Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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there are 41 students in a class , number of girls is one more than number of guys . we need to form a team of 4 students . all 4 in the team can not be from same gender . number of girls and guys in the team should not be equal . how many ways can such a team be made ? | boys = 20 and girls = 21 now the combinations are { girl , girl , girl , boy } or { boy , boy , boy , girl } so 21 c 3 * 20 c 1 + 20 c 3 * 21 c 1 = 50540 ways answer : a | a ) 50540 ways , b ) 60540 ways , c ) 50840 ways , d ) 40540 ways , e ) 51540 ways | a | add(multiply(divide(divide(factorial(subtract(41, floor(divide(41, const_2)))), factorial(subtract(4, const_1))), factorial(subtract(subtract(41, floor(divide(41, const_2))), subtract(4, const_1)))), floor(divide(41, const_2))), multiply(divide(divide(factorial(floor(divide(41, const_2))), factorial(subtract(floor(divi... | divide(n0,const_2)|subtract(n1,const_1)|factorial(#1)|floor(#0)|factorial(#3)|subtract(n0,#3)|subtract(#3,#1)|factorial(#5)|factorial(#6)|subtract(#5,#1)|divide(#7,#2)|divide(#4,#8)|factorial(#9)|divide(#10,#12)|divide(#11,#2)|multiply(#13,#3)|multiply(#14,#5)|add(#15,#16) | general |
find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 | "( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = 2 ^ 11 x 2 ^ 11 x 7 ^ 5 x 11 ^ 2 = 2 ^ 22 x 7 ^ 5 x 11 ^ 2 total number of prime factors = ( 22 + 5 + 2 ) = 29 . answer is e ." | a ) 26 , b ) 22 , c ) 25 , d ) 27 , e ) 29 | e | add(add(multiply(2, 11), 5), 2) | multiply(n1,n5)|add(n3,#0)|add(n5,#1)| | general |
pipe p can drain the liquid from a tank in 3 / 4 the time that it takes pipe q to drain it and in 3 / 3 the time that it takes pipe r to do it . if all 3 pipes operating simultaneously but independently are used to drain liquid from the tank , then pipe q drains what portion of the liquid from the tank ? | suppose q can drain in 1 hr . so , rq = 1 / 1 = 1 so , rp = 1 / [ ( 3 / 4 ) rq ] = 4 / 3 also , rp = rr / ( 3 / 3 ) = > 4 / 3 = rr / ( 3 / 3 ) = > rr = 4 / 3 let h is the time it takes to drain by running all 3 pipes simultaneously so combined rate = rc = 1 / h = 1 + 4 / 3 + 4 / 3 = 11 / 3 = 1 / ( 3 / 11 ) thus running... | a ) 9 / 29 , b ) 8 / 23 , c ) 3 / 8 , d ) 3 / 11 , e ) 3 / 4 | d | divide(multiply(3, 3), add(multiply(multiply(4, 3), const_2), multiply(3, 3))) | multiply(n0,n0)|multiply(n0,n1)|multiply(#1,const_2)|add(#2,#0)|divide(#0,#3) | physics |
xavier starts from p towards q at a speed of 40 kmph and after every 12 mins increases his speed by 20 kmph . if the distance between p and q is 56 km , then how much time does he take to cover the distance ? | "first 12 min = 40 * 12 / 60 = 8 km 2 nd 12 min = 60 * 12 / 60 = 12 km 3 rd 12 min = 80 * 12 / 60 = 16 km 4 th 12 min = 100 * 12 / 60 = 20 km total time 12.4 = 48 min b" | a ) 40 , b ) 48 , c ) 44 , d ) 36 , e ) 30 | b | add(add(add(12, 12), 12), 12) | add(n1,n1)|add(n1,#0)|add(n1,#1)| | physics |
in the next cricket world cup t - 20 , team w has decided to score 200 runs only through 4 s and 6 s . in how many ways can the team w score these 200 runs ? | team w can score a maximum of 50 fours and a minimum of 2 fours with an interval or spacing of 3 units to accommodate the 6 ' s . so the number of fours scored forms an ap 2 , 5 , 8 , . . . 50 with a common difference of 3 . number of ways of scoring 200 only through 4 ' s and 6 ' s = ( 50 - 2 ) / 3 + 1 = 17 answer : d... | a ) 13 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | d | subtract(add(const_10, divide(200, multiply(6, 4))), const_1) | multiply(n2,n3)|divide(n1,#0)|add(#1,const_10)|subtract(#2,const_1) | general |
in a class of 50 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ? | "total = english + german - both + neither - - > 50 = english + 22 - 12 + 0 - - > english = 40 - - > only english = english - both = 40 - 12 = 28 . answer : d ." | a ) 30 , b ) 10 , c ) 18 , d ) 28 , e ) 32 | d | subtract(subtract(add(50, 12), 22), 12) | add(n0,n1)|subtract(#0,n2)|subtract(#1,n1)| | other |
the perimeter of a triangle is 20 cm and the inradius of the triangle is 3 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 3 * 20 / 2 = 30 cm 2 answer : b" | a ) 22 , b ) 30 , c ) 77 , d ) 54 , e ) 23 | b | triangle_area(3, 20) | triangle_area(n0,n1)| | geometry |
a train 130 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is | "speed of train relative to man : 130 / 6 * 18 / 5 km / hr = 78 km / hr let speed of train = x therefore x + 6 = 78 x = 78 - 6 x = 72 km / hr answer : d" | a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 82 kmph | d | divide(divide(subtract(130, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
two pipes can fill a tank in 30 minutes and 20 minutes . an outlet pipe can empty the tank in 15 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 30 + v / 20 - v / 15 = v / 60 per minute the tank will be filled in 60 minutes . the answer is c ." | a ) 48 , b ) 54 , c ) 60 , d ) 66 , e ) 72 | c | subtract(add(divide(const_1, 30), divide(const_1, 20)), divide(const_1, 15)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)| | physics |
a barrel full of beer has 2 taps one midway , , which draw a litre in 6 minutes and the other at the bottom , which draws a litre in 4 minutes . the lower tap is lower normally used after the level of beer in the barrel is lower than midway . the capacity of the barrel is 36 litres . a new assistant opens the lower tap... | sol . the top tab is operational till 18 litres is drawn out . β΄ time after which the lower tap is usually open = 18 Γ 6 = 108 minutes β΄ time after which it is open now = 108 β 24 = 84 minutes β΄ litres drawn = 84 / 6 = 14 litres β΄ 18 β 14 = 4 litres were drawn by the new assistant . β΄ time = 4 Γ 4 = 16 minutes answer b | a ) 15 minutes , b ) 16 minutes , c ) 17 minutes , d ) 18 minutes , e ) none of these | b | multiply(4, subtract(divide(36, const_2), divide(subtract(multiply(divide(36, const_2), 6), 24), 6))) | divide(n3,const_2)|multiply(n1,#0)|subtract(#1,n4)|divide(#2,n1)|subtract(#0,#3)|multiply(n2,#4) | physics |
if a cube has a volume of 125 , what is the surface area of one side ? | volume of a cube = side ^ 3 125 = side ^ 3 so side = 5 area of one side of the cube = side ^ 2 = 5 ^ 2 = 25 correct answer - b | ['a ) 5', 'b ) 25', 'c ) 50', 'd ) 150', 'e ) 625'] | b | square_area(cube_edge_by_volume(125)) | cube_edge_by_volume(n0)|square_area(#0) | geometry |
