Datasets:

math_qa

like 67

a certain company ’ s profit in 1996 was 15 percent greater than its profit in 1995 , and its profit in 1997 was 35 percent greater than its profit in 1996 . the company ’ s profit in 1997 was what percent greater than its profit in 1995 ?

"profit in 1995 - 100 profit in 1996 - 115 % increae profit in 1997 in comparison to 1995 = 15 + 115 * 35 % = 55.25 correct option : e"

a ) 5 % , b ) 18 % , c ) 33 % , d ) 35 % , e ) 55.25 %

e

multiply(subtract(multiply(add(divide(35, const_100), const_1), add(const_1, divide(15, const_100))), const_1), const_100)

divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|

gain

a , b and c started a partnership business by investing rs . 12000 , rs . 16000 , rs . 20000 respectively . at the end of the year , the profit were distributed among them . if c ' s share of profit is 36000 , what is the total profit ?

"a : b : c = 12000 : 16000 : 20000 = 3 : 4 : 5 let total profit = p then p ã — 5 / 12 = 36000 p = ( 36000 ã — 12 ) / 5 = 86400 answer is c ."

a ) 70000 , b ) 60000 , c ) 86400 , d ) 90000 , e ) 50000

c

add(add(multiply(divide(12000, 20000), 36000), multiply(divide(16000, 20000), 36000)), 36000)

divide(n0,n2)|divide(n1,n2)|multiply(n3,#0)|multiply(n3,#1)|add(#2,#3)|add(n3,#4)|

gain

two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 24 seconds , what is the ratio of their speeds ?

"let the speed of the trains be x and y respectively length of train 1 = 27 x length of train 2 = 17 y relative speed = x + y time taken to cross each other = 24 s = ( 27 x + 17 y ) / ( x + y ) = 24 = ( 27 x + 17 y ) / = 24 ( x + y ) = 3 x = 7 y = x / y = 7 / 3 answer : c"

a ) 1 : 3 , b ) 3 : 1 , c ) 7 : 3 , d ) 3 : 2 , e ) 3 : 4

c

divide(subtract(27, 24), subtract(24, 17))

subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)|

physics

sarah operated her lemonade stand monday through friday over a two week period and made a total profit of 450 dollars . on hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days . each cup she sold had a total cost of 75 cents and sarah did not incur any other costs . if every day she sold exactly 32 cups and 3 of the days were hot , then what was the price of 1 cup on a hot day ?

7 regular days - - > sales = 7 * 32 * x = 224 x ; 3 hot days - - > sales = 3 * 32 * ( 1.25 x ) = 120 x ; total sales = 224 x + 120 x = 344 x . total cost = 10 * 32 * 0.75 = 240 . profit = 344 x - 240 = 450 - - > x = 2 . 1.25 x = ~ 2.50 . answer : d .

a ) $ 1.50 , b ) $ 1.88 , c ) $ 2.25 , d ) $ 2.50 , e ) $ 3.25

d

multiply(divide(add(multiply(multiply(32, divide(75, const_100)), multiply(add(const_4, const_1), const_2)), 450), add(multiply(subtract(multiply(add(const_4, const_1), const_2), 3), 32), multiply(multiply(divide(add(const_100, 25), const_100), 3), 32))), divide(add(const_100, 25), const_100))

add(const_1,const_4)|add(n1,const_100)|divide(n2,const_100)|divide(#1,const_100)|multiply(n3,#2)|multiply(#0,const_2)|multiply(#4,#5)|multiply(n4,#3)|subtract(#5,n4)|add(n0,#6)|multiply(n3,#8)|multiply(n3,#7)|add(#10,#11)|divide(#9,#12)|multiply(#13,#3)

gain

at sandi ’ s pet shop , 25 cups of bird seed are used every 5 days to feed 15 parakeets . how many cups of bird seed would be required to feed 14 parakeets for 7 days ?

let ' s go step - by - step . 25 cups are used over a 5 day period , which means 5 cups a day . this feeds 15 parakeets , which means each parakeet needs 1 / 3 of a cup every day . for 14 parakeets , we need 14 * 1 / 3 cups = 4.62 cups a day . over 7 days , we need 32.34 cups . choice b .

a ) 40.5 , b ) 32.34 , c ) 6.23 , d ) 9 , e ) 15

b

divide(multiply(multiply(14, 25), 7), multiply(5, 15))

multiply(n0,n3)|multiply(n1,n2)|multiply(n4,#0)|divide(#2,#1)

physics

there are 2 available positions and 50 candidates , one half of whom are democrats and another half are republicans . if it was decided that the positions would be filled at random , then what is the probability e that the both positions will be taken by members of just one party ?

e probability of one party having both spots : ( 1 / 2 ) * ( 24 / 49 ) = 12 / 49 ( 1 / 2 ) or ( 25 / 50 ) because it does not matter which party or which person gets the first spot . ( 24 / 49 ) because after one person from a particular party is chosen , there are 24 members of the same party left out of 49 total candidates . since this result can happen for both parties , ( 12 / 49 ) + ( 12 / 49 ) = ( 24 / 49 ) answer : d

a ) 1 / 25 , b ) 12 / 49 , c ) 1 / 4 , d ) 24 / 49 , e ) 1 / 2

d

multiply(multiply(divide(subtract(divide(50, const_2), const_1), subtract(50, const_1)), divide(divide(50, const_2), 50)), 2)

divide(n1,const_2)|subtract(n1,const_1)|divide(#0,n1)|subtract(#0,const_1)|divide(#3,#1)|multiply(#4,#2)|multiply(n0,#5)

other

a car travels from point a to point b . the average speed of the car is 60 km / hr and it travels the first half of the trip at a speed of 80 km / hr . what is the speed of the car in the second half of the trip ?

"let d be the distance and let v be the speed in the second half . the total time = t 1 + t 2 d / 60 = d / 160 + ( d / 2 ) / v 8 d / 480 - 3 d / 480 = ( d / 2 ) / v d / 96 = d / 2 v and so v = 48 km / hr the answer is d ."

a ) 40 , b ) 42 , c ) 45 , d ) 48 , e ) 50

d

divide(multiply(60, divide(multiply(80, 60), multiply(subtract(80, 60), 60))), divide(60, const_10))

divide(n0,const_10)|multiply(n0,n1)|subtract(n1,n0)|multiply(n0,#2)|divide(#1,#3)|multiply(n0,#4)|divide(#5,#0)|

general

sonika deposited rs . 8000 which amounted to rs . 11200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ?

"( 8000 * 3 * 2 ) / 100 = 480 11200 - - - - - - - - 11680 answer : c"

a ) 9680 , b ) 2277 , c ) 11680 , d ) 12068 , e ) 1212

c

add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(11200, 8000), 3), 8000)), 8000), 3), 8000)

divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)|

gain

a manufacturer sells a pair of glasses to a wholesale dealer at a profit of 18 % . the wholesaler sells the same to retailer at a profit of 20 % . the retailer in turn sells them to a customer for rs . 30.09 , there by earning a profit of 25 % . the cost price for the manufacturer is :

let the cost price for the manufacturer be rs . x . then , 125 % of 120 % of 118 % of x = 30.09 . 125 / 100 * 120 / 100 * 118 / 100 * x = 3009 / 100 177 / 100 x = 3009 / 100 = > x = 17 answer : c

a ) rs . 15 , b ) rs . 16 , c ) rs . 17 , d ) rs . 18 , e ) rs . 19

c

divide(divide(divide(add(multiply(add(const_1, const_2), divide(20, const_2)), divide(multiply(const_3, const_3), const_100)), divide(add(const_100, 25), const_100)), divide(add(20, const_100), const_100)), divide(add(const_100, 18), const_100))

add(const_1,const_2)|add(n3,const_100)|add(n1,const_100)|add(n0,const_100)|divide(n1,const_2)|multiply(const_3,const_3)|divide(#5,const_100)|divide(#1,const_100)|divide(#2,const_100)|divide(#3,const_100)|multiply(#0,#4)|add(#6,#10)|divide(#11,#7)|divide(#12,#8)|divide(#13,#9)

gain

how long will a boy take to run round a square field of side 40 meters , if he runs at the rate of 12 km / hr ?

speed = 12 km / hr = 12 * 5 / 18 = 10 / 3 m / sec distance = 40 * 4 = 160 m time taken = 160 * 3 / 10 = 48 sec answer is b

a ) 52 sec , b ) 48 sec , c ) 60 sec , d ) 25 sec , e ) 39 sec

b

divide(multiply(40, const_4), multiply(12, divide(const_1000, const_3600)))

divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)

gain

the sum of the first 50 positive even integers is 2550 . what is the sum of the even integers from 602 to 700 inclusive ?

