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 and b can do a piece of work in 40 days and 40 days respectively . they work together for 10 days and b leaves . in how many days the whole work is completed ? | "explanation : ( a + b ) ’ s 10 days work = 10 [ 1 / 40 + 1 / 40 ] = 10 [ 1 + 1 / 40 ] = 1 / 2 a complete remaining work in 1 / 2 * 40 = 20 total work = 10 + 20 = 30 days answer : option a" | a ) 30 days , b ) 35 days , c ) 40 days , d ) 45 days , e ) 50 days | a | divide(subtract(10, add(divide(10, 40), divide(10, 40))), divide(10, 40)) | divide(n2,n1)|divide(n2,n0)|add(#0,#1)|subtract(n2,#2)|divide(#3,#0)| | physics |
in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 20 overs to reach the target of 282 runs ? | "explanation : runs scored in the first 10 overs = 10 × 3.2 = 32 total runs = 282 remaining runs to be scored = 282 - 32 = 250 remaining overs = 20 run rate needed = 250 / 20 = 12.5 answer : option e" | a ) 6.25 , b ) 5.5 , c ) 7.4 , d ) 5 , e ) 12.5 | e | divide(subtract(282, multiply(10, 3.2)), 20) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
a car dealership has 40 cars on the lot , 30 % of which are silver . if the dealership receives a new shipment of 80 cars , 40 % of which are not silver , what percentage of total number of cars are silver ? | "number of silver cars before shipment = 30 % of 40 = 12 number of silver cars after shipment = 60 % of 80 = 48 total number of silver cars out of 120 = 48 + 12 = 60 , which is 50 % e is my answer ." | a ) 35 % , b ) 37.5 % , c ) 45 % , d ) 47.5 % , e ) 50 % | e | multiply(divide(add(multiply(40, divide(30, const_100)), multiply(80, divide(40, const_100))), add(40, 80)), const_100) | add(n0,n2)|divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
a person walks at a speed of 4 km / hr and runs at a speed of 8 km / hr . how many hours will the person require to cover a distance of 12 km , if the person completes half of the distance by walking and the other half by running ? | "time = 6 / 4 + 6 / 8 = 18 / 8 = 2.25 hours the answer is c ." | a ) 1.25 , b ) 1.75 , c ) 2.25 , d ) 2.75 , e ) 3.25 | c | add(divide(divide(12, const_2), 4), divide(divide(12, const_2), 8)) | divide(n2,const_2)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)| | physics |
the total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year , less the age of the tree in years ( rounded down to the nearest integer ) . during its 5 th year , the plum tree grew 50 plums . if this trend continues , how many plums will ... | "1 st year : 0 - 1 ( age ) , we take age = 0 ( as the question says that we have to ( rounded down to the nearest integer ) ) 2 ndyear : 1 - 2 ( age ) , we take age = 1 3 rd year : 2 - 3 ( age ) , we take age = 2 4 th year : 3 - 4 ( age ) , we take age = 3 5 th year : 4 - 5 ( age ) , we take age = 4 6 th year : 5 - 6 (... | a ) 40 , b ) 42 , c ) 45 , d ) 48 , e ) 54 | c | add(divide(add(subtract(subtract(5, const_1), const_1), add(50, subtract(5, const_1))), const_3), const_3) | subtract(n0,const_1)|add(n1,#0)|subtract(#0,const_1)|add(#1,#2)|divide(#3,const_3)|add(#4,const_3)| | general |
in what ratio must rice of rs . 32 per kg be mixed with rice of rs . 42 per kg so that cost of mixture is rs . 20 per kg ? | "( 20 - 42 ) / ( 32 - 20 ) = 22 / 12 = 11 : 6 answer : c" | a ) 13 : 6 , b ) 13 : 5 , c ) 11 : 6 , d ) 11 : 5 , e ) 12 : 5 | c | divide(divide(subtract(20, 42), subtract(32, 42)), subtract(const_1, divide(subtract(20, 42), subtract(32, 42)))) | subtract(n2,n1)|subtract(n0,n1)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)| | other |
in a lottery there are 10 prizes and 25 blanks . a lottery is drawn at random . what is the probability of getting a prize ? | "total draws = prizes + blanks = 10 + 25 = 35 probability of getting a prize = 10 / 35 = 2 / 7 correct option is b" | a ) 1 / 5 , b ) 2 / 7 , c ) 2 / 5 , d ) 1 / 7 , e ) 3 / 5 | b | divide(25, add(10, 25)) | add(n0,n1)|divide(n1,#0)| | probability |
if the number is divided by 3 , it reduced by 48 . the number is | "explanation : let the number be x . then , x - ( x / 3 ) = 48 = > 2 x / 3 = 48 = > x = 72 answer : option d" | a ) a ) 51 , b ) b ) 50 , c ) c ) 45 , d ) d ) 72 , e ) e ) 36 | d | divide(multiply(48, 3), subtract(3, const_1)) | multiply(n0,n1)|subtract(n0,const_1)|divide(#0,#1)| | general |
an equilateral triangle t 2 is formed by joining the mid points of the sides of another equilateral triangle t 1 . a third equilateral triangle t 3 is formed by joining the mid - points of t 2 and this process is continued indefinitely . if each side of t 1 is 80 cm , find the sum of the perimeters of all the triangles... | "we have 80 for first triangle , when we join mid - points of first triangle we get the second equilateral triangle then the length of second one is 40 and continues . so we have 80 , 40,20 , . . . we have ratio = 1 / 2 , and it is gp type . sum of infinite triangle is a / 1 - r = 80 / 1 - ( 1 / 2 ) = 160 equilateral t... | a ) 180 cm , b ) 220 cm , c ) 240 cm , d ) 270 cm , e ) 480 cm | e | add(triangle_perimeter(80, 80, 80), triangle_perimeter(80, 80, 80)) | triangle_perimeter(n5,n5,n5)|add(#0,#0)| | geometry |
if 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 26 , then n = | "2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 26 = > 4 x 2 ^ 2 n = 4 ^ 26 = 2 ^ 52 = > 2 ^ 2 x 2 ^ 2 n = 2 ^ 52 = > 2 ^ ( 2 n + 2 ) = 2 ^ 52 = > 2 n + 2 = 52 = > n = 25 so . answer will be e" | a ) 3 , b ) 6 , c ) 12 , d ) 23 , e ) 25 | e | divide(subtract(multiply(26, 2), 2), 2) | multiply(n0,n9)|subtract(#0,n0)|divide(#1,n0)| | general |
if a - b = 5 and a 2 + b 2 = 34 , find the value of ab . | "explanation : 2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 34 - 25 = 9 ab = 4.5 . answer : a" | a ) a ) 4.5 , b ) b ) 12 , c ) c ) 15 , d ) d ) 18 , e ) e ) 20 | a | divide(subtract(34, power(5, 2)), 2) | power(n0,n1)|subtract(n3,#0)|divide(#1,n1)| | general |
