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by selling 22 pens for a rupee a woman loses 10 % . how many for a rupee should he sell in order to gain 50 % ? | "1394 x 1394 = ( 1394 ) 2 = ( 1400 - 2 ) 2 = ( 1400 ) 2 + ( 6 ) 2 - ( 6 x 1400 x 6 ) = 1943236 . answer : option a" | a ) 1943236 , b ) 1981709 , c ) 18362619 , d ) 2031719 , e ) none of these | a | multiply(divide(1394, 1394), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
1394 x 1394 | b claims 2 / 7 of the profit a claims remaining 5 / 7 of the profit = > a : b = 5 / 7 : 2 / 7 = 5 : 2 let the money contributed by b = b then a : b = 16000 Γ 8 : b Γ 4 therefore , 16000 Γ 8 : b Γ 4 = 5 : 2 16000 Γ 8 Γ 2 = b Γ 4 Γ 5 16000 Γ 2 Γ 2 = b Γ 5 3200 Γ 2 Γ 2 = b b = 12800 answer is a . | a ) 12800 , b ) 13000 , c ) 11500 , d ) 12500 , e ) 12000 | a | divide(multiply(multiply(16000, 8), divide(8, 4)), multiply(4, add(4, const_1))) | add(n2,const_1)|divide(n1,n2)|multiply(n0,n1)|multiply(#1,#2)|multiply(n2,#0)|divide(#3,#4) | general |
a and b start a business jointly . a invests rs 16000 for 8 month and b remains in the business for 4 months . out of total profit , b claims 2 / 7 of the profit . how much money was contributed by b ? | we are looking for two numbers . # 1 - x # 2 - 3 x Γ’ β¬ β 2 the sum is 82 . # 1 + # 2 = 82 substituting x + 3 x Γ’ β¬ β 2 = 82 4 x Γ’ β¬ β 2 = 82 4 x = 84 x = 21 the first number is 21 , the second number is two less than three times 21 or 61 . correct answer e | a ) 7 - 19 , b ) 8 - 20 , c ) 10 - 16 , d ) 15 - 9 , e ) 21 - 61 | e | subtract(divide(add(82, const_2), add(3, const_1)), subtract(multiply(3, divide(add(82, const_2), add(3, const_1))), const_2)) | add(n1,const_2)|add(n0,const_1)|divide(#0,#1)|multiply(n0,#2)|subtract(#3,const_2)|subtract(#2,#4) | general |
the second of two numbers is two less than 3 times the first . find the numbers if there sum is 82 . | "let t be the time that john spent for his final push . thus , per the question , 4.2 t = 3.7 t + 10 + 2 - - - > 0.5 t = 12 - - - > t = 24 seconds . c is the correct answer ." | a ) 13 seconds , b ) 17 seconds , c ) 24 seconds , d ) 34 seconds , e ) 51 seconds | c | divide(add(divide(multiply(3.7, add(10, 2)), subtract(4.2, 3.7)), add(10, 2)), 4.2) | add(n0,n3)|subtract(n1,n2)|multiply(n2,#0)|divide(#2,#1)|add(#0,#3)|divide(#4,n1)| | physics |
john and steve are speed walkers in a race . john is 10 meters behind steve when he begins his final push . john blazes to the finish at a pace of 4.2 m / s , while steve maintains a blistering 3.7 m / s speed . if john finishes the race 2 meters ahead of steve , how long was john β s final push ? | "solution 32.5 let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 171 . β΄ x 16 x 9 / 12 x 1 / 100 } = 171 or x = 1425 . β΄ p . w . = rs . 1425 . answer c" | a ) rs . 1386 , b ) rs . 1764 , c ) rs . 1425 , d ) rs . 2268 , e ) none of these | c | add(divide(171, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 171) | multiply(const_3,const_4)|divide(n0,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n2,#3)|add(n2,#4)| | gain |
the true discount on a bill due 9 months hence at 16 % per annum is rs . 171 . the amount of the bill is | "let x be pair of shoes and y be pair of boots . 22 x + 16 y = 540 . . . eq 1 8 x + 32 y = 720 . . . . eq 2 . now multiply eq 1 by 2 and sub eq 2 . 44 x = 1080 8 x = 720 . 36 x = 360 = > x = 10 . sub x in eq 2 . . . . we get 80 + 32 y = 720 . . . then we get 32 y = 640 then y = 20 differenece between x and y is 10 answer : c" | a ) 12 , b ) 11 , c ) 10 , d ) 15 , e ) 16 | c | divide(subtract(540, multiply(22, divide(subtract(multiply(540, const_2), 720), subtract(multiply(22, const_2), 8)))), 16) | multiply(n2,const_2)|multiply(n0,const_2)|subtract(#0,n5)|subtract(#1,n3)|divide(#2,#3)|multiply(n0,#4)|subtract(n2,#5)|divide(#6,n1)| | general |
suzie β s discount footwear sells all pairs of shoes for one price and all pairs of boots for another price . on monday the store sold 22 pairs of shoes and 16 pairs of boots for $ 540 . on tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $ 720 . how much more do pairs of boots cost than pairs of shoes at suzie β s discount footwear ? | f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 3 * 180 / 12 = 45 ounces of water answer : a | ['a ) 45', 'b ) 50', 'c ) 55', 'd ) 60', 'e ) 63'] | a | multiply(multiply(30, const_2), divide(3, const_4)) | divide(n3,const_4)|multiply(n2,const_2)|multiply(#0,#1) | other |
in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the ` ` sport ' ' formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the ` ` sport ' ' formulation contains 3 ounces of corn syrup , how many ounces of water does it contain ? | "as two numbers are prime , only options satisfy ie option b and c and d , e but option a will not make the product of numbers i . e 35 answer : b" | a ) 8 and 4 , b ) 7 and 5 , c ) 6 and 5 , d ) 8 and 5 , e ) 6 and 5 | b | add(35, 12) | add(n0,n1)| | physics |
sum of two numbers prime to each other is 12 and their l . c . m . is 35 . what are the numbers ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 440 but x = 120 % of y = 120 y / 100 = 12 y / 10 Γ’ Λ Β΄ 12 y / 10 + y = 440 Γ’ β‘ β y [ 12 / 10 + 1 ] = 440 Γ’ β‘ β 22 y / 10 = 440 Γ’ β‘ β 22 y = 4400 Γ’ β‘ β y = 4400 / 22 = 400 / 2 = rs . 200 b )" | a ) s . 150 , b ) s . 200 , c ) s . 250 , d ) s . 350 , e ) s . 400 | b | divide(multiply(440, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)| | general |
two employees x and y are paid a total of rs . 440 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "equate the fat : 0.1 x + 0.25 * 8 = 0.2 ( x + 8 ) - - > x = 4 . answer : a ." | a ) 4 , b ) 12 , c ) 14 , d ) 16 , e ) 28 | a | divide(multiply(subtract(25, 20), 8), 10) | subtract(n2,n3)|multiply(n1,#0)|divide(#1,n0)| | general |
how many gallons of milk that is 10 percent butter - fat must be added to 8 gallons of milk that is 25 percent butterfat to obtain milk that is 20 percent butterfat ? | "820 * 8.5 6970.0 gm 6.97 kg answer : d" | a ) 6.6 kg , b ) 6.8 kg , c ) 6.7 kg , d ) 6.97 kg , e ) 7.8 kg | d | divide(multiply(8.5, 820), const_1000) | multiply(n0,n1)|divide(#0,const_1000)| | general |
a envelop weight 8.5 gm , if 820 of these envelop are sent with an advertisement mail . how much wieght ? | "l * w = 500 : area , l is the length and w is the width . 2 l + 2 w = 90 : perimeter l = 45 - w : solve for l ( 45 - w ) * w = 500 : substitute in the area equation w = 20 and l = 25 correct answer d" | a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | d | divide(subtract(divide(90, const_2), sqrt(subtract(multiply(divide(90, const_2), divide(90, const_2)), multiply(const_4, 500)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|multiply(#0,#0)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry |
the area of a rectangular field is equal to 500 square meters . its perimeter is equal to 90 meters . find the width of this rectangle . | angle between hands of a clock when the minute hand is behind the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( h β m / 5 ) + m / 2 degree when the minute hand is ahead of the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( m / 5 β h ) β m / 2 degree here h = 11 , m = 30 and minute hand is behind the hour hand . hence the angle = 30 ( h β m / 5 ) + m / 2 = 30 ( 11 β 30 / 5 ) + 30 / 2 = 30 ( 11 β 6 ) + 15 = 30 Γ 5 + 15 = 165 Β° answer is d . | a ) 35 Β° , b ) 65 Β° , c ) 45 Β° , d ) 165 Β° , e ) 95 Β° | d | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | add(const_1,const_4)|divide(const_60,const_2)|multiply(const_3,const_4)|subtract(#2,const_1)|divide(#2,#3)|multiply(#0,#3)|multiply(#4,#5)|subtract(#6,#1)|multiply(#7,#3)|divide(#8,const_2) | physics |
the angle between the minute hand and the hour hand of a clock when the time is 11.30 , is | 16 + 6 = 22 . answer is c . | a ) 7 , b ) 33 , c ) 22 , d ) 17 , e ) 25 | c | add(16, 6) | add(n0,n1)| | general |
there are 16 bees in the hive , then 6 more fly . how many bees are there in all ? | "explanation : 108 kmph = ( 108 x 5 / 18 ) m / sec = 30 m / s . answer : c" | a ) 10.8 , b ) 18 , c ) 30 , d ) 38.8 , e ) none of these | c | multiply(108, const_0_2778) | multiply(n0,const_0_2778)| | physics |
a train moves with a speed of 108 kmph . its speed in metres per second is : | "( 60 / 100 ) * x Γ’ β¬ β 20 = 88 6 x = 1080 x = 180 answer : c" | a ) 120 , b ) 300 , c ) 180 , d ) 170 , e ) 148 | c | divide(add(20, 88), divide(60, const_100)) | add(n0,n2)|divide(n1,const_100)|divide(#0,#1)| | gain |
20 is subtracted from 60 % of a number , the result is 88 . find the number ? | "( 50 / 100 ) * 250 Γ’ β¬ β ( 25 / 100 ) * 400 125 - 100 = 25 answer : b" | a ) 25 , b ) 26 , c ) 29 , d ) 39 , e ) 26 | b | subtract(multiply(250, divide(50, const_100)), multiply(divide(25, const_100), 400)) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n3,#1)|subtract(#2,#3)| | gain |
by how much is 50 % of 250 greater than 25 % of 400 . | solution required ratio = 60 : 90 = 2 : 3 answer b | a ) 1 : 3 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) none of these | b | divide(subtract(6.3, 5.7), subtract(7.2, 6.3)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1) | other |
find the ratio in which rice at rs . 7.20 a kg be mixed with rice at rs . 5.70 a kg to produce a mixture worth rs . 6.30 a kg ? | the slope of the line m is rise / run = 22 - ( - 18 ) / 15 - ( - 15 ) = 4 / 3 4 / 3 = 22 - 2 / 15 - x 60 - 4 x = 66 - 6 x = 0 the answer is c . | a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | c | add(1522, 18) | add(n1,n2) | general |
in a rectangular coordinate system , if a line passes through the points ( - 15 , - 18 ) , ( 1522 ) and ( x , 2 ) then what is the value of x ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 108 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 39.6 + 42 = 81.6 cm answer : a" | a ) 81.6 cm , b ) 85.9 cm , c ) 90 cm , d ) 92 cm , e ) 95 cm | a | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
the sector of a circle has radius of 21 cm and central angle 108 o . find its perimeter ? | "let the total solution is 150 l with 80 l water 70 l syrup . to make 35 % syrup solution , the result solution must have 97.5 l syrup and 52.5 l syrup . therefore we are taking 17.5 l of syrup from initial solution and replacing with water . using urinary method : 70 l syrup in 150 l solution 17.5 l syrup in 37.5 l solution we started by multiplying 10 now to get to the result we need to divide by 17.5 = > amount of solution to be replaced with water ( 37.5 / 17.5 ) = 2.14 . correct option : c" | a ) 1.5 , b ) 1.75 , c ) 2.14 , d ) 2.34 , e ) 2.64 | c | multiply(divide(subtract(divide(7, add(8, 7)), divide(const_2, add(const_2, const_3))), divide(7, add(8, 7))), add(8, 7)) | add(n0,n1)|add(const_2,const_3)|divide(n1,#0)|divide(const_2,#1)|subtract(#2,#3)|divide(#4,#2)|multiply(#0,#5)| | gain |
a solution contains 8 parts of water for every 7 parts of lemonade syrup . how many parts of the solution should be removed and replaced with water so that the solution will now contain 35 % lemonade syrup ? | let ' s take the approach that uses the answer choices to eliminate wasted time . 500 / 100 = 5 coffee bar per minute per machine . 20 machines = 100 per minute . 2 minutes worth = 200 coffe bar . looking at the answers it is clear . . . we can only choose ( d ) the correct answer is d . | a ) 110 , b ) 220 , c ) 330 , d ) 200 , e ) 789 | d | multiply(multiply(divide(500, 100), 2), 20) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1) | gain |
running at the same constant rate , 100 identical machines can produce a total of 500 coffee bar per minute . at this rate , how many bottles could 20 such machines produce in 2 minutes ? | "x / ( 1 / 36 ) = ( 4 / 5 ) / ( 2 / 9 ) x = 4 * 9 * 1 / 36 * 5 * 2 = 1 / 10 the answer is b ." | a ) 1 / 5 , b ) 1 / 10 , c ) 1 / 15 , d ) 1 / 20 , e ) 1 / 25 | b | divide(1, 36) | divide(n0,n1)| | other |
a certain fraction has the same ratio to 1 / 36 , as 4 / 5 does to 2 / 9 . what is this certain fraction ? | "( 5 ! * 5 ! + 6 ! * 5 ! ) / 3 = 5 ! ( 5 ! + 6 ! ) / 3 = 120 ( 120 + 720 ) / 3 = ( 120 * 840 ) / 3 = 120 * 280 units digit of the above product will be equal to 0 answer e" | a ) 4 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | e | divide(add(multiply(factorial(5), factorial(5)), multiply(factorial(5), factorial(5))), 5) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
what is the units digit of ( 5 ! * 5 ! + 6 ! * 5 ! ) / 3 ? | "- x 2 - ( k + 8 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 8 ) x - 8 = - x 2 + 6 x - 8 - ( k + 8 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 14 : solve the above for k correct answer c" | a ) 11 , b ) 12 , c ) 14 , d ) 19 , e ) 15 | c | add(8, add(4, 2)) | add(n0,n4)|add(n1,#0)| | general |
find the constant k so that : - x 2 - ( k + 8 ) x - 8 = - ( x - 2 ) ( x - 4 ) | "explanation : 6000 : 4000 : 10000 3 : 2 : 5 3 / 10 * 11000 = 3300 answer : a" | a ) 3300 , b ) 1100 , c ) 2667 , d ) 600 , e ) 4000 | a | multiply(divide(6000, add(add(6000, 4000), 10000)), 11000) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)| | gain |
a , b and c invested rs . 6000 , rs . 4000 and rs . 10000 respectively , in a partnership business . find the share of a in profit of rs . 11000 after a year ? | "both cooking and weaving = 20 - ( 5 + 8 + 5 ) = 2 so , the correct answer is b ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | subtract(subtract(subtract(20, 8), 5), 5) | subtract(n1,n4)|subtract(#0,n5)|subtract(#1,n3)| | other |
