Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
x can do a piece of work in 40 days . he works at it for 8 days and then y finished it in 16 days . how long will y take to complete the work ? | "work done by x in 8 days = 8 * 1 / 40 = 1 / 5 remaining work = 1 - 1 / 5 = 4 / 5 4 / 5 work is done by y in 16 days whole work will be done by y in 16 * 5 / 4 = 20 days answer is e" | a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | e | multiply(16, inverse(subtract(const_1, divide(8, 40)))) | divide(n1,n0)|subtract(const_1,#0)|inverse(#1)|multiply(n2,#2)| | physics |
a positive number x is multiplied by 3 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x ? | "we need to produce an equation from the information given in the problem stem . we are first given that x is multiplied by 2 and then the product is divided by 3 . this gives us : 2 x / 3 next we are given that the positive square root of the result ( which is 2 x / 3 ) is equal to x . this gives us √ ( 2 x / 3 ) = x ... | a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 2 / 3 , e ) 1 / 2 | c | divide(3, 3) | divide(n0,n1)| | general |
10 men do a work in 10 days . how many men are needed to finish the work in 4 days ? | "men required to finish the work in 4 days = 10 * 10 / 4 = 25 answer is a" | a ) 25 , b ) 20 , c ) 30 , d ) 10 , e ) 15 | a | divide(multiply(10, 10), 4) | multiply(n0,n1)|divide(#0,n2)| | physics |
a trader sells 80 meters of cloth for rs . 6900 at the profit of rs . 20 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 6900 / 80 = rs . 86 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 86 - rs . 20 = rs . 66 . answer : e" | a ) 22 , b ) 75 , c ) 90 , d ) 78 , e ) 66 | e | subtract(divide(6900, 80), 20) | divide(n1,n0)|subtract(#0,n2)| | physics |
during a certain two - week period , 80 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 4 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedies , ... | "movies : 80 % comedies . 20 % remaining genre . now in this 20 % , there are only 2 categories . action movies and drama movies . if action = x ; drama movies = 4 x . total 5 x . 5 x = 20 ; x = 4 action movies : 4 % drama movies : 16 % we can say that out of 100 z , : comedies : 80 z action : 4 z drama : 16 z now acti... | a ) 5 a , b ) 15 a , c ) 20 a , d ) 1 / 25 a , e ) 30 a | c | floor(divide(80, add(4, const_1))) | add(n1,const_1)|divide(n0,#0)|floor(#1)| | general |
( 100 x 60 ) + ( 138 x 400 ) = ? x 60 | "explanation : ? = ( 100 x 60 ) + ( 138 x 400 ) / 60 = 6000 + 55200 / 60 = 1020 answer : option d" | a ) 880 , b ) 550 , c ) 1100 , d ) 1020 , e ) 600 | d | multiply(100, power(add(const_4, const_1), const_4)) | add(const_1,const_4)|power(#0,const_4)|multiply(n0,#1)| | general |
the speed of the boat in still water in 12 kmph . it can travel downstream through 42 kms in 3 hrs . in what time would it cover the same distance upstream ? | "still water = 12 km / hr downstream = 42 / 3 = 14 km / hr upstream = > > still water = ( u + v / 2 ) = > > 12 = u + 14 / 2 = 10 km / hr so time taken in upstream = 42 / 10 = 4.2 hrs answer : c" | a ) 8 hours , b ) 6 hours , c ) 4.2 hours , d ) 5 hours , e ) 6 hours | c | divide(42, subtract(12, subtract(divide(42, 3), 12))) | divide(n1,n2)|subtract(#0,n0)|subtract(n0,#1)|divide(n1,#2)| | physics |
tickets numbered from 1 to 20 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bearsa number which is a multiple of 3 ? | "here , s = [ 1 , 2 , 3 , 4 , … . , 19 , 20 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 ] p ( e ) = n ( e ) / n ( s ) = 6 / 20 = 3 / 10 c" | a ) 1 / 5 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 7 , e ) 1 / 7 | c | divide(divide(20, 3), 20) | divide(n1,n2)|divide(#0,n1)| | general |
two equally sized jugs full of water are each emptied into two separate unequally sized empty jugs , x and y . now , jug x is 1 / 4 full , while jug y is 2 / 3 full . if water is poured from jug x into jug y until jug y is filled , what fraction of jug x then contains water ? | "suppose the water in each jug is l liters cx x ( 1 / 4 ) = l cx = 4 l liters cx is capacity of x cy x ( 2 / 3 ) = l cy = 3 l / 2 liters cy is capacity of y now , y is 3 l / 2 - l empty = l / 2 empty so , we can put only l / 2 water in jug y from jug x jug x ' s remaining water = l - l / 2 = l / 2 fraction of x which c... | a ) 0 , b ) 1 / 15 , c ) 2 / 15 , d ) 1 / 8 , e ) 2 / 10 | d | multiply(divide(1, 4), divide(subtract(1, divide(2, 3)), divide(2, 3))) | divide(n0,n1)|divide(n2,n3)|subtract(n0,#1)|divide(#2,#1)|multiply(#0,#3)| | general |
harry started a 6 - mile hike with a full 7 - cup canteen of water and finished the hike in 2 hours with 1 cup of water remaining in the canteen . if the canteen leaked at the rate of 1 cup per hour and harry drank 1 cups of water during the last mile , how many cups did he drink per mile during the first 5 miles of th... | "a . no of cups leaked during the trip = 2 cups . no of cups harry drank = 4 cups . no of cups harry drank during the first 5 miles = 4 . drink / mile = 4 / 5 answer : e" | a ) 1 / 5 , b ) 2 / 3 , c ) 1 / 3 , d ) 3 / 5 , e ) 4 / 5 | e | divide(subtract(subtract(subtract(7, 2), 1), 1), 5) | subtract(n1,n2)|subtract(#0,n3)|subtract(#1,n3)|divide(#2,n6)| | physics |
the average weight of a class of 20 boys was calculated to be 58.4 kgs and it was later found that one weight was misread as 56 kg instead of 60 kg . what is the correct weight ? | actual total weight is ( 20 x 58.4 - 56 + 60 ) = 1172 kgs actual average weight is 1172 / 20 = 58.6 kgs a | a ) 58.6 kgs , b ) 58.85 kgs , c ) 58.95 kgs , d ) 59 kgs , e ) 59.85 kgs | a | divide(add(multiply(58.4, 20), subtract(60, 56)), 20) | multiply(n0,n1)|subtract(n3,n2)|add(#0,#1)|divide(#2,n0) | general |
hillary and eddy are climbing to the summit of mt . everest from a base camp 4,700 ft from the summit . when they depart for the summit at 06 : 00 , hillary climbs at a rate of 800 ft / hr with eddy lagging behind at a slower rate of 500 ft / hr . if hillary stops 700 ft short of the summit and then descends at a rate ... | "solution : h stopped 700 ft before reaching the final point , time taken to reach 4000 ft = 4000 / 800 = 5 hrs . this means she reached there at 11 : 00 . speed difference between them is 800 - 500 = 300 ft / hr so by the time h stops they have 1500 ft of distance so now here we use relative speed formula they both ar... | a ) 14.0 , b ) 13.0 , c ) 12.0 , d ) 11.0 , e ) 10.0 | c | add(06, add(divide(multiply(divide(subtract(divide(500, const_100), const_1), divide(800, 1,000)), subtract(divide(800, 1,000), divide(500, 700))), add(divide(500, 700), const_1)), divide(subtract(divide(500, const_100), const_1), divide(800, 700)))) | divide(n4,const_100)|divide(n3,n6)|divide(n4,n6)|add(#2,const_1)|subtract(#0,const_1)|subtract(#1,#2)|divide(#4,#1)|multiply(#6,#5)|divide(#7,#3)|add(#8,#6)|add(n1,#9)| | general |
