Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
the least number which when increased by 6 each divisible by each one of 24 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 5 = 864 - 6 = 858 . answer a" | a ) 858 , b ) 859 , c ) 869 , d ) 4320 , e ) none of these | a | subtract(lcm(lcm(lcm(24, 32), 36), 54), 6) | lcm(n1,n2)|lcm(n3,#0)|lcm(n4,#1)|subtract(#2,n0)| | general |
find the mean proportional between 81 & 100 ? | "formula = √ a × b a = 81 and b = 100 √ 81 × 100 = 9 × 10 = 90 e" | a ) a ) 59 , b ) b ) 61 , c ) c ) 63 , d ) d ) 65 , e ) e ) 90 | e | divide(add(81, 100), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
if the sum of the first n positive integers is s , what is the sum of the first n positive integers divisible by 5 , in terms of s ? | "answer is e given sum of 1 2 3 . . . . . n = s sum of 5 , 10 , 15 . . . . 5 n = 5 [ sum ( 1 2 3 . . . . n ) ] = 5 s" | a ) s / 2 , b ) s , c ) 2 s , d ) 2 s + 2 , e ) 5 s | e | negate(const_2) | negate(const_2)| | general |
a man can row upstream at 30 kmph and downstream at 60 kmph , and then find the speed of the man in still water ? | "us = 30 ds = 60 m = ( 30 + 60 ) / 2 = 45 answer : e" | a ) 27 , b ) 29 , c ) 30 , d ) 20 , e ) 45 | e | divide(add(30, 60), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
if the average ( arithmetic mean ) of 2 a + 16 , 3 a - 8 is 94 , what is the value of a ? | "am of 2 a + 16 , 3 a - 8 = 2 a + 16 + 3 a - 8 / 2 = 5 a + 8 / 2 given that 5 a + 8 / 2 = 94 a = 36 answer is d" | a ) 25 , b ) 30 , c ) 28 , d ) 36 , e ) 42 | d | subtract(multiply(16, const_2), multiply(2, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
a group of students was interviewed for that if it was asked whether or not they speak french and / or english . among those who speak french , 10 speak english well , while 40 of them do not speak english . if 75 % of students do not speak french , how many students were surveyed ? | "number of students who speak french are 40 + 10 = 50 of total students , the percentage of students who do not speak french was 75 % - - > percentage of who do is 25 % 50 - - - - - - - 25 % x - - - - - - - 100 % x = 50 * 100 / 25 = 200 = number of all students answer is c" | a ) 210 , b ) 225 , c ) 200 , d ) 250 , e ) 300 | c | divide(add(10, 40), divide(subtract(const_100, 75), const_100)) | add(n0,n1)|subtract(const_100,n2)|divide(#1,const_100)|divide(#0,#2)| | gain |
a train crosses a platform of 120 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be ‘ x ’ x + 120 / 15 = x + 250 / 20 4 x + 480 = 3 x + 750 x = 270 m answer : b" | a ) a ) 150 m , b ) b ) 270 m , c ) c ) 180 m , d ) d ) 158 m , e ) e ) 350 m | b | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(120, divide(20, divide(15, const_3)))) | divide(n1,const_3)|divide(n1,#0)|divide(n3,#0)|multiply(n2,#1)|multiply(n0,#2)|subtract(#3,#4)| | physics |
two numbers are in the ratio 3 : 4 . if their l . c . m . is 36 . what is sum of the numbers ? | "explanation : let the numbers be 3 x and 4 x lcm of 3 x and 4 x = 12 x ( since lcm of 3 and 4 is 12 . hence lcm of 3 x and 4 x is 12 x ) given that lcm of 3 x and 4 x is 36 = > 12 x = 36 = > x = 36 / 12 = 3 sum of the numbers = 3 x + 4 x = 7 x = 7 x 3 = 21 answer : option a" | a ) 21 , b ) 56 , c ) 76 , d ) 84 , e ) 85 | a | add(multiply(multiply(3, const_4.0), const_100), multiply(4, 36)) | multiply(n0,const_4.0)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | other |
in what time will a train 120 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 120 / 40 = 3 sec . answer : b" | a ) 2.5 sec , b ) 3 sec , c ) 8.9 sec , d ) 6.9 sec , e ) 2.9 sec | b | divide(120, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
evaluate : 1234562 - 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 1234562 - 12 * 3 * 2 = 1234562 - 8 = 1234554 correct answer b" | a ) 32435453 , b ) 1234554 , c ) 76768786 , d ) 97979797 , e ) 75868656 | b | subtract(1234562, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
33 1 / 3 % of 360 ? | "33 1 / 3 % = 1 / 3 1 / 3 × 360 = 120 d )" | a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | d | divide(multiply(add(33, divide(1, 3)), 360), const_100) | divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)| | gain |
what is the area of a square field whose sides have a length of 14 meters ? | "14 * 14 = 196 sq m the answer is a ." | a ) 196 sq m , b ) 286 sq m , c ) 298 sq m , d ) 267 sq m , e ) 231 sq m | a | divide(square_area(14), const_2) | square_area(n0)|divide(#0,const_2)| | geometry |
x can do a piece of work in 40 days . he works at it for 8 days and then y finished it in 32 days . how long will y take to complete the work ? | "work done by x in 8 days = 8 * 1 / 40 = 1 / 5 remaining work = 1 - 1 / 5 = 4 / 5 4 / 5 work is done by y in 32 days whole work will be done by y in 32 * 5 / 4 = 40 days answer is b" | a ) 10 , b ) 40 , c ) 15 , d ) 18 , e ) 20 | b | multiply(32, inverse(subtract(const_1, divide(8, 40)))) | divide(n1,n0)|subtract(const_1,#0)|inverse(#1)|multiply(n2,#2)| | physics |
a man traveled a total distance of 1800 km . he traveled one - third of the whole trip by plane and the distance traveled by train is three - fifth of the distance traveled by bus . if he traveled by train , plane and bus , then find the distance traveled by bus ? | "explanation : total distance traveled = 1800 km . distance traveled by plane = 600 km . distance traveled by bus = x distance traveled by train = 3 x / 5 = > x + 3 x / 5 + 600 = 1800 = > 8 x / 5 = 1200 = > x = 750 km . answer : d" | a ) 239 , b ) 247 , c ) 277 , d ) 270 , e ) 898 | d | divide(multiply(divide(multiply(1800, const_2), const_3), const_3), add(const_2, const_3)) | add(const_2,const_3)|multiply(n0,const_2)|divide(#1,const_3)|multiply(#2,const_3)|divide(#3,#0)| | physics |
if k is a non - negative integer and 12 ^ k is a divisor of 856,736 then 3 ^ k - k ^ 3 = | "8 + 5 + 6 + 7 + 3 + 6 = 35 , so this number is not divisible by 3 and thus not divisible by 12 . therefore , k = 0 3 ^ k - k ^ 3 = 1 - 0 = 1 the answer is b ." | a ) 0 , b ) 1 , c ) 36 , d ) 120 , e ) 524 | b | subtract(power(3, subtract(const_1, const_1)), power(subtract(const_1, const_1), 3)) | subtract(const_1,const_1)|power(n2,#0)|power(#0,n2)|subtract(#1,#2)| | general |
