Problem
stringlengths 5
967
| Rationale
stringlengths 1
2.74k
| options
stringlengths 37
164
| correct
stringclasses 5
values | annotated_formula
stringlengths 7
1.65k
| linear_formula
stringlengths 8
925
| category
stringclasses 6
values | answer
stringclasses 5
values |
|---|---|---|---|---|---|---|---|
what is the smallest 5 digit number that is divisible by 15 , 32 , 45 , and 54 ? | 15 = 3 * 5 32 = 2 ^ 5 45 = 3 ^ 2 * 5 54 = 2 * 3 ^ 3 lcm = 2 ^ 5 * 3 ^ 3 * 5 = 4320 the smallest five - digit number that is a multiple of 4320 is 3 * 4320 = 12,960 the answer is d . | a ) 11260 , b ) 11860 , c ) 12360 , d ) 12960 , e ) 13560 | d | add(multiply(const_100, const_100), subtract(lcm(lcm(lcm(lcm(5, 15), 32), 45), 54), reminder(multiply(const_100, const_100), lcm(lcm(lcm(lcm(5, 15), 32), 45), 54)))) | lcm(n0,n1)|multiply(const_100,const_100)|lcm(n2,#0)|lcm(n3,#2)|lcm(n4,#3)|reminder(#1,#4)|subtract(#4,#5)|add(#1,#6) | general | D |
if n is a natural number , then ( 7 ( n 2 ) + 7 n ) is always divisible by : | "solution ( 7 n 2 + 7 n ) = 7 n ( n + 1 ) , which is always divisible by 7 and 14 both , since n ( n + 1 ) is always even . answer d" | a ) 3 and 7 , b ) 14 only , c ) 7 only , d ) 7 and 14 , e ) none | d | add(multiply(7, const_100), multiply(2, 7)) | multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)| | general | D |
a man swims downstream 18 km and upstream 12 km taking 3 hours each time , what is the speed of the man in still water ? | "18 - - - 3 ds = 6 ? - - - - 1 12 - - - - 3 us = 4 ? - - - - 1 m = ? m = ( 6 + 4 ) / 2 = 5 answer : b" | a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 8 | b | divide(add(divide(12, 3), divide(18, 3)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics | B |
if jake loses 32 pounds , he will weigh twice as much as his sister . together they now weigh 212 pounds . what is jake β s present weight , in pounds ? | lets say j is the weight of jack and s is the wt of his sister . if he loses 32 pounds , he s twice as heavy as his sister . j - 32 = 2 * s also , together they weight 212 pounds j + s = 212 solvong the 2 equation , we get j = 152 pounds ! d | a ) 131 , b ) 135 , c ) 139 , d ) 152 , e ) 188 | d | add(multiply(divide(subtract(212, 32), const_3), const_2), 32) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2) | general | D |
solution a is 20 % sugar and solution b is 80 % sugar . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % sugar solution ? | forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 80 - 50 ) / ( 50 - 20 ) = 1 / 1 so the volume of the two solutions will be equal . answer has to be 4 : 4 = c | a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 3 : 7 | c | multiply(divide(subtract(divide(80, const_100), divide(50, const_100)), divide(60, const_100)), const_2) | divide(n1,const_100)|divide(n4,const_100)|divide(n3,const_100)|subtract(#0,#1)|divide(#3,#2)|multiply(#4,const_2) | other | C |
if ( 1 / 2 ) ^ 16 ( 1 / 81 ) ^ k = 1 / 18 ^ 16 , then k = | "i ' m going to focus on denominator only . . ( 2 ^ 16 ) . ( ( 3 ^ 4 ) ^ k = 18 ^ 16 ( 2 ^ 16 ) . ( ( 3 ^ 4 k ) = ( 2 . 3 ^ 2 ) ^ 16 ( 2 ^ 16 ) . ( ( 3 ^ 4 k ) = ( 2 ^ 24 ) . ( 3 ^ 2 ) ^ 16 hence 4 k = 32 k = 8 answer a i hope it ' s quite clear" | a ) 8 , b ) 12 , c ) 16 , d ) 24 , e ) 36 | a | divide(multiply(16, 2), const_4) | multiply(n1,n2)|divide(#0,const_4)| | general | A |
if 16 ^ y = 2 ^ 16 , what is y ? | "16 ^ y = 2 ^ 4 y = 2 ^ 16 4 y = 16 y = 4 the answer is b ." | a ) 2 , b ) 4 , c ) 8 , d ) 10 , e ) 12 | b | divide(16, const_2) | divide(n2,const_2)| | general | B |
if 9 gallons of gasoline are added to a tank that is already filled to 3 / 4 of its capacity , the tank is then filled to 9 / 10 of its capacity . how many gallons does the tank hold ? | "let the capacity of the tank = c ( 3 / 4 ) c + 9 = ( 9 / 10 ) c = > ( 9 / 10 ) c - ( 3 / 4 ) c = 9 = > ( 3 / 20 ) c = 9 = > c = ( 9 * 20 ) / 3 = 60 number of gallons of gasoline that the tank currently holds = 3 / 4 * c + 9 = 45 + 9 = 54 answer e" | a ) 20 , b ) 24 , c ) 36 , d ) 40 , e ) 54 | e | add(multiply(divide(9, subtract(divide(9, 10), divide(3, 4))), divide(3, 4)), 9) | divide(n3,n4)|divide(n1,n2)|subtract(#0,#1)|divide(n0,#2)|multiply(#3,#1)|add(n0,#4)| | general | E |
find the smallest number of five digits exactly divisible by 16 , 24,36 and 54 . | "smallest number of five digits is 10000 . required number must be divisible by l . c . m . of 16,24 , 36,54 i . e 432 , on dividing 10000 by 432 , we get 64 as remainder . therefore , required number = 10000 + ( 432 β 64 ) = 10368 . answer is a ." | a ) 10368 , b ) 10638 , c ) 10836 , d ) 10846 , e ) none of them | a | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 24,36) | multiply(const_100,const_100)|multiply(#0,const_10)|subtract(#1,const_100)|add(n1,#2)| | general | A |
9 people decided to split the restaurant bill evenly . if the bill was $ 514.16 dollars , how much money did they 1 cent is the smallest unit ? | "this is equivalent to finding the first number that is divisible by 9 that occurs after 51416 . in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 51416 = 17 and 18 is divisible by 9 . hence , we need to add 1 to this number for it to be divisible by 9 correct option : b" | a ) $ 514.16 , b ) $ 514.17 , c ) $ 514.18 , d ) $ 514.19 , e ) $ 514.20 | b | add(514.16, divide(const_3, const_100)) | divide(const_3,const_100)|add(n1,#0)| | general | B |
the circumferences of two circles are 268 meters and 380 meters . find the difference between the areas of the larger and the smaller circles . | "let the radii of the smaller and the larger circles be s m and l m respectively . 2 β s = 268 and 2 β l = 380 s = 268 / 2 β and l = 380 / 2 β difference between the areas = β l 2 - β s 2 = β { 1902 / β 2 - 1342 / β 2 } = 1902 / β - 1342 / β = ( 190 - 134 ) ( 190 + 134 ) / β = ( 56 ) ( 324 ) / ( 22 / 7 ) = ( 18144 ) / ( 22 ) ( 7 ) = 5773.09 sq m . answer : e" | a ) 388.15 , b ) 2992 , c ) 4312.5 , d ) 2887.27 , e ) 5773.09 | e | subtract(circle_area(divide(380, multiply(const_2, const_pi))), circle_area(divide(268, multiply(const_2, const_pi)))) | multiply(const_2,const_pi)|divide(n1,#0)|divide(n0,#0)|circle_area(#1)|circle_area(#2)|subtract(#3,#4)| | geometry | E |
if the least common multiple of two prime numbers x and y is 10 , where x > y , then the value of 2 x + y is | lcm of two prime numbers will basically be multiple of those numbers . lcm = 10 = 2 * 5 , both of which are prime . now we know x > y , so x = 5 and y = 2 . 2 x + y = 2 * 5 + 2 = 10 + 2 = 12 . answer : d | a ) 7 , b ) 9 , c ) 11 , d ) 12 , e ) 21 | d | add(multiply(divide(10, const_2), 2), const_2) | divide(n0,const_2)|multiply(n1,#0)|add(#1,const_2) | general | D |
