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marts income is 50 percent more than tims income and tims income is 40 percent less than juans income . what percentage of juans income is marts income | m = ( 150 / 100 ) t t = ( 60 / 100 ) j = > m = ( 90 / 100 ) j answer d . | a ) 480 , b ) 83 . , c ) 55 , d ) 501 , e ) 90 % | e | multiply(divide(add(const_100, 50), multiply(divide(const_100, subtract(const_100, 40)), const_100)), const_100) | add(n0,const_100)|subtract(const_100,n1)|divide(const_100,#1)|multiply(#2,const_100)|divide(#0,#3)|multiply(#4,const_100) | general |
what least number should be subtracted from 13601 such that the remainder is divisible by 87 ? | "13601 Γ· 87 = 156 , remainder = 29 hence 29 is the least number which can be subtracted from 13601 such that the remainder is divisible by 87 answer is b" | a ) $ 55.55 , b ) 29 , c ) 6 2 / 3 days , d ) 6 , e ) 2 : 3 | b | reminder(13601, 87) | reminder(n0,n1)| | general |
if the complement of a certain angle is 7 times the measure of that certain angle , then what is the measure of that certain angle ? | thecomplementof angle a is the angle which , when added to angle a , gives 90 degrees . the two acute angles of a right triangle are complements , for example . the original angle is x , so the complement is 7 x , and together , these add up to 90 degrees . x + 7 x = 90 8 x = 90 x = 11.25 Β° answer = ( e ) | a ) 42415758 , b ) 80 , c ) 11.25 Β° , d ) 99 kmph , e ) 120 | c | divide(subtract(const_100, const_10), add(7, const_1)) | add(n0,const_1)|subtract(const_100,const_10)|divide(#1,#0) | geometry |
if the average ( arithmetic mean ) of x and y is 80 , and z β x = 100 , what is the average of y and z ? | "x + y / 2 = 80 = > x + y = 160 x = z - 100 . . . sub this value z - 100 + y = 160 = > z + y = 260 = > z + y / 2 = 130 answer : e" | a ) 26250 , b ) 548 , c ) 130 , d ) 4.8 cm', ' , e ) 200 | c | subtract(multiply(100, const_2), multiply(80, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
the area of a triangle is with base 2 m and height 5 m ? | "1 / 2 * 2 * 5 = 5 m 2 answer : e" | a ) 5 , b ) $ 216 , c ) 300 m , d ) rs . 15000 , e ) 49 | a | triangle_area(2, 5) | triangle_area(n0,n1)| | geometry |
the ratio of 2 numbers is 2 : 8 and their h . c . f . is 40 . their l . c . m . is ? | "let the numbers be 2 x and 8 x their h . c . f . = 40 so the numbers are 2 * 40 , 8 * 40 = 80 , 320 l . c . m . = 320 answer is d" | a ) 75 , b ) 24 metre , c ) 9 / 25 , d ) 320 , e ) 9 : 16 | d | sqrt(divide(40, add(power(8, 2), add(power(2, 2), power(2, 2))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
if n divided by 3 has a remainder of 2 , what is the remainder when 2 times n is divided by 3 ? | "as per question = > n = 3 p + 2 for some integer p hence 2 n = > 6 q + 4 = > remainder = > 1 for some integer q hence a" | a ) 1 , b ) $ 3.85 , c ) 378 , d ) 2140 , e ) 21 | a | multiply(2, 2) | multiply(n1,n2)| | general |
ramu bought an old car for rs . 38000 . he spent rs . 12000 on repairs and sold it for rs . 64900 . what is his profit percent ? | "total cp = rs . 38000 + rs . 12000 = rs . 50000 and sp = rs . 64900 profit ( % ) = ( 64900 - 50000 ) / 50000 * 100 = 29.8 % answer : e" | a ) 8 / 3 , b ) 17 th , c ) 50 , d ) 135 , e ) 29.8 % | e | multiply(divide(subtract(64900, add(38000, 12000)), add(38000, 12000)), const_100) | add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)| | gain |
two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 90 kmph and 70 kmph respectively . in what time will they cross each other completely ? | "explanation : d = 250 m + 250 m = 500 m rs = 90 + 70 = 160 * 5 / 18 = 400 / 9 t = 500 * 9 / 400 = 11.25 sec answer : option e" | a ) 36 , b ) 11.25 sec , c ) 2700 , d ) 10 , e ) 60 | b | divide(250, multiply(90, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
the mass of 1 cubic meter of a substance is 200 kilograms under certain conditions . what is the volume , in cubic centimeters , of 1 gram of this substance under these conditions ? ( 1 kilogram = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | "density is mass divided by volume . so density of the given substance will be mass / volume = 200 kg / 1 m ^ 3 = 800 kg / m ^ 3 or 1 g / 5 cm ^ 3 = 0.2 g / cm ^ 3 . next , ask yourself if 200,000 g is equivalent to 1 , 000,000 cubic centimeters then 1 g is equivalent to how many cubic centimeters ? - - > 1 g - 1 , 000,000 / 800,000 = 10 / 2 = 5 cubic centimeters . answer is e" | a ) 8 days , b ) 5.0 , c ) 1 / 3 , d ) 8 kmph , e ) s : 1300 | b | subtract(divide(power(const_100, const_3), multiply(200, 1,000)), const_2) | multiply(n1,n4)|power(const_100,const_3)|divide(#1,#0)|subtract(#2,const_2)| | geometry |
alfred buys an old scooter for $ 4700 and spends $ 800 on its repairs . if he sells the scooter for $ 6400 , his gain percent is ? | "c . p . = 4700 + 800 = $ 5500 s . p . = $ 6400 gain = 6400 - 5500 = $ 900 gain % = 900 / 5500 * 100 = 16.36 % answer is b" | a ) 3 : 2 , b ) 1 , c ) 16.36 % , d ) 4 years , e ) 226 | c | multiply(divide(subtract(6400, add(4700, 800)), add(4700, 800)), const_100) | add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)| | gain |
3251 + 587 + 369 - ? = 3007 | let 4207 - x = 3007 then x = 4207 - 3007 = 1200 answer is d | a ) 8.5 , b ) 51 days , c ) 1200 , d ) $ 20,800 , e ) 73 | c | subtract(add(add(3251, 587), 369), 3007) | add(n0,n1)|add(n2,#0)|subtract(#1,n3) | general |
the manufacturer β s suggested retail price ( msrp ) of a certain item is $ 60 . store a sells the item for 20 percent more than the msrp . the regular price of the item at store b is 30 percent more than the msrp , but the item is currently on sale for 10 percent less than the regular price . if sales tax is 5 percent of the purchase price at both stores , what is the result when the total cost of the item at store b is subtracted from the total cost of the item at store a ? | msrp = 60 price at store a = 60 β 120100 = 72 = 60 β 120100 = 72 price at store b = 60 β 130100 β 90100 = 70.2 = 60 β 130100 β 90100 = 70.2 difference = 72.0 - 70.2 = 1.8 sales tax applicable = 5 % on both = 1.8 + 0.09 = 1.89 answer = d | a ) 11.36 % , b ) 25 % , c ) 104 meters', ' , d ) $ 1.89 , e ) 2991 | d | subtract(multiply(60, divide(add(const_100, 20), const_100)), multiply(divide(subtract(const_100, 10), const_100), multiply(divide(add(const_100, 30), const_100), 60))) | add(n1,const_100)|add(n2,const_100)|subtract(const_100,n3)|divide(#0,const_100)|divide(#2,const_100)|divide(#1,const_100)|multiply(n0,#3)|multiply(n0,#5)|multiply(#4,#7)|subtract(#6,#8) | general |
