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 lady has fine gloves and hats in her closet - 18 blue , 32 red , and 25 yellow . the lights are out and it is totally dark . in spite of darkness , she can make out the difference between a hat and a glove . she takes out an item out of the closet only if she is sure that it is a glove . how many gloves must she take... | "32 r + 24 y + 1 y + 1 b + 2 b = 60 answer : a" | a ) 60 , b ) 65 , c ) 70 , d ) 75 , e ) 80 | a | add(add(25, 32), const_2) | add(n1,n2)|add(#0,const_2)| | general |
of the total amount that jill spent on a shopping trip , excluding taxes , she spent 40 percent on clothing , 30 percent on food , and 30 percent on other items . if jill paid a 4 percent tax on the clothing , no tax on the food , and an 8 percent tax on all other items , then the total tax that she paid was what perce... | "let amount spent by jill = 100 clothing = 40 , food = 30 , others = 30 tax on clothing = 1.6 tax on others = 2.4 percentage = 4 / 100 = 4 % answer : c" | a ) 2.8 % , b ) 3.6 % , c ) 4 % , d ) 5.2 % , e ) 6.0 % | c | multiply(divide(add(multiply(40, divide(4, const_100)), multiply(30, divide(8, const_100))), const_100), const_100) | divide(n3,const_100)|divide(n4,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,const_100)|multiply(#5,const_100)| | general |
a student chose a number , multiplied it by 5 , then subtracted 138 from the result and got 102 . what was the number he chose ? | "solution : let xx be the number he chose , then 5 ⋅ x − 138 = 102 5 x = 240 x = 48 answer a" | a ) 48 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | a | divide(add(102, 138), 5) | add(n1,n2)|divide(#0,n0)| | general |
in a class , 30 students pass in english and 20 students in maths , while some students among these pass in both . how many students do only english as compared to those doing only maths ? | "for doing union or intersection we would need three values . in this question the missing information in this question is total no . of students in the class . answer : e" | a ) 10 , b ) 15 , c ) 4 , d ) 12 , e ) indeterminate | e | subtract(30, 20) | subtract(n0,n1)| | other |
if 31 / 198 = 0.1565 , what is the 97 nd digit to the right of the decimal point of the fraction ? | "we are not concerned what 31 / 198 means . . we have to look at the decimal . . 0.1565 means 0.1565656 . . . . so leaving girst and second digit to the right of decimal , all odd numbered are 6 and all even numbered are 5 . . here 97 is odd , so ans is 6 c" | a ) 1 , b ) 2 , c ) 6 , d ) 7 , e ) 9 | c | add(const_2, const_3) | add(const_2,const_3)| | general |
sally has a gold credit card with a certain spending limit , and a platinum card with twice the spending limit of the gold card . currently , she has a balance on her gold card that is 1 / 3 of the spending limit on that card , and she has a balance on her platinum card that is 1 / 8 of the spending limit on that card ... | let s assume the platinum card spending limit = x gold card spending limit will be = x / 2 balance on gold card is = x / 2 * 1 / 3 = x / 6 platinum card unspent limit is = x - 1 / 8 x = 7 / 8 x so if gold card balance is transferred then the rest unspent will be 7 / 8 x - x / 6 = 17 / 24 x so the ans is c | a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 24 , d ) 19 / 30 , e ) 11 / 15 | c | subtract(1, add(multiply(inverse(3), inverse(const_2)), inverse(8))) | inverse(n3)|inverse(n1)|inverse(const_2)|multiply(#1,#2)|add(#0,#3)|subtract(n0,#4) | general |
robert spent $ 100 in buying raw materials , $ 125 in buying machinery and 10 % of the total amount he had as cash with him . what was the total amount ? | let the total amount be x then , ( 100 - 10 ) % of x = 100 + 125 90 % of x = 225 90 x / 100 = 225 x = $ 250 answer is c | a ) $ 150 , b ) $ 210 , c ) $ 250 , d ) $ 160 , e ) $ 200 | c | add(125, 125) | add(n1,n1) | gain |
a 12 % stock yields 8 % . the market value of the stock is : | "solution to obtain rs . 8 , investment = rs . 100 . to obtain rs . 12 , investment = rs . ( 100 / 8 x 12 ) = rs . 150 ∴ market value of rs . 100 stock = rs . 150 answer b" | a ) rs . 72 , b ) rs . 150 , c ) rs . 112.50 , d ) rs . 116.50 , e ) none of these | b | multiply(divide(const_100, 8), 12) | divide(const_100,n1)|multiply(n0,#0)| | gain |
there is a square with sides of 13 . what is the area of the biggest circle that can be cut out of this square ? | "area of a circle = a = ï € r ^ 2 square is 13 wide , so circle ' s diameter would be 13 , and radius would be 6.5 a = ï € 6.5 ^ 2 which is approximately 132.73 answer is a" | a ) 132.73 , b ) 231.92 , c ) 530.93 , d ) 113.1 , e ) 204.33 | a | circumface(divide(13, const_2)) | divide(n0,const_2)|circumface(#0)| | geometry |
how long does a train 130 m long running at the speed of 98 km / hr takes to cross a bridge 160 m length ? | "speed = 98 * 5 / 18 = 27 m / sec total distance covered = 130 + 160 = 290 m . required time = 290 / 22 = 13.1 sec . answer : b" | a ) 13.9 sec , b ) 13.1 sec , c ) 17.9 sec , d ) 61.9 sec , e ) 47.98 sec | b | divide(add(130, 160), multiply(98, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 5500 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a ’ s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 5500 x ( 2 / 11 ) ) = rs . 1000 . answer is a" | a ) 1000 , b ) 800 , c ) 1400 , d ) 1200 , e ) none of them | a | multiply(5500, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | divide(const_2,n0)|multiply(const_2,n0)|multiply(#0,n0)|add(#1,#2)|add(#3,n0)|divide(const_2,#4)|multiply(n1,#5)| | gain |
if { x } is the product of all even integers from 1 to x inclusive , what is the greatest prime factor of { 14 } + { 12 } ? | "soln : { 14 } + { 12 } = 14 * { 12 } + { 12 } = 15 * { 12 } answer : c" | a ) 23 , b ) 20 , c ) 15 , d ) 5 , e ) 2 | c | add(divide(add(14, 12), const_2), multiply(1, const_2)) | add(n1,n2)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | general |
if x + ( 1 / x ) = 5 , what is the value of w = x ^ 2 + ( 1 / x ) ^ 2 ? | "squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c" | a ) w = 21 , b ) w = 22 , c ) w = 23 , d ) w = 24 , e ) 27 | c | subtract(power(5, 2), 2) | power(n1,n2)|subtract(#0,n2)| | general |
a metallic sheet is of rectangular shape with dimensions 48 m x 36 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 7 m , the volume of the box ( in m 3 ) is : | "clearly , l = ( 48 - 14 ) m = 34 m , b = ( 36 - 14 ) m = 22 m , h = 8 m . volume of the box = ( 34 x 22 x 7 ) m 3 = 5236 m 3 . answer : option b" | a ) 4830 , b ) 5236 , c ) 6420 , d ) 8960 , e ) 7960 | b | volume_rectangular_prism(subtract(48, multiply(7, const_2)), subtract(36, multiply(7, const_2)), 7) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry |
if the price of a certain computer increased 30 percent from d dollars to 351 dollars , then 2 d = | "before price increase price = d after 30 % price increase price = d + ( 30 / 100 ) * d = 1.3 d = 351 ( given ) i . e . d = 351 / 1.3 = $ 270 i . e . 2 d = 2 * 270 = 540 answer : option a" | a ) 540 , b ) 570 , c ) 619 , d ) 649 , e ) 700 | a | multiply(divide(351, divide(add(const_100, 30), const_100)), 2) | add(n0,const_100)|divide(#0,const_100)|divide(n1,#1)|multiply(n2,#2)| | general |
a certain number of men can do a work in 54 days . if there were 6 men more it could be finished in 6 days less . how many men were there in the beginning ? | "explanation : m ( 54 ) = ( m + 6 ) ( 48 ) 9 m - 8 m = 48 m = 48 answer : option b" | a ) 50 , b ) 48 , c ) 70 , d ) 40 , e ) 50 | b | divide(multiply(add(54, 6), 6), 6) | add(n0,n2)|multiply(n1,#0)|divide(#1,n2)| | physics |
in traveling from a dormitory to a certain city , a student went 1 / 2 of the way by foot , 3 / 5 of the way by bus , and the remaining 4 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "whole trip = distance by foot + distance by bus + distance by car x = 1 / 2 x + 3 / 5 x + 4 x - 1 / 2 x - 3 / 5 x = 4 x = 20 km option : a" | a ) 20 , b ) 15 , c ) 40 , d ) 10 , e ) 12 | a | multiply(4, inverse(subtract(1, add(divide(1, 2), divide(3, 5))))) | divide(n0,n1)|divide(n2,n3)|add(#0,#1)|subtract(n0,#2)|inverse(#3)|multiply(n4,#4)| | physics |
a small pool filled only with water will require an additional 500 gallons of water in order to be filled to 80 % of its capacity . if pumping in these additional 500 gallons of water will increase the amount of water in the pool by 30 % , what is the total capacity of the pool in gallons ? | "since pumping in additional 500 gallons of water will increase the amount of water in the pool by 30 % , then initially the pool is filled with 1,000 gallons of water . so , we have that 1,000 + 500 = 0.8 * { total } - - > { total } = 1,875 answer : e ." | a ) 1000 , b ) 1250 , c ) 1300 , d ) 1600 , e ) 1875 | e | divide(add(divide(multiply(500, const_100), 30), 500), divide(80, const_100)) | divide(n1,const_100)|multiply(n0,const_100)|divide(#1,n3)|add(n0,#2)|divide(#3,#0)| | general |
a certain bag contains 60 balls — 22 white , 18 green , 17 yellow , 3 red , and 1 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 ) = ( 22 + 18 + 17 ) / 60 = 57 / 60 = 0.95 . answer : e ." | a ) 0.09 , b ) 0.15 , c ) 0.54 , d ) 0.85 , e ) 0.95 | e | divide(add(add(22, 18), 17), 60) | add(n1,n2)|add(n3,#0)|divide(#1,n0)| | other |
if a speaks the truth 40 % of the times , b speaks the truth 20 % of the times . what is the probability that at least one will tell the truth | "probability of a speaks truth p ( a ) = 4 / 10 ; false = 6 / 10 probability of b speaks truth p ( b ) = 2 / 10 ; false = 8 / 10 . for given qtn ans = 1 - ( neither of them tell truth ) . because a & b are independent events = 1 - [ ( 6 / 10 ) * ( 8 / 10 ) ] = 1 - 48 / 100 = 1 - 0.48 = 0.52 answer : d" | a ) 1.8 , b ) 2.0 , c ) 1.52 , d ) 0.52 , e ) 1.3 | d | multiply(divide(40, multiply(multiply(const_4, const_5), const_5)), divide(20, multiply(multiply(const_4, const_5), const_5))) | multiply(const_4,const_5)|multiply(#0,const_5)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3)| | gain |
find the number of zeroes in 58 ! ( 58 factorial ) | "no of zeroes is 58 / 5 = 11 11 / 5 = 2 11 + 2 = 13 answer : b" | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | add(const_2, const_2) | add(const_2,const_2)| | other |
find the cost of fencing around a circular field of diameter 36 m at the rate of rs . 3.50 a meter ? | "2 * 22 / 7 * 18 = 113 113 * 3 1 / 2 = rs . 395 answer : b" | a ) 438 , b ) 395 , c ) 378 , d ) 279 , e ) 222 | b | multiply(circumface(divide(36, const_2)), 3.50) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics |
a number when divided by 342 gives a remainder 47 . when the same number ift divided by 19 , what would be the remainder ? | "sol . on dividing the given number by 342 , let k be the quotient and 47 as remainder . then , number – 342 k + 47 = ( 19 x 18 k + 19 x 2 + 9 ) = 19 ( 18 k + 2 ) + 9 . the given number when divided by 19 , gives ( 18 k + 2 ) as quotient and 9 as remainder . option c" | a ) 7 , b ) 8 , c ) 9 , d ) 21 , e ) 14 | c | subtract(47, multiply(19, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general |
which is the least number that must be subtracted from 1100 so that the remainder when divided by 7 , 12 , 16 is 4 ? | "first we need to figure out what numbers are exactly divisible by 7 , 12,16 . this will be the set { lcm , lcmx 2 , lcmx 3 , . . . } lcm ( 7 , 12,16 ) = 48 * 7 = 336 the numbers which will leave remainder 4 will be { 336 + 4 , 336 x 2 + 4 , 336 x 3 + 4 , . . . } the largest such number less than or equal to 1100 is 33... | a ) 58 , b ) 68 , c ) 78 , d ) 88 , e ) 90 | d | subtract(1100, add(4, multiply(gcd(1100, lcm(lcm(7, 12), 16)), lcm(lcm(7, 12), 16)))) | lcm(n1,n2)|lcm(n3,#0)|gcd(n0,#1)|multiply(#2,#1)|add(n4,#3)|subtract(n0,#4)| | general |
two integers are in the ratio of 1 to 4 . if 12 is added to the smaller number , the ratio becomes 1 to 1 . find the larger integer . | "one option is to set up the equations and solve : if the ratio of two integers x and y is 1 to 4 , then 4 x = y , where x is the smaller integer . if adding 12 to the smaller integer makes the ratio 1 to 1 , then x + 12 = y . substituting y = 4 x into the second equation yields x + 12 = 4 x . so , x = 4 ( smaller inte... | a ) 8 , b ) 16 , c ) 32 , d ) 48 , e ) 54 | b | multiply(divide(multiply(1, 12), subtract(4, 1)), 4) | multiply(n2,n4)|subtract(n1,n4)|divide(#0,#1)|multiply(n1,#2)| | other |
the salaries of a and b together amount to $ 4000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ? | let a ' s salary is x b ' s salary = 4000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 4000 - x ) x = $ 3000 answer is d | a ) $ 1000 , b ) $ 1250 , c ) $ 2500 , d ) $ 3000 , e ) $ 1200 | d | subtract(4000, divide(4000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1))) | subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,#1)|add(#2,const_1)|divide(n0,#3)|subtract(n0,#4) | gain |
1 / 2 + [ ( 2 / 3 * 3 / 8 ) + 4 ] - 8 / 16 = | 1 / 2 + [ ( 2 / 3 * 3 / 8 ) + 4 ] - 8 / 16 = 1 / 2 + [ ( 1 / 4 ) + 4 ] - 8 / 16 = 1 / 2 + [ 17 / 4 ] - 9 / 16 = 8 / 16 + 68 / 16 - 8 / 16 = 68 / 16 = 17 / 4 d | a ) 29 / 16 , b ) 19 / 16 , c ) 15 / 16 , d ) 17 / 4 , e ) 0 | d | subtract(add(add(multiply(divide(2, 3), divide(3, 8)), 4), divide(1, 2)), divide(8, 16)) | divide(n1,n3)|divide(n3,n5)|divide(n0,n1)|divide(n5,n8)|multiply(#0,#1)|add(n6,#4)|add(#5,#2)|subtract(#6,#3) | general |