xy = 1 then what is ( 6 ^ ( x + y ) ^ 2 ) / ( 6 ^ ( x - y ) ^ 2 ) | "( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 4 6 ^ 4 = 1296 answer a" | a ) 1296 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | a | power(6, multiply(const_4, 1)) | multiply(n0,const_4)|power(n1,#0)| | general |
reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 192 as interest at the end of the loan period , what was the rate of interest ? | "let rate = r % and time = r years . then , 1200 x r x r / 100 = 192 12 r 2 = 192 r 2 = 16 r = 4 . answer : c" | a ) 3.6 , b ) 6 , c ) 4 , d ) can not be determined , e ) none of these | c | sqrt(divide(multiply(192, const_100), 1200)) | multiply(n1,const_100)|divide(#0,n0)|sqrt(#1)| | gain |
β ( 25 ) ^ 2 | "explanation β ( 25 ) ^ 2 = ? or , ? = 25 answer c" | a ) 5 , b ) 14 , c ) 25 , d ) 21 , e ) none of these | c | sqrt(power(25, 2)) | power(n0,n1)|sqrt(#0)| | general |
x can finish a work in 15 days . y can finish the same work in 20 days . y worked for 12 days and left the job . how many days does x alone need to finish the remaining work ? | "work done by x in 1 day = 1 / 15 work done by y in 1 day = 1 / 20 work done by y in 12 days = 12 / 20 = 3 / 5 remaining work = 1 β 3 / 5 = 2 / 5 number of days in which x can finish the remaining work = ( 1 / 3 ) / ( 1 / 15 ) = 5 a" | a ) 5 , b ) 3 , c ) 4 , d ) 7 , e ) 8 | a | divide(subtract(const_1, multiply(12, divide(const_1, 20))), divide(const_1, 15)) | divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|subtract(const_1,#2)|divide(#3,#1)| | physics |
the average of first 18 natural numbers is ? | "sum of 18 natural no . = 342 / 2 = 171 average = 171 / 18 = 9.5 answer : b" | a ) 5.6 , b ) 9.5 , c ) 9.1 , d ) 9.8 , e ) 5.2 | b | add(18, const_1) | add(n0,const_1)| | general |
what is the sum of 20 consecutive integers from - 9 inclusive , in a increasing order ? | "from - 9 to - 1 - - > 9 nos . zero - - > 1 number from + 1 to + 9 - - > 9 nos . when we add up nos . from - 9 to + 9 sum will be zero . total 19 nos will be added . 20 th number will be 10 . sum of these 20 nos . = 10 . d is the answer ." | a ) - 9 , b ) 9 , c ) - 10 , d ) 10 , e ) 20 | d | add(9, const_1) | add(n1,const_1)| | general |
if log 1087.5 = 2.9421 , then the number of digits in ( 875 ) 10 is ? | "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 2087.5 + 1 ) = 10 ( 2.9421 + 1 ) = 10 ( 3.9421 ) = 39.421 x = antilog ( 39.421 ) therefore , number of digits in x = 40 . answer : e" | a ) 30 , b ) 28 , c ) 27 , d ) 26 , e ) 40 | e | add(multiply(const_4, 2.9421), divide(log(const_100), log(const_10))) | log(const_100)|log(const_10)|multiply(n1,const_4)|divide(#0,#1)|add(#3,#2)| | other |
annie and sam set out together on bicycles traveling at 15 and 12 km per hour respectively . after 40 minutes , annie stops to fix a flat tire . if it takes annie 20 minutes to fix the flat tire and sam continues to ride during this time , how many minutes will it take annie to catch up with sam assuming that annie res... | "annie gains 3 km per hour ( or 1 km every 20 minutes ) on sam . after 40 minutes annie is 2 km ahead . sam rides 1 km every 5 minutes . in the next 20 minutes , sam rides 4 km so sam will be 2 km ahead . it will take annie 40 minutes to catch sam . the answer is b ." | a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) 100 | b | multiply(divide(subtract(divide(12, multiply(subtract(15, 12), divide(40, const_60))), multiply(subtract(15, 12), divide(40, const_60))), subtract(15, 12)), const_60) | divide(n2,const_60)|subtract(n0,n1)|multiply(#0,#1)|divide(n1,#2)|subtract(#3,#2)|divide(#4,#1)|multiply(#5,const_60)| | physics |
last year the range of the annual yield on equal amount of investments in 100 mutual funds was $ 10000 . if the annual yield this year for each of the 100 mutual funds has improved by 15 percent this year than it was last year , what is the range of the annual yield on mutual funds this year ? | let the lowest yield be x . therefore , highest yield is x + 10000 . now yield of each mutual fund investment is improved by 10 % . therefore the yields will remain arranged in the same order as before . or lowest yield = 1.15 x and highest = 1.15 * ( x + 10000 ) or range = highest - lowest = 1.15 * ( x + 10000 ) - 1.1... | a ) $ 12700 , b ) $ 13000 , c ) $ 23000 , d ) $ 11500 , e ) $ 33000 | d | add(10000, multiply(10000, divide(15, 100))) | divide(n3,n0)|multiply(n1,#0)|add(n1,#1) | gain |
a customer purchased a package of ground beef at a cost of $ 1.80 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 25 percent less than the package of ground beef . what was the cost per pound of the steak ? | "for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.80 . let x be the price per pound for the steak . then 0.75 x = 180 x = 180 / 0.75 = $ 2.40 the answer is d ." | a ) $ 2.00 , b ) $ 2.15 , c ) $ 2.30 , d ) $ 2.40 , e ) $ 2.65 | d | divide(1.80, add(multiply(const_0_25, const_2), multiply(const_0_33, const_1))) | multiply(const_0_25,const_2)|multiply(const_0_33,const_1)|add(#0,#1)|divide(n0,#2)| | general |
two trains a and b starting from two points and travelling in opposite directions , reach their destinations 9 hours and 4 hours respectively after meeting each other . if the train a travels at 100 kmph , find the rate at which the train b runs . | "if two objects a and b start simultaneously from opposite points and , after meeting , reach their destinations in β a β and β b β hours respectively ( i . e . a takes β a hrs β to travel from the meeting point to his destination and b takes β b hrs β to travel from the meeting point to his destination ) , then the ra... | a ) 40 , b ) 60 , c ) 150 , d ) 80 , e ) 100 | c | multiply(100, sqrt(divide(9, 4))) | divide(n0,n1)|sqrt(#0)|multiply(n2,#1)| | physics |
a soft drink company had 6000 small and 15000 big bottles in storage . if 12 % of small 14 % of big bottles have been sold , then the total bottles remaining in storage is | "6000 + 15000 - ( 0.12 * 6000 + 0.14 * 15000 ) = 18180 . answer : c ." | a ) 15360 , b ) 16010 , c ) 18180 , d ) 14930 , e ) 16075 | c | subtract(add(6000, 15000), add(multiply(6000, divide(12, const_100)), multiply(divide(14, const_100), 15000))) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)| | general |
in a box of 10 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | "first , there are 8 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 10 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 8 c 3 / 10 c 3 = 56 / 120 = 7 / 15 answer is b" | a ) 1 / 12 , b ) 7 / 15 , c ) 2 / 13 , d ) 2 / 15 , e ) 1 / 17 | b | divide(divide(factorial(subtract(10, 2)), multiply(factorial(subtract(subtract(10, 2), 3)), factorial(3))), divide(factorial(10), multiply(factorial(subtract(10, 3)), factorial(3)))) | factorial(n2)|factorial(n0)|subtract(n0,n1)|subtract(n0,n2)|factorial(#2)|factorial(#3)|subtract(#2,n2)|factorial(#6)|multiply(#5,#0)|divide(#1,#8)|multiply(#7,#0)|divide(#4,#10)|divide(#11,#9)| | general |
