options
stringlengths 37
300
| correct
stringclasses 5
values | annotated_formula
stringlengths 7
727
| problem
stringlengths 5
967
| rationale
stringlengths 1
2.74k
| program
stringlengths 10
646
|
|---|---|---|---|---|---|
a ) 42 , b ) 42.5 , c ) 44 , d ) 50 , e ) 52.5 | a | divide(add(8, 20), add(add(divide(8, 40), divide(12, 60)), divide(20, 60))) | jerry travels 8 miles at an average speed of 40 miles per hour , stops for 12 minutes , and then travels another 20 miles at an average speed of 60 miles per hour . what is jerry β s average speed , in miles per hour , for this trip ? | "total time taken by jerry = ( 8 / 40 ) * 60 minutes + 12 minutes + ( 20 / 60 ) * 60 minutes = 44 minutes average speed = total distance / total time = ( 8 + 20 ) miles / ( 44 / 60 ) hours = 28 * 60 / 44 = 42 miles per hour answer : option a" | a = 8 + 20
b = 8 / 40
c = 12 / 60
d = b + c
e = 20 / 60
f = d + e
g = a / f
|
a ) 17 , b ) 10 , c ) 15 , d ) 20 , e ) 16 | e | divide(multiply(multiply(8, divide(subtract(15, 9), 4)), 20), 15) | 8 men can dig a pit in 20 days . if a men work half as much again as a boy , then 4 men and 9 boys can dig a similar pit . fint the days for 15 boys can dig ? | 1 man = 3 / 2 boys ( 4 men + 9 boys ) = 15 boys 8 men = [ ( 3 / 2 ) + 8 ] = 12 boys now , 12 boys can dig the pit in 20 days . 15 boys can dig = 16 days . answer is option e | a = 15 - 9
b = a / 4
c = 8 * b
d = c * 20
e = d / 15
|
a ) 22.5 % , b ) 42.5 % , c ) 12.5 % , d ) 62.5 % , e ) 82.5 % | c | multiply(divide(16, 128), const_100) | a man invests in a 16 % stock at 128 . the interest obtained by him is | by investing rs 128 , income derived = rs . 16 by investing rs . 100 , income derived = = rs . 12.5 interest obtained = 12.5 % answer : c | a = 16 / 128
b = a * 100
|
a ) 3 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 7 : 3 , e ) none | d | divide(subtract(70, 63), subtract(63, 60)) | in what ratio must tea at rs . 60 per kg be mixed with tea at rs . 70 per kg so that the mixture must be worth rs . 63 per kg ? | "required ratio = 700 : 300 = 7 : 3 answer d" | a = 70 - 63
b = 63 - 60
c = a / b
|
a ) 5 , b ) 15 , c ) 40 , d ) 50 , e ) 25 | a | divide(subtract(multiply(48, 40), multiply(divide(add(const_100, 20), const_100), multiply(35, 40))), subtract(multiply(80, divide(add(const_100, 20), const_100)), 48)) | how many pounds of salt at 80 cents / lb must be mixed with 40 lbs of salt that costs 35 cents / lb so that a merchant will get 20 % profit by selling the mixture at 48 cents / lb ? | "selling price is 48 cents / lb for a 20 % profit , cost price should be 40 cents / lb ( cp * 6 / 5 = 48 ) basically , you need to mix 35 cents / lb ( salt 1 ) with 80 cents / lb ( salt 2 ) to get a mixture costing 40 cents / lb ( salt avg ) weight of salt 1 / weight of salt 2 = ( salt 2 - saltavg ) / ( saltavg - salt 1 ) = ( 80 - 40 ) / ( 40 - 35 ) = 8 / 1 we know that weight of salt 1 is 40 lbs . weight of salt 2 must be 5 lbs . answer ( a )" | a = 48 * 40
b = 100 + 20
c = b / 100
d = 35 * 40
e = c * d
f = a - e
g = 100 + 20
h = g / 100
i = 80 * h
j = i - 48
k = f / j
|
a ) 33 , b ) 72 , c ) 52 , d ) 82 , e ) 62 | c | divide(add(120, 660), multiply(54, const_0_2778)) | how long does a train 120 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length ? | explanation : t = ( 660 + 120 ) / 54 * 18 / 5 t = 52 answer : option c | a = 120 + 660
b = 54 * const_0_2778
c = a / b
|
a ) 8 / 2652 , b ) 18 / 2652 , c ) 1 / 2652 , d ) 12 / 2652 , e ) 16 / 2652 | b | divide(multiply(multiply(subtract(4, const_1), subtract(4, const_1)), const_2), multiply(52, subtract(52, const_1))) | if 2 cards are selected at random from the deck of 52 cards then what is the probability of one of the selected cards will be king and other will be 10 ? a deck of cards has a total of 52 cards , consisting of 4 suits ; ( spades ( black ) , hearts ( red ) , diamond ( red ) s , and clubs ( black ) ) ; and 13 cards including 1 king , 1 queen and 1 jack in each suit | 2 possible cases : king - 10 or 10 - king ( 4 kings and 4 10 ) . either way , the total probability = 2 ( king - 10 ) = 2 ( 4 / 52 * 4 / 51 ) = 18 / 2652 . b is the correct answer . | a = 4 - 1
b = 4 - 1
c = a * b
d = c * 2
e = 52 - 1
f = 52 * e
g = d / f
|
a ) 20 , b ) 19 , c ) 18 , d ) 17 , e ) 16 | b | subtract(multiply(10, const_2), const_1) | a and b are two multiples of 14 , and q is the set of consecutive integers between a and b , inclusive . if q contains 10 multiples of 14 , how many multiples of 7 are there in q ? | "halfway between the multiples of 14 , there will be another multiple of 7 . the total number of multiples of 7 is 10 + 9 = 19 . the answer is b ." | a = 10 * 2
b = a - 1
|
a ) 47 , b ) 46 , c ) 54 , d ) 58 , e ) 67 | e | add(add(subtract(60, 4), 7), subtract(10, 7)) | 60 persons like apple . 7 like orange and mango dislike apple . 10 like mango and apple and dislike orange . 4 like all . how many people like apple ? | "orange + mango - apple = 7 mango + apple - orange = 10 apple = 60 orange + mango + apple = 4 60 + 10 + 4 - 7 = 67 like apple answer : e" | a = 60 - 4
b = a + 7
c = 10 - 7
d = b + c
|
a ) 2287 , b ) 140 , c ) 128 , d ) 797 , e ) 123 | b | multiply(subtract(divide(300, divide(30, const_100)), 300), divide(20, const_100)) | bhanu spends 30 % of his income on petrol on scooter 20 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ? | "given 30 % ( income ) = 300 β β income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 20 % ( 700 ) = rs . 140 answer : b" | a = 30 / 100
b = 300 / a
c = b - 300
d = 20 / 100
e = c * d
|
a ) 20 , b ) 27 , c ) 28 , d ) 30 , e ) none of these | b | divide(add(add(add(add(9, multiply(9, const_2)), multiply(9, const_3)), multiply(9, const_4)), multiply(9, 5)), 5) | the average of the first 5 multiples of 9 is : | solution : required average = ( total sum of multiple of 9 ) / 5 = ( 9 + 18 + 27 + 36 + 45 ) / 5 = 27 note that , average of 9 and 45 is also 27 . and average of 18 and 36 is also 27 . answer : option b | a = 9 * 2
b = 9 + a
c = 9 * 3
d = b + c
e = 9 * 4
f = d + e
g = 9 * 5
h = f + g
i = h / 5
|
a ) 80 , b ) 160 , c ) 720 , d ) 1100 , e ) 2184 | e | multiply(divide(divide(factorial(13), factorial(subtract(13, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2)) | 13 different biology books and 8 different chemistry books lie on a shelf . in how many ways can a student pick 2 books of each type ? | "no . of ways of picking 2 biology books ( from 13 books ) = 13 c 2 = ( 13 * 12 ) / 2 = 78 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) / 2 = 28 total ways of picking 2 books of each type = 78 * 28 = 2184 ( option e )" | a = math.factorial(13)
