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 |
|---|---|---|---|---|---|---|
a certain elevator has a safe weight limit of 2,500 pounds . what is the greatest possible number of people who can safely ride on the elevator at one time with the average ( arithmetic mean ) weight of half the riders being 200 pounds and the average weight of the others being 230 pounds ? | "lets assume there are 2 x people . half of them have average weight of 200 and other half has 230 . maximum weight is = 2500 so 200 * x + 230 * x = 2500 = > 430 x = 2500 = > x is approximately equal to 6 . so total people is 2 * 6 = 12 we are not taking 12 as answer because say 11 th person has minimum of 180 weight t... | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | e | multiply(divide(multiply(const_10, 200), add(200, 230)), const_2) | add(n1,n2)|multiply(n1,const_10)|divide(#1,#0)|multiply(#2,const_2)| | general |
the ratio of two quantities is 7 to 12 . if each of the quantities is divided by 3 , what is the ratio of these 2 new quantities ? | "if both sides of a ratio are divided by the same number there is no change in the ratio . 5 : 6 means 7 x : 12 x . . . the ratio starts as 7 / ( 1 ) : 12 / ( 1 ) . so when you diveide by three it becomes 7 / ( 3 ) : 12 / ( 3 ) . . so if the xs are equal the ratio does not change . . answer : c" | a ) 3.5 : 6 , b ) 14 : 24 , c ) 7 : 12 , d ) 2.3 : 4 , e ) it can not be determined from the information given | c | divide(add(7, 3), add(12, 3)) | add(n0,n2)|add(n1,n2)|divide(#0,#1)| | other |
if 45 % of z is 72 % of y and y is 75 % of x , what percent of x is z ? | "( 45 / 100 ) z = ( 72 / 100 ) y and y = ( 75 / 100 ) x i . e . y = ( 3 / 4 ) x i . e . ( 45 / 100 ) z = ( 72 / 100 ) * ( 3 / 4 ) x i . e . z = ( 72 * 3 ) x / ( 45 * 4 ) i . e . z = ( 1.2 ) x = ( 120 / 100 ) x i . e . z is 120 % of x answer : option b" | a ) 200 , b ) 120 , c ) 100 , d ) 65 , e ) 50 | b | multiply(divide(divide(75, const_100), divide(divide(45, const_100), divide(72, const_100))), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|divide(#1,#2)|divide(#0,#3)|multiply(#4,const_100)| | gain |
a certain car can travel 40 minutes on a gallon of gasoline at 70 miles per hour . if the car had started with a full tank and had 8 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 80 miles at 60 mph ? | let , tank capacity = t gallon used fuel = ( t - 8 ) gallons distance travelled ( @ 70 miles / hr ) = 80 miles distance travelled in 1 gallon = distance travelled in 40 mins ( @ 60 miles / hr ) = ( 70 / 70 ) * 40 = 40 miles fuel used to travel 80 miles = ( 80 / 40 ) = 2 gallon i . e . used fuel = ( t - 8 ) = 2 gallon i... | a ) 15 % , b ) 30 % , c ) 25 % , d ) 20 % , e ) 40 % | d | divide(divide(multiply(40, 70), multiply(80, 60)), add(divide(multiply(40, 70), multiply(80, 60)), 8)) | multiply(n0,n1)|multiply(n3,n4)|divide(#0,#1)|add(n2,#2)|divide(#2,#3) | physics |
on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 420 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | "let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 420 ) = 2 + 2 ( extra ) = > 2 x - 840 = x = > x = 840 . answer : e" | a ) 600 , b ) 620 , c ) 500 , d ) 520 , e ) 840 | e | multiply(420, const_2) | multiply(n0,const_2)| | general |
population of a city in 20004 was 1200000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city at the end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1300650 e" | a ) 354354 , b ) 545454 , c ) 465785 , d ) 456573 , e ) 1300650 | e | multiply(1200000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6)| | gain |
two trains 111 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | t = ( 111 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 6.85 answer : c | a ) 4.85 , b ) 7.85 , c ) 6.85 , d ) 5.85 , e ) 6.15 | c | divide(add(111, 165), multiply(add(80, 65), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2) | physics |
working alone , mary can pave a driveway in 5 hours and hillary can pave the same driveway in 2 hours . when they work together , mary thrives on teamwork so her rate increases by 33.33 % , but hillary becomes distracted and her rate decreases by 50 % . if they both work together , how many hours will it take to pave t... | "initial working rates : mary = 1 / 5 per hour hillary = 1 / 2 per hour rate when working together : mary = 1 / 5 + ( 1 / 3 * 1 / 5 ) = 1 / 4 per hour hillary = 1 / 2 - ( 1 / 2 * 1 / 2 ) = 1 / 4 per hour together they work 1 / 4 + 1 / 4 = 1 / 2 per hour so they will need 2 hours to complete the driveway . the correct a... | a ) 2 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | a | inverse(add(multiply(divide(const_1, 5), add(divide(33.33, const_100), const_1)), multiply(divide(const_1, 2), divide(50, const_100)))) | divide(n2,const_100)|divide(const_1,n0)|divide(const_1,n1)|divide(n3,const_100)|add(#0,const_1)|multiply(#2,#3)|multiply(#4,#1)|add(#6,#5)|inverse(#7)| | gain |
the two lines y = x and x = - 2 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length of a cube whose surface area is equal to 6 z ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 2 , - 2 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length ... | a ) 16 , b ) 8 √ 2 , c ) 8 , d ) √ 2 , e ) ( √ 2 ) / 3 | d | sqrt(divide(multiply(2, 2), const_2)) | multiply(n0,n0)|divide(#0,const_2)|sqrt(#1)| | general |
sum of two numbers is 15 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ? | "explanation : x + y = 15 2 x – 3 y = 5 x = 10 y = 5 a )" | a ) 5 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | a | subtract(15, divide(subtract(15, divide(5, const_2)), const_2)) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,const_2)|subtract(n0,#2)| | general |
some of 50 % - intensity red paint is replaced with 20 % solution of red paint such that the new paint intensity is 30 % . what fraction of the original paint was replaced ? | 30 % is 10 % - points above 20 % and 20 % - points below 50 % . thus the ratio of 25 % - solution to 50 % - solution is 2 : 1 . 2 / 3 of the original paint was replaced . the answer is c . | a ) 1 / 30 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | c | divide(subtract(divide(30, const_100), divide(50, const_100)), subtract(divide(20, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4) | gain |
