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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how many positive integers less than 248 are multiple of 4 but not multiples of 6 ? | "244 / 4 = 61 multiples of 4 which are a multiple of 6 will be of the form 2 * 2 * 3 = 12 n where n > 0 240 / 12 = 20 61 - 20 = 41 answer : b" | a ) 20 , b ) 41 , c ) 42 , d ) 53 , e ) 64 | b | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 210 sq . feet , how many feet of fencing will be required ? | "explanation we have : l = 20 ft and lb = 210 sq . ft . so , b = 10.5 ft . length of fencing = ( l + 2 b ) = ( 20 + 21 ) ft = 41 ft . answer b" | a ) 34 , b ) 41 , c ) 68 , d ) 88 , e ) none | b | add(multiply(divide(210, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
in the coordinate plane a slope of the line k is 4 times the x - intercept of the line k . what is the y - intercept of the line k ? | as 4 y = mx + c , from 4 y = 0 m + m we get y = 1 / 4 . hence , the correct answer choice is d . | a ) - 4 , b ) 4 , c ) - 1 / 4 , d ) 1 / 4 , e ) 2 | d | divide(const_1, 4) | divide(const_1,n0)| | general |
find the smallest four - digit number which is a multiple of 112 . | "the smallest four digit number is 1000 . if 1000 is divided by 112 , the remainder is 104 . 112 - 104 = 8 , if 8 is added to 1000 , it will become the smallest four digit number and a multiple of 112 . answer : b" | a ) 896 , b ) 1008 , c ) 1120 , d ) 1024 , e ) none of these | b | multiply(divide(divide(divide(divide(112, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(112, const_1000), const_3), const_3), const_3)) | divide(n0,const_1000)|divide(#0,const_3)|divide(#1,const_3)|divide(#2,const_3)|multiply(#3,#3)| | general |
the difference between the place values of two sevens in the numerical 69758472 is | "required difference = ( 700000 - 70 ) = 699930 . answer is c" | a ) 0 , b ) 6993 , c ) 699930 , d ) 996330 , e ) none of them | c | divide(add(69758472, const_1), const_2) | add(n0,const_1)|divide(#0,const_2)| | general |
a set s = { x , - 8 , - 5 , - 3 , 3 , 6 , 9 , y } with elements arranged in increasing order . if the median and the mean of the set are the same , what is the value of | x | - | y | ? | "median of the set = ( - 3 + 3 ) / 2 = 0 as per statement , mean of the set = 0 mean of the set | y | - | x | + 18 - 16 = 0 ( where x is negative n y is positive ) | y | - | x | = - 2 so the absolute difference between two numbers is 2 answer a" | a ) 2 , b ) 0 , c ) - 1 , d ) can not be determined , e ) 1 | a | subtract(subtract(subtract(add(add(9, 8), 3), 3), 5), 8) | add(n0,n5)|add(n2,#0)|subtract(#1,n2)|subtract(#2,n1)|subtract(#3,n0)| | general |
the market value of a certain machine decreased by 30 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 10,000 , what was its market value two years later ? | "b . market value in 1982 = $ 10000 market value in 1983 = $ 10000 - ( $ 10000 x 30 / 100 ) = 10000 - 3000 = $ 7000 market value in 1984 = market value in 1983 - ( 30 % of $ 10000 ) = 7000 - 3000 = $ 4000" | a ) $ 8,000 , b ) $ 4,000 , c ) $ 3,200 , d ) $ 2,400 , e ) $ 800 | b | subtract(multiply(multiply(const_100, 30), multiply(const_2, const_4)), multiply(multiply(multiply(const_100, 30), multiply(const_2, const_4)), multiply(divide(30, const_100), const_2))) | divide(n0,const_100)|multiply(n0,const_100)|multiply(const_2,const_4)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#3,#4)|subtract(#3,#5)| | gain |
the value of log 2 16 is | "solution let log 216 = n . then , 2 n = 16 = 24 βΉ = βΊ n = 4 . answer b" | a ) 1 / 8 , b ) 4 , c ) 8 , d ) 16 , e ) 17 | b | log(divide(log(subtract(16, multiply(add(const_4, const_1), const_1000))), log(add(const_4, const_1)))) | add(const_1,const_4)|log(#0)|multiply(#0,const_1000)|subtract(n1,#2)|log(#3)|divide(#4,#1)|log(#5)| | other |
dividing by 3 β 7 and then multiplying by 5 β 6 is the same as dividing by what number ? | "say x / 3 / 7 * 5 / 6 = x * 7 / 3 * 5 / 6 = x * 35 / 18 a" | a ) 35 β 18 , b ) 16 β 5 , c ) 20 β 9 , d ) 9 β 20 , e ) 5 β 16 | a | multiply(divide(7, 3), divide(5, 6)) | divide(n1,n0)|divide(n2,n3)|multiply(#0,#1)| | general |
the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 13 . if p invested the money for 5 months , find for how much time did q invest the money ? | "7 * 5 : 5 * x = 7 : 13 x = 13 answer : c" | a ) 19 , b ) 17 , c ) 13 , d ) 10 , e ) 12 | c | multiply(multiply(divide(7, 5), divide(13, 7)), 5) | divide(n0,n1)|divide(n3,n2)|multiply(#0,#1)|multiply(n4,#2)| | gain |
excluding stoppages , the speed of a bus is 50 kmph and including stoppages , it is 45 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 5 km less . time taken to cover 5 km = ( 5 / 50 ) x 60 = 6 min answer : b" | a ) 7 min , b ) 6 min , c ) 9 min , d ) 10 min , e ) 11 min | b | multiply(const_60, divide(subtract(50, 45), 50)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
by selling 16 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ? | "80 % - - - 16 120 % - - - ? 80 / 120 * 16 = 24 answer : c" | a ) 28 , b ) 25 , c ) 24 , d ) 23 , e ) 21 | c | multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 16) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain |
the ratio of spinsters to cats is 2 to 9 . if there are 42 more cats than spinsters , how many spinsters are there ? | "let 2 x be the number of spinsters . then 9 x is the number of cats . 9 x - 2 x = 42 x = 6 and the number of spinsters is 2 ( 6 ) = 12 . the answer is d ." | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | multiply(divide(42, subtract(9, 2)), 2) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)| | other |
find the smallest number in a gp whose sum is 38 and product 1728 | "in gp : - a / r + a + ar . . . . = 38 a ( 1 + r + r ^ 2 ) = 38 r . . . . . . eq 1 product a ^ 3 = 1728 a = 12 now put this value in eq 1 12 ( 1 + r + r ^ 2 ) = 38 r r = 2 / 3 and 3 / 2 . . so sallest number is 8 answer : c" | a ) 12 , b ) 20 , c ) 8 , d ) all of these , e ) none of these | c | multiply(divide(divide(divide(divide(38, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(38, const_1000), const_3), const_3), const_3)) | divide(n0,const_1000)|divide(#0,const_3)|divide(#1,const_3)|divide(#2,const_3)|multiply(#3,#3)| | general |
