if the average ( arithmetic mean ) price of a pair of rackets is $ 9.8 , how many pairs were sold ? | "average price for a pair of rackets = $ 9.8 total cost = $ 9.8 * x = $ 490 x = 50 pairs were sold . answer : a" | a ) 50 , b ) 100 , c ) 490 , d ) 500 , e ) 980 | a | divide(490, 9.8) | divide(n0,n1)| | general |
a little girl went to the farm to buy fruits . she bought a bag of apples and grabbed a bag of oranges which was 5 times as heavy as the apples . she brought home a total of 12 pounds of fruits . how many pounds was the oranges ? | pounds of apples = x pounds of oranges = 5 x x + 5 x = 12 6 x = 12 x = 12 / 6 = 2 oranges = 5 x = 5 * 2 = 10 answer : d | a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 12 | d | divide(multiply(5, 12), add(5, const_1)) | add(n0,const_1)|multiply(n0,n1)|divide(#1,#0) | general |
a batsman scored 120 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 110 - ( 3 x 4 + 8 x 6 ) = 120 - ( 60 ) = 60 now , we need to calculate 60 is what percent of 120 . = > 60 / 120 * 100 = 50 % answer is b" | a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 75 % | b | multiply(divide(subtract(120, add(multiply(3, 8), multiply(8, 3))), 120), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
if 7 < x < 9 < y < 15 , then what is the greatest possible positive integer difference of x and y ? | "let x = 7.1 and y = 14.1 greatest possible difference = 14.1 - 7.1 = 7 answer a" | a ) 7 , b ) 4 , c ) 5 , d ) 6 , e ) 3 | a | subtract(subtract(15, 7), const_1) | subtract(n2,n0)|subtract(#0,const_1)| | general |
on the number line , the number p is twice as many units from - 2 as - 2 is from 2 . if p is less than β 2 , what is the value of p ? | distance between - 2 and 2 = 4 since number p is twice as many units from - 2 as - 2 is from 2 , therefore p can be - 10 since , p is less than - 2 , the value of p = - 10 answer b | a ) β 18 , b ) β 10 , c ) β 6 , d ) 10 , e ) 14 | b | subtract(negate(2), multiply(const_2, subtract(2, negate(2)))) | negate(n0)|subtract(n0,#0)|multiply(#1,const_2)|subtract(#0,#2) | general |
when 100 is divided by positive integer x , the remainder is 3 . what is the remainder when 197 is divided by x ? | if 100 / x leaves a reminder 3 then ( 100 - 3 ) i . e . 97 is divisible by x so ( 100 + 97 ) / x leaves a reminder rem ( 100 / x ) + rem ( 97 / x ) = > 3 + 0 = 3 answer : b | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | b | subtract(100, subtract(197, 100)) | subtract(n2,n0)|subtract(n0,#0) | general |
if 4 ^ k + 4 ^ k = ( 4 ^ 9 ) ^ ( 4 ^ 9 ) - 4 ^ k , then k = ? | 4 ^ k + 4 ^ k = ( 4 ^ 9 ) ^ 4 ^ 9 - 4 ^ k 4 * ( 4 ^ k ) = 4 ^ ( 16 * 4 ^ 9 ) = 4 ^ ( 4 ^ 2 * 4 ^ 9 ) = 4 ( 4 ^ 11 ) 4 ^ k + 1 = 4 ^ ( 4 ^ 11 ) so k + 1 = 4 ^ 11 so k = 4 ^ 11 - 1 answer is b | a ) 11 / 3 , b ) 4 ^ 11 - 1 , c ) 242 , d ) 3 ^ 10 , e ) 3 ^ 11 - 1 | b | subtract(power(4, add(9, const_2)), const_1) | add(n3,const_2)|power(n0,#0)|subtract(#1,const_1) | general |
if n is a positive integer and n ^ 2 is divisible by 212 , then the largest positive integer that must divide n is | "the question asks aboutthe largest positive integer that must divide n , not could divide n . since the least value of n for which n ^ 2 is a multiple of 72 is 12 then the largest positive integer that must divide n is 12 . complete solution of this question is given above . please ask if anything remains unclear . i ... | a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | d | multiply(sqrt(divide(212, 2)), 2) | divide(n1,n0)|sqrt(#0)|multiply(n0,#1)| | general |
a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 5.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs at ... | "the amount by which employment costs rose is equal to 0.035 ( salary costs + fringe benefit costs ) ; on the other hand the amount by which employment costs rose is equal to 0.03 * salary costs + 0.055 * fringe benefit costs ; so , 35 ( s + f ) = 30 s + 55 f - - > s = 4 f - - > f / s = 1 / 4 - - > f / ( s + f ) = 1 / ... | a ) 16.5 % , b ) 20 % , c ) 35 % , d ) 55 % , e ) 65 % | b | multiply(divide(subtract(3.5, 3), subtract(5.5, 3)), const_100) | subtract(n1,n2)|subtract(n3,n2)|divide(#0,#1)|multiply(#2,const_100)| | gain |
a man performs 1 / 2 of the total journey by rail , 1 / 3 by bus and the remaining 5 km on foot . his total journey is | "explanation : let the journey be x km then , 1 x / 2 + 1 x / 3 + 5 = x 5 x + 30 = 6 x x = 30 km answer : option d" | a ) 16 km , b ) 10 km , c ) 12 km , d ) 30 km , e ) 25 km | d | multiply(3, 5) | multiply(n3,n4)| | general |
the sum of two consecutive even integers is 26 . what are the two numbers ? | "let 2 n be the first even integer and let 2 n + 2 be the second integer 2 n + 2 n + 2 = 26 4 n + 2 = 26 4 n + 2 - 2 = 26 - 2 4 n = 24 n = 6 so the first even integer is 2 n = 2 Γ 6 = 12 and the second is 12 + 2 = 14 correct answer is a ) 12 - 14" | a ) 12 - 14 , b ) 13 - 13 , c ) 10 - 16 , d ) 9 - 17 , e ) 8 - 18 | a | add(add(power(add(add(divide(subtract(subtract(26, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(26, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(26, const_10), const_2), const_4), const_2), power(add... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
total 75 cows 25 cow gives each 2 liter milk 25 cow gives each 3 / 4 liter milk 25 cow gives each 1 / 4 liter milk this is split into 3 son per each 25 cows & 25 liter milk how ? | "25 cow 2 liter each = 50 liter 25 cow 3 / 4 liter each = 3 / 4 = 0.75 * 25 = 18.75 25 cow 1 / 4 liter each = 1 / 4 = 0.25 * 25 = 6.25 add 50 + 18.75 + 6.25 = 75 milk split into 3 son each 25 liter then 75 / 3 = 25 answer : d" | a ) 10 , b ) 12 , c ) 15 , d ) 25 , e ) 18 | d | add(add(divide(3, 4), multiply(divide(3, 4), 25)), multiply(const_0_25, 25)) | divide(n4,n5)|multiply(n1,const_0_25)|multiply(n1,#0)|add(#0,#2)|add(#3,#1)| | general |
if the wheel is 14 cm then the number of revolutions to cover a distance of 968 cm is ? | "2 * 22 / 7 * 14 * x = 968 = > x = 11 answer : e" | a ) a ) 15 , b ) b ) 10 , c ) c ) 14 , d ) d ) 12 , e ) e ) 11 | e | divide(968, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)| | physics |
two trains 210 m and 260 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 210 + 260 = 470 m . required time = 470 * 9 / 250 = 16.92 = 16.92 sec . answer : a" | a ) 16.92 sec , b ) 14.8 sec , c ) 10.68 sec , d ) 10.8 sec , e ) 20.8 sec | a | divide(add(210, 260), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