the length of a rectangular plot is 10 mtr more than its width . the cost of fencing the plot along its perimeter at the rate of rs . 6.5 mtr is rs . 1950 . the perimeter of the plot is ? | sol . let width = x , length = ( 10 + x ) perimeter = 2 ( x + ( 10 + x ) ) = 2 ( 2 x = 10 ) & 2 ( 2 x + 10 ) * 6.5 = 1950 x = 70 required perimeter = 2 ( 70 + 80 ) = 300 d | ['a ) 126', 'b ) 156', 'c ) 190', 'd ) 300', 'e ) 260'] | d | multiply(add(divide(subtract(divide(divide(1950, 6.5), const_2), const_10), const_2), add(divide(subtract(divide(divide(1950, 6.5), const_2), const_10), const_2), const_10)), const_2) | divide(n2,n1)|divide(#0,const_2)|subtract(#1,const_10)|divide(#2,const_2)|add(#3,const_10)|add(#4,#3)|multiply(#5,const_2) | geometry |
a can do a piece of work in 8 days . b can do it in 6 days . with the assistance of c they completed the work in 2 days . find in how many days can c alone do it ? | "c = 1 / 2 - 1 / 8 - 1 / 8 = 5 / 24 = > 4.8 days answer : e" | a ) 8.7 days , b ) 2.0 days , c ) 6.6 days , d ) 7.7 days , e ) 4.8 days | e | divide(multiply(8, 6), divide(subtract(multiply(8, 6), multiply(add(divide(multiply(8, 6), 8), divide(multiply(8, 6), 6)), 2)), 2)) | multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|multiply(n2,#3)|subtract(#0,#4)|divide(#5,n2)|divide(#0,#6)| | physics |
a car started running at a speed of 30 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 10 hours of the journey . | "the total distance covered by the car in the first 10 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 = sum of 10 terms in ap whose first term is 30 and last term is 48 = 10 / 2 [ 30 + 48 ] = 390 km . answer : b" | a ) 342 km , b ) 390 km , c ) 642 km , d ) 742 km , e ) 382 km | b | divide(add(add(30, multiply(2, 10)), 30), 2) | multiply(n1,n2)|add(n0,#0)|add(n0,#1)|divide(#2,n1)| | physics |
the maximum number of students among them 1230 pens and 920 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | "explanation : required number of students = h . c . f of 1230 and 920 = 10 . answer : b" | a ) 91 , b ) 10 , c ) 1001 , d ) 1911 , e ) none of these | b | gcd(1230, 920) | gcd(n0,n1)| | general |
the length of a room is 7 m and width is 4.75 m . what is the cost of paying the floor by slabs at the rate of rs . 900 per sq . metre . | "area = 7 × 4.75 sq . metre . cost for 1 sq . metre . = rs . 900 hence total cost = 7 × 4.75 × 900 = 7 × 4275 = rs . 29925 answer is e" | a ) 25650 , b ) 25750 , c ) 26550 , d ) 26750 , e ) 29925 | e | multiply(900, multiply(7, 4.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
a train 110 m long is running with a speed of 27 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 27 + 6 = 33 km / hr . = 33 * 5 / 18 = 9.1667 m / sec . time taken to pass the men = 110 / 9.1667 = 12 sec . answer : e" | a ) 7 sec , b ) 6 sec , c ) 8 sec , d ) 4 sec , e ) 12 sec | e | divide(110, multiply(add(27, 6), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
a , b and c can do a piece of work in 24 days , 30 days and 60 days respectively . they began the work together but c left 4 days before the completion of the work . in how many days was the work completed ? | one day work of a , b and c = 1 / 24 + 1 / 30 + 1 / 60 = 11 / 120 work done by a and b together in the last 4 days = 4 * ( 1 / 24 + 1 / 30 ) = 3 / 10 remaining work = 7 / 10 the number of days required for this initial work = 7 days . the total number of days required = 4 + 7 = 11 days . answer : b | a ) 10 days , b ) 11 days , c ) 13 days , d ) 21 days , e ) 31 days | b | add(divide(subtract(const_1, multiply(add(divide(const_1, 24), divide(const_1, 30)), 4)), add(divide(const_1, multiply(add(const_2, const_3), multiply(const_2, const_4))), add(divide(const_1, 24), divide(const_1, 30)))), 4) | add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|multiply(const_2,const_4)|add(#1,#2)|multiply(#0,#3)|divide(const_1,#5)|multiply(n3,#4)|add(#4,#6)|subtract(const_1,#7)|divide(#9,#8)|add(n3,#10) | physics |
machine p and machine q are each used to manufacture 550 sprockets . it takes machine p 10 hours longer to produce 550 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | "p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 550 / x = 550 / 1.1 x + 10 1.1 ( 550 ) = 550 + 11 x 11 x = 55 x = 5 the answer is a ." | a ) 5 , b ) 15 , c ) 55 , d ) 95 , e ) 125 | a | divide(subtract(550, divide(550, add(divide(10, const_100), const_1))), 10) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)| | gain |
what percent of a day is 7 hours ? | explanation : solution : required percentage = ( 7 / 100 * 24 ) = 29.2 % answer : a | a ) 29.2 % , b ) 12.5 % , c ) 13 % , d ) 11.3 % , e ) none of these | a | multiply(divide(7, multiply(multiply(const_4, const_3), const_2)), const_100) | multiply(const_3,const_4)|multiply(#0,const_2)|divide(n0,#1)|multiply(#2,const_100) | physics |
if 4 xz + yw = 5 and xw + yz = 10 , what is the value of the expression ( 2 x + y ) ( 2 z + w ) ? | "( 2 x + y ) * ( 2 z + w ) = 5 + 2 ( 10 ) = 25 answer : e" | a ) 9 . , b ) 12 . , c ) 15 . , d ) 16 . , e ) 25 | e | add(4, multiply(10, 2)) | multiply(n2,n3)|add(n0,#0)| | general |
rectangular tile each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of f... | "area of tile = 70 * 30 = 2100 area of floor = 130 * 110 = 14300 no of tiles = 14300 / 2100 = 6.8 so , the no of tile = 6 answer : b" | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | divide(multiply(110, 130), multiply(70, 30)) | multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)| | geometry |