in business , a and c invested amounts in the ratio 3 : 2 , whereas the ratio between amounts invested by a and b was 3 : 4 , if rs 30,000 was their profit , how much amount did b receive . | "explanation : a : b = 3 : 4 = 3 : 4 = > a : c = 3 : 2 = 3 : 2 = > a : b : c = 3 : 4 : 2 c share = ( 2 / 9 ) * 30000 = 6666 option c" | a ) 6666 , b ) 50 , c ) 859622 , d ) 26 , e ) 1040 | a | multiply(multiply(multiply(multiply(add(4, 3), 4), const_100), const_100), divide(4, add(add(2, 3), 4))) | 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 |
if the selling price of 100 articles is equal to the cost price of 63 articles , then the loss or gain percent is : | "let c . p . of each article be re . 1 . then , c . p . of 100 articles = rs . 100 ; s . p . of 100 articles = rs . 63 . loss % = 37 / 100 * 100 = 37 % answer : e" | a ) 87 % , b ) $ 615 , c ) 37 % , d ) 115 , e ) 3280 | c | subtract(100, 63) | subtract(n0,n1)| | gain |
rs . 1775 is divided amongst a , b , c so that 5 times a ' s share , 3 times b ' s share and 7 times c ' s share are all equal . find c ' s share ? | a + b + c = 590 5 a = 3 b = 7 c = x a : b : c = 1 / 5 : 1 / 3 : 1 / 7 = 21 : 35 : 15 15 / 71 * 1775 = rs . 375 answer : e | a ) 4.2 , b ) 2,120 , c ) 375 , d ) 28 / 55 , e ) 21.21 | c | divide(1775, add(add(divide(7, 5), divide(7, 3)), const_1)) | divide(n3,n1)|divide(n3,n2)|add(#0,#1)|add(#2,const_1)|divide(n0,#3)| | general |
n ^ ( n / 2 ) = 2 is true when n = 2 in the same way what is the value of n if n ^ ( n / 2 ) = 4 ? | "n ^ ( n / 2 ) = 4 apply log n / 2 logn = log 4 nlogn = 2 log 4 = log 4 ^ 2 = log 16 logn = log 16 now apply antilog n = 16 / n now n = 4 . answer : c" | a ) 113 , b ) rs . 145 , c ) 4 , d ) 12800 , e ) rs . 740 | c | divide(power(4, 2), 4) | power(n4,n0)|divide(#0,n4)| | general |
a team won 50 percent of its first 60 games in a particular season , and 80 percent of its remaining games . if the team won a total of 60 percent of its games that season , what was the total number of games that the team played ? | "won 50 % of the first 60 games means it won 30 games . let the number of remaining games played be x . hence the number of remaining games won is 0.8 x . so the equation for winning can be written as : 30 + 0.8 x = 0.6 ( 60 + x ) x = 30 total games played = 60 + 30 = 90 option c ." | a ) 42 , b ) rs 66.66 , c ) 90 , d ) 36 , e ) 3 / 50 | c | divide(multiply(60, divide(50, const_100)), subtract(divide(80, const_100), divide(60, const_100))) | divide(n0,const_100)|divide(n2,const_100)|divide(n3,const_100)|multiply(n1,#0)|subtract(#1,#2)|divide(#3,#4)| | gain |
if a - b = 3 and a ( power 2 ) + b ( power 2 ) = 23 , find the value of ab . | "2 ab = ( a ( power 2 ) + b ( power 2 ) - ( a - b ) ( power 2 ) = 23 - 9 = 14 ab = 7 . answer is c ." | a ) 6 days , b ) 501 , c ) 4 / 15 , d ) 21 , e ) 7 | e | divide(subtract(23, power(3, 2)), 2) | power(n0,n1)|subtract(n3,#0)|divide(#1,n1)| | general |
a person buys an article at $ 380 . at what price should he sell the article so as to make a profit of 25 % ? | "c 475 cost price = $ 380 profit = 25 % of 380 = $ 95 selling price = cost price + profit = 380 + 95 = 475" | a ) 8915 , b ) 475 , c ) 8100 , d ) 12 days , e ) 90 | b | add(380, multiply(380, divide(25, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
in certain code ' twice ' is written as ' 34 $ 5 Ξ΄ ' and ' wears ' is written as ' 4 Ξ΄ 29 % ' . how is ' seat ' written in that code ? | answer : option b | a ) 23 , b ) 2 , c ) 3 : 2 , d ) 4 , e ) 6 | a | subtract(subtract(29, 5), const_1) | subtract(n3,n1)|subtract(#0,const_1) | gain |
the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 12.5 percent , what was the annual interest rate last year ? | "12.5 = 1.1 * x x = 11.36 % answer d )" | a ) 18 feet , b ) 184 , c ) 11.36 % , d ) 12 , e ) 120 | c | divide(multiply(12.5, const_100), add(12.5, const_100)) | add(n1,const_100)|multiply(n1,const_100)|divide(#1,#0)| | gain |
two pipes a and b can fill a cistern in 10 and 15 minutes respectively . both fill pipes are opened together , but at the end of 3 minutes , β b β is turned off . how much time will the cistern take to fill ? | in one min , ( a + b ) fill the cistern = 1 β 10 + 1 β 15 = 1 β 6 th in 3 min , ( a + b ) fill the cistern = 3 β 6 = 1 β 2 th remaining part = 1 - 1 β 2 = 1 β 2 β΅ 1 β 10 th part filled by a in one min . β΄ 1 β 2 nd part filled by a in 10 Γ 1 β 2 = 5 min . β΄ total time = 3 + 5 = 8 min . answer b | a ) 22.4 hours , b ) 8 min , c ) 4 / 9 , d ) 0 % , e ) 78 | b | add(multiply(10, subtract(const_1, multiply(add(inverse(10), inverse(15)), const_3))), 3) | inverse(n0)|inverse(n1)|add(#0,#1)|multiply(#2,const_3)|subtract(const_1,#3)|multiply(n0,#4)|add(n2,#5) | physics |
a swimmer can swim in still water at 4 km / h . if the speed of the water current is 2 km / h , how many hours will the swimmer take to swim against the current for 8 km ? | "the swimmer can swim against the current at a speed of 4 - 2 = 2 km / h . the time it will take is 8 / 2 = 4 hours . the answer is d ." | a ) 4 , b ) 15 , c ) 87.3 % , d ) 27.38 % , e ) 1000 | a | divide(8, subtract(4, 2)) | subtract(n0,n1)|divide(n2,#0)| | physics |
find the simple interest on rs . 945 for 5 months at 4 paisa per month ? | "explanation : i = ( 945 * 5 * 4 ) / 100 = 189 answer : option d" | a ) 5 / 4 , b ) 7.4 , c ) 24 , d ) s . 189 , e ) 8 | d | multiply(945, divide(5, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
find a sum for first 8 prime numbers ? | required sum = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 ) = 77 note : 1 is not a prime number option d | a ) 77 , b ) 1 : 88 , c ) 40 , d ) 49 , e ) 500 | a | add(add(8, add(add(add(const_2, const_3), const_2), const_4)), add(add(add(add(add(add(add(add(add(add(add(const_2, const_3), const_2), const_4), const_2), const_4), add(add(add(add(const_2, const_3), const_2), const_4), const_2)), add(add(add(const_2, const_3), const_2), const_4)), add(add(const_2, const_3), const_2)), add(const_2, const_3)), const_3), const_2)) | add(const_2,const_3)|add(#0,const_2)|add(#1,const_4)|add(n0,#2)|add(#2,const_2)|add(#4,const_4)|add(#5,#4)|add(#6,#2)|add(#7,#1)|add(#8,#0)|add(#9,const_3)|add(#10,const_2)|add(#3,#11) | general |
a jogger running at 9 km / hr along side a railway track is 290 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 290 + 120 = 410 m . time taken = 410 / 10 = 41 sec . answer : e" | a ) $ 130 , b ) 2000 , c ) 41 sec , d ) 1 / 6 , e ) t = 121 | c | divide(add(290, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)| | general |
if you write down all the numbers from 1 to 100 , then how many times do you write 1 ? | "explanation : explanation : clearly , from 1 to 100 , there are ten numbers with 1 as the unit ' s digit - 1 , 11 , 21 , 31 , 41 , 51 , 61 , 71 , 81 , 91 ; and ten numbers with 1 as the ten ' s digit - 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 100 so , required number = 10 + 11 = 21 . answer : d" | a ) 54 Β°', ' , b ) 21 , c ) 27.6 miles , d ) $ 250 , e ) 253 | b | divide(subtract(100, 1), 1) | subtract(n1,n0)|divide(#0,n2)| | general |
