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
values |
|---|---|---|---|---|---|---|
two heavily loaded sixteen - wheeler transport trucks are 1025 kilometers apart , sitting at two rest stops on opposite sides of the same highway . driver a begins heading down the highway driving at an average speed of 90 kilometers per hour . exactly one hour later , driver b starts down the highway toward driver a ,... | "i ' ve been reading the website for a while and i ' m always keen to see different approaches so i would like to share one that works for me : short version : truck a travels for an hour . distance remaining = 1025 - 90 = 935 k ratio of speeds 9 : 8 - > 935 / 17 = 55 truck a = 90 + 55 * 9 = 585 truck b = 55 * 8 = 440 ... | a ) 90 , b ) 130 , c ) 150 , d ) 320 , e ) 145 | e | add(90, multiply(divide(subtract(1025, 90), add(90, 80)), subtract(90, 80))) | add(n1,n2)|subtract(n0,n1)|subtract(n1,n2)|divide(#1,#0)|multiply(#3,#2)|add(n1,#4)| | general |
find the sum of all 2 digit numbers divisible by 3 . | "all 2 digit numbers divisible by 3 are 12,15 , 18,21 , … … … . . 99 this is an a . p with a = 12 , and d = 3 , let it contain n terms then , 12 + ( n – 1 ) * 3 = 99 , or n = ( 99 - 12 ) / 3 + 1 = 30 required sum = 30 / 2 * ( 12 + 99 ) = 15 * 111 = 1665 answer : b" | a ) 1565 , b ) 1665 , c ) 1300 , d ) 1448 , e ) 1465 | b | multiply(add(add(const_100, const_4), subtract(multiply(const_100, const_10), 3)), divide(add(divide(subtract(subtract(multiply(const_100, const_10), 3), add(const_100, const_4)), 3), const_1), const_2)) | add(const_100,const_4)|multiply(const_10,const_100)|subtract(#1,n1)|add(#0,#2)|subtract(#2,#0)|divide(#4,n1)|add(#5,const_1)|divide(#6,const_2)|multiply(#3,#7)| | general |
2 , 3 , 6 , 0 , 10 , - 3 , 14 , ( . . . ) | "there are two series 2 , 6 , 10 , 14 , . . . ( adding 4 ) 3 , 0 , - 3 , . . . ( subtracting 3 ) hence , next term is - 3 - 3 = - 6 answer is a" | a ) - 6 , b ) 8 , c ) - 2 , d ) 4 , e ) 10 | a | subtract(negate(0), multiply(subtract(3, 6), divide(subtract(3, 6), subtract(2, 3)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
a man can row downstream at the rate of 12 km / hr and upstream at 8 km / hr . find man ' s rate in still water and the rate of current ? | "explanation : rate of still water = 1 / 2 ( 12 + 8 ) = 10 km / hr rate of current = 1 / 2 ( 12 - 8 ) = 2 km / hr answer : option d" | a ) 9,6 , b ) 6,3 , c ) 9,3 , d ) 10,2 , e ) none of these | d | divide(subtract(12, 8), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
the present population of a town is 1000 . population increase rate is 25 % p . a . find the population of town after 2 years ? | "p = 1000 r = 25 % required population of town = p ( 1 + r / 100 ) ^ t = 1000 ( 1 + 25 / 100 ) ^ 2 = 1000 ( 5 / 4 ) ^ 2 = 1563 ( approximately ) answer is c" | a ) 1000 , b ) 2000 , c ) 1563 , d ) 2153 , e ) 1245 | c | add(1000, divide(multiply(1000, 25), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 2.0 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 – 6 = 666 * 2.0 = 1332 answer : b" | a ) s . 1014 , b ) s . 1332 , c ) s . 999 , d ) s . 1085 , e ) s . 1020 | b | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 2.0), 3) | multiply(n3,const_2)|sqrt(n0)|multiply(#1,const_4)|subtract(#2,#0)|multiply(n2,#3)|multiply(#4,n1)| | physics |
two employees m and n are paid a total of $ 605 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 605 2.2 n = 605 n = 275 the answer is d ." | a ) $ 245 , b ) $ 255 , c ) $ 265 , d ) $ 275 , e ) $ 285 | d | divide(605, add(divide(120, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
the weight of 4 dogs is determined to be 25 pounds , 31 pounds , 35 pounds and 33 pounds respectively . the weight of a fifth dog is determined to be y pounds . if the average ( arithmetic mean ) weight of the first 4 dogs is the same as that of all 5 dogs what is the value of y ? | total weight of the 4 dogs = ( 25 + 31 + 35 + 33 ) = 124 avg = 124 / 4 = 31 total weight of 5 dogs = 124 + y or 4 ( 31 ) + y average of 5 dogs as per question = 31 equation : 4 ( 31 ) + y = 5 ( 31 ) , or y = 31 . choose a | a ) 31 , b ) 33 , c ) 35 , d ) 37 , e ) 39 | a | divide(add(add(25, 31), add(35, 33)), 4) | add(n1,n2)|add(n3,n4)|add(#0,#1)|divide(#2,n0) | general |
how many ounces of a 60 % salt solution must be added to 40 ounces of a 20 percent salt solution so that the resulting mixture is 40 % salt ? | "let x = ounces of 60 % salt solution to be added . 2 * 40 + . 6 x = . 4 ( 40 + x ) x = 40 amswer b" | a ) 16.67 , b ) 40 , c ) 50 , d ) 60.33 , e ) 70 | b | divide(subtract(multiply(divide(40, const_100), 40), multiply(divide(20, const_100), 40)), subtract(divide(40, const_100), divide(20, const_100))) | divide(n3,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#0,#1)|subtract(#2,#3)|divide(#5,#4)| | gain |
car a runs at the speed of 50 km / hr and reaches its destination in 8 hours . car b runs at the speed of 60 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ? | "car a travels 50 × 8 = 400 km car b travels 60 × 4 = 240 km the ratio is 400 : 240 = 40 : 24 = 5 : 3 the answer is d ." | a ) 3 : 7 , b ) 4 : 9 , c ) 13 : 7 , d ) 5 : 3 , e ) 6 : 11 | d | divide(multiply(50, 8), multiply(60, 4)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
if the cost price of 50 articles is equal to the selling price of 46 articles , then the gain or loss percent is ? | given that , cost price of 50 article is equal to selling price of 46 articles . let cost price of one article = rs . 1 selling price of 46 articles = rs . 50 but cost price of 46 articles = rs . 46 therefore , the trader made profit . \ percentage of profit = 4 / 46 * 100 = 8.69 % answer : e | a ) 22 , b ) 65 , c ) 78 , d ) 33 , e ) 8.69 | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 46), 50)), divide(multiply(const_100, 46), 50))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100) | gain |
johnny travels a total of one hour to and from school . on the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 21 miles per hour . how far is it to the school ? | "answer : c ) 6.5 miles . average speed for round trip = 2 * a * b / ( a + b ) , where a , b are speeds so , average speed was = 2 * 5 * 21 / ( 5 + 21 ) = 6.5 m / hr the distance between schoolhome should be half of that . ie . 6.5 miles answer c" | a ) 2 miles , b ) 4 miles , c ) 6.5 miles , d ) 8 miles , e ) 10 miles | c | multiply(divide(const_1, add(divide(const_1, 5), divide(const_1, 21))), const_1_6) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,const_1_6)| | physics |