1394 x 1394 = ? | "1394 x 1394 = ( 1394 ) 2 = ( 1400 - 6 ) 2 = ( 1400 ) 2 + ( 6 ) 2 - ( 2 x 1400 x 6 ) = 1960000 + 36 - 16800 = 1960036 - 16800 = 1943236 . answer : d" | a ) a ) 1951609 , b ) b ) 1951601 , c ) c ) 1951602 , d ) d ) 1943236 , e ) e ) 1951604 | d | multiply(divide(1394, 1394), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a circular ground whose diameter is 34 metres , has a 2 metre - broad garden around it . what is the area of the garden in square metres ? | req . area = ï € [ ( 19 ) 2 â € “ ( 17 ) 2 ] = 22 â „ 7 ã — ( 36 ã — 2 ) [ since a 2 - b 2 = ( a + b ) ( a - b ) ] = ( 22 ã — 36 ã — 2 ) / 7 = 226.3 sq m answer a | ['a ) 226.3', 'b ) 226.2', 'c ) 228.2', 'd ) 227', 'e ) 226'] | a | subtract(circle_area(add(divide(34, const_2), 2)), circle_area(divide(34, const_2))) | divide(n0,const_2)|add(n1,#0)|circle_area(#0)|circle_area(#1)|subtract(#3,#2) | geometry |
3889 + 12.808 - ? = 3854.002 | let 3889 + 12.808 - x = 3854.002 . then x = ( 3889 + 12.808 ) - 3854.002 = 3901.808 - 3854.002 = 47.806 . answer = c | a ) a ) 47.095 , b ) b ) 47.752 , c ) c ) 47.806 , d ) d ) 47.95 , e ) of the above | c | subtract(add(3889, 12.808), 3854.002) | add(n0,n1)|subtract(#0,n2) | general |
a sum of money is sufficient to pay a ' s wages for 20 days and b ' s wages for 30 days . the same money is sufficient to pay the wages of both for ? | "let the total money be $ x a ' s 1 day work = $ x / 20 b ' s 1 day work = $ x / 30 a + b 1 day work = $ x / 12 money is sufficient to pay the wages of both for 12 days answer is a" | a ) 12 days , b ) 15 days , c ) 20 days , d ) 18 days , e ) 22 days | a | divide(const_1, add(divide(const_1, 20), divide(const_1, 30))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | general |
a river boat leaves silver town and travels upstream to gold town at an average speed of 6 kilometers per hour . it returns by the same route at an average speed of 8 kilometers per hour . what is the average speed for the round - trip in kilometers per hour ? | "pick a number which is lcm of 8 and 6 = 24 . upstream time = 24 / 6 = 4 hrs downstream time = 24 / 8 = 3 hrs total time = 7 hrs total distance = 48 average speed = 48 / 7 = 6.8 km / hr" | a ) 6.8 , b ) 7.1 , c ) 7.2 , d ) 7.5 , e ) 8.0 | a | divide(multiply(multiply(6, 8), const_2), add(6, 8)) | add(n0,n1)|multiply(n0,n1)|multiply(#1,const_2)|divide(#2,#0)| | physics |
the sides of a square region , measured to the nearest centimeter , are 7 centimeters long . the least possible value of the actual area of the square region is | though there might be some technicalities concerning the termnearest ( as 6.5 is equidistant from both 6 and 7 ) the answer still should be a : 6.5 ^ 2 = 42.25 . answer : a | ['a ) 42.25 sq cm', 'b ) 45.25 sq cm', 'c ) 43.25 sq cm', 'd ) 50.25 sq cm', 'e ) 52.25 sq cm'] | a | power(subtract(subtract(7, const_0_25), const_0_25), const_2) | subtract(n0,const_0_25)|subtract(#0,const_0_25)|power(#1,const_2) | geometry |
a box contains 6 white , 5 red and 8 white marbles . 5 marbles are drawn from the box at random . what is the probability that both the marbles are of the same color ? | explanation : total marbles in a box = 6 black + 5 red + 8 green marbles = 19 marbles 5 marbles are drawn from 19 marbles at random . therefore , n ( s ) = 19 c 5 = 11628 ways let a be the event that 2 marbles drawn at random are of the same color . number of cases favorable to the event a is n ( a ) = 6 c 5 + 5 c 5 + ... | a ) 19 / 11628 , b ) 36 / 11628 , c ) 613 / 11628 , d ) 6113 / 11628 , e ) 63 / 11628 | e | divide(add(add(divide(factorial(6), factorial(5)), divide(factorial(5), factorial(5))), divide(factorial(8), multiply(factorial(subtract(8, 5)), factorial(5)))), divide(factorial(add(add(6, 5), 8)), multiply(factorial(5), factorial(subtract(add(add(6, 5), 8), 5))))) | add(n0,n1)|factorial(n0)|factorial(n1)|factorial(n2)|subtract(n2,n1)|add(n2,#0)|divide(#1,#2)|divide(#2,#2)|factorial(#4)|add(#6,#7)|factorial(#5)|multiply(#8,#2)|subtract(#5,n1)|divide(#3,#11)|factorial(#12)|add(#9,#13)|multiply(#2,#14)|divide(#10,#16)|divide(#15,#17) | other |
105 * 2 | "here d = 5 . 100 + 2 × × 5 / 25 = 11025 answer : a" | a ) 11025 , b ) 23889 , c ) 29088 , d ) 1999 , e ) 17710 | a | multiply(divide(105, 2), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
what will be the reminder when ( 43 ^ 43 + 43 ) is divided by 44 ? | "( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 43 ^ 43 + 1 ) will be divisible by ( 43 + 1 ) ; ( 43 ^ 43 + 1 ) + 42 when divided by 44 will give 42 as remainder . correct option : b" | a ) 32 , b ) 42 , c ) 83 , d ) 71 , e ) 92 | b | subtract(add(power(43, 43), 43), multiply(44, floor(divide(add(power(43, 43), 43), 44)))) | power(n0,n0)|add(n0,#0)|divide(#1,n3)|floor(#2)|multiply(n3,#3)|subtract(#1,#4)| | general |
maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 40 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h , what is the distance traveled by brad ? | "time taken = total distance / relative speed total distance = 40 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken = 40 / 10 = 4 hrs distance traveled by brad = brad ' s speed * time taken = 6 * 4 = 24 kms . . . answer - d" | a ) 16 , b ) 18 , c ) 20 , d ) 24 , e ) 30 | d | multiply(4, divide(40, add(4, 6))) | add(n1,n2)|divide(n0,#0)|multiply(n1,#1)| | physics |
the total of 342 of 20 paise and 25 paise make a sum of rs . 71 . the no of 20 paise coins is | "explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 342 - x ) . 0.20 * ( x ) + 0.25 ( 342 - x ) = 71 = > x = 290 . . answer : b ) 290" | a ) 238 , b ) 290 , c ) 278 , d ) 200 , e ) 288 | b | divide(subtract(multiply(342, 25), multiply(71, const_100)), subtract(25, 20)) | multiply(n0,n2)|multiply(n3,const_100)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)| | general |
a retailer buys a radio for rs 225 . his overhead expenses are rs 15 . he sellis the radio for rs 350 . the profit percent of the retailer is | "explanation : cost price = ( 225 + 15 ) = 240 sell price = 350 gain = ( 110 / 240 ) * 100 = 45.8 % . answer : b" | a ) 10 % , b ) 45.8 % , c ) 25 % , d ) 52 % , e ) none of these | b | subtract(multiply(divide(350, add(225, 15)), const_100), const_100) | add(n0,n1)|divide(n2,#0)|multiply(#1,const_100)|subtract(#2,const_100)| | gain |