michelle deposited a certain sum of money in a savings account on july 1 st , 2007 . she earns an 6 % interest compounded semiannually . the sum of money in the account on december 31 st , 2009 is approximately what percent of the initial deposit ? | "since michelle earns 6 % interest compounded semiannually , then she earns 3 % interest every 6 months . now , the simple interest earned in 5 periods ( 30 months = 5 * 6 months ) would be 3 % * 5 = 15 % . but , since the interest iscompoundedevery 6 months , then there would be interest earned on interest ( very smal... | a ) 117 % , b ) 120 % , c ) 121 % , d ) 135 % , e ) 140 % | a | multiply(power(add(1, divide(divide(6, const_100), const_2)), add(const_2, const_3)), const_100) | add(const_2,const_3)|divide(n2,const_100)|divide(#1,const_2)|add(#2,n0)|power(#3,#0)|multiply(#4,const_100)| | gain |
a company producing fruit juice changed its packaging from boxes measuring 5 x 10 x 20 centimeters to boxes measuring 6 x 10 x 20 centimeters . if the price of a box did not change and all boxes are full of juice , by approximately what percent did the price of the juice decrease ? | suppose when v = 5 x 10 x 20 = 1000 cm , the price is $ 1200 per cm price = $ 1.2 the price is $ 1200 if v = 6 x 10 x 20 = 1000 cm . per cm price = $ 1.0 so the price is decreased by $ 0.2 so the % price decreased by $ 0.2 / 1.20 = 16.67 % answer : b | a ) 12.00 % , b ) 16.67 % , c ) 18.33 % , d ) 20.00 % , e ) 21.50 % | b | multiply(subtract(const_1, divide(multiply(multiply(5, 10), 20), multiply(multiply(6, 10), 20))), const_100) | multiply(n0,n1)|multiply(n1,n3)|multiply(n2,#0)|multiply(n2,#1)|divide(#2,#3)|subtract(const_1,#4)|multiply(#5,const_100) | general |
if $ 5,000 is invested in an account at a simple annual rate of r percent , the interest is $ 250 . when $ 19,000 is invested at the same interest rate , what is the interest from the investment ? | "- > 250 / 5,000 = 5 % and 19,000 * 5 % = 950 . thus , d is the answer ." | a ) $ 700 , b ) $ 750 , c ) $ 800 , d ) $ 950 , e ) $ 900 | d | divide(multiply(250, multiply(multiply(const_2, const_100), const_100)), divide(multiply(multiply(const_2, const_100), const_100), const_4)) | multiply(const_100,const_2)|multiply(#0,const_100)|divide(#1,const_4)|multiply(n1,#1)|divide(#3,#2)| | gain |
a man can row 4.2 km / hr in still water . it takes him twice as long to row upstream as to row downstream . what is the rate of the current ? | "speed of boat in still water ( b ) = 4.2 km / hr . speed of boat with stream ( down stream ) , d = b + u speed of boat against stream ( up stream ) , u = b – u it is given upstream time is twice to that of down stream . ⇒ downstream speed is twice to that of upstream . so b + u = 2 ( b – u ) ⇒ u = b / 3 = 1.4 km / hr ... | a ) 1.9 , b ) 1.7 , c ) 1.4 , d ) 1.5 , e ) 1.1 | c | divide(subtract(multiply(4.2, const_2), 4.2), const_3) | multiply(n0,const_2)|subtract(#0,n0)|divide(#1,const_3)| | general |
divide rs . 116000 among 3 persons a , b and c such that the ratio of the shares of a and b is 3 : 4 and that of b : c is 5 : 6 . find the share of a ? | compound ratio of a : b : c a : b = 3 : 4 b : c = 5 : 6 - - - - - - - - - - a : b : c = 15 : 20 : 24 we can divide rs . 116000 in this ratio . share of a = 15 / 59 * 116000 = 29491 answer : a | a ) 29491 , b ) 28491 , c ) 39491 , d ) 49491 , e ) 59491 | a | multiply(116000, divide(multiply(3, 5), add(add(multiply(3, 5), multiply(4, 5)), multiply(4, 6)))) | multiply(n1,n4)|multiply(n3,n4)|multiply(n3,n5)|add(#0,#1)|add(#3,#2)|divide(#0,#4)|multiply(n0,#5) | other |
the diagonal of a rhombus are 40 m and 30 m . its area is : | area of the rhombus = 1 / 2 d 1 d 2 = ( 1 / 2 ã — 40 ã — 30 ) cm ( power ) 2 = 40 ã — 15 = 600 cm ( power ) 2 answer is a . | ['a ) 600', 'b ) 450', 'c ) 350', 'd ) 500', 'e ) 620'] | a | rhombus_area(40, 30) | rhombus_area(n0,n1) | geometry |
in a certain corporation , there are 300 male employees and 150 female employees . it is known that 40 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 450 employees is chosen at random , what is the probability this employee has an advanced degree or is female... | "p ( female ) = 150 / 450 = 1 / 3 p ( male with advanced degree ) = 0.4 * 300 / 450 = 120 / 450 = 4 / 15 the sum of the probabilities is 9 / 15 = 3 / 5 the answer is c ." | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 5 , d ) 7 / 10 , e ) 11 / 15 | c | add(divide(multiply(subtract(const_1, divide(40, multiply(40, 40))), 150), 450), divide(add(multiply(divide(40, multiply(40, 40)), 300), multiply(divide(40, multiply(40, 40)), 150)), 450)) | multiply(n2,n2)|divide(n3,#0)|divide(n2,#0)|multiply(n0,#2)|multiply(n1,#1)|subtract(const_1,#1)|add(#3,#4)|multiply(n1,#5)|divide(#7,n4)|divide(#6,n4)|add(#8,#9)| | other |
in a park there are two ponds with both brown ducks and green ducks . in the smaller pond there are 20 ducks and in the larger pond there are 80 ducks . if 20 % of the ducks in the smaller pond are green and 15 % of the ducks in the larger pond are green , then what percentage of ducks are green ? | "number of ducks in small pond = 20 green ducks in small pond = 20 % of 20 = 4 ducks number of ducks in large pond = 80 green ducks in large pond = 15 % of 80 = 12 ducks total number of ducks = 20 + 80 = 100 total number of green ducks = 4 + 12 = 16 ducks percentage of green ducks = 16 / 100 * 100 = 16 % answer : d" | a ) 13 % , b ) 14 % , c ) 15 % , d ) 16 % , e ) 17 % | d | multiply(divide(add(multiply(20, divide(20, const_100)), multiply(80, divide(15, const_100))), add(20, 80)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
the current of a stream runs at the rate of 5 kmph . a boat goes 6 km and back to the starting point in 2 hours , then find the speed of the boat in still water ? | "s = 5 m = x ds = x + 5 us = x - 5 6 / ( x + 5 ) + 6 / ( x - 5 ) = 2 x = 8.83 answer : c" | a ) a ) 7.63 , b ) b ) 2.6 , c ) c ) 8.83 , d ) d ) 6.69 , e ) e ) 3 | c | divide(power(5, 2), 2) | power(n0,n2)|divide(#0,n2)| | physics |
if two projectiles are launched at the same moment from 1386 km apart and travel directly towards each other at 445 km per hour and 545 km per hour respectively , how many minutes will it take for them to meet ? | "the projectiles travel a total of 990 km per hour . the time to meet is 1386 / 990 = 1.4 hours = 84 minutes the answer is c ." | a ) 80 , b ) 82 , c ) 84 , d ) 86 , e ) 88 | c | multiply(divide(1386, add(445, 545)), const_60) | add(n1,n2)|divide(n0,#0)|multiply(#1,const_60)| | physics |
find the average of all the numbers between 11 and 21 which are divisible by 2 . | "sol . average = ( 12 + 14 + 16 + 18 + 20 / 5 ) = 80 / 5 = 16 . answer c" | a ) 15 , b ) 18 , c ) 16 , d ) 22 , e ) none | c | divide(add(add(11, const_4), subtract(21, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
a train speeds past a pole in 15 sec and a platform 140 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 140 ) / 25 = x / 15 = > x = 210 m . answer : d" | a ) 238 , b ) 150 , c ) 988 , d ) 210 , e ) 171 | d | multiply(140, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 400 . find the total worth of the stock ? | let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 400 = > x = 20000 answer : b | a ) rs . 25000 , b ) rs . 20000 , c ) rs . 15000 , d ) rs . 22000 , e ) none of these | b | divide(400, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|multiply(#0,#4)|subtract(#6,#5)|divide(n3,#7) | gain |
