Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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a shopkeeper sold an article offering a discount of 10 % and earned a profit of 25 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be $ 100 . then , s . p . = $ 125 let marked price be $ x . then , 90 / 100 * x = 125 x = 12500 / 90 = $ 139 now , s . p . = $ 139 , c . p . = rs . 100 profit % = 39 % . answer : a" | a ) 39 % , b ) 52 % , c ) 15 % , d ) 21 % , e ) 28 % | a | subtract(divide(multiply(add(const_100, 25), const_100), subtract(const_100, 10)), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)| | gain |
a certain high school has 500 students . of these students , 30 are taking music , 20 are taking art , and 10 are taking both music and art . how many students are taking neither music nor art ? | "we ' re given a series of facts to work with : 1 ) a certain high school has 500 students . 2 ) of these students : x are taking music , y are taking art , and z are taking both music and art . we ' re asked how many students are taking neither music nor art ? let ' s test x = 30 y = 20 z = 10 so , we have 30 students... | a ) 430 , b ) 440 , c ) 450 , d ) 460 , e ) 470 | d | subtract(500, subtract(add(30, 20), 10)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
a ranch has both horses and ponies . exactly 3 / 10 of the ponies have horseshoes , and exactly 5 / 8 of the ponies with horseshoes are from iceland . if there are 3 more horses than ponies , what is the minimum possible combined number of horses and ponies on the ranch ? | "3 / 10 * p are ponies with horseshoes , so p is a multiple of 10 . 5 / 8 * 3 / 10 * p = 3 / 16 * p are icelandic ponies with horseshoes , so p is a multiple of 16 . the minimum value of p is 80 . then h = p + 3 = 83 . the minimum number of horses and ponies is 163 . the answer is d ." | a ) 130 , b ) 141 , c ) 152 , d ) 163 , e ) 174 | d | add(add(lcm(10, multiply(divide(10, 5), 8)), 3), lcm(10, multiply(divide(10, 5), 8))) | divide(n1,n2)|multiply(n3,#0)|lcm(n1,#1)|add(n4,#2)|add(#3,#2)| | general |
two bullet trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post . if the length of each bullet train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | "c 12 sec . speed of the first bullet train = 120 / 10 m / sec = 12 m / sec . speed of the second bullet train = 120 / 15 m / sec = 8 m / sec . relative speed = ( 12 + 8 ) = 20 m / sec . required time = ( 120 + 120 ) / 20 sec = 12 sec ." | a ) 13 sec . , b ) 15 sec . , c ) 12 sec . , d ) 17 sec . , e ) 19 sec . | c | divide(add(120, 120), add(speed(120, 10), speed(120, 15))) | add(n2,n2)|speed(n2,n0)|speed(n2,n1)|add(#1,#2)|divide(#0,#3)| | physics |
there is food for 760 men for 22 days . how many more men should join after two days so that the same food may last for 8 days more ? | 760 - - - - 22 760 - - - - 20 x - - - - - 8 x * 8 = 760 * 20 x = 1900 760 - - - - - - - 1140 answer : b | a ) 1122 , b ) 1140 , c ) 1199 , d ) 1188 , e ) 1166 | b | subtract(divide(multiply(760, subtract(22, const_2)), 8), 760) | subtract(n1,const_2)|multiply(n0,#0)|divide(#1,n2)|subtract(#2,n0) | physics |
if the price of a tv is first decreased by 20 % and then increased by 55 % , then the net change in the price will be : | "explanation : solution : let the original price be rs . 100 . new final price = 155 % of ( 80 % of 100 ) = rs . 155 / 100 * 80 / 100 * 100 = rs . 124 . . ' . increase = 24 % answer : b" | a ) 4 % increase , b ) 24 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | b | subtract(divide(multiply(subtract(const_100, 20), add(const_100, 55)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)| | general |
what is the smallest integer c for which 27 ^ c > 3 ^ 24 ? | "27 ^ c > 3 ^ 24 converting into the same bases : 27 ^ c > 27 ^ 8 therefore for the equation to hold true , c > 8 or c = 9 option c" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | divide(n2,const_3)|floor(#0)|add(#1,const_1)| | general |
in order to obtain an income of rs . 600 from 12 % stock at rs . 106 , one must make an investment of | "to obtain rs . 10 , investment = rs . 106 . to obtain rs . 600 , investment = = rs . 7632 answer : d" | a ) 5363 , b ) 6240 , c ) 2368 , d ) 7632 , e ) 7732 | d | multiply(divide(106, 12), 600) | divide(n2,n1)|multiply(n0,#0)| | gain |
speed of a boat in standing water is 16 kmph and the speed of the stream is 2 kmph . a man rows to a place at a distance of 7200 km and comes back to the starting point . the total time taken by him is : | "explanation : speed downstream = ( 16 + 2 ) = 18 kmph speed upstream = ( 16 - 2 ) = 14 kmph total time taken = 7200 / 18 + 7200 / 14 = 400 + 514.2 = 914.2 hours answer : option a" | a ) 914.2 hours , b ) 900 hours , c ) 915 hours , d ) 905 hours , e ) 915 hours | a | add(divide(7200, add(16, 2)), divide(7200, subtract(16, 2))) | add(n0,n1)|subtract(n0,n1)|divide(n2,#0)|divide(n2,#1)|add(#2,#3)| | physics |
a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 7000 . with the help of c , they completed the work in 3 days . how much is to be paid to c ? | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 3 - 7 / 24 = 1 / 24 a ' s wages : b ' s wages : c ' s wages = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 c ' s share ( for 3 days ) = rs . ( 3 * 1 / 24 * 7000 ) = rs . 875 answer = e" | a ) s . 375 , b ) s . 400 , c ) s . 600 , d ) s . 800 , e ) s . 875 | e | multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 7000), 3) | inverse(n3)|inverse(n1)|inverse(n0)|add(#1,#2)|subtract(#0,#3)|multiply(n2,#4)|multiply(n3,#5)| | physics |
what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 400 inclusive ? | "we ' re asked for the average of all of the multiples of 10 from 10 to 400 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 40 ( 10 ) = 400 so we know that there are 40 total numbers . we can now figure out the sum of those numbers with ' bunching '... | a ) 190 , b ) 195 , c ) 200 , d ) 205 , e ) 210 | d | divide(divide(multiply(add(10, 400), add(divide(subtract(400, 10), 10), const_1)), const_2), add(divide(subtract(400, 10), 10), const_1)) | add(n0,n2)|subtract(n2,n0)|divide(#1,n0)|add(#2,const_1)|multiply(#0,#3)|divide(#4,const_2)|divide(#5,#3)| | general |