a man can row upstream at 35 kmph and downstream at 45 kmph , and then find the speed of the man in still water ? | "us = 35 ds = 45 m = ( 35 + 45 ) / 2 = 40 answer : c" | a ) 32 kmph , b ) 34 kmph , c ) 40 kmph , d ) 45 kmph , e ) 65 kmph | c | divide(add(35, 45), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a 220 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | speed = ( 120 + 80 ) km / h ( because direction is opposite hence relative velocity is added ) = 500 / 9 m / s time = 9 sec let the lenght of second train is x total distance covered = 220 + x therefore , d = speed * time thus 220 + x = 500 / 9 * 9 x = 500 - 220 = 280 m answer : e | a ) 230 m , b ) 240 m , c ) 250 m , d ) 260 m , e ) 280 m | e | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 220) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0) | physics |
if 80 lamps can be lighted 5 hours per day for 10 days for rs . 21.25 , then the number of lamps which can be lighted 4 hours daily for 30 days for rs . 76.50 , is ? | explanation : ( hrs / day 4 : 5 ) : ( money 21.25 : 76.50 ) : ( days 30 : 10 ) : : 80 : x 4 * 21.25 * 30 * x = 5 * 76.50 * 10 * 80 so x = 120 answer : c | a ) 100 , b ) 110 , c ) 120 , d ) 130 , e ) none of these | c | divide(multiply(multiply(multiply(5, 76.5), 10), 80), multiply(multiply(21.25, 30), 4)) | multiply(n1,n6)|multiply(n3,n5)|multiply(n2,#0)|multiply(n4,#1)|multiply(n0,#2)|divide(#4,#3) | physics |
in a group of ducks and cows , the total number of legs are 20 more than twice the no . of heads . find the total no . of buffaloes . | "let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 20 = > 2 x = 20 = > x = 10 a" | a ) 10 , b ) 12 , c ) 13 , d ) 15 , e ) 16 | a | divide(20, const_2) | divide(n0,const_2)| | general |
the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 41 kg , then the weight of b is : | "let d sum of a , b , c is 3 * 45 = 135 and sum of a and b s 2 * 40 = 80 sum of b and c is 2 * 41 = 82 hence 80 + 82 - 135 = 27 ans = 27 answer : a" | a ) 27 kg , b ) 20 kg , c ) 26 kg , d ) 31 kg , e ) 35 kg | a | subtract(add(multiply(40, const_2), multiply(41, const_2)), multiply(45, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general |
i sold a book at a profit of 10 % . had i sold it for $ 100 more , 15 % would have been gained . find the cost price ? | "115 % of cost - 110 % of cost = $ 100 5 % of cost = $ 100 cost = 100 * 100 / 5 = $ 2000 answer is a" | a ) $ 2000 , b ) $ 2500 , c ) $ 3000 , d ) $ 3120 , e ) $ 1540 | a | divide(multiply(divide(multiply(100, 10), subtract(15, 10)), const_100), 10) | multiply(n0,n1)|subtract(n2,n0)|divide(#0,#1)|multiply(#2,const_100)|divide(#3,n0)| | gain |
an inspector rejects 0.08 % of the meters as defective . how many will he examine to reject 2 ? | "then , 0.08 % of x = 2 ( 8 / 100 ) * ( 1 / 100 ) x = 2 x = ( 2 * 100 * 100 ) / 8 = 2500 answer is e" | a ) 1200 , b ) 1240 , c ) 1852 , d ) 3200 , e ) 2500 | e | divide(multiply(2, const_100), 0.08) | multiply(n1,const_100)|divide(#0,n0)| | gain |
the average age of a class is 15.8 years . the average age of the boys in the class is 16.4 years and that of the girls is 15.4 years . what is the ratio of boys to girls in the class ? | "let number of boys = x ; let number of girls = y total numbers of students = x + y ( x + y ) × 15.8 = 16.4 x + 15.4 y 0.6 x = 0.4 y x / y = 0.4 / 0.6 = 2 / 3 answer : c" | a ) 1 : 2 , b ) 3 : 4 , c ) 2 : 3 , d ) 3 : 2 , e ) none of these | c | divide(subtract(15.8, 15.4), subtract(16.4, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general |
in how many years will a sum of money doubles itself at 25 % per annum on simple interest ? | "p = ( p * 25 * r ) / 100 r = 4 % answer : a" | a ) 4 % , b ) 8 % , c ) 10 % , d ) 12 % , e ) 15 % | a | divide(const_100, 25) | divide(const_100,n0)| | gain |
log 3 n + log 7 n what is 3 digit number n that will be whole number | no of values n can take is 1 7 ^ 3 = 343 answer : a | a ) 343 , b ) 729 , c ) 829 , d ) 929 , e ) 727 | a | power(7, 3) | power(n1,n0) | other |
at the a - street fish market a vendor sells fresh caught fish . his store is filled with 2 / 7 bass , 1 / 5 tuna , 1 / 7 trout , and the rest are catfish . if a customer just bought 1 / 7 of all the tuna and 1 / 7 of all the catfish , and a fisherman arrived to restock the bass , doubling the number of bass , what pro... | bass = 2 / 7 = 10 / 35 tuna = 1 / 5 = 7 / 35 trout = 1 / 7 = 5 / 35 catfish = 1 - ( 10 / 35 + 7 / 35 + 5 / 35 ) = 13 / 35 purchased : tuna = 7 / 35 - ( 7 / 35 ) ( 1 / 7 ) = 6 / 35 , and catfish = 13 / 35 - ( 13 / 35 ) ( 1 / 7 ) = 12 / 35 stocked : doubling the number of bass ( 2 ) ( 10 / 35 ) = 20 / 35 now : 20 + 6 + 5... | a ) 30 / 71 , b ) 4 / 7 , c ) 5 / 43 , d ) 1 / 2 , e ) 1 / 5 | c | subtract(divide(1, 5), subtract(subtract(const_1, add(add(divide(2, 7), divide(1, 5)), divide(1, 7))), divide(2, 7))) | divide(n2,n3)|divide(n0,n1)|divide(n2,n1)|add(#1,#0)|add(#3,#2)|subtract(const_1,#4)|subtract(#5,#1)|subtract(#0,#6) | general |
a and b started a business in partnership investing rs . 20000 and rs . 15000 respectively . after 6 months , c joined them with rs . 20000 . whatwill be b ' s share in total profit of rs . 20000 earned at the end of 2 years from the startingof the business ? | a : b : c = ( 20,000 x 24 ) : ( 15,000 x 24 ) : ( 20,000 x 18 ) = 4 : 3 : 3 . b ' s share = rs . 20000 x 3 / 10 = rs . 6,000 . d | a ) s . 5,000 , b ) s . 5,500 , c ) s . 5,700 , d ) s . 6,000 , e ) s . 7,500 | d | divide(multiply(20000, divide(multiply(15000, multiply(const_12, const_2)), add(add(multiply(20000, multiply(const_12, const_2)), multiply(15000, multiply(const_12, const_2))), multiply(subtract(multiply(const_12, const_2), 6), 20000)))), const_12) | multiply(const_12,const_2)|multiply(n1,#0)|multiply(n0,#0)|subtract(#0,n2)|add(#2,#1)|multiply(n0,#3)|add(#4,#5)|divide(#1,#6)|multiply(n0,#7)|divide(#8,const_12) | gain |
the product of two positive integers is 323 and their difference is 2 . what is the smaller number ? | let ' s use trial and error to find the two numbers . 20 * 18 = 360 ( too high ) 19 * 17 = 323 the answer is a . | a ) 17 , b ) 18 , c ) 16 , d ) 19 , e ) 14 | a | subtract(add(multiply(divide(323, divide(323, 2)), const_4), const_10), const_1) | divide(n0,n1)|divide(n0,#0)|multiply(#1,const_4)|add(#2,const_10)|subtract(#3,const_1) | general |
