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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if the length , breadth and the height of a cuboid are in the ratio 6 : 5 : 4 and if the total surface area is 33300 cm 2 cm 2 , then the length , breadth and height in cms , are respectively . | explanation : let length = 6 x , breadth = 5 x and height = 4 x in cm 2 ( 6 x Γ 5 x + 5 x Γ 4 x + 6 x Γ 4 x ) = 33300 148 x 2 = 33300 β x 2 = 33300 / 148 = 225 β x = 15 length = 90 cm , breadth = 75 cm and height = 60 cm correct option : d | ['a ) 90 , 85,60', 'b ) 85 , 75,60', 'c ) 90 , 75,70', 'd ) 90 , 75,60', 'e ) none'] | d | multiply(6, sqrt(divide(33300, multiply(add(multiply(6, 4), add(multiply(6, 5), multiply(5, 4))), const_2)))) | multiply(n0,n1)|multiply(n1,n2)|multiply(n0,n2)|add(#0,#1)|add(#3,#2)|multiply(#4,const_2)|divide(n3,#5)|sqrt(#6)|multiply(n0,#7) | geometry |
a train covers a distance of 14 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 14 / 10 * 60 ) km / hr = ( 84 * 5 / 18 ) m / sec = 70 / 3 m / sec . length of the train = 70 / 3 * 6 = 140 m . answer : c" | a ) m , b ) m , c ) m , d ) m , e ) m | c | divide(14, subtract(divide(14, 10), 6)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
eight identical machines can produce 360 aluminum cans per hour . if all of the machines work at the same constant rate , how many cans could 5 such machines produce in 2 hours ? | "8 machines / 360 cans = 5 machines / x cans 8 x = 1800 x = 225 ( 225 ) ( 2 hours ) = 450 cans . the answer is a ." | a ) 450 , b ) 750 , c ) 1,800 , d ) 5,900 , e ) 7,500 | a | subtract(multiply(2, 360), multiply(2, divide(multiply(5, 360), add(const_4, const_4)))) | add(const_4,const_4)|multiply(n0,n2)|multiply(n0,n1)|divide(#2,#0)|multiply(n2,#3)|subtract(#1,#4)| | physics |
one hour after yolanda started walking from x to y , a distance of 31 miles , bob started walking along the same road from y to x . if yolanda ' s walking rate was 3 miles per hour and bob Ρ ' s was 4 miles per hour , how many miles had bob walked when they met ? | "when b started walking y already has covered 3 miles out of 31 , hence the distance at that time between them was 31 - 3 = 28 miles . combined rate of b and y was 3 + 4 = 7 miles per hour , hence they would meet each other in 28 / 7 = 4 hours . in 6 hours b walked 4 * 4 = 16 miles . answer : e ." | a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 16 | e | multiply(divide(subtract(31, 3), add(3, 4)), 4) | add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n2,#2)| | physics |
two persons a and b can complete a piece of work in 30 days and 45 days respectively . if they work together , what part of the work will be completed in 5 days ? | "a ' s one day ' s work = 1 / 30 b ' s one day ' s work = 1 / 45 ( a + b ) ' s one day ' s work = 1 / 30 + 1 / 45 = 1 / 18 the part of the work completed in 5 days = 5 ( 1 / 18 ) = 5 / 18 . answer a" | a ) 5 / 18 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 9 , e ) 2 / 6 | a | multiply(5, add(divide(const_1, 30), divide(const_1, 45))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)| | physics |
a retail appliance store priced a video recorder at 20 percent above the wholesale cost of $ 200 . if a store employee applied the 25 percent employee discount to the retail price to buy the recorder , how much did the employee pay for the recorder ? | wholesale cost of video recorder = 200 $ video recorder was priced at 20 percent above 200 = 240 $ % discount given by store employee = 25 emlpoyee paid = . 75 * 240 = 180 $ answer d | a ) $ 198 , b ) $ 216 , c ) $ 220 , d ) $ 180 , e ) $ 240 | d | subtract(add(200, multiply(divide(200, const_100), 20)), multiply(divide(add(200, multiply(divide(200, const_100), 20)), const_100), 25)) | divide(n1,const_100)|multiply(n0,#0)|add(n1,#1)|divide(#2,const_100)|multiply(n2,#3)|subtract(#2,#4) | gain |
789.009 - ? + 56.84 = 215.943 | "explanation : 629.906 answer : option d" | a ) 899.015 , b ) 752.804 , c ) 714.642 , d ) 629.906 , e ) none of these | d | subtract(multiply(divide(789.009, const_100), 56.84), multiply(divide(const_1, const_3), multiply(divide(789.009, const_100), 56.84))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
an seller earns an income of re 3 on the first day of his business . on every subsequent day , he earns an income which is just thrice of that made on the previous day . on the 15 th day of business , he earns an income of : | 2 nd day he earns = 3 ( 2 β 3 ) 3 rd day he earns = 3 ( 3 β 3 ) on 15 th day he earns 3 ( 15 - 3 ) = 36 rupees answer : e | a ) 19 , b ) 22 , c ) 25 , d ) 20 , e ) 36 | e | multiply(subtract(15, 3), 3) | subtract(n1,n0)|multiply(n0,#0) | physics |
how much 80 % of 40 is greater than 10 % of 15 ? | "( 80 / 100 ) * 40 β ( 10 / 100 ) * 15 32 - 15 = 17 answer : e" | a ) 18 , b ) 99 , c ) 19 , d ) 18 , e ) 17 | e | subtract(divide(multiply(80, 40), const_100), divide(multiply(10, 15), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3)| | gain |
in a certain pond , 50 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 10 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the ... | "total fish = x percentage of second catch = ( 10 / 50 ) * 100 = 20 % so , x * 20 % = 50 x = 250 ans . c" | a ) 400 , b ) 625 , c ) 250 , d ) 2,500 , e ) 10,000 | c | divide(50, divide(10, 50)) | divide(n2,n1)|divide(n0,#0)| | gain |
a can finish a work in 12 days , b in 9 days and c in 2 days , b and c start the work but are forced to leave after 3 days . the remaining work was done by a in ? | b + c 1 day work = 1 / 9 + 1 / 12 = 7 / 36 work done by b and c in 3 days = 7 / 36 * 3 = 7 / 12 remaining work = 1 - 7 / 12 = 5 / 12 1 / 24 work is done by a in 1 day 5 / 12 work is done by a in 12 * 5 / 12 = 5 days answer is e | a ) 10 days , b ) 12 days , c ) 6 days , d ) 7 days , e ) 5 days | e | multiply(divide(const_1, add(divide(const_1, 9), divide(const_1, 2))), 3) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|divide(const_1,#2)|multiply(n3,#3) | physics |
75 boys can complete a job in 24 days . how many men need to complete the job twice in 20 days | one man can complete the job in 24 * 75 = 1800 days to complete the job twice he would need 3600 days . thus , to complete the job in 20 days , 3600 / 20 = 180 men are needed . answer : c | a ) 160 , b ) 170 , c ) 180 , d ) 190 , e ) 200 | c | divide(multiply(multiply(24, const_2), 75), 20) | multiply(n1,const_2)|multiply(n0,#0)|divide(#1,n2) | physics |
