Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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80 % of the population of a village is 32000 . the total population of the village is ? | "x * ( 80 / 100 ) = 32000 x = 400 * 100 x = 40000 answer : a" | a ) 40000 , b ) 24000 , c ) 26682 , d ) 29973 , e ) 12312 | a | multiply(divide(const_100, 80), 32000) | divide(const_100,n0)|multiply(n1,#0)| | general |
a number x is 4 times another number y . the percentage that y is less than x is | "say y = 1 and x = 4 . then y = 1 is less than x = 4 by ( 4 - 1 ) / 4 * 100 = 3 / 4 * 100 = 75 % . answer : b ." | a ) 12.5 % , b ) 75 % , c ) 80 % , d ) 11 % , e ) 1 % | b | multiply(divide(subtract(4, const_1), 4), const_100) | subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)| | general |
a trader has 1600 kg of sugar . he sells a part at 8 % profit and the rest at 12 % profit . if he gains 11 % on the whole , find the quantity sold at 12 % ? | by rule of alligation , % profit by selling part 1 % profit by selling part 2 8 12 net % profit 11 12 - 11 = 1 11 - 8 = 3 = > quantity of part 1 : quantity of part 2 = 1 : 3 given that total quantity = 1600 kg hence , quantity of part 2 ( quantity sold at 12 % profit ) = 1600 × 3 4 = 1200 b ) | a ) 1000 , b ) 1200 , c ) 1400 , d ) 1450 , e ) 1500 | b | divide(subtract(multiply(1600, 11), multiply(1600, 8)), subtract(12, 8)) | multiply(n0,n3)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2) | gain |
john spent 40 percent of his earning last month on rent and 30 percent less than what he spent on rent to purchase a new dishwasher . what percent of last month ' s earning did john have left over ? | say john ' s earning last month was $ 100 . john spent 40 percent of his earning last month on rent - - > $ 40 on rent ; 30 percent less than what he spent on rent to purchase a new dishwasher - - > $ 40 * 0.7 = $ 28 on the dishwasher . left over amount 100 - ( 40 + 28 ) = $ 32 . answer : b . | a ) 30 % , b ) 32 % , c ) 45 % , d ) 68 % , e ) 70 % | b | subtract(const_100, add(multiply(divide(subtract(const_100, 30), const_100), multiply(const_100, divide(40, const_100))), multiply(const_100, divide(40, const_100)))) | divide(n0,const_100)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#0,const_100)|multiply(#2,#3)|add(#4,#3)|subtract(const_100,#5) | gain |
10 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 4 children take to complete the work ? | "1 women ' s 1 day work = 1 / 70 1 child ' s 1 day work = 1 / 140 ( 5 women + 4 children ) ' s 1 day work = ( 5 / 10 + 4 / 140 ) = 1 / 10 5 women and 4 children will complete the work in 10 days . e" | a ) 4 , b ) 5 , c ) 7 , d ) 8 , e ) 10 | e | inverse(add(divide(5, multiply(10, 7)), divide(10, multiply(10, 14)))) | multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)| | physics |
lloyd normally works 7.5 hours per day and earns $ 4.50 per hour . for each hour he works in excess of 7.5 hours on a given day , he is paid 2.0 times his regular rate . if lloyd works 10.5 hours on a given day , how much does he earn for that day ? | "daily working hour * regular rate + overtime * increased rate 7.5 * 4.5 + 3 * 4.5 * 2.0 = 60.75 answer c" | a ) $ 33.75 , b ) $ 47.25 , c ) $ 60.75 , d ) $ 54.00 , e ) $ 70.00 | c | add(multiply(7.5, 4.50), multiply(multiply(subtract(10.5, 7.5), 2.0), 4.50)) | multiply(n0,n1)|subtract(n4,n0)|multiply(n3,#1)|multiply(n1,#2)|add(#0,#3)| | physics |
on a certain road , 20 % of the motorists exceed the posted speed limit and receive speeding tickets , but 20 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? | "suppose there are x motorists . 20 % of them exceeded the speed limit and received the ticket , i . e . x / 5 . again , suppose total no . of motorists who exceeded the speed limit are y . 20 % of y exceeded the speed limit but did n ' t received the ticket , i . e . y / 5 . it means 4 y / 5 received the ticket . henc... | a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 25 % , e ) 30 % | d | multiply(divide(20, subtract(const_100, 20)), const_100) | subtract(const_100,n1)|divide(n0,#0)|multiply(#1,const_100)| | gain |
if the remainder is 16 when the integer n is divided by 30 , what is the remainder when 2 n is divided by 15 ? | n = 30 k + 16 2 n = 2 ( 30 k + 16 ) = 4 k * 15 + 32 = 4 k * 15 + 2 * 15 + 2 = 15 j + 2 the answer is b . | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | b | subtract(multiply(16, 2), 30) | multiply(n0,n2)|subtract(#0,n1) | general |
to apply for the position of photographer at a local magazine , a photographer needs to include 4 or 5 photos in an envelope accompanying the application . if the photographer has pre - selected 7 photos representative of her work , how many choices does she have to provide the photos for the magazine ? | 7 c 4 + 7 c 5 = 35 + 21 = 56 the answer is e . | a ) 48 , b ) 50 , c ) 52 , d ) 54 , e ) 56 | e | add(divide(factorial(7), factorial(5)), multiply(7, const_2)) | factorial(n2)|factorial(n1)|multiply(n2,const_2)|divide(#0,#1)|add(#3,#2) | physics |
at 6 ′ o a clock ticks 6 times . the time between first and last ticks is 30 seconds . how long does it tick at 12 ′ o clock | explanation : for ticking 6 times , there are 5 intervals . each interval has time duration of 30 / 5 = 6 secs at 12 o ' clock , there are 11 intervals , so total time for 11 intervals = 11 × 6 = 66 secs . answer : d | a ) 47 , b ) 76 , c ) 28 , d ) 66 , e ) 11 | d | multiply(divide(30, subtract(6, const_1)), subtract(12, const_1)) | subtract(n0,const_1)|subtract(n3,const_1)|divide(n2,#0)|multiply(#2,#1) | physics |
john purchased some shirts and trousers for $ 1100 . he paid $ 550 less for the shirts than he did for the trousers . if he bought 5 shirts and the cost of a shirt is $ 20 less than that of a trouser , how many trousers did he buy ? | "given that the total purchase of two items cost 1100 . so the average purchase of one item will cost 1100 / 2 = 550 . its given as total shirt cost 100 $ less . hence total shirt cost = 550 - 275 and total trouser cost = 300 + 275 5 shirts = 275 $ = = > one shirt = 55 $ one trouser = 55 + 20 = 75 $ total trousers = 82... | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 11 | e | divide(subtract(1100, multiply(5, add(20, 20))), add(add(20, 20), 20)) | add(n3,n3)|add(n3,#0)|multiply(n2,#0)|subtract(n0,#2)|divide(#3,#1)| | general |