a train 130 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 4 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 4 = 64 km / hr . = 64 * 5 / 18 = 160 / 9 m / sec . time taken to pass the men = 130 * 9 / 160 = 7.13 sec . answer : option d" | a ) 10 , b ) 6.5 , c ) 7 , d ) 7.13 , e ) 10.2 | d | divide(130, multiply(add(60, 4), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
after giving a discount of rs . 80 the shopkeeper still gets a profit of 10 % , if the cost price is rs . 200 . find the markup % ? | "cost price = 180 s . p = 200 * 110 / 100 = 220 disc = 80 so . . . mark price = 216 + 80 = 296 . . . . . . mark up % = 261 - 200 / 200 = 61 / 200 = 0.305 or 31 % answer : a" | a ) 31 % , b ) 33 % , c ) 28 % , d ) 30 % , e ) 32 % | a | divide(subtract(add(add(200, divide(200, divide(const_100, 10))), 80), 200), 200) | divide(const_100,n1)|divide(n2,#0)|add(n2,#1)|add(n0,#2)|subtract(#3,n2)|divide(#4,n2)| | gain |
if the digits 27 in the decimal 0.00027 repeat indefinitely , what is the value of ( 10 ^ 5 - 10 ^ 3 ) ( 0.00027 ) ? | "99 * 0.27 = 26.73 approx . 27 answer : c" | a ) 0.27 , b ) 0.027 , c ) 27 , d ) 0.0027 , e ) 2.7 e - 06 | c | multiply(divide(multiply(multiply(multiply(10, 10), subtract(multiply(10, 10), const_1)), divide(27, subtract(multiply(10, 10), const_1))), const_1000), 10) | multiply(n2,n2)|subtract(#0,const_1)|divide(n0,#1)|multiply(#0,#1)|multiply(#2,#3)|divide(#4,const_1000)|multiply(n2,#5)| | general |
one ball will drop from a certain height . the height it will reach after rebounding from the floor is 50 percent of the previous height . the total travel is 200 cm when it touches the floor on third time . what is the value of the original height ? | "when ball comes down , then i have indicated the distance covered in green when ball goes up , then i have indicated the distance covered in red distance travelled uptil the ball touches the floor 3 rd time : h + 0.5 h + 0.5 h + 0.5 * 0.5 h + 0.5 * 0.5 h h + 2 * 0.5 * h + 2 * 0.25 * h = h ( 1 + 2 * 0.5 + 2 * 0.25 ) = ... | a ) 80 cm , b ) 90 cm , c ) 100 cm , d ) 120 cm , e ) 130 cm | a | divide(200, add(const_2, divide(50, const_100))) | divide(n0,const_100)|add(#0,const_2)|divide(n1,#1)| | gain |
in the coordinate plane , one of the vertices of a square is the point ( - 3 , - 4 ) . if the diagonals of that square intersect at point ( 5 , 3 ) , what is the area of that square ? | "one point ( - 3 - 4 ) , intersection ( 3,2 ) so the distance from the first point - 3 - 5 = - 8 is the midpoint of the square - - > whole side 16 , 16 * 16 = 256 e" | a ) 100 , b ) 169 , c ) 225 , d ) 324 , e ) 256 | e | multiply(multiply(3, 5), multiply(3, 5)) | multiply(n0,n2)|multiply(#0,#0)| | geometry |
in a school with 652 students , the average age of the boys is 12 years and that of the girls is 11 years . if the average age of the school is 11 years 9 months , then the number of girls in the school is | "sol . let the number of grils be x . then , number of boys = ( 652 - x ) . then , ( 11 3 / 4 × 652 ) ⇔ 11 x + 12 ( 652 - x ) ⇔ x = 7824 - 7661 ⇔ 163 . answer d" | a ) 150 , b ) 200 , c ) 250 , d ) 163 , e ) none | d | subtract(multiply(12, 652), multiply(add(11, divide(9, 12)), 652)) | divide(n4,n1)|multiply(n0,n1)|add(n2,#0)|multiply(n0,#2)|subtract(#1,#3)| | general |
( 35423 + 7164 + 41720 ) - ( 317 x 89 ) = ? | "= ( 84307 ) - 317 x 89 = 84307 - ( 317 x ( 90 - 1 ) ) = 84307 - ( 317 x 90 - 317 ) = 84307 - 28213 = 54094 answer is d" | a ) 28213 , b ) 84307 , c ) 50694 , d ) 54094 , e ) none of them | d | divide(add(35423, 7164), 41720) | add(n0,n1)|divide(#0,n2)| | general |
in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 2 / 3 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and straight | "good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 2 / 3 of the garden is green , so , since all the flowers must be either green or yellow , we know that 1 / 3 are yellow . we ' re also told there is an equal p... | a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 4 , d ) 3 / 4 , e ) 7 / 8 | a | multiply(subtract(1, divide(2, 3)), divide(1, 2)) | divide(n2,n3)|divide(n0,n1)|subtract(n2,#1)|multiply(#0,#2)| | probability |
how many words , with or without meaning , can be formed using all letters of the word beautiful using each letter exactly once ? | "the word beautiful has exactly 9 letters which are all different . therefore the number of words that can be formed = number of permutations of 9 letters taken all at a time . = p ( 9 , 9 ) = 9 ! = 9 x 8 × 7 x 6 × 5 x 4 x 3 x 2 × 1 = 362,880 answer : c" | a ) 388,620 , b ) 388,260 , c ) 362,880 , d ) 352,880 , e ) 342,880 | c | factorial(const_3) | factorial(const_3)| | general |
a shopkeeper sells his goods at cost price but uses a faulty meter that weighs 940 grams . find the profit percent . | explanation : ( 100 + g ) / ( 100 + x ) = true measure / faulty measure x = 0 true measure = 1000 faulty measure = 940 100 + g / 100 + 0 = 1000 / 940 100 + g = 100 / 94 * 100 g = 6.38 answer : d | a ) 7.38 , b ) 5.38 , c ) 4.38 , d ) 6.38 , e ) 3.38 | d | multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 940), 940), const_100) | add(const_1,const_4)|add(#0,#0)|multiply(#1,const_100)|subtract(#2,n0)|divide(#3,n0)|multiply(#4,const_100) | gain |
the average of first six multiples of 4 is : | "solution average = 4 ( 1 + 2 + 3 + 4 + 5 + 6 ) / 6 = 84 / 6 = 14 answer a" | a ) 14 , b ) 9 , c ) 12 , d ) 15 , e ) 17 | a | add(4, const_1) | add(n0,const_1)| | general |
a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 20 hour to row to a place and come back , how far is the place ? | "speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x / 12 + x / 8 = 20 2 x + 3 x = 480 x = 96 km answer is c" | a ) 24 km , b ) 30 km , c ) 96 km , d ) 12 km , e ) 15 km | c | divide(multiply(multiply(subtract(10, 2), add(10, 2)), 20), add(subtract(10, 2), add(10, 2))) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|multiply(n2,#3)|divide(#4,#2)| | physics |
during a sale , the price of a pair of shoes is marked down 20 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | "assume the price = 100 price during sale = 80 price after sale = 100 percent increase = 20 / 80 * 100 = 25 % approx . correct option : e" | a ) 9 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 25 % | e | divide(multiply(20, const_100), subtract(const_100, 20)) | multiply(n0,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