on june 1 a bicycle dealer noted that the number of bicycles in stock had decreased by 4 for each of the past 5 months . if the stock continues to decrease at the same rate for the rest of the year , how many fewer bicycles will be in stock on december 1 than were in stock on january 1 ? | "jan 1 = c feb 1 = c - 4 march 1 = c - 8 april 1 = c - 12 may 1 = c - 16 june 1 = c - 20 july 1 = c - 24 aug 1 = c - 28 sept 1 = c - 32 oct 1 = c - 36 nov 1 = c - 40 dec 1 = c - 44 difference between stock on december 1 than were in stock on january 1 will be - c - ( c - 44 ) = 44 hence answer will be ( d )" | a ) 8 , b ) 12 , c ) 20 , d ) 44 , e ) 36 | d | multiply(subtract(const_10, 1), 4) | subtract(const_10,n0)|multiply(n1,#0)| | gain |
there are 760 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | "in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 760 . hence 3 x + 5 x = 760 solve for x 8 x = 760 x = 95 number of boy... | a ) 320 , b ) 345 , c ) 425 , d ) 475 , e ) 400 | d | multiply(divide(760, 5), 3) | divide(n0,n2)|multiply(n1,#0)| | other |
a contractor is engaged for 30 days on the condition that he receives rs . 25 for eachday he works & is fined rs . 7.50 for each day is absent . he gets rs . 425 in all . for how many days was heabsent ? | "30 * 25 = 750 425 - - - - - - - - - - - 325 25 + 7.50 = 32.5 325 / 32.5 = 10 b" | a ) 5 , b ) 10 , c ) 17 , d ) 25 , e ) 30 | b | subtract(30, divide(add(multiply(30, 7.50), 425), add(25, 7.50))) | add(n1,n2)|multiply(n0,n2)|add(n3,#1)|divide(#2,#0)|subtract(n0,#3)| | physics |
if a dozen of oranges cost $ 5 , what would be the cost of 10 oranges ? | the cost of one orange = 5 / 12 = 0.416 the cost of ten oranges = 0.416 x 10 = $ 4.16 answer : b | a ) 3.56 , b ) 4.16 , c ) 4.86 , d ) 5.1 , e ) 5.2 | b | divide(multiply(10, 5), add(10, const_2)) | add(n1,const_2)|multiply(n0,n1)|divide(#1,#0) | general |
the perimeter of one square is 48 cm and that of another is 36 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? | "4 a = 48 4 a = 36 a = 12 a = 9 a 2 = 144 a 2 = 81 combined area = a 2 = 225 = > a = 15 d = 15 β 2 answer : e" | a ) 13 β 4 , b ) 13 β 2 , c ) 23 β 2 , d ) 12 β 4 , e ) 15 β 2 | e | sqrt(multiply(add(power(divide(48, const_4), const_2), power(divide(36, const_4), const_2)), const_2)) | divide(n0,const_4)|divide(n1,const_4)|power(#0,const_2)|power(#1,const_2)|add(#2,#3)|multiply(#4,const_2)|sqrt(#5)| | geometry |
light glows for every 14 seconds . how many times did it between 1 : 57 : 58 and 3 : 20 : 47 am | "the diff in sec between 1 : 57 : 58 and 3 : 20 : 47 is 4969 sec , 4969 / 14 = 354 so total 355 times light ll glow answer : e" | a ) 381 , b ) 382 , c ) 383 , d ) 384 , e ) 355 | e | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 14) | add(n6,const_2)|add(const_2,const_60)|add(n5,#1)|multiply(#2,const_60)|add(#0,#3)|divide(#4,n0)| | physics |
if 36 men can do a piece of work in 20 hours , in how mwny hours will 15 men do it ? | "explanation : let the required no of hours be x . then less men , more hours ( indirect proportion ) \ inline \ fn _ jvn \ therefore 15 : 36 : : 20 : x \ inline \ fn _ jvn \ leftrightarrow ( 15 x x ) = ( 36 x 20 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 20 } { 15 } = 48 hence ,... | a ) 22 , b ) 48 , c ) 60 , d ) 88 , e ) 72 | b | divide(multiply(36, 20), 15) | multiply(n0,n1)|divide(#0,n2)| | physics |
20 beavers , working together in a constant pace , can build a dam in 3 hours . how many hours z will it take 12 beavers that work at the same pace , to build the same dam ? | "c . 5 hrs if there were 10 beavers it qould have taken double z = 6 hrs . . so closest to that option is 5 ." | a ) 2 . , b ) z = 4 . , c ) z = 5 . , d ) 6 . , e ) 8 . | c | divide(multiply(3, 20), 12) | multiply(n0,n1)|divide(#0,n2)| | physics |
find the simple interest on rs . 580 for 11 months at 9 paisa per month ? | "i = ( 580 * 11 * 9 ) / 100 = 574 answer : a" | a ) 574 , b ) 270 , c ) 566 , d ) 266 , e ) 121 | a | multiply(580, divide(11, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
every letter in the alphabet has a number value that is equal to its place in the alphabet . thus , the letter a has a value of 1 , the letter b has a value of 2 , the letter c has a value of 3 , etc . . . the number value of a word is obtained by adding up the value of the letters in the word and then multiplying that... | ` ` rat ' ' = ( 18 + 1 + 20 ) * 3 = 117 . the answer is d . | a ) 108 , b ) 111 , c ) 114 , d ) 117 , e ) 120 | d | multiply(add(add(subtract(multiply(2, const_10), 2), 1), multiply(2, const_10)), 3) | multiply(n1,const_10)|subtract(#0,n1)|add(n0,#1)|add(#2,#0)|multiply(n2,#3) | general |
a train 100 m long crosses a platform 125 m long in 2 sec ; find the speed of the train ? | "d = 100 + 125 = 225 t = 2 s = 225 / 2 * 18 / 5 = 405 kmph answer : b" | a ) 36 , b ) 405 , c ) 200 , d ) 302 , e ) 404 | b | subtract(multiply(2, multiply(125, const_0_2778)), 100) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
0.0015 Γ· ? = 0.003 | "let 0.0015 / x = 0.003 x = 0.0015 / 0.003 = 0.5 answer : c" | a ) 0.05 , b ) 0.005 , c ) 0.5 , d ) 5 , e ) 50 | c | multiply(divide(0.0015, 0.003), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
β ( 9 ) ^ 2 | "explanation β ( 9 ) ^ 2 = ? or , ? = 9 answer c" | a ) 3 , b ) 14 , c ) 9 , d ) 21 , e ) none of these | c | sqrt(power(9, 2)) | power(n0,n1)|sqrt(#0)| | general |
the majority owner of a business received 25 % of the profit , with each of 4 partners receiving 25 % of the remaining profit . if the majority owner and two of the owners combined to receive $ 39,375 , how much profit did the business make ? | "let p be the total profit . p / 4 + 1 / 2 * ( 3 p / 4 ) = p / 4 + 3 p / 8 = 5 p / 8 = $ 39,375 p = $ 63,000 the answer is d ." | a ) $ 54,000 , b ) $ 57,000 , c ) $ 60,000 , d ) $ 63,000 , e ) $ 66,000 | d | divide(multiply(const_100, multiply(const_100, add(const_1, 4))), add(divide(25, const_100), multiply(multiply(divide(25, const_100), subtract(const_1, divide(25, const_100))), const_2))) | add(const_1,n1)|divide(n0,const_100)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#2,const_100)|multiply(#1,#3)|multiply(#5,const_2)|add(#1,#6)|divide(#4,#7)| | gain |
