Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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the captain of a cricket team of 11 members is 24 years old and the wicket keeper is 7 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 24 + 31 ) = 9 ( x - 1 ) 11 x - 9 x = 46 2 x = 46 x = 23 . the average age of the team is 23 years . the answer is c ." | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | c | divide(subtract(add(24, add(24, 7)), multiply(const_3.0, const_3.0)), const_2) | add(n1,n2)|multiply(const_3.0,const_3.0)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
in a box of 16 pens , a total of 3 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ? | "method - 1 there are 9 fine pieces of pen and 3 defective in a lot of 16 pens i . e . probability of first pen not being defective = ( 13 / 16 ) i . e . probability of second pen not being defective = ( 12 / 15 ) [ 15 pen remaining with 12 defective remaining considering that first was defective ] probability of both ... | a ) 1 / 6 , b ) 2 / 9 , c ) 6 / 20 , d ) 13 / 20 , e ) 3 / 4 | d | multiply(divide(subtract(16, 3), 16), divide(subtract(subtract(16, 3), const_1), subtract(16, const_1))) | subtract(n0,n1)|subtract(n0,const_1)|divide(#0,n0)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)| | general |
during a pizza buffet where a eats more times 2.6 than b , and b eats 4 times less than c . find the least number of times all the 3 has to eat | a eats more than b if b eats 1 times than the ratio of a and b is a : b is 2.6 : 1 or 13 : 5 and as b eat 4 times less the c the the ratio of b : c is 5 : 20 the the least number of times all three has eat is the lcm of a , b , c that is 260 . . answer : b | a ) 250 , b ) 260 , c ) 270 , d ) 280 , e ) 285 | b | lcm(divide(multiply(2.6, const_10), const_2), divide(multiply(4, const_10), const_2)) | multiply(n0,const_10)|multiply(n1,const_10)|divide(#0,const_2)|divide(#1,const_2)|lcm(#2,#3) | general |
if the mean of numbers 28 , x , 42 , 78 and 104 is 62 , what is the mean of 48 , 62 , 98 , 124 and x ? | "x is common to both the series . so , x is not going to make a difference to the average . only the remaining 4 numbers will contribute to the difference in average between the two series . sum of the 4 numbers , excluding x , of the first series is 28 + 42 + 78 + 104 = 252 sum of the 4 numbers , excluding x , of the ... | a ) 78 , b ) 58 , c ) 390 , d ) 310 , e ) 66 | a | divide(add(add(add(add(subtract(multiply(62, add(const_4, const_1)), add(add(add(28, 42), 78), 104)), 48), 62), 98), 124), add(const_4, const_1)) | add(const_1,const_4)|add(n0,n1)|add(n2,#1)|multiply(n4,#0)|add(n3,#2)|subtract(#3,#4)|add(n5,#5)|add(n6,#6)|add(n7,#7)|add(n8,#8)|divide(#9,#0)| | general |
company p had 15 percent more employees in december than it had in january . if company p had 450 employees in december , how many employees did it have in january ? | "d = number of employees in december j = number of employees in january j x 1.15 = d j x 1.15 = 450 j = 450 / 1.15 j = 45,000 / 115 = 392 thus b is the correct answer ." | a ) 391 , b ) 392 , c ) 410 , d ) 423 , e ) 445 | b | divide(450, add(const_1, divide(15, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
salley ' s internet provider offers unlimited traffic which costs $ 0.5 per day charged off at 12 a . m . the service is discontinued when client ' s debt exceeds $ 5 . if today in the morning salley makes a payment of $ 7 , for how many days , including today , will she be connected to the internet without any additio... | day 0 - payment of $ 7 is made in the morning ( after 12 am midnight ) day 1 onwards ( after 12 am midnight ) internet service will begin and will run for 7 * 2 = 14 days i . e . at the end of day 15 balance will be zero . now salley can still use the internet till her debt reaches $ 5 which means total of 5 * 2 = 10 d... | a ) 15 , b ) 24 , c ) 25 , d ) 26 , e ) 13 | c | divide(add(7, 5), 0.5) | add(n2,n3)|divide(#0,n0) | physics |
a salesman â € ™ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 700 more than that by the previous schema , his sales were worth ? | [ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 700 x = 8000 answer a | a ) s . 8,000 , b ) s . 9,000 , c ) s . 20,000 , d ) s . 10,000 , e ) s . 50,000 | a | subtract(multiply(5, const_4), const_12) | multiply(n0,const_4)|subtract(#0,const_12) | general |
5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 150 only . men â € ™ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 150 rs . 5 m + 5 m + 5 m - - - - - 150 rs . 15 m - - - - - - 150 rs . = > 1 m = 10 rs . answer : e" | a ) 6 rs , b ) 2 rs , c ) 4 rs , d ) 9 rs , e ) 10 rs | e | divide(150, multiply(const_3, 5)) | multiply(n0,const_3)|divide(n2,#0)| | general |
if a : b : : 3 : 4 , then what is ( 5 a + 6 b ) : ( a - 2 b ) ? | "a / b = 3 / 4 dividing numerator & denominator of ' ( 5 a + 6 b ) / ( a - 2 b ) ' by b , [ 5 ( a / b ) + 6 ] / [ ( a / b ) - 2 ] = [ 5 * ( 3 / 4 ) + 6 ] / [ ( 3 / 4 ) - 2 ] = - 39 / 5 answer : d" | a ) 57 : 11 , b ) - 57 : 11 , c ) 11 : 10 , d ) - 39 : 5 , e ) - 1 : 10 | d | divide(add(multiply(5, 3), multiply(4, 6)), subtract(3, multiply(2, 4))) | multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|subtract(n0,#2)|divide(#3,#4)| | general |
a began business with rs . 45000 and was joined afterwards by b with rs . 36000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ? | "45 * 12 : 36 * x = 2 : 1 x = 7.5 12 - 7.5 = 4.5 answer : c" | a ) 1.5 , b ) 6 , c ) 7.5 , d ) 8.5 , e ) 2 | c | subtract(multiply(const_4, const_3), divide(divide(multiply(45000, multiply(const_4, const_3)), 36000), 2)) | multiply(const_3,const_4)|multiply(n0,#0)|divide(#1,n1)|divide(#2,n2)|subtract(#0,#3)| | other |
in the xy - coordinate system , what is the slope of the line that goes through point ( 2 , 4 ) and is equidistant from the two points p = ( 9 , 3 ) and q = ( 5 , 9 ) ? | "first , get the middle coordinate between ( 9,3 ) and ( 5,9 ) . x = 9 + ( 5 - 9 ) / 2 = 7 y = 3 + ( 9 - 3 ) / 2 = 6 second , get the slope of ( 7,6 ) and ( 2,4 ) . m = 6 - 4 / 7 - 2 = 2 / 5 = 0.4 answer : d" | a ) 0.1 , b ) 0.2 , c ) 0.3 , d ) 0.4 , e ) 0.5 | d | divide(divide(add(4, 3), 3), divide(add(const_4.0, 4), const_2)) | add(n1,n3)|add(n0,n2)|divide(#0,n3)|divide(#1,const_2)|divide(#2,#3)| | general |