a certain sum is invested at simple interest at 15 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 900 . find the sum ? | "let the sum be rs . x . ( x * 15 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 900 = > 30 x / 100 - 24 x / 100 = 900 = > 6 x / 100 = 900 = > x = 15000 . answer : c" | a ) s . 7000 , b ) s . 9000 , c ) s . 15000 , d ) s . 17000 , e ) s . 27000 | c | divide(900, divide(multiply(subtract(15, 12), const_2), const_100)) | subtract(n0,n1)|multiply(#0,const_2)|divide(#1,const_100)|divide(n2,#2)| | gain |
a batsman scored 125 runs which included 5 boundaries and 5 sixes . what percent of his total score did he make by running between the wickets ? | "explanation : number of runs made by running , = > 125 β ( 5 Γ 4 + 5 Γ 6 ) . = > 125 - 50 = > 75 hence , the required percentage is : - = > 75 / 125 * 100 = > 60 % answer : e" | a ) 45 % , b ) 500 / 11 , c ) 600 / 11 , d ) 55 % , e ) 60 % | e | multiply(divide(subtract(125, add(multiply(5, 5), multiply(5, 5))), 125), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
a certain basketball team that has played 2 / 3 of its games has a record of 15 wins and 5 losses . what is the greatest number of the remaining games that the team can lose and still win at least 3 / 4 of all of its games ? | "15 wins , 3 losses - total 20 games played . the team has played 2 / 3 rd of all games so total number of games = 30 3 / 4 th of 30 is 22.5 so the team must win 23 games and can afford to lose at most 7 total games . it has already lost 5 games so it can lose another 2 at most . answer ( e )" | a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 2 | e | subtract(multiply(divide(3, 2), add(15, 5)), add(floor(multiply(divide(3, 4), multiply(divide(3, 2), add(15, 5)))), const_1)) | add(n2,n3)|divide(n1,n0)|divide(n1,n5)|multiply(#0,#1)|multiply(#2,#3)|floor(#4)|add(#5,const_1)|subtract(#3,#6)| | general |
on selling 13 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is : | "( c . p . of 13 balls ) - ( s . p . of 13 balls ) = ( c . p . of 5 balls ) c . p . of 8 balls = s . p . of 13 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 8 = rs . 90 . answer : option e" | a ) s . 45 , b ) s . 50 , c ) s . 55 , d ) s . 60 , e ) s . 90 | e | divide(720, subtract(13, 5)) | subtract(n0,n2)|divide(n1,#0)| | gain |
3 friends neha , sabi and mahi divide $ 1100 amongs them in such a way that if $ 5 , $ 8 and $ 4 are removed from the sums that neha , sabi and mahi received respectively , then the share of the sums that they got will be in the ratio of 2 : 8 : 6 . how much did mahi receive ? | explanatory answer let the sums of money received by a , b and c be x , y and z respectively . then x - 5 : y - 8 : z - 4 is 2 a : 8 a : 6 a when $ 5 , $ 8 and $ 4 are removed , we are removing a total of $ 17 from $ 1100 . therefore , 2 a + 8 a + 6 a = 1100 - 17 = 1093 i . e . , 16 a = 1093 or a = 68 . we know that z ... | a ) $ 122 , b ) $ 152 , c ) $ 202 , d ) $ 107 , e ) $ 102 | e | add(add(multiply(add(8, const_12), 4), const_12), const_10) | add(n3,const_12)|multiply(n4,#0)|add(#1,const_12)|add(#2,const_10) | general |
the maximum number of girls among them 2923 bags and 3239 eyeliners can be distributed in such a way that each student gets the same number of bags and eyeliners . | max no of girls = hcf of 2923 and 3239 = 79 answer : b | a ) 80 , b ) 79 , c ) 78 , d ) 81 , e ) 82 | b | gcd(2923, 3239) | gcd(n0,n1) | general |
for any integer n greater than 1 , n * denotes the product of all the integers from 1 to n , inclusive . how many prime numbers w are there between 6 * + 2 and 6 * + 6 , inclusive ? | "given that n * denotes the product of all the integers from 1 to n , inclusive so , 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 . now , notice that we can factor out 2 our of 6 ! + 2 so it can not be a prime number , we can factor out 3 our of 6 ! + 3 so it can not be a prime number , we can factor out 4 our of 6 ! + 4 so... | a ) none , b ) one , c ) two , d ) three , e ) four | a | divide(add(factorial(6), 6), add(factorial(6), 6)) | factorial(n2)|add(n2,#0)|divide(#1,#1)| | general |
in may , the groundskeeper at spring lake golf club built a circular green with an area of 95 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 ( 95 Ο ) = 380 Ο the answer is e . | ['a ) 190 Ο', 'b ) 210 Ο', 'c ) 240 Ο', 'd ) 320 Ο', 'e ) 380 Ο'] | e | multiply(power(multiply(sqrt(95), const_2), const_2), const_pi) | sqrt(n0)|multiply(#0,const_2)|power(#1,const_2)|multiply(#2,const_pi) | geometry |
i have a huge bed - sheet of 1 meter long . i want to cuts this role into pieces of 20 centimeter long cloth . how long will it take me to cut the bed sheet if each 20 centimeter cut took me 5 minutes ? | b it will take 49 cuts not 50 . so 49 * 5 = > 245 minutes | a ) 150 min , b ) 245 min , c ) 185 min , d ) 300 min , e ) 205 min | b | multiply(subtract(add(add(20, 20), const_10), const_1), 5) | add(n1,n1)|add(#0,const_10)|subtract(#1,const_1)|multiply(n3,#2) | physics |
30 men can do a work in 40 days . when should 14 men leave the work so that the entire work is completed in 40 days after they leave the work ? | "total work to be done = 30 * 40 = 1200 let 14 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 14 * 40 ) = 1200 40 p = 640 = > p = 16 days answer : b" | a ) 87 days , b ) 16 days , c ) 55 days , d ) 44 days , e ) 22 days | b | divide(subtract(multiply(30, 40), multiply(40, 14)), 40) | multiply(n0,n1)|multiply(n1,n2)|subtract(#0,#1)|divide(#2,n1)| | physics |
if the cost price of 50 articles is equal to the selling price of 32 articles , then the gain or loss percent is ? | "given that , cost price of 50 article is equal to selling price of 32 articles . let cost price of one article = rs . 1 selling price of 32 articles = rs . 50 but cost price of 32 articles = rs . 32 therefore , the trader made profit . \ percentage of profit = 18 / 32 * 100 = 56.25 % answer : b" | a ) 22 , b ) 56.25 , c ) 78 , d ) 33 , e ) 25 | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 32), 50)), divide(multiply(const_100, 32), 50))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
on dividing 23 by a number , the quotient is 7 and the remainder is 2 . find the divisor . | "d = ( d - r ) / q = ( 23 - 2 ) / 7 = 21 / 7 = 3 c" | a ) 1 , b ) 2 , c ) 3 , d ) 6 , e ) 7 | c | floor(divide(23, 7)) | divide(n0,n1)|floor(#0)| | general |
