Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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the milk level in a rectangular box measuring 56 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons ) | "6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 56 * 25 * 1 / 2 = 700 feet ^ 3 of milk must be removed , which equals to 700 * 7.5 = 5250 gallons . answer : d ." | a ) 100 , b ) 250 , c ) 750 , d ) 5250 , e ) 5635 | d | multiply(multiply(multiply(56, 25), divide(1, const_2)), 7.5) | divide(n3,const_2)|multiply(n0,n1)|multiply(#0,#1)|multiply(n4,#2)| | general |
what is the sum of all even numbers from 1 to 501 ? | "explanation : 500 / 2 = 250 250 * 251 = 62750 answer : d" | a ) 122821 , b ) 281228 , c ) 281199 , d ) 62750 , e ) 128111 | d | divide(multiply(1, 501), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
during 2005 , a company produced an average of 2,500 products per month . how many products will the company need to produce from 2006 through 2008 in order to increase its monthly average for the period from 2005 through 2008 by 300 % over its 2005 average ? | "company produced 12 * 2500 = 30,000 products in 2005 . if company produces x products from 2006 to 2008 , then total amount of product produced in 4 years ( 2005 through 2008 ) is x + 30,000 . the gives the average of ( x + 30,000 ) / 4 . this average needs to be 200 % higher than that in 2005 . in math terms , 30,000... | a ) 450,000 , b ) 475,000 , c ) 500,000 , d ) 525,000 , e ) 550,000 | a | multiply(300, divide(subtract(2008, 2005), const_2)) | subtract(n3,n0)|divide(#0,const_2)|multiply(n6,#1)| | general |
how many quarters are equal to 3 dollars ? | "3 * 4 = 12 quarters answer : c" | a ) 1 , b ) 8 , c ) 12 , d ) 9 , e ) 7 | c | multiply(3, const_4) | multiply(n0,const_4)| | general |
if k ^ 3 is divisible by 60 , what is the least possible value of integer k ? | "60 = 2 ^ 2 * 3 * 5 therefore k must include at least 2 * 3 * 5 = 30 . the answer is b ." | a ) 12 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | b | divide(divide(60, const_2), const_2) | divide(n1,const_2)|divide(#0,const_2)| | general |
the purchase price of an article is $ 48 . in order to include 20 % of cost for overhead and to provide $ 12 of net profit , the markup should be | "cost price of article = 48 $ % of overhead cost = 20 net profit = 12 $ we need to calculate % markup net profit as % of cost price = ( 12 / 48 ) * 100 = 25 % total markup should be = 25 + 20 = 45 % answer e" | a ) 15 % , b ) 25 % , c ) 35 % , d ) 40 % , e ) 45 % | e | add(multiply(divide(12, 48), const_100), 20) | divide(n2,n0)|multiply(#0,const_100)|add(n1,#1)| | gain |
pipe a can fill a tank in 10 hrs and pipe b can fill it in 8 hrs . if both the pipes are opened in the empty tank . there is an outlet pipe in 3 / 4 th of the tank . in how many hours will it be fill 3 / 4 th of that tank ? | part filled a in 1 hr = ( 1 / 10 ) part filled b in 1 hr = ( 1 / 8 ) part filled by ( a + b ) together in 1 hr = ( 1 / 10 ) + ( 1 / 8 ) = 18 / 80 so , the tank will be full in 80 / 18 hrs . time taken to fill exact 3 / 4 th of the tank = ( 80 / 18 ) * ( 3 / 4 ) = 3.20 hrs answer : d | a ) 3 hr , b ) 3.10 hr , c ) 3.15 hr , d ) 3.20 hr , e ) 3.30 hr | d | multiply(divide(3, 4), inverse(add(divide(const_1, 10), divide(const_1, 8)))) | divide(n2,n3)|divide(const_1,n0)|divide(const_1,n1)|add(#1,#2)|inverse(#3)|multiply(#0,#4) | physics |
a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 30 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 30 = 220000 + 30 = 220030 e" | a ) 122443 , b ) 154546 , c ) 165454 , d ) 186545 , e ) 220030 | e | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 30) | add(n0,n1)|subtract(n0,n1)|multiply(n2,#0)|multiply(#2,#1)|add(n3,#3)| | general |
there are some rabbits and peacocks in a zoo . the total number of their heads is 60 and total number of their legs is 192 . find the number of total rabbits ? | let the number of rabbits and peacocks be ' r ' and ' p ' respectively . as each animal has only one head , so r + p = 60 - - - ( 1 ) each rabbit has 4 legs and each peacock has 2 legs . total number of legs of rabbits and peacocks , 4 r + 2 p = 192 - - - ( 2 ) multiplying equation ( 1 ) by 2 and subtracting it from eq... | a ) 36 , b ) 77 , c ) 88 , d ) 99 , e ) 22 | a | divide(subtract(192, multiply(const_2, 60)), subtract(const_4, const_2)) | multiply(n0,const_2)|subtract(const_4,const_2)|subtract(n1,#0)|divide(#2,#1) | general |
