Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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in a party every person shakes hands with every other person . if there were a total of 136 handshakes in the party then what is the number of persons present in the party ? | "explanation : let the number of persons be n Γ’ Λ Β΄ total handshakes = nc 2 = 136 n ( n - 1 ) / 2 = 136 Γ’ Λ Β΄ n = 17 answer : c" | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | c | divide(add(sqrt(add(multiply(multiply(136, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general | C |
you hold some gold in a vault as an investment . over the past year the price of gold increases by 30 % . in order to keep your gold in the vault , you must pay 2 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year . | "( 100 % + 30 % ) * ( 100 % - 2 % ) = 130 * 0.98 = 127.4 % an increase of 27.4 % your gold holdings have increased in value by 27.4 % . the answer is b" | a ) 27.5 % , b ) 27.4 % , c ) 27.6 % , d ) 27.7 % , e ) 28.4 % | b | subtract(30, divide(30, 2)) | divide(n0,n1)|subtract(n0,#0)| | gain | B |
there are 4 runners on a track team who run an average of 4.5 seconds per 40 yards . if another runner joins their team who runs 4.3 seconds per 40 yards , what will the new average 40 yard time be ? | ( sum of the 4 times ) / 4 = 4.5 sum of the 4 times = 18 new sum = 18 + 4.3 = 22.3 new average = 22.3 / 5 = 4.46 ans : a | a ) 4.46 , b ) 4.39 , c ) 4.42 , d ) 4.49 , e ) 4.33 | a | divide(add(4.5, 4.3), const_2) | add(n1,n3)|divide(#0,const_2) | general | A |
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 | B |
how many factors does 49 ^ 2 have ? | "36 ^ 2 = 6 * 6 * 6 * 6 = 2 ^ 4 * 3 ^ 4 total factors = ( 4 + 1 ) * ( 4 + 1 ) = 6 * 6 = 36 answer e ." | a ) 2 , b ) 8 , c ) 24 , d ) 25 , e ) 36 | e | divide(49, multiply(const_10, const_2)) | multiply(const_10,const_2)|divide(n0,#0)| | other | E |
what is the remainder when ( 71 ) ( 73 ) is divided by 8 ? | "( 71 ) ( 73 ) = ( 72 - 1 ) ( 72 + 1 ) = 72 ^ 2 - 1 which is 1 less than a multiple of 8 . then the remainder will be 7 . the answer is e ." | a ) 1 , b ) 2 , c ) 4 , d ) 5 , e ) 7 | e | reminder(multiply(73, 71), 8) | multiply(n0,n1)|reminder(#0,n2)| | general | E |
sum of the numbers from 1 to 29 is | explanation : sum of first n natural numbers = 1 + 2 + 3 + . . . . . n = n ( n + 1 ) / 2 substitute n = 29 . so s 20 = 29 Γ£ β 30 / 2 = 435 correct option : c | a ) 400 , b ) 430 , c ) 435 , d ) 405 , e ) none of these | c | add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(const_10, add(add(add(add(add(add(add(1, const_2), const_4), add(add(1, const_2), const_2)), add(add(add(1, const_2), const_2), 1)), add(add(add(add(1, const_2), const_2), 1), 1)), add(add(add(add(add(1, const_2), const_2), 1), 1), 1)), add(add(add(add(add(add(1, const_2), const_2), 1), 1), 1), 1))), add(const_10, 1)), const_12), add(const_12, 1)), add(const_12, const_2)), add(const_12, add(1, const_2))), add(add(const_12, add(1, const_2)), 1)), add(add(add(const_12, add(1, const_2)), 1), 1)), add(add(add(add(const_12, add(1, const_2)), 1), 1), 1)), add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1)), add(add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1), 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(add(add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(add(add(add(add(add(add(add(const_12, add(1, const_2)), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1), 1)), 29) | add(n0,const_2)|add(n0,const_10)|add(n0,const_12)|add(const_12,const_2)|add(#0,const_4)|add(#0,const_2)|add(#0,const_12)|add(#4,#5)|add(n0,#5)|add(n0,#6)|add(#7,#8)|add(n0,#8)|add(n0,#9)|add(#10,#11)|add(n0,#11)|add(n0,#12)|add(#13,#14)|add(n0,#14)|add(n0,#15)|add(#16,#17)|add(n0,#18)|add(#19,const_10)|add(n0,#20)|add(#21,#1)|add(n0,#22)|add(#23,const_12)|add(n0,#24)|add(#25,#2)|add(n0,#26)|add(#27,#3)|add(n0,#28)|add(#29,#6)|add(n0,#30)|add(#31,#9)|add(n0,#32)|add(#33,#12)|add(n0,#34)|add(#35,#15)|add(#37,#18)|add(#38,#20)|add(#39,#22)|add(#40,#24)|add(#41,#26)|add(#42,#28)|add(#43,#30)|add(#44,#32)|add(#45,#34)|add(#46,#36)|add(n1,#47) | general | C |
in what time will a train 100 m long cross an electric pole , it its speed be 136 km / hr ? | "speed = 136 * 5 / 18 = 37.8 m / sec time taken = 100 / 37.8 = 2.6 sec . answer : a" | a ) 2.6 sec , b ) 1.9 sec , c ) 8.9 sec , d ) 6.9 sec , e ) 2.9 sec | a | divide(100, multiply(136, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | A |
a train speeds past a pole in 15 seconds and a platform 110 meters long in 25 seconds . what is the length of the train ? | "let the length of the train be x meters . the speed of the train is x / 15 . then , x + 110 = 25 * ( x / 15 ) 10 x = 1650 x = 165 meters the answer is e ." | a ) 100 m , b ) 125 m , c ) 130 m , d ) 150 m , e ) 165 m | e | multiply(110, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics | E |
equal amount of water were poured into two empty jars of different capacities , which made one jar 1 / 8 full and other jar 1 / 6 full . if the water in the jar with lesser capacity is then poured into the jar with greater capacity , what fraction of the larger jar will be filled with water ? | "same amount of water made bigger jar 1 / 8 full , then the same amount of water ( stored for a while in smaller jar ) were added to bigger jar , so bigger jar is 1 / 8 + 1 / 8 = 1 / 4 full . answer : e ." | a ) 1 / 7 , b ) 2 / 7 , c ) 1 / 2 , d ) 7 / 12 , e ) 1 / 4 | e | divide(const_2, 8) | divide(const_2,n1)| | general | E |
a store owner estimates that the average price of type a products will increase by 35 % next year and that the price of type b products will increase by 20 % next year . this year , the total amount aid for type a products was $ 3500 and the total price paid for type b products was $ 6300 . according to the store owner ' s estimate , and assuming the number of products purchased next year remains the same as that of this year , how much will be spent for both products next year ? | "cost of type a products next year = 1.35 * 3500 = 4725 cost of type b products next year = 1.2 * 6300 = 7560 total 4725 + 7560 = 12285 answer : d" | a ) $ 14,755 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 12,285 , e ) $ 17,155 | d | multiply(divide(const_3, const_4), const_1000) | divide(const_3,const_4)|multiply(#0,const_1000)| | general | D |
gold is 19 times as heavy as water and copper is 9 times as heavy as water . in what ratio should these be mixed to get an alloy 16 times as heavy as water ? | "g = 19 w c = 9 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 19 w + 9 wx = x + 1 * 16 w 19 + 9 x = 16 ( x + 1 ) x = 3 / 7 ratio of gold with copper = 1 : 3 / 7 = 7 : 3 answer is b" | a ) 1 : 2 , b ) 7 : 3 , c ) 4 : 1 , d ) 5 : 2 , e ) 6 : 5 | b | divide(subtract(16, 9), subtract(19, 16)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general | B |