a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent t of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is t = . 25 / 1 * 100 = 25 % d is the answer ." | a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | divide(n1,n0)|multiply(#0,const_100)|subtract(const_100,#1)| | general |
in a throw of a coin find the probability of getting a head ? | s = { h , t } e = { h } p ( e ) = 1 / 2 answer is a | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 2 / 3 , e ) 2 / 5 | a | divide(const_1, const_2) | divide(const_1,const_2) | probability |
a can do a piece of work in 15 days . a does the work for 5 days only and leaves the job . b does the remaining work in 12 days . in how many days b alone can do the work ? | explanation : a ’ s 5 day work = 5 * 1 / 15 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 b completes 2 / 3 work in 6 days b alone can do in x days 2 / 3 * x = 12 x = 18 days answer : option e | a ) 5 days , b ) 7 days , c ) 12 days , d ) 9 days , e ) 18 days | e | inverse(multiply(inverse(12), subtract(const_1, multiply(5, inverse(15))))) | inverse(n2)|inverse(n0)|multiply(n1,#1)|subtract(const_1,#2)|multiply(#0,#3)|inverse(#4) | physics |
evaluate : 460 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 460 - 12 * 3 * 2 = 460 - 8 = 452 correct answer e" | a ) 878 , b ) 545 , c ) 434 , d ) 442 , e ) 452 | e | subtract(460, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
in a km race , a beats bby 28 metres or 7 seconds . find a ' s timeoverthe course | clearly , b covers 28 m in 7 seconds . : . b ' s time over the course = ( 278 x 1000 ) sec = 250 seconds . : . a ' s time over the course = ( 250 - 7 - ) sec = 243 sec = 4 min . 3 sec . answer e 4 min 3 sec | a ) 4 min , b ) 3 min 3 sec , c ) 4 min 4 sec , d ) 5 min 3 sec , e ) 4 min 3 sec | e | subtract(subtract(multiply(divide(7, 28), const_1000), 7), multiply(const_2, const_100)) | divide(n1,n0)|multiply(const_100,const_2)|multiply(#0,const_1000)|subtract(#2,n1)|subtract(#3,#1) | physics |
a thief is spotted by a policeman from a distance of 200 meters . when the policeman starts the chase , the thief also starts running . if the speed of the thief be 8 km / hr and that of the policeman 10 km / hr , how far the thief will have run before he is overtaken ? | "relative speed of the policeman = ( 10 - 8 ) km / hr = 2 km / hr . time taken by police man to cover ( 200 m / 1000 ) x 1 / 2 hr = 1 / 10 hr . in 1 / 10 hrs , the thief covers a distance of 8 x 1 / 10 km = 4 / 5 km = 800 m answer is d ." | a ) 350 m , b ) 200 m , c ) 400 m , d ) 800 m , e ) none of them | d | divide(multiply(200, 8), subtract(10, 8)) | multiply(n0,n1)|subtract(n2,n1)|divide(#0,#1)| | physics |
set x consists of 10 integers and has median of 30 and a range of 20 . what is the value of the greatest possible integer that can be present in the set ? | note that both median and range do not restrict too many numbers in the set . range is only concerned with the smallest and greatest . median only cares about the middle . quick check of each option starting from the largest : ( e ) 50 range of 20 means the smallest integer will be 30 . so 20 can not lie in between and... | a ) 32 , b ) 37 , c ) c . 40 , d ) 43 , e ) 50 | e | add(30, 20) | add(n1,n2) | general |
a cistern can be filled by a tap in 5 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 5 - 1 / 9 = 4 / 45 therefore the cistern will be filled in 45 / 4 hours or 11.25 hours . answer : e" | a ) 4.5 hrs , b ) 5 hrs , c ) 6.5 hrs , d ) 7.25 hrs , e ) 11.25 hrs | e | divide(const_1, subtract(divide(const_1, 5), divide(const_1, 9))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
find the sum lend at c . i . at 5 p . c per annum will amount to rs . 650 in 2 years ? | "explanation : 650 = p ( 21 / 20 ) 2 p = 589.56 answer : d" | a ) 221 , b ) 287 , c ) 400 , d ) 589 , e ) 171 | d | divide(650, power(add(const_1, divide(5, const_100)), 2)) | divide(n0,const_100)|add(#0,const_1)|power(#1,n2)|divide(n1,#2)| | general |
there are 456 doctors and nurses in a hospital . if the ratio of the doctors to the nurses is 8 : 11 , then how many nurses are there in the hospital ? | given , the ratio of the doctors to the nurses is 8 : 11 number of nurses = 11 / 19 x 456 = 264 answer : a | a ) 264 , b ) 209 , c ) 57 , d ) 171 , e ) 181 | a | multiply(multiply(8, subtract(11, 8)), 11) | subtract(n2,n1)|multiply(n1,#0)|multiply(n2,#1) | other |
find the simple interest on rs . 70,400 at 16 2 / 3 % per annum for 9 months . | "p = rs . 70400 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 70,400 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8800 answer is d ." | a ) s . 8500 , b ) s . 8000 , c ) s . 7500 , d ) s . 8800 , e ) s . 6500 | d | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)| | gain |
the true discount on a bill due 9 months hence at 16 % per annum is rs . 153 . the amount of the bill is | "solution 32.5 let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 153 . ∴ x 16 x 9 / 12 x 1 / 100 } = 153 or x = 1275 . ∴ p . w . = rs . 1275 . answer c" | a ) rs . 1386 , b ) rs . 1764 , c ) rs . 1275 , d ) rs . 2268 , e ) none of these | c | add(divide(153, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 153) | multiply(const_3,const_4)|divide(n0,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n2,#3)|add(n2,#4)| | gain |
a voltage will influence current only if the circuit is : | answer : b | ['a ) 2', 'b ) 7', 'c ) 6', 'd ) 889', 'e ) 1'] | b | add(multiply(const_3, const_2), const_1) | multiply(const_2,const_3)|add(#0,const_1) | geometry |
a watch was sold at a loss of 10 % . if it was sold for rs . 140 more , there would have been a gain of 2 % . what is the cost price ? | "explanation : 90 % 102 % - - - - - - - - 12 % - - - - 140 100 % - - - - ? = > rs . 1166.7 answer : c" | a ) s . 1000 , b ) s . 1009 , c ) s . 1166.7 , d ) s . 1006 , e ) s . 1002 | c | divide(multiply(140, const_100), subtract(add(const_100, 2), subtract(const_100, 10))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