the events a and b are independent , the probability that event a occurs is greater than 0 , and the probability that event a occurs is twice the probability that event b occurs . the probability that at least one of events a and b occurs is 12 times the probability that both events a and b occur . what is the probabil... | "let us say probability of a occuring is a . let us say probability of b occuring is b . a = 2 b probability ( either a or b or both ) = 12 times probability ( a and b ) a * ( 1 - b ) + b * ( 1 - a ) + ab = 12 * ab substituting a = 2 b in the second equation : 2 b * ( 1 - b ) + b * ( 1 - 2 b ) + 2 b * b = 12 * 2 b * b ... | a ) 21 / 26 , b ) 25 / 26 , c ) 11 / 26 , d ) 6 / 26 , e ) 22 / 33 | d | multiply(divide(add(const_2, const_1), add(multiply(12, const_2), const_2)), const_2) | add(const_1,const_2)|multiply(n1,const_2)|add(#1,const_2)|divide(#0,#2)|multiply(#3,const_2)| | general |
a snooker tournament charges $ 40.00 for vip seats and $ 15.00 for general admission ( “ regular ” seats ) . on a certain night , a total of 320 tickets were sold , for a total cost of $ 7,500 . how many fewer tickets were sold that night for vip seats than for general admission seats ? | "let no of sits in vip enclosure is x then x * 40 + 15 ( 320 - x ) = 7500 or 25 x = 7500 - 4800 , x = 2700 / 25 = 108 vip = 108 general 212 a" | a ) 212 , b ) 200 , c ) 220 , d ) 230 , e ) 240 | a | subtract(320, divide(subtract(add(multiply(add(const_3, const_4), const_1000), multiply(add(const_2, const_3), const_100)), multiply(320, 15.00)), multiply(add(const_2, const_3), add(const_2, const_3)))) | add(const_3,const_4)|add(const_2,const_3)|multiply(n1,n2)|multiply(#0,const_1000)|multiply(#1,const_100)|multiply(#1,#1)|add(#3,#4)|subtract(#6,#2)|divide(#7,#5)|subtract(n2,#8)| | geometry |
a sum of money is to be divided among ann , bob and chloe . first , ann receives $ 4 plus one - half of what remains . next , bob receives $ 4 plus one - third of what remains . finally , chloe receives the remaining $ 32 . how much money r did bob receive ? | "notice that we need not consider ann ' s portion in the solution . we can just let k = the money remaining after ann has received her portion and go from there . our equation will use the fact that , once we remove bob ' s portion , we have $ 32 for chloe . so , we getk - bob ' s $ = 32 bob received 4 dollars plus one... | a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 52 | b | divide(multiply(32, const_2), const_3) | multiply(n2,const_2)|divide(#0,const_3)| | general |
a ’ s speed is 20 / 12 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 20 meters , b covers 12 meters in that time . so if the race i... | a ) 1 / 17 , b ) 3 / 17 , c ) 1 / 10 , d ) 6 / 20 , e ) 3 / 10 | d | divide(subtract(20, 12), 20) | subtract(n0,n1)|divide(#0,n0)| | general |
a train speeds past a pole in 15 seconds and a platform 130 meters long in 25 seconds . what is the length of the train ( in meters ) ? | "let the length of the train be x meters . the speed of the train is x / 15 . then , x + 130 = 25 * ( x / 15 ) 10 x = 1950 x = 195 meters the answer is c ." | a ) 175 , b ) 185 , c ) 195 , d ) 205 , e ) 215 | c | multiply(130, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
calculate the time it will take for a full tank to become completely empty due to a leak given that the tank could be filled in 7 hours , but due to the leak in its bottom it takes 8 hours to be filled ? | part filled without leak in 1 hour = 1 / 7 part filled with leak in 1 hour = 1 / 8 work done by leak in 1 hour = 1 / 7 â ˆ ’ 1 / 8 = 56 hours answer : d | a ) 59 hours , b ) 54 hours , c ) 59 hours , d ) 56 hours , e ) 26 hours | d | inverse(subtract(divide(const_1, 7), divide(const_1, 8))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|inverse(#2) | physics |
a is 2 times as fast as b . a alone can do the work in 20 days . if a and b working together in how many days will the work be completed ? | "a can finish 1 work in 20 days b can finish 1 / 2 work in 20 days - since a is 2 faster than b this means b can finish 1 work in 20 * 2 days = 40 days now using the awesome gmat formula when two machines work together they can finish the job in = ab / ( a + b ) = 20 * 40 / ( 20 + 40 ) = 13 days so answer is b" | a ) 23 , b ) 13 , c ) 21 , d ) 24 , e ) 25 | b | divide(const_1, add(divide(const_1, 20), divide(divide(const_1, 20), 2))) | divide(const_1,n1)|divide(#0,n0)|add(#0,#1)|divide(const_1,#2)| | physics |
company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 1.2 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last m... | "the easiest way for me is to plug in numbers . let the number of basic stereos produced be 40 , and number of delux stereos produced be 20 . total of 60 stereos . if it takes an hour to produce a basic stereo then it will take 1.2 hours to produce a deluxe stereo . 40 basic stereos = 40 hours . 20 delux stereos = 24 h... | a ) 3 / 8 , b ) 14 / 31 , c ) 7 / 15 , d ) 17 / 35 , e ) 1 / 2 | a | divide(add(3, 2), const_10) | add(n0,n1)|divide(#0,const_10)| | general |
a train leaves mumabai at 9 am at a speed of 30 kmph . after one hour , another train leaves mumbai in the same direction as that of the first train at a speed of 60 kmph . when and at what distance from mumbai do the two trains meet ? | when the second train leaves mumbai the first train covers 30 * 1 = 30 km so , the distance between first train and second train is 30 km at 10.00 am time taken by the trains to meet = distance / relative speed = 30 / ( 60 - 30 ) = 1 hours so , the two trains meet at 11 a . m . the two trains meet 1 * 60 = 60 km away f... | a ) 27 , b ) 279 , c ) 60 , d ) 278 , e ) 379 | c | multiply(divide(multiply(30, const_1), subtract(60, 30)), 60) | multiply(n1,const_1)|subtract(n2,n1)|divide(#0,#1)|multiply(n2,#2) | physics |
a = 4 ^ 15 - 625 ^ 3 and a / x is an integer , where x is a positive integer greater than 1 , such that it does not have a factor p such that 1 < p < x , then how many different values for x are possible ? | "this is a tricky worded question and i think the answer is should be d not c . . . here is my reason : the stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . the stem wants to say that x is a prime number . because any prime number has no factor grater than 1 and itse... | a ) none , b ) one , c ) two , d ) three , e ) four | c | subtract(15, multiply(3, const_4)) | multiply(n3,const_4)|subtract(n1,#0)| | general |
a envelop weight 8.5 gm , if 800 of these envelop are sent with an advertisement mail . how much wieght ? | "800 * 8.5 6800.0 gm 6.8 kg answer : b" | a ) 6.6 kg , b ) 6.8 kg , c ) 6.7 kg , d ) 6.9 kg , e ) 7.8 kg | b | divide(multiply(8.5, 800), const_1000) | multiply(n0,n1)|divide(#0,const_1000)| | general |