there are 9 cities numbered 1 to 9 . from how many cities the flight can start so as to reach the city 8 either directly or indirectly such that the path formed is divisible by 3 . | start from 1 : 1368 , start from 2 : 2358 , start from 3 : 378 , start from 4 : 48 , start from 5 : 528 , start from 6 : 678 , start from 7 : 78 , why start from 8 ? ? ? , start from 9 : 918 . so flights can start from all 8 other cities . answer : b | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | subtract(9, const_1) | subtract(n0,const_1) | general | B |
the length of the bridge , which a train 100 metres long and travelling at 45 km / hr can cross in 30 seconds , is : | "speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 100 + x ) / 30 = 25 / 2 = > 2 ( 100 + x ) = 750 = > x = 275 m . answer : option e" | a ) 230 , b ) 240 , c ) 245 , d ) 250 , e ) 275 | e | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 100) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics | E |
how many different positive integers are factors of 324 ? | "18 Γ 18 = 6 ^ 2 Γ 3 ^ 2 so total factors = ( 5 + 1 ) ( 5 + 1 ) = 36 answer : d" | a ) 9 , b ) 16 , c ) 25 , d ) 36 , e ) 49 | d | add(power(const_2, const_2), const_2) | power(const_2,const_2)|add(#0,const_2)| | other | D |
carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 5 inches by 24 inches and jordan ' s rectangle is 8 inches long , how wide is jordan ' s rectangle , in inches ? | "area of carol ' s rectangle = 24 * 5 = 120 let width of jordan ' s rectangle = w since , the areas are equal 8 w = 120 = > w = 15 answer d" | a ) 25 , b ) 23 , c ) 22 , d ) 15 , e ) 18 | d | divide(rectangle_area(5, 24), 8) | rectangle_area(n0,n1)|divide(#0,n2)| | geometry | D |
at what rate percent on simple interest will rs . 415 amount to rs . 514 in 4 years ? | "explanation : difference in amount = 514 - 415 = 99 99 = ( 415 x 4 x r ) / 100 r = 5.96 % answer is b" | a ) 4.58 % , b ) 5.96 % , c ) 6.52 % , d ) 4.98 % , e ) none of these | b | multiply(divide(divide(subtract(514, 415), 415), 4), const_100) | subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)| | gain | B |
the number 189 is equal to the sum of the cubes of two integers . what is the product of those integers ? | "4 ^ 3 + 5 ^ 3 = 189 the number is 4 * 5 = 20 c" | a ) 8 , b ) 15 , c ) 20 , d ) 27 , e ) 39 | c | multiply(floor(power(divide(189, const_2), divide(const_1, const_3))), power(subtract(189, power(floor(power(divide(189, const_2), divide(const_1, const_3))), const_3)), divide(const_1, const_3))) | divide(n0,const_2)|divide(const_1,const_3)|power(#0,#1)|floor(#2)|power(#3,const_3)|subtract(n0,#4)|power(#5,#1)|multiply(#3,#6)| | general | C |
a person has to make 146 pieces of long bar . he take $ seconds to cut piece . what is total time taken by him to make 146 pieces ? | if it is 4 seconds , as u have mistaken with shift key , then . . . . we ' ll have to make 145 cuts for 146 pieces . so 145 * 5 = 580 seconds answer : a | a ) 580 seconds , b ) 680 seconds , c ) 500 seconds , d ) 540 seconds , e ) 560 seconds | a | multiply(const_4, subtract(146, const_1)) | subtract(n0,const_1)|multiply(#0,const_4) | physics | A |
a student gets 50 % in one subject , 70 % in the other . to get an overall of 70 % how much should get in third subject . | "let the 3 rd subject % = x 50 + 70 + x = 3 * 70 120 + x = 210 x = 210 - 120 = 90 answer : a" | a ) 90 % , b ) 25 % , c ) 45 % , d ) 55 % , e ) 65 % | a | subtract(multiply(70, const_3), add(50, 70)) | add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0)| | gain | A |
at 12.00 hours , j starts to walk from his house at 6 kmph . at 13.30 , p follows him from j ' s house on his bicycle at 8 kmph . when will j be 3 km behind p ? | by the time p starts j is 1.5 hr x 6 = 9 km away from his house . j is 3 km behind when p is 3 km ahead of him . ie . , p has to cover 12 km . so he takes 12 / ( 8 - 6 ) = 6 hrs after 13.30 . so the required time is 19.30 hrs answer : b | a ) 19.39 , b ) 19.3 , c ) 19.32 , d ) 19.33 , e ) 19.36 | b | add(13.3, divide(add(multiply(6, divide(const_3, const_2)), 3), subtract(8, 6))) | divide(const_3,const_2)|subtract(n3,n1)|multiply(n1,#0)|add(n4,#2)|divide(#3,#1)|add(n2,#4) | physics | B |
how long does a train 200 m long running at the speed of 72 km / hr takes to cross a bridge 132 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 200 + 132 = 332 m . required time = 332 / 20 = 16.6 sec . answer : e" | a ) 82.1 sec , b ) 12.1 sec , c ) 16.1 sec , d ) 13.1 sec , e ) 16.6 sec | e | divide(add(200, 132), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics | E |
what is the greatest of 3 consecutive integers whose sum is 39 ? | "39 / 3 = 13 the three numbers are 12 , 13 , and 14 . the answer is a ." | a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | a | add(divide(subtract(39, 3), 3), const_2) | subtract(n1,n0)|divide(#0,n0)|add(#1,const_2)| | physics | A |
if 1 / 2 of the air in a tank is removed with each stroke of a vacuum pump , what fraction of the original amount of air has been removed after 6 strokes ? | "left after 1 st stroke = 1 / 2 left after 2 nd stroke = 1 / 2 * 1 / 2 = 1 / 4 left after 3 rd stroke = 1 / 2 * 1 / 4 = 1 / 8 left after 4 th stroke = 1 / 2 * 1 / 8 = 1 / 16 left after 5 th stroke = 1 / 2 * 1 / 16 = 1 / 32 left after 6 th stroke = 1 / 2 * 1 / 32 = 1 / 64 so removed = 1 - 1 / 64 = 63 / 64" | a ) 63 / 64 , b ) 7 / 8 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | add(add(add(add(divide(1, 2), divide(divide(1, 2), 2)), divide(divide(divide(1, 2), 2), 2)), divide(divide(divide(divide(1, 2), 2), 2), 2)), divide(divide(divide(divide(divide(1, 2), 2), 2), 2), 2)) | divide(n0,n1)|divide(#0,n1)|add(#0,#1)|divide(#1,n1)|add(#2,#3)|divide(#3,n1)|add(#4,#5)|divide(#5,n1)|add(#6,#7)| | physics | A |
how many seconds will a train 110 meters long take to cross a railway platform 165 meters long if the speed of the train is 132 kmph ? | "d = 110 + 165 = 275 s = 132 * 5 / 18 = 110 / 3 mps t = 275 * 3 / 110 = 7.5 sec b ) 7.5 sec" | a ) 6 sec , b ) 7.5 sec , c ) 8 sec , d ) 3 sec , e ) 10 sec | b | divide(add(165, 110), multiply(132, const_0_2778)) | add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)| | physics | B |
a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late . had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late . what is the speed t of the train . | "let y be the balance distance to be covered and x be the former speed . a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late so , y / ( 3 x / 4 ) - y / x = 35 / 60 4 y / 3 x - y / x = 7 / 12 y / x ( 4 / 3 - 1 ) = 7 / 12 y / x * 1 / 3 = 7 / 12 y / x = 7 / 4 4 y - 7 x = 0 . . . . . . . . 1 had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late so , ( y - 24 ) / ( 3 x / 4 ) - ( y - 24 ) / x = 25 / 60 4 ( y - 24 ) / 3 x - ( y - 24 ) / x = 5 / 12 ( y - 24 ) / x ( 4 / 3 - 1 ) = 5 / 12 ( y - 24 ) / x * 1 / 3 = 5 / 12 ( y - 24 ) * 12 = 3 x * 5 ( y - 24 ) * 4 = 5 x 4 y - 5 x = 96 . . . . . . . 2 eq 2 - eq 1 2 x = 96 x = 48 = t ans = c" | a ) a ) 45 , b ) b ) 33 , c ) c ) 48 , d ) d ) 55 , e ) e ) 61 | c | multiply(subtract(divide(multiply(24, 4), 3), 24), divide(add(subtract(35, 25), 50), subtract(35, 25))) | multiply(n2,n4)|subtract(n3,n5)|add(n0,#1)|divide(#0,n1)|divide(#2,#1)|subtract(#3,n4)|multiply(#4,#5)| | physics | C |