john makes $ 60 a week from his job . he earns a raise and now makes $ 110 a week . what is the % increase ? | "increase = ( 50 / 60 ) * 100 = ( 5 / 6 ) * 100 = 83.33 % . b" | a ) 16 % , b ) 83.33 % , c ) 17 % , d ) 17.61 % , e ) 17.56 % | b | multiply(divide(subtract(110, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a certain number of workers can do a work in 35 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ? | "number of workers = 10 * ( 35 - 10 ) / 10 = 25 answer is a" | a ) 25 , b ) 30 , c ) 28 , d ) 24 , e ) 32 | a | divide(multiply(subtract(35, 10), 10), subtract(35, subtract(35, 10))) | subtract(n0,n1)|multiply(n1,#0)|subtract(n0,#0)|divide(#1,#2)| | physics |
in the formula a = ( 2 r ) ^ 3 , if r is halved , then a is multiplied by ? | say r = 2 = > a 1 = 64 when r = 1 ; v 2 = 8 v 2 = 1 / 8 * v 1 . answer : b | a ) 64 , b ) 1 / 8 , c ) 1 , d ) 8 , e ) 1 / 64 | b | divide(const_1, power(2, 3)) | power(n0,n1)|divide(const_1,#0) | general |
the value of a machine depreciates at 20 % per annum . if its present value is rs . 1 , 50,000 , at what price should it be sold after two years such that a profit of rs . 22,000 is made ? | the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = rs . 96,000 sp such that a profit of rs . 22,000 is made = 96,000 + 24,000 = rs . 1 , 18,000 answer : c | a ) 328897 , b ) 120000 , c ) 118000 , d ) 277768 , e ) 188871 | c | add(multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100))) | add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10) | gain |
a cycle is bought for rs . 900 and sold for rs . 1180 , find the gain percent ? | "900 - - - - 280 100 - - - - ? = > 31 % answer : b" | a ) 11 , b ) 31 , c ) 99 , d ) 77 , e ) 18 | b | multiply(divide(subtract(1180, 900), 900), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
all the students of class are told to sit in circle shape . here the boy at the 8 th position is exactly opposite to 33 th boy . total number of boys in the class ? | "as half the circle shape consist of 33 - 8 = 25 boys , so total number of boys in full circle = 2 * 25 = 50 answer : b" | a ) 40 , b ) 50 , c ) 28 , d ) 38 , e ) 48 | b | multiply(subtract(33, 8), const_2) | subtract(n1,n0)|multiply(#0,const_2)| | general |
45 girls and 60 boys are present at a party . there are 17 girl - boy are in relationship among them . if a girl and a boy is selected at random , what is the probability that they will be not a couple ? | in how many ways we can select a girl and a boy from 45 lady and 60 gentlemen ? in 45 * 60 = 2,700 ways . we have a total of 17 couples so , the probability of selecting a couple is 17 / 2,700 = 17 / 2700 . probability of there will not be a couple = 1 - ( 17 / 2700 ) = 2683 / 2700 ans - b | a ) 17 / 2900 , b ) 2683 / 2700 , c ) 2496 / 2700 , d ) 2369 / 2700 , e ) 198 / 270 | b | divide(subtract(multiply(choose(45, const_1), choose(60, const_1)), choose(17, const_1)), multiply(choose(45, const_1), choose(60, const_1))) | choose(n0,const_1)|choose(n1,const_1)|choose(n2,const_1)|multiply(#0,#1)|subtract(#3,#2)|divide(#4,#3) | probability |
if p ( a ) = 2 / 15 , p ( b ) = 4 / 15 , and p ( a â ˆ ª b ) = 11 / 15 find p ( a | b ) | "p ( a | b ) = p ( a â ˆ ª b ) / p ( b ) p ( a | b ) = ( 11 / 15 ) / ( 4 / 15 ) = 11 / 4 . a" | a ) 11 / 4 , b ) 2 / 3 , c ) 3 / 2 , d ) 4 / 5 , e ) 4 / 7 | a | divide(divide(11, 15), divide(2, 15)) | divide(n4,n1)|divide(n0,n1)|divide(#0,#1)| | general |
how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 134 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 134 = 244 m . required time = 244 / 20 = 12.2 sec . answer : b" | a ) 82.1 sec . , b ) 12.2 sec , c ) 19.1 sec . , d ) 17.1 sec . , e ) 42.1 sec . | b | divide(add(110, 134), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
what percent is 70 of 280 ? | "70 / 280 = 1 / 4 1 / 4 × 100 = 25 % b" | a ) 12 % , b ) 25 % , c ) 26 % , d ) 27 % , e ) 28 % | b | multiply(divide(70, 280), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
ifaequals the sum of the even integers from 2 to 40 , inclusive , andbequals the sum of the odd integers from 1 to 39 , inclusive , what is the value of a - b ? | this is a solution from beatthegmat : even numbers : ( 40 - 2 ) / 2 + 1 = 20 even integers . ( 40 + 2 ) / 2 = 21 is the average of the even set . sum = avg * ( # of elements ) = 21 * 20 = 420 = a odd numbers : ( 39 - 1 ) / 2 + 1 = 20 odd integers . ( 39 + 1 ) / 2 = 20 is the average of the odd set . sum = avg * ( # of ... | a ) 1 , b ) 10 , c ) 19 , d ) 20 , e ) 21 | d | subtract(multiply(divide(40, const_2), add(divide(40, const_2), 1)), multiply(divide(add(39, 1), const_2), add(divide(subtract(39, 1), 2), 1))) | add(n2,n3)|divide(n1,const_2)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,const_2)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8) | general |
a train running at the speed of 90 km / hr crosses a pole in 12 seconds . find the length of the train ? | "speed = 90 * ( 5 / 18 ) m / sec = 75 / 3 m / sec length of train ( distance ) = speed * time ( 75 / 3 ) * 12 = 300 meter answer : e" | a ) 150 , b ) 872 , c ) 287 , d ) 288 , e ) 300 | e | multiply(divide(multiply(90, const_1000), const_3600), 12) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
two cars , car 1 and car 2 move towards each other from e and y respectively with respective speeds of 20 m / s and 15 m / s . after meeting each other car 1 reaches y in 10 seconds . in how many seconds does car 2 reach e starting from y ? | e - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - | - - - - - - - - - - - - - - - - - - - - - - - - - - - - y car a ( 20 mps ) - - - - - - - - - - - - - - - - - - - - - - - - - > p < - - - - - - - - - - - - - - - car b ( 15 mps ) let 2 cars meet each other at point p in t seconds . car 1 cov... | a ) 15.5 sec , b ) 8.4 sec , c ) 33.6 sec , d ) 31.11 sec , e ) 16.8 sec | d | divide(multiply(add(20, 15), divide(multiply(20, 10), 15)), 15) | add(n2,n3)|multiply(n2,n5)|divide(#1,n3)|multiply(#0,#2)|divide(#3,n3) | physics |
the radius of a semicircular plot is 25 . what is the approximate perimeter of the plot ? | perimeter of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 25 + 2 * 25 = 128.50 approximately 129 answer c | ['a ) 114', 'b ) 150', 'c ) 129', 'd ) 25', 'e ) 57'] | c | add(25, divide(circumface(25), const_2)) | circumface(n0)|divide(#0,const_2)|add(n0,#1) | geometry |