b = 13 - 2
c = math.factorial(b)
d = a / c
e = d / 2
f = math.factorial(8)
g = 8 - 2
h = math.factorial(g)
i = f / h
j = i / 2
k = e * j
|
a ) 29 / 32 , b ) 28 / 31 , c ) 30 / 31 , d ) 40 / 41 , e ) 34 / 15 | a | multiply(divide(divide(multiply(divide(add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(subtract(const_100, 20), const_100))), const_2), subtract(const_100, 20)), const_100), multiply(subtract(const_100, 20), divide(subtract(const_100, 20), const_100))), const_10) | real - estate salesman z is selling a house at a 20 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 20 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 20 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ? | "let the retail price be = x selling price of z = 0.80 x selling price of x = 0.80 * 0.80 x = 0.64 x selling price of y = ( ( 0.80 x + 0.64 x ) / 2 ) * 0.80 = 0.72 x * 0.80 = 0.58 x 0.58 x = k * 0.64 x k = 0.58 / 0.64 = 58 / 64 = 29 / 32 answer : a" | a = 100 - 20
b = 100 - 20
c = 100 - 20
d = c / 100
e = b * d
f = a + e
g = f / 2
h = 100 - 20
i = g * h
j = i / 100
k = 100 - 20
l = 100 - 20
m = l / 100
n = k * m
o = j / n
p = o * 10
|
a ) 32 , b ) 27 , c ) 25 , d ) 35 , e ) 29 | a | subtract(160, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 100 and 160 , inclusive , can not be evenly divided by 3 nor 5 ? | "the total numbers between 100 and 160 , inclusive , is 61 . 3 * 34 = 102 and 3 * 53 = 159 so the number of multiples of 3 is 20 . 5 * 20 = 100 and 5 * 32 = 160 so the number of multiples of 5 is 13 . however , the multiples of 15 have been counted twice . 15 * 7 = 105 and 15 * 10 = 150 so the number of multiples of 15 is 4 . thus the total number is 61 - 20 - 13 + 4 = 32 . the answer is a ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 160 - e
|
a ) 190 , b ) 160 , c ) 120 , d ) 151 , e ) 150 | e | multiply(divide(690, add(add(5, 7), 11)), 5) | calculate the share of y , if rs . 690 is divided among x , y and z in the ratio 5 : 7 : 11 ? | "5 + 7 + 11 = 23 690 / 23 = 30 so y ' s share = 5 * 30 = 150 answer : e" | a = 5 + 7
b = a + 11
c = 690 / b
d = c * 5
|
a ) 42.88 $ , b ) 43 % , c ) 45 % , d ) 45.5 % , e ) 46.3 % | a | subtract(52, divide(52, 6)) | you hold some gold in a vault as an investment . over the past year the price of gold increases by 52 % . in order to keep your gold in the vault , you must pay 6 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year . | ( 100 % + 52 % ) * ( 100 % - 6 % ) = 152 * 0.94 = 142.88 % an increase of 42.88 % your gold holdings have increased in value by 42.88 % . the answer is a | a = 52 / 6
b = 52 - a
|
a ) 6 % , b ) 7 % , c ) 8 % , d ) 9 % , e ) 10 % | d | multiply(divide(multiply(multiply(const_100, const_100), divide(6, 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) | a tank contains 9,000 gallons of a solution that is 6 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 9,000 gallons of a solution that is 6 % sodium chloride by volume . this means that there are 0.06 x 9,000 = 54 gallons of sodium chloride . when 3,000 gallons of water evaporate we are left with 6,000 gallons of solution . from here we can determine what percent of the 6,000 gallon solution is sodium chloride . ( sodium chloride / total solution ) x 100 = ? ( 540 / 6,000 ) x 100 = ? 9 / 100 x 100 = ? = 9 % answer is d ." | a = 100 * 100
b = 6 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
|
a ) 2600 , b ) 2500 , c ) 2900 , d ) 2800 , e ) 2700 | b | divide(divide(multiply(750, const_100), 6), 5) | a sum fetched a total simple interest of rs . 750 at the rate of 6 p . c . p . a . in 5 years . what is the sum ? | "sol . principal = rs . [ 100 * 750 / 6 * 5 ] = rs . [ 75000 / 30 ] = rs . 2500 . answer b" | a = 750 * 100
b = a / 6
c = b / 5
|
a ) a ) 4 , b ) b ) 7 , c ) c ) 9 , d ) d ) 5 , e ) e ) 2 | c | subtract(12, 3) | robert ate 12 chocolates , nickel ate 3 chocolates . how many more chocolates did robert ate than nickel ? | 12 - 3 = 9 . answer is c | a = 12 - 3
|
a ) 5 seconds , b ) 4.5 seconds , c ) 3 seconds , d ) 2.4 seconds , e ) none of these | d | divide(140, multiply(210, const_0_2778)) | in what time will a train 140 meters long cross an electric pole , if its speed is 210 km / hr | "explanation : first convert speed into m / sec speed = 210 * ( 5 / 18 ) = 58 m / sec time = distance / speed = 140 / 58 = 2.4 seconds option d" | a = 210 * const_0_2778
b = 140 / a
|
a ) 300 , b ) 360 , c ) 400 , d ) 450 , e ) 500 | e | sqrt(add(power(add(multiply(60, 5), multiply(20, 5)), const_2), power(multiply(60, 5), const_2))) | there are two cars . one is 300 miles north of the other . simultaneously , the car to the north is driven westward at 20 miles per hour and the other car is driven eastward at 60 miles per hour . how many miles apart are the cars after 5 hours ? | here , drawing a quick sketch of the ' actions ' described will end in a diagonal line that you canbuilda right triangle around : the right triangle will have a base of 400 and a height of 300 . the hidden pattern here is a 3 / 4 / 5 right triangle ( the 300 lines up with the ' 3 ' and the 400 lines up with the ' 4 ' ) . in this way , you can deduce that each side is ' 100 times ' bigger than it ' s corresponding side : 3 / 4 / 5 becomes 300 / 400 / 500 thus the distance between the two cars is the length of the hypotenuse of this larger right triangle . . . final answer : e | a = 60 * 5
b = 20 * 5
c = a + b
d = c ** 2
e = 60 * 5
f = e ** 2
g = d + f
h = math.sqrt(g)
|
a ) 3 , b ) 4 , c ) 6 , d ) 9 , e ) 12 | b | add(floor(divide(log(const_3), log(add(const_1, divide(34.1, const_100))))), const_1) | an investment compounds annually at an interest rate of 34.1 % what is the smallest investment period by which time the investment will more than triple in value ? | assume initial amount is x annual interest is 34.1 % so after 1 year the amount will become x * ( 100 + 34.1 ) / 100 = > x * 4 / 3 now we need to find n for x * ( 4 / 3 ) ^ n = 3 x ; or in other words n = 4 answer : b | a = math.log(3)
b = 34 / 1
c = 1 + b
d = math.log(c)
e = a / d
f = math.floor(e)
g = f + 1