two women started running simultaneously around a circular track of length 1800 m from the same point at speeds of 10 km / hr and 20 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? | time taken to meet for the first time anywhere on the track = length of the track / relative speed = 1800 / ( 30 + 60 ) 5 / 18 = 1800 * 18 / 90 * 5 = 72 seconds . answer : a | a ) 72 , b ) 74 , c ) 76 , d ) 78 , e ) 80 | a | divide(divide(1800, add(multiply(10, const_0_2778), multiply(20, const_0_2778))), const_3) | multiply(n1,const_0_2778)|multiply(n2,const_0_2778)|add(#0,#1)|divide(n0,#2)|divide(#3,const_3) | physics |
the average runs scored by a batsman in 20 matches is 40 . in the next 10 matches the batsman scored an average of 20 runs . find his average in all the 30 matches ? | "total score of the batsman in 20 matches = 800 . total score of the batsman in the next 10 matches = 200 . total score of the batsman in the 30 matches = 1000 . average score of the batsman = 1000 / 30 = 33.33 answer : d" | a ) 31 , b ) 46 , c ) 88 , d ) 33.33 , e ) 12 | d | divide(add(multiply(40, 20), multiply(20, 10)), add(20, 10)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general |
a goods train runs at the speed of 72 kmph and crosses a 240 m long platform in 26 seconds . what is the length of the goods train ? | "speed = ( 72 x 5 / 18 ) m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , x + 240 / 26 = 20 x + 240 = 520 x = 280 . answer : d" | a ) 230 m , b ) 270 m , c ) 643 m , d ) 280 m , e ) 270 m | d | multiply(subtract(26, divide(240, multiply(const_0_2778, 72))), multiply(const_0_2778, 72)) | multiply(n0,const_0_2778)|divide(n1,#0)|subtract(n2,#1)|multiply(#0,#2)| | physics |
if 9 a - b = 10 b + 70 = - 12 b - 2 a , what is the value of 2 a + 22 b ? | "this implies 9 a - b = 10 b + 70 , 9 a - b = - 12 b - 2 a , 10 b + 70 = - 12 b - 2 a manipulating the second equation gives us 10 b + 70 = - 12 b - 2 a = = > 2 a + 22 b = - 70 answer is b" | a ) - 4 , b ) - 70 , c ) 0 , d ) 2 , e ) 4 | b | multiply(negate(multiply(divide(70, 2), 2)), 2) | divide(n2,n4)|multiply(#0,n4)|negate(#1)|multiply(n5,#2)| | general |
how much time will a train of length 300 m moving at a speed of 72 kmph take to cross another train of length 500 m , moving at 36 kmph in the same direction ? | "the distance to be covered = sum of their lengths = 300 + 500 = 800 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 / 18 = 10 mps . time required = d / s = 800 / 10 = 80 sec . answer : b" | a ) 50 , b ) 80 , c ) 88 , d ) 76 , e ) 12 | b | divide(add(300, 500), multiply(subtract(72, 36), const_0_2778)) | add(n0,n2)|subtract(n1,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
a certain number when divided by 39 leaves a remainder 17 , what is the remainder when the same number is divided by 13 ? | "explanation : 39 + 17 = 56 / 13 = 4 ( remainder ) answer : e" | a ) 7 , b ) 8 , c ) 9 , d ) 6 , e ) 4 | e | reminder(17, 13) | reminder(n1,n2)| | general |
a tank contains 8,000 gallons of a solution that is 4 percent sodium chloride by volume . if 3,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "we start with 8,000 gallons of a solution that is 4 % sodium chloride by volume . this means that there are 0.04 x 8,000 = 320 gallons of sodium chloride . when 3,000 gallons of water evaporate we are left with 5,000 gallons of solution . from here we can determine what percent of the 5,000 gallon solution is sodium c... | a ) 3.40 % , b ) 4.40 % , c ) 5.40 % , d ) 6.40 % , e ) 7.40 % | d | multiply(divide(multiply(multiply(const_100, const_100), divide(4, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | add(const_2,const_3)|divide(n1,const_100)|multiply(const_100,const_100)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#4,const_100)|multiply(#0,#6)|add(#7,#5)|subtract(#2,#8)|divide(#3,#9)|multiply(#10,const_100)| | gain |
x , y and z , each working alone can complete a job in 2 , 4 and 6 days respectively . if all three of them work together to complete a job and earn $ 2000 , what will be z ' s share of the earnings ? | the dollars earned will be in the same ratio as amount of work done 1 day work of z is 1 / 6 ( or 2 / 12 ) 1 day work of the combined workforce is ( 1 / 2 + 1 / 4 + 1 / 6 ) = 11 / 12 z ' s contribution is 2 / 9 of the combined effort translating effort to $ = 6 / 11 * 2000 = $ 1090.90 hence : e | a ) $ 1080.90 , b ) $ 1000.90 , c ) $ 1070.90 , d ) $ 1050.90 , e ) $ 1090.90 | e | multiply(divide(2000, add(add(inverse(2), inverse(4)), inverse(6))), inverse(2)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|multiply(#5,#0) | physics |
the least number , which when divided by 12 , 15 , 20 and 54 leaves in each case a remainder of 8 is : | "required number = ( l . c . m . of 12 , 15 , 20 , 54 ) + 8 = 540 + 8 = 548 . answer : option d" | a ) 504 , b ) 536 , c ) 544 , d ) 548 , e ) 568 | d | multiply(54, const_10) | multiply(n3,const_10)| | general |
nicky and cristina are running a race . since cristina is faster than nicky , she gives him a 30 meter head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "used pluging in method say t is the time for cristina to catch up with nicky , the equation will be as under : for nicky = n = 3 * t + 30 for cristina = c = 5 * t @ t = 15 , n = 75 c = 75 right answer ans : a" | a ) 15 seconds , b ) 18 seconds , c ) 25 seconds , d ) 30 seconds , e ) 45 seconds | a | divide(30, subtract(5, 3)) | subtract(n1,n2)|divide(n0,#0)| | physics |
how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6.6 m x 22.5 cm ? | number of bricks = volume of wall / volume of bricks = 800 x 660 x 22.5 / 25 x 11.25 x 6 = = 7040 answer : e | a ) 6400 , b ) 6410 , c ) 6440 , d ) 6500 , e ) 7040 | e | divide(multiply(multiply(multiply(8, const_100), multiply(6.6, const_100)), 22.5), multiply(multiply(25, 11.25), 6)) | multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4) | physics |
the total number of students in grades 1 and 2 is 30 more than the total number of students in grades 2 and 5 . how much lesser is the number of students in grade 5 as compared to grade 1 ? | ( grade 1 + grade 2 ) - ( grade 2 + grade 5 ) = 30 grade 1 - grade 5 = 30 answer : b | a ) 20 , b ) 30 , c ) 10 , d ) 40 , e ) 15 | b | divide(add(add(add(add(add(add(add(30, 5), 2), 2), 2), const_3), const_4), const_12), 2) | add(n2,n4)|add(n1,#0)|add(n1,#1)|add(n1,#2)|add(#3,const_3)|add(#4,const_4)|add(#5,const_12)|divide(#6,n1) | general |