the contents of two vessels containing copper and tin in the ratio 2 : 3 and 5 : 7 are mixed in the ratio 3 : 5 . the resulting mixture will have copper and tin in the ratio ? | the ratio of copper and tin the new vessel = ( 2 / 5 * 3 / 8 + 5 / 12 * 5 / 8 ) : ( 3 / 5 * 3 / 8 + 7 / 12 * 5 / 8 ) = 197 / 480 : 283 / 480 = 197 : 283 answer is d | a ) 210 : 283 , b ) 312 : 433 , c ) 178 : 213 , d ) 197 : 283 , e ) 145 : 256 | d | divide(add(multiply(2, 3), multiply(5, 5)), add(multiply(3, 3), multiply(7, 5))) | multiply(n0,n1)|multiply(n2,n2)|multiply(n1,n1)|multiply(n2,n3)|add(#0,#1)|add(#2,#3)|divide(#4,#5) | other | D |
a certain number when divided by 39 leaves a remainder 19 , what is the remainder when the same number is divided by 13 ? | explanation : 39 + 19 = 58 / 13 = 6 ( remainder ) answer : d | a ) 7 , b ) 8 , c ) 9 , d ) 6 , e ) 4 | d | reminder(19, 13) | reminder(n1,n2) | general | D |
before 4 years , dog a β s age was 4 times of dog b β s age and after 4 years , dog a β s age will be 3 times of dog b β s age . what is the difference of dog a β s age and dog b β s now ? | "a - 4 = 4 ( b - 4 ) - - > a - 4 b = - 12 . . . . . . . . . . . . . 1 a + 4 = 3 ( b + 4 ) - - > a - 3 b = 8 . . . . . . . . . . . . . 2 ( 2 ) - ( 1 ) - - > b = 20 - - > a = 3 ( 24 ) = 72 a - b = 72 - 20 = 52 answer : c" | a ) 36 , b ) 42 , c ) 52 , d ) 60 , e ) 64 | c | subtract(subtract(multiply(4, add(subtract(multiply(4, 4), 4), subtract(multiply(4, 4), 4))), subtract(multiply(4, 4), 4)), add(subtract(multiply(4, 4), 4), subtract(multiply(4, 4), 4))) | multiply(n1,n2)|multiply(n0,n1)|subtract(#0,n1)|subtract(#1,n0)|add(#2,#3)|multiply(n1,#4)|subtract(#5,#3)|subtract(#6,#4)| | general | C |
the ratio of 3 numbers is 1 : 2 : 4 and the sum of their squares is 1701 . the sum of the numbers is ? | let the numbers be x , 2 x , 4 x then , x ^ 2 + 4 x ^ 2 + 16 x ^ 2 = 1701 21 x ^ 2 = 1701 x ^ 2 = 81 x = 9 answer is e | a ) a ) 10 , b ) b ) 12 , c ) c ) 15 , d ) d ) 14 , e ) e ) 9 | e | sqrt(divide(1701, add(power(4, 2), add(power(1, 2), power(2, 2))))) | power(n1,n2)|power(n2,n2)|power(n3,n2)|add(#0,#1)|add(#3,#2)|divide(n4,#4)|sqrt(#5) | other | E |
aarti can do a piece of work in 5 days . in how many days will she complete three time of work of same type ? | we have the important relation , more work , more time ( days ) a piece of work can be done in 5 days . three times of work of same type can be done in 5 x 3 = 15 days answer e | a ) 6 days , b ) 18 days , c ) 21 days , d ) 3 days , e ) 15 days | e | multiply(const_3, 5) | multiply(n0,const_3) | physics | E |
a wheel rotates 10 times every minute and moves 20 cm during each rotation . how many ems does the wheel move in 1 hour ? | explanation : number of wheel moves in 1 hour = 10 * 60 = 600 distance moved = 600 * 200 = 12000 answer : d | a ) 20000 , b ) 14000 , c ) 23000 , d ) 120000 , e ) none of these | d | multiply(multiply(const_60, 10), multiply(20, 10)) | multiply(n0,const_60)|multiply(n0,n1)|multiply(#0,#1) | physics | D |
it takes avery 3 hours to build a brick wall while tom can do it in 2 hours . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ? | "avery takes 3 hours tom takes 2 hours efficiency of avery is 1 / 3 units / hr efficiency of tom is 1 / 2 units / hr combined efficiency of tom and avery is 1 / 3 + 1 / 2 = 5 / 6 units / hr since they worked for 1 hour they completed 5 / 6 units of work and 1 / 6 units of work is left which is to be completed by tom ( since avery left ) so time taken by tom to complete the remaining work will be 1 / 6 / 1 / 2 hours = > 1 / 3 * 60 = 20 minutes . . . answer will be ( a )" | a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | a | multiply(divide(subtract(const_1, add(divide(const_1, 3), divide(const_1, 2))), divide(const_1, 2)), const_60) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|subtract(const_1,#2)|divide(#3,#1)|multiply(#4,const_60)| | physics | A |
a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 4 feet and a height of 10 feet , how far ( in feet ) did the oil level drop in the stationary tank ? | "the volume of oil pumped to the tank = the volume of oil taken away from stationary cylinder . pi * 16 * 10 = pi * h * 100 * 100 ( h is distance that the oil level dropped ) h = 160 / 10,000 = 16 / 1000 = 0.016 ft the answer is e ." | a ) 1.6 , b ) 1 , c ) 0.4 , d ) 0.16 , e ) 0.016 | e | divide(volume_cylinder(4, 10), circle_area(100)) | circle_area(n0)|volume_cylinder(n2,n3)|divide(#1,#0)| | geometry | E |
a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 100 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ? | the distance between the nest and the ditch is 100 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 100 = 3000 . d = st 3000 / 1.5 = t , i think we can take 3000 meters as 3 km , then only we get t = 2 . ( 1000 meters = 1 km ) d ) | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 7 | d | divide(divide(multiply(100, multiply(15, const_2)), const_1000), divide(15, const_10)) | divide(n1,const_10)|multiply(n1,const_2)|multiply(n0,#1)|divide(#2,const_1000)|divide(#3,#0) | physics | D |
of 30 applicants for a job , 10 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | c . 9 30 - 3 = 27 27 - 10 - 18 = - 9 then 9 are in the intersection between 4 years experience and degree . answer c | a ) 14 , b ) 13 , c ) 9 , d ) 7 , e ) 5 | c | add(subtract(add(10, 18), subtract(30, 3)), subtract(18, 10)) | add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2) | general | C |
a candidate appearing for an examination has to secure 52 % marks to pass paper i . but he secured only 45 marks and failed by 35 marks . what is the maximum mark for paper i ? | "he secured 45 marks nd fail by 35 marks so total marks for pass the examinatn = 80 let toal marks x x * 52 / 100 = 80 x = 154 answer : b" | a ) 110 , b ) 154 , c ) 130 , d ) 140 , e ) 150 | b | divide(add(45, 35), divide(52, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | gain | B |
eight identical machines can produce 360 aluminum cans per hour . if all of the machines work at the same constant rate , how many cans could 5 such machines produce in 4 hours ? | "8 machines / 360 cans = 5 machines / x cans 8 x = 1800 x = 225 ( 225 ) ( 4 hours ) = 900 cans . the answer is b ." | a ) 675 , b ) 900 , c ) 1,800 , d ) 5,900 , e ) 7,500 | b | subtract(multiply(4, 360), multiply(4, divide(multiply(5, 360), add(const_4, const_4)))) | add(const_4,const_4)|multiply(n0,n2)|multiply(n0,n1)|divide(#2,#0)|multiply(n2,#3)|subtract(#1,#4)| | physics | B |
what is the average of first 21 multiples of 7 ? | "required average = 7 ( 1 + 2 + . . . . + 21 ) / 21 ( 7 / 21 ) x ( ( 21 x 22 ) / 2 ) ( because sum of first 21 natural numbers ) = 77 c" | a ) 67 , b ) 69 , c ) 77 , d ) 79 , e ) 81 | c | multiply(divide(divide(multiply(21, add(21, const_1)), const_2), 21), 7) | add(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)|divide(#2,n0)|multiply(n1,#3)| | general | C |
64 + 5 * 12 / ( 180 / 3 ) = ? | 64 + 5 * 12 / ( 180 / 3 ) = 64 + 5 * 12 / ( 60 ) = 64 + ( 5 * 12 ) / 60 = 64 + 1 = 65 . answer : d | a ) 22 , b ) 77 , c ) 29 , d ) 65 , e ) 21 | d | add(64, divide(multiply(5, 12), divide(180, 3))) | divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2) | general | D |
