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 |
|---|---|---|---|---|---|---|
a train 250 m long is running at a speed of 27 km / hr . in what time will it pass a bridge 200 m long ? | "speed = 27 * 5 / 18 = 15 / 2 m / sec total distance covered = 250 + 200 = 450 m required time = 450 * 2 / 15 = 60 sec answer : e" | a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | e | divide(250, multiply(subtract(27, 200), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
how many different ways can 2 students be seated in a row of 6 desks , so that there is always at least one empty desk between the students ? | "10 ways to seat the students : with two empty seats between 1 empty w / one student on the left most 1 empty . . . . right most two students can be interchanged 10 x 2 = 20 d" | a ) 2 , b ) 3 , c ) 4 , d ) 20 , e ) 12 | d | permutation(subtract(6, const_1), 2) | subtract(n1,const_1)|permutation(#0,n0)| | probability |
if 75 percent of a class answered the first question on a certain test correctly , 35 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ? | "i did n ' t understand how to connect 20 percent with the whole equation . why 20 is out of the percentile diagram ( circles ) but inside of 100 ? 20 % appeared in the exam ; however did not answer question aquestion b correctly so they are out of the two circles , however as they have appeared in the exam , they have... | a ) 10 % , b ) ( 8 ) 20 % , c ) 30 % , d ) 50 % , e ) 65 % | c | subtract(add(add(75, 35), 20), const_100) | add(n0,n1)|add(n2,#0)|subtract(#1,const_100)| | other |
fresh grapes contain 90 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 30 kg of fresh grapes ? | "the weight of non - water in 30 kg of fresh grapes ( which is 100 - 90 = 10 % of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is 100 - 20 = 80 % of whole weight ) , so 30 â ˆ — 0.1 = x â ˆ — 0.8 - - > x = 3.75 answer : a" | a ) 3.75 kg , b ) 2.4 kg , c ) 2.5 kg , d ) 10 kg , e ) none of these | a | multiply(divide(divide(multiply(subtract(const_100, 90), 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 |
what least number must be added to 1056 , so that the sum is completely divisible by 23 ? | "23 ) 1056 ( 45 92 - - - 136 115 - - - 21 - - - required number = ( 23 - 21 ) = 2 . a )" | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | subtract(multiply(add(multiply(const_4, const_10), const_2), 23), 1056) | multiply(const_10,const_4)|add(#0,const_2)|multiply(n1,#1)|subtract(#2,n0)| | general |
4 ! / ( 4 - 3 ) ! = ? | 4 ! / ( 4 - 3 ) ! = 4 ! / 1 ! = 4 * 3 * 2 = 24 . hence , the correct answer is a . | a ) 24 , b ) 34 , c ) 35 , d ) 36 , e ) 39 | a | divide(factorial(4), factorial(subtract(4, 3))) | factorial(n0)|subtract(n0,n2)|factorial(#1)|divide(#0,#2)| | general |
natasha climbs up a hill , and descends along the same way she went up . it takes her 4 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 1.5 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "let the distance to the top be x , so the total distance traveled by natasha is 2 x . the total time is 4 + 2 = 6 hours the average speed = total distance / total time taken = 2 x / 6 = x / 3 the average speed of the complete journey is 1.5 km / hour x / 3 = 1.5 x = 4.5 km the average speed while climbing = distance /... | a ) 1.065 , b ) 1.125 , c ) 1.225 , d ) 1.375 , e ) 1.425 | b | divide(divide(multiply(add(4, 2), 1.5), 2), 4) | add(n0,n1)|multiply(n2,#0)|divide(#1,n1)|divide(#2,n0)| | physics |
5358 x 61 = ? | "a 5358 x 61 = 5358 x ( 60 + 1 ) = 5358 x 60 + 5358 x 1 = 321480 + 5358 = 326838 ." | a ) 326838 , b ) 323758 , c ) 323298 , d ) 273258 , e ) 327382 | a | multiply(divide(5358, 61), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
if n = 4 ^ 11 – 4 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 4 ^ 3 - 4 . now 4 ^ 3 has last digit 4 . , we subtract 4 from 4 = 0 is the answer . option a | a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | a | divide(log(4), log(power(4, 11))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 11 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos las... | "# of basic stereos was 2 / 3 of total and # of deluxe stereos was 1 / 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 11 / 5 units . total time for basic would be 10 * 1 = 10 and ... | a ) 11 / 21 , b ) 14 / 31 , c ) 7 / 15 , d ) 17 / 35 , e ) 1 / 2 | a | divide(add(3, 2), const_10) | add(n0,n1)|divide(#0,const_10)| | general |
after decreasing 24 % in the price of an article costs rs . 608 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 608 cp = 8 * 100 = > cp = 800 answer : c" | a ) 128 , b ) 277 , c ) 800 , d ) 2688 , e ) 1991 | c | divide(608, subtract(const_1, divide(24, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
in a zoo there are 30 penguins , 22 zebras , and 8 tigers with some zookeepers . if the total number of heads are 132 less than the number of feet , the number of zookeepers is | explanation : let number of zookeepers be x . then , total number of feet = 2 x 30 + 4 x 22 + 4 x 8 + 2 x = 2 x + 180 . total number of heads = 30 + 22 + 8 + x = 60 + x . therefore ( 2 x + 180 ) - 132 = ( 60 + x ) + or x = 12 . answer : a | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 22 | a | subtract(multiply(30, const_2), subtract(add(add(multiply(30, const_2), multiply(22, const_4)), multiply(8, const_4)), 132)) | multiply(n0,const_2)|multiply(n1,const_4)|multiply(n2,const_4)|add(#0,#1)|add(#3,#2)|subtract(#4,n3)|subtract(#0,#5) | general |
what is the greatest number of identical bouquets that can be made out of 21 white and 91 red tulips if no flowers are to be left out ? ( two bouquets are identical whenever the number of red tulips in the two bouquets is equal and the number of white tulips in the two bouquets is equal ) | since no flowers are to be left out , then the number of bouquets must be a factor of both 21 and 91 . for example , we can not have 2 bouquets since we can not divide 91 red tulips into 2 bouquets without one tulip left over . only answer choice which is a factor of 91 is e ( 7 ) . answer : e . | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | divide(21, const_3) | divide(n0,const_3) | general |
what least number must be added to 1049 , so that the sum is completely divisible by 25 ? | if we divide 1049 by 25 remainder is 24 25 - 24 = 1 answer : e | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 1 | e | subtract(multiply(add(multiply(const_4, const_10), const_2), 25), 1049) | multiply(const_10,const_4)|add(#0,const_2)|multiply(n1,#1)|subtract(#2,n0) | general |
