Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 40 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | "lets assume that candidate got 40 % votes and total votes is 100 . candidate won = 40 remaining = 60 to get 50 % , candidate requires 10 votes from 100 which is 10 % and 10 votes from 60 . 10 / 60 = 1 / 6 = . 166 = 16.67 % which is approx 17 % . hence the answer is d" | a ) 10 % , b ) 12 % , c ) 15 % , d ) 17 % , e ) 20 % | d | multiply(divide(subtract(divide(50, const_100), divide(40, const_100)), subtract(const_1, divide(40, const_100))), const_100) | divide(n2,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(const_1,#1)|divide(#2,#3)|multiply(#4,const_100)| | general |
a certain manufacturer of cake , muffin , and bread mixes has 100 buyers , of whom 50 purchases cake mix , 40 purchase muffin mix , and 16 purchase both cake mix and muffin mix . if a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neithe... | "c + m + b - cm - mb - cb - 2 cmb = 100 c - cake buyers , m - muffin and b - bread buyers . cm , mb , cb and cmb are intersecting regions . the question asks for people who have bought only bread mixes = b - cb - mb - 2 cmb has to be found out . 50 + 40 + b - cb - mb - 16 - 2 cmb = 100 b - cb - mb - 2 cmb = 26 hence th... | a ) 1 / 10 , b ) 13 / 50 , c ) 1 / 2 , d ) 7 / 10 , e ) 9 / 10 | b | divide(subtract(100, subtract(add(50, 40), 16)), 100) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)|divide(#2,n0)| | other |
if it is 5 : 14 in the evening on a certain day , what time in the morning was it exactly 2 , 880702 minutes earlier ? ( assume standard time in one location . ) | 5 : 14 minus 2 , 880,702 must end with 2 , the only answer choice which ends with 2 is a . answer : a . | a ) 5 : 32 , b ) 5 : 36 , c ) 5 : 48 , d ) 5 : 54 , e ) 5 : 58 | a | divide(5, add(const_60, subtract(14, reminder(add(multiply(multiply(2, const_1000), const_1000), 880702), const_60)))) | multiply(n2,const_1000)|multiply(#0,const_1000)|add(n3,#1)|reminder(#2,const_60)|subtract(n1,#3)|add(#4,const_60)|divide(n0,#5) | physics |
a man can row 6 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . how far is the place ? | "m = 6 s = 1.2 ds = 6 + 1.2 = 7.2 us = 6 - 1.2 = 4.8 x / 7.2 + x / 4.8 = 1 x = 2.88 answer : b" | a ) 2.29 , b ) 2.88 , c ) 2.2 , d ) 2.12 , e ) 2.14 | b | multiply(divide(multiply(add(6, 1.2), subtract(6, 1.2)), add(add(6, 1.2), subtract(6, 1.2))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
when positive integer n is divided by 3 , the remainder is 2 . when n is divided by 6 , 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 | d | subtract(100, reminder(const_4.0, 6)) | reminder(const_4.0,n2)|subtract(n4,#0)| | general |
a sports equipment store sold ping pong rackets for a total of $ 539 . if the average ( arithmetic mean ) price of a pair of rackets is $ 9.8 , how many pairs were sold ? | "average price for a pair of rackets = $ 9.8 total cost = $ 9.8 * x = $ 539 x = 55 pairs were sold . answer : a" | a ) 55 , b ) 100 , c ) 490 , d ) 500 , e ) 980 | a | divide(539, 9.8) | divide(n0,n1)| | general |
a retailer bought a machine at a wholesale price of $ 80 and later on sold it after a 20 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the wholesale price , what is the retail price of the machine ? | "the price after the discount was 1.2 * 80 = $ 96 let x be the retail price . 0.8 x = $ 96 x = 96 / 0.8 = $ 120 the answer is c ." | a ) $ 80 , b ) $ 100 , c ) $ 120 , d ) 135 , e ) 160 | c | divide(multiply(add(80, divide(multiply(80, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 20)) | multiply(n0,n2)|multiply(const_3,const_3)|divide(#0,const_100)|multiply(n1,#1)|add(n0,#2)|multiply(#4,const_100)|divide(#5,#3)| | gain |
there are 7 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 6 teams . . . so total no of matches = 7 x 6 = 42 . now , each match is played twice = > 42 x 2 but 2 teams play a match = > 42 x 2 / 2 = 42 . answer : d" | a ) 15 , b ) 16 , c ) 28 , d ) 42 , e ) 64 | d | divide(multiply(7, subtract(7, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
in expressing a length of 61.472 km as nearly as possible with the 3 significant digits , find the percentage error ? | explanation : error = ( 61.5 - 61.472 ) = 0.028 required percentage = 0.028 / 61.472100 = 0.045 answer : c | a ) 0.35 % , b ) 0.4 % , c ) 0.045 % , d ) 0.6 % , e ) none of these | c | multiply(divide(divide(multiply(add(const_3, const_4), const_4), const_100), 61.472), const_10) | add(const_3,const_4)|multiply(#0,const_4)|divide(#1,const_100)|divide(#2,n0)|multiply(#3,const_10) | general |
if x + y + z = 1 . then xy + yz + zx is | 1 / 2 + 1 / 2 + 0 = 1 xy = 1 / 2 * 1 / 2 = 1 / 4 answer : c | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | c | divide(1, const_4) | divide(n0,const_4) | general |
in the storage room of a certain bakery , the ratio of sugar to flour is 5 to 2 , and the ratio of flour to baking soda is 10 to 1 . if there were 60 more pounds of baking soda in the room , the ratio of flour to baking soda would be 8 to 1 . how many pounds of sugar are stored in the room ? | "sugar : flour = 5 : 2 = 25 : 10 ; flour : soda = 10 : 1 = 10 : 1 ; thus we have that sugar : flour : soda = 25 x : 10 x : 1 x . also given that 10 x / ( 1 x + 60 ) = 8 / 1 - - > x = 240 - - > sugar = 25 x = 6000 answer : e ." | a ) 600 , b ) 1200 , c ) 1500 , d ) 3600 , e ) 6000 | e | multiply(divide(5, 2), multiply(10, divide(multiply(8, 60), subtract(10, 8)))) | divide(n0,n1)|multiply(n4,n5)|subtract(n2,n5)|divide(#1,#2)|multiply(n2,#3)|multiply(#0,#4)| | general |
machine a can process 6000 envelopes in 3 hours . machines b and c working together but independently can process the same number of envelopes in 2.5 hours . if machines a and c working together but independently process 3000 envelopes in 1 hour , then how many hours would it take machine b to process 9800 envelopes . | "you can either take the amount of work done as the same as karishma has done or take the work done by each in the same time . i will do the latter 1 . work done in 1 hr by a is 2000 envelopes 2 . work done in 1 hr by a and c is 3000 envelopes 3 . so work done in 1 hr by c is 1000 envelopes 4 . work done in 1 hr by b a... | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | e | divide(9800, subtract(divide(6000, 2.5), subtract(divide(3000, 1), divide(6000, 3)))) | divide(n0,n2)|divide(n3,n4)|divide(n0,n1)|subtract(#1,#2)|subtract(#0,#3)|divide(n5,#4)| | physics |