a football player scores 5 goals in his fifth match thus increasing his average goals score by 0.2 . the total number of goals in his 5 matches would be | "while this question can be solved with a rather straight - forward algebra approach ( as the other posters have noted ) , it can also be solved by testing the answers . one of those numbers must be the total number of goals . . . from a tactical standpoint , it ' s best to test either answer b or answer d , so if the ... | a ) 14 , b ) 16 , c ) 18 , d ) 10 , e ) 21 | e | add(subtract(multiply(const_4, 5), multiply(multiply(const_4, 5), 0.2)), 5) | multiply(n0,const_4)|multiply(n2,const_4)|multiply(n1,#1)|subtract(#0,#2)|add(n0,#3)| | general |
a jar full of whisky contains 40 % alcohol . a part of this whisky is replaced by another containg 19 % alcohol and now the percentage of alcohol was found to be 22 % . what quantity of whisky is replaced ? | "let us assume the total original amount of whiskey = 10 ml - - - > 4 ml alcohol and 6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = 4 - 0.4 x new quantity of whiskey added = x ml out of which 0.19 is the alcohol . thus , the final quantity of alcohol = 4 - 0.4 x + 0.19 x - - - - > ( 4 ... | a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 4 / 5 | d | divide(subtract(40, 22), subtract(40, 19)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | gain |
a jeep takes 4 hours to cover a distance of 620 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 4 distance = 620 3 / 2 of 4 hours = 4 * 3 / 2 = 6 hours required speed = 620 / 6 = 103 kmph d )" | a ) 148 kmph , b ) 152 kmph , c ) 106 kmph , d ) 103 kmph , e ) 165 kmph | d | divide(620, multiply(divide(3, 2), 4)) | divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)| | physics |
the mean of 50 observations was 36 . it was found later that an observation 60 was wrongly taken as 23 . the corrected new mean is ? | "correct sum = ( 36 * 50 + 60 - 23 ) = 1837 . correct mean = 1837 / 50 = 36.7 answer : a" | a ) 36.7 , b ) 36.1 , c ) 36.5 , d ) 36.9 , e ) 36.3 | a | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
in a party every person shakes hands with every other person . if there were a total of 105 handshakes in the party then what is the number of persons present in the party ? | "explanation : let the number of persons be n â ˆ ´ total handshakes = nc 2 = 105 n ( n - 1 ) / 2 = 105 â ˆ ´ n = 15 answer : a" | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | a | divide(add(sqrt(add(multiply(multiply(105, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general |
at a certain organisation , the number of male members went up by 12 % in the year 2001 from year 2000 , and the number of females members went down by 4 % in the same time period . if the total membership at the organisation went up by 1.2 % from the year 2000 to 2001 , what was the ratio of male members to female mem... | "men increase by 12 % = = > 1.12 m = males in 2001 women decrease by 4 % = = > 0.96 f = women in 2001 total employees increase by 1.2 % = = > 1.012 * ( m + f ) = total number of employees in 2001 obviously ( males in 2001 ) + ( females in 2001 ) = total number of employees in 2001 1.12 m + 0.96 f = 1.012 * ( m + f ) 1.... | a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 3 , d ) 3 : 2 , e ) 2 : 1 | a | divide(subtract(multiply(add(const_1, divide(1.2, const_100)), const_1000), multiply(subtract(const_1, divide(4, const_100)), const_1000)), subtract(multiply(add(const_1, divide(12, const_100)), const_1000), multiply(add(const_1, divide(1.2, const_100)), const_1000))) | divide(n4,const_100)|divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#2,const_1)|subtract(const_1,#1)|multiply(#3,const_1000)|multiply(#5,const_1000)|multiply(#4,const_1000)|subtract(#6,#7)|subtract(#8,#6)|divide(#9,#10)| | other |
what least value should be replaced by * in 842 * 124 so the number become divisible by 9 | "explanation : trick : number is divisible by 9 , if sum of all digits is divisible by 9 , so ( 8 + 4 + 2 + * + 1 + 2 + 4 ) = 21 + * should be divisible by 9 , 21 + 6 will be divisible by 9 , so that least number is 6 . answer : option d" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | subtract(124, subtract(124, 124)) | subtract(n1,n1)|subtract(n1,#0)| | general |
there are 7 players in a bowling team with an average weight of 76 kg . if two new players join the team , one weighs 110 kg and the second weighs 60 kg , what will be the new average weight ? | "the new average will be = ( 76 * 7 + 110 + 60 ) / 9 = 78 kgs a is the answer" | a ) 78 kg . , b ) 70 kg . , c ) 75 kg . , d ) 70 kg . , e ) 72 kg . | a | divide(add(multiply(7, 76), add(110, 60)), add(7, const_2)) | add(n2,n3)|add(n0,const_2)|multiply(n0,n1)|add(#0,#2)|divide(#3,#1)| | general |
ratio between rahul and deepak is 4 : 3 , after 2 years rahul age will be 26 years . what is deepak present age . | "explanation : present age is 4 x and 3 x , = > 4 x + 2 = 26 = > x = 6 so deepak age is = 3 ( 6 ) = 18 option b" | a ) 14 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | b | divide(multiply(subtract(26, 2), 3), 4) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other |
the sum of all consecutive odd integers from − 25 to 35 , inclusive , is | "the sum of the odd numbers from - 25 to + 25 is 0 . let ' s add the remaining numbers . 27 + 29 + 31 + 33 + 35 = 5 ( 31 ) = 155 the answer is c ." | a ) 130 , b ) 135 , c ) 155 , d ) 195 , e ) 235 | c | add(add(add(add(25, const_2), add(add(25, const_2), const_2)), add(add(add(25, const_2), const_2), const_2)), 35) | add(n0,const_2)|add(#0,const_2)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(n1,#4)| | physics |
a train crosses a platform of 130 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 + 130 / 15 = x + 250 / 20 20 x + 2600 = 15 x + 3750 5 x = 1150 x = 230 m answer : e" | a ) 150 , b ) 887 , c ) 167 , d ) 197 , e ) 230 | e | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(130, 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 |