p and q are the only two applicants qualified for a short - term research project that pays 420 dollars in total . candidate p has more experience and , if hired , would be paid 50 percent more per hour than candidate q would be paid . candidate q , if hired , would require 10 hours more than candidate p to do the job ... | let q ' s hourly wage be x , then p ' s hourly wage is 1.5 x let t be the number of hours that q needs , then p needs t - 10 hours to do the job . since they both are paid an equal total amount of $ 420 : x * t = 1.5 x * ( t - 10 ) t = 30 hours and q ' s hourly wage is 420 / 30 = $ 14 p ' s hourly wage is 420 / ( t - 1... | a ) $ 3 , b ) $ 5 , c ) $ 7 , d ) $ 9 , e ) $ 11 | c | subtract(divide(420, divide(10, subtract(divide(const_3, const_2), const_1))), divide(420, add(divide(10, subtract(divide(const_3, const_2), const_1)), 10))) | divide(const_3,const_2)|subtract(#0,const_1)|divide(n2,#1)|add(n2,#2)|divide(n0,#2)|divide(n0,#3)|subtract(#4,#5) | general |
log ( a ( a ( a ) ^ 1 / 4 ) ^ 1 / 4 ) ^ 1 / 4 here base is a . | log ( a ( a ( a ^ 1 / 4 ) ^ 1 / 4 ) ^ 1 / 4 = > log ( a ( a ^ 1 + 1 / 4 ) ^ 1 / 4 ) ^ 1 / 4 = > log ( a ( a ^ 5 / 4 ) ^ 1 / 4 ) ^ 1 / 4 = > log ( a ( a ^ 5 / 16 ) ^ 1 / 4 = > log ( a ^ 1 + 5 / 16 ) ^ 1 / 4 = > log ( a ^ 21 / 16 ) ^ 1 / 4 = > log ( a ^ 21 / 64 ) = > 21 / 64 log ( a ) = > 21 / 64 answer : b | a ) 0 , b ) 21 / 64 , c ) 1 , d ) 22 / 63 , e ) 23 / 65 | b | divide(add(const_1, divide(add(const_1, divide(1, 4)), 4)), 4) | divide(n0,n1)|add(#0,const_1)|divide(#1,n1)|add(#2,const_1)|divide(#3,n1) | general |
rahul is 30 years older than his daughter mary . in 20 years rahul will be twice as old as mary . what is mary current age . | now : mary = x , rahul = x + 30 in 20 years mary = x + 20 , rahul = x + 30 + 20 or 2 ( x + 20 ) x + 30 + 20 = 2 ( x + 20 ) x + 50 = 2 x + 40 50 - 40 = 2 x - x x = 10 mary is 10 years old answer : c | a ) 12 , b ) 14 , c ) 10 , d ) 16 , e ) 18 | c | subtract(add(30, 20), multiply(20, const_2)) | add(n0,n1)|multiply(n1,const_2)|subtract(#0,#1) | general |
a rectangular field has to be fenced on three sides leaving a side of 2020 feet uncovered . if the area of the field is 680680 sq . feet , how many feet of fencing will be required ? | "explanation : area of the field = 680 sq . feet . length of the adjacent sides are 20 feet and 680 / 20 = 34 feet . required length of the fencing = 20 + 34 + 34 = 88 feet answer : option c" | a ) 95 , b ) 92 , c ) 88 , d ) 82 , e ) 80 | c | subtract(rectangle_perimeter(2020, divide(680680, 2020)), 2020) | divide(n1,n0)|rectangle_perimeter(n0,#0)|subtract(#1,n0)| | geometry |
if a lends rs . 3500 to b at 10 % per annum and b lends the same sum to c at 13 % per annum then the gain of b in a period of 3 years is ? | "( 3500 * 3 * 3 ) / 100 = > 315 answer : d" | a ) 320 , b ) 300 , c ) 310 , d ) 315 , e ) 330 | d | subtract(divide(multiply(multiply(3500, 13), 3), const_100), divide(multiply(multiply(3500, 10), 3), const_100)) | multiply(n0,n2)|multiply(n0,n1)|multiply(#0,n3)|multiply(n3,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)| | gain |
what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular 11 - sided polygon ) ? | "from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 / 8 = 1 / 4 . the answer is c ." | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 11 | c | divide(11, divide(multiply(11, subtract(11, const_3)), const_2)) | subtract(n0,const_3)|multiply(n0,#0)|divide(#1,const_2)|divide(n0,#2)| | probability |
a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 6 % . what was its profit on the items sold in february ? | "assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.94 * 150 x = 141 x hence total profit = 41 % option e" | a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 37.5 % , e ) 41 % | e | subtract(multiply(divide(subtract(const_100, 6), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100) | add(n0,const_100)|add(n1,const_100)|subtract(const_100,n2)|divide(#2,const_100)|divide(#1,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)| | gain |
the perimeter of a semi circle is 144 cm then the radius is ? | 36 / 7 r = 144 = > r = 28 answer : b | ['a ) 22', 'b ) 28', 'c ) 99', 'd ) 77', 'e ) 16'] | b | divide(144, add(const_2, const_pi)) | add(const_2,const_pi)|divide(n0,#0) | physics |
find the area of the square , one of whose diagonals is 3.8 m long ? | area of the square = 1 / 2 ( diagonal ) ^ 2 = ( 1 / 2 * 3.8 * 3.8 ) m ^ 2 = 7.22 m ^ 2 answer ( a ) | ['a ) 7.22 m ^ 2', 'b ) 3.86 m ^ 2', 'c ) 8.96 m ^ 2', 'd ) 2.68 m ^ 2', 'e ) 7.89 m ^ 2'] | a | multiply(divide(const_1, const_2), multiply(3.8, 3.8)) | divide(const_1,const_2)|multiply(n0,n0)|multiply(#0,#1) | geometry |
in an election between two candidates , the winner has a margin of 10 % of the votes polled . if 1500 people change their mind and vote for the loser , the loser would have won by a margin of 10 % of the votes polled . find the total number of votes polled in the election ? | "total # of votes cast = 100 x , out of these , let the # of votes won by the winner = w , and for the loser ( l ) = 100 x - w . given that w - ( 100 x - w ) = 10 % of total votes cast = 10 x thus , 2 w - 100 x = 10 x → → w = 55 x and l = 100 x - 55 x = 45 x . according to the new condition , if the loser had won 45 x ... | a ) 30,000 , b ) 15,450 , c ) 45,000 , d ) 60,000 , e ) 15,000 | e | divide(1500, divide(10, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
in a certain game of dice , the player ’ s score is determined as a sum of three throws of a single die . the player with the highest score wins the round . if more than one player has the highest score , the winnings of the round are divided equally among these players . if john plays this game against 22 other player... | "to guarantee that john will get some monetary payoff he must score the maximum score of 6 + 6 + 6 = 18 , because if he gets even one less than that so 17 , someone can get 18 and john will get nothing . p ( 18 ) = 1 / 6 ^ 3 = 1 / 216 . answer : b ." | a ) 41 / 50 , b ) 1 / 216 , c ) 1 / 221 , d ) 1 / 84 , e ) 1 / 42 | b | divide(const_1, power(subtract(divide(22, const_3), const_1), const_2)) | divide(n0,const_3)|subtract(#0,const_1)|power(#1,const_2)|divide(const_1,#2)| | general |
line m lies in the xy - plane . the y - intercept of line m is - 2 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 8 ) and ( 8 , - 2 ) . what is the slope of line m ? | "the midpoint of ( 2,8 ) and ( 8 , - 2 ) is ( 5,3 ) . the slope of a line through ( 0 , - 2 ) and ( 5,3 ) is ( 3 - ( - 2 ) ) / ( 5 - 0 ) = 5 / 5 = 1 the answer is d ." | a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | d | divide(add(divide(subtract(8, 2), 2), 2), divide(add(2, 8), 2)) | add(n1,n3)|subtract(n2,n4)|divide(#1,n1)|divide(#0,n1)|add(#2,n0)|divide(#4,#3)| | general |