what is the value of 3 x ^ 2 β 1.8 x + 0.3 for x = 0.3 ? | "3 x ^ 2 - 1.8 x + 0.3 for x = 0.6 = 3 ( 0.3 * 0.3 ) - 6 * 0.3 * ( 0.3 ) + 0.3 = - 3 * 0.3 * 0.3 + 0.3 = - 0.27 + 0.3 = 0.03 correct option : c" | a ) β 0.3 , b ) 0 , c ) 0.03 , d ) 1.08 , e ) 2.46 | c | subtract(multiply(divide(divide(subtract(power(3, 2), power(1.8, 0.3)), const_1000), const_1000), 3), divide(divide(subtract(power(3, 2), power(1.8, 0.3)), const_1000), const_1000)) | power(n0,n1)|power(n2,n3)|subtract(#0,#1)|divide(#2,const_1000)|divide(#3,const_1000)|multiply(n0,#4)|subtract(#5,#4)| | general |
the average of temperatures at noontime from monday to friday is 50 ; the lowest one is 42 , what is the possible maximum range of the temperatures ? | "there are 5 days so the sum of temperature can be 50 * 5 = 250 lowest is 42 . to find the maximum range we can say the temperature was the lowest for 4 of the 5 days so 4 * 42 = 168 . on the fifth day it is 250 - 168 = 281 range is therefore 128 - 42 = 86 answer e" | a ) 20 , b ) 25 , c ) 40 , d ) 45 , e ) 86 | e | subtract(subtract(multiply(50, add(const_2, const_3)), multiply(42, const_4)), 42) | add(const_2,const_3)|multiply(n1,const_4)|multiply(n0,#0)|subtract(#2,#1)|subtract(#3,n1)| | general |
a , b and c enter into a partnership and theirs shares are in the ratio 1 / 2 : 1 / 3 : 1 / 4 . after two months , a withdraws half of his capital and after 10 months , a profit of rs . 378 is divided among them . what is b β s share ? | ratio of initial investments = 1 / 2 : 1 / 3 : 1 / 4 = 6 : 4 : 3 . let their initial investments be 6 x , 2 x and 3 x respectively . a : b : c = ( 6 x * 2 + 3 x * 10 ) : ( 4 x * 12 ) : ( 3 x * 12 ) = 42 : 48 : 36 = 7 : 8 : 6 . therefore , b β s share = rs . 378 * 8 / 21 = rs . 144 . answer : b | a ) rs . 129 , b ) rs . 144 , c ) rs . 156 , d ) rs . 168 , e ) rs . 178 | b | divide(multiply(378, multiply(2, 4)), multiply(add(3, 4), 3)) | add(n3,n5)|multiply(n1,n5)|multiply(n7,#1)|multiply(n3,#0)|divide(#2,#3) | general |
on a scale that measures the intensity of a certain phenomenon , a reading of d + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of d . on that scale , the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3 ? | to solve this problem we need to examine the information in the first sentence . we are told that β a reading of d + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of d . β let β s practice this idea with some real numbers . let β s say d is 2 . this means that d + 1 = 3 . with ... | a ) 5 , b ) 50 , c ) 10 ^ 5 , d ) 5 ^ 10 , e ) 8 ^ 10 - 3 ^ 10 | c | power(10, subtract(8, 3)) | subtract(n2,n3)|power(n1,#0) | general |
if 6 x ^ 2 + x - 12 = ( hx + b ) ( cx + d ) , then | h | + | b | + | c | + | d | = for a complete solution and more practice problems , see this blog : http : / / magoosh . com / gmat / 2012 / algebra - on . . . to - factor / | 6 x ^ 2 + x - 12 = 6 x ^ 2 + 9 x - 8 x - 12 = > 3 x ( 2 x + 3 ) - 4 ( 2 x + 3 ) = > ( 2 x + 3 ) ( 3 x - 4 ) = ( hx + b ) ( cx + d ) hence h = 2 , b = c = 3 , d = - 4 so , 2 + 3 + 3 + | - 4 | = 2 + 3 + 3 + 4 = 12 answer b . | a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | b | add(add(add(2, const_3), const_3), const_4) | add(n1,const_3)|add(#0,const_3)|add(#1,const_4) | general |
4 men and 6 women finish a job in 8 days , while 3 men and 7 women finish it in 10 days . in how many days will 10 women working together finish it ? | explanation : let 1 man ' s 1 day work = x and 1 woman ' s 1 days work = y . then , 4 x + 6 y = 1 / 8 and 3 x + 7 y = 1 / 10 solving , we get y = 1 / 400 [ means work done by a woman in 1 day ] 10 women 1 day work = 10 / 400 = 1 / 40 10 women will finish the work in 40 days option b | a ) 30 days , b ) 40 days , c ) 50 days , d ) 60 days , e ) 70 days | b | divide(add(multiply(divide(subtract(multiply(10, 7), multiply(8, 6)), subtract(multiply(8, 4), multiply(10, 3))), multiply(8, 4)), multiply(8, 6)), 10) | multiply(n4,n5)|multiply(n1,n2)|multiply(n0,n2)|multiply(n3,n5)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(#6,#2)|add(#7,#1)|divide(#8,n5) | physics |
how many trailing zeros will be there after the rightmost non - zero digit in the value of 25 ! ( factorial 25 ) ? | explanatory answer 25 ! means factorial 25 whose value = 25 * 24 * 23 * 22 * . . . . * 1 when a number that has 5 as its factor is multiplied by an even number , it results in a trailing zero . ( product of 5 and 2 is 10 and any number when multiplied with 10 or a power of 10 will have one or as many zeroes as the powe... | a ) 25 , b ) 8 , c ) 6 , d ) 5 , e ) 2 | c | add(sqrt(25), const_1) | sqrt(n0)|add(#0,const_1) | other |
alice leaves her home and walks toward bob ' s house . one hour later , bob leaves his home and walks toward alice ' s house . the distance between their homes is 41 kilometers , alice ' s walking speed is 5 km / h , and bob ' s walking speed is 4 km / h . how many kilometers will alice walk before she meets bob ? | alice walks 5 km in the first hour so there are 36 km remaining . when bob starts walking , they complete a total of 9 km per hour . they will meet four hours after bob starts walking . since alice walks for 5 hours , she walks 25 km . the answer is b . | a ) 22 , b ) 25 , c ) 27 , d ) 38 , e ) 30 | b | add(multiply(divide(subtract(41, 5), add(5, 4)), 5), 5) | add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n1,#2)|add(n1,#3) | physics |
the average of 1 st 3 of 4 numbers is 6 and of the last 3 are 5 . if the sum of the first and the last number is 17 . what is the last numbers ? | "a + b + c = 18 b + c + d = 15 a + d = 17 a β d = 3 a + d = 17 2 d = 14 d = 7 answer : d" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | subtract(subtract(multiply(3, 6), add(subtract(17, 6), 3)), 6) | multiply(n1,n3)|subtract(n6,n3)|add(n1,#1)|subtract(#0,#2)|subtract(#3,n3)| | general |
a man can row 7 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ? | "m = 7 s = 1.2 ds = 8.2 us = 5.8 x / 8.2 + x / 5.8 = 1 x = 3.40 d = 3.40 * 2 = 6.79 answer : c" | a ) 6.24 km , b ) 6 km , c ) 6.79 km , d ) 5.66 km , e ) 10 km | c | multiply(divide(multiply(add(7, 1.2), subtract(7, 1.2)), add(add(7, 1.2), subtract(7, 1.2))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
how many prime numbers between 1 and 100 are factors of 67830 ? | "factor of 67830 = 2 * 3 * 5 * 7 * 17 * 19 - - - 6 prime numbers c" | a ) 4 , b ) 5 , c ) 6 , d ) 3 , e ) 2 | c | multiply(const_4, 1) | multiply(n0,const_4)| | other |