"2 + 4 + 6 + 8 + . . . + 100 = 2550 602 + 604 + . . . + 700 = 50 ( 600 ) + ( 2 + 4 + . . . + 100 ) = 30,000 + 2550 = 32,550 the answer is c ."

a ) 24,550 , b ) 28,550 , c ) 32,550 , d ) 36,550 , e ) 40,550

c

multiply(divide(add(700, 602), const_2), add(divide(subtract(700, 602), const_2), const_1))

add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3)|

general

the average runs scored by a batsman in 10 matches is 40 . in the next 10 matches the batsman scored an average of 30 runs . find his average in all the 20 matches ?

"total score of the batsman in 10 matches = 400 . total score of the batsman in the next 10 matches = 300 . total score of the batsman in the 20 matches = 700 . average score of the batsman = 700 / 20 = 35 . answer : b"

a ) 31 , b ) 35 , c ) 88 , d ) 13 , e ) 12

b

divide(add(multiply(40, 10), multiply(30, 10)), add(10, 10))

add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|

general

there are 300 giraffes at a large zoo . the number of giraffes is 3 times the number of all the other animals combined . how many more giraffes are there than non giraffes animals at the zoo ?

answer is d . total giraffes = 300 = 3 * other animals ( x ) or x = 10 . so , difference in giraffes and x = 300 - 10 = 290 .

a ) 400 , b ) 360 , c ) 300 , d ) 290 , e ) 200

d

subtract(300, divide(divide(300, 3), const_10))

divide(n0,n1)|divide(#0,const_10)|subtract(n0,#1)

general

bucket a is 1 / 4 full of sand , and bucket b , which has one - half the capacity of bucket a , is 3 / 8 full of sand . jana pours the contents of bucket b into bucket a . bucket a is now filled to what fraction of its capacity ?

let capapcity of a = x sand = x / 4 capacity of b = x / 2 sand = 3 x / 16 a total = x / 4 + 3 x / 16 = 7 x / 16 answer : b

a ) 3 / 8 , b ) 7 / 16 , c ) 1 / 2 , d ) 5 / 8 , e ) 7 / 8

b

divide(add(4, 3), multiply(8, const_2))

add(n1,n2)|multiply(n3,const_2)|divide(#0,#1)

general

if n is an integer , f ( n ) = f ( n - 1 ) - n and f ( 4 ) = 15 . what is the value of f ( 6 ) ?

since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 15 then f ( 5 ) = 15 - 5 = 10 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 10 - 6 = 4 . answer : a . questions on funtions to practice :

a ) 4 , b ) 0 , c ) 1 , d ) 2 , e ) 24

a

subtract(subtract(15, add(1, 4)), 6)

add(n0,n1)|subtract(n2,#0)|subtract(#1,n3)

general

a salesman ’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1100 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ?

"[ 1100 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 14000 answer : a"

a ) 14000 , b ) 12000 , c ) 12019 , d ) 12197 , e ) 12012

a

divide(600, divide(5, const_100))

divide(n0,const_100)|divide(n4,#0)|

general

andy solves problems 78 to 125 inclusive in a math exercise . how many problems does he solve ?

"125 - 78 + 1 = 48 ' e ' is the answer"

a ) 53 , b ) 52 , c ) 51 , d ) 50 , e ) 48

e

add(subtract(125, 78), const_1)

subtract(n1,n0)|add(#0,const_1)|

general

a fair 2 sided coin is flipped 3 times . what is the probability that tails will be the result at least twice , but not more than 3 times ?

"at least twice , but not more than 3 timesmeans exactly 2 times , 3 times the probability of getting exactly k results out of n flips is nck / 2 ^ n 3 c 2 / 2 ^ 3 = 3 / 8 option : c"

a ) 5 / 8 , b ) 3 / 4 , c ) 3 / 8 , d ) 7 / 4 , e ) 5 / 6

c

subtract(const_1, add(multiply(inverse(power(2, 3)), 3), add(inverse(power(2, 3)), inverse(power(2, 3)))))

power(n0,n1)|inverse(#0)|add(#1,#1)|multiply(n1,#1)|add(#2,#3)|subtract(const_1,#4)|

general

a ' and ' b ' are positive integers such that their lcm is 20 and their hcf is 1 . what is the addition of the maximum and minimum possible values of ' a + b ' ?

"possible values of a and b can be 5,4 ; 4,5 ( which are same for a + b ) and 1,20 ; 20,1 ( same result for a + b ) so 21 + 9 = 30 ans c"

a ) 21 , b ) 34 , c ) 30 , d ) 25 , e ) 12

c

subtract(add(20, 20), add(1, 1))

add(n0,n0)|add(n1,n1)|subtract(#0,#1)|

general

3 / 5 of all married couples have more than one child . 2 / 5 of all married couples have more than 3 children . what fraction of all married couples have 2 or 3 children ?

plug in simple numbers . take 100 couples for example . 3 / 5 of 100 couples have more than one child = 60 couples . 2 / 5 of 100 couples have more than 3 kids = 40 couples . this implies that 40 couples are a subset of 60 couples and the complement of 60 couples within those 100 couples , which equals 20 couples have either one or no kids at all . we need to find couples that have 2 or 3 kids , so essentially , it is 60 - 40 = 20 . fraction will be 20 / 100 = 1 / 5 . option a

a ) 1 / 5 , b ) 1 / 4 , c ) 7 / 20 , d ) 3 / 5 , e ) it can not be determined from the given information .

a

subtract(divide(3, 5), divide(2, 5))

divide(n0,n1)|divide(n2,n1)|subtract(#0,#1)

general

one pump drains one - half of a pond in 2 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ?

"the tricky part here , i believed is one half hour = 1 / 2 . then everything would be easy . we have the 1 st pump working rate / hour = 1 / 2 : 2 = 1 / 4 working rate of 2 pumps : 1 / 2 : 1 / 2 = 1 . working rate of 2 nd pump : 1 - 1 / 4 = 3 / 4 - - > time taken for the 2 nd pump to finish : 1 : 3 / 4 = 4 / 3 = 1.3 hours . a"

a ) 1.3 hour , b ) 1.2 hour , c ) 3 hours , d ) 5 hours , e ) 6 hours

a

divide(const_1, subtract(const_1, divide(const_1, multiply(2, const_2))))

multiply(n0,const_2)|divide(const_1,#0)|subtract(const_1,#1)|divide(const_1,#2)|

physics

if 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 18 , then n =

"2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 18 = > 4 x 2 ^ 2 n = 4 ^ 24 = 2 ^ 36 = > 2 ^ 2 x 2 ^ 2 n = 2 ^ 36 = > 2 ^ ( 2 n + 2 ) = 2 ^ 36 = > 2 n + 2 = 36 = > n = 17 so . answer will be c ."

a ) 3 , b ) 6 , c ) 17 , d ) 23 , e ) 24

c

divide(subtract(multiply(18, 2), 2), 2)

multiply(n0,n9)|subtract(#0,n0)|divide(#1,n0)|

general

a fair price shopkeeper takes 10 % profit on his goods . he lost 40 % goods during theft . his loss percent is :

explanation : suppose he has 100 items . let c . p . of each item be re . 1 . total cost = rs . 100 . number of items left after theft = 60 . s . p . of each item = rs . 1.10 total sale = 1.10 * 60 = rs . 66 hence , loss % = 34 / 100 * 100 = 34 % answer : c

a ) 72 % , b ) 42 % , c ) 34 % , d ) 12 % , e ) 22 %

c

subtract(const_100, subtract(add(const_100, 10), divide(multiply(add(const_100, 10), 40), const_100)))

add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(const_100,#3)

gain

a and b go around a circular track of length 300 m on a cycle at speeds of 34 kmph and 50 kmph . after how much time will they meet for the first time at the starting point ?