a group of people participate in some curriculum , 30 of them practice yoga , 20 study cooking , 15 study weaving , 5 of them study cooking only , 8 of them study both the cooking and yoga , 5 of them participate all curriculums . how many people study both cooking and weaving ? | "75 = ( 750 * 5 * r ) / 100 r = 2 % . answer : a" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | multiply(divide(divide(subtract(825, 750), 750), 5), const_100) | subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)| | gain |
at what rate percent on simple interest will rs . 750 amount to rs . 825 in 5 years ? | "690 = 2 * 3 * 5 * 23 so the least value forncan be 23 . e" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 23 | e | divide(divide(divide(divide(690, const_2), const_3), const_4), divide(const_10, const_2)) | divide(n1,const_2)|divide(const_10,const_2)|divide(#0,const_3)|divide(#2,const_4)|divide(#3,#1)| | general |
if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 690 , what is the least possible value of n ? | lets do it by picking up numbers . let arthur carry 2 crumbs per trip , this means amy carries 3 crumbs per trip . also let arthur make 2 trips and so amy makes 4 trips . thus total crumbs carried by arthur ( b ) = 2 x 2 = 4 , total crumbs carried by amy = 3 x 4 = 12 . 12 is 3 times 4 , so e | a ) b / 2 , b ) b , c ) 3 b / 2 , d ) 2 b , e ) 3 b | e | multiply(const_2, add(const_1, divide(const_1, const_2))) | divide(const_1,const_2)|add(#0,const_1)|multiply(#1,const_2) | general |
two ants , arthur and amy , have discovered a picnic and are bringing crumbs back to the anthill . amy makes twice as many trips and carries one and a half times as many crumbs per trip as arthur . if arthur carries a total of b crumbs to the anthill , how many crumbs will amy bring to the anthill , in terms of b ? | detailed solution let the total students be ( n + 1 ) let total badges be x let the average of β n β students be y the student who got 1 / 6 th of x = 5 y or y = x / 30 therefore β n β students got 1 / 30 th of total share each or n * x / 30 + 1 * x / 6 = x nx + 5 x = 30 x n + 5 = 30 or n = 25 total = n + 1 = 26 correct answer : b | a ) 30 , b ) 26 , c ) 11 , d ) 31 , e ) 32 | b | add(subtract(multiply(6, 5), 5), 1) | multiply(n1,n2)|subtract(#0,n2)|add(n0,#1) | general |
a certain number of badges were distributed among a class of students . the student who got 1 / 6 th of the total number of badges actually got 5 times the average number of badges the others got ! how many students were there in the class ? | "explanation : 23 23 + 2 ^ 2 = 27 23 + 3 ^ 2 = 36 23 + 4 ^ 2 = 52 23 + 5 ^ 2 = 77 answer : c" | a ) 40 , b ) 24 , c ) 77 , d ) 36 , e ) 12 | c | subtract(negate(52), multiply(subtract(27, 36), divide(subtract(27, 36), subtract(23, 27)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
23 , 27 , 36 , 52 , . . . | # of people times the # of hours : 4 * 7 = 28 - - > 4 lawyers do 28 worksin 7 hours . 3 * 14 / 3 = 14 - - > 3 assistants do 14 worksin 4 hours so , since the amount of work the assistants do is half the work the lawyers do , the time will be double , soans a | a ) 14 , b ) 10 , c ) 9 , d ) 6 , e ) 5 | a | multiply(multiply(divide(const_2, const_3), 3), 7) | divide(const_2,const_3)|multiply(n2,#0)|multiply(n1,#1) | general |
tough and tricky questions : work / rate problems . a group of 4 junior lawyers require 7 hours to complete a legal research assignment . how many hours would it take a group of 3 legal assistants to complete the same research assignment assuming that a legal assistant works at two - thirds the rate of a junior lawyer ? source : chili hot gmat | "work done by mahesh in 30 days = 20 * 1 / 30 = 2 / 3 remaining work = 1 - 2 / 3 = 1 / 3 1 / 3 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 3 = 90 days answer is a" | a ) 90 , b ) 25 , c ) 37 , d ) 41 , e ) 30 | a | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 30))), 30)) | divide(const_1,n0)|multiply(n1,#0)|subtract(const_1,#1)|divide(#2,n2)|divide(const_1,#3)| | physics |
mahesh can do a piece of work in 30 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | from ( x + 6 ) ( y - 6 ) = 0 it follows that either x = - 6 or y = 6 . thus either x ^ 2 = 36 or y ^ 2 = 36 . now , if x ^ 2 = 36 , then the least value of y ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 36 + 0 = 36 . similarly if y ^ 2 = 36 , then the least value of x ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 0 + 36 = 36 . answer : d . | a ) 0 , b ) 16 , c ) 25 , d ) 36 , e ) 49 | d | power(6, 2) | power(n0,n3) | general |
if x and y are numbers such that ( x + 6 ) ( y - 6 ) = 0 , what is the smallest possible value of x ^ 2 + y ^ 2 | let x be the larger number and y be the smaller number . x - y = 6 x + 2 ( y ) = 15 solve by substitution : y = x - 6 x + 2 ( x - 6 ) = 15 x + 2 x - 12 = 15 3 x = 27 x = 9 the larger number is 9 , so answer c is correct . | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | divide(add(15, multiply(const_2, 6)), add(const_1, const_2)) | add(const_1,const_2)|multiply(n0,const_2)|add(n1,#1)|divide(#2,#0) | general |
the difference of a larger number and a smaller number is 6 . the sum of the larger number and twice the smaller is 15 . what is the larger number ? | "explanation : length of the platform = speed of train * extra time taken to cross the platform . length of platform = 72 kmph * 12 seconds convert 72 kmph into m / sec 1 kmph = 518518 m / s ( this can be easily derived . but if you can remember this conversion , it saves a good 30 seconds ) . β΄ 72 kmph = 518 β 72518 β 72 = 20 m / sec therefore , length of the platform = 20 m / s * 12 sec = 240 meters . correct answer : a" | a ) 240 meters , b ) 360 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined | a | subtract(multiply(divide(multiply(72, const_1000), const_3600), 30), multiply(divide(multiply(72, const_1000), const_3600), 18)) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)| | physics |
a train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds . what is the length of the platform in meters ? | "16 % of 2 litres = 0.32 litres 40 % of 6 litres = 2.4 litres therefore , total quantity of alcohol is 2.72 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 27.2 % answer : d" | a ) 31 % . , b ) 71 % . , c ) 49 % . , d ) 27.2 % . , e ) 51 % . | d | multiply(divide(add(multiply(divide(16, const_100), 2), multiply(divide(40, const_100), 6)), 10), const_100) | divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100)| | general |
a vessel of capacity 2 litre has 16 % of alcohol and another vessel of capacity 6 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | 1 : 4 answer : b | ['a ) 1 : 0', 'b ) 1 : 4', 'c ) 1 : 6', 'd ) 1 : 2', 'e ) 1 : 1'] | b | divide(1, const_4) | divide(n0,const_4) | geometry |
the diameters of two spheres are in the ratio 1 : 2 what is the ratio of their surface area ? | explanation : 3 prize among 10 students can be distributed in 10 c 3 ways = 120 ways . answer : d | a ) 10 , b ) 45 , c ) 95 , d ) 120 , e ) none of these | d | add(multiply(10, 3), multiply(subtract(10, const_1), 10)) | multiply(n0,n1)|subtract(n1,const_1)|multiply(n1,#1)|add(#0,#2) | general |
