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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 ? | "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 ) 7 % , b ) 25 % , c ) s . 9360 , d ) 3 / 5 , e ) 1300 | e | 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 |
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 filled to capacity . ) | "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 answers . . . . we ' re told that there are 2 types of boxes : those that hold 12 glasses and those that hold 16 glasses . since the average number of boxes is 15 , we know that there must be at least some of each . we ' re also told that that there are 16 more of the larger boxes . this means , at the minimum , we have . . . 1 small box and 17 large boxes = 1 ( 12 ) + 17 ( 16 ) = 12 + 272 = 284 glasses at the minimum since the question asks for the total number of glasses , we can now eliminate answers a , b and c . . . . the difference in the number of boxes must be 16 though , so we could have . . . . 2 small boxes and 18 large boxes 3 small boxes and 19 large boxes etc . with every additional small box + large box that we add , we add 12 + 16 = 28 more glasses . thus , we can justadd 28 suntil we hit the correct answer . . . . 284 + 28 = 312 312 + 28 = 340 340 + 28 = 368 368 + 28 = 396 at this point , we ' ve ' gone past ' answer d , so the correct answer must be answer e . . . . . but here ' s the proof . . . . 396 + 28 = 424 424 + 28 = 452 452 + 28 = 480 final answer : e" | a ) 53.5 % , b ) 868 cm ^ 2', ' , c ) w = 480 , d ) 5 % , e ) 159 | c | multiply(multiply(16, const_2), 15) | multiply(n1,const_2)|multiply(n2,#0)| | 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 ? | "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 ) 24 . , b ) 501 , c ) 14 , d ) 12.5 , e ) 30 | b | 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 |
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 ? | "28 - - - 4 ds = 7 ? - - - - 1 16 - - - - 4 us = 4 ? - - - - 1 m = ? m = ( 7 + 4 ) / 2 = 5.5 answer : e" | a ) 5.5 , b ) 3280 , c ) 10 , d ) 1200 , e ) 45 | a | divide(add(divide(16, 4), divide(28, 4)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | 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 bob at 11 am , then how fast is bob driving ( in kilometers per hour ) ? | 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 ) 4 , b ) 6 days , c ) 80 % , d ) 20 , e ) 3 | d | subtract(divide(add(100, 100), subtract(11, 7)), 30) | add(n1,n1)|subtract(n4,n2)|divide(#0,#1)|subtract(#2,n0) | physics |
the calendar of the year 2040 can be used again in the year ? | "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 ) 1 / 4 , b ) 32 , c ) 2068 , d ) - 3 , e ) 1 / 6 | c | 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 |
how many integers from 1 to 100 exist such that each is divisible by 5 and also has 5 as a digit ? | "5 , 15,25 , 35,40 , 50,55 , 65,75 , 85,95 so there are total 11 such type of numbers . answer : c" | a ) 128 , b ) 2 : 1 , c ) 11 , d ) 5 , e ) 16.2 % | c | divide(100, const_10) | divide(n1,const_10)| | general |
in what time will a train 100 metres long cross an electic pole , if its speed be 144 km / hr ? | "sol . speed = [ 144 x 5 / 18 ] m / sec = 40 m / sec . time taken = ( 100 / 40 ) sec = 2.5 sec . answer a" | a ) 23 , b ) 80 % , c ) 9 , d ) 2.5 sec , e ) 1 / 16 | d | divide(100, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | 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 ? | "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 ) 120 km , b ) 0.2 , c ) 40 , d ) 6 , e ) 625 | a | 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 |
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 ? | "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 ) 90 % , b ) 5 , c ) 18 / 35 , d ) 43 % , e ) 140 | c | 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 |
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 | explanation : number = ( 800 * 300 * 600 ) / 8 * 3 * 2 = 3000000 answer : d | a ) 1 , b ) 10 , c ) 20 , d ) 60 kg , e ) 3000000 | e | 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 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 ? | "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 ) 2 minutes , b ) 2,000 , c ) 6 , d ) 120 , e ) 129.5 | c | divide(multiply(8, 24), add(8, 24)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | physics |