if 3 x + y = 40 , 2 x - y = 20 , for integers of x and y , 3 y ^ 2 = ? | 3 x + y = 40 2 x - y = 20 5 x = 60 x = 12 y = 4 3 y ^ 2 = 3 * 16 = 48 answer is b | a ) 2 , b ) 48 , c ) 0 , d ) 10 , e ) 3 | b | multiply(3, power(subtract(40, multiply(divide(add(40, 20), add(3, 2)), 3)), const_2)) | add(n1,n3)|add(n0,n2)|divide(#0,#1)|multiply(n0,#2)|subtract(n1,#3)|power(#4,const_2)|multiply(n0,#5) | general |
how many distinct 3 digit number less than 10 | "000,001 , 002,003 , 004,005 , 006,007 , 008,009 . 10 distinct 3 digit numbers less than 10 . answer : c" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | add(const_2, const_3) | add(const_2,const_3)| | general |
the average age of a group of persons going for picnic is 16 years . 20 new persons with an average age of 15 years join the group on the spot due to which their average age becomes 15.5 years . the number of persons initially going for picnic is ? | let the initial number of persons be x . then , 16 x + 20 * 15 = 15.5 ( x + 20 ) < = > 0.5 x = 10 < = > x = 20 . answer : b | a ) 10 , b ) 20 , c ) 37 , d ) 26 , e ) 22 | b | divide(subtract(multiply(15.5, 20), multiply(15, 20)), subtract(16, 15.5)) | multiply(n1,n3)|multiply(n1,n2)|subtract(n0,n3)|subtract(#0,#1)|divide(#3,#2) | general |
in an election between two candidates , the winner has a margin of 10 % of the votes polled . if 2000 people change their mind and vote for the loser , the loser would have won by a margin of 10 % of the votes polled . find the total number of votes polled in the election ? | "winner - looser 55 % - 45 % if 2000 people change their mind and vote for the loser : winner - looser 45 % - 55 % thus 2,000 people compose 10 % of all voters , which means that the total number of votes is 20,000 . answer : a ." | a ) 20000 , b ) 25000 , c ) 24000 , d ) 23000 , e ) 22000 | a | divide(2000, divide(10, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
if 99 < x < 199 and 10 < y < 100 , then the product xy can not be equal to : | correct answer : ( b ) determine the range of xy by multiplying the two extremes of each individual range together . the smallest value of xy must be greater than 99 * 10 . the largest value must be less than 199 * 100 . this means that 990 < xy < 19,900 . ( b ) is outside of this range , so it is not a possible produc... | a ) 19,104 , b ) 19,903 , c ) 19 , 356.732 , d ) 19,502 , e ) 19,880 | b | add(100, const_1) | add(n3,const_1) | general |
a , b , c subscribe rs . 50,000 for a business . if a subscribes rs . 4000 more than b and b rs . 5000 more than c , out of a total profit of rs . 39,000 , what will be the amount a receives ? | "total amount invested = 50000 assume that investment of c = x . then investment of b = 5000 + x , investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000 ⇒ 3 x + 14000 = 50000 ⇒ 3 x = 50000 – 14000 = 36000 ⇒ x = 36000 / 3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 inves... | a ) 14700 , b ) 14500 , c ) 14900 , d ) 16380 , e ) 14000 | d | multiply(add(divide(subtract(multiply(5000, const_10), add(add(4000, 5000), 5000)), const_3), add(4000, 5000)), divide(multiply(multiply(const_3, const_12), const_1000), multiply(5000, const_10))) | add(n1,n2)|multiply(n2,const_10)|multiply(const_12,const_3)|add(n2,#0)|multiply(#2,const_1000)|divide(#4,#1)|subtract(#1,#3)|divide(#6,const_3)|add(#0,#7)|multiply(#8,#5)| | general |
if the operation € is defined for all x and y by the equation x € y = 3 * x * y , then 3 € ( 4 € 5 ) = | "working inside out , ( 4 € 5 ) = 3 * 4 * 5 = 60 3 € 60 = 3 * 3 * 60 = 540 hence , answer is d" | a ) 80 , b ) 120 , c ) 160 , d ) 540 , e ) 660 | d | multiply(multiply(3, 3), multiply(multiply(3, 4), 5)) | multiply(n0,n1)|multiply(n0,n2)|multiply(n3,#1)|multiply(#0,#2)| | general |
a person can swim in still water at 8 km / h . if the speed of water 4 km / h , how many hours will the man take to swim back against the current for 8 km ? | "m = 8 s = 4 us = 8 - 4 = 4 d = 4 t = 8 / 4 = 2 answer : a" | a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(8, subtract(8, 4)) | subtract(n0,n1)|divide(n2,#0)| | physics |
thomas and matt has some money with them in the ratio of 3 : 4 . if their father gives rs . 30 to each of them then the ratio becomes 4 : 5 . what would be the ratio if thomas spend rs . 40 and matt spend rs . 20 from what they have now ? | let money of thomas and matt be x and y respectively . x / y = 3 / 4 = > x = 3 / 4 y ( x + 30 ) / ( y + 30 ) = 4 / 5 = > 5 x + 150 = 4 y + 120 = > 5 x = 4 y - 30 but x = 3 / 4 y 5 * 3 / 4 y = 4 y - 30 15 y = 16 y - 120 = > y = 120 = > x = 3 * 120 / 4 = > x = 90 ratio of their money , if thomas spend rs . 40 and matt sp... | a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | a | divide(subtract(multiply(multiply(30, 4), divide(3, 4)), 40), subtract(multiply(30, 4), 20)) | divide(n0,n1)|multiply(n1,n2)|multiply(#0,#1)|subtract(#1,n6)|subtract(#2,n5)|divide(#4,#3) | other |
if the integer n has exactly 5 positive divisors , including 1 and n , how many positive divisors does √ n have ? | take the example of 16 . . . it has 3 positive divisors ( 1 , 2,4 , 8,16 ) now , take the example of 4 . . . it has only 3 divisors . . so d is the ans | a ) 6 , b ) 5 , c ) 2 , d ) 3 , e ) 4 | d | subtract(5, add(1, 1)) | add(n1,n1)|subtract(n0,#0) | general |
a straight line in the xy - plane has a slope of 2 and a y - intercept of 2 . on this line , what is the x - coordinate of the point whose y - coordinate is 480 ? | "slope of 2 and a y - intercept of 2 y - coordinate is 480 y = 2 x + 2 478 = 2 x x = 239 answer : c . 239" | a ) 249 , b ) 498 , c ) 239 , d ) 823 , e ) 1,002 | c | divide(subtract(480, 2), 2) | subtract(n2,n0)|divide(#0,n0)| | general |
bruce purchased 7 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | cost of 7 kg grapes = 70 × 7 = 490 . cost of 9 kg of mangoes = 55 × 9 = 495 . total cost he has to pay = 490 + 495 = 985 a | a ) a ) 985 , b ) b ) 1050 , c ) c ) 1055 , d ) d ) 1065 , e ) e ) 1075 | a | add(multiply(7, 70), multiply(9, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1) | gain |
if the sum of two positive integers is 26 and the difference of their squares is 52 , what is the product of the two integers ? | let the 2 positive numbers x and y x + y = 26 - - 1 x ^ 2 - y ^ 2 = 52 = > ( x + y ) ( x - y ) = 52 - - 2 using equation 1 in 2 , we get = > x - y = 2 - - 3 solving equation 1 and 3 , we get x = 14 y = 12 product = 14 * 12 = 168 answer e | a ) 108 , b ) 119 , c ) 128 , d ) 135 , e ) 168 | e | multiply(divide(subtract(26, divide(52, 26)), divide(52, 26)), add(divide(subtract(26, divide(52, 26)), divide(52, 26)), divide(52, 26))) | divide(n1,n0)|subtract(n0,#0)|divide(#1,#0)|add(#2,#0)|multiply(#3,#2) | general |