in a city , 35 % of the population is composed of migrants , 20 % of whom are from rural areas . of the local population , 48 % is female while this figure for rural and urban migrants is 30 % and 40 % respectively . if the total population of the city is 728400 , what is its female population ? | explanation : total population = 728400 migrants = 35 % of 728400 = 254940 local population = ( 728400 - 254940 ) = 473460 . rural migrants = 20 % of 254940 = 50988 urban migrants = ( 254940 - 50988 ) = 203952 female population = 48 % of 473460 + 30 % of 50988 + 40 % of 203952 = 324138 answer : a | a ) 324138 , b ) 248888 , c ) 378908 , d ) 277880 , e ) 379010 | a | add(add(divide(multiply(48, subtract(728400, divide(multiply(728400, 35), const_100))), const_100), divide(multiply(30, divide(multiply(divide(multiply(728400, 35), const_100), 20), const_100)), const_100)), divide(multiply(40, subtract(divide(multiply(728400, 35), const_100), divide(multiply(divide(multiply(728400, 35... | multiply(n0,n5)|divide(#0,const_100)|multiply(n1,#1)|subtract(n5,#1)|divide(#2,const_100)|multiply(n2,#3)|divide(#5,const_100)|multiply(n3,#4)|subtract(#1,#4)|divide(#7,const_100)|multiply(n4,#8)|add(#6,#9)|divide(#10,const_100)|add(#11,#12) | gain |
a batsman scored 120 runs whichincluded 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets ? | "number of runs made by running = 110 - ( 3 x 4 + 8 x 6 ) = 120 - ( 60 ) = 60 now , we need to calculate 60 is what percent of 120 . = > 60 / 120 * 100 = 50 % b" | a ) 40 % , b ) 50 % , c ) 65 % , d ) 68 % , e ) 70 % | b | multiply(divide(add(multiply(3, const_4), multiply(add(3, 3), 8)), 120), const_100) | add(n1,n1)|multiply(n1,const_4)|multiply(n2,#0)|add(#1,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
397 x 397 + 104 x 104 + 2 x 400 x 104 = x ? | "given exp . = ( 397 ) 2 + ( 104 ) 2 + 2 x 397 x 104 = ( 397 + 104 ) 2 = ( 501 ) 2 = ( 500 + 1 ) 2 = ( 5002 ) + ( 1 ) 2 + ( 3 x 500 x 1 ) = 250000 + 1 + 1500 = 251501 e" | a ) 234341 , b ) 235633 , c ) 234677 , d ) 315656 , e ) 251501 | e | subtract(add(multiply(397, 397), multiply(104, 104)), multiply(multiply(2, 397), 104)) | multiply(n0,n0)|multiply(n2,n2)|multiply(n0,n4)|add(#0,#1)|multiply(n2,#2)|subtract(#3,#4)| | general |
the events a and b are independent , the probability that event a occurs is greater than 0 , and the probability that event a occurs is twice the probability that event b occurs . the probability that at least one of events a and b occurs is 3 times the probability that both events a and b occur . what is the probabili... | "let us say probability of a occuring is a . let us say probability of b occuring is b . a = 2 b probability ( either a or b or both ) = 3 times probability ( a and b ) a * ( 1 - b ) + b * ( 1 - a ) + ab = 3 * ab substituting a = 2 b in the second equation : 2 b * ( 1 - b ) + b * ( 1 - 2 b ) + 2 b * b = 3 * 2 b * b 3 b... | a ) 44 / 7 , b ) 9 / 8 , c ) 25 / 8 , d ) 3 / 4 , e ) 7 / 8 | d | multiply(divide(add(const_2, const_1), add(multiply(3, const_2), const_2)), const_2) | add(const_1,const_2)|multiply(n1,const_2)|add(#1,const_2)|divide(#0,#2)|multiply(#3,const_2)| | general |
john goes to his office by car at a speed of 40 kmph and reaches 8 minutes earlier . if he goes at a speed of 30 kmph , he reaches 4 mins late . what is the distance from his house to office ? | let his office time be 10 a . m . he reaches office at 8 minutes earlier when he travels at 40 kmph = > he reaches at 9 : 52 a . m he reaches his office 4 minutes late when travels at 30 kmph speed = > he reaches at 10.04 a . m the time difference is ( 10.04 – 9.52 ) = 12 minutes let the distance be d and time = distan... | a ) 18 km , b ) 20 km , c ) 22 km , d ) 24 km , e ) 30 km | d | multiply(40, divide(multiply(30, divide(add(8, 4), const_60)), subtract(40, 30))) | add(n1,n3)|subtract(n0,n2)|divide(#0,const_60)|multiply(n2,#2)|divide(#3,#1)|multiply(n0,#4) | physics |
a certain list consists of 11 different numbers . if n is in the list and n is 5 times the average ( arithmetic mean ) of the other 10 numbers in the list , then n is what fraction of the sum of the 11 numbers in the list ? | "series : a 1 , a 2 . . . . a 10 , n sum of a 1 + a 2 + . . . + a 10 = 10 * x ( x = average ) so , n = 5 * x hence , a 1 + a 2 + . . + a 10 + n = 15 x so , the fraction asked = 5 x / 15 x = 1 / 3 answer is a" | a ) 1 / 3 , b ) 2 / 5 , c ) 1 / 4 , d ) 3 / 5 , e ) 1 / 6 | a | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(10, 5), 11), 5), const_2), 5), const_3)) | divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,n1)|multiply(#3,const_2)|subtract(#4,n1)|subtract(#5,const_3)|divide(#1,#6)| | general |
compound interest of rs . 4000 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 = 4000 [ 1 + ( 5 / 100 ) ] ^ 3 - 4000 = 4000 [ ( 21 / 20 ) ^ 3 - 1 ] = 4000 ( 9261 - 8000 ) / 8000 = 4 * 1261 / 8 = 630 answer : d | a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 630 , e ) none of the above | d | subtract(multiply(4000, power(add(const_1, divide(divide(10, const_2), const_100)), multiply(add(1, divide(1, 2)), const_2))), 4000) | divide(n1,const_2)|divide(n2,n4)|add(n2,#1)|divide(#0,const_100)|add(#3,const_1)|multiply(#2,const_2)|power(#4,#5)|multiply(n0,#6)|subtract(#7,n0) | gain |
find the length of the longest pole that can be placed in an indoor stadium 24 m long , 18 m wide and 16 m high | "sqrt ( 1156 ) = 34 becoz max length is cuboid diagonal d = sqrt ( l ^ 2 + b ^ 2 + h ^ 2 ) where l lenght b breadth & h height answer : d" | a ) 31 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | d | sqrt(add(power(16, const_2), add(power(24, const_2), power(18, const_2)))) | power(n0,const_2)|power(n1,const_2)|power(n2,const_2)|add(#0,#1)|add(#3,#2)|sqrt(#4)| | physics |
one hour before john started walking from p to q , a distance of 13 miles , ann had started walking along the same road from q to p . ann walked at a constant speed of 3 miles per hour and john at 2 miles per hour . how many miles had ann walked when they met ? | "ann walks from q to p at a speed of 3 miles / hr for one hour . she covers 3 miles in 1 hour and now distance between john and ann is 13 - 3 = 10 miles . ann walks at 3 mph and john at 2 mph so their relative speed is 3 + 2 = 5 mph . they have to cover 10 miles so it will take them 10 / 5 = 2 hours to meet . in 2 hrs ... | a ) 6 miles , b ) 8,4 miles , c ) 9 miles , d ) 9,6 miles , e ) 12 miles | c | multiply(divide(13, add(3, 2)), 2) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | physics |
sheela deposits rs . 3800 in bank savings account . if this is 22 % of her monthly income . what is her monthly income in ? | "explanation : 22 % of income = rs . 3800 100 % of income = 3800 x 100 / 22 = rs . 17272 answer : d" | a ) 22000 , b ) 20000 , c ) 25123 , d ) 17272 , e ) none of these | d | divide(multiply(3800, const_100), 22) | multiply(n0,const_100)|divide(#0,n1)| | gain |