78 , 64 , 48 , 30 10 , ( . . . ) | "explanation : 78 - 14 = 64 64 - 16 = 48 48 - 18 = 30 30 - 20 = 10 10 - 22 = - 12 answer : option a" | a ) - 12 number , b ) - 14 , c ) 2 , d ) 8 , e ) 6 | a | subtract(negate(30), multiply(subtract(64, 48), divide(subtract(64, 48), subtract(78, 64)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general | A |
a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130,000 and variables costs averaging $ 8 per item . if the manufacturer β s selling price per item is $ 12 , how many items the manufacturer produce and sell to earn an annual profit of $ 140,000 ? | "let the items manufactured or sold bex 130000 + 8 x = 12 x - 140000 4 x = 270000 x = 67500 ans : e" | a ) 2,858 , b ) 8,667 , c ) 21,429 , d ) 35,000 , e ) 67,500 | e | divide(multiply(divide(divide(multiply(add(add(const_3, 8), 12), power(const_100, const_2)), subtract(12, 8)), multiply(const_4, const_2)), const_3), const_1000) | add(n1,const_3)|multiply(const_2,const_4)|power(const_100,const_2)|subtract(n2,n1)|add(n2,#0)|multiply(#4,#2)|divide(#5,#3)|divide(#6,#1)|multiply(#7,const_3)|divide(#8,const_1000)| | general | E |
the length of rectangle is thrice its breadth and its perimeter is 88 m , find the area of the rectangle ? | 2 ( 3 x + x ) = 88 l = 33 b = 11 lb = 33 * 11 = 363 answer : b | ['a ) 432 sq m', 'b ) 363 sq m', 'c ) 452 sq m', 'd ) 428 sq m', 'e ) 525 sq m'] | b | multiply(multiply(divide(88, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(88, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4) | geometry | B |
there is a square with sides of 13 . what is the area of the biggest circle that can be cut out of this square ? | "area of a circle = a = Γ― β¬ r ^ 2 square is 13 wide , so circle ' s diameter would be 13 , and radius would be 6.5 a = Γ― β¬ 6.5 ^ 2 which is approximately 132.73 answer is a" | a ) 132.73 , b ) 231.92 , c ) 530.93 , d ) 113.1 , e ) 204.33 | a | circumface(divide(13, const_2)) | divide(n0,const_2)|circumface(#0)| | geometry | A |
of 60 children , 30 are happy , 10 are sad , and 20 are neither happy nor sad . there are 17 boys and 43 girls . if there are 6 happy boys and 4 sad girls , how many boys are neither happy nor sad ? | "venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too , it becomes unwieldy . in this case , either use the table or logic . table method is shown above ; here is how you will use logic : there are 6 happy boys . there are 4 sad girls but total 10 sad children . so rest 6 sad children must be sad boys . we have 6 happy boys and 6 sad boys . total we have 17 boys . so 17 - 6 - 6 = 5 boys must be neither happy nor sad . answer ( a )" | a ) 5 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | subtract(subtract(17, subtract(10, 4)), 6) | subtract(n2,n7)|subtract(n4,#0)|subtract(#1,n6)| | other | A |
the difference between a positive proper fraction and its reciprocal is 9 / 20 . the fraction is ? | "let the required fraction be x . then 1 - x = 9 x 20 1 - x 2 = 9 x 20 20 - 20 x 2 = 9 x 20 x 2 + 9 x - 20 = 0 20 x 2 + 25 x - 16 x - 20 = 0 5 x ( 4 x + 5 ) - 4 ( 4 x + 5 ) = 0 ( 4 x + 5 ) ( 5 x - 4 ) = 0 x = 4 / 5 option a" | a ) 4 / 5 , b ) 4 / 7 , c ) 5 , d ) 5 / 7 , e ) none | a | divide(const_4, subtract(9, const_4)) | subtract(n0,const_4)|divide(const_4,#0)| | general | A |
a 220 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ? | "relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 220 ) / 9 = 500 / 9 = > x = 280 . answer : option a" | a ) 280 , b ) 270 , c ) 260 , d ) 250 , e ) 240 | a | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 220) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics | A |
the market value of a 10.5 % stock , in which an income of rs . 756 is derived by investing rs . 7000 , brokerage being 1 / 4 % , is : | "face value = rs . 7000 . dividend = 10.5 % . annual income = rs . 756 . brokerage per rs . 100 = rs . 0.25 . dividend is always paid on the face value of a share . face value * dividend / ( market value + brokerage per rs . 100 ) = annual income . = 7000 * 10.5 / 756 = market value of rs . 100 stock + brokerage per rs . 100 . = market value of rs . 100 stock + brokerage per rs . 100 = rs . 97.22 . = market value of rs . 100 stock = rs . 97.22 - re . 0.25 . = market value of rs . 100 stock = rs . 96.97 answer : a" | a ) 96.97 , b ) 114 , c ) 114.75 , d ) 124 , e ) 124.75 | a | subtract(multiply(divide(7000, 756), 10.5), divide(1, 4)) | divide(n2,n1)|divide(n3,n4)|multiply(n0,#0)|subtract(#2,#1)| | gain | A |
a bag contains 20 red jellybeans and 20 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability that all 3 jellybeans removed from the bag are blue ? | "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 3 / 20 answer : option d" | a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | d | divide(choose(20, 3), choose(add(20, 20), 3)) | add(n0,n0)|choose(n0,n2)|choose(#0,n2)|divide(#1,#2)| | probability | D |
find the l . c . m of 16 , 25 , 24 and 10 . | "explanation : 4 x 5 x 2 x 3 x 5 x 2 = 1200 answer : option d" | a ) 1000 , b ) 1400 , c ) 1320 , d ) 1200 , e ) 1650 | d | multiply(multiply(power(const_3, const_3), multiply(power(const_2, const_3), power(add(const_4, const_1), const_2))), divide(divide(divide(divide(divide(24, const_2), const_2), const_3), add(const_4, const_1)), add(const_4, const_1))) | add(const_1,const_4)|divide(n2,const_2)|power(const_2,const_3)|power(const_3,const_3)|divide(#1,const_2)|power(#0,const_2)|divide(#4,const_3)|multiply(#2,#5)|divide(#6,#0)|multiply(#7,#3)|divide(#8,#0)|multiply(#10,#9)| | physics | D |
find the cube root of 185193 . | "observe unit digit here is 3 so unit digit of the answer is 7 . now omit 193 . now find after which perfect cube 193 lies . it is 125 . so our tenth place is 5 . so answer is 57 . answer : b" | a ) 33 , b ) 57 , c ) 99 , d ) 17 , e ) 01 | b | circle_area(185193) | circle_area(n0)| | geometry | B |
if x > 0 , x / 40 + x / 20 is what percent of x ? | "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 40 + 100 / 20 = 2.5 + 5 = 7.5 7.5 is 75 % of 100 = e" | a ) 6 % , b ) 25 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | e | multiply(add(divide(const_1, 40), divide(const_1, 20)), const_100) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|multiply(#2,const_100)| | general | E |
if x is to be chosen at random from the integers between 1 to 5 , inclusive , and y is to be chosen at random from the integers between 7 and 10 , inclusive , what is the probability that x + y will be even ? | x + y will be even if x and y are both even or both odd . p ( x and y are both even ) = 2 / 5 * 2 / 4 = 1 / 5 p ( x and y are both odd ) = 3 / 5 * 2 / 4 = 3 / 10 p ( x + y is even ) = 1 / 5 + 3 / 10 = 1 / 2 the answer is a . | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | a | divide(add(multiply(const_2, const_2), multiply(const_3, const_2)), multiply(5, subtract(5, const_1))) | multiply(const_2,const_2)|multiply(const_2,const_3)|subtract(n1,const_1)|add(#0,#1)|multiply(n1,#2)|divide(#3,#4) | general | A |