fonzie , aunt bee and lapis paid $ 7000 , $ 8000 and $ 9000 respectively to buy a map to capt . plaidstache ' s treasure . when they found it , the treasure turned out to be worth a cool $ 900000 . how much of it should lapis get ? | a = 7000 b = 8000 c = 9000 a share 7 parts , b share 8 parts & c share 9 parts total 24 parts - - - - - > 900000 - - - - > 1 part - - - - - - - > 37500 c share = 9 parts - - - - - > 337500 e | a ) $ 300000 , b ) $ 320000 , c ) $ 335000 , d ) $ 336500 , e ) $ 337500 | e | multiply(9000, divide(900000, add(9000, add(7000, 8000)))) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n2,#2) | general |
for all positive integers m , [ m ] = 3 m when m is odd and [ m ] = ( 1 / 2 ) * m when m is even . what is [ 9 ] * [ 4 ] equivalent to ? | "[ 9 ] * [ 4 ] = 27 * 2 = 54 = ( 1 / 2 ) ( 108 ) = [ 108 ] the answer is a ." | a ) [ 108 ] , b ) [ 54 ] , c ) [ 36 ] , d ) [ 27 ] , e ) [ 18 ] | a | divide(multiply(divide(4, 2), multiply(3, 9)), 3) | divide(n4,n2)|multiply(n0,n3)|multiply(#0,#1)|divide(#2,n0)| | general |
two pipes p and q can fill a cistern in 10 and 15 minutes respectively . both are opened together , but at the end of 4 minutes the first is turned off . how many more minutes will it take for the cistern to fill after the first pipe is turned off ? | "let x be the total time it takes for the cistern to fill . 4 / 10 + x / 15 = 1 x / 15 = 3 / 5 x = 9 after the first pipe is turned off , it takes 5 more minutes to fill the cistern . the answer is d ." | a ) 3.5 , b ) 4 , c ) 4.5 , d ) 5 , e ) 5.5 | d | multiply(subtract(const_1, multiply(add(divide(const_1, 10), divide(const_1, 15)), 4)), 15) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n1,#4)| | physics |
express 20 mps in kmph ? | 20 * 18 / 5 = 72 kmph answer : d | a ) 23 , b ) 88 , c ) 90 , d ) 72 , e ) 21 | d | multiply(divide(20, const_1000), const_3600) | divide(n0,const_1000)|multiply(#0,const_3600) | physics |
you hold some gold in a vault as an investment . over the past year the price of gold increases by 40 % . in order to keep your gold in the vault , you must pay 5 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year . | "( 100 % + 40 % ) * ( 100 % - 5 % ) = 140 * 0.95 = 133 % an increase of 33 % your gold holdings have increased in value by 33 % . the answer is b" | a ) 35 % , b ) 33 % , c ) 36 % , d ) 38 % , e ) 40 % | b | subtract(40, divide(40, 5)) | divide(n0,n1)|subtract(n0,#0)| | gain |
a large tanker can be filled by two pipes a and b in 60 and 40 minutes respectively . how many minutes will it take to fill the tanker from empty state if b is used for half the time and a and b fill it together for the other half ? | suppose the tank is filled in x minutes . then , x / 2 ( 1 / 24 + 1 / 40 ) = 1 x / 2 * 1 / 15 = 1 = > x = 30 min answer a | a ) 30 , b ) 15 , c ) 20 , d ) 25 , e ) 50 | a | multiply(divide(const_1, add(inverse(40), add(inverse(60), inverse(40)))), const_2) | inverse(n0)|inverse(n1)|add(#0,#1)|add(#2,#1)|divide(const_1,#3)|multiply(#4,const_2) | physics |
the captain of a cricket team of 11 members is 27 years old and the wicket keeper is 1 year older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 27 + 28 ) = 9 ( x - 1 ) 11 x - 9 x = 46 2 x = 46 x = 23 . the average age of the team is 23 years . the answer is d ." | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | d | divide(subtract(add(27, add(27, 1)), multiply(const_3.0, const_3.0)), const_2) | add(n1,n2)|multiply(const_3.0,const_3.0)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
if log 1087.5 = 6.9421 , then the number of digits in ( 875 ) 10 is ? | "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 1087.5 + 1 ) = 10 ( 6.9421 + 1 ) = 10 ( 7.9421 ) = 79.421 x = antilog ( 79.421 ) therefore , number of digits in x = 80 . answer : c" | a ) 30 , b ) 28 , c ) 80 , d ) 26 , e ) 25 | c | add(multiply(const_4, 6.9421), divide(log(const_100), log(const_10))) | log(const_100)|log(const_10)|multiply(n1,const_4)|divide(#0,#1)|add(#3,#2)| | other |
two heavily loaded sixteen - wheeler transport trucks are 787 kilometers apart , sitting at two rest stops on opposite sides of the same highway . driver a begins heading down the highway driving at an average speed of 90 kilometers per hour . exactly one hour later , driver b starts down the highway toward driver a , ... | i ' ve been reading the website for a while and i ' m always keen to see different approaches so i would like to share one that works for me : short version : truck a travels for an hour . distance remaining = 787 - 90 = 697 k ratio of speeds 9 : 8 - > 697 / 17 = 41 truck a = 90 + 41 * 9 = 459 truck b = 41 * 8 = 328 de... | a ) 90 , b ) 131 , c ) 150 , d ) 320 , e ) 450 | b | add(90, multiply(divide(subtract(787, 90), add(90, 80)), subtract(90, 80))) | add(n1,n2)|subtract(n0,n1)|subtract(n1,n2)|divide(#1,#0)|multiply(#3,#2)|add(n1,#4) | general |
in a group of 800 people , 1 / 5 play at least one instrument , 64 play two or more . what is the probability that one student play exactly one instrument ? | "p ( playing 2 or more instruments ) = 64 / 800 = 2 / 25 . then , the probability of playing exactly one instrument is given by : p ( playing 1 or more instruments ) - p ( playing 2 or more instruments ) = 1 / 5 - 2 / 25 = 3 / 25 . answer a ." | a ) 3 / 25 , b ) 3 / 125 , c ) c ) 2 / 25 , d ) 3 / 25 , e ) 1 / 5 | a | divide(subtract(divide(multiply(800, 1), 5), 64), 800) | multiply(n0,n1)|divide(#0,n2)|subtract(#1,n3)|divide(#2,n0)| | general |
what is the angle between the hands of a clock when time is 8 : 30 ? | "angle between two hands = 30 h - 11 / 2 m = 30 * 8 - 30 * 11 / 2 = 240 - 165 = 75 deg answer : b" | a ) 65 deg , b ) 75 deg , c ) 45 deg , d ) 15 deg , e ) 30 deg | b | subtract(multiply(30, multiply(const_3, const_2)), 8) | multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0)| | geometry |
if the difference between compound interest ( interest compounded yearly ) and simple interest on a certain sum at the rate 20 % p . a . after 3 years is rs . 1280 then what is the principal ? | "let p is the principal , so p ( [ 120 / 100 ] ^ 3 - 1 ) - p * 20 / 100 * 3 = 1280 p [ 728 / 1000 - 3 / 5 ] = 1280 p = 1280 * 5 * 1000 / 728 * 5 - 3000 = 10000 answer : c" | a ) 7900 , b ) 8,000 , c ) 10,000 , d ) 8,300 , e ) 9,300 | c | multiply(multiply(1280, 3), 3) | multiply(n1,n2)|multiply(n1,#0)| | gain |
in the xy - coordinate system , if ( m , n ) and ( m + 4 , n + k ) are two points on the line with the equation x = 2 y + 5 , then k = | "since ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 they should satisfy m = 2 n + 5 and m + 4 = 2 * ( n + k ) + 5 . by 1 st equation we have m - 2 n = 5 and by 2 nd equation m - 2 n = 2 k + 1 - - - > 5 = 2 k + 1 - - - > k = 2 . the answer is , therefore , ( c ) ." | a ) 1 / 2 , b ) 1 , c ) 2 , d ) 5 / 2 , e ) 4 | c | divide(subtract(add(5, 4), 5), 2) | add(n0,n2)|subtract(#0,n2)|divide(#1,n1)| | general |