the sale price sarees listed for rs . 340 after successive discount is 20 % and 15 % is ? | "340 * ( 80 / 100 ) * ( 85 / 100 ) = 231 answer : b" | a ) 298 , b ) 231 , c ) 342 , d ) 876 , e ) 291 | b | subtract(subtract(340, divide(multiply(340, 20), const_100)), divide(multiply(subtract(340, divide(multiply(340, 20), const_100)), 15), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
what is 5 + 7 | d | a ) 2 , b ) 4 , c ) 10 , d ) 12 , e ) 11 | d | circle_area(divide(5, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | general |
a train crosses a platform of 100 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be â € ˜ x â € ™ x + 100 / 15 = x + 250 / 20 4 x + 400 = 3 x + 750 x = 350 m answer : e" | a ) 150 , b ) 88 , c ) 77 , d ) 62 , e ) 350 | e | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(100, divide(20, divide(15, const_3)))) | divide(n1,const_3)|divide(n1,#0)|divide(n3,#0)|multiply(n2,#1)|multiply(n0,#2)|subtract(#3,#4)| | physics |
if 0.75 : x : : 5 : 9 , then x is equal to : | "explanation : ( x * 5 ) = ( 0.75 * 9 ) x = 6.75 / 5 = 1.35 answer : d" | a ) 1.12 , b ) 1.16 , c ) 1.2 , d ) 1.35 , e ) none of these | d | divide(multiply(0.75, 9), 5) | multiply(n0,n2)|divide(#0,n1)| | general |
in the xy - plane , the point ( 3 , 1 ) is the center of a circle . the point ( 3 , - 2 ) lies inside the circle and the point ( - 2 , 1 ) lies outside the circle . if the radius r of the circle is an integer , then r = | "an easy way to solve this question will be just to mark the points on the coordinate plane . you ' ll see that the distance between the center ( 3 , 1 ) and the point inside the circle ( 3 , - 2 ) is 3 units ( both points are on x = 3 line so the distance will simply be 1 - ( - 2 ) = 3 ) so the radius must be more tha... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | add(add(2, const_2.0), 1) | add(n4,n4)|add(n1,#0)| | general |
s ( n ) is a n - digit number formed by attaching the first n perfect squares , in order , into one integer . for example , s ( 1 ) = 1 , s ( 2 ) = 14 , s ( 3 ) = 149 , s ( 4 ) = 14916 , s ( 5 ) = 1491625 , etc . how many digits t are in s ( 99 ) ? | "focus on the points where the number of digits in squares change : 1 , 2 , 3 - single digit squares . first 2 digit number is 10 . 4 , 5 , . . . 9 - two digit squares . to get 9 , the last number with two digit square , think that first 3 digit number is 100 which is 10 ^ 2 . so 9 ^ 2 must be the last 2 digit square .... | a ) t = 350 , b ) t = 353 , c ) t = 354 , d ) t = 356 , e ) 357 | b | add(add(add(add(const_60, 4), 4), multiply(add(add(const_60, 4), 4), 4)), add(multiply(3, 1), multiply(multiply(3, 2), 2))) | add(n6,const_60)|multiply(n0,n4)|multiply(n2,n4)|add(n6,#0)|multiply(n2,#2)|add(#1,#4)|multiply(n6,#3)|add(#3,#6)|add(#7,#5)| | general |
you have been given a physical balance and 7 weights of 57 , 55 , 45 , 44 , 43 , 41 and 85 kgs . keeping weights on one pan and object on the other , what is the maximum you can weigh less than 199 kgs . | "85 + 55 + 57 = 197 answer : d" | a ) 198 , b ) 195 , c ) 196 , d ) 197 , e ) 200 | d | add(add(add(55, 45), 44), 41) | add(n2,n3)|add(n4,#0)|add(n6,#1)| | general |
the average age of 7 men increases by 4 years when two women are included in place of two men of ages 26 and 30 years . find the average age of the women ? | "explanation : 26 + 30 + 7 * 4 = 84 / 2 = 42 answer : a" | a ) 42 , b ) 40 , c ) 30 , d ) 18 , e ) 11 | a | divide(add(add(26, 30), multiply(7, 4)), const_2) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|divide(#2,const_2)| | general |
what is the area of the square with the following coordinates : ( x , y ) , ( 20 , 20 ) , ( 20 , 9 ) , ( x , 9 ) ? | to find the area of a square , knowing the length of one side is sufficient . from the coordinates given , we can see that one side equals 11 . the area is 11 * 11 = 121 . the answer is a . | ['a ) 121', 'b ) 132', 'c ) 144', 'd ) 156', 'e ) 169'] | a | square_area(subtract(20, 9)) | subtract(n0,n3)|square_area(#0) | geometry |
450.271 + 369.422 ÷ 0.03 - 108.612 x 2.001 = ? | "explanation : ? = 450.271 + ( 369.422 ÷ 0.03 ) - ( 108.612 x 2.001 ) ≈ 450 + 369 / 0.03 - ( 109 x 2 ) ≈ 450 + 12300 - 218 ≈ 12532 answer : option b" | a ) 12980 , b ) 12532 , c ) 12610 , d ) 12430 , e ) 12560 | b | add(450.271, divide(multiply(369.422, 0.03), divide(108.612, 2.001))) | divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)| | general |
how many 3 - digit numerals begin with a digit that represents a prime number ? | prime digits 2 , 35 and 7 . three digit numbers _ _ _ 1 st place can be filled in 4 ways 2 nd place can be filled in 10 ways 3 rd place can be filled in 10 ways total = 4 * 10 * 10 = 400 ans : c | a ) 16 , b ) 80 , c ) 400 , d ) 180 , e ) 240 | c | add(multiply(const_100, 3), const_100) | multiply(n0,const_100)|add(#0,const_100) | general |
if leo gains 10 pounds , he will weigh 50 % more than his sister kendra . currently their combined weight is 150 pounds . what is leo ' s current weight ? | "l + k = 150 and so k = 150 - l l + 10 = 1.5 k = 1.5 ( 150 - l ) 2.5 l = 215 l = 86 the answer is b ." | a ) 80 , b ) 86 , c ) 92 , d ) 98 , e ) 104 | b | divide(subtract(multiply(150, divide(add(const_100, 50), const_100)), 10), add(divide(add(const_100, 50), const_100), const_1)) | add(n1,const_100)|divide(#0,const_100)|add(#1,const_1)|multiply(n2,#1)|subtract(#3,n0)|divide(#4,#2)| | gain |
a group of workers had agreed on a very strange payment schedule . each of them would get $ 100 for every day that they had worked , but for every day they did not work will have to return $ 25 . after 30 days they realised they did not earn any money . how many days the workers did not work ? | the workers earn 4 times more than they have to return per day for not working . so the number of days they did not work is 4 times the number of days they worked . that means 24 days of not working in 30 days . correct answer b | a ) 15 , b ) 24 , c ) 30 , d ) 20 , e ) 10 | b | subtract(30, divide(multiply(25, 30), add(100, 25))) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|subtract(n2,#2) | physics |
a pump can fill a tank with a water in 2 hours . because of a leak , it took 2 x 1 / 3 hours to fill the tank . the leak can drain all the water of the tank in | "sol . work done by the leak in 1 hour = ( 1 / 2 - 3 / 7 ) = 1 / 14 . ∴ leak will empty the tank in 14 hrs . answer d" | a ) 5 hrs , b ) 7 hrs , c ) 12 hrs , d ) 14 hrs , e ) none | d | divide(const_1, subtract(divide(const_1, 2), divide(const_1, add(2, divide(1, 3))))) | divide(const_1,n0)|divide(n2,n3)|add(n0,#1)|divide(const_1,#2)|subtract(#0,#3)|divide(const_1,#4)| | physics |