difference between two numbers is 5 , 6 times of the smaller lacks by 8 from the 4 times of the greater . find the numbers ? | explanation : x β y = 5 4 x β 6 y = 8 x = 11 y = 6 d ) | a ) 5,1 , b ) 7,3 , c ) 9,8 , d ) 11,6 , e ) 13,7 | d | subtract(subtract(8, 4), const_1) | subtract(n2,n3)|subtract(#0,const_1) | general |
angelina walked 600 meters from her home to the grocery at a constant speed . she then walked 480 meters to the gym at double the speed . she spent 40 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed , in meters per second , from the grocery to t... | "let the speed be x . . . so time taken from home to grocery = 600 / x . . the speed to gym = 2 x . . so time taken = 480 / 2 x = 240 / x . . its given 600 / x - 240 / x = 40 . . 360 / x = 40 . . x = 9 m / secs . . so grocery to gym = 2 * 9 = 18 m / s . . . d" | a ) 2 , b ) 3 , c ) 4 , d ) 9 , e ) 12 | d | multiply(divide(subtract(600, divide(480, const_2)), 40), const_2) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,n2)|multiply(#2,const_2)| | physics |
a train running at the speed of 180 km / hr crosses a pole in 36 seconds . what is the length of the train ? | "speed = ( 180 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 36 ) m = 600 m . answer : d" | a ) 450 m , b ) 500 m , c ) 550 m , d ) 600 m , e ) 650 m | d | multiply(divide(multiply(180, const_1000), const_3600), 36) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores 1 / 4 , what is the value of y / x ? | "answer is c : 1 / 5 x - y = ( x + y ) / 4 solving for y / x = 1 / 5" | a ) 3 . , b ) 2 . , c ) 1 / 5 , d ) 1 / 3 . , e ) there is n ' t enough data to answer the question . | c | add(4, divide(multiply(4, 1), 4)) | multiply(n0,n1)|divide(#0,n1)|add(n1,#1)| | general |
walking 3 / 2 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ? | "speed ratio = 1 : 3 / 2 = 2 : 3 time ratio = 3 : 2 1 - - - - - - - - 3 4 - - - - - - - - - ? 12 m . answer : a" | a ) 12 , b ) 99 , c ) 27 , d ) 28 , e ) 20 | a | multiply(4, 3) | multiply(n0,n2)| | gain |
when 5 is added to half of one - third of one - fifth of a number , the result is one - fifteenth of the number . find the number ? | "explanation : let the number be 5 + 1 / 2 [ 1 / 3 ( a / 5 ) ] = a / 15 = > 5 = a / 30 = > a = 150 answer : b" | a ) 32 , b ) 150 , c ) 60 , d ) 27 , e ) 11 | b | divide(5, divide(divide(const_1, multiply(const_3, add(const_2, 5))), const_2)) | add(const_2,n0)|multiply(#0,const_3)|divide(const_1,#1)|divide(#2,const_2)|divide(n0,#3)| | general |
in a certain game , a large bag is filled with blue , green , purple and red chips worth 1 , 5 , x and 11 points each , respectively . the purple chips are worth more than the green chips , but less than the red chips . a certain number of chips are then selected from the bag . if the product of the point values of the... | "440 = 1 * 5 * 8 * 11 the factor of 8 must come from the purple point value , so there is 1 purple chip . the answer is a ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(5, const_2) | subtract(n1,const_2)| | general |
9 business executives and 7 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ? | there are 9 business exec and in each handshake 2 business execs are involved . hence 9 c 2 = 36 also , each of 9 exec will shake hand with every 7 other chairmen for total of 63 handshake . total = 36 + 63 = 99 ans : c | a ) 144 , b ) 131 , c ) 99 , d ) 90 , e ) 45 | c | add(divide(multiply(9, subtract(9, const_1)), const_2), multiply(9, 7)) | multiply(n0,n1)|subtract(n0,const_1)|multiply(n0,#1)|divide(#2,const_2)|add(#3,#0) | geometry |
from a pack of cards , two cards are drawn one after the other , with replacement . what is the probability that the first card is a club and the second card is a king or queen ? | "p ( club ) = 1 / 4 p ( king or queen ) = 2 / 13 p ( club then a king / queen ) = 1 / 4 * 2 / 13 = 1 / 26 the answer is c ." | a ) 2 / 13 , b ) 3 / 13 , c ) 1 / 26 , d ) 1 / 52 , e ) 3 / 52 | c | multiply(divide(add(multiply(const_3, const_4), const_1), const_52), divide(const_2, const_52)) | divide(const_2,const_52)|multiply(const_3,const_4)|add(#1,const_1)|divide(#2,const_52)|multiply(#3,#0)| | probability |
a container has 40 l milk . 4 l milk is taken out and replaced with water , the process done 3 times . what is the amount of milk now left ? | explanation : in regard to the problem let after n times liquid in the container is x and liquid removed is y = > liquid in container = x = > liquid removed = y after one trial , quantity of pure liquid = x - ( 1 - y / x ) after n trials quantity of pure liquid = x - ( 1 - y / x ) n therefore , putting the values ( x =... | a ) 29.16 l , b ) 29.19 l , c ) 29.12 l , d ) 29.11 l , e ) 29.14 l | a | subtract(40, multiply(4, 3)) | multiply(n1,n2)|subtract(n0,#0) | general |
if the length of rectangle is three times of its breadth . if the area of rectangle is 6075 sq . m , then calculate the length ? | let breadth = x , length = 3 x . area of rectangle = length * breadth = 3 x ^ 2 = 6075 x ^ 2 = 2025 , x = β 2025 = 45 m length = 135 m answer b | ['a ) 145', 'b ) 135', 'c ) 75', 'd ) 140', 'e ) none of these'] | b | multiply(sqrt(divide(6075, const_3)), const_3) | divide(n0,const_3)|sqrt(#0)|multiply(#1,const_3) | geometry |
the average of 9 observations was 9 , that of the 1 st of 4 being 10 and that of the last 4 being 8 . what was the 5 th observation ? | "1 to 9 = 9 * 9 = 81 1 to 4 = 4 * 10 = 40 4 to 9 = 6 * 8 = 48 5 th = 48 + 40 = 88 β 81 = 7 answer : d" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | subtract(add(multiply(10, 4), multiply(9, 4)), multiply(9, 9)) | multiply(n3,n4)|multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general |
how much water must be added to 48 litres of milk at 1 1 β 2 litres for 20 so as to have a mixture worth 10 2 β 3 a litre ? | "c . p . of 1 litre of milk = ( 20 Γ 2 β 3 ) = 40 β 3 β΄ ratio of water and milk = 8 β 3 : 32 β 3 = 8 : 32 = 1 : 4 β΄ quantity of water to be added to 48 litres of milk = ( 1 β 4 Γ 48 ) litres = 12 litres . answer b" | a ) 10 litres , b ) 12 litres , c ) 15 litres , d ) 18 litres , e ) none of these | b | divide(20, add(1, divide(1, 2))) | divide(n1,n3)|add(n1,#0)|divide(n4,#1)| | general |