two unbiased coins are tossed . what is probability of getting at most one tail ? | "explanation : total 4 cases = [ hh , tt , th , ht ] favourable cases = [ hh , th , ht ] please note we need atmost one tail , not atleast one tail . so probability = 3 / 4 answer : d" | a ) 12 , b ) 13 , c ) 32 , d ) 34 , e ) none of these | d | negate_prob(divide(const_1, power(const_2, const_3))) | power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)| | probability |
when positive integer n is divided by 5 , the remainder is 2 . when n is divided by 11 , the remainder is 8 . what is the smallest positive integer k such that k + n is a multiple of 55 ? | "n = 5 p + 2 = 11 q + 8 n + 3 = 5 p + 5 = 11 q + 11 n + 3 is a multiple of 5 and 11 , so it is a multiple of 55 . the answer is b ." | a ) 2 , b ) 3 , c ) 6 , d ) 7 , e ) 10 | b | subtract(55, reminder(const_4.0, 11)) | reminder(const_4.0,n2)|subtract(n4,#0)| | general |
a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 10 hour to row to a place and come back , how far is the place ? | "speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x / 12 + x / 8 = 10 2 x + 3 x = 240 x = 48 km answer is e" | a ) 24 km , b ) 30 km , c ) 48 km , d ) 12 km , e ) 48 km | e | divide(multiply(multiply(subtract(10, 2), add(10, 2)), 10), add(subtract(10, 2), add(10, 2))) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|multiply(n2,#3)|divide(#4,#2)| | physics |
what is the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ? | "solution length of largest tile = h . c . f . of 1517 cm & 902 cm = 41 cm . area of each tile = ( 41 x 41 ) cm 2 β΄ required number of tiles = [ 1517 x 902 / 41 x 41 ] = 814 . answer c" | a ) 724 , b ) 804 , c ) 814 , d ) 844 , e ) none | c | divide(multiply(add(multiply(9, const_100), 2), multiply(15, const_100)), power(2, const_2)) | multiply(n2,const_100)|multiply(n0,const_100)|power(n3,const_2)|add(n3,#0)|multiply(#3,#1)|divide(#4,#2)| | physics |
the average of first 8 prime numbers is ? | "sum of first 8 prime nos . = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = 77 average = 77 / 8 = 9.625 answer : b" | a ) 10.11 , b ) 9.625 , c ) 12.11 , d ) 13.11 , e ) 14.11 | b | add(8, const_1) | add(n0,const_1)| | general |
from a group of 3 boys and 3 girls , 2 children are to be randomly selected . what is the probability that 1 boy and 1 girl will be selected ? | "the total number of ways to choose 2 children from 6 is 6 c 2 = 15 the number of ways to choose 1 boy and 1 girl is 3 * 3 = 9 p ( 1 boy and 1 girl ) = 9 / 15 the answer is d ." | a ) 1 / 10 , b ) 4 / 9 , c ) 1 / 2 , d ) 9 / 15 , e ) 2 / 3 | d | divide(multiply(choose(3, const_2), choose(3, const_2)), choose(add(3, 3), 2)) | add(n0,n0)|choose(n0,const_2)|choose(n0,const_2)|choose(#0,n2)|multiply(#1,#2)|divide(#4,#3)| | probability |
the average of first five multiples of 7 is ? | "average = 7 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 105 / 5 = 21 . answer : e" | a ) 6 , b ) 8 , c ) 9 , d ) 5 , e ) 21 | e | add(7, const_1) | add(n0,const_1)| | general |
in one hour , a boat goes 21 km / hr along the stream and 11 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : | "explanation : let the speed downstream be a km / hr and the speed upstream be b km / hr , then speed in still water = 1 / 2 ( a + b ) km / hr rate of stream = 1 / 2 ( a β b ) km / hr speed in still water = 1 / 2 ( 21 + 11 ) kmph = 16 kmph . answer : option e" | a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 16 kmph | e | stream_speed(21, 11) | stream_speed(n0,n1)| | physics |
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 / 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) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n1,#4)| | physics |
( 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 ) 1 , b ) 100 , c ) 229 , d ) 329 , e ) 337 | e | subtract(power(169, 2), power(168, 2)) | power(n0,n1)|power(n2,n1)|subtract(#0,#1) | general |
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 ) 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) | divide(n2,const_100)|multiply(n0,#0)|add(n0,#1)| | general |
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 ) a ) 4 , b ) b ) 1 , c ) c ) 2 , d ) d ) 3 , e ) e ) 5 | a | subtract(23, reminder(1054, 23)) | reminder(n0,n1)|subtract(n1,#0)| | general |
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 ) 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_... | divide(n1,n2)|divide(n3,n4)|multiply(n3,const_4)|multiply(#2,const_1000)|multiply(#0,#3)|subtract(#3,#4)|multiply(#1,#5)|subtract(#5,#6)|divide(#7,const_1000)|divide(#8,const_4)|multiply(#9,n3)| | general |
