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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the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is ? | "let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 * 3 ) = 135 - - - ( i ) a + b = ( 40 * 2 ) = 80 - - - ( ii ) b + c = ( 43 * 2 ) = 86 - - - ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 166 - - - ( iv ) subtracting ( i ) from ( iv ) , we get : b = 31 b ' s weight = 31 k... | a ) 24 kg , b ) 62 kg , c ) 34 kg , d ) 31 kg , e ) 37 kg | d | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(45, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general |
what is the units digit of 17 ^ 83 × 13 ^ 82 × 11 ^ 83 ? | "to find : the units digit of 17 ^ 83 × 13 ^ 82 × 11 ^ 87 let ' s reduce the clutter and simplify the product ( 7 ^ 83 ) ( 3 ^ 82 ) ( 1 ^ 87 ) 7 has a cyclicity of 4 : the last digit of any positive power of 7 repeats itself after every 4 th power so 7 ^ 5 has the same last digit as 7 ^ 1 , 7 ^ 9 , 7 ^ 13 thus , 7 ^ 83... | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | divide(add(multiply(factorial(17), factorial(83)), multiply(factorial(17), factorial(82))), 17) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
400 students took a mock exam in delhi , 60 % of the boys and 80 % of the girls cleared the cut off in the examination . if the total percentage of students qualifying is 65 % , how many girls appeared in the examination ? | 60 % * x + 80 % * ( 400 - x ) = 65 % * 400 ans = 100 answer : b | a ) 50 , b ) 100 , c ) 120 , d ) 140 , e ) 180 | b | divide(subtract(multiply(400, divide(65, const_100)), multiply(400, divide(60, const_100))), subtract(divide(80, const_100), divide(60, const_100))) | divide(n3,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(#2,#1)|subtract(#3,#4)|divide(#6,#5) | general |
what is the greater of the two numbers whose product is 2688 , given that the sum of the two numbers exceeds their difference by 64 ? | "let the greater and the smaller number be g and s respectively . gs = 2560 g + s exceeds g - s by 64 i . e . , g + s - ( g - s ) = 64 i . e . , 2 s = 64 = > s = 32 . g = 2688 / s = 84 . answer : a" | a ) a ) 84 , b ) b ) 104 , c ) c ) 110 , d ) d ) 72 , e ) of these | a | divide(2688, multiply(power(const_2, const_4), const_2)) | power(const_2,const_4)|multiply(#0,const_2)|divide(n0,#1)| | general |
the proportion of copper and zinc in the brass is 3 : 7 . how much zinc will there be in 100 kg of brass ? | "7 / 10 * 100 = 70 answer : a" | a ) 70 , b ) 50 , c ) 40 , d ) 270 , e ) 170 | a | multiply(divide(7, add(7, 3)), 100) | add(n0,n1)|divide(n1,#0)|multiply(n2,#1)| | general |
a contractor undertakes to do a job within 100 days and hires 10 people to do it . after 20 days , he realizes that one fourth of the work is done so he fires 2 people . in how many more days q will the work get over ? | we can also use the concept of man - days here 100 days - - > 10 men so the job includes 100 * 10 = 1000 man - days after 20 days 1 / 4 of job is completed so 1 / 4 x 1000 man - days = 250 man - days job is done now the balance job = 1000 - 250 = 750 man - days worth of job since 2 men are fired so b / l men = 8 theref... | a ) q = 60 , b ) q = 70 , c ) q = 75 , d ) 80 , e ) 100 | c | divide(multiply(divide(multiply(10, 20), const_0_25), subtract(const_1, const_0_25)), subtract(10, 2)) | multiply(n1,n2)|subtract(const_1,const_0_25)|subtract(n1,n3)|divide(#0,const_0_25)|multiply(#3,#1)|divide(#4,#2) | physics |
find the value of c from ( 8 ) ^ 3 x 9 ^ 3 ã · 679 = c . | given exp . = ( 8 ) ^ 3 x 9 ^ 3 ã · 679 = c = 512 x 729 ã · 679 = 549.7 c | a ) 249.7 , b ) 529.7 , c ) 549.7 , d ) 594.7 , e ) 119.7 | c | divide(multiply(power(8, 3), power(9, 3)), 679) | power(n0,n1)|power(n2,n1)|multiply(#0,#1)|divide(#2,n4) | general |
a man has $ 10,000 to invest . he invests $ 4000 at 5 % and $ 3500 at 4 % . in order to have a yearly income of $ 500 , he must invest the remainder at | income from $ 4000 at 5 % in one year = $ 4000 of 5 % . = $ 4000 × 5 / 100 . = $ 4000 × 0.05 . = $ 200 . income from $ 3500 at 4 % in one year = $ 3500 of 4 % . = $ 3500 × 4 / 100 . = $ 3500 × 0.04 . = $ 140 . total income from 4000 at 5 % and 3500 at 4 % = $ 200 + $ 140 = $ 340 . remaining income amount in order to ha... | a ) 6 % , b ) 6.1 % , c ) 6.2 % , d ) 6.3 % , e ) 6.4 % | e | divide(subtract(subtract(500, multiply(divide(4000, const_100), 5)), multiply(divide(3500, const_100), 4)), divide(subtract(subtract(multiply(const_100, const_100), 4000), 3500), const_100)) | divide(n1,const_100)|divide(n3,const_100)|multiply(const_100,const_100)|multiply(n2,#0)|multiply(n4,#1)|subtract(#2,n1)|subtract(n5,#3)|subtract(#5,n3)|divide(#7,const_100)|subtract(#6,#4)|divide(#9,#8) | gain |
a train speeds past a pole in 11 seconds and a platform 120 m long in 22 seconds . its length is ? | let the length of the train be x meters and its speed be y m / sec . they , x / y = 11 = > y = x / 11 x + 120 / 22 = x / 11 x = 120 m . answer : d | a ) 128 , b ) 177 , c ) 199 , d ) 120 , e ) 150 | d | multiply(120, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0) | physics |
a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year 10 % of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals 180 , while 3 / 4 of all his trees are pure fuji . how many of his trees are pure gala ? | "let f = pure fuji , g = pure gala and c - cross pollinated . c = 10 % of x where x is total trees . c = . 1 x also 3 x / 4 = f and c + f = 180 = > . 1 x + 3 / 4 x = 180 = > x = 211 211 - 180 = pure gala = 31 . a" | a ) 31 , b ) 33 , c ) 55 , d ) 77 , e ) 88 | a | subtract(divide(180, add(divide(10, const_100), divide(3, 4))), 180) | divide(n0,const_100)|divide(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,n1)| | general |
the banker â € ™ s gain on a sum due 3 years hence at 12 % per annum is rs . 90 . the banker â € ™ s discount is | "solution t . d = ( b . g x 100 / r x t ) = rs . ( 90 x 100 / 12 x 3 ) = rs . 250 . b . d = rs ( 250 + 90 ) = rs . 340 . answer a" | a ) rs . 340 , b ) rs . 840 , c ) rs . 1020 , d ) rs . 760 , e ) none | a | add(divide(multiply(90, const_100), multiply(3, 12)), 90) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2)| | gain |
in the quadratic equation x 2 - bx + 20 = 0 , if the product of two roots is 20 , what is the sum of the two roots ? | explanation : the sum of the roots of the quadratic equation ax 2 + ax + c = 0 are ( - b / a ) and the product of the roots are ( c / a ) . thus , in the equation x 2 - ax + 20 = 0 , where a = 1 , b = - a and c = 20 . we get , product of the roots = 20 sum of the roots = - ( - 12 ) / 1 = 12 answer : b | a ) 20 , b ) 12 , c ) 16 , d ) 14 , e ) 18 | b | add(divide(20, 2), 2) | divide(n1,n0)|add(n0,#0) | general |