a company conducted a survey about its two brands , a and b . x percent of respondents liked product a , ( x β 20 ) percent liked product b , 23 percent liked both products , and 23 percent liked neither product . what is the minimum number f of people surveyed by the company ? | "100 = x + x - 20 + 23 - 23 x = 60 , so , product a = 60 % , product b = 40 % , both = 23 % , neither = 23 % 23 % of the total no . of people should be an integer . so , a , bc are out . 60 % of d and 40 % of d are both integers . so , d satisfies all conditions . so , answer is d ." | a ) 46 , b ) 80 , c ) f = 90 , d ) f = 100 , e ) 200 | d | add(subtract(divide(add(add(subtract(const_100, 23), 23), 20), const_2), 20), divide(add(add(subtract(const_100, 23), 23), 20), const_2)) | subtract(const_100,n1)|add(n1,#0)|add(n0,#1)|divide(#2,const_2)|subtract(#3,n0)|add(#3,#4)| | other | D |
find the principle on a certain sum of money at 5 % per annum for 3 1 / 5 years if the amount being rs . 1740 ? | "explanation : 1740 = p [ 1 + ( 5 * 16 / 5 ) / 100 ] p = 1500 answer : option d" | a ) rs . 1000 , b ) rs . 1550 , c ) rs . 1510 , d ) rs . 1500 , e ) none of these | d | divide(1740, add(divide(multiply(divide(add(multiply(3, 5), 3), 5), 5), const_100), const_1)) | multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)| | general | D |
arjun started a business with rs . 20000 and is joined afterwards by anoop with rs . 4000 . after how many months did anoop join if the profits at the end of the year are divided equally ? | suppose anoop joined after 3 months . then , 20000 * 12 = 40000 * ( 12 β x ) = > = > x = 6 . answer : d | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(multiply(20000, const_12), multiply(4000, const_10)) | multiply(n0,const_12)|multiply(n1,const_10)|divide(#0,#1) | general | D |
what is the value of ( 2 ) ^ - 6 ? | "2 ^ - 6 = 1 / ( 2 ) ^ 6 = 1 / 64 answer : b" | a ) 1 / 62 , b ) 1 / 64 , c ) 1 / 65 , d ) 1 / 66 , e ) 1 / 67 | b | divide(multiply(add(add(const_100, const_60), const_1), 2), const_100) | add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)| | general | B |
the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 50 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 50 + 20 ) kg = 70 kg . a )" | a ) 70 kg , b ) 80 kg , c ) 85 kg , d ) 90 kg , e ) 95 kg | a | add(multiply(8, 2.5), 50) | multiply(n0,n1)|add(n2,#0)| | general | A |
a cement mixture is composed of 3 elements . by weight , 1 / 3 of the mixture is sand , 1 / 2 of the mixture is water , and the remaining 8 pounds of the mixture is gravel . what is the weight of the entire mixture in pounds ? | "let the total weight be x . sand content = ( 1 / 3 ) x water content = ( 1 / 2 ) x gravel = x - ( 1 / 3 ) x - ( 1 / 2 ) x = ( 1 / 6 ) x = 8 x = 48 then answer will be e = 48" | a ) 30 , b ) 36 , c ) 42 , d ) 46 , e ) 48 | e | divide(8, subtract(1, add(divide(1, 3), divide(1, 2)))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|subtract(n1,#2)|divide(n5,#3)| | general | E |
if n is a positive integer and the product of all the integers from 1 to n , inclusive , is a multiple of 9 , what is the least possible value of n ? | we need at least 2 factors of three , which means n must be at least 6 , so b is the correct answer . | a ) 7 , b ) 6 , c ) 9 , d ) 12 , e ) 7 | b | divide(9, const_2) | divide(n1,const_2)| | general | B |
q - 1 ) of the 84 parents who attended a meeting at a school , 25 volunteered to supervise children during the school picnic and 11 volunteered both to supervise children during the picnic and to bring refreshments to the picnic . if the number of parents who volunteered to bring refreshments was 1.5 times the number of parents who neither volunteered to supervise children during the picnic nor volunteered to bring refreshments , how many of the parents volunteered to bring refreshments ? | i used the following formula : total = group 1 + group 2 + neither - both using the information from the question stem , we have : 84 total people 25 who supervise ( group 1 ) x who neither supervise nor bring refreshments 1.5 x people who bring refreshments ( group 2 ) 11 who supervise and bring refreshments ( both ) therefore . . . 84 = 25 + 1.5 x + x - 11 solving , we get x = 28 since we want the value of those who bring refreshments and supervise : 1.5 ( 28 ) = 42 answer is e | a ) 25 , b ) 36 , c ) 38 , d ) 45 , e ) 42 | e | add(divide(subtract(multiply(subtract(84, 25), 1.5), 11), add(1.5, const_1)), 11) | add(n4,const_1)|subtract(n1,n2)|multiply(n4,#1)|subtract(#2,n3)|divide(#3,#0)|add(n3,#4) | other | E |
the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 49 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ? | "20 x + 49 = 21 ( x β 5 ) x = 56 answer : b" | a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) 59 | b | subtract(multiply(add(20, const_1), 5), 49) | add(n0,const_1)|multiply(n2,#0)|subtract(#1,n1)| | general | B |
a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 500 bolts ? | "old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 500 bolts 500 * 6 / 25 = 120 mins ans c" | a ) 36 , b ) 72 , c ) 120 , d ) 144 , e ) 180 | c | divide(500, add(divide(100, const_60), divide(150, const_60))) | divide(n0,const_60)|divide(n1,const_60)|add(#0,#1)|divide(n2,#2)| | physics | C |
two trains of length 200 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 200 + 280 = 480 m . the time required = d / s = 480 / 20 = 24 sec . answer : e" | a ) 28 , b ) 266 , c ) 990 , d ) 20 , e ) 24 | e | divide(add(200, 280), multiply(add(42, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics | E |
pascal has 96 miles remaining to complete his cycling trip . if he reduced his current speed by 4 miles per hour , the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50 % . what is his current speed w ? | "let the current speed be x miles per hour . time taken if speed is 50 % faster ( i . e . 3 x / 2 = 1.5 x ) = 96 / 1.5 x time taken if speed is reduced by 4 miles / hr ( i . e . ( x - 4 ) ) = 96 / ( x - 4 ) as per question , 96 / ( x - 4 ) - 96 / 1.5 x = 16 solving this w we get x = 8 . b ." | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 16 | b | divide(add(divide(96, 16), sqrt(add(multiply(multiply(divide(divide(96, 16), add(const_1, divide(50, const_100))), 4), 4), power(divide(96, 16), const_2)))), const_2) | divide(n0,n2)|divide(n3,const_100)|add(#1,const_1)|power(#0,const_2)|divide(#0,#2)|multiply(n1,#4)|multiply(#5,n1)|add(#6,#3)|sqrt(#7)|add(#0,#8)|divide(#9,const_2)| | physics | B |
what will be the remainder if 2 ^ 300 is divided by 4 ? | d 4 now if you go on calculating 2 ^ 300 , you will lose so much time and it might not even be feasible to carry out so long of calculations . thus we will make use of a trick here . we will calculate the remainder of each power of 2 till we come across a pattern . 2 ^ 1 divided by 4 leaves the remainder 2 . 2 ^ 2 divided by 4 leaves the remainder 0 . 2 ^ 3 divided by 4 leaves the remainder 0 . 2 ^ 4 divided by 4 leaves the remainder 0 . you can see that all the following powers of two will be divisible by 4 . therefore the remainder when 2 ^ 300 is divided by 4 will be 0 only . | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | power(2, const_2) | power(n0,const_2) | general | D |
