Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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a worker is paid a regular rate of rs . 10 for completing a survey . the worker completes 100 surveys per week . for any survey involving the use of her cellphone , she is paid at a rate of that is 30 % higher than her regular rate . if she completed 60 surveys involving the use of her cellphone , how much did she get ... | "amount earned using her cell phone = 60 * 13 = 780 earned for remaining surveys = 40 * 10 = 400 total earning = 1180 answer : b" | a ) 180 , b ) 1180 , c ) 280 , d ) 580 , e ) 2080 | b | add(multiply(10, 100), multiply(60, multiply(10, divide(30, 100)))) | divide(n2,n1)|multiply(n0,n1)|multiply(n0,#0)|multiply(n3,#2)|add(#1,#3)| | gain |
if the price of sugar rises from rs . 2 per kg to rs . 5 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by | "sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 2 = x Γ 5 = x = 40 kg . β΄ reduction in consumption = 60 % . answer e" | a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 60 % | e | multiply(subtract(const_1, divide(multiply(const_1, 2), 5)), const_100) | multiply(n0,const_1)|divide(#0,n1)|subtract(const_1,#1)|multiply(#2,const_100)| | general |
what is the 124 th digit to the right of the decimal point in the decimal form of 24 / 44 ? | "to solve this question , we first have to use some long division . this long division allows us to get 24 / 44 in decimal form , which is 0.545454 β¦ where β 54 β is repeating . we can see that the 1 st , 3 rd , 5 th digit to the right of the decimal point is a 5 and that the 2 nd , 4 th , 6 th digit to the right of th... | a ) 8 , b ) 9 , c ) 6 , d ) 4 , e ) 2 | d | divide(24, 44) | divide(n1,n2)| | general |
a palindrome is a number that reads the same forward and backward , such as 145 . how many odd , 4 - digit numbers are palindromes ? | "a palindrome is a number that reads the same forward and backward . examples of four digit palindromes are 1221 , 4334 , 2222 etc you basically get to choose the first two digits and you repeat them in opposite order . say , you choose 45 as your first two digits . the next two digits are 54 and the number is 4554 . a... | a ) 40 , b ) 45 , c ) 80 , d ) 90 , e ) 2500 | c | divide(power(const_10, divide(4, const_2)), const_2) | divide(n1,const_2)|power(const_10,#0)|divide(#1,const_2)| | general |
when n is divided by 25 , the remainder is 4 . what is the remainder when n + 15 is divided by 5 ? | "assume n = 29 remainder ( n / 24 ) = 4 n + 15 = 44 remainder ( 44 / 5 ) = 4 option c" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | subtract(4, reminder(15, 5)) | reminder(n2,n3)|subtract(n1,#0)| | general |
find the 25 % of rs . 500 . | "explanation : 25 % of 500 = > 25 / 100 * 500 = rs . 125 answer : d" | a ) s . 50 , b ) s . 70 , c ) s . 100 , d ) s . 125 , e ) s . 60 | d | multiply(500, divide(25, const_100)) | divide(n0,const_100)|multiply(n1,#0)| | gain |
what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 200 inclusive ? | "this question can be solved with the average formula and ' bunching . ' we ' re asked for the average of all of the multiples of 10 from 10 to 200 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 20 ( 10 ) = 200 so we know that there are 40 total nu... | a ) 190 , b ) 105 , c ) 200 , d ) 205 , e ) 210 | b | divide(divide(multiply(add(10, 200), add(divide(subtract(200, 10), 10), const_1)), const_2), add(divide(subtract(200, 10), 10), const_1)) | add(n0,n2)|subtract(n2,n0)|divide(#1,n0)|add(#2,const_1)|multiply(#0,#3)|divide(#4,const_2)|divide(#5,#3)| | general |
the ages of two persons differ by 12 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "explanation : let their ages be x and ( x + 12 ) years . 5 ( x - 5 ) = ( x + 12 - 5 ) or 4 x = 32 or x = 8 . their present ages are 20 years and 8 years option b" | a ) 20,20 , b ) 20,8 , c ) 25,15 , d ) 30,10 , e ) none of these | b | subtract(add(divide(multiply(12, 5), subtract(5, const_1)), 5), 12) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general |
if n is the smallest integer such that 54 times n is the square of an integer , what is the value of n ? | "54 can written as = 2 * 3 * 3 * 3 - - > 2 ^ 1 * 3 ^ 3 - - - ( 1 ) so for 54 * n to be a square of an integer , the integer should have even powers to the prime numbers it composed of . here 2 already has even power - > so n has to be 2 to make the power of 1 in ( 1 ) even option b is correct" | a ) 2 , b ) 1 , c ) 6 , d ) 12 , e ) 24 | b | divide(divide(divide(divide(divide(divide(54, const_2), const_2), const_2), const_2), const_3), const_3) | divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_3)|divide(#4,const_3)| | geometry |
two trains a and b are 150 m and 150 m long and are moving at one another at 54 km / hr and 36 km / hr respectively . arun is sitting on coach b 1 of train a . calculate the time taken by arun to completely cross train b . | "detailed solution speed of a = 54 β 1000 / 60 β 60 = 15 m / s speed of b = 36 β 1000 / 60 β 60 = 10 m / s relative speed = s 1 + s 2 = 15 + 10 m / s = 25 m / s the length that needs to be crossed = length of train b = 150 m . therefore time taken = 150 / 25 = 6 s . what is the time taken for trains to completely cross... | a ) 10 s , b ) 6 s , c ) 4 s , d ) 8 s , e ) 12 s | e | divide(add(150, 150), add(divide(multiply(54, const_1000), const_3600), divide(multiply(36, const_1000), const_3600))) | add(n0,n1)|multiply(n2,const_1000)|multiply(n3,const_1000)|divide(#1,const_3600)|divide(#2,const_3600)|add(#3,#4)|divide(#0,#5)| | physics |
if 4 a = 5 b = 30 , then 40 ab = | "4 a * 5 b = 30 * 30 = 900 20 ab = 900 i . e . 30 ab = 1800 answer : option b" | a ) 900 , b ) 1800 , c ) 2250 , d ) 1650 , e ) 1250 | b | multiply(40, multiply(divide(30, 4), divide(30, 5))) | divide(n2,n0)|divide(n2,n1)|multiply(#0,#1)|multiply(n3,#2)| | general |
if a train , travelling at a speed of 18 kmph , crosses a pole in 5 sec , then the length of train is ? | "b b = 18 * 5 / 18 * 5 = 25 m" | a ) 281 m , b ) 25 m , c ) 50 m , d ) 125 m , e ) 75 m | b | multiply(multiply(18, const_0_2778), 5) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
a train speeds past a pole in 15 seconds and a platform 110 meters long in 25 seconds . what is the length of the train ? | "let the length of the train be x meters . the speed of the train is x / 15 . then , x + 110 = 25 * ( x / 15 ) 10 x = 1650 x = 165 meters the answer is e ." | a ) 100 m , b ) 125 m , c ) 130 m , d ) 150 m , e ) 165 m | e | multiply(110, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