if 2 x + y = 7 and x + 2 y = 5 , then ( 2 x + 2 y ) / 3 = | "2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 ( 2 * 3 + 2 * 1 ) / 3 = ( 6 + 3 ) / 3 = 9 / 3 = 3 a" | a ) 3 , b ) 4 / 3 , c ) 17 / 5 , d ) 18 / 5 , e ) 4 | a | divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 2) | multiply(n0,n1)|multiply(n0,n0)|subtract(#0,n3)|subtract(#1,const_1)|divide(#2,#3)|multiply(n0,#4)|subtract(n1,#5)|add(#4,#6)|divide(#7,n4)| | general |
johnny travels a total of one hour to and from school . on the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 25 miles per hour . how far is it to the school ? | answer : c ) 6.6 miles . average speed for round trip = 2 * a * b / ( a + b ) , where a , b are speeds so , average speed was = 2 * 5 * 25 / ( 5 + 25 ) = 6.6 m / hr the distance between schoolhome should be half of that . ie . 6.6 miles answer c | a ) 2 miles , b ) 4 miles , c ) 6.6 miles , d ) 8 miles , e ) 10 miles | c | multiply(divide(const_1, add(divide(const_1, 5), divide(const_1, 25))), const_1_6) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,const_1_6) | physics |
what is the units digit of ( 12 ^ 3 ) ( 15 ^ 4 ) ( 31 ^ 7 ) ? | "the units digit of 12 ^ 3 is the units digit of 2 ^ 3 which is 8 . the units digit of 15 ^ 4 is the units digit of 5 ^ 4 which is 5 . the units digit of 31 ^ 7 is the units digit of 1 ^ 7 which is 1 . the units digit of 8 * 5 * 1 is 0 . the answer is a ." | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | a | divide(add(multiply(factorial(12), factorial(3)), multiply(factorial(12), factorial(4))), 12) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 12 % ? | "let marked price = rs . 100 . then , c . p . = rs . 64 , s . p . = rs . 88 gain % = 24 / 64 * 100 = 37.5 % . answer : a" | a ) 37.5 % , b ) 37.6 % , c ) 38.5 % , d ) 17.5 % , e ) 37.2 % | a | multiply(subtract(divide(subtract(const_100, 12), 64), const_1), const_100) | subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)| | gain |
the average of 25 results is 18 . the average of first 12 of those is 14 and the average of last 12 is 17 . what is the 13 th result ? | "solution : sum of 1 st 12 results = 12 * 14 sum of last 12 results = 12 * 17 13 th result = x ( let ) now , 12 * 14 + 12 * 17 + x = 25 * 18 or , x = 78 . answer : option d" | a ) 74 , b ) 75 , c ) 69 , d ) 78 , e ) 45 | d | subtract(subtract(multiply(25, 18), multiply(12, 17)), multiply(12, 14)) | multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)| | general |
an woman sitting in a train which is travelling at 20 kmph observes that a goods train travelling in a opposite direction , takes 15 seconds to pass him . if the goods train is 300 m long , find its speed . | relative speed = ( 300 / 15 ) m / s = ( 300 / 15 ) * ( 18 / 5 ) = 72 kmph speed of goods train = 72 - 20 = 52 kmph answer is a | a ) 52 , b ) 54 , c ) 56 , d ) 58 , e ) 60 | a | subtract(speed(speed(300, 15), const_0_2778), 20) | speed(n2,n1)|speed(#0,const_0_2778)|subtract(#1,n0) | physics |
a number consists of 3 digits whose sum is 10 . the middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed . the number is | solution let the number be x . then 2 x = 10 or x = 5 . so , the number is either 253 or 352 . since the number increases on reversing the digits , so the hundred ' s digit is smaller than the units digit . hence , required number = 253 . answer c | a ) 145 , b ) 185 , c ) 253 , d ) 370 , e ) none | c | add(add(multiply(const_100, subtract(divide(10, const_2), 3)), multiply(divide(10, const_2), 10)), 3) | divide(n1,const_2)|multiply(n1,#0)|subtract(#0,n0)|multiply(#2,const_100)|add(#3,#1)|add(n0,#4) | general |
what least number must be subtracted from 7538 so that remaining no . is divisible by 14 | "explanation : on dividing 7538 by 14 we get the remainder 6 , so 6 should be subtracted option c" | a ) 3 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | c | subtract(7538, multiply(floor(divide(7538, 14)), 14)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
in a survey of parents , exactly 5 / 6 of the mothers and 3 / 4 of the fathers held full - time jobs . if 60 percent of the parents surveyed were women , what percent of the parents did not hold full - time jobs ? | "fathers without full - time jobs are 1 / 4 * 2 / 5 = 2 / 20 of all the parents surveyed . mothers without full - time jobs are 1 / 6 * 3 / 5 = 1 / 10 of all the parents surveyed . the percent of parents without full - time jobs is 2 / 20 + 1 / 10 = 1 / 5 = 20 % the answer is d ." | a ) 12 % , b ) 15 % , c ) 18 % , d ) 20 % , e ) 25 % | d | add(subtract(subtract(const_100, 60), multiply(divide(3, 4), subtract(const_100, 60))), subtract(60, multiply(divide(5, 6), 60))) | divide(n2,n3)|divide(n0,n1)|subtract(const_100,n4)|multiply(#0,#2)|multiply(n4,#1)|subtract(#2,#3)|subtract(n4,#4)|add(#5,#6)| | general |
on a scale of map , 0.4 cm represents 5.3 km . if the distance between the points on the map is 64 cm , the actual distance between these points is : | explanation : let the actual distance be x km . then , more distance on the map , more is the actual distance ( direct proportion ) = > 0.4 : 64 : : 5.3 : x = > 0.4 x = 64 x 5.3 = > x = 64 x 5.3 / 0.4 = > x = 848 answer : d | a ) 9 km , b ) 72.5 km , c ) 190.75 km , d ) 848 km , e ) none of these | d | multiply(divide(5.3, 0.4), 64) | divide(n1,n0)|multiply(n2,#0) | physics |
the ratio w , by volume of soap to alcohol to water in a 76 litre solution is 2 : 50 : 100 . the solution is then altered by adding more soap , alcohol and water . after this alteration the ratio , by volume of soap to water in the solution doubles whereas the ratio , by volume of soap to water remains the same as befo... | i guess it should be the ratio w , by volume ofsoaptowaterin the solutiondoubleswhereas the ratio , by volume ofalocoholtowaterremains thesameas before 2 : 50 : 100 = > 1 : 25 : 50 . if we add all the parts , we get 76 liters so we have 1 liters of soap , 25 liters of alcohol and 50 liters of water . now as per the que... | a ) 79 litres , b ) 78 litres , c ) 77 litres , d ) 152 liters , e ) 304 litres | c | add(76, const_1) | add(n0,const_1) | general |
the sum of the numbers is 98 . if the ratio between the first and the second be 2 : 3 and that between the second and third be 5 : 8 , then find the second number ? | given ratios 2 : 3 5 : 8 10 : 15 : 24 the second number = 98 / ( 10 + 15 + 24 ) * 15 = 30 answer is b | a ) 24 , b ) 30 , c ) 36 , d ) 42 , e ) 50 | b | divide(multiply(98, lcm(3, 5)), add(multiply(8, 3), add(multiply(5, 2), multiply(5, 3)))) | lcm(n2,n3)|multiply(n1,n3)|multiply(n2,n3)|multiply(n2,n4)|add(#1,#2)|multiply(n0,#0)|add(#4,#3)|divide(#5,#6) | other |
