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 |
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if the ratio of a to b is 11 to 3 and the ratio of b to c is 1 to 5 , what is the ratio of a to c ? | "a : b = 11 : 3 - - 1 b : c = 1 : 5 = > b : c = 3 : 15 - - 2 from 1 and 2 , we get a : c = 11 : 15 answer b" | a ) 4 / 15 , b ) 11 / 15 , c ) 2 / 5 , d ) 4 / 5 , e ) 7 / 6 | b | divide(multiply(11, 1), multiply(3, 5)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other |
what should be the least number to be added to the 2496 number to make it divisible by 5 ? | "answer : 4 option : d" | a ) 12 , b ) 17 , c ) 18 , d ) 4 , e ) 26 | d | subtract(5, reminder(2496, 5)) | reminder(n0,n1)|subtract(n1,#0)| | general |
1 ⁄ 5 of a number is equal to 5 ⁄ 8 of the second number . if 35 is added to the first number then it becomes 4 times of second number . what is the value of the second number ? | let x be the first number and y be the second number . 1 ⁄ 5 x = 5 ⁄ 8 y \ x ⁄ y = 25 ⁄ 8 . . . . . . ( i ) x + 35 = 4 y or , 25 ⁄ 8 y + 35 = 4 y \ y = 40 answee c | a ) 125 , b ) 70 , c ) 40 , d ) 25 , e ) none of these | c | divide(multiply(35, 8), subtract(multiply(8, 4), multiply(5, 5))) | multiply(n3,n4)|multiply(n3,n5)|multiply(n1,n1)|subtract(#1,#2)|divide(#0,#3) | general |
when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 75 . | "first , let us say i have a number n which is divisible by 5 and by 7 . we all agree that it will be divisible by 35 , the lcm of 5 and 7 . now , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 1 , we can say the number is of the form n = 5 a + 1 e . g . 5 + 1 ,... | a ) 3 , b ) 4 , c ) 12 , d ) 32 , e ) 35 | e | subtract(75, reminder(3, 7)) | reminder(n3,n2)|subtract(n4,#0)| | general |
in what time will a train 130 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 130 / 40 = 3.25 sec . answer : b" | a ) 2.5 sec , b ) 3.25 sec , c ) 5 sec , d ) 12.5 sec , e ) 6 sec | b | divide(130, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
donald duck can can swim his pool downstream ( with the pool current helping time ) in exact 40 seconds and upstream ( against the pool current ) in a pool in exact 60 seconds . the length of pool is 2 kilometers . how long donald duck can cover distance of one side at a still pool ( with no current ) . | d 48 minutes . donald duck ' s speed = x km / seconds pool current speed = y km / seconds 2 / ( x + y ) = 40 2 / ( x - y ) = 60 solving the simultaneous equations gives x = 1 / 24 therefore , to cover 2 km will take 2 / x = 48 seconds | a ) 40 minutes , b ) 45 minutes , c ) 38 minutes , d ) 48 minutes , e ) 30 minutes | d | divide(multiply(2, const_1000), divide(add(divide(multiply(2, const_1000), 40), divide(multiply(2, const_1000), 60)), 2)) | multiply(n2,const_1000)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|divide(#3,n2)|divide(#0,#4) | physics |
a number is increased by 11 % and then reduced by 10 % . after these operations , the number : | "let the original number be 100 . then , the new number = 100 × 1.1 × 0.9 = 99 i . e . the number decreases by 1 % . answer b" | a ) does not change , b ) decreases by 1 % , c ) increases by 1 % , d ) increases by 0.1 % , e ) none of these | b | divide(multiply(add(11, const_100), subtract(const_100, 11)), const_100) | add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)| | other |
an plane covers a certain distance at a speed of 240 kmph in 5 hours . to cover the same distance in 5 / 3 hours , it must travel at a speed of ? | we know distance = speed * time = 240 * 5 = 1200 km now to cover 1200 km in 5 / 3 hrs , speed = distance / time = 1200 / ( 5 / 3 ) = 720 kmph ans - b | a ) 100 , b ) 720 , c ) 1100 , d ) 940 , e ) 860 | b | multiply(240, divide(5, divide(5, 3))) | divide(n1,n3)|divide(n1,#0)|multiply(n0,#1) | physics |
the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 4 . if the sum of their ages is 132 , what is the difference between the ages of patrick and monica ? | ages of p and mi = 3 x : 5 x ages of mi and mo = 3 x : 4 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 20 x sum = 44 x = 132 x = 3 difference if ages of pa and mo = 20 x - 9 x = 11 x = 11 * 3 = 33 answer e | a ) 27 , b ) 48 , c ) 45 , d ) 72 , e ) 33 | e | subtract(multiply(multiply(4, 5), divide(132, add(add(multiply(3, const_3), multiply(const_3, 5)), multiply(4, 5)))), multiply(multiply(3, const_3), divide(132, add(add(multiply(3, const_3), multiply(const_3, 5)), multiply(4, 5))))) | multiply(n0,const_3)|multiply(n1,const_3)|multiply(n1,n3)|add(#0,#1)|add(#3,#2)|divide(n4,#4)|multiply(#5,#2)|multiply(#5,#0)|subtract(#6,#7) | general |
two vessels contains equal number of mixtures milk and water in the ratio 5 : 2 and 6 : 1 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ? | "the ratio of milk and water in the new vessel is = ( 5 / 7 + 6 / 7 ) : ( 2 / 7 + 1 / 7 ) = 11 / 7 : 3 / 7 = 11 : 3 answer is d" | a ) 1 : 3 , b ) 9 : 13 , c ) 5 : 11 , d ) 11 : 3 , e ) 15 : 4 | d | divide(add(multiply(5, divide(add(6, 1), add(5, 2))), 6), add(multiply(2, divide(add(6, 1), add(5, 2))), 1)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n2,#3)|add(n3,#4)|divide(#5,#6)| | other |
ashok secured average of 80 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 80 therefore total marks in 6 subjects = 80 * 6 = 480 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 480 – 370 = 110 answer : d" | a ) 38 , b ) 27 , c ) 99 , d ) 110 , e ) 80 | d | subtract(multiply(80, 6), multiply(74, 5)) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)| | general |
out of 410 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ? | n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 410 - 50 = 360 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 360 = 140 . answer is d | a ) 120 , b ) 150 , c ) 100 , d ) 140 , e ) 220 | d | subtract(add(175, 325), subtract(410, 50)) | add(n1,n2)|subtract(n0,n3)|subtract(#0,#1) | other |