"time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 300 / ( 36 * 5 / 18 ) , 300 / ( 54 * 5 / 18 ) } = lcm ( 30 , 20 ) = 60 sec . answer : b"

a ) 120 sec , b ) 60 sec , c ) 186 sec , d ) 167 sec , e ) 168 sec

b

divide(300, subtract(multiply(50, const_0_2778), multiply(34, const_0_2778)))

multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)|

physics

in cliff ’ s impressive rock collection , there are half as many igneous rocks as sedimentary rocks . of the igneous rocks , 3 / 4 are shiny and the rest are matte , while 1 / 5 of the sedimentary rocks are shiny . if there are 30 shiny igneous rocks , how many total rocks does cliff have ?

"we can start with the known quantity and then go on to find the others . shiny igneous ricks are 30 . these are ( 3 / 4 ) of total igneous rocks . ( 2 / 3 ) * total igneous rocks = 30 total igneous rocks = 30 * ( 4 / 3 ) = 40 total sedimentary rocks = 2 * total igneous rocks = 2 * 40 = 80 total number of rocks = 40 + 80 = 120 answer ( d )"

a ) 30 , b ) 45 , c ) 60 , d ) 120 , e ) 135

d

divide(multiply(divide(multiply(multiply(3, 4), 30), 3), 4), 3)

multiply(n0,n1)|multiply(n4,#0)|divide(#1,n0)|multiply(n1,#2)|divide(#3,n0)|

general

two assembly line inspectors , smithy and glen , inspect widgets as they come off the assembly line . if smithy inspects every fifth widget , starting with the fifth , and glen inspects every third , starting with the third , how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector ?

widgets inspected by smithy : ( ( 95 - 5 ) / 5 ) + 1 = 18 + 1 = 19 widgets inspected by glen : ( ( 96 - 3 ) / 3 ) + 1 = 31 + 1 = 32 widgets inspected by both : ( ( 90 - 15 ) + 1 = 6 total : 19 + 32 - 6 = 45 hence , widgets not inspected : 98 - 45 = 53 option b

a ) 50 , b ) 53 , c ) 55 , d ) 60 , e ) 65

b

subtract(98, subtract(add(divide(98, divide(const_10, const_2)), divide(98, const_3)), divide(98, multiply(divide(const_10, const_2), const_3))))

divide(const_10,const_2)|divide(n0,const_3)|divide(n0,#0)|multiply(#0,const_3)|add(#2,#1)|divide(n0,#3)|subtract(#4,#5)|subtract(n0,#6)

other

what is the sum of the numbers between 1 and 12 , inclusive ?

"sol . add the numbers between 1 and 12 . answer is d , 78 ."

a ) 99 , b ) 65 , c ) 79 , d ) 78 , e ) 80

d

divide(multiply(1, 12), const_4)

multiply(n0,n1)|divide(#0,const_4)|

general

the speed of a car is 120 km in the first hour and 70 km in the second hour . what is the average speed of the car ?

"s = ( 120 + 70 ) / 2 = 95 kmph answer : c"

a ) 72 kmph , b ) 75 kmph , c ) 95 kmph , d ) 80 kmph , e ) 82 kmph

c

divide(add(120, 70), const_2)

add(n0,n1)|divide(#0,const_2)|

physics

in a class of 60 children , 30 % children can speak only english , 20 % hindi and english both and the rest of the children can speak only hindi . how many children can speak hindi ?

number of students who speak only english = 30 % of 60 = 18 ∵ number of students who speak hindi and english = 20 % of 60 = 12 ∵ number of students who speak only hindi = ( 60 – 30 = ) 30 ∵ no . of students who speak hindi = 30 + 12 = 42 answer a

a ) 42 , b ) 36 , c ) 30 , d ) 48 , e ) none of these

a

add(subtract(60, 30), divide(multiply(20, 60), const_100))

multiply(n0,n2)|subtract(n0,n1)|divide(#0,const_100)|add(#2,#1)

other

25 % of a number is more than 20 % of 650 by 190 . find the number ?

"( 25 / 100 ) * x â € “ ( 20 / 100 ) * 650 = 190 1 / 4 x = 320 x = 1280 answer : e"

a ) 288 , b ) 278 , c ) 800 , d ) 267 , e ) 1280

e

divide(multiply(add(divide(multiply(650, 20), const_100), 190), const_100), 25)

multiply(n1,n2)|divide(#0,const_100)|add(n3,#1)|multiply(#2,const_100)|divide(#3,n0)|

gain

george ' s car calls for 8 ounces of oil for each cylinder used . his car ' s engine has 6 cylinders . if 16 ounces of oil has already been put into the engine used , then how many additional ounces of oil are needed ?

number of ounces per cylinder 8 number of cylinders 6 ounces needed 48 ounces used 16 answer d

['a ) 16', 'b ) 48', 'c ) 64', 'd ) 32', 'e ) 8']

d

subtract(multiply(8, 6), 16)

multiply(n0,n1)|subtract(#0,n2)

geometry

a car traveled 35 % of the way from town x to town y at an average speed of 35 mph . the car traveled at an average speed of v mph for the remaining part of the trip . the average speed for the entire trip was 50 mph . what is v in mph ?

"assume total distance = 100 miles time taken for 35 miles = 35 / 35 = 1 hour time taken for the rest of the 65 miles = 65 / v hours . average speed = 50 therefore the total time needed = 2 hours . 2 = 1 + 65 / v hence v = 65 mph answer : b"

a ) 45 , b ) 65 , c ) 55 , d ) 75 , e ) 85

b

divide(subtract(const_100, 35), divide(35, 35))

divide(n0,n0)|subtract(const_100,n0)|divide(#1,#0)|

general

if the compound interest on a certain sum of money for 2 years at 10 % per annum be rs . 993 , what would be the simple interest ?

"let p = principal a - amount we have a = p ( 1 + r / 100 ) 3 and ci = a - p atq 993 = p ( 1 + r / 100 ) 3 - p ? p = 3000 / - now si @ 10 % on 3000 / - for 2 yrs = ( 3000 x 10 x 2 ) / 100 = 600 / - answer : a ."

a ) rs . 600 , b ) rs . 890 , c ) rs . 895 , d ) rs . 900 , e ) none

a

divide(multiply(multiply(multiply(multiply(const_3.0, const_100), 10), 10), 2), const_100)

multiply(const_3.0,const_100)|multiply(n1,#0)|multiply(n1,#1)|multiply(n0,#2)|divide(#3,const_100)|

gain

hammers and wrenches are manufactured at a uniform weight per hammer and a uniform weight per wrench . if the total weight of two hammers and two wrenches is one - third that of 8 hammers and 5 wrenches , then the total weight of one wrench is how many times that of one hammer ?

x be the weight of a hammer and y be the weight of a wrench . ( 2 x + 2 y ) = 1 / 3 * ( 8 x + 5 y ) 3 ( 2 x + 2 y ) = ( 8 x + 5 y ) 6 x + 6 y = 8 x + 5 y y = 2 x ans - e

a ) 1 / 2 , b ) 2 / 3 , c ) 1 , d ) 3 / 2 , e ) 2

e

divide(subtract(8, multiply(const_3, const_2)), subtract(multiply(const_3, const_2), 5))

multiply(const_2,const_3)|subtract(n0,#0)|subtract(#0,n1)|divide(#1,#2)

general

tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 163 per kg , the price of the third variety per kg will be

"solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 163 / 22.50 = 1 = â € º x - 163 = 22.50 = â € º x = 185.50 . hence , price of the third variety = rs . 175.50 per kg . answer c"

a ) rs . 169.50 , b ) rs . 1700 , c ) rs . 185.50 , d ) rs . 180 , e ) none

c

divide(subtract(multiply(163, add(add(1, 1), 2)), add(126, 126)), 2)

add(n1,n1)|add(n0,n0)|add(n3,#0)|multiply(n4,#2)|subtract(#3,#1)|divide(#4,n3)|

other

in an examination , 25 % of total students failed in hindi , 40 % failed in english and 35 % in both . the percentage of these who passed in both the subjects is :

"pass percentage = 100 - ( 25 + 40 - 35 ) = 100 - 70 = 30 answer : c"

a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 %

c

subtract(const_100, subtract(add(25, 40), 35))

add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)|

general

the total circumference of two circles is 49 . if the first circle has a circumference that is exactly twice the circumference of the second circle , then what is the approximate sum of their two radii ?