man can row upstream at 25 kmph and downstream at 35 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 35 m = ( 35 + 25 ) / 2 = 30 answer : c" | a ) 12 , b ) 22 , c ) 30 , d ) 88 , e ) 13 | c | divide(add(25, 35), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
according to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . how many 10 - ounce cans of the concentrate are required to prepare 200 6 - ounce servings of orange juice ? | "orange juice concentrate : water : : 1 : 3 total quantity of orange juice = 200 * 6 = 1200 oz so orange juice concentrate : water : : 300 oz : 900 oz no . of 10 oz can = 300 oz / 10 oz = 30 answer c , 30 cans" | a ) 25 , b ) 34 , c ) 30 , d ) 67 , e ) 100 | c | divide(multiply(6, 200), divide(200, const_10)) | divide(n2,const_10)|multiply(n2,n3)|divide(#1,#0)| | general |
when positive integer n is divided by positive integer j , the remainder is 14 . if n / j = 213.28 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 213.28 here 213 is the quotient . given that remainder = 14 so , 213.28 = 213 + 14 / j so , j = 50" | a ) 30 , b ) 35 , c ) 40 , d ) 50 , e ) 60 | d | divide(14, subtract(213.28, add(const_100, add(multiply(const_4, const_10), const_2)))) | multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3)| | general |
the diagonals of a rhombus are 30 cm and 12 cm . find its area ? | "1 / 2 * 30 * 12 = 180 answer : c" | a ) 176 , b ) 190 , c ) 180 , d ) 278 , e ) 292 | c | rhombus_area(30, 12) | rhombus_area(n0,n1)| | geometry |
if the cost price of 50 articles is equal to the selling price of 30 articles , then the gain or loss percent is ? | "percentage of profit = 20 / 30 * 100 = 67 % answer : e" | a ) 16 , b ) 127 , c ) 12 , d ) 18 , e ) 67 | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 30), 50)), divide(multiply(const_100, 30), 50))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 percent of the distribution lies within one standard deviation d of the mean , what percent w of the distribution is less than m + d ? | "d the prompt says that 68 % of the population lies between m - d and m + d . thus , 32 % of the population is less than m - d or greater than m + d . since the population is symmetric , half of this 32 % is less than m - d and half is greater than m + d . thus , w = ( 68 + 16 ) % or ( 100 - 16 ) % of the population is... | a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | d | subtract(const_100, divide(subtract(const_100, 68), const_2)) | subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)| | general |
a man whose speed is 4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 2.5 kmph , find his average speed for the total journey ? | explanation : speed of man = 4.5 kmph speed of stream = 2.5 kmph speed in downstream = 7 kmph speed in upstream = 2 kmph average speed = ( 2 x 7 x 2 ) / 9 = 3.1 kmph . answer : a | a ) 3.1 kmph , b ) 5 kmph , c ) 6 kmph , d ) 4 kmph , e ) 7 kmph | a | divide(add(4.5, subtract(4.5, 2.5)), const_2) | subtract(n0,n1)|add(n0,#0)|divide(#1,const_2) | general |
nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw . if the total weight of one screw and one nail is half that of 6 screws and one nail , then the total weight of 3 screws , and 3 nails is how many times that of 1.5 screws and 4.5 nails ? | "let the weight of nail be n and that of screw be s . . so s + w = 1 / 2 * ( 6 s + 1 n ) . . . or 1 n = 4 s . . lets see the weight of 3 s and 3 n = 3 s + 3 * 4 s = 15 s . . and weight of 1.5 s and 4.5 n = 1.5 s + 4.5 * 1.5 s = 8.25 s . . ratio = 15 s / 8.25 s = 15 / 8.25 = 20 / 11 e" | a ) a . 4 , b ) b . 5 / 2 , c ) c . 2 , d ) d . 3 / 2 , e ) e . 20 / 11 | e | divide(add(multiply(3, const_4), 3), const_10) | multiply(n1,const_4)|add(n1,#0)|divide(#1,const_10)| | general |
a camera lens filter kit containing 5 filters sells for $ 52.50 . if the filters are purchased individually , 2 of them are priced at $ 6.45 each , 2 at $ 11.05 each , 1 at $ 15.50 . the amount saved by purchasing the kit is what percent of the total price of the 5 filters purchased individually ? | "cost of kit = $ 52.50 if filters are purchased individually - $ 6.45 * 2 + $ 11.05 * 2 + $ 15.50 = $ 50.50 amount saved = $ 52.50 - $ 50.50 = $ 2.50 required % age = ( $ 2.50 / $ 52.50 ) * 100 = 4.76 % so , the correct answer is b ." | a ) 3.76 % , b ) 4.76 % , c ) 5.76 % , d ) 6.76 % , e ) 7.76 % | b | divide(multiply(subtract(52.50, add(15.50, add(multiply(2, 6.45), multiply(2, 11.05)))), const_100), 52.50) | multiply(n2,n3)|multiply(n2,n5)|add(#0,#1)|add(n7,#2)|subtract(n1,#3)|multiply(#4,const_100)|divide(#5,n1)| | general |
the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 2 students ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . ∴ required average = 266 / 2 = 133 years . answer a" | a ) 133 years , b ) 177 years , c ) 119 years , d ) 212 years , e ) none | a | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)| | general |
if the sides of a cube are in the ratio 4 : 3 . what is the ratio of their diagonals ? | "a 1 : a 2 = 4 : 3 d 1 : d 2 = 4 : 3 answer : a" | a ) 4 : 3 , b ) 4 : 8 , c ) 4 : 1 , d ) 4 : 2 , e ) 4 : 7 | a | divide(4, 3) | divide(n0,n1)| | geometry |
if 638521 is to be divisible by 5 , what is the least whole number that should be added to ? | "a number is divisible by 5 if the last digit is either 0 or 5 . here , 638521 = 1 ( last digit is neither 0 or 5 ) . the next multiple of 5 i . e to make the last digit 0 or 5 add 4 4 must be added to 638521 to make it divisible by 5 d" | a ) 8 , b ) 7 , c ) 5 , d ) 4 , e ) 3 | d | add(floor(5), const_1) | floor(n1)|add(#0,const_1)| | general |
the manager at a health foods store mixes a unique superfruit juice cocktail that costs $ 1399.45 per litre to make . the cocktail includes mixed fruit juice and a ç ai berry juice , which cost $ 262.85 per litre and $ 3104.35 per litre , respectively . the manager has already opened 37 litres of the mixed fruit juice ... | "262.85 ( 37 ) + 3 , 104.35 x = 1 , 399.45 ( 37 + x ) solve the equation . 262.85 ( 37 ) + 3 , 104.35 x = 1 , 399.45 ( 37 + x ) 9725.45 + 3 , 104.35 x = 51779.65 + 1 , 399.45 x 9725.45 + 1 , 704.9 x = 51779.65 1 , 704.9 x = 42054.2 x ≈ 24.67 answer is b ." | a ) 17 litres , b ) 24.67 litres , c ) 11 litres , d ) 24.07 litres , e ) 38.67 litres | b | divide(subtract(multiply(37, 1399.45), multiply(37, 262.85)), subtract(3104.35, 1399.45)) | multiply(n0,n3)|multiply(n1,n3)|subtract(n2,n0)|subtract(#0,#1)|divide(#3,#2)| | general |