what is the remainder when 50 ! is divided by 16 ^ 8 ? ? | "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 ) 1000 , b ) 157 / 160 , c ) 6 , d ) 0 , e ) 2 | d | reminder(multiply(16, 50), 8) | multiply(n0,n1)|reminder(#0,n2)| | general |
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 ? | "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 ) 15 days , b ) 3 , c ) 120 , d ) 18 , e ) 700 | c | add(45, 30) | add(n1,n2)| | physics |
if the cost price of 140 pencils is equal to the selling price of 100 pencils , the gain percent is | "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 ) 629 , b ) 40 , c ) 15 litres , d ) 70 , e ) 150 | b | divide(const_100, divide(100, subtract(140, 100))) | subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1)| | gain |
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 ? | "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 ) 1 / 5 , b ) 6 cm', ' , c ) 44 , d ) 0.18 , e ) 13 | c | subtract(60, divide(subtract(140, 60), 5)) | subtract(n0,n1)|divide(#0,n3)|subtract(n1,#1)| | general |
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 : | "( 3 x - 9 ) : ( 5 x - 9 ) = 5 : 2 x = 1 = > 3 x = 3 answer : a" | a ) 6 : 40 , b ) 3 , c ) 31 , d ) 4 , e ) 15 days | b | 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 |
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 ? | 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 ) / 8 = 1 / 5 of the mixture . answer : c . | a ) 27.5 % , b ) 10749 , c ) 41 , d ) 14 , e ) 1 / 5 | e | divide(add(const_1, divide(3, 5)), add(5, 3)) | add(n0,n1)|divide(n0,n1)|add(#1,const_1)|divide(#2,#0) | 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 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 ) q = 1600 , b ) 19 , c ) 75 % decrease , d ) 9 / 25 , e ) 0.075 days | 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 |
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 ? | let the cost price of book = x selling price of book = 24 $ markup % = 20 ( 120 / 100 ) x = 24 = > x = 20 answer e | a ) 40 min , b ) $ 20.00 , c ) 160 , d ) 12 , e ) 40 % | b | subtract(24, multiply(divide(20, const_100), 24)) | divide(n0,const_100)|multiply(n1,#0)|subtract(n1,#1) | 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 ? | "the profit 0 f 2009 in terms of 2008 = 0.9 * 15 / 10 * 100 = 135 % c" | a ) 12 kmph , b ) 2000 , c ) 135 % , d ) $ 864 , e ) rs . 937.5 | 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 |
find the simple interest on $ 10000 at 6 % per annum for 12 months ? | "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 ) 32 kmph , b ) $ 600 , c ) 40 % , d ) 1652 , e ) 10749 | b | multiply(10000, divide(6, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
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 ? | "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 liters then the water ratio is increse 1 parts 1 part - - - - - > 10 liters the quantity of milk in the original mixture is = 5 : 2 = 5 + 2 = 7 7 parts - - - - - > 70 liters ( answer is = 70 ) short cut method - 2 : for only milk problems milk : water 5 : 2 5 : 3 milk ratio same but water ratio 1 part incress per 10 liters 1 part of ratio - - - - - - - > 10 liters 7 part of ratio - - - - - - - > 70 liters c )" | a ) 50 , b ) 6 liters , c ) 720 , d ) 9.5 gallons , e ) $ 8 | a | 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 |
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 ? | 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 / 3168 people total you get ( a ) 40 % | a ) 1000 , b ) 23 , c ) 40 % , d ) $ 1.96 , e ) 3 | c | 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 |
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 ? | "in such a case there is always a loss loss % = ( 13 / 10 ) ^ 2 = 120 / 71 = 1.69 % answer is a" | a ) 64.3 , b ) 266 cm 2 , c ) 720 , d ) 2 minutes , 00 seconds , e ) 1.69 % | e | 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 |