jaime earned enough money by selling seashells at 35 cents each to buy several used paperback books at 55 cents each . if he spent all of the money he earned selling seashells to buy the books , what is the least number of seashells he could have sold ? | let ' s test answer d : 22 seashells . . . . with 22 seashells , jamie would have 22 ( 35 ) = 770 cents . this would allow him to buy 14 books for 770 cents total , with no money left over . this is an exact match for what we were told , so this must be the answer . final answer : [ reveal ] spoiler : d | a ) 5 , b ) 11 , c ) 17 , d ) 22 , e ) 30 | d | add(divide(lcm(35, 55), 35), const_10) | lcm(n0,n1)|divide(#0,n0)|add(#1,const_10) | general |
a manufacturer produces a certain men ' s athletic shoe in integer sizes from 8 to 17 . for this particular shoe , each unit increase in size corresponds to a 1 / 5 - inch increase in the length of the shoe . if the largest size of this shoe is 40 % longer than the smallest size , how long , in inches , is the shoe in ... | let x be the length of the size 8 shoe . then 0.4 x = 9 / 5 x = 4.5 inches the size 15 shoe has a length of 4.5 + 7 / 5 = 5.9 inches the answer is b . | a ) 5.4 , b ) 5.9 , c ) 6.3 , d ) 6.7 , e ) 7.5 | b | add(divide(multiply(divide(1, 5), subtract(17, 8)), divide(40, const_100)), multiply(subtract(15, 8), divide(1, 5))) | divide(n2,n3)|divide(n4,const_100)|subtract(n1,n0)|subtract(n5,n0)|multiply(#0,#2)|multiply(#0,#3)|divide(#4,#1)|add(#6,#5) | general |
the average salary / head of allthe workers in a workshop is rs . 850 , if the average salary / head of 7 technician is rs . 1000 and the average salary / head of the rest is rs . 780 , the total no . of workers in the work - shop is ? | "let the total number of workers be y . so sum of salary for all workers = sum of salary of 7 technician + sum of salary for other y - 7 workers . 7 x 1000 + 780 ( y - 7 ) = 850 y ⇒ 7000 + 780 y - 5460 = 850 y ⇒ 70 y = 1540 ∴ y = 22 so total number of workers = 22 c" | a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | c | divide(subtract(multiply(7, 1000), multiply(7, 780)), subtract(850, 780)) | multiply(n1,n2)|multiply(n1,n3)|subtract(n0,n3)|subtract(#0,#1)|divide(#3,#2)| | general |
a man buys a cycle for rs . 1600 and sells it at a loss of 20 % . what is the selling price of the cycle ? | "s . p . = 80 % of rs . 1600 = 80 / 100 x 1600 = rs . 1280 answer : e" | a ) 1410 , b ) 1420 , c ) 1430 , d ) 1440 , e ) 1280 | e | divide(multiply(subtract(const_100, 20), 1600), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain |
15 people went to a hotel for combine dinner party 10 of them spent rs . 50 each on their dinner and rest spent 4 more than the average expenditure of all the 15 . what was the total money spent by them . | "solution : let average expenditure of 15 people be x . then , 15 x = 10 * 50 + 5 * ( x + 4 ) ; or , 15 x = 10 * 50 + 5 x + 20 ; or , x = 52 ; so , total money spent = 52 * 15 = rs . 780 . answer : option c" | a ) 1628.4 , b ) 1534 , c ) 780 , d ) 1496 , e ) none of these | c | multiply(divide(add(multiply(10, 50), multiply(subtract(15, 10), 4)), subtract(15, subtract(15, 10))), 15) | multiply(n1,n2)|subtract(n0,n1)|multiply(n3,#1)|subtract(n0,#1)|add(#0,#2)|divide(#4,#3)|multiply(n0,#5)| | general |
what is the maximum number of pieces of birthday cake of size 5 ” by 5 ” that can be cut from a cake 15 ” by 15 ” ? | "the prompt is essentially asking for the maximum number of 5 x 5 squares that can be cut from a larger 15 by 15 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 3 ' pieces ' each , we have ( 3 ) ( 3 ) = 9 total smaller squares ( at maximum ) . e" | a ) 5 , b ) 10 , c ) 16 , d ) 20 , e ) 9 | e | divide(multiply(15, 15), multiply(5, 5)) | multiply(n2,n2)|multiply(n0,n0)|divide(#0,#1)| | physics |
the width of a rectangular hall is ½ of its length . if the area of the hall is 450 sq . m , what is the difference between its length and breadth ? | "let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 450 = x * 1 x / 2 x ² = 900 x = 30 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 30 / 2 = 15 m answer : d" | a ) 8 m , b ) 10 m , c ) 12 m , d ) 15 m , e ) 17 m | d | divide(sqrt(divide(450, divide(const_1, const_2))), const_2) | divide(const_1,const_2)|divide(n0,#0)|sqrt(#1)|divide(#2,const_2)| | geometry |
if p / q = 4 / 5 , then the value of 1 / 7 + { ( 2 q - p ) / ( 2 q + p ) } is ? | "answer given exp . = 4 / 7 + { ( 2 q - p ) / ( 2 q + p ) } dividing numerator as well as denominator by q , exp = 1 / 7 + { 2 - p / q ) / ( 2 + p / q ) } = 1 / 7 + { ( 2 - 4 / 5 ) / ( 2 + 4 / 5 ) } = 1 / 7 + 6 / 14 = 1 / 7 + 3 / 7 = 4 / 7 correct option : a" | a ) 4 / 7 , b ) 34 , c ) 1 , d ) 2 , e ) 3 | a | add(divide(1, 7), divide(subtract(2, divide(4, 5)), add(2, divide(4, 5)))) | divide(n2,n3)|divide(n0,n1)|add(n4,#1)|subtract(n4,#1)|divide(#3,#2)|add(#0,#4)| | general |
when a natural number n is successively divided by 1020 , 3040 . the remainders are 5060 , 7080 . what will be the sum of the remainders if the order of the division is reversed ? | 10 20 30 40 50 60 70 80 leave the top right - most number 8 start with bottom right - most number 5 80 * 30 + 70 = 2470 2470 * 20 + 60 = 49460 49460 * 10 + 50 = 494650 this is the number required now , do the successive division in the reverse order the sum of the remainders is 28 hence , the correct option is c | a ) 16 , b ) 14 , c ) 28 , d ) 11 , e ) 9 | c | add(multiply(add(add(const_4, const_3), add(const_1, const_2)), const_2), multiply(floor(divide(5060, 1020)), floor(divide(7080, 3040)))) | add(const_3,const_4)|add(const_1,const_2)|divide(n2,n0)|divide(n3,n1)|add(#0,#1)|floor(#2)|floor(#3)|multiply(#4,const_2)|multiply(#5,#6)|add(#7,#8) | general |
in the land of oz only one or two - letter words are used . the local language has 68 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ? | "the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa , bb , . . . the number of 1 letter words ( x ) that can be made from 68 letters is 68 ; the number of 2 letter words ( xx ) that can be made from 68 letters is 68 * 68 , since each x can take 68 values... | a ) 65 , b ) 66 , c ) 67 , d ) 131 , e ) 136 | e | add(add(const_1, 68), 68) | add(n0,const_1)|add(n0,#0)| | general |