a train covers a distance at a speed of 150 kmph in 3 hours . to cover the same distance in 1 hours , it must travel at a speed of ? | distance = 150 * 3 = 450 km required speed = 450 / 1 = 450 km / hr answer is e | a ) 320 , b ) 250 , c ) 560 , d ) 400 , e ) 450 | e | divide(multiply(150, 3), 1) | multiply(n0,n1)|divide(#0,n2) | physics |
find the quotient : 0.63 / 9 | "63 / 9 = 7 . dividend contains 2 places decimal . 0.63 / 9 = 0.7 answer is e ." | a ) 0.5 , b ) 0.3 , c ) 0.9 , d ) 0.1 , e ) 0.7 | e | divide(0.63, 9) | divide(n0,n1)| | general |
the ratio of boys to girls in a class is 2 : 3 . the career preference of the students in the class are to be represented in a circle graph . if the area of the graph allocated to each career preference is to be proportional to the number of students who have that career preference , how many degrees of the circle shou... | let the common ratio be x . . so b = 2 x and g = 3 x and total = 5 x 1 / 3 of m = 2 x / 3 and 2 / 3 of f = 2 x . . total preferring that carrer = 2 x / 3 + 2 x = 8 x / 3 . . now 5 x = 360 , so x = 360 / 5 = 72 . . so x * 8 / 3 = 72 * 8 / 3 = 192 ans 192 e | ['a ) a ) 160 degree', 'b ) b ) 168 degree', 'c ) c ) 191 degree', 'd ) d ) 272 degree', 'e ) e ) 192 degree'] | e | multiply(add(divide(divide(2, add(2, 3)), const_3), divide(multiply(divide(const_3, add(2, 3)), const_2), const_3)), const_360) | add(n0,n1)|divide(n0,#0)|divide(const_3,#0)|divide(#1,const_3)|multiply(#2,const_2)|divide(#4,const_3)|add(#3,#5)|multiply(#6,const_360) | geometry |
x , y , and z are different prime numbers . the product x ^ 2 * y * z is divisible by how many different positive numbers ? | the exponents of x ^ 2 * y * z are 2 , 1 , and 1 . the number of factors is ( 2 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 12 the answer is e . | a ) 4 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | e | subtract(power(2, const_4), const_4) | power(n0,const_4)|subtract(#0,const_4) | general |
due to construction , the speed limit along an 10 - mile section of highway is reduced from 55 miles per hour to 25 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 10 miles stretch = 10 * 60 / 55 = 10 * 12 / 11 = 10.9 new time in minutes to cross 10 miles stretch = 10 * 60 / 25 = 10 * 12 / 5 = 24 time difference = 13.1 ans : b" | a ) a ) 6.24 , b ) b ) 13.1 , c ) c ) 10 , d ) d ) 15 , e ) e ) 24 | b | max(multiply(subtract(add(55, 10), const_1), subtract(divide(10, 25), divide(10, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics |
a 120 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | "s = 120 / 6 * 18 / 5 = 72 kmph answer : d" | a ) 229 , b ) 108 , c ) 278 , d ) 72 , e ) 112 | d | multiply(divide(120, 6), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
peter takes a loan of $ 100000 with 12 % annual interest : the interest is paid once , at the end of the year . martha takes a loan of $ 100000 with 12 % annual interest , compounding monthly at the end of each month . at the end of one full year , compared to peter ' s loan interest , approximately how much more does ... | peters interest = $ 100000 * 0.12 = $ 12000 or $ 1000 each month . martha β s interest , 12 % / 12 = 1 % each month : for the 1 st month = $ 100000 * 0.01 = $ 1000 ; for the 2 nd month = $ 1000 + 1 % of 1000 = $ 1010 , so we would have interest earned on interest ( very small amount ) ; for the 3 rd month = $ 1010 + 1 ... | a ) $ 68.25 , b ) zero , c ) $ 6825.00 , d ) $ 6.82 , e ) $ 682.50 | e | add(multiply(const_0_25, const_1000), add(multiply(multiply(multiply(100000, divide(divide(12, 12), const_100)), divide(divide(12, 12), const_100)), add(multiply(multiply(100000, divide(divide(12, 12), const_100)), divide(divide(12, 12), const_100)), const_1)), multiply(const_3, const_100))) | divide(n1,n1)|multiply(const_100,const_3)|multiply(const_0_25,const_1000)|divide(#0,const_100)|multiply(n0,#3)|multiply(#3,#4)|add(#5,const_1)|multiply(#6,#5)|add(#7,#1)|add(#8,#2) | general |
the price of a jacket is reduced by 25 % . during a special sale the price of the jacket is reduced another 15 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 25 % , therefore bringing down the price to $ 75 . 3 ) again it is further discounted by 10 % , therefore bringing down the price to $ 63.75 . 4 ) now 63.75 has to be added byx % in order to equal the original price . 63.75 + ( x % ) 63.75 = ... | a ) 32.5 , b ) 35 , c ) 48 , d ) 56.8 , e ) 67.5 | d | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(15, const_100)))), subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(15, const_100))))) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain |
3 numbers are randomly selected , with replacement , from the set of integers { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . if the first number selected is w , the second number selected is x , and the third number is y , what is the probability that w < x < y ? | solution : total possible ways = 10 * 10 * 10 = 1000 case 1 : w = 0 . if x = 1 , then y can have 8 ways . if x = 2 , then y can have 7 ways and so on . so , no . of ways = 8 + 7 + . . + 1 case 2 : w = 1 . if x = 2 , then y can have 7 ways . if x = 3 , then y can have 6 ways and so on . so , no . of ways = 7 + 6 + . . +... | a ) 3 / 40 , b ) 28 / 243 , c ) 3 / 25 , d ) 33 / 100 , e ) 64 / 125 | c | divide(1, 8) | divide(n2,n9) | physics |
two trains a and b are 200 m and 150 m long and are moving at one another at 54 km / hr and 36 km / hr respectively . arun is sitting on coach b 1 of train a . calculate the time taken by arun to completely cross train b . | "detailed solution speed of a = 54 β 1000 / 60 β 60 = 15 m / s speed of b = 36 β 1000 / 60 β 60 = 10 m / s relative speed = s 1 + s 2 = 15 + 10 m / s = 25 m / s the length that needs to be crossed = length of train b = 150 m . therefore time taken = 150 / 25 = 6 s . what is the time taken for trains to completely cross... | a ) 10 s , b ) 6 s , c ) 14 s , d ) 8 s , e ) 12 s | c | divide(add(200, 150), add(divide(multiply(54, const_1000), const_3600), divide(multiply(36, const_1000), const_3600))) | add(n0,n1)|multiply(n2,const_1000)|multiply(n3,const_1000)|divide(#1,const_3600)|divide(#2,const_3600)|add(#3,#4)|divide(#0,#5)| | physics |
if 20 % of a class averages 80 % on a test , 50 % of the class averages 60 % on the test , and the remainder of the class averages 40 % on the test , what is the overall class average ? | 2 students scored 80 % 5 students scored 60 % 3 students scored 40 % ( 2 ) ( 80 ) + 5 ( 60 ) + 3 ( 40 ) = 160 + 300 + 120 = 580 580 / 10 students = 58 - - > 58 % average answer : c | a ) 80 % , b ) 74 % , c ) 58 % , d ) 56 % , e ) 50 % | c | divide(add(add(multiply(20, 80), multiply(50, 60)), multiply(subtract(const_100, add(20, 50)), 40)), const_100) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(const_100,#0)|multiply(n4,#4)|add(#3,#5)|divide(#6,const_100) | general |
