Problem stringlengths 5 628 | Rationale stringlengths 1 2.74k | options stringlengths 37 137 | correct stringclasses 5
values | annotated_formula stringlengths 6 848 | linear_formula stringlengths 7 357 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
a batsman makes a score of 76 runs in the 17 th inning and thus increases his average by 3 . find his average after 17 th inning . | "find the pattern of the remainders after each power : ( 2 ^ 1 ) / 7 remainder 2 ( 2 ^ 2 ) / 7 remainder 4 ( 2 ^ 3 ) / 7 remainder 1 - - > this is where the cycle ends ( 2 ^ 4 ) / 7 remainder 2 - - > this is where the cycle begins again ( 2 ^ 5 ) / 7 remainder 4 continuing the pattern to ( 2 ^ 14 ) / 7 gives us a remai... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | reminder(power(2, 14), 7) | power(n0,n1)|reminder(#0,n2)| | general |
find the remainder of the division ( 2 ^ 14 ) / 7 . | "1 / 18 + 1 / 30 = 8 / 90 = 4 / 45 45 / 4 = 11 ¼ * 2 = 22 ½ days answer : b" | a ) 21 ½ days , b ) 22 ½ days , c ) 23 ½ days , d ) 12 ½ days , e ) none of these | b | add(divide(const_1, 18), divide(const_1, 30)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)| | physics |
a can do a job in 18 days and b can do it in 30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "let the additional invested amount for 8 % interest be x ; equation will be ; 2600 + 0.05 * 2600 + x + 0.08 x = 2600 + x + 0.06 ( 2600 + x ) 0.05 * 2600 + 0.08 x = 0.06 x + 0.06 * 2600 0.02 x = 2600 ( 0.06 - 0.05 ) x = 2600 * 0.01 / 0.02 = 1300 ans : ` ` c ' '" | a ) 1200 , b ) 3000 , c ) 1300 , d ) 3600 , e ) 2400 | c | divide(subtract(multiply(divide(6, const_100), 2600), multiply(2600, divide(5, const_100))), subtract(divide(8, const_100), divide(6, const_100))) | divide(n3,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(#2,#0)|subtract(#3,#4)|divide(#6,#5)| | general |
barbata invests $ 2600 in the national bank at 5 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 6 % of her entire investment ? | "most test takers would recognize thesystemof equations in this prompt and just do algebra to get to the solution ( and that ' s fine ) . the wording of the prompt and the ' spread ' of the answer choices actually provide an interesting ' brute force ' shortcut that you can take advantage of to eliminate the 4 wrong an... | a ) 96 , b ) 240 , c ) w = 256 , d ) w = 384 , e ) w = 480 | e | multiply(multiply(16, const_2), 15) | multiply(n1,const_2)|multiply(n2,#0)| | general |
at a restaurant , glasses are stored in two different - sized boxes . one box contains 12 glasses , and the other contains 16 glasses . if the average number of glasses per box is 15 , and there are 16 more of the larger boxes , what is the total number of glasses w at the restaurant ? ( assume that all boxes are fille... | "sum s = n / 2 { 2 a + ( n - 1 ) d } = 11 / 2 { 2 * 40 + ( 11 - 1 ) * 1 } = 11 * 45 = 495 = x number of even number = ( 50 - 40 ) / 2 + 1 = 6 = y x + y = 495 + 6 = 501 d" | a ) 171 , b ) 281 , c ) 391 , d ) 501 , e ) 613 | d | add(multiply(divide(add(40, 50), const_2), add(subtract(50, 40), const_1)), add(divide(subtract(50, 40), const_2), const_1)) | add(n0,n1)|subtract(n1,n0)|add(#1,const_1)|divide(#1,const_2)|divide(#0,const_2)|add(#3,const_1)|multiply(#2,#4)|add(#5,#6)| | general |
if x is equal to the sum of the integers from 40 to 50 , inclusive , and y is the number of even integers from 40 to 50 , inclusive , what is the value of x + y ? | "28 - - - 4 ds = 7 ? - - - - 1 16 - - - - 4 us = 4 ? - - - - 1 m = ? m = ( 7 + 4 ) / 2 = 5.5 answer : e" | a ) 6.5 , b ) 8.6 , c ) 7.5 , d ) 9.2 , e ) 5.5 | e | divide(add(divide(16, 4), divide(28, 4)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
a man swims downstream 28 km and upstream 16 km taking 4 hours each time , what is the speed of the man in still water ? | alice and bob complete 200 km / 4 hours = 50 km / hour bob ' s speed is 50 - 30 = 20 km / hour the answer is a . | a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | a | subtract(divide(add(100, 100), subtract(11, 7)), 30) | add(n1,n1)|subtract(n4,n2)|divide(#0,#1)|subtract(#2,n0) | physics |
alice and bob drive at constant speeds toward each other on a highway . alice drives at a constant speed of 30 km per hour . at a certain time they pass by each other , and then keep driving away from each other , maintaining their constant speeds . if alice is 100 km away from bob at 7 am , and also 100 km away from b... | "explanation : given year 2040 when divided by 4 , leaves a remainder 0 . note : when remainder is 0 , 28 is added to the given year to get the result . so , 2040 + 28 = 2068 answer : e" | a ) 2063 , b ) 2061 , c ) 2111 , d ) 2191 , e ) 2068 | e | add(multiply(subtract(multiply(const_4, const_4), const_2), const_2), 2040) | multiply(const_4,const_4)|subtract(#0,const_2)|multiply(#1,const_2)|add(n0,#2)| | gain |
the calendar of the year 2040 can be used again in the year ? | "5 , 15,25 , 35,40 , 50,55 , 65,75 , 85,95 so there are total 11 such type of numbers . answer : c" | a ) 10 , b ) 12 , c ) 11 , d ) 20 , e ) 25 | c | divide(100, const_10) | divide(n1,const_10)| | general |
how many integers from 1 to 100 exist such that each is divisible by 5 and also has 5 as a digit ? | "sol . speed = [ 144 x 5 / 18 ] m / sec = 40 m / sec . time taken = ( 100 / 40 ) sec = 2.5 sec . answer a" | a ) 2.5 sec , b ) 4.25 sec , c ) 5 sec , d ) 12.5 sec , e ) none | a | divide(100, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
in what time will a train 100 metres long cross an electic pole , if its speed be 144 km / hr ? | "speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x / 12 + x / 8 = 25 2 x + 3 x = 600 x = 120 km answer is d" | a ) 24 km , b ) 30 km , c ) 48 km , d ) 120 km , e ) 15 km | d | divide(multiply(multiply(subtract(10, 2), add(10, 2)), 25), add(subtract(10, 2), add(10, 2))) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|multiply(n2,#3)|divide(#4,#2)| | physics |
a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 25 hour to row to a place and come back , how far is the place ? | "the total number of ways to choose 4 children from 8 is 8 c 4 = 70 the number of ways to choose 2 boys and 2 girls is 4 c 2 * 4 c 2 = 6 * 6 = 36 p ( 2 boys and 2 girls ) = 36 / 70 = 18 / 35 the answer is d ." | a ) 12 / 29 , b ) 14 / 31 , c ) 16 / 33 , d ) 18 / 35 , e ) 20 / 37 | d | divide(multiply(choose(4, const_2), choose(4, const_2)), choose(add(4, 4), 4)) | add(n0,n0)|choose(n0,const_2)|choose(n0,const_2)|choose(#0,n2)|multiply(#1,#2)|divide(#4,#3)| | probability |