what is the greatest of 3 consecutive integers whose sum is 30 ? | "30 / 3 = 10 the three numbers are 9 , 10 , and 11 . the answer is d ." | a ) 170 , b ) 96 , c ) s . 200 , d ) 11 , e ) 7 | d | add(divide(subtract(30, 3), 3), const_2) | subtract(n1,n0)|divide(#0,n0)|add(#1,const_2)| | physics |
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 ? | 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 ) 72 , b ) 120 , c ) 67 % , d ) 28.57 % , e ) 2 : 3 | 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 |
if the sides of a triangle are 20 cm , 12 cm and 16 cm , what is its area ? | "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 ) 2056 , b ) 12 , c ) 7 , d ) 96 , e ) 360 | d | divide(multiply(12, 16), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
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 ? | "d = 100 + 150 = 250 s = 36 * 5 / 18 = 10 mps t = 250 / 10 = 25 sec . answer : c" | a ) 4 , b ) 2 , c ) 16 , d ) 25 , e ) 14 | d | divide(add(150, 100), multiply(36, const_0_2778)) | add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)| | physics |
34.94 + 240.016 + 23.98 = ? | 34.94 240.016 + 23.98 - - - - - - - - 298.936 answer is a . | a ) 24 % , b ) 28000 , c ) 5 / 4 , d ) 2068 , e ) 298.936 | e | add(add(34.94, 240.016), 23.98) | add(n0,n1)|add(n2,#0) | 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 ? | total cost of items : 2000 / - amount paid : 3000 / - balance receivable : 3000 - 2000 = 1000 / - answer is b | a ) 60 , b ) 158 , c ) 1000 , d ) 55 , e ) 1 / 13 | c | subtract(3000, 2000) | subtract(n3,n2) | 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 . | "correct average = 35 x 72 + ( 86 - 36 ) / 35 ≈ 72 + 1.43 = 73.43 answer d" | a ) q = 35 , b ) 1540 , c ) $ 333.33 , d ) 73.43 , e ) 243 | d | divide(subtract(multiply(35, 72), subtract(86, 36)), 35) | multiply(n0,n2)|subtract(n4,n3)|subtract(#0,#1)|divide(#2,n0)| | 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 ? | 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 ) 2 / 5 , b ) 165 , c ) 466 , d ) 43.33 % , e ) e = 121 | e | add(lcm(10, lcm(6, 8)), const_1) | lcm(n2,n3)|lcm(n4,#0)|add(#1,const_1) | general |
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 charges the cost price of the weight that he claims , find his profit percentage . | "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 ) 129 , b ) 2 , c ) 11 , d ) 28 m , e ) 82 % | e | 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 |
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 | "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 ) 23 , b ) 40000 , c ) 15 , d ) 28 , e ) 2 / 9 | c | 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 |
how many numbers from 2 to 13 are exactly divisible by 2 ? | "2 / 2 = 1 and 13 / 2 = 6 6 - 1 = 5 5 + 1 = 6 numbers . answer : e" | a ) 15 / 2 minutes , b ) 21 , c ) rs 20 , d ) 37.5 % , 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 |
in a class of 37 students 26 play football and play 20 long tennis , if 17 play above , many play neither ? | "26 + 20 - 17 = 29 37 - 29 = 8 play neither answer is b" | a ) 16 / 625 , b ) s . 189 , c ) 6 , d ) 8 , e ) 2 : 3 | d | subtract(37, subtract(add(26, 20), 17)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
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 . | "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 ) 56.25 $ , b ) 192 , c ) 21 , d ) 110 , e ) 4.1 hr | e | divide(86, add(15, 6)) | add(n0,n1)|divide(n2,#0)| | physics |
on dividing a number by 357 , we get 38 as remainder . on dividing the same number by 17 , what will be the remainder ? | "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 ) 131.6 , b ) 81 , c ) 4 , d ) 20 minutes , e ) 8 days | c | 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 |
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 ? | 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 ) 172 , b ) 245 , c ) $ 20 , d ) 84 , e ) 405 | d | 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 |