a salesman Γ’ β¬ β’ 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 . 100 more than that by the previous schema , his sales were worth ? | "[ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 100 x = 32000 answer : c" | a ) s . 14,000 , b ) s . 12,000 , c ) s . 32,000 , d ) s . 40,000 , e ) s . 50,000 | c | subtract(multiply(5, const_4), const_12) | multiply(n0,const_4)|subtract(#0,const_12)| | general |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 24 seconds . the ratio of their speeds is ? | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 24 = = > 27 x + 17 y = 24 x + 24 y = = > 3 x = 7 y = = > x / y = 7 / 3 . answer : c" | a ) 3 / 6 , b ) 3 / 2 , c ) 7 / 3 , d ) 3 / 5 , e ) 5 / 2 | c | divide(subtract(27, 24), subtract(24, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics |
32 % 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 ( e ) 80 think of 100 people total : from the first fact , 32 of these are women with fair hair . from the second fact , these 30 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 ) * 32 equals th... | a ) 25 , b ) 30 , c ) 50 , d ) 55 , e ) 80 | e | multiply(divide(32, 40), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
while visiting a small town in the united states , i lost my overcoat in a bus . when i reported the matter to the bus company i was asked the number of the bus . though i did not remember the exact number i did remember that the bus number bad a certain peculiarity about it . the number plate showed the bus number as ... | b 196 the only numbers that can be turned upside down and still read as a number are 0 , 1 , 6 , 8 and 9 . the numbers 0 , 1 and 8 remain 0 , 1 and 8 when turned over , but 6 becomes 9 and 9 becomes 6 . therefore the possible numbers on the bus were 9 , 16 , 81 , 100 , 169 or 196 . however , the number 196 is the only ... | a ) 127 , b ) 196 , c ) 116 , d ) 140 , e ) 234 | b | add(subtract(100, const_4), const_100) | subtract(n1,const_4)|add(#0,const_100) | geometry |
a car traveled from san diego to san francisco at an average speed of 51 miles per hour . if the journey back took twice as long , what was the average speed of the trip ? | "let the time taken be = x one way distance = 51 x total distance traveled = 2 * 51 x = 102 x total time taken = x + 2 x = 3 x average speed = 102 x / 3 x = 34 answer : b" | a ) 24 . , b ) 34 . , c ) 36 . , d ) 42 . , e ) 44 . | b | inverse(add(inverse(51), divide(inverse(51), const_2))) | inverse(n0)|divide(#0,const_2)|add(#1,#0)|inverse(#2)| | physics |
in what time will a train 150 metres long cross an electric pole , if its speed be 122 km / hr ? | "solution speed = ( 122 x 5 / 18 ) m / sec = 33.88 m / sec time taken = ( 150 / 33.88 ) sec = 4.42 sec . answer d" | a ) 2.42 , b ) 6.42 , c ) 5.42 , d ) 4.42 , e ) 3.42 | d | divide(150, multiply(122, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
what is the sum of the greatest common factor and the lowest common multiple of 16 and 36 ? | "prime factorization of the given numbers 16 = 2 ^ 4 36 = 2 ^ 2 * 3 * 2 greatest common factor = 2 ^ 2 = 4 lowest common multiple = 2 ^ 4 * 3 ^ 2 = 144 sum = 4 + 144 = 148 answer a" | a ) 148 , b ) 222 , c ) 144 , d ) 157 , e ) 164 | a | divide(multiply(16, 36), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
x is a positive integer less than 100 . when x is divided by 7 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ? | "the nubmer which when divided by 7 leaves remainder 1 should be of the form 7 k + 1 this number when divided by 3 leaves remainder 2 . so , ( 7 k + 1 ) - 2 should be divisible by 3 or 7 k - 1 should be divisible by 3 . we now put the values of k starting from 0 to find first number divisible by 3 we find 1 st number a... | a ) 14 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | a | subtract(add(multiply(reminder(7, 100), 3), reminder(3, 100)), reminder(1, 100)) | reminder(n1,n0)|reminder(n3,n0)|reminder(n2,n0)|multiply(n3,#0)|add(#3,#1)|subtract(#4,#2)| | general |
a student has to obtain 33 % of the total marks to pass . he got 125 marks and failed by 40 marks . the maximum mark is | "the student got 125 marks and still failed by 40 marks = 125 + 40 = 165 therefore 33 % of the total marks = 165 100 % of the total = 165 * 100 / 33 = 55 max mark = 500 answer b" | a ) 600 , b ) 500 , c ) 400 , d ) 300 , e ) 800 | b | divide(add(125, 40), divide(33, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | general |
the number of boxes in a warehouse can be divided evenly into 9 equal shipments by boat or 24 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ? | "answer is the lcm of 9 and 24 = 72 answer c" | a ) 27 , b ) 33 , c ) 72 , d ) 81 , e ) 162 | c | multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 9) | multiply(const_2,const_2)|multiply(#0,const_2)|multiply(#1,const_3)|multiply(n0,#2)| | general |
an outlet pipe empties a tank which is full in 5 hours . if the inlet pipe is kept open , which lets water in at the rate of 8 litres / min then outlet pipe would take 3 hours longer . find the capacity of the tank . | "let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 8 litres / min , or 8 * 60 = 480 liters / hour ; net outflow rate when both pipes operate would be x - 480 liters / hour . capacity of the tank = x * 5 hours = ( x - 480 ) * ( 5 + 3 ) hours 5 x = ( x - 480 ) * 8 - - > x = 1280 - - > capacity = 5 x ... | a ) 8600 litres , b ) 200 litres , c ) 12800 litres , d ) 6400 litres , e ) 13200 litres | d | multiply(divide(multiply(multiply(8, const_60), add(5, 3)), 3), 5) | add(n0,n2)|multiply(n1,const_60)|multiply(#0,#1)|divide(#2,n2)|multiply(n0,#3)| | physics |
a can do a piece of work in 3 hours ; b and c together can do it in 2 hours , which a and b together can do it in 2 hours . how long will c alone take to do it ? | "a ' s 1 hour work = 1 / 3 ; ( b + c ) ' s 1 hour work = 1 / 2 ; ( a + b ) ' s 1 hour work = 1 / 2 ( a + b + c ) ' s 1 hour work = ( 1 / 3 + 1 / 2 ) = 5 / 6 c ' s 1 hour work = ( 5 / 6 - 1 / 2 ) = 1 / 3 c alone will take 3 hours to do the work . answer : e" | a ) 2 hours , b ) 5 hours , c ) 6 hours , d ) 4 hours , e ) 3 hours | e | divide(const_1, subtract(divide(const_1, 2), subtract(divide(const_1, 2), divide(const_1, 3)))) | divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4)| | physics |