an alloy is to contain steel and iron in the ratio 5 : 2 . the iron required to be melted with 35 kg of steel is ? | let the required quantity of steel be x kg 5 : 2 : : 35 : x 5 x = 2 * 35 x = 14 kg answer is b | a ) 12 kg , b ) 14 kg , c ) 15 kg , d ) 20 kg , e ) 18 kg | b | multiply(divide(35, 5), 2) | divide(n2,n0)|multiply(n1,#0) | other |
a man complete a journey in 30 hours . he travels first half of the journey at the rate of 20 km / hr and second half at the rate of 10 km / hr . find the total journey in km . | 0.5 x / 20 + 0.5 x / 10 = 30 - - > x / 20 + x / 10 = 60 - - > 3 x = 20 x 60 - - > x = ( 60 x 20 ) / 3 = 400 km . answer : d | a ) 220 km , b ) 224 km , c ) 230 km , d ) 400 km , e ) 234 km | d | multiply(const_2, divide(multiply(multiply(20, 10), 30), add(20, 10))) | add(n1,n2)|multiply(n1,n2)|multiply(n0,#1)|divide(#2,#0)|multiply(#3,const_2) | physics |
a man sitting in a train which is traveling at 60 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . ? | "relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) * ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 60 ) kmph = 52 kmph . answer : a" | a ) 52 kmph , b ) 58 kmph , c ) 62 kmph , d ) 65 kmph , e ) 75 kmph | a | subtract(multiply(divide(280, 9), const_3_6), 60) | divide(n2,n1)|multiply(#0,const_3_6)|subtract(#1,n0)| | physics |
0.0006 ? = 0.03 | "explanation : required answer = 0.0006 / 0.03 = 0.06 / 3 = 0.02 . answer : option d" | a ) 2 , b ) 20 , c ) 0.2 , d ) 0.02 , e ) none of these | d | multiply(divide(0.0006, 0.03), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a train 125 m long passes a man , running at 4 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 4 ) km / hr . x - 4 = 45 = = > x = 49 km / hr . answer : b" | a ) 11 , b ) 49 , c ) 88 , d ) 65 , e ) 22 | b | divide(divide(subtract(125, multiply(multiply(4, const_0_2778), 4)), 10), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n2)|divide(#3,const_0_2778)| | physics |
it would take one machine 9 hours to complete a large production order and another machine 8 hours to complete the same order . how many hours would it take both machines , working simultaneously at their respective constant rates , to complete the order ? | the rate of the first machine is 1 / 9 job per hour ; the rate of the second machine is 1 / 8 job per hour ; thus , the combined rate of the machines is 1 / 9 + 1 / 8 = 17 / 72 job per hour , which means that it takes 1 / ( 17 / 72 ) = 72 / 17 hours both machines to do the job . answer : a . | a ) 72 / 17 , b ) 17 / 72 , c ) 14 / 24 , d ) 24 / 56 , e ) 34 / 56 | a | divide(multiply(9, 8), add(9, 8)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0) | physics |
3034 − ( 1002 ÷ 20.04 ) = ? | "explanation : = 3034 − ( 1002 / 2004 × 100 ) = 3034 − 50 = 2984 option d" | a ) 1964 , b ) 1984 , c ) 2964 , d ) 2984 , e ) none of these | d | subtract(3034, divide(1002, 20.04)) | divide(n1,n2)|subtract(n0,#0)| | general |
a number is doubled and 9 is added . if the resultant is trebled , it becomes 57 . what is that number ? | "let the number be x . then , 3 ( 2 x + 9 ) = 57 2 x = 10 = > x = 5 answer : a" | a ) 5 , b ) 6 , c ) 8 , d ) 7 , e ) 4 | a | divide(subtract(57, multiply(const_3, 9)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
a room is 5 meters 76 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room . | "let us calculate both the length and width of the room in centimeters . length = 5 meters and 76 centimeters = 576 cm width = 4 meters and 32 centimeters = 432 cm as we want the least number of square tiles required , it means the length of each square tile should be as large as possible . further , the length of each... | a ) 12 , b ) 15 , c ) 17 , d ) 137 , e ) 147 | b | divide(multiply(add(multiply(5, const_100), 76), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(5, const_100), 76), add(multiply(4, const_100), 32)), gcd(add(multiply(5, const_100), 76), add(multiply(4, const_100), 32)))) | multiply(n0,const_100)|multiply(n2,const_100)|add(n1,#0)|add(n3,#1)|gcd(#2,#3)|multiply(#2,#3)|multiply(#4,#4)|divide(#5,#6)| | general |
a train sets off at 2 : 00 pm at the speed of 70 km / h . another train starts at 3 : 00 pm in the same direction at the rate of 80 km / h . at what time will the second train catch the first train ? | in one hour the first train travels 70 km . the second train catches the first train at a rate of 80 km / h - 70 km / h = 10 km / h . the second train will catch the first train in 70 / 10 = 7 hours , so at 10 : 00 pm . the answer is a . | a ) 10 : 00 , b ) 10 : 30 , c ) 11 : 00 , d ) 11 : 30 , e ) 12 : 00 | a | add(divide(70, subtract(80, 70)), 3) | subtract(n5,n2)|divide(n2,#0)|add(n3,#1) | physics |
a boat can travel with a speed of 15 km / hr in still water . if the speed of the stream is 5 km / hr . find the time taken by the boat to go 68 km downstream ? | "solution speed downstream = ( 15 + 5 ) km / hr = 20 km / hr . time taken to travel 68 km downstream = ( 68 / 20 ) hrs = 3 hrs 24 minutes . answer d" | a ) 2 hours , b ) 3 hours , c ) 4 hours , d ) 3 hours 24 minutes , e ) none | d | divide(68, add(15, 5)) | add(n0,n1)|divide(n2,#0)| | physics |
mark and ann together were allocated n boxes of cookies to sell for a club project . mark sold 9 boxes less than n and ann sold 2 boxes less than n . if mark and ann have each sold at least one box of cookies , but together they have sold less than n boxes , what is the value of n ? | "if n = 10 mark sold 1 box and ann sold 8 boxes total 9 < 10 answer : b" | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | b | add(9, const_1) | add(n0,const_1)| | general |
135 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ? | "milk = 3 / 5 * 135 = 81 liters water = 54 liters 81 : ( 54 + p ) = 3 : 4 162 + 3 p = 324 = > p = 54 54 liters of water are to be added for the ratio become 3 : 4 . answer : d" | a ) 12 liters , b ) 32 liters , c ) 41 liters , d ) 54 liters , e ) 34 liters | d | multiply(divide(135, add(3, 2)), 2) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | general |