there are 3 prizes to be distributed among 10 students . if no students gets more than one prize , then this can be done in ? | "72 + 82 + x = 3 * 75 x = 71 the answer is b ." | a ) 69 % , b ) 71 % , c ) 73 % , d ) 75 % , e ) 77 % | b | subtract(multiply(const_3, 75), add(72, 82)) | add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0)| | general |
a student got 72 % in math and 82 % in history . to get an overall average of 75 % , how much should the student get in the third subject ? | let b ' s capital be rs . x . { 3500 \ 12 } / { 7 x } = { 2 } / { 3 } = > x = 9000 . answer : d | a ) rs . 9228 , b ) rs . 9129 , c ) rs . 9120 , d ) rs . 9000 , e ) rs . 1922 | d | divide(multiply(multiply(3500, const_12), 3), multiply(subtract(const_12, 5), 2)) | multiply(n0,const_12)|subtract(const_12,n1)|multiply(n3,#0)|multiply(n2,#1)|divide(#2,#3) | other |
a starts business with rs . 3500 and after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b ' s contribution in the capital ? | "c . p . of 1 litre of milk = ( 20 Γ 2 β 3 ) = 40 β 3 β΄ ratio of water and milk = 8 β 3 : 32 β 3 = 8 : 32 = 1 : 4 β΄ quantity of water to be added to 60 litres of milk = ( 1 β 4 Γ 60 ) litres = 15 litres . answer c" | a ) 10 litres , b ) 12 litres , c ) 15 litres , d ) 18 litres , e ) none of these | c | divide(20, add(1, divide(1, 2))) | divide(n1,n3)|add(n1,#0)|divide(n4,#1)| | general |
how much water must be added to 60 litres of milk at 1 1 β 2 litres for 20 so as to have a mixture worth 10 2 β 3 a litre ? | "the number of silver cars is 0.15 * 40 + 0.45 * 80 = 42 the percentage of cars which are silver is 42 / 120 = 35 % the answer is b ." | a ) 30 % , b ) 35 % , c ) 40 % , d ) 45 % , e ) 50 % | b | multiply(divide(add(multiply(40, divide(15, const_100)), multiply(80, divide(45, const_100))), add(40, 80)), const_100) | add(n0,n2)|divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
a car dealership has 40 cars on the lot , 15 % of which are silver . if the dealership receives a new shipment of 80 cars , 45 % of which are not silver , what percentage of total number of cars are silver ? | "factors of 24 - 1 , 2 , 3 , 4 , 6 , 8 , 12,24 factors of 27 - 1 , 3 , 9,27 comparing both , we have 6 factors of 24 which are not factors of 27 - 2,4 , 6,8 , 12,24 answer : b" | a ) 2 , b ) 6 , c ) 4 , d ) 1 , e ) 5 | b | divide(27, 24) | divide(n1,n0)| | other |
how many of the positive factors of 24 are not factors of 27 | "speed on return trip = 120 % of 60 = 72 km / hr . average speed of trip = 60 + 72 / 2 = 132 / 2 = 66 km / hr answer : d" | a ) 33 , b ) 77 , c ) 48 , d ) 66 , e ) 21 | d | divide(add(multiply(60, add(const_1, divide(20, const_100))), 60), const_2) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|add(n0,#2)|divide(#3,const_2)| | general |
a person travels from p to q a speed of 60 km / hr and returns by increasing his speed by 20 % . what is his average speed for both the trips ? | "a + b 5 days work = 5 * 1 / 10 = 1 / 2 remaining work = 1 - 1 / 2 = 1 / 2 1 / 2 work is done by a in 5 days whole work will be done by a in 5 * 2 = 10 days answer is a" | a ) 10 , b ) 15 , c ) 20 , d ) 5 , e ) 30 | a | divide(multiply(5, 10), subtract(10, 5)) | multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | physics |
a and b can together finish a work in 10 days . they worked together for 5 days and then b left . after another 5 days , a finished the remaining work . in how many days a alone can finish the job ? | "area = l * w = ( l ) * ( l - 10 ) = 200 trial and error : 19 * 9 = 171 ( too low ) 20 * 10 = 200 the length is 20 meters . the answer is a ." | a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | a | add(10, add(const_0_25, add(const_0_33, divide(divide(200, 10), const_2)))) | divide(n1,n0)|divide(#0,const_2)|add(#1,const_0_33)|add(#2,const_0_25)|add(n0,#3)| | geometry |
a rectangular field has a length 10 meters more than it is width . if the area of the field is 200 , what is the length ( in meters ) of the rectangular field ? | "drawing two balls of same color from seven green balls can be done in β· c β ways . similarly from eight white balls two can be drawn in ways . 7 / 15 answer : e" | a ) 7 / 18 , b ) 7 / 19 , c ) 7 / 11 , d ) 7 / 12 , e ) 7 / 15 | e | add(multiply(divide(8, add(7, 8)), divide(subtract(8, const_1), subtract(add(7, 8), const_1))), multiply(divide(7, add(7, 8)), divide(subtract(7, const_1), subtract(add(7, 8), const_1)))) | add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#4,#7)|add(#8,#9)| | other |
a bag contains 7 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is - . | "explanation : ( men 4 : 9 ) : ( hrs / day 10 : 6 ) : : 1600 : x hence 4 * 10 * x = 9 * 6 * 1600 or x = 9 * 6 * 1600 / 4 * 10 = 2160 answer : d" | a ) rs 840 , b ) rs 1320 , c ) rs 1620 , d ) rs 2160 , e ) none of these | d | multiply(divide(multiply(9, 6), multiply(4, 10)), 1600) | multiply(n3,n4)|multiply(n0,n1)|divide(#0,#1)|multiply(n2,#2)| | physics |
if 4 men working 10 hours a day earn rs . 1600 per week , then 9 men working 6 hours a day will earn how much per week ? | "let the number be x . ( 5 / 4 ) * x = ( 4 / 5 ) * x + 9 25 x = 16 x + 180 9 x = 180 x = 20 the answer is c ." | a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | c | divide(divide(multiply(multiply(9, divide(4, 5)), divide(4, 5)), subtract(const_1, multiply(divide(4, 5), divide(4, 5)))), divide(4, 5)) | divide(n0,n1)|multiply(n4,#0)|multiply(#0,#0)|multiply(#0,#1)|subtract(const_1,#2)|divide(#3,#4)|divide(#5,#0)| | general |
a student was asked to find 4 / 5 of a number . but the student divided the number by 4 / 5 , thus the student got 9 more than the correct answer . find the number . | cp per kg of mixture = [ 24 ( 150 ) + 36 ( 125 ) ] / ( 24 + 36 ) = rs . 135 sp = cp [ ( 100 + profit % ) / 100 ] = 135 * [ ( 100 + 40 ) / 100 ] = rs . 189 . answer : c | a ) rs . 135 , b ) rs . 162 , c ) rs . 189 , d ) rs . 198 , e ) none of these | c | add(divide(add(multiply(24, 150), multiply(36, 125)), add(36, 24)), multiply(divide(add(multiply(24, 150), multiply(36, 125)), add(36, 24)), divide(40, const_100))) | add(n0,n2)|divide(n4,const_100)|multiply(n0,n1)|multiply(n2,n3)|add(#2,#3)|divide(#4,#0)|multiply(#5,#1)|add(#5,#6) | gain |
raman mixed 24 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | "j + k = 284 and so k = 284 - j j - 8 = 2 k j - 8 = 2 ( 284 - j ) 3 j = 576 j = 192 the answer is e ." | a ) 176 , b ) 180 , c ) 184 , d ) 188 , e ) 192 | e | add(multiply(divide(subtract(284, 8), const_3), const_2), 8) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)| | general |