from a group of 4 boys and 4 girls , 4 children are to be randomly selected . what is the probability that 2 boys and 2 girls will be selected ? | explanation : number = ( 800 * 300 * 600 ) / 8 * 3 * 2 = 3000000 answer : d | a ) 9800000 , b ) 1000000 , c ) 7500000 , d ) 3000000 , e ) none of these | d | divide(multiply(multiply(multiply(const_4.0, const_100), multiply(3, const_100)), multiply(const_4.0, const_100)), multiply(multiply(8, 3), 3)) | multiply(n2,const_100)|multiply(n1,const_100)|multiply(const_4.0,n1)|multiply(#0,#1)|multiply(n1,#2)|multiply(#3,#0)|divide(#5,#4)| | physics |
a wooden box of dimensions 8 m x 3 m x 6 m is to carry rectangularboxes of dimensions 8 cm x 3 cm x 2 cm . the maximum number ofboxes that can be carried in the wooden box , is | "a rate = 1 / 8 b rate = 1 / 24 ( a + b ) rate = ( 1 / 8 ) + ( 1 / 24 ) = 1 / 6 a & b finish the work in 6 days correct option is e" | a ) 3 , b ) 5 , c ) 4 , d ) 2 , e ) 6 | e | divide(multiply(8, 24), add(8, 24)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | physics |
a can do a work in 8 days . b can do the same work in 24 days . if both a & b are working together in how many days they will finish the work ? | "16 raise to 8 = 2 raise to 32 , now highest power of 2 divisible by 50 ! is 25 + 12 + 6 + 3 + 1 = 47 since 2 raise to 47 is divisible , 2 raise to 32 also will be divisible answer : a" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | reminder(multiply(16, 50), 8) | multiply(n0,n1)|reminder(#0,n2)| | general |
what is the remainder when 50 ! is divided by 16 ^ 8 ? ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 1250 + 1250 = 2500 m . required time = 2500 * 6 / 125 = 120 sec . answer : e" | a ) 228 , b ) 278 , c ) 48 , d ) 27 , e ) 120 | e | add(45, 30) | add(n1,n2)| | physics |
two goods trains each 1250 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "let c . p . of each pencil be re . 1 . then , c . p . of 100 pencils = rs . 100 ; s . p . of 100 pencils = rs . 140 . gain % = 40 / 100 * 100 = 40 % answer : e" | a ) 36 , b ) 37 , c ) 38 , d ) 39 , e ) 40 | e | divide(const_100, divide(100, subtract(140, 100))) | subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1)| | gain |
if the cost price of 140 pencils is equal to the selling price of 100 pencils , the gain percent is | "mother + daughter + child = 140 kg daughter + child = 60 kg mother = 140 - 60 = 80 kg child = 1 / 5 th of mother = ( 1 / 5 ) * 80 = 16 kg so now daughter = 140 - ( mother + child ) = 140 - ( 80 + 16 ) = 44 kg answer : a" | a ) 44 , b ) 47 , c ) 48 , d ) 49 , e ) 50 | a | subtract(60, divide(subtract(140, 60), 5)) | subtract(n0,n1)|divide(#0,n3)|subtract(n1,#1)| | general |
mother , her daughter and her grand child weighs 140 kg . daughter and her daughter ( child ) weighs 60 kg . child is 1 / 5 th of her grand mother . what is the age of the daughter ? | "( 3 x - 9 ) : ( 5 x - 9 ) = 5 : 2 x = 1 = > 3 x = 3 answer : a" | a ) a ) 3 , b ) b ) 98 , c ) c ) 34 , d ) d ) 35 , e ) e ) 62 | a | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | other |
two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 5 : 2 . the first number is : | we have total of 8 parts : 3 parts of sand and 5 parts of cement . in order there to be half sand and half cement ( 4 parts of sand and 4 parts of cement ) , we should remove 1 part of cement . with 1 part of cement comes 3 / 5 parts of sand , so we should remove 1 + 3 / 5 = 8 / 5 part of the mixture , which is ( 8 / 5... | a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 5 , d ) 1 / 7 , e ) 1 / 8 | c | divide(add(const_1, divide(3, 5)), add(5, 3)) | add(n0,n1)|divide(n0,n1)|add(#1,const_1)|divide(#2,#0) | general |
a mixture of sand and cement contains , 3 parts of sand and 5 parts of cement . how much of the mixture must be substituted with sand to make the mixture half sand and half cement ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 65 5 x = 35 x = 7 . x + 12 = 7 + 12 = 19 b" | a ) 17 , b ) 19 , c ) 16 , d ) 18 , e ) 21 | b | divide(add(add(add(add(const_2.0, const_4), add(3, const_4)), add(const_4, const_4)), 65), 5) | add(const_2.0,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)| | general |
the sum of ages of 5 children born 3 years different each is 65 yrs . what is the age of the elder child ? | let the cost price of book = x selling price of book = 24 $ markup % = 20 ( 120 / 100 ) x = 24 = > x = 20 answer e | a ) $ 14.40 , b ) $ 14.00 , c ) $ 10.00 , d ) $ 9.60 , e ) $ 20.00 | e | subtract(24, multiply(divide(20, const_100), 24)) | divide(n0,const_100)|multiply(n1,#0)|subtract(n1,#1) | gain |
a bookseller sells his books at a 20 % markup in price . if he sells a book for $ 24.00 , how much did he pay for it ? | "the profit 0 f 2009 in terms of 2008 = 0.9 * 15 / 10 * 100 = 135 % c" | a ) 80 % , b ) 105 % , c ) 135 % , d ) 124.2 % , e ) 138 % | c | multiply(divide(multiply(15, subtract(const_1, divide(10, const_100))), 10), const_100) | divide(n3,const_100)|subtract(const_1,#0)|multiply(n4,#1)|divide(#2,n1)|multiply(#3,const_100)| | gain |
in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 10 percent , but profits were 15 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "p = $ 10000 r = 6 % t = 12 / 12 years = 1 year s . i . = p * r * t / 100 = 10000 * 6 * 1 / 100 = $ 600 answer is c" | a ) $ 410 , b ) $ 500 , c ) $ 600 , d ) $ 710 , e ) $ 1000 | c | multiply(10000, divide(6, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
find the simple interest on $ 10000 at 6 % per annum for 12 months ? | "milk : water = 5 : 2 5 x : 2 x + 10 = 5 : 3 3 [ 5 x ] = 5 [ 2 x + 10 ] 15 x = 10 x + 50 15 x - 10 x = 50 x = 10 the quantity of milk in the original mixture is = 5 : 2 = 5 + 2 = 7 7 x = 70 short cut method : milk : water = 5 : 2 after adding 10 liters of water milk : water = 5 : 3 milk is same but water increse 10 lit... | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | divide(multiply(10, divide(const_2.0, const_3.0)), subtract(divide(3, add(5, 2)), multiply(divide(2, add(5, 2)), divide(2, 5)))) | add(n4,n1)|divide(n3,n4)|divide(n0,#0)|divide(n1,#0)|multiply(n2,#1)|multiply(#3,#1)|subtract(#2,#5)|divide(#4,#6)| | general |