how many two - digit whole numbers yield a remainder of 3 when divided by 10 and also yield a remainder of 3 when divided by 4 ? options : | numbers giving remainder 3 when divided by 10 = 13,23 , 33,43 , 53,63 , 73,83 , 93 numbers giving remainder 3 when divided by 4 = alternate numbers ( coloured in red ) = 4 correct option : d | a ) one , b ) two , c ) three , d ) four , e ) five | d | divide(lcm(10, 4), add(const_3, const_2)) | add(const_2,const_3)|lcm(n1,n3)|divide(#1,#0) | general |
on a race track a maximum of 5 horses can race together at a time . there are a total of 25 horses . there is no way of timing the races . what is the minimum number t of races we need to conduct to get the top 3 fastest horses ? | "t = 7 is the correct answer . good solution buneul . b" | a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | b | add(divide(25, 5), const_2) | divide(n1,n0)|add(#0,const_2)| | general |
from the beginning to the end of 2007 , the price of a stock rose 20 percent . in 2008 , it dropped 25 percent . in 2009 , it rose 30 percent . what percent of the stock â € ™ s 2007 starting price was the price of the stock at the end of 2009 ? | "assume a value at the beginning of 2007 . as this is a % question , assume p = 100 . at the end of 2007 it becmae = 1.2 * 100 = 120 at the end of 2008 it decreased by 25 % = 120 * . 75 = 90 at the end of 2009 it increased by 30 % = 90 * 1.3 = 117 thus ratio = 117 / 100 = 1.17 ( in % terms = 117 % ) . thus b is the cor... | a ) 80 , b ) 117 , c ) 95 , d ) 100 , e ) 108 | b | multiply(multiply(multiply(const_100, divide(add(const_100, 20), const_100)), divide(subtract(const_100, 25), const_100)), divide(add(const_100, 30), const_100)) | add(n5,const_100)|add(n1,const_100)|subtract(const_100,n3)|divide(#0,const_100)|divide(#2,const_100)|divide(#1,const_100)|multiply(#5,const_100)|multiply(#4,#6)|multiply(#3,#7)| | gain |
a rectangular tiled patio is composed of 40 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 40 tiles , and it will still be rectangular . how many rows are in the tile patio before th... | "r * c = 70 and ( 6 + 4 ) ( c - 2 ) = 70 - - > r = 6 and c = 6 . answer : b ." | a ) 5 , b ) 6 , c ) 10 , d ) 13 , e ) 28 | b | divide(40, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(40, 2))))), 2)) | multiply(n0,n1)|negate(n2)|power(n2,n1)|multiply(n2,#0)|add(#3,#2)|sqrt(#4)|add(#1,#5)|divide(#6,n1)|divide(n0,#7)| | geometry |
set s consists of integers { 6 , 7 , 10 , 12 , 15 } . if integer n is included in the set , the average ( arithmetic mean ) of set s will increase by 40 % . what is the value of integer n ? | "the average of the numbers in set s is 10 . if we increase the mean by 40 % , the new mean is 14 . thus , on average , 5 numbers increase by 4 . therefore n = 14 + 20 = 34 the answer is c ." | a ) 26 , b ) 30 , c ) 34 , d ) 38 , e ) 42 | c | add(add(10, 12), 12) | add(n2,n3)|add(n3,#0)| | general |
a girl was asked to multiply a certain number by 43 . she multiplied it by 34 and got his answer less than the correct one by 1224 . find the number to be multiplied . | "let the required number be x . then , 43 x – 34 x = 1224 or 9 x = 1224 or x = 136 . required number = 136 . answer : d" | a ) 130 , b ) 132 , c ) 134 , d ) 136 , e ) 138 | d | divide(1224, subtract(43, 34)) | subtract(n0,n1)|divide(n2,#0)| | general |
what will be the difference between simple and compound interest @ 20 % per annum on a sum of $ 500 after 2 years ? | "s . i . = 500 * 20 * 2 / 100 = $ 200 c . i . = 500 * ( 1 + 20 / 100 ) ^ 2 - 500 = $ 220 difference = 220 - 200 = $ 20 answer is b" | a ) $ 10 , b ) $ 20 , c ) $ 25 , d ) $ 30 , e ) $ 40 | b | subtract(subtract(multiply(500, power(add(divide(20, const_100), const_1), 2)), 500), multiply(multiply(500, divide(20, const_100)), 2)) | divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|multiply(n2,#2)|power(#1,n2)|multiply(n1,#4)|subtract(#5,n1)|subtract(#6,#3)| | gain |
martins made a part payment of $ 650 toward buying a brand new washing machine , which represents 15 % of the total cost of the washing machine , how much remains to be paid ? | explanation : let ' s start with what the total price of the washing machine would be . if 15 % is equal to $ 650 then 100 % equals $ x . we just have to multiply $ 650 by 6.67 to get total amount = $ 4333.3 . out of this amount we then need to deduct the amount already paid which was $ 650 so we have $ 4333.3 - $ 650 ... | a ) $ 1116.3 , b ) $ 3583.3 , c ) $ 3283.3 , d ) $ 3683.3 , e ) $ 2683.3 | d | subtract(divide(650, divide(15, const_100)), 650) | divide(n1,const_100)|divide(n0,#0)|subtract(#1,n0) | general |
a small , rectangular park has a perimeter of 560 feet and a diagonal measurement of 300 feet . what is its area , in square feet ? | "you can avoid a lot of work in this problem by recognizing that , with the info provided , the diagonal forms a triangle inside the rectangle with sides that have a 3 : 4 : 5 ratio . diagonal = 200 2 x + 2 y = 560 , or x + y = 280 a ^ 2 + b ^ 2 = c ^ 2 for each the sides of the triangle using the ratio 3 : 4 : 5 for s... | a ) 43,200 , b ) 19,600 , c ) 20,000 , d ) 20,400 , e ) 20,800 | a | multiply(multiply(divide(300, add(const_2, const_3)), const_3), multiply(divide(300, add(const_2, const_3)), const_4)) | add(const_2,const_3)|divide(n1,#0)|multiply(#1,const_3)|multiply(#1,const_4)|multiply(#2,#3)| | geometry |
in the storage room of a certain bakery , the ratio of sugar to flour is 5 to 6 , and the ratio of flour to baking soda is 10 to 1 . if there were 60 more pounds of baking soda in the room , the ratio of flour to baking soda would be 8 to 1 . how many pounds of sugar are stored in the room ? | "sugar : flour = 5 : 6 = 25 : 30 ; flour : soda = 10 : 1 = 30 : 3 ; thus we have that sugar : flour : soda = 25 x : 30 x : 3 x . also given that 30 x / ( 3 x + 60 ) = 8 / 1 - - > x = 80 - - > sugar = 25 x = 2,000 . answer : e ." | a ) 600 , b ) 1200 , c ) 1500 , d ) 1600 , e ) 2000 | e | multiply(divide(5, 6), multiply(10, divide(multiply(8, 60), subtract(10, 8)))) | divide(n0,n1)|multiply(n4,n5)|subtract(n2,n5)|divide(#1,#2)|multiply(n2,#3)|multiply(#0,#4)| | general |
