options
stringlengths
37
300
correct
stringclasses
5 values
annotated_formula
stringlengths
7
727
problem
stringlengths
5
967
rationale
stringlengths
1
2.74k
program
stringlengths
10
646
a ) 1 / 8 , b ) 1 / 2 , c ) 2 / 4 , d ) 1 / 4 , e ) 1 / 4
b
multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 10)), 3)), 10)
two pipes p and q can fill a cistern in 12 and 10 minutes respectively . both are opened together , but at the end of 3 minutes the first is turned off . how much longer will the cistern take to fill ?
"3 / 12 + x / 10 = 1 x = 8 1 / 2 answer : b"
a = 1 / 12 b = 1 / 10 c = a + b d = c * 3 e = 1 - d f = e * 10
a ) 1 , b ) 100 , c ) 229 , d ) 329 , e ) 337
e
subtract(power(169, 2), power(168, 2))
( 169 ) ^ 2 - ( 168 ) ^ 2 =
using the formula : ( a + 1 ) ^ 2 - a ^ 2 = 2 a + 1 so , answer = 168 * 2 + 1 = 336 + 1 = 337 = answer = e
a = 169 ** 2 b = 168 ** 2 c = a - b
a ) 40 - 42 , b ) 39 - 41 , c ) 38 - 40 , d ) 37 - 39 , e ) 36 - 37
d
add(multiply(33.50, divide(15, const_100)), 33.50)
a meal cost $ 33.50 and there was no tax . if the tip was more than 10 pc but less than 15 pc of the price , then the total amount paid should be :
"10 % ( 33.5 ) = 3.35 15 % ( 33.5 ) = 5.025 total amount could have been 33.5 + 3.35 and 33.5 + 5.025 = > could have been between 36.85 and 38.525 = > approximately between 37 and 39 answer is d ."
a = 15 / 100 b = 33 * 50 c = b + 33
a ) a ) 4 , b ) b ) 1 , c ) c ) 2 , d ) d ) 3 , e ) e ) 5
a
subtract(23, reminder(1054, 23))
what least number should be added to 1054 , so that the sum is completely divisible by 23
"explanation : ( 1054 / 23 ) gives remainder 19 19 + 4 = 23 , so we need to add 4 answer : option a"
a = 23 - reminder
a ) 2,000 , b ) 3,000 , c ) 6,000 , d ) 8,000 , e ) 9,000
e
multiply(divide(divide(subtract(subtract(multiply(multiply(1, const_4), const_1000), multiply(multiply(multiply(1, const_4), const_1000), divide(1, 5))), multiply(subtract(multiply(multiply(1, const_4), const_1000), multiply(multiply(multiply(1, const_4), const_1000), divide(1, 5))), divide(1, 4))), const_1000), const_...
in a recent head - to - head run - off election , 15,000 absentee ballets were cast . 1 / 5 of the absentee ballets were thrown out and 1 / 4 of the remaining absentee ballets were cast for candidate a . how many absentee votes did candidate b receive ?
"4 / 5 * 3 / 4 ( total absentee votes ) = 3 / 5 ( total votes ) = 3 / 5 * 15000 = 9000 answer is e"
a = 1 * 4 b = a * 1000 c = 1 * 4 d = c * 1000 e = 1 / 5 f = d * e g = b - f h = 1 * 4 i = h * 1000 j = 1 * 4 k = j * 1000 l = 1 / 5 m = k * l n = i - m o = 1 / 4 p = n * o q = g - p r = q / 1000 s = r / 4 t = s * 1
a ) 10 , b ) 12 , c ) 15 , d ) 11 , e ) 20
d
add(divide(90, subtract(10, 1)), 1)
the operation is defined for all integers a and b by the equation ab = ( a - 1 ) ( b - 1 ) . if y 10 = 90 , what is the value of y ?
"ab = ( a - 1 ) ( b - 1 ) y 10 = ( y - 1 ) ( 10 - 1 ) = 90 - - > y - 1 = 10 - - > y = 11 answer : d"
a = 10 - 1 b = 90 / a c = b + 1
a ) 10.30 a . m , b ) 11 a . m , c ) 10 a . m , d ) 12 noon , e ) none of these
b
add(divide(add(330, 75), add(60, 75)), 8)
the distance between two cities a and b is 330 km . a train starts from a at 8 a . m . and travel towards b at 60 km / hr . another train starts from b at 9 a . m . and travels towards a at 75 km / hr . at what time will they meet ?
"explanation : assume that they meet x hours after 8 a . m . then , train 1 , starting from a , travelling towards b , travels x hours till the trains meet β‡’ distance travelled by train 1 in x hours = speed Γ— time = 60 x then , train 2 , starting from b , travelling towards a , travels ( x - 1 ) hours till the trains m...
a = 330 + 75 b = 60 + 75 c = a / b d = c + 8
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 12
e
divide(multiply(multiply(multiply(const_4, 2), 2), 1), multiply(multiply(2, 2), 1))
a card game called β€œ high - low ” divides a deck of 52 playing cards into 2 types , β€œ high ” cards and β€œ low ” cards . there are an equal number of β€œ high ” cards and β€œ low ” cards in the deck and β€œ high ” cards are worth 2 points , while β€œ low ” cards are worth 1 point . if you draw cards one at a time , how many ways...
"great question ravih . this is a permutations problem ( order matters ) with repeating elements . given thatlowcards are worth 1 pt andhigh cards 2 pts , and you must draw 3 low cards , we know that you must also draw 1 high card . the formula for permutations problems with repeating elements isn ! / a ! b ! . . . whe...
a = 4 * 2 b = a * 2 c = b * 1 d = 2 * 2 e = d * 1 f = c / e
a ) 1 hour , b ) 1.2 hour , c ) 3 hours , d ) 1.25 hours , e ) 6 hours
d
divide(const_1, subtract(const_1, divide(const_1, multiply(2.5, const_2))))
one pump drains one - half of a pond in 2.5 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 : 5 / 2 = 1 / 5 working rate of 2 pumps : 1 / 2 : 1 / 2 = 1 . working rate of 2 nd pump : 1 - 1 / 5 = 4 / 5 - - > time taken for the 2 nd pump to finish : 1 : 4 / 5 = 5 / 4 = 1...
a = 2 * 5 b = 1 / a c = 1 - b d = 1 / c
a ) 35 , b ) 45 , c ) 55 , d ) 60 , e ) 50
e
subtract(subtract(add(add(divide(multiply(400, 1), 2), divide(multiply(400, 5), 8)), divide(multiply(400, 3), 4)), multiply(divide(multiply(400, 3), 8), 2)), 400)
a high school has 400 students 1 / 2 attend the arithmetic club , 5 / 8 attend the biology club and 3 / 4 attend the chemistry club . 3 / 8 attend all 3 clubs . if every student attends at least one club how many students attend exactly 2 clubs .
a - club has 200 members ( 1 / 2 of 400 ) b - club has 250 members ( 5 / 8 of 400 ) c - club has 300 members ( 3 / 4 of 400 ) we can create an equation to solve this : 200 + 250 + 150 = n + x + 2 y where n is the number of students , x is the number of students in two clubs , and y is the number of students in three cl...
a = 400 * 1 b = a / 2 c = 400 * 5 d = c / 8 e = b + d f = 400 * 3 g = f / 4 h = e + g i = 400 * 3 j = i / 8 k = j * 2 l = h - k m = l - 400
a ) 15 , b ) 7 , c ) 10 , d ) 12 , e ) 14
a
subtract(30, const_4)
in a group of cows and hens , the number of legs are 30 more than twice the number of heads . the number of cows is
"explanation : let the number of cows be x and the number of hens be y . then , 4 x + 2 y = 2 ( x + y ) + 30 4 x + 2 y = 2 x + 2 y + 30 2 x = 30 x = 15 answer : a"
a = 30 - 4
a ) 650 meter , b ) 555 meter , c ) 500 meter , d ) 458 meter , e ) none of these
c
subtract(multiply(divide(multiply(78, const_1000), const_3600), const_60), 800)
a train , 800 meter long is running with a speed of 78 km / hr . it crosses a tunnel in 1 minute . what is the length of the tunnel ?
explanation : let length of tunnel is x meter distance = 800 + x meter time = 1 minute = 60 seconds speed = 78 km / hr = 78 * 5 / 18 m / s = 65 / 3 m / s distance = speed * time = > 800 + x = 65 / 3 βˆ— 60 = > 800 + x = 20 βˆ— 65 = 1300 = > x = 1300 βˆ’ 800 = 500 so the length of the tunnel is 500 meters . option c
a = 78 * 1000 b = a / 3600 c = b * const_60 d = c - 800
a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28
c
add(10, add(const_3, const_4))
how many digits are in ( 8 Γ— 10 ^ 14 ) ( 10 Γ— 10 ^ 10 ) ?