a number consists of 3 digit whose sum is 10 . the middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed . the number is : | explanation : let the middle digit be x . then , 2 x = 10 or x = 5 . so , the number is either 253 or 352 . since the number increases on reversing the digits , so the hundred ' s digit is smaller than the unit ' s digit . hence , required number = 253 . answer is b | a ) 145 , b ) 253 , c ) 370 , d ) 352 , e ) 300 | b | add(add(multiply(subtract(divide(add(10, const_2), const_4), const_1), const_100), multiply(add(divide(add(10, const_2), const_4), subtract(divide(add(10, const_2), const_4), const_1)), const_10)), divide(add(10, const_2), const_4)) | add(n1,const_2)|divide(#0,const_4)|subtract(#1,const_1)|add(#1,#2)|multiply(#2,const_100)|multiply(#3,const_10)|add(#4,#5)|add(#6,#1) | general |
what amount does an investor receive if the investor invests $ 5000 at 10 % p . a . compound interest for two years , compounding done annually ? | "a = ( 1 + r / 100 ) ^ n * p ( 1.1 ) ^ 2 * 5000 = 1.21 * 5000 = 6050 the answer is c ." | a ) $ 5850 , b ) $ 5950 , c ) $ 6050 , d ) $ 6150 , e ) $ 6250 | c | multiply(5000, power(add(const_1, divide(10, const_100)), const_2)) | divide(n1,const_100)|add(#0,const_1)|power(#1,const_2)|multiply(n0,#2)| | gain |
of the 3,600 employees of company x , 1 / 4 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ? | "let ' s see , the way i did it was 1 / 4 are clerical out of 3600 so 900 are clerical 900 reduced by 1 / 4 is 900 * 1 / 4 so it reduced 225 people , so there is 675 clerical people left but since 225 people left , it also reduced from the total of 3600 so there are 3375 people total since 675 clerical left / 3375 peop... | a ) 25 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | c | multiply(divide(multiply(divide(1, 4), subtract(1, divide(1, 4))), add(multiply(divide(1, 4), subtract(1, divide(1, 4))), subtract(1, divide(1, 4)))), const_100) | divide(n1,n2)|divide(n3,n4)|subtract(n3,#1)|subtract(n3,#0)|multiply(#0,#2)|add(#4,#3)|divide(#4,#5)|multiply(#6,const_100)| | general |
there are 12 teams in a certain league and each team plays each of the other teams exactly once . what is the total number of games played ? | "12 c 2 = 66 the answer is b ." | a ) 55 , b ) 66 , c ) 77 , d ) 88 , e ) 99 | b | divide(multiply(12, subtract(12, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
tough and tricky questions : statistics . set x consists of prime numbers { 3 , 11 , 7 , a , 17 , 19 } . if integer y represents the product of all elements in set x and if 11 y is an even number , what is the range of set x ? | since 11 y = even therefore y has to beevensince 11 is a odd integer ( even * odd = even ) similarly , y is the product of all integers in set x but all integers in set x are odd except the unknown a and since x contains only prime numbers , a has to equal to 2 . . . ( 2 is the only even prime number and the product of... | a ) 14 , b ) 16 , c ) 17 , d ) 20 , e ) 26 | c | subtract(19, const_2) | subtract(n4,const_2) | general |
in a simultaneous throw of a pair of dice , find the probability of getting a total more than 4 | "total number of cases = 3 * 3 = 9 favourable cases = [ ( 2,3 ) , ( 3,2 ) , ( 3,3 ) ] = 3 so probability = 3 / 9 = 1 / 3 answer is e" | a ) 1 / 2 , b ) 7 / 12 , c ) 5 / 13 , d ) 5 / 12 , e ) 1 / 3 | e | divide(subtract(4, multiply(const_2, const_3)), 4) | multiply(const_2,const_3)|subtract(n0,#0)|divide(#1,n0)| | general |
what is the total number of integers between 100 and 250 that are divisible by 5 ? | "105 , 110 , 115 , . . . , 240,245 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 245 - 105 ) / ( 5 ) + 1 = 140 / 5 + 1 = 28 + 1 = 29 answer is c" | a ) 27 , b ) 28 , c ) 29 , d ) 30 , e ) 25 | c | add(divide(subtract(250, 100), 5), const_1) | subtract(n1,n0)|divide(#0,n2)|add(#1,const_1)| | general |
what profit percent is made by selling an article at a certain price , if by selling at 4 / 5 rd of that price , there would be a loss of 20 % ? | "sp 2 = 4 / 5 sp 1 cp = 100 sp 2 = 80 4 / 5 sp 1 = 80 sp 1 = 100 100 - - - 100 = > 0 % answer : a" | a ) 0 % , b ) 29 % , c ) 70 % , d ) 27 % , e ) 28 % | a | subtract(divide(subtract(const_100, 20), divide(4, 5)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain |
on a sum of money , the simple interest for 2 years is rs . 320 , while the compound interest is rs . 340 , the rate of interest being the same in both the cases . the rate of interest is | explanation : - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - solution 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -... | a ) 15 % , b ) 14.25 % , c ) 12.5 % , d ) 10.5 % , e ) 11.5 % | c | divide(multiply(const_100, subtract(subtract(340, divide(320, 2)), divide(320, 2))), divide(320, 2)) | divide(n1,n0)|subtract(n2,#0)|subtract(#1,#0)|multiply(#2,const_100)|divide(#3,#0) | gain |
if x , y , and z are positive real numbers such that x ( y + z ) = 62 , y ( z + x ) = 82 , and z ( x + y ) = 100 , then xyz is | "xy + xz = 62 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 ) yz + yx = 82 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 ) xz + zy = 100 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 3 ) re - writing equation 3 as follows ... | a ) 120 , b ) 50 , c ) 60 , d ) 70 , e ) 90 | a | subtract(100, subtract(82, 62)) | subtract(n1,n0)|subtract(n2,#0)| | general |
of the families in city x in 1992 , 30 percent owned a personal computer . the number of families in city x owning a computer in 1999 was 50 percent greater than it was in 1992 , and the total number of families in city x was 5 percent greater in 1999 than it was in 1994 . what percent of the families in city x owned a... | "say a 100 families existed in 1992 then the number of families owning a computer in 1992 - 30 number of families owning computer in 1999 = 30 * 150 / 100 = 45 number of families in 1999 = 105 the percentage = 45 / 105 * 100 = 42.85 % . answer : b" | a ) 43 % , b ) 42.85 % , c ) 45.25 % , d ) 46.23 % , e ) 41.66 % | b | multiply(const_100, divide(divide(multiply(add(50, const_100), 30), const_100), add(const_100, 5))) | add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | general |
of the votes cast on a certain proposal , 70 more were in favor of the proposal than were against it . if the number of votes against the proposal was 40 percent of the total vote , what was the total number of votes cast ? ( each vote cast was either in favor of the proposal or against it . ) | "let x be the total number of votes cast . 0.6 x = 0.4 x + 70 0.2 x = 70 x = 350 the answer is a ." | a ) 350 , b ) 375 , c ) 400 , d ) 425 , e ) 450 | a | divide(70, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | divide(n1,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n0,#2)| | gain |