company workers decided to raise rs . 3 lakhs by equal contribution from each . had they contributed rs . 50 each extra , the contribution would have been rs . 3.25 lakhs . how many workers were they ? | explanation : n * 50 = ( 325000 - 300000 ) = 25000 n = 25000 / 50 = 500 option c | a ) 300 , b ) 400 , c ) 500 , d ) 600 , e ) 700 | c | divide(subtract(multiply(3.25, multiply(const_100, const_1000)), multiply(3, multiply(const_100, const_1000))), 50) | multiply(const_100,const_1000)|multiply(n2,#0)|multiply(n0,#0)|subtract(#1,#2)|divide(#3,n1) | general |
the average of 7 numbers is 20 . if each number be multiplied by 5 . find the average of new set of numbers ? | "explanation : average of new numbers = 20 * 5 = 100 answer : option a" | a ) a ) 100 , b ) b ) 122 , c ) c ) 120 , d ) d ) 125 , e ) e ) 145 | a | multiply(20, 5) | multiply(n1,n2)| | general |
right triangle pqr is the base of the prism in the figure above . if pq = pr = Γ’ Λ Ε‘ 5 and the height of the prism is 10 , what is the volume of the prism ? | volume of prism = area of base * height = 1 / 2 * ( square root of 5 ) * ( square root of 5 ) * 10 = 25 answer : e | ['a ) 5', 'b ) 10', 'c ) 15', 'd ) 20', 'e ) 25'] | e | multiply(triangle_area(sqrt(5), sqrt(5)), 10) | sqrt(n0)|triangle_area(#0,#0)|multiply(n1,#1) | geometry |
each of the positive integers a and c is a 4 - digit integer . if each of the digits 0 through 9 appears in one of these 3 integers , what is the maximum possible value of the sum of a and c ? | according to the stem we should use the digits 0 through 9 to construct 2 four - digit integers , so that their sum is as big as possible . to maximize the sum , maximize the thousands digits of a and c , so make them 9 and 8 . next , maximize hundreds digits . make them 7 and 6 . next maximize 10 ' s digit place by 5 ... | a ) 18695 , b ) 18325 , c ) 18365 , d ) 18395 , e ) 18485 | d | add(add(add(add(multiply(9, const_1000), multiply(add(3, 4), const_100)), multiply(add(4, const_1), const_10)), 3), add(add(add(multiply(multiply(4, const_2), const_1000), multiply(multiply(3, const_2), const_100)), multiply(4, const_10)), const_2)) | add(n0,n3)|add(n0,const_1)|multiply(n2,const_1000)|multiply(n0,const_2)|multiply(n3,const_2)|multiply(n0,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|multiply(#3,const_1000)|multiply(#4,const_100)|add(#2,#6)|add(#8,#9)|add(#10,#7)|add(#11,#5)|add(n3,#12)|add(#13,const_2)|add(#14,#15) | general |
the length of a rectangle is doubled while its width is doubled . what is the % change in area ? | "the original area is l * w the new area is 2 l * 2 w = 4 * l * w = l * w + 3 * l * w the area increased by 300 % . the answer is b ." | a ) 250 % , b ) 300 % , c ) 500 % , d ) 650 % , e ) 700 % | b | multiply(subtract(multiply(const_2, const_3), const_1), const_10) | multiply(const_2,const_3)|subtract(#0,const_1)|multiply(#1,const_10)| | geometry |
riya and priya set on a journey . riya moves eastward at a speed of 18 kmph and priya moves westward at a speed of 24 kmph . how far will be priya from riya after 15 minutes | total eastward distance = 18 kmph * 1 / 4 hr = 4.5 km total westward distance = 24 kmph * 1 / 4 hr = 6 km total distn betn them = 4.5 + 6 = 10.5 m ans 11 km answer : b | a ) 25 kms , b ) 11 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms | b | multiply(speed(add(18, 24), const_60), 15) | add(n0,n1)|speed(#0,const_60)|multiply(n2,#1) | physics |
36 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 36 : : 12 : x working hours / day 6 : 5 30 x 6 x x = 36 x 5 x 12 x = ( 36 x 5 x 12 ) / ( 30 x 6 ) x = 12 answer a" | a ) 12 , b ) 16 , c ) 13 , d ) 18 , e ) 19 | a | divide(multiply(multiply(36, 12), 5), multiply(30, 6)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)| | physics |
a shopkeeper buys two articles for rs . 1000 each and then sells them , making 10 % profit on the first article and 10 % loss on second article . find the net profit or loss percent ? | profit on first article = 10 % of 1000 = 100 . this is equal to the loss he makes on the second article . that , is he makes neither profit nor loss . answer : c | a ) 200 , b ) 278 , c ) 100 , d ) 202 , e ) 270 | c | multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 10)), 1000), multiply(divide(const_100, add(const_100, 10)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 10)), 1000), multiply(divide(const_100, add(const_100, 10)), 1000))), const_100) | add(n1,const_100)|add(n0,n0)|subtract(const_100,n1)|divide(const_100,#2)|divide(const_100,#0)|multiply(n0,#3)|multiply(n0,#4)|add(#5,#6)|subtract(#7,#1)|multiply(#8,const_100)|divide(#9,#7)|multiply(#10,const_100) | gain |
a can run 4 times as fast as b and gives b a start of 63 m . how long should the race course be so that a and b might reach in the same time ? | "speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 63 m he gains 63 * ( 4 / 3 ) i . e 84 m answer : e" | a ) 70 m , b ) 60 m , c ) 80 m , d ) 65 m , e ) 84 m | e | add(multiply(4, divide(divide(63, 4), subtract(4, const_1))), 63) | divide(n1,n0)|subtract(n0,const_1)|divide(#0,#1)|multiply(n0,#2)|add(n1,#3)| | physics |
if c is 25 % of a and 10 % of b , what percent of a is b ? | "c is 25 % of a - - > c = a / 4 ; c is 10 % of b - - > c = b / 10 ; thus a / 4 = b / 10 - - > b = 5 / 2 * a = 2.5 a . therefore , b is 250 % of a . answer : e" | a ) 2.5 % , b ) 15 % , c ) 25 % , d ) 35 % , e ) 250 % | e | multiply(divide(divide(25, const_100), divide(10, const_100)), const_100) | divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)| | gain |
car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 40 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? | "car y began travelling after 72 minutes or 1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so , 35 ( t + 1.2 ) = 40 t t = 8.4 distance traveled by car x from the time car y began traveling until both cars stopped is 35 x 8.4 = 294 ... | a ) 205 , b ) 220 , c ) 240 , d ) 247 , e ) 294 | e | multiply(35, divide(multiply(divide(72, const_60), 35), subtract(40, 35))) | divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3)| | physics |
what is the ratio between perimeters of two squares one having 2.5 times the diagonal then the other ? | "d = 2.5 d d = d a β 2 = 2.5 d a β 2 = d a = 2.5 d / β 2 a = d / β 2 = > 2.5 : 1 answer : c" | a ) 4 : 5 , b ) 1 : 3 , c ) 2.5 : 1 , d ) 3.5 : 1 , e ) 3 : 2 | c | divide(2.5, divide(2.5, 2.5)) | divide(n0,n0)|divide(n0,#0)| | geometry |