the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the four walls of the room at rs . 3 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each ? | "area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 3 = rs . 2718 answer : e" | a ) s . 4528 , b ) s . 4520 , c ) s . 4527 , d ) s . 4530 , e ) s . 2718 | e | multiply(subtract(subtract(multiply(multiply(const_2, 12), add(15, 25)), multiply(6, 3)), multiply(3, multiply(4, 3))), 3) | add(n0,n1)|multiply(n2,const_2)|multiply(n4,n5)|multiply(n5,n6)|multiply(#0,#1)|multiply(n5,#3)|subtract(#4,#2)|subtract(#6,#5)|multiply(n3,#7)| | geometry |
the length of the bridge , which a train 160 meters long and travelling at 45 km / hr can cross in 30 seconds , is ? | "speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 160 + x ) / 30 = 25 / 2 = = > 2 ( 160 + x ) = 750 = = > x = 215 m . answer : c" | a ) 766 m , b ) 156 m , c ) 215 m , d ) 156 m , e ) 156 m | c | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 160) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
a car that moves at an average speed of 60 kmph , reaches its destination on time . when its average speed becomes 50 kmph , then it reaches its destination 45 minutes late . find the length of journey . | sol . difference between timings = 45 min = 3 / 4 hr . let the length of journey be x km . then , x / 50 - x / 60 = 1 / 4 Γ’ β‘ β 6 x - 5 x = 225 Γ’ β‘ β x = 225 km . answer b | a ) 189 km , b ) 225 km , c ) 255 km , d ) 199 km , e ) none | b | multiply(divide(multiply(50, divide(45, const_60)), subtract(60, 50)), 60) | divide(n2,const_60)|subtract(n0,n1)|multiply(n1,#0)|divide(#2,#1)|multiply(n0,#3) | physics |
one pipe can fill a tank in 15 hour . but because of hole in a tank , this tank fill in 20 hour . so in how much time this hole will empty the full tank ? | in 1 hour hole empty = [ 1 / 15 - 1 / 20 ] = 1 / 60 so total time taken to empty the full tank through hole = 60 hour . answer d | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | d | inverse(subtract(divide(const_1, 15), divide(const_1, 20))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|inverse(#2) | physics |
the ratio of the volumes of two cubes is 125 : 1728 . what is the ratio of their total surface areas ? | "ratio of the sides = Γ’ Β³ Γ’ Λ Ε‘ 125 : Γ’ Β³ Γ’ Λ Ε‘ 1728 = 5 : 12 ratio of surface areas = 5 ^ 2 : 12 ^ 2 = 25 : 144 answer : c" | a ) 81 : 121 , b ) 9 : 11 , c ) 25 : 144 , d ) 27 : 121 , e ) none of these | c | power(divide(125, 1728), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry |
moli buys 3 ribbons , 7 clips and 1 soap for rs . 120 exactly . at the same place it would cost rs . 164 for 4 ribbons , 10 clips and one soap . how much would it cost for one ribbon , one clip and one soap ? | explanation : let us consider , ribbon is denoted by r clips is denoted by c soaps is denoted by s now according to the question , = > 3 r + 7 c + s = 120 - - - - - - - - ( 1 ) = > 4 r + 10 c + s = 164 - - - - - - - - ( 2 ) ( 1 ) - ( 2 ) = > ( 3 r - 4 r ) + ( 7 c - 10 c ) + ( s - s ) = 120 - 164 = > - r - 3 c = - 44 mu... | a ) 36 , b ) 34 , c ) 38 , d ) 32 , e ) 31 | d | subtract(subtract(164, 120), add(10, const_2)) | add(n6,const_2)|subtract(n4,n3)|subtract(#1,#0) | general |
the speed of a train is 90 kmph . what is the distance covered by it in 5 minutes ? | "90 * 5 / 60 = 7.5 kmph answer : a" | a ) 7.5 , b ) 66 , c ) 77 , d ) 52 , e ) 42 | a | multiply(divide(5, const_60), 90) | divide(n1,const_60)|multiply(n0,#0)| | physics |
a rainstorm increased the amount of water stored in state j reservoirs from 124 billion gallons to 140 billion gallons . if the storm increased the amount of water in the reservoirs to 80 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the... | "let total capacity be x we know 140 = 0.80 x x = 140 / 0.80 = 175 prior to storm , we had 124 bn gallons 175 - 124 = 51 answer : a" | a ) 51 , b ) 48 , c ) 55 , d ) 63 , e ) 65 | a | divide(divide(multiply(140, const_100), 80), const_2) | multiply(n1,const_100)|divide(#0,n2)|divide(#1,const_2)| | general |
if ( m - 8 ) is a factor of m ^ 2 - km - 24 , then k = | "( m - 8 ) ( m - a ) = m ^ 2 - km - 24 a = - 3 k = 8 + a = 5 = b" | a ) 3 , b ) 5 , c ) 6 , d ) 11 , e ) 16 | b | subtract(8, divide(24, 8)) | divide(n2,n0)|subtract(n0,#0)| | general |
a box contains either blue or red flags . the total number of flags in the box is an even number . a group of children are asked to pick up two flags each . if all the flags are used up in the process such that 60 % of the children have blue flags , and 70 % have red flags , what percentage of children have flags of bo... | solution : let the total number of flags be 100 ( even number ) let the total number of ' blue ' flags alone be ' a ' let the total number of ' red ' flags alone be ' b ' let the total number of ' both ' flags be ' c ' we have given , total number of blue flags = 60 % = 60 = a + c total number of red flags = 70 % = 70 ... | a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 30 % | e | subtract(add(60, 70), const_100) | add(n0,n1)|subtract(#0,const_100) | general |
a dog breeder currently has 9 breeding dogs . 6 of the dogs have exactly 1 littermate , and 3 of the dogs have exactly 2 littermates . if 2 dogs are selected at random , what is the probability w that both selected dogs are not littermates ? | we have three pairs of dogs for the 6 with exactly one littermate , and one triplet , with each having exactly two littermates . so , in fact there are two types of dogs : those with one littermate - say a , and the others with two littermates - b . work with probabilities : choosing two dogs , we can have either one d... | a ) 1 / 6 , b ) 2 / 9 , c ) 5 / 6 , d ) 7 / 9 , e ) 8 / 9 | c | divide(const_5, 6) | divide(const_5,n1) | other |
how many 4 - digit positive integers are there , where each digit is positive , and no 4 adjacent digits are same ? | "first digit . . 9 posibilities second digit , 8 possibilities third digit , 7 possibilities fourth digit , 6 possibilities . 9 * 8 * 7 * 6 = 3024 . b" | a ) 1236 , b ) 3024 , c ) 4096 , d ) 4608 , e ) 6561 | b | multiply(multiply(add(4, 4), add(4, 4)), multiply(add(4, 4), multiply(4, 4))) | add(n1,n0)|add(n0,n0)|multiply(n1,n1)|multiply(#0,#0)|multiply(#1,#2)|multiply(#3,#4)| | general |