"he question simplifies to ( 8 Γ— 10 ^ 14 ) ( 10 ^ 11 ) = > 8 * 10 ^ 25 = > will contain 25 zeros + 1 digit 8 = > 26 ans c"
a = 3 + 4 b = 10 + a
a ) 75 , b ) 180 , c ) 198 , d ) 216 , e ) 252
a
divide(multiply(add(subtract(90, 10), subtract(90, 70)), 30), 40)
richard walks along sunrise boulevard daily . he starts walking at 07 : 00 from block 10 and walks to block 90 where he turns around and walks back to block 70 , where he stops at 07 : 30 . the blocks along the boulevard are numbered sequentially ( 1 , 2,3 ) , and each block measures 40 meters . what is richard ' s spe...
total distance from 10 to 90 = 80 + from 90 to 70 = 20 so the dist is 100 Γ— 30 ( per block dist ) speed = 3000 mts / 40 min = 75 m / min a is the answer
a = 90 - 10 b = 90 - 70 c = a + b d = c * 30 e = d / 40
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
b
subtract(5, reminder(4, 8))
when n is divided by 32 , the remainder is 5 . what is the remainder when 4 n is divided by 8 ?
"let n = 5 ( leaves a remainder of 5 when divided by 32 ) 4 n = 4 ( 5 ) = 20 , which leaves a remainder of 4 when divided by 8 . answer b"
a = 5 - reminder
a ) a ) 25 , b ) b ) 40 , c ) c ) 32 , d ) d ) 50 , e ) e ) 28
d
divide(subtract(10, add(const_2, const_3)), subtract(divide(const_1, const_2), divide(const_2, add(const_2, const_3))))
a person ' s present age is two - fifth of the age of his mother . after 10 years , he will be one - half of the age of his mother . how old is the mother at present ?
let the mother ' s present age be x years then the person ' s present age = 2 x / 5 ( 2 x / 5 ) + 10 = 1 / 2 ( x + 10 ) 2 ( 2 x + 50 ) = 5 ( x + 10 ) x = 50 answer is d
a = 2 + 3 b = 10 - a c = 1 / 2 d = 2 + 3 e = 2 / d f = c - e g = b / f
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
e
divide(multiply(add(4, 3), const_2), 5)
| x + 3 | – | 4 - x | = | 5 + x | how many solutions will this equation have ?
"you have | x + 3 | - | 4 - x | = | 8 + x | first , look at the three values independently of their absolute value sign , in other words : | x + 3 | - | 4 - x | = | 8 + x | ( x + 3 ) - ( 4 - x ) = ( 8 + x ) now , you ' re looking at x < - 8 , so x is a number less than - 8 . let ' s pretend x = - 10 here to make things...
a = 4 + 3 b = a * 2 c = b / 5
a ) 90 , b ) 120 , c ) 150 , d ) 180 , e ) 160
a
multiply(divide(const_60, 40), 1)
if the population of a certain country increases at the rate of one person every 40 seconds , by how many persons does the population increase in 1 hour ?
"answer = 1.5 * 60 = 90 answer is a"
a = const_60 / 40 b = a * 1
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
b
divide(add(multiply(13, add(5, 2)), 2), add(multiply(13, add(5, 2)), 2))
when positive integer t is divided by 13 , the remainder is 2 . when n is divided by 8 , the remainder is 5 . how many such values are less than 180 ?
the equation that can be formed t is 13 x + 2 = 8 y + 5 . . 13 x - 3 = 8 y . . . as we can see x can take only odd values as the rhs will always be even . . also x can take values till 13 as 13 * 14 > 180 . . now we have to substitue x as 1 , 35 , 79 , 1113 . . . once we find 7 fitting in , any other value need not be ...
a = 5 + 2 b = 13 * a c = b + 2 d = 5 + 2 e = 13 * d f = e + 2 g = c / f
a ) 636 , b ) 640 , c ) 647 , d ) 651 , e ) 675
d
divide(multiply(add(multiply(11, const_100), 47), add(multiply(7, const_100), 77)), multiply(subtract(47, add(multiply(const_2, const_4), const_2)), subtract(47, add(multiply(const_2, const_4), const_2))))
a room 11 m 47 cm long and 7 m 77 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor .
explanation : area of the room = ( 1147 x 777 ) cm 2 . size of largest square tile = h . c . f . of 1147 cm and 777 cm = 37 cm . area of 1 tile = ( 37 x 37 ) cm 2 . number of tiles required = ( 1147 Γ— 777 ) / ( 37 Γ— 37 ) = 651 answer : option d
a = 11 * 100 b = a + 47 c = 7 * 100 d = c + 77 e = b * d f = 2 * 4 g = f + 2 h = 47 - g i = 2 * 4 j = i + 2 k = 47 - j l = h * k m = e / l
a ) 42 , b ) 44 , c ) 46 , d ) 36 , e ) 50
d
divide(180, subtract(divide(180, 30), divide(60, 60)))
the distance from city a to city b is 180 miles . while driving from city a to city b , cara drives at a constant speed of 30 miles per hour . dan leaves city a 60 minutes after cara . what is the minimum constant speed in miles per hour that dan must exceed in order to arrive in city b before cara ?
"the time it takes cara to drive to city b is 180 / 30 = 6 hours . dan needs to take less than 5 hours for the trip . dan needs to exceed a constant speed of 180 / 5 = 36 miles per hour . the answer is d ."
a = 180 / 30 b = 60 / 60 c = a - b d = 180 / c
['a ) 20 cm and 21 cm', 'b ) 25 cm and 23 cm', 'c ) 27 cm and 23 cm', 'd ) 30 cm and 34 cm', 'e ) 28 cm and 36 cm']
c
divide(add(multiply(divide(250, 10), const_2), 10), const_2)
difference of two parallel sides of atrapezium is 4 cm . perpendicular distance between them is 10 cm . if the area of the trapezium is 250 cm ^ 2 . find the lengths of the parallel side ?
let the two parallel sides of the trapezium be a cm and b cm . then , a - b = 4 - - - - - - ( 1 ) and , ( 1 / 2 ) x ( a + b ) x 10 = 475 = > ( a + b ) = ( ( 250 x 2 ) / 10 ) = > a + b = 50 - - - - - - - ( 2 ) solving 1 and 2 , we get : a = 27 , b = 23 so , the two parallel sides are 27 cm and 23 cm . c
a = 250 / 10 b = a * 2 c = b + 10 d = c / 2
a ) 30 , b ) 5 , c ) 10 , d ) 36 , e ) 25
b
sqrt(25)
the length of the longest tape in cm which can be used to measure exactly , the length 100 cm ; 2 m 25 cm ; and 7 m 80 cm is :
the three lengths in cm are 100 , 225 & 780 . hcf of 100 , 225 & 780 is 5 . hence , the answer is 5 cm . answer : b
a = math.sqrt(25)
a ) 6 , b ) 12 , c ) 16 , d ) 18 , e ) 24
b
divide(subtract(90, const_10), const_10)
how many positive factors do 180 and 90 have in common ?