a mixture contains milk and water in the ratio 5 : 2 . on adding 10 liters of water , the ratio of milk to water becomes 5 : 3 . the quantity of milk in the original mixture is ? | let ' s see , the way i did it was 12 / 25 are clerical out of 3600 so 1728 are clerical 1728 reduced by 1 / 4 is 1728 * 1 / 4 so it reduced 432 people , so there is 1296 clerical people left but since 432 people left , it also reduced from the total of 3600 so there are 3168 people total since 1296 clerical left / 316... | a ) 40 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | a | multiply(divide(multiply(divide(12, 25), subtract(1, divide(1, 4))), add(multiply(divide(12, 25), subtract(1, divide(1, 4))), subtract(const_1, divide(12, 25)))), const_100) | divide(n1,n2)|divide(n3,n4)|subtract(n3,#1)|subtract(const_1,#0)|multiply(#0,#2)|add(#4,#3)|divide(#4,#5)|multiply(#6,const_100) | general |
of the 3,600 employees of company x , 12 / 25 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ? | "in such a case there is always a loss loss % = ( 13 / 10 ) ^ 2 = 120 / 71 = 1.69 % answer is a" | a ) 1.69 % , b ) 2.56 % , c ) 3.12 % , d ) 4.65 % , e ) 5.12 % | a | multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 13)), 675958), multiply(divide(const_100, subtract(const_100, 13)), 675958)), add(675958, 675958)), add(multiply(divide(const_100, add(const_100, 13)), 675958), multiply(divide(const_100, subtract(const_100, 13)), 675958))), const_100) | add(n1,const_100)|add(n0,n0)|subtract(const_100,n1)|divide(const_100,#0)|divide(const_100,#2)|multiply(n0,#3)|multiply(n0,#4)|add(#5,#6)|subtract(#7,#1)|divide(#8,#7)|multiply(#9,const_100)| | gain |
a man two flats for $ 675958 each . on one he gains 13 % while on the other he loses 13 % . how much does he gain or lose in the whole transaction ? | "30 / 3 = 10 the three numbers are 9 , 10 , and 11 . the answer is d ." | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | d | add(divide(subtract(30, 3), 3), const_2) | subtract(n1,n0)|divide(#0,n0)|add(#1,const_2)| | physics |
what is the greatest of 3 consecutive integers whose sum is 30 ? | solution : sum of pass students of first , second and third class , = ( 45 % of 10 ) + ( 60 % of 15 ) + ( 80 % of 25 ) = 4.5 + 9 + 20 = 33.5 total students appeared , = 10 + 15 + 25 = 50 pass average , = 33.5 * 100 / 50 = 67 % . answer : option c | a ) 74 % , b ) 75 % , c ) 67 % , d ) 72 % , e ) none | c | divide(multiply(add(add(divide(multiply(10, 45), const_100), divide(multiply(15, 60), const_100)), divide(multiply(25, 80), const_100)), const_100), add(add(10, 15), 25)) | add(n1,n3)|multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(n5,#0)|divide(#1,const_100)|divide(#2,const_100)|divide(#3,const_100)|add(#5,#6)|add(#8,#7)|multiply(#9,const_100)|divide(#10,#4) | general |
students of 3 different classes appeared in common examination . pass average of 10 students of first class was 45 % , pass average of 15 students of second class was 60 % and pass average of 25 students of third class was 80 % then what will be the pass average of all students of 3 classes ? | "the triangle with sides 20 cm , 12 cm and 16 cm is right angled , where the hypotenuse is 20 cm . area of the triangle = 1 / 2 * 12 * 16 = 96 cm 2 answer : option d" | a ) 70 , b ) 79 , c ) 85 , d ) 96 , e ) 92 | d | divide(multiply(12, 16), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
if the sides of a triangle are 20 cm , 12 cm and 16 cm , what is its area ? | "d = 100 + 150 = 250 s = 36 * 5 / 18 = 10 mps t = 250 / 10 = 25 sec . answer : c" | a ) 2 , b ) 28 , c ) 25 , d ) 99 , e ) 12 | c | divide(add(150, 100), multiply(36, const_0_2778)) | add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)| | physics |
how many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ? | 34.94 240.016 + 23.98 - - - - - - - - 298.936 answer is a . | a ) 298.936 , b ) 298.694 , c ) 289.496 , d ) 289.469 , e ) 298.964 | a | add(add(34.94, 240.016), 23.98) | add(n0,n1)|add(n2,#0) | general |
34.94 + 240.016 + 23.98 = ? | total cost of items : 2000 / - amount paid : 3000 / - balance receivable : 3000 - 2000 = 1000 / - answer is b | a ) 650 , b ) 1000 , c ) 1500 , d ) 800 , e ) 750 | b | subtract(3000, 2000) | subtract(n3,n2) | general |
meera purchased two 3 items from a shop . total price for 3 items is rs . 2000 / - she have given rs . 3000 / - what is the balance amount meera got ? | "correct average = 35 x 72 + ( 86 - 36 ) / 35 ≈ 72 + 1.43 = 73.43 answer d" | a ) 73.41 , b ) 74.31 , c ) 72.43 , d ) 73.43 , e ) can not be determined | d | divide(subtract(multiply(35, 72), subtract(86, 36)), 35) | multiply(n0,n2)|subtract(n4,n3)|subtract(#0,#1)|divide(#2,n0)| | general |
a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 72 . if the marks secured by reema was written as 36 instead of 86 then find the correct average marks up to two decimal places . | or u can just use the answer choices here . since the answers are already arranged in ascending order , the first number which gives remainder e as 1 for all three is the correct answer . in the given question , the first number which gives a remainder of 1 for 68 and 10 is 121 . c | a ) 21 , b ) 41 , c ) e = 121 , d ) 241 , e ) 481 | c | add(lcm(10, lcm(6, 8)), const_1) | lcm(n2,n3)|lcm(n4,#0)|add(#1,const_1) | general |
what is the smallest integer e greater than 1 that leaves a remainder of 1 when divided by any of the integers 6 , 8 , and 10 ? | "anyways , one can infer that he ' steals ' 30 % from suppliers and then charges 40 % extra to customers so basically 1.3 * 1.4 = 1.82 given that 1 is start point , we get 21 % more hence answer is b" | a ) 28 % , b ) 82 % , c ) 24.33 % , d ) 29.109 % , e ) 78 % | b | subtract(multiply(divide(add(const_100, 40), const_100), add(const_100, 30)), const_100) | add(n0,const_100)|add(n1,const_100)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,const_100)| | gain |
a trader cheats both his supplier and customer by using faulty weights . when he buys from the supplier , he takes 30 % more than the indicated weight . when he sells to his customer , he gives the customer a weight such that 40 % of that is added to the weight , the weight claimed by the trader is obtained . if he cha... | "milk quantity = 3 / 4 * 60 = 45 water quantity = 60 - 45 = 15 new ratio of m : w = 45 : 15 + x = 3 : 2 45 + 3 x = 90 x = 15 answer is b" | a ) 1 , b ) 15 , c ) 7 , d ) 5 , e ) 12 | b | multiply(subtract(divide(multiply(divide(3, add(3, 1)), 60), divide(3, add(3, 2))), 60), divide(add(const_10, 1), const_10)) | add(const_10,n2)|add(n1,n2)|add(n1,n4)|divide(#0,const_10)|divide(n1,#1)|divide(n1,#2)|multiply(n0,#4)|divide(#6,#5)|subtract(#7,n0)|multiply(#3,#8)| | general |