one pipe can fill a pool 1.5 times faster than a second pipe . when both pipes are opened , they fill the pool in four hours . how long would it take to fill the pool if only the slower pipe is used ? | "say the rate of the slower pipe is r pool / hour , then the rate of the faster pipe would be 1.5 r = 3 r / 2 . since when both pipes are opened , they fill the pool in four hours , then their combined rate is 1 / 4 pool / hour . thus we have that r + 3 r / 2 = 1 / 4 - - > r = 1 / 10 pool / hour - - > time is reciproca... | a ) 11.25 , b ) 11.52 , c ) 1.25 , d ) 9 , e ) 10 | e | divide(inverse(divide(inverse(add(const_3, const_2)), add(const_1, 1.5))), 1.5) | add(const_2,const_3)|add(n0,const_1)|inverse(#0)|divide(#2,#1)|inverse(#3)|divide(#4,n0)| | physics |
3 photographers , lisa , mike and norm , take photos of a wedding . the total of lisa and mikes photos is 60 less than the sum of mike ' s and norms . if norms photos number 10 more than twice the number of lisa ' s photos , then how many photos did norm take ? | l + m = m + n - 60 / n = 2 l + 10 60 = m + n - l - m 60 = n - l 60 = 2 l + 10 - l 50 = l 2 ( 50 ) + 10 = 110 d | a ) 40 , b ) 50 , c ) 60 , d ) 110 , e ) 80 | d | subtract(multiply(60, const_2), 10) | multiply(n1,const_2)|subtract(#0,n2) | general |
the area of a rectangle is 360 sq . m . if its length is increased by 10 m and its width is decreased by 6 m , then its area does not change . find the perimeter of the original rectangle . | let l and b are the length and width of the original rectangle . since , l * b = 360 sq . m - ( 1 ) now , a / c to the problem ( l + 10 ) * ( b - 6 ) = 360 sq . m therefore , b = 6 ( 1 + l / 10 ) - ( 2 ) l * l + 10 l - 600 = 0 { by using equations 1 and 2 } thereafter , ( l - 20 ) ( l + 30 ) = 0 hence , l = 20 m b = 18... | ['a ) 74 m', 'b ) 75 m', 'c ) 76 m', 'd ) 77 m', 'e ) 78 m'] | c | rectangle_perimeter(multiply(6, const_3), divide(360, multiply(6, const_3))) | multiply(n2,const_3)|divide(n0,#0)|rectangle_perimeter(#1,#0) | geometry |
the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 39 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 39 / 3 = 13 kmph answer : b" | a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 16 kmph | b | subtract(39, divide(multiply(39, const_2), const_3)) | multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)| | physics |
in measuring the sides of a rectangle , one side is taken 7 % in excess , and the other 6 % in deficit . find the error percent in the area calculated from these measurements . | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 61 / 57 ) x ( 47 / 50 ) y = ( 867 / 862 ) ( xy ) error in measurement = ( 867 / 862 ) xy - xy = ( 5 / 862 ) xy error percentage = [ ( 5 / 862 ) xy ( 1 / xy ) 100 ] % = ( 29 / 50 ) % = 0.58 % . answer is e ." | a ) 0.11 % , b ) 0.7 % , c ) 0.4 % , d ) 0.6 % , e ) 0.58 % | e | subtract(subtract(7, 6), divide(multiply(7, 6), const_100)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)| | geometry |
how many squares are there between 2011 to 2300 ? ? ? ? | nos . are 2025 , 2116 , 2209 answers is 3 answer : a | ['a ) 3', 'b ) 4', 'c ) 5', 'd ) 6', 'e ) 7'] | a | subtract(floor(sqrt(2300)), floor(sqrt(2011))) | sqrt(n1)|sqrt(n0)|floor(#0)|floor(#1)|subtract(#2,#3) | geometry |
the average age of a group of 10 students is 14 years . if 5 more students join the group , the average age rises by 1 year . the average age of the new students is : | explanation : total age of the 10 students = 10 × 14 = 140 total age of 15 students including the newly joined 5 students = 15 × 15 = 225 total age of the new students = 225 − 140 = 85 average age = 85 / 5 = 17 years answer : d | a ) 22 , b ) 38 , c ) 11 , d ) 17 , e ) 91 | d | divide(subtract(multiply(add(14, 1), add(14, 1)), multiply(14, 10)), 5) | add(n1,n3)|multiply(n0,n1)|multiply(#0,#0)|subtract(#2,#1)|divide(#3,n2) | general |
average expenditure of a person for the first 3 days of a week is rs . 310 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is : | "explanation : assumed mean = rs . 310 total excess than assumed mean = 4 × ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 / 7 = rs . 40 therefore , average expenditure for 7 days = rs . 310 + rs . 40 = rs . 350 correct option : a" | a ) 350 , b ) 370 , c ) 390 , d ) 430 , e ) none | a | add(310, divide(multiply(4, subtract(420, 310)), add(3, 4))) | add(n0,n2)|subtract(n3,n1)|multiply(n2,#1)|divide(#2,#0)|add(n1,#3)| | general |
if the perimeter of a rectangular garden is 800 m , its length when its breadth is 100 m is ? | "2 ( l + 100 ) = 800 = > l = 300 m answer : c" | a ) 286 m , b ) 899 m , c ) 300 m , d ) 166 m , e ) 187 m | c | subtract(divide(800, const_2), 100) | divide(n0,const_2)|subtract(#0,n1)| | physics |
two equal sums of money were invested , one at 4 % and the other at 4.5 % . at the end of 7 years , the simple interest received from the latter exceeded to that received from the former by 31.50 . each sum was : | difference of s . i . = √ 31.50 let each sum be x . then x × 4 1 / 2 × 7 / 100 − x × 4 × 7 / 100 = 31.50 or 7 ⁄ 100 x × 1 ⁄ 2 = 63 ⁄ 2 or x = 900 answer d | a ) 1200 , b ) 600 , c ) 750 , d ) 900 , e ) none of these | d | divide(multiply(divide(31.5, subtract(4.5, 4)), const_100), 7) | subtract(n1,n0)|divide(n3,#0)|multiply(#1,const_100)|divide(#2,n2) | gain |
shekar scored 76 , 65 , 82 , 67 and 75 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | "explanation : average = ( 76 + 65 + 82 + 67 + 75 ) / 5 = 365 / 5 = 73 hence average = 73 answer : e" | a ) 65 , b ) 69 , c ) 75 , d ) 85 , e ) 73 | e | divide(add(add(add(add(76, 65), 82), 67), 75), add(const_1, const_4)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
tickets for all but 100 seats in a 10000 - seat stadium were sold . of the tickets sold , 30 % were sold at half price and the remaining tickets were sold at the full price of $ 2 . what was the total revenue from ticket sales ? | 10000 seats - - > full price : half price = 7000 : 3000 price when all seats are filled = 14000 + 3000 = 17000 100 seats are unsold - - > loss due to unfilled seats = 30 + 2 * 70 = 170 revenue = 17000 - 170 = 16830 answer : a | a ) $ 16,830 , b ) $ 17,820 , c ) $ 18,000 , d ) $ 19,800 , e ) $ 21,780 | a | floor(divide(add(multiply(2, subtract(subtract(10000, 100), divide(multiply(subtract(10000, 100), 30), 100))), divide(multiply(subtract(10000, 100), 30), 100)), const_1000)) | subtract(n1,n0)|multiply(n2,#0)|divide(#1,n0)|subtract(#0,#2)|multiply(n3,#3)|add(#2,#4)|divide(#5,const_1000)|floor(#6) | general |
a and b started a business investing rs . 10,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "a : b = 10000 : 20000 = 1 : 2 answer : d" | a ) 3 : 2 , b ) 9 : 2 , c ) 18 : 20 , d ) 1 : 2 , e ) 18 : 4 | d | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | multiply(const_3,const_3)|add(#0,const_1)|multiply(#1,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3)| | gain |