"the number of common factors will be same as number of factors of the highest common factor ( hcf ) hcf of 180 and 90 is 90 number of factors of 90 = 12 answer : b"
a = 90 - 10 b = a / 10
a ) 14000 , b ) 14400 , c ) 16129 , d ) 14600 , e ) 14700
c
add(10000, multiply(divide(multiply(10000, 27), const_100), 2))
the population of a town is 10000 . it increases annually at the rate of 27 % p . a . what will be its population after 2 years ?
"formula : 10000 Γ— 127 / 100 Γ— 127 / 100 = 16129 answer : c"
a = 10000 * 27 b = a / 100 c = b * 2 d = 10000 + c
a ) 28 % , b ) 40 % , c ) 85.5 % , d ) 70 % , e ) 72 %
c
add(multiply(divide(divide(5, const_100), subtract(1, divide(1, 10))), const_100), 2)
the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 5 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ?
"original price = 100 day 1 discount = 10 % , price = 100 - 10 = 90 day 2 discount = 10 % , price = 90 - 9 = 81 day 3 discount = 5 % , price = 81 - 4.05 = 76.95 which is 76.95 / 90 * 100 of the sale price on day 1 = ~ 8 answer c"
a = 5 / 100 b = 1 / 10 c = 1 - b d = a / c e = d * 100 f = e + 2
a ) 20 , b ) 23 , c ) 32 , d ) 19 , e ) 29
b
divide(subtract(add(30, add(30, 5)), multiply(5, 5)), const_2)
the captain of a cricket team of 11 members is 30 years old and the wicket keeper is 5 years younger . 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 of the team ?
"let the average of the whole team be x years . 11 x - ( 30 + 25 ) = 9 ( x - 1 ) = 11 x - 9 x = 46 = 2 x = 46 = > x = 23 so , average age of the team is 23 years . answer : b"
a = 30 + 5 b = 30 + a c = 5 * 5 d = b - c e = d / 2
a ) s . 1200 , b ) s . 1300 , c ) s . 600 , d ) s . 800 , e ) s . 1200
c
divide(300, multiply(divide(5, const_100), 10))
a sum was put at simple interest at a certain rate for 10 years . had it been put at 5 % higher rate , it would have fetched rs . 300 more . what was the sum ?
"at 5 % more rate , the increase in s . i for 10 years = rs . 300 ( given ) so , at 5 % more rate , the increase in si for 1 year = 300 / 10 = rs . 30 / - i . e . rs . 30 is 5 % of the invested sum so , 1 % of the invested sum = 30 / 5 therefore , the invested sum = 30 Γ— 100 / 5 = rs . 600 answer : c"
a = 5 / 100 b = a * 10 c = 300 / b
a ) 2000 , b ) 3000 , c ) 3500 , d ) 3800 , e ) 4000
b
multiply(630, divide(const_100, subtract(36, 15)))
in an exam 49 % candidates failed in english and 36 % failed in hindi and 15 % failed in both subjects . if the total number of candidates who passed in english alone is 630 . what is the total number of candidates appeared in exam ?
not fail in english = 51 % not fail in hindi = 64 % not fail in both = 30 % ( 49 + 36 - 15 ) pass in english only = 51 - 30 = 21 21 / 100 * x = 630 x = 3000 answer : b
a = 36 - 15 b = 100 / a c = 630 * b
a ) 4 / 15 , b ) 7 / 30 , c ) 11 / 30 , d ) 5 / 18 , e ) 7 / 18
c
add(divide(1, 5), divide(1, 6))
each of the three people individually can complete a certain job in 4 , 5 , and 6 hours , respectively . what is the lowest fraction of the job that can be done in 1 hour by 2 of the people working together at their respective rates ?
"the two slowest people work at rates of 1 / 5 and 1 / 6 of the job per hour . the sum of these rates is 1 / 5 + 1 / 6 = 11 / 30 of the job per hour . the answer is c ."
a = 1 / 5 b = 1 / 6 c = a + b
a ) 187 , b ) 197 , c ) 207 , d ) 136 , e ) 227
d
subtract(multiply(11, 13), add(const_10, const_1))
find the smallest number which when divided by 11 and 13 leaves respective remainders of 4 and 6 .
"let ' n ' is the smallest number which divided by 11 and 13 leaves respective remainders of 4 and 6 . required number = ( lcm of 11 and 13 ) - ( common difference of divisors and remainders ) = ( 143 ) - ( 7 ) = 136 . answer : d"
a = 11 * 13 b = 10 + 1 c = a - b
a ) 5 seconds , b ) 4.5 seconds , c ) 3 seconds , d ) 2 seconds , e ) none of these
d
divide(100, multiply(180, const_0_2778))
in what time will a train 100 meters long cross an electric pole , if its speed is 180 km / hr
"explanation : first convert speed into m / sec speed = 180 * ( 5 / 18 ) = 50 m / sec time = distance / speed = 100 / 50 = 2 seconds answer : d"
a = 180 * const_0_2778 b = 100 / a
a ) 18 sec , b ) 7.5 sec , c ) 15 sec , d ) 20 sec , e ) none of these
b
divide(subtract(multiply(12, 15), 15), divide(multiply(12, 15), 9))
a train consists of 12 boggies , each boggy 15 metres long . the train crosses a telegraph post in 9 seconds . due to some problem , two boggies were detached . the train now crosses a telegraph post in
"length of train = 12 Γ£ β€” 15 = 180 m . then , speed of train = 180 Γ’  β€ž 9 = 20 m / s now , length of train = 10 Γ£ β€” 15 = 150 m Γ’ Λ† Β΄ required time = 150 Γ’  β€ž 20 = 7.5 sec . answer b"
a = 12 * 15 b = a - 15 c = 12 * 15 d = c / 9 e = b / d
a ) 6 / 11 , b ) 8 / 15 , c ) 9 / 20 , d ) 11 / 25 , e ) 15 / 28
e
multiply(divide(subtract(8, 2), 8), divide(subtract(subtract(8, 2), const_1), subtract(8, const_1)))
in a box of 8 pens , a total of 2 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ?
"p ( neither pen is defective ) = 6 / 8 * 5 / 7 = 15 / 28 the answer is e ."
a = 8 - 2 b = a / 8 c = 8 - 2 d = c - 1 e = 8 - 1 f = d / e g = b * f
a ) 8000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000
c
divide(power(divide(500, const_2), const_2), subtract(512.50, 500))
if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 500 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 512.50 . what is the value of x ?
"simple way to solve this question is to use options . from si , we know that x * y = 25,000 . now , put the value of x = 5000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 5000 + 5 % of 5000 = 5250 . 2 nd year , amount = 5250 + 5 % of 5250 = 5512.50 . we can see after 2 years interest = 551...
a = 500 / 2 b = a ** 2 c = 512 - 50 d = b / c
a ) a ) 5 , b ) b ) 2 , c ) c ) 17 / 5 , d ) d ) 18 / 5 , e ) e ) 4
a
divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 5)
if 2 x + y = 7 and x + 2 y = 5 , then 5 xy / 3 = ?
"2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 ( 5 * 3 * 1 ) / 3 = 15 / 3 = 5 a"
a = 7 * 2 b = a - 5 c = 2 * 2 d = c - 1 e = b / d f = 7 * 2 g = f - 5 h = 2 * 2 i = h - 1 j = g / i k = 2 * j l = 7 - k m = e + l n = m / 5
a ) 8100 , b ) 3388 , c ) 7767 , d ) 2009 , e ) 3645
e
add(add(2880, multiply(divide(1, 8), 2880)), multiply(divide(1, 8), add(2880, multiply(divide(1, 8), 2880))))
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 . 2880 ?