what no . should be subtracted from x ^ 3 + 4 x 2 β 7 x + 12 x ^ 3 + 4 x ^ 2 β 7 x + 12 , if it is to be perfectly divisible by x + 3 x + 3 ? | according to remainder theorem when dfracf ( x ) x + adfracf ( x ) x + a , then the remainder is f ( β a ) f ( β a ) . in this case , as x + 3 x + 3 divides x 3 + 4 x 2 β 7 x + 12 β kx 3 + 4 x 2 β 7 x + 12 β k perfectly ( kk being the number to be subtracted ) , the remainder is 0 when the value of xx is substituted by... | a ) 38 , b ) 42 , c ) 46 , d ) 49 , e ) 62 | b | add(subtract(add(power(negate(3), 3), multiply(4, power(negate(3), 2))), multiply(7, negate(3))), 12) | negate(n0)|multiply(n3,#0)|power(#0,n2)|power(#0,n0)|multiply(n1,#2)|add(#4,#3)|subtract(#5,#1)|add(n4,#6) | general |
76 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | "x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 76 = 33.25 answer : c" | a ) 13.25 , b ) 38.25 , c ) 33.25 , d ) 34.25 , e ) 31.25 | c | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 76), const_2) | add(const_3,const_4)|add(#0,const_2)|divide(#0,#1)|multiply(n0,#2)|divide(#3,const_2)| | general |
in a division sum , the quotient is 40 , the divisor 72 and the remainder 64 , find the dividend ? | explanation : 40 * 72 + 64 = 2944 answer : b | a ) 2674 , b ) 2944 , c ) 2464 , d ) 2924 , e ) 2994 | b | add(multiply(40, 72), 64) | multiply(n0,n1)|add(n2,#0) | general |
a store sells chairs and tables . if the price of 3 chairs and 1 table is 60 % of the price of 1 chair and 3 tables , and the price of 1 table and 1 chair is $ 80 , what is the price , in dollars , of 1 table ? ( assume that every chair has the same price and every table has the same price . ) | "let c = chair ; t = table 3 c + 1 t = 0.6 ( 1 c + 3 t ) or c ( 3 - 0.6 ) = 1.8 t - 1 t or 2.4 c = 0.8 t therefore c = 0.8 / 2.4 t = 1 / 3 t ic + 1 t = 80 or 1 / 3 t + 1 t = 80 therefore t = 80 * 3 / 4 = 60 d" | a ) 68 , b ) 84 , c ) 72 , d ) 60 , e ) 64 | d | divide(subtract(multiply(80, 3), multiply(80, divide(60, const_100))), add(divide(60, const_100), 1)) | divide(n2,const_100)|multiply(n7,n0)|add(#0,n1)|multiply(n7,#0)|subtract(#1,#3)|divide(#4,#2)| | gain |
the probability that a man will be alive for 10 more yrs is 1 / 4 & the probability that his wife will alive for 10 more yrs is 1 / 3 . the probability that none of them will be alive for 10 more yrs , is | sol . required probability = pg . ) x p ( b ) = ( 1 β d x ( 1 β i ) = : x 1 = 1 / 2 ans . ( a ) | a ) 1 / 2 , b ) 1 , c ) 2 / 3 , d ) 3 / 5 , e ) 3 / 8 | a | multiply(subtract(1, divide(1, 4)), subtract(1, divide(1, 3))) | divide(n1,n2)|divide(n1,n5)|subtract(n1,#0)|subtract(n1,#1)|multiply(#2,#3) | general |
how many positive even integers less than 100 contain digits 4 or 7 ? | "one digit numbers : 4 two digit numbers : 4 at tens place : 40 , 42,44 , 46,48 7 at tens place : 70 , 72,74 , 76,78 4 at units place : 14 , 24,34 , 54,64 , 84,94 ( duplicates removed ) if 7 is at units place , the number cant be even total : 1 + 5 + 5 + 7 = 18 answer c" | a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | c | multiply(4, const_2) | multiply(n1,const_2)| | general |
a car traveled from san diego to san francisco at an average speed of 48 miles per hour . if the journey back took twice as long , what was the average speed of the trip ? | "let the time taken be = x one way distance = 48 x total distance traveled = 2 * 48 x = 96 x total time taken = x + 2 x = 3 x average speed = 96 x / 3 x = 32 answer : b" | a ) 24 . , b ) 32 . , c ) 36 , d ) 42 . , e ) 44 . | b | inverse(add(inverse(48), divide(inverse(48), const_2))) | inverse(n0)|divide(#0,const_2)|add(#1,#0)|inverse(#2)| | physics |
amount of bacteria present time amount 1 : 00 p . m . 8.0 grams 4 : 00 p . m . x grams 7 : 00 p . m . 12.4 grams data for a certain biology experiment are given in the table above . if the amount of bacteria present increased by the same fraction during each of the two 3 - hour periods shown , how many grams of bacteri... | the question says that bacteria increased by same fraction , not by same amount in 2 intervals of 3 hours . let x represent the amount of bacteria present at 4 : 00 pm . since the fractional increase must remain constant from 1 to 4 pm as it is from 4 pm to 7 pm : fractional increase from 1 pm to 4 pm = x / 8.0 fractio... | a ) 12.0 , b ) 12.1 , c ) 12.2 , d ) 10.0 , e ) 10.4 | d | sqrt(multiply(8, 12.4)) | multiply(n2,n7)|sqrt(#0) | general |
find the ratio in which rice at rs . 7.20 a kg be mixed with rice at rs . 5.60 a kg to produce a mixture worth rs . 6.30 a kg ? | "solution required ratio = 70 : 90 = 7 : 9 answer d" | a ) 1 : 3 , b ) 2 : 3 , c ) 3 : 4 , d ) 7 : 9 , e ) none of these | d | divide(subtract(6.30, 5.60), subtract(7.20, 6.30)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | other |
i bought two books ; for rs . 300 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | "x * ( 85 / 100 ) = ( 300 - x ) 119 / 100 x = 175 answer : a" | a ) 175 , b ) 280 , c ) 189 , d ) 278 , e ) 268 | a | divide(multiply(300, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | add(n2,const_100)|subtract(const_100,n1)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)| | gain |
p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 130 days , in how many days can r alone complete the remaining work ? | "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 25... | a ) 50 days , b ) 83.6 days , c ) 100 days , d ) 150 days , e ) 80 days | b | subtract(multiply(const_4, 50), multiply(divide(130, const_100), 60)) | divide(n4,const_100)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#1,#2)| | physics |
a confectioner decides to sell all of his pastry due to the coming holiday . his pastry goods are equally divided among a group of 15 regular customers . if only 49 customers come to the bakery , each one will receive 6 less pastry goods . how much pastry does the confectioner needs to sell ? | "pastry is divided in 15 customers equally . so , total number of pastry must be a multiple of 15 only option a satisfies the condition , and hence is the answer" | a ) 450 . , b ) 459 . , c ) 432 . , d ) 502 . , e ) 522 . | a | multiply(divide(multiply(49, 6), subtract(49, 15)), 15) | multiply(n1,n2)|subtract(n1,n0)|divide(#0,#1)|multiply(n0,#2)| | general |
for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 8 + 3 and * 8 + 8 , inclusive ? | "generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 8 would be divisible by all numbers from 1 to 8 . = > * 8 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 8 is divisible by 3 ) in fact adding anyprimenumber between 1 to 8 to * 8 will definitely be ... | a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(subtract(add(multiply(multiply(multiply(8, 3), const_2), const_4), 8), add(multiply(multiply(multiply(8, 3), const_2), const_4), 3)), 1) | multiply(n1,n2)|multiply(#0,const_2)|multiply(#1,const_4)|add(n1,#2)|add(n2,#2)|subtract(#3,#4)|subtract(#5,n0)| | general |