two equilateral triangle of side 12 cm are placed one on top another , such a 6 pionted star is formed if the six vertices lie on a circle what is the area of the circle not enclosed by the star ? | answer : a | ['a ) 21', 'b ) 88937', 'c ) 269', 'd ) 279', 'e ) 2701'] | a | multiply(multiply(divide(sqrt(const_3), const_2), 12), const_2) | sqrt(const_3)|divide(#0,const_2)|multiply(n0,#1)|multiply(#2,const_2) | geometry |
the average age of a class is 15.8 years . the average age of boys in the class is 16.4 years while that of girls is 15.4 years . what is the ratio of boys to girls in the class ? | let x = number of girls y = number of boys sum of girls ages / x = 15.4 sum of boys ages / y = 16.4 the sum of the girls ' ages is 15.4 x , and boys ' sum is 16.4 y ( sum of girls ages + sum of boys ages ) / ( x + y ) = 15.8 ( 15.4 x + 16.4 y ) / ( x + y ) = 15.8 15.4 x + 16.4 y = 15.8 ( x + y ) 15.4 x + 16.4 y = 15.8 ... | a ) 1 : 2 , b ) 3 : 4 , c ) 3 : 5 , d ) 2 : 3 , e ) none of these | d | divide(subtract(15.8, 15.4), subtract(16.4, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1) | general |
subtracting 7 % of a from a is equivalent to multiplying a by how much ? | "answer let a - 7 % of a = ab . ⇒ ( 93 x a ) / 100 = ab ∴ b = 0.93 correct option : a" | a ) 0.93 , b ) 9.4 , c ) 0.094 , d ) 94 , e ) none | a | divide(subtract(const_100, 7), const_100) | subtract(const_100,n0)|divide(#0,const_100)| | general |
a number divided by 84 leaves remainder 57 what is the remainder when same number divided by 12 | "add 84 + 57 = 141 now 141 divided by 12 so we get 9 as reaminder answer : c" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | divide(subtract(57, 12), subtract(57, 12)) | subtract(n1,n2)|divide(#0,#0)| | general |
a children ' s gift store sells gift certificates in denominations of $ 3 and $ 5 . the store sold ' m ' $ 3 certificates and ' n ' $ 5 certificates worth $ 93 on a saturday afternoon . if ' m ' and ' n ' are natural numbers , how many different values can ' m ' take ? | explanatory answer step 1 : key data 1 . total value of all certificates sold = $ 93 . 2 . certificates sold were in denominations of $ 3 and $ 5 . 3 . both ' m ' and ' n ' are natural numbers . step 2 : approach the value of all certificates sold , 93 is divisible by 3 . so , a maximum of 31 $ 3 certificates and no $ ... | a ) 5 , b ) 7 , c ) 6 , d ) 31 , e ) 18 | c | floor(divide(divide(93, 3), 5)) | divide(n4,n0)|divide(#0,n1)|floor(#1) | physics |
how long will a boy take to run round a square field of side 45 meters , if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 45 * 4 = 180 m time taken = 180 * 2 / 5 = 72 sec answer is a" | a ) 72 sec , b ) 45 sec , c ) 1 min , d ) 32 sec , e ) 25 sec | a | divide(multiply(45, const_4), multiply(9, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain |
what is the greatest positive integer w such that 3 ^ w is a factor of 9 ^ 10 ? | "what is the greatest positive integer w such that 3 ^ w is a factor of 9 ^ 10 ? 9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 d . 20" | a ) 5 , b ) w = 9 , c ) w = 10 , d ) w = 20 , e ) w = 30 | d | multiply(subtract(9, 10), 10) | subtract(n1,n2)|multiply(n2,#0)| | other |
in a group of 100 people , 60 like volleyball , 50 like hockey . how many like both volleyball and hockey ? | "make a venn diagram , and enter your data . let the number of people who like both volleyball and hockey be x 60 - x + x + 50 - x = 100 x = 10 so number who like both volleyball and hockey = 10 answer c" | a ) 20 , b ) 15 , c ) 10 , d ) 5 , e ) 25 | c | add(add(50, subtract(60, 50)), 50) | subtract(n1,n2)|add(n2,#0)|add(n2,#1)| | other |
triathlete dan runs along a 2 - mile stretch of river and then swims back along the same route . if dan runs at a rate of 10 miles per hour and swims at a rate of 8 miles per hour , what is his average rate for the entire trip in miles per minute ? | "dan travels 4 miles round trip . running part : ( 2 / 10 = 1 / 5 * 60 = 12 minutes ) swimming part : ( 2 / 8 = 1 / 4 * 60 = 15 minutes ) 4 miles in ( 12 + 15 ) minutes 4 / 27 = 1 / 7 mile per minute answer : 1 / 7 mile per minute a" | a ) 1 / 7 , b ) 2 / 15 , c ) 3 / 15 , d ) 1 / 4 , e ) 3 / 8 | a | divide(add(2, 2), add(multiply(divide(2, 8), const_60), multiply(divide(2, 10), const_60))) | add(n0,n0)|divide(n0,n2)|divide(n0,n1)|multiply(#1,const_60)|multiply(#2,const_60)|add(#3,#4)|divide(#0,#5)| | physics |
a square has a side 5 centimeters shorter than the side of a second square . the area of the larger square is four times the area of the smaller square . find the side of each square | let x be the side of the smaller square and y be the side of the larger square . the statement ` ` a square has a side 5 centimeters shorter than the side of a second square ' ' may be formulated by x = y - 5 the area of the smaller square is equal to x 2 = ( y - 5 ) 2 and the area of the larger square is equal to y 2 ... | ['a ) 5', 'b ) 6', 'c ) 7', 'd ) 8', 'e ) 9'] | a | square_edge_by_area(square_area(5)) | square_area(n0)|square_edge_by_area(#0) | physics |
a train 360 m long is running at a speed of 48 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 48 * 5 / 18 = 40 / 3 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 3 / 40 = 37.5 sec answer : c" | a ) 40 sec , b ) 11 sec , c ) 37.5 sec , d ) 19 sec , e ) 10 sec | c | divide(360, multiply(subtract(48, 140), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
a train covers a distance of 100 km in 1 hour . if its speed is decreased by 30 km / hr , the time taken by the car to cover the same distance will be ? | "speed = 100 / 1 = 100 km / hr new speed = 100 - 30 = 70 km / hr time taken = 100 / 70 = 1 hr 25 min answer is b" | a ) 1 hr , b ) 1 hr 25 min , c ) 50 min , d ) 1 hr 30 min , e ) 1 hr 45 min | b | divide(100, subtract(divide(100, 1), 30)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