if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 284 pounds . what is jake β s present weight , in pounds ? | "they together completed 4 / 10 work in 4 days . balance 6 / 10 work will be completed by arun alone in 40 * 6 / 10 = 24 days . answer : e" | a ) 16 days . , b ) 17 days . , c ) 18 days . , d ) 19 days . , e ) 24 days . | e | subtract(40, multiply(divide(40, 10), 4)) | divide(n2,n0)|multiply(n1,#0)|subtract(n2,#1)| | physics |
arun and tarun can do a work in 10 days . after 4 days tarun went to his village . how many days are required to complete the remaining work by arun alone . arun can do the work alone in 40 days . | "50 * 23 = 1150 1150 - 1145 = 5 answer : d" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 8 | d | subtract(multiply(add(multiply(const_4, const_10), const_2), 23), 1145) | multiply(const_10,const_4)|add(#0,const_2)|multiply(n1,#1)|subtract(#2,n0)| | general |
what least number must be added to 1145 , so that the sum is completely divisible by 23 ? | percentage change in area = ( β 30 β 20 + ( 30 Γ 20 ) / 100 ) % = β 44 % i . e . , area is decreased by 44 % answer : c | a ) 24 % , b ) 30 % , c ) 44 % , d ) 54 % , e ) 64 % | c | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 30), subtract(const_100, 20))), const_100) | multiply(const_100,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(#1,#2)|subtract(#0,#3)|divide(#4,const_100) | gain |
a towel , when bleached , lost 30 % of its length and 20 % of its breadth . what is the percentage decrease in area ? | "solution let the son ' s present age be x years . then , man ' s present age = ( x + 30 ) years . then Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ ( x + 30 ) + 2 = 2 ( x + 2 ) Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ x + 32 = 2 x + 4 x = 28 . answer b" | a ) 14 years , b ) 28 years , c ) 20 years , d ) 22 years , e ) none | b | divide(subtract(30, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
a man is 30 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is | "let x be the discount on pony jeans . then 0.22 - x is the discount on fox jeans . 3 ( 0.22 - x ) ( 15 ) + 2 x ( 18 ) = 8.73 9.9 - 45 x + 36 x = 8.73 9 x = 1.17 x = 0.13 the answer is e ." | a ) 9 % , b ) 10 % , c ) 11 % , d ) 12 % , e ) 13 % | e | multiply(subtract(divide(22, const_100), divide(subtract(8.73, multiply(divide(22, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100) | divide(n6,const_100)|multiply(n1,n5)|multiply(n0,n4)|multiply(#0,#1)|subtract(#2,#1)|subtract(n2,#3)|divide(#5,#4)|subtract(#0,#6)|multiply(#7,const_100)| | gain |
fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 8.73 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discount rates is 22 percent , what is the discount rate on pony jeans ? | "original price = 100 cp = 80 s = 80 * ( 180 / 100 ) = 112 100 - 144 = 44 % answer : e" | a ) 18 % , b ) 13 % , c ) 12 % , d ) 32 % , e ) 44 % | e | multiply(subtract(divide(divide(multiply(subtract(const_100, 20), add(const_100, 80)), const_100), const_100), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
a trader bought a car at 20 % discount on its original price . he sold it at a 80 % increase on the price he bought it . what percent of profit did he make on the original price ? | "800 * 15 = 1000 * x x = 12 answer : e" | a ) 11.5 , b ) 12.5 , c ) 10.5 , d ) 11 , e ) 12 | e | divide(multiply(15, 800), add(800, 200)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
800 men have provisions for 15 days . if 200 more men join them , for how many days will the provisions last now ? | as you see 5 ! till 10 ! each unit digit is zero . so 1 ! + 2 ! + 3 ! + 4 ! = 33 so unit digit 3 + 0 = 3 n = 3 n ^ n = 3 ^ 3 = 27 so last digit is 7 . answer : e | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | power(subtract(10, 3), const_1) | subtract(n3,n2)|power(#0,const_1) | general |
n = 1 ! + 2 ! + 3 ! . . . . . + 10 ! . what is the last digit of n ^ n ? | i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 250 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 250 ) / 100 + ( 1.8 * 700 ) / 100 = 14.6 gms % of liquid x in resultant mixture = ( 14.6 / 1000 ) * 100 = 1.46 % a | a ) 1.46 % , b ) 1.93 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(250, 0.8), multiply(700, 1.8)), const_1000) | multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,const_1000) | gain |
by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 250 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "old set s - total is avg * no of elements = 6.2 * 10 = 62 if one number is increased by 3 then total increased to 62 + 3 = 65 new avg - 65 / 10 = 6.5 . hence answer is a ." | a ) 6.5 , b ) 6.7 , c ) 6.8 , d ) 6.85 , e ) 6.9 | a | divide(add(multiply(10, 6.2), 3), 10) | multiply(n0,n1)|add(n2,#0)|divide(#1,n0)| | general |
set s contains exactly 10 numbers and has an average ( arithmetic mean ) of 6.2 . if one of the numbers in set s is increased by 3 , while all other numbers remain the same , what is the new average of set s ? | "the amount a gets for managing = 20 % of rs . 11000 = 2200 remaining profit = 11000 β 2200 = 8800 this is to be divided in the ratio 2 : 9 . share of b = 9 / 11 of 8800 = 7200 answer b" | a ) 3500 , b ) 7200 , c ) 6800 , d ) 4800 , e ) none of these | b | add(divide(multiply(subtract(11000, divide(multiply(20, 11000), const_100)), add(const_2, const_3)), add(add(const_2, const_3), add(const_2, const_4))), divide(multiply(20, 11000), const_100)) | add(const_2,const_3)|add(const_2,const_4)|multiply(n2,n3)|add(#0,#1)|divide(#2,const_100)|subtract(n3,#4)|multiply(#0,#5)|divide(#6,#3)|add(#7,#4)| | gain |
a is a working partner and b is a sleeping partner in a business . a puts in 20,000 and b 90,000 . a gets 20 % of the profit for managing the business , and the rest is divided in proportion to their capitals . find the share of b in profit of 11000 . | "both cooking and weaving = 25 - ( 6 + 8 + 7 ) = 4 so , the correct answer is d ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | subtract(subtract(subtract(25, 8), 7), 6) | subtract(n1,n4)|subtract(#0,n5)|subtract(#1,n3)| | other |
a group of people participate in some curriculum , 30 of them practice yoga , 25 study cooking , 15 study weaving , 6 of them study cooking only , 8 of them study both the cooking and yoga , 7 of them participate all curriculums . how many people study both cooking and weaving ? | "a quick note on doubling . when you double a length you have 2 * l 1 . when you double all lengths of a rectangle you have ( 2 * l 1 ) ( 2 * l 2 ) = a . an increase of 2 ^ 2 or 4 . when you double all lengths of a rectangular prism you have ( 2 * l 1 ) ( 2 * l 2 ) ( 2 * l 3 ) = v . an increase of 2 ^ 3 or 8 . this leads to the basic relationship : line : 2 * original size rectangle : 4 * original size rectangular prism : 8 * original size answer is d" | a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) 100 | d | multiply(power(const_2, const_3), 10) | power(const_2,const_3)|multiply(n0,#0)| | geometry |
a carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet . if the carpenter were to make a similar sandbox twice as long , twice as wide , and twice as high as the first sandbox , what would be the capacity , in cubic feet , of the second sandbox ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 25 + 21 = 46 t = 12 d = 46 * 12 = 552 answer : c" | a ) 457 km , b ) 444 km , c ) 552 km , d ) 645 km , e ) 453 km | c | add(multiply(divide(60, subtract(21, 25)), 25), multiply(divide(60, subtract(21, 25)), 21)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 25 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "8 ( 8 Γ 3 ) Γ· 2 = 12 ( 12 Γ 3 ) Γ· 2 = 18 ( 18 Γ 3 ) Γ· 2 = 27 ( 27 Γ 3 ) Γ· 2 = 40.5 ( 40.5 Γ 3 ) Γ· 2 = 60.75 answer is b ." | a ) 62 , b ) 60.75 , c ) 60.5 , d ) 60.25 , e ) 60 | b | subtract(negate(27), multiply(subtract(12, 18), divide(subtract(12, 18), subtract(8, 12)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
8 , 12 , 18 , 27 , 40.5 , ( . . . ) | "let total capacity be x we know 160 = 0.80 x x = 160 / 0.80 = 200 prior to storm , we had 130 bn gallons 200 - 130 = 70 answer : b" | a ) 75 , b ) 70 , c ) 65 , d ) 85 , e ) 90 | b | divide(divide(multiply(160, const_100), 80), const_2) | multiply(n1,const_100)|divide(#0,n2)|divide(#1,const_2)| | general |
a rainstorm increased the amount of water stored in state j reservoirs from 130 billion gallons to 160 billion gallons . if the storm increased the amount of water in the reservoirs to 80 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the storm ? | "explanation : a : b = 2 : 5 = 2 : 5 = > a : c = 2 : 1 = 2 : 1 = > a : b : c = 6 : 5 : 1 b share = ( 5 / 12 ) * 15000 = 6250 option e" | a ) 1950 , b ) 6895 , c ) 4879 , d ) 8126 , e ) 6250 | e | multiply(multiply(multiply(multiply(add(5, 2), 5), const_100), const_100), divide(5, add(add(1, 2), 5))) | add(n0,n1)|add(n0,n3)|add(n3,#0)|multiply(#1,n3)|divide(n3,#2)|multiply(#3,const_100)|multiply(#5,const_100)|multiply(#4,#6)| | gain |
in business , a and c invested amounts in the ratio 2 : 1 , whereas the ratio between amounts invested by a and b was 2 : 5 , if rs 15,000 was their profit , how much amount did b receive . | "1 women ' s 1 day work = 1 / 70 1 child ' s 1 day work = 1 / 140 ( 8 women + 12 children ) ' s 1 day work = ( 8 / 70 + 12 / 140 ) = 1 / 5 8 women and 4 children will complete the work in 5 days . b" | a ) 4 , b ) 5 , c ) 7 , d ) 8 , e ) 2 | b | inverse(add(divide(8, multiply(10, 7)), divide(10, multiply(10, 14)))) | multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)| | physics |
10 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 8 women and 12 children take to complete the work ? | "find the pattern of the remainders after each power : ( 3 ^ 1 ) / 7 remainder 3 ( 3 ^ 2 ) / 7 remainder 2 ( 3 ^ 3 ) / 7 remainder 6 ( 3 ^ 4 ) / 7 remainder 4 ( 3 ^ 5 ) / 7 remainder 5 ( 3 ^ 6 ) / 7 remainder 1 - - > this is where the cycle ends ( 3 ^ 7 ) / 7 remainder 3 - - > this is where the cycle begins again ( 3 ^ 8 ) / 7 remainder 2 continuing the pattern to ( 3 ^ 50 ) / 7 gives us a remainder of 2 final answer : c ) 2" | a ) 5 , b ) 3 , c ) 2 , d ) 1 , e ) 7 | c | reminder(power(3, 50), 7) | power(n0,n1)|reminder(#0,n2)| | general |
find the remainder of the division ( 3 ^ 50 ) / 7 . | first , let ' s graph the lines y = 2 and x = 2 at this point , we need to find the points where the line y = 10 - x intersects the other two lines . for the vertical line , we know that x = 2 , so we ' ll plug x = 2 into the equation y = 10 - x to get y = 10 - 2 = 8 perfect , when x = 2 , y = 8 , so one point of intersection is ( 28 ) for the horizontal line , we know that y = 2 , so we ' ll plug y = 2 into the equation y = 10 - x to get 2 = 10 - x . solve to get : x = 8 so , when y = 2 , x = 8 , so one point of intersection is ( 82 ) now add these points to our graph and sketch the line y = 10 - x at this point , we can see that we have the following triangle . the base has length 6 and the height is 6 area = ( 1 / 2 ) ( base ) ( height ) = ( 1 / 2 ) ( 6 ) ( 6 ) = 18 answer : e | a ) a ) 8 , b ) b ) 10 , c ) c ) 12 , d ) d ) 14 , e ) e ) 18 | e | multiply(subtract(subtract(10, 2), 2), multiply(multiply(const_2, const_0_25), subtract(subtract(10, 2), 2))) | multiply(const_0_25,const_2)|subtract(n2,n0)|subtract(#1,n0)|multiply(#0,#2)|multiply(#3,#2) | general |
what is the area inscribed by the lines y = 2 , x = 2 , y = 10 - x on an xy - coordinate plane ? | "from the details given in the problem principle = p = $ 8,000 and r = 9 % or 0.09 expressed as a decimal . as the annual interest is to be calculated , the time period t = 1 . plugging these values in the simple interest formula , i = p x t x r = 8,000 x 1 x 0.09 = 720.00 annual interest to be paid = $ 720 answer : c" | a ) 680 , b ) 700 , c ) 720 , d ) 730 , e ) 750 | c | divide(multiply(multiply(multiply(9, const_100), sqrt(const_100)), 9), const_100) | multiply(n1,const_100)|sqrt(const_100)|multiply(#0,#1)|multiply(n1,#2)|divide(#3,const_100)| | gain |
alex takes a loan of $ 8,000 to buy a used truck at the rate of 9 % simple interest . calculate the annual interest to be paid for the loan amount . | "this one took me bout 3 1 / 2 min . just testin numbers and what not . first notice that n is positive . save time by noticing thati worked out one solution where n = 0 only to find that thats not an option : p . 1 - 7 stands for ^ 1 thru 7 1 : 7 * 1 = 7 2 : 7 * 7 = 9 3 : 7 * 9 = 3 4 : 7 * 3 = 1 5 : 7 * 1 = 7 6 : 7 * 7 = 9 7 : 7 * 9 = 3 pattern repeats every @ 5 . notice every ^ 4 or multiple of 4 is always going to be 1 . this is just for future notice for similar problems . so 7 ^ 4 n + 3 - - - > if n = 1 then its ( ( 7 ^ 7 ) * 6 ) ) / 10 which can say is going to be 3 * 8 - - > 18 / 10 - - > r = 8 now from here if id double check just to make sure . 7 ^ 4 ( 2 ) + 3 * 6 ^ 2 - - - > 7 ^ 11 * 36 or we can just say again 7 ^ 11 * 6 ( b / c we are only interested in the units digit ) . since ^ 12 is going to be 1 that means ^ 11 is going to be 3 ( as taken from our pattern ) so again 3 * 6 = 18 / 10 - - - > r = 8 . c or j in this problem ." | a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | c | reminder(multiply(multiply(const_3, const_3), 6), 10) | multiply(const_3,const_3)|multiply(n3,#0)|reminder(#1,n4)| | general |
if n is a positive integer , what is the remainder when ( 4 ^ ( 4 n + 3 ) ) ( 6 ^ n ) is divided by 10 ? | let initial price be 1000 price in day 1 after 10 % discount = 900 price in day 2 after 20 % discount = 720 price in day 3 after 40 % discount = 432 so , price in day 3 as percentage of the sale price on day 1 will be = 432 / 900 * 100 = > 48 % answer will definitely be ( b ) | a ) 28 % , b ) 48 % , c ) 64.8 % , d ) 70 % , e ) 72 % | b | add(multiply(divide(divide(40, const_100), subtract(1, divide(1, 10))), const_100), 2) | divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4) | gain |