the captain of a cricket team of 11 members is 27 years old and the wicket keeper is 1 year older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 27 + 28 ) = 9 ( x - 1 ) 11 x - 9 x = 46 2 x = 46 x = 23 . the average age of the team is 23 years . the answer is d ." | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | d | divide(subtract(add(27, add(27, 1)), multiply(const_3.0, const_3.0)), const_2) | add(n1,n2)|multiply(const_3.0,const_3.0)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general | D |
in a coconut grove , ( x + 2 ) trees yield 60 nuts per year , x trees yield 120 nuts per year and ( x β 2 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 2 ) Γ 60 + x Γ 120 + ( x β 2 ) Γ 180 / ( x + 2 ) + x + ( x β 2 ) = 100 β 360 x β 240 / 3 x = 100 β 60 x = 240 β x = 4 answer e" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | divide(subtract(multiply(180, 2), multiply(60, 2)), subtract(add(add(60, 120), 180), multiply(100, const_3))) | add(n1,n2)|multiply(n0,n4)|multiply(n0,n1)|multiply(n5,const_3)|add(n4,#0)|subtract(#1,#2)|subtract(#4,#3)|divide(#5,#6)| | general | E |
addison high school β s senior class has 300 boys and 240 girls . if 30 % of the boys did not attend the college then 70 % of the total class attended college , what percentage of the girls class did not attend college ? | "% of boy attending the college = 100 - 30 = 70 % number of boys attending the college : 0.70 * 300 = 210 total people = 300 + 240 = 540 total number of students attending the college : 0.70 * 540 = 378 number of girls attending the college = 378 - 210 = 168 number of girls not attending the college = 240 - 168 = 72 % of girls not attending = 72 / 240 = 30 % . e is the correct answer" | a ) 70 , b ) 45 , c ) 50 , d ) 46 , e ) 30 | e | multiply(divide(add(multiply(divide(30, const_100), 300), multiply(subtract(const_1, divide(70, const_100)), 240)), add(300, 240)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|subtract(const_1,#2)|multiply(n1,#4)|add(#3,#5)|divide(#6,#0)|multiply(#7,const_100)| | general | E |
if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 210 games will be played during the tournament . what is the number of participants ? | "let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 210 n * ( n - 1 ) = 420 = 21 * 20 ( trial and error ) the answer is d ." | a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | d | divide(add(sqrt(add(multiply(multiply(210, 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 | D |
s is a set of 100 consecutive multiples of 8 . if the smallest number in s is 108 , then the greatest number in s is | "last term = first term + ( total no . of terms - 1 ) consecutive difference s is a set of 100 consecutive multiples of 8 . if the smallest number in s is 108 , then the greatest number in s is first term = 102 ; total terms = 100 ; difference = 3 108 + ( 99 ) 8 = 900 ans e" | a ) 998 , b ) 297 , c ) 999 , d ) 902 , e ) 900 | e | add(108, multiply(subtract(100, const_1), 8)) | subtract(n0,const_1)|multiply(n1,#0)|add(n2,#1)| | general | E |
of the two square fields , the area of the one is 1 hectare , while anothe one is broader by 1 % . there differences in area is : | area of one square field = 10000 m ( power ) 2 10000 Γ 1 = 10000 side of this field = β 10000 m = 100 m side of another square = 101 m difference of areas = [ 101 ( power ) 2 - 100 ( power ) 2 ] m ( power ) 2 [ 101 + 100 ] [ 101 - 100 ] m ( power ) 2 ( 201 ) ( 1 ) m 2 = 201 m ( power ) 2 answer is a . | ['a ) 201 m ( power ) 2', 'b ) 220 m ( power ) 2', 'c ) 211 m ( power ) 2', 'd ) 219 m ( power ) 2', 'e ) 205 m ( power ) 2'] | a | subtract(square_area(add(1, sqrt(multiply(const_10, const_1000)))), multiply(const_10, const_1000)) | multiply(const_10,const_1000)|sqrt(#0)|add(n0,#1)|square_area(#2)|subtract(#3,#0) | geometry | A |
at a certain paint store forest green is made by mixing 4 parts blue paint with 3 parts yellow paint . verdant green is made by mixing 4 parts yellow paint with 3 parts blue paint . how many liters of yellow paint must be added to 7 liters of forest green to change it to verdant green ? | 7 liter of forset green have 4 liter of blue and 3 liter of yellow suppose we add x liter of yellow to make it a verdant green so the ratio of blue to yellow in verdant green is ΒΎ so the equation is blue / yellow = 4 / ( 3 + x ) = ΒΎ 9 + 3 x = 16 = > x = 7 / 3 answer : c | a ) 5 / 3 , b ) 2 / 7 , c ) 7 / 3 , d ) 1 / 7 , e ) 1 / 4 | c | subtract(multiply(divide(4, 3), 4), 3) | divide(n0,n1)|multiply(n0,#0)|subtract(#1,n1) | general | C |
in a garden , 26 trees are planted at equal distances along a yard 800 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ? | "26 trees have 25 gaps between them . length of each gap = 800 / 25 = 32 i . e . , distance between two consecutive trees = 32 answer is c ." | a ) 10 , b ) 8 , c ) 32 , d ) 14 , e ) 16 | c | divide(800, subtract(26, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | physics | C |
two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 3 litres from vessel p is mixed with 4 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ? | "quantity of alcohol in vessel p = 62.5 / 100 * 3 = 15 / 8 litres quantity of alcohol in vessel q = 87.5 / 100 * 4 = 7 / 2 litres quantity of alcohol in the mixture formed = 15 / 8 + 7 / 2 = 43 / 8 = 5.375 litres as 7 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 5.375 : 1.625 = 179 : 54 . answer : d" | a ) 179 : 2 , b ) 178 : 3 , c ) 179 : 1 , d ) 179 : 54 , e ) 179 : 45 | d | divide(add(divide(multiply(62.5, 3), const_100), divide(multiply(87.5, 4), const_100)), add(subtract(3, divide(multiply(62.5, 3), const_100)), subtract(4, divide(multiply(87.5, 4), const_100)))) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|subtract(n2,#2)|subtract(n3,#3)|add(#5,#6)|divide(#4,#7)| | other | D |
last year sandy saved 10 % of her annual salary . this year , she made 10 % more money than last year , and she saved 15 % of her salary . the amount saved this year was what percent of the amount she saved last year ? | "let last year ' s salary be x . last year , sandy save 0.1 x this year , sandy saved 0.15 * 1.1 x = 0.165 x 0.165 x / 0.1 x = 1.65 = 165 % the answer is c ." | a ) 142 % , b ) 150 % , c ) 165 % , d ) 176 % , e ) 188 % | c | multiply(divide(multiply(divide(15, const_100), add(const_100, 10)), divide(multiply(10, const_100), const_100)), const_100) | add(n1,const_100)|divide(n2,const_100)|multiply(n0,const_100)|divide(#2,const_100)|multiply(#0,#1)|divide(#4,#3)|multiply(#5,const_100)| | gain | C |