two trains of length 160 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 160 + 280 = 440 m . the time required = d / s = 440 / 20 = 22 sec . answer : e" | a ) 28 , b ) 266 , c ) 990 , d ) 20 , e ) 22 | e | divide(add(160, 280), multiply(add(42, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
the parameter of a square is equal to the perimeter of a rectangle of length 20 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | "let the side of the square be a cm . parameter of the rectangle = 2 ( 20 + 14 ) = 68 cm parameter of the square = 68 cm i . e . 4 a = 68 a = 17 diameter of the semicircle = 17 cm circimference of the semicircle = 1 / 2 ( ∏ ) ( 17 ) = 1 / 2 ( 22 / 7 ) ( 17 ) = 374 / 14 = 26.71 cm to two decimal places answer : c" | a ) 77.14 cm , b ) 47.14 cm , c ) 26.71 cm , d ) 94.94 cm , e ) 23.57 cm | c | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(20, 14)), const_2)), const_2) | rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)| | geometry |
if x > 0 , x / 10 + x / 25 is what percent of x ? | "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 10 + 100 / 25 = 10 + 4 = 14 14 is 14 % of 100 = b" | a ) 6 % , b ) 14 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | b | multiply(add(divide(const_1, 10), divide(const_1, 25)), const_100) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|multiply(#2,const_100)| | general |
evaluate : 30 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 30 - 12 * 3 * 2 = 30 - 8 = 22 correct answer e" | a ) 62 , b ) 52 , c ) 32 , d ) 12 , e ) 22 | e | subtract(30, multiply(multiply(12, const_2.0), 2)) | multiply(n1,const_2.0)|multiply(n3,#0)|subtract(n0,#1)| | general |
one day , raviraj left home and cycled 20 km southwards , turned right and cycled 10 km and turned right and cycled 20 km and turned left and cycled 20 km . how many kilometres will he have to cycle to reach his home straight ? | raviraj starts from home at a , moves 20 km in south upto b . then he turns right and moves 10 km upto c , then he turns right and moves 20 km upto d , then he turns lefts and moves 20 km upto e . so from image it is clear that , if he moves straight then he will have to move ad + de , ad = bc = 10 km so , he will have... | a ) 50 , b ) 30 , c ) 40 , d ) 60 , e ) 80 | b | add(20, 10) | add(n0,n1) | physics |
a train running at a speed of 36 km / h passes an electric pole in 14 seconds . in how many seconds will the whole train pass a 370 - meter long platform ? | let the length of the train be x meters . when a train crosses an electric pole , the distance covered is its own length x . speed = 36 km / h = 36000 m / 3600 s = 10 m / s x = 14 * 10 = 140 m . the time taken to pass the platform = ( 140 + 370 ) / 10 = 51 seconds the answer is d . | a ) 45 , b ) 47 , c ) 49 , d ) 51 , e ) 53 | d | divide(add(multiply(multiply(36, const_0_2778), 14), 370), multiply(36, const_0_2778)) | multiply(n0,const_0_2778)|multiply(n1,#0)|add(n2,#1)|divide(#2,#0) | physics |
a train is 360 meter long is running at a speed of 50 km / hour . in what time will it pass a bridge of 140 meter length . | "explanation : speed = 50 km / hr = 50 * ( 5 / 18 ) m / sec = 125 / 9 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 / 125 / 9 = 36 seconds answer : b" | a ) 26 seconds , b ) 36 seconds , c ) 46 seconds , d ) 56 seconds , e ) none of these | b | divide(add(360, 140), divide(multiply(50, const_1000), const_3600)) | add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)| | physics |
if 50 % of x is 30 less than 25 % of 1500 , then x is ? | "50 % of x = x / 2 ; 25 % of 1500 = 25 / 100 * 1500 = 375 given that , x / 2 = 375 - 30 = > x / 2 = 345 = > x = 690 . answer : a" | a ) 690 , b ) 860 , c ) 560 , d ) 875 , e ) 880 | a | divide(subtract(multiply(1500, divide(25, const_100)), 30), divide(50, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#2,n1)|divide(#3,#1)| | general |
an investor can sell her microtron stock for 36 $ per share and her dynaco stock for 44 $ per share , if she sells 300 shares altogether , some of each stock , at an average price per share of 40 $ , how many shares of dynaco stock has she sold ? | w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) = ( 44 - 40 ) / ( 40 - 36 ) = 4 / 4 = 1 / 1 = number of microtron stocks / number of dynaco stocks so for every 1 microtron stock , she sold 1 dynaco stock . so out of 300 total stocks , ( 1 / 1 ) th i . e . 300 / 2 = 150 must be dynaco stock . answer ( e ) | a ) 52 , b ) 75 , c ) 92 , d ) 136 , e ) 150 | e | divide(multiply(300, divide(40, subtract(44, 36))), divide(add(36, 44), subtract(44, 36))) | add(n0,n1)|subtract(n1,n0)|divide(n3,#1)|divide(#0,#1)|multiply(n2,#2)|divide(#4,#3) | general |
a man is 35 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "let ' s son age is x , then father age is x + 35 . = > 2 ( x + 2 ) = ( x + 35 + 2 ) = > 2 x + 4 = x + 37 = > x = 33 years answer : c" | a ) 78 years , b ) 22 years , c ) 33 years , d ) 66 years , e ) 65 years | c | divide(subtract(35, 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 jogger running at 9 km / hr along side a railway track is 120 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 = 120 + 120 = 360 m . time taken = 240 / 10 = 24 sec . answer : option d" | a ) 89 , b ) 20 , c ) 36 , d ) 24 , e ) 34 | d | divide(add(120, 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 |
the width of a rectangular hall is ½ of its length . if the area of the hall is 200 sq . m , what is the difference between its length and breadth ? | let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 200 = x * 1 x / 2 x ² = 400 x = 20 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 20 / 2 = 10 m answer : b | ['a ) 8 m', 'b ) 10 m', 'c ) 12 m', 'd ) 15 m', 'e ) 17 m'] | b | divide(sqrt(divide(200, divide(const_1, const_2))), const_2) | divide(const_1,const_2)|divide(n0,#0)|sqrt(#1)|divide(#2,const_2) | geometry |
a work which could be finished in 10 days was finished 3 days earlier after 10 more men joined . the number of men employed was ? | "x - - - - - - - 10 ( x + 10 ) - - - - 6 x * 10 = ( x + 10 ) 6 x = 15 \ answer : a" | a ) 15 , b ) 20 , c ) 88 , d ) 71 , e ) 11 | a | divide(multiply(multiply(3, const_2), 10), subtract(10, multiply(3, const_2))) | multiply(n1,const_2)|multiply(n2,#0)|subtract(n0,#0)|divide(#1,#2)| | physics |