the speed of a car is 20 km in the first hour and 30 km in the second hour . what is the average speed of the car ? | "s = ( 20 + 30 ) / 2 = 25 kmph answer : a" | a ) 25 kmph , b ) 85 kmph , c ) 34 kmph , d ) 23 kmph , e ) 45 kmph | a | divide(add(20, 30), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
how many shots of 1 cm radius can be prepared from a sphere of 4 cm radius ? | "4 / 3 Γ― β¬ * 4 * 4 * 4 = 4 / 3 Γ― β¬ * 1 * 1 * 1 * x x = 64 answer : d" | a ) 33 , b ) 88 , c ) 27 , d ) 64 , e ) 99 | d | multiply(power(4, const_2), 4) | power(n1,const_2)|multiply(n1,#0)| | physics |
in a class , 7 students like to play basketball and 8 like to play cricket . 3 students like to play on both basketball and cricket . how many students like to play basketball or cricket or both ? | "draw a venn diagram yourself ! b + c - bc = number of students that play either basketball or cricket 7 + 8 - 3 = 12 a )" | a ) 12 , b ) 15 , c ) 16 , d ) 18 , e ) 22 | a | subtract(add(7, 8), 3) | add(n0,n1)|subtract(#0,n2)| | other |
p has $ 42 more than what q and r together would have had if both b and c had 1 / 8 of what p has . how much does p have ? | "p = ( 2 / 8 ) * p + 42 ( 6 / 8 ) * p = 42 p = 56 the answer is d ." | a ) $ 44 , b ) $ 48 , c ) $ 52 , d ) $ 56 , e ) $ 60 | d | divide(42, subtract(1, multiply(divide(1, 8), const_2))) | divide(n1,n2)|multiply(#0,const_2)|subtract(n1,#1)|divide(n0,#2)| | general |
one hour after yolanda started walking from x to y , a distance of 24 miles , bob started walking along the same road from y to x . if yolanda ' s walking rate was 3 miles per hour and bob Ρ ' s was 4 miles per hour , how many miles had bob walked when they met ? | "when b started walking y already has covered 3 miles out of 24 , hence the distance at that time between them was 24 - 3 = 21 miles . combined rate of b and y was 3 + 4 = 7 miles per hour , hence they would meet each other in 21 / 7 = 3 hours . in 6 hours b walked 3 * 4 = 12 miles . answer : e ." | a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 12 | e | multiply(divide(subtract(24, 3), add(3, 4)), 4) | add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n2,#2)| | physics |
the cost of an article is decreased by 10 % . if the original cost is $ 100 , find the decrease cost . | "original cost = $ 100 decrease in it = 10 % of $ 100 = 10 / 100 Γ£ β 100 = 1000 / 100 = $ 10 therefore , decrease cost = $ 100 - $ 10 = $ 90 answer : e" | a ) 33 , b ) 11 , c ) 68 , d ) 36 , e ) 90 | e | divide(multiply(100, 10), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain |
how long will a boy take to run round a square field of side 35 metres . if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 35 * 4 = 140 m time taken = 140 * 2 / 5 = 56 sec . answer is a" | a ) 56 sec , b ) 1 minute , c ) 45 sec , d ) 58 sec , e ) 50 sec | a | divide(multiply(35, const_4), multiply(9, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain |
a can do a piece of work in 20 days , b can do a piece of work in 12 days , both are working together for 3 days after a left the job , then how many days b requires to complete the remaining work ? | first take lcm of ( 20,12 ) = 60 ; total work = 60 ; a ' s one day work = 60 / 20 = 3 ; b ' s one day work = 60 / 12 = 5 ; one day work of both a + b = 3 + 5 = 8 ; total work completed in 3 days = 8 * 3 = 24 ; remaining work is = 60 - 24 = 36 ; remaining work completed by b in = 36 / 12 = 3 days . answer = c | a ) 6 days , b ) 5 days , c ) 3 days , d ) 7 days , e ) 12 days | c | floor(add(inverse(multiply(subtract(inverse(add(divide(const_1, 20), divide(const_1, 12))), 3), divide(const_1, 12))), const_1)) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|inverse(#2)|subtract(#3,n2)|multiply(#0,#4)|inverse(#5)|add(#6,const_1)|floor(#7) | physics |
the average ( arithmetic mean ) of the integers from 200 to 800 , inclusive , is how much greater than the average of the integers from 50 to 300 , inclusive ? | "for an ap the mean or average of series is average of first and last term . so , average of numbers between 200 to 800 , inclusive = ( 200 + 800 ) / 2 = 500 average of numbers between 50 to 300 , inclusive = ( 50 + 300 ) / 2 = 175 difference = 500 - 175 = 425 answer is d" | a ) 450 , b ) 375 , c ) 400 , d ) 425 , e ) 300 | d | subtract(divide(add(200, 800), const_2), divide(add(50, 300), const_2)) | add(n0,n1)|add(n2,n3)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)| | general |
the sum of the first 50 positive even integers is 2550 . what is the sum r of even integers from 102 to 200 inclusive ? | "my solution is : first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : r = 2550 + ( 100 * 50 ) = 7550 . b" | a ) 5100 , b ) 7550 , c ) 10100 , d ) 15500 , e ) 20100 | b | multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1)) | add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3)| | general |
bipin is 6 times old as alok . bipin ' s age will be twice of chandan ' s age after 10 years . if chandan ' s 7 th birthday was celebrated 3 years ago , what is alok ' s present age ? | let the present ages of bipin , alok , chandan are b , a , c respectively given conditions are 1 ) b = 6 a 2 ) ( b + 10 ) = 2 * ( c + 10 ) 3 ) chandans present age = c = 10 years therefore b = 30 so a = 5 answer : d | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | divide(add(multiply(10, const_2), 10), 6) | multiply(n1,const_2)|add(n1,#0)|divide(#1,n0) | general |
a big container is 30 % full with water . if 36 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container ? | "a big container is 30 % full with water and after 36 liters of water is added , the container becomes 75 % full . hence these 36 liters account for 45 % of the container , which means that the capacity of it is 36 / 0.45 = 80 liters . or : if the capacity of the container is x liters then : 0.3 x + 36 = 0.75 x - - > x... | a ) 36 liters , b ) 40 liters , c ) 45 liters , d ) 54 liters , e ) 80 liters | e | divide(36, subtract(divide(3, 4), divide(30, const_100))) | divide(n2,n3)|divide(n0,const_100)|subtract(#0,#1)|divide(n1,#2)| | general |
find the cost of fencing around a circular field of diameter 42 m at the rate of rs . 5 a meter ? | "2 * 22 / 7 * 14 = 131.95 131.95 * 5 = rs . 659.75 answer : b" | a ) 655.9 , b ) 659.75 , c ) 650.57 , d ) 655.9 , e ) 600.3 | b | multiply(circumface(divide(42, const_2)), 5) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics |
length of a rectangular plot is 60 mtr more than its breadth . if the cost of fencing the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 60 ) metres . perimeter = 5300 / 26.5 m = 200 m . 2 [ ( x + 60 ) + x ] = 200 2 x + 60 = 100 2 x = 40 x = 20 . hence , length = x + 60 = 80 m e" | a ) 46 m , b ) 60 m , c ) 58 m , d ) 78 m , e ) 80 m | e | divide(add(divide(5300, 26.50), multiply(const_2, 60)), const_4) | divide(n2,n1)|multiply(n0,const_2)|add(#0,#1)|divide(#2,const_4)| | physics |
exactly 36 % of the numbers in set a are even multiples of 3 . if 40 % of the even integers in set a are not multiples of 3 , what percent of the numbers in set a are not even integers ? | everything is correct except the red part with a typo : it should be n = 0.6 a - - > even numbers are 60 % of a - - > 40 % of a are not even integers . answer : d . | a ) 76 % , b ) 60 % , c ) 50 % , d ) 40 % , e ) 24 % | d | subtract(const_100, multiply(divide(divide(36, const_100), divide(subtract(const_100, 40), const_100)), const_100)) | divide(n0,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#0,#2)|multiply(#3,const_100)|subtract(const_100,#4) | gain |
the number of timeshare condos available at sunset beach is 3 / 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 320 , what is the difference between the number of condos available at sunset beach and the number of condos avai... | "let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 / 5 x we know , x + 3 / 5 x = 320 hence , x = 200 . so , number of timeshare condos available at playa del mar = 200 the difference between the number of condos available at sunset beach a... | a ) 80 , b ) 90 , c ) 120 , d ) 150 , e ) 240 | a | add(divide(multiply(320, 3), 5), multiply(3, 5)) | multiply(n0,n2)|multiply(n0,n1)|divide(#0,n1)|add(#2,#1)| | general |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 15 cm | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 15 ) = 285 cm 2 answer : c" | a ) 178 cm 2 , b ) 179 cm 2 , c ) 285 cm 2 , d ) 167 cm 2 , e ) 197 cm 2 | c | quadrilateral_area(15, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
if the operation @ is defined for all a and b by the equation a @ b = ( a ^ 2 * b ) * 3 , then 2 @ ( 3 @ - 1 ) = | "work within the parenthesis first so solve ( 3 @ - 1 ) first ( 3 @ - 1 ) = ( ( 3 ^ 2 ) * - 1 ) * 3 = ( 9 * - 1 ) * 3 = - 9 * 3 = - 27 now take - 3 plug back into equation and solve the rest 2 @ ( - 27 ) = ( ( 2 ^ 2 ) * - 3 ) * 3 = ( 4 * - 27 ) * 3 = - 324 so - 324 is the answer . . . . this question is merely testing ... | a ) - 324 , b ) 223 , c ) 324 , d ) - 223 , e ) - 410 | a | subtract(2, 3) | subtract(n2,n1)| | general |
a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 7 inches wide on each side , the total area would have been ( m + 264 ) square inches . what is the perimeter of the photograph , in inches ? | "let x and y be the width and length of the photograph . ( x + 2 ) ( y + 2 ) = m and so ( 1 ) xy + 2 x + 2 y + 4 = m ( x + 14 ) ( y + 14 ) = m and so ( 2 ) xy + 14 x + 14 y + 196 = m + 264 let ' s subtract equation ( 1 ) from equation ( 2 ) . 12 x + 12 y + 192 = 264 2 x + 2 y = 12 , which is the perimeter of the photog... | a ) 8 , b ) 12 , c ) 16 , d ) 20 , e ) 24 | b | divide(subtract(264, subtract(power(multiply(7, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | multiply(n1,const_2)|multiply(n0,const_2)|power(#0,const_2)|power(#1,const_2)|subtract(#2,#3)|subtract(n2,#4)|divide(#5,const_2)| | geometry |
a picnic attracts 240 persons . there are 60 more men than women , and 60 more adults than children . how many men are at this picnic ? | "adult + children = 240 let , children = y then , adult = y + 60 i . e . y + ( y + 60 ) = 240 i . e . y = 90 i . e . adult = 90 + 20 = 110 adults include only men and women i . e . men + women = 110 let women , w = x then men , m = x + 60 i . e . x + ( x + 60 ) = 2 x + 60 = 110 i . e . x = 25 i . e . men , m = 25 + 60 ... | a ) 240 , b ) 75 , c ) 110 , d ) 130 , e ) 85 | e | add(divide(subtract(add(divide(subtract(240, 60), const_2), 60), 60), const_2), 60) | subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)|subtract(#2,n1)|divide(#3,const_2)|add(n1,#4)| | general |
the price for a loaf of bread in april was $ 1.17 , in may the price was $ 1.30 , and in june the price was $ 1.05 . if 2 / 3 as much bread was sold in april as in may , and twice as much was sold in june as in april , what was the average price for a loaf of bread sold during the 3 - month period ? | let x be the number of loaves sold in may . then 2 x / 3 loaves were sold in april and 4 x / 3 loaves were sold in june . the average price was ( 2 x / 3 ) ( 117 ) + 130 x + ( 4 x / 3 ) ( 105 ) / ( 2 x / 3 + x + 4 x / 3 ) = ( 78 + 130 + 140 ) / ( 3 ) = 348 / 3 = $ 1.16 the answer is d . | a ) $ 1.13 , b ) $ 1.14 , c ) $ 1.15 , d ) $ 1.16 , e ) $ 1.17 | d | divide(add(add(1.17, 1.3), multiply(const_2, 1.05)), add(add(const_1, const_1), const_2)) | add(n0,n1)|add(const_1,const_1)|multiply(n2,const_2)|add(#0,#2)|add(#1,const_2)|divide(#3,#4) | general |
a person travels equal distances with speeds of 9 km / hr , 10 km / hr and 11 km / hr and takes a total time of 47 minutes . the total distance is ? | "let the total distance be 3 x km . then , x / 9 + x / 10 + x / 11 = 47 / 60 29 x / 96 = 47 / 60 = > x = 2.59 total distance = 3 * 2.59 = 7.78 km . answer : c" | a ) 6 km , b ) 3 km , c ) 8 km , d ) 9 km , e ) 2 km | c | multiply(multiply(divide(divide(47, const_60), add(add(divide(const_1, 9), divide(const_1, 10)), divide(const_1, 11))), const_3), const_1000) | divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3)|multiply(#7,const_1000)| | physics |
if the sides of a triangle are 26 cm , 22 cm and 10 cm , what is its area ? | "the triangle with sides 26 cm , 22 cm and 10 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 / 2 * 22 * 10 = 110 cm 2 answer : option b" | a ) 120 , b ) 110 , c ) 130 , d ) 140 , e ) 150 | b | divide(multiply(22, 10), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
the average mark of the students of a class in a particular exam is 72 . if 5 students whose average mark in that exam is 40 are excluded , the average mark of the remaining will be 92 . find the number of students who wrote the exam ? | "let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 92 ( x - 5 ) 72 x - ( 5 * 40 ) = 92 ( x - 5 ) 260 = 20 x = > x = 13 answer : b" | a ) 12 , b ) 13 , c ) 14 , d ) 16 , e ) 15 | b | divide(subtract(multiply(92, 5), multiply(5, 40)), subtract(92, 72)) | multiply(n1,n3)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)| | general |