p can do a work in 15 days and q cando the same work in 20 days . if they can work together for 4 days , what is the fraction of work left ? | amount of work p can do in 1 day = 1 / 15 amount of work q can do in 1 day = 1 / 20 amount of work p and q can do in 1 day = 1 / 15 + 1 / 20 = 7 / 60 amount of work p and q can together do in 4 days = 4 Γ ( 7 / 60 ) = 7 / 15 fraction of work left = 1 β 7 / 15 = 8 / 15 c | a ) 6 / 15 , b ) 7 / 13 , c ) 8 / 15 , d ) 9 / 17 , e ) 2 / 13 | c | divide(subtract(const_60, multiply(divide(const_60, 15), add(divide(const_60, 15), divide(const_60, 20)))), const_60) | divide(const_60,n0)|divide(const_60,n1)|add(#0,#1)|multiply(#2,#0)|subtract(const_60,#3)|divide(#4,const_60) | physics |
how many of the integers between 25 and 95 are even ? | "number start between 25 to 95 is 70 numbers half of them is even . . which is 35 answer : a" | a ) 35 , b ) 90 , c ) 11 , d ) 10 , e ) 9 | a | divide(subtract(95, 25), const_2) | subtract(n1,n0)|divide(#0,const_2)| | general |
the sale price shirts listed for rs . 400 after successive discount is 10 % and 5 % is ? | 400 * ( 90 / 100 ) * ( 95 / 100 ) = 342 c | a ) 280 , b ) 290 , c ) 342 , d ) 250 , e ) 253 | c | subtract(400, add(divide(multiply(400, 10), const_100), divide(multiply(5, 400), const_100))) | multiply(n0,n1)|multiply(n0,n2)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|subtract(n0,#4) | gain |
village x has a population of 78000 , which is decreasing at the rate of 1200 per year . village y has a population of 42000 , which is increasing at the rate of 800 per year . in how many years will the population of the two villages be equal ? | "let the population of two villages be equal after p years then , 78000 - 1200 p = 42000 + 800 p 2000 p = 36000 p = 18 answer is d ." | a ) 15 , b ) 19 , c ) 11 , d ) 18 , e ) 13 | d | divide(subtract(78000, 42000), add(800, 1200)) | add(n1,n3)|subtract(n0,n2)|divide(#1,#0)| | general |
if a whole number n is divided by 4 , we will get 3 as remainder . what will be the remainder if 2 n is divided by 4 ? | explanation : let n Γ· 4 = p , remainder = 3 = > n = 4 p + 3 2 n = 2 ( 4 p + 3 ) = 8 p + 6 = 8 p + 4 + 2 = 4 ( 2 p + 1 ) + 2 hence , if 2 n is divided by 4 , we will get 2 as remainder . answer : c | a ) 4 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | c | reminder(multiply(2, add(multiply(4, const_1), 3)), 4) | multiply(n0,const_1)|add(n1,#0)|multiply(n2,#1)|reminder(#2,n0) | general |
in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 4000 , the number of valid votes that the other candidate got , was : | "number of valid votes = 80 % of 4000 = 3200 . valid votes polled by other candidate = 45 % of 4000 = ( 45 / 100 ) x 4000 = 1800 answer = e" | a ) 3500 , b ) 1200 , c ) 1650 , d ) 3700 , e ) 1800 | e | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 4000) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)| | gain |
a invested $ 300 in a business after 6 months b invested $ 200 in the business . end of the year if they got $ 100 as profit . find a shares ? | "a : b = 300 * 12 : 200 * 6 a : b = 3 : 1 a ' s share = 100 * 3 / 4 = $ 75 answer is b" | a ) $ 100 , b ) $ 75 , c ) $ 20 , d ) $ 120 , e ) $ 50 | b | multiply(100, subtract(const_1, divide(divide(200, const_2), add(300, divide(200, const_2))))) | divide(n2,const_2)|add(n0,#0)|divide(#0,#1)|subtract(const_1,#2)|multiply(n3,#3)| | gain |
a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 7.50 . if the company β s latest order required a truck with at least 336,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | "number of boxes = total volume / volume of one box = 336,000 / ( 8 * 10 * 12 ) = 350 one box costs 7.50 , so 350 box will cost = 350 * 7.5 = 2625 d is the answer" | a ) $ 960 , b ) $ 1,350 , c ) $ 1,725 , d ) $ 2,625 , e ) $ 2,250 | d | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, 10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 7.50), const_1000) | multiply(const_100,const_3)|multiply(n0,n1)|multiply(n0,n1)|add(#0,#1)|multiply(n2,#2)|add(#3,const_4)|multiply(#5,const_1000)|divide(#6,#4)|multiply(n3,#7)|divide(#8,const_1000)| | general |
the sum of 77 consecutive integers is 7777 . what is the greatest integer in the set ? | "let x be the first integer in the set . then x + 76 is the largest integer . the sum is : x + ( x + 1 ) + ( x + 2 ) + . . . + ( x + 76 ) = 77 x + 76 * 77 / 2 = 77 ( x + 38 ) then x + 38 = 101 x = 63 the largest integer in the set is 63 + 76 = 139 the answer is a ." | a ) 139 , b ) 141 , c ) 143 , d ) 145 , e ) 147 | a | add(add(power(add(add(divide(subtract(subtract(77, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(77, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(77, const_10), const_2), const_4), const_2), power(add... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
a group of boy scouts and girls scouts is going on a rafting trip . 80 % of the scouts arrived with signed permission slips . if 40 % of the scouts were boy scouts and 75 % of the boy scouts arrived with signed permission slips , then what percentage of the girl scouts arrived with signed permission slips ? round to th... | "40 % were boy scouts so 60 % ( 100 - 40 = 60 ) were girl scouts . # of boy scouts with permission slips signed + # of girl scouts with permission slips signed = total # with permission slip signed ( 75 % of 40 % of the total going ) + ( ? % of 60 % of the total going ) = 80 % of the total going we can let the ` ` tota... | a ) 79 , b ) 81 , c ) 82 , d ) 83 , e ) 85 | d | multiply(divide(subtract(80, divide(multiply(40, 75), const_100)), subtract(const_100, 40)), const_100) | multiply(n1,n2)|subtract(const_100,n1)|divide(#0,const_100)|subtract(n0,#2)|divide(#3,#1)|multiply(#4,const_100)| | gain |
a sum fetched a total simple interest of rs . 4016.25 at the rate of 1 % p . a . in 3 years . what is the sum ? | "principal = ( 100 * 4016.25 ) / ( 1 * 3 ) = rs . 133875 . answer : d" | a ) 122762 , b ) 132877 , c ) 122882 , d ) 133875 , e ) 132887 | d | divide(divide(multiply(4016.25, const_100), 1), 3) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain |
pipe a can fill a tank in 12 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ? | "let the pipe b be closed after x minutes . 30 / 12 - x / 24 = 1 = > x / 24 = 30 / 12 - 1 = 3 / 2 = > x = 3 / 2 * 24 = 36 . answer : e" | a ) 18 , b ) 27 , c ) 98 , d ) 27 , e ) 36 | e | multiply(subtract(divide(30, 12), const_1), 24) | divide(n2,n0)|subtract(#0,const_1)|multiply(n1,#1)| | physics |
sandy is younger than molly by 18 years . if the ratio of their ages is 7 : 9 , how old is sandy ? | "let sandy ' s age be 7 x and let molly ' s age be 9 x . 9 x - 7 x = 18 x = 9 sandy is 63 years old . the answer is d ." | a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70 | d | multiply(divide(7, subtract(9, 7)), 18) | subtract(n2,n1)|divide(n1,#0)|multiply(n0,#1)| | other |
score interval - - - - - - - - - - - - - - - - number of scores 50 - 59 - - - - - - - - - - - - - - - - - - - - - - - - - - 2 60 - 69 - - - - - - - - - - - - - - - - - - - - - - - - - - 4 70 - 79 - - - - - - - - - - - - - - - - - - - - - - - - - - 10 80 - 89 - - - - - - - - - - - - - - - - - - - - - - - - - - 27 90 - 9... | total scores = 2 + 4 + 10 + 27 + 18 = 61 , which is odd , therefore the median is the floor ( 61 / 2 ) + 1 = 31 th score . and the 31 th score is in the 80 - 89 range , because 50 - 79 only reference 28 scores . c | a ) 30 , b ) 32 , c ) 31 th ( 80 - 89 ) , d ) 34 , e ) 36 | c | subtract(70, 79) | subtract(n6,n7) | general |
a side of beef lost 20 percent of its weight in processing . if the side of beef weighed 640 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 80 / 100 ) * x = 640 = > x = ( 640 * 100 ) / 80 = 800 answer c" | a ) 191 , b ) 355 , c ) 800 , d ) 840 , e ) 900 | c | divide(multiply(640, const_100), subtract(const_100, 20)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds . the ratio of their speeds is ? | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 23 = = > 27 x + 17 y = 23 x + 23 y = = > 4 x = 6 y = = > x / y = 3 / 2 . answer : b" | a ) 3 / 7 , b ) 3 / 2 , c ) 3 / 5 , d ) 3 / 1 , e ) 3 / 3 | b | divide(subtract(27, 23), subtract(23, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics |
the cost price of 16 articles is the same as the selling price of 12 articles . find the loss / profit percentages . | "explanation : the gain is 4 out of 12 articles . therefore , gain percentage = 4 x 100 / 12 = 100 / 3 = 33 1 / 3 % answer : option c" | a ) 30 % , b ) 32.50 % , c ) 33 1 / 3 % , d ) 40 % , e ) 50 % | c | divide(multiply(16, const_4), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)| | gain |
selling an kite for rs . 30 , a shop keeper gains 30 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ? | "explanation : marked price = rs . 30 c . p . = 100 / 130 * 30 = rs . 23.07 sale price = 90 % of rs . 30 = rs . 27 required gain % = 3.92 / 23.07 * 100 = 17 % . answer : d" | a ) 8 % , b ) 10 % , c ) 11 % , d ) 17 % , e ) 20 % | d | multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(30, const_100))), divide(multiply(30, const_100), add(30, const_100))), const_100) | add(n1,const_100)|divide(n0,const_100)|multiply(n0,const_100)|subtract(const_100,n2)|divide(#2,#0)|multiply(#1,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)| | gain |
the manufacturing cost of a shoe is rs . 220 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains | "explanation : total cost of a watch = 220 + ( 500 / 100 ) = 225 . gain = 20 % = > sp = 1.2 cp = 1.2 x 225 = 270 answer : d" | a ) s 222 , b ) s 216 , c ) s 220 , d ) s 270 , e ) s 217 | d | add(add(220, divide(500, 100)), multiply(divide(20, 100), add(220, divide(500, 100)))) | divide(n1,n2)|divide(n3,n2)|add(n0,#0)|multiply(#2,#1)|add(#2,#3)| | gain |
a grocer has a sale of rs . 5124 , rs . 5366 , rs . 5808 , rs . 5399 and rs . 6124 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 5400 ? | "total sale for 5 months = rs . ( 5124 + 5366 + 5808 + 5399 + 6124 ) = rs . 27821 . required sale = rs . [ ( 5400 x 6 ) - 27821 ] = rs . ( 32400 - 27821 ) = rs . 4579 . answer : b" | a ) 4079 , b ) 4579 , c ) 5579 , d ) 5679 , e ) 5779 | b | subtract(multiply(add(5, const_1), 5400), add(add(add(add(5124, 5366), 5808), 5399), 6124)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general |
one bacterium splits into 8 bacteria of the next generation . but due to environment , only 50 % of one generation can produced the next generation . if the seventh generation number is 4096 million , what is the number in first generation ? | solution : let the number of bacteria in the 1 st generation be x , then number of bacteria in 2 nd , 3 rd , 4 th . . . . . generation would be 8 ( x / 2 ) , 8 ( 4 x / 2 ) , 8 ( 16 x / 2 ) . . . . and so on . as x , 4 x , 16 x , 64 x . . . . . it is in gp with common ratio 4 . hence , 7 th term of gp , x ( 4 ) 6 = 4096... | a ) 1 million , b ) 2 million , c ) 4 million , d ) 8 million , e ) none of these | a | divide(power(divide(8, const_2), subtract(8, const_2)), power(divide(8, const_2), subtract(8, const_2))) | divide(n0,const_2)|subtract(n0,const_2)|power(#0,#1)|divide(#2,#2) | other |
xavier , yvonne , and zelda each try independently to solve a problem . if their individual probabilities for success are 1 / 4 , 1 / 3 and 5 / 8 , respectively , what is the probability that xavier and yvonne , but not zelda , will solve the problem ? | "p ( xavier will solve ) = 1 / 4 p ( yvonne will solve ) = 1 / 2 p ( zelda will not solve ) = 1 - 5 / 8 = 3 / 8 . now , we need to multiply all this ps to find an answer : p = ( 1 / 4 ) * ( 1 / 3 ) * ( 3 / 8 ) = 1 / 32 . ans . a ." | a ) 1 / 32 , b ) 7 / 8 , c ) 9 / 64 , d ) 5 / 64 , e ) 3 / 64 | a | multiply(multiply(divide(1, 4), divide(1, 3)), subtract(1, divide(5, 8))) | divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|multiply(#0,#1)|subtract(n0,#2)|multiply(#3,#4)| | general |
a class has a ratio of 3 : 6 : 7 of children with red hair , blonde hair and black hair respectively . if the class has 9 kids with red hair , how many kids are there with black hair ? | "since there is a 3 : 7 ratio between red haired children and black - haired children , if there are 9 children with red hair , then there must be 3 times the amount of black - haired children . therefore , are 21 children with black hair . answer : d )" | a ) 7 , b ) 6 , c ) 9 , d ) 21 , e ) 63 | d | multiply(add(add(7, 6), 3), 3) | add(n1,n2)|add(n0,#0)|multiply(n0,#1)| | other |