the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each . the radius of the sphere is : | "explanation : curved surface area of sphere = 4 / Ο r 2 surface area of cylinder = 2 Ο rh = > 4 / Ο r 2 = 2 Ο rh = > r 2 = 6 β 12 / 2 = > r 2 = 36 = > r = 6 note : diameter of cylinder is 12 so radius is taken as 6 . option b" | a ) 4 cm , b ) 6 cm , c ) 8 cm , d ) 10 cm , e ) 9 cm | b | sqrt(divide(multiply(multiply(const_pi, multiply(12, divide(12, const_2))), const_2), multiply(const_pi, const_4))) | divide(n0,const_2)|multiply(const_4,const_pi)|multiply(n0,#0)|multiply(#2,const_pi)|multiply(#3,const_2)|divide(#4,#1)|sqrt(#5)| | geometry |
during the first week of performances of a certain play , 5400 tickets were sold , all at reduced price . during the remaining weeks of performances , 5 times as many tickets were sold at full price as were sold at reduced price . if the total number of tickets e sold was 25200 , how many of them were sold at full pric... | given : the question tells us about the number of tickets sold in the first week at reduced price = 5400 . it also tells us that during the remaining weeks , the number of tickets sold at full price was 5 times the no . of tickets sold at reduced price . the total tickets sold was 25200 , and we are asked to find the n... | a ) 21000 , b ) 19800 , c ) 16500 , d ) 15800 , e ) 8700 | c | divide(multiply(subtract(25200, 5400), 5), add(const_2, const_4)) | add(const_2,const_4)|subtract(n2,n0)|multiply(n1,#1)|divide(#2,#0) | general |
a cricket bat is sold for $ 900 , making a profit of $ 100 . the profit percentage would be | "100 / ( 900 - 100 ) = 100 / 800 = 0.125 = 12.5 % . answer : b" | a ) 24 % , b ) 12.5 % , c ) 30 % , d ) 36 % , e ) 40 % | b | multiply(divide(100, subtract(900, 100)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain |
a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 279 metres long in 20 seconds . the length of the train ( in metres ) is : | "explanation : let the length of train be l m . acc . to question ( 279 + l ) / 20 = l / 8 2232 + 8 l = 20 l l = 2232 / 12 = 186 m answer a" | a ) 186 , b ) 176 , c ) 175 , d ) 96 , e ) none of these | a | multiply(divide(279, subtract(20, 8)), 8) | subtract(n2,n0)|divide(n1,#0)|multiply(n0,#1)| | physics |
mary ' s income is 60 % more than tim ' s income and tim ' s income is 10 % less than juan ' s income . what % of juan ' s income is mary ' s income . | "even i got 96 % j = 100 t = 100 * 0.9 = 90 m = 90 * 1.6 = 144 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 144 * 100 / 100 = 144 ans : a" | a ) 144 % , b ) b . 120 % , c ) 96 % , d ) 80 % , e ) 64 % | a | add(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(60, const_100))) | divide(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|add(#2,#1)| | general |
let s be the set of all positive integers that , when divided by 8 , have a remainder of 5 . what is the 75 th number in this set ? | "the set s = { 5 , 13 , 21 , 29 , . . . . . . . . . . . . . . . . . . . . . } 1 st number = 8 * 0 + 5 = 5 2 nd number = 8 * 1 + 5 = 13 3 rd number = 8 * 2 + 5 = 21 75 th number = 8 * ( 75 - 1 ) + 5 = 597 answer = a" | a ) 597 , b ) 608 , c ) 613 , d ) 616 , e ) 621 | a | add(multiply(subtract(75, const_1), 8), 5) | subtract(n2,const_1)|multiply(n0,#0)|add(n1,#1)| | general |
the perimeter of a square is equal to the radius of a circle having area 39424 sq cm , what is the area of square ? | let the side of square = a cm and radius of circle = r cm given , 4 a = r also area of square = pi * r ^ 2 = 39424 , r ^ 2 = 12544 or r = 112 so a = r / 4 = 112 / 4 = 28 hence area of square = a ^ 2 = 28 ^ 2 = 784 cm ^ 2 answer : c | ['a ) 1225 sq cm', 'b ) 441 sq cm', 'c ) 784 sq cm', 'd ) can β t say', 'e ) none of these'] | c | square_area(square_edge_by_perimeter(sqrt(divide(39424, const_pi)))) | divide(n0,const_pi)|sqrt(#0)|square_edge_by_perimeter(#1)|square_area(#2) | geometry |
a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 65 percent of books that were loaned out are returned and there are 54 books in the spec... | "the total number of books is 75 . let x be the number of books which were loaned out . 65 % of books that were loaned out are returned . 35 % of books that were loaned out are not returned . now , there are 54 books , thus the number of un - returned books is 75 - 54 = 21 books . 0.35 x = 21 x = 60 the answer is e ." | a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 60 | e | divide(subtract(75, 54), subtract(const_1, divide(65, const_100))) | divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)| | gain |
rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 86 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 86 5 x / 100 + 90 β 6 x / 100 = 86 x / 100 = 4 = > x = 400 . answer : e" | a ) 388 , b ) 2998 , c ) 277 , d ) 500 , e ) 400 | e | multiply(add(5, 6), const_100) | add(n1,n2)|multiply(#0,const_100)| | gain |
what is the smallest no . that should be added to 236 to make it exactly divisible by 9 ? | "if a number is divisible by 9 , the sum of its digits must be a multiple of 9 . here , 2 + 3 + 6 = 11 , the next multiple of 9 is 18 . 7 must be added to 236 to make it divisible by 9 d" | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | add(subtract(9, reminder(236, 9)), const_2) | reminder(n0,n1)|subtract(n1,#0)|add(#1,const_2)| | general |
each of the positive integers a , b , and c is a 3 - digit integer . if each of the digits 1 through 9 appears in one of these 3 integers , what is the maximum possible value of the sum of a , b , and c ? | according to the stem we should use the digits 1 through 9 to construct 3 three - digit integers , so that their sum is as big as possible . to maximize the sum , maximize the hundreds digits of a , b , and c , so make them 9 , 8 , and 7 . next , minimize tens digits . make them 6 , 5 , and 4 . use the remaining digits... | a ) 2445 , b ) 2331 , c ) 2874 , d ) 2801 , e ) 2256 | e | subtract(subtract(add(add(multiply(multiply(add(const_2, const_3), const_3), const_10), multiply(subtract(multiply(9, const_3), const_3), const_100)), add(const_3, const_3)), const_100), const_100) | add(const_2,const_3)|add(const_3,const_3)|multiply(n2,const_3)|multiply(#0,const_3)|subtract(#2,const_3)|multiply(#3,const_10)|multiply(#4,const_100)|add(#5,#6)|add(#7,#1)|subtract(#8,const_100)|subtract(#9,const_100) | general |