a man can do a piece of work in 4 days , but with the help of his son he can do it in 3 days . in what time can the son do it alone ? | "explanation : in this type of question , where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as son ' s one day work = ( 1 / 3 − 1 / 4 ) = ( 4 − 3 ) / 12 = 1 / 12 so son will do whole work in 12 days answer : a" | a ) 12 days , b ) 6 days , c ) 5 days , d ) 4 days , e ) none of these | a | divide(multiply(4, 3), subtract(4, 3)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics |
beginning in town a , biker bob rides his bike 10 miles west , 5 miles north , 5 miles east , and then 15 miles north , to town b . how far apart are town a and town b ? ( ignore the curvature of the earth . ) | "using pythagoras we have one side i , e total distance traveled in north direction = 15 + 5 = 20 m other being the base ie distance traveled west - distance traveled eat = 10 - 5 = 5 m now this third side or the distance between town a and town b = 20 ^ 2 + 5 ^ 2 = sq root 425 = 20.62 mile answer : b" | a ) 22 miles , b ) 20.62 miles , c ) 25 miles , d ) 26 miles , e ) 28 miles | b | sqrt(add(power(add(5, 15), const_2), power(5, const_2))) | add(n1,n3)|power(n2,const_2)|power(#0,const_2)|add(#2,#1)|sqrt(#3)| | physics |
a , b and c have rs . 400 between them , a and c together have rs . 300 and b and c rs . 150 . how much does c have ? | "a + b + c = 400 a + c = 300 b + c = 150 - - - - - - - - - - - - - - a + b + 2 c = 450 a + b + c = 400 - - - - - - - - - - - - - - - - c = 50 answer : b" | a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 100 | b | subtract(add(300, 150), 400) | add(n1,n2)|subtract(#0,n0)| | general |
0.0002152 / 0.000205 x 12.05 = ? | "explanation : ? = 0.0002152 / 0.000205 x 12.05 = 12.6 answer : option c" | a ) 19.64 , b ) 14.64 , c ) 12.6 , d ) 15.64 , e ) 13.64 | c | divide(0.0002152, 0.000205) | divide(n0,n1)| | general |
city x has a population 8 times as great as the population of city y , which has a population twice as great as the population of city z . what is the ratio of the population of city x to the population of city z ? | "x = 8 y , y = 2 * z x : y , y : z 8 : 1 , 2 : 1 16 : 2 , 2 : 1 so , x : z = 16 : 1 ( e )" | a ) 1 : 8 , b ) 1 : 4 , c ) 2 : 1 , d ) 4 : 1 , e ) 16 : 1 | e | multiply(8, const_2) | multiply(n0,const_2)| | general |
in digging a pond 20 m * 10 m * 8 m the volumes of the soil extracted will be ? | "20 * 10 * 8 = 1600 answer : b" | a ) 3387 , b ) 1600 , c ) 2866 , d ) 2787 , e ) 1121 | b | multiply(multiply(20, 10), 8) | multiply(n0,n1)|multiply(n2,#0)| | general |
a man took loan from a bank at the rate of 12 % p . a . s . i . after 10 years he had to pay rs . 1500 interest only for the period . the principal amount borrowed by him was ? | "principal = ( 100 * 1500 ) / ( 12 * 10 ) = rs . 1250 answer : c" | a ) 1502 , b ) 1900 , c ) 1250 , d ) 1750 , e ) 2000 | c | divide(divide(multiply(1500, const_100), 10), 12) | multiply(n2,const_100)|divide(#0,n1)|divide(#1,n0)| | gain |
in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 8 people all together . how many people can sit in a bus ? | "right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 8 total people can seat = 81 + 8 = 89 answer : e" | a ) 52 , b ) 49 , c ) 95 , d ) 88 , e ) 89 | e | add(multiply(add(15, subtract(15, 3)), 3), 8) | subtract(n0,n1)|add(n0,#0)|multiply(n1,#1)|add(n3,#2)| | general |
if a town of 45,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ? | 1 % is quite small and hence the answer is a ) | a ) a . 46000 , b ) b . 47000 , c ) c . 48000 , d ) d . 49000 , e ) e . 50000 | a | multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3)) | add(const_3,const_4)|add(#0,const_3)|multiply(n2,#0)|multiply(#2,const_100)|multiply(#1,#3)| | gain |
a computer wholesaler sells 10 different computers and each is priced differently . if the wholesaler chooses 3 computers for display at a trade show , what is the probability ( all things being equal ) that the two most expensive computers will be among the 3 chosen for display ? | since , two of the choices are prefixed , we are free to choose 1 from the rest of the 8 avilable . so 8 c 1 is the numerator . total no of ways we can choose 3 from 10 is 10 c 3 , which is the denominator . so , the probability : 8 c 1 / 10 c 3 = 1 / 15 ans is d . | a ) 15 / 56 , b ) 3 / 28 , c ) 1 / 28 , d ) 1 / 15 , e ) 1 / 168 | d | multiply(3, multiply(divide(subtract(3, const_1), 10), divide(const_1, subtract(10, const_1)))) | subtract(n1,const_1)|subtract(n0,const_1)|divide(#0,n0)|divide(const_1,#1)|multiply(#2,#3)|multiply(n1,#4) | general |
how many internal diagonals does a octagon ( eight sided polygon ) have ? | number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 8 no . of diagonals = 8 ( 8 - 3 ) / 2 = 20 ans e | ['a ) 90', 'b ) 85', 'c ) 70', 'd ) 35', 'e ) 20'] | e | multiply(subtract(multiply(const_2, const_4), const_3), divide(multiply(const_2, const_4), const_2)) | multiply(const_2,const_4)|divide(#0,const_2)|subtract(#0,const_3)|multiply(#1,#2) | geometry |
on thursday mabel handled 90 transactions . anthony handled 10 % more transactions than mabel , cal handled 2 / 3 rds of the transactions that anthony handled , and jade handled 17 more transactions than cal . how much transactions did jade handled ? | "solution : mabel handled 90 transactions anthony handled 10 % more transactions than mabel anthony = 90 + 90 × 10 % = 90 + 90 × 0.10 = 90 + 9 = 99 cal handled 2 / 3 rds of the transactions than anthony handled cal = 2 / 3 × 99 = 66 jade handled 16 more transactions than cal . jade = 66 + 17 = 83 jade handled = 83 tran... | a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 84 | d | add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), 3), 17) | add(n1,const_100)|divide(n0,const_100)|multiply(#0,#1)|multiply(n2,#2)|divide(#3,n3)|add(n4,#4)| | general |