the h . c . f of two numbers is 19 and their l . c . m is 4263 . if one of the numbers is 399 , then the other is ? | "other number = ( 19 * 4263 ) / 399 = 203 . answer : b" | a ) 202 , b ) 203 , c ) 204 , d ) 205 , e ) 206 | b | multiply(19, 399) | multiply(n0,n2)| | physics |
in a restaurant , the profit is 110 % of the cost . if the cost increases by 12 % but the selling price remains constant , approximately what percentage of the selling price is the profit ? | explanation : let c . p . = rs . 100 . then , profit = rs . 110 , s . p . = rs . 210 . new c . p . = 112 % of rs . 100 = rs . 112 new s . p . = rs . 210 . profit = rs . ( 210 - 112 ) = rs . 98 . required percentage = ( 98 / 210 * 100 ) % = 47 % appox answer : b | a ) 30 % , b ) 47 % , c ) 90 % , d ) 100 % , e ) none of these | b | multiply(const_100, divide(subtract(add(multiply(const_100, divide(110, const_100)), const_100), multiply(const_100, add(const_1, divide(12, const_100)))), add(multiply(const_100, divide(110, const_100)), const_100))) | divide(n0,const_100)|divide(n1,const_100)|add(#1,const_1)|multiply(#0,const_100)|add(#3,const_100)|multiply(#2,const_100)|subtract(#4,#5)|divide(#6,#4)|multiply(#7,const_100) | gain |
at a loading dock , each worker on the night crew loaded 3 / 4 as many boxes as each worker on the day crew . if the night crew has 3 / 4 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "let x be the number of workers on the day crew . let y be the number of boxes loaded by each member of the day crew . then the number of boxes loaded by the day crew is xy . the number of boxes loaded by the night crew is ( 3 x / 4 ) ( 3 y / 4 ) = 9 xy / 16 the total number of boxes is xy + 9 xy / 16 = 25 xy / 16 the ... | a ) 3 / 5 , b ) 4 / 5 , c ) 11 / 15 , d ) 16 / 25 , e ) 19 / 25 | d | divide(multiply(4, 4), add(multiply(4, 4), multiply(3, 3))) | multiply(n1,n3)|multiply(n0,n2)|add(#0,#1)|divide(#0,#2)| | physics |
walking 6 / 5 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ? | "speed ratio = 1 : 6 / 5 = 5 : 6 time ratio = 6 : 5 1 - - - - - - - - 6 4 - - - - - - - - - ? 24 m . answer : d" | a ) 22 , b ) 99 , c ) 27 , d ) 24 , e ) 20 | d | multiply(4, 6) | multiply(n0,n2)| | gain |
how many seconds will a 500 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 66 km / hr ? | "speed of train relative to man = 66 - 3 = 63 km / hr . = 63 * 5 / 18 = 35 / 2 m / sec . time taken to pass the man = 500 * 2 / 35 = 28.57 sec . approx . 29 sec answer : c" | a ) 11 , b ) 30 , c ) 29 , d ) 88 , e ) 61 | c | divide(500, multiply(subtract(66, 3), const_0_2778)) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
what number times ( 1 ⁄ 4 ) ^ 2 will give the value of 4 ^ 3 ? | "x * ( 1 / 4 ) ^ 2 = 4 ^ 3 x = 4 ^ 2 * 4 ^ 3 = 4 ^ 5 = 1024 the answer is e ." | a ) 4 , b ) 16 , c ) 64 , d ) 256 , e ) 1024 | e | multiply(power(const_2.0, 4), power(4, 3)) | power(n2,n1)|power(n1,n4)|multiply(#0,#1)| | general |
a vendor bought toffees at 6 for a rupee . how many for a rupee must he sell to gain 25 % ? | "c . p . of 6 toffees = re . 1 s . p . of 6 toffees = 150 % of re . 1 = rs . 3 / 2 for rs . 3 / 2 , toffees sold = 6 . for re . 1 , toffees sold = 6 x 3 / 2 = 9 . answer b" | a ) 2 , b ) 9 , c ) 6 , d ) 3 , e ) 5 | b | multiply(6, add(const_1, divide(25, const_100))) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)| | gain |
one night 19 percent of the female officers on a police force were on duty . if 152 police officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let x be the number of female police officers on the police force . the number of female police officers on duty was 76 . 0.19 x = 76 x = 400 the answer is c ." | a ) 300 , b ) 350 , c ) 400 , d ) 450 , e ) 500 | c | divide(divide(152, const_2), divide(19, const_100)) | divide(n1,const_2)|divide(n0,const_100)|divide(#0,#1)| | gain |
a car ferry can hold up to 70 tons of cargo . what is the greatest number of vehicles that the ferry can carry if half the vehicles are cars with an average ( arithmetic mean ) weight of 0.75 tons and half of the vehicles are trucks with an average ( arithmetic mean ) weight of 6 tons ? | "the weight of one car and one truck is 6.75 tons . 70 / 6.75 = 10 plus a remainder the ferry could carry 10 cars and 10 trucks for a total of 20 vehicles . the answer is a ." | a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | a | add(divide(70, add(0.75, 6)), divide(70, add(0.75, 6))) | add(n1,n2)|divide(n0,#0)|add(#1,#1)| | general |
a train passes a platform in 35 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km / hr , the length of the platform is | "explanation : speed of the train = 54 km / hr = ( 54 × 10 ) / 35 m / s = 15 m / s length of the train = speed × time taken to cross the man = 15 × 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) / 15 = > ( 300 + l ) / 15 = 35 = > 300 + l = 15 × 35 = 525 = > l = 525 - 300 = ... | a ) 280 meter , b ) 240 meter , c ) 225 meter , d ) 200 meter , e ) none of these | c | multiply(multiply(const_0_2778, 54), subtract(35, 20)) | multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)| | physics |
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 50 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 . 50 % of 100 = 50 pounds of decaffeinated coffee 3 . wt have 130 pounds of decaffeinated out of 500 pounds , that means 130 / 500 * 100 % = 26 % . the correct answer is a ." | a ) 26 % , b ) 30 % , c ) 32 % , d ) 34 % , e ) 40 % | a | multiply(divide(add(multiply(divide(20, 100), 400), multiply(100, divide(50, 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 |
praveen starts business with rs . 3220 and after 5 months , hari joins with praveen as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is hari ’ s contribution in the capital ? | "let hari ’ s capital be rs . x . then , 3220 * 12 / 7 x = 2 / 3 = > 14 x = 115920 = > x = 8280 . answer : c" | a ) s . 7500 , b ) s . 8000 , c ) s . 8280 , d ) s . 9000 , e ) s . 6000 | c | divide(divide(3220, subtract(const_1, divide(5, const_12))), divide(2, 3)) | divide(n1,const_12)|divide(n2,n3)|subtract(const_1,#0)|divide(n0,#2)|divide(#3,#1)| | other |