if the sales tax be reduced from 4 ( 1 / 3 ) % to 2 ( 1 / 2 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 8000 ? | "explanation : required difference = ( 4 ( 1 / 3 ) of rs . 8000 ) - ( 2 ( 1 / 2 ) of rs . 8000 ) = ( 13 / 3 – 5 / 2 ) % of rs . 8000 = ( 11 / 6 ) x ( 1 / 100 ) x 8000 = rs . 146.66 answer : b" | a ) 146.69 , b ) 146.66 , c ) 146.62 , d ) 146.61 , e ) 146.6 | b | subtract(multiply(add(divide(add(4, divide(1, 4)), const_100), 1), divide(8000, add(divide(add(divide(1, 2), 1), const_100), 1))), 8000) | divide(n1,n0)|divide(n1,n5)|add(n0,#0)|add(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|add(n1,#4)|add(n1,#5)|divide(n6,#7)|multiply(#6,#8)|subtract(#9,n6)| | general |
how many positive 3 - digit integers are divisible by both 3 and 7 ? | a number to be divisible by both 3 and 7 should be divisible by the least common multiple of 3 and 7 so by 21 . multiples of 21 between 100 and 999 , inclusive is ( last - first ) / multiple + 1 = ( 987 - 105 ) / 21 + 1 = 42 + 1 = 43 answer : a . | a ) 43 , b ) 45 , c ) 47 , d ) 50 , e ) 52 | a | subtract(floor(divide(const_1000, lcm(3, 7))), floor(divide(const_100, lcm(3, 7)))) | lcm(n0,n2)|divide(const_1000,#0)|divide(const_100,#0)|floor(#1)|floor(#2)|subtract(#3,#4) | general |
the average of 50 numbers id 62 . if two numbers , namely 45 and 55 are discarded , the average of the remaining numbers is : | "explanation : total of 50 numbers = ( 50 × 62 ) = 3100 total of 48 numbers = ( 3100 - ( 45 + 55 ) ] = 3000 required average = 3000 / 48 = 62.5 answer : e" | a ) 22.3 , b ) 33.2 , c ) 22.2 , d ) 51.3 , e ) 62.5 | e | divide(subtract(multiply(50, 62), add(45, 55)), subtract(50, const_2)) | add(n2,n3)|multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,#0)|divide(#3,#2)| | general |
in what time will a train 100 meters long cross an electric pole , if its speed is 144 km / hr | "first convert speed into m / sec speed = 144 * ( 5 / 18 ) = 40 m / sec time = distance / speed = 100 / 40 = 2.5 seconds answer : d" | a ) 8.5 seconds , b ) 2.8 seconds , c ) 3.5 seconds , d ) 2.5 seconds , e ) 2.6 seconds | d | divide(100, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
what is the unit digit of ( 6 ! * 6 ! / 6 ! * 3 ! ) ? | ( 6 ! * 6 ! / 6 ! * 3 ! ) = ( 6 ! / 3 ! ) = 720 / 6 = 120 units digit of the above product will be equal to 0 answer a | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | divide(multiply(factorial(6), factorial(6)), multiply(factorial(6), factorial(3))) | factorial(n0)|factorial(n3)|multiply(#0,#0)|multiply(#0,#1)|divide(#2,#3) | general |
two vessels contains equal number of mixtures milk and water in the ratio 5 : 3 and 6 : 2 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ? | "the ratio of milk and water in the new vessel is = ( 5 / 8 + 6 / 8 ) : ( 3 / 8 + 2 / 8 ) = 11 / 8 : 5 / 8 = 11 : 5 answer is a" | a ) 11 : 5 , b ) 9 : 13 , c ) 5 : 11 , d ) 11 : 3 , e ) 15 : 4 | a | divide(add(multiply(5, divide(add(6, 2), add(5, 3))), 6), add(multiply(3, divide(add(6, 2), add(5, 3))), 2)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n2,#3)|add(n3,#4)|divide(#5,#6)| | other |
mona and donald fly to rome for the weekend . they take cash only in notes of $ 10 and notes of € 10 . mona carries 3 times the amount of euros donald carries . she also carries as many dollars as donald carries . the number of € 10 notes they take is double the number of $ 10 notes they take . if donald carries a tota... | let e 10 = x no . d 10 = y no . donald is having x + y notes mona carries 3 x + y again x = 2 y or donald x + y = 39 or 3 y = 39 y = 13 ; x = 26 , total notes they carry = 104 + 26 = 130 a | a ) 130 , b ) 80 , c ) 100 , d ) 120 , e ) 150 | a | add(multiply(39, 3), divide(39, 3)) | divide(n5,n2)|multiply(n2,n5)|add(#0,#1) | general |
in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 34 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ? | "working backward from the day it ' s covered : day 34 : fully covered day 33 : half covered so 33 days answer : a" | a ) 33 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47 | a | subtract(34, const_1) | subtract(n0,const_1)| | physics |
the water level in a reservoir has been dropping at the rate of 14 inches per day . exactly 5 days ago , the water level was at h inches . what will be the water level exactly 4 days from now if the rate at which the level is dropping remains the same ? | "drop = 14 inches / day 5 days ago = h , means now it ' s equal h - 70 and in 4 days = h - 70 - 56 = h - 126 answer e" | a ) h − 12 , b ) h − 56 , c ) h − 14 , d ) h + 14 , e ) h - 126 | e | multiply(14, divide(4, 5)) | divide(n2,n1)|multiply(n0,#0)| | gain |
how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 850 meters in length ? | "t = ( 850 + 165 ) / 54 * 18 / 5 t = 67.7 answer : e" | a ) 28 , b ) 27 , c ) 55 , d ) 18 , e ) 67.7 | e | divide(add(165, 850), multiply(54, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
in the faculty of reverse - engineering , 350 second year students study numeric methods , 250 second year students study automatic control of airborne vehicles and 100 second year students study them both . how many students are there in the faculty if the second year students are approximately 25 % of the total ? | "total number of students studying both are 350 + 250 - 100 = 500 ( subtracting the 100 since they were included in the both the other numbers already ) . so 25 % of total is 500 , so 100 % is 2000 answer is d" | a ) 1800 , b ) 1900 , c ) 2100 , d ) 2000 , e ) 2200 | d | add(350, 250) | add(n0,n1)| | general |
how many numbers from 39 to 79 are exactly divisible by 11 ? | 39 / 11 = 1 and 79 / 11 = 7 = = > 7 - 3 = 4 numbers answer : c | a ) 5 , b ) 7 , c ) 4 , d ) 11 , e ) 12 | c | add(divide(subtract(multiply(floor(divide(79, 11)), 11), multiply(add(floor(divide(39, 11)), const_1), 11)), 11), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1) | general |
in the coordinate plane , points ( x , 1 ) and ( 12 , y ) are on line k . if line k passes through the origin and has slope 1 / 2 , then x + y = | "line k passes through the origin and has slope 1 / 2 means that its equation is y = 1 / 2 * x . thus : ( x , 1 ) = ( 2 , 1 ) and ( 12 , y ) = ( 12,6 ) - - > x + y = 2 + 6 = 8 . answer : c ." | a ) 4.5 , b ) 7 , c ) 8 , d ) 11 , e ) 12 | c | multiply(multiply(12, 2), divide(1, 2)) | divide(n0,n3)|multiply(n1,n3)|multiply(#0,#1)| | general |
in a rectangular coordinate system , what is the area of a rhombus whose vertices have the coordinates ( 0 , 3.5 ) , ( 8 , 0 ) , ( 0 , - 3.5 ) , ( - 8 , 0 ) ? | "area of rhombus = 1 / 2 * d 1 * d 2 length of 1 st diagonal = 8 + 8 = 16 length of 2 nd diagonal = 3.5 + 3.5 = 7 area = 1 / 2 * 16 * 7 = 56 a is the answer" | a ) 56 , b ) 88 , c ) 112 , d ) 116 , e ) 120 | a | rhombus_area(multiply(8, const_2), multiply(3.5, const_2)) | multiply(n2,const_2)|multiply(n1,const_2)|rhombus_area(#0,#1)| | geometry |