a carpenter worked alone for 1 day on a job that would take him 7 more days to finish . he and another carpenter completed the job in 4 more days . how many days would it have taken the second carpenter to do the complete job working alone ? | "a carpenter worked only 1 day on something that takes him 7 more days . means ; carpenter finishes his work in 8 days . let his buddy finish the same task in x days . respective rates per day : 1 / 8 and 1 / x to complete 1 work : first guy worked for 5 days @ rate = 1 / 8 per day . second one worked for 4 days @ rate... | a ) 16 , b ) 7 , c ) 9 , d ) 14 , e ) 24 | a | divide(multiply(4, add(1, 7)), const_2) | add(n0,n1)|multiply(n2,#0)|divide(#1,const_2)| | physics |
in a certain company 20 % of the men and 40 % of the women attended the annual company picnic . if 55 % of all the employees are men . what % of all the employee went to the picnic ? | "total men in company 55 % means total women in company 45 % ( assume total people in company 100 % ) no of men employees attended picnic = 55 x ( 20 / 100 ) = 11 no of women employees attend picnic = 45 x ( 40 / 100 ) = 18 total percentage of employees attend the picnic = 11 + 18 = 29 % answer : a" | a ) 29 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 37 % | a | multiply(add(multiply(divide(55, const_100), divide(20, const_100)), multiply(divide(subtract(const_100, 55), const_100), divide(40, const_100))), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n2)|divide(#3,const_100)|multiply(#0,#1)|multiply(#4,#2)|add(#5,#6)|multiply(#7,const_100)| | gain |
fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 7 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 18... | "you know that fox jeans costs $ 15 , and pony jeans costs $ 18 , you also know that 3 pairs of fox jeans and 2 pairs of pony jeans were purchased . so 3 ( 15 ) = 45 - fox 2 ( 18 ) = 36 - pony the total discount discount is $ 3 and you are asked to find the percent discount of pony jeans , so 45 ( 18 - x ) / 100 + 36 (... | a ) 9 % , b ) 10 % , c ) 11 % , d ) 12 % , e ) 15 % | e | multiply(subtract(divide(18, const_100), divide(subtract(7, multiply(divide(18, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100) | divide(n6,const_100)|multiply(n1,n5)|multiply(n0,n4)|multiply(#0,#1)|subtract(#2,#1)|subtract(n2,#3)|divide(#5,#4)|subtract(#0,#6)|multiply(#7,const_100)| | gain |
how many of the positive factors of 42 are not factors of 56 ? | "factors of 42 - 1 , 2,3 , 6,7 , 14,42 factors of 56 - 1,2 , 4,7 , 8,14 , 28,56 comparing both , we have three factors of 42 which are not factors of 56 - 3 , 6,42 . the answer is 3 . pls check your options again - d $ e are same . answer : c" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(56, 42) | divide(n1,n0)| | other |
if taxi fares were $ 10.00 for the first 1 / 5 mile and $ 1.00 for each 1 / 5 mile there after , then the taxi fare for a 10 - mile ride was | "in 10 miles , initial 1 / 5 mile charge is $ 10 rest of the distance = 10 - ( 1 / 5 ) = 49 / 5 rest of the distance charge = 49 ( 1.00 ) = $ 49.00 ( as the charge is 1.00 for every 1 / 5 mile ) = > total charge for 4 miles = 10 + 49.00 = 59.00 answer is d" | a ) $ 55.10 , b ) $ 59.20 , c ) $ 59.40 , d ) $ 59.00 , e ) $ 55.00 | d | add(10.00, multiply(subtract(divide(10.00, divide(1, 5)), 1), 1.00)) | divide(n1,n2)|divide(n0,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)| | general |
two cubes of tire sides ratio 2 : 3 . find its cube volumes ratio ? | a ( power 3 ) : b ( power 3 ) = 2 ( power 3 ) : 3 ( power 3 ) = 8 : 27 answer is d . | ['a ) 2 : 27', 'b ) 3 : 27', 'c ) 1 : 27', 'd ) 8 : 27', 'e ) 5 : 27'] | d | divide(volume_cube(2), volume_cube(3)) | volume_cube(n0)|volume_cube(n1)|divide(#0,#1) | geometry |
the market value of a 10.5 % stock , in which an income of rs . 756 is derived by investing rs . 7000 , brokerage being 1 / 4 % , is : | "face value = rs . 7000 . dividend = 10.5 % . annual income = rs . 756 . brokerage per rs . 100 = rs . 0.25 . dividend is always paid on the face value of a share . face value * dividend / ( market value + brokerage per rs . 100 ) = annual income . = 7000 * 10.5 / 756 = market value of rs . 100 stock + brokerage per rs... | a ) 96.97 , b ) 114 , c ) 114.75 , d ) 124 , e ) 124.75 | a | subtract(multiply(divide(7000, 756), 10.5), divide(1, 4)) | divide(n2,n1)|divide(n3,n4)|multiply(n0,#0)|subtract(#2,#1)| | gain |
a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 3600 . with the help of c , they completed the work in 3 days . how much is to be paid to c ? | c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 3 - 7 / 24 = 1 / 24 a ' s wages : b ' s wages : c ' s wages = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 c ' s share ( for 3 days ) = rs . ( 3 * 1 / 24 * 3600 ) = rs . 450 answer = b | a ) s . 375 , b ) s . 450 , c ) s . 600 , d ) s . 800 , e ) s . 850 | b | multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 3600), 3) | inverse(n3)|inverse(n1)|inverse(n0)|add(#1,#2)|subtract(#0,#3)|multiply(n2,#4)|multiply(n3,#5) | physics |
in town x , 60 percent of the population are employed , and 15 percent of the population are employed males . what percent of the employed people in town x are females ? | "total employed people 60 % , out of which 15 are employed males , hence 45 % are employed females . ( employed females ) / ( total employed people ) = 45 / 60 = 3 / 4 = 75 % answer : b ." | a ) 16 % , b ) 75 % , c ) 32 % , d ) 40 % , e ) 52 % | b | multiply(divide(subtract(60, 15), 60), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
according to a recent student poll , 5 / 6 out of 30 members of the finance club are interested in a career in investment banking . if two students are chosen at random , what is the probability that at least one of them is interested in investment banking ? | 25 students are interested , 5 are not interested prob = 1 - 5 c 2 / 30 c 2 = 1 - ( 5 * 4 / ( 30 * 29 ) ) = 1 - 20 / 580 = 1 - 1 / 29 = 28 / 29 answer : d | a ) 1 / 14 , b ) 4 / 49 , c ) 2 / 7 , d ) 28 / 29 , e ) 13 / 14 | d | divide(subtract(choose(30, const_2), choose(subtract(30, multiply(30, divide(5, 6))), const_2)), choose(30, const_2)) | choose(n2,const_2)|divide(n0,n1)|multiply(n2,#1)|subtract(n2,#2)|choose(#3,const_2)|subtract(#0,#4)|divide(#5,#0) | gain |
for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 80 ? | "4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 80 4 n < 73 n < 18 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 18 ) combining the two - 1.5 < n < 18.25 n = 1 to 18 ( 18 integers ) and... | a ) 24 , b ) 26 , c ) 20 , d ) 22 , e ) 18 | c | subtract(floor(divide(subtract(80, 7), 4)), floor(divide(subtract(1, 7), 4))) | subtract(n3,n1)|subtract(n2,n1)|divide(#0,n0)|divide(#1,n0)|floor(#2)|floor(#3)|subtract(#4,#5)| | general |