how many integers from 22 to 160 , inclusive , are divisible by 3 but not divisible by 7 ? | "we should find # of integers divisible by 3 but not by 3 * 7 = 21 . # of multiples of 21 in the range from 22 to 160 , inclusive is ( 147 - 42 ) / 21 + 1 = 6 ; 46 - 6 = 40 . answer : c ." | a ) 30 , b ) 35 , c ) 40 , d ) 50 , e ) 60 | c | divide(160, const_10) | divide(n1,const_10)| | general |
8 identical machines , working alone and at their constant rates , take 6 hours to complete a job lot . how long would it take for 4 such machines to perform the same job ? | let each machine do 1 unit of work for 1 hour 8 machines - - > 8 units of work in 1 hour for 6 hours = 8 * 6 = 48 units of total work is done . now this 48 units of total work must be done by 4 machines 4 units of work ( 4 machines ) - - - > 1 hour for 48 units of work 4 * 12 - - - > 1 * 12 hours c 12 hours | a ) 2.25 hours , b ) 8.75 hours , c ) 12 hours , d ) 14.25 hours , e ) 16 hours | c | divide(multiply(6, 8), 4) | multiply(n0,n1)|divide(#0,n2) | physics |
a candidate got 35 % of the votes and lost to the rival by 2100 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % = 2100 10 % = 700 100 % of the votes = 7000 answer : a" | a ) 7000 , b ) 7200 , c ) 7400 , d ) 7600 , e ) 7800 | a | divide(2100, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain |
if you write down all the numbers from 1 to 25 , then how many times do you write 3 ? | "explanation : explanation : clearly , from 1 to 25 , there are ten numbers with 3 as the unit ' s digit - 3 , 13 , 23 so , required number = 3 answer : b" | a ) a ) 11 , b ) b ) 3 , c ) c ) 2 , d ) d ) 1 , e ) e ) 6 | b | divide(subtract(25, 1), 3) | subtract(n1,n0)|divide(#0,n2)| | general |
what is the square root of 81 , divided by 3 ? | "square root is a number times itself square root of 81 = 6 , 9 / 3 = 3 ( d ) 3" | a ) 9 , b ) 36 , c ) 122 , d ) 3 , e ) 1 | d | divide(sqrt(81), 3) | sqrt(n0)|divide(#0,n1)| | other |
the mean of 30 values was 250 . it was detected on rechecking that one value 165 was wrongly copied as 135 for the computation of the mean . find the correct mean . | "corrected mean = 250 × 30 − 135 + 165 / 30 = 7500 − 135 + 165 / 30 = 7530 / 30 = 251 answer d" | a ) 151 , b ) 149 , c ) 152 , d ) 251 , e ) none of the above | d | divide(add(multiply(30, 250), subtract(165, 135)), 30) | multiply(n0,n1)|subtract(n2,n3)|add(#0,#1)|divide(#2,n0)| | general |
a cyclist rides a bicycle 9 km at an average speed of 11 km / hr and again travels 11 km at an average speed of 9 km / hr . what is the average speed for the entire trip ? | "distance = 20 km time = 9 / 11 + 11 / 9 = ( 81 + 121 ) / 99 = 202 / 99 hours average speed = ( 20 * 99 ) / 202 = 9.8 km / h the answer is c ." | a ) 9.3 , b ) 9.6 , c ) 9.8 , d ) 10.1 , e ) 10.3 | c | divide(add(9, 11), add(divide(9, 11), divide(11, 9))) | add(n0,n1)|divide(n0,n1)|divide(n1,n0)|add(#1,#2)|divide(#0,#3)| | general |
a man can row upstream at 26 kmph and downstream at 30 kmph , and then find the speed of the man in still water ? | "us = 26 ds = 30 m = ( 30 + 26 ) / 2 = 28 answer : b" | a ) 29 , b ) 28 , c ) 30 , d ) 32 , e ) 23 | b | divide(add(26, 30), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
in a group of 60 people , 35 have visited iceland and 23 have visited norway . if 31 people have visited both iceland and norway , how many people have visited neither country ? | this is an example of a standard overlapping sets question . it has no ' twists ' to it , so you ' ll likely find using the overlapping sets formula to be a fairly easy approach . if you ' re not familiar with it , then here is the formula : 60 = 35 + 23 - 31 + ( # in neither group ) = 33 the prompt gives you all of th... | a ) 30 , b ) 31 , c ) 32 , d ) 33 , e ) 34 | d | subtract(60, subtract(add(35, 23), 31)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1) | other |
a reduction of 5 % in the price of oil enables a house wife to obtain 10 kgs more for rs . 5000 , what is the reduced price for kg ? | "5000 * ( 5 / 100 ) = 250 - - - - 10 ? - - - - 1 = > rs . 25 answer : c" | a ) 23 , b ) 27 , c ) 25 , d ) 35 , e ) 15 | c | divide(divide(multiply(5000, 5), const_100), 10) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain |
if a 2 - b 2 = 10 and a * b = 8 , find a 4 + b 4 . | "a 2 - b 2 = 10 : given a 4 + b 4 - 2 a 2 b 2 = 102 : square both sides and expand . a * b = 8 : given a 2 b 2 = 82 : square both sides . a 4 + b 4 - 2 ( 64 ) = 100 : substitute a 4 + b 4 = 228 correct answer b" | a ) 200 , b ) 228 , c ) 182 , d ) 102 , e ) 128 | b | add(power(10, 2), multiply(power(8, 2), 2)) | power(n3,n0)|power(n2,n0)|multiply(#0,n0)|add(#2,#1)| | general |
find the sum of first 80 odd numbers | "explanation : n 2 = 802 = 6400 answer : option a" | a ) 6400 , b ) 8000 , c ) 5000 , d ) 4980 , e ) 4960 | a | add(divide(divide(80, divide(divide(divide(divide(divide(80, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(80, divide(divide(divide(divide(divide(80, const_2), const_2), const_2), const_2), const_2)), const_2)))) | divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(n0,#4)|divide(#5,const_2)|sqrt(#6)|add(#7,const_1)|add(#8,#6)| | general |
of 58 players on a cricket team , 37 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | total = 58 thrower = 37 rest = 58 - 37 = 21 left handed = 21 / 3 = 7 right handed = 14 if all thrower are right handed then total right handed is 37 + 14 = 51 so a . 51 is the right answer | a ) 51 , b ) 55 , c ) 59 , d ) 71 , e ) 92 | a | add(37, multiply(subtract(const_1, divide(const_1, const_3)), subtract(58, 37))) | divide(const_1,const_3)|subtract(n0,n1)|subtract(const_1,#0)|multiply(#2,#1)|add(n1,#3) | general |