during the first week of performances of a certain play , 5400 tickets were sold , all at reduced price . during the remaining weeks of performances , 5 times as many tickets were sold at full price as were sold at reduced price . if the total number of tickets r sold was 25200 , how many of them were sold at full pric... | given : the question tells us about the number of tickets sold in the first week at reduced price = 5400 . it also tells us that during the remaining weeks , the number of tickets sold at full price was 5 times the no . of tickets sold at reduced price . the total tickets sold was 25200 , and we are asked to find the n... | a ) 21000 , b ) 19800 , c ) 16500 , d ) 15800 , e ) 8700 | c | multiply(divide(subtract(25200, 5400), add(5, const_1)), 5) | add(n1,const_1)|subtract(n2,n0)|divide(#1,#0)|multiply(n1,#2) | general |
income and expenditure of a person are in the ratio 10 : 4 . if the income of the person is rs . 19000 , then find his savings ? | "let the income and the expenditure of the person be rs . 10 x and rs . 4 x respectively . income , 10 x = 19000 = > x = 1900 savings = income - expenditure = 10 x - 4 x = 6 x = 6 ( 1900 ) so , savings = rs . 11400 . answer : d" | a ) 79754 , b ) 87689 , c ) 59875 , d ) 11400 , e ) 87648 | d | subtract(19000, multiply(divide(4, 10), 19000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other |
a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 2000 . the difference between the profit shares of a and c is ? | "explanation : ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 2000 = > 5 parts out of 15 parts is rs . 2000 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 2000 ) = rs . 800 answer : d" | a ) 621 , b ) 276 , c ) 236 , d ) 800 , e ) 211 | d | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 2000) | divide(n2,n1)|divide(n0,n1)|subtract(#0,#1)|multiply(n3,#2)| | gain |
if a card is drawn from a well shuffled deck of cards , what is the probability of drawing a black card or a face card ? | "p ( b á ´ œ f ) = p ( b ) + p ( f ) - p ( b â ˆ © f ) , where b denotes black cards and f denotes face cards . p ( b á ´ œ f ) = 26 / 52 + 12 / 52 - 6 / 52 = 8 / 13 answer : d" | a ) 7 / 13 , b ) 15 / 26 , c ) 9 / 12 , d ) 8 / 13 , e ) 29 / 52 | d | add(divide(const_3, const_52), divide(divide(const_52, const_4), const_52)) | divide(const_3,const_52)|divide(const_52,const_4)|divide(#1,const_52)|add(#0,#2)| | probability |
a cistern 4 m long and 2 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is : | "area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 2 x 1.25 + 4 x 1.25 ) + 4 x 2 ] m 2 = 23 m 2 . answer : a" | a ) 23 m 2 , b ) 20 m 2 , c ) 52 m 2 , d ) 24 m 2 , e ) 55 m 2 | a | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 2), multiply(add(divide(25, const_100), 1), 4))), multiply(2, 4)) | divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)| | physics |
jane makes toy bears . when she works with an assistant , she makes 62 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? | "let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 64.8 bears per 36 hours a week or 1.8 bears / hr ( [ 40 bears * 1.62 ] / [ 40 hrs * . 90 ] ) . [ ( 1.8 - 1 ) / 1 ] * 100 % = 80 % answer : b" | a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | b | multiply(divide(10, subtract(subtract(const_100, 62), 10)), const_100) | subtract(const_100,n0)|subtract(#0,n1)|divide(n1,#1)|multiply(#2,const_100)| | physics |
find the average of all numbers between 5 and 37 which are divisible by 4 | "explanation : average = ( 4 + 8 + 12 + 16 + 20 + 24 + 28 + 32 + 36 ) / 4 ) = 180 / 4 = 45 option c" | a ) 25 , b ) 20 , c ) 45 , d ) 30 , e ) 35 | c | divide(add(add(add(multiply(4, const_3), add(4, multiply(4, const_2))), multiply(4, const_4)), multiply(add(const_4, const_1), 4)), 4) | add(const_1,const_4)|multiply(n2,const_2)|multiply(n2,const_3)|multiply(n2,const_4)|add(n2,#1)|multiply(n2,#0)|add(#4,#2)|add(#6,#3)|add(#7,#5)|divide(#8,n2)| | general |
uncle bruce is baking chocolate chip cookies . he has 36 ounces of dough ( with no chocolate ) and 18 ounces of chocolate . how many ounces of chocolate are left over if he uses all the dough but only wants the cookies to consist of 20 % chocolate ? | "answer is e . x / x + 36 = 1 / 5 x = 9 18 - 9 = 9" | a ) 3 , b ) 6 , c ) 7.2 , d ) 7.8 , e ) 9 | e | multiply(divide(20, const_100), 20) | divide(n2,const_100)|multiply(n2,#0)| | gain |
a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 50 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease o... | "assume the broker invested $ 100 . after year 1 ( 50 % increase in wealth ) : $ 100 * 0.5 = $ 50 increase + $ 100 = $ 150 after year 2 ( 30 % decrease in wealth ) : $ 150 * 0.3 = $ 45 ; $ 150 - $ 45 decrease = $ 105 net increase = $ 105 / $ 100 - 1 = 5 % answer is b ." | a ) − 5 % , b ) 5 % , c ) 15 % , d ) 20 % , e ) 80 % | b | multiply(subtract(multiply(add(const_1, divide(50, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
the sum of the squares of three numbers is 225 , while the sum of their products taken two at a time is 200 . their sum is : | "x ^ + y ^ 2 + z ^ 2 = 225 xy + yz + zx = 200 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 225 + ( 2 * 200 ) ( x + y + z ) ^ 2 = 625 so x + y + z = 25 answer : d" | a ) 10 , b ) 30 , c ) 40 , d ) 25 , e ) none of these | d | add(multiply(sqrt(divide(subtract(225, 200), const_2)), const_100), sqrt(subtract(225, divide(subtract(225, 200), const_2)))) | subtract(n0,n1)|divide(#0,const_2)|sqrt(#1)|subtract(n0,#1)|multiply(#2,const_100)|sqrt(#3)|add(#4,#5)| | general |
a train travels from new york to chicago , a distance of approximately 480 miles , at an average rate of 40 miles per hour and arrives in chicago at 6 : 00 in evening , chicago time . at what hour in the morning , new york time , did the train depart for chicago ? ( note : chicago time is one hour earlier than new york... | "6 : 00 in evening in chicago = 7 : 00 in evening in new york . so , the train was in chicago 7 : 00 in the evening , new york time . the trip took t = d / r = 480 / 40 = 12 hours . therefore , the train depart from new york at 7 : 00 - 12 hours = 7 : 00 in the morning , new york time . answer : e ." | a ) 3 : 00 , b ) 4 : 00 , c ) 5 : 00 , d ) 6 : 00 , e ) 7 : 00 | e | subtract(multiply(add(6, const_1), const_3), divide(480, 40)) | add(n2,const_1)|divide(n0,n1)|multiply(#0,const_3)|subtract(#2,#1)| | physics |