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | divide(630, divide(add(12, 6), const_2)) | the cross - section of a water channel is a trapezium in shape . if the channel is 12 meters wide at the top and 6 meters wide at the bottom and the area of cross - section is 630 square meters , what is the depth of the channel ( in meters ) ? | "1 / 2 * d * ( 12 + 6 ) = 630 d = 70 the answer is c ." | a = 12 + 6
b = a / 2
c = 630 / b
|
a ) 246 , b ) 448 , c ) 1408 , d ) 710 , e ) 624 | e | divide(multiply(const_1000, 1), multiply(add(const_3, divide(add(multiply(const_3, const_4), const_2), power(add(multiply(const_4, const_2), const_2), const_2))), 0.51)) | a bicycle wheel has a diameter of 0.51 m . how many complete revolutions does it make in 1 km ? | 1 revolution = 3.14 * diameter . number of revolutions in 1 km = 1000 m / ( 3.14 * 0.51 m ) = 624.5 . hence 624 complete revolutions . answer e | a = 1000 * 1
b = 3 * 4
c = b + 2
d = 4 * 2
e = d + 2
f = e ** 2
g = c / f
h = 3 + g
i = h * 0
j = a / i
|
a ) 9 : 10 , b ) 17 : 19 , c ) 23 : 27 , d ) 16 : 17 , e ) 15 : 23 | d | sqrt(divide(1536, 1734)) | triangle atriangle b are similar triangles with areas 1536 units square and 1734 units square respectively . the ratio of there corresponding height would be | "let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 1536 / 1734 ( x ^ 2 / y ^ 2 ) = ( 16 * 16 * 6 ) / ( 17 * 17 * 7 ) ( x ^ 2 / y ^ 2 ) = 16 ^ 2 / 17 ^ 2 x / y = 16 / 17 ans = d" | a = 1536 / 1734
b = math.sqrt(a)
|
a ) 2 , b ) 3 , c ) 9 , d ) 5 , e ) 8 | d | divide(subtract(const_1, multiply(12, divide(const_1, 16))), divide(const_1, 20)) | x can finish a work in 20 days . y can finish the same work in 16 days . y worked for 12 days and left the job . how many days does x alone need to finish the remaining work ? | "work done by x in 1 day = 1 / 20 work done by y in 1 day = 1 / 16 work done by y in 12 days = 12 / 16 = 3 / 4 remaining work = 1 β 3 / 4 = 1 / 4 number of days in which x can finish the remaining work = ( 1 / 4 ) / ( 1 / 20 ) = 5 d" | a = 1 / 16
b = 12 * a
c = 1 - b
d = 1 / 20
e = c / d
|
a ) 15 , b ) 20 , c ) 23 , d ) 24 , e ) 25 | e | add(multiply(10, const_2), multiply(subtract(20.5, multiply(10, const_2)), 10)) | if the average of 10 consecutive integers is 20.5 then the 10 th integer is : - | the average falls between the 5 th and 6 th integers , integer 5 = 20 , integer 6 = 21 . counting up to the tenth integer we get 25 . answer : e | a = 10 * 2
b = 10 * 2
c = 20 - 5
d = c * 10
e = a + d
|
a ) 19 , b ) 21 , c ) 23 , d ) 25 , e ) 27 | c | divide(add(multiply(2, 30), 55), add(2, 3)) | sandy gets 3 marks for each correct sum and loses 2 marks for each incorrect sum . sandy attempts 30 sums and obtains 55 marks . how many sums did sandy get correct ? | "let x be the correct sums and ( 30 - x ) be the incorrect sums . 3 x - 2 ( 30 - x ) = 55 5 x = 115 x = 23 the answer is c ." | a = 2 * 30
b = a + 55
c = 2 + 3
d = b / c
|
a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 52 | b | divide(multiply(32, const_2), const_3) | a sum of money is to be divided among ann , bob and chloe . first , ann receives $ 4 plus one - half of what remains . next , bob receives $ 4 plus one - third of what remains . finally , chloe receives the remaining $ 32 . how much money m did bob receive ? | "notice that we need not consider ann ' s portion in the solution . we can just let k = the money remaining after ann has received her portion and go from there . our equation will use the fact that , once we remove bob ' s portion , we have $ 32 for chloe . so , we getk - bob ' s $ = 32 bob received 4 dollars plus one - third of what remained once bob receives $ 4 , the amount remaining is k - 4 dollars . so , bob gets a 1 / 3 of that as well . 1 / 3 of k - 4 is ( k - 4 ) / 3 so altogether , bob receives 4 + ( k - 4 ) / 3 so , our equation becomes : k - [ 4 + ( k - 4 ) / 3 ] = 32 simplify to get : k - 4 - ( k - 4 ) / 3 = 32 multiply both sides by 3 to get : 3 k - 12 - k + 4 = 96 simplify : 2 k - 8 = 96 solve : k = 52 plug this k - value intok - bob ' s $ = 32 to get : 52 - bob ' s $ = 32 so , bob ' s $ m = 20 answer : b" | a = 32 * 2
b = a / 3
|
a ) 252.5 , b ) 364.5 , c ) 367.5 , d ) 375 , e ) 345 | e | multiply(divide(multiply(add(divide(15, const_100), const_1), 30), 10), const_100) | last year elaine spent 10 % of her annual earnings on rent . this year she earned 15 % more than last year and she spent 30 % of her annual earnings on rent . the amount she spent on rent this year is what percent of the amount spent on rent last year ? | "for this it is easiest to use simple numbers . let ' s assume that elaine ' s annual earnings last year were $ 100 . she would ' ve spent $ 10 of this on rent . this year she earned 15 % more , or $ 115 . she would ' ve spent 30 % of this on rent , or $ 34.5 do $ 34.5 / $ 10 this will give you 345 % e is the correct answer ." | a = 15 / 100
b = a + 1
c = b * 30
d = c / 10
e = d * 100
|
a ) 18 , b ) 27 , c ) 36 , d ) 45 , e ) 54 | c | multiply(divide(7, subtract(9, 7)), 8) | sandy is younger than molly by 8 years . if their ages are in the respective ratio of 7 : 9 , how old is molly ? | "s = m - 8 s / m = 7 / 9 9 s = 7 m 9 ( m - 8 ) = 7 m m = 36 the answer is c ." | a = 9 - 7
b = 7 / a
c = b * 8
|
a ) - 4 , b ) - 2 , c ) 5 , d ) 13 , e ) 22 | c | divide(add(16, 4), add(3, const_1)) | when x is multiplied by 3 , the result is 4 more than the result of subtracting x from 16 . what is the value of x ? | "the equation that can be formed is : 3 x - 4 = 16 - x or , 4 x = 20 or , x = 5 . answer : c" | a = 16 + 4
b = 3 + 1
c = a / b
|
a ) 72 , b ) 80 , c ) 120 , d ) 150 , e ) 100 | b | divide(add(36, multiply(36, divide(2, 3))), subtract(const_1, divide(25, const_100))) | in a certain school , 25 % of students are below 8 years of age . the number of students above 8 years of age is 2 / 3 of the number of students of 8 years of age which is 36 . what is the total number of students in the school ? | "explanation : let the number of students be x . then , number of students above 8 years of age = ( 100 - 25 ) % of x = 75 % of x . 75 % of x = 36 + 2 / 3 of 36 75 / 100 x = 60 x = 80 . answer : option b" | a = 2 / 3
b = 36 * a
c = 36 + b
d = 25 / 100
e = 1 - d
f = c / e
|