a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 30 percent . | 1 hr = 20 percent . thus 15 min = 5 percent . now to charge 30 percent 90 min . answer : b | a ) 145 , b ) 90 , c ) 175 , d ) 160 , e ) 130 | b | multiply(divide(30, 20), const_60) | divide(n2,n1)|multiply(#0,const_60)| | physics | B |
a meeting has to be conducted with 4 managers . find the number of ways in which the managers may be selected from among 8 managers , if there are 2 managers who refuse to attend the meeting together . | "the total number of ways to choose 4 managers is 8 c 4 = 70 we need to subtract the number of groups which include the two managers , which is 6 c 2 = 15 . 70 - 15 = 55 the answer is b ." | a ) 48 , b ) 55 , c ) 58 , d ) 62 , e ) 70 | b | subtract(choose(8, 4), choose(subtract(8, 2), 2)) | choose(n1,n0)|subtract(n1,n2)|choose(#1,n2)|subtract(#0,#2)| | probability | B |
in an examination , 25 % of total students failed in hindi , 40 % failed in english and 35 % in both . the percentage of these who passed in both the subjects is : | "pass percentage = 100 - ( 25 + 40 - 35 ) = 100 - 70 = 30 answer : c" | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | c | subtract(const_100, subtract(add(25, 40), 35)) | add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)| | general | C |
the difference between the place values of 6 and 1 in the number 527631 is | sol . = ( place value of 6 ) β ( place value of 1 ) = ( 600 - 1 ) = 599 answer e | a ) 596 , b ) 587 , c ) 533 , d ) 568 , e ) 599 | e | subtract(multiply(const_10, 6), 6) | multiply(n0,const_10)|subtract(#0,n0)| | general | E |
the edge of a cube is 5 a cm . find its surface ? | "6 a 2 = 6 * 5 a * 5 a = 150 a 2 answer : a" | a ) 150 a 2 cm 2 , b ) 170 a 2 cm 2 , c ) 125 a 2 cm 2 , d ) 175 a 2 cm 2 , e ) 120 a 2 cm 2 | a | surface_cube(5) | surface_cube(n0)| | geometry | A |
the average of first 25 prime numbers is ? | "explanation : average = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 / 9 = 100 / 9 = 11.11 ( approx ) answer is e" | a ) 9.9 , b ) 8.8 , c ) 5.6 , d ) 11.9 , e ) 11.11 | e | add(25, const_1) | add(n0,const_1)| | general | E |
find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 . | "( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = 2 ^ 11 x 2 ^ 11 x 7 ^ 5 x 11 ^ 2 = 2 ^ 22 x 7 ^ 5 x 11 ^ 2 total number of prime factors = ( 22 + 5 + 2 ) = 29 . answer is a ." | a ) 29 , b ) 27 , c ) 28 , d ) 24 , e ) 21 | a | add(add(multiply(2, 11), 5), 2) | multiply(n1,n5)|add(n3,#0)|add(n5,#1)| | general | A |
find the area of a parallelogram with base 25 cm and height 15 cm ? | "area of a parallelogram = base * height = 25 * 15 = 375 cm 2 answer : d" | a ) 295 cm 2 , b ) 385 cm 2 , c ) 275 cm 2 , d ) 375 cm 2 , e ) 285 cm 2 | d | multiply(25, 15) | multiply(n0,n1)| | geometry | D |
the sector of a circle has radius of 7 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 7 ) + 2 ( 7 ) = 16.5 + 14 = 30.5 cm answer : d" | a ) 91.5 cm , b ) 92.2 cm , c ) 28.9 cm , d ) 30.5 cm , e ) 98.2 cm | d | multiply(multiply(const_2, divide(multiply(subtract(7, const_3), const_2), add(const_4, const_3))), 7) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics | D |
a circular dartboard of radius 1 foot is at a distance of 20 feet from you . you throw a dart at it and it hits the dartboard at some point q in the circle . what is the probability that q is closer to the center of the circle than the periphery ? | radius of inner dartboard / overall radius of dartboard ( ( r / 2 ) ^ 2 ) / ( r ^ 2 ) = 1 / 4 = 0.25 answer : c | a ) 0.75 , b ) 1 , c ) 0.25 , d ) 0.5 , e ) 0.58 | c | divide(multiply(divide(1, const_2), divide(1, const_2)), multiply(const_1, const_1)) | divide(n0,const_2)|multiply(const_1,const_1)|multiply(#0,#0)|divide(#2,#1) | probability | C |
two numbers are in the ratio 3 : 4 . if their l . c . m . is 96 . what is sum of the numbers ? | "explanation : let the numbers be 3 x and 4 x lcm of 3 x and 4 x = 12 x ( since lcm of 3 and 4 is 12 . hence lcm of 3 x and 4 x is 12 x ) given that lcm of 3 x and 4 x is 96 = > 12 x = 96 = > x = 96 / 12 = 8 sum of the numbers = 3 x + 4 x = 7 x = 7 x 8 = 56 answer : option b" | a ) 48 , b ) 56 , c ) 76 , d ) 84 , e ) 85 | b | add(multiply(multiply(3, const_4.0), const_100), multiply(4, 96)) | multiply(n0,const_4.0)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | other | B |
a merchant sells an item at a 20 % discount , but still makes a gross profit of 25 percent of the cost . what percent of the cost would the gross profit on the item have been if it had been sold without the discount ? | "original sp = x cost = c current selling price = . 8 x ( 20 % discount ) . 8 x = 1.5 c ( 25 % profit ) x = 1.25 / . 8 * c x = 12.5 / 8 c original selling price is 1.5625 c which is 56.25 % profit answer c" | a ) 20 % , b ) 40 % , c ) 56.25 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 25)) | subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1)| | gain | C |
the sale price shirts listed for rs . 400 after successive discount is 10 % and 5 % is ? | 400 * ( 90 / 100 ) * ( 95 / 100 ) = 342 c | a ) 280 , b ) 290 , c ) 342 , d ) 250 , e ) 253 | c | subtract(400, add(divide(multiply(400, 10), const_100), divide(multiply(5, 400), const_100))) | multiply(n0,n1)|multiply(n0,n2)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|subtract(n0,#4) | gain | C |
what is the distance covered by a car if it travels with a speed of 160 kmh for 5 hours ? | distance = time x speed 5 x 160 = 800 answer : c | a ) 350 km , b ) 400 km , c ) 800 km , d ) 600 km , e ) 650 km | c | multiply(160, 5) | multiply(n0,n1) | physics | C |
how many seconds will it take for a car that is traveling at a constant rate of 90 miles per hour to travel a distance of 22 yards ? ( 1 mile = 1,160 yards ) | speed = 90 miles / hr = 44 yard / s distance = 22 yards time = distance / speed = 22 / 44 = 0.5 sec ans - b | a ) 0.8 , b ) 0.5 , c ) 1.0 , d ) 1.1 , e ) 1.2 | b | divide(22, multiply(divide(90, const_3600), multiply(subtract(add(90, 90), const_4), const_10))) | add(n0,n0)|divide(n0,const_3600)|subtract(#0,const_4)|multiply(#2,const_10)|multiply(#1,#3)|divide(n1,#4) | physics | B |
{ 5 , 6 , 3 , 8 , 4 } if we want to get the median value of the set to be 10 which value should be inserted in the set of 5 values ? | median is the middle number of a sorted distribution . in case of even number of items , median will be the average of middle two values . the question asks fora new number that will cause the median of new set at 10 . mentionable that given set is already sorted and number of items of new set will be 5 + 1 = 6 . to get median 10 , the sum of 3 th 4 th term should be 20 i . e . 3 + 7 . so to keep the median 10 , the new item should be > = 7 answer is b | a ) 5 , b ) 7 , c ) 15 , d ) 12 , e ) 13 | b | add(add(add(add(add(add(subtract(add(5, 6), 10), const_1), subtract(add(5, 6), 10)), subtract(add(5, 6), 10)), subtract(add(5, 6), 10)), subtract(add(5, 6), 10)), subtract(add(5, 6), 10)) | add(n0,n1)|subtract(#0,n5)|add(#1,const_1)|add(#2,#1)|add(#3,#1)|add(#4,#1)|add(#5,#1)|add(#6,#1) | general | B |