in a dairy farm , 30 cows eat 30 bags of husk in 30 days . in how many days one cow will eat one bag of husk ? | "explanation : one bag of husk = 30 cows per day ⇒ 30 × 1 × 30 = 1 × 30 × x for one cow = 30 days answer : e" | a ) 1 , b ) 40 , c ) 20 , d ) 26 , e ) 30 | e | multiply(divide(30, 30), 30) | divide(n0,n0)|multiply(n0,#0)| | physics |
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 56 hours . if harry and anni... | "annie earned 40 x + 16 ( 2 x ) = 72 x let h be the number of hours that harry worked . harry earned 30 x + 1.5 x ( h - 30 ) = 72 x ( 1.5 x ) ( h ) = 87 x h = 58 hours the answer is d ." | a ) 52 , b ) 54 , c ) 56 , d ) 58 , e ) 60 | d | add(divide(subtract(add(40, 2), 30), 1.5), 30) | add(n2,n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)| | general |
car dealership x , the total profit from sales increased by 9 percent over the previous years , while the number of cars sold decreased by 9 percent over the previous years . approximately what was the average percent increase in profit per car over the previous year . | "let profit be x and cars sold be y originally profit / car = x / y now it is 1.09 x / 0.91 y = 109 / 91 ( x / y ) increase in profit per car = ( ( 18 / 91 ) ( x / y ) ) / ( x / y ) * 100 = 19.78 % answer = b" | a ) 18 % , b ) 19.78 % , c ) 22 % , d ) 23 % , e ) 25 % | b | multiply(divide(subtract(divide(add(const_100, 9), subtract(const_10, const_3)), divide(const_100, const_10)), divide(const_100, const_10)), const_100) | add(n0,const_100)|divide(const_100,const_10)|subtract(const_10,const_3)|divide(#0,#2)|subtract(#3,#1)|divide(#4,#1)|multiply(#5,const_100)| | general |
when positive integer n is divided by 2 , the remainder is 2 . when n is divided by 7 , the remainder is 5 . how many values less than 100 can n take ? | "a quick approac to this q is . . the equation we can form is . . 3 x + 2 = 7 y + 5 . . 3 x - 3 = 7 y . . . 3 ( x - 1 ) = 7 y . . . so ( x - 1 ) has to be a multiple of 7 as y then will take values of multiple of 3 . . here we can see x can be 1 , 8,15 , 22,29 so 5 values till 100 is reached as ( 29 - 1 ) * 3 = 84 and ... | a ) 0 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(100, reminder(5, 7)) | reminder(n3,n2)|subtract(n4,#0)| | general |
two goods trains each 250 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 250 + 250 = 500 m . required time = 500 * 6 / 125 = 24 sec . answer : option a" | a ) 24 , b ) 45 , c ) 48 , d ) 51 , e ) 44 | a | add(45, 30) | add(n1,n2)| | physics |
if x / y = 2 / 7 , then ( x + y ) / ( x - y ) = ? | "any x and y satisfying x / y = 2 / 7 should give the same value for ( x + y ) / ( x - y ) . say x = 2 and y = 7 , then ( x + y ) / ( x - y ) = ( 2 + 7 ) / ( 2 - 7 ) = - 9 / 5 . answer : d ." | a ) 5 , b ) 1 / 5 , c ) - 1 / 6 , d ) - 9 / 5 , e ) - 5 | d | divide(add(2, 7), subtract(2, 7)) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | general |
two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 70 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 70 + 30 = 100 km / hr . 100 * 5 / 18 = 250 / 9 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 9 / 250 = 36 sec answer : b" | a ) 22 , b ) 36 , c ) 48 , d ) 99 , e ) 23 | b | add(70, 30) | add(n1,n2)| | physics |
p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 140 days , in how many days can r alone complete the remaining work ? | "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 25... | a ) 50 days , b ) 102.7 days , c ) 100 days , d ) 150 days , e ) 80 days | b | subtract(multiply(const_4, 50), multiply(divide(140, const_100), 60)) | divide(n4,const_100)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#1,#2)| | physics |
a train 560 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 560 / 16 * 18 / 5 = 126 kmph answer : a" | a ) 126 kmph , b ) 77 kmph , c ) 54 kmph , d ) 71 kmph , e ) 88 kmph | a | multiply(const_3_6, divide(560, 16)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
a man invests rs . 3,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | "explanation : interest on rs . 3000 at 5 % per annum = ( 3000 × 5 × 1 ) / 100 = rs . 150 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x × 8 × 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 3000 + x ) × 6 × 1 / 100 . = > 150 + 2 x / 25 = ( 3000 + x ) × 6 × ... | a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 2000 , e ) none of these | c | multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6))) | multiply(n2,const_1000)|subtract(n3,n1)|subtract(n2,n3)|divide(#1,#2)|multiply(#3,#0)| | gain |
k is 50 % faster than l . if l starts at 9 a . m . and k starts at 10 a . m . l travels at a speed of 50 km / hr . if l and k are 300 kms apart , the time when they meet when they travel in opposite direction is ? | explanation : let ' t ' be the time after which they met since l starts . given k is 50 % faster than l 50 t + 1.5 * 50 ( t - 1 ) = 300 50 t + 75 t = 300 + 75 t = 375 / 125 = 3 hrs past the time that l starts so they meet at ( 9 + 3 ) hrs = 12 : 00 noon . answer : a | a ) 12 pm , b ) 22 pm , c ) 72 pm , d ) 92 pm , e ) 23 pm | a | add(divide(subtract(300, 50), add(multiply(add(const_1, divide(50, const_100)), 50), 50)), 10) | divide(n0,const_100)|subtract(n4,n0)|add(#0,const_1)|multiply(n0,#2)|add(n0,#3)|divide(#1,#4)|add(n2,#5) | physics |
how many prime numbers between 1 and 100 are factors of 10010 ? | "factor of 10010 = 2 * 5 * 7 * 11 * 13 - - - 5 prime numbers a" | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | a | multiply(const_4, 1) | multiply(n0,const_4)| | other |
one lady has 2 children , one of her child is boy , what is the probability of having both are boys | n ( s ) = one of her child is boy out of 2 children = bb , bg = 2 n ( e ) = both are boys = bb = 1 so p = n ( e ) / n ( s ) = 1 / 2 answer : a | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 3 / 4 | a | divide(const_1, 2) | divide(const_1,n0) | other |
7 / 10 of the population of the country of venezia lives in montague province , while the rest lives in capulet province . in the upcoming election , 80 % of montague residents support romeo , while 70 % of capulet residents support juliet ; each resident of venezia supports exactly one of these two candidates . rounde... | "total population = 100 ( assume ) . 7 / 10 * 100 = 70 people from montague . 3 / 10 * 100 = 30 people from capulet . 0.2 * 70 = 14 people from montague support juliet . 0.7 * 30 = 21 people from capulet support juliet . the probability that a juliet supporter chosen at random resides in capulet is 21 / ( 14 + 21 ) = ~... | a ) 28 % , b ) 41 % , c ) 60 % , d ) 72 % , e ) 78 % | c | multiply(divide(multiply(divide(70, const_100), divide(subtract(10, 7), 10)), add(multiply(divide(subtract(const_100, 80), const_100), divide(7, 10)), multiply(divide(70, const_100), divide(subtract(10, 7), 10)))), const_100) | divide(n3,const_100)|divide(n0,n1)|subtract(n1,n0)|subtract(const_100,n2)|divide(#2,n1)|divide(#3,const_100)|multiply(#0,#4)|multiply(#5,#1)|add(#7,#6)|divide(#6,#8)|multiply(#9,const_100)| | gain |