three cubes of metal whose edges are 5 , 12 and 15 cm respectively , are melted and one new cube is made . find the edge of the new cube ? | "53 + 123 + 153 = a 3 = > a = 17.3 answer : c" | a ) 28 , b ) 77 , c ) 17.3 , d ) 18 , e ) 99 | c | power(add(add(volume_cube(5), volume_cube(12)), volume_cube(15)), const_0_33) | volume_cube(n0)|volume_cube(n1)|volume_cube(n2)|add(#0,#1)|add(#3,#2)|power(#4,const_0_33)| | physics |
a rectangular pig farm has a fence along three sides and a wall along the fourth side . the fenced side opposite the wall is twice the length of each of the other two fenced sides . if the area of the rectangular region is 1250 square feet , what is the total length of the fence , in feet ? | two sides each = x the third = 2 x and the wall length is thus 2 x too x * 2 x = 2 x ^ 2 = 1250 ie x ^ 2 = 625 ie x = 25 l = 50 w = 25 total length of fence = 2 * 25 + 50 = 100 my answer is c | ['a ) 25', 'b ) 50', 'c ) 100', 'd ) 150', 'e ) 200'] | d | rectangle_perimeter(sqrt(divide(1250, const_2)), multiply(sqrt(divide(1250, const_2)), const_2)) | divide(n0,const_2)|sqrt(#0)|multiply(#1,const_2)|rectangle_perimeter(#2,#1) | geometry |
the ratio of pens to pencils is 5 to 6 . there are 6 more pencils than pens . how many pencils are there ? | "let the number of pens be 5 x and the number of pencils be 6 x . 6 x - 5 x = 6 x = 6 the number of pencils is 36 . the answer is b ." | a ) 32 , b ) 36 , c ) 40 , d ) 44 , e ) 48 | b | add(multiply(6, 5), 6) | multiply(n0,n2)|add(n2,#0)| | other |
there is a clock which shows false timing ie it shows 10 minutes less for each 1 hour . . . . if the clock shows 12 noon now how many minutes will be lost when it shows 3 pm ? | for 1 hour , 10 minutes is slower , then for 06 miuntes - 1 minutes is slower 30 minutes - 5 minutes actual time false time 12 12 01 pm 12.50 pm ( 10 minutes slower ) 02 pm 01.40 pm ( 12.50 + 60 min - 10 minutes slower ) 03 pm 02.30 pm ( 01.40 + 60 min - 10 minutes slower ) 03.36 pm 03.00 pm ( 02.30 + 36 min - 6 minute... | a ) 16 minutes , b ) 26 minutes , c ) 36 minutes , d ) 46 minutes , e ) 56 minutes | c | multiply(subtract(divide(multiply(3, const_60), subtract(const_60, 10)), 3), const_60) | multiply(n3,const_60)|subtract(const_60,n0)|divide(#0,#1)|subtract(#2,n3)|multiply(#3,const_60) | physics |
in a box of 7 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | "first , there are 5 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 7 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 5 c 3 / 7 c 3 = 10 / 35 = 2 / 7 answer is d" | a ) 1 / 2 , b ) 1 / 5 , c ) 2 / 3 , d ) 2 / 7 , e ) 1 / 7 | d | divide(divide(factorial(subtract(7, 2)), multiply(factorial(subtract(subtract(7, 2), 3)), factorial(3))), divide(factorial(7), multiply(factorial(subtract(7, 3)), factorial(3)))) | factorial(n2)|factorial(n0)|subtract(n0,n1)|subtract(n0,n2)|factorial(#2)|factorial(#3)|subtract(#2,n2)|factorial(#6)|multiply(#5,#0)|divide(#1,#8)|multiply(#7,#0)|divide(#4,#10)|divide(#11,#9)| | general |
the mean of 50 observations was 36 . it was found later that an observation 90 was wrongly taken as 23 . the corrected new mean is ? | "correct sum = ( 36 * 50 + 90 - 23 ) = 1867 . correct mean = 1825 / 50 = 37.3 answer : a" | a ) 37.3 , b ) 36.1 , c ) 36.5 , d ) 36.9 , e ) 36.3 | a | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
how many kilograms of pure water is to be added to 100 kilograms of a 30 % saline solution to make it a 10 % saline solution . | let x be the weights , in kilograms , of pure water to be added . let y be the weight , in kilograms , of the 10 % solution . hence x + 100 = y x = 200 kilograms . answer b | a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | b | subtract(divide(multiply(divide(30, const_100), 100), divide(10, const_100)), 100) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,n0) | other |
there is a escalator and 2 persons move down it . a takes 50 steps and b takes 75 steps while the escalator is moving down . given that the time taken by a to take 1 step is equal to time taken by b to take 3 steps . find the no . of steps in the escalator while it is staionary ? | if a takes 1 step in one second , then b takes 3 steps in one second . if a takes t 1 seconds to take 50 steps , then b takes 150 steps in t 1 seconds . for b , to take 150 steps he requires t 1 seconds , then to take 75 steps he requires t 1 / 2 seconds . now use relative velocity l - v * t 1 = 50 ( for first person )... | a ) 60 , b ) 70 , c ) 80 , d ) 90 , e ) 100 | e | add(75, divide(75, 3)) | divide(n2,n4)|add(n2,#0) | general |
at what price must an article costing rs . 57 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 = 57 sp = 57 * ( 125 / 100 ) = 71.25 mp * ( 95 / 100 ) = 71.25 mp = 75 answer : e" | a ) 62.5 , b ) 69.5 , c ) 62.3 , d ) 62.1 , e ) 75 | e | divide(multiply(add(57, divide(multiply(57, 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 taxi fares were $ 1.00 for the first 1 / 5 mile and $ 0.20 for each 1 / 5 mile there after , then the taxi fare for a 3 - mile ride was | "in 3 miles , initial 1 / 5 mile charge is $ 1 rest of the distance = 3 - ( 1 / 5 ) = 14 / 5 rest of the distance charge = 14 ( 0.2 ) = $ 2.8 ( as the charge is 0.2 for every 1 / 5 mile ) = > total charge for 3 miles = 1 + 2.8 = 3.8 answer is c ." | a ) $ 1.56 , b ) $ 2.40 , c ) $ 3.80 , d ) $ 4.20 , e ) $ 2.80 | c | add(1.00, multiply(subtract(divide(1.00, divide(1, 5)), 1), 0.20)) | divide(n1,n2)|divide(n0,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)| | general |
if 3 ^ x * 4 ^ y = 59,049 and x – y = 10 , then x = ? | "since 59,049 is an odd number , it is not a multiple of 4 . this means that y must equal zero . since x - y = 10 and y = 0 , then x = 10 . the answer is a ." | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | a | multiply(10, const_1) | multiply(n3,const_1)| | general |
a train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 18 seconds . what is the length of the train in meters ? | "train overtakes a bike means that we are talking about total length of the train . ( train ' s head is close to bike when it started and its tail crosses the bike when it overtakes the bike ) relative speed = 100 - 64 = 36 km / h = 36000 m / h time = 18 seconds distance = speed * time 36000 * 18 / 3600 = 180 meters . ... | a ) 180 meters , b ) 1111 meters , c ) 1777 meters , d ) 1822 meters , e ) none of these | a | multiply(multiply(subtract(100, 64), const_0_2778), 18) | subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(n2,#1)| | physics |