working simultaneously and independently at an identical constant rate , 20 machines of a certain type can produce a total of x units of product p in 4 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 3 x units of product p in 8 days ? | "the rate of 20 machines is rate = job / time = x / 4 units per day - - > the rate of 1 machine 1 / 20 * ( x / 4 ) = x / 80 units per day ; now , again as { time } * { combined rate } = { job done } then 8 * ( m * x / 80 ) = 3 x - - > m = 30 . answer : a ." | a ) 11.36 % , b ) 60000 , c ) 30 , d ) 120 , e ) $ 12.48 | c | multiply(multiply(20, 3), divide(4, 8)) | divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)| | general |
the sum of first n consecutive odd integers is n ^ 2 . what is the sum of all odd integers between 13 and 31 inclusive . | we ' re dealing with a sequence of consecutive odd integers : 13 to 31 , inclusive . we ' re asked for the sum of this group . 1 ) start with the sum of the smallest and the biggest : 13 + 31 = 44 2 ) now look at the ' next smallest ' and the ' next biggest ' : 15 + 29 = 44 now we have proof that there is no middle term . we have 5 bunches of 44 5 ( 44 ) = 220 a | a ) 1812.5 , b ) 14 % , c ) 19 , d ) 6.25 , e ) 220 | e | subtract(power(divide(add(31, const_1), 2), const_2), power(divide(add(subtract(13, 2), const_1), 2), 2)) | add(n2,const_1)|subtract(n1,n0)|add(#1,const_1)|divide(#0,n0)|divide(#2,n0)|power(#3,const_2)|power(#4,n0)|subtract(#5,#6) | general |
find the l . c . m of 15 , 18 , 28 and 30 . | "explanation : 2 x 3 x 5 x 3 x 14 = 1260 answer : option b" | a ) 20 , b ) 1260 , c ) 250 m , d ) 14 : 00 , e ) 45 | b | multiply(multiply(power(const_3, const_3), multiply(power(const_2, const_3), power(add(const_4, const_1), const_2))), divide(divide(divide(divide(divide(28, const_2), const_2), const_3), add(const_4, const_1)), add(const_4, const_1))) | add(const_1,const_4)|divide(n2,const_2)|power(const_2,const_3)|power(const_3,const_3)|divide(#1,const_2)|power(#0,const_2)|divide(#4,const_3)|multiply(#2,#5)|divide(#6,#0)|multiply(#7,#3)|divide(#8,#0)|multiply(#10,#9)| | physics |
if o is the center of the circle in the figure above and the area of the unshaded sector is 5 , what is the area of the shaded region ? | 60 / 360 = 1 / 6 1 / 6 of total area = 5 5 / 6 of total area = 5 * 5 = 25 answer : d | a ) 41 , b ) 1 : 4', ' , c ) 9.89 % , d ) rs . 800 , e ) 25', ' | e | power(5, const_2) | power(n0,const_2) | geometry |
the smallest value of n , for which 2 n is not a prime number , is | "( 2 Γ£ β 1 ) = 2 . ( 2 Γ£ β 2 ) = 4 . ( 2 Γ£ β 3 ) = 6 . ( 2 Γ£ β 4 ) = 8 . which is not prime , n = 2 , 3,4 . answer : a" | a ) 16 , b ) $ 2,350 , c ) 2 , 3,4 , d ) 12000 , e ) 8 / 45 | c | add(2, 2) | add(n0,n0)| | general |
if a mixture is 5 β 9 alcohol by volume and 4 β 9 water by volume , what is the ratio of the volume of alcohol to the volume of water in this mixture ? | "should be a sub - 600 level q . . volume = { 5 / 9 } / { 4 / 9 } = 5 / 4 c" | a ) 80 %', ' , b ) 6.24 , c ) 5 / 4 , d ) rs . 90 , e ) 786 | c | divide(divide(5, 9), divide(4, 9)) | divide(n0,n1)|divide(n2,n1)|divide(#0,#1)| | general |
two spherical balls lie on the ground touching . if one of the balls has a radius of 8 cm , and the point of contact is 12 cm above the ground , what is the radius of the other ball ( in centimeters ) ? | "a straight line will join the two centers and the point of contact , thus making similar triangles . 4 / 8 = ( r - 12 ) / r 4 r = 8 r - 96 r = 24 the answer is b ." | a ) 7.2 hr , b ) 7 / 3 , c ) 1 / 12 , d ) 25 % , e ) 24 | e | add(add(const_4.0, 12), 8) | add(const_4.0,n1)|add(#0,n0)| | physics |
s = { 12 , 35 , 69 } t = { 45 , 67 , 13 } what is the probability that x chosen from s and y chosen from t will result x * y = even | p : the probability that x * y is even , then p = 1 - p ( x * y is odd ) p ( x * y odd ) = p ( x odd ) * p ( y odd ) = 4 / 6 * 4 / 6 = 16 / 36 = 4 / 9 and p = 1 - 4 / 9 = 5 / 9 option : a | a ) 1 / 81 , b ) 5 / 9 , c ) 17 , d ) 4 , e ) 11.25 sec | b | divide(add(divide(12, const_4), const_2), multiply(divide(12, const_4), divide(12, const_4))) | divide(n0,const_4)|add(#0,const_2)|multiply(#0,#0)|divide(#1,#2) | general |
a certain music store stocks 800 cellos and 600 violas . of these instruments , there are 110 cello - viola pairs , such that a cello and a viola were both made with wood from the same tree ( each tree can make at most one viola and one cello , so there are no pairs other than these 90 ) . if one viola and one cello are chosen at random , what is the probability that the two instruments are made with wood from the same tree ? | "solution provided by stanford 2012 is correct : 110 / 800 choosing one of the cellos which has a pair viola , 1 / 600 choosing the viola which is the pair of chosen cello - - > p = 110 / 800 * 1 / 600 = 311 / 48,000 . answer : a ." | a ) 15 , b ) 2160 , c ) 1 [ 1 / 5 ] , d ) 2 , e ) 11 / 48,000 | e | multiply(divide(110, 800), divide(const_1, 600)) | divide(n2,n0)|divide(const_1,n1)|multiply(#0,#1)| | other |
a certain roller coaster has 3 cars , and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster . if a certain passenger is to ride the roller coaster 3 times , what is the probability that the passenger will ride in each of the 3 cars ? | "probability = ( favorable cases ) / ( total number of cases ) total number of ways in which a person can ride car = 3 * 3 * 3 = 27 ( in first ride he has 3 options to sit , in second right again he has 3 seats available to sit and so on ) number of favorable cases , i . e . , when he rides on different cars ; he can choose seat car in 3 ways in his 1 st ride . he can choose seat car in 2 ways in his 2 nd ride . he can choose seat car in 1 ways in his 3 rd ride . so , 3 * 2 * 1 = 6 ways thus , probability of choosing different seats = 6 / 27 = 2 / 9 answer : c" | a ) 36 , b ) 15 , c ) $ 300 , d ) 2 / 9 , e ) 44 | d | multiply(factorial(3), power(divide(1, 3), 3)) | divide(n1,n0)|factorial(n0)|power(#0,n0)|multiply(#1,#2)| | general |
of the 100 people in a room , 4 / 5 are women . if 3 / 5 of the people are married , what is the maximum number of women in the room who could be unmarried ? | "women = 4 / 5 * 100 = 80 married = 3 / 5 * 100 = 60 unmarried = 40 max ( un - married women ) = 40 e" | a ) 5 cm , b ) s . 15000 , c ) 120 , d ) 40 , e ) rs . 3800 | d | multiply(100, divide(4, 5)) | divide(n1,n2)|multiply(n0,#0)| | general |
one copy machine can make 25 copies a minute , and a second copy machine makes 35 copies a minute . if the two copiers work together , how long would it take them to make 4,800 copies ? | "total work done by both machines in a minute = 25 + 35 = 60 copies total number of copies required = 6000 time = 4800 / 60 = 80 mins answer e" | a ) 136.8 , b ) 6 , c ) 80 minutes , d ) 155 , e ) 60 | c | divide(power(35, const_3), add(25, 35)) | add(n0,n1)|power(n1,const_3)|divide(#1,#0)| | physics |