a man whose bowling average is 12.4 , takes 6 wickets for 26 runs and there by decreases his average by 0.4 . the number of wickets taken by him before his last match is ? | "12.4 * x + 26 = ( 6 + x ) 12 solve equation x = 115 answer : c" | a ) 113 , b ) 114 , c ) 115 , d ) 116 , e ) 117 | c | divide(subtract(multiply(floor(12.4), 6), 26), subtract(12.4, floor(12.4))) | floor(n0)|multiply(n1,#0)|subtract(n0,#0)|subtract(#1,n2)|divide(#3,#2)| | general |
of the 200 stamps in a collection , 90 are foreign and 70 are more than 10 years old . if 20 stamps are both foreign and more than 10 years old , how many stamps are neither foreign nor more than 10 years old ? | 20 stamps are both foreign and more than 10 years old . 70 stamps are foreign only . 50 stamps are 10 years old only . the number of remaining stamps is 200 - ( 20 + 70 + 50 ) = 60 the answer is a . | a ) 60 , b ) 80 , c ) 100 , d ) 130 , e ) 150 | a | subtract(200, subtract(add(90, 70), 20)) | add(n1,n2)|subtract(#0,n4)|subtract(n0,#1) | other |
the average age of 20 students in a class is 5 years . if teacher ' s age is also included then average increases 1 year then find the teacher ' s age ? | "total age of 50 students = 20 * 5 = 100 total age of 51 persons = 20 * 6 = 120 age of teacher = 120 - 100 = 20 years answer is c" | a ) 28 , b ) 24 , c ) 20 , d ) 22 , e ) 18 | c | subtract(add(add(multiply(20, 5), 1), 20), multiply(20, 5)) | multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)| | general |
how many zeroes are there at the end of the number n , if n = 80 ! + 160 ! ? | "the number of zeroes at the end of 80 ! will be less than the number of zeroes at the end of 160 ! hence it is sufficient to calculate the number of zeroes at the end of 80 ! the number of zeroes = [ 80 / 5 ] + [ 80 / 25 ] + [ 80 / 125 ] = 16 + 3 + 0 = 19 the answer is b ." | a ) 15 , b ) 19 , c ) 23 , d ) 27 , e ) 31 | b | add(divide(80, add(const_4, const_1)), const_2) | add(const_1,const_4)|divide(n0,#0)|add(#1,const_2)| | general |
the principal that amounts to rs . 4913 in 3 years at 6 1 / 4 % per annum c . i . compounded annually , is ? | "principal = [ 4913 / ( 1 + 25 / ( 4 * 100 ) ) 3 ] = 4913 * 16 / 17 * 16 / 17 * 16 / 17 = rs . 4096 . answer : d" | a ) 288 , b ) 267 , c ) 200 , d ) 4096 , e ) 2771 | d | divide(4913, power(add(1, divide(add(6, divide(1, 4)), const_100)), 3)) | divide(n3,n4)|add(n2,#0)|divide(#1,const_100)|add(#2,n3)|power(#3,n1)|divide(n0,#4)| | gain |
a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 3 and b : c = 2 : 5 . if the total runs scored by all of them are 75 , the runs scored by c are ? | "a : b = 2 : 3 b : c = 2 : 5 a : b : c = 4 : 6 : 15 15 / 25 * 75 = 45 answer : c" | a ) 11 , b ) 18 , c ) 45 , d ) 17 , e ) 12 | c | multiply(divide(75, add(add(divide(2, 3), divide(5, 2)), 2)), 5) | divide(n0,n1)|divide(n3,n0)|add(#0,#1)|add(#2,n0)|divide(n4,#3)|multiply(n3,#4)| | general |
if n is a positive integer and the product of all the integers from 1 to n , inclusive , is a multiple of 9 , what is the least possible value of n ? | we need at least 2 factors of three , which means n must be at least 6 , so b is the correct answer . | a ) 7 , b ) 6 , c ) 9 , d ) 12 , e ) 7 | b | divide(9, const_2) | divide(n1,const_2)| | general |
β 289 - β 625 Γ· β 25 is equal to ? | answer β 289 - β 625 Γ· β 25 = β 17 x 17 - β 25 x 25 Γ· β 5 x 5 = 17 - 25 Γ· 5 = 17 - 5 = 12 correct option : c | a ) 17 , b ) 15 , c ) 12 , d ) - 8 / 5 , e ) none | c | subtract(sqrt(289), divide(sqrt(625), sqrt(sqrt(625)))) | sqrt(n0)|sqrt(n1)|sqrt(#1)|divide(#1,#2)|subtract(#0,#3) | general |
the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students β grades was 83 , and the average of the female students β grades was 92 , how many female students took the test ? | total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 83 thus m = 664 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 664 + 92 f ) / ( 8 + f ) = 9... | a ) 8 , b ) 9 , c ) 10 , d ) 21 , e ) 28 | e | divide(subtract(multiply(90, 8), multiply(83, 8)), subtract(92, 90)) | multiply(n0,n1)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2) | general |
what is the compound interest on rs . 9500 at 6 % p . a . compounded half - yearly for 1 1 / 2 years . | "compound interest : a = p ( 1 + r / n ) nt a = 10 , 380.91 c . i . > > 10 , 380.91 - 9500 > > rs . 880.91 answer : c" | a ) 810.91 , b ) 840.91 , c ) 880.91 , d ) 887.91 , e ) 980.91 | c | multiply(9500, subtract(power(divide(add(divide(6, const_2), const_100), const_100), multiply(1, const_2)), const_1)) | divide(n1,const_2)|multiply(n2,const_2)|add(#0,const_100)|divide(#2,const_100)|power(#3,#1)|subtract(#4,const_1)|multiply(n0,#5)| | gain |
a and b β s salaries together amount to rs . 4,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is b β s salary ? | "( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 1000 4 b = 1000 = > b = 1000 answer b" | a ) 2777 , b ) 1000 , c ) 2789 , d ) 2776 , e ) 2881 | b | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | add(const_2,const_3)|divide(n2,const_100)|divide(n1,const_100)|multiply(#0,const_4)|subtract(const_1,#1)|subtract(const_1,#2)|add(#5,#4)|multiply(#3,const_2)|multiply(#7,const_100)|multiply(#8,#4)|divide(#9,#6)| | gain |
the population of a bacteria culture doubles every 6 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 500,000 bacteria | "this one ' s easy . 1000 * 2 ^ t = 500,000 2 ^ t = 500 now gauging , since 2 ^ 8 = 256 , then 2 ^ 9 = 512 so t = 9 but be careful , ' t ' is in time intervals of 6 minutes so answer is 9 * 6 = 54 minutes answer ( d )" | a ) 10 , b ) 12 , c ) 14 , d ) 54 , e ) 18 | d | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 6) | add(const_1,const_4)|multiply(#0,n1)|multiply(#1,const_100)|divide(#2,n1)|log(#3)|multiply(n0,#4)| | general |
can you solve it ? 2 + 3 = 8 , 3 + 7 = 27 , 4 + 5 = 32 , 5 + 8 = 60 , 6 + 7 = 72 , 7 + 8 = ? ? | 2 + 3 = 2 * [ 3 + ( 2 - 1 ) ] = 8 3 + 7 = 3 * [ 7 + ( 3 - 1 ) ] = 27 4 + 5 = 4 * [ 5 + ( 4 - 1 ) ] = 32 5 + 8 = 5 * [ 8 + ( 5 - 1 ) ] = 60 6 + 7 = 6 * [ 7 + ( 6 - 1 ) ] = 72 therefore 7 + 8 = 7 * [ 8 + ( 7 - 1 ) ] = 98 x + y = x [ y + ( x - 1 ) ] = x ^ 2 + xy - x correct answer is d ) 98 | a ) 68 , b ) 78 , c ) 88 , d ) 98 , e ) 56 | d | multiply(add(subtract(7, const_1), 8), 7) | subtract(n4,const_1)|add(n2,#0)|multiply(n4,#1) | general |
if the remainder is 10 when positive integer n is divided by 14 , what is the remainder when n is divided by 7 ? | "assume x is quotient here , n = 14 x + 7 - - - - - - - - - - ( 1 ) and n = 7 x + ? we can also write equation ( 1 ) as : n = ( 14 x + 7 ) + 3 . ie 7 ( 2 x + 1 ) + 3 ie the first term is perfectly divisible by 7 . so , the remainder left is 3 . so , answer ( d ) is right choice ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | divide(14, 7) | divide(n1,n2)| | general |