a walks at 10 kmph and 5 hours after his start , b cycles after him at 20 kmph . how far from the start does b catch up with a ? | "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 5 hours . x / 10 - x / 20 = 5 x = 100 km answer is a" | a ) 100 km , b ) 150 km , c ) 50 km , d ) 120 km , e ) 200 km | a | multiply(5, 20) | multiply(n1,n2)| | physics |
evaluate : 28 % of 400 + 45 % of 250 | explanation : 28 % of 400 + 45 % of 250 = ( 28 / 100 * 400 + 45 / 100 * 250 ) = ( 112 + 112.5 ) = 224.5 answer b | a ) 220.3 , b ) 224.5 , c ) 190.3 , d ) 150 , e ) none of these | b | add(divide(multiply(400, 28), const_100), divide(multiply(45, 250), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3) | gain |
a rope of which a calf is tied is increased from 10 m to 20 m , how much additional grassy ground shall it graze ? | "Ο ( 202 β 102 ) = 942.86 answer : e" | a ) 540 , b ) 128 , c ) 100 , d ) 942 , e ) 942.86 | e | multiply(subtract(power(20, const_2), power(10, const_2)), divide(add(multiply(10, const_2), const_2), add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_2)|power(n1,const_2)|power(n0,const_2)|add(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#6,#5)| | general |
a card is drawn from a pack of 52 cards . the probability of getting a queen of club or a king of heart is : | "here , n ( s ) = 52 . let e = event of getting a queen of club or a king of heart . then , n ( e ) = 2 . p ( e ) = n ( e ) / n ( s ) = 2 / 52 = 1 / 26 answer should be a" | a ) 1 / 26 , b ) 1 / 52 , c ) 2 / 62 , d ) 2 / 26 , e ) 3 / 26 | a | divide(subtract(52, multiply(const_4, const_4)), 52) | multiply(const_4,const_4)|subtract(n0,#0)|divide(#1,n0)| | probability |
a man can row 8 kmph in still water . when the river is running at 1.8 kmph , it takes him 2 hour to row to a place and back . what is the total distance traveled by the man ? | "m = 8 s = 1.8 ds = 9.8 us = 6.2 x / 9.8 + x / 6.2 = 1 x = 3.8 d = 2.88 * 2 = 7.6 answer : c" | a ) 8.2 km , b ) 6.7 km , c ) 7.6 km , d ) 7.4 km , e ) 6.3 km | c | multiply(divide(multiply(add(8, 1.8), subtract(8, 1.8)), add(add(8, 1.8), subtract(8, 1.8))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
a cube of side 5 cm is painted on all its side . if it is sliced into 1 cubic centimer cubes , how many 1 cubic centimeter cubes will have exactly one of their sides painted ? | explanatory answer when a 5 cc cube is sliced into 1 cc cubes , we will get 5 * 5 * 5 = 125 cubes of 1 cubic centimeter . in each side of the larger cube , the smaller cubes on the edges will have more than one of their sides painted . therefore , the cubes which are not on the edge of the larger cube and that lie on t... | ['a ) 9', 'b ) 61', 'c ) 98', 'd ) 54', 'e ) 64'] | d | add(divide(surface_cube(5), const_3), const_4) | surface_cube(n0)|divide(#0,const_3)|add(#1,const_4) | physics |
a person buys an article at $ 800 . at what price should he sell the article so as to make a profit of 12 % ? | "a 896 cost price = $ 800 profit = 12 % of 800 = $ 96 selling price = cost price + profit = 800 + 96 = 896" | a ) 896 , b ) 890 , c ) 990 , d ) 789 , e ) 740 | c | add(800, multiply(800, divide(12, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
if m = 3 ^ n , what is the greatest value of n for which m is a factor of 25 ! | "we should find the highest power of 3 in 25 ! : 25 / 3 + 25 / 3 ^ 2 = 8 + 2 = 10 ( take only the quotient into account ) . . so the highest power of 3 in 25 ! is 10 . answer : b ." | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | b | add(floor(divide(25, 3)), floor(divide(25, power(3, const_2)))) | divide(n1,n0)|power(n0,const_2)|divide(n1,#1)|floor(#0)|floor(#2)|add(#3,#4)| | other |
jane makes toy bears . when she works with an assistant , she makes 44 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane β s output of toy bears per hour by what percent ? | "let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 57.6 bears per 36 hours a week or 1.6 bears / hr ( [ 40 bears * 1.44 ] / [ 40 hrs * . 90 ] ) . [ ( 1.6 - 1 ) / 1 ] * 100 % = 60 % answer : b" | a ) 20 % , b ) 60 % , c ) 100 % , d ) 180 % , e ) 200 % | b | multiply(divide(10, subtract(subtract(const_100, 44), 10)), const_100) | subtract(const_100,n0)|subtract(#0,n1)|divide(n1,#1)|multiply(#2,const_100)| | physics |
two trains are moving in the same direction at 108 kmph and 36 kmph . the faster train crosses the slower train in 17 seconds . find the length of the faster train in meters . | relative speed = ( 108 - 36 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in 17 sec = 17 * 20 = 340 m . the length of the faster train = 340 m . answer : b | a ) 270 m , b ) 340 m , c ) 310 m , d ) 350 m , e ) 327 m | b | multiply(divide(subtract(108, 36), const_3_6), 17) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1) | physics |
how many seconds will a 500 metre long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "solution speed of train relative to man = ( 63 - 3 ) km / hr = 60 km / hr = ( 60 x 5 / 18 ) m / sec = 50 / 3 m / sec time taken to pass the man = ( 500 x 3 / 50 ) sec = 30 sec answer b" | a ) 25 , b ) 30 , c ) 40 , d ) 45 , e ) 55 | b | divide(500, multiply(const_0_2778, subtract(63, 3))) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
if 5 < x < 8 < y < 13 , then what is the greatest possible positive integer difference of x and y ? | "let x = 5.1 and y = 12.1 greatest possible difference = 12.1 - 5.1 = 7 answer e" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | subtract(subtract(13, 5), const_1) | subtract(n2,n0)|subtract(#0,const_1)| | general |
in a garden , 26 trees are planted at equal distances along a yard 500 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ? | "26 trees have 25 gaps between them . length of each gap = 500 / 25 = 20 i . e . , distance between two consecutive trees = 20 answer is a ." | a ) 20 , b ) 8 , c ) 12 , d ) 14 , e ) 16 | a | divide(500, subtract(26, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | physics |
a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 2200 in all from both of them as interest . the rate of interest per annum is : | "explanation : let the rate of interest per annum be r % simple interest for rs . 5000 for 2 years at rate r % per annum + simple interest for rs . 3000 for 4 years at rate r % per annum = rs . 2200 β 5000 Γ r Γ 2 / 100 + 3000 Γ r Γ 4 / 100 = 2200 β 100 r + 120 r = 2200 β 220 r = 2200 β r = 10 i . e , rate = 10 % . ans... | a ) 5 % , b ) 10 % , c ) 7 % , d ) 8 % , e ) 9 % | b | multiply(divide(2200, add(multiply(5000, 2), multiply(3000, 4))), const_100) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(n4,#2)|multiply(#3,const_100)| | gain |