suresh borrowed some money at the rate of 12 % p . a . for the first 3 years , 9 % p . a . for the next 5 years and 13 % p . a . for the period beyond 8 years . if the total interest paid by him at the end of 11 years is rs . 8160 , how much money did he borrow ? | using the formula s . i = p * t * r / 100 ( x * 12 * 3 / 100 ) + ( x * 9 * 5 / 100 ) + ( x * 13 * 3 / 100 ) = 8160 x = 6800 i . e money borrowed is rs . 6800 answer : c | a ) s . 8500 , b ) s . 6000 , c ) s . 6800 , d ) s . 9000 , e ) s . 7000 | c | divide(8160, add(add(multiply(divide(12, const_100), 3), multiply(divide(9, const_100), 5)), multiply(divide(13, const_100), subtract(8, 5)))) | divide(n0,const_100)|divide(n2,const_100)|divide(n4,const_100)|subtract(n5,n3)|multiply(n1,#0)|multiply(n3,#1)|multiply(#2,#3)|add(#4,#5)|add(#7,#6)|divide(n7,#8) | gain |
in township k , 1 / 4 of the housing units are equiped with cable tv . if 1 / 10 of the housing units , including 1 / 3 of those that are equiped with cable tv , are equipped with videocassette recorders , what fraction of the housing units have neither cable tv nor videocassette recorders ? | "1 / 4 - - cable tv ( this includes some data from video cassette recorder ) 1 / 10 - - video cassette recorder including 1 / 3 ( equiped with cable tv ) i . e . 1 / 3 ( 1 / 4 ) = 1 / 12 therefore only video cassette recorder = 1 / 10 - 1 / 12 = 1 / 60 total = 1 / 4 + 1 / 60 + neither cable tv nor videocassette recorde... | a ) 23 / 30 , b ) 11 / 15 , c ) 7 / 10 , d ) 1 / 6 , e ) 2 / 15 | b | divide(add(10, 4), multiply(10, const_2)) | add(n1,n3)|multiply(n3,const_2)|divide(#0,#1)| | other |
if both 5 ^ 2 and 3 ^ 4 are factors of the number a * 4 ^ 3 * 6 ^ 2 * 13 ^ 11 , then what is the smallest possible value of a ? | "the number a must include at least 3 ^ 2 * 5 ^ 2 = 225 the answer is d ." | a ) 80 , b ) 120 , c ) 150 , d ) 225 , e ) 300 | d | add(add(add(add(add(multiply(multiply(5, 2), const_2.0), multiply(multiply(5, 2), 2)), multiply(multiply(5, 2), 2)), 2), const_4), const_4) | multiply(n0,n7)|multiply(n4,#0)|add(#1,#1)|add(#2,#1)|add(n7,#3)|add(#4,const_4)|add(#5,const_4)| | other |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling alongthe oundary of the park at the speed of 12 km / hr completes one round in 8 min , then the area of the park ( in sq . m ) is ? | "perimeter = distance covered in 8 min . = 12000 x 8 m = 1600 m . 60 let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1600 or x = 160 . length = 480 m and breadth = 320 m . area = ( 480 x 320 ) m 2 = 153600 m c" | a ) 7889890 m , b ) 2334356 m , c ) 1536001 m , d ) 1232431 m , e ) 2435467 m | c | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
what will be the result of 12 ^ 4 + 3 + 2 ^ 9 + 3 - 8 / 2 ? | "the multiplication ( e . g . 12 ^ 4 ) and division ( e . g . 8 / 2 ) is done first and then the sum ( e . g . 48 + 3 ) and subtraction ( e . g . 4 - 4 ) , and the answer is 69 , option d ." | a ) 58 , b ) 14 , c ) 145 , d ) 69 , e ) 74 | d | subtract(multiply(add(divide(subtract(multiply(12, 4), 3), 2), 9), 3), 12) | multiply(n0,n1)|subtract(#0,n2)|divide(#1,n3)|add(n4,#2)|multiply(n5,#3)|subtract(#4,n0)| | general |
in a certain game of dice , the player β s score is determined as a sum of two throws of a single die . the player with the highest score wins the round . if more than one player has the highest score , the winnings of the round are divided equally among these players . if jim plays this game against 21 other players ,... | toguaranteethat jim will get some monetary payoff he must score the maximum score of 6 + 6 = 12 , because if he gets even one less than that so 11 , someone can get 12 and jim will get nothing . p ( 12 ) = 1 / 6 ^ 2 = 1 / 36 . answer : d | a ) 41 / 50 , b ) 1 / 221 , c ) 1 / 216 , d ) 1 / 36 , e ) 1 / 42 | d | divide(const_1, power(subtract(divide(21, const_3), const_1), const_2)) | divide(n0,const_3)|subtract(#0,const_1)|power(#1,const_2)|divide(const_1,#2) | general |
how many times will the digit 6 be written when listing the integers from 1 to 1000 ? | "many approaches are possible . for example : consider numbers from 0 to 999 written as follows : 1 . 000 2 . 001 3 . 002 4 . 003 . . . . . . . . . 1000 . 999 we have 1000 numbers . we used 3 digits per number , hence used total of 3 * 1000 = 3000 digits . now , why should any digit have preferences over another ? we u... | a ) 150 , b ) 280 , c ) 310 , d ) 420 , e ) 300 | e | multiply(multiply(multiply(1, const_10), const_10), const_3) | multiply(n1,const_10)|multiply(#0,const_10)|multiply(#1,const_3)| | general |
an assembly line produces 20 cogs per hour until an initial order of 60 cogs is completed . the speed of the assembly line is then immediately increased so that it can produce 60 cogs per hour until another 60 cogs are produced . what is the overall average output , in cogs per hour , for the assembly line during this ... | "the time to produce the first 60 cogs is 60 / 20 = 3 hours . the time to produce the next 60 cogs is 60 / 60 = 1 hour . the average output is 120 cogs / 4 hours = 30 cogs per hour . the answer is b ." | a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | b | divide(multiply(60, const_2), add(divide(60, 20), const_1)) | divide(n1,n0)|multiply(n1,const_2)|add(#0,const_1)|divide(#1,#2)| | general |
what is 120 % of 13 / 24 of 960 ? | "120 % * 13 / 24 * 360 = 1.2 * 13 * 40 = 624 the answer is e ." | a ) 420 , b ) 484 , c ) 526 , d ) 578 , e ) 624 | e | divide(multiply(120, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
3 candidates in an election and received 1136 , 7636 and 11628 votes respectively . what % of the total votes did the winningcandidate got in that election ? | "total number of votes polled = ( 1136 + 7636 + 11628 ) = 20400 so , required percentage = 11628 / 20400 * 100 = 57 % c" | a ) 45 % , b ) 56 % , c ) 57 % , d ) 70 % , e ) 72 % | c | multiply(divide(11628, add(add(1136, 7636), 11628)), const_100) | add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)| | gain |
the average of first 80 non - zero positive numbers is : | "explanation : sum of first n natural numbers = n ( n + 1 ) / 2 so , average of first n natural numbers = n ( n + 1 ) / 2 n = ( n + 1 ) / 2 required average = ( 80 + 1 ) / 2 = 81 / 2 = 40.5 . answer : d" | a ) 22.25 , b ) 31.25 , c ) 35 , d ) 40.5 , e ) none of these | d | add(80, const_1) | add(n0,const_1)| | general |