a number when divided by 899 gives a remainder 63 . what remainder will be obtained by dividing the same number by 29 | 63 / 29 , thereforerequired number is : 5 , correct answer ( b ) | a ) 8 , b ) 5 , c ) 7 , d ) 6 , e ) 9 | b | subtract(63, multiply(29, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general |
a and b can do a piece of work in 10 days , while b and c can do the same work in 15 days and c and a in 25 days . they started working together , after 4 days a left . after another 4 days b left . in how many days c can finish the remaining work ? | "let the rates of a , b and c be a , b , and c respectively . a and b can do a piece of work in 10 days : a + b = 1 / 10 ; b and c can do the same work in 15 days : b + c = 1 / 15 ; c and a can do the same work in 25 days : c + a = 1 / 25 . sum the above 3 equations : 2 ( a + b + c ) = 31 / 150 - - > a + b + c = 31 / 3... | a ) 16 , b ) 32 , c ) 64 , d ) 96 , e ) none of these | d | divide(subtract(const_1, multiply(divide(add(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 25)), const_2), 4)), subtract(divide(add(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 25)), const_2), divide(const_1, 15))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|multiply(n3,#5)|subtract(#5,#1)|subtract(const_1,#6)|divide(#8,#7)| | physics |
17 men take 21 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ? | "3 w = 2 m 17 m - - - - - - 21 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 17 * 21 * 8 = 14 * x * 6 x = 34 answer : a" | a ) 34 , b ) 87 , c ) 30 , d ) 99 , e ) 77 | a | add(floor(divide(multiply(multiply(21, 8), multiply(17, 3)), multiply(multiply(21, 2), 6))), const_1) | multiply(n1,n2)|multiply(n0,n5)|multiply(n4,n6)|multiply(#0,#1)|multiply(n3,#2)|divide(#3,#4)|floor(#5)|add(#6,const_1)| | physics |
if xy = 9 , x / y = 36 , for positive numbers x and y , y = ? | very easy question . 2 variables and 2 easy equations . xy = 9 - - - > x = 9 / y - ( i ) x / y = 36 - - - > replacing ( i ) here - - - > 9 / ( y ^ 2 ) = 36 - - - > y ^ 2 = 9 / 36 = 1 / 4 - - - > y = 1 / 2 or - 1 / 2 the question states that x and y are positive integers . therefore , y = 1 / 2 is the answer . answer a ... | a ) 1 / 2 , b ) 2 , c ) 1 / 3 , d ) 3 , e ) 1 / 6 | a | sqrt(divide(9, 36)) | divide(n0,n1)|sqrt(#0) | general |
in one year , the population , of a village increased by 20 % and in the next year , it decreased by 20 % . if at the end of 2 nd year , the population was 9600 , what was it in the beginning ? | x * 120 / 100 * 80 / 100 = 9600 x * 0.96 = 9600 x = 9600 / 0.96 = > 10000 answer : a | a ) 10000 , b ) 8000 , c ) 1988 , d ) 1277 , e ) 2081 | a | divide(divide(9600, subtract(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))) | divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|divide(n3,#2)|divide(#3,#1) | general |
two numbers are in the ratio of 15 : 11 . if their h . c . f . is 13 , find the numbers . | let the required numbers be 15 . x and llx . then , their h . c . f . is x . so , x = 13 . the numbers are ( 15 x 13 and 11 x 13 ) i . e . , 195 and 143 . answer is a . | a ) 195,143 , b ) 185,133 , c ) 175,123 , d ) 165,113 , e ) none of them | a | multiply(15, 13) | multiply(n0,n2) | other |
a bag contains 7 red , 5 blue and 4 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 7 c 216 c 2 = 7 c 216 c 2 = 21 / 120 = 7 / 40 c" | a ) 2 / 15 , b ) 2 / 21 , c ) 7 / 40 , d ) 3 / 29 , e ) 4 / 27 | c | divide(choose(7, 2), choose(add(add(7, 5), 4), 2)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)| | other |
a boat can travel with a speed of 22 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 216 km downstream | "explanation : speed of the boat in still water = 22 km / hr speed of the stream = 5 km / hr speed downstream = ( 22 + 5 ) = 27 km / hr distance travelled downstream = 216 km time taken = distance / speed = 216 / 27 = 8 hours . answer : option e" | a ) 5 hours , b ) 4 hours , c ) 3 hours , d ) 2 hours , e ) 8 hours | e | divide(216, add(22, 5)) | add(n0,n1)|divide(n2,#0)| | physics |
a certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio w of revenue from car sales ... | a . . i have probably solved this question 3 - 4 times by now . . remember the answer . . 1 : 2 | a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3 | a | divide(subtract(add(const_100, 7), add(const_100, 1)), subtract(add(const_100, 1), subtract(const_100, 11))) | add(n3,const_100)|add(n6,const_100)|subtract(const_100,n1)|subtract(#0,#1)|subtract(#1,#2)|divide(#3,#4)| | other |
in 1979 approximately 1 / 3 of the 37.3 million airline passengers traveling to or from the united states used kennedy airport . if the number of such passengers that used miami airport was 1 / 2 the number that used kennedy airport and 5 times the number that used logan airport , approximately how many millions of the... | "number of passengers using kennedy airport = 37 / 3 = ~ 12.43 passengers using miami airport = 12.43 / 2 = ~ 6.2 passengers using logan airport = 6.2 / 5 = ~ 1.24 so c" | a ) 18.6 , b ) 9.3 , c ) 1.2 , d ) 3.1 , e ) 1.6 | c | divide(divide(37.3, 3), multiply(5, 2)) | divide(n3,n2)|multiply(n5,n6)|divide(#0,#1)| | general |
the average of 5 numbers is 6.8 . if one of the numbers is multiplied by a factor of 3 , the average of the numbers increases to 12.8 . what number is multiplied by 3 ? | "the average of 5 numbers is 6.8 the sum of 5 numbers will be 6.8 x 5 = 34 the average of 5 number after one of the number is multiplied by 3 is 12.8 the sum of the numbers will now be 12.8 x 5 = 64 so the sum has increased by 64 - 34 = 30 let the number multiplied by 3 be n then , 3 n = n + 30 or 2 n = 30 or n = 15 an... | a ) 15.0 , b ) 13.0 , c ) 13.9 , d ) 10.0 , e ) 6.0 | a | subtract(multiply(12.8, 5), multiply(6.8, 5)) | multiply(n0,n3)|multiply(n0,n1)|subtract(#0,#1)| | general |
in covering a distance of 18 km , abhay takes 2 hours more than sameer . if abhay doubles his speed , then he would take 1 hour less than sameer . abhay ' s speed is : | let abhay ' s speed be x km / hr . then , 18 / x - 18 / 2 x = 3 6 x = 18 x = 3 km / hr . answer : option a | a ) 3 kmph , b ) 6 kmph , c ) 6.25 kmph , d ) 7.5 kmph , e ) 7.8 kmph | a | divide(subtract(18, divide(18, 2)), add(1, 2)) | add(n1,n2)|divide(n0,n1)|subtract(n0,#1)|divide(#2,#0) | physics |
a portion of the 50 % solution of chemicals was replaced with an equal amount of 60 % solution of chemicals . as a result , 55 % solution of chemicals resulted . what part of the original solution was replaced ? | "this is a weighted average question . say x % of the solution was replaced - - > equate the amount of chemicals : 0.5 ( 1 - x ) + 0.6 * x = 0.55 - - > x = 1 / 2 . answer : e ." | a ) 3 / 4 , b ) 1 / 4 , c ) 3 / 4 , d ) 2 / 5 , e ) 1 / 2 | e | divide(subtract(50, 55), subtract(50, 60)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | gain |