a ) 48 , b ) 36 , c ) 26 , d ) 11 , e ) 18 | b | divide(600, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 600 meter long train moving with a speed of 63 km / hr take to cross a man walking with a speed of 3 km / hr in the direction of the train ? | "explanation : here distance d = 600 mts speed s = 63 - 3 = 60 kmph = 60 x 5 / 18 m / s time t = = 36 sec . answer : b" | a = 63 - 3
b = a * const_0_2778
c = 600 / b
|
a ) 100 m , b ) 120 m , c ) 140 m , d ) 160 m , e ) 180 cm | e | divide(const_100.0, subtract(divide(const_100.0, 10), 9)) | a train covers a distance of 12 km in 10 minutes . if it takes 9 seconds to pass a telegraph post , then the length of the train is | "explanation : speed = 12 / 10 x 60 km / hr = 72 x 5 / 18 m / sec = 20 m / sec . length of the train = ( speed x time ) = ( 20 x 9 ) m = 180 m answer : option e" | a = 100 / 0
b = a - 9
c = 100 / 0
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 60 / 7 | c | divide(5600, subtract(divide(6000, 2.5), subtract(divide(3000, 1), divide(6000, 3)))) | machine a can process 6000 envelopes in 3 hours . machines b and c working together but independently can process the same number of envelopes in 2.5 hours . if machines a and c working together but independently process 3000 envelopes in 1 hour , then how many hours would it take machine b to process 5600 envelopes . | "you can either take the amount of work done as the same as karishma has done or take the work done by each in the same time . i will do the latter 1 . work done in 1 hr by a is 2000 envelopes 2 . work done in 1 hr by a and c is 3000 envelopes 3 . so work done in 1 hr by c is 1000 envelopes 4 . work done in 1 hr by b and c is 2400 envelopes 5 . so work done in 1 hr by b is 1400 envelopes 6 . so to process 5600 envelopes b will take 5600 / 1400 hrs = 4 hrs so the answer is choice c" | a = 6000 / 2
b = 3000 / 1
c = 6000 / 3
d = b - c
e = a - d
f = 5600 / e
|
a ) 1 / 16 , b ) 3 / 32 , c ) 5 / 32 , d ) 3 / 64 , e ) 7 / 64 | d | multiply(4, power(divide(const_1, const_2), 4)) | if the probability of rain on any given day in chicago during the summer is 25 % , independent of what happens on any other day , what is the probability of having exactly 3 rainy days from july 4 through july 7 inclusive ? | "one possible case is : rainy - rainy - rainy - not rainy . the probability of this case is 1 / 4 * 1 / 4 * 1 / 4 * 3 / 4 = 3 / 256 the number of possible cases is 4 c 3 = 4 . p ( exactly 3 rainy days ) = 4 * 3 / 256 = 3 / 64 the answer is d ." | a = 1 / 2
b = a ** 4
c = 4 * b
|
a ) 36 , b ) 45 , c ) 52 , d ) 44 , e ) 39 | a | subtract(multiply(4, const_10), 4) | what is the difference between the place value and face value of 4 in the numeral 46 ? | difference between the place value and face value of 4 = 40 - 4 = 36 answer is a | a = 4 * 10
b = a - 4
|
a ) 50 % , b ) 40 % , c ) 20 % , d ) 30 % , e ) 10 % | e | divide(multiply(1,000, const_100), add(add(multiply(const_2, const_1000), const_100), 1,000)) | in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ 1,000 higher than the average rental price for all one - bedroom apartments , and if the average rental price for all two - bedroom apartments is $ 9,000 higher that m , then what percentage of apartments in the building are two - bedroom apartments ? | "ratio of 2 bedroom apartment : 1 bedroom apartment = 1000 : 9000 - - - - - > 1 : 9 let total number of apartments be x no . of 2 bedroom apartment = ( 1 / 10 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 / 10 ) * 100 - - - > 10 % answer : e" | a = 1 * 0
b = 2 * 1000
c = b + 100
d = c + 1
e = a / d
|
a ) 33 , b ) 77 , c ) 48 , d ) 99 , e ) 34.5 | e | divide(add(multiply(30, add(const_1, divide(30, const_100))), 30), const_2) | a person travels from p to q a speed of 30 km / hr and returns by increasing his speed by 30 % . what is his average speed for both the trips ? | "speed on return trip = 130 % of 30 = 39 km / hr . average speed of trip = 30 + 39 / 2 = 69 / 2 = 34.5 km / hr answer : e" | a = 30 / 100
b = 1 + a
c = 30 * b
d = c + 30
e = d / 2
|
a ) 0 , b ) - 2 , c ) - 25 , d ) - 49 , e ) - 51 | d | add(add(negate(26), const_1), add(add(negate(26), const_1), const_1)) | the sum of all the integers f such that - 26 < f < 24 is | "easy one - - 25 , - 24 , - 23 , - 22 , . . . . . . - 1,0 , 1 , 2 . . . . , 22 , 23 cancel everyhitng and we ' re left with - - 25 and - 24 f = - 49 . d is the answer ." | a = negate + (
b = a + 1
|
a ) 69 , b ) 72 , c ) 75 , d ) 78 , e ) 80 | a | divide(add(add(multiply(70, 6), multiply(60, 4)), 30), add(6, 4)) | the average expenditure of a labourer for 6 months was 70 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | "income of 6 months = ( 6 Γ 70 ) β debt = 420 β debt income of the man for next 4 months = 4 Γ 60 + debt + 30 = 270 + debt β΄ income of 10 months = 690 average monthly income = 690 Γ· 10 = 69 answer a" | a = 70 * 6
b = 60 * 4
c = a + b
d = c + 30
e = 6 + 4
f = d / e
|
a ) 30 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | a | divide(add(102, 138), 8) | a student chose a number , multiplied it by 8 , then subtracted 138 from the result and got 102 . what was the number he chose ? | "solution : let xx be the number he chose , then 8 β
x β 138 = 102 8 x = 240 x = 30 answer a" | a = 102 + 138
b = a / 8
|
a ) 37 , b ) 36 , c ) 22 , d ) 90 , e ) 28 | b | add(add(divide(192, add(add(const_1, 5), 10)), divide(192, add(add(const_1, 5), 10))), divide(192, add(add(const_1, 5), 10))) | a man has rs . 192 in the denominations of one - rupee notes , 5 - rupee notes and 10 - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? | let number of notes of each denomination be x . then , x + 5 x + 10 x = 192 16 x = 192 x = 12 . hence , total number of notes = 3 x = 36 . answer : b | a = 1 + 5
b = a + 10
c = 192 / b
d = 1 + 5
e = d + 10
f = 192 / e
g = c + f
h = 1 + 5
i = h + 10
j = 192 / i
k = g + j
|
a ) 6 , b ) 12 , c ) 14 , d ) 16 , e ) 28 | d | divide(multiply(subtract(40, 20), 8), 10) | how many gallons of milk that is 10 percent butter - fat must be added to 8 gallons of milk that is 40 percent butterfat to obtain milk that is 20 percent butterfat ? | "equate the fat : 0.1 x + 0.4 * 8 = 0.2 ( x + 8 ) - - > x = 16 . answer : d ." | a = 40 - 20
b = a * 8
c = b / 10
|