if x ^ 2 + 1 / x ^ 2 = 2 , what is the value of x ^ 4 + 1 / x ^ 4 ? | important : i notice that if we square x Β² , we get x β΄ , and if we square 1 / x Β² , we get 1 / x β΄ , so let ' s see what happens if we take the equation x Β² + 1 / x Β² = 2 andsquareboth sides : ( x Β² + 1 / x Β² ) Β² = 4 so , ( x Β² + 1 / x Β² ) ( x Β² + 1 / x Β² ) = 4 expand to get : x β΄ + 1 + 1 + 1 / x β΄ = 4 simplify : x β΄ ... | a ) 10 , b ) 2 , c ) 4 , d ) 16 , e ) 18 | b | subtract(power(2, const_2), const_2) | power(n0,const_2)|subtract(#0,const_2) | general |
a certain debt will be paid in 52 installments from january 1 to december 31 of a certain year . each of the first 12 payments is to be $ 410 ; each of the remaining payments is to be $ 65 more than each of the first 12 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the yea... | "total number of installments = 52 payment per installment for the first 12 installments = 410 payment per installment for the remaining 32 installments = 410 + 65 = 475 average = ( 12 * 410 + 40 * 475 ) / 52 = 460 answer c" | a ) 443 , b ) 450 , c ) 460 , d ) 468 , e ) 475 | c | divide(add(multiply(12, 410), multiply(add(410, 65), subtract(52, 12))), 52) | add(n4,n5)|multiply(n3,n4)|subtract(n0,n3)|multiply(#0,#2)|add(#1,#3)|divide(#4,n0)| | general |
at what rate percent of simple interest will a sum of money double itself in 22 years ? | "let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 22 ) = 100 / 22 = 4.54 % answer : b" | a ) 6.54 % , b ) 4.54 % , c ) 8 . 2 % , d ) 4.94 % , e ) 5.54 % | b | divide(const_100, 22) | divide(const_100,n0)| | gain |
consider a square of diagonal length d . let another square be formed with d as its side . find the ratio of the area of the bigger square to that of the smaller square . | d ^ 2 / ( 1 / 2 ) * d ^ 2 = 2 answer : a | ['a ) 2', 'b ) 3', 'c ) 4', 'd ) 16', 'e ) 41'] | a | divide(power(const_1, const_2), power(sqrt(divide(const_1, add(const_1, const_1))), const_2)) | add(const_1,const_1)|power(const_1,const_2)|divide(const_1,#0)|sqrt(#2)|power(#3,const_2)|divide(#1,#4) | geometry |
find the missing value : ? - 1936248 = 1635773 | "let x - 1936248 = 1635773 . then , x = 1635773 + 1936248 = 3572021 answer is a ." | a ) 3572021 , b ) 3570221 , c ) 3572012 , d ) 3752012 , e ) none of them | a | divide(1635773, divide(1936248, const_100)) | divide(n0,const_100)|divide(n1,#0)| | general |
the tax on a commodity is diminished by 14 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 86 * 115 = 9890 - - - - - - - - - - - 10000 - - - - - - - - - - - 110 100 - - - - - - - - - - - ? = > 1.1 % decrease answer : e" | a ) 2.1 % , b ) 8.1 % , c ) 5.1 % , d ) 6.1 % , e ) 1.1 % | e | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(14, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general |
the sum of the ages of 4 children born at the intervals of 5 years each is 58 years . what is the age of the youngest child ? | "explanation : let x = the youngest child . each of the other four children will then be x + 5 , x + 10 , x + 15 we know that the sum of their ages is 58 . so , x + ( x + 5 ) + ( x + 10 ) + ( x + 15 ) = 58 x = 7 the youngest child is 4 years old . answer : a" | a ) 7 , b ) 8 , c ) 10 , d ) none of these , e ) can not be determined | a | divide(subtract(divide(58, divide(4, 5)), multiply(subtract(4, const_1), 5)), 5) | divide(n0,n1)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,n1)| | general |
in a party every person shakes hands with every other person . if there were a total of 136 handshakes in the party then what is the number of persons present in the party ? | "explanation : let the number of persons be n Γ’ Λ Β΄ total handshakes = nc 2 = 136 n ( n - 1 ) / 2 = 136 Γ’ Λ Β΄ n = 17 answer : c" | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | c | divide(add(sqrt(add(multiply(multiply(136, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general |
set a consists of all the prime numbers between 12 and 32 . what is the range of set a ? | "the range of a set of data is the difference between the highest and lowest values in the set in this set , highest number = 31 lowest number = 13 range = highest - lowest = 31 - 13 = 18 option b" | a ) 12 , b ) 18 , c ) 16.6 , d ) 17 , e ) 23 | b | subtract(subtract(32, const_2), add(add(12, const_4), const_1)) | add(n0,const_4)|subtract(n1,const_2)|add(#0,const_1)|subtract(#1,#2)| | general |
caleb spends $ 74.50 on 50 hamburgers for the marching band . if single burgers cost $ 1.00 each and double burgers cost $ 1.50 each , how many double burgers did he buy ? | "solution - lets say , single hamburgersxand double hamburgersy given that , x + y = 50 and 1 x + 1.5 y = 74.50 . by solving the equations y = 49 . ans e ." | a ) 5 , b ) 10 , c ) 20 , d ) 40 , e ) 49 | e | divide(subtract(74.50, 50), subtract(1.50, 1.00)) | subtract(n0,n1)|subtract(n3,n2)|divide(#0,#1)| | general |
the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 40 degrees . if the temperature on monday was 42 degrees . find the temperature on friday ? | "m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 40 = 160 m = 42 tu + w + th = 192 - 42 = 150 f = 160 β 150 = 10 answer : c" | a ) 8 , b ) 9 , c ) 10 , d ) 12 , e ) 16 | c | subtract(42, subtract(multiply(48, const_4), multiply(40, const_4))) | multiply(n0,const_4)|multiply(n1,const_4)|subtract(#0,#1)|subtract(n2,#2)| | general |
if eight machines working at the same rate can do 3 / 4 of a job in 30 minutes , how many minutes would it take two machines working at the same rate to do 3 / 5 of the job ? | "using the std formula m 1 d 1 h 1 / w 1 = m 2 d 2 h 2 / w 2 substituting the values we have 8 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 16 / 3 = 10 / 3 * x x = 8 / 5 hour so 96 minutes answer : d" | a ) 45 , b ) 60 , c ) 75 , d ) 96 , e ) 100 | d | divide(divide(3, 5), multiply(divide(divide(divide(3, 4), 5), 30), const_2)) | divide(n0,n4)|divide(n0,n1)|divide(#1,n4)|divide(#2,n2)|multiply(#3,const_2)|divide(#0,#4)| | physics |
if the sum of the 4 th term and the 12 th term of an arithmetic progression is 8 , what is the sum of the first 15 terms of the progression ? | "let a = first term , d = common difference . 4 th term = a + 3 d 12 th term = a + 11 d a + 3 d + a + 11 d = 8 2 a + 14 d = 8 a + 7 d = 4 . sum of the first 15 terms = 15 / 2 ( 2 a + 14 d ) = 15 ( a + 7 d ) = 15 ( 4 ) = 60 . ans ( a )" | a ) 60 , b ) 120 , c ) 160 , d ) 240 , e ) 840 | a | multiply(divide(15, const_2), 8) | divide(n3,const_2)|multiply(n2,#0)| | general |