among 600 students , 56 % study sociology , 44 % study mathematics and 40 % study biology . if 30 % of students study both mathematics and sociology , what is the largest possible number of students who study biology but do not study either mathematics or sociology ? | "i would just like to add a bit of explanation after the step where you calculate that the number of students studying both m and s = 180 using your analysis : we see that the total number of students who study either maths or sociology = 264 + 336 - 180 = 420 so , in the image we know that the number of students in th... | a ) 30 , b ) 90 , c ) 120 , d ) 172 , e ) 180 | e | subtract(600, subtract(add(multiply(divide(600, const_100), 56), multiply(divide(600, const_100), 44)), multiply(divide(600, const_100), 30))) | divide(n0,const_100)|multiply(n1,#0)|multiply(n2,#0)|multiply(n4,#0)|add(#1,#2)|subtract(#4,#3)|subtract(n0,#5)| | other |
if a farmer sells 15 of his goats , his stock of feed will last for 4 more days than planned , but if he buys 20 more goats , he will run out of feed 3 days earlier than planned . if no goats are sold or bought , the farmer will be exactly on schedule . how many goats does the farmer have ? | "say farmer has n goat and he is good for d days . : - we have 3 equations given in question : - ( n - 15 ) * d + 4 = ( n + 20 ) * ( d - 3 ) = n * d solving these : ( you can solve 1 st and 3 rd and 2 nd and 3 rd together ) we get : 20 d - 3 n = 60 4 n - 15 d = 60 = > n = 60 ans d it is !" | a ) 12 , b ) 24 , c ) 48 , d ) 60 , e ) 55 | d | multiply(subtract(multiply(3, 4), 20), 20) | multiply(n1,n3)|subtract(#0,n2)|multiply(n2,#1)| | general |
15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 3 hrs . each would it take for 21 women if 3 women do as much work as 2 men ? | "let 1 man does 1 unit / hr of work 15 m in 21 days of 8 hrs will do ( 15 * 21 * 8 ) units 3 w = 2 m 1 w = ( 2 / 3 ) units / hr 21 w with 3 hrs a day will take ( 15 * 21 * 8 ) / ( 21 * 3 * ( 2 / 3 ) ) days = > 60 days answer : a" | a ) 60 , b ) 20 , c ) 19 , d ) 29 , e ) 39 | a | divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 3), divide(2, 3))) | divide(n6,n5)|multiply(n0,n1)|multiply(n1,n3)|multiply(n2,#1)|multiply(#0,#2)|divide(#3,#4)| | physics |
if a tire rotates at 400 revolutions per minute when the car is traveling 72 km / h , what is the circumference of the tire ? | "400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 72,000 m : c is the circumference c = 3 meters correct answer e" | a ) 7 meters , b ) 9 meters , c ) 8 meters , d ) 5 meters , e ) 3 meters | e | multiply(divide(72, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000) | multiply(const_2,const_3)|multiply(#0,const_10)|multiply(n0,#1)|divide(n1,#2)|multiply(#3,const_1000)| | physics |
a , b , c subscribe rs . 50000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35000 , c receives : | let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . c ' s share = rs . ( 35000 x 12 / 50 ) = rs . 8,400 . d | a ) s . 14,000 , b ) s . 14,200 , c ) s . 4,400 , d ) s . 8,400 , e ) s . 4,800 | d | divide(subtract(divide(multiply(divide(divide(subtract(50000, add(add(4000, 5000), 5000)), const_3), 50000), 35000), const_4), const_100), const_10) | add(n1,n2)|add(n2,#0)|subtract(n0,#1)|divide(#2,const_3)|divide(#3,n0)|multiply(n3,#4)|divide(#5,const_4)|subtract(#6,const_100)|divide(#7,const_10) | general |
the sum of number of boys and girls in a school is 900 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ? | we have x + x % of 900 = 900 x + x / 100 * 900 = 900 10 * x = 900 x = 90 answer is d | a ) 50 , b ) 40 , c ) 60 , d ) 90 , e ) 70 | d | divide(900, add(divide(900, const_100), const_1)) | divide(n0,const_100)|add(#0,const_1)|divide(n0,#1) | general |
one half of a two digit number exceeds its one fourth by 3 . what is the sum of the digits of the number ? | explanation : x / 2 – x / 4 = 3 = > x = 12 1 + 2 = 3 a | a ) 3 , b ) 9 , c ) 11 , d ) 12 , e ) 15 | a | add(reminder(multiply(3, const_4), const_10), const_1) | multiply(n0,const_4)|reminder(#0,const_10)|add(#1,const_1) | general |
a student chose a number , multiplied it by 2 , then subtracted 200 from the result and got 110 . what was the number he chose ? | "solution : let x be the number he chose , then 2 * x * 200 = 110 2 x = 310 x = 155 correct answer c" | a ) 168 , b ) 100 , c ) 155 , d ) 160 , e ) 200 | c | divide(add(110, 200), 2) | add(n1,n2)|divide(#0,n0)| | general |
20 beavers , working together in a constant pace , can build a dam in 3 hours . how many hours will it take 12 beavers that work at the same pace , to build the same dam ? | "total work = 20 * 3 = 60 beaver hours 12 beaver * x = 60 beaver hours x = 60 / 12 = 5 answer : d" | a ) 2 . , b ) 4 . , c ) 5 . , d ) 6 , e ) 8 . | d | divide(multiply(3, 20), 12) | multiply(n0,n1)|divide(#0,n2)| | physics |
the average expenditure of a labourer for 6 months was 90 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | income of 6 months = ( 6 × 90 ) – debt = 540 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 810 average monthly income = 810 ÷ 10 = 81 answer e | a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 81 | e | divide(add(add(multiply(90, 6), multiply(60, 4)), 30), add(6, 4)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(n4,#3)|divide(#4,#0) | general |
an aeroplane covers a certain distance at a speed of 240 kmph in 5 hours . to cover the same distance in 1 hours , it must travel at a speed of : | "explanation : distance = ( 240 x 5 ) = 1200 km . speed = distance / time speed = 1200 / ( 5 / 3 ) km / hr . required speed = ( 1200 x 3 / 5 ) km / hr = 720 km / hr . answer is d" | a ) 300 kmph , b ) 360 kmph , c ) 600 kmph , d ) 720 kmph , e ) 750 kmph | d | divide(divide(multiply(240, 5), add(const_1, divide(const_2, const_3))), const_2) | divide(const_2,const_3)|multiply(n0,n1)|add(#0,const_1)|divide(#1,#2)|divide(#3,const_2)| | physics |
the length of rectangle is thrice its breadth and its perimeter is 48 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 48 l = 18 b = 6 lb = 18 * 6 = 108 answer : b" | a ) 432 , b ) 108 , c ) 252 , d ) 992 , e ) 212 | b | multiply(multiply(divide(48, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(48, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4)| | geometry |