the ratio of the present age of sandy to that of molly is 7 : 2 . sixteen years from now , the ratio of the ages of sandy to molly will be 5 : 2 . what was sandy ' s age 6 years ago ? | let the present age of sandy be 7 x years and that of molly be 2 x years . ( 7 x + 16 ) / ( 2 x + 16 ) = 5 / 2 4 x = 48 x = 12 six years ago , sandy ' s age was 7 ( 12 ) - 6 = 78 the answer is d . | a ) 66 , b ) 70 , c ) 74 , d ) 78 , e ) 82 | d | subtract(divide(multiply(subtract(multiply(5, add(const_10, 6)), multiply(2, add(const_10, 6))), 7), add(2, 2)), 6) | add(n4,const_10)|add(n1,n1)|multiply(n2,#0)|multiply(n1,#0)|subtract(#2,#3)|multiply(n0,#4)|divide(#5,#1)|subtract(#6,n4) | general |
evaluate : 50 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 50 - 12 * 3 * 2 = 50 - 8 = 42 correct answer e" | a ) 62 , b ) 52 , c ) 32 , d ) 12 , e ) 42 | e | subtract(50, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
a grocer has 400 pounds of coffee in stock , 20 percent of which is decaffeinated . if the grocer buys another 100 pounds of coffee of which 60 percent is decaffeinated , what percent , by weight , of the grocer ’ s stock of coffee is decaffeinated ? | "1 . 20 % of 400 = 80 pounds of decaffeinated coffee 2 . 60 % of 100 = 60 pounds of decaffeinated coffee 3 . wt have 140 pounds of decaffeinated out of 500 pounds , that means 140 / 500 * 100 % = 28 % . the correct answer is a ." | a ) 28 % , b ) 30 % , c ) 32 % , d ) 34 % , e ) 40 % | a | multiply(divide(add(multiply(divide(20, 100), 400), multiply(100, divide(60, 100))), add(400, 100)), 100) | add(n0,n2)|divide(n1,n2)|divide(n3,n2)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,n2)| | gain |
the area of a parallelogram is 1764 sq m and its base is nine times size of the corresponding altitude . then the length of the base is ? | "9 x * x = 1764 = > x = 14 answer : d" | a ) 21 , b ) 9 , c ) 11 , d ) 14 , e ) 7 | d | sqrt(divide(1764, const_2)) | divide(n0,const_2)|sqrt(#0)| | geometry |
at what rate percent per annum will a sum of money double in 8 years . | "explanation : let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) / ( p x 8 ) ] % = 12.5 % per annum . answer : a ) 12.5 %" | a ) 12.5 % , b ) 12.9 % , c ) 18.5 % , d ) 11.5 % , e ) 12.3 % | a | divide(const_100, 8) | divide(const_100,n0)| | gain |
paper charge is rs . 60 per kg . how much expenditure would be there to cover a cube of edge 10 m with a paper , if one kg of paper covers 20 sq . m . area ? | explanation : total surface area of the cube = 6 x ( 10 ) 2 = 600 sq . m 1 kg covers 20 sq . m area , for 600 we need = 600 / 20 = 30 kg paper . expenditure = rate x quantity = rs . 60 x 30 kg = rs . 1800 answer is d | ['a ) rs . 2250', 'b ) rs . 3600', 'c ) rs . 2700', 'd ) rs . 1800', 'e ) none of these'] | d | multiply(divide(multiply(divide(60, 10), multiply(10, 10)), 20), 60) | divide(n0,n1)|multiply(n1,n1)|multiply(#0,#1)|divide(#2,n2)|multiply(n0,#3) | geometry |
a shopkeeper sells his goods at cost price but uses a weight of 800 gm instead of kilogram weight . what is his profit percentage ? | "100 ( 1000 - 800 ) / 800 = 25 % profit percentage = 25 % answer a" | a ) 25 % , b ) 18 % , c ) 20 % , d ) 40 % , e ) 45 % | a | multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 800), 800), 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 |
how many odd numbers between 10 and 900 are the squares of integers ? | "the square of an odd number is an odd number : 10 < odd < 1,000 10 < odd ^ 2 < 1,000 3 . something < odd < 31 . something ( by taking the square root ) . so , that odd number could be any odd number from 5 to 31 , inclusive : 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 , 29 , and 31 . 13 numbers . answer : ... | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | add(10, const_4) | add(n0,const_4)| | geometry |
what is the value of x ^ 2 yz − xyz ^ 2 , if x = 2 , y = 1 , and z = 2 ? | "4 * 1 * 2 - ( 2 * 1 * 4 ) = 8 + 8 = 0 ans : c" | a ) 20 , b ) 24 , c ) 0 , d ) 32 , e ) 48 | c | multiply(multiply(multiply(negate(2), 1), 1), subtract(negate(2), 1)) | negate(n0)|multiply(n3,#0)|subtract(#0,n3)|multiply(n3,#1)|multiply(#3,#2)| | general |
by selling 18 pencils for a rupee a man loses 10 % . how many for a rupee should he sell in order to gain 35 % ? | 90 % - - - 18 135 % - - - ? 90 / 135 * 18 = 12 answer : e | a ) 11 , b ) 13 , c ) 15 , d ) 18 , e ) 12 | e | multiply(divide(const_1, multiply(add(const_100, 35), divide(const_1, subtract(const_100, 10)))), 18) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4) | gain |
how many pieces of 0.35 meteres can be cut from a rod 12.5 meteres long | "explanation : we need so simple divide 42.5 / 0.85 , = ( 1250 / 35 ) = 35 option c" | a ) 30 , b ) 40 , c ) 35 , d ) 60 , e ) 70 | c | divide(12.5, 0.35) | divide(n1,n0)| | physics |
in a class , 7 students like to play basketball and 5 like to play cricket . 3 students like to play on both basketball and cricket . how many students like to play basketball or cricket or both ? | "draw a venn diagram yourself ! b + c - bc = number of students that play either basketball or cricket 7 + 5 - 3 = 9 d )" | a ) 12 , b ) 15 , c ) 16 , d ) 9 , e ) 22 | d | subtract(add(7, 5), 3) | add(n0,n1)|subtract(#0,n2)| | other |
a jar contains only red , yellow , and orange marbles . if there are 3 red , 6 yellow , and 4 orange marbles , and 3 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ? | "i started by finding the 2 probabilities , without calculation , like this : p ( yyr ) p ( yry ) p ( ryy ) i calculated the first one and ended in 1 / 22 . i looked at the answer choices at this point and saw answer d : 3 / 22 . this helped me realise that for the 3 possible orderings the probabbility is the same . so... | a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 5 / 22 | e | divide(multiply(choose(6, 2), choose(3, 1)), choose(add(add(6, 4), 3), 3)) | add(n1,n2)|choose(n1,n4)|choose(n0,n5)|add(n0,#0)|multiply(#1,#2)|choose(#3,n0)|divide(#4,#5)| | probability |
jerry travels 8 miles at an average speed of 40 miles per hour , stops for 12 minutes , and then travels another 20 miles at an average speed of 60 miles per hour . what is jerry ’ s average speed , in miles per hour , for this trip ? | "total time taken by jerry = ( 8 / 40 ) * 60 minutes + 12 minutes + ( 20 / 60 ) * 60 minutes = 44 minutes average speed = total distance / total time = ( 8 + 20 ) miles / ( 44 / 60 ) hours = 28 * 60 / 44 = 42 miles per hour answer : option a" | a ) 42 , b ) 42.5 , c ) 44 , d ) 50 , e ) 52.5 | a | divide(add(8, 20), add(add(divide(8, 40), divide(12, 60)), divide(20, 60))) | add(n0,n3)|divide(n0,n1)|divide(n2,n4)|divide(n3,n4)|add(#1,#2)|add(#4,#3)|divide(#0,#5)| | physics |