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 ? | "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 ) 1 / 4 , b ) 10 , c ) 52 , d ) 14 : 00 , e ) 81.6 cm | c | 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 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 ? | "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 ) 10 % , b ) 3 / 11 , c ) 30 hr , d ) 1 / 8 , e ) 9560 | 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 |
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 ? | "soln : - 13 x = 45 - - > 87 / 77 * x = 45 - - > x = 45 * 77 / 87 = 677 / 17 = ~ 40 . answer : c ." | a ) 40 , b ) 60 , c ) 0.1 , d ) 149 , e ) $ 320 | a | multiply(divide(const_100, add(const_100, 13)), 45) | add(n4,const_100)|divide(const_100,#0)|multiply(n7,#1)| | 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 ? | "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 the total number of married students = 30 , implying 10 married females . therefore 20 single females ; = 2 / 3 of total females . answer : d" | a ) rs . 9000 , b ) 321.05 $ , c ) 2 / 3 , d ) 1380 , e ) 5 cm | c | divide(const_10, 30) | divide(const_10,n1)| | gain |
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 spends driving to the meeting ? | "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 : b ." | a ) 32 % , b ) 2972 , c ) 3 : 2 , d ) 875 , e ) 3 / 7 | c | 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 |
if ( a - b - c + d = 18 ) and ( a + b - c - d = 4 ) , what is the value of ( b - d ) ^ 2 ? | "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 ) 265 miles . , b ) 49 . , c ) 30 kmph , d ) 15 π , e ) 4264 | b | power(subtract(4, divide(add(18, 4), 2)), 2) | add(n0,n1)|divide(#0,n2)|subtract(n1,#1)|power(#2,n2)| | 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 ? | 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 / 60 = 10 . e | a ) $ 900 , b ) 8 , c ) 2 , d ) - 9 , 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 |
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 ? | "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 ) 650 , b ) 0.4 , c ) 540 , d ) 48 , e ) 66 | d | 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 |
45 men can complete a piece of work in 18 days . in how many days will 27 men complete the same work ? | "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 } } x = 30 so 30 days will be required to get work done by 27 men . answer : c" | a ) 53.5 % , b ) 30 , c ) 4991 , d ) 85 , e ) 71 % | b | divide(multiply(18, 45), 27) | multiply(n0,n1)|divide(#0,n2)| | 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 ? | "( 27 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 60 60 – 27 = 33 answer : a" | a ) 4.37 sec , b ) 34 , c ) 33 , d ) 33.33 % , e ) 40 | c | 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 |
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 ? | "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 small as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers , the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use 2 and 7 in the hundreds places and reserve them for the tens places . now what are the options ? try and make a pair with ( 3 * * and 4 * * ) . make the 3 * * number as large as possible and make the 4 * * number as small as possible . 376 and 425 ( difference is 49 ) or try and make a pair with ( 5 * * and 6 * * ) . make the 5 * * number as large as possible and make the 6 * * number as small as possible . we get 574 and 623 ( difference is 49 ) b" | a ) $ 144 , b ) 11 / 3 , c ) 49 , d ) 60 , e ) 298.936 | c | 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 |
54671 - 14456 - 35466 = ? | "e if we calculate we will get 4749" | a ) 4749 , b ) 26250 , c ) $ . 90 , d ) 79860 , e ) 5 | a | 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 |
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 ? | "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 ) 625 , b ) 45 , c ) 84 % , d ) 60 , e ) 11.36 % | b | 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 |
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 ? | "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 ) 14', ' , b ) 4 / 7 , c ) 58 , d ) 78 + ( 4 / , e ) 9.1 litres | b | 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 |
on dividing 21 by a number , the quotient is 10 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 21 - 1 ) / 10 = 20 / 10 = 2 b" | a ) 170 m , b ) 135 ° , c ) 1 , d ) 13 % , e ) 2 | e | floor(divide(21, 10)) | divide(n0,n1)|floor(#0)| | general |