one computer can upload 100 megabytes worth of data in 4 seconds . two computers , including this one , working together , can upload 1300 megabytes worth of data in 20 seconds . how long would it take for the second computer , working on its own , to upload 100 megabytes of data ? | "since the first computer can upload 100 megabytes worth of data in 4 seconds then in 4 * 5 = 20 seconds it can upload 5 * 100 = 500 megabytes worth of data , hence the second computer in 20 seconds uploads 1300 - 500 = 800 megabytes worth of data . the second computer can upload 100 megabytes of data in 2.5 seconds . ... | a ) 6 , b ) 2.5 , c ) 9 , d ) 11 , e ) 13 | b | subtract(divide(1300, 20), divide(100, 4)) | divide(n2,n3)|divide(n0,n1)|subtract(#0,#1)| | physics |
the “ length of integer x ” refers to the number of prime factors , not necessarily distinct , that x has . ( if x = 60 , the length of x would be 4 because 60 = 2 × 2 × 3 × 5 . ) what is the greatest possible length of integer z if z < 500 ? | "to maximize the length of z , we should minimize its prime base . the smallest prime is 2 and since 2 ^ 8 = 256 < 500 , then the greatest possible length of integer z is 8 . the answer is c ." | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | c | log(power(2, const_10)) | power(n3,const_10)|log(#0)| | general |
after decreasing 50 % in the price of an article costs rs . 620 . find the actual cost of an article ? | "cp * ( 50 / 100 ) = 620 cp = 12.4 * 100 = > cp = 1240 answer : b" | a ) 1400 , b ) 1240 , c ) 1200 , d ) 1100 , e ) 1500 | b | divide(620, subtract(const_1, divide(50, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
a cube has a volume of 27 cubic feet . if a similar cube is twice as long , twice as wide , and twice as high , then the volume , in cubic feet of such cube is ? | "volume = 27 = side ^ 3 i . e . side of cube = 3 new cube has dimensions 6 , 6 , and 6 as all sides are twice of teh side of first cube volume = 6 * 6 * 6 = 216 square feet answer : option a" | a ) 216 , b ) 48 , c ) 64 , d ) 80 , e ) 100 | a | volume_cube(multiply(const_2, cube_edge_by_volume(27))) | cube_edge_by_volume(n0)|multiply(#0,const_2)|volume_cube(#1)| | geometry |
if the compound interest on a certain sum of money for 7 years at 10 % per annum be rs . 993 , what would be the simple interest ? | "let p = principal a - amount we have a = p ( 1 + r / 100 ) 3 and ci = a - p atq 993 = p ( 1 + r / 100 ) 3 - p ? p = 3000 / - now si @ 10 % on 3000 / - for 7 yrs = ( 3000 x 10 x 7 ) / 100 = 2100 / - answer : d ." | a ) rs . 880 , b ) rs . 890 , c ) rs . 895 , d ) rs . 2100 , e ) none | d | divide(multiply(multiply(multiply(multiply(const_3.0, const_100), 10), 10), 7), const_100) | multiply(const_3.0,const_100)|multiply(n1,#0)|multiply(n1,#1)|multiply(n0,#2)|divide(#3,const_100)| | gain |
for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 4 + 3 and * 4 + 4 , inclusive ? | "generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 4 would be divisible by all numbers from 1 to 4 . = > * 4 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 4 is divisible by 3 ) in fact adding anyprimenumber between 1 to 4 to * 4 will definitely be ... | a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(subtract(add(multiply(multiply(multiply(4, 3), const_2), const_4), 4), add(multiply(multiply(multiply(4, 3), const_2), const_4), 3)), 1) | multiply(n1,n2)|multiply(#0,const_2)|multiply(#1,const_4)|add(n1,#2)|add(n2,#2)|subtract(#3,#4)|subtract(#5,n0)| | general |
in the hillside summer camp there are 50 children . 85 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 5 % of the total number of children in the camp . how many more boys must she bring to make that happen ? | given there are 50 students , 84 % of 50 = 42 boys and remaining 8 girls . now here 84 % are boys and 16 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 5 or 8 % . . if we add 50 to existing 45 then the count will be 92 and the girls number will be 8 as it . ... | a ) 20 . , b ) 45 . , c ) 50 . , d ) 30 . , e ) 25 . | c | add(multiply(divide(subtract(const_100, 5), const_100), 50), multiply(divide(5, const_100), 50)) | divide(n2,const_100)|subtract(const_100,n2)|divide(#1,const_100)|multiply(n0,#0)|multiply(n0,#2)|add(#4,#3) | general |
4 , 5 , 7 , 11 , 19 , ( . . . ) | "4 4 × 2 - 3 = 5 5 × 2 - 3 = 7 7 × 2 - 3 = 11 11 × 2 - 3 = 19 19 × 2 - 3 = 35 answer is c" | a ) 32 , b ) 22 , c ) 35 , d ) 27 , e ) 28 | c | subtract(negate(11), multiply(subtract(5, 7), divide(subtract(5, 7), subtract(4, 5)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
a set s = { x , - 8 , - 5 , - 2 , 2 , 6 , 9 , y } with elements arranged in increasing order . if the median and the mean of the set are the same , what is the value of | x | - | y | ? | "median of the set = ( - 2 + 2 ) / 2 = 0 as per statement , mean of the set = 0 mean of the set | y | - | x | + 17 - 15 = 0 ( where x is negative n y is positive ) | y | - | x | = - 2 so the absolute difference between two numbers is 2 answer c" | a ) 1 , b ) 0 , c ) 2 , d ) - 1 , e ) can not be determined | c | subtract(subtract(subtract(add(add(9, 8), 2), 2), 5), 8) | add(n0,n5)|add(n2,#0)|subtract(#1,n2)|subtract(#2,n1)|subtract(#3,n0)| | general |
a farmer used 1,034 acres of land for beans , wheat , and corn in the ratio of 5 : 2 : 4 , respectively . how many acres q were used for corn ? | consider 5 x acres of land used for bean consider 2 x acres of land used for wheat consider 4 x acres of land used for corn total given is 1034 acres 11 x = 1034 x = 94 land used for corn q = 4 * 94 = 376 correct option - c | a ) 188 , b ) 258 , c ) 376 , d ) 470 , e ) 517 | c | multiply(divide(add(multiply(const_1000, const_1), add(multiply(const_10, const_3), 4)), add(add(5, 2), 4)), 4) | add(n1,n2)|multiply(const_10,const_3)|multiply(const_1,const_1000)|add(n3,#1)|add(n3,#0)|add(#3,#2)|divide(#5,#4)|multiply(n3,#6) | other |
a sum fetched a total simple interest of $ 4016.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ? | "e 8925 principal = $ 100 x 4016.25 / 9 x 5 = $ 401625 / 45 = $ 8925 ." | a ) $ 8829 , b ) $ 2840 , c ) $ 6578 , d ) $ 7782 , e ) $ 8925 | e | divide(divide(multiply(4016.25, const_100), 9), 5) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain |
a train running at 1 / 2 of its own speed reached a place in 8 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 / 2 * 8 = 4 hrs time saved = 8 - 4 = 4 hrs answer : c | a ) 8 hrs , b ) 10 hrs , c ) 4 hrs , d ) 15 hrs , e ) 6 hrs | c | multiply(divide(1, 2), 8) | divide(n0,n1)|multiply(n2,#0) | physics |