8 men can dig a pit in 20 days . if a men work half as much again as a boy , then 4 men and 9 boys can dig a similar pit . fint the days for 15 boys can dig ? | 1 man = 3 / 2 boys ( 4 men + 9 boys ) = 15 boys 8 men = [ ( 3 / 2 ) + 8 ] = 12 boys now , 12 boys can dig the pit in 20 days . 15 boys can dig = 16 days . answer is option e | a ) 17 , b ) 10 , c ) 15 , d ) 20 , e ) 16 | e | divide(multiply(multiply(8, divide(subtract(15, 9), 4)), 20), 15) | subtract(n4,n3)|divide(#0,n2)|multiply(n0,#1)|multiply(n1,#2)|divide(#3,n4) | physics |
a man invests in a 16 % stock at 128 . the interest obtained by him is | by investing rs 128 , income derived = rs . 16 by investing rs . 100 , income derived = = rs . 12.5 interest obtained = 12.5 % answer : c | a ) 22.5 % , b ) 42.5 % , c ) 12.5 % , d ) 62.5 % , e ) 82.5 % | c | multiply(divide(16, 128), const_100) | divide(n0,n1)|multiply(#0,const_100) | gain |
in what ratio must tea at rs . 60 per kg be mixed with tea at rs . 70 per kg so that the mixture must be worth rs . 63 per kg ? | "required ratio = 700 : 300 = 7 : 3 answer d" | a ) 3 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 7 : 3 , e ) none | d | divide(subtract(70, 63), subtract(63, 60)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | other |
how many pounds of salt at 80 cents / lb must be mixed with 40 lbs of salt that costs 35 cents / lb so that a merchant will get 20 % profit by selling the mixture at 48 cents / lb ? | "selling price is 48 cents / lb for a 20 % profit , cost price should be 40 cents / lb ( cp * 6 / 5 = 48 ) basically , you need to mix 35 cents / lb ( salt 1 ) with 80 cents / lb ( salt 2 ) to get a mixture costing 40 cents / lb ( salt avg ) weight of salt 1 / weight of salt 2 = ( salt 2 - saltavg ) / ( saltavg - salt ... | a ) 5 , b ) 15 , c ) 40 , d ) 50 , e ) 25 | a | divide(subtract(multiply(48, 40), multiply(divide(add(const_100, 20), const_100), multiply(35, 40))), subtract(multiply(80, divide(add(const_100, 20), const_100)), 48)) | add(n3,const_100)|multiply(n1,n4)|multiply(n1,n2)|divide(#0,const_100)|multiply(#3,#2)|multiply(n0,#3)|subtract(#1,#4)|subtract(#5,n4)|divide(#6,#7)| | gain |
how long does a train 120 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length ? | explanation : t = ( 660 + 120 ) / 54 * 18 / 5 t = 52 answer : option c | a ) 33 , b ) 72 , c ) 52 , d ) 82 , e ) 62 | c | divide(add(120, 660), multiply(54, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1) | physics |
if 2 cards are selected at random from the deck of 52 cards then what is the probability of one of the selected cards will be king and other will be 10 ? a deck of cards has a total of 52 cards , consisting of 4 suits ; ( spades ( black ) , hearts ( red ) , diamond ( red ) s , and clubs ( black ) ) ; and 13 cards inclu... | 2 possible cases : king - 10 or 10 - king ( 4 kings and 4 10 ) . either way , the total probability = 2 ( king - 10 ) = 2 ( 4 / 52 * 4 / 51 ) = 18 / 2652 . b is the correct answer . | a ) 8 / 2652 , b ) 18 / 2652 , c ) 1 / 2652 , d ) 12 / 2652 , e ) 16 / 2652 | b | divide(multiply(multiply(subtract(4, const_1), subtract(4, const_1)), const_2), multiply(52, subtract(52, const_1))) | subtract(n4,const_1)|subtract(n1,const_1)|multiply(#0,#0)|multiply(n1,#1)|multiply(#2,const_2)|divide(#4,#3) | probability |
a and b are two multiples of 14 , and q is the set of consecutive integers between a and b , inclusive . if q contains 10 multiples of 14 , how many multiples of 7 are there in q ? | "halfway between the multiples of 14 , there will be another multiple of 7 . the total number of multiples of 7 is 10 + 9 = 19 . the answer is b ." | a ) 20 , b ) 19 , c ) 18 , d ) 17 , e ) 16 | b | subtract(multiply(10, const_2), const_1) | multiply(n1,const_2)|subtract(#0,const_1)| | physics |
60 persons like apple . 7 like orange and mango dislike apple . 10 like mango and apple and dislike orange . 4 like all . how many people like apple ? | "orange + mango - apple = 7 mango + apple - orange = 10 apple = 60 orange + mango + apple = 4 60 + 10 + 4 - 7 = 67 like apple answer : e" | a ) 47 , b ) 46 , c ) 54 , d ) 58 , e ) 67 | e | add(add(subtract(60, 4), 7), subtract(10, 7)) | subtract(n0,n3)|subtract(n2,n1)|add(n1,#0)|add(#2,#1)| | general |
bhanu spends 30 % of his income on petrol on scooter 20 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ? | "given 30 % ( income ) = 300 ⇒ ⇒ income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 20 % ( 700 ) = rs . 140 answer : b" | a ) 2287 , b ) 140 , c ) 128 , d ) 797 , e ) 123 | b | multiply(subtract(divide(300, divide(30, const_100)), 300), divide(20, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(n2,#1)|subtract(#2,n2)|multiply(#0,#3)| | gain |
the average of the first 5 multiples of 9 is : | solution : required average = ( total sum of multiple of 9 ) / 5 = ( 9 + 18 + 27 + 36 + 45 ) / 5 = 27 note that , average of 9 and 45 is also 27 . and average of 18 and 36 is also 27 . answer : option b | a ) 20 , b ) 27 , c ) 28 , d ) 30 , e ) none of these | b | divide(add(add(add(add(9, multiply(9, const_2)), multiply(9, const_3)), multiply(9, const_4)), multiply(9, 5)), 5) | multiply(n1,const_2)|multiply(n1,const_3)|multiply(n1,const_4)|multiply(n0,n1)|add(n1,#0)|add(#4,#1)|add(#5,#2)|add(#6,#3)|divide(#7,n0) | general |
13 different biology books and 8 different chemistry books lie on a shelf . in how many ways can a student pick 2 books of each type ? | "no . of ways of picking 2 biology books ( from 13 books ) = 13 c 2 = ( 13 * 12 ) / 2 = 78 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) / 2 = 28 total ways of picking 2 books of each type = 78 * 28 = 2184 ( option e )" | a ) 80 , b ) 160 , c ) 720 , d ) 1100 , e ) 2184 | e | multiply(divide(divide(factorial(13), factorial(subtract(13, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2)) | factorial(n0)|factorial(n1)|subtract(n0,n2)|subtract(n1,n2)|factorial(#2)|factorial(#3)|divide(#0,#4)|divide(#1,#5)|divide(#6,n2)|divide(#7,n2)|multiply(#8,#9)| | other |