tony alone can paint a wall in 7 days and his friend roy alone can paint the same wall in 9 days . in how many days they can paint the wall working together ? round off the answer to the nearest integer . | sol . use formula ( xy / x + y ) so nearest value for 3.93 = 4 answer : option b | a ) 2 , b ) 4 , c ) 6 , d ) 7 , e ) 9 | b | divide(multiply(7, 9), add(7, 9)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0) | physics |
when the number 9 y 30012 is exactly divisible by 11 , what is the smallest whole number that can replace y ? | "the given number = 9 y 30012 sum of the odd places = 2 + 0 + 3 + 9 = 14 sum of the even places = 1 + 0 + y ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) 14 - ( 1 + y ) = divisible by 11 13 οΏ½ y = divisible by 11 . y must be 2 , to make given number divisible by 11 . b" | a ) 1 , b ) 2 , c ) 5 , d ) 6 , e ) 8 | b | subtract(9, 9) | subtract(n0,n0)| | general |
a worker ' s daily wage is increased by 40 % and the new wage is $ 35 per day . what was the worker ' s daily wage before the increase ? | "let x be the daily wage before the increase . 1.4 x = $ 35 x = $ 25 the answer is d ." | a ) $ 20 , b ) $ 21 , c ) $ 24 , d ) $ 25 , e ) $ 28 | d | divide(35, add(const_1, divide(40, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | physics |
for the past n days , the average ( arithmetic mean ) daily production at a company was 60 units . if today ' s production of 90 units raises the average to 65 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 60 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 60 n + 90 = 65 * ( n + 1 ) - - > n = 5 . or as 30 extra units increased the average for n + 1 days by 5 u... | a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 5 | e | subtract(divide(subtract(90, 60), subtract(65, 60)), const_1) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,#1)|subtract(#2,const_1)| | general |
a 7 - digit combination lock on a safe has zero exactly 3 times , does not have the digit 1 at all . what is the probability that exactly 3 of its digits are odd ? | remaining number of digits after three ' 0 ' digits = 4 constraint - no more ' 0 ' and no more ' 1 ' well , these four digits can be odd or even 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 ways to choose 3 odd digits out of total 4 digits = 4 c 3 = 4 4 * 1 / 16 = 1 / 4 d is the answer hi ! since we are talking about remaini... | a ) 9 / 16 , b ) 1 / 2 , c ) 1 / 3 , d ) 1 / 4 , e ) 1 / 6 | d | divide(1, subtract(7, 3)) | subtract(n0,n1)|divide(n2,#0) | general |
area of a square is 1 / 2 hectare . the diagonal of the square is ? | area = 1 / 2 hectare = 10000 / 2 m 2 = 5000 m 2 again area = 1 / 2 x ( diagonal ) 2 so 1 / 2 x ( diagonal ) 2 = 5000 m 2 diagonal 2 = 10000 diagonal = 100 answer : b | ['a ) 250 meter', 'b ) 100 meter', 'c ) 50 β 2 meter', 'd ) 50 meter', 'e ) 35 meter'] | b | sqrt(multiply(const_2, multiply(divide(1, 2), multiply(const_100, const_100)))) | divide(n0,n1)|multiply(const_100,const_100)|multiply(#0,#1)|multiply(#2,const_2)|sqrt(#3) | geometry |
the ratio of the cost price and the selling price is 4 : 5 . the profit percent is : | "c c . p . = rs . 4 x . then , s . p . = rs . 5 x gain = ( 5 x - 4 x ) = rs . x gain % = ( x * 100 ) / 4 x = 25 % ." | a ) 67 % , b ) 45 % , c ) 25 % , d ) 70 % , e ) 50 % | c | multiply(subtract(divide(5, 4), const_1), const_100) | divide(n1,n0)|subtract(#0,const_1)|multiply(#1,const_100)| | gain |
pipe a and pipe b fill water into a tank of capacity 5000 litres , at a rate of 200 l / min and 50 l / min . pipe c drains at a rate of 25 l / min . pipe a is open for 1 min and closed , then pipe b is open for 2 min and closed . further the pipe c is opened and drained for another 2 min . this process is repeated unti... | tank capacity : 5000 l , 1 st - 200 l / min for 1 min , volume filled : 200 l 2 nd - 100 l / min for 2 min , volume filled : 100 l 3 rd ( water draining ) : 25 l / min * 2 : 50 l total : ( 200 + 100 ) - 50 = 250 l filled for 1 cycle number of 250 in 5000 l tank : 1000 / 250 = 20 time taken to fill : 20 * total time = 2... | a ) 120 , b ) 100 , c ) 140 , d ) 80 , e ) 90 | b | multiply(add(add(1, 2), 2), divide(5000, subtract(add(200, multiply(50, const_2)), multiply(25, 2)))) | add(n4,n5)|multiply(n2,const_2)|multiply(n3,n5)|add(n5,#0)|add(n1,#1)|subtract(#4,#2)|divide(n0,#5)|multiply(#3,#6) | physics |
a can do a piece of work in 10 days and b in 20 days . they began the work together but 5 days before the completion of the work , a leaves . the work was completed in ? | b ( x Γ’ β¬ β 5 ) / 10 + x / 20 = 1 x = 10 days | a ) 15 days , b ) 10 days , c ) 12 days , d ) 19 days , e ) 13 days | b | add(divide(subtract(const_1, add(multiply(5, divide(const_1, 10)), multiply(5, divide(const_1, 20)))), divide(const_1, 20)), 5) | divide(const_1,n0)|divide(const_1,n1)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)|add(n2,#6) | physics |
a car gets 20 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 130 kilometers ? | "each 20 kilometers , 1 gallon is needed . we need to know how many 20 kilometers are there in 130 kilometers ? 130 / 20 = 6.5 * 1 gallon = 6.5 gallons correct answer c" | a ) 8.5 gallons , b ) 7.5 gallons , c ) 6.5 gallons , d ) 5.5 gallons , e ) 4.5 gallons | c | divide(130, 20) | divide(n1,n0)| | physics |
the diameter of a garden roller is 1.4 m and it is 2 m long . how much area will it cover in 6 revolutions ? ( use Γ― β¬ = 22 Γ’ Β β 7 ) | "required area covered in 5 revolutions = 6 Γ£ β 2 Γ― β¬ rh = 6 Γ£ β 2 Γ£ β 22 Γ’ Β β 7 Γ£ β 0.7 Γ£ β 2 = 52.8 m 2 answer c" | a ) 40 m 2 , b ) 44 m 2 , c ) 52.8 m 2 , d ) 36 m 2 , e ) none of these | c | multiply(6, multiply(multiply(multiply(2, divide(22, 7)), divide(1.4, 2)), 2)) | divide(n0,n1)|divide(n3,n4)|multiply(#1,n1)|multiply(#0,#2)|multiply(n1,#3)|multiply(n2,#4)| | physics |
at what price must an article costing rs . 200 be marked in order that after deducting 10 % from the list price . it may be sold at a profit of 20 % on the cost price ? | "cp = 200 sp = 200 * ( 120 / 100 ) = 240 mp * ( 90 / 100 ) = 240 mp = 266.67 answer : c" | a ) 62.5 , b ) 62.0 , c ) 266.67 , d ) 62.1 , e ) 62.2 | c | divide(multiply(add(200, divide(multiply(200, 20), const_100)), const_100), subtract(const_100, 10)) | multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)| | gain |