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 , b ) 12 , c ) 15 , d ) 11 , e ) 20 | d | add(divide(90, subtract(10, 1)), 1) | subtract(n2,n0)|divide(n3,#0)|add(n0,#1)| | general |
how many 4 x 4 x 4 cubes could fit in a 9 x 13 x 16 box ? | the answer is c ) 29 . 9 x 13 x 16 gives the box an area of 1,872 . if you divide 1,872 by 64 ( the total of 4 x 4 x 4 ) , you get 29.25 . this means 29 full boxes will fit in the larger box . | a ) 22 , b ) 28 , c ) 29 , d ) 31 , e ) 32 | c | volume_rectangular_prism(9, 13, 16) | volume_rectangular_prism(n3,n4,n5)| | geometry |
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 ) 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) | add(n0,n4)|add(n2,n4)|divide(#0,#1)|add(n1,#2)| | physics |
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... | "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 ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 12 | e | divide(multiply(multiply(multiply(const_4, 2), 2), 1), multiply(multiply(2, 2), 1)) | multiply(n5,const_4)|multiply(n1,n5)|multiply(n1,#0)|multiply(n3,#1)|multiply(n3,#2)|divide(#4,#3)| | general |
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 ) 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)))) | multiply(n0,const_2)|divide(const_1,#0)|subtract(const_1,#1)|divide(const_1,#2)| | physics |
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 ) 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) | multiply(n0,n1)|multiply(n0,n3)|multiply(n0,n5)|divide(#0,n2)|divide(#1,n4)|divide(#2,n6)|divide(#2,n4)|add(#3,#4)|multiply(n2,#6)|add(#7,#5)|subtract(#9,#8)|subtract(#10,n0) | general |
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 ) 15 , b ) 7 , c ) 10 , d ) 12 , e ) 14 | a | subtract(30, const_4) | subtract(n0,const_4)| | general |
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 ) 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) | multiply(n1,const_1000)|divide(#0,const_3600)|multiply(#1,const_60)|subtract(#2,n0) | physics |
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 ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | c | add(10, add(const_3, const_4)) | add(const_3,const_4)|add(#0,n1)| | general |
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 ) 75 , b ) 180 , c ) 198 , d ) 216 , e ) 252 | a | divide(multiply(add(subtract(90, 10), subtract(90, 70)), 30), 40) | subtract(n3,n2)|subtract(n3,n4)|add(#0,#1)|multiply(n6,#2)|divide(#3,n9) | physics |
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 ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | subtract(5, reminder(4, 8)) | reminder(n2,n3)|subtract(n1,#0)| | general |
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 ) 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)))) | add(const_2,const_3)|divide(const_1,const_2)|divide(const_2,#0)|subtract(n0,#0)|subtract(#1,#2)|divide(#3,#4) | general |
| 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 ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | divide(multiply(add(4, 3), const_2), 5) | add(n0,n1)|multiply(#0,const_2)|divide(#1,n2)| | general |
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 ) 90 , b ) 120 , c ) 150 , d ) 180 , e ) 160 | a | multiply(divide(const_60, 40), 1) | divide(const_60,n0)|multiply(n1,#0)| | physics |
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 ) 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)) | add(n1,n3)|multiply(n0,#0)|add(n1,#1)|divide(#2,#2) | general |
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 ) 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)))) | multiply(n0,const_100)|multiply(n2,const_100)|multiply(const_2,const_4)|add(n1,#0)|add(n3,#1)|add(#2,const_2)|multiply(#3,#4)|subtract(n1,#5)|multiply(#7,#7)|divide(#6,#8) | physics |
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 ) 42 , b ) 44 , c ) 46 , d ) 36 , e ) 50 | d | divide(180, subtract(divide(180, 30), divide(60, 60))) | divide(n0,n1)|divide(n2,n2)|subtract(#0,#1)|divide(n0,#2)| | physics |
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 ) 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) | divide(n2,n1)|multiply(#0,const_2)|add(n1,#1)|divide(#2,const_2) | physics |