ram get 450 marks in his exam which is 90 % of total marks . what is the total marks ? | x * ( 90 / 100 ) = 450 x = 5 * 100 x = 500 answer : d | a ) 475 , b ) 600 , c ) 550 , d ) 500 , e ) 525 | d | add(450, multiply(add(const_3, const_2), const_10)) | add(const_2,const_3)|multiply(#0,const_10)|add(n0,#1) | general |
all the faces of a cube are painted with blue colour . then it is cut into 125 small equal cubes . how many small cubes will be formed having only one face coloured ? | no . of small cubes will have only one face painted = ( x - 2 ) 2 * 6 here x = side of the small cube = 5 therfore ( x - 2 ) 2 * 6 = 54 answer : a | a ) 54 , b ) 8 , c ) 16 , d ) 24 , e ) 34 | a | divide(multiply(power(subtract(power(125, divide(const_1, const_3)), const_2), const_2), const_60), const_10) | divide(const_1,const_3)|power(n0,#0)|subtract(#1,const_2)|power(#2,const_2)|multiply(#3,const_60)|divide(#4,const_10) | geometry |
in a certain country 1 / 7 of 8 = 5 . assuming the same proportion , what would be the value of 1 / 5 of 40 ? | "d 35" | a ) 54 , b ) 25 , c ) 45 , d ) 35 , e ) 52 | d | multiply(divide(7, 8), 40) | divide(n1,n2)|multiply(n6,#0)| | general |
mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 30 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | let me try a simpler one . lets assume that candidate got 30 % votes and total votes is 100 . candidate won = 30 remaining = 70 to get 50 % , candidate requires 20 votes from 100 which is 20 % and 20 votes from 70 . 20 / 70 = 2 / 7 = . 285 = 28.5 % which is approx 29 % . hence the answer is d . | a ) 10 % , b ) 12 % , c ) 15 % , d ) 29 % , e ) 20 % | d | multiply(divide(subtract(divide(50, const_100), divide(30, const_100)), subtract(const_1, divide(30, const_100))), const_100) | divide(n2,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(const_1,#1)|divide(#2,#3)|multiply(#4,const_100) | general |
if f ( x ) = x ^ 4 - 4 x ^ 3 - 2 x ^ 2 + 5 x , then f ( - 1 ) = | f ( - 1 ) = ( - 1 ) ^ 4 - 4 ( - 1 ) ^ 3 - 2 ( - 1 ) ^ 2 + 5 ( - 1 ) = 1 + 4 - 2 - 5 = - 2 the answer is b . | a ) - 4 , b ) - 2 , c ) - 1 , d ) 1 , e ) 2 | b | add(subtract(subtract(power(negate(1), const_4), multiply(4, power(negate(1), const_3))), multiply(2, power(negate(1), const_2))), multiply(5, negate(1))) | negate(n6)|multiply(n5,#0)|power(#0,const_4)|power(#0,const_3)|power(#0,const_2)|multiply(n0,#3)|multiply(n3,#4)|subtract(#2,#5)|subtract(#7,#6)|add(#1,#8) | general |
in the junior basketball league there are 18 teams , 2 / 3 of them are bad and Β½ are rich . what ca n ' t be the number of teams that are rich and bad ? | otal teams = 18 bad teams = ( 2 / 3 ) * 18 = 12 rich teams = 9 so maximum value that the both rich and bad can take will be 9 . so e = 10 can not be that value . answer : e | a ) 4 . , b ) 6 . , c ) 7 . , d ) 8 . , e ) 10 | e | add(multiply(18, divide(const_1, const_2)), const_1) | divide(const_1,const_2)|multiply(n0,#0)|add(#1,const_1) | general |
working alone , john finishes cleaning half the house in a third of the time it takes nick to clean the entire house alone . john alone cleans the entire house in 6 hours . how many hours will it take nick and john to clean the entire house if they work together ? | answer is 3.6 hours . john does the complete house in 6 hours while nick does it in 9 hours . 1 / ( 1 / 6 + 1 / 9 ) = 3.6 answer is e | a ) 1.5 , b ) 2 , c ) 2.4 , d ) 3 , e ) 3.6 | e | inverse(add(inverse(6), inverse(multiply(divide(const_3, const_2), 6)))) | divide(const_3,const_2)|inverse(n0)|multiply(n0,#0)|inverse(#2)|add(#1,#3)|inverse(#4) | physics |
the radius of a cylinder is 2 r units and height is 3 r units . find the curved surface ? | "explanation : 2 * Ο * 2 r * 3 r = 12 Ο r 2 answer : b" | a ) 18 Ο r 2 , b ) 12 Ο r 2 , c ) 3 Ο r 2 , d ) 6 Ο r 2 , e ) 9 Ο r 2 | b | multiply(circumface(2), 3) | circumface(n0)|multiply(n1,#0)| | geometry |
on the richter scale , which measures the total amount of energy released during an earthquake , a reading of x - 1 indicates one - tenth the released energy as is indicated by a reading of x . on that scale , the frequency corresponding to a reading of 7 is how many times as great as the frequency corresponding to a r... | "if richter scale reading goes from x - 1 to x it will be 10 if richter scale reading goes from 5 to 6 it will be 10 similarly if richter scale reading goes from 6 to 7 it will be 10 so it will from 5 to 7 i . e 6,7 = 10 * 10 = 10 ^ 2 answer is a" | a ) 10 ^ 2 , b ) 10 ^ 3 , c ) 10 ^ 4 , d ) 10 ^ 5 , e ) 10 ^ 6 | a | power(const_10, subtract(7, 5)) | subtract(n1,n2)|power(const_10,#0)| | general |
the total of 344 of 20 paise and 25 paise make a sum of rs . 71 . the no of 20 paise coins is | explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 344 - x ) . 0.20 * ( x ) + 0.25 ( 344 - x ) = 71 = > x = 300 . . answer : e ) 300 | a ) 238 , b ) 277 , c ) 278 , d ) 200 , e ) 300 | e | divide(subtract(multiply(344, 25), multiply(71, const_100)), subtract(25, 20)) | multiply(n0,n2)|multiply(n3,const_100)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2) | general |
a man buys a cycle for rs . 800 and sells it at a loss of 15 % . what is the selling price of the cycle ? | "s . p . = 85 % of rs . 800 = rs . 85 / 100 x 800 = rs . 680 answer : c" | a ) s . 1090 , b ) s . 1160 , c ) s . 680 , d ) s . 520 , e ) s . 700 | c | divide(multiply(subtract(const_100, 15), 800), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain |
a math teacher has 26 cards , each of which is in the shape of a geometric figure . half of the cards are rectangles , and a third of the cards are rhombuses . if 8 cards are squares , what is the maximum possible number of cards that re circles . | "a square is a special kind of rhombus ( sides are perpendicular ) a square is a special kind of rectangles ( sides with same length ) among the 26 cards with have : 15 rectangles 10 rhombus 8 squares among the 15 rectangles , there could be 8 special ones ( with sides of same length ) that are squares . that lets at l... | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | subtract(subtract(divide(divide(multiply(26, 8), const_10), const_2), const_0_25), const_0_25) | multiply(n0,n1)|divide(#0,const_10)|divide(#1,const_2)|subtract(#2,const_0_25)|subtract(#3,const_0_25)| | geometry |