"let r = radius of smaller circle . let r = radius of larger circle therefore : 2 π r + 2 π r = 49 where 2 r = r thus : 2 π r + 4 π r = 49 6 π r = 49 r = approx 2.6 π r + 2 r π = 49 3 π r = 49 r = approx 5.2 r + r = approx 7.8 answer : a"

a ) 7.8 , b ) 6.0 , c ) 6.7 , d ) 9.7 , e ) 18.0

a

divide(add(divide(divide(49, const_3), const_3), divide(multiply(divide(49, const_3), const_2), const_3)), const_2)

divide(n0,const_3)|divide(#0,const_3)|multiply(#0,const_2)|divide(#2,const_3)|add(#1,#3)|divide(#4,const_2)|

general

a certain pair of used shoes can be repaired for $ 10.50 and will last for 1 year . a pair of the same kind of shoes can be purchased new for $ 30.00 and will last for 2 years . the average cost per year of the new shoes is what percent greater than the cost of repairing the used shoes ?

"1 ) cost of repairing = 10.5 ( for one year ) , therefore for 2 years it would be $ 21 . 2 ) cost of new pair which will last for 2 years is $ 30.00 percentage change formula = ( final value - initial value ) / ( initial value ) * 100 . in this case the final value would be the price of new shoe initial value would be the cost of repairing the old shoe . i . e ( 30 - 21 ) / ( 21 ) * 100 = 42.85 % . ans is a"

a ) 42.85 , b ) 44 , c ) 48 , d ) 38 , e ) 40

a

multiply(const_100, divide(subtract(30.00, multiply(10.50, 2)), multiply(10.50, 2)))

multiply(n0,n3)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)|

general

a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 6 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ?

"net part filled in 1 hour = ( 1 / 4 - 1 / 6 ) = 1 / 12 the cistern will be filled in 12 / 1 hrs i . e . , 12 hrs . answer : d"

a ) 17.5 , b ) 7.4 , c ) 7.9 , d ) 12.0 , e ) 7.1

d

divide(const_1, subtract(divide(const_1, 4), divide(const_1, 6)))

divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)|

physics

a and b invests rs . 8000 and rs . 9000 in a business . after 4 months , a withdraws half of his capital and 2 months later , b withdraws one - third of his capital . in what ratio should they share the profits at the end of the year ?

"a : b ( 8000 * 4 ) + ( 4000 * 8 ) : ( 9000 * 6 ) + ( 6000 * 6 ) 64000 : 90000 32 : 45 . answer : a"

a ) 32 : 45 , b ) 32 : 47 , c ) 32 : 45 , d ) 32 : 41 , e ) 32 : 42

a

divide(add(multiply(8000, 4), multiply(divide(9000, const_3), multiply(2, 4))), add(multiply(9000, multiply(2, const_3)), multiply(subtract(9000, divide(9000, const_3)), multiply(2, const_3))))

divide(n1,const_3)|multiply(n0,n2)|multiply(n3,n2)|multiply(n3,const_3)|multiply(#0,#2)|multiply(n1,#3)|subtract(n1,#0)|add(#1,#4)|multiply(#3,#6)|add(#5,#8)|divide(#7,#9)|

gain

a runs twice as fast as b and gives b a start of 42 m . how long should the racecourse be so that a and b might reach in the same time ?

"ratio of speeds of a and b is 2 : 1 b is 42 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 42 m is 42 / 1 = 42 m so the total time taken by a and b to reach = 2 * 42 = 84 m answer : c"

a ) 75 m . , b ) 80 m . , c ) 84 m . , d ) 100 m . , e ) none of the above

c

multiply(42, const_2)

multiply(n0,const_2)|

physics

in a certain town , the ratio of ny yankees fans to ny mets fans is 3 : 2 , and the ratio of ny mets fans to boston red sox fans is 4 : 5 . if there are 330 baseball fans in the town , each of whom is a fan of exactly one of those three teams , how many ny mets fans are there in this town ?

"the ratio of yankees : mets : red sox = 6 : 4 : 5 the mets fans are 4 / 15 of the population . ( 4 / 15 ) * 330 = 88 the answer is b ."

a ) 72 , b ) 88 , c ) 95 , d ) 120 , e ) 135

b

multiply(divide(330, add(add(multiply(3, 2), 4), 5)), 4)

multiply(n0,n1)|add(n2,#0)|add(n3,#1)|divide(n4,#2)|multiply(n2,#3)|

other

what is the 26 th digit to the right of the decimal point in the decimal form of 6 / 11 ?

"6 / 11 = 0.54545454 . . . the even numbered positions in the decimal expansion are all 4 . the answer is b ."

a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7

b

divide(6, 11)

divide(n1,n2)|

general

what is the area of square field whose side of length 13 m ?

"13 * 13 = 169 sq m answer : a"

a ) 169 , b ) 196 , c ) 266 , d ) 288 , e ) 261

a

square_area(13)

square_area(n0)|

geometry

a can complete a certain job in 9 days . b is 80 % more efficient than a . in how many days can b complete the same job ?

"let , total work unit = 90 units a can finish in 9 days = 90 unit work i . e . a can finish in 1 days = 10 unit work i . e . b can finish in 1 days = 10 + ( 80 / 100 ) * 10 = 18 unit work days in which b will complete the work alone = 90 / 18 = 5 days answer : option a"

a ) 5 , b ) 6.25 , c ) 7 , d ) 7.5 , e ) 4.8

a

divide(multiply(9, const_60.0), const_100)

multiply(n0,const_60.0)|divide(#0,const_100)|

gain

a metallic sheet is of rectangular shape with dimensions 50 m x 30 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 5 m , the volume of the box ( in m cube ) is :

"explanation : l = ( 50 - 10 ) m = 40 m , [ because 5 + 5 = 10 ] b = ( 30 - 10 ) m = 20 m , h = 5 m . volume of the box = ( 40 x 20 x 5 ) m cube = 4000 m cube . option e"

a ) 4120 m cube , b ) 4140 m cube , c ) 5140 m cube , d ) 5120 m cube , e ) none of these

e

volume_rectangular_prism(subtract(50, multiply(5, const_2)), subtract(30, multiply(5, const_2)), 5)

multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)|

geometry

in what time a sum of money double itself at 2 % per annum simple interest ?

p = ( p * 2 * r ) / 100 r = 50 % answer : e

a ) 60 % , b ) 58 % , c ) 55 % , d ) 40 % , e ) 50 %

e

divide(multiply(const_100, const_100), multiply(2, const_100))

multiply(const_100,const_100)|multiply(n0,const_100)|divide(#0,#1)

gain

a , b , c can complete a piece of work in 18 , 6,12 days . working together , they complete the same work in how many days ?