in the junior basketball league there are 15 teams , 2 / 3 of them are bad and ½ are rich . what ca n ' t be the number of teams that are rich and bad ? | "total teams = 16 bad teams = ( 2 / 3 ) * 15 = 10 rich teams = 8 so maximum value that the both rich and bad can take will be 8 . so e = 9 can not be that value . ans d ." | a ) 4 . , b ) 6 . , c ) 7 . , d ) 9 . , e ) 5 . | d | add(multiply(15, divide(const_1, 2)), const_1) | divide(const_1,n1)|multiply(n0,#0)|add(#1,const_1)| | general |
a movie buff buys movies on dvd and on blu - ray in a ratio of 17 : 4 . if she returns 4 blu - ray movies , that ratio would change to 9 : 2 . if she buys movies on no other medium , what was the original number of movies purchased ? | if u can just keep an eye on the options 99 is the only multiple of 9 in options given . . so you can mark it wid in seconds . now coming to the process m ( d ) = 17 x and b ( d ) = 4 x now from the next line the new eqn becomes 17 x / ( 4 x - 4 ) = 9 / 2 solving it 34 x = 36 x - 36 x = 18 which means m ( d ) = 306 and... | a ) 22 , b ) 28 , c ) 77 , d ) 99 , e ) 378 | e | multiply(divide(multiply(4, 9), 2), add(17, 4)) | add(n0,n1)|multiply(n1,n3)|divide(#1,n4)|multiply(#0,#2) | other |
in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 20 percent , but profits were 15 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "let ' s test : 1998 revenues = $ 100 profits = $ 10 next we ' re told that , in 1999 , revenue fell by 20 % , but profits were 15 % of revenues . using the numbers from 1998 ( above ) , we end up with . . . 1999 revenues = $ 80 profits = $ 12 we ' re asked to compare the profits in 1999 to the profits in 1998 ( as a p... | a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | c | multiply(divide(multiply(subtract(const_1, divide(20, const_100)), divide(15, const_100)), divide(10, const_100)), const_100) | divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(#0,#3)|divide(#4,#2)|multiply(#5,const_100)| | gain |
a cube has two of its faces painted half red and half white . the other faces are completely painted white . what is the ratio between the red painted areas and the white painted areas of the cube ? | "let x be the area of each face of the cube . the area painted red is 2 ( x / 2 ) = x the area painted white is 2 ( x / 2 ) + 4 x = 5 x the ratio of red to white is x : 5 x which is 1 : 5 . the answer is a ." | a ) 1 : 5 , b ) 3 : 6 , c ) 1 : 2 , d ) 2 : 9 , e ) 1 : 3 | a | divide(multiply(multiply(add(const_1, const_4), divide(const_1, const_2)), const_2), multiply(add(multiply(add(const_1, const_4), divide(const_1, const_2)), const_1), const_2)) | add(const_1,const_4)|divide(const_1,const_2)|multiply(#0,#1)|add(#2,const_1)|multiply(#2,const_2)|multiply(#3,const_2)|divide(#4,#5)| | geometry |
if athul rows 16 km upstream and 24 km down steam taking 4 hours each , then the speed of the stream | speed upstream = 16 / 4 = 4 kmph speed down stream = 24 / 4 = 6 kmph speed of stream = ½ ( 6 - 4 ) = 1 kmph answer : a | a ) 1 kmph , b ) 2 kmph , c ) 1.5 kmph , d ) 12 kmph , e ) 15 kmph | a | divide(subtract(divide(24, 4), divide(16, 4)), const_2) | divide(n1,n2)|divide(n0,n2)|subtract(#0,#1)|divide(#2,const_2) | physics |
the average age of 30 students in a class is 5 years . if teacher ' s age is also included then average increases 1 year then find the teacher ' s age ? | "total age of 50 students = 30 * 5 = 150 total age of 51 persons = 31 * 6 = 186 age of teacher = 186 - 150 = 36 years answer is e" | a ) 59 , b ) 55 , c ) 61 , d ) 45 , e ) 36 | e | subtract(add(add(multiply(30, 5), 1), 30), multiply(30, 5)) | multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)| | general |
a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 11000 . what is the share of c ? | "explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 × ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) × 11000 ] = rs . 3000 answer : option b" | a ) rs . 2250 , b ) rs . 3000 , c ) rs . 6750 , d ) rs . 5625 , e ) none of these | b | multiply(11000, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1))) | divide(n1,n0)|multiply(n0,#0)|add(#0,#1)|add(#2,const_1)|inverse(#3)|multiply(n3,#4)| | gain |
a batsman scored 120 runs which included 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets | "number of runs made by running = 110 - ( 3 x 4 + 8 x 6 ) = 120 - ( 60 ) = 60 now , we need to calculate 60 is what percent of 120 . = > 60 / 120 * 100 = 50 % b" | a ) 30 % , b ) 50 % , c ) 60 % , d ) 80 % , e ) 90 % | b | multiply(divide(subtract(120, add(multiply(3, 8), multiply(8, 3))), 120), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
worker a takes 8 hours to do a job . worker b takes 10 hours to do a job . how long should it take both a and b , working together to do same job | explanation : in this type of questions , first we need to calculate 1 hours work , then their collective work as a ' s 1 hour work is 1 / 8 b ' s 1 hour work is 1 / 10 ( a + b ) ' s 1 hour work = 1 / 8 + 1 / 10 = 9 / 40 so both will finish the work in 40 / 9 hours = 449 answer : d | a ) 49 , b ) 249 , c ) 349 , d ) 449 , e ) none of these | d | inverse(add(inverse(8), inverse(10))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2) | physics |
4.036 divided by 0.04 gives : | "= 4.036 / 0.04 = 403.6 / 4 = 100.9 answer is d ." | a ) 10.09 , b ) 1.06 , c ) 10.06 , d ) 100.9 , e ) 100.6 | d | divide(4.036, 0.04) | divide(n0,n1)| | general |
find the constant k so that : - x 2 - ( k + 7 ) x - 8 = - ( x - 2 ) ( x - 4 ) | "- x 2 - ( k + 7 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 7 ) x - 8 = - x 2 + 6 x - 8 - ( k + 7 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 13 : solve the above for k correct answer c" | a ) - 11 , b ) - 12 , c ) - 13 , d ) - 14 , e ) - 15 | c | add(7, add(4, 2)) | add(n0,n4)|add(n1,#0)| | general |
a car takes 4.5 hours to travel from a to b , which is 360 miles apart . what is the average speed of the car ? | average speed = 360 / 4.5 hours = 360 / 270 = 1.33 miles per min = 80 miles per hour answer : d | a ) 40 , b ) 50 , c ) 60 , d ) 80 , e ) 85 | d | divide(360, 4.5) | divide(n1,n0) | physics |