"2880 * 9 / 8 * 9 / 8 = 3645 answer : e"
a = 1 / 8 b = a * 2880 c = 2880 + b d = 1 / 8 e = 1 / 8 f = e * 2880 g = 2880 + f h = d * g i = c + h
a ) 1000 meters , b ) 1050 meters , c ) 1200 meters , d ) 1250 meters , e ) none of these
d
multiply(multiply(5, divide(15, const_60)), const_1000)
a man is walking at the rate of 5 km / hr crosses a bridge in 15 minutes . the length of the bridge is
"explanation : we need to get the answer in meters . so we will first of change distance from km / hour to meter / sec by multiplying it with 5 / 18 and also change 15 minutes to seconds by multiplying it with 60 . answer : d"
a = 15 / const_60 b = 5 * a c = b * 1000
a ) 19 , b ) 25 , c ) 77 , d ) 88 , e ) 11
a
subtract(multiply(divide(const_100, 840), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100)
a dishonest dealer professes to sell goods at the cost price but uses a weight of 840 grams per kg , what is his percent ?
"840 - - - 160 100 - - - ? = > 19 % answer : a"
a = 100 / 840 b = 3 + 2 c = b * 2 d = 100 * c e = a * d f = e - 100
a ) 105 , b ) 95 , c ) 90 , d ) 99 , e ) 100
d
divide(divide(subtract(500, multiply(multiply(9, const_0_2778), 9)), 9), const_0_2778)
a train 500 m long passes a man , running at 9 km / hr in the same direction in which the train is going , in 20 seconds . the speed of the train is :
"speed of the train relative to man = 500 / 20 m / sec = 25 m / sec . = 25 x 18 / 5 km / hr = 90 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 9 ) km / hr . x - 9 = 90 = 99 km / hr . answer : d"
a = 9 * const_0_2778 b = a * 9 c = 500 - b d = c / 9 e = d / const_0_2778
a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 4
a
multiply(divide(subtract(4800, 4400), 4800), const_100)
the cost price of a radio is rs . 4800 and it was sold for rs . 4400 , find the loss % ?
"4800 - - - - 400 100 - - - - ? = > 8 % answer : a"
a = 4800 - 4400 b = a / 4800 c = b * 100
a ) - 3 , b ) - 1 , c ) - 1 / 3 , d ) 0 , e ) undefined
a
divide(add(divide(subtract(4, 8), 2), 5), divide(add(2, 6), 2))
line m lies in the xy - plane . the y - intercept of line m is - 5 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 4 ) and ( 6 , - 8 ) . what is the slope of line m ?
"ans : a solution : line m goes through midpoint of ( 2 , 4 ) and ( 6 , - 8 ) . midpoint is ( 4 , - 2 ) as we can see that the y axis of intercept point is ( 0 , - 5 ) means line m is parallel to x axis slope m = - 3 ans : a"
a = 4 - 8 b = a / 2 c = b + 5 d = 2 + 6 e = d / 2 f = c / e
a ) 7,000 , b ) 24,000 , c ) 40,000 , d ) 100,000 , e ) 158,400
e
multiply(const_4, const_10)
a certain machine produces 1,100 units of product p per hour . working continuously at this constant rate , this machine will produce how many units of product p in 6 days ?
since 6 days consist of 24 * 6 hours the total is 144 hours . since every hour the machine produces 1100 units of product p the total product during 144 hours is 144 * 1100 = 158,400 . correct option : e
a = 4 * 10
a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 900
b
divide(150, subtract(const_1, divide(3, 4)))
linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 150 , what were her original savings ?
"if linda spent 3 / 4 of her savings on furniture , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 150 . so 1 / 4 of her savings is $ 150 . so her original savings are 4 times $ 150 = $ 600 correct answer b"
a = 3 / 4 b = 1 - a c = 150 / b
a ) 12 days , b ) 14 days , c ) 16 days , d ) 18 days , e ) 20 days
c
divide(multiply(8, 10), 5)
if 8 people can eat a gallon of ice cream in 10 days , how long would it take 5 people to eat a gallon of ice cream ?
8 * 10 = 5 * x x = 16 answer : c
a = 8 * 10 b = a / 5
a ) 9.46 , b ) 2.03 , c ) 50.4 , d ) 14.65 , e ) 12.85
a
divide(subtract(multiply(8, 2.4), multiply(5, 1.4)), 1.5)
8 x 2.4 - 5 x 1.4 / 1.5 = ?
"given expression = ( 19.2 - 5 x 1.4 ) / 1.5 = 14.2 / 1.5 = 14.2 / 1.5 = 9.46 answer is a"
a = 8 * 2 b = 5 * 1 c = a - b d = c / 1
a ) 2.5 sec , b ) 2.8 sec , c ) 0.5 sec , d ) 2.3 sec , e ) 1.5 sec
c
divide(20, multiply(144, const_0_2778))
in what time will a train 20 m long cross an electric pole , it its speed be 144 km / hr ?
"speed = 144 * 5 / 18 = 40 m / sec time taken = 20 / 40 = 0.5 sec . answer : c"
a = 144 * const_0_2778 b = 20 / a
a ) 50 , b ) 40 , c ) 60 , d ) 100 , e ) 70
c
divide(150, add(divide(150, const_100), const_1))
the sum of number of boys and girls in a school is 150 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ?
"we have x + x % of 150 = 150 x + x / 100 * 150 = 150 5 / 2 * x = 150 x = 150 * 2 / 5 = 60 answer is c"
a = 150 / 100 b = a + 1 c = 150 / b
a ) 10 s , b ) 6 s , c ) 4 s , d ) 8 s , e ) 13 s
e
divide(add(175, 150), add(divide(multiply(54, const_1000), const_3600), divide(multiply(36, const_1000), const_3600)))
two trains a and b are 175 m and 150 m long and are moving at one another at 54 km / hr and 36 km / hr respectively . arun is sitting on coach b 1 of train a . calculate the time taken by arun to completely cross train b .
detailed solution speed of a = 54 βˆ— 1000 / 60 βˆ— 60 = 15 m / s speed of b = 36 βˆ— 1000 / 60 βˆ— 60 = 10 m / s relative speed = s 1 + s 2 = 15 + 10 m / s = 25 m / s the length that needs to be crossed = length of train b = 150 m . therefore time taken = 150 / 25 = 6 s . what is the time taken for trains to completely cross ...
a = 175 + 150 b = 54 * 1000 c = b / 3600 d = 36 * 1000 e = d / 3600 f = c + e g = a / f
a ) 30 , b ) 40 , c ) 60 , d ) 50 , e ) none
a
divide(subtract(multiply(12, 120), multiply(90, 15)), subtract(15, 12))
a train traveling with constant speed crosses a 90 m long platform in 12 seconds and a 120 m long platform in 15 seconds . find the length of the train and its speed .
let the length of the train be x m and its speed be y m / sec . then , x / y = 12 = > y = x / 12 ( x + 90 ) / 12 = x + 120 / 15 = > x = 30 m . answer : a
a = 12 * 120 b = 90 * 15 c = a - b d = 15 - 12 e = c / d
a ) 150 , b ) 140 , c ) 160 , d ) 170 , e ) 180
c
subtract(multiply(120, const_2), multiply(100, const_2))
if the average ( arithmetic mean ) of a and b is 100 , and c – a = 120 , what is the average of b and c ?
"a + b / 2 = 100 = > a + b = 200 a = c - 120 . . . sub this value c - 120 + b = 200 = > c + b = 320 = > c + b / 2 = 160 answer : c"
a = 120 * 2 b = 100 * 2 c = a - b
a ) 28 sec , b ) 23 sec , c ) 24 sec , d ) 25 sec , e ) 26 sec
a
divide(add(150, 130), multiply(36, const_0_2778))
how many seconds will a train 130 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ?
"d = 130 + 150 = 280 s = 36 * 5 / 18 = 10 mps t = 280 / 10 = 28 sec a"
a = 150 + 130 b = 36 * const_0_2778 c = a / b
a ) $ 37.80 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 39.50 , e ) $ 40.60
b
add(multiply(35, divide(40, const_100)), 35)
a farmer spent $ 35 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 20 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have s...