the ratio 6 : 5 expressed as a percent equals | "solution 6 : 5 = 6 / 5 = ( 6 / 5 x 100 ) % . = 120 % . answer c" | a ) 12.5 % , b ) 40 % , c ) 120 % , d ) 125 % , e ) none | c | multiply(divide(6, 5), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
find the area of a parallelogram with base 20 cm and height 16 cm ? | "area of a parallelogram = base * height = 20 * 16 = 320 cm 2 answer : b" | a ) 198 cm 2 , b ) 320 cm 2 , c ) 279 cm 2 , d ) 128 cm 2 , e ) 297 cm 2 | b | multiply(20, 16) | multiply(n0,n1)| | geometry |
a part of certain sum of money is invested at 10 % per annum and the rest at 16 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | 16 : 10 = 8 : 5 answer : a | a ) 8 : 5 , b ) 4 : 9 , c ) 4 : 3 , d ) 4 : 1 , e ) 4 : 2 | a | multiply(divide(16, const_100), 10) | divide(n1,const_100)|multiply(n0,#0) | gain |
the h . c . f of two numbers is 12 and their l . c . m is 6600 . if one of the numbers is 288 , then the other is ? | "other number = ( 12 * 6600 ) / 288 = 275 . answer : e" | a ) 255 , b ) 260 , c ) 265 , d ) 270 , e ) 275 | e | multiply(12, 288) | multiply(n0,n2)| | physics |
the unit digit in the product ( 624 * 708 * 913 * 463 ) is : | "explanation : unit digit in the given product = unit digit in ( 4 * 8 * 3 * 3 ) = 8 answer : d" | a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | d | subtract(multiply(multiply(multiply(624, 708), 913), 463), subtract(multiply(multiply(multiply(624, 708), 913), 463), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general |
in the faculty of reverse - engineering , 226 second year students study numeric methods , 450 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 80 % of the total ? | "answer is b : 678 solution : total number of students studying both are 450 + 226 - 134 = 542 ( subtracting the 134 since they were included in the both the other numbers already ) . so 80 % of total is 542 , so 100 % is approx . 678 ." | a ) 515 . , b ) 678 . , c ) 618 . , d ) 644 . , e ) 666 . | b | add(226, 450) | add(n0,n1)| | general |
the volume of the sphere q is ( 37 / 64 ) % less than the volume of sphere p and the volume of sphere r is ( 19 / 27 ) % less than that of sphere q . by what is the surface area of sphere r less than the surface area of sphere p ? | explanation : let the volume of sphere p be 64 parts . therefore volume of sphere q = > 64 − ( 37 / 64 ) % of 64 . = > 64 − 37 = 27 parts . the volume of r is : - = > 27 − ( 19 / 27 ) × 27 . = > 27 − 19 = 8 parts . volume ratio : = > p : q : r = 64 : 27 : 8 . radius ratio : = > p : q : r = 4 : 3 : 2 . the surface area ... | ['a ) 77.77 %', 'b ) 75 %', 'c ) 67.5 %', 'd ) 87.5 %', 'e ) none of these'] | b | multiply(divide(subtract(divide(64, power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3))), divide(64, power(64, divide(const_1, const_3)))), const_100) | divide(const_1,const_3)|power(n1,#0)|divide(n1,#1)|subtract(#2,#1)|divide(#3,#2)|multiply(#4,const_100) | geometry |
what is the radius of a circle that has a circumference of 3.14 meters ? | circumference of a circle = 2 π r . given , circumference = 3.14 meters . therefore , 2 π r = circumference of a circle or , 2 π r = 3.14 . or , 2 ? 3.14 r = 3.14 , [ putting the value of pi ( π ) = 3.14 ] . or , 6.28 r = 3.14 . or , r = 3.14 / 6.28 . or , r = 0.5 . answer : 0.5 meter . correct answer e | ['a ) 2.5', 'b ) 2', 'c ) 1.5', 'd ) 1', 'e ) 0.5'] | e | divide(3.14, multiply(const_2, const_pi)) | multiply(const_2,const_pi)|divide(n0,#0) | physics |
the difference between a two - digit number and the number obtained by interchanging the digit is 36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ? | since the number is greater than the number obtained on reversing the digits , so the ten ' s digit is greater than the unit ' s digit . let the ten ' s and unit ' s digits be 2 x and x respectively . then , ( 10 * 2 x + x ) - ( 10 x + 2 x ) = 36 9 x = 36 x = 4 required difference = ( 2 x + x ) - ( 2 x - x ) = 2 x = 8 ... | a ) 8 , b ) 15 , c ) 14 , d ) 12 , e ) 10 | a | subtract(add(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2)), subtract(multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2), divide(36, subtract(subtract(add(multiply(2... | multiply(n2,const_10)|add(n1,#0)|subtract(#1,const_10)|subtract(#2,n2)|divide(n0,#3)|multiply(n2,#4)|add(#4,#5)|subtract(#5,#4)|subtract(#6,#7) | general |
a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ? | "first ratio = 3 / 20 = 15 % second ratio = ( 3 + 12 ) / ( 9 + 108 ) = 15 / 117 = 12,8 % decrease = 2,2 % 2,2 % / 15 % = 15 % approximately . answer : b" | a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 56 % | b | multiply(divide(3, 20), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the difference of two numbers is 11 and one fifth of their sum is 9 . the numbers are : | "x − y = 11 , x + y = 5 × 9 x − y = 11 , x + y = 45 , y = 17 , x = 28 answer : d" | a ) 31 , 20 , b ) 30 , 19 , c ) 29 , 18 , d ) 28 , 17 , e ) none | d | add(subtract(multiply(divide(const_10, const_2), 9), divide(add(11, multiply(divide(const_10, const_2), 9)), const_2)), divide(const_10, const_2)) | divide(const_10,const_2)|multiply(n1,#0)|add(n0,#1)|divide(#2,const_2)|subtract(#1,#3)|add(#0,#4)| | general |
an internet recently hired 8 new network , in addvertisement 20 network already employed . all new network cam from university a . in addition 75 % of computer addvertisement came from same university a . what fraction of original 20 network addvertisement came from same univerity a ? pls help to solve | new networks = 8 already employed networks = 20 total networks = 28 computer advt . from uni . a = 28 * ( 75 / 100 ) = 21 out 28 networks 21 came from uni . a among these 21 we know 8 are newly hired so 21 - 8 = 13 networks among previously employed networks came from uni . a answer is : 13 / 20 answer : a | a ) 13 / 20 , b ) 15 / 20 , c ) 14 / 20 , d ) 12 / 20 , e ) 11 / 20 | a | divide(subtract(multiply(add(8, 20), divide(75, const_100)), 8), 20) | add(n0,n1)|divide(n2,const_100)|multiply(#0,#1)|subtract(#2,n0)|divide(#3,n1) | gain |
dividing by 3 ⁄ 9 and then multiplying by 5 ⁄ 6 is the same as dividing by what number ? | "say x / 3 / 9 * 5 / 6 = x * 9 / 3 * 5 / 6 = x * 5 / 2 d" | a ) 31 ⁄ 5 , b ) 16 ⁄ 5 , c ) 20 ⁄ 9 , d ) 5 / 2 , e ) 5 ⁄ 16 | d | multiply(divide(9, 3), divide(5, 6)) | divide(n1,n0)|divide(n2,n3)|multiply(#0,#1)| | general |