there are 20 balls which are red , blue or green . if 7 balls are green and the sum of red balls and green balls is less than 13 , at most how many red balls are there ? | "explanation : given : red + green + blue = 20 green = 7 solution : r + g < 13 therefore r + 7 < 13 so r < 6 ans : at most 5 red balls b . 5 hence ( b ) is correct . answer : b" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | subtract(7, const_1) | subtract(n1,const_1)| | general |
an amount of rs . 100000 is invested in two types of shares . the first yields an interest of 9 % p . a and the second , 11 % p . a . if the total interest at the end of one year is 9 1 / 2 % , then the amount invested at 11 % was ? | "let the sum invested at 9 % be rs . x and that invested at 11 % be rs . ( 100000 - x ) . then , ( x * 9 * 1 ) / 100 + [ ( 100000 - x ) * 11 * 1 ] / 100 = ( 100000 * 19 / 2 * 1 / 100 ) ( 9 x + 1100000 - 11 x ) = 950000 x = 75000 sum invested at 9 % = rs . 75000 sum invested at 11 % = rs . ( 100000 - 75000 ) = rs . 2500... | a ) 23777 , b ) 25000 , c ) 29977 , d ) 26777 , e ) 19871 | b | divide(subtract(multiply(100000, divide(add(9, divide(1, 2)), const_100)), multiply(100000, divide(9, const_100))), subtract(divide(11, const_100), divide(9, const_100))) | divide(n4,n5)|divide(n1,const_100)|divide(n2,const_100)|add(n1,#0)|multiply(n0,#1)|subtract(#2,#1)|divide(#3,const_100)|multiply(n0,#6)|subtract(#7,#4)|divide(#8,#5)| | gain |
at what price must an book costing $ 47.50 be marked in order that after deducting 8 % from the list price . it may be sold at a profit of 25 % on the cost price ? | c $ 62.50 cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 92 / 100 ) = 59.375 mp = 64.5 b | a ) 72.5 , b ) 64.5 , c ) 62.5 , d ) 82.5 , e ) 60.5 | b | multiply(divide(divide(multiply(47.5, add(const_100, 25)), const_100), subtract(const_100, 8)), const_100) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100) | gain |
exactly 36 % of the numbers in set s are even multiples of 3 . if 40 % of the even integers in set s are not multiples of 3 , what percent of the numbers in set s are not even integers ? | let s be the total number of elements in set s . we know that 0.36 * s is the number of even multiples of three . let n be the number of even numbers . we know 0.40 * n are even numbers not multiple of three . this also means , 0.60 * n are even numbers that are multiples of three . therefore : 0.06 * n = 0.36 * s n = ... | a ) 76 % , b ) 60 % , c ) 50 % , d ) 40 % , e ) 24 % | d | subtract(const_100, multiply(divide(36, subtract(const_100, 40)), const_100)) | subtract(const_100,n2)|divide(n0,#0)|multiply(#1,const_100)|subtract(const_100,#2) | gain |
the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 48 years , was replaced by a new person . find the age of the new person ? | "initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 48 + q = > q = 18 answer : a" | a ) 18 , b ) 56 , c ) 12 , d ) 17 , e ) 14 | a | subtract(48, multiply(10, 3)) | multiply(n0,n1)|subtract(n2,#0)| | general |
a glucose solution contains 10 grams of glucose per 100 cubic centimeters of solution . if 45 cubic centimeters of the solution were poured into an empty container , how many grams of glucose would be in the container ? | "we are given that a glucose solution contains 10 grams of glucose per 100 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of glucose is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of glucose would be in the container whe... | a ) 3.00 , b ) 5.00 , c ) 4.50 , d ) 5.50 , e ) 6.75 | c | multiply(45, divide(10, 100)) | divide(n0,n1)|multiply(n2,#0)| | physics |
find the two digit number , such that the ratio is 7 / 4 of original number to the number formed by reversing the digits . | here ratio is 7 / 4 . so lets check one by one . for 1 : 7 * 1 / 4 * 1 = 7 / 4 for 2 : 7 * 2 / 4 * 2 = 14 / 8 , which does n ' t satisfy the conditions . for 3 : 7 * 3 / 4 * 3 = 21 / 12 answer : b | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | b | subtract(multiply(7, 4), 7) | multiply(n0,n1)|subtract(#0,n0) | other |
the owner of a furniture shop charges his customer 10 % more than the cost price . if a customer paid rs . 7350 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 7350 ( 100 / 110 ) = rs . 6681 . answer : c" | a ) rs . 6289 , b ) rs . 6298 , c ) rs . 6681 , d ) rs . 6725 , e ) rs . 6708 | c | divide(7350, add(const_1, divide(10, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
a can do a piece of work in 4 days . b can do it in 5 days . with the assistance of c they completed the work in 2 days . find in how many days can c alone do it ? | "c = 1 / 2 - 1 / 4 - 1 / 5 = 1 / 20 = > 20 days answer : b" | a ) 87 days , b ) 20 days , c ) 16 days , d ) 19 days , e ) 36 days | b | divide(multiply(4, 5), divide(subtract(multiply(4, 5), multiply(add(divide(multiply(4, 5), 4), divide(multiply(4, 5), 5)), 2)), 2)) | multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|multiply(n2,#3)|subtract(#0,#4)|divide(#5,n2)|divide(#0,#6)| | physics |
what least no . must be subtracted from 10154 so that remaining no . is divisible by 30 ? | "explanation : on dividing 10154 by 30 we get the remainder 14 , so 14 should be subtracted option c" | a ) 10 , b ) 12 , c ) 14 , d ) 8 , e ) 9 | c | subtract(10154, multiply(floor(divide(10154, 30)), 30)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
the length of the bridge , which a train 140 meters long and travelling at 45 km / hr can cross in 30 seconds , is : | "speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 140 + x ) / 30 = 25 / 2 = = > 2 ( 140 + x ) = 750 = = > x = 235 m . answer : option e" | a ) 230 , b ) 244 , c ) 245 , d ) 238 , e ) 235 | e | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 140) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
the present population of a town is 240 . population increase rate is 10 % p . a . find the population of town after 1 years ? | "p = 240 r = 10 % required population of town = p * ( 1 + r / 100 ) ^ t = 240 * ( 1 + 10 / 100 ) = 240 * ( 11 / 10 ) = 264 answer is e" | a ) 100 , b ) 120 , c ) 200 , d ) 220 , e ) 264 | e | add(240, divide(multiply(240, 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