in what time will a train 320 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 320 / 40 = 8.0 sec . answer : b" | a ) 2.5 sec , b ) 8.0 sec , c ) 3.5 sec , d ) 2.9 sec , e ) 9.5 sec | b | divide(320, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
among 200 students , 56 % study sociology , 44 % study mathematics and 40 % study biology . if 30 % of students study both mathematics and sociology , what is the largest possible number of students who study biology but do not study either mathematics or sociology ? | "i would just like to add a bit of explanation after the step where you calculate that the number of students studying both m and s = 60 using your analysis : we see that the total number of students who study either maths or sociology = 88 + 112 - 60 = 140 so , in the image we know that the number of students in the z... | a ) 60 , b ) 90 , c ) 120 , d ) 172 , e ) 188 | a | subtract(200, subtract(add(multiply(divide(200, const_100), 56), multiply(divide(200, const_100), 44)), multiply(divide(200, const_100), 30))) | divide(n0,const_100)|multiply(n1,#0)|multiply(n2,#0)|multiply(n4,#0)|add(#1,#2)|subtract(#4,#3)|subtract(n0,#5)| | other |
a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 5 hour to row to a place and come back , how far is the place ? | speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x / 12 + x / 8 = 5 2 x + 3 x = 120 x = 24 km answer is a | a ) 24 km , b ) 30 km , c ) 48 km , d ) 12 km , e ) 15 km | a | divide(multiply(multiply(subtract(10, 2), add(10, 2)), 5), add(subtract(10, 2), add(10, 2))) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|multiply(n2,#3)|divide(#4,#2) | physics |
what is the cost of leveling the field in the form of parallelogram at the rate of rs . 70 / 10 sq . metre , whose base & perpendicular distance from the other side being 84 m & 24 m respectively ? | "area of the parallelogram = length of the base * perpendicular height = 84 * 24 = 2016 m . total cost of levelling = rs . 2016 b" | a ) s . 2400 , b ) s . 2016 , c ) s . 1400 , d ) s . 3480 , e ) s . 2000 | b | multiply(multiply(84, 24), divide(70, 10)) | divide(n0,n1)|multiply(n2,n3)|multiply(#0,#1)| | physics |
during a certain two - week period , 70 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 5 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 , ... | "total movies = 100 . comedies = 70 . action + drama = 30 . since there were 5 times as many dramas as action movies , then action + 5 * action = 30 - - > action = a = 5 . comedies = 70 = 14 a . answer : d ." | a ) a / 14 , b ) 5 a / 7 , c ) 7 a / 5 , d ) 14 a , e ) 35 a | d | floor(divide(70, add(5, const_1))) | add(n1,const_1)|divide(n0,#0)|floor(#1)| | general |
country x taxes each of its citizens an amount equal to 11 percent of the first $ 40,000 of income , plus 20 percent of all income in excess of $ 40,000 . if a citizen of country x is taxed a total of $ 8,000 , what is her income ? | "equation is correct , so math must be a problem . 0.11 * 40,000 + 0.2 * ( x - 40,000 ) = 8,000 - - > 4,400 + 0.2 x - 8,000 = 8,000 - - > 0.2 x = 11,600 - - > x = 58,000 . answer : d ." | a ) $ 40,000 , b ) $ 56,000 , c ) $ 64,000 , d ) $ 58,000 , e ) $ 80,000 | d | add(multiply(multiply(const_4, const_10), const_1000), divide(subtract(multiply(multiply(const_4, const_2), const_1000), multiply(divide(11, const_100), multiply(multiply(const_4, const_10), const_1000))), divide(20, const_100))) | divide(n0,const_100)|divide(n2,const_100)|multiply(const_2,const_4)|multiply(const_10,const_4)|multiply(#2,const_1000)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(#4,#6)|divide(#7,#1)|add(#8,#5)| | general |
list k consists of 14 consecutive integers . if - 4 is the least integer in list k , what is the range of the positive integers in list k ? | answer = d = 8 if least = - 4 , then largest = 9 range = 9 - 1 = 8 | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 12 | d | subtract(subtract(subtract(14, 4), const_1), const_1) | subtract(n0,n1)|subtract(#0,const_1)|subtract(#1,const_1) | general |
a person travels equal distances with speeds of 3 km / hr , 4 km / hr and 5 km / hr and takes a total time of 40 minutes . the total distance is ? | "e 3 km let the total distance be 3 x km . then , x / 3 + x / 4 + x / 5 = 40 / 60 47 x / 60 = 40 / 60 = > x = 0.85 . total distance = 3 * 0.85 = 2.55 km ." | a ) 1 km , b ) 2 km , c ) 3 km , d ) 4 km , e ) 2.55 km | e | multiply(multiply(divide(divide(40, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5))), const_3), const_1000) | divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3)|multiply(#7,const_1000)| | physics |
the average of 20 results is 65 and the average of other 65 results is 20 . what is the average of all the results ? | "answer sum of 85 result = sum of 20 result + sum of 65 result . = 20 x 65 + 65 x 20 = 2600 correct option : a" | a ) 30 , b ) 25 , c ) 48 , d ) 50 , e ) none | a | divide(add(multiply(20, 65), multiply(65, 20)), add(20, 65)) | add(n0,n1)|multiply(n0,n1)|add(#1,#1)|divide(#2,#0)| | general |
by investing rs . 1620 in 8 % stock , michael earns rs . 135 . the stock is then quoted at | "to earn rs . 135 , investment = rs . 1620 . to earn rs . 8 , investment = rs . 96 . market value of rs . 100 stock = rs . 96 . answer : b" | a ) 46 , b ) 96 , c ) 35 , d ) 87 , e ) 13 | b | multiply(multiply(divide(1620, 135), divide(8, const_100)), const_100) | divide(n0,n2)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)| | gain |
in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 192 votes . find the total number of valid votes ? | "let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 192 = > 4 x / 10 = 192 = > 4 x = 1920 = > x = 480 . answer : d" | a ) 430 , b ) 438 , c ) 436 , d ) 480 , e ) 422 | d | divide(192, divide(subtract(70, subtract(const_100, 70)), const_100)) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)| | gain |
how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 20 - percent solution ? | "how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 20 - percent solution ? 3 % of a 50 liter solution is 1.5 l . so you are trying to determine how many liters must a solution be for the 1.5 l to represent 20 % of the solution . set up an inequality and solve for x : 1.... | a ) 45 , b ) 33 1 / 3 , c ) 27 , d ) 16 2 / 3 , e ) 15 | a | subtract(50, multiply(divide(50, const_100), 20)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