"a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 35 = $ 14 on chicken feed , thus he spent the remaining 35 - 14 = $ 21 on goat feed . now , since he bought chicken feed at a 20 % discount then the original price of it was x * 0.8 = $ 14 - - > x = $ 17.5 . therefore if the farmer had paid full price for...
a = 40 / 100 b = 35 * a c = b + 35
a ) 45 , b ) 38 , c ) 72 , d ) 74 , e ) 78
c
add(multiply(subtract(18, const_1), 5), 4)
find the 18 th term of an arithmetic progression whose first term is 4 and the common difference is 5 .
"n th term of a . p = a + ( n - 1 ) * d = 4 + ( 18 - 1 ) * 4 , = 4 + 68 = 72 . answer : c"
a = 18 - 1 b = a * 5 c = b + 4
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8
c
divide(18, add(2, const_1))
if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 18 - m ) then m = ?
"2 m = 18 - m 3 m = 18 m = 6 the answer is c ."
a = 2 + 1 b = 18 / a
a ) a ) 125 , b ) b ) 119 , c ) c ) 153 , d ) d ) 721 , e ) e ) 159
b
divide(multiply(850, subtract(const_100, add(add(44, 32), 10))), const_100)
in a school of 850 boys , 44 % of muslims , 32 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ?
"44 + 32 + 10 = 86 % 100 – 84 = 14 % 850 * 14 / 100 = 119 answer : b"
a = 44 + 32 b = a + 10 c = 100 - b d = 850 * c e = d / 100
a ) 35 , b ) 37 , c ) 39 , d ) 41 , e ) 42
e
add(multiply(const_2, 10), add(12, 10))
in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 12 years older than b , the present age of b is
"explanation : let b ' s age = x years . then , as age = ( x + 12 ) years . ( x + 12 + 10 ) = 2 ( x β€” 10 ) hence x = 42 . present age of b = 42 years answer : option e"
a = 2 * 10 b = 12 + 10 c = a + b
a ) 3 / 10 , b ) 7 / 10 , c ) 9 / 10 , d ) 9 / 20 , e ) 13 / 20
d
divide(add(multiply(divide(1, 5), const_12), multiply(divide(const_12, 2), divide(1, 2))), const_12)
drum x is 1 / 2 full of oil and drum y , which has twice the capacity of drum x , is 1 / 5 full of oil . if all of the oil in drum x is poured into drum y , then drum y will be filled to what capacity ?
"( 1 / 2 ) x = ( 1 / 4 ) y ( 1 / 4 ) y + ( 1 / 5 ) y = ( 9 / 20 ) y the answer is d ."
a = 1 / 5 b = a * 12 c = 12 / 2 d = 1 / 2 e = c * d f = b + e g = f / 12
a ) 10 , b ) 9 , c ) 15 , d ) 6 , e ) 12
c
multiply(divide(120, add(add(2, 6), 8)), 2)
the average age of 3 boys is 120 years and their ages are in proportion 2 : 6 : 8 . what is the age in years of the youngest boy ?
2 x + 6 x + 8 x = 120 x = 7.5 2 x = 15 answer : c
a = 2 + 6 b = a + 8 c = 120 / b d = c * 2
a ) s . 6000 , b ) s . 9000 , c ) s . 9900 , d ) s . 9990 , e ) s . 10000
a
add(subtract(42000, divide(multiply(25000, 5), 3)), subtract(divide(multiply(42000, 4), 7), 25000))
for 2 consecutive yrs , my incomes are in the ratio of 4 : 7 and expenses in the ratio of 3 : 5 . if my income in the 2 nd yr is rs . 42000 & my expenses in the first yr in rs . 25000 , my total savings for the two - year is
"sol . income in first year = * x 42000 = rs . 24000 expenses in second year = \ x 21000 = rs . 35000 total savings = total income - total expenses = ( 42000 + 24000 ) - ( 25000 + 35000 ) = 66000 - 60000 = rs . 6000 a"
a = 25000 * 5 b = a / 3 c = 42000 - b d = 42000 * 4 e = d / 7 f = e - 25000 g = c + f
a ) 8 , b ) 10 , c ) 9 , d ) 6 , e ) 5
d
divide(subtract(78, 54), subtract(54, 50))
rahul played weel in this season . his current batting averagge is 50 . if he score 78 runs in today match . his batting average will become 54 . how many matches had he played in this season .
"50 x + 78 = 54 ( x + 1 ) = > 4 x = 24 = > x = 6 answer : d"
a = 78 - 54 b = 54 - 50 c = a / b
a ) 1 / 8 , b ) 2 / 8 , c ) 3 / 8 , d ) 4 / 8 , e ) 5 / 8
e
add(divide(5, subtract(add(5, 12), 1)), divide(multiply(2, 2), subtract(add(5, 12), 1)))
in 2 bags , there are to be put together 5 red and 12 white balls , neither bag being empty . how must the balls be divided so as to give a person who draws 1 ball from either bag - the greatest chance of drawing a red ball ?
"greatest chance . 1 / 2 * 1 + 1 / 2 * 4 / 16 = 5 / 8 answer : e"
a = 5 + 12 b = a - 1 c = 5 / b d = 2 * 2 e = 5 + 12 f = e - 1 g = d / f h = c + g
a ) 5 minutes , b ) 6 minutes , c ) 8 minutes , d ) 9 minutes , e ) 10 minutes
b
divide(subtract(12, divide(12, divide(add(4, 12), subtract(12, 4)))), const_1)
a man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed , what is the time interval between consecutive buses ?
"let ' s say the distance between the buses is d . we want to determine interval = \ frac { d } { b } , where b is the speed of bus . let the speed of cyclist be c . every 12 minutes a bus overtakes cyclist : \ frac { d } { b - c } = 12 , d = 12 b - 12 c ; every 4 minutes cyclist meets an oncoming bus : \ frac { d } { ...
a = 4 + 12 b = 12 - 4 c = a / b d = 12 / c e = 12 - d f = e / 1
a ) s . 2900 , b ) s . 3570 , c ) s . 4500 , d ) s . 4550 , e ) s . 2500
a
subtract(multiply(8000, const_4), subtract(multiply(8900, const_4), 6500))
the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8900 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ?
"sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8900 = 35600 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 3600 salary of may is rs . 6500 salary of jan...
a = 8000 * 4 b = 8900 * 4 c = b - 6500 d = a - c
a ) 5 , b ) 8 , c ) 9 , d ) 10 , e ) 15
e
add(add(multiply(3, 4), multiply(5, 4)), subtract(multiply(4, 4), multiply(5, 3)))
( 3 x + 4 ) ( 2 x - 5 ) = ax ^ 2 + kx + n . what is the value of a - n + k ?
"expanding we have 6 x ^ 2 - 15 x + 8 x - 20 6 x ^ 2 - 7 x - 20 taking coefficients , a = 6 , k = - 7 , n = - 20 therefore a - n + k = 6 - ( - 20 ) - 11 = 26 - 11 = 15 the answer is e ."
a = 3 * 4 b = 5 * 4 c = a + b d = 4 * 4 e = 5 * 3 f = d - e g = c + f
a ) 24 % , b ) 25 % , c ) 27 % , d ) 28 % , e ) 35 %
c
divide(multiply(subtract(add(multiply(divide(multiply(280, 45), const_100), divide(add(const_100, 20), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280)
a shopkeeper has 280 kg of apples . he sells 45 % of these at 20 % profit and remaining 60 % at 30 % profit . find his % profit on total .