a batsman makes a score of 74 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 74 = total runs scored in 17 innings = > 16 ( x - 3 ) + 74 = 17 x = > 16 x - 48 + 74 = 17 x = > x = 26 answer is c" | a ) 25 , b ) 31 , c ) 26 , d ) 29 , e ) 39 | c | add(subtract(74, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
a 5 - digit number divisible by 3 is to be formed using numerical 0 , 1 , 2 , 3 , 4 and 5 without repetition . the total number q of ways this can be done is : | we should determine which 5 digits from given 6 , would form the 5 digit number divisible by 3 . we have six digits : 0 , 1 , 2 , 3 , 4 , 5 . their sum = 15 . for a number to be divisible by 3 the sum of the digits must be divisible by 3 . as the sum of the six given numbers is 15 ( divisible by 3 ) only 5 digits good ... | a ) 122 , b ) 210 , c ) 216 , d ) 217 , e ) 225 | c | subtract(multiply(factorial(5), const_2), factorial(4)) | factorial(n0)|factorial(n6)|multiply(#0,const_2)|subtract(#2,#1) | general |
the red triangle is equilateral with a side of 23 centimetres . its perimeter is | solution : all three sides of an equilateral triangle are equal . thus , its perimeter will be 23 × 3 = 69 cm answer b | ['a ) 68 cm', 'b ) 69 cm', 'c ) 70 cm', 'd ) 71 cm', 'e ) none'] | b | triangle_perimeter(23, 23, 23) | triangle_perimeter(n0,n0,n0) | geometry |
the average height of 35 boys in a class was calculated as 180 cm . it has later found that the height of one of the boys in the class was wrongly written as 156 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . | "calculated average height of 35 boys = 180 cm . wrong total height of 35 boys = 180 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 156 cm . correct total height of 35 boys = 180 * 35 cm - 156 cm + 106 cm = 180 * 35 cm - 156 cm + 106 cm / 35 = 180 cm - 50 / 35 cm = 180 cm - 1.42 c... | a ) 168.58 cm , b ) 158.58 cm , c ) 179.29 cm , d ) 168.58 cm , e ) 178.58 cm | e | floor(divide(add(subtract(multiply(35, 180), 156), 106), 35)) | multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|floor(#3)| | general |
in a group of 100 cars , 37 cars do not have air conditioning . if at least 51 cars have racing stripes , what is the greatest number of cars that could have air conditioning but not racing stripes ? | "lets assume ac = 63 ( includesonly ac carsandcars with ac and racing stripes ) lets assume rs ( racing stripes ) > = 51 ( includescars with ac and racing stripesandonly racing stripes ) . now since we want to maximize ( only ac ) we have to see to it thatcars with ac and racing stripesis minimal ( assume 0 ) but since... | a ) 45 , b ) 47 , c ) 48 , d ) 49 , e ) 53 | d | subtract(100, 51) | subtract(n0,n2)| | other |
the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability r that the next couple the maitre ' d seats will not order dessert ? | could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . r = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c o... | a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 % | b | multiply(subtract(const_1, divide(divide(60, multiply(multiply(const_2, const_5), multiply(const_2, const_5))), subtract(const_1, divide(20, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))))), multiply(multiply(const_2, const_5), multiply(const_2, const_5))) | multiply(const_2,const_5)|multiply(#0,#0)|divide(n0,#1)|divide(n1,#1)|subtract(const_1,#3)|divide(#2,#4)|subtract(const_1,#5)|multiply(#1,#6) | probability |
john purchased 1365 large bottles at $ 1.89 per bottle and 720 small bottles at $ 1.42 per bottle . what was the approximate average price paid per bottle ? | ( 1365 * 1.89 + 720 * 1.42 ) / ( 1365 + 720 ) = ~ 1.73 option ( d ) | a ) $ 1.63 , b ) $ 1.64 , c ) $ 1.68 , d ) $ 1.73 , e ) $ 1.76 | d | divide(add(multiply(1365, 1.89), multiply(720, 1.42)), add(1365, 720)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0) | general |
a train 100 meters long completely crosses a 300 meters long bridge in 15 seconds . what is the speed of the train is ? | "s = ( 100 + 300 ) / 45 = 400 / 15 * 18 / 5 = 96 answer : b" | a ) 32 kmph , b ) 96 kmph , c ) 34 kmph , d ) 43 kmph , e ) 40 kmph | b | divide(divide(add(100, 300), const_1000), divide(15, const_3600)) | add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)| | physics |
( 500 + 200 + 100 ) ã — 4 ã — ( 3 + 4 ) = ? | "( 500 + 200 + 100 ) ã — 4 ã — ( 3 + 4 ) = ? or , ? = 800 ã — 4 ã — 7 = 22400 answer c" | a ) 52000 , b ) 32400 , c ) 22400 , d ) 22800 , e ) 24200 | c | subtract(divide(multiply(4, add(4, const_1)), const_2), divide(multiply(subtract(500, const_1), 500), const_2)) | add(n3,const_1)|subtract(n0,const_1)|multiply(n3,#0)|multiply(n0,#1)|divide(#2,const_2)|divide(#3,const_2)|subtract(#4,#5)| | general |
jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent e ? | "c . let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . e = [ ( 2 - 1 ) / 1 ] * 100 % = 100 % . c" | a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(10, subtract(subtract(const_100, 80), 10)), const_100) | subtract(const_100,n0)|subtract(#0,n1)|divide(n1,#1)|multiply(#2,const_100)| | physics |
if the wheel is 10 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 10 * x = 1056 = > x = 16.8 answer : a" | a ) 16.8 , b ) 26 , c ) 14 , d ) 12 , e ) 91 | a | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 10)) | multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)| | physics |
the perimeter of a semi circle is 126 cm then the radius is ? | "36 / 7 r = 126 = > r = 24.5 answer : d" | a ) 22 , b ) 28 , c ) 98 , d ) 24.5 , e ) 13 | d | divide(126, add(const_2, const_pi)) | add(const_2,const_pi)|divide(n0,#0)| | physics |
two cyclist start from the same places in opposite directions . one is going towards north at 10 kmph and the other is going towards south 15 kmph . what time will they take to be 50 km apart ? | "to be ( 10 + 15 ) km apart , they take 1 hour to be 50 km apart , they take 1 / 25 * 50 = 2 hrs answer is b" | a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 5 hrs , e ) 6 hrs | b | divide(50, add(10, 15)) | add(n0,n1)|divide(n2,#0)| | physics |