a ) $ 11.73 , b ) $ 12.78 , c ) $ 13.80 , d ) $ 14.00 , e ) $ 15.87 | b | multiply(multiply(divide(211.00, add(const_100, 15)), const_100), divide(const_1, 15)) | the price of lunch for 15 people was $ 211.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ? | "take the initial price before the gratuity is 100 the gratuity is calculated on the final price , so as we assumed the final bill before adding gratuity is 100 so gratuity is 15 % of 100 is 15 so the total price of meals is 115 so the given amount i . e 211 is for 115 then we have to calculate for 100 for 115 211 for 100 x so by cross multiplication we get 115 x = 100 * 211 = > x = 100 * 211 / 110 by simplifying we get x as 191.81 which is the price of lunch before gratuity so the gratuity is 19.19 so as the question ask the average price person excluding gratuity is 191.81 / 15 = 12.78 so our answer is b )" | a = 100 + 15
b = 211 / 0
c = b * 100
d = 1 / 15
e = c * d
|
a ) 20 , b ) 30 , c ) 15 , d ) 25 , e ) 20 | a | add(add(divide(subtract(500, 100), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(20, const_2)) | how many times digit 20 is used while writing numbers from 100 to 500 ? | "in 100 to 200 there are 5 20 ' s in 200 to 300 there are 5 20 ' s in 300 to 400 there are 5 20 ' s in 400 to 500 there are 5 20 ' s so , total 20 is 20 ' s correct option : a" | a = 500 - 100
b = a / 10
c = 10 + 1
d = 10 + 1
e = c * d
f = b + e
g = 20 * 2
h = f + g
|
a ) 5.2 , b ) 3.5 , c ) 5.3 , d ) 5.9 , e ) 5.1 | b | add(6, const_1) | the average of first 6 natural numbers is ? | "sum of 6 natural no . = 42 / 2 = 21 average = 21 / 6 = 3.5 answer : b" | a = 6 + 1
|
a ) 9 % , b ) 17 % , c ) 25 % , d ) 50 % , e ) 100 % | b | multiply(divide(divide(20, const_100), divide(add(20, const_100), const_100)), const_100) | a part - time employee β s hourly wage was increased by 20 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? | "let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 / hour and worked 100 hours / week so , the total weekly income was $ 100 / week after the 20 % wage increase , the employee makes $ 1.20 / hour we want the employee ' s income to remain at $ 100 / week . so , we want ( $ 1.20 / hour ) ( new # of hours ) = $ 100 divide both sides by 1.20 to get : new # of hours = 100 / 1.20 β 83 hours so , the number of hours decreases from 100 hours to ( approximately ) 83 hours . this represents a 17 % decrease ( approximately ) . answer : b" | a = 20 / 100
b = 20 + 100
c = b / 100
d = a / c
e = d * 100
|
['a ) a ) 10', 'b ) b ) 12', 'c ) c ) 15', 'd ) d ) 14', 'e ) e ) 9'] | c | sqrt(divide(4725, add(power(4, const_2), add(power(1, const_2), power(2, const_2))))) | the ratio of three numbers is 1 : 2 : 4 and the sum of their squares is 4725 . the sum of the numbers is ? | let the numbers be x , 2 x , 4 x then , x ^ 2 + 4 x ^ 2 + 16 x ^ 2 = 4725 21 x ^ 2 = 4725 x ^ 2 = 225 x = 15 answer is c | a = 4 ** 2
b = 1 ** 2
c = 2 ** 2
d = b + c
e = a + d
f = 4725 / e
g = math.sqrt(f)
|
a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | c | add(subtract(add(add(multiply(const_3, const_4), add(const_1, add(multiply(const_3, const_4), const_4))), const_2), add(const_1, add(multiply(const_3, const_4), const_4))), const_1) | madhav ranks seventeenth in a class of thirtyone . what is his rank from the last ? | ( rank from top ) + ( rank from bottom ) - 1 = total students 17 + x - 1 = 31 x = 15 answer : c | a = 3 * 4
b = 3 * 4
c = b + 4
d = 1 + c
e = a + d
f = e + 2
g = 3 * 4
h = g + 4
i = 1 + h
j = f - i
k = j + 1
|
['a ) 64', 'b ) 32', 'c ) 8', 'd ) 4', 'e ) 2'] | c | sqrt(power(power(17, divide(const_1, const_2)), const_3)) | what positive number , when squared , is equal to the cube of the positive square root of 17 ? | let the positive number be x x ^ 2 = ( ( 17 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 64 = > x = 8 answer c | a = 1 / 2
b = 17 ** a
c = b ** 3
d = math.sqrt(c)
|
a ) rs . 72 , b ) rs . 60 , c ) rs . 75 , d ) rs . 76 , e ) none | a | multiply(multiply(75, add(const_1, divide(20, const_100))), subtract(const_1, divide(20, const_100))) | the cost of an article was rs . 75 . the cost was first increased by 20 % and later on it was reduced by 20 % . the present cost of the article is : | solution : initial cost = rs . 75 after 20 % increase in the cost , it becomes , ( 75 + 20 % of 75 ) = rs . 90 now , cost is decreased by 20 % , so cost will become , ( 90 - 20 % of 90 ) = rs . 72 . so , present cost is rs . 72 . mind calculation method : 75 - - - - - 20 % β - - β 90 - - - - - 20 % β - - - - - β 72 . answer : option a | a = 20 / 100
b = 1 + a
c = 75 * b
d = 20 / 100
e = 1 - d
f = c * e
|
a ) 4.85 , b ) 7.85 , c ) 4.51 , d ) 5.85 , e ) 6.15 | c | divide(add(111, 165), multiply(add(100, 120), const_0_2778)) | two trains 111 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 100 km and the other at the rate of 120 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 111 + 165 ) / ( 100 + 120 ) * 18 / 5 t = 4.51 answer : c" | a = 111 + 165
b = 100 + 120
c = b * const_0_2778
d = a / c
|
a ) 11 , b ) 14 , c ) 13 , d ) 15 , e ) 10 | e | divide(const_100, 10) | in how many years will a sum of money be doubled given that the annual interest on it is 10 % ? | p = ( p * 10 * r ) / 100 r = 10 % answer : e | a = 100 / 10
|
a ) 2 / 5 , b ) 2 / 25 , c ) 5 / 2 , d ) 5 / 4 , e ) 4 / 25 | e | divide(power(2, const_2), power(5, const_2)) | rectangle a has sides a and b , and rectangle b has sides c and d . if a / c = b / d = 2 / 5 , what is the ratio of rectangle a β s area to rectangle b β s area ? | "the area of rectangle a is ab . c = 5 a / 2 and d = 5 b / 2 . the area of rectangle b is cd = 25 ab / 4 . the ratio of rectangle a ' s area to rectangle b ' s area is ab / ( 25 ab / 4 ) = 4 / 25 . the answer is e ." | a = 2 ** 2
b = 5 ** 2
c = a / b
|
a ) 21 , 21 , b ) 9 , 21 , c ) 21 , 9 , d ) 9 , 9 , e ) 20 , 8 | c | divide(add(divide(48, 4), divide(600, 20)), const_2) | if a boat is rowed downstream for 600 km in 20 hours and upstream for 48 km in 4 hours , what is the speed of the boat and the river ? | "explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x Γ’ β¬ β y x = ( ds + us ) / 2 y = ( ds Γ’ β¬ β us ) / 2 in the above problem ds = 30 ; us = 12 x = ( 30 + 12 ) / 2 = 42 / 2 = 21 km / hr y = ( 30 - 12 ) / 2 = 18 / 2 = 9 km / hr answer : c" | a = 48 / 4
b = 600 / 20
c = a + b
d = c / 2
|
a ) 333 sec , b ) 190 sec , c ) 176 sec , d ) 2687 sec , e ) 1876 sec | b | divide(add(1200, 700), divide(1200, 120)) | a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 700 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 1900 m t = d / s t = 1900 / 10 t = 190 sec answer : b" | a = 1200 + 700
b = 1200 / 120
c = a / b
|