17 times a positive integer is more than its square by 60 , then the positive integer is | "explanation : let the number be x . then , 17 x = x 2 + 60 = > x 2 - 17 x + 60 = 0 = > ( x - 12 ) ( x - 5 ) = 0 = > x = 5 or 12 answer : option d" | a ) 13 , b ) 10 , c ) 11 , d ) 5 , e ) 14 | d | divide(add(17, subtract(power(17, const_2), multiply(60, const_4))), const_2) | multiply(n1,const_4)|power(n0,const_2)|subtract(#1,#0)|add(n0,#2)|divide(#3,const_2)| | general |
a certain junior class has 1000 students and a certain senior class has 800 students . among these students , there are 40 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ? | "let ' s see pick 40 / 1000 first then we can only pick 1 other pair from the 800 so total will be 40 / 800 * 1000 simplify and you get 1 / 20000 answer is c" | a ) 3 / 40000 , b ) 1 / 3600 , c ) 1 / 20000 , d ) 1 / 60 , e ) 1 / 15 | c | divide(1, const_3) | divide(n3,const_3)| | probability |
a card is drawn from a pack of 52 cards . the probability of getting a queen of club or a king of heart is | "explanation : total number of cases = 52 favourable cases = 2 probability = 2 / 56 = 1 / 26 answer is c" | a ) 1 / 13 , b ) 2 / 13 , c ) 1 / 26 , d ) 1 / 52 , e ) 1 / 23 | c | divide(subtract(52, multiply(const_4, const_4)), 52) | multiply(const_4,const_4)|subtract(n0,#0)|divide(#1,n0)| | probability |
frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 480 square feet . how many feet offence does frank use ? | area = length x breadth 480 = 40 x breadth so , breadth = 12 units fencing required is - breadth + breadth + length 12 + 12 + 40 = > 64 feet answer must be ( e ) 64 | ['a ) 14', 'b ) 47', 'c ) 54', 'd ) 180', 'e ) 64'] | e | add(add(divide(480, 40), divide(480, 40)), 40) | divide(n1,n0)|add(#0,#0)|add(n0,#1) | geometry |
the height of the wall is 6 times its width and length of the wall is 7 times its height . if the volume of the wall be 6804 cu . m . its width is | "explanation : let width = x then , height = 6 x and length = 42 x 42 x ã — 6 x ã — x = 6804 x = 3 answer : a" | a ) 3 m , b ) 5 m , c ) 6 m , d ) 7 m , e ) 8 m | a | power(divide(6804, multiply(multiply(6, 7), 6)), divide(const_1, const_3)) | divide(const_1,const_3)|multiply(n0,n1)|multiply(n0,#1)|divide(n2,#2)|power(#3,#0)| | physics |
90 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 90 = 35 answer : b | a ) 66 , b ) 35 , c ) 42 , d ) 27 , e ) 11 | b | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 90), const_2) | add(const_3,const_4)|add(#0,const_2)|divide(#0,#1)|multiply(n0,#2)|divide(#3,const_2) | general |
a train running at 1 / 3 of its own speed reached a place in 9 hours . how much time could be saved if the train would have run at its own speed ? | "time taken if run its own speed = 1 / 3 * 9 = 3 hrs time saved = 9 - 3 = 6 hrs answer : e" | a ) 8 hrs , b ) 10 hrs , c ) 12 hrs , d ) 15 hrs , e ) 6 hrs | e | multiply(divide(1, 3), 9) | divide(n0,n1)|multiply(n2,#0)| | physics |
the radius of a semi circle is 7 cm then its perimeter is ? | "radius = 7 cm then diameter = 14 cm 1 / 2 * 22 / 7 * 14 + 14 = 36 answer : b" | a ) 32.8 , b ) 36 , c ) 32.1 , d ) 32.2 , e ) 32.9 | b | add(divide(circumface(7), const_2), multiply(7, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | physics |
two persons raj and ramu started working for a company in similar jobs on january 1 , 1991 . raj ' s initial monthly salary was rs 400 , which increases by rs 40 after every year . ramu ' s initial monthly salary was rs 500 which increases by rs 20 after every 6 months . if these arrangements continue till december 31 ... | explanation : raj ' s salary as on 1 jan 1991 is rs 400 per month . his increment in his month salary is rs 40 per annum his total salary from 1 jan 1991 to 31 st dec 2000 i . e . in ten years = > 12 [ 2 ( 400 ) + ( 10 − 1 ) 40 ] × 10 / 2 . = > rs 69600 . ramu ' s salary as on jan 1 st 1991 is rs 550 and his half yearl... | a ) rs 108000 , b ) rs 144000 , c ) rs 132000 , d ) rs 152400 , e ) none of these | d | add(multiply(multiply(add(multiply(const_2, 400), multiply(40, subtract(const_10, const_1))), const_12), add(const_2, const_3)), multiply(multiply(add(multiply(const_2, 500), multiply(subtract(20, const_1), 20)), 6), const_10)) | add(const_2,const_3)|multiply(n2,const_2)|multiply(n4,const_2)|subtract(const_10,const_1)|subtract(n5,const_1)|multiply(n3,#3)|multiply(n5,#4)|add(#1,#5)|add(#2,#6)|multiply(#7,const_12)|multiply(n6,#8)|multiply(#0,#9)|multiply(#10,const_10)|add(#11,#12) | general |
at a certain college , 80 percent of the total number of students are freshmen . if 60 percent of the fresh - men are enrolled in the school of liberal arts and , of these , 50 percent are psychology majors , what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal ar... | let ' s say there is a total of 100 students at this college . 80 percent of the total number of students are freshmen . # of freshmen = 80 % of 100 = 80 60 percent of the fresh - men are enrolled in the school of liberal arts . . . number of liberal arts freshmen = 60 % of 80 = 48 . . . and , of these , 50 percent are... | a ) 25 % , b ) 20 % , c ) 22 % , d ) 21 % , e ) 24 % | e | divide(multiply(divide(multiply(80, 60), const_100), 50), const_100) | multiply(n0,n1)|divide(#0,const_100)|multiply(n2,#1)|divide(#2,const_100) | general |
difference of two numbers is 1670 . if 7.5 % of the number is 12.5 % of the other number , find the number ? | "let the numbers be x and y . then , 7.5 % of x = 12.5 % of y x = 125 * y / 75 = 5 * y / 3 . now , x - y = 1670 5 * y / 3 – y = 1670 2 * y / 3 = 1670 y = [ ( 1670 * 3 ) / 2 ] = 2505 . one number = 2505 , second number = 5 * y / 3 = 4175 . answer e ." | a ) 2660 , 1000 , b ) 3660 , 2000 , c ) 3000 , 4160 , d ) 2490 , 4150 , e ) 2505 , 4175 | e | multiply(divide(7.5, 12.5), divide(1670, subtract(const_1, divide(7.5, 12.5)))) | divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)|multiply(#0,#2)| | gain |