the amount of principal rs . 8000 at compound interest at the ratio of 5 % p . a . for 6 years is | "c . i = p ( 1 + r / 100 ) ^ n = 8000 ( 1 + 5 / 100 ) ^ 6 = rs 10721 answer : c" | a ) s . 9621 , b ) s . 6921 , c ) s . 10721 , d ) s . 6261 , e ) s . 6361 | c | multiply(8000, power(add(const_1, divide(5, const_100)), 6)) | divide(n1,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)| | gain |
angelo and isabella are both salespersons . in any given week , angelo makes $ 590 in base salary plus 8 percent of the portion of his sales above $ 1,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of m... | let the weekly sales of both = x 590 + ( x − 1000 ) 8 / 100 = 10 / 100 x x = 25500 answer : c | a ) 23,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | c | floor(divide(divide(subtract(590, multiply(const_1000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), const_1000)) | divide(n1,const_100)|divide(n3,const_100)|multiply(#0,const_1000)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|divide(#5,const_1000)|floor(#6) | general |
the product of the squares of two positive integers is 1600 . how many pairs of positive integers satisfy this condition ? | "ans : e - 4 pairs ( x ˆ 2 ) ( y ˆ 2 ) = 1600 [ square root both sides ] xy = 40 40 = 1 x 40 , 4 x 10 , 40 x 1 , 10 x 4 , 5 x 8 , 8 x 5 , 20 x 2 , 2 x 20 cancel the repeats this leaves us with exactly 4 options . hence , e" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(add(const_2, const_3), const_2) | add(const_2,const_3)|subtract(#0,const_2)| | geometry |
what percent is 20 % of 25 % ? | required percentage = 20 % / 25 % * 100 = 20 / 25 * 100 = 80 % answer is e | a ) 20 % , b ) 30 % , c ) 50 % , d ) 60 % , e ) 80 % | e | multiply(divide(20, 25), const_100) | divide(n0,n1)|multiply(#0,const_100) | gain |
walking at 40 % 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 40 % of speed meand 2 / 5 s takes 5 / 2 t . it takes 24 minutes extra to cover the distance . then 5 / 2 t = t + 24 5 t = 2 t + 48 3 t = 48 t = 16 . option a is correct | a ) 16 , b ) 36 , c ) 42 , d ) 48 , e ) 54 | a | divide(24, subtract(divide(const_1, divide(40, const_100)), const_1)) | divide(n0,const_100)|divide(const_1,#0)|subtract(#1,const_1)|divide(n1,#2) | physics |
the area of the largest circle that can be drawn inside a square of side 70 cm in length is : | radius of required circle = 35 cm its area = 22 / 7 × 35 × 35 110 × 35 = 3850 cm ( power ) 2 answer is c . | ['a ) 3600', 'b ) 2646', 'c ) 3850', 'd ) 4215', 'e ) 5205'] | c | circle_area(divide(70, const_2)) | divide(n0,const_2)|circle_area(#0) | geometry |
there are 660 male and female participants in a meeting . half the female participants and one - quarter of the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "female = x male = 660 - x x / 2 + 660 - x / 4 = 1 / 3 * ( 660 ) = 220 x = 220 x / 2 = 110 is supposed to be the answer m is missing something correct option b" | a ) 75 , b ) 110 , c ) 125 , d ) 175 , e ) 225 | b | divide(subtract(multiply(divide(660, const_3), const_4), 660), const_2) | divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)| | general |
the speed of a train is 70 kmph . what is the distance covered by it in 10 minutes ? | "70 * 10 / 60 = 11.66 kmph answer : b" | a ) 10.66 kmph , b ) 11.66 kmph , c ) 21.66 kmph , d ) 12.66 kmph , e ) 11.96 kmph | b | multiply(divide(10, const_60), 70) | divide(n1,const_60)|multiply(n0,#0)| | physics |
what is the ratio e of the surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled ? | one side surface area of a cube = x * x = x ^ 2 total 6 sides = 6 x ^ 2 as for the rectangular , height ( h ) and width ( w ) are same as cube , x . only length = 2 x . l x h = 2 x * x = 2 x ^ 2 - - - - > 4 sides = 2 x ^ 2 * 4 = 8 x ^ 2 w * h = x * x = x ^ 2 - - - - - - > 2 sides = x ^ 2 * 2 = 2 x ^ 2 total 6 sides = 8... | ['a ) 1 / 4', 'b ) 3 / 8', 'c ) 1 / 2', 'd ) 3 / 5', 'e ) 2'] | d | divide(multiply(multiply(const_1, const_1), multiply(const_2, const_3)), add(multiply(multiply(const_1, const_1), const_2), multiply(multiply(const_1, const_2), const_4))) | multiply(const_1,const_1)|multiply(const_2,const_3)|multiply(const_1,const_2)|multiply(#0,#1)|multiply(#0,const_2)|multiply(#2,const_4)|add(#4,#5)|divide(#3,#6) | geometry |
if the operation @ is defined for all a and b by the equation a @ b = ( a ^ b ) / 2 , then ( 3 @ 2 ) = ? | work within the parenthesis first so solve ( 3 @ 2 ) first ( 3 @ 2 ) = ( 3 ^ 2 ) / 2 = 9 / 2 so 9 / 2 is the answer . . . . this question is merely testing order of operations remember pemdas answer : c | a ) 4 , b ) 2 , c ) 9 / 2 , d ) - 9 / 2 , e ) - 8 / 2 | c | divide(power(3, 2), 2) | power(n1,n0)|divide(#0,n0) | general |
a boat can travel with a speed of 10 km / hr in still water . if the speed of the stream is 8 km / hr , find the time taken by the boat to go 54 km downstream . | "speed of boat in still water = 10 km / hr speed of the stream = 8 km / hr speed downstream = ( 10 + 8 ) = 18 km / hr time taken to travel 68 km downstream = 54 â „ 18 = 17 â „ 4 = 3 hours answer is c" | a ) 4.25 hr , b ) 5.25 hr , c ) 3 hr , d ) 2.25 hr , e ) 2.50 hr | c | divide(54, add(10, 8)) | add(n0,n1)|divide(n2,#0)| | physics |
there is enough provisions for 9600 women in an army camp for 400 days . if there were 4800 men less , how long will the provision last ? | we have , m 1 d 1 = m 2 d 2 9600 * 400 = 4800 * d 2 d 2 = 9600 * 400 / 4800 = 800 days . answer : b | a ) 799 , b ) 800 , c ) 801 , d ) 802 , e ) 803 | b | divide(multiply(9600, 400), 4800) | multiply(n0,n1)|divide(#0,n2) | physics |
if a * b * c = ( √ ( a + 2 ) ( b + 3 ) ) / ( c + 1 ) , find the value of 6 * 15 * 3 . | "6 * 15 * 3 = ( √ ( 6 + 2 ) ( 15 + 3 ) ) / ( 3 + 1 ) = ( √ 8 * 18 ) / 4 = ( √ 144 ) / 4 = 12 / 4 = 3 . answer is d ." | a ) 8 , b ) 5 , c ) 11 , d ) 3 , e ) 4 | d | divide(sqrt(multiply(add(6, 2), add(15, 3))), add(3, 1)) | add(n0,n3)|add(n1,n4)|add(n2,n5)|multiply(#0,#1)|sqrt(#3)|divide(#4,#2)| | general |
shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 60 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 60 minutes after shannon . if... | "nice question + 1 we have that x / 24 - ( 60 - x ) / 2 r = 40 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 48 hence answer is e" | a ) 20 , b ) 34 , c ) 38 , d ) 40 , e ) 48 | e | divide(subtract(multiply(const_3, 60), 60), const_3) | multiply(n0,const_3)|subtract(#0,n0)|divide(#1,const_3)| | physics |