a ) 1 / 12 , b ) 1 / 10 , c ) 1 / 8 , d ) 1 / 42 , e ) 5 / 9 | d | multiply(multiply(multiply(divide(add(const_4, const_1), add(9, const_1)), divide(subtract(add(const_4, const_1), const_1), subtract(add(9, const_1), const_1))), divide(subtract(subtract(add(const_4, const_1), const_1), const_1), subtract(subtract(add(9, const_1), const_1), const_1))), divide(subtract(subtract(subtract(add(const_4, const_1), const_1), const_1), const_1), subtract(subtract(subtract(add(9, const_1), const_1), const_1), const_1))) | each of the integers from 0 to 9 , inclusive , is written on a separate slip of blank paper and the ten slips are dropped into a hat . if 4 of the slips are the drawn , without replacement , what is the probability that all 4 have a even number written on it ? | key is that there is no replacement , so each successive choice will become more skewed towards picking a neg ( i . e . the pool of positives decreases , while the pool of negatives stay the same ) p ( + on 1 st pick ) = 5 / 10 p ( + on 2 nd pick ) = 4 / 9 p ( + on 3 rd pick ) = 3 / 8 p ( + on 4 rd pick ) = 2 / 7 5 / 10 * 4 / 9 * 3 / 8 * 2 / 7 = 1 / 42 d | a = 4 + 1
b = 9 + 1
c = a / b
d = 4 + 1
e = d - 1
f = 9 + 1
g = f - 1
h = e / g
i = c * h
j = 4 + 1
k = j - 1
l = k - 1
m = 9 + 1
n = m - 1
o = n - 1
p = l / o
q = i * p
r = 4 + 1
s = r - 1
t = s - 1
u = t - 1
v = 9 + 1
w = v - 1
x = w - 1
y = x - 1
z = u / y
A = q * z
|
a ) 630 km , b ) 540 km , c ) 276 km , d ) 178 km , e ) 176 km | a | multiply(add(20, 25), divide(70, subtract(25, 20))) | two trains start at same time from two stations and proceed towards each other at the rate of 20 km / hr and 25 km / hr respectively . when they meet , it is found that one train has traveled 70 km more than the other . what is the distance between the two stations ? | "explanation : let us assume that trains meet after ' x ' hours distance = speed * time distance traveled by two trains = 20 x km and 25 x km resp . as one train travels 70 km more than the other , 25 x Γ’ β¬ β 20 x = 70 5 x = 70 x = 14 hours as the two trains are moving towards each other , relative speed = 20 + 25 = 45 km / hr therefore , total distance = 45 * 14 = 630 km . answer : a" | a = 20 + 25
b = 25 - 20
c = 70 / b
d = a * c
|
a ) 21 , b ) 32 , c ) 30 , d ) 41 , e ) 37 | a | divide(add(add(10, 12), multiply(10, 2)), const_2) | the average age of 10 men increases by 2 years when two women are included in place of two men of ages 10 and 12 years . find the average age of the women ? | "10 + 12 + 10 * 2 = 42 / 2 = 21 answer : a" | a = 10 + 12
b = 10 * 2
c = a + b
d = c / 2
|
a ) 8400 , b ) 14703 , c ) 37798 , d ) 13777 , e ) 14778 | a | multiply(35000, divide(divide(subtract(50000, add(add(4000, 5000), 5000)), const_3), 50000)) | a , b , c subscribe rs . 50000 for a business , a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35000 , c receives : | explanation : let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 < = > 3 x = 36000 < = > x = 12000 . a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . c ' s share = rs . ( 35000 * 12 / 50 ) = rs . 8,400 . answer : a ) | a = 4000 + 5000
b = a + 5000
c = 50000 - b
d = c / 3
e = d / 50000
f = 35000 * e
|
a ) 4 , b ) 8 , c ) 10 , d ) 15 , e ) 24 | a | max(multiply(subtract(add(55, 6), const_1), subtract(divide(6, 35), divide(6, 55))), const_4) | due to construction , the speed limit along an 6 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | 6 / 35 - 6 / 55 = 6 / 5 * ( 11 - 7 ) / 77 = 6 / 5 * 4 / 77 * 60 min = 6 * 12 * 4 / 77 = 288 / 77 ~ 3.7 answer - a | a = 55 + 6
b = a - 1
c = 6 / 35
d = 6 / 55
e = c - d
f = b * e
g = max(f)
|
a ) 276.92 , b ) 267.92 , c ) 266.27 , d ) 900 , e ) 237.8 | a | multiply(divide(600, add(add(600, 800), 1200)), 900) | a , b and c invested rs . 600 , rs . 800 and rs . 1200 respectively , in a partnership business . find the share of b in profit of rs . 900 after a year ? | "explanation : 600 : 800 : 1200 3 : 4 : 6 4 / 13 * 900 = 276.92 answer : a" | a = 600 + 800
b = a + 1200
c = 600 / b
d = c * 900
|
a ) 36 , b ) 42 , c ) 45 , d ) 48 , e ) 51 | b | divide(multiply(subtract(1, divide(1, 3)), 21), subtract(1, subtract(1, divide(1, 3)))) | a driver would have reduced the time it took to drive from home to the store by 1 / 3 if the average speed had been increased by 21 miles per hour . what was the actual average speed , in miles per hour , when the driver drove from home to the store ? | "let r be the original speed and let t be the original time . since the distance remains the same ( we ' re just changing the rate and time ) , any increase in rate or time is met with a decrease in the other term . decreasing the time by 1 / 3 would give us : d = ( r ) ( t ) = ( 2 t / 3 ) ( x * r ) x = 3 / 2 since ( 2 t / 3 ) ( 3 r / 2 ) = ( r ) ( t ) = d 3 r / 2 = r + 21 r / 2 = 21 r = 42 the answer is b ." | a = 1 / 3
b = 1 - a
c = b * 21
d = 1 / 3
e = 1 - d
f = 1 - e
g = c / f
|
a ) 50 days , b ) 92.4 days , c ) 100 days , d ) 150 days , e ) 80 days | b | subtract(multiply(const_4, 50), multiply(divide(135, const_100), 60)) | p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 135 days , in how many days can r alone complete the remaining work ? | "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 250 days and hence r alone in 385 days p works for 60 days - - > 60 / 200 work done = > 30 % q works for 135 days - - > 135 / 250 work done = > 54 % 24 % work left . . . r alone will take 24 % * 385 = 92.4 days answer is ( b )" | a = 4 * 50
b = 135 / 100
c = b * 60
d = a - c
|
a ) 1 , b ) 4 , c ) 5 , d ) 0 , e ) 3 | c | divide(add(15, 10), subtract(7, const_2)) | find a number such that when 15 is subtracted from 7 times the number , the result is 10 more than twice the number . | let the number be x . then , 7 x - 15 = 2 x + 10 = > 5 x = 25 = > x = 5 . hence , the required number is 5 . answer is c . | a = 15 + 10
b = 7 - 2
c = a / b
|
a ) a ) 1000 , b ) b ) 1055 , c ) c ) 965 , d ) d ) 1075 , e ) e ) 1080 | c | add(multiply(8, 70), multiply(9, 45)) | harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 45 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 70 Γ 8 = 560 . cost of 9 kg of mangoes = 45 Γ 9 = 405 . total cost he has to pay = 560 + 405 = 965 . c )" | a = 8 * 70
b = 9 * 45
c = a + b
|