evaluate : 11 + sqrt ( - 4 + 6 Γ 4 Γ· 3 ) = ? | "according to order of operations , inner brackets first where 6 Γ 4 Γ· 3 is first calculated since it has a multiplication and a division . 6 Γ 4 Γ· 3 = 24 Γ· 3 = 8 hence 11 + sqrt ( - 4 + 6 Γ 4 Γ· 3 ) = 11 + sqrt ( - 4 + 8 ) = 11 + sqrt ( 4 ) = 11 + 2 = 13 correct answer d ) 13" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | d | add(11, sqrt(subtract(divide(multiply(6, 4), 3), 4))) | multiply(n1,n2)|divide(#0,n4)|subtract(#1,n1)|sqrt(#2)|add(n0,#3)| | general |
the lcm and hcf of two numbers are 1000 and 100 respectively . find the larger of the two numbers if their sum is 1100 . | "there are 2 approaches in solving this . methode 1 . hcf * lcm = the actual number . 1000 * 100 = 100000 so the answer which we are looking for has to be a factor of 100000 . so among the options shortlist the answers by eliminating those numbers which is not divisible by 100000 . and then take the highest number as t... | a ) 1000 , b ) 10 , c ) 567 , d ) 57 , e ) 678 | a | divide(multiply(1000, 100), 1100) | multiply(n0,n1)|divide(#0,n2)| | physics |
ramesh purchased a refrigerator for rs . 12500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 15 % if no discount was offered ? | "price at which the tv set is bought = rs . 12,500 discount offered = 20 % marked price = 12500 * 100 / 80 = rs . 15625 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 15625 + 375 = rs . 16000 the price at which the tv should be sold to get a profit of 15 % if no di... | a ) 17608 , b ) 17606 , c ) 17604 , d ) 18400 , e ) 117601 | d | divide(multiply(add(const_100, 15), add(divide(multiply(12500, const_100), subtract(const_100, 20)), add(125, 250))), const_100) | add(n4,const_100)|add(n2,n3)|multiply(n0,const_100)|subtract(const_100,n1)|divide(#2,#3)|add(#1,#4)|multiply(#0,#5)|divide(#6,const_100)| | gain |
in a school 10 % of the boys are same in number as 1 / 6 th of the girls . what is the ratio of boys to the girls in the school ? | "10 % of b = 1 / 6 g 10 b / 100 = g / 6 b = 5 g / 3 b / g = 5 / 3 b : g = 5 : 3 answer is a" | a ) 5 : 3 , b ) 2 : 3 , c ) 1 : 4 , d ) 3 : 7 , e ) 2 : 5 | a | divide(divide(10, const_100), divide(1, 6)) | divide(n0,const_100)|divide(n1,n2)|divide(#0,#1)| | other |
the number of positive integer solutions for the equation x + y + z + t = 15 is | the number of positive integer solutions for the equatio fx 1 + x 2 + β― + xn = k ( k - 1 ) c ( n - 1 ) - where k is the number and n is number of variable in the equation . 15 - 1 c 4 - 1 = 14 c 3 = 364 answer : d | a ) 654 , b ) 584 , c ) 384 , d ) 364 , e ) 394 | d | divide(factorial(subtract(15, const_1)), multiply(factorial(subtract(const_4, const_1)), factorial(subtract(subtract(15, const_1), subtract(const_4, const_1))))) | subtract(n0,const_1)|subtract(const_4,const_1)|factorial(#0)|factorial(#1)|subtract(#0,#1)|factorial(#4)|multiply(#3,#5)|divide(#2,#6) | general |
in a certain pond , 50 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the ... | "if x is total number of fish in the pond : 4 = 50 / x * 100 = > x = 1250 so answer is c" | a ) 400 , b ) 625 , c ) 1250 , d ) 2500 , e ) 10 000 | c | divide(50, divide(2, 50)) | divide(n2,n1)|divide(n0,#0)| | gain |
if the cost price of 12 pencils is equal to the selling price of 8 pencils , the gain percent is : | c 50 % let c . p . of each pencil be $ 1 . then , c . p . of 8 pencils = $ 8 ; s . p . of 8 pencils = $ 12 . gain % = 4 / 8 * 100 = 50 % | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | divide(const_100, divide(8, subtract(12, 8))) | subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1) | gain |
a and b together can do a work in 9 days . if a alone can do it in 18 days . in how many days can b alone do it ? | "1 / 9 β 1 / 18 = 1 / 18 = > 18 answer : c" | a ) 10 , b ) 99 , c ) 18 , d ) 55 , e ) 21 | c | add(inverse(subtract(divide(const_1, 9), divide(const_1, 18))), divide(const_2, add(const_2, const_3))) | add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|divide(const_2,#0)|subtract(#1,#2)|inverse(#4)|add(#3,#5)| | physics |
solve below question 6 x - 1 = - 13 | 1 . subtract 1 from both sides : 6 x - 1 + 1 = - 13 + 1 2 . simplify both sides : 6 x = - 12 3 . divide both sides by 6 : 4 . simplify both sides : x = - 2 b | a ) - 8 , b ) - 2 , c ) - 5 , d ) - 4 , e ) 1 | b | divide(negate(add(13, 1)), 6) | add(n1,n2)|negate(#0)|divide(#1,n0)| | general |
if n = ( 32 ) ^ 43 + ( 43 ) ^ 33 what is the units digit of n ? | "first of all , the units digit of ( 32 ) ^ 43 is the same as that of 3 ^ 43 and the units digit of ( 43 ) ^ 33 is the same as that of 3 ^ 33 . so , we need to find the units digit of 3 ^ 43 + 3 ^ 33 . next , the units digit of 3 in positive integer power repeats in blocks of four { 3 , 9 , 7 , 1 } : 3 ^ 1 = 3 ( the un... | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | e | subtract(reminder(add(power(reminder(32, const_10), reminder(43, const_4)), power(reminder(43, const_10), const_4)), const_100), multiply(const_2, const_3)) | multiply(const_2,const_3)|reminder(n0,const_10)|reminder(n1,const_4)|reminder(n1,const_10)|power(#1,#2)|power(#3,const_4)|add(#4,#5)|reminder(#6,const_100)|subtract(#7,#0)| | general |
if 125 ( 5 ^ y ) = 1 then y = | 5 ^ x = 1 / 125 5 ^ x = 1 / 5 ^ 3 5 ^ x = 5 ^ - 3 x = - 3 e | a ) β 2 , b ) β 1 , c ) 0 , d ) 1 , e ) - 3 | e | divide(log(divide(1, 125)), log(5)) | divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1) | general |
a certain store sells all maps at one price and all books at another price . on monday the store sold 12 maps and 8 books for a total of $ 38.00 , and on tuesday the store sold 20 maps and 12 books for a total of $ 60.00 . at this store , how much less does a map sell for than a book ? | "12 x + 8 y = 38 20 x + 12 y = 60 multiply eq 1 by 5 and eq 2 by 3 60 x + 40 y = 38 * 5 60 x + 36 y = 60 * 3 subtracting 1 from 2 4 y = 190 - 180 y = 2.5 therefore x = 1.5 difference in price = 1 d" | a ) $ 0.25 , b ) $ 0.50 , c ) $ 0.75 , d ) $ 1.00 , e ) $ l . 25 | d | divide(subtract(60.00, negate(multiply(12, divide(subtract(multiply(38.00, const_4), multiply(60.00, const_2)), subtract(multiply(8, const_4), multiply(12, const_2)))))), 20) | multiply(n2,const_4)|multiply(n5,const_2)|multiply(n1,const_4)|multiply(n4,const_2)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(n4,#6)|negate(#7)|subtract(n5,#8)|divide(#9,n3)| | general |