the profits of qrs company rose 50 % from march to april , then dropped 20 % from april to may , then rose 50 % from may to june . what was the percent increase for the whole quarter , from march to june ? | "assume 100 in march , then 150 in april as 50 % increase , then 120 in may as 20 % decrease from april , and then 180 in june which is 150 % of 120 . so overall increase is from 100 to 180 is 80 % answer e" | a ) 15 % , b ) 32 % , c ) 40 % , d ) 62 % , e ) 80 % | e | multiply(const_100, subtract(multiply(add(const_1, divide(50, const_100)), multiply(add(const_1, divide(50, const_100)), subtract(const_1, divide(20, const_100)))), const_1)) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#2)|multiply(#4,#5)|multiply(#3,#6)|subtract(#7,const_1)|multiply(#8,const_100)| | gain |
a certain bag contains 100 balls Γ’ β¬ β 20 white , 30 green , 10 yellow , 37 red , and 3 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 20 + 30 + 10 ) / 100 = 60 / 100 = 0.6 . answer is c" | a ) 0.9 , b ) 0.75 , c ) 0.6 , d ) 0.8 , e ) 0.5 | c | divide(subtract(100, add(37, 3)), 100) | add(n4,n5)|subtract(n0,#0)|divide(#1,n0)| | other |
what is the next number : 2 , 17 , 257 , __ | "4 ^ 0 + 1 = 2 4 ^ 2 + 1 = 17 4 ^ 4 + 1 = 257 4 ^ 6 + 1 = 4097 the answer is c ." | a ) 2977 , b ) 3837 , c ) 4097 , d ) 4847 , e ) 5387 | c | subtract(subtract(subtract(multiply(257, const_10), const_100), 2), 2) | multiply(n2,const_10)|subtract(#0,const_100)|subtract(#1,n0)|subtract(#2,n0)| | general |
calculate the effect changes in dimension of a rectangle will have on its area , if length is increased by 15 % and its breadth is decreased by 10 % ? | "let l and b be 100 each 100 * 100 = 10000 l increase by 15 % = 115 b decrease by 10 % = 90 115 * 90 = 10350 3.5 percent increase answer : e" | a ) 3.5 percent increase , b ) 9.5 percent increase , c ) 1.5 percent increase , d ) 3.9 percent increase , e ) 3.5 percent increase | e | multiply(divide(subtract(multiply(add(const_100, 15), subtract(const_100, 10)), multiply(const_100, const_100)), multiply(const_100, const_100)), const_100) | add(n0,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#3,#1)|divide(#4,#1)|multiply(#5,const_100)| | geometry |
in order to obtain an income of rs . 450 from 10 % stock at rs . 108 , one must make an investment of | "to obtain rs . 10 , investment = rs . 108 . to obtain rs . 450 , investment = = rs . 4860 . answer : c" | a ) 5363 , b ) 6240 , c ) 4860 , d ) 1987 , e ) 2732 | c | multiply(divide(108, 10), 450) | divide(n2,n1)|multiply(n0,#0)| | gain |
in a ratio which is equal to 4 : 6 , if the antecedent is 20 , then the consequent is ? | "we have 4 / 6 = 20 / x 4 x = 20 * 6 x = 30 consequent = 30 answer is c" | a ) 72 , b ) 85 , c ) 30 , d ) 51 , e ) 45 | c | add(20, divide(20, const_2)) | divide(n2,const_2)|add(n2,#0)| | other |
the area of a rectangular plot is 20 times its breadth . if the difference between the length and the breadth is 10 metres , what is its breadth ? | "l Γ b = 20 Γ b β΄ l = 20 m and l β b = 10 β΄ b = 20 β 10 = 10 m answer a" | a ) 10 metres , b ) 5 metres , c ) 7.5 metres , d ) data inadequate , e ) none of these | a | subtract(20, 10) | subtract(n0,n1)| | geometry |
78654 * 9999 | "explanation : 78654 * ( 10000 - 1 ) = 786540000 - 78654 = 786461346 option d" | a ) 784598562 , b ) 765858254 , c ) 784569825 , d ) 786461346 , e ) none of these | d | multiply(divide(78654, 9999), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the weight of a glass of jar is 40 % of the weight of the jar filled with coffee beans . after some of the beans have been removed , the weight of the jar and the remaining beans is 60 % of the original total weight . what fraction part of the beans remain in the jar ? | "let weight of jar filled with beans = 100 g weight of jar = 40 g weight of coffee beans = 60 g weight of jar and remaining beans = 60 g weight of remaining beans = 20 g fraction remaining = 20 / 60 = 1 / 3 answer is b ." | a ) 1 / 5 , b ) 1 / 3 , c ) 2 / 5 , d ) 1 / 2 , e ) 2 / 3 | b | divide(subtract(60, 40), subtract(const_100, 40)) | subtract(n1,n0)|subtract(const_100,n0)|divide(#0,#1)| | gain |
one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 75,000 as total savings , how much has he saved in public provident fund ? | "let savings in n . s . c and p . p . f . be rs . x and rs . ( 175000 - x ) respectively . then , = 1 / 3 x = 1 / 2 ( 175000 - x ) = x / 3 + x / 2 = 87500 = 5 x / 6 = 87500 = x = 87500 x 6 / 5 = 105000 savings in public provident fund = rs . ( 175000 - 105000 ) = rs . 70000 answer is d ." | a ) 80000 , b ) 60000 , c ) 50000 , d ) 70000 , e ) 90000 | d | multiply(add(multiply(multiply(const_100, const_10), const_100), subtract(multiply(multiply(const_100, const_10), const_100), multiply(multiply(const_2, const_100), const_100))), divide(1, add(divide(const_3, const_2), 1))) | divide(const_3,const_2)|multiply(const_10,const_100)|multiply(const_100,const_2)|add(n0,#0)|multiply(#1,const_100)|multiply(#2,const_100)|divide(n0,#3)|subtract(#4,#5)|add(#4,#7)|multiply(#8,#6)| | general |
how many positive integers less than 120 are there such that they are multiples of 13 or multiples of 12 but not both ? | "for 13 : 13 . . . 117 = 13 * 9 = 117 for 12 : 12 . . . 120 = 12 * 10 = 120 but there is one integer 13 * 12 . so n = ( 9 ) + ( 10 ) = 19 b" | a ) 13 , b ) 19 , c ) 15 , d ) 16 , e ) 17 | b | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
a rectangular metallic piece of paper that covers exactly the area of a cube . the length of the piece of paper is 96 inches and the width is 72 inches . what is the volume of the cube in cubic feet is 1 feet is 12 inches ? | l = 96 / 12 = 8 ft w = 72 / 12 = 6 ft area of paper = 48 area of cube = 8 * side ^ 2 side of cube = 4 v of cube = 64 | ['a ) a 64', 'b ) b 22', 'c ) c 27', 'd ) d 40', 'e ) e 50'] | a | divide(96, divide(divide(divide(volume_cube(divide(divide(sqrt(rectangle_area(96, 72)), const_3), const_2)), 12), 12), 12)) | rectangle_area(n0,n1)|sqrt(#0)|divide(#1,const_3)|divide(#2,const_2)|volume_cube(#3)|divide(#4,n3)|divide(#5,n3)|divide(#6,n3)|divide(n0,#7) | geometry |