r is the set of positive even integers less than 101 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "r is the set of positive even integers less than 101 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 2,4 , 6,8 , 10,12 . . . s = 4,16 , 36,64 . . . numbers : 4 , 16 , 36 , 64 , and 100 are even integers ( less than 101 ) that are in both sets . ... | a ) none , b ) two , c ) four , d ) five , e ) seven | d | subtract(const_4, const_1) | subtract(const_4,const_1)| | physics |
a fill pipe can fill 1 / 4 of cistern in 16 minutes in how many minutes , it can fill 3 / 4 of the cistern ? | "1 / 4 of the cistern can fill in 16 min 3 / 4 of the cistern can fill in = 16 * 4 * 3 / 4 = 48 min answer is a" | a ) 48 min , b ) 36 min , c ) 25 min , d ) 30 min , e ) 50 min | a | divide(16, 1) | divide(n2,n0)| | physics |
how many three digit numbers e are divisible by 78 or 91 ? | "the answer will be 19 . explanation : 78 = 2 * 3 * 13 now multiples of 78 , 156 . . . . 780 , now 1000 - 780 = 220 only two more muktiples of 78 can exists . so total number of 3 digit multiples of 78 are 9 + 2 = 11 91 = 13 * 7 - - total number of three digit multiples - - 9 no remember we have a common multiples as w... | a ) 17 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | b | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(78, 78)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general |
calculate the effect changes in dimension of a rectangle will have on its area , if length is increased by 35 % and its breadth is decreased by 21 % ? | "let l and b be 100 each 100 * 100 = 10000 l increase by 35 % = 135 b decrease by 21 % = 79 135 * 79 = 10665 6.65 % increase answer : c" | a ) 4.65 % increase , b ) 5.65 % increase , c ) 6.65 % increase , d ) 6.65 % decrease , e ) 7.65 % increase | c | multiply(divide(subtract(multiply(add(const_100, 35), subtract(const_100, 21)), multiply(const_100, const_100)), multiply(const_100, const_100)), const_100) | add(n0,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#3,#1)|divide(#4,#1)|multiply(#5,const_100)| | geometry |
fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 3 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 18... | "you know that fox jeans costs $ 15 , and pony jeans costs $ 18 , you also know that 3 pairs of fox jeans and 2 pairs of pony jeans were purchased . so 3 ( 15 ) = 45 - fox 2 ( 18 ) = 36 - pony the total discount discount is $ 3 and you are asked to find the percent discount of pony jeans , so 45 ( 18 - x ) / 100 + 36 (... | a ) 9 % , b ) 56.6 % , c ) 11 % , d ) 12 % , e ) 15 % | b | multiply(subtract(divide(18, const_100), divide(subtract(3, multiply(divide(18, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100) | divide(n6,const_100)|multiply(n1,n5)|multiply(n0,n4)|multiply(#0,#1)|subtract(#2,#1)|subtract(n2,#3)|divide(#5,#4)|subtract(#0,#6)|multiply(#7,const_100)| | gain |
a and b began business with rs . 2000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of b . | "( 2 * 8 + 1 * 4 ) : ( 4 * 8 + 5 * 4 ) 5 : 13 13 / 18 * 630 = 455 answer : d" | a ) 240 , b ) 288 , c ) 277 , d ) 455 , e ) 361 | d | multiply(divide(630, add(add(multiply(2000, 8), multiply(subtract(2000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(2000, 8), multiply(subtract(2000, 1000), subtract(const_12, 8)))) | add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)| | gain |
if 70 % of a number exceeds 28 % of it by 39.9 , then find the number ? | "use the elimination method to find the correct option . of all the options only 95 fits 70 % of 95 = 66.5 28 % of 95 = 26.6 66.5 - 26.6 = 39.9 required number is 95 . answer : d" | a ) 65 , b ) 67 , c ) 100 , d ) 95 , e ) 33 | d | add(divide(39.9, divide(subtract(70, 28), const_100)), 28) | subtract(n0,n1)|divide(#0,const_100)|divide(n2,#1)|add(n1,#2)| | gain |
40 onions on a scale weigh 7.68 kg . when 5 onions are removed from the scale , the average weight of the 35 onions is 190 grams . what is the average weight ( in grams ) of the 5 onions which were removed ? | "35 * 190 = 6650 . the other 5 onions weigh a total of 1030 grams . the average weight is 1030 / 5 = 206 grams . the answer is d ." | a ) 200 , b ) 202 , c ) 204 , d ) 206 , e ) 208 | d | divide(subtract(multiply(7.68, const_1000), multiply(35, 190)), 5) | multiply(n1,const_1000)|multiply(n3,n4)|subtract(#0,#1)|divide(#2,n2)| | general |
evaluate : 6402.5 + 640.25 + 64.025 + 6.4025 + 0.64025 | "6402.5 640.25 64.025 6.4025 + 0.64025 - - - - - - - - - - - - - - - 7113.81775 answer is d ." | a ) 6819.59775 , b ) 6981.59775 , c ) 6918.59775 , d ) 7113.81775 , e ) 6891.59775 | d | subtract(6402.5, multiply(multiply(640.25, 64.025), 6.4025)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
mudit ' s age 16 years hence will be thrice his age 4 years ago . find mudit ' s present age ? | explanation : let mudit ' s present age be ' m ' years . m + 16 = 3 ( m - 4 ) = > 2 m = 28 = > m = 14 years . answer : a | a ) 14 , b ) 19 , c ) 27 , d ) 18 , e ) 15 | a | divide(add(16, multiply(const_3, 4)), const_2) | multiply(n1,const_3)|add(n0,#0)|divide(#1,const_2) | general |
the compound interest earned on a sum for the second and the third years are $ 1400 and $ 1540 respectively . what is the rate of interest ? | 1540 - 1400 = 140 is the rate of interest on $ 1400 for one year . the rate of interest = ( 100 * 140 ) / ( 1400 ) = 10 % the answer is e . | a ) 2 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 10 % | e | divide(multiply(subtract(1540, 1400), const_100), 1400) | subtract(n1,n0)|multiply(#0,const_100)|divide(#1,n0) | gain |
on a certain planet where people usually tend to live for more than a century , the ratio of present ages of father and son is 7 : 4 . 10 years later , the ratio of their ages will be 5 : 3 . what is the sum of their present ages ? | the ratio of present ages of father and son is 7 : 4 , so suppose the present ages are 7 x and 4 x , respectively . 10 years later , the ages of father and son , respectively , shall be 7 x + 10 and 4 x + 10 . the ratio of ages 10 years later is given to be 5 : 3 . this means ( 7 x + 10 ) / ( 4 x + 10 ) = 5 / 3 solving... | a ) 220 , b ) 200 , c ) 140 , d ) 80 , e ) 60 | a | add(multiply(7, subtract(multiply(5, 10), multiply(3, 10))), multiply(4, subtract(multiply(5, 10), multiply(3, 10)))) | multiply(n2,n3)|multiply(n2,n4)|subtract(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(#3,#4) | general |