if rs . 1440 / - are divided among a , b and c so that a receives 1 / 3 rd as much as b and b receives 1 / 4 th as much as c . the amount b received is ? | a : b : c = 1 : 3 : 12 total parts = 16 b ' s share is = 3 parts 16 - - - - - > 1440 1 - - - - - > 90 3 - - - - - > 270 ( b ' s share is 270 ) b ) | a ) 250 , b ) 270 , c ) 310 , d ) 330 , e ) 350 | b | divide(divide(1440, add(add(divide(1, multiply(3, 4)), divide(1, 4)), 1)), 4) | divide(n1,n4)|multiply(n2,n4)|divide(n1,#1)|add(#2,#0)|add(n1,#3)|divide(n0,#4)|divide(#5,n4) | general | B |
a man traveled a total distance of 1800 km . he traveled one - third of the whole trip by plane and the distance traveled by train is one - half of the distance traveled by bus . if he traveled by train , plane and bus , how many kilometers did he travel by bus ? | "total distance traveled = 1800 km . distance traveled by plane = 600 km . distance traveled by bus = x distance traveled by train = x / 2 x + x / 2 + 600 = 1800 3 x / 2 = 1200 x = 800 km the answer is b ." | a ) 700 , b ) 800 , c ) 900 , d ) 1000 , e ) 1100 | b | divide(multiply(divide(multiply(1800, const_2), const_3), const_3), add(const_2, const_3)) | add(const_2,const_3)|multiply(n0,const_2)|divide(#1,const_3)|multiply(#2,const_3)|divide(#3,#0)| | physics | B |
a man goes downstream at 14 kmph , and upstream 8 kmph . the speed of the stream is | speed of the stream = 1 / 2 ( 14 - 8 ) kmph = 3 kmph . correct option : e | a ) 0 kmph , b ) 4 kmph , c ) 16 kmph , d ) 2.5 kmph , e ) 3 kmph | e | divide(subtract(14, 8), const_2) | subtract(n0,n1)|divide(#0,const_2) | physics | E |
what is the least number . which should be added to 221 to make it a perfect square ? | "221 + 4 = 225 15 ^ 2 answer : c" | a ) 5 , b ) 8 , c ) 4 , d ) 6 , e ) 7 | c | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 221) | add(const_12,const_4)|add(#0,const_2)|divide(#1,const_100)|multiply(#2,#2)|subtract(#3,n0)| | general | C |
a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 600 , what is the reduced price for kg ? | "a 600 * ( 25 / 100 ) = 150 - - - - 5 ? - - - - 1 = > rs . 30" | a ) 30 , b ) 60 , c ) 70 , d ) 80 , e ) 20 | a | divide(divide(multiply(600, 25), const_100), 5) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain | A |
in a single throw of a die , what is probability of getting a number greater than 4 ? | "e = { 5,6 } n ( e ) = 2 p ( s ) = n ( e ) / n ( s ) = 2 / 6 = 1 / 3 ans : e" | a ) 2 / 3 , b ) 3 / 5 , c ) 4 / 7 , d ) 8 / 9 , e ) 1 / 3 | e | divide(const_2, add(4, const_2)) | add(n0,const_2)|divide(const_2,#0)| | probability | E |
calculate the circumference of a circular field whose radius is 8 centimeters . | "circumference c is given by c = 2 Ο r = 2 Ο * 8 = 16 Ο cm answer : d" | a ) 17 Ο cm , b ) 18 Ο cm , c ) 1 Ο cm , d ) 16 Ο cm , e ) 6 Ο cm | d | circumface(8) | circumface(n0)| | physics | D |
matt and peter can do together a piece of work in 20 days . after they have worked together for 10 days matt stops and peter completes the remaining work in 10 days . in how many days peter complete the work separately . | "together they complete the job in 20 days means they complete 10 / 20 of the job after 10 days . peter completes the remaining ( 10 / 20 ) of the job in 10 days which means that the whole job ( 1 ) can be completed in x days . < = > 8 / 20 - > 10 < = > x = 10 / ( 10 / 20 ) = 20 b" | a ) 21 , b ) 20 , c ) 22 , d ) 23 , e ) 24 | b | divide(10, subtract(const_1, multiply(10, divide(const_1, 20)))) | divide(const_1,n0)|multiply(n1,#0)|subtract(const_1,#1)|divide(n2,#2)| | physics | B |
if t = 5 / 9 * ( k - 32 ) , and if t = 50 , then what is the value of k ? | "k - 32 = 9 t / 5 k = 9 t / 5 + 32 k = 9 ( 50 ) / 5 + 32 = 122 the answer is c ." | a ) 116 , b ) 119 , c ) 122 , d ) 125 , e ) 128 | c | add(divide(multiply(50, 9), 5), 32) | multiply(n1,n3)|divide(#0,n0)|add(n2,#1)| | general | C |
how many of the positive divisors q of 120 are also multiples of 4 not including 120 ? | "4 , 8,12 , 20,24 , 40,60 . ( 7 ) is the answer other way : factors of 120 = 2 ^ 3 * 3 * 5 separate 2 ^ 2 ( which means 4 ) now , calculate the number of other factors . q = 2 * 3 * 5 = total positive factors are 2 * 2 * 2 = 8 this 8 factors include 120 so subtract 1 from 8 ans is 7 = d" | a ) 3 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | d | divide(divide(divide(120, 4), const_2), const_3) | divide(n0,n1)|divide(#0,const_2)|divide(#1,const_3)| | general | D |
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of hydrogen are there in 171 grams of water ? | "( 2 / 18 ) * 171 = 19 grams the answer is e ." | a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | e | multiply(2, divide(171, add(2, 16))) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other | E |
a certain bacteria colony doubles in size every day for 21 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat β s limit ? | "if there is one bacteria colony , then it will reach the limit of its habitat in 21 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 20 times . answer : c . similar questions to practice : hope it helps ." | a ) 6.33 , b ) 7.5 , c ) 20 , d ) 15 , e ) 19 | c | subtract(21, divide(21, 21)) | divide(n0,n0)|subtract(n0,#0)| | physics | C |
the sum of all solutions for x in the equation x ^ 2 β 8 x + 21 = | x β 4 | + 3 is equal to : | "x ^ 2 - 8 x + 18 = | x - 4 | rhs can be - ve or + ve x ^ 2 - 9 x + 22 = 0 x ^ 2 - 7 x + 14 = 0 x = 11 , 7,2 we test all 3 values in original equation , all ok . thus , sum = 11 + 7 + 2 = 20 ans ( c )" | a ) β 7 , b ) 7 , c ) 20 , d ) 12 , e ) 14 | c | multiply(3, 4) | multiply(n3,n4)| | general | C |
a alone can complete a work in 16 days and b alone can do in 12 days . star Ι΅ ng with a , they work on alternate days . the total work will be completed in | explana Ι΅ on : a ' s 1 day work = 1 / 16 b ' s 1 day work = 1 / 12 as they are working on alternate day ' s so their 2 days work = ( 1 / 16 ) + ( 1 / 12 ) = 7 / 48 [ here is a small technique , total work done will be 1 , right , then mul Ι΅ ply numerator Ι΅ ll denominator , as 7 * 6 = 42 , 7 * 7 = 49 , as 7 * 7 is more than 48 , so we will consider 7 * 6 , means 6 pairs ] work done in 6 pairs = 6 * ( 7 / 48 ) = 7 / 8 remaining work = 1 - 7 / 8 = 1 / 8 on 13 th day it will a turn , then remaining work = ( 1 / 8 ) - ( 1 / 16 ) = 1 / 16 on 14 th day it is b turn , 1 / 12 work done by b in 1 day 1 / 16 work will be done in ( 12 * 1 / 16 ) = 3 / 4 day so total days = 1334 answer : c | a ) 1314 , b ) 1312 , c ) 1334 , d ) 1344 , e ) none of these | c | multiply(const_100, multiply(inverse(add(inverse(12), inverse(16))), const_2)) | inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2)|multiply(#3,const_2)|multiply(#4,const_100) | physics | C |