a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 20 gallons of water and 40 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ? | "nope , 150 . i can only get it by following pr ' s backsolving explanation . i hate that . original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 40 gallons salt is added and 20 gallons of water is added . so salt now becomes - > ( 0.2 x + 40 ) and water - - > ( 0.6 x + 20 ) amount of salt is 33.33 % of total . so amount of water is 66.66 % . so salt is half of the volume of water . so ( 0.2 x + 40 ) = ( 0.6 x + 20 ) / 2 = > 0.4 x + 80 = 0.6 x + 20 = > 0.2 x = 60 solving , x = 300 answer : a" | a ) 140 , b ) $ 5 , c ) 11 and 9 , d ) 300 , e ) 6666 | d | divide(subtract(multiply(40, const_2), 20), subtract(subtract(subtract(1, divide(20, const_100)), multiply(subtract(1, divide(20, const_100)), divide(1, const_4))), multiply(const_2, divide(20, const_100)))) | divide(n0,const_100)|divide(n4,const_4)|multiply(n2,const_2)|multiply(#0,const_2)|subtract(#2,n1)|subtract(n4,#0)|multiply(#1,#5)|subtract(#5,#6)|subtract(#7,#3)|divide(#4,#8)| | general |
a rectangular lawn of length 200 m by 120 m has two roads running along its center , one along the length and the other along the width . if the width of the roads is 5 m what is the area w covered by the two roads ? | "area covered by road along the length = 5 * 200 = 1000 square meter area covered by road along the width = 5 * 120 = 600 square meter common area in both roads ( where the roads intersect ) = square with side 5 meter = 5 * 5 = 25 total area of the roads w = 1000 + 600 - 25 = 1575 answer : option c" | a ) 1575 , b ) 60 , c ) 14 kmph , d ) 45 , e ) 7200 | a | add(rectangle_area(200, 5), rectangle_area(120, 5)) | rectangle_area(n0,n2)|rectangle_area(n1,n2)|add(#0,#1)| | geometry |
each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , annie is paid x dollars per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week annie worked a total of 53 hours . if harry and annie were paid the same amount last week , how many hours did harry work last week ? | "annie earned 40 x + 13 ( 2 x ) = 66 x let h be the number of hours that harry worked . harry earned 30 x + 1.5 x ( h - 30 ) = 66 x ( 1.5 x ) ( h ) = 81 x h = 54 hours the answer is d ." | a ) 63 , b ) 3334 , c ) 48.68077 , d ) 137 / 216', ' , e ) 54 | e | add(divide(subtract(add(40, 2), 30), 1.5), 30) | add(n2,n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)| | general |
the diagonals of a rhombus are 18 cm and 22 cm . find its area ? | 1 / 2 * 18 * 22 = 198 answer : c | a ) 18 : 5 , b ) 4 , c ) 0 , d ) 7 / 15 , e ) 198', ' | e | rhombus_area(18, 22) | rhombus_area(n0,n1) | geometry |
if 20 liters of chemical x are added to 80 liters of a mixture that is 25 % chemical x and 75 % chemical y , then what percentage of the resulting mixture is chemical x ? | the amount of chemical x in the solution is 20 + 0.25 ( 80 ) = 40 liters . 40 liters / 100 liters = 40 % the answer is d . | a ) 103.4 % , b ) 10 % , c ) 24 . , d ) 2 : 1 , e ) 40 % | e | add(20, multiply(divide(25, const_100), 80)) | divide(n2,const_100)|multiply(n1,#0)|add(n0,#1) | general |
a number increased by 20 % gives 600 . the number is | "formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 20 % = 120 % 120 % - - - - - - - > 600 ( 120 Γ 5 = 600 ) 100 % - - - - - - - > 500 ( 100 Γ 5 = 500 ) d )" | a ) 22 1 / 2 days , b ) 15 sec . , c ) 19 , d ) 10 v 2', ' , e ) 500 | e | divide(600, add(const_1, divide(20, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
when a laptop is sold for rs . 49,000 , the owner loses 30 % . at what price must that laptop be sold in order to gain 30 % ? | 70 : 49000 = 130 : x x = ( 49000 x 130 ) / 70 = 91000 . hence , s . p . = rs . 91,000 . answer : option e | a ) 66 , b ) 2.9 , c ) 91000 , d ) $ 320 , e ) 7 | c | multiply(divide(multiply(multiply(multiply(add(const_3, const_4), add(const_3, const_4)), const_100), multiply(add(const_3, const_2), const_2)), subtract(const_100, 30)), add(const_100, 30)) | add(n1,const_100)|add(const_3,const_4)|add(const_2,const_3)|subtract(const_100,n1)|multiply(#1,#1)|multiply(#2,const_2)|multiply(#4,const_100)|multiply(#6,#5)|divide(#7,#3)|multiply(#0,#8) | gain |
if a and b together can finish a work in 16 days . a can finish same work alone in 24 days then b alone can finish same work alone in how many days ? | ( a + b ) work in 1 day = 1 / 16 , a work in 1 day = 1 / 24 b work in 1 day = [ 1 / 16 - 1 / 24 ] = 1 / 48 . b alone can finish same work in 48 days . answer b | a ) 4 , b ) 48 days , c ) 400,200 , d ) s . 340.90 , e ) 660 | b | inverse(subtract(inverse(16), inverse(24))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2) | physics |
if 5 a + 7 b = m , where a and b are positive integers , what is the largest possible value of m for which exactly one pair of integers ( a , b ) makes the equation true ? | 5 * a 1 + 7 * b 1 = m 5 * a 2 + 7 * b 2 = m 5 * ( a 1 - a 2 ) = 7 * ( b 2 - b 1 ) since we are dealing with integers we can assume that a 1 - a 2 = 7 * q and b 2 - b 1 = 5 * q where q is integer , so whenever we get a pair for ( a ; b ) we can find another one by simply adding 7 to a and subtracting 5 from b or vice versa , subtracting 7 from a and adding 5 to b . lets check how it works for our numbers , starting from the largest : e ) 74 = 5 * 12 + 7 * 2 ( a 1 = 12 , b 1 = 2 ) , subtract 7 from a and add 5 to b respectively , so a 2 = 5 and b 2 = 7 , second pair - bad d ) 70 = 5 * 7 + 7 * 5 ( a 1 = 7 , b 1 = 5 ) , if we add 7 toawe will have to subtract 5 from b but b ca n ' t be 0 , so - no pair , if we subtract 7 froma , we ' ll get a = 0 which also is n ' t allowed - no pair , thus this is the only pair for ( a ; b ) that works , good ! , thus d is the answer | a ) 6 , b ) 70 , c ) 85 , d ) 49 . , e ) 7.7 % . | b | add(multiply(7, add(const_3, const_4)), multiply(5, const_4)) | add(const_3,const_4)|multiply(n0,const_4)|multiply(n1,#0)|add(#2,#1) | general |
the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 15 % ? | "let marked price = rs . 100 . then , c . p . = rs . 64 , s . p . = rs . 85 gain % = 21 / 64 * 100 = 32.8 % . answer : a" | a ) 9 , b ) 8 , c ) 32.8 % , d ) 44 , e ) $ 20 | c | multiply(subtract(divide(subtract(const_100, 15), 64), const_1), const_100) | subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)| | gain |
the calendar of the year 2028 can be used again in the year ? | "explanation : given year 2028 when divided by 4 , leaves a remainder 0 . note : when remainder is 0 , 28 is added to the given year to get the result . so , 2028 + 28 = 2056 answer : c" | a ) 2056 , b ) 21 , c ) 100 % , d ) 100 , e ) 700 | a | add(multiply(subtract(multiply(const_4, const_4), const_2), const_2), 2028) | multiply(const_4,const_4)|subtract(#0,const_2)|multiply(#1,const_2)|add(n0,#2)| | gain |