a small , rectangular park has a perimeter of 560 feet and a diagonal measurement of 100 feet . what is its area , in square feet ? | "you can avoid a lot of work in this problem by recognizing that , with the info provided , the diagonal forms a triangle inside the rectangle with sides that have a 3 : 4 : 5 ratio . diagonal = 200 2 x + 2 y = 560 , or x + y = 280 a ^ 2 + b ^ 2 = c ^ 2 for each the sides of the triangle using the ratio 3 : 4 : 5 for sides , and knowing c = 100 , you can deduce the following a = 60 b = 80 60 x 80 = 4,800 a is the answer ." | a ) 4,800 , b ) 19,600 , c ) 20,000 , d ) 20,400 , e ) 20,800 | a | multiply(multiply(divide(100, add(const_2, const_3)), const_3), multiply(divide(100, add(const_2, const_3)), const_4)) | add(const_2,const_3)|divide(n1,#0)|multiply(#1,const_3)|multiply(#1,const_4)|multiply(#2,#3)| | geometry | A |
what is the remainder when 9 ^ 1 + 9 ^ 2 + 9 ^ 3 + . . . + 9 ^ 9 is divided by 9 ? | "notice that in the brackets we have the sum of 9 odd multiples of 9 , which yields remainder of 0 upon division by 9 . answer : c" | a ) 2 , b ) 3 , c ) 0 , d ) 5 , e ) none of the above | c | divide(subtract(power(9, 1), power(9, 1)), 9) | power(n0,n1)|power(n2,n1)|subtract(#0,#1)|divide(#2,n4)| | general | C |
what is the rate percent when the simple interest on rs . 9100 amount to rs . 455 in 3 years ? | "455 = ( 4100 * 3 * r ) / 100 r = 3.7 % answer : a" | a ) 3.7 % , b ) 5.7 % , c ) 6.7 % , d ) 8.7 % , e ) 9.7 % | a | divide(multiply(const_100, 455), multiply(9100, 3)) | multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)| | gain | A |
during a sale , the price of a pair of shoes is marked down 13 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | "assume the price = 100 price during sale = 87 price after sale = 100 percent increase = 13 / 87 * 100 = 15 % approx . correct option : d" | a ) 9 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 90 % | d | divide(multiply(13, const_100), subtract(const_100, 13)) | multiply(n0,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain | D |
a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 55 paise per sq m is | explanation : area to be plastered = [ 2 ( l + b ) Γ£ β h ] + ( l Γ£ β b ) = [ 2 ( 25 + 12 ) Γ£ β 6 ] + ( 25 Γ£ β 12 ) = 744 sq m cost of plastering = 744 Γ£ β ( 55 / 100 ) = rs . 409.20 answer : c | a ) rs . 209.20 , b ) rs . 309.20 , c ) rs . 409.20 , d ) rs . 509.20 , e ) none of these | c | multiply(divide(55, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | divide(n3,const_100)|multiply(n0,n2)|multiply(n1,n2)|multiply(n0,n1)|multiply(#1,const_2)|multiply(#2,const_2)|add(#4,#5)|add(#6,#3)|multiply(#7,#0) | physics | C |
a certain article of clothing was discounted during a special sale to 2 / 5 of its original retail price . when the clothing did n ' t sell , it was discounted even further to 1 / 2 of its original retail price during a second sale . by what percent did the price of this article of clothing decrease from the first sale to the second sale ? | "say the original retail price of the item was $ 200 . the price after the first sale = 3 / 5 * $ 200 = $ 120 . the price after the second sale = 1 / 2 * $ 200 = $ 100 . the percent change from the first sale to the second = ( 120 - 100 ) / 120 = 1 / 3 = 16.66 % . answer : d ." | a ) 50 % , b ) 33.33 % , c ) 25 % , d ) 16.66 % , e ) 12.5 % | d | multiply(subtract(divide(2, 5), multiply(divide(2, 5), divide(1, 2))), const_100) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|subtract(#0,#2)|multiply(#3,const_100)| | gain | D |
what is the greatest number of identical bouquets that can be made out of 21 white and 91 red tulips if no flowers are to be left out ? ( two bouquets are identical whenever the number of red tulips in the two bouquets is equal and the number of white tulips in the two bouquets is equal ) | since no flowers are to be left out , then the number of bouquets must be a factor of both 21 and 91 . for example , we can not have 2 bouquets since we can not divide 91 red tulips into 2 bouquets without one tulip left over . only answer choice which is a factor of 91 is e ( 7 ) . answer : e . | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | divide(21, const_3) | divide(n0,const_3) | general | E |
ajay and balu together can do a piece of work in 12 days . balu and charan together can do the same work in 16 days . after ajay has been working at it for 5 days and balu for 7 days , charan finishes it in 13 days . in how many days will charan alone be able to do the work ? | let the total work be 48 units . ajay and balu β s one day work = 48 / 12 = 4 units . balu and charan β s one day work = 48 / 16 = 3 units . ajay β s one day work = 4 - balu β s one day work . balu β s one day work = 3 - charan β s one day work . therefore , ajay β s one day work = 1 + charan β s one day work . also given , 5 x ajay β s one day work + 7 x balu β s one day work + 13 x charan β s one day work = 48 equating all the equations formed , we get charan β s one day work is 2 units . hence , charan can finish the whole work in 48 / 2 = 24 days . answer : b | a ) 16 days , b ) 24 days , c ) 36 days , d ) 48 days , e ) 58 days | b | multiply(inverse(subtract(subtract(const_1, multiply(5, inverse(12))), multiply(subtract(7, 5), inverse(16)))), subtract(13, const_2)) | inverse(n0)|inverse(n1)|subtract(n3,n2)|subtract(n4,const_2)|multiply(n2,#0)|multiply(#1,#2)|subtract(const_1,#4)|subtract(#6,#5)|inverse(#7)|multiply(#8,#3) | physics | B |
find the ratio between 1 hr 30 minutes and 2 hrs 40 minutes . | solution : the ratio = 1 hr 30 min : 2 hr 40 min . 90 min : 160 min = 9 : 16 . answer is a | a ) 9 : 16 , b ) 5 : 9 , c ) 12 : 17 , d ) 13 : 24 , e ) none of these | a | divide(add(multiply(1, const_60), 30), add(multiply(2, const_60), 40)) | multiply(n0,const_60)|multiply(n2,const_60)|add(n1,#0)|add(n3,#1)|divide(#2,#3) | other | A |
every student of a certain school must take one and only one elective course . in last year , 1 / 2 of the students took p . e . as an elective , 1 / 3 of the students took theatre as an elective , and all of the other students took music . in this year , 1 / 3 of the students who took p . e . and 1 / 4 of the students who took theatre left school , other students did not leave , and no fresh student come in . what fraction of all students took p . e . or music ? | lets pick smart numbers . total number of students : 12 p . e . ( 1 / 2 ) : 6 theatre ( 1 / 3 ) : 4 music ( 1 / 6 ) : 2 after leaving school p . e . : 4 theatre : 3 music : 2 new total number of students : 9 p . e . and music : 6 answer 2 / 3 or a | a ) 2 / 3 , b ) 1 / 4 , c ) 7 / 9 , d ) 1 / 5 , e ) 8 / 11 | a | divide(add(subtract(const_1, add(divide(1, 2), divide(1, 3))), multiply(subtract(const_1, divide(1, 3)), divide(1, 2))), add(add(subtract(const_1, add(divide(1, 2), divide(1, 3))), multiply(subtract(const_1, divide(1, 3)), divide(1, 2))), multiply(subtract(const_1, divide(1, 4)), divide(1, 3)))) | divide(n0,n1)|divide(n0,n3)|divide(n0,n7)|add(#0,#1)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#0,#4)|multiply(#1,#5)|subtract(const_1,#3)|add(#6,#8)|add(#9,#7)|divide(#9,#10) | general | A |
1.20 can be expressed in terms of percentage as | "explanation : while calculation in terms of percentage we need to multiply by 100 , so 1.20 * 100 = 120 answer : option a" | a ) 120 % , b ) 0.120 % , c ) 1.20 % , d ) 0.209 % , e ) none of these | a | multiply(1.20, const_100) | multiply(n0,const_100)| | general | A |