a combustion reaction forms carbon dioxide . a carbon dioxide molecule contains one carbon and two oxygen atoms . if , over a period of 10 minutes , a combustion reaction creates 10,000 molecules of carbon dioxide then approximately how many more atoms of oxygen than carbon are created on average per minute ? | solution : 10,000 carbon dioxide molecules are created over a period of 10 minutes . therefore 10,000 / 10 = 1,000 carbon dioxide molecules are created on average per minute each carbon dioxide molecule contains one carbon atom and two oxygen atoms . so 1,000 carbon dioxide molecules contain 1 × 1,000 = 1,000 carbon at... | a ) 1,000 , b ) 500 , c ) 250 , d ) 50 , e ) 0 | a | divide(divide(multiply(subtract(multiply(10, const_2), 10), const_1000), 10), const_1000) | multiply(n0,const_2)|subtract(#0,n0)|multiply(#1,const_1000)|divide(#2,n0)|divide(#3,const_1000) | general |
find the average of all the numbers between 6 and 34 which is divisible by 5 | "avg = ( 10 + 15 + 20 + 25 + 30 ) / 5 = 100 / 5 = > 20 answer e" | a ) 15 , b ) 18 , c ) 13 , d ) 19 , e ) 20 | e | divide(add(add(6, const_4), subtract(34, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
. 3 + 33 + 333 + 3.33 = ? | . 3 33 333 3.33 - - - - - - - - - - 369.63 - - - - - - - - - - answer is a | a ) 369.63 , b ) 372.33 , c ) 702.33 , d ) 702 , e ) none of them | a | add(add(add(33, 333), 3.33), divide(const_3, const_10)) | add(n1,n2)|divide(const_3,const_10)|add(n3,#0)|add(#2,#1) | general |
an woman swims downstream 45 km and upstream 25 km taking 5 hours each time ; what is the speed of the current ? | "45 - - - 5 ds = 9 ? - - - - 1 25 - - - - 5 us = 5 ? - - - - 1 s = ? s = ( 9 - 5 ) / 2 = 2 answer : a" | a ) 2 , b ) 1 , c ) 3 , d ) 4 , e ) 2.1 | a | divide(subtract(divide(45, 5), divide(25, 5)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
find a sum for first 6 prime number ' ss ? | required sum = ( 2 + 3 + 5 + 7 + 11 + 13 ) = 41 note : 1 is not a prime number option e | a ) 25 , b ) 28 , c ) 30 , d ) 35 , e ) 41 | e | subtract(multiply(add(6, 6), const_4), add(6, const_1)) | add(n0,n0)|add(n0,const_1)|multiply(#0,const_4)|subtract(#2,#1) | general |
a computer factory produces 8400 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | "number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 8400 / ( 28 * 24 * 2 ) = 6.25 answer e" | a ) 2.25 . , b ) 3.125 . , c ) 4.5 . , d ) 5.225 . , e ) 6.25 . | e | divide(divide(8400, 28), multiply(subtract(28, const_4), const_2)) | divide(n0,n2)|subtract(n2,const_4)|multiply(#1,const_2)|divide(#0,#2)| | physics |
what least number should be subtracted from 3,381 so that the remainder when divided by 9 , 11 , and 17 will leave in each case the same remainder 8 ? | "the lcm of 9 , 11 and 17 is 1,683 . the next multiple is 2 * 1,683 = 3,366 . 3,366 + { remainder } = 3,366 + 8 = 3,374 , which is 7 less than 3,381 . answer : b ." | a ) 6 , b ) 7 , c ) 8 , d ) 12 , e ) 15 | b | subtract(3,381, add(multiply(multiply(multiply(9, 11), 17), const_2), 8)) | multiply(n1,n2)|multiply(n3,#0)|multiply(#1,const_2)|add(n4,#2)|subtract(n0,#3)| | general |
a trader marked the selling price of an article at 40 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of : | sol . let c . p . = rs . 100 . then , marked price = rs . 140 , s . p . = rs . 99 . ∴ discount % = [ 11 / 140 * 100 ] % = 7.8 % answer b | a ) 10 % , b ) 7.8 % , c ) 11 % , d ) 12.5 % , e ) none | b | multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), const_1), add(const_100, 40))) | add(const_2,const_3)|add(n0,const_100)|multiply(#0,const_2)|add(#2,const_1)|divide(#3,#1)|multiply(#4,const_100) | gain |
diana is painting statues . she has 7 / 16 of a gallon of paint remaining . each statue requires 1 / 16 gallon of paint . how many statues can she paint ? | "number of statues = all the paint ÷ amount used per statue = 7 / 16 ÷ 1 / 16 = 7 / 16 * 16 / 1 = 7 / 1 = 7 answer is a ." | a ) 7 , b ) 20 , c ) 28 , d ) 14 , e ) 19 | a | divide(divide(7, 16), divide(1, 16)) | divide(n0,n1)|divide(n2,n3)|divide(#0,#1)| | general |
in an election a candidate who gets 60 % of the votes is elected by a majority of 1300 votes . what is the total number of votes polled ? | "let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 60 ) % of x = 40 % of x 60 % of x - 40 % of x = 1300 20 x / 100 = 1300 x = 1300 * 100 / 20 = 6500 answer is e" | a ) a ) 4500 , b ) b ) 5200 , c ) c ) 6900 , d ) d ) 7520 , e ) e ) 6500 | e | divide(1300, divide(subtract(60, subtract(const_100, 60)), const_100)) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)| | gain |
fresh grapes contain 60 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 30 kg of fresh grapes ? | "from the question we know : 30 kg * 60 % = 18 kg of water in the fresh grapes 30 kg - 18 kg of water = 12 kg of non - water mass we are looking for the weight of the dry grapes ( x ) . since the question tells us that 20 % of the weight of the dry graps is water and we know that 12 kg is non - water mass we can set up... | a ) 10 kg , b ) 15 kg , c ) 20 kg , d ) 25 kg , e ) 40 kg | b | multiply(divide(divide(multiply(subtract(const_100, 60), 30), const_100), subtract(const_100, 20)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|multiply(n2,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | gain |
without stoppages , a train travels certain distance with an average speed of 80 km / h , and with stoppages , it covers the same distance with an average speed of 60 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 20 km less . time taken to cover 20 km = 20 ⁄ 80 h = 1 ⁄ 4 h = 1 ⁄ 4 × 60 min = 15 min answer a" | a ) 15 , b ) 18 , c ) 10 , d ) 16 , e ) none of these | a | subtract(const_60, multiply(divide(60, 80), const_60)) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
if 5 / ( 4 + 1 / x ) = 1 , then x = | "the expression 5 / ( 4 + 1 / x ) = 1 should have been equal to something . if 5 / ( 4 + 1 / x ) = 1 = > 5 x / ( 4 x + 1 ) = 1 = > 5 x = 4 x + 1 = > x = 1 correct option : b" | a ) 3 , b ) 1 , c ) 1 / 3 , d ) - 1 / 3 , e ) - 3 | b | divide(1, subtract(5, 4)) | subtract(n0,n1)|divide(n2,#0)| | general |
the average age of a class of 22 students is 21 years . the average increased by 1 when the teacher ' s age also included . what is the age of the teacher ? | "total age = 22 * 21 tot age of all stu + age of the teacher = 23 * 22 age of the teacher = 23 * 22 - 22 * 21 = 44 answer a" | a ) 44 , b ) 48 , c ) 45 , d ) 43 , e ) 50 | a | subtract(multiply(add(22, 1), add(21, 1)), multiply(22, 21)) | add(n0,n2)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