in 60 litres mixture milk and water are in the ratio 3 : 1 . after adding how many liters of water its ratio will become 3 : 2 | "2 / 2 = 1 and 13 / 2 = 6 6 - 1 = 5 5 + 1 = 6 numbers . answer : e" | a ) a ) 2 , b ) b ) 3 , c ) c ) 5 , d ) d ) 7 , e ) e ) 6 | e | add(divide(subtract(multiply(floor(divide(13, 2)), 2), multiply(add(floor(divide(2, 2)), const_1), 2)), 2), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general |
how many numbers from 2 to 13 are exactly divisible by 2 ? | "26 + 20 - 17 = 29 37 - 29 = 8 play neither answer is b" | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | subtract(37, subtract(add(26, 20), 17)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
in a class of 37 students 26 play football and play 20 long tennis , if 17 play above , many play neither ? | "speed of boat in still water = 15 km / hr speed of the stream = 6 km / hr speed downstream = ( 15 + 6 ) = 21 km / hr time taken to travel 86 km downstream = 86 â „ 16 = 17 â „ 4 = 4.1 hours answer is a" | a ) 4.1 hr , b ) 5.25 hr , c ) 8.25 hr , d ) 2.25 hr , e ) 2.50 hr | a | divide(86, add(15, 6)) | add(n0,n1)|divide(n2,#0)| | physics |
a boat can travel with a speed of 15 km / hr in still water . if the speed of the stream is 6 km / hr , find the time taken by the boat to go 86 km downstream . | "let x be the number and y be the quotient . then , x = 357 * y + 38 = ( 17 * 21 * y ) + ( 17 * 2 ) + 4 = 17 * ( 21 y + 2 ) + 4 . required number = 4 . answer is a" | a ) 4 , b ) 5 , c ) 8 , d ) 7 , e ) 2 | a | multiply(subtract(divide(power(38, const_2), 357), floor(divide(power(38, const_2), 357))), 357) | power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)| | general |
on dividing a number by 357 , we get 38 as remainder . on dividing the same number by 17 , what will be the remainder ? | given 30 % ( income ) = 300 ⇒ ⇒ income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 12 % ( 700 ) = rs . 84 answer : c | a ) 62 , b ) 140 , c ) 84 , d ) 60 , e ) 123 | c | multiply(subtract(divide(300, divide(30, const_100)), 300), divide(12, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(n2,#1)|subtract(#2,n2)|multiply(#0,#3) | gain |
bhanu spends 30 % of his income on petrol on scooter 12 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ? | "let the length of the train be x meters . when a train crosses an electric pole , the distance covered is its own length x . speed = 36 km / h = 36000 m / 3600 s = 10 m / s x = 15 * 10 = 150 m . the time taken to pass the platform = ( 150 + 370 ) / 10 = 52 seconds the answer is d ." | a ) 46 , b ) 48 , c ) 50 , d ) 52 , e ) 54 | d | divide(add(multiply(multiply(36, const_0_2778), 15), 370), multiply(36, const_0_2778)) | multiply(n0,const_0_2778)|multiply(n1,#0)|add(n2,#1)|divide(#2,#0)| | physics |
a train running at a speed of 36 km / h passes an electric pole in 15 seconds . in how many seconds will the whole train pass a 370 - meter long platform ? | "work done by the tank in 1 hour = ( 1 / 3 - 3 1 / 3 ) = 1 / 30 leak will empty the tank in 30 hrs . answer : c" | a ) 17 hr , b ) 19 hr , c ) 30 hr , d ) 14 hr , e ) 16 hr | c | inverse(subtract(divide(1, 3), inverse(divide(add(multiply(3, 3), 1), 3)))) | divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)| | physics |
a pump can fill a tank with water in 3 hours . because of a leak , it took 3 1 / 3 hours to fill the tank . the leak can drain all the water of the tank in ? | "soln : - 13 x = 45 - - > 87 / 77 * x = 45 - - > x = 45 * 77 / 87 = 677 / 17 = ~ 40 . answer : c ." | a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | c | multiply(divide(const_100, add(const_100, 13)), 45) | add(n4,const_100)|divide(const_100,#0)|multiply(n7,#1)| | gain |
from january 1 , 2015 , to january 1 , 2017 , the number of people enrolled in health maintenance organizations increased by 13 percent . the enrollment on january 1 , 2017 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 2015 ? | "let the universal set be x = { all students in the graduate physics course } , such that n ( x ) = 100 it will contain 2 mutually exclusive sets ; m ( all male students ) & f ( all female students ) , where n ( m ) = 70 , n ( f ) = 30 now 2 / 7 of all male students are married , implying their number = 20 . however th... | a ) 2 / 7 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 7 | d | divide(const_10, 30) | divide(const_10,n1)| | gain |
in a graduate physics course , 70 percent of the students are male and 30 percent of the students are married . if two - sevenths of the male students are married , what fraction of the female students is single ? | "say the rate of bob is 3 mph and he covers 6 miles then he needs 6 / 3 = 2 hours to do that . now , in this case the rate of ann would be 3 + 3 * 1 / 3 = 4 mph and the distance she covers would be 6 * 2 = 12 miles , so she needs 12 / 4 = 3 hours for that . the ratio r of ann ' s time to bob ' s time is 3 : 2 . answer ... | a ) 8 : 3 , b ) 3 : 2 , c ) 4 : 3 , d ) 2 : 3 , e ) 3 : 8 | b | divide(const_2, add(const_1, divide(const_1, const_3))) | divide(const_1,const_3)|add(#0,const_1)|divide(const_2,#1)| | general |
ann and bob drive separately to a meeting . ann ' s average driving speed is greater than bob ' s avergae driving speed by one - third of bob ' s average driving speed , and ann drives twice as many miles as bob . what is the ratio r of the number of hours ann spends driving to the meeting to the number of hours bob sp... | "eq 1 : a - b - c + d = 18 eq 2 : a + b - c - d = 4 ( 1 ) subtract eq 1 from eq 2 a - b - c + d = 18 - a + b - c - d = 4 - - - - - - - - - - - - - - - - - - - - - - - - - 2 b + 2 d = 14 ( 2 ) simplify - b + d = 7 b - d = - 7 ( b - d ) ^ 2 = ( - 7 ) ^ 2 = 49 my answer : a" | a ) 49 . , b ) 8 . , c ) 12 . , d ) 16 . , e ) 64 . | a | power(subtract(4, divide(add(18, 4), 2)), 2) | add(n0,n1)|divide(#0,n2)|subtract(n1,#1)|power(#2,n2)| | general |
if ( a - b - c + d = 18 ) and ( a + b - c - d = 4 ) , what is the value of ( b - d ) ^ 2 ? | given that the total purchase of two items cost 800 . so the average purchase of one item will cost 800 / 2 = 400 . its given as total shirt cost 400 $ less . hence total shirt cost = 400 - 200 and total trouser cost = 400 + 200 5 shirts = 200 $ = = > one shirt = 40 $ one trouser = 40 + 20 = 60 $ total trousers = 600 /... | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 10 | e | divide(subtract(800, 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 |