a certain candy manufacturer reduced the weight of candy bar m by 40 percent buy left the price unchanged . what was the resulting percent increase in the price per ounce of candy bar m ? | "assume 1 oz candy cost $ 1 before . now price remain same $ 1 but weight of candy reduces to 0.6 oz new price of candy = 1 / 0.6 = 1.6666 price increase 66.66 % a" | a ) 66.66 , b ) 62.5 , c ) 58 , d ) 60 , e ) 62 | a | subtract(multiply(divide(const_1, subtract(const_1, divide(40, const_100))), const_100), const_100) | divide(n0,const_100)|subtract(const_1,#0)|divide(const_1,#1)|multiply(#2,const_100)|subtract(#3,const_100)| | gain |
weights of two friends ram and shyam are in the ratio 1 : 5 . if ram ' s weight is increased by 10 % and total weight of ram and shyam become 82.8 kg , with an increases of 15 % . by what percent did the weight of shyam has to be increased ? | solution : given ratio of ram and shayam ' s weight = 1 : 5 hence , ( x - 15 ) / ( 15 - 10 ) = 1 / 5 or , x = 16 % . answer : option d | a ) 19 % , b ) 10 % , c ) 21 % , d ) 16 % , e ) none | d | add(15, multiply(subtract(15, 10), divide(1, 5))) | divide(n0,n1)|subtract(n4,n2)|multiply(#0,#1)|add(n4,#2) | gain |
the average of first 20 prime numbers is ? | "explanation : average = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 / 8 = 77 / 8 = 9.6 ( approx ) answer is d" | a ) 9.5 , b ) 8.6 , c ) 9.3 , d ) 9.6 , e ) 9.8 | d | add(20, const_1) | add(n0,const_1)| | general |
what is the value of x ^ 2 yz − xyz ^ 2 , if x = − 2 , y = 1 , and z = 4 ? | "4 * 1 * 4 - ( - 2 * 1 * 16 ) = 16 + 32 = 48 ans : e" | a ) 20 , b ) 24 , c ) 30 , d ) 32 , e ) 48 | e | multiply(multiply(multiply(negate(2), 1), 4), subtract(negate(2), 1)) | negate(n0)|multiply(n4,#0)|subtract(#0,n4)|multiply(n3,#1)|multiply(#3,#2)| | general |
a train of 35 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ? | "d = 35 * 60 + 1500 = 3600 m t = 3600 / 60 * 18 / 5 = 216 sec = 3.6 mins answer : e" | a ) 6 , b ) 3 , c ) 4 , d ) 9 , e ) 3.6 | e | add(divide(multiply(add(35, const_1), 60), const_1000), 1.5) | add(n0,const_1)|multiply(n1,#0)|divide(#1,const_1000)|add(n4,#2)| | physics |
on july 1 of last year , total employees at company e was decreased by 10 percent . without any change in the salaries of the remaining employees , the average ( arithmetic mean ) employee salary was 10 percent more after the decrease in the number of employees than before the decrease . the total of the combined salar... | "the total number of employees = n the average salary = x total salary to all emplyoees = xn after the total number of employees = n - 0.1 n = 0.9 n the average salary = x + 10 % of x = 1.1 x total salary to all emplyoees = 0.9 n ( 1.1 x ) total salary after as a % of total salary before q = [ 0.9 n ( 1.1 x ) ] / xn = ... | a ) 90 % , b ) 99 % , c ) 100 % , d ) 101 % , e ) 110 % | b | multiply(10, 10) | multiply(n1,n1)| | general |
the difference of two numbers is 11 and one - fifth of their sum is 9 . find the numbers | if two numbers are x , y x - y = 11 - - ( i ) & ( x + y ) / 5 = 9 - - ( ii ) solving ( i ) & ( ii ) , x = 28 , y = 17 answer : b | a ) 28,22 , b ) 28,17 , c ) 27,17 , d ) 26,17 , e ) 28,19 | b | add(subtract(multiply(divide(const_10, const_2), 9), divide(add(11, multiply(divide(const_10, const_2), 9)), const_2)), divide(const_10, const_2)) | divide(const_10,const_2)|multiply(n1,#0)|add(n0,#1)|divide(#2,const_2)|subtract(#1,#3)|add(#0,#4) | general |
a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 18000 left with him . find salary . . | "[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 18000 , = [ 1 - 9 / 10 ] * total salary = $ 18000 , total salary = $ 18000 * 10 = $ 180000 , correct answer ( c )" | a ) $ 10800 , b ) $ 18000 , c ) $ 180000 , d ) $ 1800 , e ) none | c | divide(18000, subtract(1, add(add(divide(1, 5), divide(1, 10)), divide(3, 5)))) | divide(n0,n1)|divide(n0,n3)|divide(n4,n1)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general |
bert left the house with n dollars . he spent 1 / 4 of this at the hardware store , then $ 9 at the dry cleaners , and then half of what was left at the grocery store . when he got home , he had $ 12 left in his pocket . what was the value of n ? | "started to test answer b if he had 44 , then he spent 11 at hardware store now he was left with 33 $ he spent 9 dollars on cleaning , thus he remained with 24 $ he then spent 1 / 2 of 24 , or 12 , and was left with 12 . hence , the only option that can be right is b" | a ) $ 36 , b ) $ 44 , c ) $ 52 , d ) $ 60 , e ) $ 68 | b | multiply(multiply(inverse(subtract(1, divide(1, 4))), add(divide(9, const_2), 12)), const_2) | divide(n2,const_2)|divide(n0,n1)|add(n3,#0)|subtract(n0,#1)|inverse(#3)|multiply(#2,#4)|multiply(#5,const_2)| | general |
if the wheel is 14 cm then the number of revolutions to cover a distance of 3520 cm is ? | "2 * 22 / 7 * 14 * x = 3520 = > x = 40 answer : e" | a ) 22 , b ) 28 , c ) 17 , d ) 12 , e ) 40 | e | divide(3520, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)| | physics |
what will be the compound interest on a sum of rs . 15000 after 3 years at the rate of 12 % p . a . ? | "explanation : amount = [ 15000 * ( 1 + 12 / 100 ) 3 ] = 15000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 35123.20 c . i . = ( 35123.20 - 15000 ) = rs . 6073.92 answer : e" | a ) s . 10123.22 , b ) s . 5823.20 , c ) s . 9123.20 , d ) s . 7256.20 , e ) s . 6073.92 | e | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
if 13 = 13 w / ( 1 - w ) , then ( w ) 2 = | 13 - 13 w = 13 w 26 w = 13 w = 1 / 2 2 w * 2 = 1 / 2 * 2 = 1 answer : c | a ) 1 / 4 , b ) 1 / 2 , c ) 1 , d ) 2 , e ) 3 | c | multiply(divide(13, add(13, 13)), 2) | add(n0,n0)|divide(n0,#0)|multiply(n3,#1) | general |