a man speaks truth 3 out of 4 times . he throws a die and reports it to be a 6 . what is the probability of it being a 6 ? | explanation : there are two cases 1 ) he is telling truth that the die reports 6 , its probability = 3 / 4 * 1 / 6 = 1 / 8 2 ) he is telling lie that the die reports 6 , its probability = 1 / 4 * 5 / 6 = 5 / 24 so required probability = ( 1 / 8 ) / ( 1 / 8 ) + ( 5 / 24 ) = ( 1 / 8 ) / ( 1 / 3 ) = 3 / 8 hencer ( d ) is the correct answer answer : d | a ) 45 , b ) 3 / 8 , c ) 12.50 % , d ) 28 m , e ) 2 / 7 | b | divide(multiply(divide(3, 4), divide(const_1, 6)), add(multiply(divide(3, 4), divide(const_1, 6)), multiply(divide(const_1, const_4), divide(const_5, 6)))) | divide(n0,n1)|divide(const_1,n2)|divide(const_1,const_4)|divide(const_5,n2)|multiply(#0,#1)|multiply(#2,#3)|add(#4,#5)|divide(#4,#6) | probability |
a fair coin is tossed 11 times . what is the probability of getting more heads than tails in 11 tosses ? | "on each toss , the probability of getting a head is 1 / 2 and the probability of getting a tail is 1 / 2 . there is no way to get the same number of heads and tails on an odd number of tosses . there will either be more heads or more tails . then there must be more heads on half of the possible outcomes and more tails on half of the possible outcomes . p ( more heads ) = 1 / 2 the answer is a ." | a ) $ 6 , b ) 2 / 7 , c ) 36 , d ) 37.5 % , e ) 1 / 2 | e | divide(add(add(add(choose(11, const_2), choose(11, const_3)), choose(11, const_4)), choose(11, 11)), power(const_2, 11)) | choose(n0,const_2)|choose(n0,const_3)|choose(n0,const_4)|choose(n0,n0)|power(const_2,n0)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,#4)| | probability |
on a sum of money , simple interest for 2 years is rs 660 and compound interest is rs 696.30 , the rate of interest being the same in both cases . | explanation : difference between c . i and s . i for 2 years = 36.30 s . i . for one year = 330 . s . i . on rs 330 for one year = 36.30 so r % = \ frac { 100 * 36.30 } { 330 * 1 } = 11 % answer : d | a ) 58 , b ) 6 hours , c ) 11 % , d ) 10 , e ) 1 / 25 | c | multiply(divide(multiply(subtract(696.3, 660), 2), 660), const_100) | subtract(n2,n1)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_100) | gain |
an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 19404 / - in 3 years at the same rate of interest . find the rate percentage ? | explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 19404 / - - 17640 / - = rs . 1764 / - rate of interest = ( 1764 / 17640 ) Γ ( 100 / 1 ) = > 10 % answer : option d | a ) 10 % , b ) 12 , c ) 6 , d ) 150 , e ) 1 | a | multiply(divide(subtract(19404, 17640), 17640), const_100) | subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100) | general |
a farmer spent $ 35 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 30 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have spent on the chicken feed and goat feed combined ? | "a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 35 = $ 14 on chicken feed , thus he spent the remaining 35 - 14 = $ 21 on goat feed . now , since he bought chicken feed at a 30 % discount then the original price of it was x * 0.7 = $ 14 - - > x = $ 20 . therefore if the farmer had paid full price for both the chicken feed and the goat feed , then he would he have spent 20 + 21 = $ 41 . answer : e ." | a ) 5 , b ) 22.37 , c ) 40 , d ) 19 , e ) $ 41 | e | add(multiply(35, divide(40, const_100)), 35) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
a man is 22 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 22 ) + 2 = 2 ( x + 2 ) x + 24 = 2 x + 4 x = 20 . answer : c" | a ) 18 , b ) 3 , c ) 20 years , d ) 1 , e ) 6 o kmph | c | divide(subtract(22, 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 is twice as good a work man as b and together they finish the work in 10 days . in how many days a alone can finish the work ? | "wc = 2 : 1 2 x + x = 1 / 10 = > x = 1 / 30 2 x = 1 / 15 a can do the work in 15 days . answer : b" | a ) 2400 , b ) 2 / 42 , c ) 15 , d ) $ 40000 , e ) 0 | c | inverse(divide(inverse(10), add(const_2, const_1))) | add(const_1,const_2)|inverse(n0)|divide(#1,#0)|inverse(#2)| | physics |
pumps a , b , and c operate at their respective constant rates . pumps a and b , operating simultaneously , can fill a certain tank in 1 / 2 hours ; pumps a and c , operating simultaneously , can fill the tank in 1 / 2 hours ; and pumps b and c , operating simultaneously , can fill the tank in 2 hours . how many hours does it take pumps a , b , and c , operating simultaneously , to fill the tank . | "a + b = 1 / 2 ; a + c = 1 / 2 , b + c = 2 ; add then 2 * ( a + b + c ) = 3 a + b + c = 3 / 2 hrs e" | a ) 1717.85 , b ) 5 / 6 , c ) 625 , d ) β 12 , e ) 9 | b | divide(subtract(add(const_3.0, 2), add(divide(2, 1), add(2, 2))), 1) | add(const_3.0,n1)|add(n0,n0)|divide(const_3.0,n3)|add(#1,#2)|subtract(#0,#3)|divide(#4,n3)| | physics |
if y > 0 , ( 10 y ) / 20 + ( 3 y ) / 10 is what percent of y ? | "can be reduced to y / 2 + 3 y / 10 = 4 y / 5 = 80 % e" | a ) 0 , b ) $ 216 , c ) 0.7 % , d ) 80 % , e ) 4 | d | multiply(const_100, add(divide(10, 20), divide(3, 10))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)| | general |
the ages of two persons differ by 16 years . if 6 years ago , the elder one be 3 times as old as the younger one , find their present ages . | "explanation : sol . let the age of the younger person be xx years then , age of the elder person = ( x + 16 ) ( x + 16 ) years Γ’ Λ Β΄ Γ’ Λ Β΄ 3 ( x Γ’ Λ β 6 ) = ( x + 16 Γ’ Λ β 6 ) 3 ( x - 6 ) = ( x + 16 - 6 ) Γ’ β‘ β 3 x Γ’ Λ β 18 = x + 10 Γ’ β‘ β 3 x - 18 = x + 10 Γ’ β‘ β 2 x = 28 Γ’ β‘ β 2 x = 28 Γ’ β‘ β x = 14 Γ’ β‘ β x = 14 hence , their present age are 14 years and 30 years . answer is d" | a ) 12 , b ) $ 500 , c ) 14 years and 30 years , d ) 18 , e ) 11.1 % | c | subtract(add(divide(multiply(16, 6), subtract(6, const_1)), 6), 16) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general |
let the number which when multiplied by 11 is increased by 300 . | solution let the number be x . then , 11 x - x = 300 βΉ = βΊ 10 x = 300 x βΉ = βΊ 30 . answer e | a ) 30 , b ) 21.21 , c ) 252 , d ) 2 , e ) 12 hours | a | divide(300, subtract(11, const_1)) | subtract(n0,const_1)|divide(n1,#0) | general |
in a division , a student took 78 as divisor instead of 36 . his answer was 24 . the correct answer is - | "x / 78 = 24 . x = 24 * 78 . so correct answer would be , ( 24 * 78 ) / 36 = 52 . answer : d" | a ) 315 , b ) 40 , c ) $ 7717.50 , d ) 52 , e ) 8 | d | divide(multiply(78, 24), 36) | multiply(n0,n2)|divide(#0,n1)| | general |