yesterday it took robert 4 hours to drive from city a to city b . today it took robert 3.5 hours to drive back from city Π² to city a along the same route . if he had saved 15 minutes in both trips , the speed for the round trip would be 60 miles per hour . what is the distance between city a and city b ? | "2 d / 60 = 7 ( because time = 4 + 3.5 - 0.5 hrs ) = > d = 210 answer - d" | a ) 90 , b ) 120 , c ) 150 , d ) 210 , e ) 300 | d | divide(multiply(60, subtract(add(4, 3.5), const_1)), const_2) | add(n0,n1)|subtract(#0,const_1)|multiply(n3,#1)|divide(#2,const_2)| | physics |
a shopkeeper loses 15 % , if an article is sold for rs . 153 . what should be the selling price of the article to gain 20 % ? | given that sp = rs . 153 and loss = 15 % cp = [ 100 ( sp ) ] / ( 100 - l % ) = ( 100 * 153 ) / 85 = rs . 180 . to get 20 % profit , new sp = [ ( 100 + p % ) cp ] / 100 = ( 180 * 120 ) / 100 = rs . 216 answer : e | a ) s . 147 , b ) s . 248 , c ) s . 244 , d ) s . 229 , e ) s . 216 | e | add(divide(153, subtract(const_1, divide(15, const_100))), multiply(divide(153, subtract(const_1, divide(15, const_100))), divide(20, const_100))) | divide(n0,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n1,#2)|multiply(#3,#1)|add(#3,#4) | gain |
the sum of two numbers is 45 . sum of their quotient and reciprocal is 2.05 , find the product of the numbers . | sol : let a , b be the numbers . a + b = 45 ab + baab + ba = 2.05 β a 2 + b 2 ab β a 2 + b 2 ab = 2.05 β ( a + b ) 2 β 2 abab = 2.05 β ( a + b ) 2 β 2 abab = 2.05 β ( a + b ) 2 β ( a + b ) 2 = 2.05 ab + 2 ab = 4.05 ab β β ab = 4524 . 054524.05 = 500 answer : b | a ) 339 , b ) 500 , c ) 288 , d ) 277 , e ) 112 | b | divide(multiply(45, 45), add(2.05, const_2)) | add(n1,const_2)|multiply(n0,n0)|divide(#1,#0) | general |
in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what q percent of all the people in smithtown are left - handed women ? | "looking at the ratio we can take total number of people q = 20 . . ans 5 / 20 or 25 % c" | a ) 50 % , b ) 40 % , c ) 25 % , d ) 20 % , e ) 10 % | c | multiply(divide(subtract(multiply(2, divide(add(3, 1), add(3, 2))), subtract(3, multiply(3, divide(add(3, 1), add(3, 2))))), add(3, 1)), const_100) | add(n0,n1)|add(n0,n3)|divide(#0,#1)|multiply(n3,#2)|multiply(n0,#2)|subtract(n0,#4)|subtract(#3,#5)|divide(#6,#0)|multiply(#7,const_100)| | general |
company p had 15 percent more employees in december than it had in january . if company p had 500 employees in december , how many employees did it have in january ? | "d = number of employees in december j = number of employees in january j x 1.15 = d j x 1.15 = 500 j = 500 / 1.15 j = 50,000 / 115 = 435 thus b is the correct answer ." | a ) 391 , b ) 435 , c ) 410 , d ) 423 , e ) 445 | b | divide(500, add(const_1, divide(15, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
at a certain fruit stand , the price of each apple is 50 cents and the price of each orange is 60 cents . mary selects a total of 10 apples and oranges from the fruit stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 56 cents . how many oranges must mary put back so that the average price o... | "let number of apples = a number of oranges = b a + b = 10 - - - 1 . 56 = ( . 5 a + . 6 b ) / 10 = > 56 = 5 a + 6 b - - - - 2 solving 1 and 2 , we get a = 4 b = 6 let the number of oranges put back = c 52 * ( 10 - c ) = 50 * 4 + 60 ( 6 - c ) = > c = 5 answer e" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(subtract(multiply(56, 10), multiply(52, 10)), subtract(60, 52)) | multiply(n2,n4)|multiply(n2,n5)|subtract(n1,n5)|subtract(#0,#1)|divide(#3,#2)| | general |
mary can do a piece of work in 28 days . rosy is 40 % more efficient than mary . the number of days taken by rosy to do the same piece of work is ? | "ratio of times taken by mary and rosy = 140 : 100 = 14 : 10 suppose rosy takes x days to do the work . 14 : 10 : : 28 : x = > x = 20 days . hence , rosy takes 20 days to complete the work . answer : c" | a ) 22 , b ) 24 , c ) 20 , d ) 25 , e ) 27 | c | divide(28, add(const_1, divide(40, const_100))) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | physics |
company x provides bottled water to its employees in 5 liter bottles , $ 5 each . at least how many 5 l bottles must company x buy monthly , so that new contract with $ 40 fixed monthly rate and $ 2 for 20 l bottle each paid off ? ( assume that no other costs apply ) | let the no . of 5 liter bottles be x , so the no . of 20 l bottles will be x / 4 ( to equate the vol . ) since the total cost will be equal , 5 x = 40 + 2 * x / 4 so x = 8.88 or 9 . answer is ( b ) . | a ) 4 , b ) 9 , c ) 12 , d ) 20 , e ) 25 | b | divide(40, subtract(5, divide(const_1, const_2))) | divide(const_1,const_2)|subtract(n0,#0)|divide(n3,#1) | general |
a reduction of 15 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | "800 * ( 15 / 100 ) = 120 - - - - 5 ? - - - - 1 = > rs . 24 answer : c" | a ) 12 , b ) 16 , c ) 24 , d ) 28 , e ) 32 | c | divide(divide(multiply(800, 15), const_100), 5) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain |
a certain number when divided by 39 leaves a remainder 18 , what is the remainder when the same number is divided by 13 ? | "explanation : 39 + 18 = 57 / 13 = 5 ( remainder ) answer : e" | a ) 7 , b ) 8 , c ) 9 , d ) 6 , e ) 5 | e | reminder(18, 13) | reminder(n1,n2)| | general |
the membership of a committee consists of 4 english teachers , 2 mathematics teachers , and 2 social studies teachers . if 2 committee members are to be selected at random to write the committee β s report , what is the probability that the two members selected will both be english teachers ? | "probability of first member an english teacher = 3 / 8 probability of second member an english teacher = 2 / 8 probability of both being english teacher = 3 / 8 x 2 / 7 = 3 / 28 ( e )" | a ) 2 / 3 , b ) 1 / 3 , c ) 2 / 9 , d ) 1 / 12 , e ) 3 / 28 | e | multiply(divide(4, add(add(4, 2), 2)), divide(2, subtract(add(add(4, 2), 2), const_1))) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|subtract(#1,const_1)|divide(n2,#3)|multiply(#2,#4)| | probability |