by selling 88 pens , a trader gains the cost of 22 pens . find his gain percentage ? | let the cp of each pen be rs . 1 . cp of 88 pens = rs . 88 profit = cost of 22 pens = rs . 22 profit % = 22 / 88 * 100 = 25 % answer : a | a ) 25 % , b ) 26 % , c ) 27 % , d ) 28 % , e ) 29 % | a | multiply(divide(22, 88), const_100) | divide(n1,n0)|multiply(#0,const_100) | gain |
the population of a bacteria colony doubles every day . if it was started 9 days ago with 2 bacteria and each bacteria lives for 13 days , how large is the colony today ? | "9 days ago - 2 8 days ago - 4 7 days ago - 8 6 days ago - 16 5 days ago - 32 4 days ago - 64 3 days ago - 128 2 days ago - 256 yesterday - 512 today - 1024 ans : b" | a ) 512 , b ) 1024 , c ) 768 , d ) 2048 , e ) 4096 | b | subtract(power(2, add(9, const_1)), const_1) | add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)| | physics |
due to construction , the speed limit along an 9 - mile section of highway is reduced from 55 miles per hour to 35 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 ? | "9 / 35 - 9 / 55 = 9 / 5 * ( 11 - 7 ) / 77 = 9 / 5 * 4 / 77 * 60 min = 9 * 12 * 4 / 77 = 432 / 77 ~ 5.6 answer - b" | a ) 5 , b ) 7 , c ) 10 , d ) 15 , e ) 24 | b | max(multiply(subtract(add(55, 9), const_1), subtract(divide(9, 35), 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 |
if k is a non - negative integer and 18 ^ k is a divisor of 624,938 then 6 ^ k - k ^ 6 = | "6 + 2 + 4 + 9 + 3 + 8 = 32 , so this number is not divisible by 3 and thus not divisible by 18 . therefore , k = 0 6 ^ k - k ^ 6 = 1 - 0 = 1 the answer is b ." | a ) 0 , b ) 1 , c ) 36 , d ) 118 , e ) 420 | b | subtract(power(6, subtract(const_1, const_1)), power(subtract(const_1, const_1), 6)) | subtract(const_1,const_1)|power(n2,#0)|power(#0,n2)|subtract(#1,#2)| | general |
in the first m games of a team ' s season , the ratio of the team ' s wins to its losses was 1 : 3 . in the subsequent n games , the ratio of the team ´ s wins to losses was 2 : 3 . if m : n = 4 : 5 , what was the ratio of the team ' s wins to its losses for all m + n games ? | m = 4 / 9 of total games n = 5 / 9 of total games wins = 1 / 4 * 4 / 9 + 2 / 5 * 5 / 9 = 1 / 9 + 2 / 9 = 1 / 3 losses = 1 - 1 / 3 = 2 / 3 the ratio of wins to losses is 1 : 2 . the answer is c . | a ) 7 : 18 , b ) 3 : 7 , c ) 1 : 2 , d ) 2 : 5 , e ) 4 : 9 | c | divide(add(1, 2), add(3, 3)) | add(n0,n2)|add(n1,n1)|divide(#0,#1) | general |
two trains start at same time from two stations and proceed towards each other at the rate of 20 km / hr and 25 km / hr respectively . when they meet , it is found that one train has traveled 55 km more than the other . what is the distance between the two stations ? | "explanation : let us assume that trains meet after ' x ' hours distance = speed * time distance traveled by two trains = 20 x km and 25 x km resp . as one train travels 55 km more than the other , 25 x â € “ 20 x = 55 5 x = 55 x = 11 hours as the two trains are moving towards each other , relative speed = 20 + 25 = 45... | a ) 540 km , b ) 495 km , c ) 276 km , d ) 178 km , e ) 176 km | b | multiply(add(20, 25), divide(55, subtract(25, 20))) | add(n0,n1)|subtract(n1,n0)|divide(n2,#1)|multiply(#0,#2)| | general |
the compound ratio of 5 : 6 , 3 : 2 and 4 : 5 ? | "5 / 6 * 3 / 2 * 4 / 5 = 1 / 1 1 : 1 answer : d" | a ) 1 : 8 , b ) 1 : 6 , c ) 1 : 2 , d ) 1 : 1 , e ) 1 : 7 | d | divide(divide(multiply(5, 3), multiply(6, 2)), divide(multiply(3, 4), multiply(2, 5))) | multiply(n0,n2)|multiply(n1,n3)|multiply(n2,n4)|multiply(n3,n5)|divide(#0,#1)|divide(#2,#3)|divide(#4,#5)| | other |
in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 17 to the thrice of the remainder . the dividend is | "divisor = ( 6 * 3 ) + 17 = 35 5 * quotient = 35 quotient = 7 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 7 ) + 6 = 146 . b )" | a ) 74 , b ) 146 , c ) 86 , d ) 92 , e ) 98 | b | add(multiply(add(multiply(6, const_3), 17), divide(add(multiply(6, const_3), 17), 5)), 6) | multiply(n0,const_3)|add(n2,#0)|divide(#1,n1)|multiply(#1,#2)|add(n0,#3)| | general |
if albert ’ s monthly earnings rise by 36 % , he would earn $ 495 . if , instead , his earnings rise by only 25 % , how much ( in $ ) would he earn this month ? | "= 495 / 1.36 ∗ 1.25 = 454 = 454 answer is c" | a ) 643 , b ) 652 , c ) 454 , d ) 460 , e ) 490 | c | multiply(divide(495, add(const_1, divide(36, const_100))), add(const_1, divide(25, const_100))) | divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n1,#3)|multiply(#2,#4)| | gain |
lilly has 10 fish and rosy has 9 fish . in total , how many fish do they have in all ? | "10 + 9 = 19 the answer is d ." | a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | d | add(10, 9) | add(n0,n1)| | general |
after giving a discount of rs . 100 the shopkeeper still gets a profit of 30 % , if the cost price is rs . 1000 . find the markup % ? | "cost price = 1000 s . p = 1000 * 130 / 100 = 1300 disc = 100 so . . . mark price = 1300 + 100 = 1400 . . . . . . mark up % = 1400 - 1000 / 1000 = 400 / 1000 = . 40 ( or ) 40 % answer : c" | a ) . 50 , b ) . 45 , c ) . 40 , d ) . 48 , e ) . 52 | c | divide(subtract(add(add(1000, divide(1000, divide(const_100, 30))), 100), 1000), 1000) | divide(const_100,n1)|divide(n2,#0)|add(n2,#1)|add(n0,#2)|subtract(#3,n2)|divide(#4,n2)| | gain |
of the goose eggs laid at a certain pond , 1 / 3 hatched and 4 / 5 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year . if 120 geese survived the first year and if no more than one goose hatched from each egg , how many ... | "let x be the number of eggs that were laid . ( 2 / 5 ) ( 4 / 5 ) ( 1 / 3 ) x = 120 ( 8 / 75 ) x = 120 x = 1125 the answer is c ." | a ) 1025 , b ) 1075 , c ) 1125 , d ) 1175 , e ) 1225 | c | divide(divide(divide(120, subtract(const_1, divide(3, 5))), divide(3, 5)), divide(1, 3)) | divide(n1,n5)|divide(n2,n3)|divide(n0,n1)|subtract(const_1,#0)|divide(n6,#3)|divide(#4,#1)|divide(#5,#2)| | general |