john purchased some shirts and trousers for $ 800 . he paid $ 400 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 ? | "total age of all students = 24 ã — 23 total age of all students + age of the teacher = 25 ã — 24 age of the teacher = 25 ã — 24 â ˆ ’ 24 ã — 23 = 24 ( 25 â ˆ ’ 23 ) = 24 ã — 2 = 48 answer is c ." | a ) 40 , b ) 41 , c ) 48 , d ) 52 , e ) 43 | c | subtract(multiply(add(24, 1), add(23, 1)), multiply(24, 23)) | add(n0,n2)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
the average age of a class of 24 students is 23 years . the average increased by 1 when the teacher ' s age also included . what is the age of the teacher ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 27 : 45 : : 18 : x [ please pay attention , we have written 27 : 45 rather than 45 : 27 , in indirect proportion , if you get it then chain rule is clear to you : ) ] { \ color { blue } x = \ frac { 45 \ times 18 } { 27... | a ) 24 , b ) 77 , c ) 30 , d ) 25 , e ) 13 | c | divide(multiply(18, 45), 27) | multiply(n0,n1)|divide(#0,n2)| | physics |
45 men can complete a piece of work in 18 days . in how many days will 27 men complete the same work ? | "( 27 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 60 60 – 27 = 33 answer : a" | a ) 33 , b ) 66 , c ) 88 , d ) 100 , e ) 281 | a | subtract(divide(multiply(divide(multiply(27, 8), 30), 50), 6), 27) | multiply(n0,n1)|divide(#0,n2)|multiply(n3,#1)|divide(#2,n4)|subtract(#3,n0)| | physics |
27 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "you have 6 digits : 2 , 3 , 4 , 5 , 6 , 7 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as sma... | a ) 59 , b ) 49 , c ) 58 , d ) 113 , e ) 131 | b | subtract(subtract(const_100, multiply(subtract(7, 2), const_10)), const_1) | subtract(n5,n1)|multiply(#0,const_10)|subtract(const_100,#1)|subtract(#2,const_1)| | general |
n and m are each 3 - digit integers . each of the numbers 2 , 3 , 4,5 , 6 , and 7 is a digit of either n or m . what is the smallest possible positive difference between n and m ? | "e if we calculate we will get 4749" | a ) 2449 , b ) 5449 , c ) 6749 , d ) 6449 , e ) 4749 | e | subtract(multiply(divide(54671, const_100), 14456), multiply(divide(const_1, const_3), multiply(divide(54671, const_100), 14456))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
54671 - 14456 - 35466 = ? | "let 1 man does 1 unit / hr of work 15 m in 21 days of 8 hrs will do ( 15 * 21 * 8 ) units 3 w = 2 m 1 w = ( 2 / 3 ) units / hr 21 w with 4 hrs a day will take ( 15 * 21 * 8 ) / ( 21 * 4 * ( 2 / 3 ) ) days = > 45 days answer : e" | a ) 30 , b ) 20 , c ) 15 , d ) 25 , e ) 45 | e | divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 4), divide(2, 3))) | divide(n6,n5)|multiply(n0,n1)|multiply(n1,n3)|multiply(n2,#1)|multiply(#0,#2)|divide(#3,#4)| | physics |
15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 4 hrs . each would it take for 21 women if 3 women do as much work as 2 men ? | "the total number of ways to choose 6 children from 8 is 8 c 6 = 28 the number of ways to choose 3 boys and 3 girls is 4 c 3 * 4 c 3 = 4 * 4 = 16 p ( 3 boys and 3 girls ) = 16 / 28 = 4 / 7 the answer is d ." | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 5 , d ) 4 / 7 , e ) 5 / 9 | d | divide(multiply(choose(4, const_2), choose(4, const_2)), choose(add(4, 4), 6)) | add(n0,n0)|choose(n0,const_2)|choose(n0,const_2)|choose(#0,n2)|multiply(#1,#2)|divide(#4,#3)| | probability |
from a group of 4 boys and 4 girls , 6 children are to be randomly selected . what is the probability that 3 boys and 3 girls will be selected ? | "d = ( d - r ) / q = ( 21 - 1 ) / 10 = 20 / 10 = 2 b" | a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 7 | b | floor(divide(21, 10)) | divide(n0,n1)|floor(#0)| | general |
on dividing 21 by a number , the quotient is 10 and the remainder is 1 . find the divisor . | side of square = √ 36 = 6 m . length = 6 m and breadth = 3 m area of rectangle = 6 * 3 = 18 sq . m answer a | ['a ) 18', 'b ) 20', 'c ) 27', 'd ) 32', 'e ) 25'] | a | multiply(sqrt(36), divide(sqrt(36), const_2)) | sqrt(n0)|divide(#0,const_2)|multiply(#1,#0) | geometry |
if length of a rectangle is equal to side of a square and breadth of rectangle is half of length . if area of square is 36 sq . m . calculate the area of rectangle ? | "money paid in cash = rs . 1000 balance payment = ( 70000 - 1000 ) = rs . 69000 answer : c" | a ) 22678 , b ) 26699 , c ) 69000 , d ) 19000 , e ) 26711 | c | subtract(70000, 1000) | subtract(n0,n2)| | gain |
the price of a t . v . set worth rs . 70000 is to be paid in 20 installments of rs . 1000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "0.8 * ( 1.2 * 200 ) = $ 192 the answer is a ." | a ) $ 192 , b ) $ 198 , c ) $ 200 , d ) $ 208 , e ) $ 216 | a | subtract(add(200, divide(multiply(200, 20), const_100)), divide(multiply(add(200, divide(multiply(200, 20), const_100)), 20), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|subtract(#2,#4)| | general |
the original price of a suit is $ 200 . the price increased 20 % , and after this increase , the store published a 20 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 20 % off the increased price , how much did these consumers pay for the suit ? | "998 x 998 = ( 998 ) 2 = ( 1000 - 2 ) 2 = ( 1000 ) 2 + ( 2 ) 2 - ( 2 x 1000 x 2 ) = 1000000 + 4 - 4000 = 1000004 - 4000 = 996004 . c )" | a ) 996000 , b ) 1000000 , c ) 996004 , d ) 4000 , e ) 996008 | c | multiply(divide(998, 998), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
998 x 998 = ? | "distance traveled in 2 hours = 2 * 55 = 110 m distance traveled in 3 hours = 3 * 80 = 240 m total distance covered = 240 + 110 = 350 m total time = 2 + 3 = 5 h hence avg speed = total distance covered / total time taken = 350 / 5 = 70 mph answer : c" | a ) 60 mph , b ) 56.67 mph , c ) 70 mph , d ) 64 mph , e ) 66.67 mph | c | add(divide(add(multiply(80, 3), multiply(55, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | add(n0,n2)|divide(const_100,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#2,#3)|subtract(#1,const_0_33)|divide(#4,#0)|add(#6,#5)| | physics |
steve traveled the first 2 hours of his journey at 55 mph and the remaining 3 hours of his journey at 80 mph . what is his average speed for the entire journey ? | let dravid scored point = x then dhoni scored = x + 30 shewag scored = 2 * ( x + 30 ) = 2 x + 60 as given , x + x + 30 + 2 x + 60 = 150 points 4 x + 90 = 150 x = 150 - 90 / 4 = 15 so dhoni scored = x + 30 i . e ) 15 + 30 = 45 answer : e | a ) 50 , b ) 52 , c ) 35 , d ) 40 , e ) 45 | e | divide(add(150, 30), add(add(const_2, const_1), const_1)) | add(n0,n2)|add(const_1,const_2)|add(#1,const_1)|divide(#0,#2) | general |