a ) 2000 , b ) 2500 , c ) 3000 , d ) 3500 , e ) 4000 | c | subtract(multiply(const_2, 4050), divide(add(multiply(const_2, 4050), subtract(multiply(const_2, 5250), multiply(const_2, 4200))), const_2)) | the average monthly income of a and b is rs . 4050 . the average monthly income of b and c is rs . 5250 and the average monthly income of a and c is rs . 4200 . what is the monthly income of a ? | let monthly income of a = x monthly income of b = y monthly income of c = z x + y = 2 Γ 4050 . . . . ( equation 1 ) y + z = 2 Γ 5250 . . . . ( equation 2 ) z + x = 2 Γ 4200 . . . . ( equation 3 ) ( equation 1 ) + ( equation 3 ) - ( equation 2 ) = > x + y + x + z - ( y + z ) = ( 2 Γ 4050 ) + ( 2 Γ 4200 ) - ( 2 Γ 5250 ) = > 2 x = 2 ( 4050 + 4200 - 5250 ) = > x = 3000 i . e . , monthly income of a = 3000 answer : c | a = 2 * 4050
b = 2 * 4050
c = 2 * 5250
d = 2 * 4200
e = c - d
f = b + e
g = f / 2
h = a - g
|
a ) 21 % , b ) 25 % , c ) 69 % , d ) 31 % , e ) 19 % | e | subtract(subtract(add(40, const_100), divide(multiply(add(40, const_100), 15), const_100)), const_100) | a merchant marks his goods up by 40 % and then offers a discount of 15 % on the marked price . what % profit does the merchant make after the discount ? | let the original price be 100 . after 40 % markup , price = 140 after 15 % discount on this marked - up price , price = 140 - ( 15 % of 140 ) = 140 - 21 = 119 final profit = ( ( 119 - 100 ) / 100 ) * 100 = 19 % option e | a = 40 + 100
b = 40 + 100
c = b * 15
d = c / 100
e = a - d
f = e - 100
|
a ) rs . 700 , b ) rs . 708.50 , c ) rs . 732.50 , d ) rs . 742.50 , e ) none of these | d | divide(multiply(subtract(multiply(100, 70), multiply(subtract(100, multiply(2.5, const_2)), subtract(70, multiply(2.5, const_2)))), 90), const_100) | a rectangular grassy plot 100 m . by 70 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 90 paise per sq . metre . | "explanation : area of the plot = ( 100 x 70 ) m 2 = 7000 m 2 area of the plot excluding the path = [ ( 100 - 5 ) * ( 70 - 5 ) ] m 2 = 6175 m 2 . area of the path = ( 7000 - 6175 ) m 2 = 825 m 2 . cost of gravelling the path = rs . 825 * ( 90 / 100 ) = rs . 742.50 answer : option d" | a = 100 * 70
b = 2 * 5
c = 100 - b
d = 2 * 5
e = 70 - d
f = c * e
g = a - f
h = g * 90
i = h / 100
|
a ) 20.23 , b ) 20.13 , c ) 30.93 , d ) 20.93 , e ) 10.93 | d | multiply(divide(75, add(add(divide(4, 3), divide(5, 4)), const_1)), 5) | a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 4 : 3 and b : c = 4 : 5 . if the total runs scored by all of them are 75 , the runs scored by b are ? | "a : b = 4 : 3 b : c = 4 : 5 a : b : c = 16 : 12 : 15 12 / 43 * 75 = 20.93 answer : d" | a = 4 / 3
b = 5 / 4
c = a + b
d = c + 1
e = 75 / d
f = e * 5
|
a ) rs . 15500 , b ) rs . 16500 , c ) rs . 17500 , d ) rs . 18500 , e ) rs . 19500 | b | multiply(800, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . find the cost of paying the floor by slabs at the rate of rs . 800 per sq . metre . | "area = 5.5 Γ 3.75 sq . metre . cost for 1 sq . metre . = rs . 800 hence total cost = 5.5 Γ 3.75 Γ 800 = 5.5 Γ 3000 = rs . 16500 answer : b" | a = 5 * 5
b = 800 * a
|
a ) 30.5 , b ) 48.5 , c ) 50.4 , d ) 62.6 , e ) 57.6 | e | subtract(add(divide(multiply(2, 38), subtract(38, const_1)), 38), 2) | the ages of 2 persons differ by 38 years . if 12 years ago the elder one be 6 times as old as the younger one , find the present age of elder person . | "age of the younger person = x age of the elder person = x + 38 6 ( x - 12 ) = x + 38 - 12 x = 19.6 age of elder person = 19.6 + 38 = 57.6 answer is e" | a = 2 * 38
b = 38 - 1
c = a / b
d = c + 38
e = d - 2
|
a ) 18 , b ) 29 , c ) 37 , d ) 44 , e ) 69 | e | multiply(add(2, const_1), subtract(25, 2)) | the product x of two prime numbers is between 15 and 70 . if one of the prime numbers is greater than 2 but less than 6 and the other is greater than 13 but less than 25 , then x = | option bc can be ruled out as they themselves are prime numbers 18 = 2 * 9 = 3 * 6 > > ignore 44 = 2 * 22 = 4 * 11 > > ignore 69 = 3 * 23 > > answer answer = e | a = 2 + 1
b = 25 - 2
c = a * b
|
a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(2020, 2020), add(divide(2020, 2020), divide(2020, 2020))) | we define that k @ j is the product of j number from k in increasing order for positive integers k , j . for example , 6 @ 4 = 6 * 7 * 8 * 9 . if a = 2020 and b = 2120 , what is the value r of the ratio a / b ? | "r - > a / b = 20 * 21 * β¦ β¦ * 39 / 21 * 22 * β¦ . * 39 * 40 = 20 / 40 = 1 / 2 . therefore , the answer is a ." | a = 2020 / 2020
b = 2020 / 2020
c = 2020 / 2020
d = b + c
e = a / d
|
a ) 25000 , b ) 2028 , c ) 2775 , d ) 5496 , e ) 6851 | a | divide(5000, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | a candidate got 40 % of the votes polled and he lost to his rival by 5000 votes . how many votes were cast ? | "40 % - - - - - - - - - - - l 60 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 20 % - - - - - - - - - - 5000 100 % - - - - - - - - - ? = > 25000 answer : a" | a = 40 / 100
b = 1 - a
c = 40 / 100
d = b - c
e = 5000 / d
|
a ) s . 130 , b ) s . 140 , c ) s . 150 , d ) s . 224 , e ) s . 280 | d | divide(560, add(divide(150, const_100), const_1)) | two employees a and b are paid a total of rs . 560 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 560 but x = 150 % of y = 150 y / 100 = 15 y / 10 β΄ 15 y / 10 + y = 560 β y [ 15 / 10 + 1 ] = 560 β 25 y / 10 = 560 β 25 y = 5600 β y = 5600 / 25 = rs . 224 d )" | a = 150 / 100
b = a + 1
c = 560 / b
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | divide(add(multiply(factorial(17), factorial(83)), multiply(factorial(17), factorial(82))), 17) | what is the units digit of 17 ^ 83 Γ 13 ^ 82 Γ 11 ^ 87 ? | "cyclicity of 7 is 4 - - > 7 , 9 , 3 , 1 cyclicity of 3 is 4 - - > 3 , 9 , 7 , 1 11 raised to any positive power has a units digit 1 17 ^ 83 Γ 13 ^ 82 Γ 11 ^ 87 = units digit of 3 * 9 * 1 = 7 answer : d" | a = math.factorial(17)
b = math.factorial(83)
c = a * b
d = math.factorial(17)
e = math.factorial(82)
f = d * e
g = c + f
h = g / 17
|
a ) 360 , b ) 420 , c ) 510 , d ) 320 , e ) 140 | e | lcm(lcm(10, 14), 20) | find the lowest common multiple of 10 , 14 and 20 . | "lcm = 2 * 2 * 5 * 7 = 140 . answer is e" | a = math.lcm(10, 14)
b = math.lcm(a, 20)
|
a ) 2984 , b ) 3029 , c ) 2982 , d ) 2981 , e ) none of these | b | subtract(3034, divide(1002, 200.4)) | 3034 - ( 1002 / 200.4 ) = ? | "3034 - 5 = 3029 answer : b" | a = 1002 / 200
b = 3034 - a
|