if the operation β¬ is defined for all x and y by the equation x β¬ y = 2 * x * y , then 9 β¬ ( 4 β¬ 5 ) = | "working inside out , ( 4 β¬ 5 ) = 2 * 4 * 5 = 40 9 β¬ 40 = 2 * 9 * 40 = 720 hence , answer is b" | a ) 80 , b ) 720 , c ) 760 , d ) 840 , e ) 860 | b | multiply(multiply(2, 9), multiply(multiply(2, 4), 5)) | multiply(n0,n1)|multiply(n0,n2)|multiply(n3,#1)|multiply(#0,#2)| | general |
jack and jill are marathon runners . jack can finish a marathon ( 43 km ) in 4.5 hours and jill can run a marathon in 4.3 hours . what is the ratio of their average running speed ? ( jack : jill ) | "average speed of jack = distance / time = 43 / ( 9 / 2 ) = 86 / 9 average speed of jill = 43 / ( 4.2 ) = 10 ratio of average speed of jack to jill = ( 86 / 9 ) / 10 = 86 / 90 = 43 / 45 answer b" | a ) 14 / 15 , b ) 43 / 45 , c ) 4 / 5 , d ) 5 / 4 , e ) can not be determined | b | divide(divide(43, 4.5), divide(43, 4.3)) | divide(n0,n1)|divide(n0,n2)|divide(#0,#1)| | physics |
two pipes a and b can fill a tank in 6 and 12 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by a in 1 min . = 1 / 6 part filled by b in 1 min . = 1 / 12 part filled by ( a + b ) in 1 min . = 1 / 6 + 1 / 12 = 1 / 4 . both the pipes can fill the tank in 4 minutes . answer : a" | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | a | add(multiply(6, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 6), divide(const_1, 12)), const_2)), 6), const_60)) | divide(const_1,n0)|divide(const_1,n1)|multiply(n0,const_100)|add(#0,#1)|multiply(#3,const_2)|subtract(const_1,#4)|multiply(n0,#5)|multiply(#6,const_60)|add(#2,#7)| | physics |
if a no . when divided by 44 , gives 432 as quotient and 0 as remainder . what will be the remainder when dividing the same no . by 34 | "p Γ£ Β· 44 = 432 = > p = 432 * 44 = 19008 p / 34 = 19008 / 34 = 559 , remainder = 2 a" | a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | reminder(multiply(432, 44), 34) | multiply(n0,n1)|reminder(#0,n3)| | general |
a car traveled 420 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 6 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "let the speed in highway be h mpg and in city be c mpg . h = c + 6 h miles are covered in 1 gallon 420 miles will be covered in 420 / h . similarly c miles are covered in 1 gallon 336 miles will be covered in 336 / c . both should be same ( as car ' s fuel capacity does not change with speed ) = > 336 / c = 420 / h = ... | a ) 14 , b ) 16 , c ) 21 , d ) 24 , e ) 27 | d | divide(336, divide(subtract(420, 336), 6)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics |
65 bags of nuts are to be divided among 13 students . each bag contains 15 nuts . how many nuts will each student receive ? | 65 Γ· 13 = 5 bags per student 5 x 15 = 75 nuts per student correct answer a | a ) 75 , b ) 65 , c ) 43 , d ) 52 , e ) 78 | a | multiply(divide(65, 13), 15) | divide(n0,n1)|multiply(n2,#0) | general |
if x = 1 - 4 t and y = 2 t - 2 , then for what value of t does x = y ? | we are given x = 1 β 4 t and y = 2 t β 2 , and we need to determine the value for t when x = y . we should notice that both x and y are already in terms of t . thus , we can substitute 1 β 4 t for x and 2 t β 2 for y in the equation x = y . this gives us : 1 β 4 t = 2 t β 2 3 = 6 t 1 / 2 = t the answer is b . | a ) 5 / 2 , b ) 1 / 2 , c ) 2 / 3 , d ) 2 / 5 , e ) 0 | b | divide(add(1, 2), add(2, 4)) | add(n0,n2)|add(n1,n2)|divide(#0,#1) | general |
the sum of two numbers is 528 and their h . c . f is 33 . the number of pairs of numbers satisfying the above conditions is : | "let the required numbers be 33 a and 33 b . then , 33 a + 33 b = 528 = > a + b = 16 . now , co - primes with sum 16 are ( 1 , 15 ) , ( 3 , 13 ) , ( 5 , 11 ) and ( 7 , 9 ) . required numbers are ( 33 * 1 , 33 * 15 ) , ( 33 * 3 , 33 * 13 ) , ( 33 * 5 , 33 * 11 ) , ( 33 * 7 , 33 * 9 ) . the number of such pairs is 4 . an... | a ) 4 , b ) 5 , c ) 6 , d ) 6 , e ) 1 | a | multiply(divide(add(528, 33), add(const_1, const_1)), subtract(divide(add(528, 33), add(const_1, const_1)), 33)) | add(n0,n1)|add(const_1,const_1)|divide(#0,#1)|subtract(#2,n1)|multiply(#2,#3)| | general |
what is the last digit in ( 7 ^ 95 - 3 ^ 58 ) ? | explanation : unit digit in 795 = unit digit in [ ( 74 ) 23 x 73 ] = unit digit in [ ( unit digit in ( 2401 ) ) 23 x ( 343 ) ] = unit digit in ( 123 x 343 ) = unit digit in ( 343 ) = 3 unit digit in 358 = unit digit in [ ( 34 ) 14 x 32 ] = unit digit in [ unit digit in ( 81 ) 14 x 32 ] = unit digit in [ ( 1 ) 14 x 32 ]... | a ) 5 , b ) 6 , c ) 4 , d ) 9 , e ) 5.5 | c | divide(divide(multiply(const_4, const_4), const_2), const_2) | multiply(const_4,const_4)|divide(#0,const_2)|divide(#1,const_2) | general |
how many cuboids of length 5 m , width 4 m and height 3 m can be farmed from a cuboid of 16 m length , 10 m width and 12 m height . | ( 16 Γ£ β 10 Γ£ β 12 ) / ( 5 Γ£ β 4 Γ£ β 3 ) = 32 answer is a . | a ) 32 , b ) 28 , c ) 36 , d ) 40 , e ) 50 | a | divide(multiply(multiply(16, 10), 12), multiply(multiply(5, 4), 3)) | multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#2,#3) | physics |
one night 10 percent of the female officers on a police force were on duty . if 200 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 200 number of female officers on duty on that night = 200 / 2 = 100 ( 10 / 100 ) * f = 100 = > f = 1000 answer e" | a ) 90 , b ) 180 , c ) 270 , d ) 500 , e ) 1,000 | e | divide(divide(200, const_2), divide(10, const_100)) | divide(n1,const_2)|divide(n0,const_100)|divide(#0,#1)| | gain |
a furniture dealer purchased a desk for $ 140 and then set the selling price equal to the purchase price plus a markup that was 30 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ? | "purchase price = 140 selling price = x 140 + 0.3 * x = x 0.7 * x = 140 x = 200 profit = 200 - 140 = 60 answer : b" | a ) $ 40 , b ) $ 60 , c ) $ 80 , d ) $ 90 , e ) $ 100 | b | divide(multiply(subtract(divide(140, subtract(const_1, divide(30, const_100))), 140), const_100), 140) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|multiply(#3,const_100)|divide(#4,n0)| | gain |
find 96 Γ Γ 97 | "here both numbers are less than 100 . so they are deficient of - 4 and - 3 compared with 100 . so answer : b" | a ) 91 / 198 , b ) 91 / 12 , c ) 91 / 13 , d ) 91 / 10 , e ) 91 / 11 | b | divide(96, 97) | divide(n0,n1)| | general |