on dividing 15968 by a certain number , the quotient is 89 and the remainder is 37 . find the divisor . | "divisor = ( dividend – remainder / quotient ) = ( 15968 - 37 ) / 89 = 179 answer b 179" | a ) 170 , b ) 179 , c ) 37 , d ) 67 , e ) 32 | b | divide(subtract(15968, 37), 89) | subtract(n0,n2)|divide(#0,n1)| | general |
58 , 52 , 46 , 40 , 34 , . . . ? | "each number is 6 less than the previous number . 58 - 6 = 52 52 - 6 = 46 46 - 6 = 40 40 - 6 = 34 34 - 6 = 28 answer : c" | a ) 20 , b ) 22 , c ) 28 , d ) 27 , e ) 32 | c | subtract(negate(40), multiply(subtract(52, 46), divide(subtract(52, 46), subtract(58, 52)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
if a coin is flipped , the probability that the coin will land tails is 1 / 2 . if the coin is flipped 4 times , what is the probability that it will land tails up on the first 2 flips and not on the last 2 flips ? | ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 16 answer : d | a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 2 , d ) 1 / 16 , e ) 1 / 32 | d | inverse(power(2, 4)) | power(n1,n2)|inverse(#0)| | probability |
if rs . 460 amount to rs . 540 in 4 years , what will it amount to in 6 years at the same rate % per annum ? | "80 = ( 460 * 4 * r ) / 100 r = 4.34 % i = ( 460 * 6 * 4.34 ) / 100 = 120 460 + 120 = 580 answer : d" | a ) rs . 575 , b ) rs . 595 , c ) rs . 590 , d ) rs . 580 , e ) rs . 585 | d | subtract(multiply(subtract(540, 460), 6), subtract(540, 460)) | subtract(n1,n0)|multiply(n3,#0)|subtract(#1,#0)| | gain |
a train passes a station platform in 50 sec and a man standing on the platform in 15 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 15 = 225 m . let the length of the platform be x m . then , ( x + 225 ) / 50 = 15 = > x = 525 m . answer : d" | a ) 227 , b ) 500 , c ) 550 , d ) 525 , e ) 171 | d | multiply(15, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
a trained covered x km at 70 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km . | "total time taken = x / 70 + 2 x / 20 hours = 4 x / 35 hours average speed = 3 x / ( 4 x / 35 ) = 26.25 kmph answer : c" | a ) 22.25 , b ) 99 , c ) 26.25 , d ) 66 , e ) 887 | c | divide(multiply(70, 3), add(divide(70, 70), divide(multiply(2, 70), 20))) | divide(n0,n0)|multiply(n0,n3)|multiply(n0,n1)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)| | general |
a hall is 15 m long and 12 m broad . if the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls , the volume of the hall is : | answer : c ) 1200 | a ) 3348 , b ) 3898 , c ) 1200 , d ) 2881 , e ) 7881 | c | volume_rectangular_prism(15, 12, divide(multiply(const_2, rectangle_area(12, 15)), add(multiply(12, const_2), multiply(15, const_2)))) | multiply(n1,const_2)|multiply(n0,const_2)|rectangle_area(n0,n1)|add(#0,#1)|multiply(#2,const_2)|divide(#4,#3)|volume_rectangular_prism(n0,n1,#5)| | geometry |
a salesman sold twice as much pears in the afternoon than in the morning . if he sold $ 420 kilograms of pears that day , how many kilograms did he sell in the afternoon ? | "3 x = 420 x = 140 therefore , the salesman sold 140 kg in the morning and 2 ⋅ 140 = 280 kg in the afternoon . so answer is d ." | a ) 120 , b ) 180 , c ) 240 , d ) 280 , e ) 320 | d | multiply(divide(420, const_3), const_2) | divide(n0,const_3)|multiply(#0,const_2)| | other |
the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 1000 per sq . metre . | solution area of the floor = ( 5.5 × 3.75 ) m 2 = 20.625 m 2 ∴ cost of paving = rs . ( 1000 × 20.625 ) = 20625 . answer d | a ) rs . 15000 , b ) rs . 15550 , c ) rs . 15600 , d ) rs . 20625 , e ) none of these | d | multiply(1000, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0) | physics |
the probability that a can solve the problem is 2 / 3 and b can solve it is 3 / 4 . if both of them attemp the problem , then what is the probability that the problem get solved ? | explanation : the event is defined as : a solves the problem and b does not solve the problem or a does n ' t solve the problem and b solves the problem or a solves the problem and b solves the problem numerically this is equivalent to : ( 2 / 3 ) x ( 1 / 4 ) + ( 1 / 3 ) x ( 3 / 4 ) + ( 2 / 3 ) x ( 3 / 4 ) = 11 / 12 an... | a ) 9 / 12 , b ) 10 / 12 , c ) 11 / 12 , d ) 1 , e ) 2 | c | subtract(const_1, multiply(subtract(const_1, divide(2, 3)), subtract(const_1, divide(3, 4)))) | divide(n0,n1)|divide(n1,n3)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|subtract(const_1,#4) | other |
a and b finish the job in 15 days . while a , b and c can finish it in 11 days . c alone will finish the job in | explanation : 11 = ( 15 * x ) / ( 15 + x ) 165 + 11 x = 15 x 4 x = 165 x = 41.25 answer : option a | a ) 41.25 days , b ) 30.25 days , c ) 60.25 days , d ) 71.25 days , e ) 51.25 days | a | divide(multiply(11, 15), subtract(15, 11)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1) | physics |
on a map , 1 inch represents 28 miles . how many x xxinches would be necessary to represent a distance of 383.6 miles ? | "x inches necessary to represent a distance of 383.6 miles = 383.6 / 28 = 13.7 answer c" | a ) 5.2 , b ) 7.4 , c ) 13.7 , d ) 21.2 , e ) 28.7 | c | divide(383.6, 28) | divide(n2,n1)| | physics |
tim has 13 10 - dollar bills , 11 5 - dollar bills , and 17 one - dollar bills . if tim needs to pay exactly $ 128 , what is the least number of bills he will need to use ? | 128 is the total sum of money . as we have 13 10 dollar bills so the closest we can get to 128 using the 10 dollar bills is by using 12 * 10 dollar bils 128 - 120 = 8 so now we need to get 8 from either 5 dollar bills or 1 dollar bills 8 - 5 ( use only 1 5 dollar bill ) = 3 so we can get 3 from using 3 $ 1 bills . henc... | a ) 68 , b ) 66 , c ) 565 , d ) 16 , e ) 35 | d | add(add(subtract(13, const_1), const_1), const_3) | subtract(n0,const_1)|add(#0,const_1)|add(#1,const_3) | general |
if in a race of 130 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by : | "explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 130 * 5 ) / 25 = 26 m hence , a beats b by 26 m . answer b" | a ) 20 m , b ) 26 m , c ) 11 m , d ) 10 m , e ) 15 m | b | multiply(divide(subtract(25, 20), 25), 130) | subtract(n2,n1)|divide(#0,n2)|multiply(n0,#1)| | physics |
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