the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 20 percent , and on day 3 , it is discounted an additional 40 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "sol . required probability = pg . ) x p ( b ) = ( 1 β d x ( 1 β i ) = : x 1 = 2 / 5 ans . ( e )" | a ) 1 / 2 , b ) 1 , c ) 2 / 3 , d ) 3 / 4 , e ) 2 / 5 | e | multiply(subtract(1, divide(1, 3)), subtract(1, divide(1, 5))) | divide(n1,n2)|divide(n1,n5)|subtract(n1,#0)|subtract(n1,#1)|multiply(#2,#3)| | general |
the probability that a man will be alive for 10 more yrs is 1 / 3 & the probability that his wife will alive for 10 more yrs is 2 / 5 . the probability that none of them will be alive for 10 more yrs , is | "the diagonal of the rectangle will be the diameter of the circle . and perimeter = 2 * pi * r ans : e" | a ) 2.5 Ο , b ) 5 Ο , c ) 10 Ο , d ) 12.5 Ο , e ) 15 Ο | e | circumface(divide(sqrt(add(power(9, const_2), power(12, 12))), 12)) | power(n0,const_2)|power(n1,n1)|add(#0,#1)|sqrt(#2)|divide(#3,n1)|circumface(#4)| | geometry |
a 9 by 12 rectangle is inscribed in circle . what is the circumference of the circle ? | "no of dogs = 45 long fur = 28 brown = 17 neither long fur nor brown = 8 therefore , either long fur or brown = 45 - 8 = 37 37 = 28 + 17 - both both = 8 answer d" | a ) 26 , b ) 19 , c ) 11 , d ) 8 , e ) 6 | d | subtract(add(28, 17), subtract(45, 8)) | add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)| | other |
each of the dogs in a certain kennel is a single color . each of the dogs in the kennel either has long fur or does not . of the 45 dogs in the kennel , 28 have long fur , 17 are brown , and 8 are neither long - furred nor brown . how many long - furred dogs are brown ? | must be twice on heads and twice on tails 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 answer : c | a ) 1 / 8 , b ) 1 / 4 , c ) 1 / 16 , d ) 1 / 32 , e ) 1 / 2 | c | divide(const_1, power(const_2, 4)) | power(const_2,n0)|divide(const_1,#0) | general |
when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 4 times , what is the probability that it will land twice on heads and twice tails ? | "amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 7 / 100 ) 2 ] = rs . 14.49 effective rate = ( 114.49 - 100 ) = 14.49 % answer : d" | a ) 16.06 % , b ) 16.07 % , c ) 16.08 % , d ) 14.49 % , e ) 16.19 % | d | add(add(divide(14, const_2), divide(14, const_2)), divide(multiply(divide(14, const_2), divide(14, const_2)), const_100)) | divide(n0,const_2)|add(#0,#0)|multiply(#0,#0)|divide(#2,const_100)|add(#1,#3)| | gain |
the effective annual rate of interest corresponding to a nominal rate of 14 % per annum payable half - yearly is ? | "juan ' s income = 100 ( assume ) ; tim ' s income = 50 ( 50 percent less than juan ' s income ) ; mary ' s income = 80 ( 60 percent more than tim ' s income ) . thus , mary ' s income ( 80 ) is 80 % of juan ' s income ( 100 ) . answer : d ." | a ) 124 % , b ) 120 % , c ) 96 % , d ) 80 % , e ) 64 % | d | multiply(multiply(subtract(const_1, divide(50, const_100)), add(const_1, divide(60, const_100))), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)| | general |
mary ' s income is 60 percent more than tim ' s income , and tim ' s income is 50 percent less than juan ' s income . what percent of juan ' s income is mary ' s income ? | a little reflection will show that chalk marks will touch the ground together for the first time after the wheels have passed over a distance which is the lcm of 2 2 / 5 metres and 3 3 / 7 metres . lcm of 12 / 5 metres and 24 / 7 metres = 24 metres . answer is e | a ) 18 metres , b ) 16 metres , c ) 38 metres , d ) 42 metres , e ) 24 metres | e | add(multiply(7, 3), 3) | multiply(n3,n5)|add(n3,#0) | general |
the circumferences of the fore and hind - wheels of a carriage are 2 2 / 5 and 3 3 / 7 meters respectively . a chalk mark is put on the point of contact of each wheel with the ground at any given moment . how far will the carriage have travelled so that their chalk marks may be again on the ground at the same time ? | "explanation : a β s one day work = 1 / 5 b β s one day work = 1 / 4 ( a + b ) β s one day work = 1 / 5 + 1 / 4 = 9 / 20 = > time = 20 / 9 = 2 2 / 9 days answer : option d" | a ) 6 / 11 , b ) 8 / 11 , c ) 7 / 9 , d ) 2 / 9 , e ) 10 / 11 | d | divide(const_1, add(divide(const_1, 5), divide(const_1, 4))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
a can do a piece of work in 5 days and b can do it in 4 days how long will they both work together to complete the work ? | "5 ^ 5 Γ 5 ^ x = ( 125 ) ^ 4 5 ^ ( 5 + x ) = 5 ^ 12 since they have the same base we can just set the exponents equal to each other : ( 5 + x ) = 12 x = 7 ans . e ) 7" | a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 7 | e | divide(subtract(multiply(const_2, 5), 125), const_3) | multiply(n1,const_2)|subtract(#0,n3)|divide(#1,const_3)| | general |
if 5 ^ 5 Γ 5 ^ x = ( 125 ) ^ 4 , then what is the value of x ? | working formula . . . initial concentration * initial volume = final concentration * final volume . let x is the part removed from 100 lts . 30.25 % ( 1 - x / 100 ) ^ 2 = 25 % * 100 % ( 1 - x / 100 ) ^ 2 = 25 / 30.25 - - - - - - > ( 1 - x / 100 ) ^ 2 = ( 5 / 5.5 ) ^ 2 100 - x = 500 / 5.5 x = 9.1 . . . ans a | a ) 9.1 litres , b ) 10 litres , c ) 8 litres , d ) 12 litres , e ) 6 litres | a | multiply(100, subtract(const_1, sqrt(divide(25, 30.25)))) | divide(n2,n1)|sqrt(#0)|subtract(const_1,#1)|multiply(n0,#2) | general |
a 100 - litre mixture of milk and water contains 30.25 litres of milk . ' x ' litres of this mixture is removed and replaced with an equal quantum of water . if the process is repeated once , then the concentration of the milk stands reduced at 25 % . what is the value of x ? | "x = ( y / 4 ) - ( 2 / 5 ) , and so y = 4 x + 8 / 5 . the slope is 4 . ( n + 12 - n ) / ( m + p - m ) = 3 p = 3 the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(12, 4) | divide(n0,n1)| | general |
m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 12 ) both lie on the line defined by the equation x = ( y / 4 ) - ( 2 / 5 ) , what is the value of p ? | "1000 * 15 = 1300 * x x = 11.5 answer : a" | a ) 11.5 , b ) 12.5 , c ) 12.6 , d ) 12.2 , e ) 12.1 | a | divide(multiply(15, 1000), add(1000, 300)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
1000 men have provisions for 15 days . if 300 more men join them , for how many days will the provisions last now ? | this can be solve in two steps and without any complex calculation . given : equation of line a as y = ( 4 / 3 ) x - 100 . so the line intercept the axes at ( 0 , - 100 ) and ( 750 ) . this can be considered a right angle triangle with right angle at ( 00 ) . so base = 100 , height = 75 and hypotenuse = 125 ( by pythagoras triplet ) so a perpendicular from the ( 00 ) to hypotenuse will be the answer . area of triangle = 0.5 * 100 * 75 = 0.5 * 125 * x = > x = 60 ; so answer is 60 = c | a ) 48 , b ) 50 , c ) 60 , d ) 75 , e ) 100 | c | divide(multiply(100, 3), sqrt(add(power(4, const_2), power(3, const_2)))) | multiply(n1,n2)|power(n0,const_2)|power(n1,const_2)|add(#1,#2)|sqrt(#3)|divide(#0,#4) | general |