31 of the scientists that attended a certain workshop were wolf prize laureates , and 14 of these 31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize , the number of scientists that had received the nobel prize was 3 greater than the number of scientists that had not received the nobel prize . if 50 of the scientists attended that workshop , how many of them were nobel prize laureates ? | "lets solve by creating equation . . w = 31 . . total = 50 . . not w = 50 - 31 = 19 . . now let people who were neither be x , so out of 19 who won nobel = x + 3 . . so x + x + 3 = 19 or x = 8 . . so who won nobel but not wolf = x + 3 = 11 . . but people who won both w and n = 13 . . so total who won n = 11 + 14 = 25 . . c" | a ) a ) 11 , b ) b ) 18 , c ) c ) 25 , d ) d ) 29 , e ) d ) 36 | c | add(add(3, divide(subtract(subtract(50, 31), 3), const_2)), 14) | subtract(n4,n0)|subtract(#0,n3)|divide(#1,const_2)|add(n3,#2)|add(n1,#3)| | physics | C |
in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 2 ounces of corn syrup , how many ounces of water does it contain ? | "standard : fl : corn s : water = 1 : 12 : 30 sport : fl : corn s : water = 3 : 12 : 180 this simplifies to 1 : 4 : 60 if the large bottle has a capacity of x ounces , then 4 x / 65 = 2 . so , x = 32.5 ounces . water = ( 60 / 65 ) * ( 65 / 2 ) = = 30 ounces . ans b" | a ) 15 , b ) 30 , c ) 45 , d ) 60 , e ) 90 | b | multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 2) | divide(n0,n1)|divide(n0,n2)|divide(#1,const_2)|multiply(#0,const_3)|divide(#3,#2)|multiply(n3,#4)| | other | B |
a train requires 10 seconds to pass a pole while it requires 35 seconds to cross a stationary train which is 500 mtrs long . find the speed of the train . | "in 10 s the train crosses the pole and in 35 sec the train crosses one more stationary train in 25 sec the train travels a distance of 500 mtrs speed = 500 / 25 = 20 m / s = 20 ( 3600 / 1000 ) = 20 * 18 / 5 = 72 kmph answer : c" | a ) 68 kmph , b ) 70 kmph , c ) 72 kmph , d ) 60 kmph , e ) 50 kmph | c | multiply(divide(500, subtract(35, 10)), const_3_6) | subtract(n1,n0)|divide(n2,#0)|multiply(#1,const_3_6)| | physics | C |
18 beavers , working together in a constant pace , can build a dam in 8 hours . how many hours will it take 36 beavers that work at the same pace , to build the same dam ? | "total work = 18 * 8 = 144 beaver hours 36 beaver * x = 144 beaver hours x = 144 / 36 = 4 answer : b" | a ) 2 . , b ) 4 . , c ) 5 . , d ) 6 , e ) 8 . | b | divide(multiply(8, 18), 36) | multiply(n0,n1)|divide(#0,n2)| | physics | B |
what is the area of a square field whose diagonal of length 10 m ? | "d 2 / 2 = ( 10 * 10 ) / 2 = 50 answer : a" | a ) 50 , b ) 70 , c ) 60 , d ) 20 , e ) 25 | a | divide(square_area(10), const_2) | square_area(n0)|divide(#0,const_2)| | geometry | A |
a baker makes chocolate cookies and peanut cookies . his recipes allow him to make chocolate cookie in batches of 7 and peanut cookies in batches of 6 . if he makes exactly 99 cookies , what is the minimum number of chocolate chip cookies he makes ? | "7 c + 6 p = 99 we need to maximize p to minimize c so that the eq is also satisfied try substitution for cp to solve so that eqn is satisfied the least value of c for which equation gets satisfied is 5 i . e . 7 * 3 + 6 * 13 = 21 + 78 = 99 hence c is the answer" | a ) 7 , b ) 14 , c ) 21 , d ) 28 , e ) 35 | c | multiply(divide(subtract(99, reminder(99, add(7, 6))), add(7, 6)), 7) | add(n0,n1)|reminder(n2,#0)|subtract(n2,#1)|divide(#2,#0)|multiply(n0,#3)| | general | C |
in a restaurant , the profit is 120 % of the cost . if the cost increases by 12 % but the selling price remains constant , approximately what percentage of the selling price is the profit ? | "explanation : let c . p . = rs . 100 . then , profit = rs . 120 , s . p . = rs . 220 . new c . p . = 112 % of rs . 100 = rs . 112 new s . p . = rs . 220 . profit = rs . ( 220 - 112 ) = rs . 108 . required percentage = ( 108 / 220 * 100 ) % = 49 % appox answer : b" | a ) 30 % , b ) 49 % , c ) 90 % , d ) 100 % , e ) none of these | b | multiply(const_100, divide(subtract(add(multiply(const_100, divide(120, const_100)), const_100), multiply(const_100, add(const_1, divide(12, const_100)))), add(multiply(const_100, divide(120, const_100)), const_100))) | divide(n0,const_100)|divide(n1,const_100)|add(#1,const_1)|multiply(#0,const_100)|add(#3,const_100)|multiply(#2,const_100)|subtract(#4,#5)|divide(#6,#4)|multiply(#7,const_100)| | gain | B |
a person travels equal distances with speeds of 3 km / hr , 6 km / hr , 9 km / hr . and takes a total time of 11 minutes . find the total distance ? | "let the each distance be x km total distance = 3 x then total time , ( x / 3 ) + ( x / 6 ) + ( x / 9 ) = 11 / 60 x = 0.3 total distance = 3 * 0.3 = 0.9 km = 900 meters correct option is c" | a ) 1 km , b ) 500 mts , c ) 900 mts , d ) 2 km , e ) 250 mts | c | multiply(multiply(divide(divide(11, const_60), add(add(divide(const_1, 3), divide(const_1, 6)), divide(const_1, 9))), const_3), const_1000) | divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3)|multiply(#7,const_1000)| | physics | C |
calculate the sum of first 89 natural numbers . | "solution we know that ( 1 + 2 + 3 + . . . . . + 89 ) = n ( n + 1 ) / 2 therefore ( 1 + 2 + 3 + . . . . + 89 ) = ( 89 Γ 90 / 2 ) = 4005 . answer e" | a ) 2017 , b ) 2647 , c ) 2147 , d ) 2045 , e ) 4005 | e | multiply(add(89, const_1), divide(89, const_2)) | add(n0,const_1)|divide(n0,const_2)|multiply(#0,#1)| | general | E |
jill invests $ 10000 in an account that pays an annual rate of 3.96 % , compounding semi - annually . approximately how much e does she have in her account after two years ? | "ps . i guess one can use simple interest to solve cause the answer choices are quite spread between you can easily arrive at something near 8 % hence b the answer" | a ) $ 10079.44 , b ) e = $ 10815.83 , c ) $ 12652.61 , d ) $ 14232.14 , e ) $ 20598.11 | b | multiply(10000, power(add(const_1, divide(divide(3.96, const_100), const_2)), const_4)) | divide(n1,const_100)|divide(#0,const_2)|add(#1,const_1)|power(#2,const_4)|multiply(n0,#3)| | gain | B |
if in a box of dimensions 6 m * 5 m * 4 m smaller boxes of dimensions 60 cm * 50 cm * 40 cm are kept in it , then what will be the maximum number of the small boxes that can be kept in it ? | 6 * 5 * 4 = 60 / 100 * 50 / 100 * 40 / 100 * x 1 = 1 / 10 * 1 / 10 * 1 / 10 * x = > x = 1000 answer : b | a ) 1098 , b ) 1000 , c ) 1628 , d ) 1098 , e ) 1094 | b | divide(multiply(multiply(6, 5), 4), multiply(multiply(divide(60, const_100), divide(50, const_100)), divide(40, const_100))) | divide(n5,const_100)|divide(n3,const_100)|divide(n4,const_100)|multiply(n0,n1)|multiply(n2,#3)|multiply(#1,#2)|multiply(#0,#5)|divide(#4,#6) | physics | B |
what is the smallest integer y for which 27 ^ y > 3 ^ 24 ? | "27 ^ y > 3 ^ 24 converting into the same bases : 27 ^ y > 27 ^ 8 therefore for the equation to hold true , y > 8 or y = 9 option c" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | divide(n2,const_3)|floor(#0)|add(#1,const_1)| | general | C |