"a + b + c 1 day work = 1 / 18 + 1 / 6 + 1 / 12 = 11 / 36 a , b , c together will complete the job in 36 / 11 days answer is d"

a ) 2 , b ) 4 / 5 , c ) 7 / 9 , d ) 36 / 11 , e ) 24 / 7

d

divide(const_1, add(add(divide(const_1, 18), divide(const_1, add(const_4, const_2))), divide(const_1, multiply(const_2, add(const_4, const_2)))))

add(const_2,const_4)|divide(const_1,n0)|divide(const_1,#0)|multiply(#0,const_2)|add(#1,#2)|divide(const_1,#3)|add(#4,#5)|divide(const_1,#6)|

physics

if the horse can run 10 miles / hour , how many days would it take for a horse to run from new york city to washington . d . c . ? ( 240 miles from nyc to d . c . )

the horse runs 10 miles / hour , thus it would take the horse 240 miles / 10 miles / hour to = 24 hours to run from nyc to d . c . there are 24 hours in a day , so days it take would be 24 hours / 24 hours / day = 1 day . correct option : 1

a ) 0.6 , b ) 0.8 , c ) 1 , d ) 1.2 , e ) 1.4

c

divide(divide(240, 10), divide(240, 10))

divide(n1,n0)|divide(#0,#0)

physics

solve below question 2 x + 1 = - 21

"2 x + 1 = - 21 x = - 11 a"

a ) - 11 , b ) - 9 , c ) 9 , d ) 8 , e ) - 7

a

divide(negate(add(21, 1)), 2)

add(n1,n2)|negate(#0)|divide(#1,n0)|

general

solution for 2.12 + . 004 + . 345

2.12 + . 004 + . 345 0 = 0 - 2.12 - 0.004 - 0.345 0 = - 2.469 answer : c

a ) 2.91 , b ) 2.911 , c ) 2.469 , d ) 2.586 , e ) 2.499

c

add(add(2.12, divide(4, const_1000)), divide(345, const_1000))

divide(n1,const_1000)|divide(n2,const_1000)|add(n0,#0)|add(#2,#1)

general

in a throw of dice what is the probability of ge æ « ng number greater than 3

"explanation : number greater than 3 is 4 , 5 , 6 , so only 3 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 3 / 6 = 1 / 2 answer : a"

a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these

a

divide(subtract(const_6, 3), const_6)

subtract(const_6,n0)|divide(#0,const_6)|

probability

a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 1300 , what is the reduced price for kg ?

"1300 * ( 25 / 100 ) = 325 - - - - 5 ? - - - - 1 = > rs . 65 answer : b"

a ) s . 40 , b ) s . 65 , c ) s . 49 , d ) s . 41 , e ) s . 42

b

divide(divide(multiply(1300, 25), const_100), 5)

multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|

gain

how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1 / 2 m board ?

"30 * 33 / 2 = 5 / 2 * 2 * x = > x = 99 answer b"

a ) 90 , b ) 99 , c ) 97 , d ) 95 , e ) 65

b

divide(multiply(30, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2))

divide(n1,n0)|add(n5,#0)|add(n0,#0)|multiply(n4,#1)|multiply(n0,#2)|divide(#3,#4)|

general

20 liters of a mixture is created by mixing liquid p and liquid q in the ratio 3 : 2 . how many liters of liquid q must be added to make the ratio 4 : 3 ?

"let x be the amount of liquid q to be added . ( 2 / 5 ) * 20 + x = ( 3 / 7 ) * ( 20 + x ) 280 + 35 x = 300 + 15 x 20 x = 20 x = 1 the answer is a ."

a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9

a

subtract(divide(multiply(divide(multiply(add(4, 3), divide(multiply(20, 3), add(3, 2))), 4), 3), add(4, 3)), divide(multiply(20, 2), add(3, 2)))

add(n3,n4)|add(n1,n2)|multiply(n0,n1)|multiply(n0,n2)|divide(#2,#1)|divide(#3,#1)|multiply(#0,#4)|divide(#6,n3)|multiply(n4,#7)|divide(#8,#0)|subtract(#9,#5)|

general

a sum amounts to rs . 5292 in 2 years at the rate of 5 % p . a . if interest was compounded yearly then what was the principal ?

"ci = 5292 , r = 5 , n = 2 ci = p [ 1 + r / 100 ] ^ 2 = p [ 1 + 5 / 100 ] ^ 2 5292 = p [ 21 / 20 ] ^ 2 5292 [ 20 / 21 ] ^ 2 4800 answer : d"

a ) s . 4000 , b ) s . 5000 , c ) s . 4500 , d ) s . 4800 , e ) s . 5800

d

divide(5292, power(add(divide(5, const_100), const_1), 2))

divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)|

gain

express a speed of 56 kmph in meters per second ?

"56 * 5 / 18 = 15.56 mps answer : d"

a ) 10.56 mps , b ) 17.56 mps , c ) 97.56 mps , d ) 15.56 mps , e ) 18.56 mps

d

multiply(const_0_2778, 56)

multiply(n0,const_0_2778)|

physics

in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 45 liters and they are all empty , how much money total will it cost to fuel all cars ?

"total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 45 ) = 21 + 351 = > 372 hence answer will be ( b ) 372"

a ) 320 $ , b ) 372 $ , c ) 420 $ , d ) 450 $ , e ) 480 $

b

multiply(multiply(0.65, 45), 12)

multiply(n1,n3)|multiply(n2,#0)|

general

a group of men decided to do a work in 12 days , but 6 of them became absent . if the rest of the group did the work in 14 days , find the original number of men ?

"original number of men = 6 * 14 / ( 14 - 12 ) = 42 answer is b"

a ) 15 , b ) 42 , c ) 44 , d ) 45 , e ) 18

b

divide(multiply(6, 14), subtract(14, 12))

multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|

physics

a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 234 for it , the cost price of the cricket bat for a is :

"125 % of 120 % of a = 234 125 / 100 * 120 / 100 * a = 234 a = 234 * 2 / 3 = 156 . answer a"

a ) 156 , b ) 120 , c ) 130 , d ) 160 , e ) 210

a

divide(234, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))))

divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)|

gain

a rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing . if the poles of the fence are kept 5 metres apart , how many poles will be needed ?

"solution perimeter of the plot = 2 ( 90 + 50 ) = 280 m . ∴ number of poles = [ 280 / 5 ] = 56 m answer b"

a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) none of these

b

divide(rectangle_perimeter(90, 50), 5)

rectangle_perimeter(n0,n1)|divide(#0,n2)|

physics

in town x , 64 percent of the population are employed , and 49 percent of the population are employed males . what percent of the employed people in town x are females ?

"answer b male - employed - 64 % female - employed - 15 % total employed 64 % means total unemployed = 36 % therefore d and e are not correct because female unemployed has to be less than 36 % female - unemployed = 32 % male unemployed = 4 % 48 % + 4 % = 52 % 16 % + 32 % = 48 % plug in a and b in place of 32 % and the sum is not 100 % 15 / 64 = 20 b"

a ) 16 % , b ) 20 % , c ) 32 % , d ) 40 % , e ) 52 %

b

multiply(divide(subtract(64, 49), 64), const_100)

subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|

gain

the hiker walking at a constant rate of 6 miles per hour is passed by a motor - cyclist traveling in the same direction along the same path at 30 miles per hour . the motor - cyclist stops to wait for the hiker 12 minutes after passing her , while the hiker continues to walk at her constant rate , how many minutes must the motor - cyclist wait until the hiker catches up ?

for the 12 minutes the motor - cyclist continues to overtake the hiker , she is going at 24 miles per hour faster than the hiker . once the motor - cyclist stops , the hiker is going at 6 miles per hour while the motor - cyclist is at rest so the amount of time the hiker will take to cover the distance between them is going to be in the ratio of the relative speeds . 24 / 6 * 12 or 48 minutes answer is ( d )

a ) 30 , b ) 36 , c ) 40 , d ) 48 , e ) 60

d

multiply(30, const_1_6)

multiply(n1,const_1_6)

physics

the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 90 metres ?

"speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 90 m = 90 / 5 = 18 seconds . answer : e"

a ) 16 seconds , b ) 76 seconds , c ) 26 seconds , d ) 12 seconds , e ) 18 seconds

e

divide(90, multiply(add(15, 3), const_0_2778))

add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)|

physics

a train consists of 12 boggies , each boggy 15 metres long . the train crosses a telegraph post in 9 seconds . due to some problem , one boggies were detached . the train now crosses a telegraph post in

"length of train = 12 ã — 15 = 180 m . then , speed of train = 180 ⠁ „ 9 = 20 m / s now , length of train = 11 ã — 15 = 165 m â ˆ ´ required time = 165 ⠁ „ 20 = 8.25 sec . answer d"

a ) 18 sec , b ) 12 sec , c ) 15 sec , d ) 8.25 sec , e ) none of these

d

divide(subtract(multiply(12, 15), 15), divide(multiply(12, 15), 9))

multiply(n0,n1)|divide(#0,n2)|subtract(#0,n1)|divide(#2,#1)|

physics

a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 2640 in all from both of them as interest . the rate of interest per annum is ?