12 + 6 = 792 10 + 2 = 110 1 + 9 = 9 2 + 7 = 16 11 + 4 = ? ? solve it ? | x + y = x [ y + ( x - 1 ) ] = x ^ 2 + xy - x 12 + 6 = 12 [ 6 + ( 12 - 1 ) ] = 792 10 + 2 = 10 [ 2 + ( 10 - 1 ) ] = 110 1 + 9 = 1 [ 9 + ( 1 - 1 ) ] = 9 2 + 7 = 2 [ 7 + ( 2 - 1 ) ] = 16 11 + 4 = 11 [ 4 + ( 11 - 1 ) ] = 154 answer : e | a ) 100 , b ) 120 , c ) 190 , d ) 160 , e ) 154 | e | add(multiply(add(11, 4), 10), 4) | add(n12,n13)|multiply(n3,#0)|add(n13,#1) | general |
at a certain high school , the senior class is 3 times the size of the junior class . if 1 / 3 of the seniors and 3 / 4 of the juniors study japanese , what fraction of the students in both classes study japanese ? | start by deciding on a number of students to represent the number of students in the senior class . for this example i will choose 300 students . that would make the number of students in the junior class 100 . then we can find out how many students are taking japanese in each grade and add them together . ( 1 / 3 ) * ... | a ) 1 / 2 , b ) 3 / 8 , c ) 5 / 16 , d ) 1 / 4 , e ) 7 / 16 | e | divide(add(const_1, divide(3, 4)), add(3, 1)) | add(n0,n1)|divide(n0,n4)|add(#1,const_1)|divide(#2,#0) | general |
a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 4 feet . how many feet does the squirrel travels if the post is 12 feet tall and 3 feet in circumference ? | "total circuit = 12 / 4 = 3 total feet squirrel travels = 3 * 3 = 9 feet answer : a" | a ) 9 feet , b ) 12 feet , c ) 13 feet , d ) 15 feet , e ) 18 feet | a | multiply(divide(12, 4), 3) | divide(n1,n0)|multiply(n2,#0)| | geometry |
evaluate 2 + 22 + 222 + 2.22 | "2 + 22 + 222 + 2.22 = 248.22 option d" | a ) 269.72 , b ) 268.22 , c ) 248.21 , d ) 248.22 , e ) 268.21 | d | divide(2, divide(22, 2)) | divide(n1,n0)|divide(n0,#0)| | general |
the least number which when increased by 7 each divisible by each one of 24 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 7 = 864 - 7 = 857 . answer a" | a ) 857 , b ) 859 , c ) 869 , d ) 4320 , e ) none of these | a | subtract(lcm(lcm(lcm(24, 32), 36), 54), 7) | lcm(n1,n2)|lcm(n3,#0)|lcm(n4,#1)|subtract(#2,n0)| | general |
what is the area of a square field whose diagonal of length 12 m ? | "d 2 / 2 = ( 12 * 12 ) / 2 = 72 answer : b" | a ) 288 , b ) 72 , c ) 200 , d ) 112 , e ) 178 | b | divide(square_area(12), const_2) | square_area(n0)|divide(#0,const_2)| | geometry |
at a contest with 3,500 participants , 1 / 2 of the people are aged 18 to 22 . next year , the number of people aged 18 to 22 will increase by 1 / 7 . after this change , what percentage of the total 3,500 people will the 18 - to 22 - year - olds represent ? | "i just wanted to mention a couple of things here : * this is a pure ratio question ; the number 3,500 is completely irrelevant , and you can ignore it if you like . when we increase something by 1 / 7 , we are multiplying it by 1 + 1 / 7 = 8 / 7 , so the answer here must be ( 1 / 2 ) * ( 8 / 7 ) = 4 / 7 = 57.14 % . an... | a ) 33 % , b ) 40 % , c ) 55 % , d ) 57.14 % , e ) 66.14 % | d | multiply(divide(multiply(divide(3,500, 2), add(1, divide(1, 7))), 3,500), const_100) | divide(n1,n8)|divide(n0,n2)|add(#0,n1)|multiply(#2,#1)|divide(#3,n0)|multiply(#4,const_100)| | general |
a recipe requires 2 1 / 2 ( mixed number ) cups of flour 2 3 / 4 ( mixed number ) cups of sugar and 1 1 / 3 ( mixed number ) cups of milk to make one cake . victor has 15 cups if flour , 16 cups of sugar and 8 cups of milk . what is the greatest number of cakes jerry can make using this recipe ? | "less work up front : go through each item and see what the greatest number of cakes you can make with each . the lowest of these will be the right answer . flour : 15 cups , we need 2.5 cups each . just keep going up the line to see how many cakes we can make : that means i can make 2 cakes with 5 cups , so 6 cakes ov... | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | min(min(divide(15, add(2, divide(1, 2))), floor(divide(16, add(divide(3, 4), 2)))), divide(8, add(divide(1, 3), 1))) | divide(n1,n4)|divide(n1,n0)|divide(n4,n5)|add(n1,#0)|add(n0,#1)|add(n0,#2)|divide(n11,#3)|divide(n9,#4)|divide(n10,#5)|floor(#8)|min(#7,#9)|min(#6,#10)| | general |
daniel went to a shop and bought things worth rs . 25 , out of which 30 paise went on sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ? | "total cost of the items he purchased = rs . 25 given that out of this rs . 25 , 30 paise is given as tax = > total tax incurred = 30 paise = rs . 30 / 100 let the cost of the tax free items = x given that tax rate = 6 % β΄ ( 25 β 30 / 100 β x ) 6 / 100 = 30 / 100 β 6 ( 25 β 0.3 β x ) = 30 β ( 25 β 0.3 β x ) = 5 β x = 2... | a ) 19.7 , b ) 20 , c ) 21.3 , d ) 21.5 , e ) 22 | a | subtract(subtract(25, divide(30, const_100)), divide(30, 6)) | divide(n1,const_100)|divide(n1,n2)|subtract(n0,#0)|subtract(#2,#1)| | gain |
the sum of the present age of henry and jill is 40 . what is their present ages if 11 years ago henry was twice the age of jill ? | "let the age of jill 11 years ago be x , age of henry be 2 x x + 11 + 2 x + 11 = 40 x = 6 present ages will be 17 and 23 answer : c" | a ) and 27 , b ) and 24 , c ) and 23 , d ) and 29 , e ) of these | c | subtract(40, divide(add(40, 11), const_3)) | add(n0,n1)|divide(#0,const_3)|subtract(n0,#1)| | general |
if log 2 = 0.3010 and log 3 = 0.4771 , the value of log 5 512 is : | "log 5 512 = log 512 / log 5 = log 2 ^ 9 / log ( 10 / 2 ) = 9 log 2 / log 10 - log 2 = ( 9 * 0.3010 ) / 1 - 0.3010 = 2.709 / 0.699 = 2709 / 699 = 3.876 answer c" | a ) 2.87 , b ) 2.965 , c ) 3.876 , d ) 3.9 , e ) 4.526 | c | multiply(0.3010, divide(0.3010, 0.4771)) | divide(n1,n3)|multiply(n1,#0)| | other |
a is half good a work man as b and together they finish a job in 18 days . in how many days working alone b finish the job ? | "c 27 wc = 1 : 2 2 x + x = 1 / 18 = > x = 1 / 54 2 x = 1 / 27 = > 27 days" | a ) 23 , b ) 22 , c ) 27 , d ) 36 , e ) 48 | c | multiply(18, divide(const_3, const_2)) | divide(const_3,const_2)|multiply(n0,#0)| | physics |