"if the total quantity was 100 then 45 x 20 % + 60 x 30 % = 27 this profit will remain same for any total quantity unless the % of products remains the same . hence ' c ' is the answer"
a = 280 * 45 b = a / 100 c = 100 + 20 d = c / 100 e = b * d f = 280 * 60 g = f / 100 h = 100 + 30 i = h / 100 j = g * i k = e + j l = k - 280 m = l * 100 n = m / 280
a ) $ 44,000 , b ) $ 54,000 , c ) $ 64,000 , d ) $ 84,000 , e ) $ 104,000
c
divide(multiply(const_100, multiply(const_100, add(const_1, 4))), add(divide(25, const_100), multiply(multiply(divide(25, const_100), subtract(const_1, divide(25, const_100))), const_2)))
the majority owner of a business received 25 % of the profit , with each of 4 partners receiving 25 % of the remaining profit . if the majority owner and two of the owners combined to receive $ 40,000 , how much profit did the business make ?
"let p be the total profit . p / 4 + 1 / 2 * ( 3 p / 4 ) = p / 4 + 3 p / 8 = 5 p / 8 = 40000 p = $ 64,000 the answer is c ."
a = 1 + 4 b = 100 * a c = 100 * b d = 25 / 100 e = 25 / 100 f = 25 / 100 g = 1 - f h = e * g i = h * 2 j = d + i k = c / j
a ) 4 % , b ) 10 % , c ) 83.33 % , d ) 90.33 % , e ) 20 %
c
multiply(divide(subtract(divide(subtract(const_100, 70), const_100), subtract(divide(40, const_100), multiply(divide(70, const_100), divide(50, const_100)))), divide(subtract(const_100, 70), const_100)), const_100)
70 % of the employees of a company are men . 50 % of the men in the company speak french and 40 % of the employees of the company speak french . what is % of the women in the company who do not speak french ?
"no of employees = 100 ( say ) men = 70 women = 30 men speaking french = 0.5 * 70 = 35 employees speaking french = 0.4 * 100 = 40 therefore women speaking french = 40 - 35 = 5 and women not speaking french = 30 - 5 = 25 % of women not speaking french = 25 / 30 * 100 = 83.33 % answer c"
a = 100 - 70 b = a / 100 c = 40 / 100 d = 70 / 100 e = 50 / 100 f = d * e g = c - f h = b - g i = 100 - 70 j = i / 100 k = h / j l = k * 100
a ) 10 , b ) 37 , c ) 27 , d ) 18 , e ) 19
a
subtract(add(const_10, 1), 1)
a and b are partners in a business . a contributes 1 / 4 of the capital for 15 months and b received 2 / 3 of the profit . for how long b ' s money was used ?
let the total profit be rs . z . then , b ' s share = rs . 2 z / 3 , a ' s share = rs . ( z - 2 z / 3 ) = rs . z / 3 . a : b = z / 3 : 2 z / 3 = 1 : 2 let the total capital be rs , x and suppose b ' s money was used for x months . then . ( 1 ( x ) / 4 * 15 ) / ( 3 x ) / 4 * y ) = 1 / 2 < = > y = ( 15 * 2 / 3 ) = 10 . t...
a = 10 + 1 b = a - 1
a ) 125 , b ) 126 , c ) 130 , d ) 148 , e ) 161
e
add(add(add(multiply(3, 15), multiply(3, 16)), multiply(3, 19)), 11)
3 * 15 + 3 * 16 + 3 * 19 + 11 = ?
"3 * 15 + 3 * 16 + 3 * 19 + 11 = 45 + 48 + 57 + 11 = 161 the answer is e ."
a = 3 * 15 b = 3 * 16 c = a + b d = 3 * 19 e = c + d f = e + 11
a ) 17 hr , b ) 19 hr , c ) 10 hr , d ) 14 hr , e ) 16 hr
d
inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 3), 1), 3))))
a pump can fill a tank with water in 2 hours . because of a leak , it took 2 1 / 3 hours to fill the tank . the leak can drain all the water of the tank in ?
"work done by the tank in 1 hour = ( 1 / 2 - 1 / 3 ) = 1 / 14 leak will empty the tank in 14 hrs . answer : d"
a = 1 / 2 b = 2 * 3 c = b + 1 d = c / 3 e = 1/(d) f = a - e g = 1/(f)
a ) 110 , b ) 1100 , c ) 9900 , d ) 10000 , e ) 12100
c
multiply(multiply(subtract(const_10, const_1), multiply(add(6, 4), subtract(16, 5))), const_10)
right triangle pqr is to be constructed in the xy - plane so that the right angle is at p and pr is parallel to the x - axis . the x - and y - coordinates of p , q and r are to be integers that satisfy the inequalities - 4 < = x < = 5 and 6 < = y < = 16 . how many different triangles with these properties could be cons...
"we know that p , q , r can only lie on a grid 10 x 11 let ' s start by fixing p ' x - coordinate : we have 10 possible places for it then we choose x - coordinate for r : since pr | | x - axis , r ' s x - coordinate can not coincide with p ' s x - coord , so we have only 9 possible choices left . then , we fix p ' s (...
a = 10 - 1 b = 6 + 4 c = 16 - 5 d = b * c e = a * d f = e * 10
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22
d
divide(multiply(multiply(5, 4), 3), 3)
sand is poured into a box so that the box is being filled at the rate of 3 cubic feet per hour . if the empty rectangular box is 5 feet long , 4 feet wide , and 3 feet deep , approximately how many hours does it take to fill the box ?
"the volume the box is : length * width * depth = 5 * 4 * 3 = 60 cubic feet . 60 cubic feet / 3 cubic feet per hour = 20 hours . it will take 20 hours to fill the box . the answer is d ."
a = 5 * 4 b = a * 3 c = b / 3
a ) 150 , b ) 180 , c ) 210 , d ) 240 , e ) 270
d
multiply(divide(270, add(4, 5)), 8)
jack and christina are standing 270 feet apart on a level surface . their dog , lindy , is standing next to christina . at the same time , they all begin moving toward each other . jack walks in a straight line toward christina at a constant speed of 4 feet per second and christina walks in a straight line toward jack ...
"the relative speed of jack and christina is 4 + 5 = 9 feet per second . the distance between them is 210 feet , hence they will meet in ( time ) = ( distance ) / ( relative speed ) = 270 / 7 = 30 seconds . for all this time lindy was running back and forth , so it covered ( distance ) = ( speed ) * ( time ) = 8 * 30 =...
a = 4 + 5 b = 270 / a c = b * 8
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18
d
multiply(divide(20, 10), divide(50, 10))
how many liters of water must be added to 20 liters of milk and water containing 10 % water to make it 50 % water ?
"by rule of alligation : 50 % - 10 % = 40 % 100 % - 50 % = 50 % quantity of pure water : quantity of the mixture = 4 : 5 there are 20 liters of mixture , so we need to add 16 liters of pure water . the answer is d ."
a = 20 / 10 b = 50 / 10 c = a * b
a ) 400 , b ) 200 , c ) 100 , d ) 150 , e ) 500
a
multiply(3200, divide(3, inverse(subtract(divide(const_1, 3), add(divide(const_1, 6), divide(const_1, 8))))))
man 1 alone can do a piece of work in 6 days and man 2 alone in 8 days . man 1 and man 2 undertook to do it for rs . 3200 . with the help of man 3 , they completed the work in 3 days . how much is to be paid to man 3 ?
man 3 1 day work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 3 - 7 / 24 = 1 / 24 man 1 : man 2 : man 3 = 1 / 6 : 1 / 8 : 1 / 24 = = > 4 : 3 : 1 man 3 share = 3 * 1 / 24 * 3200 = rs 400 answer a
a = 1 / 3 b = 1 / 6 c = 1 / 8 d = b + c e = a - d f = 1/(e) g = 3 / f h = 3200 * g
a ) 16.5 kg , b ) 66.5 kg , c ) 26.5 kg , d ) 56.5 kg , e ) 86.5 kg
b
divide(add(68, add(65, const_1)), const_2)
in arun ' s opinion , his weight is greater than 65 kg but leas than 72 kg . his brother does not agree with arun and he thinks that arun ' s weight is greater than 60 kg but less than 70 kg . his mother ' s view is that his weight can not be greater than 68 kg . if all of them are correct in their estimation , what is...