the ratio of flour to water to sugar in a recipe is 11 : 8 : 1 . the ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar . if the new recipe calls for 4 cups of water , how much sugar is required ? | "the ratio of flour to water is 22 : 8 = 11 : 4 . the ratio of flour to sugar is 5.5 : 1 = 11 : 2 . the new ratio of flour to water to sugar is 11 : 4 : 2 if we need 4 cups of water , then we need 2 cups of sugar . the answer is d ." | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | d | multiply(divide(4, 8), multiply(const_4, 1)) | divide(n3,n1)|multiply(n2,const_4)|multiply(#0,#1)| | other |
craig sells major appliances . for each appliance he sells , craig receives a commission of $ 50 plus 10 percent of the selling price . during one particular week craig sold 6 appliances for selling prices totaling $ 3620 . what was the total of craig ' s commissions for that week ? | craig receives a commission of $ 50 on each appliance plus a 10 % commission on total sales , thus his commission was 6 * 50 + 0.1 * 3620 = 662 . answer : e . | a ) $ 412 , b ) $ 526 , c ) $ 585 , d ) $ 605 , e ) $ 662 | e | add(divide(3620, 10), multiply(50, 6)) | divide(n3,n1)|multiply(n0,n2)|add(#0,#1) | general |
if 8 ^ x = 2 ^ 9 , what is x ? | 2 ^ 3 x = 2 ^ 9 3 x = 9 , therefore x = 3 answer b | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(9, const_3) | divide(n2,const_3) | general |
a certain list consists of 21 different numbers . if n is in the list and n is 3 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction of the sum of the 21 numbers in the list ? | "series : a 1 , a 2 . . . . a 20 , n sum of a 1 + a 2 + . . . + a 20 = 20 * x ( x = average ) so , n = 3 * x hence , a 1 + a 2 + . . + a 20 + n = 23 x so , the fraction asked = 3 x / 23 x = 3 / 23 answer e" | a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 5 , d ) 4 / 21 , e ) 3 / 23 | e | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 3), 21), const_4.0), const_2), 3), const_3)) | divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,const_4.0)|multiply(#3,const_2)|subtract(#4,n1)|subtract(#5,const_3)|divide(#1,#6)| | general |
i chose a number and divide it by 8 . then i subtracted 100 from the result and got 6 . what was the number i chose ? | "solution : let x be the number i chose , then x / 8 â ˆ ’ 100 = 6 x / 8 = 106 x = 848 answer e" | a ) 600 , b ) 700 , c ) 800 , d ) 900 , e ) 848 | e | multiply(add(100, 6), 8) | add(n1,n2)|multiply(n0,#0)| | general |
the population of a bacteria colony doubles every day . if it was started 7 days ago with 2 bacteria and each bacteria lives for 12 days , how large is the colony today ? | 2 ^ 7 ( 2 ) = 2 ^ 8 = 256 the answer is b . | a ) 128 , b ) 256 , c ) 512 , d ) 1024 , e ) 2048 | b | subtract(power(2, add(7, const_1)), const_1) | add(n0,const_1)|power(n1,#0)|subtract(#1,const_1) | physics |
light glows for every 32 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am . | "time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 / 32 ) = 15582 times . answer : c" | a ) 180 times , b ) 381 times , c ) 155 times , d ) 392 times , e ) 150 times | c | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 32) | add(n6,const_2)|add(const_2,const_60)|add(n5,#1)|multiply(#2,const_60)|add(#0,#3)|divide(#4,n0)| | general |
if the height of a right cone c is 3 and the diameter of the base of c is 8 , what is the distance from the apex of c to any point on the circumference of the base of c ? | height of cone = 3 radius of base = 4 so the distance from the apex of c to any point on the circumference of the base of c will be a right angled triangle hence the distance is sqrt ( 3 ^ 2 + 4 ^ 2 ) = 5 answer : b | ['a ) 3 √ 2', 'b ) 5', 'c ) 5 √ 2', 'd ) √ ( 67 )', 'e ) √ ( 73 )'] | b | sqrt(add(power(divide(8, const_2), const_2), power(3, const_2))) | divide(n1,const_2)|power(n0,const_2)|power(#0,const_2)|add(#2,#1)|sqrt(#3) | geometry |
a can copy 50 papers in 10 hrs , while a & b can copy 70 papers in 10 hrs . how many hours are required for b to copy 26 papers ? | a can copy 50 papers in 10 hrs , while a & b can copy 70 papers in 10 hrs . it means b can copy 20 papers in 10 hrs . then time taken by b to copy 26 papers = 26 * 10 / 20 = 13 hours answer : c | a ) 11 hours , b ) 12 hours , c ) 13 hours , d ) 14 hours , e ) 18 hours | c | divide(multiply(26, 10), subtract(70, 50)) | multiply(n1,n4)|subtract(n2,n0)|divide(#0,#1) | physics |
45 % of 816 - 23 % of ? = 240.7 | "d 550 ( 816 * 45 ) / 100 - ( ? * 23 ) / 100 = 240.7 = > 367.2 - ( ? * 23 ) / 100 = 240.7 = > ( ? * 23 ) / 100 = 367.2 - 240.7 = > ? = ( 126.5 * 100 ) / 23 = 550" | a ) 330 , b ) 440 , c ) 240 , d ) 550 , e ) 230 | d | divide(multiply(45, 816), 23) | multiply(n0,n1)|divide(#0,n2)| | general |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 121 hours and 100 hours respectively . the ratio of their speeds is : | "let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 100 : â ˆ š 121 = 10 : 11 . answer e" | a ) 2 : 3 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none of these | e | divide(sqrt(100), sqrt(121)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics |
shanmukham buys good worth rs . 6650 . he gets a rebate of 6 % on it . after getting the rebate , he pays sales tax @ 10 % . find the amount he will have to pay for the goods | explanation : rebate = 6 % of rs . 6650 = rs . 6 / 100 x 6650 = rs . 399 . sales tax = 10 % of rs . ( 6650 399 ) = rs . 10 / 100 x 6251 = rs . 625.10 final amount = rs . ( 6251 + 625.10 ) = rs . 6876.10 answer : b | a ) 6999.2 , b ) 6876.1 , c ) 6654 , d ) 7000 , e ) none of these | b | add(subtract(6650, divide(multiply(6650, 6), const_100)), divide(multiply(subtract(6650, divide(multiply(6650, 6), const_100)), const_10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(#2,const_10)|divide(#3,const_100)|add(#4,#2) | gain |