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 ? | side of square = √ 36 = 6 m . length = 6 m and breadth = 3 m area of rectangle = 6 * 3 = 18 sq . m answer a | a ) 500 , b ) 18', ' , c ) 87 % , d ) 445 , e ) 50 % | b | multiply(sqrt(36), divide(sqrt(36), const_2)) | sqrt(n0)|divide(#0,const_2)|multiply(#1,#0) | geometry |
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 ? | "money paid in cash = rs . 1000 balance payment = ( 70000 - 1000 ) = rs . 69000 answer : c" | a ) 69000 , b ) 3 , c ) 61 cm , d ) 8 and 20 , e ) 855 | a | subtract(70000, 1000) | subtract(n0,n2)| | gain |
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 ? | "0.8 * ( 1.2 * 200 ) = $ 192 the answer is a ." | a ) 400 , b ) $ 192 , c ) 25', ' , d ) 250 m , e ) 90 | b | 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 |
998 x 998 = ? | "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 ) 996004 , b ) 9 , c ) 7.36 kg , d ) 5 : 3 , e ) 5 : 11 | a | multiply(divide(998, 998), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
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 ? | "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 ) $ . 90 , b ) 14.63 , c ) 20 % , d ) 70 mph , e ) 36 | d | 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 |
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 ? | 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 ) 2 / 3 , b ) 1528 , c ) 1 , d ) 35 , 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 |
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 he cover while the sun was shining on him ? | 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 12 t . therefore , total distance covered while the sun was shining over him is 2 / 14 = 1 / 7 . answer : d | a ) 3,300 , b ) 120 , c ) 480 cm 2 , d ) 82 , e ) 1 / 7 | e | 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 |
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 ? | 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 ) 3 / 7 , b ) 8 : 1 , c ) $ 1260 , d ) 19 / 27 , e ) 16 : 15 | c | 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), const_100))), divide(multiply(1200, const_100), multiply(10, 2))) | 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 35 % of a number is 12 less than 50 % of that number , then the number is ? | "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 ) 5.75 , b ) 120 , c ) $ 555.55 , d ) 80 , e ) 18 days | d | divide(12, divide(subtract(50, 35), const_100)) | subtract(n2,n0)|divide(#0,const_100)|divide(n1,#1)| | gain |
find the number of different prime factors of 441 | "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 ) 33 1 / 3 % , c ) 2 , d ) 80 % , e ) 200 | c | add(const_2, const_2) | add(const_2,const_2)| | other |
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 % ? | 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 - 30000 = 20000 first part = 20000 answer : a | a ) 28000 , b ) 2220 , c ) 54 °', ' , d ) 38 years , e ) 20000 | e | 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 |
the average of 10 consecutive odd numbers is 22 . what is the sum of the first 3 numbers ? | "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 ) 9 , b ) $ 700 , c ) 129 , d ) 20 , e ) 45 | e | 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(const_1, const_2), const_1), const_1), const_1)) | 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 |
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 ? | "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 ) 35.6 % , c ) 36 , d ) 62 , e ) 89 | a | subtract(multiply(245, const_2), 112) | multiply(n0,const_2)|subtract(#0,n1)| | 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 ? | 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 ) 150 km , b ) 6 , c ) 16 , d ) none , e ) 18 | d | multiply(divide(subtract(5, 4), 5), const_100) | subtract(n2,n0)|divide(#0,n2)|multiply(#1,const_100) | 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 ? | "answer = a please refer diagram below 85 - 10 = 75" | a ) 1 , b ) 20 , c ) 36 , d ) 6 , e ) 75 | e | subtract(85, 10) | subtract(n0,n1)| | general |