if m is the set of all consecutive multiples of 9 between 100 and 500 , what is the median of m ? | since 100 / 9 = 11.11 , 500 / 9 = 55.55 , m can be written as { 9 * 12 , 9 * 13 , . . . . , 9 * 54 , 9 * 55 } . the number of elements of m is , therefore , 44 ( = 55 - 12 + 1 ) . since the number of elements is even number the median is the average of the 22 nd element ( = 9 * 33 ) and 23 rd element ( = 9 * 34 ) . so ... | a ) 300 , b ) 301 , c ) 301.5 , d ) 302.5 , e ) 306 | c | divide(add(add(multiply(9, floor(divide(100, 9))), multiply(divide(floor(divide(subtract(500, 100), 9)), const_2), 9)), add(add(multiply(9, floor(divide(100, 9))), multiply(divide(floor(divide(subtract(500, 100), 9)), const_2), 9)), 9)), const_2) | divide(n1,n0)|subtract(n2,n1)|divide(#1,n0)|floor(#0)|floor(#2)|multiply(n0,#3)|divide(#4,const_2)|multiply(n0,#6)|add(#5,#7)|add(n0,#8)|add(#8,#9)|divide(#10,const_2) | physics |
a 300 meter long train crosses a platform in 42 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 42 = 50 / 3 3 ( x + 300 ) = 2100 è x = 400 m . answer : e" | a ) 99 , b ) 289 , c ) 350 , d ) 882 , e ) 400 | e | subtract(multiply(divide(300, 18), 42), 300) | divide(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
shop offered 20 % offer for every shirt , smith bought a shirt at rs . 780 . and what was the shop ' s original selling price ? | "sp * ( 80 / 100 ) = 780 sp = 9.75 * 100 = > cp = 975 answer : e" | a ) 500 , b ) 550 , c ) 600 , d ) 700 , e ) 975 | e | divide(780, subtract(const_1, divide(20, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
two positive integers differ by 4 , and sum of their reciprocals is 1 . then one of the numbers is | algebraic approach : let n be the smaller integer = > 1 / n + 1 / ( n + 4 ) = 1 or ( ( n + 4 ) + n ) / n ( n + 4 ) = 1 or n ^ 2 + 4 n = 2 n + 4 or n = 2 as n can not be - negative solve for n = > n = 2 . hence , a | a ) 2 , b ) b ) 1 , c ) c ) 5 , d ) d ) 21 , e ) e ) 28 | a | add(1, const_1) | add(n1,const_1) | general |
john has rs 360 for his expenses . if he exceeds his days by 4 days he must cut down daily expenses by rs 3 . the number of days of john ' s tour program is | let john under takes a tour of x days . then , expenses for each day = 360 x 360 x + 4 = 360 x − 3 x = 20 and − 24 hence , x = 20 days . c | a ) 10 , b ) 15 , c ) 20 , d ) 24 , e ) 27 | c | divide(subtract(sqrt(add(multiply(multiply(multiply(360, 4), 4), 3), power(multiply(3, 4), const_2))), multiply(3, 4)), multiply(const_2, 3)) | multiply(n0,n1)|multiply(n1,n2)|multiply(n2,const_2)|multiply(n1,#0)|power(#1,const_2)|multiply(n2,#3)|add(#5,#4)|sqrt(#6)|subtract(#7,#1)|divide(#8,#2) | physics |
three numbers are in the ratio 3 : 4 : 5 and their l . c . m . is 1200 . their h . c . f is ? | "let the numbers be 3 x , 4 x and 5 x their l . c . m . = 60 x 60 x = 1200 x = 20 the numbers are 3 * 20 , 4 * 20 , 5 * 20 hence required h . c . f . = 20 answer is a" | a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | a | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | other |
the price of food in the area where the sims family lives is scheduled to increase by 50 % next year . since the sims family can not afford an increase in their food bill , how much will they have to reduce consumption to keep their cost the same ? | solution : let the current food expense be represented by rs . 100 . the cost of food rises 50 % . so , to buy same amount of food , they need to increase their expense , = ( 100 + 50 % of 100 ) = rs . 150 . but , they want to keep food expense the same , so they have to cut rs . by 50 to keep it to rs . = 100 . the % ... | a ) 17 % , b ) 25.6 % , c ) none % , d ) 33.3 % , e ) 50 % | d | multiply(subtract(const_1, divide(const_1, divide(add(const_100, 50), const_100))), const_100) | add(n0,const_100)|divide(#0,const_100)|divide(const_1,#1)|subtract(const_1,#2)|multiply(#3,const_100) | general |
find the simple interest on $ 10000 at 5 % per annum for 12 months ? | "p = $ 10000 r = 5 % t = 12 / 12 years = 1 year s . i . = p * r * t / 100 = 10000 * 5 * 1 / 100 = $ 500 answer is b" | a ) $ 410 , b ) $ 500 , c ) $ 650 , d ) $ 710 , e ) $ 1000 | b | multiply(10000, divide(5, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
if the given two numbers are respectively 8 % and 16 % of a third number , then what percentage is the first of the second ? | "here , l = 8 and m = 16 therefore , first number = l / m x 100 % of second number = 8 / 16 x 100 % of second number = 50 % of second number answer : d" | a ) 20 % , b ) 25 % , c ) 18 % , d ) 50 % , e ) none of these | d | multiply(divide(divide(8, const_100), divide(16, const_100)), const_100) | divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)| | gain |
if you keep rolling a pair of dice together till a sum of 5 or 7 is obtained , then what is the probability that a sum of 5 comes before a sum of 7 ? | this can be solved in a generic and complex way but let us not go into all that . there can be four ways through which the pair of dice results in a sum of 5 . there can be six ways through which the pair of dice can result in a sum of 7 . now , we want the probability of the pair of dice resulting in a sum of 5 before... | a ) 5 / 5 , b ) 4 / 5 , c ) 3 / 5 , d ) 2 / 5 , e ) 1 / 5 | d | divide(const_4, add(const_4, const_6)) | add(const_4,const_6)|divide(const_4,#0) | probability |
a pupil ' s marks were wrongly entered as 73 instead of 63 . due to the average marks for the class got increased by half . the number of pupils in the class is ? | "let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 73 - 63 ) = > x / 2 = 10 = > x = 20 . answer : c" | a ) 18 , b ) 82 , c ) 20 , d ) 27 , e ) 29 | c | multiply(subtract(73, 63), const_2) | subtract(n0,n1)|multiply(#0,const_2)| | general |
in the coordinate plane , one of the vertices of a square is the point ( - 3 , - 4 ) . if the diagonals of that square intersect at point ( 4 , 2 ) , what is the area of that square ? | "one point ( - 3 - 4 ) , intersection ( 4,2 ) so the distance from the first point - 3 - 4 = - 7 is the midpoint of the square - - > whole side 14 , 14 * 14 = 196 a" | a ) 196 , b ) 125 , c ) 140 , d ) 128 , e ) 134 | a | multiply(multiply(3, 4), multiply(3, 4)) | multiply(n0,n2)|multiply(#0,#0)| | geometry |
a sports equipment store sold ping pong rackets for a total of $ 588 . 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 = $ 588 x = 60 pairs were sold . answer : b" | a ) 50 , b ) 60 , c ) 490 , d ) 500 , e ) 980 | b | divide(588, 9.8) | divide(n0,n1)| | general |