the equation of line a is y = 4 / 3 * x - 100 . what is the smallest possible distance in the xy - plane from the point with coordinates ( 0 , 0 ) to any point on line a ? | "within one standard deviation of the average age means 31 + / - 7 25 - - 31 - - 37 number of dif . ages - 25 26 27 28 29 30 31 32 33 34 35 36 37 total = 13 b" | a ) 8 , b ) 13 , c ) 15 , d ) 18 , e ) 30 | b | add(multiply(6, const_2), const_1) | multiply(n1,const_2)|add(#0,const_1)| | general |
the average age of applicants for a new job is 31 , with a standard deviation of 6 . the hiring manager is only willing to accept applications whose age is within one standard deviation of the average age . what is the maximum number of different ages of the applicants ? | "3 w = 2 m 20 m - - - - - - 21 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 20 * 21 * 8 = 14 * x * 6 x = 40 answer : c" | a ) 32 , b ) 87 , c ) 40 , d ) 99 , e ) 77 | c | add(floor(divide(multiply(multiply(21, 8), multiply(20, 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 |
20 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 ? | "1 sd from the mean is adding and subtrating the amount if standard deviation from the mean one time . 2 sd from the mean is adding and subtracting twice . 1 sd from the mean ranges from 25 to 55 , where 55 is within sd above the mean and 25 within 1 sd below the mean 2 sd = 15 twice = 30 from the the mean , which is 55 to 25 , where 55 is within 2 sd above the mean and 30 is within 2 sd below the mean . answer = a" | a ) 20 , b ) 31 , c ) 45 , d ) 90 , e ) 89 | a | multiply(2, 15) | multiply(n1,n2)| | general |
the class mean score on a test was 40 , and the standard deviation was 15 . if jack ' s score was within 2 standard deviations of the mean , what is the lowest score he could have received ? | machine b produces 100 part in 30 minutes . machine a produces 100 parts thrice as fast as b , so machine a produces 100 parts in 30 / 3 = 10 minutes . now , machine a produces 100 parts in 10 minutes which is 100 / 10 = 10 parts / minute . 10 parts x a total of 6 minutes = 60 d | a ) 20 , b ) 80 , c ) 40 , d ) 60 , e ) 50 | d | multiply(multiply(divide(100, 30), const_3), 6) | divide(n0,n2)|multiply(#0,const_3)|multiply(n3,#1) | gain |
machine a produces 100 parts thrice as fast as machine b does . machine b produces 100 parts in 30 minutes . if each machine produces parts at a constant rate , how many parts does machine a produce in 6 minutes ? | "machine y produces 20 percent more widgets in an hour than machine x does in an hour . so if machine x produces 100 widgets , then machine y produces 120 widgets . ratio of 120 / 100 = 6 / 5 . this is their speed of work ( y : x ) . i . e . speed of their work ( x : y ) = 5 / 6 now , time is inversely proportional to speed . hence the ratio of the time spent ( x : y ) = 6 / 5 let us assume that they spend 6 x and 5 x hours . given that 6 x - 5 x = 80 so , x = 80 . hence 6 x = 6 * 80 = 480 hours . hence x takes 120 hours to produce 1080 widgets . so , in 1 hour , it can produce ( 1 * 1080 ) / 480 = 2.25 hence option ( e ) ." | a ) 100 , b ) 65 , c ) 25 , d ) 11 , e ) 2.25 | e | divide(1080, multiply(divide(80, const_10), 80)) | divide(n0,const_10)|multiply(n0,#0)|divide(n1,#1)| | general |
machine x takes 80 hours longer than machine y to produce 1080 widgets . machine y produces 20 percent more widgets in an hour than machine x does in an hour . how many widgets per hour does machine x produce | "solution : total number of students studying both are 423 + 226 - 134 = 515 ( subtracting the 134 since they were included in the both the other numbers already ) . so 80 % of total is 515 , so 100 % is approx . 644 . answer is d : 644" | a ) 515 , b ) 545 , c ) 618 . , d ) 644 . , e ) 666 | d | add(226, 423) | add(n0,n1)| | general |
in the faculty of reverse - engineering , 226 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 80 % of the total ? | "answer : a ( 5 x 9 - 3 x 4 ) / 2 = 16.5" | a ) 16.5 , b ) 10 , c ) 8 , d ) 9.5 , e ) none of these | a | divide(subtract(multiply(5, 9), multiply(3, 4)), 2) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
the average of 5 quantities is 9 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ? | "relative speed = 60 + 30 = 90 km / hr . 90 * 5 / 18 = 25 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 / 25 = 40 sec . answer : c" | a ) 12 sec , b ) 24 sec , c ) 40 sec , d ) 60 sec , e ) 62 sec | c | add(60, 30) | add(n1,n2)| | physics |
two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 60 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "the total height was 45 cm too much . the average height should be reduced by 45 cm / 30 = 1.5 cm the answer is b ." | a ) 176.5 cm , b ) 175.5 cm , c ) 174.5 cm , d ) 173.5 cm , e ) 172.5 cm | b | divide(subtract(multiply(30, 177), subtract(151, 106)), 30) | multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#2,n0)| | general |
the average height of 30 students in a class was calculated as 177 cm . it has later found that the height of one of the students in the class was incorrectly written as 151 cm whereas the actual height was 106 cm . what was the actual average height of the students in the class ? | "rest time = number of rest Γ£ β time for each rest = 5 Γ£ β 6 = 30 minutes total time to cover 6 km = ( 6 Γ’ Β β 10 Γ£ β 60 ) minutes + 30 minutes = 66 minutes answer c" | a ) 48 min . , b ) 50 min . , c ) 66 min . , d ) 55 min . , e ) none of these | c | add(multiply(divide(6, 10), speed(const_60, const_1)), multiply(const_4, 6)) | divide(n1,n0)|multiply(n1,const_4)|speed(const_60,const_1)|multiply(#0,#2)|add(#3,#1)| | physics |
a man is walking at a speed of 10 km per hour . after every kilometre , he takes rest for 6 minutes . how much time will be take to cover a distance of 6 kilometres ? | given that two trains are of same length i . e . . 150 mtrs first train passes the km stone in 15 seconds . here we have time and distance so speed = 150 / 15 = 10 m / s we need to find out the second train speed . suppose the speed of the 2 nd train is x m / s relative speed of two trains is ( 10 + x ) = = > ( 150 + 150 ) / ( 10 + x ) = 8 = = > ( 300 ) / ( 10 + x ) = 8 = = > 300 = 80 + 8 x = = > 300 - 80 = 8 x = = > 220 = 8 x : - x = 55 / 2 m / s convert m / s into km / ph ( 55 / 2 ) * ( 18 / 5 ) = 99 kmph answer : d | a ) 60 kmph , b ) 66 kmph , c ) 72 kmph , d ) 99 kmph , e ) 89 kmph | d | multiply(divide(subtract(add(150, 150), multiply(divide(150, 15), 8)), 8), const_3_6) | add(n0,n0)|divide(n0,n1)|multiply(n2,#1)|subtract(#0,#2)|divide(#3,n2)|multiply(#4,const_3_6) | physics |