a and b began business with rs . 6000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of a . | "( 6 * 8 + 5 * 4 ) : ( 4 * 8 + 5 * 4 ) 17 : 13 17 / 30 * 630 = 357 answer : d" | a ) 240 , b ) 288 , c ) 277 , d ) 357 , e ) 361 | d | multiply(divide(630, add(add(multiply(6000, 8), multiply(subtract(6000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(6000, 8), multiply(subtract(6000, 1000), subtract(const_12, 8)))) | add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)| | gain |
a , b and c rents a pasture for rs . 950 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 950 = 295 answer : a" | a ) 295 , b ) 199 , c ) 676 , d ) 156 , e ) 122 | a | multiply(divide(950, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|multiply(#5,#1)| | general |
in what ratio p : q should the mixture p of milk and water in the ratio of 2 : 1 be mixed with another mixture q of milk and water in the ratio 1 : 5 so that the resultant mixture contains equal quantities of milk and water ? | "( 2 / 3 ) * p + ( 1 / 6 ) * q = ( 1 / 3 ) * p + ( 5 / 6 ) * q 2 p = 4 q p / q = 2 / 1 the answer is b ." | a ) 3 : 2 , b ) 2 : 1 , c ) 4 : 3 , d ) 3 : 5 , e ) 3 : 1 | b | add(subtract(add(1, 5), add(2, const_2.0)), const_1) | add(n1,n3)|add(n0,const_2.0)|subtract(#0,#1)|add(#2,const_1)| | general |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 350 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 350 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . ... | a ) 1,108 , b ) 2,900 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 350) | divide(n6,n5)|multiply(n4,#0)| | general |
ifaequals the sum of the even integers from 2 to 80 , inclusive , andbequals the sum of the odd integers from 1 to 79 , inclusive , what is the value of a - b ? | "this is a solution from beatthegmat : even numbers : ( 80 - 2 ) / 2 + 1 = 40 even integers . ( 80 + 2 ) / 2 = 41 is the average of the even set . sum = avg * ( # of elements ) = 41 * 40 = 1640 = a odd numbers : ( 79 - 1 ) / 2 + 1 = 40 odd integers . ( 79 + 1 ) / 2 = 40 is the average of the odd set . sum = avg * ( # o... | a ) 1 , b ) 10 , c ) 19 , d ) 40 , e ) 21 | d | subtract(multiply(divide(80, 2), add(divide(80, 2), 1)), multiply(divide(add(79, 1), 2), add(divide(subtract(79, 1), 2), 1))) | add(n2,n3)|divide(n1,n0)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,n0)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8)| | general |
if 31 % of a number is 46.5 , the number is ? | "answer let 31 % of a = 46.5 ⇒ ( 31 x a ) / 100 = 46.5 ∴ a = ( 46.5 x 100 ) / 31 = 150 correct option : a" | a ) 150 , b ) 155 , c ) 160 , d ) 165 , e ) none | a | divide(divide(46.5, const_100), divide(31, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)| | gain |
if an average hard drive had a capacity of 0.6 tb in 2000 , and average hard drive capacities double every 5 years , what will be the average hard drive capacity in 2050 ? | "0.6 * 2 ^ 10 = 0.6 * 1024 = 614.4 the answer is d ." | a ) 256 , b ) 512 , c ) 768 , d ) 6144 , e ) 1536 | d | multiply(add(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(0.6, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), 0.6), const_10) | multiply(n0,const_2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(#3,const_2)|multiply(#4,const_2)|multiply(#5,const_2)|multiply(#6,const_2)|multiply(#7,const_2)|multiply(#8,const_2)|add(n0,#9)|multiply(#10,const_10)| | general |
two digits in father ' s age are the same as the digit in the son ' s age , but in reverse order . in 5 years father will be twice as old as son will be then . what is the difference in their current age ? | father ' s age = 10 x + y so son ' s age = 10 y + x . . after 5 years , 10 x + y + 5 = 2 * ( 10 y + x + 5 ) . . . . so 5 + 19 y = 8 x . . . check for odd values of y , y = 1 satisfies the eqn with x = 3 . . . so ages are 31 and 13 and ans as found correctly by u is a . . 18 | a ) 18 , b ) 16 , c ) 34 , d ) 25 , e ) 12 | a | subtract(multiply(5, const_4), const_2) | multiply(n0,const_4)|subtract(#0,const_2) | general |
a train running at the speed of 60 km / hr crosses a pole in 18 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 18 ) m = 300 m . answer : c" | a ) 187 m , b ) 278 m , c ) 300 m , d ) 150 m , e ) 267 m | c | multiply(divide(multiply(60, const_1000), const_3600), 18) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
the average monthly income of a and b is rs . 5050 . the average monthly income of b and c is rs . 6250 and the average monthly income of a and c is rs . 5200 . what is the monthly income of a ? | "explanation : let monthly income of a = a monthly income of b = b monthly income of c = c a + b = 2 × 5050 . . . . ( equation 1 ) b + c = 2 × 6250 . . . . ( equation 2 ) a + c = 2 × 5200 . . . . ( equation 3 ) ( equation 1 ) + ( equation 3 ) - ( equation 2 ) = > a + b + a + c - ( b + c ) = ( 2 × 5050 ) + ( 2 × 5200 ) ... | a ) 2000 , b ) 3000 , c ) 4000 , d ) 5000 , e ) 6200 | c | subtract(multiply(const_2, 5050), divide(add(multiply(const_2, 5050), subtract(multiply(const_2, 6250), multiply(const_2, 5200))), const_2)) | multiply(n0,const_2)|multiply(n1,const_2)|multiply(n2,const_2)|subtract(#1,#2)|add(#0,#3)|divide(#4,const_2)|subtract(#0,#5)| | general |
x , y , and z are different prime numbers . the product x ^ 3 * y ^ 2 * z ^ 1 is divisible by how many different positive numbers ? | "the exponents of x ^ 3 * y ^ 2 * z ^ 1 are 3 , 2 , and 1 . the number of factors is ( 3 + 1 ) ( 2 + 1 ) ( 1 + 1 ) = 24 the answer is a ." | a ) 24 , b ) 30 , c ) 36 , d ) 42 , e ) 48 | a | subtract(power(const_2.0, const_4), const_4) | power(const_2.0,const_4)|subtract(#0,const_4)| | general |
a man is 28 years older than his son . in four years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 28 . = > 2 ( x + 4 ) = ( x + 28 + 4 ) = > 2 x + 8 = x + 32 = > x = 24 years answer : option d" | a ) 21 years , b ) 22 years , c ) 23 years , d ) 24 years , e ) 25 years | d | divide(subtract(28, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