a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 20 miles per gallon , and from town b to town c , the car averaged 24 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through ... | "step 1 ) took lcm of 20 and 24 . . came as 120 . step 2 ) 120 distance between b to c . . . do 120 / 24 hence 5 gallons used step 3 ) twice distance . . hence 120 * 2 = 240 . . . do as above . . 240 / 20 = 7 gallons used step 4 ) total gallons . . 5 + 7 = 12 gallons step ) total miles = 120 + 240 = 360 miles hence . .... | a ) 40 , b ) 30 , c ) 42 , d ) 33 , e ) 35 | b | divide(add(multiply(24, const_10), divide(multiply(24, const_10), const_2)), add(divide(multiply(24, const_10), 20), divide(divide(multiply(24, const_10), const_2), 24))) | multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)| | general |
a bowl of fruit contains 10 apples and 23 oranges . how many oranges must be removed so that 50 % of the pieces of fruit in the bowl will be apples ? | "number of apples = 10 number of oranges = 23 let number of oranges that must be removed so that 50 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 10 + ( 23 - x ) = 33 - x 10 / ( 33 - x ) = 5 / 10 = > 20 = 33 - x = > x = 13 answer b" | a ) 3 , b ) 13 , c ) 14 , d ) 17 , e ) 20 | b | divide(subtract(multiply(add(10, 23), divide(50, const_100)), 10), divide(50, const_100)) | add(n0,n1)|divide(n2,const_100)|multiply(#0,#1)|subtract(#2,n0)|divide(#3,#1)| | gain |
given that p is a positive even integer with a positive units digit , if the units digit of p ^ 3 minus the units digit of p ^ 2 is equal to 0 , what is the units digit of p + 5 ? | "p is a positive even integer with a positive units digit - - > the units digit of p can be 2 , 4 , 6 , or 8 - - > in order the units digit of p ^ 3 - p ^ 2 to be 0 , the units digit of p ^ 3 and p ^ 2 must be the same . thus the units digit of p can be 0 , 1 , 5 or 6 . intersection of values is 6 , thus the units digi... | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | e | add(add(3, 3), 5) | add(n0,n0)|add(n3,#0)| | general |
if a / x = 1 / 3 and a / y = 1 / 4 , then ( x + y ) = | ratio 1 : 3 a = x ratio 2 : 4 a = y x + y = 7 a answer is d | a ) 2 a , b ) a / 2 , c ) 6 a , d ) 7 a , e ) 9 a | d | add(4, 3) | add(n1,n3) | general |
for the past n days , the average ( arithmetic mean ) daily production at a company was 70 units . if today ' s production of 90 units raises the average to 75 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 70 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 70 n + 90 = 75 * ( n + 1 ) - - > n = 3 . answer : e ." | a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 3 | e | subtract(divide(subtract(90, 70), subtract(75, 70)), const_1) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,#1)|subtract(#2,const_1)| | general |
a is twice as good a work man as b and together they finish the work in 6 days . in how many days a alone can finish the work ? | "wc = 2 : 1 2 x + x = 1 / 6 = > x = 1 / 18 2 x = 1 / 9 a can do the work in 9 days . answer : b" | a ) 3 , b ) 9 , c ) 7 , d ) 2 , e ) 6 | b | inverse(divide(inverse(6), add(const_2, const_1))) | add(const_1,const_2)|inverse(n0)|divide(#1,#0)|inverse(#2)| | physics |
three cubes of metal whose edges are 9 , 12 and 21 cm respectively , are melted and one new cube is made . find the edge of the new cube ? | "93 + 123 + 213 = a 3 = > a = 22.7 answer : b" | a ) 28 , b ) 22.7 , c ) 66 , d ) 18 , e ) 99 | b | power(add(add(volume_cube(9), volume_cube(12)), volume_cube(21)), const_0_33) | volume_cube(n0)|volume_cube(n1)|volume_cube(n2)|add(#0,#1)|add(#3,#2)|power(#4,const_0_33)| | physics |
a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 2 , and the ratio of soap to salt is five times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is halved by water evaporation . at that ... | "water : soap = 3 : 2 soap : salt = 15 : 2 = > for 15 soap , salt = 2 = > for 2 soap , salt = ( 2 / 15 ) * 2 = 4 / 15 so , water : soap : salt = 3 : 2 : 4 / 15 = 45 : 30 : 4 after open container , water : soap : salt = 22.5 : 30 : 4 so , water : salt = 22.5 : 4 = 45 : 8 e" | a ) 1 : 1 , b ) 2 : 3 , c ) 3 : 2 , d ) 9 : 4 , e ) 45 : 8 | e | divide(multiply(multiply(2, 3), 3), multiply(multiply(2, 2), 2)) | multiply(n0,n1)|multiply(n1,n1)|multiply(n0,#0)|multiply(n1,#1)|divide(#2,#3)| | other |
if a @ b = a * b ^ ( 1 / 2 ) then 4 @ 9 = ? self made | 4 * ( 9 ) ^ 1 / 2 = 4 * 3 = 12 b is the answer | a ) 2 , b ) 12 , c ) 3 , d ) 8 , e ) 4 | b | multiply(4, power(9, divide(1, 2))) | divide(n0,n1)|power(n3,#0)|multiply(n2,#1) | general |
population of a city in 20004 was 1000000 . if in 2005 there isan increment of 25 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city atthe end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 25 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1178125 d" | a ) 976374 , b ) 979923 , c ) 980241 , d ) 1178125 , e ) 1083875 | d | multiply(1000000, multiply(multiply(add(const_1, divide(25, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6)| | gain |
the difference between simple interest and compound interest of a certain sum of money at 20 % per annum for 2 years is $ 48 . then the sum is | "20 % of ( year 1 interest ) = 20 % of ( 20 % of principal ) = 48 principal = $ 1200 answer ( c )" | a ) $ 600 , b ) $ 1000 , c ) $ 1200 , d ) $ 1500 , e ) $ 2000 | c | subtract(subtract(multiply(power(add(divide(divide(2, const_2), const_100), const_1), const_2), 20), 20), divide(multiply(20, 2), const_100)) | divide(n1,const_2)|multiply(n0,n1)|divide(#0,const_100)|divide(#1,const_100)|add(#2,const_1)|power(#4,const_2)|multiply(n0,#5)|subtract(#6,n0)|subtract(#7,#3)| | gain |
the equal amounts of money are deposited in two banks each at 15 % per annum for 3.5 years and 5 years respectively . if the difference between their interests is rs . 144 , find the each sum ? | ( p * 5 * 15 ) / 100 - ( p * 3.5 * 15 ) / 100 = 144 75 p / 100 â € “ 52.5 p / 100 = 144 22.5 p = 144 * 100 = > p = rs . 640 answer a | a ) 640 , b ) 500 , c ) 3000 , d ) 1000 , e ) 2000 | a | divide(144, divide(subtract(multiply(15, 5), multiply(15, 3.5)), const_100)) | multiply(n0,n2)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,const_100)|divide(n3,#3) | general |
the number of stamps that p and q had were in the ratio of 5 : 3 respectively . after p gave q 10 stamps , the ratio of the number of p ' s stamps to the number of q ' s stamps was 9 : 7 . as a result of the gift , p had how many more stamps than q ? | "p started with 5 k stamps and q started with 3 k stamps . ( 5 k - 10 ) / ( 3 k + 10 ) = 9 / 7 35 k - 27 k = 160 k = 20 p has 5 ( 20 ) - 10 = 90 stamps and q has 3 ( 20 ) + 10 = 70 stamps . the answer is a ." | a ) 20 , b ) 30 , c ) 40 , d ) 60 , e ) 90 | a | divide(add(multiply(10, 9), multiply(10, 7)), add(9, 7)) | add(n3,n4)|multiply(n2,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)| | other |