from january 1 , 2015 , to january 1 , 2017 , the number of people enrolled in health maintenance organizations increased by 11 percent . the enrollment on january 1 , 2017 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 2015 ? | "soln : - 11 x = 45 - - > 10 / 9 * x = 45 - - > x = 45 * 9 / 10 = 1500 / 37 = ~ 41 . answer : d ." | a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | d | multiply(divide(const_100, add(const_100, 11)), 45) | add(n4,const_100)|divide(const_100,#0)|multiply(n7,#1)| | gain |
when you multiply a number by 4 and then subtract 3 , the result is the same as if you first subtracted 7 from the same number and then multiplied by 9 . what is the number ? | let x equal the number and then change the words into an equation and solve for x : 4 x – 3 = 9 ( x – 7 ) 4 x – 3 = 9 x – 63 60 = 5 x 12 = x correct answer d ) 12 | a ) a ) 23 , b ) b ) 13 , c ) c ) 19 , d ) d ) 12 , e ) e ) 10 | d | divide(subtract(multiply(9, 7), 3), subtract(9, 4)) | multiply(n2,n3)|subtract(n3,n0)|subtract(#0,n1)|divide(#2,#1) | general |
a man can row his boat with the stream at 25 km / h and against the stream in 13 km / h . the man ' s rate is ? | "ds = 25 us = 13 s = ? s = ( 25 - 13 ) / 2 = 6 kmph answer : c" | a ) 1 kmph , b ) 2 kmph , c ) 6 kmph , d ) 8 kmph , e ) 3 kmph | c | divide(subtract(25, 13), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
an alloy of zinc and copper contains the metals in the ratio 5 : 3 . the quantity of zinc to be added to 6 kg of the alloy so that the ratio of the metal may be 3 : 1 is : | explanation : in sixteen kg of alloy 10 kg of zinc and 6 kg copper is present to make the ratio 3 : 1 we must add 8 kg of zinc to make it 18 kg of zinc and 6 kg copper so answer is 8 . answer : c | a ) 2 , b ) 9 , c ) 8 , d ) 7 , e ) 4 | c | subtract(multiply(multiply(divide(3, add(5, 3)), add(const_10, 6)), 3), multiply(divide(5, add(5, 3)), add(const_10, 6))) | add(n2,const_10)|add(n0,n1)|divide(n1,#1)|divide(n0,#1)|multiply(#0,#2)|multiply(#0,#3)|multiply(n1,#4)|subtract(#6,#5) | other |
what is the cp of rs 100 stock at 5 discount , with 1 / 5 % brokerage ? | "explanation : use the formula , cp = 100 â € “ discount + brokerage % cp = 100 - 5 + 1 / 5 95.2 thus the cp is rs 95.2 . answer : c" | a ) 96.9 , b ) 96.3 , c ) 95.2 , d ) 96.7 , e ) 96.21 | c | add(subtract(100, 5), divide(1, 5)) | divide(n2,n3)|subtract(n0,n1)|add(#0,#1)| | gain |
a monkey start climbing up a tree 21 ft tall . each hour it hops 3 ft and slips back 2 ft . how much time would it take the monkey to reach the top . | "if monkey hops 3 ft and slips back 2 ft in a hour , it means the monkey hops ( 3 ft - 2 ft ) = 1 ft / hr . similarly in 18 hrs it wil be 18 ft . but since the height of the tree is 21 ft , so if the monkey hops up the tree in the next hr i . e 19 th hr then it reaches at the top of the tree . hence it takes 19 hrs for... | a ) 15 hrs , b ) 18 hrs , c ) 19 hrs , d ) 17 hrs , e ) 16 hrs | c | subtract(divide(21, subtract(3, 2)), 2) | subtract(n1,n2)|divide(n0,#0)|subtract(#1,n2)| | physics |
john makes $ 50 a week from his job . he earns a raise andnow makes $ 75 a week . what is the % increase ? | "increase = ( 25 / 50 ) * 100 = ( 1 / 2 ) * 100 = 50 % . c" | a ) 15 % , b ) 16.66 % , c ) 50 % , d ) 19 % , e ) 21 % | c | multiply(divide(subtract(75, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 6400 ? | "6400 * 9 / 8 * 9 / 8 = 8100 answer : a" | a ) 8100 , b ) 3388 , c ) 7767 , d ) 2009 , e ) 22888 | a | add(add(6400, multiply(divide(1, 8), 6400)), multiply(divide(1, 8), add(6400, multiply(divide(1, 8), 6400)))) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)| | general |
oil cans x and y are right circular cylinders and the height and radius of y are each thrice those of x . if the oil in can x filled to capacity sells for $ 4 , how much does the oil in y sell for if y is only 1 / 3 rd filled ? | "formula for vol of a cyl is pi * r ^ 2 * h so vy = 27 * vy y when 1 / 3 rd filled will cost 9 times x so ans is c" | a ) $ 30 , b ) $ 25 , c ) $ 36 , d ) $ 40 , e ) $ 28 | c | multiply(power(4, 1), 4) | power(n0,n1)|multiply(n0,#0)| | general |
an alloy weighing 60 ounces is 20 percent copper . how many ounces of copper must be added to create an alloy that is 60 percent copper ? | "an alloy of 60 oz which is 20 % copper means there is 12 oz of gold . to get to an alloy that is 60 % copper , let ' s use this expression : ( 12 + x ) / ( 60 + x ) = 0.60 with x representing the amount of pure copper that must be added to get to 60 % . the expression we are using represents the new total weight of pu... | a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | c | divide(multiply(divide(60, const_100), 60), divide(20, const_100)) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|divide(#2,#1)| | gain |
by selling 16 pens for a rupee a woman loses 5 % . how many for a rupee should he sell in order to gain 18 % ? | d 95 % - - - 16 118 % - - - ? 95 / 118 * 16 = 13 | a ) 12 , b ) 14 , c ) 45 , d ) 13 , e ) 65 | d | inverse(add(divide(divide(const_100, subtract(const_100, 5)), 16), divide(multiply(divide(divide(const_100, subtract(const_100, 5)), 16), 18), const_100))) | subtract(const_100,n1)|divide(const_100,#0)|divide(#1,n0)|multiply(n2,#2)|divide(#3,const_100)|add(#2,#4)|inverse(#5) | gain |
a and b complete a work in 30 days . a alone can do it in 40 days . if both together can do the work in how many days ? | "1 / 30 + 1 / 40 = 0.0583 days answer : d" | a ) 1.0875 days , b ) 0.1873 days , c ) 0.0673 days , d ) 0.0583 days , e ) 0.0773 days | d | inverse(add(inverse(30), inverse(40))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)| | physics |
an art gallery has only paintings and sculptures . currently , 1 / 3 of the pieces of art are displayed , and 1 / 6 of the pieces on display are sculptures . if 1 / 3 of the pieces not on display are paintings , and 1200 sculptures are not on display , how many pieces of art does the gallery have ? | "too many words and redundant info there . ( i ) 1 / 3 of the pieces of art are displayed , hence 2 / 3 of the pieces of art are not displayed . ( ii ) 1 / 6 of the pieces on display are sculptures , hence 5 / 6 of the pieces on display are paintings . ( iii ) 1 / 3 of the pieces not on display are paintings , hence 2 ... | a ) 360 , b ) 2700 , c ) 540 , d ) 640 , e ) 720 | b | divide(divide(1200, subtract(1, divide(1, 3))), subtract(1, divide(1, 3))) | divide(n0,n1)|subtract(n0,#0)|divide(n6,#1)|divide(#2,#1)| | general |
a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 12600 after a year ? | "6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12600 = 3780 answer : e" | a ) 3630 , b ) 3637 , c ) 3630 , d ) 3631 , e ) 3780 | e | multiply(divide(6300, add(add(6300, 4200), 10500)), 12600) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)| | gain |