what is the smallest positive integer k such that the product of 450 x k is a perfect square ? | "a perfect square , is just an integer that can be written as the square of some other integer . for example 16 = 4 ^ 2 , is a perfect square . now , 450 = 3 ^ 2 * 5 ^ 2 * 3 , so if k = 3 then 450 k = ( 3 * 5 * 3 ) ^ 2 , which is a perfect square ( basically the least positive value of k must complete only the power of 7 to even power as powers of other primes are already even ) . answer : a ." | a ) 3 , b ) 9 , c ) 15 , d ) 25 , e ) 63 | a | add(const_3, const_4) | add(const_3,const_4)| | general | A |
a solution is 90 % glycerin . if there are 4 gallons of the solution , how much water , in gallons must be addded to make a 75 % glycerin solution | in 4 gallons of the solution there are 0.9 β 4 = 3.80 gallons of glycerin . we want to add ww gallons of water to 4 gallons of solution so that these 3.6 gallons of glycerin to be 75 % of new solution : 0.9 β 4 = 0.75 ( 4 + w ) - - > w = 0.8 answer : e . | a ) 1.8 , b ) 1.4 , c ) 1.2 , d ) 1.0 , e ) 0.8 | e | subtract(divide(multiply(divide(90, const_100), 4), divide(75, const_100)), 4) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|divide(#2,#1)|subtract(#3,n1) | general | E |
find the cost of fencing around a circular field of diameter 32 m at the rate of rs . 1.50 a meter ? | "2 * 22 / 7 * 16 = 100.6 100.6 * 1 1 / 2 = rs . 150.9 answer : c" | a ) 287 , b ) 132 , c ) 150.9 , d ) 158 , e ) 267 | c | multiply(circumface(divide(32, const_2)), 1.50) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics | C |
a man rows his boat 95 km downstream and 60 km upstream , taking 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 95 / ( 2 ) = 48 kmph speed upstream = d / t = 60 / ( 2 ) = 30 kmph the speed of the stream = ( 48 - 30 ) / 2 = 9 kmph answer : c" | a ) 76 kmph , b ) 6 kmph , c ) 9 kmph , d ) 8 kmph , e ) 4 kmph | c | divide(subtract(divide(95, 2), divide(60, 2)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics | C |
a certain basketball team that has played 2 / 3 of its games has a record of 18 wins and 2 losses . what is the greatest number of the remaining games that the team can lose and still win at least 3 / 4 of all of its games ? | "18 wins , 2 losses - total 20 games played . the team has played 2 / 3 rd of all games so total number of games = 30 3 / 4 th of 30 is 22.5 so the team must win 23 games and can afford to lose at most 7 total games . it has already lost 2 games so it can lose another 5 at most . answer ( c )" | a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | c | subtract(multiply(divide(3, 2), add(18, 2)), add(floor(multiply(divide(3, 4), multiply(divide(3, 2), add(18, 2)))), const_1)) | add(n2,n3)|divide(n1,n0)|divide(n1,n5)|multiply(#0,#1)|multiply(#2,#3)|floor(#4)|add(#5,const_1)|subtract(#3,#6)| | general | C |
a wooden cube whose edge length is 5 inches is composed of smaller cubes with edge lengths of one inch . the outside surface of the large cube is painted red and then it is split up into its smaller cubes . if one cube is randomly selected from the small cubes , what is the probability that the cube will have at least one red face ? | "there are a total of 5 * 5 * 5 = 125 cubes . all the exterior cubes will have at least one face painted red . the interior is formed by 3 * 3 * 3 = 27 cubes . the number of cubes with at least one side painted red is 125 - 27 = 98 cubes the probability that a cube has at least one side painted red is 98 / 125 which is 78.4 % the answer is c ." | a ) 70.6 % , b ) 74.5 % , c ) 78.4 % , d ) 82.1 % , e ) 86.9 % | c | multiply(const_100, subtract(const_1, divide(volume_cube(multiply(const_1, const_4)), volume_cube(5)))) | multiply(const_1,const_4)|volume_cube(n0)|volume_cube(#0)|divide(#2,#1)|subtract(const_1,#3)|multiply(#4,const_100)| | geometry | C |
the tax on a commodity is diminished by 10 % and its consumption increased by 25 % . the effect on revenue is ? | "100 * 100 = 10000 90 * 125 = 11250 - - - - - - - - - - - 10000 - - - - - - - - - - - 1250 100 - - - - - - - - - - - ? = > 12.5 % decrease answer : e" | a ) 12 % decrease , b ) 18 % decrease , c ) 19 % decrease , d ) 13 % decrease , e ) 12.5 % increase | e | subtract(const_100, multiply(multiply(add(const_1, divide(25, const_100)), subtract(const_1, divide(10, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general | E |
if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 55 minutes ? | "since the population increases at the rate of 1 person every 15 seconds , it increases by 4 people every 60 seconds , that is , by 4 people every minute . thus , in 55 minutes the population increases by 55 x 4 = 220 people . answer . e ." | a ) 80 , b ) 100 , c ) 150 , d ) 180 , e ) 220 | e | multiply(divide(const_60, 15), 55) | divide(const_60,n0)|multiply(n1,#0)| | physics | E |
the original price of a suit is $ 200 . the price increased 30 % , and after this increase , the store published a 30 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 30 % off the increased price , how much did these consumers pay for the suit ? | "given the foregoing discussion , it may be obvious now the trap - mistake answer is ( c ) . even if you can β t remember the correct thing to do , at the very least , learn to spot the trap ! the multiplier for a 30 % increases is 1 + 0.30 = 1.3 , and the multiplier for a 30 % decrease is 1 β 0.30 = 0.70 , so the combined change is 1.3 * 0.7 = 0.91 , 91 % percent of the original , or a 9 % decreases . now , multiply $ 200 * 0.91 = $ 182 . answer = ( a ) ." | a ) $ 182 , b ) $ 191 , c ) $ 200 , d ) $ 209 , e ) $ 219 | a | subtract(add(200, divide(multiply(200, 30), const_100)), divide(multiply(add(200, divide(multiply(200, 30), const_100)), 30), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|subtract(#2,#4)| | general | A |
find the circumference and area of radius 6 cm . | area of circle = Ο r Β² = 22 / 7 Γ 6 Γ 6 cm Β² = 113 cm Β² answer : a | ['a ) 113 cm Β²', 'b ) 144 cm Β²', 'c ) 154 cm Β²', 'd ) 184 cm Β²', 'e ) 194 cm Β²'] | a | circle_area(6) | circle_area(n0) | geometry | A |
the average age of 3 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is | "total age increased = ( 3 * 2 ) years = 6 years . sum of ages of two new men = ( 21 + 23 + 6 ) years = 50 years average age of two new men = ( 50 / 2 ) years = 25 years . answer : c" | a ) 22 , b ) 30 , c ) 25 , d ) 38 , e ) 27 | c | add(divide(add(21, 23), const_2), multiply(const_1, 3)) | add(n1,n2)|multiply(n0,const_1)|divide(#0,const_2)|add(#2,#1)| | general | C |
a man , a woman and a boy can together complete a piece of work in 3 days . if a man alone can do it in 6 days and a boy alone in 9 days , how long will a woman take to complete the work ? | explanation : ( 1 man + 1 woman + 1 boy ) β s 1 day β s work = 1 / 3 1 man β s 1 day work = 1 / 6 1 boy β s 1 day β s work = 1 / 9 ( 1 man + 1 boy ) β s 1 day β s work = 1 / 6 + 1 / 9 = 5 / 18 therefore , 1 woman β s 1 day β s work = 1 / 3 β 5 / 18 = 1 / 18 therefore , the woman alone can finish the work in 18 days . answer : option e | a ) 12 days , b ) 15 days , c ) 16 days , d ) 24 days , e ) 18 days | e | inverse(subtract(inverse(3), add(inverse(6), inverse(9)))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#1,#2)|subtract(#0,#3)|inverse(#4) | physics | E |