if y is the smallest positive integer such that 31,360 multiplied by y is the square of an integer , then y must be | "31360 = 8 * 8 * 7 * 7 * 5 * 2 , so we need one 5 and one 2 to make it a square of a number . so 5 * 2 = 10 ans : b" | a ) 21 , b ) 10 , c ) 61 , d ) 72 , e ) 14 | b | multiply(add(const_2, const_3), const_2) | add(const_2,const_3)|multiply(#0,const_2)| | geometry |
solution y is 30 percent liquid x and 70 percent water . if 3 kilograms of water evaporate from 8 kilograms of solution y and 3 kilograms of solution y are added to the remaining 6 kilograms of liquid , what percent of this new solution is liquid x ? | in 8 kilograms of solution y there are 0.3 * 8 = 2.4 kilograms of solution x ; after 3 kilograms of water are replaced by 3 kilograms of solution y , to the existing 2.4 kilograms of solution x , 0.3 * 3 = 0.9 kilograms of solution x are added , so in the new solution of 8 kilograms there are 2.4 + 0.9 = 3.3 kilograms ... | a ) 35 % , b ) 37 % , c ) 41.3 % , d ) 42 % , e ) 45 % | c | multiply(const_100, divide(add(multiply(8, divide(30, const_100)), multiply(3, divide(30, const_100))), add(add(multiply(8, divide(30, const_100)), multiply(3, divide(30, const_100))), add(subtract(multiply(8, divide(70, const_100)), 3), multiply(3, divide(70, const_100)))))) | divide(n0,const_100)|divide(n1,const_100)|multiply(n3,#0)|multiply(n2,#0)|multiply(n2,#1)|multiply(n3,#1)|add(#2,#3)|subtract(#5,n2)|add(#4,#7)|add(#6,#8)|divide(#6,#9)|multiply(#10,const_100) | gain |
a train 480 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 480 / 16 * 18 / 5 = 108 kmph answer : b" | a ) 22 kmph , b ) 108 kmph , c ) 54 kmph , d ) 71 kmph , e ) 88 kmph | b | multiply(const_3_6, divide(480, 16)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
a boat can travel with a speed of 13 km / hr in still water . if the speed of the stream is 4 km / hr . find the time taken by the boat to go 80 km downstream ? | "solution speed downstream = ( 13 + 4 ) km / hr = 17 km / hr . time taken to travel 80 km downstream = ( 80 / 17 ) hrs = 4 hrs 42 minutes . answer e" | a ) 2 hours , b ) 3 hours , c ) 4 hours , d ) 5 hours , e ) none | e | divide(80, add(13, 4)) | add(n0,n1)|divide(n2,#0)| | physics |
a library has an average of 510 visitors on sunday and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is ? | "since the month begins with a sunday , so there will be five sundays in the month . required average = [ ( 510 * 5 ) + ( 240 * 25 ) ] / 30 = 8550 / 30 = 285 . answer : d" | a ) 187 , b ) 279 , c ) 280 , d ) 285 , e ) 262 | d | divide(add(multiply(add(const_4, const_1), 510), multiply(multiply(add(const_4, const_1), add(const_4, const_1)), 240)), 30) | add(const_1,const_4)|multiply(n0,#0)|multiply(#0,#0)|multiply(n1,#2)|add(#1,#3)|divide(#4,n2)| | general |
the sum of present age of abe and the age before 7 years is 35 . find the present age of abe . what will be his age after 7 years ? | present age = x before 7 yrs , y = x - 7 after 7 yrs , z = x + 7 by the qn , x + ( x - 7 ) = 35 2 x - 7 = 35 2 x = 35 + 7 x = 42 / 2 x = 21 z = x + 7 = 21 + 7 = 28 answer : d | a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | d | add(divide(add(35, 7), const_2), 7) | add(n0,n1)|divide(#0,const_2)|add(n0,#1) | general |
the two lines y = x and x = - 6 intersect on the coordinate plane . what is the value of the area of the figure formed by the intersecting lines and the x - axis ? | "the point of intersection is ( - 6 , - 6 ) . the triangle has a base of length 6 and a height of 6 . area = ( 1 / 2 ) * base * height = ( 1 / 2 ) * 6 * 6 = 18 the answer is d ." | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | divide(multiply(6, 6), const_2) | multiply(n0,n0)|divide(#0,const_2)| | general |
what distance will be covered by a bus moving at 36 kmph in 30 seconds ? | "36 kmph = 36 * 5 / 18 = 10 mps d = speed * time = 10 * 30 = 300 m . answer : d" | a ) 287 , b ) 600 , c ) 289 , d ) 300 , e ) 207 | d | multiply(multiply(36, const_0_2778), 30) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
if 2.5 of a number is equal to 0.75 of another number , the ratio of the numbers is : | "2.5 a = 0.75 b - > a / b = 0.75 / 2.5 = 75 / 250 = 3 / 10 : . a : b = 3 : 10 answer : c" | a ) 2 : 3 , b ) 3 : 4 , c ) 3 : 10 , d ) 20 : 3 , e ) 30 : 7 | c | divide(multiply(0.75, const_100), multiply(2.5, const_100)) | multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)| | other |
the average weight of 29 students is 28 kg . by the admission of a new student , the average weight is reduced to 27.5 kg . the weight of the new student is | "exp . the total weight of 29 students = 29 * 28 the total weight of 30 students = 30 * 27.5 weight of the new student = ( 30 * 27.5 – 29 * 28 ) = 825 - 812 = 13 answer : d" | a ) 22 kg , b ) 21.6 kg , c ) 22.4 kg , d ) 13 kg , e ) none of these | d | subtract(multiply(add(29, const_1), 27.5), multiply(29, 28)) | add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)| | general |
if 4 tic equals 3 tacs and 6 tacs equal 4 tocs , what is the ratio of one tic to one toc ? | "4 tic = 3 * tac and 6 * tac = 4 * toc ; 24 * tic = 18 * tac and 18 * tac = 12 * toc - - > 24 * tic = 12 * toc - - > tic / toc = 24 / 12 = 2 / 1 . answer : e ." | a ) 15 / 2 , b ) 6 / 5 , c ) 5 / 6 , d ) 3 / 10 , e ) 2 / 1 | e | add(divide(4, 4), divide(3, 4)) | divide(n3,n0)|divide(n1,n0)|add(#0,#1)| | other |
the speed of a boat in still water is 30 kmph . what is the speed of the stream if the boat can cover 80 km downstream or 40 km upstream in the same time ? | x = the speed of the stream ( 30 + x ) / ( 30 - x ) = 2 / 1 30 + x = 60 - 2 x 3 x = 30 x = 10 km / hour if the speed of the stream is 10 km / hour , then the ' downstream ' speed of the boat is 30 + 10 = 40 km / hour and the ' upstream ' speed of the boat is 30 - 10 = 20 km / hour . in that way , if the boat traveled f... | a ) 11 kmph , b ) 10 kmph , c ) 12 kmph , d ) 16 kmph , e ) 15 kmph | b | divide(30, add(const_1, const_2)) | add(const_1,const_2)|divide(n0,#0) | physics |