niki covers a distance in 45 min , if she drives at a speed of 60 km / h on an average . find the speed at which she must drive at to reduce the time of the journey by 50 % . | given speed = 60 kmph . it means niki covered 60 km in 60 min ( 1 hour ) . so , in 45 min he will cover 45 km . 50 % time reduced = 45 - 50 % of 45 = 22.5 min . thus , niki needs to cover 45 km in 22.5 min ( 0.37 hour ) . speed * time = distance . speed * 0.37 = 45 speed = 121.62 kmph . answer : option b | a ) 101.02 kmph , b ) 121.62 kmph , c ) 100.62 kmph , d ) 111.00 kmph , e ) 121.00 kmph | b | multiply(divide(const_1, divide(50, const_100)), 60) | divide(n2,const_100)|divide(const_1,#0)|multiply(n1,#1) | general |
a man complete a journey in 20 hours . he travels first half of the journey at the rate of 10 km / hr and second half at the rate of 15 km / hr . find the total journey in km . | "0.5 x / 10 + 0.5 x / 15 = 20 - - > x / 10 + x / 15 = 40 - - > 5 x = 30 x 40 - - > x = ( 30 x 40 ) / 5 = 240 km . answer : b ." | a ) 220 km , b ) 240 km , c ) 230 km , d ) 232 km , e ) 234 km | b | multiply(const_2, divide(multiply(multiply(10, 15), 20), add(10, 15))) | add(n1,n2)|multiply(n1,n2)|multiply(n0,#1)|divide(#2,#0)|multiply(#3,const_2)| | physics |
a satellite is composed of 24 modular units , each of which is equipped with a set of sensors , some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is 1 / 4 the total number of upgraded sensors on the entire satellite , wh... | "let x be the number of upgraded sensors on the satellite . the number of non - upgraded sensors per unit is x / 4 . the number of non - upgraded sensors on the whole satellite is 24 ( x / 4 ) = 6 x . the fraction of sensors which have been upgraded is x / ( x + 6 x ) = x / 7 x = 1 / 7 the answer is d ." | a ) 5 / 6 , b ) 1 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) 1 / 24 | d | divide(24, add(24, multiply(24, multiply(divide(1, 4), 24)))) | divide(n1,n2)|multiply(n0,#0)|multiply(n0,#1)|add(n0,#2)|divide(n0,#3)| | general |
20 beavers , working together in a constant pace , can build a dam in 3 hours . how many b hours will it take 12 beavers that work at the same pace , to build the same dam ? | "c . 5 hrs if there were 10 beavers it could have taken double b = 6 hrs . . so closest to that option is 5 . c" | a ) 2 . , b ) 4 . , c ) b = 5 . , d ) b = 6 . , e ) b = 8 . | c | divide(multiply(3, 20), 12) | multiply(n0,n1)|divide(#0,n2)| | physics |
| x + 3 | β | 2 - x | = | 8 + x | how many solutions will this equation have ? | "you have | x + 3 | - | 4 - x | = | 8 + x | first , look at the three values independently of their absolute value sign , in other words : | x + 3 | - | 4 - x | = | 8 + x | ( x + 3 ) - ( 4 - x ) = ( 8 + x ) now , you ' re looking at x < - 8 , so x is a number less than - 8 . let ' s pretend x = - 10 here to make things... | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(multiply(add(2, const_3.0), const_2), 8) | add(n0,const_4.0)|multiply(#0,const_2)|divide(#1,n2)| | general |
from january 1 , 1991 , to january 1 , 1993 , the number of people enrolled in health maintenance organizations increased by 10 percent . the enrollment on january 1 , 1993 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 1991 ? | "1.10 x = 45 - - > 11 / 10 * x = 45 - - > x = 45 * 10 / 11 = 450 / 11 = ~ 41 . answer : d ." | a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | d | multiply(divide(const_100, add(const_100, 10)), 45) | add(n4,const_100)|divide(const_100,#0)|multiply(n7,#1)| | gain |
if a 2 cm cube is cut into 1 cm cubes , then what is the percentage increase in the surface area of the resulting cubes ? | "the area a of the large cube is 2 * 2 * 6 = 24 square cm . the area of the 8 small cubes is 8 * 6 = 48 = 2 a , an increase of 100 % . the answer is b ." | a ) 50 % , b ) 100 % , c ) 150 % , d ) 200 % , e ) 250 % | b | multiply(const_100, divide(multiply(surface_cube(1), surface_cube(2)), surface_cube(2))) | surface_cube(n1)|surface_cube(n0)|multiply(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | geometry |
a group of 4 investment bankers and 5 clients recently frequented the bonbon ribs restaurant . the total bill for the meal , including 20 % gratuity , came to $ 756 . on average , how much did the meal of each individual cost before gratuity ? | 4 ibs and 5 clients - so total 9 people the bill $ 756 includes 20 % gratuity . . . so the actual cost of dinner was $ 630 now , the cost per person will be $ 630 / 9 which is $ 70 option b | a ) $ 160 , b ) $ 70 , c ) $ 90 , d ) $ 80 , e ) $ 55 | b | divide(divide(756, divide(add(20, const_100), const_100)), add(4, 5)) | add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(n3,#2)|divide(#3,#1) | general |
a salesman ' s income consists of a commission and a base salary of $ 370 per week . over the past 5 weeks , his weekly income totals have been $ 406 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) commission be per week over the next two weeks so that his average weekly income is $ 500 ov... | "total weekly income over 5 weeks = $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070 for avg weekly income to be $ 500 over 7 weeks , we need total weekly income over 7 weeks = $ 3500 now , $ 3500 - $ 2070 = $ 1430 from this , we subtract base salary for 2 weeks i . e $ 370 * 2 = $ 740 therefore , commission = $ 1430 - $... | a ) $ 150 , b ) $ 345 , c ) $ 365 , d ) $ 715 , e ) $ 730 | b | subtract(divide(subtract(multiply(500, 7), add(add(add(406, 413), add(436, 420)), 395)), const_2), 370) | add(n2,n3)|add(n4,n5)|multiply(n7,n8)|add(#0,#1)|add(n6,#3)|subtract(#2,#4)|divide(#5,const_2)|subtract(#6,n0)| | general |
a car is running at a speed of 60 kmph . what distance will it cover in 8 sec ? | "speed = 60 kmph = 60 * 5 / 18 = 17 m / s distance covered in 8 sec = 17 * 8 = 136 m answer is b" | a ) 100 m , b ) 136 m , c ) 180 m , d ) 200 m , e ) 250 m | b | multiply(divide(60, const_3_6), 8) | divide(n0,const_3_6)|multiply(n1,#0)| | physics |
without stoppages , a train travels certain distance with an average speed of 300 km / h , and with stoppages , it covers the same distance with an average speed of 200 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 100 km less . time taken to cover 100 km = 100 Γ’ Β β 300 h = 1 Γ’ Β β 3 h = 1 Γ’ Β β 3 Γ£ β 60 min = 20 min answer a" | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | a | subtract(const_60, multiply(divide(200, 300), const_60)) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