ratio between rahul and deepak is 4 : 3 , after 22 years rahul age will be 26 years . what is deepak present age | "explanation : present age is 4 x and 3 x , = > 4 x + 22 = 26 = > x = 1 so deepak age is = 3 ( 1 ) = 3 answer : option b" | a ) 10 , b ) 3 , c ) 5 , d ) 7 , e ) 8 | b | divide(multiply(subtract(26, 22), 3), 4) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other | B |
in an examination 35 % of the students passed and 260 failed . how many students appeared for the examination ? | "let the number of students appeared be x then , 65 % of x = 260 65 x / 100 = 260 x = 260 * 100 / 65 = 400 answer is b" | a ) a ) 540 , b ) b ) 400 , c ) c ) 700 , d ) d ) 650 , e ) e ) 840 | b | divide(multiply(260, const_100), subtract(const_100, 35)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain | B |
there are 60 doors marked with numbers 1 to 60 . there are 60 individuals marked 1 to 60 . an operation on a door is defined as changing the status of the door from open to closed or vice versa . all the doors are closed to start with . one at a time , one randomly chosen individual goes and operates the doors . the individual however operates only those doors which are a multiple of the number he / she is carrying . for example , the individual marked with number 5 operates the doors marked with 5 , 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , and 60 . if every individual in the group gets one turn , then how many doors are open at the end ? | if a door is closed at the start , it requires an odd number of people to operate to be open at the end . only the perfect squares have an odd number of factors . the doors which are open at the end are : 1 , 4 , 9 , 16 , 25 , 36 , 49 for a total of 7 doors . the answer is d . | a ) 0 , b ) 2 , c ) 5 , d ) 7 , e ) 9 | d | add(divide(60, const_10), 1) | divide(n0,const_10)|add(n1,#0) | general | D |
how is 5 % expressed as a decimal fraction ? | "5 / 100 = 0.05 answer : b" | a ) 0.5 , b ) 0.05 , c ) 0.005 , d ) 0.0005 , e ) 5 | b | divide(5, const_100) | divide(n0,const_100)| | gain | B |
what will be the compound interest on a sum of rs . 21,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 21000 * ( 1 + 12 / 100 ) 3 ] = 21000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 29503.49 c . i . = ( 29503.49 - 21000 ) = rs . 8503.49 answer : a" | a ) rs . 8503.49 , b ) rs . 9720 , c ) rs . 10123.20 , d ) rs . 10483.20 , e ) none | a | 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 | A |
a man buys a cycle for rs . 1800 and sells it at a loss of 10 % . what is the selling price of the cycle ? | "s . p . = 90 % of rs . 1800 = 90 / 100 x 1800 = rs . 1620 answer : b" | a ) 1410 , b ) 1620 , c ) 1430 , d ) 1440 , e ) 1540 | b | divide(multiply(subtract(const_100, 10), 1800), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain | B |
an investor can sell her microtron stock for 36 $ per share and her dynaco stock for 56 $ per share , if she sells 300 shares altogether , some of each stock , at an average price per share of 40 $ , how many shares of dynaco stock has she sold ? | "w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) = ( 56 - 40 ) / ( 40 - 36 ) = 16 / 4 = 4 / 1 = number of microtron stocks / number of dynaco stocks so for every 4 microtron stock , she sold 1 dynaco stock . so out of 300 total stocks , ( 1 / 5 ) th i . e . 300 / 5 = 60 must be dynaco stock . answer ( c )" | a ) 52 , b ) 75 , c ) 60 , d ) 136 , e ) 184 | c | divide(multiply(300, divide(40, subtract(56, 36))), divide(add(36, 56), subtract(56, 36))) | add(n0,n1)|subtract(n1,n0)|divide(n3,#1)|divide(#0,#1)|multiply(n2,#2)|divide(#4,#3)| | general | C |
if n is the smallest integer such that 432 times n is the square of an integer , what is the value of n ? | the prime factorization of a square has to have even powers of all its prime factors . if the original number has a factor , say of 7 , then when it β s squared , the square will have a factor of 7 ^ 2 . another way to say that is : any positive integer all of whose prime factors have even powers must be a perfect square of some other integer . look at the prime factorization of 432 432 = ( 2 ^ 4 ) * ( 3 ^ 3 ) the factor of 2 already has an even power β - that β s all set . the factor of 3 currently has an odd power . if n = 3 , then 432 * n would have an even power of 2 and an even power of 3 ; therefore , it would be a perfect square . thus , n = 3 is a choice that makes 432 * n a perfect square . answer : b . | ['a ) 2', 'b ) 3', 'c ) 6', 'd ) 12', 'e ) 24'] | b | divide(divide(divide(divide(divide(divide(432, const_2), const_2), const_2), const_2), const_3), const_3) | divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_3)|divide(#4,const_3) | geometry | B |
a merchant purchased a jacket for $ 54 and then determined a selling price that equalled the purchase price of the jacket plus a markup that was 40 percent of the selling price . during a sale , the merchant discounted the selling price by 20 percent and sold the jacket . what was the merchant β s gross profit on this sale ? | "actual cost = $ 54 sp = actual cost + mark up = actual cost + 40 % sp = 54 * 100 / 60 on sale sp = 80 / 100 ( 54 * 100 / 60 ) = 72 gross profit = $ 18 answer is e" | a ) $ 0 , b ) $ 3 , c ) $ 4 , d ) $ 12 , e ) $ 18 | e | subtract(multiply(divide(54, subtract(const_1, divide(40, const_100))), subtract(const_1, divide(20, const_100))), 54) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|divide(n0,#2)|multiply(#4,#3)|subtract(#5,n0)| | gain | E |
the current of a stream runs at the rate of 4 kmph . a boat goes 6 km and back to the starting point in 4 hours , then find the speed of the boat in still water ? | "s = 4 m = x ds = x + 4 us = x - 4 6 / ( x + 4 ) + 6 / ( x - 4 ) = 4 x = 5.77 answer : d" | a ) a ) 7 , b ) b ) 2 , c ) c ) 8.73 , d ) d ) 5.77 , e ) e ) 3 | d | divide(power(4, 4), 4) | power(n0,n2)|divide(#0,n2)| | physics | D |
a and b enter into a partnership . a contributed rs . 5000 for 8 months and b rs . 6000 for 5 months . find a ' s share in a total profit of rs . 8400 . | explanation : ratio of capitals of a and b = ratio of their investments = 5000 Γ 8 : 6000 Γ 5 = 40000 : 30000 = 4 : 3 . : a ' s share = rs . [ ( 4 / 7 ) Γ 8400 ] = rs . 4800 answer : option a | a ) rs . 4800 , b ) rs . 2400 , c ) rs . 3600 , d ) rs . 5600 , e ) none of these | a | multiply(8400, divide(multiply(5000, 8), add(multiply(5000, 8), multiply(6000, 5)))) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#0,#2)|multiply(n4,#3) | gain | A |
if the sum of the first n positive integers is s , what is the sum of the first n negative even integers , in terms of s ? | answer is c given sum of - 1 - 2 - 3 . . . . . - n = s sum of - 2 , - 4 , - 6 . . . . - 2 n = 2 [ sum ( - 1 - 2 - 3 . . . . n ) ] = 2 s | a ) s / 2 , b ) s , c ) 2 s , d ) 2 s + 2 , e ) 4 s | c | negate(const_2) | negate(const_2) | general | C |
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