of 3 numbers , the third is twice the first and the second is 4 times the first . if their average is 77 , the smallest of the 3 numbers is : | explanation : let first number be x . so , 2 nd no . = 4 x & 3 rd no . = 2 x . so , x + 4 x + 2 x = 77 Γ 3 = 231 7 x = 231 x = 231 / 7 hence , smallest number x = 33 answer : e | a ) 18 , b ) 25 , c ) 20 , d ) 32 , e ) 33 | e | divide(multiply(77, 3), add(add(const_1, const_2), 4)) | add(const_1,const_2)|multiply(n0,n2)|add(n1,#0)|divide(#1,#2) | general |
the slant height of a cone is 21 cm and radius of the base is 10 cm , find the curved surface of the cone . | "Ο * 21 * 10 = 660 answer : c" | a ) 330 , b ) 770 , c ) 660 , d ) 100 , e ) 600 | c | multiply(multiply(const_pi, 10), 21) | multiply(n1,const_pi)|multiply(n0,#0)| | geometry |
a store reported total sales of $ 400 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase in sales ? | "new value β old value ) / old value x 100 we are given : february sales this year = 400 million february sales last year = 320 million we need to determine the percent increase between sales from last year to sales this year . thus , the new value = 400 million and the old value = 320 million . let β s plug them into ... | a ) 2 % , b ) 17 % , c ) 25 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(400, 320), 320), const_100) | subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)| | general |
two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 20 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ? | "same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 10... | a ) 3 : 5 , b ) 9 : 25 , c ) 5 : 2 , d ) 25 : 9 , e ) can not be determined from the information provided | c | divide(circumface(divide(50, const_2)), circumface(divide(20, const_2))) | divide(n1,const_2)|divide(n0,const_2)|circumface(#0)|circumface(#1)|divide(#2,#3)| | physics |
in a company , 56 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ? | "the percent of employees who are unionized and men is 0.7 * 0.6 = 42 % the percent of employees who are unionized and women is 60 - 42 = 18 % 44 % of all employees are women , so non - union women are 44 % - 18 % = 26 % 40 % of all employees are non - union . the percent of non - union employees who are women is 26 % ... | a ) 45 % , b ) 50 % , c ) 55 % , d ) 60 % , e ) 65 % | e | multiply(const_100, divide(subtract(subtract(const_100, 56), subtract(60, multiply(60, divide(70, const_100)))), subtract(const_100, 60))) | divide(n2,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(n1,#0)|subtract(n1,#3)|subtract(#1,#4)|divide(#5,#2)|multiply(#6,const_100)| | gain |
in assembling a bluetooth device , a factory uses one of two kinds of modules . one module costs $ 15 and the other one , that is cheaper , costs $ 10 . the factory holds a $ 140 worth stock of 13 modules . how many of the modules in the stock are of the cheaper kind ? | "so the number of $ 10 modules must be 11 so that the leftover 2 modules are of $ 15 which will give a total value $ 140 . 11 * 10 + 2 * 15 = 110 + 30 = 140 answer : d" | a ) 20 , b ) 25 , c ) 30 , d ) 11 , e ) 35 | d | divide(subtract(140, 15), 10) | subtract(n2,n0)|divide(#0,n1)| | other |
in a two digit number it is known that its unit ' s digit exceeds its 10 ' s digit by 4 and that the product of the given number and the sum of its digits is equal to 208 , what is the number ? | xy is 2 digit no . y - x = 4 ; ( 10 x + y ) * ( 2 x + 4 ) = 208 ; sub y = x + 4 ; we get 26 answer : c | a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | c | add(add(divide(subtract(sqrt(add(power(add(multiply(add(const_10, const_1), 4), multiply(4, const_2)), const_2), multiply(multiply(4, multiply(add(const_10, const_1), const_2)), subtract(208, multiply(const_4, const_4))))), add(multiply(add(const_10, const_1), 4), multiply(4, const_2))), multiply(add(const_10, const_1)... | add(const_1,const_10)|multiply(const_4,const_4)|multiply(n1,const_2)|multiply(#0,const_2)|multiply(n1,#0)|subtract(n2,#1)|add(#4,#2)|multiply(n1,#3)|multiply(#7,#5)|power(#6,const_2)|add(#8,#9)|sqrt(#10)|subtract(#11,#6)|divide(#12,#4)|add(n1,#13)|multiply(n0,#13)|add(#14,#15) | general |
if a fills a tank in 30 mins , and b fills the same tank in 20 mins , and c emptys it in 40 min , if a , b , c works together how much time will the tank take to fill ? | 1 / 30 + 1 / 20 - 1 / 40 = 7 / 120 so total time = 120 / 7 = 17.14 mins answer : c | a ) 15.14 mins , b ) 16.14 mins , c ) 17.14 mins , d ) 18.14 mins , e ) 19.14 mins | c | divide(const_1, subtract(add(divide(const_1, 30), divide(const_1, 20)), divide(const_1, 40))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|divide(const_1,#4) | physics |
an article is bought for rs . 560 and sold for rs . 340 , find the loss percent ? | 560 - - - - 220 100 - - - - ? = > 39 % answer : d | a ) 36 % , b ) 38 % , c ) 16 % , d ) 39 % , e ) 12 % | d | subtract(const_100, divide(multiply(340, const_100), 560)) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1) | gain |
jennifer had $ 180 to spend on herself . she spent 1 / 5 of the money on a sandwich , 1 / 6 for a ticket to a museum , and 1 / 2 of it on a book . how much money does jennifer have left over ? | 1 / 5 x $ 180 = $ 36 for sandwich 1 / 6 x $ 180 = $ 30 for museum 1 / 2 x $ 180 = $ 90 for book $ 36 + $ 30 + $ 90 = $ 156 spent $ 180 - $ 156 = $ 24 left over correct answer e | a ) $ 4 , b ) $ 14 , c ) $ 5 , d ) $ 15 , e ) $ 24 | e | subtract(180, add(add(multiply(180, divide(1, 5)), multiply(180, divide(1, 6))), multiply(180, divide(1, 2)))) | divide(n1,n2)|divide(n1,n4)|divide(n1,n6)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|add(#3,#4)|add(#6,#5)|subtract(n0,#7) | general |
a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 12 miles per gallon , and from town b to town c , the car averaged 16 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through ... | "ans is c given d _ ab = 2 * d _ bc let d _ ab = d and d _ bc = x so d = 2 x for average miles per gallon = ( d + x ) / ( ( d / 12 ) + ( x / 16 ) ) = 14.4 ( formula avg speed = total distance / total time )" | a ) 13 , b ) 13.5 , c ) 14.4 , d ) 14.5 , e ) 15 | c | divide(add(multiply(16, const_10), divide(multiply(16, const_10), const_2)), add(divide(multiply(16, const_10), 12), divide(divide(multiply(16, const_10), const_2), 16))) | multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)| | general |
the greater of two numbers is twice the less , and the sum of the numbers is 96 . what are the numbers ? | let x = the less number 2 x = the greater number the sum of the numbers is 96 x + 2 x = 96 3 x = 96 x = 32 2 x = 64 option a | a ) 64 , b ) 68 , c ) 70 , d ) 79 , e ) 80 | a | multiply(const_2, divide(96, add(const_2, const_1))) | add(const_1,const_2)|divide(n0,#0)|multiply(#1,const_2) | general |