a fruit drink is made of grapefruit , lemon , and orange juice , where 25 percent of the drink is grapefruit juice and 35 percent is lemon juice . if the drink is made with 20 ounces of orange juice , how many ounces is the drink total ? | let the total number of ounces in the drink be x % of orange = 25 % % of watermelon = 35 % % of grape = 100 % - 60 % = 40 % but this number is given as 20 ounces so 40 % of x = 20 and x = ( 20 ) ( 100 / 40 ) = ( 20 ) ( 5 / 2 ) x = 50 therefore there a total of 50 ounces in the drink . correct answer - c | a ) 60 , b ) 80 , c ) 50 , d ) 100 , e ) 40 | c | divide(20, divide(subtract(const_100, add(25, 35)), const_100)) | add(n0,n1)|subtract(const_100,#0)|divide(#1,const_100)|divide(n2,#2) | general |
` ` how old are you , alchemerion ? ' ' asked one of the wizards appearances the wizards answer with a riddle , ` ` i am still very young as wizards go . i am only 3 times my son ' s age . my father is 40 year more than twice of my age . together the 3 of us are a 1240 year old ' ' . how old is alchemerion | let alchemerion be a let son be s let father be f a = 3 * s f = 40 + ( 2 * a ) therefore 1240 = ( s + f + a ) 1240 = ( a / 3 ) + ( 40 + ( 2 * a ) ) + a a = 360 answer : d | a ) 300 , b ) 320 , c ) 340 , d ) 360 , e ) 380 | d | multiply(divide(subtract(1240, 40), const_10), 3) | subtract(n3,n1)|divide(#0,const_10)|multiply(n0,#1) | general |
in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 20 percent , but profits were 16 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,128 r = x / 100 * 0.1 r answer e" | a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 128 % | e | multiply(divide(multiply(subtract(const_1, divide(20, const_100)), divide(16, const_100)), divide(10, const_100)), const_100) | divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(#0,#3)|divide(#4,#2)|multiply(#5,const_100)| | gain |
if a person walks at 14 km / hr instead of 10 km / hr , he would have walked 20 km more . the actual distance travelled by him is | let the actual distance travelled be x km . then x / 10 = ( x + 20 ) / 14 - - > 14 x = 10 x + 200 - - > 4 x = 200 - - > x = 50 km . answer : a . | a ) 50 km , b ) 56 km , c ) 70 km , d ) 80 km , e ) 90 km | a | multiply(10, divide(20, subtract(14, 10))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1) | general |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 300 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 300 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . ... | a ) 1,108 , b ) 2,500 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 300) | divide(n6,n5)|multiply(n4,#0)| | general |
find the 25 % of rs . 300 . | "explanation : 25 % of 300 = > 25 / 100 * 300 = rs . 75 answer : e" | a ) s . 50 , b ) s . 70 , c ) s . 100 , d ) s . 80 , e ) s . 75 | e | multiply(300, divide(25, const_100)) | divide(n0,const_100)|multiply(n1,#0)| | gain |
a reduction of 50 % in the price of apples would enable a man to obtain 50 more for rs . 50 , what is reduced price per dozen ? | "c 50 * ( 50 / 100 ) = 25 - - - 50 ? - - - 12 = > rs . 6" | a ) 1 , b ) 3 , c ) 6 , d ) 7 , e ) 10 | c | multiply(divide(divide(multiply(50, 50), const_100), 50), const_12) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|multiply(#2,const_12)| | gain |
a 1000 liter tank , half - full of water is being filled from a pipe with a flow rate of 1 kiloliter every 2 minutes . at the same time , the tank is losing water from two drains at a rate of 1 kiloliter every 4 minutes and every 6 minutes . how many minutes does it take to fill the tank completely ? | "in : we have : 1,000 / 2 min = 500 litres per minute out : we have : 1,000 / 4 + 1,000 / 6 then do : in - out to figure out the net inflow per minute ( you get 83.3 ) . then divide the total number of litres you need ( 500 by that net inflow to get the minutes ) - 6 min . answer a ." | a ) 6 , b ) 12 , c ) 18 , d ) 24 , e ) 48 | a | divide(multiply(add(add(6, 4), 2), divide(1000, 2)), const_1000) | add(n4,n5)|divide(n0,n2)|add(n2,#0)|multiply(#2,#1)|divide(#3,const_1000)| | physics |
working simultaneously and independently at an identical constant rate , 4 machines of a certain type can produce a total of x units of product p in 6 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 2 x units of product p in 3 days ? | "4 machines do x units in 6 days we have x / 6 = > rate of the 4 machines we know that we need to have 2 x units in 3 days therefore , we need to get to 2 x / 3 rate of the machines . rate of one machine is x / 6 * 1 / 4 = x / 24 . now , we need to know how many machines need to work simultaneously , to get 2 x done in... | a ) 24 , b ) 18 , c ) 16 , d ) 12 , e ) 8 | c | multiply(multiply(4, 2), divide(6, 3)) | divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)| | general |
a bag of potatoes weighs 36 lbs divided by half of its weight . how much does the bag of potatoes weight ? | sol . 36 Γ· 3 = 12 . answer : e | a ) 20 lb , b ) 30 lb , c ) 10 lb , d ) 15 lb , e ) 36 lb | e | divide(36, const_1) | divide(n0,const_1) | general |
p , q and r together earn rs . 1800 in 9 days . p and r can earn rs . 600 in 5 days . q and r in 7 days can earn rs . 910 . how much amount does r can earn per day ? | explanation : amount earned by p , q and r in 1 day = 1800 / 9 = 200 - - - ( 1 ) amount earned by p and r in 1 day = 600 / 5 = 120 - - - ( 2 ) amount earned by q and r in 1 day = 910 / 7 = 130 - - - ( 3 ) ( 2 ) + ( 3 ) - ( 1 ) = > amount earned by p , q and 2 r in 1 day - amount earned by p , q and r in 1 day = 120 + 1... | a ) s . 50 , b ) s . 70 , c ) s . 90 , d ) s . 100 , e ) s . 120 | a | subtract(add(divide(600, 5), divide(910, 7)), divide(1800, 9)) | divide(n2,n3)|divide(n5,n4)|divide(n0,n1)|add(#0,#1)|subtract(#3,#2) | physics |
jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume , respectively . if these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid , approximately how many liters of the 2 percent solution will be required ? | "use weighted average : 2 % and 12 % solutions mix to give 5 % solution . w 1 / w 2 = ( a 2 - avg ) / ( avg - a 1 ) = ( 12 - 5 ) / ( 5 - 2 ) = 7 / 3 you need 7 parts of 2 % solution and 3 parts of 12 % solution to get 10 parts of 5 % solution . if total 5 % solution is actually 60 litres , you need 7 * 6 = 42 litres of... | a ) 18 , b ) 20 , c ) 24 , d ) 36 , e ) 42 | e | multiply(const_3, divide(60, const_10)) | divide(n2,const_10)|multiply(#0,const_3)| | gain |
in a kilo meter race , if a gives b a 40 m start , a wins by 19 s . but if a gives b a 30 s start , b wins by 40 m . find the time taken by b to run 5,000 m ? | when b is given a 40 m start then b runs 960 m in t secs and a runs 1000 m in t - 19 secs . when b is given a 30 second start then b runs 1000 m in t secs and a runs 960 m in t - 30 secs . assume that the speed of a ( va ) and b ( vb ) is constant in both situations . then vb = 960 Γ· t = 1000 Γ· t , and va = 1000 Γ· ( t ... | a ) 150 s , b ) 450 s , c ) 750 s , d ) 825 s , e ) 925 s | c | divide(multiply(const_1000, add(const_1, const_4)), divide(divide(40, const_2), const_3)) | add(const_1,const_4)|divide(n0,const_2)|divide(#1,const_3)|multiply(#0,const_1000)|divide(#3,#2) | physics |