in a basketball game , dhoni scored 30 points more than dravid , but only half as many points as shewag . if the 3 players scored a combined total of 150 points , how many points did dhoni score ? | if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 / 4 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or... | a ) 1 / 5 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 7 , e ) 1 / 3 | d | divide(1, divide(add(add(2.8, add(2.8, 2.8)), add(2.8, 2.8)), const_2)) | add(n1,n1)|add(n1,#0)|add(#1,#0)|divide(#2,const_2)|divide(n0,#3) | general |
on a partly cloudy day , milton decides to walk back from work . when it is sunny , he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction of the total distance did ... | rate = 10 % time = 2 years s . i . = $ 1200 principal = 100 * 1200 / 10 * 2 = $ 6000 amount = 6000 ( 1 + 10 / 100 ) ^ 2 = $ 7260 c . i . = 7260 - 6000 = $ 1260 answer is a | a ) $ 1260 , b ) $ 1520 , c ) $ 1356 , d ) $ 1440 , e ) $ 1210 | a | subtract(add(divide(multiply(add(divide(multiply(1200, const_100), multiply(10, 2)), divide(multiply(divide(multiply(1200, const_100), multiply(10, 2)), 10), const_100)), 10), const_100), add(divide(multiply(1200, const_100), multiply(10, 2)), divide(multiply(divide(multiply(1200, const_100), multiply(10, 2)), 10), con... | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|multiply(n0,#2)|divide(#3,const_100)|add(#2,#4)|multiply(n0,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2) | gain |
if the sample interest on a sum of money 10 % per annum for 2 years is $ 1200 , find the compound interest on the same sum for the same period at the same rate ? | "let the number be x . then , 50 % of x - 35 % of x = 12 50 / 100 x - 35 / 100 x = 12 x = ( 12 * 100 ) / 15 = 80 . answer : d" | a ) 40 , b ) 50 , c ) 60 , d ) 80 , e ) 70 | d | divide(12, divide(subtract(50, 35), const_100)) | subtract(n2,n0)|divide(#0,const_100)|divide(n1,#1)| | gain |
if 35 % of a number is 12 less than 50 % of that number , then the number is ? | "explanation : l . c . m of 441 = 3 x 3 x 7 x 7 3 , 7 number of different prime factors is 2 . answer : option b" | a ) 4 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | b | add(const_2, const_2) | add(const_2,const_2)| | other |
find the number of different prime factors of 441 | let first parrt is x and second part is y then x + y = 50000 - - - - - - - - - - eq 1 total profit = profit on x + profit on y 8000 = ( x * 10 * 1 ) / 100 + ( y * 20 * 1 ) / 100 80000 = x + 2 y - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - eq 2 80000 = 50000 + y so y = 30000 then x = 50000 - 300... | a ) 20000 , b ) 40000 , c ) 50000 , d ) 60000 , e ) 70000 | a | divide(subtract(divide(multiply(50000, 20), const_100), 8000), divide(10, const_100)) | divide(n1,const_100)|multiply(n0,n2)|divide(#1,const_100)|subtract(#2,n3)|divide(#3,#0) | gain |
rs 50000 is divided into two parts one part is given to a person with 10 % interest and another part is given to a person with 20 % interest . at the end of first year he gets profit 8000 find money given by 10 % ? | "22 = ( n + n + 2 + n + 4 + . . . + ( n + 18 ) ) / 10 22 = ( 10 n + ( 2 + 4 + . . . + 18 ) ) / 10 220 = 10 n + 2 ( 1 + 2 + . . . + 9 ) 220 = 10 n + 2 ( 9 ) ( 10 ) / 2 220 = 10 n + 90 220 - 90 = 10 n 130 = 10 n n = 13 so the first three numbers are 13 , 15 , 17 13 + 15 + 17 = 45 option b" | a ) 13 , b ) 45 , c ) 17 , d ) 220 , e ) 90 | b | add(divide(subtract(multiply(22, 10), add(add(add(add(const_1, add(add(add(add(add(const_1, const_2), const_1), const_1), const_1), const_1)), const_1), const_1), add(add(add(add(const_1, add(add(add(add(add(const_1, const_2), const_1), const_1), const_1), const_1)), const_1), const_1), const_1))), 10), add(add(add(add... | add(const_1,const_2)|multiply(n0,n1)|add(#0,const_1)|add(#2,const_1)|add(#3,const_1)|add(#4,const_1)|add(#5,const_1)|add(#6,const_1)|add(#7,const_1)|add(#8,const_1)|add(#8,#9)|subtract(#1,#10)|divide(#11,n0)|add(#4,#12)| | general |
the average of 10 consecutive odd numbers is 22 . what is the sum of the first 3 numbers ? | "easy solution : n = dq 1 + 245 2 n = 2 dq 1 + 490 - ( 1 ) 2 n = dq 2 + 112 - ( 2 ) as ( 1 ) = ( 2 ) = 2 n d * ( q 2 - 2 q 1 ) = 378 d * some integer = 378 checking all options only ( a ) syncs with it . answer a" | a ) 378 , b ) 365 , c ) 380 , d ) 456 , e ) 460 | a | subtract(multiply(245, const_2), 112) | multiply(n0,const_2)|subtract(#0,n1)| | general |
a number when divided by a certain divisor left remainder 245 , when twice the number was divided by the same divisor , the remainder was 112 . find the divisor ? | given 50 % of 8 employees including 4 who are capable of doing task . 60 % of 5 employeees = 50 / 100 * 4 = 4 employees = = = > 4 employees who are capable of doing the task and no one employee who is not capable . percentage of employees assigned who are not capable answer : e | a ) 43.33 % , b ) 33.33 % , c ) 13.33 % , d ) 38.33 % , e ) none | e | multiply(divide(subtract(5, 4), 5), const_100) | subtract(n2,n0)|divide(#0,n2)|multiply(#1,const_100) | general |
4 out of 8 employees are capable of doing a certain task . sixty percent of the 5 employees , including the 4 who are capable , are assigned to a project involving this task . what percentage of employees assigned to the project are not capable ? | "answer = a please refer diagram below 85 - 10 = 75" | a ) 75 , b ) 80 , c ) 85 , d ) 90 , e ) 95 | a | subtract(85, 10) | subtract(n0,n1)| | general |
85 white and black tiles will be used to form a 10 x 10 square pattern . if there must be at least one black tile in every row and at least one white tile in every column , what is the maximum difference between the number of black and white tiles that can be used ? | "let the side of the square be a cm . parameter of the rectangle = 2 ( 16 + 14 ) = 60 cm parameter of the square = 60 cm i . e . 4 a = 60 a = 15 diameter of the semicircle = 15 cm circimference of the semicircle = 1 / 2 ( ∏ ) ( 15 ) = 1 / 2 ( 22 / 7 ) ( 15 ) = 330 / 14 = 23.57 cm to two decimal places answer : option e... | a ) 34 , b ) 35 , c ) 56 , d ) 67 , e ) 23.57 | e | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(16, 14)), const_2)), const_2) | rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)| | geometry |
the parameter of a square is equal to the perimeter of a rectangle of length 16 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places | "let height of the building be x meters 40.25 : 28.75 : : 17.5 < = > 40.25 x x = 28.75 x 17.5 x = 28.75 x 17.5 / 40.25 x = 12.5 answer : option b" | a ) 10 m , b ) 12.5 m , c ) 17.5 m , d ) 21.25 m , e ) none | b | multiply(28.75, divide(17.5, 40.25)) | divide(n0,n1)|multiply(n2,#0)| | physics |