if taxi fares were $ 1.00 for the first 1 / 5 mile and $ 0.45 for each 1 / 5 mile there after , then the taxi fare for a 3 - mile ride was | "in 3 miles , initial 1 / 5 mile charge is $ 1 rest of the distance = 3 - ( 1 / 5 ) = 14 / 5 rest of the distance charge = 14 ( 0.45 ) = $ 6.3 ( as the charge is 0.45 for every 1 / 5 mile ) = > total charge for 3 miles = 1 + 6.3 = 7.3 answer is e ." | a ) $ 1.56 , b ) $ 2.40 , c ) $ 3.80 , d ) $ 4.20 , e ) $ 7.30 | e | add(1.00, multiply(subtract(divide(1.00, divide(1, 5)), 1), 0.45)) | divide(n1,n2)|divide(n6,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)| | general |
a and b together can do a work in 3 days . a alone can do it in 12 days . what time b will take to do the work alone ? | "explanation : a and b 1 day ' s work = 1 / 3 a alone can do 1 day ' s work = 1 / 12 what time b will take to do the work alone ? b = ( a + b ) - a = ( 1 / 3 - ( 1 / 12 ) = 4 days answer : option a" | a ) 4 days , b ) 8 days , c ) 12 days , d ) 10 days , e ) 5 days | a | add(inverse(subtract(divide(const_1, 3), divide(const_1, 12))), divide(const_2, add(const_2, const_3))) | add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|divide(const_2,#0)|subtract(#1,#2)|inverse(#4)|add(#3,#5)| | physics |
the distance from city a to city b is 180 miles . while driving from city a to city b , cara drives at a constant speed of 30 miles per hour . dan leaves city a 120 minutes after cara . what is the minimum constant speed in miles per hour that dan must exceed in order to arrive in city b before cara ? | "the time it takes cara to drive to city b is 180 / 30 = 6 hours . dan needs to take less than 4 hours for the trip . dan needs to exceed a constant speed of 180 / 4 = 45 miles per hour . the answer is b ." | a ) 42 , b ) 45 , c ) 46 , d ) 48 , e ) 50 | b | divide(180, subtract(divide(180, 30), divide(120, const_60))) | divide(n0,n1)|divide(n2,const_60)|subtract(#0,#1)|divide(n0,#2)| | physics |
of the 600 residents of clermontville , 35 % watch the television show island survival , 40 % watch lovelost lawyers and 50 % watch medical emergency . if all residents watch at least one of these 3 shows and 18 % watch exactly 2 of these shows , then how many clermontville residents c watch all of the shows ? | oa is d . 100 = a + b + c - ab - ac - bc + abc , which is the same as the following formula 100 = a + b + c + ( - ab - ac - bc + abc + abc + abc ) - 2 abc . the term between parantheses value 18 % so the equation to resolve is 100 = 35 + 40 + 50 - 18 - 2 abc therefore the value of abc is 3.5 % of 600 , c is 21 . d is t... | a ) 150 , b ) 108 , c ) 42 , d ) 21 , e ) - 21 | d | divide(subtract(subtract(add(add(multiply(divide(35, const_100), 600), multiply(divide(50, const_100), 600)), multiply(divide(40, const_100), 600)), 600), multiply(divide(18, const_100), 600)), 2) | divide(n1,const_100)|divide(n3,const_100)|divide(n2,const_100)|divide(n5,const_100)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|multiply(n0,#3)|add(#4,#5)|add(#8,#6)|subtract(#9,n0)|subtract(#10,#7)|divide(#11,n6) | gain |
find the lowest positive integer that can be divided by all the even numbers between 14 and 21 . | the even numbers between 14 and 21 are 1416 , 1820 . find the lcm of these numbers lcm = 2 ^ 4 * 7 * 3 ^ 2 * 5 the answer is b ) | a ) 4581 , b ) 5040 , c ) 3682 , d ) 1231 , e ) 1563 | b | multiply(lcm(lcm(const_10, add(add(divide(14, const_2), const_1), const_1)), divide(14, const_2)), add(divide(14, const_2), const_1)) | divide(n0,const_2)|add(#0,const_1)|add(#1,const_1)|lcm(#2,const_10)|lcm(#0,#3)|multiply(#1,#4) | general |
by combining number of small cubes , a larger cube is produced . the ratio between volume of larger cube and small cube is 1000 : 8 then how many small cubes are required to produce a larger cube ? | the ratio is a 13 : a 23 = 1000 : 8 then number small cubes requiered is 1000 / 8 = 125 so the answre is d | ['a ) 100', 'b ) 2', 'c ) 5', 'd ) 125', 'e ) 80'] | d | divide(1000, 8) | divide(n0,n1) | geometry |
jacob is now 12 years younger than michael . if 3 years from now michael will be twice as old as jacob , how old will jacob be in 4 years ? | "jacob = x years , mike = x + 12 years 3 years from now , 2 ( x + 3 ) = x + 15 2 x + 6 = x + 15 x = 9 x + 4 = 13 years answer e" | a ) 3 , b ) 7 , c ) 15 , d ) 21 , e ) 13 | e | add(4, subtract(add(12, 3), multiply(const_2, 3))) | add(n0,n1)|multiply(n1,const_2)|subtract(#0,#1)|add(n2,#2)| | general |
the area of a triangle will be when a = 2 m , b = 5 m , c = 7 m , a , b , c being lengths of respective sides ? | "s = ( 2 + 5 + 7 ) / 2 = 7 answer : b" | a ) 2 , b ) 7 , c ) 5 , d ) 3 , e ) 4 | b | divide(add(add(2, 5), 7), 5) | add(n0,n1)|add(n2,#0)|divide(#1,n1)| | geometry |
in a certain circle there are 10 points . what is the number of the triangles connecting 2 points of the 10 points ? | "imo : b here we have to select 2 points out of 10 points . order is not important so the answer will be 10 c 2 = 45 answer b" | a ) 40 , b ) 45 , c ) 108 , d ) 120 , e ) 132 | b | divide(factorial(10), multiply(factorial(divide(const_10, const_2)), factorial(2))) | divide(const_10,const_2)|factorial(n0)|factorial(n1)|factorial(#0)|multiply(#3,#2)|divide(#1,#4)| | geometry |
the length of the bridge , which a train 140 m long and traveling at 45 km / hr can cross in 30 sec is ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec . time = 30 sec let the length of bridge be x meters . then , ( 140 + x ) / 30 = 25 / 2 x = 235 m . answer : option a" | a ) 235 , b ) 240 , c ) 245 , d ) 250 , e ) 255 | a | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 140) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
a 8000 liter tank , half - full of water is being filled from a pipe with a flow rate of 1 kiloliter every 2 minutes . at the same time , the tank is losing water from two drains at a rate of 1 kiloliter every 4 minutes and every 6 minutes . how many minutes does it take to fill the tank completely ? | in : we have : 1,000 / 2 min = 500 litres per minute out : we have : 1,000 / 4 + 1,000 / 6 then do : in - out to figure out the net inflow per minute ( you get 83.3 ) . then divide the total number of litres you need ( 4,000 by that net inflow to get the minutes ) - 48 min . answer e . | a ) 8 , b ) 12 , c ) 18 , d ) 24 , e ) 48 | e | divide(multiply(add(add(6, 4), 2), divide(8000, 2)), const_1000) | add(n4,n5)|divide(n0,n2)|add(n2,#0)|multiply(#2,#1)|divide(#3,const_1000) | physics |
julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ 120 and $ 124 from the simple interest account and the compound interest account respectively . if the interest rates for both account... | "$ 120 for 2 years = $ 60 per year . extra $ 4 yearned with the compound interest is the percent yearned on percent . so , $ 4 is yearned on $ 60 , which means that the interest = 15 % . this on the other hand means that half of the savings = 60 * 15 = $ 900 . twice of that = $ 1,800 . answer : e ." | a ) 600 , b ) 720 , c ) 1080 , d ) 1200 , e ) 1800 | e | divide(120, divide(multiply(const_2, subtract(124, 120)), 120)) | subtract(n1,n0)|multiply(#0,const_2)|divide(#1,n0)|divide(n0,#2)| | gain |