if a is a positive integer , and if the units digit of a ^ 2 is 1 and the units digit of ( a + 1 ) ^ 2 is 4 , what is the units digit of ( a + 2 ) ^ 2 ? | "if the units digit of a ^ 2 is 1 , then the units digit of a is either 1 or 9 . if the units digit of ( a + 1 ) ^ 2 is 4 , then the units digit of a + 1 is either 2 or 8 . to satisfy both conditions , the units digit of a must be 1 . then a + 2 has the units digit of 3 , thus the units digit of ( a + 2 ) ^ 2 will be 9... | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | e | power(add(multiply(1, 2), 2), 2) | multiply(n0,n1)|add(n0,#0)|power(#1,n0)| | general |
the difference between simple interest and compound interest on $ 1200 for one year at 10 % per annum reckoned half - yearly is : | "s . i = $ [ ( 1200 * 10 * 1 ) / 100 ] = $ 120 . c . i = $ [ 1200 * ( 1 + ( 5 / 100 ) ^ 2 ) - 1200 ] = $ 123 . difference = $ ( 123 - 120 ) = $ 3 . answer ( d )" | a ) $ 5 , b ) $ 8 , c ) $ 9 , d ) $ 3 , e ) $ 7 | d | subtract(subtract(multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 1200), 1200), divide(multiply(1200, 10), const_100)) | divide(n1,const_2)|multiply(n0,n1)|divide(#0,const_100)|divide(#1,const_100)|add(#2,const_1)|power(#4,const_2)|multiply(n0,#5)|subtract(#6,n0)|subtract(#7,#3)| | gain |
a train 110 meters long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train . find the speed of the train . | explanation : let the speed of the train be x kmph . speed of the train relative to man = ( x + 5 ) kmph = ( x + 5 ) × 5 / 18 m / sec . therefore 110 / ( ( x + 5 ) × 5 / 18 ) = 6 < = > 30 ( x + 5 ) = 1980 < = > x = 61 speed of the train is 61 kmph . answer : option d | a ) 45 kmph , b ) 50 kmph , c ) 55 kmph , d ) 61 kmph , e ) 70 kmph | d | subtract(divide(110, multiply(6, const_0_2778)), 5) | multiply(n1,const_0_2778)|divide(n0,#0)|subtract(#1,n2) | physics |
the price of a bushel of corn is currently $ 3.20 , and the price of a peck of wheat is $ 5.80 . the price of corn is increasing at a constant rate of 5 x 5 x cents per day while the price of wheat is decreasing at a constant rate of 2 √ ∗ x − x 2 ∗ x − x cents per day . what is the approximate price when a bushel of c... | "let yy be the # of days when these two bushels will have the same price . first let ' s simplify the formula given for the rate of decrease of the price of wheat : 2 √ ∗ x − x = 1.41 x − x = 0.41 x 2 ∗ x − x = 1.41 x − x = 0.41 x , this means that the price of wheat decreases by 0.41 x 0.41 x cents per day , in yy day... | a ) $ 4.50 , b ) $ 5.10 , c ) $ 5.30 , d ) $ 5.50 , e ) $ 5.60 | e | add(3.20, multiply(divide(subtract(5.80, 3.20), add(5, subtract(sqrt(5), 2))), 5)) | sqrt(n3)|subtract(n1,n0)|subtract(#0,n4)|add(n2,#2)|divide(#1,#3)|multiply(n2,#4)|add(n0,#5)| | gain |
find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number | "f the number is x , then , x + 17 = 60 / x x 2 + 17 x - 60 = 0 ( x + 20 ) ( x - 3 ) = 0 x = 3 , - 20 , so x = 3 ( as 3 is positive ) answer : e" | a ) 17 , b ) 15 , c ) 12 , d ) 8 , e ) 3 | e | add(const_2, gcd(60, 17)) | gcd(n0,n1)|add(#0,const_2)| | general |
the length of a rectangular garden is three times its width . if the area of the rectangular garden is 588 square meters , then what is the width of the rectangular garden ? | "let x be the width of the garden . 3 x ^ 2 = 588 x ^ 2 = 196 x = 14 the answer is c ." | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | c | sqrt(divide(588, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry |
a train 360 m long is running at a speed of 72 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 1 / 20 = 25 sec answer : c" | a ) 40 sec , b ) 29 sec , c ) 25 sec , d ) 27 sec , e ) 34 sec | c | divide(360, multiply(subtract(72, 140), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
if 20 typists can type 48 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "20 typists can type 48 letters , so 30 typists can type = 48 * 30 / 20 48 * 30 / 20 letters can be typed in 20 mins . in 60 mins typist can type = 48 * 30 * 60 / 20 * 20 = 216 d is the answer" | a ) 63 , b ) 72 , c ) 144 , d ) 216 , e ) 400 | d | multiply(divide(multiply(48, const_3), 20), 30) | multiply(n1,const_3)|divide(#0,n0)|multiply(n3,#1)| | physics |
a car traveling at a certain constant speed takes 15 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 60 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "60 * t = 1 km = > t = 1 / 60 km / h v * ( t + 15 / 3600 ) = 1 v ( 1 / 60 + 15 / 3600 ) = 1 v ( 75 / 3600 ) = 1 v = 48 km / h the answer is a ." | a ) 48 , b ) 50 , c ) 52 , d ) 54 , e ) 56 | a | divide(1, divide(add(multiply(const_3600, divide(1, 60)), 15), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
p and q started a business investing rs . 96,000 and rs . 28,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 96000 : 28000 = 24 : 7 answer : b" | a ) 17 : 6 , b ) 24 : 7 , c ) 25 : 6 , d ) 21 : 3 , e ) 17 : 4 | b | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | add(n2,const_3)|add(#0,const_3)|multiply(n2,#0)|multiply(#1,#2)|multiply(#2,const_3)|add(#0,#3)|add(#0,#4)|divide(#5,#6)| | gain |
a train 90 m long running at 56 kmph crosses a platform in 18 sec . what is the length of the platform ? | "d = 56 * 5 / 18 = 25 = 280 – 90 = 190 answer : a" | a ) 190 , b ) 200 , c ) 180 , d ) 881 , e ) 271 | a | subtract(multiply(18, multiply(56, const_0_2778)), 90) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
a card is drawn from a pack of 52 cards . the probability of ge ƫ ng a queen of club or a king of heart is | "explanation : total number of cases = 52 favourable cases = 2 probability = 2 / 56 = 1 / 26 answer : c" | a ) 1 / 13 , b ) 2 / 13 , c ) 1 / 26 , d ) 1 / 52 , e ) none of these | c | divide(subtract(52, multiply(const_4, const_4)), 52) | multiply(const_4,const_4)|subtract(n0,#0)|divide(#1,n0)| | probability |
this topic is locked . if you want to discuss this question please re - post it in the respective forum . matt and peter can do together a piece of work in 20 days . after they have worked together for 12 days matt stops and peter completes the remaining work in 14 days . in how many days peter complete the work separa... | together they complete the job in 20 days means they complete 12 / 20 of the job after 12 days . peter completes the remaining ( 8 / 20 ) of the job in 14 days which means that the whole job ( 1 ) can be completed in x days . < = > 8 / 20 - > 14 < = > x = 14 / ( 8 / 20 ) = 35 thus the answer is d . | a ) 26 days , b ) 27 days , c ) 23 days , d ) 35 days , e ) 24 days | d | add(multiply(divide(const_1, const_2), 14), multiply(14, const_2)) | divide(const_1,const_2)|multiply(n2,const_2)|multiply(n2,#0)|add(#2,#1) | physics |
we bought a total of 80 books at the store . math books cost $ 4 and history books cost $ 5 . the total price was $ 373 . how many math books did we buy ? | "m + h = 80 h = 80 - m 4 m + 5 h = 373 4 m + 5 * ( 80 - m ) = 373 m = 27 the answer is b ." | a ) 23 , b ) 27 , c ) 31 , d ) 35 , e ) 39 | b | subtract(80, subtract(373, multiply(80, 4))) | multiply(n0,n1)|subtract(n3,#0)|subtract(n0,#1)| | general |