"let the rate be r % p . a . then , ( 5000 * r * 2 ) / 100 + ( 3000 * r * 4 ) / 100 = 2640 100 r + 120 r = 2640 r = 12 % answer : b"

a ) 16 % , b ) 12 % , c ) 74 % , d ) 10 % , e ) 45 %

b

multiply(divide(2640, add(multiply(5000, 2), multiply(3000, 4))), const_100)

multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(n4,#2)|multiply(#3,const_100)|

gain

tickets numbered from 1 to 26 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bears a number which is a multiple of 3 ?

"here , s = [ 1 , 2 , 3 , 4 , … . , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 ] p ( e ) = n ( e ) / n ( s ) = 8 / 26 = 4 / 13 the answer is d ."

a ) 1 / 5 , b ) 2 / 5 , c ) 3 / 10 , d ) 4 / 13 , e ) 1 / 7

d

divide(divide(26, 3), 26)

divide(n1,n2)|divide(#0,n1)|

general

in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 56 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ?

"since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 56 , so 0.14 * { total } = 56 - - > { total } = 400 . answer : b ."

a ) 70 , b ) 400 , c ) 150 , d ) 35 , e ) 350

b

divide(56, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100)))

divide(n0,const_100)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#2,#0)|divide(n2,#3)|

gain

the number 219 can be written as sum of the squares of 3 different positive integers . what is the sum of these 3 different integers ?

"sum of the squares of 3 different positive integers = 219 13 ^ 2 + 7 ^ 2 + 1 ^ 2 = 219 now , sum of these 3 different integers = 13 + 7 + 1 = 21 ans - b"

a ) 20 , b ) 21 , c ) 19 , d ) 17 , e ) 15

b

multiply(const_2, sqrt(divide(219, const_2)))

divide(n0,const_2)|sqrt(#0)|multiply(#1,const_2)|

geometry

a certain list consists of 21 different numbers . if n is in the list and n is 7 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction of the sum of the 21 numbers in the list ?

"series : a 1 , a 2 . . . . a 20 , n sum of a 1 + a 2 + . . . + a 20 = 20 * x ( x = average ) so , n = 7 * x hence , a 1 + a 2 + . . + a 20 + n = 27 x so , the fraction asked = 7 x / 27 x = 7 / 27 answer a"

a ) 7 / 27 , b ) 1 / 6 , c ) 1 / 5 , d ) 4 / 21 , e ) 5 / 21

a

divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 7), 21), 7), const_2), 7), const_3))

divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,n1)|multiply(#3,const_2)|subtract(#4,n1)|subtract(#5,const_3)|divide(#1,#6)|

general

in a boat there are 8 men whose average weight is increased by 1 kg when 1 man of 60 kg is replaced by a new man . what is weight of new comer ?

solution : member in group * age increased = difference of replacement or , 8 * 1 = new comer - man going out or , new comer = 8 + 60 ; or , new comer = 68 years . answer : option c

a ) 70 , b ) 66 , c ) 68 , d ) 69 , e ) 72

c

add(60, 8)

add(n0,n3)

general

the ratio of a and b is 5 : 3 . after 2 years their ratio is 3 : 2 . what is the age of b ?

the ratio of a and b is 5 x and 3 x years . after 2 years , their ratio will be 3 and 2 . ( i . e ) ( 5 x + 2 ) / ( 3 x + 2 ) = 3 / 2 10 x + 4 = 9 x + 6 x = 2 b ' s age = 3 x = 3 ( 2 ) = 6 years answer is option c

a ) 5 , b ) 3 , c ) 6 , d ) 7 , e ) 8

c

subtract(divide(multiply(multiply(5, 2), 3), 2), multiply(3, 3))

multiply(n0,n2)|multiply(n1,n1)|multiply(n1,#0)|divide(#2,n2)|subtract(#3,#1)

other

a train leaves delhi at 7 a . m . at a speed of 30 kmph . another train leaves at 2 p . m . at a speed of 40 kmph on the same day and in the same direction . how far from delhi , will the two trains meet ?

"d = 30 * 7 = 210 rs = 40 – 30 = 10 t = 210 / 10 = 21 d = 40 * 21 = 840 km answer : b"

a ) 229 , b ) 840 , c ) 600 , d ) 888 , e ) 121

b

multiply(divide(multiply(30, add(const_3, 2)), subtract(40, 30)), 40)

add(n2,const_3)|subtract(n3,n1)|multiply(n1,#0)|divide(#2,#1)|multiply(n3,#3)|

physics

if the probability of rain on any given day in chicago during the summer is 50 % , 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 case is rainy - rainy - rainy - not rainy . the probability of this case is ( 1 / 2 ) ^ 4 = 1 / 16 the number of possible cases is 4 c 3 = 4 . p ( exactly 3 rainy days ) = 4 * 1 / 16 = 1 / 4 the answer is a .

a ) 1 / 4 , b ) 1 / 8 , c ) 3 / 8 , d ) 3 / 16 , e ) 5 / 16

a

multiply(4, power(divide(const_1, const_2), 4))

divide(const_1,const_2)|power(#0,n2)|multiply(n2,#1)

general

eight cubes , each with a volume of 512 cm ^ 3 , are joined to form one large cube . what is the surface area of the large cube ?

volume = a x a x a = a ^ 3 volume of cube = 8 x 512 = 4096 cm ^ 3 4096 = a ^ 3 a = 16 surface area of cube = 6 a ^ 2 = 6 x 16 ^ 2 = 1536 answer is b

['a ) 4096 sq cm', 'b ) 1536 sq cm', 'c ) 1024 sq cm', 'd ) 2048 sq cm', 'e ) 512 sq cm']

b

surface_cube(cube_edge_by_volume(multiply(multiply(const_2, const_4), 512)))

multiply(const_2,const_4)|multiply(n0,#0)|cube_edge_by_volume(#1)|surface_cube(#2)

geometry

the length of a rectangle is twice its breadth . if its lengthis decreased by 5 cm & breadth is increased by 5 cm , the area of the rectangle is increased by 75 sq . cm . what is the length of the rectangle ?

let breadth = x cm then , length = 2 x cm area = x × 2 x = 2 x 2 sq . cm . new length = ( 2 x − 5 ) cm new breadth = ( x + 5 ) cm new area = ( 2 x − 5 ) ( x + 5 ) sq . cm . given that , new area = initial area + 75 sq . cm . ⇒ ( 2 x − 5 ) ( x + 5 ) = 2 x 2 + 75 ⇒ 2 x 2 + 10 x − 5 x − 25 = 2 x 2 + 75 ⇒ 5 x − 25 = 75 ⇒ 5 x = 75 + 25 = 100 ⇒ x = 1005 = 20 cm length = 2 x = 2 × 20 = 40 cm c

a ) 25 cm , b ) 30 cm , c ) 40 cm , d ) 45 cm , e ) 60 cm

c

multiply(const_2, divide(add(75, multiply(5, 5)), subtract(multiply(const_2, 5), 5)))

multiply(n0,n0)|multiply(n0,const_2)|add(n2,#0)|subtract(#1,n0)|divide(#2,#3)|multiply(#4,const_2)

geometry

every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 51200 ?

51200 * 9 / 8 * 9 / 8 = 64800 . answer : e

a ) 81000 , b ) 81028 , c ) 27772 , d ) 29889 , e ) 64800

e

add(add(51200, multiply(divide(1, 8), 51200)), multiply(divide(1, 8), add(51200, multiply(divide(1, 8), 51200))))

divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)|

general

set s contains exactly 10 numbers and has an average ( arithmetic mean ) of 6.2 . if one of the numbers in set s is increased by 4 , while all other numbers remain the same , what is the new average of set s ?