a flagstaff 17.5 m high casts a shadow of length 40.25 m . the height of the building , which casts a shadow of length 28.75 m under similar conditions will be : | x 1 hour ' s work = 1 / 4 ; y + z ' s hour ' s work = 1 / 3 x + y + z ' s 1 hour ' s work = 1 / 4 + 1 / 3 = 7 / 12 y ' s 1 hour ' s work = ( 7 / 12 - 1 / 2 ) = 1 / 12 . y alone will take 12 hours to do the work . c | a ) 5 hours , b ) 10 hours , c ) 12 hours , d ) 24 hours , e ) 15 hours | c | inverse(subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 4)))) | divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|inverse(#4) | physics |
x can do a piece of work in 4 hours ; y and z together can do it in 3 hours , while x and z together can do it in 2 hours . how long will y alone take to do it ? | "6 % of a 50 liter solution is 3 l which is 10 % of the solution at the end . the solution at the end must be 30 l . we need to evaporate 20 liters . the answer is a ." | a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | a | subtract(50, multiply(divide(50, const_100), 10)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
how many liters of water must be evaporated from 50 liters of a 6 - percent sugar solution to get a 10 - percent solution ? | "total men in company 40 % means total women in company 60 % ( assume total people in company 100 % ) no of men employees attended picnic = 40 x ( 20 / 100 ) = 8 no of women employees attended picnic = 60 x ( 40 / 100 ) = 24 total percentage of employees attended the picnic = 8 + 24 = 32 % answer : a" | a ) 32 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 37 % | a | multiply(add(multiply(divide(40, const_100), divide(20, const_100)), multiply(divide(subtract(const_100, 40), const_100), divide(40, const_100))), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n2)|divide(#3,const_100)|multiply(#0,#1)|multiply(#4,#2)|add(#5,#6)|multiply(#7,const_100)| | gain |
in a certain company 20 % of the men and 40 % of the women attended the annual company picnic . if 40 % of all the employees are men . what % of all the employee went to the picnic ? | "exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 7 log 7 ( 8 ) = 8 correct answer e" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 8 | e | divide(log(multiply(7, 7)), log(const_10)) | log(const_10)|multiply(n0,n0)|log(#1)|divide(#2,#0)| | other |
7 log 7 ( 8 ) = ? | "due to stoppages , it covers 14 km less . time taken to cover 14 km = ( 14 / 84 x 60 ) min = 10 min answer : b" | a ) 12 min , b ) 10 min , c ) 15 min , d ) 14 min , e ) 13 min | b | multiply(const_60, divide(subtract(84, 70), 84)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
excluding stoppages , the speed of a bus is 84 kmph and including stoppages , it is 70 kmph . for how many minutes does the bus stop per hour ? | 40 / 160 × 100 = 25 % answer : e | a ) 35 % , b ) 40 % , c ) 45 % , d ) 50 % , e ) 25 % | e | multiply(divide(40, 160), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
40 is what percent of 160 ? | "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 57 ) = 465.60 hence answer will be ( e )" | a ) 320.50 $ , b ) 380.50 $ , c ) 425.50 $ , d ) 450.50 $ , e ) 465.60 $ | e | multiply(multiply(0.65, 57), 12) | multiply(n1,n3)|multiply(n2,#0)| | general |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 57 liters and they are all empty , how much money total will it cost to fuel all cars ? | answer : c ) 5940 | a ) 5942 , b ) 2887 , c ) 5940 , d ) 2888 , e ) 28881 | c | multiply(multiply(multiply(multiply(const_2, const_2), multiply(multiply(const_3, const_3), const_3)), divide(divide(divide(135, const_3), const_3), const_3)), divide(22, const_2)) | divide(n0,const_2)|divide(n3,const_3)|multiply(const_2,const_2)|multiply(const_3,const_3)|divide(#1,const_3)|multiply(#3,const_3)|divide(#4,const_3)|multiply(#2,#5)|multiply(#6,#7)|multiply(#0,#8) | physics |
the l . c . m of 22 , 54 , 108 , 135 and 198 is | "280 * ( 88 / 100 ) * ( 92 / 100 ) = 226 answer : b" | a ) 288 , b ) 226 , c ) 250 , d ) 230 , e ) 262 | b | subtract(subtract(280, divide(multiply(280, 12), const_100)), divide(multiply(subtract(280, divide(multiply(280, 12), const_100)), 8), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
the sale price sarees listed for rs . 280 after successive discount is 12 % and 8 % is ? | relative speed = ( 1120 / 12 ) m / s = ( 1120 / 12 ) * ( 18 / 5 ) = 336 kmph speed of goods train = 336 - 40 = 296 kmph answer is b | a ) 295 , b ) 296 , c ) 297 , d ) 298 , e ) 299 | b | subtract(divide(divide(1120, 12), const_0_2778), 40) | divide(n2,n1)|divide(#0,const_0_2778)|subtract(#1,n0) | physics |
a girl sitting in a train which is travelling at 40 kmph observes that a goods train travelling in a opposite direction , takes 12 seconds to pass him . if the goods train is 1120 m long , find its speed . | "this is how i used to calculate which i think works pretty well : if you let the average of the 20 other numbers equal a , can you write this equation for sum of the list ( s ) n + 20 a = s the question tells us that n = 4 a plug this back into the first equation and you get that the sum is 24 a 4 a + 20 a = 24 a ther... | a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 5 , d ) 4 / 21 , e ) 5 / 21 | b | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 4), 21), 4), const_2), 4), const_3)) | divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,n1)|multiply(#3,const_2)|subtract(#4,n1)|subtract(#5,const_3)|divide(#1,#6)| | general |
a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction t of the sum of the 21 numbers in the list ? | "= ( 107 ) ^ 2 + ( 93 ) ^ 2 = ( 100 + 7 ) ^ 2 + ( 100 - 7 ) ^ 2 = 2 x [ ( 100 ) ^ 2 + 7 ^ 2 ] = 2 [ 10000 + 49 ] = 2 x 10049 = 20098 answer is c" | a ) 19578 , b ) 19418 , c ) 20098 , d ) 21908 , e ) none of them | c | multiply(107, power(107, 93)) | power(n1,n2)|multiply(n0,#0)| | general |
107 x 107 + 93 x 93 = ? | "1 percent for 3 years = 66 1 percent for 1 year = 22 = > 100 percent = 2200 answer : c" | a ) 2000 , b ) 2100 , c ) 2200 , d ) 2300 , e ) 2400 | c | multiply(divide(66, 3), const_100) | divide(n2,n0)|multiply(#0,const_100)| | gain |
a sum was put at simple interest at certain rate for 3 years . had it been put at 1 % higher rate it would have fetched rs . 66 more . the sum is : a . rs . 2,400 b . rs . 2,100 c . rs . 2,200 d . rs . 2,480 | "( place value of 6 ) - ( face value of 6 ) = ( 6000 - 6 ) = 5994 answer : option c" | a ) 973 , b ) 6973 , c ) 5994 , d ) 6084 , e ) none of these | c | subtract(multiply(const_10, 6), 6) | multiply(n0,const_10)|subtract(#0,n0)| | general |