find the value of ( √ 1.1 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) is | "( √ 1.1 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) = > 2.879 answer is b" | a ) 195 / 63 , b ) 2.879 , c ) 155 / 63 , d ) 125 / 63 , e ) 185 / 63 | b | add(divide(sqrt(1.1), sqrt(0.81)), divide(sqrt(1.44), sqrt(0.49))) | sqrt(n0)|sqrt(n1)|sqrt(n2)|sqrt(n3)|divide(#0,#1)|divide(#2,#3)|add(#4,#5)| | general |
if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 6 years the amount to be paid is ? | "a = 7500 ( 26 / 25 ) ^ 6 = 9490 answer : d" | a ) 3377 , b ) 2688 , c ) 2688 , d ) 9490 , e ) 1268 | d | multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100))) | add(n1,const_100)|divide(#0,const_100)|multiply(#1,#1)|multiply(n0,#2)| | gain |
if x ^ 2 + y ^ 2 = 15 and xy = 3 , then ( x − y ) ^ 2 = | but you can not take xy + 3 to mean xy = - 3 . . only if xy + 3 = 0 , it will mean xy = - 3 . . rest your solution is perfect and you will get your correct answer as 15 - 2 * 3 = 9 . . answer b | a ) 8 , b ) 9 , c ) 14 , d ) 17 , e ) 20 | b | power(3, 2) | power(n3,n0) | general |
the perimeter of a triangle is 48 cm and the in radius of the triangle is 2.5 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the in radius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 48 / 2 = 60 cm 2 answer : c" | a ) 76 , b ) 88 , c ) 60 , d ) 55 , e ) 35 | c | triangle_area(2.5, 48) | triangle_area(n0,n1)| | geometry |
if h = { 1 , 7 , 18 , 20 , 29 , 33 } , how much less is the mean of the numbers in h than the median of the numbers in h ? | this is a good question to understand the difference between mean and median . mean : average of all the numbers . ( sum of all the elements divided by the number of elements ) median : arrange the elements of the set in increasing order . if the number of terms is odd , the middle term is the median . if the number of... | a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0 | a | subtract(29, add(add(7, 1), 20)) | add(n0,n1)|add(n3,#0)|subtract(n4,#1) | general |
what will be the cost of house to paint which area equal to 484 sq ft , if the price per foot of building is rs . 20 | let the side of the square plot be a ft . a 2 = 484 = > a = 22 length of the fence = perimeter of the plot = 4 a = 88 ft . cost of building the fence = 88 * 20 = rs . 1760 . answer : b | a ) 1800 , b ) 1760 , c ) 1400 , d ) 2600 , e ) 3600 | b | multiply(multiply(sqrt(484), const_4), 20) | sqrt(n0)|multiply(#0,const_4)|multiply(n1,#1) | geometry |
x varies inversely as square of y . given that y = 2 for x = 1 . the value of x for y = 6 will be equal to ? | "given x = k / y 2 , where k is a constant . now , y = 2 and x = 1 gives k = 4 . x = 4 / y 2 = > x = 4 / 62 , when y = 6 = > x = 4 / 36 = 1 / 9 . answer : d" | a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 0 , d ) 1 / 9 , e ) 1 / 5 | d | divide(multiply(1, power(2, const_2)), power(6, const_2)) | power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1)| | general |
two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability r that jane will be on the same committee as albert ? | "total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability r = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d" | a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 % | d | multiply(divide(multiply(choose(const_4, const_1), const_2), choose(6, 3)), multiply(multiply(const_5, const_5), const_4)) | choose(const_4,const_1)|choose(n0,n1)|multiply(const_5,const_5)|multiply(#0,const_2)|multiply(#2,const_4)|divide(#3,#1)|multiply(#5,#4)| | probability |
there are , in a certain league , 50 teams , and each team face another team for a total of 4 times . how many games are played in the season ? | by using the formula , t [ n ( n - 1 ) / 2 ] , where t = no . of games between two teams and n = total no . of teams , we get : 4900 option d . | a ) 6660 , b ) 3600 , c ) 2400 , d ) 4900 , e ) 5400 | d | multiply(multiply(subtract(50, const_1), const_4), divide(50, const_2)) | divide(n0,const_2)|subtract(n0,const_1)|multiply(#1,const_4)|multiply(#0,#2) | general |
a cycle is bought for rs . 1000 and sold for rs . 2000 , find the gain percent ? | "1000 - - - - 1000 100 - - - - ? = > 100 % answer : c" | a ) 11 , b ) 20 , c ) 100 , d ) 77 , e ) 12 | c | multiply(divide(subtract(2000, 1000), 1000), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a 360 m long train is running at a speed of 55 km / hr . it crossed a platform of length 520 m in ? | "speed = 55 km / hr ( to convert km / hr in to m / s ) = 55 x 5 / 18 m / s distance = 360 m + 520 m ( if questions is about train crossing a post you need to consider only the length of train , ) = 880 m time = distance / speed = 880 x 18 / ( 5 x 55 ) = 57.6 sec ans is : a" | a ) 57.6 sec , b ) 20.2 sec , c ) 31.8 sec , d ) 50.4 sec , e ) none of the above | a | divide(add(360, 520), multiply(55, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
5 x 1.6 - 2 x 1.4 / 1.3 = ? | "given expression = ( 8 - 2.8 ) / 1.3 = 5.2 / 1.3 = 52 / 1.3 = 4 . answer is d ." | a ) 3 , b ) 0 , c ) 5 , d ) 4 , e ) 2 | d | divide(subtract(multiply(5, 1.6), multiply(2, 1.4)), 1.3) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
the speed of a car is 98 km in the first hour and 70 km in the second hour . what is the average speed of the car ? | "s = ( 98 + 70 ) / 2 = 84 kmph answer : e" | a ) 76 kmph , b ) 75 kmph , c ) 87 kmph , d ) 56 kmph , e ) 84 kmph | e | divide(add(98, 70), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 10 days , working 12 hours a day , the number of men required ? | "that is , 1 work done = 12 × 8 × 10 then , 12 8 × 10 = ? × 12 × 10 ? ( i . e . no . of men required ) = 12 × 8 × 10 / 12 × 10 = 8 days d )" | a ) 10 days , b ) 9 days , c ) 6 days , d ) 8 days , e ) 7 days | d | divide(multiply(multiply(12, 10), 8), multiply(10, 12)) | multiply(n0,n2)|multiply(n3,n4)|multiply(n1,#0)|divide(#2,#1)| | physics |
divide rs . 1301 between a and b , so that the amount of a after 7 years is equal to the amount of b after 9 years , the interest being compounded at 4 % per annum . | explanation : let the two parts be rs . x and rs . ( 1301 - x ) . 625 x = 676 ( 1301 - x ) 1301 x = 676 x 1301 x = 676 . so , the parts are rs . 676 and rs . ( 1301 - 676 ) i . e rs . 676 and rs . 625 answer : a ) rs . 625 | a ) 625 , b ) 367 , c ) 288 , d ) 209 , e ) 198 | a | divide(1301, add(const_1, divide(power(add(const_1, divide(4, const_100)), 9), power(add(const_1, divide(4, const_100)), 7)))) | divide(n3,const_100)|add(#0,const_1)|power(#1,n2)|power(#1,n1)|divide(#2,#3)|add(#4,const_1)|divide(n0,#5) | gain |
heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 40 litres of tomato juice ? | "answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . â ‡ ’ â ‡ ’ 10 % ( 40 ) = 80 % ( x ) â ‡ ’ â ‡ ’ x = 5 liters . answer : c" | a ) 2.8 liters . , b ) 2.5 liters . , c ) 5 liters . , d ) 2.6 liters . , e ) 2.1 liters . | c | divide(multiply(divide(subtract(const_100, 90), const_100), 40), divide(subtract(const_100, 20), const_100)) | subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)| | gain |
david works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 4 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? | first notice the number of months for which he worked - march to sept i . e . 7 months avg of monthly totals in tips for months other than august = x tips in august = 4 x total tips for all months = 6 * x + 4 x = 10 x tips for august / total tips for all months = 4 x / 10 x = 2 / 5 answer : b | a ) 3 / 5 , b ) 2 / 5 , c ) 4 / 5 , d ) 1 / 5 , e ) 6 / 5 | b | divide(4, add(subtract(const_10, 4), 4)) | subtract(const_10,n0)|add(n0,#0)|divide(n0,#1) | general |
a train 330 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in : | "relative speed = ( 25 + 2 ) km / hr = 27 km / hr = ( 27 × 5 / 18 ) m / sec = 15 / 2 m / sec . time taken by the train to pass the man = ( 330 × 2 / 15 ) sec = 44 sec answer : e" | a ) 30 sec , b ) 32 sec , c ) 36 sec , d ) 38 sec , e ) 44 sec | e | multiply(const_3600, divide(divide(330, const_1000), add(25, 2))) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)| | physics |
walking at 4 / 5 of her normal speed , a worker is 15 minutes later than usual in reaching her office . the usual time ( in minutes ) taken by her to cover the distance between her home and her office is | "let v be her normal speed and let t be her normal time . d = ( 4 / 5 ) v * ( t + 15 ) since the distance is the same we can equate this to a regular day which is d = v * t v * t = ( 4 / 5 ) v * ( t + 15 ) t / 5 = 12 t = 60 the answer is d ." | a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 65 | d | multiply(4, 15) | multiply(n0,n2)| | physics |
a cylinder of height h is 8 / 9 of water . when all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder , the new cylinder is 3 / 5 full . the height of the new cylinder is what percent of h ? | basically we can disregard the radius is 25 % information , as we are only asked about the height of the original and the new cylinder . this is becausethe new cylinder is 3 / 5 fullmeans the same as that it ' s height is 3 / 5 . original cylinder 8 / 9 new cylinder 3 / 5 so 3 / 5 / 8 / 9 = 3 / 5 * 9 / 8 = 12 / 15 = 4 ... | a ) 25 % , b ) 50 % , c ) 68 % , d ) 80 % , e ) 100 % | c | multiply(divide(divide(3, 5), divide(8, 9)), const_100) | divide(n3,n4)|divide(n0,n1)|divide(#0,#1)|multiply(#2,const_100) | gain |
mary passed a certain gas station on a highway while traveling west at a constant speed of 40 miles per hour . then , 15 minutes later , paul passed the same gas station while traveling west at a constant speed of 60 miles per hour . if both drivers maintained their speeds and both remained on the highway for at least ... | "d = rt m : r = 40 mph , t = t + 1 / 4 hr d = 40 ( t + 1 / 4 ) p : r = 60 , t = t d = 60 t since they went the same distance : 40 t + 40 / 4 = 60 t 20 t = 40 / 4 t = 1 / 2 or 1 / 2 hr a" | a ) 1 / 2 , b ) 2 , c ) 1 , d ) 4 , e ) 5 | a | multiply(const_60, divide(multiply(40, divide(15, const_60)), subtract(60, 40))) | divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(#3,const_60)| | physics |
a can do a job in 15 days and b in 30 days . if they work on it together for 4 days , then the fraction of the work that is left is ? | "a ' s 1 day work = 1 / 15 b ' s 1 day work = 1 / 30 a + b 1 day work = 1 / 15 + 1 / 30 = 3 / 30 = 1 / 10 a + b 4 days work = 1 / 10 * 4 = 2 / 5 remaining work = 1 - 2 / 5 = 3 / 5 answer is b" | a ) 2 / 15 , b ) 3 / 5 , c ) 3 / 11 , d ) 1 / 12 , e ) 6 / 13 | b | subtract(const_1, multiply(4, add(divide(const_1, 15), divide(const_1, 30)))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics |
the population of a city increases by 9 % per year but due to migration it decrease by 1 % per years . what will be the percentage increase in population in 3 years ? | "actual increase in population = 8 % let , earlier population = 100 then the population after 3 years = 100 ( 1 + 8 / 100 ) ^ 3 = 125.97 ∴ required percentage = 25.97 % answer : c" | a ) 9 % , b ) 9.27 % , c ) 25.97 % , d ) 12 % , e ) none of these | c | add(divide(multiply(add(const_100, add(divide(multiply(add(const_100, subtract(9, 1)), subtract(9, 1)), const_100), subtract(9, 1))), subtract(9, 1)), const_100), add(divide(multiply(add(const_100, subtract(9, 1)), subtract(9, 1)), const_100), subtract(9, 1))) | subtract(n0,n1)|add(#0,const_100)|multiply(#1,#0)|divide(#2,const_100)|add(#3,#0)|add(#4,const_100)|multiply(#5,#0)|divide(#6,const_100)|add(#4,#7)| | general |
a cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.5 m and its walls are 5 cm thick . the thickness of the bottom is : | "explanation : let the thickness of the bottom be x cm . then , [ ( 330 - 10 ) × ( 260 - 10 ) × ( 150 - x ) ] = 8000 × 1000 = > 320 × 250 × ( 150 - x ) = 8000 × 1000 = > ( 150 - x ) = 8000 × 1000 / 320 = 100 = > x = 50 cm = 5 dm . answer : b" | a ) 90 cm , b ) 5 dm , c ) 1 m , d ) 1.1 cm , e ) none of these | b | subtract(multiply(multiply(3.3, 2.6), 1.5), divide(8000, const_1000)) | divide(n0,const_1000)|multiply(n1,n2)|multiply(n3,#1)|subtract(#2,#0)| | physics |
if k is the greatest positive integer such that 3 ^ k is a divisor of 18 ! then k = | "18 / 3 = 6 18 / 9 = 2 6 + 2 = 8 k = 8 answer : e" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | add(const_1, divide(18, 3)) | divide(n1,n0)|add(#0,const_1)| | general |
if 18888 – n is divisible by 11 , and 0 < n < 11 , what is n ? | 18,888 / 11 = 1717 with a remainder of 1 . we need to subtract the remainder to get a multiple of 11 . the answer is a . | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | gcd(18888, 11) | gcd(n0,n1) | general |