"old set s - total is avg * no of elements = 6.2 * 10 = 62 if one number is increased by 4 then total increased to 62 + 4 = 66 new avg - 66 / 10 = 6.6 . hence answer is a ."

a ) 6.6 , b ) 6.7 , c ) 6.8 , d ) 6.85 , e ) 6.9

a

divide(add(multiply(10, 6.2), 4), 10)

multiply(n0,n1)|add(n2,#0)|divide(#1,n0)|

general

during a sale of 20 % on everything in a store , a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies . the store still makes a profit of 18 % on this sale . what is the mark up percentage on each candy ?

"hi akhil , i can share the way i solved it . . let ' s say marked price = m so , there ' s a discount of 20 % on m so , new s . p . = 80 % of m now , the child convinces the owner to sell 20 candies for the price of 14 candies . let ' s say each candy after discount is 1 $ . so , s . p . of 20 candies = 20 $ . the child bought it for 14 $ so , he got a discount of 6 / 20 * 100 = 30 % so , the latest s . p . = 70 % of 80 % of m = 0.7 * 0.8 m now , we are given that the shopkeeper still makes a profit of 12 % . so we have , 0.7 * 0.8 * m = 1.12 c . p so , we get , m = 2 c . p . i . e . marked price was kept 100 % above c . p . d"

a ) 100 % , b ) 80 % , c ) 75 % , d ) 66 + 2 / 3 % , e ) 55 %

d

multiply(subtract(divide(add(divide(18, const_100), const_1), multiply(subtract(const_1, divide(20, const_100)), divide(subtract(20, 14), 20))), const_1), const_100)

divide(n3,const_100)|divide(n0,const_100)|subtract(n1,n2)|add(#0,const_1)|divide(#2,n1)|subtract(const_1,#1)|multiply(#4,#5)|divide(#3,#6)|subtract(#7,const_1)|multiply(#8,const_100)|

gain

how many numbers are divisible by 4 between 1 and 100 .

"numbers divisible by 4 till 100 = ( 100 / 4 ) = 25 but we should not consider 100 as we are asked to find the number between 1 to 100 which are divisible by 4 . so answer is 24 numbers . option : a"

a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 25

a

divide(100, multiply(4, 1))

multiply(n0,n1)|divide(n2,#0)|

general

if a person walks at 14 km / hr instead of 10 km / hr , he would have walked 20 km more . what is the actual distance travelled by him ?

"assume that the person would have covered x km if travelled at 10 km / hr ⇒ speed = distancetime = x 10 . . . . . ( equation 1 ) give that the person would have covered ( x + 20 ) km if travelled at 14 km / hr ⇒ speed = distancetime = ( x + 20 ) 14 . . . . . ( equation 2 ) from equations 1 and 2 , x 10 = ( x + 20 ) 1414 x = 10 x + 2004 x = 200 x = 2004 = 50 answer : d"

a ) 33 , b ) 11 , c ) 38 , d ) 50 , e ) 12

d

multiply(10, divide(20, subtract(14, 10)))

subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)|

general

arjun started a business with rs . 40,000 and is joined afterwards by anoop with rs . 80 , 000 . after how many months did anoop join if the profits at the end of the year are divided equally ?

"suppose anoop joined after 3 months . then , 40000 * 12 = 80000 * ( 12 – x ) = > x = 6 . answer : d"

a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7

d

divide(subtract(multiply(multiply(multiply(const_4, const_100), const_100), const_12), multiply(multiply(multiply(const_2, const_100), const_100), const_12)), multiply(multiply(80, const_100), const_10))

multiply(const_100,const_4)|multiply(const_100,const_2)|multiply(n1,const_100)|multiply(#0,const_100)|multiply(#1,const_100)|multiply(#2,const_10)|multiply(#3,const_12)|multiply(#4,const_12)|subtract(#6,#7)|divide(#8,#5)|

general

5555 × 9999 = ?

"e 55544445 5555 × 9999 = 5555 ( 10000 - 1 ) = 5555 × 10000 - 5555 × 1 = 55550000 - 5555 = 55544445"

a ) 55500005 , b ) 55511115 , c ) 55522225 , d ) 55533335 , e ) 55544445

e

multiply(divide(5555, 9999), const_100)

divide(n0,n1)|multiply(#0,const_100)|

general

1397 x 1397 = ?

"1397 x 1397 = ( 1397 ) 2 = ( 1400 - 3 ) 2 = ( 1400 ) 2 + ( 3 ) 2 - ( 2 x 1400 x 3 ) = 1960000 + 9 - 8400 = 1960009 - 8400 = 1951609 . b )"

a ) 1951607 , b ) 1951609 , c ) 1951610 , d ) 1951611 , e ) 1951613

b

multiply(divide(1397, 1397), const_100)

divide(n0,n1)|multiply(#0,const_100)|

general

if 5 < x < 11 and y = x + 6 , what is the greatest possible integer value of x + y ?

"x + y = x + x + 6 = 2 x + 6 we need to maximize this value and it needs to be an integer . 2 x is an integer when the decimal of x is . 0 or . 5 the largest such value is 10.5 then x + y = 10.5 + 16.5 = 27 . the answer is c ."

a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29

c

add(add(6, const_10), const_10)

add(n2,const_10)|add(#0,const_10)|

general

pipe p can fill a tank in 8 hours , pipe q in 12 hours and pipe r in 20 hours . if all the pipes are open , in how many hours will the tank be filled ?

explanation : part filled by ( p + q + r ) in 1 hour = ( 1 / 8 + 1 / 12 + 1 / 20 ) = 31 / 120 all the three pipes together will fill the tank = 120 / 31 = 3.8 hours answer c

a ) 2 hours , b ) 2.4 hours , c ) 3.8 hours , d ) 3.5 hours , e ) 4.5 hours

c

inverse(add(divide(const_1, 20), add(divide(const_1, 8), divide(const_1, 12))))

divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|inverse(#4)

physics

a worker ' s take - home pay last year was the same each month , and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was 5 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she saved last year was from her take - home pay , what fraction of her take - home pay did she save each month ?

"let x be the fraction of her take - home pay that the worker saved . let p be the monthly pay . 12 xp = 5 ( 1 - x ) p 12 xp = 5 p - 5 xp 17 xp = 5 p x = 5 / 17 the answer is c ."

a ) 2 / 11 , b ) 3 / 14 , c ) 5 / 17 , d ) 7 / 20 , e ) 11 / 30

c

inverse(add(divide(const_12, 5), const_1))

divide(const_12,n0)|add(#0,const_1)|inverse(#1)|

general

a trader mixes 80 kg of tea at 15 per kg with 20 kg of tea at cost price of 20 per kg . in order to earn a profit of 35 % , what should be the sale price of the mixed tea ?

"c . p . of mixture = 80 × 15 + 20 × 20 / 80 + 20 = 16 ∴ s . p . = ( 100 + 35 ) / 100 × 16 = 21.6 answer b"

a ) 23.75 , b ) 21.6 , c ) 20 , d ) 19.2 , e ) none of these

b

add(divide(multiply(divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)), 35), const_100), divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)))

add(n0,n2)|multiply(n0,n1)|multiply(n2,n2)|add(#1,#2)|divide(#3,#0)|multiply(n4,#4)|divide(#5,const_100)|add(#6,#4)|

gain

if the sum of two numbers is 55 and the h . c . f and l . c . m of these numbers are 5 and 120 respectively , then the sum of the reciprocal of the numbers is equal to :

"let the numbers be a and b . then , a + b = 55 and ab = 5 * 120 = 600 . required sum = 1 / a + 1 / b = ( a + b ) / ab = 55 / 600 = 11 / 120 . answer : c"

a ) 55 / 601 , b ) 601 / 55 , c ) 11 / 120 , d ) 120 / 11 , e ) 120 / 13

c

divide(add(5, 120), multiply(5, 120))

add(n1,n2)|multiply(n1,n2)|divide(#0,#1)|

general

a person bought 129 glass bowls at a rate of rs . 11 per bowl . he sold 118 of them at rs . 15 and the remaining broke . what is the percentage gain for a ?