car a runs at the speed of 58 km / hr & reaches its destination in 8 hr . car b runs at the speed of 70 km / h & reaches its destination in 4 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 58 ã — 8 = 464 km distance travelled by car b = 70 ã — 4 = 280 km ratio = 464 / 280 = 29 : 18 c" | a ) 11 : 6 , b ) 12 : 7 , c ) 29 : 18 , d ) 15 : 6 , e ) 13 : 6 | c | divide(multiply(58, 8), multiply(70, 4)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum y of all the even integers between 99 and 301 ? | "i would solve it in a different way . first of all , total number of even integers between 99 and 301 are , ( 301 - 99 ) / 2 = 202 / 2 = 101 average = ( 301 + 99 ) / 2 = 400 / 2 = 200 sum y = average * total count = 200 * 101 = 20,200 answer is b" | a ) 10,100 , b ) 20,200 , c ) 22,650 , d ) 40,200 , e ) 45,150 | b | add(divide(subtract(subtract(301, 1), add(99, 1)), 2), 1) | add(n2,n0)|subtract(n3,n0)|subtract(#1,#0)|divide(#2,n1)|add(n0,#3)| | general |
in what ratio p : q should the mixture p of milk and water in the ratio of 4 : 1 be mixed with another mixture q of milk and water in the ratio 2 : 3 so that the resultant mixture contains equal quantities of milk and water ? | "( 4 / 5 ) * p + ( 2 / 5 ) * q = ( 1 / 5 ) * p + ( 3 / 5 ) * q 3 p = q p / q = 1 / 3 the answer is a ." | a ) 1 : 3 , b ) 2 : 3 , c ) 1 : 4 , d ) 3 : 4 , e ) 3 : 5 | a | add(subtract(add(1, 3), add(4, 2)), const_1) | add(n1,n3)|add(n0,n2)|subtract(#0,#1)|add(#2,const_1)| | general |
if n is an integer and 101 n ^ 2 is less than or equal to 12100 , what is the greatest possible value of n ? | "101 * n ^ 2 < = 12100 n ^ 2 < = 12100 / 101 which will be less than 121 since 12100 / 100 = 121 which is the square of 9 next closest value of n where n ^ 2 < = 121 is 10 ans d" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | d | floor(sqrt(divide(12100, 101))) | divide(n2,n0)|sqrt(#0)|floor(#1)| | general |
how many minutes is it before 12 noon if 20 minutes ago it was 3 times as many minutes after 9 am ? | a 40 minutes , or 11.20 am | a ) 40 , b ) 39 , c ) 42 , d ) 55 , e ) 29 | a | divide(subtract(subtract(multiply(12, const_60), multiply(9, const_60)), 20), add(3, const_1)) | add(n2,const_1)|multiply(n0,const_60)|multiply(n3,const_60)|subtract(#1,#2)|subtract(#3,n1)|divide(#4,#0) | physics |
a bag holds 4 red marbles and 5 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be red ? | "probability of selecting first red marble = 4 / 9 probability of selecting second red marble without replacement = 3 / 8 final probability = 4 / 9 * 3 / 8 = 1 / 6 the correct answer is b ." | a ) 1 / 10 , b ) 1 / 6 , c ) 3 / 10 , d ) 2 / 5 , e ) 1 / 4 | b | divide(choose(const_2.0, const_2.0), choose(add(4, 5), 4)) | add(const_2.0,const_3.0)|choose(n0,n0)|choose(#0,n0)|divide(#1,#2)| | other |
the sum of four consecutive even numbers is 140 . what would be the smallest number ? | "let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 140 = > x = 18 smallest number is : 2 ( x - 2 ) = 32 answer : a" | a ) 32 , b ) 88 , c ) 70 , d ) 123 , e ) 12 | a | add(add(power(add(add(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), power(... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
a number is doubled and 15 is added . if the resultant is multiplied by 5 , it becomes 125 . what is that number ? | "solution let the number be x . then , 5 ( 2 x + 15 ) ‹ = › 125 ‹ = › 2 x + 15 = 25 ‹ = › 2 x = 10 x = 5 . answer d" | a ) 5 , b ) 10 , c ) 8 , d ) none of these , e ) can not be determined | d | divide(subtract(5, multiply(const_3, 15)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
a train is 360 meter long is running at a speed of 75 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 75 km / hr = 75 * ( 5 / 18 ) m / sec = 125 / 6 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 * ( 6 / 125 ) = 24 seconds answer : e" | a ) 87 , b ) 69 , c ) 40 , d ) 72 , e ) 24 | e | divide(add(360, 140), divide(multiply(75, const_1000), const_3600)) | add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)| | physics |
if k ^ 3 is divisible by 120 , what is the least possible value of integer k ? | "k ^ 3 = 120 * x = 2 ^ 3 * 3 * 5 * x the factors of k must at minimum include 2 * 3 * 5 = 30 . the answer is b ." | a ) 15 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | b | divide(divide(120, const_2), const_2) | divide(n1,const_2)|divide(#0,const_2)| | general |
population of a city in 20004 was 1000000 . if in 2005 there isan increment of 20 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city atthe end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 20 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1131000 a" | a ) 1131000 , b ) 979923 , c ) 980241 , d ) 2356677 , e ) 1083875 | a | multiply(1000000, multiply(multiply(add(const_1, divide(20, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6)| | gain |
sony and johnny caught 40 fishes . sony caught 4 times as many as johnny . how many fishes did johnny catch ? | s + j = 40 s = 4 * j j = 8 answer : e | a ) 12 , b ) 16 , c ) 34 , d ) 38 , e ) 8 | e | divide(40, add(4, const_1)) | add(n1,const_1)|divide(n0,#0) | general |
two digits in brother ' s age are the same as the digit in the sister ' s age , but in reverse order . in 20 9 years brother will be twice as old as sister will be then . what is the difference in their current age ? | brother ' s age = 10 x + y so sister ' s age = 10 y + x . . after 29 years , 10 x + y + 29 = 2 * ( 10 y + x + 29 ) . . . . so 29 + 19 y = 8 x . . . check for odd values of y , y = 1 satisfies the eqn with x = 6 . . . so ages are 61 and 16 and ans as found correctly by u is e . . 45 | a ) 34 , b ) 50 , c ) 32 , d ) 28 , e ) 45 | e | add(add(20, 20), subtract(9, const_4)) | add(n0,n0)|subtract(n1,const_4)|add(#0,#1) | general |