if a * b * c = ( β ( a + 2 ) ( b + 3 ) ) / ( c + 1 ) , find the value of 6 * 15 * 7 . | "6 * 15 * 3 = ( β ( 6 + 2 ) ( 15 + 3 ) ) / ( 7 + 1 ) = ( β 8 * 18 ) / 8 = ( β 144 ) / 8 = 12 / 8 = 1.5 answer is e" | a ) 8 , b ) 5 , c ) 11 , d ) 3 , e ) 1.5 | e | divide(sqrt(multiply(add(6, 2), add(15, 3))), add(7, 1)) | add(n0,n3)|add(n1,n4)|add(n2,n5)|multiply(#0,#1)|sqrt(#3)|divide(#4,#2)| | general |
if 3 x = 6 y = z , what is x + y , in terms of z ? | "3 x = 6 y = z x = z / 3 and y = z / 6 x + y = z / 3 + z / 6 = z / 2 answer is a" | a ) z / 2 , b ) 2 z , c ) z / 3 , d ) 3 z / 5 , e ) z / 9 | a | divide(subtract(divide(multiply(3, const_100), const_2), const_2), add(divide(multiply(3, const_100), const_2), const_2)) | multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)| | general |
a salesman β s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1400 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ? | "[ 1400 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 20000 answer : c" | a ) 12028 , b ) 12000 , c ) 20000 , d ) 12197 , e ) 12012 | c | divide(600, divide(5, const_100)) | divide(n0,const_100)|divide(n4,#0)| | general |
how many bricks each measuring 21 cm x 10 cm x 8 cm , will be needed to build a wall 9 m x 5 m x 18.5 m | "explanation : no . of bricks = volume of the wall / volume of 1 brick = ( 900 x 500 x 18.5 ) / ( 21 x 10 x 8 ) = 4955 answer : a" | a ) 4955 , b ) 4899 , c ) 4650 , d ) 7200 , e ) none of these | a | divide(multiply(multiply(9, 5), 18.5), divide(divide(multiply(multiply(21, 10), 8), const_100), const_100)) | multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#3,const_100)|divide(#4,const_100)|divide(#2,#5)| | physics |
at what price must an article costing rs . 47.50 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "explanation : cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 95 / 100 ) = 59.375 mp = 62.5 answer : a" | a ) 62.5 , b ) 62.8 , c ) 62.1 , d ) 62.9 , e ) 32.5 | a | divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 5)) | multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)| | gain |
during a certain two - week period , 64 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 5 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedies , ... | total movies = 100 . comedies = 64 . action + drama = 36 . since there were 5 times as many dramas as action movies , then action + 5 * action = 36 - - > action = a = 6 . comedies = 60 = 10 a . a | a ) 10 a , b ) 12 a , c ) 14 a , d ) 16 a , e ) 18 a | a | floor(divide(64, add(5, const_1))) | add(n1,const_1)|divide(n0,#0)|floor(#1) | general |
find the greatest 4 digit number which leaves respective remainders of 2 and 5 when divided by 15 and 24 . a . 9974 | explanation : since the difference between the divisors and the respective remainders is not constant , back substitution is the convenient method . none of the given numbers is satisfying the condition . answer : e | a ) 388 , b ) 282 , c ) 378 , d ) 292 , e ) 281 | e | subtract(subtract(multiply(15, 24), multiply(5, 15)), const_4) | multiply(n3,n4)|multiply(n2,n3)|subtract(#0,#1)|subtract(#2,const_4) | general |
p has $ 35 more than what q and r together would have had if both b and c had 1 / 7 of what p has . how much does p have ? | "p = ( 2 / 7 ) * p + 35 ( 5 / 7 ) * p = 35 p = 49 the answer is c ." | a ) $ 45 , b ) $ 47 , c ) $ 49 , d ) $ 51 , e ) $ 53 | c | divide(35, subtract(1, multiply(divide(1, 7), const_2))) | divide(n1,n2)|multiply(#0,const_2)|subtract(n1,#1)|divide(n0,#2)| | general |
a can do a work in 20 days and b in 30 days . if they work on it together for 4 days , then the fraction of the work that is left is : | "ans is : b a ' s 1 day ' s work = 1 / 20 b ' s 1 day ' s work = 1 / 30 ( a + b ) ' s 1 day ' s work = ( 1 / 20 + 1 / 30 ) = 1 / 12 ( a + b ) ' s 4 day ' s work = ( 1 / 12 * 4 ) = 1 / 3 therefore , remaining work = ( 1 - 1 / 3 ) = 2 / 3" | a ) 1 / 3 , b ) 2 / 3 , c ) 4 / 3 , d ) 5 / 3 , e ) 7 / 3 | b | subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 30)), 4)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics |
3 candidates in an election and received 1136 , 7636 and 10628 votes respectively . what % of the total votes did the winning candidate gotin that election ? | "total number of votes polled = ( 1136 + 7636 + 10628 ) = 19400 so , required percentage = 10628 / 19400 * 100 = 54.8 % c" | a ) 40 % , b ) 55 % , c ) 54.8 % , d ) 60 % , e ) 62 % | c | multiply(divide(10628, add(add(1136, 7636), 10628)), const_100) | add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)| | gain |
a is the hundreds digit of the 3 digit integer x , b is the tens digit of x , and c is the units digit of x . 4 a = 2 b = c , and a > 0 . what is the difference between the two greatest possible values of x ? tip : dont stop till you have exhausted all answer choices to arrive at the correct one . | ratio of a : b : c = 1 : 2 : 4 two possible greatest single digit values for c are 8 and 4 if c is 8 , then x = 248 if c is 4 , then x = 124 difference = 248 - 124 = 124 a is the answer | a ) 124 , b ) 297 , c ) 394 , d ) 421 , e ) 842 | a | add(multiply(multiply(4, 3), const_10), 4) | multiply(n0,n1)|multiply(#0,const_10)|add(n1,#1) | general |
if albert β s monthly earnings rise by 27 % , he would earn $ 567 . if , instead , his earnings rise by only 26 % , how much ( in $ ) would he earn this month ? | "= 567 / 1.27 β 1.26 = 562 = 562 answer is c" | a ) 643 , b ) 652 , c ) 562 , d ) 578 , e ) 693 | c | multiply(divide(567, add(const_1, divide(27, const_100))), add(const_1, divide(26, const_100))) | divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n1,#3)|multiply(#2,#4)| | gain |
indu gave bindu rs . 1250 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "1250 = d ( 100 / 4 ) 2 d = 2 answer : b" | a ) 5 , b ) 2 , c ) 9 , d ) 5 , e ) 1 | b | subtract(subtract(multiply(1250, power(add(const_1, divide(4, const_100)), 2)), 1250), multiply(multiply(1250, divide(4, const_100)), 2)) | divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)| | gain |
when n is divided by 24 , the remainder is 4 . what is the remainder when 4 n is divided by 8 ? | "let n = 4 ( leaves a remainder of 4 when divided by 24 ) 4 n = 4 ( 4 ) = 16 , which leaves a remainder of 0 when divided by 8 . answer d" | a ) 3 , b ) 4 , c ) 5 , d ) 0 , e ) 7 | d | subtract(4, reminder(4, 8)) | reminder(n2,n3)|subtract(n1,#0)| | general |