rs . 6000 is lent out in two parts . one part is lent at 7 % p . a simple interest and the other is lent at 9 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | "let the amount lent at 7 % be rs . x amount lent at 9 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 7 / 100 x + 9 / 100 ( 6000 - x ) = 540 - 2 x / 100 = > 540 - 1 / 50 x = 450 = > x = 4500 amount lent at 10 % = 1500 required ratio = 4500 : 1500 = 9 : 3 answer : a" | a ) 9 : 3 , b ) 9 : 5 , c ) 5 : 8 , d ) 5 : 4 , e ) 9 : 2 | a | divide(divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(9, 7)), divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(9, 7))) | multiply(n3,const_100)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|divide(#4,#4)| | gain |
a man sitting in a train which is travelling at 15 kmph observes that a goods train , travelling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed ? | "solution relative speed = ( 280 / 9 ) m / sec = ( 280 / 9 x 18 / 5 ) = 112 kmph . speed of the train = ( 112 - 15 ) kmph = 97 kmph . answer c" | a ) 52 kmph . , b ) 62 kmph . , c ) 97 kmph . , d ) 80 kmph . , e ) none | c | subtract(multiply(divide(280, 9), const_3_6), 15) | divide(n2,n1)|multiply(#0,const_3_6)|subtract(#1,n0)| | physics |
there are 35 student in a school . 20 if them speek hindi 19 of them speak eng 9 of them speak both , then how many student neithe speak eng nor speak hindi | total number of students = 35 no of hindi speaking students ( h ) = 20 no of english speaking students ( e ) = 19 h intersection e = 9 thus h union e = 20 + 19 - 9 = 30 thus number of student neithe speak eng nor speak hindi = 35 - 30 = 5 answer : e | a ) 28 , b ) 30 , c ) 32 , d ) 35 , e ) none of these | e | subtract(35, subtract(add(20, 19), 9)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1) | other |
a student chose a number , multiplied it by 2 , then subtracted 152 from the result and got 102 . what was the number he chose ? | "solution : let x be the number he chose , then 2 * x * 152 = 102 2 x = 254 x = 127 correct answer a" | a ) 127 , b ) 100 , c ) 129 , d ) 160 , e ) 200 | a | divide(add(102, 152), 2) | add(n1,n2)|divide(#0,n0)| | general |
the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 50 feet wide by 150 feet long by 10 feet deep and is currently at 80 % capacity , how long will it take to drain the pool ? | volume of pool = 50 * 150 * 10 cu . ft , 80 % full = 50 * 150 * 10 * 0.8 cu . ft water is available to drain . draining capacity = 60 cu . ft / min therefore time taken = 50 * 150 * 10 * 0.8 / 60 min = 1000 min a | a ) 1000 min , b ) 1200 min , c ) 1300 min , d ) 1400 min , e ) 1600 min | a | divide(multiply(divide(80, const_100), multiply(multiply(50, 150), 10)), 60) | divide(n4,const_100)|multiply(n1,n2)|multiply(n3,#1)|multiply(#0,#2)|divide(#3,n0) | gain |
if a is an integer greater than 9 but less than 26 and b is an integer greater than 14 but less than 31 , what is the range of a / b ? | "range of a / b = max ( a / b ) - min ( a / b ) to get max ( a / b ) = > max ( a ) / min ( b ) = 25 / 15 to get min ( a / b ) = > min ( a ) / max ( b ) = 10 / 30 range = 25 / 15 - 10 / 30 = 4 / 3" | a ) 1 , b ) 1 / 2 , c ) 5 / 6 , d ) 2 / 3 , e ) 4 / 3 | e | subtract(divide(subtract(26, const_1), add(14, const_1)), divide(add(9, const_1), subtract(31, const_1))) | add(n2,const_1)|add(n0,const_1)|subtract(n1,const_1)|subtract(n3,const_1)|divide(#2,#0)|divide(#1,#3)|subtract(#4,#5)| | general |
if the cost price is 40 % of selling price . then what is the profit percent . | "explanation : let the s . p = 100 then c . p . = 40 profit = 60 profit % = ( 60 / 40 ) * 100 = 150 % . answer : a" | a ) 150 % , b ) 120 % , c ) 130 % , d ) 200 % , e ) none of these | a | multiply(divide(subtract(const_100, 40), 40), const_100) | subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
when positive integer x is divided by positive integer y , the remainder is 12 . if x / y = 75.12 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 12 - - > x = qy + 12 ; x / y = 75.12 - - > x = 75 y + 0.12 y ( so q above equals to 75 ) ; 0.12 y = 12 - - > y = 100 . answer : e ." | a ) 84 , b ) 98 , c ) 51 , d ) 65 , e ) 100 | e | divide(12, subtract(75.12, floor(75.12))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general |
the cost to park a car in a certain parking garage is $ 20.00 for up to 2 hours of parking and $ 1.75 for each hour in excess of 2 hours . what is the average ( arithmetic mean ) cost per hour to park a car in the parking garage for 9 hours ? | "total cost of parking for 9 hours = 20 $ for the first 2 hours and then 1.75 for ( 9 - 2 ) hours = 20 + 7 * 1.75 = 32.25 thus the average parking price = 32.25 / 9 = 3.58 $ a is the correct answer ." | a ) $ 3.58 , b ) $ 1.67 , c ) $ 2.25 , d ) $ 2.37 , e ) $ 2.50 | a | divide(add(20.00, multiply(1.75, subtract(9, 2))), 9) | subtract(n4,n1)|multiply(n2,#0)|add(n0,#1)|divide(#2,n4)| | general |
the unit digit in the product ( 611 * 704 * 912 * 261 ) is : | "explanation : unit digit in the given product = unit digit in ( 1 * 4 * 2 * 1 ) = 2 answer : a" | a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | a | subtract(multiply(multiply(multiply(611, 704), 912), 261), subtract(multiply(multiply(multiply(611, 704), 912), 261), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general |
two trains are moving in opposite directions with speed of 150 km / hr and 90 km / hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds | "explanation : relative speed = 150 + 90 = 240 km / hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 / 240 hr = 1 / 120 hr = 3600 / 120 seconds = 30 seconds answer : option d" | a ) 56 , b ) 48 , c ) 47 , d ) 30 , e ) 25 | d | multiply(divide(add(1.10, 0.9), add(150, 90)), const_3600) | add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(#2,const_3600)| | physics |
a train 60 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 60 * 3 / 55 = 3 sec . answer e" | a ) 7 , b ) 6 , c ) 8 , d ) 2 , e ) 3 | e | divide(60, multiply(add(60, 6), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
if the sum and difference of two numbers are 20 and 10 respectively , then the difference of their square is : | "let the numbers be x and y . then , x + y = 20 and x - y = 8 x 2 - y 2 = ( x + y ) ( x - y ) = 20 * 10 = 200 . answer : d" | a ) 12 , b ) 28 , c ) 160 , d ) 200 , e ) 18 | d | subtract(power(divide(add(20, 10), const_2), const_2), power(subtract(20, divide(add(20, 10), const_2)), const_2)) | add(n0,n1)|divide(#0,const_2)|power(#1,const_2)|subtract(n0,#1)|power(#3,const_2)|subtract(#2,#4)| | general |