a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 215 m long ? | ": speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 215 = 575 m required time = 575 * 2 / 25 = 46 sec answer : b" | a ) 40 , b ) 46 , c ) 88 , d ) 77 , e ) 21 | b | divide(360, multiply(subtract(45, 215), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
a and b complete a work in 5 days . a alone can do it in 10 days . if both together can do the work in how many days ? | "1 / 5 + 1 / 10 = 3 / 10 10 / 3 = 3.33 days answer : a" | a ) 3.33 days , b ) 3.99 days , c ) 2.99 days , d ) 2.98 days , e ) 2.44 days | a | inverse(add(inverse(5), inverse(10))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)| | physics |
when tom works alone he chops 2 lb . salad in 3 minutes , and when tammy works alone she chops 3 lb . salad in 2 minutes . they start working together , and after some time finish chopping 65 lb . of salad . of those 80 lb . , the salad quantity chopped by tom is what percent lesser than the quantifying chopped by tamm... | "tom chops 4 lbs in 6 minutes tammy chops 9 lbs in 6 minutes so in the same amount of time , tom chops 125 % less than tammy , since 9 is 125 % greater than 4 . so 125 % is the answer . note that the actual time does n ' t matter . if you multiply the time each work by x , you ' ll multiply the work each does by x , an... | a ) 44 % , b ) 100 % , c ) 105 % , d ) 225 % , e ) 125 % | e | multiply(divide(subtract(divide(3, 2), divide(2, 3)), divide(2, 3)), const_100) | divide(n1,n0)|divide(n0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | physics |
suresh can complete a job in 15 hours . ashutosh alone can complete the same job in 35 hours . suresh works for 9 hours and then the remaining job is completed by ashutosh . how many hours will it take ashutosh to complete the remaining job alone ? | "the part of job that suresh completes in 9 hours = 9 Γ’ Β β 15 = 3 Γ’ Β β 5 remaining job = 1 - 3 Γ’ Β β 5 = 2 Γ’ Β β 5 remaining job can be done by ashutosh in 2 Γ’ Β β 5 Γ£ β 35 = 14 hours answer c" | a ) 4 , b ) 5 , c ) 14 , d ) 12 , e ) none of these | c | multiply(subtract(const_1, divide(9, 15)), 35) | divide(n2,n0)|subtract(const_1,#0)|multiply(n1,#1)| | physics |
a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 9 days , a finished the remaining work . in how many days a alone can finish the job ? | "a + b 10 days work = 10 * 1 / 40 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 3 / 4 work is done by a in 9 days whole work will be done by a in 9 * 4 / 3 = 12 days answer is a" | a ) 12 , b ) 25 , c ) 60 , d ) 30 , e ) 20 | a | divide(multiply(9, 40), subtract(40, 10)) | multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | physics |
two trains of length 90 m and 95 m are 250 m apart . they start moving towards each other on parallel tracks , at speeds 64 kmph and 92 kmph . after how much time will the trains meet ? | "they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 64 + 92 ) * 5 / 18 = 43.3 mps . the time required = d / s = 250 / 43.3 = 58 / 10 sec . answer : e" | a ) 53 / 10 , b ) 55 / 10 , c ) 56 / 10 , d ) 57 / 10 , e ) 58 / 10 | e | divide(250, multiply(add(64, 92), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
what is the greatest 6 - digit number when divided by 6 , 7 , 8 , 9 , and 10 leaves a remainder of 4 , 5 , 6 , 7 , and 8 respectively ? | when you divide a positive integer by 10 , the remainder will just be the units digit . we know the remainder is 8 when we divide by 10 , so d is the only possible answer . | a ) 456780 , b ) 678910 , c ) 997479 , d ) 997918 , e ) 997920 | d | divide(multiply(const_1000, const_1000), 10) | multiply(const_1000,const_1000)|divide(#0,n5)| | general |
in one alloy there is 12 % chromium while in another alloy it is 10 % . 15 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 15 + 35 = 50 kg alloy is 0.12 * 15 + 0.10 * 35 = 5.3 kg , so the percentage is 5.3 / 50 * 100 = 10.6 % . answer : e ." | a ) 8.8 % , b ) 9 % , c ) 9.2 % , d ) 8.6 % , e ) 10.6 % | e | multiply(divide(add(divide(multiply(12, 15), const_100), divide(multiply(10, 35), const_100)), add(15, 35)), const_100) | add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
a baseball card decreased in value 10 % in its first year and 10 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | "consider the initial value of the baseball card as $ 100 after first year price = 100 * 0.9 = 90 after second year price = 90 * 0.9 = 81 final decrease = [ ( 100 - 81 ) / 100 ] * 100 = 19 % correct answer - a" | a ) 19 % , b ) 30 % , c ) 32 % , d ) 36 % , e ) 72 % | a | subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(10, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | gain |
at a certain restaurant , the average ( arithmetic mean ) number of customers served for the past x days was 80 . if the restaurant serves 120 customers today , raising the average to 90 customers per day , what is the value of x ? | "withoutusing the formula , we can see that today the restaurant served 30 customers above the average . the total amount above the average must equal total amount below the average . this additional 30 customers must offset the β deficit β below the average of 90 created on the x days the restaurant served only 80 cus... | a ) 2 , b ) 3 , c ) 9 , d ) 15 , e ) 30 | b | subtract(divide(subtract(120, 90), subtract(90, 80)), divide(subtract(120, const_100), const_100)) | subtract(n1,n2)|subtract(n2,n0)|subtract(n1,const_100)|divide(#0,#1)|divide(#2,const_100)|subtract(#3,#4)| | general |
bhanu spends 30 % of his income on petrol on scooter 10 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ? | "given 30 % ( income ) = 300 β β income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 10 % ( 700 ) = rs . 70 answer : b" | a ) 2287 , b ) 70 , c ) 128 , d ) 797 , e ) 120 | b | multiply(subtract(divide(300, divide(30, const_100)), 300), divide(10, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(n2,#1)|subtract(#2,n2)|multiply(#0,#3)| | gain |
how many 4 digit numbers are there , if it is known that the first digit is odd , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 1 can be used only once ? | "4 options for the first digit : 1 , 3 , 5 , 7 , 9 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 5 * 5 * 4 * 4 = 400 numbers with one 1 - s , 1 ... | a ) 20 , b ) 150 , c ) 384 , d ) 300 , e ) 320 | c | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | add(n0,const_1)|multiply(n0,n0)|multiply(#0,#0)|multiply(#2,#1)|multiply(#2,n0)|subtract(#3,#4)| | physics |
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