0.9999 + 0.1111 = ? | "0.9999 + 0.1111 = 0.9999 + 0.111 + 0.0001 = ( 0.9999 + 0.0001 ) + 0.111 = 1 + 0.111 = 1.111 d" | a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.111 , e ) 1.1111 | d | multiply(divide(0.9999, 0.1111), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
exactly 18 % of the reporters for a certain wire service cover local politics in country x . if 40 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ? | "let ' s assume there are 100 reporters - - > 18 reporters cover local politics . now , as 40 % of the reporters who cover all politics do not cover local politics then the rest 60 % of the reporters who cover politics do cover local politics , so if there are x reporters who cover politics then 60 % of them equal to 1... | a ) 20 % , b ) 42 % , c ) 44 % , d ) 70 % , e ) 84 % | d | multiply(subtract(const_1, divide(18, subtract(const_100, 40))), const_100) | subtract(const_100,n1)|divide(n0,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | gain |
when 200 is divided by positive integer x , the remainder is 5 . what is the remainder when 395 is divided by x ? | "if 200 / x leaves a reminder 5 then ( 200 - 5 ) i . e . 195 is divisible by x so ( 200 + 195 ) / x leaves a reminder rem ( 200 / x ) + rem ( 195 / x ) = > 5 + 0 = 5 answer : e" | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 5 | e | subtract(200, subtract(395, 200)) | subtract(n2,n0)|subtract(n0,#0)| | general |
xy = 1 then what is ( 7 ^ ( x + y ) ^ 2 ) / ( 7 ^ ( x - y ) ^ 2 ) | "( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 4 7 ^ 4 = 2401 answer a" | a ) 2401 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | a | power(7, multiply(const_4, 1)) | multiply(n0,const_4)|power(n1,#0)| | general |
if a ( a + 2 ) = 80 and b ( b + 2 ) = 80 , where a β b , then a + b = | i . e . if a = 8 then b = - 10 or if a = - 10 then b = 8 but in each case a + b = - 10 + 8 = - 2 answer : e | a ) - 3 , b ) - 4 , c ) - 6 , d ) - 8 , e ) - 2 | e | add(divide(80, const_10), divide(80, divide(80, const_10))) | divide(n1,const_10)|divide(n1,#0)|add(#0,#1) | general |
a man sitting in a train which is traveling at 40 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . ? | "relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) * ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 40 ) kmph = 72 kmph . answer : e" | a ) 50 kmph , b ) 58 kmph , c ) 62 kmph , d ) 65 kmph , e ) 72 kmph | e | subtract(multiply(divide(280, 9), const_3_6), 40) | divide(n2,n1)|multiply(#0,const_3_6)|subtract(#1,n0)| | physics |
evaluate : 45 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 45 - 12 * 3 * 2 = 45 - 8 = 37 correct answer c" | a ) 62 , b ) 52 , c ) 37 , d ) 12 , e ) 22 | c | subtract(45, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 55 paise per sq m is | explanation : area to be plastered = [ 2 ( l + b ) Γ£ β h ] + ( l Γ£ β b ) = [ 2 ( 25 + 12 ) Γ£ β 6 ] + ( 25 Γ£ β 12 ) = 744 sq m cost of plastering = 744 Γ£ β ( 55 / 100 ) = rs . 409.20 answer : c | a ) rs . 209.20 , b ) rs . 309.20 , c ) rs . 409.20 , d ) rs . 509.20 , e ) none of these | c | multiply(divide(55, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | divide(n3,const_100)|multiply(n0,n2)|multiply(n1,n2)|multiply(n0,n1)|multiply(#1,const_2)|multiply(#2,const_2)|add(#4,#5)|add(#6,#3)|multiply(#7,#0) | physics |
let p be a prime number greater than 2 and let n = 14 p . how many even numbers divide n ? | answer : a . there is exactly one . the prime factorization of 14 is 14 = 2 * 7 , so n = 2 * 7 * p = 7 * 2 * p = 7 * 2 p , so 2 p is the only even number that divides n . | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(2, const_1) | subtract(n0,const_1) | general |
compound interest of rs . 5000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) . | "10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 5000 [ 1 + ( 5 / 100 ) ] ^ 3 - 5000 = 5000 [ ( 21 / 20 ) ^ 3 - 1 ] = 5000 ( 9261 - 8000 ) / 8000 = 5 * 1261 / 8 = 788 answer : d" | a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 788 , e ) none of the above | d | subtract(multiply(5000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 5000) | divide(n1,n4)|divide(n2,n4)|add(n2,#1)|divide(#0,const_100)|add(#3,n2)|multiply(#2,n4)|power(#4,#5)|multiply(n0,#6)|subtract(#7,n0)| | gain |
a train starts full of passengers . at the first station , it drops one - third of the passengers and takes 280 more . at the second station , it drops one - half of the new total and takes 12 more . on arriving at the third station , it is found to have 248 passengers . find the number of passengers in the beginning . | let x be no . of passengers at beginning at , 1 st station is drops 1 / 3 rd so remaining 2 / 3 rd are in train only plus 280 i . e . , 2 x / 3 + 280 at 2 nd stop 1 / 2 of new total and 12 more i . e . , ( 2 x / 3 + 280 ) / 2 + 12 = 248 on solving above equation we get x as 288 answer : d | a ) 240 , b ) 248 , c ) 280 , d ) 288 , e ) 290 | d | divide(multiply(subtract(multiply(subtract(248, 12), const_2), 280), const_3), const_2) | subtract(n2,n1)|multiply(#0,const_2)|subtract(#1,n0)|multiply(#2,const_3)|divide(#3,const_2) | general |
the credit card and a global payment processing companies have been suffering losses for some time now . a well known company recently announced its quarterly results . according to the results , the revenue fell to $ 52.0 billion from $ 69.0 billion , a year ago . by what percent did the revenue fall ? | "$ 69 - $ 52 = 17 $ ( 17 / 69 ) * 100 = 24.6 % answer : b" | a ) 20.8 , b ) 24.6 , c ) 31.8 , d ) 32.5 , e ) 33.8 | b | multiply(divide(subtract(69.0, 52.0), 69.0), const_100) | subtract(n1,n0)|divide(#0,n1)|multiply(#1,const_100)| | gain |
what will be the compound interest on rs . 35000 after 3 years at the rate of 12 % per annum | "explanation : ( 35000 Γ ( 1 + 12 / 100 ) 3 ) = > 35000 Γ 28 / 25 Γ 28 / 25 Γ 28 / 25 = > 49172.48 so compound interest will be 49172.48 - 35000 = rs 14172.48 option d" | a ) rs 13123.20 , b ) rs 14103.48 , c ) rs 12172.48 , d ) rs 14172.48 , e ) none of these | d | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
in what ratio must wheat at rs . 3.20 pe rkg be mixed with wheat at rs . 2.90 per kg so that the mixture be worth rs . 3.08 per kg | if 2 ingredients are mixed , then the required ratio is given by , the rule of allegation c . p of a unit quantity of cheaper ( c ) c . p . of a unit quantity of dearer ( d ) mean price ( m ) d - m m - c ο required ratio is cheaper quantity : dearer quantity = ( d - m ) : ( m - c ) answer with explanation : given i . c... | a ) 3 : 4 , b ) 3 : 2 , c ) 3 : 5 , d ) 3 : 7 , e ) 2 : 3 | b | divide(subtract(3.08, 2.9), subtract(3.2, 3.08)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1) | other |