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 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 ) 23.57 , b ) 80 % , c ) s . 1665 , d ) 1.2 , e ) 200 | a | 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 |
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 : | "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 ) 2 , b ) 12.5 m , c ) 11 , d ) 7 , e ) 650 | b | multiply(28.75, divide(17.5, 40.25)) | divide(n0,n1)|multiply(n2,#0)| | 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 ? | 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 ) 70 , b ) 3 : 2 , c ) 18 , d ) 120 , e ) 12 hours | e | 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 |
how many liters of water must be evaporated from 50 liters of a 6 - percent sugar solution to get a 10 - percent solution ? | "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 ) 331 times , b ) 12 , c ) 20 , d ) 11 / 3 , e ) 100 | c | subtract(50, multiply(divide(50, const_100), 10)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | 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 ? | "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 ) 74 , b ) 32.5 % , c ) 32 % , d ) 6 , e ) 39 | c | 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 |
7 log 7 ( 8 ) = ? | "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 ) 3400 , b ) 4 , c ) 2 , d ) 8 , e ) 50 % | d | divide(log(multiply(7, 7)), log(const_10)) | log(const_10)|multiply(n0,n0)|log(#1)|divide(#2,#0)| | other |
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 ? | "due to stoppages , it covers 14 km less . time taken to cover 14 km = ( 14 / 84 x 60 ) min = 10 min answer : b" | a ) 50 % , b ) 81 , c ) 868 cm ^ 2', ' , d ) 10 min , e ) 122 | d | multiply(const_60, divide(subtract(84, 70), 84)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
40 is what percent of 160 ? | 40 / 160 × 100 = 25 % answer : e | a ) 80 minutes , b ) 40 , c ) 25 % , d ) 14 , e ) 20 minutes | c | multiply(divide(40, 160), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
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 ? | "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 57 ) = 465.60 hence answer will be ( e )" | a ) 10,000 , b ) 1.9 , c ) 4091 , d ) 10 , e ) 465.60 $ | e | multiply(multiply(0.65, 57), 12) | multiply(n1,n3)|multiply(n2,#0)| | general |
the l . c . m of 22 , 54 , 108 , 135 and 198 is | answer : c ) 5940 | a ) 7500 , b ) 30 , c ) 5940 , d ) 2 minutes , e ) $ 22000 | 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 sale price sarees listed for rs . 280 after successive discount is 12 % and 8 % is ? | "280 * ( 88 / 100 ) * ( 92 / 100 ) = 226 answer : b" | a ) 2 : 1 , b ) 57.5 , c ) 20 , d ) 94 , e ) 226 | e | 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 |
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 . | relative speed = ( 1120 / 12 ) m / s = ( 1120 / 12 ) * ( 18 / 5 ) = 336 kmph speed of goods train = 336 - 40 = 296 kmph answer is b | a ) 59.8 % , b ) 50 km / hr , c ) s . 19,800 , d ) 296 , e ) s . 9360 | d | subtract(divide(divide(1120, 12), const_0_2778), 40) | divide(n2,n1)|divide(#0,const_0_2778)|subtract(#1,n0) | physics |
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 ? | "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 therefore fraction t of n to the total would be 4 a / 24 a or 1 / 6 answer b" | a ) 63 % , b ) $ 290 , c ) 1 / 6 , d ) 0.7 , e ) 33 | c | 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 |
107 x 107 + 93 x 93 = ? | "= ( 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 ) 6000 , b ) 60 liters , c ) 20098 , d ) 113 , e ) 0.036 | c | multiply(107, power(107, 93)) | power(n1,n2)|multiply(n0,#0)| | general |
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 | "1 percent for 3 years = 66 1 percent for 1 year = 22 = > 100 percent = 2200 answer : c" | a ) 2200 , b ) 500 , c ) 8 , d ) 35 , e ) 6666 | a | multiply(divide(66, 3), const_100) | divide(n2,n0)|multiply(#0,const_100)| | gain |
the difference between the place value and the face value of 6 in the numeral 856973 is | "( place value of 6 ) - ( face value of 6 ) = ( 6000 - 6 ) = 5994 answer : option c" | a ) 14 , b ) 1235 , c ) 9 , d ) rs . 27500 , e ) 5994 | e | subtract(multiply(const_10, 6), 6) | multiply(n0,const_10)|subtract(#0,n0)| | general |