a cube has a volume of 64 cubic feet . if a similar cube is twice as long , twice as wide , and twice as high , then the volume , in cubic feet of such cube is ? | volume = 64 = side ^ 3 i . e . side of cube = 4 new cube has dimensions 8 , 8 , and 8 as all sides are twice of teh side of first cube volume = 8 * 8 * 8 = 512 square feet answer : option e | ['a ) 24', 'b ) 48', 'c ) 64', 'd ) 80', 'e ) 512'] | e | volume_cube(multiply(const_2, cube_edge_by_volume(64))) | cube_edge_by_volume(n0)|multiply(#0,const_2)|volume_cube(#1) | geometry |
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 ) 30 , b ) 5 , c ) 10 , d ) 36 , e ) 25 | b | sqrt(25) | sqrt(n2) | physics |
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 ) 6 , b ) 12 , c ) 16 , d ) 18 , e ) 24 | b | divide(subtract(90, const_10), const_10) | subtract(n1,const_10)|divide(#0,const_10)| | other |
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 ) 14000 , b ) 14400 , c ) 16129 , d ) 14600 , e ) 14700 | c | add(10000, multiply(divide(multiply(10000, 27), const_100), 2)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,n2)|add(n0,#2)| | gain |
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 ) 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) | divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4)| | gain |
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 ) 20 , b ) 23 , c ) 32 , d ) 19 , e ) 29 | b | divide(subtract(add(30, add(30, 5)), multiply(5, 5)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
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 ) s . 1200 , b ) s . 1300 , c ) s . 600 , d ) s . 800 , e ) s . 1200 | c | divide(300, multiply(divide(5, const_100), 10)) | divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)| | gain |
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 ) 2000 , b ) 3000 , c ) 3500 , d ) 3800 , e ) 4000 | b | multiply(630, divide(const_100, subtract(36, 15))) | subtract(n1,n2)|divide(const_100,#0)|multiply(n3,#1) | general |
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 ) 4 / 15 , b ) 7 / 30 , c ) 11 / 30 , d ) 5 / 18 , e ) 7 / 18 | c | add(divide(1, 5), divide(1, 6)) | divide(n3,n1)|divide(n3,n2)|add(#0,#1)| | physics |
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 ) 187 , b ) 197 , c ) 207 , d ) 136 , e ) 227 | d | subtract(multiply(11, 13), add(const_10, const_1)) | add(const_1,const_10)|multiply(n0,n1)|subtract(#1,#0)| | general |
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 ) 5 seconds , b ) 4.5 seconds , c ) 3 seconds , d ) 2 seconds , e ) none of these | d | divide(100, multiply(180, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
a train consists of 12 boggies , each boggy 15 metres long . the train crosses a telegraph post in 9 seconds . due to some problem , 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 ) 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)) | multiply(n0,n1)|divide(#0,n2)|subtract(#0,n1)|divide(#2,#1)| | physics |
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 ) 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))) | subtract(n0,n1)|subtract(n0,const_1)|divide(#0,n0)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)| | general |
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 ) 8000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000 | c | divide(power(divide(500, const_2), const_2), subtract(512.50, 500)) | divide(n0,const_2)|subtract(n1,n0)|power(#0,const_2)|divide(#2,#1)| | gain |
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 ) 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) | multiply(n0,n1)|multiply(n0,n0)|subtract(#0,n3)|subtract(#1,const_1)|divide(#2,#3)|multiply(n0,#4)|subtract(n1,#5)|add(#4,#6)|divide(#7,n4)| | general |
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 ) 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)))) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)| | general |
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 ) 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) | divide(n1,const_60)|multiply(n0,#0)|multiply(#1,const_1000)| | gain |
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 ) 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) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)| | gain |
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 ) 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) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
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 ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 4 | a | multiply(divide(subtract(4800, 4400), 4800), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