the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 11000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | "answer average salary = 8000 + 5000 + 11000 + 7000 + 9000 / 5 = rs . 8000 correct option : b" | a ) rs . 7000 , b ) rs . 8000 , c ) rs . 8500 , d ) rs . 9000 , e ) none | b | divide(add(add(add(add(8000, 5000), 11000), 7000), 9000), add(const_4, const_1)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
a car averages 75 mph for the first 4 hours of a trip and averages 60 mph for each additional hour . the average speed for the entire trip was 70 mph . how many hours long is the trip ? | "let the time for which car averages 60 mph = t 70 * ( t + 4 ) = 75 * 4 + 60 t = > 10 t = 20 = > t = 2 total duration of the trip = 4 + 2 = 6 answer a" | a ) 6 , b ) 78 , c ) 8 , d ) 9 , e ) 10 | a | add(divide(subtract(multiply(70, 4), multiply(75, 4)), subtract(60, 70)), 4) | multiply(n3,n1)|multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)|add(n1,#4)| | general |
linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 210 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 210 . so 1 / 4 of her savings is $ 210 . so her original savings are 4 times $ 210 = $ 840 correct answer a" | a ) $ 840 , b ) $ 800 , c ) $ 1000 , d ) $ 700 , e ) $ 1500 | a | divide(210, subtract(const_1, divide(3, 4))) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)| | general |
the duplicate ratio of 1 : 4 is ? | "1 ^ 2 : 4 ^ 2 = 2 : 16 = 1 : 8 answer : c" | a ) 1 : 2 , b ) 1 : 4 , c ) 1 : 8 , d ) 1 : 18 , e ) 1 : 13 | c | divide(power(1, const_2), power(4, const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,#1)| | other |
find the cost of fencing around a circular field of diameter 20 m at the rate of rs . 1.50 a meter ? | "2 * 22 / 7 * 10 = 62.8 62.8 * 1 1 / 2 = rs . 94.2 answer : b" | a ) 287 , b ) 94.2 , c ) 156 , d ) 158 , e ) 267 | b | multiply(circumface(divide(20, const_2)), 1.50) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics |
a 60 kg metal bar made of alloy of tin and silver lost 6 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ? | "you can simply use this formula to avoid confusion : w 1 / w 2 = ( a 2 - aavg ) / ( avg - a 1 ) here is how you will find the values of a 1 an a 2 . we have an overall loss ( average loss ) . the average loss is 6 kg when 60 kg alloy is immersed . this is a loss of ( 6 / 60 ) * 100 = 10 % . this is aavg the loss of ti... | a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3 | e | divide(divide(subtract(6, multiply(divide(0.375, 5), 60)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(60, divide(subtract(6, multiply(divide(0.375, 5), 60)), subtract(divide(1.375, 10), divide(0.375, 5))))) | divide(n5,n4)|divide(n3,n2)|multiply(n0,#0)|subtract(#1,#0)|subtract(n1,#2)|divide(#4,#3)|subtract(n0,#5)|divide(#5,#6)| | other |
a walks at 10 kmph and 10 hours after his start , b cycles after him at 20 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 10 hours . x / 10 - x / 20 = 10 x = 200 km answer is e" | a ) 100 km , b ) 150 km , c ) 50 km , d ) 120 km , e ) 200 km | e | multiply(10, 20) | multiply(n1,n2)| | physics |
the sum of the non - prime numbers between 60 and 70 , non - inclusive , is | "sum of consecutive integers from 61 to 69 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 61 + 69 ) / 2 * 9 = 65 * 9 = 585 sum of non - prime numbers b / w 60 and 70 , non inclusive = = = > 585 - 128 ( i . e . , 61 + 67 , being the prime # s in the range ) = 457 answer : c" | a ) 387 , b ) 429 , c ) 457 , d ) 499 , e ) 536 | c | add(add(add(add(add(add(add(60, const_1), add(add(60, const_1), const_1)), add(add(add(60, const_1), const_1), const_2)), add(add(add(add(60, const_1), const_1), const_2), const_1)), add(add(add(add(add(60, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(60, const_1), const_1), const_2), cons... | add(n0,const_1)|add(#0,const_1)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(#3,const_1)|add(#4,#5)|add(#5,const_1)|add(#6,#7)|add(#7,const_1)|add(#8,#9)|add(#9,const_1)|add(#10,#11)| | general |
how many seconds will a 500 meter long train moving with a speed of 63 km / hr , take to cross a man walking with a speed of 3 km / hr in the direction of the train ? | distance = 500 m speed = 63 - 3 km / hr = 60 km / hr = 600 / 36 m / s = 50 / 3 m / s time taken = distance / speed = 500 / ( 50 / 3 ) = 30 s answer is d . | a ) 15 s , b ) 20 s , c ) 25 s , d ) 30 s , e ) 35 s | d | divide(500, multiply(subtract(63, 3), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1) | physics |
evaluate 50 % of 450 + 45 % of 500 | "explanation : = ( 50 / 100 ) * 450 + ( 45 / 100 ) * 500 = 450 option b" | a ) 232 , b ) 450 , c ) 252 , d ) 262 , e ) 272 | b | divide(50, divide(450, 50)) | divide(n1,n0)|divide(n0,#0)| | gain |
ifaequals the sum of the even integers from 2 to 20 , inclusive , andbequals the sum of the odd integers from 1 to 19 , inclusive , what is the value of b - a ? | "yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / 2 here a = sum of even ints from 2 to 20 , inclusive number of even ints = 10 , first int + last int = 2 + 20 = 22 a = 10 * 22 / 2 = 110 b = sum of odd ints from 1 to 19 ... | a ) - 10 , b ) 10 , c ) - 19 , d ) 20 , e ) 21 | a | subtract(multiply(divide(20, 2), add(divide(20, 2), 1)), multiply(divide(add(19, 1), 2), add(divide(subtract(19, 1), 2), 1))) | add(n2,n3)|divide(n1,n0)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,n0)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8)| | general |
the angle of elevation of a ladder leaning against a wall is 60 ΒΊ and the foot of the ladder is 4.6 m away from the wall . the length of the ladder is : | "let ab be the wall and bc be the ladder . then , acb = 60 ΒΊ and ac = 4.6 m . ac / bc = cos 60 ΒΊ = 1 / 2 bc = 2 x ac = ( 2 x 4.6 ) m = 9.2 m . answer : option d" | a ) 2.3 m , b ) 4.6 m , c ) 7.8 m , d ) 9.2 m , e ) 10.1 | d | multiply(4.6, const_2) | multiply(n1,const_2)| | gain |