the radius and height of a right circular cone are in the ratio 3 : 4 . if its volume is 96 β cm Β³ , what is its slant height ? | sol . let the radius and the height of the cone be 3 x and 4 x respectively . then , 1 / 3 * β * ( 3 x ) Β² * 4 x = 96 β β 36 x Β³ = ( 96 * 3 ) β x Β³ = [ 96 * 3 / 36 ] = 8 β x = 2 & ther 4 ; radius = 6 cm , height = 8 cm . slant height = β 6 Β² + 8 Β² cm = β 100 cm = 10 cm . answer b | ['a ) 8 cm', 'b ) 10 cm', 'c ) 12 cm', 'd ) 14 cm', 'e ) none'] | b | divide(divide(96, const_3), const_pi) | divide(n2,const_3)|divide(#0,const_pi) | geometry |
rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 84 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 84 5 x / 100 + 90 β 6 x / 100 = 84 x / 100 = 6 = > x = 600 answer : d" | a ) 299 , b ) 466 , c ) 578 , d ) 600 , e ) 677 | d | multiply(add(5, 6), const_100) | add(n1,n2)|multiply(#0,const_100)| | gain |
an engine moves at the speed of 60 kmph without any coaches attached to it . speed of the train reduces at the rate that varies directly as the square root of the number of coaches attached . when 4 coaches are attached speed decreases to 48 kmph . what will be the speed of train when 36 coaches are attached . | "1 . no . of coaches = 4 sqr root = 2 speed decreases by 12 12 = k * 2 k = 6 no . of coaches = 36 swr root = 6 decrease = 6 * 6 = 36 new speed = 60 - 36 = 24 a" | a ) 24 , b ) 28 , c ) 20 , d ) 32 , e ) 40 | a | subtract(60, multiply(sqrt(36), divide(subtract(60, 48), sqrt(4)))) | sqrt(n1)|sqrt(n3)|subtract(n0,n2)|divide(#2,#0)|multiply(#3,#1)|subtract(n0,#4)| | physics |
you and your friend spent a total of $ 15 for lunch . your friend spent $ 5 more than you . how much did your friend spend on their lunch ? | "my lunch = l , my friends lunch = l + 5 ( l ) + ( l + 5 ) = 15 l + l + 5 - 5 = 15 - 5 2 l = 10 l = 5 my friends lunch l + 5 = 5 + 5 = $ 10 , the answer is e" | a ) $ 9 , b ) $ 3 , c ) $ 4 , d ) $ 6 , e ) $ 10 | e | add(divide(subtract(15, 5), const_2), 5) | subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)| | general |
the rear β most end of a 66 foot truck exits a 330 foot tunnel exactly 6 seconds after the front β most end of the truck entered the tunnel . if the truck traveled the entire tunnel at a uniform speed , what is the speed of the truck in miles per hour ( 1 mile = 5,280 feet ) ? | total length will be 330 + 66 = 396 time = 6 secs speed = 396 / 6 = 66 feet / sec conversion to miles per hour = 66 * 3600 / 5280 = 45 . answer c | a ) 225 , b ) 90 , c ) 45 , d ) 37.5 , e ) 27 | c | divide(multiply(divide(add(330, 66), 6), const_3600), add(multiply(add(const_4, const_1), const_1000), subtract(multiply(const_3, const_100), multiply(const_2, const_10)))) | add(n0,n1)|add(const_1,const_4)|multiply(const_100,const_3)|multiply(const_10,const_2)|divide(#0,n2)|multiply(#1,const_1000)|subtract(#2,#3)|add(#5,#6)|multiply(#4,const_3600)|divide(#8,#7) | physics |
the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 800 per sq . metre . | "solution area of the floor = ( 5.5 Γ 3.75 ) m 2 = 20.625 m 2 β΄ cost of paving = rs . ( 800 Γ 20.625 ) = 16500 . answer d" | a ) rs . 15000 , b ) rs . 15550 , c ) rs . 15600 , d ) rs . 16500 , e ) none of these | d | multiply(800, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
2 men and 2 women are lined up in a row . what is the number of cases where they stand with each other in turn ? ( the number of cases in which men ( or women ) do not stand next to each other ) | the list should be wmwmw . hence , from women 2 ! and men 2 ! , we get ( 2 ! ) ( 2 ! ) = 4 . therefore , the correct answer is a . | a ) 4 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | a | multiply(factorial(2), factorial(2)) | factorial(n0)|factorial(n1)|multiply(#0,#1)| | general |
a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 60 km / hr respectively . b starts six hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "the table you made does n ' t make sense to me . all three meet at the same point means the distance they cover is the same . we know their rates are 30 , 40 and 60 . say the time taken by b is t hrs . then a takes 6 + t hrs . and we need to find the time taken by k . distance covered by a = distance covered by b 30 *... | a ) 3 , b ) 4.5 , c ) 4 , d ) d ) 12 , e ) e ) 5 | d | subtract(40, 30) | subtract(n1,n0)| | physics |
how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 70 m long and 16 1 / 2 m board ? | "70 * 33 / 2 = 5 / 2 * 2 * x = > x = 231 answer : e" | a ) 99 , b ) 18 , c ) 16 , d ) 10 , e ) 231 | e | divide(multiply(70, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | divide(n1,n0)|add(n5,#0)|add(n0,#0)|multiply(n4,#1)|multiply(n0,#2)|divide(#3,#4)| | general |
what is the rate percent when the simple interest on rs . 1200 amount to rs . 160 in 4 years ? | "160 = ( 1200 * 4 * r ) / 100 r = 3.3 % answer : a" | a ) 3.3 % , b ) 7 % , c ) 9 % , d ) 2 % , e ) 4 % | a | divide(multiply(const_100, 160), multiply(1200, 4)) | multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)| | gain |
the simple form of the ratio 2 / 3 : 2 / 5 is ? | "2 / 3 : 2 / 5 = 10 : 6 = 5 : 3 answer : d" | a ) 5 : 0 , b ) 5 : 3 , c ) 5 : 1 , d ) 5 : 2 , e ) 5 : 7 | d | divide(2, 5) | divide(n2,n3)| | other |
on a map , 1 inch represents 28 miles . how many inches would be necessary to represent a distance of 383.6 miles ? | "28 miles represented by 1 inch 1 mile represented by ( 1 / 28 ) inch 383.6 miles represented by ( 1 / 28 ) * 383.6 approximately = 14 we should use estimation here as the answer options are not close . ( 28 * 10 = 280 and 28 * 20 = 560 , hence we select the only option between 10 and 20 ) answer c" | a ) 5.2 , b ) 7.4 , c ) 13.7 , d ) 21.2 , e ) 28.7 | c | divide(383.6, 28) | divide(n2,n1)| | physics |
the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.5 years , the ratio of the number of boys to the number of girls in the class is | "explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.5 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.5 ) < = > k = 0.3 / 0.6 = 1 / 2 . required ratio = 1 / 2 : 1 = 1 : 2 . answer : a" | a ) 1 : 2 , b ) 2 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 2 : 5 | a | divide(subtract(15.8, 15.5), subtract(16.4, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general |
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