the difference between the place value and the face value of 6 in the numeral 856973 is | ": ( 60 / 100 ) * x + 180 = x 2 x = 900 x = 450 answer : e" | a ) 300 , b ) 277 , c ) 266 , d ) 99 , e ) 450 | e | divide(180, divide(180, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain |
60 % of a number is added to 180 , the result is the same number . find the number ? | ordering the data from least to greatest , we get : $ 1.61 , $ 1.75 , $ 1.79 , $ 1.82 , $ 1.96 , $ 2.09 , $ 2.11 the median gasoline price is $ 1.82 . ( there were 3 states with higher gasoline prices and 3 with lower prices . ) b | a ) $ 1 , b ) $ 1.82 , c ) $ 1.92 , d ) $ 2.13 , e ) $ 2.15 | b | min(divide(add(add(add(add(add(add(1.75, 1.61), 1.79), 2.11), 1.96), 2.09), 1.82), 7), 1.82) | add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|add(n6,#3)|add(n7,#4)|divide(#5,n0)|min(n7,#6) | general |
andrew travelling to 7 cities . gasoline prices varied from city to city . $ 1.75 , $ 1.61 , $ 1.79 , $ 2.11 , $ 1.96 , $ 2.09 , $ 1.82 . what is the median gasoline price ? | "5 ^ x = 1 / 625 5 ^ x = 1 / 5 ^ 4 5 ^ x = 5 ^ - 4 x = - 4 b" | a ) – 2 , b ) - 4 , c ) 0 , d ) - 1 , e ) 2 | b | divide(log(divide(1, 625)), log(5)) | divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)| | general |
if 625 ( 5 ^ x ) = 1 then x = | "2 years answer : a" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | floor(add(divide(log(const_2), log(add(const_1, divide(45, const_100)))), const_1)) | divide(n0,const_100)|log(const_2)|add(#0,const_1)|log(#2)|divide(#1,#3)|add(#4,const_1)|floor(#5)| | general |
find the least number of complete years in which a sum of money put out at 45 % compound interest will be more than double of itself ? | still water = 12 km / hr downstream = 45 / 3 = 15 km / hr upstream = > > still water = ( u + v / 2 ) = > > 12 = u + 15 / 2 = 9 km / hr so time taken in upstream = 45 / 9 = 5 hrs answer : d | a ) 8 hours , b ) 6 hours , c ) 4 hours , d ) 5 hours , e ) 6 hours | d | divide(45, subtract(12, subtract(divide(45, 3), 12))) | divide(n1,n2)|subtract(#0,n0)|subtract(n0,#1)|divide(n1,#2) | physics |
the speed of the boat in still water in 12 kmph . it can travel downstream through 45 kms in 3 hrs . in what time would it cover the same distance upstream ? | "a + b + c = 3 * 55 = 165 a + b + c + d = 4 * 60 = 240 - - - - ( i ) so , d = 75 & e = 75 + 3 = 78 b + c + d + e = 58 * 4 = 232 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 240 – 232 = 8 a = e + 8 = 78 + 8 = 86 answer : d" | a ) 56 , b ) 65 , c ) 75 , d ) 86 , e ) 90 | d | subtract(multiply(60, const_4), subtract(multiply(58, const_4), add(3, subtract(multiply(60, const_4), multiply(55, 3))))) | multiply(n1,const_4)|multiply(n3,const_4)|multiply(n0,n2)|subtract(#0,#2)|add(n2,#3)|subtract(#1,#4)|subtract(#0,#5)| | general |
the avg weight of a , b & c is 55 kg . if d joins the group , the avg weight of the group becomes 60 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 58 kg . what is the weight of a ? | "s = 120 / 12 * 18 / 5 = 36 kmph answer : c" | a ) 16 kmph , b ) 88 kmph , c ) 36 kmph , d ) 18 kmph , e ) 19 kmph | c | multiply(const_3_6, divide(120, 12)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
a train 120 m in length crosses a telegraph post in 12 seconds . the speed of the train is ? | "from the info that the maximum sides of the cubes is 4 , we know that the gcf of 12 ( = 2 ^ 2 * 3 ) andbis 4 ( = 2 ^ 2 ) , sob = 2 ^ x , where x > = 2 . from the second premise , we know that the lcm of 12 ( 2 ^ 2 * 3 ) andbis 32 ( 2 ^ 5 ) , sob = 2 ^ 5 combining 2 premises shows the answer is d ( 32 ) ." | a ) 8 , b ) 16 , c ) 24 , d ) 32 , e ) 48 | d | sqrt(subtract(power(divide(32, 4), const_2), power(12, const_2))) | divide(n2,n1)|power(n0,const_2)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)| | geometry |
a small table has a length of 12 inches and a breadth of b inches . cubes are placed on the surface of the table so as to cover the entire surface . the maximum side of such cubes is found to be 4 inches . also , a few such tables are arranged to form a square . the minimum length of side possible for such a square is ... | explanation : hint : a ' s one day work = 1 / 3 b ' s one day work = 1 / 5 c ' s one day work = 1 / 10 ( a + b + c ) ' s one day work = 1 / 3 + 1 / 5 + 1 / 10 = 1 / 1.5 hence , a , b & c together will take 1.5 days to complete the work . answer is a | a ) 1.5 days , b ) 4.5 days , c ) 7 days , d ) 9.8 days , e ) 9 days | a | add(subtract(3, const_2), divide(5, 10)) | divide(n1,n2)|subtract(n0,const_2)|add(#0,#1) | physics |
if ' a ' completes a piece of work in 3 days , which ' b ' completes it in 5 days and ' c ' takes 10 days to complete the same work . how long will they take to complete the work , if they work together ? | "in 105 liters of drink a , there are 60 liters of milk and 45 liters of juice . with 60 liters of milk , we need a total of 80 liters of juice to make drink b . we need to add 35 liters of juice . the answer is d ." | a ) 14 , b ) 21 , c ) 28 , d ) 35 , e ) 42 | d | subtract(divide(multiply(multiply(divide(4, add(4, 3)), 105), 4), 3), multiply(divide(3, add(4, 3)), 105)) | add(n0,n1)|divide(n0,#0)|divide(n1,#0)|multiply(n4,#1)|multiply(n4,#2)|multiply(n0,#3)|divide(#5,n1)|subtract(#6,#4)| | general |
a certain drink of type a is prepared by mixing 4 parts milk with 3 parts fruit juice . another drink of type b is prepared by mixing 4 parts of fruit juice and 3 parts of milk . how many liters of fruit juice must be added to 105 liters of drink a to convert it to drink b ? | "let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) / ( p x 9 ) ] % = 11.1 % per annum . answer : d" | a ) 12.5 % , b ) 13.5 % , c ) 11.5 % , d ) 11.1 % , e ) 21.5 % | d | divide(const_100, 9) | divide(const_100,n0)| | gain |
at what rate percent per annum will a sum of money double in 9 years . | "according to order of operations , 12 ÷ 3 × 2 ( division and multiplication ) is done first from left to right 12 ÷ 3 × 2 = 4 × 2 = 8 hence 30 - 12 ÷ 3 × 2 = 30 - 8 = 22 correct answer is b ) 22" | a ) 11 , b ) 22 , c ) 33 , d ) 44 , e ) 55 | b | subtract(30, multiply(multiply(12, const_2.0), 2)) | multiply(n1,const_2.0)|multiply(n3,#0)|subtract(n0,#1)| | general |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.