an athlete takes 10 seconds to run 100 m . what is his avg . speed in miles per hour ? | his average speed is 10 m / s . which is 36 km / hr . but 36 km = 22.37 miles . the average speed of the athlete is 22.37 mph answer : a | a ) 30 , b ) 40 , c ) 22.37 , d ) 352 , e ) 8 kmph | c | divide(multiply(divide(100, const_1000), const_0_6), divide(10, const_3600)) | divide(n1,const_1000)|divide(n0,const_3600)|multiply(#0,const_0_6)|divide(#2,#1) | physics |
chris age after 13 years will be 5 times his age 5 years back . what is the present age of chris ? | "chris present age = x after 13 years = x + 13 5 years back = x - 5 x + 13 = 5 ( x - 5 ) x = 12 answer is e" | a ) 12 , b ) 7 , c ) 29 , d ) 15000 , e ) 16 | a | subtract(divide(add(multiply(5, 5), 13), subtract(5, const_1)), subtract(divide(add(multiply(5, 5), 13), subtract(5, const_1)), 5)) | multiply(n1,n1)|subtract(n1,const_1)|add(n0,#0)|divide(#2,#1)|subtract(#3,n1)|subtract(#3,#4)| | general |
a pair of articles was bought for $ 50 at a discount of 50 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 50 / 2 = $ 25 let m . p = $ x 50 % of x = 25 x = 25 * . 5 = $ 12.50 answer is b" | a ) 41 sec , b ) 21 years , c ) 16 , d ) $ 12.50 , e ) 112 | d | divide(multiply(subtract(const_100, 50), divide(50, const_2)), const_100) | divide(n0,const_2)|subtract(const_100,n1)|multiply(#0,#1)|divide(#2,const_100)| | gain |
a tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride . the proportion of sodium chloride in the tank is 40 % by volume and the capacity of the tank is 24 gallons . if the water evaporates from the tank at the rate of 0.5 gallons per hour , and the amount of sodium chloride stays the same , what will be the concentration of water in the mixture in 6 hours ? | the number of gallons in the tank is ( 1 / 4 ) 24 = 6 gallons the amount of sodium chloride is 0.4 ( 6 ) = 2.4 gallons at the start , the amount of water is 0.6 ( 6 ) = 3.6 gallons after 6 hours , the amount of water is 3.6 - 0.5 ( 6 ) = 0.6 gallons the concentration of water is 0.6 / ( 2.4 + 0.6 ) = 2 / 10 = 20 % the answer is b . | a ) 20 % , b ) 30 , c ) 9 , d ) 144 , e ) 80 % | a | multiply(divide(subtract(divide(multiply(6, subtract(const_100, 40)), const_100), multiply(0.5, 6)), subtract(6, multiply(0.5, 6))), const_100) | multiply(n2,n3)|subtract(const_100,n0)|multiply(n3,#1)|subtract(n3,#0)|divide(#2,const_100)|subtract(#4,#0)|divide(#5,#3)|multiply(#6,const_100) | gain |
in what proportion must rice at rs 3.10 per kg be mixed with rice at rs 3.75 per kg , so that the mixture be worth rs 3.25 a kg ? | c . p of 1 kg of cheaper rice = rs 3.10 c . p of 1 kg of expensive rice = rs 3.75 the mixture be worth for 1 kg = rs 3.25 by the alligation rule : quantity of cheaper rice / quantity of expensive rice = ( 3.75 - 3.25 ) / ( 3.25 - 3.10 ) = ( 0.50 ) / ( 0.15 ) = 10 / 3 c | a ) 6 , b ) $ 280 , c ) 60.75 , d ) 0.05 , e ) 10 / 3 | e | divide(subtract(3.75, 3.25), subtract(3.25, 3.1)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1) | general |
the distance between 2 cities a and b is 1000 km . a train starts from a at 12 p . m . and travels towards b at 100 km / hr . another starts from b at 1 p . m . and travels towards a at 150 km / hr . at what time do they meet ? | "suppose they meet x hrs after 12 p . m . distance moved by first in x hrs + distance moved by second in ( x - 1 ) hrs = 1000 100 x + 150 ( x - 1 ) = 1000 x = 4.60 = 5 hrs they meet at 10 + 5 = 5 p . m . answer is c" | a ) 1 / 7 , b ) rs . 98.56 , c ) 5 pm . , d ) 72.33 , e ) 3 | c | add(divide(add(2, 1), add(12, 1)), 1000) | add(n0,n4)|add(n2,n4)|divide(#0,#1)|add(n1,#2)| | physics |
in a certain quiz that consists of 10 questions , each question after the first is worth 4 points more than the preceding question . if the 10 questions on the quiz are worth a total of 300 points , how many points is the third question worth ? | "x x + 4 x + 8 x + 12 x + 16 x + 20 x + 24 x + 28 x + 32 x + 36 10 x + 180 = 300 10 x = 120 x = 12 3 rd question = x + 8 = 12 + 8 = 20 answer d" | a ) 20 , b ) four , c ) 3 / 5 , d ) 165 Β° , e ) 786858 | a | add(divide(300, 10), subtract(subtract(10, 4), const_1)) | divide(n3,n2)|subtract(n0,n1)|subtract(#1,const_1)|add(#0,#2)| | general |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 256 sq . feet , how many feet of fencing will be required ? | "given that length and area , so we can find the breadth . length x breadth = area 20 x breadth = 256 breadth = 12.8 feet area to be fenced = 2 b + l = 2 ( 12.8 ) + 20 = 45.6 feet answer : a" | a ) 45.6 , b ) 24.78 , c ) 18 , d ) $ 1260 , e ) 27 | a | add(multiply(divide(256, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
there are 180 doctors and nurses at a hospital . if the ratio of doctors to nurses is 2 to 3 , how many nurses are at the hospital ? | "the number of nurses at the hospital is ( 3 / 5 ) * 180 = 108 . the answer is c ." | a ) 2 : 3 , b ) 11 , c ) 48.7 $ , d ) 31 , e ) 108 | e | multiply(divide(180, add(2, 3)), 3) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | other |
a reduction of 40 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | "800 * ( 40 / 100 ) = 320 - - - - 5 ? - - - - 1 = > rs . 64 answer : c" | a ) 7 % , b ) 305 , c ) 2 , d ) 64 , e ) 3 : 2 | d | divide(divide(multiply(800, 40), const_100), 5) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain |
a tank contains 6,500 gallons of a solution that is 5 percent sodium chloride by volume . if 2,500 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "we start with 6,500 gallons of a solution that is 5 % sodium chloride by volume . this means that there are 0.05 x 6,500 = 325 gallons of sodium chloride . when 2,500 gallons of water evaporate we are left with 4,000 gallons of solution . from here we can determine what percent of the 4,000 gallon solution is sodium chloride . ( sodium chloride / total solution ) x 100 = ? ( 325 / 4,000 ) x 100 = ? 0.0812 x 100 = ? = 8.12 % answer is d ." | a ) 15 , b ) 8.12 % , c ) 6 , d ) 1800 , e ) 75 kg | b | multiply(divide(multiply(multiply(const_100, const_100), divide(5, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | add(const_2,const_3)|divide(n1,const_100)|multiply(const_100,const_100)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#4,const_100)|multiply(#0,#6)|add(#7,#5)|subtract(#2,#8)|divide(#3,#9)|multiply(#10,const_100)| | gain |
the l . c . m . of 2 numbers is 72 . the numbers are in the ratio 2 : 3 . find their sum ? | "let the numbers be 2 x and 3 x l . c . m . = 6 x 6 x = 72 x = 12 the numbers are = 24 and 36 required sum = 24 + 36 = 60 answer is d" | a ) 25 , b ) 4 , c ) none , d ) 60 , e ) 6 | d | add(multiply(divide(2, multiply(72, 2)), 2), multiply(divide(2, multiply(72, 2)), 72)) | multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n1,#1)|add(#2,#3)| | other |