if it takes a machine 1 ⁄ 3 minute to produce one item , how many items will it produce in 2 hours ? | "1 item takes 1 / 3 min so it takes 120 min to produce x x / 3 = 120 the x = 360 answer : e" | a ) 1 ⁄ 3 , b ) 4 ⁄ 3 , c ) 80 , d ) 120 , e ) 360 | e | divide(multiply(2, const_60), divide(1, 3)) | divide(n0,n1)|multiply(n2,const_60)|divide(#1,#0)| | physics |
a basket of 1430 apples is divided equally among a group of apple lovers . if 45 people join the group , each apple lover would receive 9 apples less . how many s apples did each person get before 45 people joined the feast ? | "before solving it algebraically , let us prime factorize 1430 = 2 * 5 * 11 * 13 . since number of apples per person * total persons s = 1430 , the answer should be a factor of 1430 . only c is . and that ' s your answer . c" | a ) 20 . , b ) 21 . , c ) 22 . , d ) 23 . , e ) 24 . | c | add(divide(1430, add(divide(1430, add(add(const_10, const_10), const_2)), 45)), 9) | add(const_10,const_10)|add(#0,const_2)|divide(n0,#1)|add(n1,#2)|divide(n0,#3)|add(n2,#4)| | general |
a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 7700 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a ’ s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 7700 x ( 2 / 11 ) ) = rs . 1400 . answer is c" | a ) 1100 , b ) 800 , c ) 1400 , d ) 1200 , e ) none of them | c | multiply(7700, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | divide(const_2,n0)|multiply(const_2,n0)|multiply(#0,n0)|add(#1,#2)|add(#3,n0)|divide(const_2,#4)|multiply(n1,#5)| | gain |
if [ [ x ] ] = x ^ 2 + 2 x + 4 , what is the value of [ [ 5 ] ] ? | "these functions questions might look intimidating , but they just test your knowledge about how well you can substitute values [ [ x ] ] = x ^ 2 + 2 x + 4 [ [ 5 ] ] = 5 ^ 2 + 2 * 5 + 4 = 39 . option a" | a ) 39 , b ) 9 , c ) 15 , d ) 19 , e ) 25 | a | add(add(power(5, const_2.0), multiply(2, 2)), 4) | multiply(n3,n3)|power(n0,n0)|add(#0,#1)|add(n2,#2)| | general |
sobha ' s father was 38 years of age when she was born while her mother was 36 years old when her brother 4 years younger to her was born . what is the difference between the ages of her parents ? | age of sobha ' s father when sobha was born = 38 age of sobha ' s mother when sobha was born = 36 − 4 = 32 required difference of age = 38 − 32 = 6 answer is b . | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 2 | b | subtract(add(38, 4), 36) | add(n0,n2)|subtract(#0,n1) | general |
a train travels from albany to syracuse , a distance of 100 miles , at the average rate of 50 miles per hour . the train then travels back to albany from syracuse . the total travelling time of the train is 5 hours and 24 minutes . what was the average rate of speed of the train on the return trip to albany ? | "50 * t = 100 therefore t = 100 / 50 = 2 on return speed * ( 5.4 - 2 ) = 100 therefore t = 100 / 3.2 = 31.25 a" | a ) 31.25 , b ) 30 , c ) 32 , d ) 34 , e ) 36 | a | speed(100, subtract(add(5, divide(24, const_60)), divide(100, 50))) | divide(n3,const_60)|divide(n0,n1)|add(n2,#0)|subtract(#2,#1)|speed(n0,#3)| | physics |
if the average ( arithmetic mean ) of ( 2 a + 16 ) and ( 3 a - 8 ) is 74 , what is the value of a ? | "( ( 2 a + 16 ) + ( 3 a - 8 ) ) / 2 = ( 5 a + 8 ) / 2 = 74 a = 28 the answer is c ." | a ) 25 , b ) 30 , c ) 28 , d ) 36 , e ) 42 | c | divide(subtract(multiply(74, 2), subtract(16, 8)), add(2, 3)) | add(n0,n2)|multiply(n0,n4)|subtract(n1,n3)|subtract(#1,#2)|divide(#3,#0)| | general |
the ratio of the arithmetic mean of two numbers to one of the numbers is 5 : 7 . what is the ratio of the smaller number to the larger number ? | "for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 5 : 7 , thus the smaller number must have a ratio of 3 . the ratio of the smaller number to the larger number is 3 : 7 . the answer is c ." | a ) 2 : 3 , b ) 2 : 5 , c ) 3 : 7 , d ) 4 : 9 , e ) 5 : 6 | c | multiply(subtract(divide(5, 7), divide(const_1, const_2)), const_2) | divide(n0,n1)|divide(const_1,const_2)|subtract(#0,#1)|multiply(#2,const_2)| | other |
find number which is 60 % less than 120 . | "explanation : 60 % less is 40 % of the given number therefore , 40 % of 120 is 48 . answer : d" | a ) 18 , b ) 22 , c ) 28 , d ) 48 , e ) 98 | d | divide(multiply(120, 60), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain |
in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 182 votes . find the total number of valid votes ? | "let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 182 = > 4 x / 10 = 182 = > 4 x = 1820 = > x = 455 . answer : a" | a ) 455 , b ) 570 , c ) 480 , d ) 520 , e ) 550 | a | divide(182, divide(subtract(70, subtract(const_100, 70)), const_100)) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)| | gain |
find the number which is nearest to 457 and is exactly divisible by 11 . | "solution on dividing 457 by 11 , remainder is 6 . required number is either 451 or 462 nearest to 456 is = 462 . answer d" | a ) 450 , b ) 451 , c ) 460 , d ) 462 , e ) none | d | add(457, subtract(11, reminder(457, 11))) | reminder(n0,n1)|subtract(n1,#0)|add(n0,#1)| | general |
what is the remainder when 1201 × 1202 × 1205 × 1210 is divided by 6 ? | "the remainders when dividing each number by six are : 1 , 2 , 5 , and 4 . the product is 1 * 2 * 5 * 4 = 40 the remainder when dividing 40 by 6 is 4 . the answer is d ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | reminder(multiply(1202, 1201), 1205) | multiply(n0,n1)|reminder(#0,n2)| | general |
a cycle is bought for rs . 450 and sold for rs . 520 , find the gain percent ? | "450 - - - - 70 100 - - - - ? = > 15.55 % answer : e" | a ) 18 , b ) 14 , c ) 20 , d ) 15 , e ) 15.55 | e | multiply(divide(subtract(520, 450), 450), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration r of the acid in the mixture ? | "do not need any computation 30 % - - - - - - - - - - - 21 % - - - - - - - - - 12 % if volume of both sol . were equal the concentration r would be 21 % = 1 / 5 , but 12 % is more than 3 times only possibility is 1 / 6 d" | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5 | d | divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)| | general |
12.1212 + 17.0005 - 9.1103 = ? | solution given expression = ( 12.1212 + 17.0005 ) - 9.1103 = ( 29.1217 - 9.1103 ) = 20.0114 . answer d | a ) 20.0015 , b ) 20.0105 , c ) 20.0115 , d ) 20.0114 , e ) none | d | subtract(add(12.1212, 17.0005), 9.1103) | add(n0,n1)|subtract(#0,n2) | general |