pipe p can fill a tank in 3 hours , pipe q in 9 hours and pipe r in 24 hours . if all the pipes are open , in how many hours will the tank be filled ? | "explanation : part filled by ( p + q + r ) in 1 hour = ( 1 / 3 + 1 / 9 + 1 / 24 ) = 35 / 72 all the three pipes together will fill the tank = 72 / 35 = 2.1 hours answer b" | a ) 2 hours , b ) 2.1 hours , c ) 3 hours , d ) 3.5 hours , e ) 4.5 hours | b | inverse(add(divide(const_1, 24), add(divide(const_1, 3), divide(const_1, 9)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|inverse(#4)| | physics |
a boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes . how long will it take to go 8 km in stationary water ? | "speed ( upstream ) = 2 / 1 = 2 kmhr speed ( downstream ) = 1 / ( 10 / 60 ) = 6 kmhr speed in still water = 1 / 2 ( 2 + 6 ) = 4 kmhr time taken in stationary = 8 / 4 = 2 hrs answer : a" | a ) 2 hours , b ) 1 hour , c ) 1 hour 15 min , d ) 1 hour 30 min , e ) 1 hour 10 min | a | divide(8, divide(add(multiply(divide(1, 10), const_60), divide(2, 2)), const_2)) | divide(n0,n0)|divide(n2,n3)|multiply(#1,const_60)|add(#0,#2)|divide(#3,const_2)|divide(n4,#4)| | physics |
one woman and one man can build a wall together in five hours , but the woman would need the help of two girls in order to complete the same job in the same amount of time . if one man and one girl worked together , it would take them nine hours to build the wall . assuming that rates for men , women and girls remain c... | solution : let work done by man , women and girl per hour be m , w , g respectively . then , m + w = 1 / 5 - - > ( 1 ) , w + 2 g = 1 / 5 - - > ( 2 ) and m + g = 1 / 9 - - > ( 3 ) . no . of hours it would take forone woman , one man , and one girl , working together , to build the wall , n = 1 / m + w + g from ( 1 ) and... | a ) 55 / 7 , b ) 1 , c ) 135 / 32 , d ) 222 / 37 , e ) 252 / 58 | c | inverse(add(divide(subtract(divide(add(inverse(add(add(const_4, const_1), const_4)), multiply(inverse(add(const_4, const_1)), const_2)), const_2), inverse(add(const_4, const_1))), divide(const_3, const_2)), inverse(add(const_4, const_1)))) | add(const_1,const_4)|divide(const_3,const_2)|add(#0,const_4)|inverse(#0)|inverse(#2)|multiply(#3,const_2)|add(#4,#5)|divide(#6,const_2)|subtract(#7,#3)|divide(#8,#1)|add(#9,#3)|inverse(#10) | physics |
a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 35 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | "explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 β 35 = rs . 170 cp on 40 metres = 170 x 40 = rs . 6800 profit earned on 40 metres cloth = rs . 8200 β rs . 6800 = rs . 1400 . answer : option d" | a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 1400 , e ) none of these | d | multiply(35, 40) | multiply(n0,n2)| | gain |
a reduction of 25 % in the price of apples would enable a man to obtain 36 more for rs . 25 , what is reduced price per dozen ? | "a 25 * ( 25 / 100 ) = 6 - - - 36 ? - - - 12 = > rs . 2" | a ) 2 , b ) 3 , c ) 6 , d ) 7 , e ) 10 | a | multiply(divide(divide(multiply(25, 25), const_100), 36), const_12) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|multiply(#2,const_12)| | gain |
the ratio between the present age of p and q is 6 : 7 . if q is 4 years old than p , what will be the ratio of the ages of p and q after 4 years ? | let p and q ' s age is 6 x and 7 x years respectively . 7 x - 6 x = 4 x = 4 required ratio = ( 6 x + 4 ) : ( 7 x + 4 ) = 28 : 32 = 7 : 8 option a is answer | a ) 8 : 7 , b ) 7 : 8 , c ) 8 : 6 , d ) 6 : 7 , e ) 6 : 8 | a | divide(add(4, add(4, divide(multiply(6, 4), subtract(7, 6)))), add(4, divide(multiply(6, 4), subtract(7, 6)))) | multiply(n0,n2)|subtract(n1,n0)|divide(#0,#1)|add(n2,#2)|add(n2,#3)|divide(#4,#3) | other |
what is the greatest value of positive integer x such that 2 ^ x is a factor of 100 ^ 60 ? | "put in simple words , we need to find the highest power of 2 in 100 ^ 60 100 = 2 ^ 2 * 5 ^ 2 therefore 100 ^ 60 = ( 2 ^ 2 * 5 ^ 2 ) ^ 60 = 2 ^ 120 * 5 ^ 120 answer : c" | a ) 100 , b ) 110 , c ) 120 , d ) 105 , e ) 115 | c | subtract(100, subtract(power(2, divide(100, 60)), const_1)) | divide(n1,n2)|power(n0,#0)|subtract(#1,const_1)|subtract(n1,#2)| | general |
if both 5 ^ 3 and 3 ^ 3 are factors of n x ( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ^ 2 ) , what is the smallest possible positive value of n ? | "( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ) has two appearances of 3 ( in 12 ^ 2 ) and two appearances of 5 ( in 10 ^ 2 ) . thus n must include at least 3 * 5 = 15 the answer is a ." | a ) 15 , b ) 45 , c ) 75 , d ) 125 , e ) 150 | a | add(add(add(add(add(multiply(multiply(5, 2), 3), multiply(multiply(5, 2), 3)), multiply(multiply(5, 2), 3)), 2), const_4), const_4) | multiply(n0,n7)|multiply(n1,#0)|add(#1,#1)|add(#2,#1)|add(n7,#3)|add(#4,const_4)|add(#5,const_4)| | other |
in one hour , a boat goes 8 km along the stream and 2 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 8 + 2 ) kmph . = 5 kmph . answer b" | a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | b | divide(add(8, 2), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
the sale price sarees listed for rs . 200 after successive discount is 20 % and 10 % is ? | "200 * ( 80 / 100 ) * ( 90 / 100 ) = 144 answer : d" | a ) 187 , b ) 124 , c ) 172 , d ) 144 , e ) 175 | d | subtract(subtract(200, divide(multiply(200, 20), const_100)), divide(multiply(subtract(200, divide(multiply(200, 20), const_100)), 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
sandy is younger than molly by 20 years . if the ratio of their ages is 7 : 9 , how old is sandy ? | "let sandy ' s age be 7 x and let molly ' s age be 9 x . 9 x - 7 x = 20 x = 10 sandy is 70 years old . the answer is e ." | a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70 | e | multiply(divide(7, subtract(9, 7)), 20) | subtract(n2,n1)|divide(n1,#0)|multiply(n0,#1)| | other |
5 times a number equals one 100 . the number is | since five times the number equals one hundred , then the number equals one hundred divided by five . therefore the number is 100 / 5 = 20 . correct answer e | a ) 50 , b ) 10 , c ) 15 , d ) 25 , e ) 20 | e | divide(100, 5) | divide(n1,n0) | general |