if a farmer sells 15 of his chickens , his stock of feed will last for 4 more days than planned , but if he buys 25 more chickens , he will run out of feed 3 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ? | "say farmer has n chicken and he is good for d days . : - we have 3 equations given in question : - ( n - 15 ) * d + 4 = ( n + 25 ) * ( d - 3 ) = n * d solving these : ( you can solve 1 st and 3 rd and 2 nd and 3 rd together ) we get : 25 d - 3 n = 75 4 n - 15 d = 60 = > n ~ 49 ans e it is !" | a ) 15 / 8 ohms', ' , b ) 169 , c ) 758 , d ) 40 litres , e ) 49 | e | multiply(add(const_2, 4), multiply(3, 4)) | add(const_2,n1)|multiply(n1,n3)|multiply(#0,#1)| | general |
what is the max number of rectangular boxes , each measuring 4 inches by 6 inches by 10 inches , that can be packed into a rectangular packing box measuring 16 inches by 18 inches by 30 inches , if all boxes are aligned in the same direction ? | the 4 inch side should be aligned to the 16 inch side ( 4 layer ) 6 inch side should be aligned to the 18 inch side . ( 3 layer ) and 10 inch side should be aligned to the 30 inch side . ( 3 layer ) maximum number of rectangles = 4 * 3 * 3 = 36 answer is a | a ) 2700 , b ) 720 , c ) 2 ^ 11 - 1 , d ) 36', ' , e ) 600 | d | divide(multiply(multiply(16, 18), 30), multiply(multiply(4, 6), 10)) | multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#2,#3) | geometry |
after decreasing 80 % in the price of an article costs rs . 320 . find the actual cost of an article ? | "cp * ( 20 / 100 ) = 320 cp = 16 * 100 = > cp = 1600 answer : e" | a ) 9 , b ) 90 , c ) 5 , d ) 1600 , e ) 22.8 | d | divide(320, subtract(const_1, divide(80, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
how many seconds will a 500 m long train take to cros a man walking with a speed of 3 kmph in the direction of the moving train if the speed of the train is 63 kmph | time = distance ( here length of the train ) / relative speed ( 63 - 3 ) thus time = 500 / 60 * 5 / 18 = 500 * 18 / 60 * 5 = 30 seconds answer : b | a ) 30 , b ) 84.6 % , c ) 210 , d ) 4487 , e ) 72.33 | a | divide(500, divide(subtract(63, 3), const_3_6)) | subtract(n2,n1)|divide(#0,const_3_6)|divide(n0,#1) | physics |
a batsman had a certain average of runs for 16 innings . in the 17 th innings , he made a score of 87 runs thereby increasing his average by 3 . what is his average after 17 innings ? | explanation : assume his initial average = xx his total runs after 16 innings = 16 xx after scoring 87 runs his average got increased by 3 to xx + 3 so his total runs after 17 innings = 17 Γ ( xx + 3 ) but it was given that the difference in the total scores after 16 innings and 17 innings = 87 therefore 17 Γ ( x + 3 ) β 16 x = 87 β x = 3617 Γ ( x + 3 ) β 16 x = 87 β x = 36 his new average = 36 + 3 = 39 answer : a | a ) 108 kmph , b ) 62 , c ) 1600 , d ) 5 , e ) 39 | e | add(subtract(87, multiply(17, 3)), 3) | multiply(n1,n3)|subtract(n2,#0)|add(n3,#1) | general |
if a card is drawn from a well shuffled pack of cards , the probability of drawing a spade or a king is - . | "explanation : p ( s α΄ k ) = p ( s ) + p ( k ) - p ( s β© k ) , where s denotes spade and k denotes king . p ( s α΄ k ) = 13 / 52 + 4 / 52 - 1 / 52 = 4 / 13 answer : b" | a ) 34 , b ) 2 , c ) 500 , d ) 4 / 13 , e ) 50 | d | add(divide(const_3, const_52), divide(divide(const_52, const_4), const_52)) | divide(const_3,const_52)|divide(const_52,const_4)|divide(#1,const_52)|add(#0,#2)| | probability |
the membership of a committee consists of 3 english teachers , 4 mathematics teachers , and 3 social studies teachers . if 2 committee members are to be selected at random to write the committee β s report , what is the probability that the two members selected will both be english teachers ? | "probability of first member an english teacher = 3 / 10 probability of second member an english teacher = 2 / 10 probability of both being english teacher = 3 / 9 x 2 / 8 = 3 / 5 ( d )" | a ) 10 , b ) 75 % , c ) 1 / 12 , d ) rs . 6720 , e ) 15 % | c | multiply(divide(3, add(add(3, 4), 3)), divide(3, subtract(add(add(3, 4), 3), const_1))) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|subtract(#1,const_1)|divide(n2,#3)|multiply(#2,#4)| | probability |
a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 1 / 5 % his yearly income diminishes by rs . 61.50 . his capital is | explanation : capital = rs . x , then 4 / 5 x = 61.5 x = 76.88 answer : c ) rs . 76.88 | a ) 76.88 , b ) 4 : 5 , c ) 1575 , d ) 6.8 % , e ) 80 % | a | divide(61.5, divide(const_4, 5)) | divide(const_4,n3)|divide(n4,#0) | gain |
thirty percent of the members of a swim club have passed the lifesaving test . among the members who havenotpassed the test , 19 have taken the preparatory course and 30 have not taken the course . how many members are there in the swim club ? | "30 % of the members have passed the test , thus 70 % have not passed the test . we also know that 30 + 19 = 49 members have not passed the test , thus 0.7 * total = 49 - - > total = 70 . answer : b ." | a ) 720 , b ) 1 , c ) 18 % , d ) 70 , e ) 10 / 3 | d | divide(add(19, 30), divide(subtract(const_100, 30), const_100)) | add(n0,n1)|subtract(const_100,n1)|divide(#1,const_100)|divide(#0,#2)| | gain |
the decimal 0.1 is how many times greater than the decimal ( 0.01 ) ^ 3 ? | 0.1 = 10 ^ - 1 ( 0.01 ) ^ 3 = ( 10 ^ - 2 ) ^ 3 = 10 ^ - 6 10 ^ 5 * 10 ^ - 6 = 10 ^ - 1 the answer is c . | a ) 36 min , b ) 77.77 meters , c ) 91000 , d ) 85 , e ) 10 ^ 5 | e | divide(0.1, power(0.01, 3)) | power(n1,n2)|divide(n0,#0) | general |
the current in a river is 5 mph . a boat can travel 20 mph in still water . how far up the river can the boat travel if the round trip is to take 10 hours ? | "upstream speed = 20 - 5 = 15 mph downstream speed = 20 + 5 = 25 mph d / 15 + d / 25 = 10 hours solving for d we get d = 93,75 answer : c" | a ) 93,75 miles , b ) 180 , c ) 3 / 11 , d ) 500 cm 2 , e ) 7 | a | divide(multiply(multiply(subtract(20, 5), add(20, 5)), 10), add(add(20, 5), subtract(20, 5))) | add(n0,n1)|subtract(n1,n0)|add(#0,#1)|multiply(#0,#1)|multiply(n2,#3)|divide(#4,#2)| | physics |
let f ( x , y ) be defined as the remainder when ( x β y ) ! is divided by x . if x = 16 , what is the maximum value of y for which f ( x , y ) = 0 ? | "the question is finding y such that ( 16 - y ) ! is a multiple of 16 . that means we need to have 2 ^ 4 in ( 16 - y ) ! 6 ! is the smallest factorial number with 2 ^ 4 as a factor . 16 - y = 6 y = 10 the answer is b ." | a ) 28 % , b ) 4 , c ) 525 , d ) 10 , e ) 30 | d | subtract(16, multiply(const_2, divide(divide(16, const_2), add(const_1, const_4)))) | add(const_1,const_4)|divide(n0,const_2)|divide(#1,#0)|multiply(#2,const_2)|subtract(n0,#3)| | general |