"cp = 129 * 11 = 1419 and sp = 118 * 15 = 1770 gain % = 100 * ( 1770 - 1419 ) / 1419 = 2900 / 129 answer : b"

a ) 40 , b ) 2900 / 129 , c ) 2430 / 70 , d ) 2791 / 11 , e ) 2719 / 8

b

multiply(divide(subtract(multiply(118, 15), multiply(129, 11)), multiply(129, 11)), const_100)

multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)|

gain

a cube of edge 12 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 15 cm , find the rise in water level ?

"increase in volume = volume of the cube = 12 * 12 * 12 cm ^ 3 rise in water level = volume / area = 12 * 12 * 12 / 20 * 15 = 5.76 cm answer is a"

a ) 5.76 cm , b ) 8.25 cm , c ) 11.25 cm , d ) 15.12 cm , e ) 20.62 cm

a

divide(volume_cube(12), multiply(20, 15))

multiply(n1,n2)|volume_cube(n0)|divide(#1,#0)|

geometry

two cars are driving toward each other . the first car is traveling at a speed of 100 km / h , which is 20 % slower than the second car ' s speed . if the distance between the cars is 1125 km , how many hours will it take until the two cars meet ?

"the speed of the first car is 100 km / h . the speed of the second car is 100 / 0.8 = 125 km / h . the two cars complete a total of 225 km each hour . the time it takes the cars to meet is 1125 / 225 = 5 hours . the answer is d ."

a ) 4 , b ) 4.5 , c ) 4.8 , d ) 5 , e ) 5.5

d

divide(add(multiply(100, const_4), multiply(divide(multiply(const_4, 100), add(const_1, const_4)), const_4)), add(100, divide(multiply(const_4, 100), add(const_1, const_4))))

add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|add(n0,#2)|multiply(#2,const_4)|add(#1,#4)|divide(#5,#3)|

physics

a train 140 m long running at 84 kmph crosses a platform in 16 sec . what is the length of the platform ?

"d = 84 * 5 / 18 = 16 = 373 â € “ 140 = 233 answer : c"

a ) 287 , b ) 298 , c ) 233 , d ) 726 , e ) 267

c

subtract(multiply(16, multiply(84, const_0_2778)), 140)

multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|

physics

by travelling at 70 kmph , a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ?

let the time taken to reach the destination be 3 x hours . total distance = 70 * 3 x = 210 x km he covered 2 / 3 * 210 x = 140 x km in 1 / 3 * 3 x = x hours so , the remaining 70 x km , he has to cover in 2 x hours . required speed = 70 x / 2 x = 35 kmph . answer : d

a ) 25 kmph , b ) 20 kmph , c ) 28 kmph , d ) 35 kmph , e ) 32 kmph

d

divide(subtract(multiply(70, const_3), divide(multiply(multiply(70, const_3), const_2), const_3)), subtract(const_3, const_1))

multiply(n0,const_3)|subtract(const_3,const_1)|multiply(#0,const_2)|divide(#2,const_3)|subtract(#0,#3)|divide(#4,#1)

physics

the cost of registration at a professional association meeting was $ 50 per person ; a lunch for registrants only was available for an additional $ 22 per person . if the number of registrants who paid for lunch was 50 more than the number who did not , and if receipts for registration and lunch totaled $ 82,900 , how many people paid just for registration at the meeting ?

"hope this might be useful to you . let the number of people who have opted only to register = x now since the registration cost is 50 $ per person , the total amount sums to = 50 x $ as per the information given in the question , the number of registrants who paid for lunch was 50 more than the number who did not . that means , total number of people who registered and paid for lunch = 50 + x . for the people who registered for lunch the cost is 50 $ ( for the event registration ) + 22 $ ( for lunch ) = 72 $ . total amount in this case sums to = 72 ( 50 + x ) = 3600 + 72 x now , total amount received was 82900 . thus , from the above data , 50 x + 3600 + 72 x = 82900 122 x = 82900 - 3600 122 x = 79300 x = 650 hence the correct ans is b"

a ) 700 , b ) 650 , c ) 1,300 , d ) 1,500 , e ) 1,800

b

multiply(const_1, const_1)

multiply(const_1,const_1)|

general

the unit digit in the product ( 594 * 897 * 929 * 462 ) is :

"explanation : unit digit in the given product = unit digit in ( 4 * 7 * 9 * 2 ) = 4 answer : d"

a ) 2 , b ) 5 , c ) 6 , d ) 4 , e ) 10

d

subtract(multiply(multiply(multiply(594, 897), 929), 462), subtract(multiply(multiply(multiply(594, 897), 929), 462), add(const_4, const_4)))

add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)|

general

how many times will the digit 8 be written when listing the integers from 1 to 1000 ?

"many approaches are possible . for example : consider numbers from 0 to 999 written as follows : 1 . 000 2 . 001 3 . 002 4 . 003 . . . . . . . . . 1000 . 999 we have 1000 numbers . we used 3 digits per number , hence used total of 3 * 1000 = 3000 digits . now , why should any digit have preferences over another ? we used each of 10 digits equal # of times , thus we used each digit ( including 8 ) 3000 / 10 = 300 times . answer : c ."

a ) 164 , b ) 297 , c ) 300 , d ) 345 , e ) 482

c

multiply(multiply(multiply(1, const_10), const_10), const_3)

multiply(n1,const_10)|multiply(#0,const_10)|multiply(#1,const_3)|

general

a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 63 km / hr . what is the length of the platform ?

"speed = 63 * 5 / 18 = 17.5 m / sec . length of the train = 17.5 * 20 = 350 m . let the length of the platform be x m . then , ( x + 350 ) / 36 = 17.5 = > x = 280 m . answer : e"

a ) 240 , b ) 288 , c ) 277 , d ) 226 , e ) 280

e

multiply(20, multiply(63, const_0_2778))

multiply(n2,const_0_2778)|multiply(n1,#0)|

physics

machine a and machine b are each used to manufacture 330 sprockets . it takes machine a 10 hours longer to produce 330 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ?

"machine b : takes x hours to produce 330 sprockets machine a : takes ( x + 10 ) hours to produce 330 sprockets machine b : in 1 hour , b makes 330 / x sprockets machine a : in 1 hour , a makes 330 / ( x + 10 ) sprockets equating : 1.1 ( 330 / ( x + 10 ) ) = 330 / x 484 / ( x + 10 ) = 330 / x 363 x = 330 x + 3300 33 x = 3300 x = 100 a makes 330 / ( 110 ) = 3 sprockets per hour answer : e"

a ) 5 , b ) 4 , c ) 6 , d ) 8 , e ) 3

e

divide(330, divide(multiply(multiply(10, 330), divide(add(const_100, 10), const_100)), subtract(multiply(330, divide(add(const_100, 10), const_100)), 330)))

add(n1,const_100)|multiply(n0,n1)|divide(#0,const_100)|multiply(#2,#1)|multiply(n0,#2)|subtract(#4,n0)|divide(#3,#5)|divide(n0,#6)|

gain

if the ratio of the sum of the first 6 terms of a g . p . to the sum of the first 3 terms of the g . p . is 217 , what is the common ratio of the g . p ?

"217 = ( a 1 + a 2 + a 3 + a 4 + a 5 + a 6 ) / ( a 1 + a 2 + a 3 ) factorize the same terms 217 = 1 + ( a 4 + a 5 + a 6 ) / ( a 1 + a 2 + a 3 ) write every term with respect to r a 1 = a 1 a 2 = a 1 * r ^ 1 a 3 = a 1 * r ^ 2 . . . . . . . . . 217 = 1 + ( a 1 ( r ^ 3 + r ^ 4 + r ^ 5 ) ) / ( a 1 ( 1 + r ^ 1 + r ^ 2 ) ) 216 = ( r ^ 3 ( 1 + r ^ 1 + r ^ 2 ) ) / ( ( 1 + r ^ 1 + r ^ 2 ) ) 216 = r ^ 3 r = 6 d"

a ) 3 , b ) 1 / 3 , c ) 2 , d ) 6 , e ) 1 / 6

d

power(subtract(217, const_1), divide(const_1, 3))

divide(const_1,n1)|subtract(n2,const_1)|power(#1,#0)|

other