the manufacturing cost of a shoe is rs . 200 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains | "explanation : total cost of a watch = 200 + ( 500 / 100 ) = 205 . gain = 20 % = > sp = 1.2 cp = 1.2 x 205 = 246 answer : b" | a ) s 222 , b ) s 246 , c ) s 220 , d ) s 210 , e ) s 217 | b | add(add(200, divide(500, 100)), multiply(divide(20, 100), add(200, divide(500, 100)))) | divide(n1,n2)|divide(n3,n2)|add(n0,#0)|multiply(#2,#1)|add(#2,#3)| | gain |
find the compound interest on $ 10,000 in 2 years at 4 % per annum , the interest being compounded half - yearly | "principal = $ 10000 ; rate = 2 % per half - year ; time = 2 years = 4 half - years . amount = $ [ 10000 * ( 1 + ( 2 / 100 ) ) 4 ] = $ ( 10000 * ( 51 / 50 ) * ( 51 / 50 ) * ( 51 / 50 ) * ( 51 / 50 ) ) = $ 10824.32 . : . c . i . = $ ( 10824.32 - 10000 ) = $ 824.32 . answer c ." | a ) 824 , b ) 824.36 , c ) 824.32 , d ) 824.24 , e ) 824.26 | c | subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 10,000), 10,000) | divide(n2,const_100)|divide(#0,n1)|add(#1,const_1)|power(#2,n2)|multiply(n0,#3)|subtract(#4,n0)| | gain |
in how many years will a sum of money doubles itself at 30 % per annum on simple interest ? | "p = ( p * 30 * r ) / 100 r = 3 % answer : b" | a ) 7 % , b ) 3 % , c ) 5 % , d ) 8 % , e ) 2 % | b | divide(const_100, 30) | divide(const_100,n0)| | gain |
a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 25 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 ( 25 / 100 ) = rs . 186 answer : a" | a ) rs . 186 , b ) rs . 258 , c ) rs . 486 , d ) rs . 586 , e ) none of these | a | multiply(divide(25, 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 |
a collection of books went on sale , and 2 / 3 of them were sold for $ 2.00 each . if none of the 36 remaining books were sold , what was the total amount received for the books that were sold ? | since 2 / 3 of the books in the collection were sold , 1 / 3 were not sold . the 36 unsold books represent 1 / 3 of the total number of books in the collection , and 2 / 3 of the total number of books equals 2 ( 36 ) or 72 . the total proceeds of the sale was 72 ( $ 2.00 ) or $ 144 . the best answer is therefore a . an... | a ) $ 144 , b ) $ 135 , c ) $ 90 , d ) $ 60 , e ) $ 54 | a | multiply(2, divide(multiply(36, divide(2, 3)), divide(const_1, const_3))) | divide(n0,n1)|divide(const_1,const_3)|multiply(n3,#0)|divide(#2,#1)|multiply(n0,#3) | general |
in some quantity of ghee , 50 % is pure ghee and 50 % is vanaspati . if 20 kg of pure ghee is added , then the strength of vanaspati ghee becomes 30 % . the original quantity was ? | "let the original quantity be x then , vanaspati ghee in xkg = 50 x / 100 kg = x / 2 kg ( x / 2 ) / ( x + 20 ) = 30 / 100 x / ( 2 x + 40 ) = 3 / 10 x = 30 answer is c" | a ) 25 , b ) 15 , c ) 30 , d ) 45 , e ) 20 | c | divide(multiply(30, 20), subtract(50, 30)) | multiply(n2,n3)|subtract(n1,n3)|divide(#0,#1)| | gain |
the roof of an apartment building is rectangular and its length is 4 times longer than its width . if the area of the roof is 1024 feet squared , what is the difference between the length and the width of the roof ? | "let the width = x x * 4 x = 1024 x ^ 2 = 256 x = 16 length = 4 * 16 = 64 difference = 64 - 16 = 48 d is the answer" | a ) 38 . , b ) 40 . , c ) 42 . , d ) 48 . , e ) 46 . | d | subtract(multiply(sqrt(divide(1024, 4)), 4), sqrt(divide(1024, 4))) | divide(n1,n0)|sqrt(#0)|multiply(#1,n0)|subtract(#2,#1)| | geometry |
at what price must an article costing rs . 38 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 38 sp = 38 * ( 125 / 100 ) = 38 mp * ( 95 / 100 ) = 47.5 mp = 50 answer : b" | a ) 52.5 , b ) 50 , c ) 62.3 , d ) 62.1 , e ) 62.4 | b | divide(multiply(add(38, divide(multiply(38, 25), const_100)), const_100), subtract(const_100, 5)) | multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)| | gain |
if the given two numbers are respectively 6 % and 18 % of a third number , then what percentage is the first of the second ? | "here , l = 6 and m = 18 therefore , first number = l / m x 100 % of second number = 6 / 18 x 100 % of second number = 33 1 / 3 % of second number answer : d" | a ) 20 % , b ) 25 % , c ) 18 % , d ) 33 1 / 3 % , e ) none of these | d | multiply(divide(divide(6, const_100), divide(18, const_100)), const_100) | divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)| | gain |
a 4 % stock yield 5 % . the market value of the stock is ? | "for an income of rs . 5 , investment = rs . 100 for an income of rs . 4 , investment = rs . ( 100 / 5 ) x 4 = rs . 80 answer : b" | a ) rs . 125 , b ) rs . 80 , c ) rs . 99 , d ) rs . 109 , e ) rs . 101 | b | multiply(divide(const_100, 5), 4) | divide(const_100,n1)|multiply(n0,#0)| | gain |
a train is 288 meter long is running at a speed of 29 km / hour . in what time will it pass a bridge of 101 meter length ? | "speed = 29 km / hr = 29 * ( 5 / 18 ) m / sec = 145 / 18 m / sec total distance = 288 + 101 = 389 meter time = distance / speed = 389 * ( 18 / 145 ) = 48.29 seconds . answer : a" | a ) 48.29 , b ) 50 , c ) 26.8 , d ) 45.29 , e ) 30 | a | divide(add(288, 101), divide(multiply(29, const_1000), const_3600)) | add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)| | physics |
rani bought more apples than oranges . she sells apples at βΉ 23 apiece and makes 15 % profit . she sells oranges at βΉ 10 apiece and makes 25 % profit . if she gets βΉ 653 after selling all the apples and oranges , find her profit percentage c . | "given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the val... | a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(1... | add(n1,const_2)|divide(n3,const_100)|multiply(n2,const_2)|add(#2,const_1)|add(#1,const_1)|divide(n2,#4)|multiply(#3,#2)|multiply(#0,#5)|add(#6,#7)|subtract(n4,#8)|divide(#9,#8)|multiply(#10,const_100)| | gain |
john had a stock of 1100 books in his bookshop . he sold 75 on monday , 50 on tuesday , 64 on wednesday , 78 on thursday and 135 on friday . what percentage of the books were not sold ? | "let n be the total number of books sold . hence n = 75 + 50 + 64 + 78 + 135 = 402 let m be the books not sold m = 1100 - n = 1100 - 402 = 698 percentage books not sold / total number of books = 698 / 1100 = 0.634 = 63.4 % correct answer b" | a ) 50.4 % , b ) 63.4 % , c ) 66.5 % , d ) 34.9 % , e ) 43.5 % | b | multiply(divide(subtract(1100, add(add(add(75, 50), add(64, 78)), 135)), 1100), const_100) | add(n1,n2)|add(n3,n4)|add(#0,#1)|add(n5,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)| | gain |
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