two tests had the same maximum mark . the pass percentages in the first and the second test were 40 % and 45 % respectively . a candidate scored 249 marks in the second test and failed by 66 marks in that test . find the pass mark in the first test ? | "let the maximum mark in each test be m . the candidate failed by 66 marks in the second test . pass mark in the second test = 249 + 66 = 315 45 / 100 m = 315 pass mark in the first test = 40 / 100 m = 40 / 45 * 315 = 280 . answer : d" | a ) 4 / 15 , b ) 280 , c ) 40 , d ) 10 , e ) rs . 1236 | b | add(249, 66) | add(n2,n3)| | gain |
the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 14 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | "sum of 10 numbers = 402 corrected sum of 10 numbers = 402 β 13 + 31 β 14 = 406 hence , new average = 406 β 10 = 40.6 answer c" | a ) 40.6 , b ) 90 % , c ) 32 % , d ) 1076 , e ) 3 | a | divide(subtract(add(multiply(40.2, 10), add(13, 14)), 31), 10) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#2,n4)|divide(#3,n0)| | general |
in 1990 the budgets for projects q and v were $ 500,000 and $ 780,000 , respectively . in each of the next 10 years , the budget for q was increased by $ 30,000 and the budget for v was decreased by $ 10,000 . in which year was the budget for q equal to the budget for v ? | "let the no of years it takes is x . 500 + 30 x = 780 - 10 x - - > 40 x = 280 and x = 7 . thus , it happens in 1997 . e ." | a ) 1997 , b ) 982.14', ' , c ) 54.2 % , d ) 45 , e ) 4 | a | add(1990, multiply(10, multiply(const_2, const_3))) | multiply(const_2,const_3)|multiply(n3,#0)|add(n0,#1)| | general |
how many numbers between 100 and 798 are divisible by 2 , 3 , and 7 together ? | explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 798 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 798 : 798 / 42 = 19 hence , 42 * 19 = 798 is the last number divisible by 42 within 798 hence , total numbers divisible by 2 , 3 , 7 together are ( 19 Γ’ β¬ β 2 ) = 17 answer : d | a ) 20 , b ) rs . 1001 , c ) 447.5 , d ) 6 , e ) 17 | e | subtract(divide(798, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7))) | multiply(n2,n3)|multiply(n4,#0)|divide(n1,#1)|divide(n0,#1)|subtract(#2,#3) | general |
a car traveled 340 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 6 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "let the speed in highway be h mpg and in city be c mpg . h = c + 6 h miles are covered in 1 gallon 340 miles will be covered in 340 / h . similarly c miles are covered in 1 gallon 336 miles will be covered in 336 / c . both should be same ( as car ' s fuel capacity does not change with speed ) = > 336 / c = 340 / h = > 336 / c = 340 / ( c + 6 ) = > 336 c + 336 * 6 = 340 c = > c = 336 * 6 / 4 = 504 answer a ." | a ) 70 mph , b ) 11 , c ) 240 meters , d ) 39 % , e ) 504 | e | divide(336, divide(subtract(340, 336), 6)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics |
1850 men have provisions for 15 days . if 150 more men join them , for how many days will the provisions last now ? | "1850 * 15 = 2000 * x x = 13.9 answer : c" | a ) 15000 , b ) 87 % , c ) 13.9 , d ) 47 , e ) 91000 | c | divide(multiply(15, 1850), add(1850, 150)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
the product of the squares of two positive integers is 100 . how many pairs of positive integers satisfy this condition ? | "ans : c - 2 pairs ( x Λ 2 ) ( y Λ 2 ) = 100 [ square root both sides ] xy = 10 10 = 1 x 10 , 10 x 1 , 2 x 5 , 5 x 2 cancel the repeats this leaves us with exactly 2 options . hence , c" | a ) 2 , b ) 1 / 81 , c ) 0.1388 , d ) 9 , e ) 16.41 | a | subtract(add(const_2, const_3), const_2) | add(const_2,const_3)|subtract(#0,const_2)| | geometry |
a number increased by 30 % gives 780 . the number is | "formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 30 % = 130 % 130 % - - - - - - - > 780 ( 130 Γ 6 = 780 ) 100 % - - - - - - - > 600 ( 100 Γ 6 = 600 ) d )" | a ) 2 , b ) 600 , c ) 32 , d ) 208 , e ) 18 years | b | divide(780, add(const_1, divide(30, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
if a > x > y > z on the number line , y is halfway between x and z , and x is halfway between w and z , then ( y - x ) / ( y - a ) = | let y - z = t - - - > since y is halfway between x and z and x > y we have x - y = t . moreover x - z = ( x - y ) + ( y - z ) = 2 t . similarly since x is halfway between w and z , we have a - x = 2 t . so y - x = - t , y - a = - 3 t . - - - > ( y - x ) / ( y - a ) = 1 / 3 . the answer is ( b ) . | a ) 39.8 , b ) 0.4 , c ) rs . 600 , d ) 6 , e ) 1 / 3 | e | divide(const_1, subtract(add(const_2, const_2), const_1)) | add(const_2,const_2)|subtract(#0,const_1)|divide(const_1,#1) | general |
3 friends james , david and charlie divide $ 1230 amongs them in such a way that if $ 5 , $ 10 and $ 15 are removed from the sums that james , david and charlie received respectively , then the share of the sums that they got will be in the ratio of 9 : 10 : 11 . how much did charlie receive ? | a + b + c = 1230 given ratio 9 : 10 : 11 let us say the shares of a , b , c deducting 5 , 1015 be a , b , c a + b + c = 1230 - 30 = 1200 = 30 k c share = ( 1200 x 30 ) / 60 = 600 c = charlie share = 600 + 15 = 615 option e | a ) 12 , b ) $ 325 , c ) 2 , d ) $ 615 , e ) 18 : 5 | d | add(add(add(add(add(multiply(11, divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11))), 15), divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11))), divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11))), divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11))), divide(subtract(1230, add(add(5, 10), 15)), add(add(9, 10), 11))) | add(n2,n3)|add(n3,n5)|add(n4,#0)|add(n7,#1)|subtract(n1,#2)|divide(#4,#3)|multiply(n7,#5)|add(n4,#6)|add(#7,#5)|add(#8,#5)|add(#9,#5)|add(#10,#5) | general |
if 2 + 3 = 31 ; 3 + 5 = 134 ; 5 + 7 = 368 then 7 + 11 = ? | "2 ^ 2 + 3 ^ 3 = 4 + 27 = 31 3 ^ 2 + 5 ^ 3 = 9 + 125 = 134 5 ^ 2 + 7 ^ 3 = 25 + 343 = 368 and 7 ^ 2 + 11 ^ 3 = 49 + 1331 = 1380 answer : d" | a ) 6 , b ) 18 , c ) 24 , d ) 9 , e ) 1380 | e | add(power(7, 3), power(11, 2)) | power(n9,n3)|power(n10,n0)|add(#0,#1)| | general |
a rectangular grass field is 70 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 70 + 55 + 2.5 * 2 ) 2 * 2.5 = > 650 650 * 2 = rs . 1300 answer : b" | a ) 11 , b ) 7 , c ) 36 , d ) 75 , e ) s . 1300 | e | multiply(subtract(rectangle_area(add(70, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(70, 55)), 2) | multiply(n2,n3)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5)| | geometry |
calculate the area of a triangle , if the sides of are 39 cm , 36 cm and 15 cm , what is its area ? | "the triangle with sides 39 cm , 36 cm and 15 is right angled , where the hypotenuse is 39 cm . area of the triangle = 1 / 2 * 36 * 15 = 270 cm 2 answer : e" | a ) 135 , b ) 100 , c ) 15 % , d ) 7 , e ) 270 cm 2 | e | multiply(divide(36, const_2), 15) | divide(n1,const_2)|multiply(n2,#0)| | geometry |