real - estate salesman z is selling a house at a 20 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 20 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 20 percent discount .... | "let the retail price be = x selling price of z = 0.80 x selling price of x = 0.80 * 0.80 x = 0.64 x selling price of y = ( ( 0.80 x + 0.64 x ) / 2 ) * 0.80 = 0.72 x * 0.80 = 0.58 x 0.58 x = k * 0.64 x k = 0.58 / 0.64 = 58 / 64 = 29 / 32 answer : a" | a ) 29 / 32 , b ) 28 / 31 , c ) 30 / 31 , d ) 40 / 41 , e ) 34 / 15 | a | multiply(divide(divide(multiply(divide(add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(subtract(const_100, 20), const_100))), const_2), subtract(const_100, 20)), const_100), multiply(subtract(const_100, 20), divide(subtract(const_100, 20), const_100))), const_10) | subtract(const_100,n1)|subtract(const_100,n0)|subtract(const_100,n2)|divide(#0,const_100)|multiply(#3,#1)|add(#4,#1)|divide(#5,const_2)|multiply(#6,#2)|divide(#7,const_100)|divide(#8,#4)|multiply(#9,const_10)| | general |
how many integers between 100 and 160 , inclusive , can not be evenly divided by 3 nor 5 ? | "the total numbers between 100 and 160 , inclusive , is 61 . 3 * 34 = 102 and 3 * 53 = 159 so the number of multiples of 3 is 20 . 5 * 20 = 100 and 5 * 32 = 160 so the number of multiples of 5 is 13 . however , the multiples of 15 have been counted twice . 15 * 7 = 105 and 15 * 10 = 150 so the number of multiples of 15... | a ) 32 , b ) 27 , c ) 25 , d ) 35 , e ) 29 | a | subtract(160, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
calculate the share of y , if rs . 690 is divided among x , y and z in the ratio 5 : 7 : 11 ? | "5 + 7 + 11 = 23 690 / 23 = 30 so y ' s share = 5 * 30 = 150 answer : e" | a ) 190 , b ) 160 , c ) 120 , d ) 151 , e ) 150 | e | multiply(divide(690, add(add(5, 7), 11)), 5) | add(n1,n2)|add(n3,#0)|divide(n0,#1)|multiply(n1,#2)| | general |
you hold some gold in a vault as an investment . over the past year the price of gold increases by 52 % . in order to keep your gold in the vault , you must pay 6 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year . | ( 100 % + 52 % ) * ( 100 % - 6 % ) = 152 * 0.94 = 142.88 % an increase of 42.88 % your gold holdings have increased in value by 42.88 % . the answer is a | a ) 42.88 $ , b ) 43 % , c ) 45 % , d ) 45.5 % , e ) 46.3 % | a | subtract(52, divide(52, 6)) | divide(n0,n1)|subtract(n0,#0) | gain |
a tank contains 9,000 gallons of a solution that is 6 percent sodium chloride by volume . if 3,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "we start with 9,000 gallons of a solution that is 6 % sodium chloride by volume . this means that there are 0.06 x 9,000 = 54 gallons of sodium chloride . when 3,000 gallons of water evaporate we are left with 6,000 gallons of solution . from here we can determine what percent of the 6,000 gallon solution is sodium ch... | a ) 6 % , b ) 7 % , c ) 8 % , d ) 9 % , e ) 10 % | d | multiply(divide(multiply(multiply(const_100, const_100), divide(6, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | add(const_2,const_3)|divide(n1,const_100)|multiply(const_100,const_100)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#4,const_100)|multiply(#0,#6)|add(#7,#5)|subtract(#2,#8)|divide(#3,#9)|multiply(#10,const_100)| | gain |
a sum fetched a total simple interest of rs . 750 at the rate of 6 p . c . p . a . in 5 years . what is the sum ? | "sol . principal = rs . [ 100 * 750 / 6 * 5 ] = rs . [ 75000 / 30 ] = rs . 2500 . answer b" | a ) 2600 , b ) 2500 , c ) 2900 , d ) 2800 , e ) 2700 | b | divide(divide(multiply(750, const_100), 6), 5) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain |
robert ate 12 chocolates , nickel ate 3 chocolates . how many more chocolates did robert ate than nickel ? | 12 - 3 = 9 . answer is c | a ) a ) 4 , b ) b ) 7 , c ) c ) 9 , d ) d ) 5 , e ) e ) 2 | c | subtract(12, 3) | subtract(n0,n1)| | general |
in what time will a train 140 meters long cross an electric pole , if its speed is 210 km / hr | "explanation : first convert speed into m / sec speed = 210 * ( 5 / 18 ) = 58 m / sec time = distance / speed = 140 / 58 = 2.4 seconds option d" | a ) 5 seconds , b ) 4.5 seconds , c ) 3 seconds , d ) 2.4 seconds , e ) none of these | d | divide(140, multiply(210, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
there are two cars . one is 300 miles north of the other . simultaneously , the car to the north is driven westward at 20 miles per hour and the other car is driven eastward at 60 miles per hour . how many miles apart are the cars after 5 hours ? | here , drawing a quick sketch of the ' actions ' described will end in a diagonal line that you canbuilda right triangle around : the right triangle will have a base of 400 and a height of 300 . the hidden pattern here is a 3 / 4 / 5 right triangle ( the 300 lines up with the ' 3 ' and the 400 lines up with the ' 4 ' )... | a ) 300 , b ) 360 , c ) 400 , d ) 450 , e ) 500 | e | sqrt(add(power(add(multiply(60, 5), multiply(20, 5)), const_2), power(multiply(60, 5), const_2))) | multiply(n2,n3)|multiply(n1,n3)|add(#0,#1)|power(#0,const_2)|power(#2,const_2)|add(#4,#3)|sqrt(#5) | physics |
an investment compounds annually at an interest rate of 34.1 % what is the smallest investment period by which time the investment will more than triple in value ? | assume initial amount is x annual interest is 34.1 % so after 1 year the amount will become x * ( 100 + 34.1 ) / 100 = > x * 4 / 3 now we need to find n for x * ( 4 / 3 ) ^ n = 3 x ; or in other words n = 4 answer : b | a ) 3 , b ) 4 , c ) 6 , d ) 9 , e ) 12 | b | add(floor(divide(log(const_3), log(add(const_1, divide(34.1, const_100))))), const_1) | divide(n0,const_100)|log(const_3)|add(#0,const_1)|log(#2)|divide(#1,#3)|floor(#4)|add(#5,const_1) | gain |
the cross - section of a water channel is a trapezium in shape . if the channel is 12 meters wide at the top and 6 meters wide at the bottom and the area of cross - section is 630 square meters , what is the depth of the channel ( in meters ) ? | "1 / 2 * d * ( 12 + 6 ) = 630 d = 70 the answer is c ." | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | divide(630, divide(add(12, 6), const_2)) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)| | physics |
a bicycle wheel has a diameter of 0.51 m . how many complete revolutions does it make in 1 km ? | 1 revolution = 3.14 * diameter . number of revolutions in 1 km = 1000 m / ( 3.14 * 0.51 m ) = 624.5 . hence 624 complete revolutions . answer e | a ) 246 , b ) 448 , c ) 1408 , d ) 710 , e ) 624 | e | divide(multiply(const_1000, 1), multiply(add(const_3, divide(add(multiply(const_3, const_4), const_2), power(add(multiply(const_4, const_2), const_2), const_2))), 0.51)) | multiply(n1,const_1000)|multiply(const_3,const_4)|multiply(const_2,const_4)|add(#1,const_2)|add(#2,const_2)|power(#4,const_2)|divide(#3,#5)|add(#6,const_3)|multiply(n0,#7)|divide(#0,#8) | physics |