what is the cost of leveling the field in the form of parallelogram at the rate of rs . 50 / 10 sq . metre , whose base & perpendicular distance from the other side being 94 m & 54 m respectively ? | "area of the parallelogram = length of the base * perpendicular height = 94 * 54 = 5076 m . total cost of levelling = rs . 5076 e" | a ) s . 6420 , b ) s . 3563 , c ) s . 2400 , d ) s . 5480 , e ) s . 5076 | e | multiply(multiply(94, 54), divide(50, 10)) | divide(n0,n1)|multiply(n2,n3)|multiply(#0,#1)| | physics |
what is the minimum value of | x - 4 | + | x + 8 | + | x - 5 | ? | "a can not be the answer as all the three terms are in modulus and hence the answer will be non negative . | x - 4 | > = 0 - - > minimum occurs at x = 4 | x + 8 | > = 0 - - > minimum occurs at x = - 8 | x - 5 | > = 0 - - > minimum occurs at x = 5 x = - 8 - - > result = 12 + 0 + 13 = 25 . also any negative value will pu... | a ) - 13 , b ) 13 , c ) 5 , d ) 7 , e ) 8 | b | add(8, 5) | add(n1,n2)| | general |
divide rs . 1600 among a , b and c so that a receives 1 / 4 as much as b and c together and b receives 3 / 4 as a and c together . a ' s share is ? | "a + b + c = 1600 a = 1 / 4 ( b + c ) ; b = 3 / 4 ( a + c ) a / ( b + c ) = 1 / 4 a = 1 / 5 * 1600 = > 320 answer : c" | a ) 350 , b ) 375 , c ) 320 , d ) 530 , e ) 230 | c | divide(1600, const_3) | divide(n0,const_3)| | general |
the population of a town is 10000 . it increases annually at the rate of 15 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 10000 Γ£ β 115 / 100 Γ£ β 115 / 100 = 13225 c" | a ) 13300 , b ) 13350 , c ) 13225 , d ) 13500 , e ) 13600 | c | add(10000, multiply(divide(multiply(10000, 15), const_100), 2)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,n2)|add(n0,#2)| | gain |
a man ' s speed with the current is 16 km / hr and the speed of the current is 3.2 km / hr . the man ' s speed against the current is | "man ' s rate in still water = ( 16 - 3.2 ) km / hr = 12.8 km / hr . man ' s rate against the current = ( 12.8 - 3.2 ) km / hr = 9.6 km / hr . answer : b" | a ) 9 , b ) 9.6 , c ) 11 , d ) 12 , e ) 13 | b | subtract(subtract(16, 3.2), 3.2) | subtract(n0,n1)|subtract(#0,n1)| | gain |
a card game called β high - low β divides a deck of 52 playing cards into 2 types , β high β cards and β low β cards . there are an equal number of β high β cards and β low β cards in the deck and β high β cards are worth 2 points , while β low β cards are worth 1 point . if you draw cards one at a time , how many ways... | "great question ravih . this is a permutations problem ( order matters ) with repeating elements . given thatlowcards are worth 1 pt andhigh cards 2 pts , and you must draw 3 low cards , we know that you must also draw 1 high card . the formula for permutations problems with repeating elements isn ! / a ! b ! . . . whe... | a ) 1 , b ) 20 , c ) 3 , d ) 4 , e ) 5 | b | divide(multiply(multiply(multiply(const_4, 3), 2), 1), multiply(multiply(3, 2), 1)) | multiply(n5,const_4)|multiply(n1,n5)|multiply(n1,#0)|multiply(n3,#1)|multiply(n3,#2)|divide(#4,#3)| | general |
a card is drawn from a pack of 52 cards . the probability of not getting a face card ? | clearly in the 52 cards other than face cards = 52 - 16 = 36 probability of not getting a face card = 36 / 52 = 9 / 13 correct option is a | a ) 9 / 13 , b ) 5 / 12 , c ) 3 / 13 , d ) 6 / 17 , e ) 1 / 13 | a | divide(subtract(52, multiply(const_4, const_4)), 52) | multiply(const_4,const_4)|subtract(n0,#0)|divide(#1,n0) | probability |
what number is 64 more than two - fourth of itself ? | "2 / 4 x + 64 = x that means 64 = 2 / 4 x x = 64 * 2 = 128 a is the answer" | a ) 128 , b ) 225 , c ) 288 , d ) 324 , e ) 336 | a | divide(64, subtract(const_1, divide(const_1, const_4))) | divide(const_1,const_4)|subtract(const_1,#0)|divide(n0,#1)| | general |
in 10 years , a will be twice as old as b was 10 years ago . if a is now 6 years older than b the present age of b is | "let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 6 = > 2 b - b - 6 = 30 b = 36 so the present age of b is 36 years answer : b" | a ) 37 , b ) 36 , c ) 39 , d ) 40 , e ) 41 | b | add(add(multiply(const_2, 10), 10), 6) | multiply(n0,const_2)|add(n0,#0)|add(n2,#1)| | general |
a train running at the speed of 56 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 56 * 5 / 18 = 140 / 9 m / sec length of the train = speed * time = 140 / 9 * 9 = 140 m answer : a" | a ) 140 m , b ) 786 m , c ) 566 m , d ) 546 m , e ) 445 m | a | multiply(divide(multiply(56, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
the cross - section of a cannel is a trapezium in shape . if the cannel is 11 m wide at the top and 9 m wide at the bottom and the area of cross - section is 960 sq m , the depth of cannel is ? | "1 / 2 * d ( 11 + 9 ) = 960 d = 96 answer : e" | a ) 69 , b ) 92 , c ) 94 , d ) 49 , e ) 96 | e | divide(divide(divide(960, divide(add(11, 9), const_2)), 9), const_2) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)| | physics |
suppose 6 monkeys take 6 minutes to eat 6 bananas . how many monkeys would it take to eat 18 bananas in 18 minutes ? | "one monkey takes 6 min to eat 1 banana , so in 18 mins 1 monkey will eat 3 bananas , so for 18 bananas in 18 min we need 18 / 3 = 6 monkeys answer : d" | a ) 9 , b ) 10 , c ) 11 , d ) 6 , e ) 13 | d | divide(18, divide(18, 6)) | divide(n3,n0)|divide(n3,#0)| | physics |
6 ) a marketing firm determined that , of 240 households surveyed , 80 used neither brand a nor brand b soap . 60 used only brand a soap and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 household surveyed used both brands of soap ? | "160 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 160 = > 4 x = 100 = > x = 25 answer - c" | a ) a ) 15 , b ) b ) 20 , c ) c ) 25 , d ) d ) 40 , e ) e ) 45 | c | divide(subtract(subtract(240, 80), 60), const_4) | subtract(n1,n2)|subtract(#0,n3)|divide(#1,const_4)| | other |
if 20 % of a number is equal to three - fifth of another number , what is the ratio of first number to the second number ? | "let 20 % of a = 3 / 5 b . then , 20 a / 100 = 3 b / 5 = > 1 a / 5 = 3 b / 5 a / b = ( 3 / 5 * 5 / 1 ) = 3 / 1 a : b = 3 : 1 . answer : a" | a ) 3 : 1 , b ) 2 : 1 , c ) 1 : 3 , d ) 1 : 2 , e ) 2 : 3 | a | divide(divide(const_1, const_4), divide(20, const_100)) | divide(const_1,const_4)|divide(n0,const_100)|divide(#0,#1)| | general |