a student chose a number , multiplied it by 2 , then subtracted 138 from the result and got 104 . what was the number he chose ? | "let xx be the number he chose , then 2 â ‹ … x â ˆ ’ 138 = 104 2 â ‹ … x â ˆ ’ 138 = 104 x = 121 answer : b" | a ) 123 , b ) 121 , c ) 277 , d ) 267 , e ) 120 | b | divide(add(104, 138), 2) | add(n1,n2)|divide(#0,n0)| | general |
one side of a rectangle is 3 cm shorter than the other side . if we increase the length of each side by 1 cm , then the area of the rectangle will increase by 18 cm 2 . find the lengths of all sides . | "let x be the length of the longer side x > 3 , then the other side ' s length is x − 3 cm . then the area is s 1 = x ( x - 3 ) cm 2 . after we increase the lengths of the sides they will become ( x + 1 ) and ( x − 3 + 1 ) = ( x − 2 ) cm long . hence the area of the new rectangle will be a 2 = ( x + 1 ) ⋅ ( x − 2 ) cm ... | a ) 10 and 3 , b ) 7 and 10 , c ) 10 and 7 , d ) 3 and 10 , e ) 10 and 10 | c | subtract(add(divide(18, 2), 1), 3) | divide(n2,n3)|add(#0,n1)|subtract(#1,n0)| | geometry |
john want to buy a $ 100 trouser at the store , but he think it â € ™ s too expensive . finally , it goes on sale for $ 75 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it â € ™ s 100 â € “ 75 = 25 . the â € œ original â € is our starting point ; in this case , it â € ™ s 100 . ( 25 / 100 ) * 100 = ( 0.25 ) * 100 = 25 % . d" | a ) 20 % , b ) 30 % , c ) 40 % , d ) 25 % , e ) 60 % | d | subtract(100, 75) | subtract(n0,n1)| | general |
an investment compounds annually at an interest rate of 30 % what is the smallest investment period by which time the investment will more than double in value ? | "1 year : 100 / 3 = 33.33 approx $ 34 : total : 134 2 nd year : 134 / 3 = 45 : total : 134 + 45 = 179 3 rd year : 179 / 3 = 60 : total : 179 + 60 = 239 > 2 ( 100 ) ; 3 years ; answer : a" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | add(floor(divide(log(const_3), log(add(const_1, divide(30, const_100))))), const_1) | divide(n0,const_100)|log(const_3)|add(#0,const_1)|log(#2)|divide(#1,#3)|floor(#4)|add(#5,const_1)| | gain |
what is the remainder when ( 55 ) ( 57 ) is divided by 8 ? | ( 55 ) ( 57 ) = ( 56 - 1 ) ( 56 + 1 ) = 56 ^ 2 - 1 which is 1 less than a multiple of 8 . then the remainder will be 7 . the answer is e . | a ) 1 , b ) 2 , c ) 4 , d ) 5 , e ) 7 | e | reminder(multiply(57, 55), 8) | multiply(n0,n1)|reminder(#0,n2) | general |
in 1950 , richard was 4 times as old as robert . in 1955 , richard was 3 times as old as robert . in which year was richard 1.25 as old as robert ? | in 1950 : ri = 4 ro - - - - - - - - - - - - - - eq 1 in 1955 : ri + 5 = 3 ( ro + 5 ) - - - - - - - - - eq 2 thus in 1950 , solving eq 1 and eq 2 ro = 10 , ri = 40 now for each year we can calculate : 1960 : ri = 50 , ro = 20 1965 : ri = 55 , ro = 25 2060 : ri = 120 , ro = 150 thus ans : e | a ) 1960 , b ) 1965 , c ) 1970 , d ) 2050 , e ) 2060 | e | add(1950, add(1.25, add(const_4, const_3))) | add(const_3,const_4)|add(n4,#0)|add(n0,#1) | general |
a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord q , in feet ? | because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . q = pi * r = 30 . round pi = 3 , then r = 10... | a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5 | d | divide(30, const_3) | divide(n0,const_3) | general |
a person crosses a 300 m long street in 4 minutes . what is his speed in km per hour ? | "distance = 300 meter time = 4 minutes = 4 x 60 seconds = 240 seconds speed = distance / time = 300 / 240 = 1.25 m / s = 1.25 ã — 18 / 5 km / hr = 4.5 km / hr answer : b" | a ) 1.6 , b ) 4.5 , c ) 8.2 , d ) 6.5 , e ) 2.9 | b | divide(divide(300, const_1000), divide(multiply(4, const_60), const_3600)) | divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)| | physics |
jerry and michelle play a card game . in the beginning of the game they have an equal number of cards . each player , at her turn , gives the other a third of her cards . michelle plays first , giving jerry a third of her cards . jerry plays next , and michelle follows . then the game ends . jerry ended up with 28 more... | "gamemichelle jerry initially 54 54 assume after game 1 36 72 after game 2 60 48 after game 3 40 68 now merry has 28 cards more than michelle . this option gives us exactly what number of cards they had initially . so the answer is d" | a ) 51 , b ) 52 , c ) 53 , d ) 54 , e ) 56 | d | subtract(multiply(28, const_3), 28) | multiply(n0,const_3)|subtract(#0,n0)| | general |
if albert ’ s monthly earnings rise by 26 % , he would earn $ 693 . if , instead , his earnings rise by only 20 % , how much ( in $ ) would he earn this month ? | = 693 / 1.26 ∗ 1.2 = 660 = 660 answer is c | a ) 643 , b ) 652 , c ) 660 , d ) 690 , e ) 693 | c | multiply(divide(693, add(const_1, divide(26, const_100))), add(const_1, divide(20, const_100))) | divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n1,#3)|multiply(#2,#4) | gain |
what is the normal price of an article sold at $ 36 after two successive discounts of 10 % and 20 % ? | "0.8 * 0.9 * cost price = $ 36 cost price = $ 50 the answer is c ." | a ) $ 45 , b ) $ 48 , c ) $ 50 , d ) $ 54 , e ) $ 56 | c | divide(36, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100))) | subtract(const_100,n1)|subtract(const_100,n2)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n0,#4)| | gain |
if | x + 3 | = 5 , what is the sum of all the possible values of x ? | "there will be two cases x + 3 = 5 or x + 3 = - 5 = > x = 2 or x = - 8 sum of both the values will be - 8 + 2 = - 6 answer : d" | a ) - 16 , b ) - 12 , c ) - 8 , d ) - 6 , e ) 12 | d | add(3, 5) | add(n0,n1)| | general |
in a group of cows and hens , the number of legs are 14 more than twice the number of heads . the number of cows is : | "let no of cows be x , no of hens be y . so heads = x + y legs = 4 x + 2 y now , 4 x + 2 y = 2 ( x + y ) + 14 2 x = 14 x = 7 . answer : c" | a ) 5 , b ) 6 , c ) 7 , d ) 10 , e ) 12 | c | subtract(14, const_4) | subtract(n0,const_4)| | general |
a and b invests rs . 3000 and rs . 6000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ? | "( 3 * 6 + 6 * 6 ) : ( 6 * 12 ) 18 : 24 = > 3 : 4 . answer : e" | a ) 9 : 6 , b ) 3 : 8 , c ) 3 : 1 , d ) 9 : 9 , e ) 3 : 4 | e | divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(6000, add(6, 6))) | add(n2,n2)|multiply(n0,n2)|multiply(n0,const_2)|multiply(n2,#2)|multiply(n1,#0)|add(#1,#3)|divide(#5,#4)| | gain |