triangle atriangle b are similar triangles with areas 1536 units square and 1734 units square respectively . the ratio of there corresponding height would be | "let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 1536 / 1734 ( x ^ 2 / y ^ 2 ) = ( 16 * 16 * 6 ) / ( 17 * 17 * 7 ) ( x ^ 2 / y ^ 2 ) = 16 ^ 2 / 17 ^ 2 x / y = 16 / 17 ans = d" | a ) 9 : 10 , b ) 17 : 19 , c ) 23 : 27 , d ) 16 : 17 , e ) 15 : 23 | d | sqrt(divide(1536, 1734)) | divide(n0,n1)|sqrt(#0)| | geometry |
x can finish a work in 20 days . y can finish the same work in 16 days . y worked for 12 days and left the job . how many days does x alone need to finish the remaining work ? | "work done by x in 1 day = 1 / 20 work done by y in 1 day = 1 / 16 work done by y in 12 days = 12 / 16 = 3 / 4 remaining work = 1 – 3 / 4 = 1 / 4 number of days in which x can finish the remaining work = ( 1 / 4 ) / ( 1 / 20 ) = 5 d" | a ) 2 , b ) 3 , c ) 9 , d ) 5 , e ) 8 | d | divide(subtract(const_1, multiply(12, divide(const_1, 16))), divide(const_1, 20)) | divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|subtract(const_1,#2)|divide(#3,#1)| | physics |
if the average of 10 consecutive integers is 20.5 then the 10 th integer is : - | the average falls between the 5 th and 6 th integers , integer 5 = 20 , integer 6 = 21 . counting up to the tenth integer we get 25 . answer : e | a ) 15 , b ) 20 , c ) 23 , d ) 24 , e ) 25 | e | add(multiply(10, const_2), multiply(subtract(20.5, multiply(10, const_2)), 10)) | multiply(n0,const_2)|subtract(n1,#0)|multiply(n0,#1)|add(#0,#2) | general |
sandy gets 3 marks for each correct sum and loses 2 marks for each incorrect sum . sandy attempts 30 sums and obtains 55 marks . how many sums did sandy get correct ? | "let x be the correct sums and ( 30 - x ) be the incorrect sums . 3 x - 2 ( 30 - x ) = 55 5 x = 115 x = 23 the answer is c ." | a ) 19 , b ) 21 , c ) 23 , d ) 25 , e ) 27 | c | divide(add(multiply(2, 30), 55), add(2, 3)) | add(n0,n1)|multiply(n1,n2)|add(n3,#1)|divide(#2,#0)| | general |
a sum of money is to be divided among ann , bob and chloe . first , ann receives $ 4 plus one - half of what remains . next , bob receives $ 4 plus one - third of what remains . finally , chloe receives the remaining $ 32 . how much money m did bob receive ? | "notice that we need not consider ann ' s portion in the solution . we can just let k = the money remaining after ann has received her portion and go from there . our equation will use the fact that , once we remove bob ' s portion , we have $ 32 for chloe . so , we getk - bob ' s $ = 32 bob received 4 dollars plus one... | a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 52 | b | divide(multiply(32, const_2), const_3) | multiply(n2,const_2)|divide(#0,const_3)| | general |
last year elaine spent 10 % of her annual earnings on rent . this year she earned 15 % more than last year and she spent 30 % of her annual earnings on rent . the amount she spent on rent this year is what percent of the amount spent on rent last year ? | "for this it is easiest to use simple numbers . let ' s assume that elaine ' s annual earnings last year were $ 100 . she would ' ve spent $ 10 of this on rent . this year she earned 15 % more , or $ 115 . she would ' ve spent 30 % of this on rent , or $ 34.5 do $ 34.5 / $ 10 this will give you 345 % e is the correct a... | a ) 252.5 , b ) 364.5 , c ) 367.5 , d ) 375 , e ) 345 | e | multiply(divide(multiply(add(divide(15, const_100), const_1), 30), 10), const_100) | divide(n1,const_100)|add(#0,const_1)|multiply(n2,#1)|divide(#2,n0)|multiply(#3,const_100)| | gain |
sandy is younger than molly by 8 years . if their ages are in the respective ratio of 7 : 9 , how old is molly ? | "s = m - 8 s / m = 7 / 9 9 s = 7 m 9 ( m - 8 ) = 7 m m = 36 the answer is c ." | a ) 18 , b ) 27 , c ) 36 , d ) 45 , e ) 54 | c | multiply(divide(7, subtract(9, 7)), 8) | subtract(n2,n1)|divide(n1,#0)|multiply(n0,#1)| | other |
when x is multiplied by 3 , the result is 4 more than the result of subtracting x from 16 . what is the value of x ? | "the equation that can be formed is : 3 x - 4 = 16 - x or , 4 x = 20 or , x = 5 . answer : c" | a ) - 4 , b ) - 2 , c ) 5 , d ) 13 , e ) 22 | c | divide(add(16, 4), add(3, const_1)) | add(n1,n2)|add(const_1,n0)|divide(#0,#1)| | general |
in a certain school , 25 % of students are below 8 years of age . the number of students above 8 years of age is 2 / 3 of the number of students of 8 years of age which is 36 . what is the total number of students in the school ? | "explanation : let the number of students be x . then , number of students above 8 years of age = ( 100 - 25 ) % of x = 75 % of x . 75 % of x = 36 + 2 / 3 of 36 75 / 100 x = 60 x = 80 . answer : option b" | a ) 72 , b ) 80 , c ) 120 , d ) 150 , e ) 100 | b | divide(add(36, multiply(36, divide(2, 3))), subtract(const_1, divide(25, const_100))) | divide(n3,n4)|divide(n0,const_100)|multiply(n6,#0)|subtract(const_1,#1)|add(n6,#2)|divide(#4,#3)| | general |
how many seconds will a 600 meter long train moving with a speed of 63 km / hr take to cross a man walking with a speed of 3 km / hr in the direction of the train ? | "explanation : here distance d = 600 mts speed s = 63 - 3 = 60 kmph = 60 x 5 / 18 m / s time t = = 36 sec . answer : b" | a ) 48 , b ) 36 , c ) 26 , d ) 11 , e ) 18 | b | divide(600, multiply(subtract(63, 3), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
a train covers a distance of 12 km in 10 minutes . if it takes 9 seconds to pass a telegraph post , then the length of the train is | "explanation : speed = 12 / 10 x 60 km / hr = 72 x 5 / 18 m / sec = 20 m / sec . length of the train = ( speed x time ) = ( 20 x 9 ) m = 180 m answer : option e" | a ) 100 m , b ) 120 m , c ) 140 m , d ) 160 m , e ) 180 cm | e | divide(const_100.0, subtract(divide(const_100.0, 10), 9)) | divide(const_100.0,n1)|subtract(#0,n2)|divide(const_100.0,#1)| | physics |
the radius of a cylindrical vessel is 7 cm and height is 1 cm . find the whole surface of the cylinder ? | "r = 7 h = 1 2 π r ( h + r ) = 2 * 22 / 7 * 7 ( 8 ) = 352 answer : a" | a ) 352 , b ) 771 , c ) 440 , d ) 767 , e ) 1981 | a | surface_cylinder(7, 1) | surface_cylinder(n0,n1)| | geometry |