the sum of the ages of 5 children born at the intervals of 3 years each is 80 years . what is the age of the youngest child ? | "let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 80 5 x = 50 = > x = 10 . age of youngest child = x = 10 years . answer : e" | a ) 2 years , b ) 4 years , c ) 6 years , d ) 8 years , e ) 10 years | e | divide(subtract(divide(80, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | divide(n0,n1)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,n1)| | general |
a runs 1 2 β 3 times as fast as b . if a gives b a start of 80 m , how far must the winning post be so that a and b might reach it at the same time ? | speed of a : speed of b = 5 β 3 : 1 = 5 : 3 i . e . , in a race of 5 m , a runs 5 m and b runs 3 m i . e . , a gains 2 m over b in a race of 5 m = > a gains 80 m over b in a race of 52 Γ 80 = 200 m hence , winning point should be 200 m away from the starting point . answer : a | a ) 200 m , b ) 250 m , c ) 450 m , d ) 350 m , e ) 270 m | a | divide(multiply(80, divide(add(multiply(1, 3), 2), 3)), subtract(divide(add(multiply(1, 3), 2), 3), const_1)) | multiply(n0,n2)|add(n1,#0)|divide(#1,n2)|multiply(n3,#2)|subtract(#2,const_1)|divide(#3,#4) | general |
in a division , a student took 42 as divisor instead of 36 . his answer was 24 . the correct answer is - | "x / 42 = 24 . x = 24 * 42 . so correct answer would be , ( 24 * 42 ) / 36 = 38 . answer : e" | a ) 42 , b ) 32 , c ) 48 , d ) 28 , e ) 38 | e | divide(multiply(42, 24), 36) | multiply(n0,n2)|divide(#0,n1)| | general |
hyderabad chennai mumbai delhi bangalore trivandram ; hyderabad chennai mumbai delhi bangalore trivandram in the table above , what is the least number of table entries that are needed to show the mileage between each city and each of the other 5 cities ? | easy way to go about this problem is we have 6 * 6 = 36 enteries in table the least number of enteries would be ( 36 - 6 ) / 2 since 6 enteries represent the distances between same points . alternatively this can be solved as combination problem . correct answer d | a ) 20 , b ) 18 , c ) 14 , d ) 15 , e ) 16 | d | multiply(5, const_3) | multiply(n0,const_3) | general |
the speed of a boat in still water in 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 24 minutes is : | "speed downstream = ( 20 + 5 ) = 25 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time Γ speed = 2 / 5 Γ 25 = 10.0 km answer is e ." | a ) 10.6 , b ) 10.2 , c ) 10.8 , d ) 10.4 , e ) 10.0 | e | multiply(add(20, 5), divide(24, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
the average weight of 8 person ' s increases by 3.5 kg when a new person comes in place of one of them weighing 65 kg . what is the weight of the new person ? | "explanation : total increase in weight = 8 Γ£ β 3.5 = 28 if x is the weight of the new person , total increase in weight = x Γ’ Λ β 65 = > 28 = x - 65 = > x = 28 + 65 = 93 answer : option a" | a ) 93 kg , b ) 50 kg , c ) 85 kg , d ) 80 kg , e ) 60 kg | a | add(multiply(8, 3.5), 65) | multiply(n0,n1)|add(n2,#0)| | general |
a river 5 m deep and 19 m wide is flowing at the rate of 4 kmph the amount of water that runs into the sea per minute is ? | "( 4000 * 5 * 19 ) / 60 = 6333 m 3 answer : b" | a ) 4500 , b ) 6333 , c ) 6576 , d ) 6345 , e ) 2882 | b | divide(multiply(multiply(5, 19), multiply(4, const_1000)), multiply(const_1, const_60)) | multiply(n0,n1)|multiply(n2,const_1000)|multiply(const_1,const_60)|multiply(#0,#1)|divide(#3,#2)| | physics |
if the number is divided by 3 , it reduced by 24 . the number is | "explanation : let the number be x . then , x - ( x / 3 ) = 24 = > 2 x / 3 = 24 = > x = 36 answer : option e" | a ) a ) 51 , b ) b ) 50 , c ) c ) 45 , d ) d ) 40 , e ) e ) 36 | e | divide(multiply(24, 3), subtract(3, const_1)) | multiply(n0,n1)|subtract(n0,const_1)|divide(#0,#1)| | general |
at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 120 . at the same place it would cost $ 158.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ? | "let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 120 4 b + 10 s + c = 158.5 subtracting the first equation from the second gives us b + 3 s = 38.5 now if we subtract the new equation two times from first or 3 times fro... | a ) $ 21 , b ) $ 27 , c ) $ 31 , d ) $ 43 , e ) it can not be determined | d | subtract(add(120, subtract(158.50, 120)), multiply(subtract(158.50, 120), 3)) | subtract(n3,n2)|add(n2,#0)|multiply(#0,n0)|subtract(#1,#2)| | general |
the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 40 / kg . how much did a kg of coffee cost in june ? | "let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.8 x . in july the price of 1 / 2 kg ( 400 gm ) of tea and 1 / 2 kg ( 400 gm ) of coffee ( equal quantities ) = 40 1.2 x ( 1 / 2 ) + 0.8 x ( 1 / 2 ) = 40 = > x = 40 thus proved . . . option a ." | a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | a | divide(40, multiply(subtract(const_1, divide(20, const_100)), add(divide(20, const_100), const_1))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4)| | general |
solve below question 4 x + 1 = 37 | 1 . subtract 1 from both sides : 4 x + 1 - 1 = 37 - 1 2 . simplify both sides : 4 x = 36 3 . divide both sides by 4 : 4 . simplify both sides : x = 9 d | a ) - 8 , b ) - 9 , c ) - 5 , d ) 9 , e ) 1 | d | divide(negate(add(37, 1)), 4) | add(n1,n2)|negate(#0)|divide(#1,n0)| | general |
thabo owns exactly 220 books , and each book is either paperback fiction , paperback nonfiction , or hardcover nonfiction . if he owns 20 more paperback nonfiction books than hardcover nonfiction books , and twice as many paperback fiction books as paperback nonfiction books , how many hardcover books nonfiction books ... | "i think we can use double - matrix method and solve using only one variable . our goal is to find the number of hardcover nonfiction books . let that number be x . we are given that all 140 books are either paperback fiction , paperback nonfiction , or hardcover nonfiction . this implies that number of hardcover ficti... | a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | d | subtract(divide(add(220, 20), const_4), 20) | add(n0,n1)|divide(#0,const_4)|subtract(#1,n1)| | general |
10 men do a work in 10 days . how many men are needed to finish the work in 5 days ? | "men required to finish the work in 5 days = 10 * 10 / 5 = 20 answer is b" | a ) 50 , b ) 20 , c ) 30 , d ) 10 , e ) 15 | b | divide(multiply(10, 10), 5) | multiply(n0,n1)|divide(#0,n2)| | physics |
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