when a number is divided by 6 & then multiply by 12 the answer is 8 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 8 = > 2 x = 8 = > x = 4.0 e" | a ) 4.5 , b ) 5 , c ) 5.5 , d ) 5.8 , e ) 4 | e | multiply(divide(8, 12), 6) | divide(n2,n1)|multiply(n0,#0)| | general |
in one year , the population , of a village increased by 30 % and in the next year , it decreased by 30 % . if at the end of 2 nd year , the population was 13650 , what was it in the beginning ? | "x * 130 / 100 * 70 / 100 = 13650 x * 0.91 = 13650 x = 13650 / 0.91 = > 15000 answer : c" | a ) 7787 , b ) 8000 , c ) 15000 , d ) 1277 , e ) 2081 | c | divide(divide(13650, subtract(const_1, divide(30, const_100))), add(const_1, divide(30, const_100))) | divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|divide(n3,#2)|divide(#3,#1)| | general |
three table runners have a combined area of 208 square inches . by overlapping the runners to cover 80 % of a table of area 175 square inches , the area that is covered by exactly two layers of runner is 24 square inches . what is the area of the table that is covered with three layers of runner ? | total = a + b + c - ( sum of exactly 2 - group overlaps ) - 2 * ( all three ) + neither 80 % * 175 = 208 - 24 - 2 * ( all three ) + 0 2 * ( all three ) = 208 - 24 - 140 all three = 22 answer : a | ['a ) 22 square inches', 'b ) 20 square inches', 'c ) 24 square inches', 'd ) 28 square inches', 'e ) 30 square inches'] | a | divide(subtract(subtract(208, 24), multiply(175, divide(80, const_100))), const_2) | divide(n1,const_100)|subtract(n0,n3)|multiply(n2,#0)|subtract(#1,#2)|divide(#3,const_2) | geometry |
calculate the amount that an investor needs to be invest to earn $ 460 in interest in 12 months if the investor plans to invest x dollars in a savings account that pays interest at an annual rate of 9 % compounded semi - annually ? | "the approach is substitution , our interest requirement is 460 after 12 months , 2 compounding period . calculate the compound interest on each option and find out the one that yields 460 in 12 months 5,000 yielded $ 460 using the formula a = p ( 1 + r / n ) nt hence answer is d" | a ) 7,000 , b ) 4,000 , c ) 6,000 , d ) 5,000 , e ) 8,000 | d | divide(multiply(const_100, 460), 9) | multiply(n0,const_100)|divide(#0,n2)| | gain |
the least number , which when divided by 48,60 , 72,108 and 140 leaves 38,50 , 62,98 and 130 as remainder respectively , is : | "solution here ( 48 - 38 ) = 10 , ( 60 - 50 ) = 10 , ( 72 - 62 ) = 10 , ( 108 - 98 ) = 10 & ( 140 - 130 ) = 10 . so , required number = ( l . c . m . of 48 , 60,72 , 108,140 ) - 10 = 15120 - 10 = 15110 . answer b" | a ) 11115 , b ) 15110 , c ) 15120 , d ) 15210 , e ) 12510 | b | multiply(38,50, const_10) | multiply(n3,const_10)| | general |
company c sells a line of 25 products with an average retail price of $ 1,200 . if none of these products sells for less than $ 400 , and exactly 12 of the products sell for less than $ 1,000 , what is the greatest possible selling price of the most expensive product ? | "the average price of 25 products is $ 1,200 means that the total price of 25 products is 25 * 1,200 = $ 30,000 . next , since exactly 12 of the products sell for less than $ 1,000 , then let ' s make these 12 items to be at $ 400 each ( min possible ) . now , the remaining 12 items can not be priced less than $ 1,000 ... | a ) 10,000 , b ) 11.0 , c ) 12,000 , d ) 13,200 , e ) 14,000 | d | subtract(multiply(multiply(const_12, const_100), 25), add(multiply(400, 12), multiply(subtract(subtract(25, 12), const_1), 1,000))) | multiply(const_100,const_12)|multiply(n2,n3)|subtract(n0,n3)|multiply(n0,#0)|subtract(#2,const_1)|multiply(#4,n4)|add(#1,#5)|subtract(#3,#6)| | general |
how many bricks , each measuring 80 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 80 x 11.25 x 6 ) = 2000 answer : b" | a ) 6400 , b ) 2000 , c ) 5500 , d ) 7400 , e ) 3000 | b | divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(80, 11.25), 6)) | multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)| | physics |
a man goes downstream at 13 kmph , and upstream 8 kmph . the speed of the stream is | "speed of the stream = 1 / 2 ( 13 - 8 ) kmph = 2.5 kmph . correct option : d" | a ) 0 kmph , b ) 4 kmph , c ) 16 kmph , d ) 2.5 kmph , e ) 26 kmph | d | divide(subtract(13, 8), const_2) | subtract(n0,n1)|divide(#0,const_2)| | physics |
drum x is 1 / 2 full of oil and drum y , which has twice the capacity of drum x , is 1 / 2 full of oil . if all of the oil in drum x is poured into drum y , then drum y will be filled to what capacity ? | "( 1 / 2 ) x = ( 1 / 4 ) y ( 1 / 4 ) y + ( 1 / 2 ) y = ( 3 / 4 ) y the answer is b ." | a ) 2 / 3 , b ) 3 / 4 , c ) 4 / 5 , d ) 5 / 6 , e ) 7 / 8 | b | divide(add(multiply(divide(1, 2), const_12), multiply(divide(const_12, 2), divide(1, 2))), const_12) | divide(n0,n3)|divide(const_12,n1)|divide(n0,n1)|multiply(#0,const_12)|multiply(#1,#2)|add(#3,#4)|divide(#5,const_12)| | general |
the average weight of 8 people increases by 2.5 kg when a new person comes in place of one of them weighing 45 kg . what is the weight of the new person ? | the total weight increase = ( 8 x 2.5 ) kg = 20 kg weight of new person = ( 45 + 20 ) kg = 65 kg the answer is d . | a ) 75 kg , b ) 85 kg , c ) 95 kg , d ) 65 kg , e ) 55 kg | d | add(multiply(2.5, 8), 45) | multiply(n0,n1)|add(n2,#0) | general |
average of 10 matches is 32 , how many runs one should should score to increase his average by 3 runs . | "explanation : average after 11 innings should be 35 so , required score = ( 11 * 35 ) - ( 10 * 32 ) = 385 - 320 = 65 answer : option a" | a ) a ) 65 , b ) b ) 76 , c ) c ) 78 , d ) d ) 80 , e ) e ) 88 | a | subtract(multiply(add(32, 3), add(10, const_1)), multiply(10, 32)) | add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
how many internal diagonals does a hexagon ( six sided polygon ) have ? | "number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 6 no . of diagonals = 6 ( 6 - 3 ) / 2 = 9 ans d" | a ) 5 , b ) 6 , c ) 8 , d ) 9 , e ) 10 | d | multiply(subtract(multiply(const_2, const_4), const_3), divide(multiply(const_2, const_4), const_2)) | multiply(const_2,const_4)|divide(#0,const_2)|subtract(#0,const_3)|multiply(#1,#2)| | geometry |
running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 3 w widgets . | "i am getting 12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + 2 days . 1 / ( y + 2 ) + 1 / y = 5 / 12 ( 5 / 12 is because ( 5 / 4 ) w widgets are done in 3 days . so , x widgets will be done in 12 / 5 days or 5 / 12 th of a widget ... | a ) 4 , b ) 6 , c ) 18 , d ) 10 , e ) 12 | c | multiply(multiply(3, 2), 3) | multiply(n0,n4)|multiply(n3,#0)| | general |