after decreasing 56 % in the price of an article costs rs . 4400 . find the actual cost of an article ? | "cp * ( 44 / 100 ) = 4400 cp = 100 * 100 = > cp = 10000 answer : e" | a ) 12000 , b ) 15000 , c ) 1200 , d ) 1000 , e ) 10000 | e | divide(4400, subtract(const_1, divide(56, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
for each of her sales , a saleswoman receives a commission equal to 20 percent of the first $ 500 of the total amount of the sale , plus 45 percent of the total amount in excess of $ 500 . if the total amount of one of her sales was $ 800 , the saleswoman ’ s commission was approximately what percent of the total amoun... | total sales = 800 comission = ( 20 / 100 ) * 500 + ( 45 / 100 ) * 300 = 100 + 135 = 235 % comission = ( 235 / 800 ) * 100 = 29.4 ~ 29 % answer is d | a ) 22 % , b ) 24 % , c ) 25 % , d ) 29 % , e ) 28 % | d | multiply(divide(add(multiply(divide(20, const_100), 500), multiply(divide(45, const_100), subtract(800, 500))), 800), const_100) | divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n1)|multiply(n1,#0)|multiply(#1,#2)|add(#3,#4)|divide(#5,n4)|multiply(#6,const_100) | general |
a shop owner sells 50 mtr of cloth and gains sp of 10 mtrs . find the gain % ? | here , selling price of 10 m cloth is obtained as profit . profit of 10 m cloth = ( s . p . of 50 m cloth ) – ( c . p . of 50 m cloth ) selling price of 40 m cloth = selling price of 50 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 40 m cloth = rs . 4000 and s . p . of 40 m cloth = rs . rs .... | a ) 25 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | a | multiply(divide(10, subtract(50, 10)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100) | gain |
find compound interest on rs . 7500 at 4 % per annum for 2 years , compounded annually . | "solution amount = rs [ 7500 x ( 1 + 4 / 100 ) ² ] = rs . ( 7500 x 26 / 25 x 26 / 25 ) = rs . 8112 . c . i = rs ( 8112 - 7500 ) = rs . 612 . answer c" | a ) rs . 512 , b ) rs . 552 , c ) rs . 612 , d ) rs . 622 , e ) none | c | subtract(add(add(7500, divide(multiply(7500, 4), const_100)), divide(multiply(add(7500, divide(multiply(7500, 4), const_100)), 4), const_100)), 7500) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#2,#4)|subtract(#5,n0)| | gain |
a number is 4 less than 6 times the other number . if the sum of both the number is 38 , what are the numbers ? | that numbers are x , 6 x - 4 so sum = x + 6 x - 4 = 38 7 x = 42 = = > x = 6 another no is 32 answer is 6 , 32 answer : a | a ) 632 , b ) 731 , c ) 533 , d ) 434 , e ) none of the above | a | add(multiply(subtract(38, subtract(38, 6)), const_100), subtract(38, 6)) | subtract(n2,n1)|subtract(n2,#0)|multiply(#1,const_100)|add(#2,#0) | general |
two trains of length 250 m and 120 m are 50 m apart . they start moving towards each other on parallel tracks , at speeds 64 kmph and 42 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 + 42 ) * 5 / 18 = 7 * 5 = 29.4 mps . the time required = d / s = 50 / 29.4 = 17 / 10 sec . answer : b" | a ) 15 / 10 , b ) 17 / 10 , c ) 16 / 10 , d ) 18 / 10 , e ) 16 / 10 | b | divide(50, multiply(add(64, 42), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
the area of sector of a circle whose radius is 12 metro and whose angle at the center is 38 ° is ? | "38 / 360 * 22 / 7 * 12 * 12 = 47.7 m 2 answer : c" | a ) 30 m 2 , b ) 40 m 2 , c ) 47.7 m 2 , d ) 50 m 2 , e ) 55 m 2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(38, divide(const_3600, const_10))) | add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|power(n0,const_2)|add(#2,const_2)|divide(n1,#1)|divide(#4,#0)|multiply(#6,#3)|multiply(#5,#7)| | geometry |
if the l . c . m of two numbers is 750 and their product is 18750 , find the h . c . f of the numbers . | "h . c . f = ( product of the numbers ) / ( their l . c . m ) = 18750 / 750 = 25 . answer : d" | a ) 50 , b ) 30 , c ) 125 , d ) 25 , e ) none of these | d | divide(18750, 750) | divide(n1,n0)| | physics |
the population of a bacteria culture doubles every 2 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 200,000 bacteria | "the question basically asks how many minutes it takes for a population to increase by factor 200 ( 200,000 / 1,000 = 200 ) . now you know that every two minutes the population doubles , i . e . is multiplied by 2 . so the equation becomes : 2 ^ x > = 200 , where x represents the number of times the population doubles ... | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | d | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 2) | add(const_1,const_4)|multiply(#0,n1)|multiply(#1,const_100)|divide(#2,n1)|log(#3)|multiply(n0,#4)| | general |
the marks obtained by polly and sandy are in the ratio 3 : 5 and those obtained by sandy and willy are in the ratio of 3 : 2 . the marks obtained by polly and willy are in the ratio of . . . ? | "polly : sandy = 3 : 5 = 9 : 15 sandy : willy = 3 : 2 = 15 : 10 polly : sandy : willy = 9 : 15 : 10 polly : willy = 9 : 10 the answer is d ." | a ) 3 : 4 , b ) 5 : 6 , c ) 7 : 8 , d ) 9 : 10 , e ) 11 : 12 | d | divide(multiply(3, 3), multiply(5, 2)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other |
if the largest 3 digit number is subtracted from the smallest 5 digit number , then the remainder is | solution required remainder = ( 10000 - 999 ) = 9001 answer c | a ) 1 , b ) 9000 , c ) 9001 , d ) 90001 , e ) none | c | subtract(multiply(add(multiply(5, const_100), multiply(5, const_100)), const_10), subtract(add(multiply(5, const_100), multiply(5, const_100)), const_1)) | multiply(n1,const_100)|add(#0,#0)|multiply(#1,const_10)|subtract(#1,const_1)|subtract(#2,#3) | general |
x and y started a business investing rs . 50000 and rs 40000 respectively . in what ratio the profit earned after 2 years be divided between x and y respectively ? | x : y = 50000 : 40000 = 5 : 4 answer : c | a ) 3 : 2 , b ) 9 : 2 , c ) 5 : 4 , d ) 1 : 4 , e ) 18 : 4 | c | divide(50000, 40000) | divide(n0,n1) | gain |
a man can row downstream at 26 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 26 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 36 = > x = 18 , y = 8 . answer : d" | a ) 2 , 9 , b ) 4 , 9 , c ) 8 , 9 , d ) 18 , 8 , e ) 7 , 9 | d | divide(divide(add(26, 10), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