tom and linda stand at point a . linda begins to walk in a straight line away from tom at a constant rate of 3 miles per hour . one hour later , tom begins to jog in a straight line in the exact opposite direction at a constant rate of 8 miles per hour . if both tom and linda travel indefinitely , what is the positive ... | "d is the answer . . . . d = ts where d = distance , t = time and s = speed to travel half distance , ( 2 + 3 t ) = 8 t = = > t = 2 / 5 = = > 24 minutes to travel double distance , 2 ( 2 + 3 t ) = 8 t = = > 2 = = > 120 minutes difference , 96 minutes d" | a ) 60 , b ) 72 , c ) 84 , d ) 96 , e ) 108 | d | multiply(subtract(divide(multiply(const_2, const_2), subtract(8, multiply(const_2, 3))), divide(const_2, subtract(8, 3))), const_60) | multiply(const_2,const_2)|multiply(n0,const_2)|subtract(n1,n0)|divide(const_2,#2)|subtract(n1,#1)|divide(#0,#4)|subtract(#5,#3)|multiply(#6,const_60)| | physics |
35 - [ 23 - { 15 - x } ] = 12 Γ 2 Γ· 1 / 2 | explanation : 35 - [ 23 - { 19 - ( 15 - x ) } ] = 12 Γ 2 Γ 2 = 48 = > 35 - 23 + ( 19 - 15 + x ) = 48 = > 12 + 4 + x = 48 = > x = 48 - ( 4 + 12 ) = 32 answer : option b | a ) 34 , b ) 32 , c ) 17 , d ) 27 , e ) 28 | b | subtract(35, const_3) | subtract(n0,const_3) | general |
amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.2 km ? | "c 300 m distance covered by amar = 18 / 4.8 ( 1.6 km ) = 3 / 8 ( 1200 ) = 300 m answer is c" | a ) 600 m , b ) 200 m , c ) 300 m , d ) 400 m , e ) 100 m | c | divide(multiply(18, multiply(1.2, const_1000)), 48) | multiply(n2,const_1000)|multiply(n0,#0)|divide(#1,n1)| | physics |
if n is an integer , f ( n ) = f ( n - 1 ) - n and f ( 4 ) = 20 . what is the value of f ( 6 ) ? | "since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 20 then f ( 5 ) = 20 - 5 = 15 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 15 - 6 = 9 . answer : a . questions on funtions to practice :" | a ) 9 , b ) 0 , c ) 1 , d ) 2 , e ) 4 | a | subtract(subtract(20, add(1, 4)), 6) | add(n0,n1)|subtract(n2,#0)|subtract(#1,n3)| | general |
what is the remainder when 14,453 Γ 15,654 Γ 16,788 is divided by 5 ? | "only the unit ' s digit of the product will decide the remainder when divided by 5 . hence , 3 * 4 * 8 = will give units digit as 5 so , whatever be the number , if it ends in 6 , the remainder after dividing with 5 will be 1 . optionb" | a ) 2 , b ) 1 , c ) 5 , d ) 4 , e ) 3 | b | reminder(multiply(15,654, 14,453), 16,788) | multiply(n0,n1)|reminder(#0,n2)| | general |
a thief goes away with a santro car at a speed of 40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 50 kmph when will the owner over take the thief from the start ? | "d 20 hours | - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - | 50 40 d = 20 rs = 50 β 40 = 10 t = 20 / 10 = 2 hours" | a ) 22 hours , b ) 21 hours , c ) 23 hours , d ) 20 hours , e ) 28 hours | d | subtract(divide(multiply(divide(const_1, const_2), 40), subtract(50, 40)), divide(const_1, const_2)) | divide(const_1,const_2)|subtract(n1,n0)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,#0)| | physics |
what is the smallest 5 digit number that is divisible by 15 , 32 , 45 , and 54 ? | 15 = 3 * 5 32 = 2 ^ 5 45 = 3 ^ 2 * 5 54 = 2 * 3 ^ 3 lcm = 2 ^ 5 * 3 ^ 3 * 5 = 4320 the smallest five - digit number that is a multiple of 4320 is 3 * 4320 = 12,960 the answer is d . | a ) 11260 , b ) 11860 , c ) 12360 , d ) 12960 , e ) 13560 | d | add(multiply(const_100, const_100), subtract(lcm(lcm(lcm(lcm(5, 15), 32), 45), 54), reminder(multiply(const_100, const_100), lcm(lcm(lcm(lcm(5, 15), 32), 45), 54)))) | lcm(n0,n1)|multiply(const_100,const_100)|lcm(n2,#0)|lcm(n3,#2)|lcm(n4,#3)|reminder(#1,#4)|subtract(#4,#5)|add(#1,#6) | general |
excluding stoppages , the average speed of a bus is 60 km / hr and including stoppages , the average speed of the bus is 45 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 60 km without stoppages and 45 km with stoppages . stoppage time = time take to travel ( 60 - 45 ) km i . e 15 km at 60 km / hr . stoppage time = 15 / 60 hrs = 15 min . answer : a" | a ) 15 , b ) 88 , c ) 77 , d ) 20 , e ) 99 | a | subtract(multiply(const_1, const_60), multiply(divide(45, 60), const_60)) | divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)| | general |
a rectangular courty 3.78 metres long and 5.25 metres wide is to be paved exactly with square tiles , all of the same size . what is the largest size of the tile which could be used for the purpose ? | solution largest size of the tile . h . c . f of 378 cm and 525 cm = 21 cms . answer b | ['a ) 14 cms', 'b ) 21 cms', 'c ) 42 cms', 'd ) none of these', 'e ) can not be determined'] | b | multiply(divide(divide(divide(divide(multiply(const_100, 3.78), const_3), const_3), const_3), const_2), const_3) | multiply(n0,const_100)|divide(#0,const_3)|divide(#1,const_3)|divide(#2,const_3)|divide(#3,const_2)|multiply(#4,const_3) | geometry |
working together , wayne and his son can shovel the entire driveway in three hours . if wayne can shovel five times as fast as his son can , how many hours would it take for his son to shovel the entire driveway on his own ? | "w : the time for wyane to do the job s : the time for his son to do the job we have 1 / w + 1 / s = 1 / 5 and w = 5 s then we have 1 / ( 5 * s ) + 1 / s = 1 / 5 < = > 6 / ( 5 * s ) = 1 / 5 < = > s = 6 ans : b" | a ) 4 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | b | multiply(multiply(add(inverse(multiply(const_2, const_4)), const_1), const_3), multiply(const_2, const_4)) | multiply(const_2,const_4)|inverse(#0)|add(#1,const_1)|multiply(#2,const_3)|multiply(#3,#0)| | physics |
the diagonal of a rhombus are 80 m and 120 m . its area is : | "area of the rhombus = 1 / 2 d 1 d 2 = ( 1 / 2 Γ£ β 80 Γ£ β 120 ) cm ( power ) 2 = 80 Γ£ β 60 = 4800 cm ( power ) 2 answer is c ." | a ) 4500 , b ) 4000 , c ) 4800 , d ) 4600 , e ) 4320 | c | rhombus_area(80, 120) | rhombus_area(n0,n1)| | geometry |
3 bodies x , y and z start moving around a circular track of length 960 m from the same point simultaneously in the same direction at speeds of 12 m / s , 20 m / s and 36 m / s respectively . when will they meet for the first time after they started moving ? | if they all meet after t seconds , it means they covered the distances 12 t , 20 t , and 36 t respectively . since they all arrive to the same spot , it means that the differences taken pairwise between the distances must be positive integer multiples of the length of the track , which is 960 m . so , 8 t , 16 t , and ... | a ) 240 seconds , b ) 120 seconds , c ) 60 seconds , d ) 180 seconds , e ) 100 seconds | b | divide(960, subtract(20, 12)) | subtract(n3,n2)|divide(n1,#0) | physics |
a part of certain sum of money is invested at 10 % per annum and the rest at 12 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "12 : 10 = 6 : 5 answer : e" | a ) 4 : 2 , b ) 4 : 8 , c ) 4 : 3 , d ) 4 : 0 , e ) 6 : 5 | e | multiply(divide(12, const_100), 10) | divide(n1,const_100)|multiply(n0,#0)| | gain |