"let arun ' s weight be x kg . according to arun , 65 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 68 . the values satisfying all the above conditions are 66 and 67 . required average = ( 66 + 67 ) / 2 = 66.5 kg answer : b"
a = 65 + 1 b = 68 + a c = b / 2
a ) 2,000 , b ) 2,500 , c ) 4,000 , d ) 5,000 , e ) 6,000
b
multiply(divide(multiply(divide(500, 2), divide(500, 2)), subtract(555, 500)), 2)
shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $ 500 as interest . he invested the remaining in a bond that paid compound interest , interest being compounded annually , for the same 2 years at the same rate of interest and received $ 555 as interest . what was the v...
"so , we know that shawn received 20 % of the amount he invested in a year . we also know that in one year shawn received $ 250 , thus 0.2 x = $ 250 - - > x = $ 1,250 . since , he invested equal sums in his 2 bonds , then his total savings before investing was 2 * $ 1,250 = $ 2,500 . answer : b"
a = 500 / 2 b = 500 / 2 c = a * b d = 555 - 500 e = c / d f = e * 2
a ) 34778 , b ) 26888 , c ) 2899 , d ) 17600 , e ) 21725
e
divide(multiply(add(const_100, 10), add(divide(multiply(15500, const_100), subtract(const_100, 20)), add(125, 250))), const_100)
ramesh purchased a refrigerator for rs . 15500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 10 % if no discount was offered ?
"price at which the tv set is bought = rs . 15,500 discount offered = 20 % marked price = 15500 * 100 / 80 = rs . 19375 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 19375 + 375 = rs . 19750 the price at which the tv should be sold to get a profit of 10 % if no di...
a = 100 + 10 b = 15500 * 100 c = 100 - 20 d = b / c e = 125 + 250 f = d + e g = a * f h = g / 100
a ) 900 , b ) 8000 , c ) 1100 , d ) 111110 , e ) 14
d
add(divide(subtract(multiply(floor(divide(1000000, 9)), 9), multiply(add(floor(divide(10, 9)), const_1), 9)), 9), const_1)
how many numbers from 10 to 1000000 are exactly divisible by 9 ?
"10 / 9 = 1 and 1000000 / 9 = 111111 = = > 111111 - 1 = 111110 . answer : d"
a = 1000000 / 9 b = math.floor(a) c = b * 9 d = 10 / 9 e = math.floor(d) f = e + 1 g = f * 9 h = c - g i = h / 9 j = i + 1
a ) 150 , b ) 872 , c ) 287 , d ) 288 , e ) 212
a
multiply(divide(multiply(60, const_1000), const_3600), 9)
a train running at the speed of 60 km / hr crosses a pole in 9 seconds . find the length of the train ?
"speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 9 = 150 meter answer : a"
a = 60 * 1000 b = a / 3600 c = b * 9
a ) 46 m , b ) 66 m , c ) 26 m , d ) 56 m , e ) 32 m
e
divide(multiply(add(30, 50), const_2), 5)
a rectangular plot measuring 30 meters by 50 meters is to be enclosed by wire fencing . if the poles of the fence are kept 5 meters apart . how many poles will be needed ?
"perimeter of the plot = 2 ( 30 + 50 ) = 160 m no of poles = 160 / 5 = 32 m answer : e"
a = 30 + 50 b = a * 2 c = b / 5
a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 5 , d ) 4 / 21 , e ) 5 / 21
b
divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 4), 21), 4), const_2), 4), const_3))
a certain list consists of 21 different numbers . if n is in the list and n is 4 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 = 4 * x hence , a 1 + a 2 + . . + a 20 + n = 24 x so , the fraction asked = 4 x / 24 x = 1 / 6 ; answer : b ."
a = 1 * 1 b = 20 / 4 c = b + 21 d = c / 4 e = d * 2 f = e - 4 g = f - 3 h = a / g
a ) 42 , b ) 43 , c ) 44 , d ) 45 , e ) 58
e
add(subtract(80, multiply(12, 2)), 2)
a batsman in his 12 th innings makes a score of 80 and thereby increases his average by 2 runs . what is his average after the 12 th innings if he had never been β€˜ not out ’ ?
"let β€˜ x ’ be the average score after 12 th innings β‡’ 12 x = 11 Γ— ( x – 2 ) + 80 ∴ x = 58 answer e"
a = 12 * 2 b = 80 - a c = b + 2
a ) rs . 859.09 , b ) rs . 349.09 , c ) rs . 449.09 , d ) rs . 749.09 , e ) rs . 849.09
e
multiply(4500, subtract(power(divide(add(divide(5, const_2), const_100), const_100), multiply(3, const_2)), const_1))
what is the compound interest on rs . 4500 at 5 % p . a . compounded half - yearly for 3 1 / 2 years .
"compound interest : a = p ( 1 + r / n ) nt a = 5 , 349.09 c . i . > > 5 , 349.09 - 4500 > > rs . 849.09 answer : e"
a = 5 / 2 b = a + 100 c = b / 100 d = 3 * 2 e = c ** d f = e - 1 g = 4500 * f
a ) 120 , b ) 155 , c ) 180 , d ) 336 , e ) 456
a
add(const_100, multiply(5, const_4))
danny can divide his herd into 5 equal parts and also to 6 equal parts , but not to 9 equal parts . what could be the number of cows danny has in his herd ?
danny can divide his herd into 5 equal parts and also to 6 equal parts = > no of cows = multiple of 5 and 6 = multiple of 30 . only option a and c qualify . now 2 nd condition : but not to 9 equal parts : so it should not be a multiple of 9 . answer : a
a = 5 * 4 b = 100 + a
a ) 8 days , b ) 10 days , c ) 12 days , d ) 7 days , e ) 5 days
a
inverse(add(subtract(divide(const_1, 12), subtract(divide(const_1, 8), divide(const_1, 6))), subtract(divide(const_1, 8), divide(const_1, 6))))
a and b can do a work in 8 days , b and c can do it in 12 days ; a , b and c together can finish it in 6 days . a and c together will do it in ?
"a + b + c 1 day work = 1 / 6 a + b 1 day work = 1 / 8 b + c 1 day work = 1 / 12 a + c 1 day work = 2 * 1 / 6 - 1 / 8 + 1 / 12 = 1 / 3 - 5 / 24 = 3 / 24 = 1 / 8 a and c together will do the work in 8 days . answer is a"
a = 1 / 12 b = 1 / 8 c = 1 / 6 d = b - c e = a - d f = 1 / 8 g = 1 / 6 h = f - g i = e + h j = 1/(i)
a ) 29 , b ) 776 , c ) 66 , d ) 16 , e ) 99
d
subtract(multiply(40, divide(90, const_100)), multiply(divide(4, 5), 25))
how much is 90 % of 40 is greater than 4 / 5 of 25 ?
"( 90 / 100 ) * 40 – ( 4 / 5 ) * 25 36 - 20 = 16 answer : d"
a = 90 / 100 b = 40 * a c = 4 / 5 d = c * 25 e = b - d
a ) 210 , b ) 252 , c ) 250 , d ) 300 , e ) 420
c
divide(multiply(multiply(25, 42), 60), multiply(multiply(7, 6), 6))
a grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches . if the measurement of each soap box is 7 inches by 6 inches by 6 inches , then what is the maximum number of soap boxes that can be placed in each carton ?