abcd is a square where ab = â ˆ š 6000 . let x be a point on ab and y be a point on cd such that ax = cy . compute the area of trapezoid axyd . | "note that trapezoids axy d and bxy c are congruent , so the area of axy d is always 6000 / 2 = 3000 . correct answer e" | a ) 3008 , b ) 2002 , c ) 1008 , d ) 2016 , e ) 3000 | e | divide(6000, const_2) | divide(n0,const_2)| | geometry |
mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8410 / - . find out the amount that he borrowed . | explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8410 = p + ( 54 p / 100 ) 8410 = ( 10... | a ) s . 5266 , b ) s . 5269 , c ) s . 5461 , d ) s . 5218 , e ) s . 52192 | c | divide(8410, add(const_1, divide(multiply(6, 9), const_100))) | multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n3,#2) | gain |
the perimeter of a semi circle is 108 cm then the radius is ? | "36 / 7 r = 108 = > r = 21 answer : b" | a ) 17 , b ) 21 , c ) 19 , d ) 11 , e ) 12 | b | divide(108, add(const_2, const_pi)) | add(const_2,const_pi)|divide(n0,#0)| | physics |
the length of a rectangle is one - sixth of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 1296 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | given that the area of the square = 1225 sq . units = > side of square = √ 1296 = 36 units the radius of the circle = side of the square = 35 units length of the rectangle = 2 / 5 * 35 = 14 units given that breadth = 10 units area of the rectangle = lb = 36 * 10 = 360 sq . units answer : b | ['a ) 560 sq . units', 'b ) 360 sq . units', 'c ) 260 sq . units', 'd ) 460 sq . units', 'e ) 160 sq . units'] | b | multiply(sqrt(1296), 10) | sqrt(n0)|multiply(n1,#0) | geometry |
the sum of first n consecutive odd integers is n ^ 2 . what is the sum of all odd integers between 13 and 35 inclusive . | "we ' re dealing with a sequence of consecutive odd integers : 13 to 35 , inclusive . we ' re asked for the sum of this group . 1 ) start with the sum of the smallest and the biggest : 13 + 35 = 48 2 ) now look at the ' next smallest ' and the ' next biggest ' : 15 + 33 = 48 now we have proof that there is no middle te... | a ) 288 , b ) 364 , c ) 410 , d ) 424 , e ) 450 | a | subtract(power(divide(add(35, const_1), 2), 2), power(divide(add(subtract(13, 2), const_1), 2), 2)) | add(n2,const_1)|subtract(n1,n0)|add(#1,const_1)|divide(#0,n0)|divide(#2,n0)|power(#3,n0)|power(#4,n0)|subtract(#5,#6)| | general |
the function f ( n ) is defined as the product of all the consecutive positive integers between 4 and n ^ 2 , inclusive , whereas the function g ( n ) is defined as the product of the squares of all the consecutive positive integers between 1 and n , inclusive . the exponent on 2 in the prime factorization of f ( 3 ) /... | f ( 3 ) / g ( 3 ) = product ( 1 to 3 ^ 2 ) / 1.2 ^ 2.3 ^ 2 = 1 . 2.3 . 4.5 . 6.7 . 8.9 / 1 . 4.9 = 1 . 2.3 . ( 2 ^ 2 ) . 5 . ( 2.3 ) . 7 . ( 2 ^ 3 ) . 9 / 1 . ( 2 ^ 2 ) . 9 = 1 . ( 2 ^ 7 ) . 3.5 . 7.9 / 1 . ( 2 ^ 2 ) . 9 loof for 2 ^ 7 / 2 ^ 2 = 2 ^ 5 - - - - exponent 4 answer : d | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | power(2, 2) | power(n1,n1) | general |
in a shower , 5 cm of rain falls . the volume of water that falls on 1.5 hectares of ground is : | solution area = ( 1.5 × 10000 ) m 2 = 45000 m 2 . depth = m = m . ∴ volume = ( area × depth ) = m 3 = 750 m 3 . answer b | ['a ) 75 cu . m', 'b ) 750 cu . m', 'c ) 7500 cu . m', 'd ) 75000 cu . m', 'e ) none of these'] | b | multiply(divide(5, const_100), multiply(multiply(1.5, const_1000), const_10)) | divide(n0,const_100)|multiply(n1,const_1000)|multiply(#1,const_10)|multiply(#0,#2) | physics |
for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 200 ? | from first constraint 4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 200 4 n < 193 n < 48 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 48 ) combining the two - 1.5 < n < 48.25 n = 1 to... | a ) 48 , b ) 49 , c ) 50 , d ) 51 , e ) 52 | c | subtract(floor(divide(subtract(200, 7), 4)), floor(divide(subtract(1, 7), 4))) | subtract(n3,n1)|subtract(n2,n1)|divide(#0,n0)|divide(#1,n0)|floor(#2)|floor(#3)|subtract(#4,#5) | general |
in the first 10 overs of a cricket game , the run rate was only 2.1 . what should be the rate in the remaining 30 overs to reach the target of 282 runs ? | "required run rate = [ 282 - ( 2.1 * 10 ) ] / 30 = 261 / 30 = 8.70 answer : d" | a ) 6.25 , b ) 8.28 , c ) 6.11 , d ) 8.7 , e ) 8.11 | d | divide(subtract(282, multiply(10, 2.1)), 30) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
5216 x 51 = ? | "normal way of multiplication may take time . here are one alternative . 5216 x 51 = ( 5216 x 50 ) + 5216 = ( 5216 x 100 ) / 2 + 5216 = 521600 / 2 + 5216 = 260800 + 5216 = 266016 answer is b" | a ) 212016 , b ) 266016 , c ) 266436 , d ) 216314 , e ) 216318 | b | multiply(divide(5216, 51), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a small pool filled only with water will require an additional 300 gallons of water in order to be filled to 70 % of its capacity . if pumping in these additional 300 gallons of water will increase the amount of water in the pool by 30 % , what is the total capacity of the pool in gallons ? | "300 gallons of water increases capacity by 30 % that means 30 % is 300 gallons , so 100 % would be = 300 * 100 / 30 = 1000 gallons now 1000 + 300 gallons is 70 % capacity of tank . so 100 % capacity would be = 1300 * 100 / 70 = 1857.14 e is the answer" | a ) 1000 , b ) 1250 , c ) 1300 , d ) 1600 , e ) 1857.14 | e | divide(add(divide(multiply(300, const_100), 30), 300), divide(70, const_100)) | divide(n1,const_100)|multiply(n0,const_100)|divide(#1,n3)|add(n0,#2)|divide(#3,#0)| | general |
a car takes 6 hours to cover a distance of 720 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distance = 720 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 720 / 9 = 80 kmph answer c ." | a ) 60 , b ) 50 , c ) 80 , d ) 70 , e ) 65 | c | divide(720, divide(multiply(6, 3), 2)) | multiply(n0,n2)|divide(#0,n3)|divide(n1,#1)| | physics |
by selling 15 pencils for a rupee a man loses 15 % . how many for a rupee should he sell in order to gain 15 % ? | "85 % - - - 15 115 % - - - ? 15 / 115 * 15 = 11 answer : c" | a ) 8 , b ) 9 , c ) 11 , d ) 89 , e ) 81 | c | multiply(divide(const_1, multiply(add(const_100, 15), divide(const_1, subtract(const_100, 15)))), 15) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain |
the consumption of diesel per hour of a bus varies directly as square of its speed . when the bus is travelling at 50 kmph its consumption is 1 litre per hour . if each litre costs $ 50 and other expenses per hous is $ 50 , then what would be the minimum expenditure required to cover a distance of 500 km ? | "50 kmph consumption is 1 lt / hr so 500 km will take 10 hrs and the consumption is 10 lt for entire distance . 1 lt costs $ 50 so 10 lt costs $ 500 extra expenses for 1 hr - $ 50 10 hrs - $ 500 total expense - $ 500 + $ 500 = $ 1000 answer : d" | a ) 800 , b ) 950 , c ) 900 , d ) 1000 , e ) 1250 | d | add(multiply(divide(500, 50), 50), multiply(divide(500, 50), 50)) | divide(n4,n0)|multiply(n2,#0)|multiply(n3,#0)|add(#1,#2)| | physics |
a man can row at 5 kmph in still water . if the velocity of current is 2 kmph and it takes him 1 hour to row to a place and come back , how far is the place ? | "explanation : speed in still water = 5 kmph speed of the current = 1 kmph speed downstream = ( 5 + 2 ) = 7 kmph speed upstream = ( 5 - 2 ) = 3 kmph let the required distance be x km total time taken = 1 hour â ‡ ’ x / 7 + x / 3 = 1 â ‡ ’ 3 x + 7 x = 21 â ‡ ’ 10 x = 21 â ‡ ’ x = 2.1 km answer : option a" | a ) 2.1 km , b ) 2.9 km , c ) 3.4 km , d ) 4.4 km , e ) 5.6 km | a | divide(multiply(subtract(5, 2), const_3), 5) | subtract(n0,n1)|multiply(#0,const_3)|divide(#1,n0)| | physics |
a number is said to be prime saturated if the product of all the different positive prime factors of r is less than the square root of r . what is the greatest two digit prime saturated integer ? | clearly d a number is said to be prime saturated if the product of all the different positive prime factors of r is less than the square root of r . 96 has more number of smaller prime factor thats the clue ! ! = d | a ) 99 , b ) 98 , c ) 97 , d ) 96 , e ) 95 | d | add(multiply(const_2, const_3), subtract(const_100, const_10)) | multiply(const_2,const_3)|subtract(const_100,const_10)|add(#0,#1) | other |
a small table has a length of 12 inches and a breadth of b inches . cubes are placed on the surface of the table so as to cover the entire surface . the maximum side of such cubes is found to be 4 inches . also , a few such tables are arranged to form a square . the minimum length of side possible for such a square is ... | "from the info that the maximum sides of the cubes is 4 , we know that the gcf of 12 ( = 2 ^ 2 * 3 ) andbis 4 ( = 2 ^ 2 ) , sob = 2 ^ x , where x > = 2 . from the second premise , we know that the lcm of 12 ( 2 ^ 2 * 3 ) andbis 24 ( 2 ^ 3 * 3 ) , sob = 2 ^ 3 or 2 ^ 3 * 3 ( 8 or 24 ) . combining 2 premises shows the ans... | a ) 8 , b ) 16 , c ) 24 , d ) 32 , e ) 48 | a | sqrt(subtract(power(divide(24, 4), const_2), power(12, const_2))) | divide(n2,n1)|power(n0,const_2)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)| | geometry |
if money is invested at r percent interest , compounded annually , the amount of investment will double in approximately 70 / r years . if pat ' s parents invested $ 7000 in a long term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of investment 18 years later , whe... | "since investment doubles in 70 / r years then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 7,000 * 2 ) * 2 = $ 28,000 ( after 9 years investment will become $ 7,000 * 2 = $ 14,0... | a ) $ 28000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | a | multiply(multiply(7000, const_2), const_2) | multiply(n1,const_2)|multiply(#0,const_2)| | general |
a , b and c completed a piece of work , a worked for 6 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was rs . 1480 ? | "3 x 4 x 5 x 6 9 4 18 x + 36 x + 20 x = 1480 74 x = 1480 = > x = 20 5 x = 100 rs . answer : c" | a ) 228 , b ) 217 , c ) 100 , d ) 288 , e ) 772 | c | multiply(5, divide(1480, add(add(multiply(6, 3), multiply(9, 4)), multiply(4, 5)))) | multiply(n0,n3)|multiply(n1,n2)|multiply(n2,n5)|add(#0,#1)|add(#3,#2)|divide(n6,#4)|multiply(n5,#5)| | physics |
if 5400 mn = k ^ 4 , where m , n , and k are positive integers , what is the least possible value of m + 2 n ? | "k ^ 4 = 5400 ∗ mn k ^ 4 = 3 ^ 3.2 ^ 3.5 ^ 2 . mn in order to make rhs a perfect power of 4 , we require it to be multiplied by 3 , 2 & 5 ^ 2 mn = 3 . 2.5 ^ 2 mn = 150 = 10 * 15 ( least possible ) answer = 10 + 15 * 2 = 40 answer : e" | a ) 11 , b ) 18 , c ) 20 , d ) 25 , e ) 40 | e | multiply(4, power(const_3, 2)) | power(const_3,n2)|multiply(n1,#0)| | general |
if the area of a square with sides of length 4 centimeters is equal to the area of a rectangle with a width of 4 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 4 ^ 2 = l * 4 = > l = 16 / 4 = 4 answer d" | a ) 14 , b ) 8 , c ) 12 , d ) 4 , e ) 18 | d | divide(power(4, const_2), 4) | power(n0,const_2)|divide(#0,n1)| | geometry |
what is the average ( arithmetic mean ) of the numbers 21 , 22 , 23 , 23 , 24 , and 25 ? | "{ 21 , 22 , 23 , 23 , 24 , 25 } = { 23 - 2,23 - 1 , 23 , 23,23 + 1,23 + 2 } - - > the average = 23 . answer : e ." | a ) 23.0 , b ) 22.5 , c ) 21 , d ) 24 , e ) 23 | e | divide(divide(multiply(add(21, 23), add(divide(subtract(23, 21), 21), const_1)), const_2), add(divide(subtract(23, 21), 21), const_1)) | add(n0,n2)|subtract(n2,n0)|divide(#1,n0)|add(#2,const_1)|multiply(#0,#3)|divide(#4,const_2)|divide(#5,#3)| | general |
the average age of 28 students in a group is 16 years . when teacher â € ™ s age is included to it , the average increases by one . what is the teacher â € ™ s age in years ? | "sol . age of the teacher = ( 29 ã — 17 â € “ 28 ã — 16 ) years = 45 years . answer a" | a ) 45 , b ) 36 , c ) 41 , d ) 51 , e ) 48 | a | add(28, const_1) | add(n0,const_1)| | general |
a and b are two stations 390 km apart . a train starts from a at 10 a . m . and travels towards b at 65 kmph . another train starts from b at 11 a . m . and travels towards a at 35 kmph . at what time do they meet ? | suppose they meet x hours after 10 a . m . then , ( distance moved by first in x hrs ) + [ distance moved by second in ( x - 1 ) hrs ] = 390 . 65 x + 35 ( x - 1 ) = 390 = > 100 x = 425 = > x = 17 / 4 so , they meet 4 hrs . 15 min . after 10 a . m i . e . , at 2.15 p . m . answer a | a ) 2.15 pm , b ) 1.15 pm , c ) 4.15 pm , d ) 3.15 pm , e ) 12.15 pm | a | subtract(add(divide(add(390, 35), add(65, 35)), 10), add(10, const_2)) | add(n0,n4)|add(n2,n4)|add(n1,const_2)|divide(#0,#1)|add(n1,#3)|subtract(#4,#2) | physics |
in a group of cows and chickens , the number of legs was 16 more than twice the number of heads . the number of cows was : | "let the number of cows be x and their legs be 4 x . let the number of chicken be y and their legs be 2 x . total number of legs = 4 x + 2 y . total number of heads = x + y . the number of legs was 16 more than twice the number of heads . therefore , 2 × ( x + y ) + 16 = 4 x + 2 y . or , 2 x + 2 y + 16 = 4 x + 2 y . or... | a ) 8 , b ) 7 , c ) 10 , d ) 12 , e ) 14 | a | divide(16, subtract(const_4, const_2)) | subtract(const_4,const_2)|divide(n0,#0)| | general |