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 ) - 3 , b ) - 1 , c ) - 1 / 3 , d ) 0 , e ) undefined | a | divide(add(divide(subtract(4, 8), 2), 5), divide(add(2, 6), 2)) | add(n1,n3)|subtract(n2,n4)|divide(#1,n1)|divide(#0,n1)|add(#2,n0)|divide(#4,#3)| | general |
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 ) 7,000 , b ) 24,000 , c ) 40,000 , d ) 100,000 , e ) 158,400 | e | multiply(const_4, const_10) | multiply(const_10,const_4) | physics |
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 ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 900 | b | divide(150, subtract(const_1, divide(3, 4))) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)| | general |
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 ) 12 days , b ) 14 days , c ) 16 days , d ) 18 days , e ) 20 days | c | divide(multiply(8, 10), 5) | multiply(n0,n1)|divide(#0,n2) | physics |
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 ) 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) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
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 ) 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)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
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 ) 50 , b ) 40 , c ) 60 , d ) 100 , e ) 70 | c | divide(150, add(divide(150, const_100), const_1)) | divide(n0,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
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 ... | 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))) | add(n0,n1)|multiply(n2,const_1000)|multiply(n3,const_1000)|divide(#1,const_3600)|divide(#2,const_3600)|add(#3,#4)|divide(#0,#5) | physics |
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 ) 30 , b ) 40 , c ) 60 , d ) 50 , e ) none | a | divide(subtract(multiply(12, 120), multiply(90, 15)), subtract(15, 12)) | multiply(n1,n2)|multiply(n0,n3)|subtract(n3,n1)|subtract(#0,#1)|divide(#3,#2) | physics |
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 ) 150 , b ) 140 , c ) 160 , d ) 170 , e ) 180 | c | subtract(multiply(120, const_2), multiply(100, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
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 ) 28 sec , b ) 23 sec , c ) 24 sec , d ) 25 sec , e ) 26 sec | a | divide(add(150, 130), multiply(36, const_0_2778)) | add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)| | physics |
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 ) $ 37.80 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 39.50 , e ) $ 40.60 | b | add(multiply(35, divide(40, const_100)), 35) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
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 ) 45 , b ) 38 , c ) 72 , d ) 74 , e ) 78 | c | add(multiply(subtract(18, const_1), 5), 4) | subtract(n0,const_1)|multiply(n2,#0)|add(n1,#1)| | general |
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 ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | divide(18, add(2, const_1)) | add(n0,const_1)|divide(n3,#0)| | general |
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 ) 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) | add(n1,n2)|add(n3,#0)|subtract(const_100,#1)|multiply(n0,#2)|divide(#3,const_100)| | gain |
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 ) 35 , b ) 37 , c ) 39 , d ) 41 , e ) 42 | e | add(multiply(const_2, 10), add(12, 10)) | add(n0,n3)|multiply(n0,const_2)|add(#0,#1)| | general |
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 ) 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) | divide(n0,n3)|divide(const_12,n1)|divide(n0,n1)|multiply(#0,const_12)|multiply(#1,#2)|add(#3,#4)|divide(#5,const_12)| | general |
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 ) 10 , b ) 9 , c ) 15 , d ) 6 , e ) 12 | c | multiply(divide(120, add(add(2, 6), 8)), 2) | add(n2,n3)|add(n4,#0)|divide(n1,#1)|multiply(n2,#2) | general |
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 ) 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)) | multiply(n4,n7)|multiply(n1,n6)|divide(#0,n3)|divide(#1,n2)|subtract(n6,#2)|subtract(#3,n7)|add(#4,#5)| | general |
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 ) 8 , b ) 10 , c ) 9 , d ) 6 , e ) 5 | d | divide(subtract(78, 54), subtract(54, 50)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | general |
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 ) 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))) | add(n1,n2)|multiply(n0,n0)|subtract(#0,n3)|divide(n1,#2)|divide(#1,#2)|add(#3,#4)| | general |