the maximum numbers of students among them 781 pens and 710 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is | "olution required number of students . = h . c . f of 781 and 710 . â € ¹ = â € º 71 . answer a" | a ) 71 , b ) 910 , c ) 1001 , d ) 1911 , e ) none | a | subtract(781, 710) | subtract(n0,n1)| | general |
what is the measure of the radius of the circle inscribed in a triangle whose sides measure 8 , 15 and 27 units ? | "sides are 8 , 15 and 27 . . . thus it is right angle triangle since 27 ^ 2 = 8 ^ 2 + 15 ^ 2 therefore , area = 1 / 2 * 15 * 8 = 60 we have to find in - radius therefore , area of triangle = s * r . . . . where s = semi - perimeter and r = in - radius now s = semi - perimeter = 27 + 15 + 8 / 2 = 30 thus , 60 = 30 * r a... | a ) 8.5 units , b ) 6 units , c ) 3 units , d ) 5 units , e ) 2 units | e | divide(triangle_area_three_edges(8, 15, 27), divide(triangle_perimeter(8, 15, 27), const_2)) | triangle_area_three_edges(n0,n1,n2)|triangle_perimeter(n0,n1,n2)|divide(#1,const_2)|divide(#0,#2)| | geometry |
the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 3 years hence ? | explanation : a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 3 years , their ages will be 43 and 23 . sum of their ages = 43 + 23 = 66 . d ) | a ) 58 , b ) 59 , c ) 62 , d ) 66 , e ) 74 | d | add(add(multiply(divide(60, 3), const_2), const_3), add(divide(60, 3), const_3)) | divide(n0,n1)|add(#0,const_3)|multiply(#0,const_2)|add(#2,const_3)|add(#3,#1) | general |
two trains 120 m and 190 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 120 + 190 = 310 m . required time = 310 * 9 / 250 = 11.6 = 11.6 sec . answer : b" | a ) 10.9 sec , b ) 11.6 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 20.8 sec | b | divide(add(120, 190), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
krishan and nandan jointly started a business . krishan invested 4 times as nandan did and invested his money for trible time as compared to nandan . if the gain is proportional to the money invested and the time for which the money is invested and the total gain was rs . 26000 , find the nandan ' s earning . | 4 : 1 3 : 1 - - - - - - 12 : 1 13 - - - - - rs . 26000 1 - - - - - ? = > rs . 26000 / 13 = rs . 2000 answer : e | a ) 4000 , b ) 1000 , c ) 5000 , d ) 3000 , e ) 2000 | e | divide(26000, add(multiply(const_3, const_4), const_1)) | multiply(const_3,const_4)|add(#0,const_1)|divide(n1,#1) | gain |
what is the 21 th digit to the right of the decimal point in the decimal expansion of 1 / 37 ? | "1 / 37 = 0.027027 . . . so , we have a repeating cycle of 027 . every third digit ( 3 rd , 6 th , 9 th , . . . ) to the right of the decimal point is 7 , thus 21 st digit to the right of the decimal point is also 7 . answer : d ." | a ) 0 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | d | divide(1, 37) | divide(n1,n2)| | general |
two brothers ram and ravi appeared for an exam . the probability of selection of ram is 2 / 7 and that of ravi is 1 / 5 . find the probability that both of them are selected . | "let a be the event that ram is selected and b is the event that ravi is selected . p ( a ) = 2 / 7 p ( b ) = 1 / 5 let c be the event that both are selected . p ( c ) = p ( a ) x p ( b ) as a and b are independent events : = 2 / 7 x 1 / 5 = 2 / 35 answer : a" | a ) 2 / 35 , b ) 2 / 3 , c ) 7 / 35 , d ) 5 / 7 , e ) 7 / 5 | a | multiply(divide(2, 7), divide(1, 5)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)| | general |
if a rectangular billboard has an area of 104 square feet and a perimeter of 42 feet , what is the length of each of the longer sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 104 and a perimeter of 42 . we ' re asked for the length of one of the longer sides of the rectangle . since the answers are all integers , and the area is 104 , the shorter side will almost certainly be ... | a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | d | divide(subtract(divide(42, const_2), sqrt(subtract(power(divide(42, const_2), const_2), multiply(const_4, 104)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry |
a cricket bat is sold for $ 900 , making a profit of $ 75 . the profit percentage would be | "75 / ( 900 - 75 ) = 75 / 825 = 0.9 bit more than 9 % . answer : a" | a ) 9 % , b ) 12 % , c ) 10 % , d ) 36 % , e ) 40 % | a | multiply(divide(75, subtract(900, 75)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain |
if 4 ^ ( 2 x + 2 ) = 16 ^ ( 3 x − 1 ) , what is the value of x ? | 2 ^ ( 4 x + 4 ) = 2 ^ ( 12 x - 4 ) 4 x + 4 = 12 x − 4 x = 1 answer b | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(4, add(2, 2)) | add(n1,n1)|divide(n0,#0) | general |
the product of three consecutive numbers is 720 . then the sum of the smallest two numbers is ? | "product of three numbers = 720 720 = 8 * 9 * 10 so , the three numbers are 8 , 9 and 10 . and sum of smallest of these two = 8 + 9 = 17 . answer : option c" | a ) 11 , b ) 15 , c ) 17 , d ) 38 , e ) 56 | c | multiply(power(const_2, const_4.0), factorial(720)) | factorial(const_4)|power(const_2,n0)|multiply(#0,#1)| | general |
you collect pens . suppose you start out with 25 . mike gives you another 22 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 19 pens . how many pens do you have at the end ? | "solution start with 25 pens . mike gives you 22 pens : 25 + 22 = 47 pens . cindy doubles the number of pens you have : 47 ã — 2 = 94 pens . sharon takes 19 pens from you : 94 - 19 = 75 pens . so you have 75 at the end . correct answer : d" | a ) 39 , b ) 40 , c ) 41 , d ) 75 , e ) 43 | d | subtract(multiply(add(22, 25), const_2), 19) | add(n0,n1)|multiply(#0,const_2)|subtract(#1,n2)| | general |
the population of a town increased from 1 , 45,000 to 2 , 30,500 in a decade . the average percent increase of population per year is : | "explanation : increase in 10 years = ( 230500 - 145000 ) = 85500 . increase % = ( 85500 / 145000 x 100 ) % = 58 % . required average = ( 58 / 10 ) % = 5.8 % . answer : option b" | a ) 4.37 % , b ) 5.8 % , c ) 6.8 % , d ) 8.75 % , e ) none | b | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,co... | general |
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