a is a working partner and b is a sleeping partner in the business . a puts in rs . 5000 and b rs . 1000 , a receives 10 % of the profit for managing the business the rest being divided in proportion of their capitals . out of a total profit of rs . 9600 , money received by a is ? | "5000 : 1000 = > 5 : 1 9600 * 10 / 100 = 960 9600 - 960 = 8640 8640 * 5 / 6 = 7200 + 960 = 8160 answer : d" | a ) 1978 , b ) 2707 , c ) 7728 , d ) 8160 , e ) 7291 | d | subtract(9600, multiply(multiply(9600, subtract(const_1, divide(10, const_100))), divide(1000, add(5000, 1000)))) | add(n0,n1)|divide(n2,const_100)|divide(n1,#0)|subtract(const_1,#1)|multiply(n3,#3)|multiply(#2,#4)|subtract(n3,#5)| | gain |
how much 90 % of 40 is greater than 80 % of 30 ? | "( 90 / 100 ) * 40 – ( 80 / 100 ) * 30 36 - 24 = 12 answer : a" | a ) 12 , b ) 14 , c ) 16 , d ) 17 , e ) 18 | a | subtract(divide(multiply(90, 40), const_100), divide(multiply(80, 30), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3)| | gain |
an auction house charges a commission of 20 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ? | "say the price of the house was $ x , then 0.2 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 190,000 ( 20 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : b ." | a ) $ 115,000 , b ) $ 190,000 , c ) $ 215,000 , d ) $ 240,000 , e ) $ 365,000 | b | add(multiply(20, 10), 10) | multiply(n0,n2)|add(n2,#0)| | general |
in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 3 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | "given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 3 ) = 36 / 12 = 3 / 1 . answer : e ." | a ) 2 / 1 , b ) 4 / 1 , c ) 2 / 3 , d ) 3 / 4 , e ) 3 / 1 | e | divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 3)) | add(n1,n2)|divide(n0,#0)|add(n3,#1)|subtract(n0,#1)|divide(#3,#2)| | general |
the compound interest earned by sunil on a certain amount at the end of two years at the rate of 8 % p . a . was rs . 2828.80 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned ? | "let the sum be rs . p p { [ 1 + 8 / 100 ] 2 - 1 } = 2828.80 p ( 8 / 100 ) ( 2 + 8 / 100 ) = 2828.80 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 2828.80 / ( 0.08 ) ( 2.08 ) = 1360 / 0.08 = 17000 principal + interest = rs . 19828.80 answer : c" | a ) 19828.29 , b ) 19828.2 , c ) 19828.8 , d ) 19828.88 , e ) 19828.21 | c | add(divide(2828.80, subtract(power(add(const_1, divide(8, const_100)), const_2), const_1)), 2828.80) | divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|divide(n1,#3)|add(n1,#4)| | gain |
a man took loan from a bank at the rate of 13 % p . a . s . i . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was ? | "principal = ( 100 * 5400 ) / ( 13 * 3 ) = rs . 13846 answer : c" | a ) rs . 1846 , b ) rs . 18846 , c ) rs . 13846 , d ) rs . 2000 , e ) rs . 14846 | c | divide(divide(multiply(5400, const_100), 3), 13) | multiply(n2,const_100)|divide(#0,n1)|divide(#1,n0)| | gain |
a sum amounts to rs . 4410 in 2 years at the rate of 5 % p . a . if interest was compounded yearly then what was the principal ? | ci = 4410 , r = 5 , n = 2 ci = p [ 1 + r / 100 ] ^ 2 = p [ 1 + 5 / 100 ] ^ 2 4410 = p [ 21 / 20 ] ^ 2 4410 [ 20 / 21 ] ^ 2 4000 answer : a | a ) rs . 4000 , b ) rs . 5000 , c ) rs . 4500 , d ) rs . 4800 , e ) rs . 5800 | a | divide(4410, power(add(divide(5, const_100), const_1), const_2)) | divide(n2,const_100)|add(#0,const_1)|power(#1,const_2)|divide(n0,#2) | gain |
a man can row downstream at 18 kmph and upstream at 8 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 18 - - - ( 1 ) and x - y = 8 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 26 = > x = 13 , y = 5 . answer : b" | a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 9 | b | divide(divide(add(18, 8), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
the compound ratio of 2 / 3 , 6 / 7 , 1 / 5 and 1 / 8 is given by ? | "2 / 3 * 6 / 7 * 1 / 5 * 1 / 8 = 12 / 1400 = 3 / 350 answer : d" | a ) 7 / 98 , b ) 1 / 48 , c ) 1 / 98 , d ) 3 / 350 , e ) 1 / 68 | d | inverse(multiply(2, 7)) | multiply(n0,n3)|inverse(#0)| | other |
how many of the positive factors of 14 are not factors of 21 ? | "factors of 14 - 1 , 2 , 7 , 14 , factors of 21 - 1 , 3 , 7 , 21 , comparing both , we have three factors of 14 which are not factors of 21 - 2,14 the answer is c" | a ) 4 , b ) 3 , c ) 1 , d ) 2 , e ) 0 | c | divide(21, 14) | divide(n1,n0)| | other |
how much water must be added to 60 litres of milk at 1.5 litres for rs . 20 so as to have a mixture worth rs . 32 / 3 a litre ? | c . p . of 1 litre of milk = rs . 20 × 2 / 3 = rs . 40 / 3 . c . p . of 1 litre of water = rs 0 . mean price = rs . 32 / 3 . by the rule of alligation , we have : c . p of 1 litre c . p of 1 litre of water of milk ( 0 ) ( rs . 40 / 3 ) \ / mean price ( rs . 32 / 3 ) / \ 40 / 3 − 32 / 3 32 / ( 3 − 0 ) 8 / 3 32 / 3 the r... | a ) 10 litres , b ) 12 litres , c ) 15 litres , d ) 18 litres , e ) 21 litres | c | divide(60, divide(divide(32, 3), subtract(divide(20, 1.5), divide(32, 3)))) | divide(n3,n4)|divide(n2,n1)|subtract(#1,#0)|divide(#0,#2)|divide(n0,#3) | general |
a cricket bat is sold for $ 850 , making a profit of $ 255 . the profit percentage would be | "255 / ( 850 - 255 ) = 255 / 595 = 51 / 119 = 43 % . answer : e ." | a ) 24 % , b ) 25 % , c ) 30 % , d ) 36 % , e ) 43 % | e | multiply(divide(255, subtract(850, 255)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain |
a car covers a distance of 624 km in 2 3 / 5 hours . find its speed ? | 624 / 2 3 / 5 = 240 kmph answer : e | a ) 104 kmph , b ) 194 kmph , c ) 109 kmph , d ) 174 kmph , e ) 240 kmph | e | divide(624, add(2, divide(3, 5))) | divide(n2,n3)|add(n1,#0)|divide(n0,#1) | physics |
three walls have wallpaper covering a combined area of 300 square meters . by overlapping the wallpaper to cover a wall with an area of 180 square meters , the area that is covered by exactly two layers of wallpaper is 30 square meters . what is the area that is covered with three layers of wallpaper ? | "300 - 180 = 120 sq m of the wallpaper overlaps ( in either two layers or three layers ) if 36 sq m has two layers , 120 - 30 = 90 sq m of the wallpaper overlaps in three layers . 90 sq m makes two extra layers hence the area over which it makes two extra layers is 45 sq m . answer ( d ) ." | a ) 5 square meters , b ) 36 square meters , c ) 42 square meters , d ) 45 square meters , e ) 120 square meters | d | divide(subtract(subtract(300, 180), 30), const_2) | subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2)| | geometry |
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