kelly and chris are moving into a new city . both of them love books and thus packed several boxes with books . if chris packed 60 % of the total number of boxes , what was the ratio of the number of boxes kelly packed to the number of boxes chris packed ? | the ratio of the number of boxes kelly packed to the number of boxes chris packed = 40 / 60 = 2 / 3 answer : b | a ) 4 / 3 , b ) 2 / 3 , c ) 1 / 3 , d ) 3 / 4 , e ) 1 / 4 | b | divide(subtract(const_100, 60), 60) | subtract(const_100,n0)|divide(#0,n0) | other |
there are 6 more women than there are men on a local co - ed softball team . if there are a total of 24 players on the team , what is the ratio of men to women ? | "w = m + 6 w + m = 24 m + 6 + m = 24 2 m = 18 m = 9 w = 15 ratio : 9 : 15 ans : b" | a ) 10 / 16 , b ) 9 / 15 , c ) 4 / 16 , d ) 6 / 10 , e ) 4 / 10 | b | divide(divide(subtract(24, 6), add(const_1, const_1)), add(divide(subtract(24, 6), add(const_1, const_1)), 6)) | add(const_1,const_1)|subtract(n1,n0)|divide(#1,#0)|add(n0,#2)|divide(#2,#3)| | general |
two cars start at the same time from opposite ends of a highway that is 60 miles long . one car is riding at 13 mph and the second car is riding at 17 mph . how long after they begin will they meet ? | "as cars are moving in opposite directions their speeds will be added . so their relative speeds : 17 + 13 = 30 mph total distance to be covered = 60 miles . time taken would be : 60 miles / 30 mph = 2.0 hours e is the answer ." | a ) 0.75 , b ) 1 , c ) 1.25 , d ) 1.5 , e ) 2.0 | e | divide(60, add(13, 17)) | add(n1,n2)|divide(n0,#0)| | physics |
( x + 6 ) is a factor in x ^ 2 - mx - 42 . what is the value of m ? | i solved the second degree equation and found it like this : x ^ 2 - mx - 42 = 0 ( x - 7 ) ( x + 6 ) = 0 x = 7 or x = - 6 substituting both values for x in the equation we find : x ^ 2 - mx - 42 = > ( - 6 ) ^ 2 - m ( - 6 ) = 42 = > 36 + 6 m = 42 = > 6 m = 42 - 36 = 6 = > m = 1 and with 7 , using a similar process we en... | a ) 2 , b ) 2.2 , c ) 1 , d ) 4 , e ) 5 | c | subtract(divide(42, 6), 6) | divide(n2,n0)|subtract(#0,n0) | general |
local kennel has cats and dogs in the ratio of 6 : 8 . if there are 10 fewer cats than dogs , how many dogs are in the kennel ? | "lets work with the data given to us . we know that there ratio of cats to dogs is 6 : 8 or cats 6 dogs 8 we can write number of cats as 6 x and number of dogs as 8 x and we know that 8 x - 6 x = 10 ( therefore 2 x = 10 = > x = 5 ) then # of dogs = 8 x 5 = 40 answer is e" | a ) 30 , b ) 35 , c ) 45 , d ) 50 , e ) 40 | e | multiply(10, 8) | multiply(n1,n2)| | other |
a rectangular field has a length 10 meters more than it is width . if the area of the field is 144 , what is the length ( in meters ) of the rectangular field ? | "area = l * w = ( l ) * ( l - 10 ) = 171 trial and error : 20 * 10 = 200 ( too high ) 19 * 9 = 171 ( too high ) 18 * 8 = 144 the length is 18 meters . the answer is b ." | a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | b | add(10, add(const_0_25, add(const_0_33, divide(divide(144, 10), const_2)))) | divide(n1,n0)|divide(#0,const_2)|add(#1,const_0_33)|add(#2,const_0_25)|add(n0,#3)| | geometry |
add 15 % of 25 and 12 % of 45 . | 15 % of 25 + 12 % of 45 25 * 15 / 100 + 45 * 12 / 100 3.8 + 5.4 = 9.2 answer a | a ) 9.2 , b ) 10.5 , c ) 11.5 , d ) 12.3 , e ) 15 | a | add(divide(multiply(15, 25), const_100), divide(multiply(12, 45), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3) | gain |
there are cats got together and decided to kill the mice of 999919 . each cat kills equal number of mice and each cat kills more number of mice than cats there were . then what are the number of cats ? | "999919 can be written as 1000000 β 81 = 10002 β 92 ie of the form a 2 - b 2 = ( a + b ) ( a - b ) = ( 1000 + 9 ) * ( 1000 - 9 ) = ( 1009 ) * ( 991 ) given that number of cats is less than number if mice . so number of cats is 991 and number of mice were 1009 answer c" | a ) 941,1009 , b ) 991,1001 , c ) 991,1009 , d ) 791,1009 , e ) 931,1009 | c | divide(999919, add(multiply(const_100, const_10), add(const_3, const_2))) | add(const_2,const_3)|multiply(const_10,const_100)|add(#0,#1)|divide(n0,#2)| | general |
a crate measures 4 feet by 8 feet by 12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius , in feet , of the pillar with the largest volume that could still... | "to fit the cylinder with largest radius inside this cuboid , we should make the base of the crate as wide as possible so we will take the base as 12 feet by 8 feet now since the limiting number in the base is 8 feet ; therefore a cylinder { we can visualise that a cylinder ' s width is its diameter } can only fit insi... | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | divide(divide(multiply(multiply(8, 12), 4), 12), 8) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,n2)|divide(#2,n1)| | geometry |
an equilateral triangle t 2 is formed by joining the mid points of the sides of another equilateral triangle t 1 . a third equilateral triangle t 3 is formed by joining the mid - points of t 2 and this process is continued indefinitely . if each side of t 1 is 45 cm , find the sum of the perimeters of all the triangles... | "we have 45 for first triangle , when we join mid - points of first triangle we get the second equilateral triangle then the length of second one is 22.5 and continues . so we have 45 , 22.5 , 11.25 , . . . we have ratio = 1 / 2 , and it is gp type . sum of infinite triangle is a / 1 - r = 45 / 1 - ( 1 / 2 ) = 90 equil... | a ) 180 cm , b ) 220 cm , c ) 240 cm , d ) 270 cm , e ) 300 cm | d | add(triangle_perimeter(45, 45, 45), triangle_perimeter(45, 45, 45)) | triangle_perimeter(n5,n5,n5)|add(#0,#0)| | geometry |
if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 120 , how much more remains to be paid ? | "10 / 100 p = 120 > > p = 120 * 100 / 10 = 1200 1200 - 120 = 1080 answer : e" | a ) $ 880 , b ) $ 990 , c ) $ 1,000 , d ) $ 1,100 , e ) $ 1,080 | e | subtract(multiply(120, divide(const_100, 10)), 120) | divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)| | general |
a telephone company needs to create a set of 3 - digit area codes . the company is entitled to use only digits 2 , 4 and 6 , which can be repeated . if the product of the digits in the area code must be even , how many different codes can be created ? | "total # of codes possible is 3 * 3 * 3 = 27 . oit of those 27 codes answer : a" | a ) 27 , b ) 22 , c ) 24 , d ) 26 , e ) 30 | a | subtract(power(3, 3), const_1) | power(n0,n0)|subtract(#0,const_1)| | general |
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