a circular well with a diameter of 6 metres , is dug to a depth of 24 metres . what is the volume of the earth dug out ? | "solution volume = Ο r 2 h βΉ = βΊ ( 22 / 7 Γ 3 Γ 3 Γ 24 ) m 3 βΉ = βΊ 678.6 m 3 . answer a" | a ) 678.6 m 3 , b ) 36 m 3 , c ) 40 m 3 , d ) 44 m 3 , e ) none | a | volume_cylinder(divide(6, const_2), 24) | divide(n0,const_2)|volume_cylinder(#0,n1)| | geometry |
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 ) 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) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)|divide(n0,#2)|subtract(n0,#3)|divide(#4,const_1)| | physics |
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 ) 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)) | multiply(n0,const_4)|multiply(n1,const_4)|subtract(#1,n2)|subtract(#0,#2)| | general |
( 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 ) 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))) | multiply(n0,n1)|multiply(n1,n3)|multiply(n1,n1)|multiply(n0,n3)|add(#0,#1)|subtract(#2,#3)|add(#4,#5)| | general |
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 ) 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) | add(n2,const_100)|add(n4,const_100)|multiply(n0,n1)|multiply(n0,n3)|divide(#2,const_100)|divide(#0,const_100)|divide(#3,const_100)|divide(#1,const_100)|multiply(#4,#5)|multiply(#6,#7)|add(#8,#9)|subtract(#10,n0)|multiply(#11,const_100)|divide(#12,n0)| | gain |
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 ) $ 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))) | add(const_1,n1)|divide(n0,const_100)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#2,const_100)|multiply(#1,#3)|multiply(#5,const_2)|add(#1,#6)|divide(#4,#7)| | gain |
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 ) 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) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|subtract(#0,#5)|subtract(#4,#6)|divide(#7,#4)|multiply(#8,const_100)| | gain |
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 , b ) 37 , c ) 27 , d ) 18 , e ) 19 | a | subtract(add(const_10, 1), 1) | add(n0,const_10)|subtract(#0,n0) | gain |
3 * 15 + 3 * 16 + 3 * 19 + 11 = ? | "3 * 15 + 3 * 16 + 3 * 19 + 11 = 45 + 48 + 57 + 11 = 161 the answer is e ." | a ) 125 , b ) 126 , c ) 130 , d ) 148 , e ) 161 | e | add(add(add(multiply(3, 15), multiply(3, 16)), multiply(3, 19)), 11) | multiply(n0,n1)|multiply(n0,n3)|multiply(n0,n5)|add(#0,#1)|add(#3,#2)|add(n6,#4)| | general |
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 ) 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)))) | divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)| | physics |
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 ) 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) | add(n0,n2)|subtract(n3,n1)|subtract(const_10,const_1)|multiply(#0,#1)|multiply(#3,#2)|multiply(#4,const_10)| | general |
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 ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | d | divide(multiply(multiply(5, 4), 3), 3) | multiply(n1,n2)|multiply(n3,#0)|divide(#1,n0)| | geometry |
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 ) 150 , b ) 180 , c ) 210 , d ) 240 , e ) 270 | d | multiply(divide(270, add(4, 5)), 8) | add(n1,n2)|divide(n0,#0)|multiply(n3,#1)| | physics |
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 ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | d | multiply(divide(20, 10), divide(50, 10)) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)| | gain |
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 ) 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)))))) | divide(const_1,n7)|divide(const_1,n1)|divide(const_1,n3)|add(#1,#2)|subtract(#0,#3)|inverse(#4)|divide(n7,#5)|multiply(n6,#6) | physics |
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 ) 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) | add(n0,const_1)|add(n4,#0)|divide(#1,const_2)| | general |
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 ) 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) | divide(n1,n0)|subtract(n3,n1)|multiply(#0,#0)|divide(#2,#1)|multiply(n0,#3)| | gain |
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 ) 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) | add(n4,const_100)|add(n2,n3)|multiply(n0,const_100)|subtract(const_100,n1)|divide(#2,#3)|add(#1,#4)|multiply(#0,#5)|divide(#6,const_100)| | gain |
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 ) 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) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general |
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