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
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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_4), 1) | 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 meet β distance travelled by train 2 in ( x - 1 ) hours = speed Γ time = 75 ( x - 1 ) total distance travelled = distance travelled by train 1 + distance travelled by train 2 = > 330 = 60 x + 75 ( x - 1 ) = > 12 x + 15 ( x - 1 ) = 66 = > 12 x + 15 x - 15 = 66 = > 27 x = 66 + 15 = 81 = > 3 x = 9 = > x = 3 hence the trains meet 3 hours after 8 a . m . , i . e . at 11 a . m . answer : option b" | 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 high - low β divides a deck of 52 playing cards into 2 types , β high β cards and β low β cards . there are an equal number of β high β cards and β low β cards in the deck and β high β cards are worth 2 points , while β low β cards are worth 1 point . if you draw cards one at a time , how many ways can you draw β high β and β low β cards to earn 5 points if you must draw exactly 2 β low β cards ? | "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 ! . . . where n represents the number of elements in the group and a , b , etc . represent the number of times that repeating elements are repeated . here there are 4 elements and thelowcard is repeated 2 times . as a result , the formula is : 4 ! / 2 ! which represents ( 4 * 3 * 2 * 1 ) / ( 2 * 1 ) which simplifies to just 4 , giving you answer e ." | 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 = 1.25 hours . d" | 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 clubs . the question provides y for us ( 150 ) . 750 = 400 + x + 300 x = 50 e | 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 β 60 = > 800 + x = 20 β 65 = 1300 = > x = 1300 β 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 speed in meters per minute ? | 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 | β | 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 bit easier to understand . when x = - 10 i . ) ( x + 3 ) ( - 10 + 3 ) ( - 7 ) ii . ) ( 4 - x ) ( 4 - [ - 10 ] ) ( double negative , so it becomes positive ) ( 4 + 10 ) ( 14 ) iii . ) ( 8 + x ) ( 8 + - 10 ) ( - 2 ) in other words , when x < - 8 , ( x + 3 ) and ( 8 + x ) are negative . to solve problems like this , we need to check for the sign change . here is how i do it step by step . i . ) | x + 3 | - | 4 - x | = | 8 + x | ii . ) ignore absolute value signs ( for now ) and find the values of x which make ( x + 3 ) , ( 4 - x ) and ( 8 + x ) = to zero as follows : ( x + 3 ) x = - 3 ( - 3 + 3 ) = 0 ( 4 - x ) x = 4 ( 4 - 4 ) = 0 ( 8 + x ) x = - 8 ( 8 + - 8 ) = 4 e" | 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 checked as every 4 th value will give us answer so next value will be 15 . . ans 1 . . b | 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)
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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
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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 β 9 = 20 m / s now , length of train = 10 Γ£ β 15 = 150 m Γ’ Λ Β΄ required time = 150 Γ’ Β β 20 = 7.5 sec . answer b" | a = 12 * 15
b = a - 15
c = 12 * 15
d = c / 9
e = b / d
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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 = 5512.50 - 5000 = 512.50 hence , it satisfies the question . hence c is the correct answer" | 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
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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
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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 β 1000 / 60 β 60 = 15 m / s speed of b = 36 β 1000 / 60 β 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 each other ? the length that needs to be crossed = 175 + 150 = 325 m . time taken = 325 / 25 = 13 s . correct answer e . | 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 β 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 spent on the chicken feed and goat feed combined ? | "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 both the chicken feed and the goat feed , then he would he have spent 17.5 + 21 = $ 38.5 . answer : b ." | 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 β 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 } { b + c } = 4 , d = 4 b + 4 c ; d = 12 b - 12 c = 4 b + 4 c , - - > b = 2 c , - - > d = 12 b - 6 b = 6 b . interval = \ frac { d } { b } = \ frac { 6 b } { b } = 6 answer : b ( 6 minutes ) ." | 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 january = rs . 2900 answer : a" | 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 . thus , b ' s money was used for 10 months . answer : a ) 10 months | 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 constructed ? | "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 ( and r ' s too ) y - coordinate : 11 possible choices . finally , after we fixed p and r , q ' s x - coordinate must be the same as p ' s ( because p is the right angle and pr | | x - axis ) , so the only choice is for y - coordinate , and we have 10 possible choices left . so , total number of different triangles : 10 * 9 * 11 * 10 = 9,900 answer : c" | 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 at a constant speed of 5 feet per second . lindy runs at a constant speed of 8 feet per second from christina to jack , back to christina , back to jack , and so forth . what is the total distance , in feet , that lindy has traveled when the three meet at one place ? | "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 = 240 feet . answer : d ." | 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 the average of diferent probable weights of arun ? | "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 value of his total savings before investing in these two bonds ? | "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 discount was offered = 19750 * 110 / 100 = rs . 21725 answer : e" | 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 β ? | "let β x β be the average score after 12 th innings β 12 x = 11 Γ ( x β 2 ) + 80 β΄ 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 β ( 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 60 we get = 5 x 5 x 10 = 250 so the question is why this particular arrangement , in order to maximize number of soaps we need to minimize the space wasted and this is the only config where we dont waste any space so we can expect the maximum number the answer is ( c )" | 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(multiply(multiply(const_3, const_4), const_2), multiply(const_100, multiply(const_2, add(const_2, const_3)))), multiply(multiply(const_4, multiply(const_2, add(const_2, const_3))), multiply(const_100, multiply(const_2, add(const_2, const_3))))))) | 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 = g * D
|
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 3.6 seconds . answer : b . | 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
|
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