machine a can process 6000 envelopes in 3 hours . machines b and c working together but independently can process the same number of envelopes in 2.5 hours . if machines a and c working together but independently process 3000 envelopes in 1 hour , then how many hours would it take machine b to process 5600 envelopes . | "you can either take the amount of work done as the same as karishma has done or take the work done by each in the same time . i will do the latter 1 . work done in 1 hr by a is 2000 envelopes 2 . work done in 1 hr by a and c is 3000 envelopes 3 . so work done in 1 hr by c is 1000 envelopes 4 . work done in 1 hr by b a... | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 60 / 7 | c | divide(5600, subtract(divide(6000, 2.5), subtract(divide(3000, 1), divide(6000, 3)))) | divide(n0,n2)|divide(n3,n4)|divide(n0,n1)|subtract(#1,#2)|subtract(#0,#3)|divide(n5,#4)| | physics |
if the probability of rain on any given day in chicago during the summer is 25 % , independent of what happens on any other day , what is the probability of having exactly 3 rainy days from july 4 through july 7 inclusive ? | "one possible case is : rainy - rainy - rainy - not rainy . the probability of this case is 1 / 4 * 1 / 4 * 1 / 4 * 3 / 4 = 3 / 256 the number of possible cases is 4 c 3 = 4 . p ( exactly 3 rainy days ) = 4 * 3 / 256 = 3 / 64 the answer is d ." | a ) 1 / 16 , b ) 3 / 32 , c ) 5 / 32 , d ) 3 / 64 , e ) 7 / 64 | d | multiply(4, power(divide(const_1, const_2), 4)) | divide(const_1,const_2)|power(#0,n2)|multiply(n2,#1)| | general |
what is the difference between the place value and face value of 4 in the numeral 46 ? | difference between the place value and face value of 4 = 40 - 4 = 36 answer is a | a ) 36 , b ) 45 , c ) 52 , d ) 44 , e ) 39 | a | subtract(multiply(4, const_10), 4) | multiply(n0,const_10)|subtract(#0,n0) | general |
in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the b... | "ratio of 2 bedroom apartment : 1 bedroom apartment = 1000 : 9000 - - - - - > 1 : 9 let total number of apartments be x no . of 2 bedroom apartment = ( 1 / 10 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 / 10 ) * 100 - - - > 10 % answer : e" | a ) 50 % , b ) 40 % , c ) 20 % , d ) 30 % , e ) 10 % | e | divide(multiply(1,000, const_100), add(add(multiply(const_2, const_1000), const_100), 1,000)) | multiply(n0,const_100)|multiply(const_1000,const_2)|add(#1,const_100)|add(n0,#2)|divide(#0,#3)| | general |
a person travels from p to q a speed of 30 km / hr and returns by increasing his speed by 30 % . what is his average speed for both the trips ? | "speed on return trip = 130 % of 30 = 39 km / hr . average speed of trip = 30 + 39 / 2 = 69 / 2 = 34.5 km / hr answer : e" | a ) 33 , b ) 77 , c ) 48 , d ) 99 , e ) 34.5 | e | divide(add(multiply(30, add(const_1, divide(30, const_100))), 30), const_2) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|add(n0,#2)|divide(#3,const_2)| | general |
the sum of all the integers f such that - 26 < f < 24 is | "easy one - - 25 , - 24 , - 23 , - 22 , . . . . . . - 1,0 , 1 , 2 . . . . , 22 , 23 cancel everyhitng and we ' re left with - - 25 and - 24 f = - 49 . d is the answer ." | a ) 0 , b ) - 2 , c ) - 25 , d ) - 49 , e ) - 51 | d | add(add(negate(26), const_1), add(add(negate(26), const_1), const_1)) | negate(n0)|add(#0,const_1)|add(#1,const_1)|add(#1,#2)| | general |
the average expenditure of a labourer for 6 months was 70 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | "income of 6 months = ( 6 × 70 ) – debt = 420 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 690 average monthly income = 690 ÷ 10 = 69 answer a" | a ) 69 , b ) 72 , c ) 75 , d ) 78 , e ) 80 | a | divide(add(add(multiply(70, 6), multiply(60, 4)), 30), add(6, 4)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(n4,#3)|divide(#4,#0)| | general |
a student chose a number , multiplied it by 8 , then subtracted 138 from the result and got 102 . what was the number he chose ? | "solution : let xx be the number he chose , then 8 ⋅ x − 138 = 102 8 x = 240 x = 30 answer a" | a ) 30 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | a | divide(add(102, 138), 8) | add(n1,n2)|divide(#0,n0)| | general |
a man has rs . 192 in the denominations of one - rupee notes , 5 - rupee notes and 10 - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? | let number of notes of each denomination be x . then , x + 5 x + 10 x = 192 16 x = 192 x = 12 . hence , total number of notes = 3 x = 36 . answer : b | a ) 37 , b ) 36 , c ) 22 , d ) 90 , e ) 28 | b | add(add(divide(192, add(add(const_1, 5), 10)), divide(192, add(add(const_1, 5), 10))), divide(192, add(add(const_1, 5), 10))) | add(n1,const_1)|add(n2,#0)|divide(n0,#1)|add(#2,#2)|add(#3,#2) | general |
how many gallons of milk that is 10 percent butter - fat must be added to 8 gallons of milk that is 40 percent butterfat to obtain milk that is 20 percent butterfat ? | "equate the fat : 0.1 x + 0.4 * 8 = 0.2 ( x + 8 ) - - > x = 16 . answer : d ." | a ) 6 , b ) 12 , c ) 14 , d ) 16 , e ) 28 | d | divide(multiply(subtract(40, 20), 8), 10) | subtract(n2,n3)|multiply(n1,#0)|divide(#1,n0)| | general |
the price of lunch for 15 people was $ 211.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ? | "take the initial price before the gratuity is 100 the gratuity is calculated on the final price , so as we assumed the final bill before adding gratuity is 100 so gratuity is 15 % of 100 is 15 so the total price of meals is 115 so the given amount i . e 211 is for 115 then we have to calculate for 100 for 115 211 for ... | a ) $ 11.73 , b ) $ 12.78 , c ) $ 13.80 , d ) $ 14.00 , e ) $ 15.87 | b | multiply(multiply(divide(211.00, add(const_100, 15)), const_100), divide(const_1, 15)) | add(n0,const_100)|divide(const_1,n0)|divide(n1,#0)|multiply(#2,const_100)|multiply(#1,#3)| | general |
how many times digit 20 is used while writing numbers from 100 to 500 ? | "in 100 to 200 there are 5 20 ' s in 200 to 300 there are 5 20 ' s in 300 to 400 there are 5 20 ' s in 400 to 500 there are 5 20 ' s so , total 20 is 20 ' s correct option : a" | a ) 20 , b ) 30 , c ) 15 , d ) 25 , e ) 20 | a | add(add(divide(subtract(500, 100), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(20, const_2)) | add(const_1,const_10)|multiply(n0,const_2)|subtract(n2,n1)|divide(#2,const_10)|multiply(#0,#0)|add(#3,#4)|add(#5,#1)| | general |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.