if log 1087.5 = 3.9421 , 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 ( 3.9421 + 1 ) = 10 ( 4.9421 ) = 49.421 x = antilog ( 49.421 ) therefore , number of digits in x = 50 . answer : d" | a ) 30 , b ) 28 , c ) 27 , d ) 50 , e ) 25 | d | add(multiply(const_4, 3.9421), divide(log(const_100), log(const_10))) | log(const_100)|log(const_10)|multiply(n1,const_4)|divide(#0,#1)|add(#3,#2)| | other |
if $ 5,000 is invested in an account that earns 12 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ? | "solution amount ( ci ) = p + ( 1 + r / n ) ^ nt = 5000 + ( 1 + 0.12 / 2 ) ^ 2 = 5412 amount ( si ) = p + ptr / 100 = 5000 + ( 5000 * 1 * 12 / 100 ) = 5400 difference = 5412 - 5400 = 12 $ c" | a ) $ 4 , b ) $ 8 , c ) $ 12 , d ) $ 16 , e ) $ 432 | c | multiply(12, const_1) | multiply(n1,const_1)| | gain |
the sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 180 m , its area is ? | "5 x + 12 x + 13 x = 180 = > x = 6 a = 30 , b = 72 , c = 78 s = ( 30 + 72 + 78 ) / 2 = 90 answer : a" | a ) 90 , b ) 882 , c ) 277 , d ) 261 , e ) 281 | a | multiply(180, divide(180, add(add(5, 12), 13))) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n3,#2)| | geometry |
if 2 ^ 4 , 2 ^ 3 , and 11 ^ 3 are factors of the product of 1,452 and w , where w is a positive integer , what is the smallest possible value of w ? | i will go with a ( pending elements to match is 2 ^ 2 * 2 ^ 2 * 11 ^ 1 = 176 | a ) 176 , b ) 288 , c ) 363 , d ) 396 , e ) 484 | a | multiply(multiply(multiply(power(2, 2), 2), divide(11, 2)), 2) | divide(n4,n0)|power(n0,n0)|multiply(n2,#1)|multiply(#0,#2)|multiply(n0,#3)| | general |
what is the sum of 80 consecutive integers from - 39 inclusive , in a increasing order ? | "from - 39 to - 1 - - > 39 nos . zero - - > 1 number from + 1 to + 39 - - > 39 nos . when we add up nos . from - 39 to + 39 sum will be zero . total 79 nos will be added . 80 th number will be 40 . sum of these 80 nos . = 40 . d is the answer ." | a ) - 29 , b ) 39 , c ) - 30 , d ) 40 , e ) 60 | d | add(39, const_1) | add(n1,const_1)| | general |
the profit earned by selling an article for rs . 832 is equal to the loss incurred when the same article is sold for rs . 448 . what should be the sale price for making 70 % profit ? | "let c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 170 % of rs . 640 = 170 / 100 * 640 = rs . 1088 . answer : b" | a ) 277 , b ) 1088 , c ) 277 , d ) 266 , e ) 121 | b | multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(70, const_100))) | divide(n2,const_100)|subtract(n0,n1)|add(#0,const_1)|divide(#1,const_2)|subtract(n0,#3)|multiply(#2,#4)| | gain |
a man has some hens and cows . if the number of heads be 46 and the number of feet equals 136 , then the number of hens will be | "explanation : let number of hens = h and number of cows = c number of heads = 46 = > h + c = 46 - - - ( equation 1 ) number of feet = 136 = > 2 h + 4 c = 136 = > h + 2 c = 68 - - - ( equation 2 ) ( equation 2 ) - ( equation 1 ) gives 2 c - c = 68 - 46 = > c = 22 substituting the value of c in equation 1 , we get h + 2... | a ) 22 , b ) 24 , c ) 26 , d ) 20 , e ) 28 | b | divide(subtract(multiply(46, const_4), 136), const_2) | multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)| | general |
find the smallest number which when divided by 13 and 17 leaves respective remainders of 3 and 7 . | "let ' n ' is the smallest number which divided by 13 and 17 leaves respective remainders of 3 and 7 . required number = ( lcm of 13 and 17 ) - ( common difference of divisors and remainders ) = ( 221 ) - ( 10 ) = 211 . answer : b" | a ) 187 , b ) 211 , c ) 207 , d ) 219 , e ) 227 | b | subtract(multiply(13, 17), add(const_10, const_1)) | add(const_1,const_10)|multiply(n0,n1)|subtract(#1,#0)| | general |
a ' s speed is 20 / 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that b beats a by 25 % of the length of the race ? | "first calculate the distance , b has covered with his speed in the time , in which a reached 60 % of the race . then add the remaining distance as head start for b to win the race . its best to apply ratios concept here . since a ' s speed is 20 / 13 of b , therefore , b ' s speed is 13 / 20 of a distance covered by b... | a ) 44 % , b ) 48 % , c ) 52 % , d ) 42 % , e ) 61 % | e | multiply(subtract(const_1, multiply(divide(subtract(const_100, 25), const_100), divide(13, 20))), const_100) | divide(n1,n0)|subtract(const_100,n2)|divide(#1,const_100)|multiply(#2,#0)|subtract(const_1,#3)|multiply(#4,const_100)| | general |
find the constant k so that : - x 2 - ( k + 11 ) x - 8 = - ( x - 2 ) ( x - 4 ) | "- x 2 - ( k + 11 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 11 ) x - 8 = - x 2 + 6 x - 8 - ( k + 11 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 17 : solve the above for k correct answer c" | a ) 11 , b ) 12 , c ) 17 , d ) 14 , e ) 15 | c | add(11, add(4, 2)) | add(n0,n4)|add(n1,#0)| | general |
what is the maximum number of pieces of birthday cake of size 10 ” by 10 ” that can be cut from a cake 20 ” by 20 ” ? | "the prompt is essentially asking for the maximum number of 10 x 10 squares that can be cut from a larger 20 by 20 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 2 ' pieces ' each , we have ( 2 ) ( 2 ) = 4 total smaller squares ( at maximum ) . a" | a ) 4 , b ) 10 , c ) 16 , d ) 20 , e ) 25 | a | divide(multiply(20, 20), multiply(10, 10)) | multiply(n2,n2)|multiply(n0,n0)|divide(#0,#1)| | physics |
if the sides of a triangle are 65 cm , 60 cm and 25 cm , what is its area ? | "the triangle with sides 65 cm , 60 cm and 25 cm is right angled , where the hypotenuse is 65 cm . area of the triangle = 1 / 2 * 60 * 25 = 750 cm 2 answer : b" | a ) 120 cm 2 , b ) 750 cm 2 , c ) 216 cm 2 , d ) 197 cm 2 , e ) 275 cm 2 | b | divide(multiply(60, 25), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
a man walking at a rate of 15 km / hr crosses a bridge in 30 minutes . the length of the bridge is ? | "speed = 15 * 5 / 18 = 15 / 18 m / sec distance covered in 30 minutes = 15 / 18 * 30 * 60 = 1500 m answer is d" | a ) 1250 m , b ) 1110 m , c ) 950 m , d ) 1500 m , e ) 1300 m | d | multiply(divide(multiply(15, const_1000), const_60), 30) | multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)| | gain |
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