albert buys 4 horses and 9 cows for rs . 13,400 . if he sells the horses at 10 % profit and the cows at 20 % profit , then he earns a total profit of rs . 1880 . the cost of a horse is ? | explanation : let c . p . of each horse be rs . x and c . p . of each cow be rs . y . then , 4 x + 9 y = 13400 - - ( i ) and , 10 % of 4 x + 20 % of 9 y = 1880 2 / 5 x + 9 / 5 y = 1880 = > 2 x + 9 y = 9400 - - ( ii ) solving ( i ) and ( ii ) , we get : x = 2000 and y = 600 . cost price of each horse = rs . 2000 . answe... | a ) 1299 , b ) 2788 , c ) 2000 , d ) 2981 , e ) 2881 | c | multiply(20, const_100) | multiply(n4,const_100) | gain |
when positive integer x is divided by positive integer y , the remainder is 8 . if x / y = 76.4 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 8 - - > x = qy + 8 ; x / y = 76.4 - - > x = 76 y + 0.4 y ( so q above equals to 76 ) ; 0.4 y = 8 - - > y = 20 . answer : a ." | a ) 20 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | a | divide(8, subtract(76.4, floor(76.4))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general |
a family pays $ 850 per year for an insurance plan that pays 80 percent of the first $ 1,000 in expenses and 100 percent of all medical expenses thereafter . in any given year , the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total . | "upfront payment for insurance plan = 850 $ family needs to pay 20 % of first 1000 $ in expense = 200 $ total amount paid by family when medical expenses are equal to or greater than 1000 $ = 850 + 200 = 1050 $ total amount paid by insurance plan for first 1000 $ = 850 $ total amount paid by family will equal amount pa... | a ) $ 1,000 , b ) $ 1,250 , c ) $ 1,400 , d ) $ 1,800 , e ) $ 2,200 | b | subtract(1,000, 850) | subtract(n2,n0)| | general |
in a group of 120 people , 90 have an age of more 30 years , and the others have an age of less than 20 years . if a person is selected at random from this group , what is the probability the person ' s age is less than 20 ? | "number of people whose age is less than 20 is given by 120 - 90 = 30 probability p that a person selected at random from the group is less than 20 is gieven by 30 / 120 = 0.25 correct answer c" | a ) 60 / 120 , b ) 15 / 120 , c ) 30 / 120 , d ) 80 / 120 , e ) 40 / 120 | c | divide(subtract(120, 90), 120) | subtract(n0,n1)|divide(#0,n0)| | general |
the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 10 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "original price = 100 day 1 discount = 10 % , price = 100 - 10 = 90 day 2 discount = 10 % , price = 90 - 9 = 81 day 3 discount = 10 % , price = 81 - 8.1 = 72.9 which is 72.9 / 90 * 100 of the sale price on day 1 = ~ 81 % answer e" | a ) 28 % , b ) 40 % , c ) 64.8 % , d ) 70 % , e ) 81 % | e | add(multiply(divide(divide(10, const_100), subtract(1, divide(1, 10))), const_100), 2) | divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4)| | gain |
given that 268 x 74 = 19432 , find the value of 2.68 x . 74 . | "solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 × . 74 = 1.9432 answer a" | a ) 1.9432 , b ) 1.0025 , c ) 1.5693 , d ) 1.0266 , e ) none | a | multiply(divide(268, const_100), divide(74, const_100)) | divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)| | general |
solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 300 milliliters of solution x to create a solution that is 22 percent alcohol by volume ? | "22 % is 12 % - points higher than 10 % but 8 % - points lower than 30 % . thus there should be 2 parts of solution x for 3 parts of solution y . we should add 450 ml of solution y . the answer is d ." | a ) 300 , b ) 350 , c ) 400 , d ) 450 , e ) 500 | d | multiply(divide(subtract(22, 10), subtract(30, 22)), 300) | subtract(n3,n0)|subtract(n1,n3)|divide(#0,#1)|multiply(n2,#2)| | general |
find the area of circle whose radius is 7 m ? | the area of circle = pie * r ^ 2 = 22 / 7 * 7 * 7 = 154 sq m answer : e | ['a ) 121 sq m', 'b ) 184 sq m', 'c ) 174 sq m', 'd ) 124 sq m', 'e ) 154 sq m'] | e | circle_area(7) | circle_area(n0) | geometry |
walking at 5 / 6 of its usual speed , a train is 10 minutes too late . find its usual time to cover the journey . | "new speed = 5 / 6 of the usual speed new time taken = 6 / 5 of the usual time taken so , ( 6 / 5 of the usual time ) - ( usual time ) = 10 min 1 / 5 of the usual time = 10 min usual time = 10 * 5 = 50 min correct option is a" | a ) 50 min , b ) 20 min , c ) 1 hour , d ) 30 min , e ) 45 min | a | divide(10, subtract(inverse(divide(5, 6)), const_1)) | divide(n0,n1)|inverse(#0)|subtract(#1,const_1)|divide(n2,#2)| | physics |
in the rectangle below , the line mn cuts the rectangle into two regions . find x the length of segment nb so that the area of the quadrilateral mnbc is 40 % of the total area of the rectangle . | solution we first note that mc = 20 - 5 = 15 the quadrilateral mnbc is a trapezoid and its area a is given by a = ( 1 / 2 ) × 10 × ( x + mc ) = 5 ( x + 15 ) 40 % of the area of the rectangle is equal to 40 % × ( 20 × 10 ) = ( 40 / 100 ) × 200 = 80 since the area of mnbc is equal to 40 % the area of the rectangle , we c... | a ) 1 meter , b ) 2 meter , c ) 3 meter , d ) 4 meter , e ) 5 meter | a | subtract(subtract(divide(40, const_12), const_0_33), const_2) | divide(n0,const_12)|subtract(#0,const_0_33)|subtract(#1,const_2) | geometry |
two trains of equal lengths take 15 sec and 20 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 15 = 8 m / sec . speed of the second train = 120 / 20 = 6 m / sec . relative speed = 8 + 6 = 14 m / sec . required time = ( 120 + 120 ) / 14 = 17 sec . answer : c" | a ) 16 sec , b ) 12 sec , c ) 17 sec , d ) 21 sec , e ) 23 sec | c | divide(multiply(120, const_2), add(speed(120, 20), speed(120, 15))) | multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)| | physics |
3,7 , 12,18 , 25,33 , . . . . . . . . . . . . . . 7 th terms | "3 + 4 = 7 7 + 5 = 12 12 + 6 = 18 18 + 7 = 25 25 + 8 = 33 33 + 9 = 42 answer : b" | a ) 43 , b ) 42 , c ) 63 , d ) 65 , e ) 78 | b | subtract(negate(7), multiply(subtract(12,18, 25,33), divide(subtract(12,18, 25,33), subtract(3,7, 12,18)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.