"however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = 25 * 42 * 60 dimensions of the soap = 6 * 6 * 7 placing the 7 inch side along 42 inch side we get 6 soaps in a line and in a similar way 5 along 25 and 6 along...
a = 25 * 42 b = a * 60 c = 7 * 6 d = c * 6 e = b / d
a ) s . 7375 , b ) s . 8379 , c ) s . 9875 , d ) s . 10875 , e ) s . 11875
e
multiply(multiply(multiply(const_4, multiply(const_2, add(const_2, const_3))), multiply(const_100, multiply(const_2, add(const_2, const_3)))), divide(multiply(subtract(multiply(multiply(const_3, const_4), const_2), add(const_2, const_3)), multiply(const_100, multiply(const_2, add(const_2, const_3)))), add(multiply(mult...
suresh and ramesh started a business investing rs . 24,000 and rs . 40,000 respectively . out of total profit of rs . 19,000 , what is ramesh ' s share ?
explanation : ratio of suresh and ramesh ' s share = 24,000 : 40,000 = 3 : 5 ramesh ' s share = rs . ( 19000 x 5 / 8 ) = rs . 11875 answer : e
a = 2 + 3 b = 2 * a c = 4 * b d = 2 + 3 e = 2 * d f = 100 * e g = c * f h = 3 * 4 i = h * 2 j = 2 + 3 k = i - j l = 2 + 3 m = 2 * l n = 100 * m o = k * n p = 3 * 4 q = p * 2 r = 2 + 3 s = 2 * r t = 100 * s u = q * t v = 2 + 3 w = 2 * v x = 4 * w y = 2 + 3 z = 2 * y A = 100 * z B = x * A C = u + B D = o / C E = ...
a ) rs . 8000 , b ) rs . 4000 , c ) rs . 5000 , d ) rs . 6000 , e ) rs . 24000
e
divide(60, multiply(divide(5, const_100), divide(5, const_100)))
if difference between compound interest and simple interest on a sum at 5 % p . a . for 2 years is rs . 60 , then what is the sum ?
"c . i - s . i = 60 si = ( x * 2 * 5 ) / 100 = x / 10 ci = { x * ( 1 + 5 / 100 ) ^ 2 - x } = 41 x / 400 ci - si = 41 x / 400 - x / 10 = 60 x = 24000 answer : e"
a = 5 / 100 b = 5 / 100 c = a * b d = 60 / c
a ) $ 21,000 , b ) $ 18,000 , c ) $ 15,00 , d ) $ 4,500 , e ) $ 15,000
e
divide(multiply(divide(subtract(360, multiply(divide(6, const_100), 1000)), subtract(divide(10, const_100), divide(6, const_100))), const_2), const_1000)
salesperson a ' s compensation for any week is $ 360 plus 6 percent of the portion of a ' s total sales above $ 1000 for that week . salesperson b ' s compensation for any week is 10 percent of a ' s total sales for that week . for what amount of total weekly sales would both salepeople earn the same compensation ?
sometime , setting up an equation is an easy way to go with : 360 + 0.06 ( x - 1000 ) = 0.1 x x = 15,000 ans : e
a = 6 / 100 b = a * 1000 c = 360 - b d = 10 / 100 e = 6 / 100 f = d - e g = c / f h = g * 2 i = h / 1000
a ) 15 / 16 , b ) 7 / 8 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16
b
add(add(add(add(divide(1, 2), divide(divide(1, 2), 2)), divide(divide(divide(1, 2), 2), 2)), divide(divide(divide(divide(1, 2), 2), 2), 2)), divide(divide(divide(divide(divide(1, 2), 2), 2), 2), 2))
if 1 / 2 of the air in a tank is removed with each stroke of a vacuum pump , what fraction of the original amount of air has been removed after 3 strokes ?
"left after 1 st stroke = 1 / 2 left after 2 nd stroke = 1 / 2 * 1 / 2 = 1 / 4 left after 3 rd stroke = 1 / 2 * 1 / 4 = 1 / 8 so removed = 1 - 1 / 8 = 7 / 8 = b"
a = 1 / 2 b = 1 / 2 c = b / 2 d = a + c e = 1 / 2 f = e / 2 g = f / 2 h = d + g i = 1 / 2 j = i / 2 k = j / 2 l = k / 2 m = h + l n = 1 / 2 o = n / 2 p = o / 2 q = p / 2 r = q / 2 s = m + r
a ) 400 , b ) 420 , c ) 480 , d ) 500 , e ) 600
d
divide(300, divide(60, const_100))
if it is assumed that 60 percent of those who receive a questionnaire by mail will respond and 300 responses are needed , what is the minimum number of questionnaires that should be mailed ?
"minimum no of mail to be sent for getting 300 responses at 60 % = 300 / 0.6 = 500 option d"
a = 60 / 100 b = 300 / a
a ) 7.1 , b ) 8.2 , c ) 4.6 , d ) 6.0 , e ) 12.8
d
divide(divide(1000, const_1000), divide(multiply(10, const_60), const_3600))
a person crosses a 1000 m long street in 10 minutes . what is his speed in km per hour ?
"distance = 1000 meter time = 10 minutes = 10 x 60 seconds = 600 seconds speed = distance / time = 1000 / 600 = 1.67 m / s = 1.67 Γ£ β€” 18 / 5 km / hr = 6.0 km / hr answer : d"
a = 1000 / 1000 b = 10 * const_60 c = b / 3600 d = a / c
a ) 15.8 sec , b ) 14.9 sec , c ) 12.4 sec , d ) 16.8 sec , e ) 11.8 sec
d
divide(add(110, 170), multiply(60, const_0_2778))
how long does a train 110 m long travelling at 60 kmph takes to cross a bridge of 170 m in length ?
"d 16.8 sec d = 110 + 170 = 280 m s = 60 * 5 / 18 = 50 / 3 t = 280 * 3 / 50 = 16.8 sec"
a = 110 + 170 b = 60 * const_0_2778 c = a / b
a ) 2 second , b ) 3.6 second , c ) 5 second , d ) 5.5 second , e ) 6 second
b
divide(add(100, 200), divide(multiply(add(100, 200), const_1000), const_3600))
two trains from opposite directions are to cross each other . the length of two trains are 100 meter and 200 meter respectively . the speed of first train is 100 km / hour and second train 200 km / hour . in how much time will they cross each other ?
the crossing moment starts when the head of first compartment of two trains meet or touch . the process of crossing each other continues till 100 + 200 meter = 300 meter is covered . the speed will be 100 km / hour + 200 km / hour = 300 km / hour . 300 x 1000 meter covered in 3600 seconds . so , 300 meter covered in 3....
a = 100 + 200 b = 100 + 200 c = b * 1000 d = c / 3600 e = a / d
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 10
a
add(add(const_1, const_1), const_2)
let n be the greatest number that will divide 1305 , 4665 and 6905 , leaving the same remainder in each case . then sum of the digits in n is :
"explanation : n = h . c . f . of ( 4665 - 1305 ) , ( 6905 - 4665 ) and ( 6905 - 1305 ) = h . c . f . of 3360 , 2240 and 5600 = 1120 . sum of digits in n = ( 1 + 1 + 2 + 0 ) = 4 answer is a"
a = 1 + 1 b = a + 2
a ) 32 , b ) 24 , c ) 26 , d ) 28 , e ) 30
a
add(add(multiply(subtract(12, const_1), 2), divide(10, 2)), divide(10, 2))
in a garden , there are 10 rows and 12 columns of mango trees . the distance between the two trees is 2 metres and a distance of five metre is left from all sides of the boundary of the garden . what is the length of the garden ?
between the 12 mango trees , there are 11 gaps and each gap has 2 meter length also , 5 meter is left from all sides of the boundary of the garden . hence , length of the garden = ( 11 Γ£ β€” 2 ) + 5 + 5 = 32 meter answer is a .
a = 12 - 1 b = a * 2 c = 10 / 2 d = b + c e = 10 / 2 f = d + e
a ) 15060000 , b ) 0.001506 , c ) 0.01506 , d ) 1.506 e - 05 , e ) none of these
d
multiply(divide(15.06, 0.000001), const_100)
15.06 * 0.000001 = ?
"explanation : clearly after decimal 8 digits should be there . option d"
a = 15 / 6 b = a * 100