60 % of a number is added to 180 , the result is the same number . find the number ? | ": ( 60 / 100 ) * x + 180 = x 2 x = 900 x = 450 answer : e" | a ) 2400 , b ) 2200 , c ) 447.5 , d ) 450 , e ) 2 | d | divide(180, divide(180, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain |
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 ? | 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 ) 95.07 square meter', ' , b ) $ 23400 , c ) 855220 , d ) $ 1.82 , e ) 6 | d | 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 |
if 625 ( 5 ^ x ) = 1 then x = | "5 ^ x = 1 / 625 5 ^ x = 1 / 5 ^ 4 5 ^ x = 5 ^ - 4 x = - 4 b" | a ) 25 , b ) - 4 , c ) rs . 90 , d ) 36', ' , e ) 975 | b | divide(log(divide(1, 625)), log(5)) | divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)| | 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 ? | "2 years answer : a" | a ) 22.5 % , b ) 9 , c ) 2 , d ) 2.6 units , e ) 300 | c | 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 |
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 ? | 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 ) 2 kmph , b ) 440 , c ) 5 hours , d ) s . 5100 , e ) 1750 m | c | divide(45, subtract(12, subtract(divide(45, 3), 12))) | divide(n1,n2)|subtract(#0,n0)|subtract(n0,#1)|divide(n1,#2) | physics |
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 ? | "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 ) 300 , b ) $ 11.20 , c ) 84.6 % , d ) 3 , e ) 86 | e | 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 |
a train 120 m in length crosses a telegraph post in 12 seconds . the speed of the train is ? | "s = 120 / 12 * 18 / 5 = 36 kmph answer : c" | a ) 2 , b ) rs . 20985 , c ) 36 kmph , d ) 32 square meters , e ) 321.05 $ | c | multiply(const_3_6, divide(120, 12)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
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 32 inches . find b . | "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 ) 36 , b ) 17 , c ) 5 / 16 , d ) 32 , e ) 53.5 % | 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 |
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 ? | 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 ) $ 94,000 , b ) 2 , c ) 15 sec , d ) 5 / 12 , e ) 1.5 days | e | add(subtract(3, const_2), divide(5, 10)) | divide(n1,n2)|subtract(n0,const_2)|add(#0,#1) | physics |
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 ? | "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 ) 180 m , b ) 35 , c ) 27 sec , d ) 12 cm', ' , e ) $ 214.16 | b | 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 |
at what rate percent per annum will a sum of money double in 9 years . | "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 ) 700 , b ) 2700 , c ) 11.5 , d ) 25 π , e ) 11.1 % | e | divide(const_100, 9) | divide(const_100,n0)| | gain |
evaluate : 30 - 12 ÷ 3 × 2 = | "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 ) 305 , b ) 391 , c ) 25 % , d ) 22 , e ) 29 | d | subtract(30, multiply(multiply(12, const_2.0), 2)) | multiply(n1,const_2.0)|multiply(n3,#0)|subtract(n0,#1)| | general |
if n is a prime number greater than 11 , what is the remainder when n ^ 2 is divided by 12 ? | "there are several algebraic ways to solve this question , but the easiest way is as follows : since we can not have two correct answers just pick a prime greater than 11 , square it and see what would be the remainder upon division of it by 12 . n = 17 - - > n ^ 2 = 289 - - > remainder upon division 289 by 12 is 1 . answer : b ." | a ) - 5 , b ) 40 , c ) 150 , d ) 88 , e ) 1 | e | subtract(power(add(11, 2), 2), multiply(12, const_4)) | add(n0,n1)|multiply(n2,const_4)|power(#0,n1)|subtract(#2,#1)| | general |
if a * b = 2 a - 3 b + ab , then 3 * 5 + 5 * 3 is equal to : | "explanation : 3 * 5 + 5 * 3 = ( 2 * 3 - 3 * 5 + 3 * 5 ) + ( 2 * 5 - 3 * 3 + 5 * 3 ) = ( 6 + 10 - 9 + 15 ) = 22 . answer : a ) 22" | a ) 30 , b ) 22 , c ) 21.48 hours , d ) 6 cm', ' , e ) 1 | b | add(multiply(2, 3), multiply(3, 5)) | multiply(n0,n1)|multiply(n1,n3)|add(#0,#1)| | general |