a 340 - liter solution of kola is made from 75 % water , 5 % concentrated kola and the rest is made from sugar . if 3.2 liters of sugar , 12 liter of water and 6.8 liters of concentrated kola were added to the solution , what percent of the solution is made from sugar ? | "denominator : 340 + 12 + 3.2 + 6.8 = 362 numerator : 340 ( 1 - . 75 - . 05 ) + 3.2 340 ( 0.2 ) + 3.2 68 + 3.2 71.2 ratio : 71.2 / 362 = 0.196 answer : c" | a ) 6 % . , b ) 7.5 % . , c ) 19.6 % . , d ) 10.5 % . , e ) 11 % . | c | multiply(divide(add(subtract(subtract(340, multiply(340, divide(75, const_100))), multiply(340, divide(5, const_100))), 3.2), add(add(add(340, 3.2), 12), 6.8)), const_100) | add(n0,n3)|divide(n1,const_100)|divide(n2,const_100)|add(n4,#0)|multiply(n0,#1)|multiply(n0,#2)|add(n5,#3)|subtract(n0,#4)|subtract(#7,#5)|add(n3,#8)|divide(#9,#6)|multiply(#10,const_100)| | gain |
the average mark of the students of a class in a particular exam is 90 . if 2 students whose average mark in that exam is 45 are excluded , the average mark of the remaining will be 95 . find the number of students who wrote the exam ? | "let the number of students who wrote the exam be x . total marks of students = 90 x . total marks of ( x - 2 ) students = 95 ( x - 2 ) 90 x - ( 2 * 45 ) = 95 ( x - 2 ) 100 = 5 x = > x = 20 answer : c" | a ) 10 , b ) 40 , c ) 20 , d ) 30 , e ) 25 | c | divide(subtract(multiply(95, 2), multiply(2, 45)), subtract(95, 90)) | multiply(n1,n3)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)| | general |
a , b and c invest in the ratio of 3 : 4 : 5 . the percentage of return on their investments are in the ratio of 6 : 5 : 4 . find the total earnings , if b earns rs . 250 more than a : | "explanation : a b c investment 3 x 4 x 5 x rate of return 6 y % 5 y % 4 y % return \ inline \ frac { 18 xy } { 100 } \ inline \ frac { 20 xy } { 100 } \ inline \ frac { 20 xy } { 100 } total = ( 18 + 20 + 20 ) = \ inline \ frac { 58 xy } { 100 } b ' s earnings - a ' s earnings = \ inline \ frac { 2 xy } { 100 } = 250 ... | a ) 2348 , b ) 7250 , c ) 2767 , d ) 1998 , e ) 2771 | b | multiply(add(add(multiply(3, 6), multiply(4, 5)), multiply(5, 4)), divide(250, subtract(multiply(4, 5), multiply(3, 6)))) | multiply(n0,n3)|multiply(n1,n2)|add(#0,#1)|subtract(#1,#0)|add(#2,#1)|divide(n6,#3)|multiply(#4,#5)| | general |
when positive integer n is divided by 3 , the remainder is 1 . when n is divided by 11 , the remainder is 9 . what is the smallest positive integer k such that k + n is a multiple of 33 ? | "n = 3 p + 1 = 11 q + 9 n + 2 = 3 p + 3 = 11 q + 11 n + 2 is a multiple of 3 and 11 , so it is a multiple of 33 . the answer is a ." | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | subtract(33, reminder(9, 11)) | reminder(n3,n2)|subtract(n4,#0)| | general |
a copy machine , working at a constant rate , makes 35 copies per minute . a second copy machine , working at a constant rate , makes 75 copies per minute . working together at their respective rates , how many copies do the two machines make in half an hour ? | "together the two machines make 35 + 75 = 110 copies per minute . so , in half an hour they will make 110 * 30 = 3,300 copies . answer : c ." | a ) 90 , b ) 2,700 , c ) 3,300 , d ) 5,400 , e ) 324,000 | c | divide(multiply(add(35, 75), const_60), const_2) | add(n0,n1)|multiply(#0,const_60)|divide(#1,const_2)| | physics |
the lenght of a room is 5.5 m and width is 4 m . find the cost of paving the floor by slabs at the rate of rs . 750 per sq . metre . | "area of the floor = ( 5.5 ã — 4 ) m 2 = 22 m 2 . cost of paving = rs . ( 750 ã — 22 ) = rs . 16500 . answer : option c" | a ) s . 15,550 , b ) s . 15,600 , c ) s . 16,500 , d ) s . 17,600 , e ) s . 17,900 | c | multiply(750, multiply(5.5, 4)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
a dishonest dealer professes to sell goods at the cost price but uses a weight of 950 grams per kg , what is his percent ? | 950 - - - 50 100 - - - ? = > 5.26 % answer : d | a ) 22 % , b ) 25 % , c ) 77 % , d ) 5.26 % , e ) 12 % | d | subtract(multiply(divide(const_100, 950), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100) | gain |
find the compound interest on rs . 25000 in 2 years at 4 % per annum . the interest being compounded half yearly . | "explanation : given : principal = rs . 25000 , rate = 4 % per half year , time = 2 years = 4 half years therefore , amount = p ( 1 + ( r / 2 ) / 100 ) 2 n amount = rs . [ 25000 * ( 1 + 2 / 100 ) 4 ] = rs . ( 25000 * 51 / 50 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 27060.804 therefore , c . i . = rs . ( 27060.804 – 25000... | a ) rs 2060.808 , b ) rs 2060.801 , c ) rs 2060.804 , d ) rs 2060.802 , e ) rs 2060.805 | c | subtract(multiply(power(add(const_1, divide(divide(2, const_4), const_100)), const_3), multiply(multiply(multiply(const_4, const_4), const_100), sqrt(const_100))), multiply(multiply(multiply(const_4, const_4), const_100), sqrt(const_100))) | divide(n1,const_4)|multiply(const_4,const_4)|sqrt(const_100)|divide(#0,const_100)|multiply(#1,const_100)|add(#3,const_1)|multiply(#4,#2)|power(#5,const_3)|multiply(#6,#7)|subtract(#8,#6)| | gain |
ravi covers a distance of 900 mtrs in 180 secs . find his speed in kmph ? | given distance d = 900 meters and time = 180 seconds speed = distance / time = 900 / 180 = 5 m / s . but speed is asked in kmph , to convert 5 mps to kmph , multiply it by 18 / 5 . speed in kmph = 5 x 18 / 5 = 18 kmph b | a ) 10 kmph , b ) 18 kmph , c ) 20 kmph , d ) 26 kmph , e ) 28 kmph | b | multiply(divide(900, 180), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6) | physics |
find the value of x from the below equation ? : 3 x ^ 2 - 5 x + 2 = 0 | "a = 3 , b = - 5 , c = 2 x 1,2 = ( 5 â ± â ˆ š ( ( - 5 ) ^ 2 - 4 ã — 3 ã — 2 ) ) / ( 2 ã — 3 ) = ( 5 â ± â ˆ š ( 25 - 24 ) ) / 6 = ( 5 â ± 1 ) / 6 x 1 = ( 5 + 1 ) / 6 = 6 / 6 = 1 x 2 = ( 5 - 1 ) / 6 = 4 / 6 = 2 / 3 a" | a ) 2 / 3 , b ) - 1 , c ) 0 , d ) - 2 / 3 , e ) 2 | a | divide(subtract(2, sqrt(subtract(power(2, 3), multiply(5, 5)))), 3) | multiply(n2,n2)|power(n1,n0)|subtract(#1,#0)|sqrt(#2)|subtract(n1,#3)|divide(#4,n0)| | general |
n ^ ( n / 2 ) = 4 is true when n = 4 in the same way what is the value of n if n ^ ( n / 2 ) = 8 ? | n ^ ( n / 2 ) = 8 apply log n / 2 logn = log 8 nlogn = 2 log 8 = log 8 ^ 2 = log 64 logn = log 64 now apply antilog n = 64 / n now n = 8 . answer : b | a ) 4 , b ) 8 , c ) 2 , d ) 6 , e ) 10 | b | divide(power(8, 2), 8) | power(n4,n0)|divide(#0,n4)| | general |
if n = 2 ^ 0.25 and n ^ b = 8 , b must equal | "25 / 100 = 1 / 4 n = 2 ^ 1 / 4 n ^ b = 2 ^ 3 ( 2 ^ 1 / 4 ) ^ b = 2 ^ 3 b = 12 answer : c" | a ) 3 / 80 , b ) 3 / 5 , c ) 12 , d ) 5 / 3 , e ) 80 / 3 | c | divide(log(8), log(power(2, 0.25))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
[ ( 3.242 x 15 ) / 100 ] = ? | "answer multiplying 3.242 x 15 = 4.863 now divide 4.863 by 100 so , 4.863 ÷ 100 = 0.04863 ∴ shift the decimal two places to the left as 100 correct option : e" | a ) 0.045388 , b ) 4.5388 , c ) 453.88 , d ) 473.88 , e ) 0.04863 | e | divide(divide(multiply(3.242, 15), 100), const_10) | multiply(n0,n1)|divide(#0,n2)|divide(#1,const_10)| | general |
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