a man owns 1 / 3 of market reserch beauro buzness , and sells 3 / 5 of his shares for 2000 rs , what is the value of buzness ? | "if value of business = x total sell ( 1 x / 3 ) ( 3 / 5 ) = 2000 - > x = 10000 answer : c" | a ) 40000 , b ) 60000 , c ) 10000 , d ) 25000 , e ) 50000 | c | divide(2000, multiply(divide(1, 3), divide(3, 5))) | divide(n0,n1)|divide(n1,n3)|multiply(#0,#1)|divide(n4,#2)| | general |
a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 100 that a person who randomly guesses the answer to each question will be a winner ? | "a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 1000 that a person who randomly guesses the answer to each question will be a winner ? a . ... | a ) 5 , b ) 10 , c ) 50 , d ) 100 , e ) 1000 | c | multiply(const_1000, divide(1, 100)) | divide(n0,n1)|multiply(#0,const_1000)| | general |
15 binders can bind 1400 books in 21 days . how many binders will be required to bind 1600 books in 20 days ? | "binders books days 15 1400 21 x 1600 20 x / 15 = ( 1600 / 1400 ) * ( 21 / 20 ) = > x = 18 answer : b" | a ) 14 , b ) 18 , c ) 24 , d ) 28 , e ) none of these | b | divide(divide(1600, 20), divide(1400, multiply(15, 21))) | divide(n3,n4)|multiply(n0,n2)|divide(n1,#1)|divide(#0,#2)| | physics |
what is the smallest number which when increased by 3 is divisible by 18 , 70 , 100 and 84 ? | "when increased by 3 , the number must include at least 2 ^ 2 * 3 ^ 2 * 5 ^ 2 * 7 = 6300 the answer is d ." | a ) 5307 , b ) 5647 , c ) 5927 , d ) 6297 , e ) 6557 | d | add(lcm(lcm(18, 70), lcm(100, 84)), 3) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|add(n0,#2)| | general |
find the compound ratio of ( 1 : 2 ) , ( 1 : 3 ) and ( 3 : 5 ) is | "required ratio = 1 / 2 * 1 / 3 * 3 / 5 = 1 / 10 = 1 : 10 answer is a" | a ) 1 : 10 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 3 : 2 | a | multiply(divide(1, 2), multiply(divide(1, 2), divide(1, 2))) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2)| | other |
the arithmetic mean and standard deviation of a certain normal distribution are 16.5 and 1.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | "the value which isexactlytwo sd less than the mean is : mean - 2 * sd = 16.5 - 2 * 1.5 = 13.5 . answer : e ." | a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 13.5 | e | subtract(16.5, multiply(2, 1.5)) | multiply(n1,n2)|subtract(n0,#0)| | general |
find the value of 3 / 11 × [ ( 1 + 1 / 3 ) ( 1 + 1 / 4 ) … … … … . ( 1 + 1 / 120 ) ] is ? | "3 / 11 × [ ( 1 + 1 / 3 ) ( 1 + 1 / 4 ) . . . . . . . ( 1 + 1 / 120 ) ] 3 / 11 × [ 4 / 3 + 5 / 4 × . . . . . . . . 121 / 120 ] 3 / 11 × 4 / 3 × 5 / 4 × . . . . 121 / 120 121 / 11 = 11 . a )" | a ) 11 , b ) 12 , c ) 14 , d ) 15 , e ) 16 | a | divide(add(120, 1), 11) | add(n2,n10)|divide(#0,n1)| | general |
two cars are driving towards one another . the first car is traveling at a speed of 120 km / h , which is 28 % faster than the second car . if the distance between the cars is 855 km , how long will it takes the cars to meet ( in hours ) ? | let s be the speed of the slower car . then the speed of the faster is 1.28 s 1.28 s = 120 s = 120 / 1.28 cumulative speed / rate = s + 1.28 s = 2.28 s time taken to meet = 855 / 2.28 s = 855 / 213.75 = 4 hrs answer : d | a ) 2.5 . , b ) 3 . , c ) 3.5 , d ) 4 . , e ) 4.5 . | d | divide(855, add(120, subtract(120, multiply(divide(120, const_100), 28)))) | divide(n0,const_100)|multiply(n1,#0)|subtract(n0,#1)|add(n0,#2)|divide(n2,#3) | physics |
according to the directions on a packet of smoothie mix , 1 3 - ounce packet of smoothie mix is to be combined with 10 ounces of water to make a smoothie . how many 3 - ounce packets of smoothie mix are required to prepare 150 12 - ounce smoothies ? | "this question was n ' t particularly grueling , but i think it ' s the first where i had the opportunity to solve it via theory andinspectionthat many on this board suggest as strategy on the gmat . it actually came to me by accident . basically , if we thought that the 3 packets of powder were included in the 12 ounc... | a ) 120 , b ) 150 , c ) 180 , d ) 240 , e ) 600 | b | add(150, multiply(3, const_10)) | multiply(n1,const_10)|add(n4,#0)| | general |
which fraction should be added to : 1 / ( 2 ã — 3 ) + 1 / ( 3 ã — 4 ) + 1 / ( 4 ã — 5 ) + . . . . . . . + 1 / ( 23 ã — 24 ) so that the result is equal to 1 ? | the given expression = 1 / ( 2 ã — 3 ) + 1 / ( 3 ã — 4 ) + 1 / ( 4 ã — 5 ) + . . . . . . + 1 / ( 23 ã — 24 ) = ( 1 / 2 - 1 / 3 ) + ( 1 / 3 - 1 / 4 ) + ( 1 / 4 - 1 / 5 ) + . . . . . . . . . . + ( 1 / 23 - 1 / 24 ) = 1 / 2 - 1 / 24 = 11 / 24 thus , the fraction we should add = 1 - 11 / 24 = 13 / 24 the answer is c . | a ) 7 / 24 , b ) 11 / 24 , c ) 13 / 24 , d ) 17 / 24 , e ) 19 / 24 | c | divide(add(3, const_10), 24) | add(n2,const_10)|divide(#0,n11) | general |
of the land owned by a farmer , 90 percent was cleared for planting . of the cleared land , 80 percent was planted with barley and 10 percent of the cleared land was planted with potato . if the remaining 90 acres of cleared land was planted with tomato , how many acres did the farmer own ? | 80 % of 90 % = 72 % , 10 % 0 f 90 % = 9 % so the remaining 90 - 72 - 9 = 9 % = 90 acres or 10 % of 90 % = 9 % - - > 90 / 9 * 100 = 1000 acres answer ( e ) | a ) 850 , b ) 950 , c ) 1025 , d ) 975 , e ) 1000 | e | divide(90, multiply(divide(subtract(subtract(const_100, 80), 10), const_100), divide(90, const_100))) | divide(n0,const_100)|subtract(const_100,n1)|subtract(#1,n2)|divide(#2,const_100)|multiply(#3,#0)|divide(n3,#4)| | gain |
what is the remainder when 14,452 × 15,652 × 16,781 is divided by 5 ? | "only the unit ' s digit of the product will decide the remainder when divided by 5 . hence , 2 * 2 * 1 = will give units digit as 4 so , whatever be the number , if it ends in 4 , the remainder after dividing with 5 will be 4 . optione" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | reminder(multiply(15,652, 14,452), 16,781) | multiply(n0,n1)|reminder(#0,n2)| | general |
if 4 xz + yw = 6 and xw + yz = 12 , what is the value of the expression ( 2 x + y ) ( 2 z + w ) ? | "( 2 x + y ) * ( 2 z + w ) = 6 + 2 ( 12 ) = 30 answer : b" | a ) 9 . , b ) 30 . , c ) 15 . , d ) 16 . , e ) 18 . | b | add(4, multiply(12, 2)) | multiply(n2,n3)|add(n0,#0)| | general |