find a sum for 1 st 7 prime number ' s ? | "required sum = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 ) = 58 note : 1 is not a prime number option c" | a ) 25 , b ) 28 , c ) 58 , d ) 34 , e ) 36 | c | add(add(add(add(add(add(add(add(const_2, const_3), add(const_2, const_3)), add(add(const_2, const_3), const_2)), add(7, const_2)), add(add(7, const_2), const_2)), add(add(add(7, const_2), const_2), const_4)), add(add(add(add(7, const_2), const_2), const_4), const_2)), add(add(add(add(add(7, const_2), const_2), const_4)... | add(const_2,const_3)|add(n1,const_2)|add(#0,#0)|add(#0,const_2)|add(#1,const_2)|add(#2,#3)|add(#4,const_4)|add(#5,#1)|add(#6,const_2)|add(#7,#4)|add(#8,const_4)|add(#9,#6)|add(#11,#8)|add(#12,#10)| | general |
a and b can do a piece of work in 9 days . with the help of c they finish the work in 6 days . c alone can do that piece of work in ? | "c = 1 / 6 â € “ 1 / 9 = 1 / 18 = > 18 days answer : c" | a ) 33 , b ) 878 , c ) 18 , d ) 88 , e ) 11 | c | inverse(subtract(6, divide(6, 9))) | divide(n1,n0)|subtract(n1,#0)|inverse(#1)| | physics |
the length of a rectangular garden is three times its width . if the area of the rectangular garden is 507 square meters , then what is the width of the rectangular garden ? | let x be the width of the garden . 3 x ^ 2 = 507 x ^ 2 = 169 x = 13 the answer is d . | ['a ) 10', 'b ) 11', 'c ) 12', 'd ) 13', 'e ) 14'] | d | sqrt(divide(507, const_3)) | divide(n0,const_3)|sqrt(#0) | geometry |
due to construction , the speed limit along an 9 - mile section of highway is reduced from 55 miles per hour to 30 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 9 miles stretch = 9 * 60 / 55 = 9 * 12 / 11 = 9.81 new time in minutes to cross 9 miles stretch = 9 * 60 / 30 = 9 * 2 / 1 = 18 time difference = 8.19 ans : e" | a ) a ) 5.61 , b ) b ) 8 , c ) c ) 10 , d ) d ) 15 , e ) e ) 8.19 | e | max(multiply(subtract(add(55, 9), const_1), subtract(divide(9, 30), divide(9, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics |
220 % of a number x is 44 . what is 44 % of x . | solution : given , 220 % of x = 44 or , x = 20 . thus , 44 % of 20 = ( 44 * 20 ) / 100 = 8.8 answer : option a | a ) 8.8 , b ) 8.9 , c ) 6.6 , d ) 7.7 , e ) none | a | subtract(add(add(divide(220, 44), 44), const_4), 44) | divide(n0,n1)|add(n1,#0)|add(#1,const_4)|subtract(#2,n1) | gain |
what is the smallest of 6 consecutive odd integers whose average ( arithmetic mean ) is b + 2 ? | since the numbers are consecutive odd integers , mean = median = 3 rd integer + 4 th integer / 2 and 1 st integer = 3 rd integer - 4 let ' s say 3 rd integer = n and 4 th integer = n + 2 2 n + 2 / 2 = b + 2 n = b + 1 1 st integer = b + 1 - 4 = b - 3 a is the answer | a ) b - 3 , b ) b - 6 , c ) b - 1 , d ) b , e ) b + 1 | a | subtract(const_2, subtract(6, const_1)) | subtract(n0,const_1)|subtract(const_2,#0) | general |
a motorcyclist goes from bombay to pune , a distance of 180 kms at an average of 20 kmph speed . another man starts from bombay by car 2 â ½ hours after the first , and reaches pune â ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | "t = 180 / 20 = 9 h t = 9 - 3 = 6 time ratio = 9 : 6 = 3 : 2 speed ratio = 2 : 3 answer : b" | a ) 1 : 2 , b ) 2 : 3 , c ) 1 : 9 , d ) 1 : 5 , e ) 1 : 1 | b | divide(divide(180, divide(180, 20)), divide(180, subtract(divide(180, 20), const_3))) | divide(n0,n1)|divide(n0,#0)|subtract(#0,const_3)|divide(n0,#2)|divide(#1,#3)| | physics |
on thursday mabel handled 90 transactions . anthony handled 10 % more transactions than mabel , cal handled 2 / 3 rds of the transactions that anthony handled , and jade handled 19 more transactions than cal . how much transactions did jade handled ? | "solution : mabel handled 90 transactions anthony handled 10 % more transactions than mabel anthony = 90 + 90 × 10 % = 90 + 90 × 0.10 = 90 + 9 = 99 cal handled 2 / 3 rds of the transactions than anthony handled cal = 2 / 3 × 99 = 66 jade handled 19 more transactions than cal . jade = 66 + 19 = 85 jade handled = 85 tran... | a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 85 | e | add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), 3), 19) | add(n1,const_100)|divide(n0,const_100)|multiply(#0,#1)|multiply(n2,#2)|divide(#3,n3)|add(n4,#4)| | general |
in a garden , there are 10 rows and 12 columns of mango trees . the distance between the two trees is 2 metres and a distance of six metres is left from all sides of the boundary of the garden . what is the length of the garden ? | "between the 12 mango trees , there are 11 gaps and each gap has 2 meter length also , 6 meter is left from all sides of the boundary of the garden . hence , length of the garden = ( 11 ã — 2 ) + 6 + 6 = 34 meter answer is b ." | a ) 32 , b ) 34 , c ) 26 , d ) 28 , e ) 30 | b | add(add(multiply(subtract(12, const_1), 2), divide(10, 2)), divide(10, 2)) | divide(n0,n2)|subtract(n1,const_1)|multiply(n2,#1)|add(#0,#2)|add(#3,#0)| | physics |
a garrison of 150 men has provisions for 31 days . at the end of 16 days , a reinforcement arrives , and it is now found that the provisions will last only for 5 days more . what is the reinforcement ? | "150 - - - - 31 150 - - - - 15 x - - - - - 5 x * 5 = 150 * 15 x = 450 150 - - - - - - - 300 answer : e" | a ) 150 , b ) 450 , c ) 400 , d ) 600 , e ) 300 | e | subtract(divide(subtract(multiply(150, 31), multiply(150, 16)), 5), 150) | multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n3)|subtract(#3,n0)| | physics |
the lenght of a room is 5.5 m and width is 4 m . find the cost of paving the floor by slabs at the rate of rs . 800 per sq . metre . | area of the floor = ( 5.5 × 4 ) m 2 = 22 m 2 . cost of paving = rs . ( 800 × 22 ) = rs . 17600 . answer : option d | a ) rs . 15,550 , b ) rs . 15,600 , c ) rs . 16,500 , d ) rs . 17,600 , e ) rs . 17,900 | d | multiply(800, multiply(5.5, 4)) | multiply(n0,n1)|multiply(n2,#0) | physics |
employees of a certain company are each to receive a unique 8 - digit identification code consisting of the digits 0 , 1 , 2 , 3 , 4 , 5 , 6 , and 7 such that no digit is used more than once in any given code . in valid codes , the second digit in the code is exactly twice the first digit . how many valid codes are the... | "there are 6 ! ways to make codes starting with 12 . there are 6 ! ways to make codes starting with 24 . there are 6 ! ways to make codes starting with 36 . the number of codes is 3 * 6 ! = 2160 . the answer is c ." | a ) 1220 , b ) 1840 , c ) 2160 , d ) 2480 , e ) 2760 | c | multiply(multiply(2, 3), 2) | multiply(n3,n4)|multiply(n3,#0)| | general |
if x is 13 percent greater than 80 , then x = | "13 % of 80 = ( 80 * 0.13 ) = 10.4 11 % greater than 80 = 80 + 10.4 = 90.4 answer is clearly b ." | a ) 88.8 , b ) 90.4 , c ) 88.0 , d ) 70.9 , e ) 71.2 | b | add(80, multiply(divide(13, const_100), 80)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.