after getting 2 successive discounts , a shirt with a list price of rs 150 is available at rs 105 . if the second discount is 12.55 , find the first discount . | "let the first discount be x % then , 87.5 % of ( 100 - x ) % of 150 = 105 = 87.5 / 100 * ( 100 - x ) / 100 * 450 = 150 = > 105 = > 100 - x = ( 105 * 100 * 100 ) / ( 150 * 87.5 ) = 80 x = ( 100 - 80 ) = 20 first discount = 20 % answer is e ." | a ) 22 % , b ) 18 % , c ) 24 % , d ) 17 % , e ) 20 % | e | divide(multiply(subtract(150, divide(multiply(105, const_100), subtract(const_100, 12.55))), const_100), 150) | multiply(n2,const_100)|subtract(const_100,n3)|divide(#0,#1)|subtract(n1,#2)|multiply(#3,const_100)|divide(#4,n1)| | gain |
a rectangular plot measuring 60 meters by 50 meters is to be enclosed by wire fencing . if the poles of the fence are kept 5 meters apart . how many poles will be needed ? | perimeter of the plot = 2 ( 60 + 50 ) = 220 m no of poles = 220 / 5 = 44 m answer : a | ['a ) 44 m', 'b ) 66 m', 'c ) 26 m', 'd ) 56 m', 'e ) 25 m'] | a | divide(multiply(add(60, 50), const_2), 5) | add(n0,n1)|multiply(#0,const_2)|divide(#1,n2) | physics |
the arithmetic mean and standard deviation of a certain normal distribution are 15 and 1.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | the value which isexactlytwo sd less than the mean is : mean - 2 * sd = 15 - 2 * 1.5 = 12 . answer : d . | a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | d | subtract(15, multiply(2, 1.5)) | multiply(n1,n2)|subtract(n0,#0) | general |
in how many years will a sum of money doubles itself at 12 % per annum on simple interest ? | "p = ( p * 12 * r ) / 100 r = 8 % answer : d" | a ) 15 % , b ) 20 % , c ) 10 % , d ) 8 % , e ) 22 % | d | divide(const_100, 12) | divide(const_100,n0)| | gain |
20 % of a number is more than 20 % of 650 by 190 . find the number ? | "( 20 / 100 ) * x â € “ ( 20 / 100 ) * 650 = 190 1 / 5 x = 320 x = 1600 answer : d" | a ) 288 , b ) 278 , c ) 800 , d ) 1600 , e ) 121 | d | divide(multiply(add(divide(multiply(650, 20), const_100), 190), const_100), 20) | multiply(n1,n2)|divide(#0,const_100)|add(n3,#1)|multiply(#2,const_100)|divide(#3,n0)| | gain |
it will take 18 days for cameron to complete a certain task alone . he worked for 9 days before she was joined by sandra . both of them completed the remaining task in 3 and half days . how many days will it take both of them to complete the entire job ? | explanation : cameron and sandra completed half work in 3.5 days = > they can complete whole work in 7 days answer : option b | a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 9 | b | divide(add(3, divide(const_1, const_2)), divide(9, 18)) | divide(const_1,const_2)|divide(n1,n0)|add(n2,#0)|divide(#2,#1) | physics |
if greg buys 5 shirts , 4 trousers and 2 ties , the total cost is $ 80 . if greg buys 7 shirts , 4 trousers and 2 ties , the total cost is $ 70 . how much will it cost him to buy 2 trousers , 3 shirts and 1 ties ? | "solution : 5 x + 4 y + 2 z = 80 7 x + 4 y + 2 z = 70 adding both the equations = 12 x + 8 y + 4 z = 150 3 x + 2 y + 1 z = 37.5 ans a" | a ) $ 37.5 , b ) $ 64 , c ) $ 75 , d ) $ 96 , e ) can not be determined | a | divide(add(70, 80), const_2.0) | add(n3,n7)|divide(#0,n5)| | general |
what is the sum of all the numbers between 1 and 16 , inclusive ? | "all you have to do is add 1 + 2 + 3 + 4 . . . + 14 + 15 + 16 , which is 190 . final answer : a" | a ) 190 , b ) 200 , c ) 185 , d ) 166 , e ) 213 | a | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 1), add(const_2, const_4)) | add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)| | general |
ramesh purchased a refrigerator for rs . 12500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 20 % if no discount was offered ? | "price at which the tv set is bought = rs . 12,500 discount offered = 20 % marked price = 12500 * 100 / 80 = rs . 15625 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 15625 + 375 = rs . 16000 the price at which the tv should be sold to get a profit of 20 % if no di... | a ) 19200 , b ) 17606 , c ) 17604 , d ) 17600 , e ) 117601 | a | divide(multiply(add(const_100, 20), add(divide(multiply(12500, const_100), subtract(const_100, 20)), add(125, 250))), const_100) | add(n4,const_100)|add(n2,n3)|multiply(n0,const_100)|subtract(const_100,n1)|divide(#2,#3)|add(#1,#4)|multiply(#0,#5)|divide(#6,const_100)| | gain |
the distance between delhi and mathura is 100 kms . a starts from delhi with a speed of 50 kmph at 7 a . m . for mathura and b starts from mathura with a speed of 45 kmph at 8 p . m . from delhi . when will they meet ? | "d = 100 – 50 = 50 rs = 45 + 50 = 95 t = 90 / 45 = 0.5 hours 8 a . m . + 0.5 = 8 1 / 2 a . m . . answer : d" | a ) 11 , b ) 77 , c ) 8 3 / 4 , d ) 8 1 / 2 , e ) 8 | d | add(8, divide(100, add(50, 45))) | add(n1,n3)|divide(n0,#0)|add(n4,#1)| | physics |
the length and breadth of a rectangular floor are 16.25 metre and 12.75 metre respectively . find how many minimum number of square tiles would be required to cover it completely ? | explanation : since we require minimum number of square tiles , the size of the tile is given as the h . c . f . of two sides of the room . the h . c . f . of 1625 cm & 1275 cm . is 25 cm . hence , we get , required number = ( 1625 * 1275 ) / ( 25 * 25 ) = 3315 answer : d | ['a ) 375', 'b ) 2570', 'c ) 2800', 'd ) 3315', 'e ) none of these'] | d | divide(rectangle_area(multiply(16.25, const_100), multiply(12.75, const_100)), multiply(add(add(const_10, const_10), add(const_3, const_2)), add(add(const_10, const_10), add(const_3, const_2)))) | add(const_10,const_10)|add(const_2,const_3)|multiply(n0,const_100)|multiply(n1,const_100)|add(#0,#1)|rectangle_area(#2,#3)|multiply(#4,#4)|divide(#5,#6) | geometry |
a man can row upstream at 25 kmph and downstream at 37 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 37 m = ( 37 + 25 ) / 2 = 31 answer : c" | a ) 86 , b ) 67 , c ) 31 , d ) 15 , e ) 17 | c | divide(add(25, 37), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
walking at 50 % of his usual speed a man takes 24 minutes more to cover a distance . what is his usual time to cover this distance ? | "speed is inversly proprtional to time walking at 50 % of speed meand 1 / 2 s takes 2 t . it takes 24 minutes extra to cover the distance . then 2 t = t + 24 t = 24 . option a is correct" | a ) 24 , b ) 36 , c ) 42 , d ) 48 , e ) 54 | a | divide(24, subtract(divide(const_1, divide(50, const_100)), const_1)) | divide(n0,const_100)|divide(const_1,#0)|subtract(#1,const_1)|divide(n1,#2)| | physics |
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