if one positive integer is greater than another positive integer by 3 , and the difference of their cubes is 63 , what is their sum ? | 1 ^ 3 = 1 2 ^ 3 = 8 3 ^ 3 = 27 4 ^ 3 = 64 the two numbers are 1 and 4 . the answer is a . | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | a | add(divide(add(power(3, const_3), sqrt(add(power(power(3, const_3), const_2), power(subtract(63, power(3, const_3)), const_2)))), multiply(power(3, const_2), const_2)), subtract(divide(add(power(3, const_3), sqrt(add(power(power(3, const_3), const_2), power(subtract(63, power(3, const_3)), const_2)))), multiply(power(3... | power(n0,const_3)|power(n0,const_2)|multiply(#1,const_2)|power(#0,const_2)|subtract(n1,#0)|power(#4,const_2)|add(#3,#5)|sqrt(#6)|add(#0,#7)|divide(#8,#2)|subtract(#9,n0)|add(#9,#10) | general |
in a recent election , geoff received 0.5 percent of the 5,000 votes cast . to win the election , a candidate needed to receive more than x % of the vote . if geoff needed exactly 2,571 more votes to win the election , what is the value of x ? | "word problems are tricky in somehow more than other problem because you have the additional step to translate . breaking the problem : geoff how many votes he receives ? ? 25 votes he needs 2571 more votes so : 25 + 2571 = 2596 now what ' s the problem wants ? ? a x % . . . . . . . . 2596 is what % of total votes 5000... | a ) 50 , b ) 54 , c ) 52 , d ) 60 , e ) 63 | c | add(divide(const_100, const_2), 0.5) | divide(const_100,const_2)|add(n0,#0)| | gain |
a bus trip of 360 miles would have taken 1 hour less if the average speed v for the trip had been greater by 5 miles per hour . what was the average speed v , in miles per hour , for the trip ? | "the time is the distance / speed . the time difference is 1 hour . 360 / v - 360 / ( v + 5 ) = 1 360 ( v + 5 ) - 360 v = ( v ) ( v + 5 ) 1800 = ( v ) ( v + 5 ) 40 * 45 = ( v ) ( v + 5 ) v = 40 mph the answer is b ." | a ) 35 , b ) 40 , c ) 45 , d ) 50 , e ) 55 | b | divide(subtract(sqrt(add(multiply(multiply(360, 5), const_4), power(5, const_2))), 5), const_2) | multiply(n0,n2)|power(n2,const_2)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n2)|divide(#5,const_2)| | physics |
the area of sector of a circle whose radius is 12 metro and whose angle at the center is 41 ° is ? | "41 / 360 * 22 / 7 * 12 * 12 = 51.5 m 2 answer : c" | a ) 49 m 2 , b ) 50 m 2 , c ) 51.5 m 2 , d ) 53 m 2 , e ) 59 m 2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(41, divide(const_3600, const_10))) | add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|power(n0,const_2)|add(#2,const_2)|divide(n1,#1)|divide(#4,#0)|multiply(#6,#3)|multiply(#5,#7)| | geometry |
10 % of employees are women with fair hair . 40 % of fair - haired employees are women . what percent of employees have fair hair ? | "i came up with ( a ) 25 think of 100 people total : from the first fact , 10 of these are women with fair hair . from the second fact , these 10 women make up 40 % of the total fair haired population . we can then make a ratio of 60 : 40 fair haired men to fair haired women . this means that ( 60 / 40 ) * 10 equals th... | a ) 25 , b ) 30 , c ) 50 , d ) 55 , e ) 60 | a | multiply(divide(10, 40), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
a rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ? | area = ( l + b – d ) d ( 80 + 60 – 10 ) 10 = > 1300 m 2 1300 * 3 = rs . 3900 answer : d | ['a ) rs . 3919', 'b ) rs . 3910', 'c ) rs . 3997', 'd ) rs . 3900', 'e ) rs . 3902'] | d | multiply(multiply(subtract(add(80, 60), 10), 10), 3) | add(n0,n1)|subtract(#0,n2)|multiply(n2,#1)|multiply(n3,#2) | geometry |
it takes 50 identical printing machines 8 hours to print 500,000 posters . how long would it take 40 of these machines to print the same total ? | "50 machines finish the job in 8 hours . we do not have to take care of the 500 ' 000 posters , it does not matter . 40 machines means , 20 % less of capacity and therefore 20 % longer to work for a certain job . 8 * 1.2 = 9 hr , 36 min answer a . . hi reto , since you are looking time for 40 machines it is 25 % less s... | a ) 10 hr , b ) 12 hr , c ) 12 hr 24 min , d ) 12 hr , 40 min , e ) 13 hr , 20 min | a | add(divide(50, 8), divide(500,000, 40)) | divide(n0,n1)|divide(n2,n3)|add(#0,#1)| | physics |
a man bought 40 shares of rs . 60 at 5 discount , the rate of dividend being 1212 % the rate of interest obtained is | explanation : face value of a share = rs . 60 he bought each share at rs . 60 - rs . 5 = rs . 55 number of shares = 40 dividend = 12 1 / 2 % = 25 / 2 % dividend per share = 60 × 25 / 2 × 100 = rs . 7.5 total dividend = ( 40 × 7.5 ) ie , he got a dividend of ( 40 × 7.5 ) for an investment of rs . ( 40 × 55 ) interest ob... | a ) 13.64 % , b ) 15.5 % , c ) 14 % , d ) 14.25 % , e ) 14.95 % | a | divide(multiply(multiply(40, 60), divide(1212, const_100)), multiply(40, subtract(60, 5))) | divide(n3,const_100)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#1)|multiply(n0,#2)|divide(#3,#4) | gain |
a equilateral triangle having one side 5 m . in this triangle there is a square of side 2 m . calculate the percentage of area outside square in that triangle . | area of triangle = â ˆ š 3 a 2 / 4 . = 1.732 * 25 / 4 = 10.825 m 2 . area of square = 2 * 2 = 4 m 2 . rest area after square = 10.825 - 4 = 6.825 m 2 . % rest area after square = 6.825 / 10.825 * 100 = 63.04 % answer c | ['a ) 68', 'b ) 54', 'c ) 63.04', 'd ) 62', 'e ) 60'] | c | multiply(divide(subtract(multiply(divide(sqrt(const_3), const_4), power(5, const_2)), square_area(2)), multiply(divide(sqrt(const_3), const_4), power(5, const_2))), const_100) | power(n0,const_2)|sqrt(const_3)|square_area(n1)|divide(#1,const_4)|multiply(#3,#0)|subtract(#4,#2)|divide(#5,#4)|multiply(#6,const_100) | geometry |
the average of numbers 0.54204 , 0.54203 , 0.54202 and 0.54201 is ? | "answer average = ( 0.54204 + 0.54203 + 0.54202 + 0.54201 ) / 4 = 2.1681 / 4 = 0.542025 correct option : d" | a ) 0.54202 , b ) 0.54204 , c ) 0.542022 , d ) 0.542025 , e ) none | d | divide(add(multiply(0.54204, 0.54203), multiply(0.54204, 0.54202)), 0.54204) | multiply(n0,n1)|multiply(n0,n2)|add(#0,#1)|divide(#2,n0)| | general |
0.007 x 0.45 = ? | "7 x 45 = 315 sum of decimal places = 5 therefore , 0.007 x 0.45 = 0.00315 answer is a ." | a ) 0.00315 , b ) 0.0315 , c ) 0.315 , d ) 0.0351 , e ) none of them | a | multiply(divide(0.007, 0.45), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