the length of a rectangular floor is more than its breadth by 200 % . if rs . 529 is required to paint the floor at the rate of rs . 3 / sq m , what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 529 / 3 = 176.33 sq m l b = 176.33 i . e . , l * l / 3 = 176.33 l ^ 2 = 529 = > l = 23 e" | a ) 12 , b ) 18 , c ) 20 , d ) 22 , e ) 23 | e | multiply(sqrt(divide(divide(529, 3), const_3)), const_3) | divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)| | gain |
walking at 5 / 6 th of its usual speed a cab is 8 mnts late . find its usual time to cover the journey ? | "new speed = 5 / 6 th of usual speed new time = 6 / 5 th of usual time 6 / 5 ut - ut = 8 m ut / 5 = 8 m ut = 40 m answer is c" | a ) 25 m , b ) 45 m , c ) 40 m , d ) 50 m , e ) 62 m | c | multiply(5, 8) | multiply(n0,n2)| | physics |
a and b can do a piece of work in 4 days . b and c can do it in 12 days and a and c in 16 days . working together they will complete the work in how many days ? | "a + b = 1 / 4 b + c = 1 / 12 c + a = 1 / 16 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 ( a + b + c ) = 1 / 4 + 1 / 12 + 1 / 16 = 19 / 48 a + b + c = 19 / 96 96 / 19 = 5.05 days answer : b" | a ) 4.05 days , b ) 5.05 days , c ) 14.76 days , d ) 8.05 days , e ) 15.05 days | b | inverse(divide(add(inverse(16), add(inverse(4), inverse(12))), const_2)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|inverse(#5)| | physics |
a certain machine produces 780 units of product p per hour . working continuously at this constant rate , this machine will produce how many units of product p in 4 days ? | "since 7 days consist of 24 * 4 hours the total is 96 hours . since every hour the machine produces 780 units of product p the total product during 96 hours is 96 * 780 = 74,880 . correct option : d" | a ) 7,000 , b ) 24,000 , c ) 74,880 , d ) 100,000 , e ) 168,000 | d | multiply(const_4, const_10) | multiply(const_10,const_4)| | physics |
the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 39 ? | "first , let ' s translate the expression 21 # 39 , using the definition given : 21 # 39 = 21 × 24 × 27 × 30 × 33 × 36 × 39 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 39 = 3 ^ 7 ( 7 × 8 × 9 × 10 × 11 × 12 × 13 ) now let ' s replace each consecutive integer with its ... | a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | d | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | add(const_2,const_2)|add(n0,#0)|add(#1,#0)|add(#2,#1)| | general |
the cost of 3 pens and 5 pencils is rs . 200 . also the cost of one pen and one pencil is in the ratio of 5 : 1 respectively . what is the cost of one dozen pens ? | "explanation : let the cost of one pen is ‘ 5 x ’ and pencil is ‘ x ’ 3 x 5 x + 5 x = rs . 200 15 x + 5 x = rs . 200 x = 200 / 20 = 10 : . cost of 1 pen = 5 x = 5 x 10 = 50 : . cost of 12 pens , i . e . ( one dozen ) = 50 x 12 = rs . 600 answer : option e" | a ) rs . 200 , b ) rs . 250 , c ) rs . 300 , d ) rs . 150 , e ) rs . 600 | e | multiply(multiply(3, const_4), divide(200, add(3, 1))) | add(n0,n4)|multiply(n0,const_4)|divide(n2,#0)|multiply(#2,#1)| | other |
the captain of a cricket team of 11 members is 25 years old and the wicket keeper is 5 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 25 + 30 ) = 9 ( x - 1 ) 11 x - 9 x = 46 2 x = 46 x = 23 . the average age of the team is 23 years . the answer is c ." | a ) 20 , b ) 21 , c ) 23 , d ) 25 , e ) 26 | c | divide(subtract(add(25, add(25, 5)), multiply(const_3.0, const_3.0)), const_2) | add(n1,n2)|multiply(const_3.0,const_3.0)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 72 % percent carbonated water , what percent of the volume of the mixture is p ? | 72 % is 8 % - points below 80 % and 17 % - points above 55 % . so the ratio of solution p to solution q is 17 : 8 . mixture p is 17 / 25 = 68 % of the volume of mixture pq . the answer is e . | a ) 48 % , b ) 54 % , c ) 60 % , d ) 64 % , e ) 68 % | e | multiply(divide(subtract(divide(72, const_100), divide(55, const_100)), add(subtract(divide(72, const_100), divide(55, const_100)), subtract(divide(80, const_100), divide(72, const_100)))), const_100) | divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#0)|add(#3,#4)|divide(#3,#5)|multiply(#6,const_100) | gain |
there are 3 girls and 6 boys . how many kids are there ? | d | a ) 10 , b ) 19 , c ) 20 , d ) 9 , e ) 12 | d | divide(3, 6) | divide(n0,n1)| | general |
two numbers n and 16 have lcm = 48 and gcf = 8 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 16 * n = 48 * 8 n = 48 * 8 / 16 = 24 correct answer c" | a ) 11 , b ) 32 , c ) 24 , d ) 48 , e ) 50 | c | divide(multiply(48, 8), 16) | multiply(n1,n2)|divide(#0,n0)| | physics |
if y > 0 , ( 2 y ) / 5 + ( 3 y ) / 10 is what percent of y ? | can be reduced to 2 y / 5 + 3 y / 10 = 2 y / 5 = 70 % d | a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 80 % | d | multiply(const_100, add(divide(2, 5), divide(3, 10))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100) | general |
convert 200 inches into centimeter ? | "1 inch = 2.54 cm 200 inches = 200 * 2.54 = 508 cm answer is c" | a ) 512 cm , b ) 520 cm , c ) 508 cm , d ) 510 cm , e ) 554 cm | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 200), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
if an object travels at seven feet per second , how many feet does it travel in one hour ? | "speed = 7 feet per second . 1 hour = 60 x 60 seconds = 3600 seconds . total no of feet traveled in 1 hour = 3600 x 7 = 25200 answer d" | a ) 30 , b ) 300 , c ) 720 , d ) 25200 , e ) 18000 | d | multiply(multiply(const_3, const_60), const_60) | multiply(const_3,const_60)|multiply(#0,const_60)| | physics |
evaluate : 2344 - 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 2344 - 12 * 3 * 2 = 2344 - 8 = 2337 correct answer a" | a ) 2337 , b ) 1227 , c ) 3327 , d ) 5457 , e ) 7677 | a | subtract(2344, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
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