a 360 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "explanation : as trains are running in opposite directions so their relative speed will get added so , relative speed = 120 + 80 = 200 kmph = 200 * ( 5 / 18 ) = 500 / 9 m / sec let the length of other train is x meter then x + 360 / 9 = 500 / 9 = > x + 360 = 500 = > x = 140 so the length of the train is 140 meters option c" | a ) 220 meter , b ) 225 meter , c ) 140 meter , d ) 235 meter , e ) none of these | c | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 360) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics | C |
a train 2000 m long can cross an electric pole in 25 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 2000 / 25 s = 80 m / sec speed = 80 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 324 kmph answer : b" | a ) 322 , b ) 324 , c ) 326 , d ) 328 , e ) 330 | b | divide(divide(2000, const_1000), divide(25, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics | B |
. for the fifa world cup , paul the octopus has been predicting the winner of each match with amazing success . it is rumored that in a match between 2 teams a and b , paul picks a with the same probability as a β s chances of winning . let β s assume such rumors to be true and that in a match between ghana and bolivia ; ghana the stronger team has a probability of 2 / 3 of winning the game . what is the probability that paul will correctly pick the winner of the ghana - bolivia game ? | the probability that paul correctly picks the winner = ( a ' s chances of winning ) x ( pauls picking the winner corectly ) + ( a ' s chances of loosing ) x ( paul picks wrongly ) = 23 Γ 23 + 13 Γ 13 = 59 answer : a | a ) 59 , b ) 88 , c ) 77 , d ) 26 , e ) 18 | a | add(add(add(multiply(const_10, add(const_3, const_2)), 3), 3), 3) | add(const_2,const_3)|multiply(#0,const_10)|add(n2,#1)|add(n2,#2)|add(n2,#3) | general | A |
in what time will a train 60 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 60 / 40 = 1.5 sec . answer : e" | a ) 2.5 sec , b ) 2.8 sec , c ) 7.5 sec , d ) 2.3 sec , e ) 1.5 sec | e | divide(60, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | E |
217 * 217 + 183 * 183 | "explanation : 1 / 2 Γ 2 ( a 2 + b 2 ) = 1 / 2 Γ [ ( a + b ) 2 + ( a β b ) 2 ] = 1 / 2 Γ [ ( 217 + 183 ) 2 + ( 217 β 183 ) 2 ] = 80578 option c" | a ) 70578 , b ) 82578 , c ) 80578 , d ) 80568 , e ) 89045 | c | multiply(217, 217) | multiply(n0,n1)| | general | C |
radha bought a watch for rs . 144 and got a percentage of profit equal to the cost price of the watch . what is the cost price of the watch ? | sp = 144 cp = x profit % = x c . p . = ( 100 / ( 100 + gain % ) ) * s . p . x = ( 100 / 100 + x ) * 144 x ^ 2 + 100 x = 14400 x ^ 2 + 180 x - 80 x - 14400 = 0 ( x + 180 ) ( x - 80 ) = 0 x = - 180 x = 80 answer : d | a ) rs . 72 , b ) rs . 78 , c ) rs . 80 , d ) rs . 90 , e ) rs . 92 | d | add(divide(subtract(sqrt(add(multiply(const_100, const_100), multiply(multiply(const_100, 144), const_4))), const_100), const_2), const_10) | multiply(const_100,const_100)|multiply(n0,const_100)|multiply(#1,const_4)|add(#0,#2)|sqrt(#3)|subtract(#4,const_100)|divide(#5,const_2)|add(#6,const_10) | gain | D |
an integer n between 1 and 100 , inclusive , is to be chosen at random . what is the probability that n ( n + 1 ) will be divisible by 5 ? | "n ( n + 1 ) to be divisible by 3 either n or n + 1 must be a multiples of 3 . in each following group of numbers : { 1 , 2 , 3 , 4 , 5 } , { 6 , 7 , 8 , 9 , 10 } , . . . , { 96 , 97 , 98 , 99,100 } there is exactly 1 numbers out of 5 satisfying the above condition . for example in { 1 , 2 , 3 , 4 , 5 } n can be : 4 or 5 . thus , the overall probability is 2 / 5 . answer : d ." | a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 2 / 5 | e | multiply(divide(divide(100, 5), 100), const_2.0) | divide(n1,n3)|divide(#0,n1)|multiply(const_2.0,#1)| | general | E |
working at constant rate , pump x pumped out half of the water in a flooded basement in 7 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constant rate , to pump out all of the water that was pumped out of the basement ? | "rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 34 hours c" | a ) a . 10 , b ) b . 12 , c ) c . 34 , d ) d . 18 , e ) e . 24 | c | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(3, 7)), const_1), const_1), const_1), const_1), 7), const_1), const_1), const_1), const_1), const_1), const_1) | add(n0,n1)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)|add(#3,const_1)|add(#4,const_1)|add(#5,n0)|add(#6,const_1)|add(#7,const_1)|add(#8,const_1)|add(#9,const_1)|add(#10,const_1)|add(#11,const_1)| | physics | C |
working alone , mary can pave a driveway in 3 hours and hillary can pave the same driveway in 6 hours . when they work together , mary thrives on teamwork so her rate increases by 10 % , but hillary becomes distracted and her rate decreases by 20 % . if they both work together , how many hours will it take to pave the driveway ? | "initial working rates : mary = 1 / 3 per hour hillary = 1 / 6 per hour rate when working together : mary = 1 / 3 + ( 1 / 10 * 1 / 3 ) = 3 / 8 per hour hillary = 1 / 6 - ( 1 / 5 * 1 / 6 ) = 2 / 15 per hour together they work 3 / 8 + 2 / 15 = 1 / 2 per hour so they will need 2 hours to complete the driveway . the correct answer is a ." | a ) 2 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | a | inverse(add(multiply(divide(const_1, 3), add(divide(10, const_100), const_1)), multiply(divide(const_1, 6), divide(20, 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 | A |
walking at 40 % of his usual speed a man takes 24 minutes more to cover a distance . what is his usual time to cover this distance ? | speed is inversly proprtional to time walking at 40 % of speed meand 2 / 5 s takes 5 / 2 t . it takes 24 minutes extra to cover the distance . then 5 / 2 t = t + 24 5 t = 2 t + 48 3 t = 48 t = 16 . option a is correct | a ) 16 , b ) 36 , c ) 42 , d ) 48 , e ) 54 | a | divide(24, subtract(divide(const_1, divide(40, const_100)), const_1)) | divide(n0,const_100)|divide(const_1,#0)|subtract(#1,const_1)|divide(n1,#2) | physics | A |
in a box of 9 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | first , there are 7 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 9 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 7 c 3 / 9 c 3 = 35 / 84 = 5 / 12 answer is a | a ) 5 / 12 , b ) 1 / 15 , c ) 2 / 13 , d ) 2 / 15 , e ) 1 / 17 | a | divide(divide(factorial(subtract(9, 2)), multiply(factorial(subtract(subtract(9, 2), 3)), factorial(3))), divide(factorial(9), multiply(factorial(subtract(9, 3)), factorial(3)))) | factorial(n2)|factorial(n0)|subtract(n0,n1)|subtract(n0,n2)|factorial(#2)|factorial(#3)|subtract(#2,n2)|factorial(#6)|multiply(#5,#0)|divide(#1,#8)|multiply(#7,#0)|divide(#4,#10)|divide(#11,#9) | general | A |