one fourth of one third of two fifth of a number is 15 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 15 then x = 15 * 30 = 450 40 % of 450 = 180 answer : option c" | a ) 140 , b ) 150 , c ) 180 , d ) 200 , e ) 220 | c | divide(multiply(divide(15, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100)| | gain |
a certain candy manufacturer reduced the weight of candy bar m by 25 percent buy left the price unchanged . what was the resulting percent increase in the price per ounce of candy bar m ? | "assume 1 oz candy cost $ 1 before . now price remain same $ 1 but weight of candy reduces to 0.75 oz new price of candy = 1 / 0.75 = 1.3333 price increase 33.33 % e" | a ) 20 , b ) 30 , c ) 34 , d ) 35 , e ) 33.33 | e | subtract(multiply(divide(const_1, subtract(const_1, divide(25, const_100))), const_100), const_100) | divide(n0,const_100)|subtract(const_1,#0)|divide(const_1,#1)|multiply(#2,const_100)|subtract(#3,const_100)| | gain |
apple costs l rupees per kilogram for first 30 kgs and q rupees per kilogram for each additional kilogram . if the price of 33 kilograms is 11.67 and for 36 kgs of apples is 12.48 then the cost of first 10 kgs of apples is | "ans : by framing equations we get 30 l + 3 q = 11.67 30 l + 6 q = 12.48 eliminate q by multiplying the first equation by 2 and subtracting second equation from the first then we get l = 0.362 cost of 10 kgs of apples = 0.362 x 10 = 3.62 answer : d" | a ) 3.69 , b ) 3.66 , c ) 3.6 , d ) 3.62 , e ) 3.61 | d | multiply(divide(subtract(11.67, multiply(subtract(33, 30), divide(subtract(12.48, 11.67), subtract(36, 33)))), 30), 10) | subtract(n4,n2)|subtract(n3,n1)|subtract(n1,n0)|divide(#0,#1)|multiply(#3,#2)|subtract(n2,#4)|divide(#5,n0)|multiply(n5,#6)| | other |
every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 76800 ? | 76800 * 9 / 8 * 9 / 8 = 97200 . answer : b | a ) 81000 , b ) 97200 , c ) 27772 , d ) 29889 , e ) 22312 | b | add(add(76800, multiply(divide(1, 8), 76800)), multiply(divide(1, 8), add(76800, multiply(divide(1, 8), 76800)))) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)| | general |
a walks at 30 kmph and 30 hours after his start , b cycles after him at 48 kmph . how far from the start does b catch up with a ? | "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 30 hours . x / 30 - x / 48 = 30 x = 2400 km answer is d" | a ) 1200 km , b ) 1500 km , c ) 2000 km , d ) 2400 km , e ) 3600 km | d | multiply(30, 48) | multiply(n1,n2)| | physics |
kamal obtained 96 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "sol . average = 96 + 65 + 82 + 67 + 85 / 5 ) = ( 395 / 5 ) = 79 . answer d" | a ) 65 , b ) 69 , c ) 72 , d ) 79 , e ) none | d | divide(add(add(add(add(96, 65), 82), 67), 85), add(const_1, const_4)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 32 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 32 5 x = 30 = > x = 6 hence , b ' s age = 2 x = 12 years . answer : e" | a ) 17 years , b ) 19 years , c ) 29 years , d ) 10 years , e ) 12 years | e | divide(multiply(subtract(32, const_2), const_2), add(const_4, const_1)) | add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)| | general |
evaluate : 6702.5 + 670.25 + 67.025 + 6.7025 + 0.67025 | "6702.5 670.25 67.025 6.7025 + 0.67025 - - - - - - - - - - - - - - - 7447.14775 answer is a ." | a ) 7447.14775 , b ) 6981.59775 , c ) 6918.59775 , d ) 6198.59775 , e ) 6891.59775 | a | subtract(6702.5, multiply(multiply(670.25, 67.025), 6.7025)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
a cylindrical can has a radius of 4 centimeters and a height of 24 centimeters . what is the area , in square centimeters , of a rectangular label that completely covers the curved surface of the can without over - lapping ? | "they are asking the lateral surface area of the cylinder . the lateral surface area = 2 * pi * r * h = 2 * pi * 4 * 24 = 192 pi answer is d ." | a ) 16 pi , b ) 64 pi , c ) 96 pi , d ) 192 pi , e ) 576 pi | d | multiply(multiply(4, 24), multiply(const_2, const_pi)) | multiply(n0,n1)|multiply(const_2,const_pi)|multiply(#0,#1)| | geometry |
it is the new year and mandy has made a resolution to lose weight this year . she plans to exercise and do yoga . for exercise she plans to workout at the gym and ride her bicycle in the ratio of 2 : 3 everyday . she will also do yoga in the ratio , yoga : exercise = 2 : 3 . if she rides her bike for 12 minutes , how m... | the ratio is 2 : 3 = gym : ride , so ( 12 ) ( 3 / 2 ) = 18 minutes at the gym , and 18 + 12 = 30 minutes exercise , so ( 2 / 3 ) ( 30 ) = 20 minutes yoga . answer : e | a ) 10 min . , b ) 41 min . , c ) 17 min . , d ) 23 min . , e ) 20 min . | e | divide(multiply(12, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3)))) | add(n0,n1)|divide(n1,#0)|multiply(n4,#1)|multiply(#1,#1)|divide(#2,#3) | physics |
the probability that a man will be alive for 10 more yrs is 3 / 4 & the probability that his wife will alive for 10 more yrs is 3 / 7 . the probability that none of them will be alive for 10 more yrs , is | "sol . required probability = pg . ) x p ( b ) = ( 1 — d x ( 1 — i ) = : x 1 = 1 / 7 ans . ( c )" | a ) 1 / 2 , b ) 1 , c ) 1 / 7 , d ) 3 / 4 , e ) 2 | c | multiply(subtract(3, divide(3, 4)), subtract(3, divide(3, 7))) | divide(n1,n2)|divide(n1,n5)|subtract(n1,#0)|subtract(n1,#1)|multiply(#2,#3)| | general |
a jar of 264 marbles is divided equally among a group of marble - players . if 2 people join the group , each one would receive 1 marble less . how many people are there in the group today ? | "only 22 and 24 divide 264 perfectly , assume initially , there are 22 people , , each will get 12 marbles two people join , then 24 will get 11 marble each . hence answer e = 24" | a ) 20 , b ) 21 . , c ) 22 , d ) 23 , e ) 24 | e | divide(subtract(sqrt(add(multiply(multiply(264, 2), const_4), power(2, 2))), 2), 2) | multiply(n0,n1)|power(n1,n1)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n1)|divide(#5,n1)| | general |
two numbers are in the ratio 3 : 4 . if the sum of numbers is 63 , find the numbers . | "sum of the terms of the ratio = 3 + 4 = 7 sum of numbers = 63 therefore , first number = 3 / 7 × 63 = 27 second number = 4 / 7 × 63 = 36 therefore , the two numbers are 27 and 36 . answer is a" | a ) 27 and 36 . , b ) 25 and 30 . , c ) 23 and 44 . , d ) 63 and 12 . , e ) 12 and 36 . | a | add(multiply(multiply(3, 63), const_100), multiply(4, 63)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | other |