if a , b , c , d , e and f are integers and ( ab + cdef ) < 0 , then what is the maximum number s of integers that can be negative ? | "minimuum should be 1 maximum should be 4 : 1 out of a or b to make the multiplication negative 3 out of c , d , e or f to make the multiplication negative . negative + negative < 0 answer : c maximum will be 5 . . you dont require both the multiplicatin to be negative for entire equation to be negative . . . any one a... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(divide(const_10, const_2), 0) | divide(const_10,const_2)|add(n0,#0)| | general |
a monkey ascends a greased pole 14 metres high . he ascends 2 metres in first minute and slips down 1 metre in the alternate minute . in which minute , he reaches the top ? | "in 2 minutes , he ascends = 1 metre â ˆ ´ 12 metres , he ascends in 24 minutes . â ˆ ´ he reaches the top in 25 th minute . answer a" | a ) 25 th , b ) 22 nd , c ) 23 rd , d ) 24 th , e ) none of these | a | subtract(multiply(2, 14), 1) | multiply(n0,n1)|subtract(#0,n2)| | physics |
due to a reduction of 25 / 4 % in the price of sugar , a man is able to buy 1 kg more for rs . 120 . find the original rate of sugar . | rate reduced by 6.25 % , so amount rs 120 reduced by = ( 6.25 / 100 ) * 120 = 7.5 so in rs 7.5 extra sugar obtained is 1 kg i . e . new rate is rs . 7.5 per kg thus original rate = [ 7.5 / ( 100 – 6.25 ) ] / 100 = 7.5 / 0.9375 = rs . 8 answer : option b | a ) 7 , b ) 8 , c ) 8.5 , d ) 9 , e ) 9.5 | b | divide(multiply(divide(divide(25, 4), const_100), 120), divide(subtract(const_100, divide(25, 4)), const_100)) | divide(n0,n1)|divide(#0,const_100)|subtract(const_100,#0)|divide(#2,const_100)|multiply(n3,#1)|divide(#4,#3) | gain |
on a test average ( arithmetic mean ) test score for 9 students is 80 . what must be 10 th student ' s score that average score for 10 students to be 82 ? | "( 9 * 80 + x ) / 10 = 82 x = ( 10 * 82 ) - ( 9 * 80 ) x = 820 - 720 total score required 820 - 720 = 100 correct answer is a" | a ) 100 , b ) 90 , c ) 80 , d ) 70 , e ) 60 | a | subtract(multiply(82, 10), multiply(9, 80)) | multiply(n2,n4)|multiply(n0,n1)|subtract(#0,#1)| | general |
a man can row 4 kmph in still water . when the river is running at 1.8 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | "m = 4 s = 1.8 ds = 5.8 us = 2.2 x / 5.8 + x / 2.2 = 1 x = 1.59 d = 1.59 * 2 = 3.19 answer : b" | a ) 7.13 , b ) 6.19 , c ) 4.12 , d ) 3.19 , e ) 4.21 | b | multiply(divide(multiply(add(4, 1.8), subtract(4, 1.8)), add(add(4, 1.8), subtract(4, 1.8))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
the annual interest rate earned by an investment increased by 5 percent from last year to this year . if the annual interest rate earned by the investment this year was 26.25 percent , what was the annual interest rate last year ? | "let i = interest rate i ( this year ) = i ( last year ) + 0.05 i ( last year ) = 1.05 i ( last year ) 26.25 = 1.05 x i ( last year ) i ( last year ) = 26.25 / 1.05 = 2625 / 105 = 25 % answer : a" | a ) 25 % , b ) 24 % , c ) 23 % , d ) 22 % , e ) 21 % | a | divide(multiply(26.25, const_100), add(26.25, const_100)) | add(n1,const_100)|multiply(n1,const_100)|divide(#1,#0)| | gain |
a certain farmer pays $ 30 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is 360 feet by 1210 feet ? ( 43,560 square feet = 1 acre ) | basically the question an error . 1 acre = 43,560 square feet and if it is then the answer is 300 ( d ) | ['a ) $ 5,330', 'b ) $ 3,360', 'c ) $ 1,350', 'd ) $ 300', 'e ) $ 150'] | d | multiply(30, divide(multiply(360, 1210), divide(multiply(360, 1210), const_10))) | multiply(n1,n2)|divide(#0,const_10)|divide(#0,#1)|multiply(n0,#2) | geometry |
how many multiples of 7 are there between 21 and 343 , inclusive ? | "7 * 3 = 21 7 * 49 = 343 total multiples = ( 49 - 3 ) + 1 = 47 include 21 and 343 = 47 answer is b" | a ) 48 , b ) 47 , c ) 46 , d ) 45 , e ) 44 | b | add(divide(subtract(343, 21), 7), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1600 employees , then the no . of female employees who are computer literate ? | "solution : total employees , = 1600 female employees , 60 % of 1600 . = ( 60 * 1600 ) / 100 = 960 . then male employees , = 640 50 % of male are computer literate , = 320 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1600 ) / 100 = 992 computer literate . thus , female computer lite... | a ) 690 , b ) 674 , c ) 672 , d ) 960 , e ) none | c | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1600) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#2)|multiply(#1,#3)|subtract(#0,#4)|multiply(n3,#5)| | gain |
shop offered 20 % offer for every shirt , smith bought a shirt at rs . 560 . and what was the shop ' s original selling price ? | "sp * ( 80 / 100 ) = 560 sp = 70 * 100 = > cp = 700 answer : d" | a ) 500 , b ) 550 , c ) 600 , d ) 700 , e ) 750 | d | divide(560, subtract(const_1, divide(20, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
the present worth of $ 3600 due in 2 years at 20 % per annum compound interest is ? | present worth = 3600 / ( 1 + 20 / 100 ) ^ 2 = 3600 * 5 / 6 * 5 / 6 = $ 2500 answer is e | a ) $ 1200 , b ) $ 1850 , c ) $ 2010 , d ) $ 3200 , e ) $ 2500 | e | divide(3600, power(add(divide(20, const_100), const_1), 2)) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2) | gain |
if the wheel is 19 cm then the number of revolutions to cover a distance of 1156 cm is ? | 2 * 22 / 7 * 19 * x = 1156 = > x = 9.7 answer : d | a ) 11 , b ) 14 , c ) 18 , d ) 10 , e ) 19 | d | divide(1156, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 19)) | 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 |
harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 60 per kg . how much amount did he pay to the shopkeeper ? | cost of 8 kg grapes = 70 × 8 = 560 . cost of 9 kg of mangoes = 60 × 9 = 540 . total cost he has to pay = 560 + 540 = 1100 . e ) | a ) a ) 1055 , b ) b ) 1075 , c ) c ) 1065 , d ) d ) 1070 , e ) e ) 1100 | e | add(multiply(8, 70), multiply(9, 60)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1) | gain |
the commuter rail between scottsdale and sherbourne is 200 km of track , and one train makes a round trip in 5 hours . if harsha boards the train at the forest grove station , which is located one fifth of the track ' s length out from scottsdale , how long will it take her to get to sherbourne ? | distance between scottsdale and sherbourne = 200 km time for a one - way trip between scottsdale and sherbourne : 5 / 2 = 2.5 hours distance between forest grove and scottsdale : 200 / 5 = 40 km distance between forest grove and sherbourne : 40 x 4 = 160 km time between scottsdale and forest grove : 2.5 / 5 = . 5 hours... | a ) . 5 hours , b ) 1 hour , c ) 1.5 hours , d ) 2 hours , e ) 2.5 hours | d | divide(200, multiply(divide(200, 5), divide(5, const_2))) | divide(n0,n1)|divide(n1,const_2)|multiply(#0,#1)|divide(n0,#2) | physics |
excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 35 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 10 / 45 * 60 = 13 answer : option d" | a ) 73 , b ) 83 , c ) 12 , d ) 13 , e ) 28 | d | subtract(const_60, multiply(const_60, divide(35, 45))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
if the price of a certain bond on may 1 st was 2 / 3 the price of the bond on june 1 st and the price of the bond on july 1 st was 25 % greater than the price of the bond on may 1 st . then the price of the bond on june 1 st st was what percent of the average ( arithmetic mean ) price of the bond on may 1 st and july 1... | "the price on june 1 st = 12 ( assume ) ; the price on may 1 st = 2 / 3 * 12 = 8 ; the price on july 1 st = 8 * 1.25 = 10 . the average price of the bond on may 1 st and july 1 st = ( 8 + 10 ) / 2 = 9 . the price of the bond on june 1 st ( 12 ) is 4 / 3 times ( 134 % ) the average price of the bond on may 1 st and july... | a ) 50 % , b ) 75 % , c ) 120 % , d ) 133 1 / 3 % , e ) 150 % | d | add(multiply(25, const_0_33), add(const_100, const_0_33)) | add(const_0_33,const_100)|multiply(n5,const_0_33)|add(#0,#1)| | general |
to asphalt 1 km road , 30 men spent 12 days working 8 hours per day . how many days , 20 men will spend to asphalt a road of 2 km working 10 hours a day ? | "man - hours required to asphalt 1 km road = 30 * 12 * 8 = 2880 man - hours required to asphalt 2 km road = 2880 * 2 = 5760 man - hours available per day = 20 * 10 = 200 therefore number of days = 5760 / 200 = 28.8 days ans = a" | a ) 28.8 , b ) 30 , c ) 32 , d ) 34 , e ) 36 | a | divide(multiply(multiply(multiply(30, 12), 8), 2), multiply(20, 10)) | multiply(n1,n2)|multiply(n4,n6)|multiply(n3,#0)|multiply(n5,#2)|divide(#3,#1)| | physics |
two good train each 900 m long , are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . | "sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 900 + 900 ) m = 1000 m . required time = ( 1800 x 6 / 125 ) sec = 86.4 sec . answer c" | a ) 12.2 sec , b ) 24.9 sec , c ) 86.4 sec , d ) 60.2 sec , e ) none | c | multiply(multiply(900, inverse(multiply(add(45, 30), const_0_2778))), const_2) | add(n1,n2)|multiply(#0,const_0_2778)|inverse(#1)|multiply(n0,#2)|multiply(#3,const_2)| | physics |
a charitable association sold an average of 66 raffle tickets per member . among the female members , the average was 70 raffle tickets . the male to female ratio of the association is 1 : 2 . what was the average number e of tickets sold by the male members of the association | given that , total average e sold is 66 , male / female = 1 / 2 and female average is 70 . average of male members isx . ( 70 * f + x * m ) / ( m + f ) = 66 - > solving this equation after substituting 2 m = f , x = 58 . ans c . | a ) 50 , b ) 56 , c ) 58 , d ) 62 , e ) 66 | c | subtract(multiply(66, add(1, 2)), multiply(70, 2)) | add(n2,n3)|multiply(n1,n3)|multiply(n0,#0)|subtract(#2,#1) | general |
uncle bruce is baking chocolate chip cookies . he has 72 ounces of dough ( with no chocolate ) and 50 ounces of chocolate . how much chocolate is left over if he uses all the dough but only wants the cookies to consist of 40 % chocolate ? | "first , you must find the total weight of the mixture given that 80 % of it will be dough . 60 % * total = 72 = > ( 6 / 10 ) total = 72 = > total = 720 / 6 = > total = 120 oz , from there , you must find 10 % of the total 40 oz of the mixture . 40 % * total = > ( 4 / 10 ) ( 120 ) = 48 oz choclate used , not forgetting... | a ) 4 , b ) 6 , c ) 8 , d ) 2 , e ) 10 | d | multiply(divide(40, const_100), 40) | divide(n2,const_100)|multiply(n2,#0)| | gain |
107 ã — 107 + 93 ã — 93 = ? | "explanation : ( a + b ) 2 + ( a â ˆ ’ b ) 2 = 2 ( a 2 + b 2 ) ( reference : basic algebraic formulas ) 1072 + 932 = ( 100 + 7 ) 2 + ( 100 â ˆ ’ 7 ) 2 = 2 ( 1002 + 72 ) = 2 ( 10000 + 49 ) = 20098 . answer : option b" | a ) 20198 , b ) 20098 , c ) 20000 , d ) 20088 , e ) 20085 | b | add(subtract(power(107, const_2), 107), subtract(power(93, const_2), 93)) | power(n0,const_2)|power(n2,const_2)|subtract(#0,n0)|subtract(#1,n2)|add(#2,#3)| | general |
how many integers from 100 to 800 , inclusive , remains the value unchanged when the digits were reversed ? | "question is asking for palindrome first digit possibilities - 1 through 7 = 7 8 is not possible here because it would result in a number greater than 8 ( i . e 808 , 818 . . ) second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 7 * 10... | a ) 50 , b ) 60 , c ) 80 , d ) 70 , e ) 90 | d | divide(800, const_10) | divide(n1,const_10)| | general |
a part - time employee whose hourly wage was increased by 50 percent decided to reduce the number of hours worked per week so that the employee ' s total weekly income would remain unchanged . by what percent should the number of hours worked be reduced ? | "let original hourly wage be x and let the no of hours worked be y total wage will be = x * y after the increment the wage will be = 1.5 x now we need to find number of hours worked so that x * y = 1.5 x * z i . e z = 1 / 1.5 y = 2 / 3 y % decrease = ( y - 2 / 3 y ) / y * 100 = 100 / 3 = 33.33 % . thus my answer is e .... | a ) 9 , b ) 13 , c ) 17 , d ) 20 , e ) 33.33 | e | multiply(divide(const_1, multiply(divide(add(50, const_100), const_100), 50)), const_100) | add(n0,const_100)|divide(#0,const_100)|multiply(#1,n0)|divide(const_1,#2)|multiply(#3,const_100)| | general |
pascal has 96 miles remaining to complete his cycling trip . if he reduced his current speed by 4 miles per hour , the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50 % . what is his current speed w ? | "let the current speed be x miles per hour . time taken if speed is 50 % faster ( i . e . 3 x / 2 = 1.5 x ) = 96 / 1.5 x time taken if speed is reduced by 4 miles / hr ( i . e . ( x - 4 ) ) = 96 / ( x - 4 ) as per question , 96 / ( x - 4 ) - 96 / 1.5 x = 16 solving this w we get x = 8 . b ." | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 16 | b | divide(add(divide(96, 16), sqrt(add(multiply(multiply(divide(divide(96, 16), add(const_1, divide(50, const_100))), 4), 4), power(divide(96, 16), const_2)))), const_2) | divide(n0,n2)|divide(n3,const_100)|add(#1,const_1)|power(#0,const_2)|divide(#0,#2)|multiply(n1,#4)|multiply(#5,n1)|add(#6,#3)|sqrt(#7)|add(#0,#8)|divide(#9,const_2)| | physics |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.