a , b and c can do a piece of work in 24 days , 30 days and 40 days respectively . they began the work together but c left 10 days before the completion of the work . in how many days was the work completed ? | "one day work of a , b and c = 1 / 24 + 1 / 30 + 1 / 40 = 1 / 10 work done by a and b together in the last 10 days = 10 * ( 1 / 24 + 1 / 30 ) = 3 / 4 remaining work = 1 / 4 the number of days required for this initial work = 2.5 days . the total number of days required = 10 + 2.5 = 12.5 days . answer : a" | a ) 12.5 days , b ) 16 days , c ) 18 days , d ) 11 days , e ) 38 days | a | add(divide(subtract(const_1, multiply(add(divide(const_1, 24), divide(const_1, 30)), 10)), add(divide(const_1, multiply(add(const_2, const_3), multiply(const_2, 10))), add(divide(const_1, 24), divide(const_1, 30)))), 10) | add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|multiply(const_2,n3)|add(#1,#2)|multiply(#0,#3)|divide(const_1,#5)|multiply(n3,#4)|add(#4,#6)|subtract(const_1,#7)|divide(#9,#8)|add(n3,#10)| | physics |
if a 2 + b 2 + c 2 = 213 and ab + bc + ca = 6 , then a + b + c is | "by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 213 and ab + bc + ca = 6 , ( a + b + c ) ^ 2 = 213 + 2 ( 6 ) = 225 = 15 ^ 2 therefore : a + b + c = 15 answer : c" | a ) 14 , b ) 20 , c ) 15 , d ) 32 , e ) 64 | c | sqrt(add(213, multiply(6, 2))) | multiply(n4,n0)|add(n3,#0)|sqrt(#1)| | general |
if a - b = 6 and a 2 + b 2 = 48 , find the value of ab . | "2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 48 - 36 = 12 ab = 6 . answer : d" | a ) 8 , b ) 10 , c ) 12 , d ) 6 , e ) 19 | d | divide(subtract(48, power(6, 2)), 2) | power(n0,n1)|subtract(n3,#0)|divide(#1,n1)| | general |
a certain article of clothing was discounted during a special sale to 2 / 5 of its original retail price . when the clothing did n ' t sell , it was discounted even further to 1 / 2 of its original retail price during a second sale . by what percent did the price of this article of clothing decrease from the first sale... | "say the original retail price of the item was $ 200 . the price after the first sale = 3 / 5 * $ 200 = $ 120 . the price after the second sale = 1 / 2 * $ 200 = $ 100 . the percent change from the first sale to the second = ( 120 - 100 ) / 120 = 1 / 3 = 16.66 % . answer : d ." | a ) 50 % , b ) 33.33 % , c ) 25 % , d ) 16.66 % , e ) 12.5 % | d | multiply(subtract(divide(2, 5), multiply(divide(2, 5), divide(1, 2))), const_100) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|subtract(#0,#2)|multiply(#3,const_100)| | gain |
a committee is reviewing a total of 30 x black - and - white films and 6 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction of the selected films are in color ? | "say x = y = 10 . in this case we would have : 30 x = 300 black - and - white films ; 6 y = 60 color films . y / x % = 10 / 10 % = 1 % of the black - and - white films , so 3 black - and - white films and all 60 color films , thus total of 63 films were selected . color films thus compose 60 / 63 = 20 / 21 of the selec... | a ) 22 / 21 , b ) 20 / 21 , c ) 23 / 21 , d ) 25 / 21 , e ) 28 / 21 | b | divide(6, add(divide(30, const_100), 6)) | divide(n0,const_100)|add(n1,#0)|divide(n1,#1)| | general |
after paying a 10 percent tax on all income over $ 3000 , a person had a net income of $ 12000 . what was the income before taxes ? | let x be the income over 3000 then ( x - x / 10 ) + 3000 = 12000 = > x = 10000 therefore income before taxes = 10000 + 3000 = 13000 option ( b ) | a ) $ 13300 , b ) $ 13000 , c ) $ 12900 , d ) $ 10000 , e ) $ 9000 | b | divide(subtract(12000, multiply(divide(10, const_100), 3000)), subtract(const_1, divide(10, const_100))) | divide(n0,const_100)|multiply(n1,#0)|subtract(const_1,#0)|subtract(n2,#1)|divide(#3,#2) | gain |
sam invested rs . 8000 @ 10 % per annum for one year . if the interest is compounded half - yearly , then the amount received by sam at the end of the year will be ? | "p = rs . 8000 ; r = 10 % p . a . = 5 % per half - year ; t = 1 year = 2 half - year amount = [ 8000 * ( 1 + 5 / 100 ) 2 ] = ( 8000 * 21 / 20 * 21 / 20 ) = rs . 8820.00 answer : c" | a ) 3200.0 , b ) 11520.0 , c ) 8820.0 , d ) 7354.0 , e ) 16537.11 | c | multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 8000) | divide(n1,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)| | gain |
half a number plus 5 is 11 . what is the number ? | "solution let x be the number . always replace ` ` is ' ' with an equal sign ( 1 / 2 ) x + 5 = 11 ( 1 / 2 ) x + 5 - 5 = 11 - 5 ( 1 / 2 ) x = 6 2 Γ ( 1 / 2 ) x = 6 Γ 2 x = 12 answer c" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | multiply(subtract(11, 5), const_2) | subtract(n1,n0)|multiply(#0,const_2)| | general |
a can complete a certain job in 19 days . b is 90 % more efficient than a . in how many days can b complete the same job ? | "let , total work unit = 190 units a can finish in 19 days = 190 unit work i . e . a can finish in 1 days = 10 unit work i . e . b can finish in 1 days = 10 + ( 90 / 100 ) * 10 = 19 unit work days in which b will complete the work alone = 190 / 19 = 10 days answer : option e" | a ) 6 , b ) 6.25 , c ) 7 , d ) 7.5 , e ) 10 | e | divide(multiply(19, 90), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain |
if 20 men can build a water fountain 56 metres long in 7 days , what length of a similar water fountain can be built by 35 men in 3 days ? | "explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 7 : 3 : : 56 : x therefore ( 20 x 7 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 140 = 42 hence , the required length is 42 m . answer : b" | a ) 40 m , b ) 42 m , c ) 47 m , d ) 49 m , e ) 50 m | b | multiply(divide(56, multiply(20, 7)), multiply(35, 3)) | multiply(n0,n2)|multiply(n3,n4)|divide(n1,#0)|multiply(#2,#1)| | physics |
jaime earned enough money by selling seashells at 20 cents each to buy several used paperback books at 55 cents each . if he spent all of the money he earned selling seashells to buy the books , what is the least number of seashells he could have sold ? | "let ' s test answer c : 22 seashells . . . with 22 seashells , jamie would have 22 ( 20 ) = 440 cents . this would allow him to buy 8 books for 440 cents total , so this is the correct answer . c" | a ) 5 , b ) 10 , c ) 22 , d ) 25 , e ) 30 | c | add(divide(lcm(20, 55), 20), const_10) | lcm(n0,n1)|divide(#0,n0)|add(#1,const_10)| | general |
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