a man spends 2 / 5 of his salary on house rent , 3 / 10 of his salary on food , and 1 / 8 of his salary on conveyance . if he has rs 1400 left with him , find his expenditure on food and conveyance | "if his salary is x , then x - x * ( 2 / 5 + 3 / 10 + 1 / 8 ) = 1400 x ( 1 - 33 / 40 ) = 1400 x = 1400 * 40 / 7 = rs 8000 expenditure on food and conveyance = 8000 * 3 / 10 + 8000 * 1 / 8 = 2400 + 1000 = 3400 answer : b" | a ) 3200 , b ) 3400 , c ) 3600 , d ) 3800 , e ) 4000 | b | add(multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(3, 10)), multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(1, 8))) | divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|add(#0,#1)|add(#3,#2)|subtract(n4,#4)|divide(n6,#5)|multiply(#6,#1)|multiply(#6,#2)|add(#7,#8)| | general |
a certain bus driver is paid a regular rate of $ 14 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 976 in total compens... | "for 40 hrs = 40 * 14 = 560 excess = 976 - 560 = 416 for extra hours = . 75 ( 14 ) = 10.5 + 14 = 24.5 number of extra hrs = 416 / 24.5 = 18.82 = 19 approx . total hrs = 40 + 19 = 59 answer c 59" | a ) 54 , b ) 51 , c ) 59 , d ) 55 , e ) 52 | c | add(40, divide(subtract(976, multiply(14, 40)), divide(multiply(14, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general |
a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 3 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? | "1 / x - 1 / 20 = - 1 / 24 x = 120 120 * 3 = 360 answer : a" | a ) 360 , b ) 487 , c ) 481 , d ) 729 , e ) 268 | a | multiply(24, 20) | multiply(n0,n2)| | physics |
the ratio of the incomes of a and b is 2 : 3 and the ratio of their expenditure is 1 : 2 . if at the end of the year , each saves $ 500 then the income of a is ? | "let the income of a and b be $ 2 x and $ 3 x let their expenditures be $ y and $ 2 y 2 x - y = 500 - - - - - - - 1 ) 3 x - 2 y = 500 - - - - - - - 2 ) from 1 ) and 2 ) x = 500 a ' s income = 2 x = 2 * 500 = $ 1000 answer is a" | a ) $ 1000 , b ) $ 1500 , c ) $ 1200 , d ) $ 900 , e ) $ 1100 | a | multiply(divide(2, 2), 500) | divide(n0,n3)|multiply(n4,#0)| | other |
the compound and the simple interests on a certain sum at the same rate of interest for two years are rs . 11730 and rs . 10200 respectively . find the sum | explanation : the simple interest for the first year is 10200 / 2 is rs . 5100 and compound interest for first year also is rs . 5100 . the compound interest for second year on rs . 5100 for one year so rate of the interest = ( 100 * 1530 ) / ( 5100 * 1 ) = 30 % p . a . so p = ( 100 * 10200 ) / ( 30 * 2 ) = rs . 17000 ... | a ) rs . 17037 , b ) rs . 17000 , c ) rs . 17276 , d ) rs . 170287 , e ) rs . 171881 | b | divide(multiply(const_100, 10200), multiply(divide(multiply(subtract(11730, 10200), const_100), divide(10200, const_2)), const_2)) | divide(n1,const_2)|multiply(n1,const_100)|subtract(n0,n1)|multiply(#2,const_100)|divide(#3,#0)|multiply(#4,const_2)|divide(#1,#5) | gain |
the sum of two numbers is 8 and the sum of their squares is 120 . find the product of the two numbers . | let a and b be the two numbers ( a + b ) ^ 2 = a ^ 2 + 2 ab + b ^ 2 given ( a + b ) = 8 a ^ 2 + b ^ 2 = 120 so , 8 ^ 2 = 120 + 2 ab 2 ab = 120 - 64 2 ab = 56 ab = 28 ans e | a ) 25 , b ) 24 , c ) 23 , d ) 21 , e ) 28 | e | divide(subtract(120, power(8, const_2)), const_2) | power(n0,const_2)|subtract(n1,#0)|divide(#1,const_2)| | general |
if the cost price of 121 chocolates is equal to the selling price of 77 chocolates , the gain percent is : | "explanation : solution : let c . p . of each chocolate be re . 1 . then , c . p . of 77 chocolates = rs . 77 ; s . p . of 77 chocolates = rs . 121 . . ' . gain % = 44 * 100 / 77 = 57.14 % answer : c" | a ) 56.14 % , b ) 57.44 % , c ) 57.14 % , d ) 56.34 % , e ) 57.87 % | c | divide(const_100, divide(77, subtract(121, 77))) | subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1)| | gain |
a library has an average of 500 visitors on sundays and 140 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "explanation : since the month begins with a sunday , so there will be five sundays in the month , required average = ( 500 * 5 + 140 * 25 ) / 30 = 6000 / 30 = 200 answer : b ) 200" | a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 550 | b | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 500), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 140)), 30) | add(const_3,const_4)|divide(n2,#0)|floor(#1)|add(#2,const_1)|multiply(n0,#3)|subtract(n2,#3)|multiply(n1,#5)|add(#4,#6)|divide(#7,n2)| | general |
a man can row his boat with the stream at 6 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 6 us = 4 s = ? s = ( 6 - 4 ) / 2 = 1 kmph answer : a" | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 7 | a | divide(subtract(6, 4), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
a man buys 100 lts of liquid which contains 25 % of the liquid and the rest is water . he then mixes it with 90 lts of another mixture with 30 % of liquid . what is the % of water in the new mixture ? | 25 % in 100 lts is 25 . so water = 100 - 25 = 75 lts . 30 % of 90 lts = 27 . so water in 2 nd mixture = 90 - 27 = 63 lts . now total quantity = 100 + 90 = 190 lts . total water in it will be 75 + 63 = 138 lts . % of water = ( 100 * 138 ) / 190 = 72.63 answer : c | a ) 81 , b ) 75 , c ) 72.63 , d ) 73 , e ) 76 | c | divide(add(multiply(100, subtract(const_100, 25)), multiply(90, subtract(const_100, 30))), add(100, 90)) | add(n0,n2)|subtract(const_100,n1)|subtract(const_100,n3)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0) | gain |
a train running at the speed of 60 km / hr crosses a pole in 3 seconds . find the length of the train ? | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 3 = 50 meter answer : e" | a ) 150 meter , b ) 170 meter , c ) 156 meter , d ) 168 meter , e ) 50 meter | e | multiply(divide(multiply(60, const_1000), const_3600), 3) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
the sale price sarees listed for rs . 750 after successive discount is 20 % and 15 % is ? | "750 * ( 80 / 100 ) * ( 85 / 100 ) = 570 answer : b" | a ) 227 , b ) 570 , c ) 342 , d ) 680 , e ) 230 | b | subtract(subtract(750, divide(multiply(750, 20), const_100)), divide(multiply(subtract(750, divide(multiply(750, 20), const_100)), 15), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
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