a can give b 300 meters start and c 600 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "a runs 1000 m while b runs 700 m and c runs 400 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 400 ) / 700 = 571.43 m . b can give c = 1000 - 571.43 = 428.57 m . answer : c" | a ) 111.12 , b ) 111.67 , c ) 428.57 , d ) 111.11 , e ) 101.12 | c | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 600)), subtract(multiply(const_100, const_10), 300))) | multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4)| | physics | C |
a 360 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | "s = 360 / 6 * 18 / 5 = 216 kmph answer : d" | a ) 229 , b ) 108 , c ) 278 , d ) 216 , e ) 112 | d | multiply(divide(360, 6), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics | D |
in one alloy there is 12 % chromium while in another alloy it is 8 % . 15 kg of the first alloy was melted together with 30 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 15 + 30 = 45 kg alloy is 0.12 * 15 + 0.08 * 30 = 4.2 kg , so the percentage is 4.2 / 45 * 100 = 9.33 % . answer : b ." | a ) 9.4 % , b ) 9.33 % , c ) 9.6 % , d ) 9.8 % , e ) 10 % | b | multiply(divide(add(divide(multiply(12, 15), const_100), divide(multiply(8, 30), const_100)), add(15, 30)), const_100) | add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain | B |
working together , wayne and his son can shovel the entire driveway in three hours . if wayne can shovel four times as fast as his son can , how many hours would it take for his son to shovel the entire driveway on his own ? | "w : the time for wyane to do the job s : the time for his son to do the job we have 1 / w + 1 / s = 1 / 4 and w = 4 s then we have 1 / ( 4 * s ) + 1 / s = 1 / 4 < = > 5 / ( 4 * s ) = 1 / 4 < = > s = 5 ans : c" | a ) 4 , b ) 6 , c ) 5 , d ) 9 , e ) 12 | c | multiply(multiply(add(inverse(multiply(const_2, const_4)), const_1), const_3), multiply(const_2, const_4)) | multiply(const_2,const_4)|inverse(#0)|add(#1,const_1)|multiply(#2,const_3)|multiply(#3,#0)| | physics | C |
if Γ’ β¬ Ε * Γ’ β¬ Β is called Γ’ β¬ Ε + Γ’ β¬ Β , Γ’ β¬ Ε / Γ’ β¬ Β is called Γ’ β¬ Ε * Γ’ β¬ Β , Γ’ β¬ Ε - Γ’ β¬ Β is called Γ’ β¬ Ε / Γ’ β¬ Β , Γ’ β¬ Ε + Γ’ β¬ Β is called Γ’ β¬ Ε - Γ’ β¬ Β . 120 / 40 Γ’ β¬ β 30 * 20 - 5 = ? | "explanation : given : 120 / 40 Γ’ β¬ β 30 * 20 - 5 = ? substituting the coded symbols for mathematical operations , we get , 120 * 40 / 30 + 20 / 5 = ? 120 * 1.33 + 4 = ? 159.6 + 4 = 163.6 answer : d" | a ) 178.5 , b ) 156.6 , c ) 160.4 , d ) 163.6 , e ) 164.1 | d | add(multiply(divide(30, 20), divide(120, 40)), 5) | divide(n2,n3)|divide(n0,n1)|multiply(#0,#1)|add(n4,#2)| | general | D |
the average salary of all the workers in a workshop is $ 1000 . the average salary of 5 technicians is $ 5000 and the average salary of the rest is $ 500 . the total number of workers in the shop is ? | "let the total number of workers be x 1000 x = 5000 * 5 + 500 ( x - 5 ) x = 25 answer is a" | a ) 25 , b ) 20 , c ) 15 , d ) 30 , e ) 18 | a | divide(subtract(multiply(5, 5000), multiply(500, 5)), subtract(1000, 500)) | multiply(n1,n2)|multiply(n1,n3)|subtract(n0,n3)|subtract(#0,#1)|divide(#3,#2)| | general | A |
a certain drink of type a is prepared by mixing 4 parts milk with 3 parts fruit juice . another drink of type b is prepared by mixing 4 parts of fruit juice and 3 parts of milk . how many liters of fruit juice must be added to 42 liters of drink a to convert it to drink b ? | in 42 liters of drink a , there are 24 liters of milk and 18 liters of juice . with 24 liters of milk , we need a total of 32 liters of juice to make drink b . we need to add 14 liters of juice . the answer is d . | a ) 7 , b ) 9 , c ) 12 , d ) 14 , e ) 18 | d | subtract(divide(multiply(multiply(divide(4, add(4, 3)), 42), 4), 3), multiply(divide(3, add(4, 3)), 42)) | add(n0,n1)|divide(n0,#0)|divide(n1,#0)|multiply(n4,#1)|multiply(n4,#2)|multiply(n0,#3)|divide(#5,n1)|subtract(#6,#4) | general | D |
balls of equal size are arranged in rows to form an equilateral triangle . the top most row consists of one ball , the 2 nd row of two balls and so on . if 789 balls are added , then all the balls can be arranged in the shape of square and each of the sides of the square contain 8 balls less than the each side of the triangle did . how many balls made up the triangle ? | "as expected , this question boils down to 2 equation , consider total number of balls in triangle = t and number of balls in last row = x . 1 + 2 + 3 + . . . + x = t x ( x + 1 ) / 2 = t - - - - ( a ) as mentioned in the question , side of a square will be ( x - 8 ) and total number of balls in square will be ( t + 789 ) ( x - 8 ) ^ 2 = t + 789 - - - - - ( b ) now the hardest part of the question will be to solve these 2 equations and this looks like time consuming but the easy way will be plug and play . also , we ' ve to find a value of t ( from 5 optiosn given below ) which can make a square of a a number . one we know this , it will be a cake walk . we can see that option a fits this criteria in eq ( b ) . add - 1711 + 789 = 2500 = 50 ^ 2 = ( x - 8 ) ^ 2 hence , x = 58 . cross check by putting in eq ( a ) = x ( x + 1 ) / 2 = t = > 58 * 59 / 2 = 1711 hence , answer is a ." | a ) 1711 , b ) 2209 , c ) 2878 , d ) 1210 , e ) 1560 | a | multiply(subtract(789, multiply(const_4, const_100)), add(multiply(subtract(789, multiply(const_4, const_100)), 2), const_1)) | multiply(const_100,const_4)|subtract(n1,#0)|multiply(n0,#1)|add(#2,const_1)|multiply(#3,#1)| | general | A |
90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 20 % of the total school population of x students , what is x ? | 90 = x / 100 * 20 / 100 * x = > x ^ 2 = 9 * 10000 / 2 = > x = 212 c | a ) 125 , b ) 150 , c ) 212 , d ) 250 , e ) 500 | c | sqrt(divide(multiply(90, const_100), divide(20, const_100))) | divide(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|sqrt(#2) | gain | C |
sum of two numbers prime to each other is 8 and their l . c . m . is 15 . what are the numbers ? | "as two numbers are prime , only options satisfy ie option a and b and c but option d will not make the product of numbers i . e 15 answer : a" | a ) 3 and 5 , b ) 3 and 4 , c ) 4 and 5 , d ) 5 and 10 , e ) 6 and 8 | a | add(15, 8) | add(n0,n1)| | physics | A |
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 135 grams of water ? | "( 16 / 18 ) * 144 = 120 grams the answer is a ." | a ) 120 , b ) 116 , c ) 112 , d ) 108 , e ) 104 | a | multiply(2, divide(135, add(2, 16))) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other | A |
roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of 6 km / hr and 3 km / hr respectively . after 4 hours what will be the distance between them ? | distance = relative speed * time = ( 6 + 3 ) * 4 = 36 km [ they are travelling in the opposite direction , relative speed = sum of the speeds ] . answer : e | a ) 22 km , b ) 20 km , c ) 65 km , d ) 18 km , e ) 36 km | e | multiply(add(6, 3), 4) | add(n0,n1)|multiply(n2,#0) | physics | E |
a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 16 sec . what is the length of the platform ? | "speed = 300 / 16 = 75 / 4 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 75 / 4 = > x = 731.25 m answer : b ( 431.25 )" | a ) 287 m , b ) 431.25 m , c ) 267 m , d ) 287.25 m , e ) 656 m | b | subtract(multiply(speed(300, 16), 39), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics | B |
what will be the compound interest on a sum of rs . 25,000 after 3 years at the rate of 12 p . c . p . a ? | "= rs . ( 25000 x ( 1 + 12 / 100 ) Β³ = rs . ( 25000 x 28 / 25 x 28 / 25 x 28 / 25 ) = rs . 35123.20 . c . i = rs ( 35123.20 - 25000 ) = rs . 10123.20 answer : b" | a ) rs . 10111.00 , b ) rs . 10123.20 , c ) rs . 10123.00 , d ) rs . 10100.00 , e ) rs . 10110.00 | b | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain | B |
logx 16 = 0.8 then what is the value of x ? | x ^ 0.8 = 16 x = ( 2 ^ 4 ) ^ ( 10 / 8 ) x = 2 ^ 5 x = 32 answer : a | a ) 32 , b ) 34 , c ) 35 , d ) 36 , e ) 37 | a | divide(multiply(add(multiply(0.8, const_4), 0.8), 16), const_2) | multiply(n1,const_4)|add(n1,#0)|multiply(n0,#1)|divide(#2,const_2) | general | A |
the ages of two persons differ by 20 years . if 5 years ago , the older one be 5 times as old as the younger one , then their present ages , in years , are | "let the ages be β x β and β y β years now . then x - y = 20 . . . ( i ) ( x - 5 ) = 5 ( y - 5 ) . . . ( ii ) from ( i ) and ( ii ) , 20 + y - 5 = 5 y - 25 y = 10 and x = 30 . answer : b" | a ) 25 , 5 , b ) 30 , 10 , c ) 35 , 15 , d ) 50 , 30 , e ) none of these | b | subtract(add(divide(multiply(20, 5), subtract(5, const_1)), 5), 20) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general | B |
the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 13 percent , what was the annual interest rate last year ? | 13 = 1.1 * x x = 11.81 % answer d ) | a ) 1 % , b ) 1.1 % , c ) 9.1 % , d ) 11.81 % , e ) 11.50 % | d | divide(multiply(13, const_100), add(13, const_100)) | add(n1,const_100)|multiply(n1,const_100)|divide(#1,#0) | gain | D |
the total circumference of two circles is 88 . if the first circle has a circumference that is exactly twice the circumference of the second circle , then what is the approximate sum of their two radii ? | "let r = radius of smaller circle . let r = radius of larger circle therefore : 2 Ο r + 2 Ο r = 88 where 2 r = r thus : 2 Ο r + 4 Ο r = 88 6 Ο r = 88 r = approx 4.7 Ο r + 2 r Ο = 88 3 Ο r = 88 r = approx 9.3 r + r = approx 14.0 answer : e" | a ) 5.7 , b ) 6.0 , c ) 6.7 , d ) 9.7 , e ) 14.0 | e | divide(add(divide(divide(88, const_3), const_3), divide(multiply(divide(88, const_3), const_2), const_3)), const_2) | divide(n0,const_3)|divide(#0,const_3)|multiply(#0,const_2)|divide(#2,const_3)|add(#1,#3)|divide(#4,const_2)| | general | E |
how many integers between 1 and 10 ^ 17 are such that the sum of their digits is 2 ? | "the integers with a sum of 2 are : 2 , 20 , 200 , . . . , 2 * 10 ^ 16 and there are 17 integers in this list . also , these integers have a sum of 2 : 11 101 , 110 1001 , 1010 , 1100 etc . . . the number of integers in this list is 1 + 2 + . . . + 16 thus , the total number of integers is 1 + 2 + . . . + 16 + 17 = 17 * 18 / 2 = 153 the answer is c ." | a ) 133 , b ) 143 , c ) 153 , d ) 163 , e ) 173 | c | subtract(10, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general | C |
a train 270 m long , running with a speed of 108 km / hr will pass a tree in | "sol . speed = ( 108 x 5 / 18 ) m / sec . = 30 m / sec . time taken = ( 270 x 1 / 30 ) sec = 9 sec answer b" | a ) 12 sec , b ) 9 sec , c ) 16 sec , d ) 20 sec , e ) none | b | multiply(divide(270, multiply(108, const_1000)), const_3600) | multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)| | physics | B |
what is the sum of natural numbers between 20 and 100 | "a = first number l = last number sn = n / 2 [ a + l ] between 20 and 100 numbers = 81 = > 100 - 20 = 80 + 1 = 81 sn = 81 / 2 Γ 120 = 81 Γ 60 = 4860 b )" | a ) 4800 , b ) 4860 , c ) 5000 , d ) 5500 , e ) 5550 | b | add(100, const_1) | add(n1,const_1)| | general | B |
find the simple interest on rs . 5000 at 6 % per annum for the period from 5 th feb to 19 th april , 2015 . | "explanation : given : 1 ) principal = rs . 5000 2 ) rate of interest = 6 % 3 ) time = 5 th feb to 19 th april , 2015 first find the time period 5 th feb to 19 th april , 2015 feb = 28 β 5 = 23 days march = 31 days april = 19 days total days = 23 + 31 + 19 = 73 days convert days into years , by dividing it by 365 time = 73 / 365 = 1 / 5 simple interest = ( p Γ r Γ t ) / 100 = [ 5000 Γ 6 Γ ( 1 / 5 ) ] / 100 = rs . 60 simple interest = rs . 60 answer is c" | a ) rs . 40 , b ) rs . 50 , c ) rs . 60 , d ) rs . 70 , e ) none of these | c | multiply(multiply(multiply(add(multiply(multiply(multiply(5, 19), const_100), const_100), multiply(multiply(multiply(19, 19), const_100), multiply(add(19, 5), 5))), divide(add(multiply(6, 19), 5), 19)), divide(multiply(19, 19), multiply(5, multiply(5, 19)))), divide(const_1, const_100)) | add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)| | gain | C |
in a kilometer race , a beats b by 16 meters or 10 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 10 ) / 16 = 625 sec . time taken by a = 625 - 10 = 615 sec . answer : e" | a ) 167 sec , b ) 190 sec , c ) 176 sec , d ) 716 sec , e ) 615 sec | e | subtract(divide(multiply(const_1, const_1000), divide(16, 10)), 10) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics | E |
edward ' s home have 3 taps , one tap , working alone at its maximum constant rate , can fill a certain sink in 270 seconds . a second tap working alone at its maximum constant rate , can fill a certain sink in 274 seconds . a third tap , working alone at its maximum constant rate , can fill the same sink in 276 seconds . if all taps work together at their respective maximum constant rates , the time it will take to fill the sink is closest to | tap 1 : 270 secs tap 2 : 274 secs tap 3 : 276 secs considering the average of these 3 taps : 273.33 secs . so 1 tap can fill the tank in 273.33 secs , so 3 taps can fill the tank in 273.33 / 3 = 91.11 secs . closest answer is a . | a ) 91 seconds , b ) 102 seconds , c ) 177 seconds , d ) 200 seconds , e ) 270 seconds | a | divide(add(add(divide(270, 3), divide(274, 3)), divide(276, 3)), 3) | divide(n1,n0)|divide(n2,n0)|divide(n3,n0)|add(#0,#1)|add(#3,#2)|divide(#4,n0) | physics | A |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.