a ) 100 , b ) 200 , c ) 120 , d ) 40 , e ) 20 | a | divide(subtract(divide(multiply(2, const_100), const_2), const_2), add(divide(multiply(2, const_100), const_2), const_2)) | if 2 x = 3 y = 10 , then 6 xy = ? | "2 x = 10 ; x = 5 3 y = 10 ; y = 10 / 3 multiply : 6 xy = 6 * 5 * 10 / 3 = 100 answer : a ." | a = 2 * 100
b = a / 2
c = b - 2
d = 2 * 100
e = d / 2
f = e + 2
g = c / f
|
a ) 90 cm , b ) 6 dm , c ) 1 m , d ) 1.1 cm , e ) none of these | b | subtract(multiply(multiply(3.3, 2.6), 1.6), divide(8000, const_1000)) | a cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.6 m and its walls are 5 cm thick . the thickness of the bottom is : | explanation : let the thickness of the bottom be x cm . then , [ ( 330 - 10 ) Γ ( 260 - 10 ) Γ ( 160 - x ) ] = 8000 Γ 1000 = > 320 Γ 250 Γ ( 160 - x ) = 8000 Γ 1000 = > ( 160 - x ) = 8000 Γ 1000 / 320 = 100 = > x = 60 cm = 6 dm . answer : b | a = 3 * 3
b = a * 1
c = 8000 / 1000
d = b - c
|
a ) 15.27 , b ) 16.27 , c ) 17.27 , d ) 19.27 , e ) 15.83 | e | multiply(divide(add(multiply(const_2, 10), const_2), add(const_3, const_4)), multiply(multiply(5.5, 5.5), divide(multiply(const_1, const_60), multiply(const_100, const_3_6)))) | the length of minute hand of a clock is 5.5 cm . what is the area covered by this in 10 minutes | "area of circle is pi * r ^ 2 but in 10 minutes area covered is ( 10 / 60 ) * 360 = 60 degree so formula is pi * r ^ 2 * ( angle / 360 ) = 3.14 * ( 5.5 ^ 2 ) * ( 60 / 360 ) = 15.83 cm ^ 2 answer : e" | a = 2 * 10
b = a + 2
c = 3 + 4
d = b / c
e = 5 * 5
f = 1 * const_60
g = 100 * const_3_6
h = f / g
i = e * h
j = d * i
|
a ) 180 , b ) 184 , c ) 188 , d ) 192 , e ) 196 | e | add(multiply(divide(subtract(290, 8), const_3), const_2), 8) | if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 290 pounds . what is jake β s present weight , in pounds ? | "j + k = 290 and so k = 290 - j j - 8 = 2 k j - 8 = 2 ( 290 - j ) 3 j = 588 j = 196 the answer is e ." | a = 290 - 8
b = a / 3
c = b * 2
d = c + 8
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | subtract(40, 30) | a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 80 km / hr respectively . b starts two hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "in 2 hours , a travels 60 km . b can catch a at a rate of 10 km / hr , so b catches a 6 hours after b starts . so a and b both travel a distance of 240 km . c needs 3 hours to travel 240 km , so c leaves 5 hours after a . the answer is d ." | a = 40 - 30
|
a ) 15 % , b ) 17 % , c ) 24 % , d ) 30 % , e ) 33 % | e | multiply(divide(subtract(divide(40, const_100), multiply(divide(25, const_100), divide(40, const_100))), subtract(const_1, multiply(divide(25, const_100), divide(40, const_100)))), const_100) | in february wilson β s earnings were 40 percent of his family β s total income . in march wilson earned 25 percent less than in february . if the rest of his family β s income was the same in both months , then , in march , wilson β s earnings were approximately what percent x of his family β s total income ? | "lets suppose the total family income in feb = 100 x wilson ' s earning in feb = 40 % of 100 x = 40 x earnings of remaining family in feb = 100 x - 40 x = 60 x wilson ' s earning in march = 75 % of wilson ' s feb earnings = 75 % of 40 x = 30 x earnings of remaining family in march = earnings of remaining family in feb = 60 x thus wilson ' s earning as % of total family income in march x = 30 x / ( 30 + 60 ) x = 30 x / 90 x = 33.33 % thus answer is e" | a = 40 / 100
b = 25 / 100
c = 40 / 100
d = b * c
e = a - d
f = 25 / 100
g = 40 / 100
h = f * g
i = 1 - h
j = e / i
k = j * 100
|
a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | d | subtract(add(23, 16), const_1) | at a garage sale , all of the items were sold at different prices . if the price of a radio sold at the garage sale was both the 16 th highest price and the 23 rd lowest price among the prices of the items sold , how many items were sold at the garage sale ? | "there were 15 items sold at a higher price than the radio and 22 items sold at a lower price than the radio . including the radio , there were 15 + 22 + 1 = 38 items sold . the answer is d ." | a = 23 + 16
b = a - 1
|
a ) 28 , b ) 23 , c ) 22 , d ) 21 , e ) 19.5 | a | multiply(divide(subtract(52, 3), add(3, 4)), 4) | one hour after yolanda started walking from x to y , a distance of 52 miles , bob started walking along the same road from y to x . if yolanda ' s walking rate was 3 miles per hour and bob Ρ ' s was 4 miles per hour , how many miles had bob walked when they met ? | "when b started walking y already has covered 3 miles out of 52 , hence the distance at that time between them was 52 - 3 = 49 miles . combined rate of b and y was 3 + 4 = 7 miles per hour , hence they would meet each other in 49 / 7 = 7 hours . in 6 hours b walked 7 * 4 = 28 miles . answer : a ." | a = 52 - 3
b = 3 + 4
c = a / b
d = c * 4
|
a ) 1300 , b ) 1992 , c ) 9921 , d ) 2798 , e ) 2789 | a | multiply(multiply(divide(13, multiply(10, 2)), const_100), multiply(10, 2)) | find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 13 of money ? | "p = 13 ( 100 / 10 ) 2 = > p = 1300 answer : a" | a = 10 * 2
b = 13 / a
c = b * 100
d = 10 * 2
e = c * d
|
a ) 87 days , b ) 10 days , c ) 55 days , d ) 44 days , e ) 20 days | e | divide(subtract(multiply(30, 40), multiply(40, 10)), 40) | 30 men can do a work in 40 days . when should 10 men leave the work so that the entire work is completed in 40 days after they leave the work ? | "total work to be done = 30 * 40 = 1200 let 10 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 10 * 40 ) = 1200 40 p = 800 = > p = 20 days answer : e" | a = 30 * 40
b = 40 * 10
c = a - b
d = c / 40
|
a ) 16 sec , b ) 12 sec , c ) 17 sec , d ) 21 sec , e ) 23 sec | b | divide(multiply(120, const_2), add(speed(120, 15), speed(120, 10))) | two trains of equal lengths take 10 sec and 15 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 / 10 = 12 m / sec . speed of the second train = 120 / 5 = 8 m / sec . relative speed = 12 + 8 = 20 m / sec . required time = ( 120 + 120 ) / 20 = 12 sec . answer : b" | a = 120 * 2
b = speed + (
c = a / b
|
a ) 84 , b ) 105 , c ) 54 , d ) 42 , e ) 70 | a | multiply(power(const_2, const_2), multiply(7, const_3)) | jill ' s favorite number is even and has some repeating prime factors . john is guessing jills favorite number , and the only hint she gives is that 7 is a prime factor . what is johns best guess ? | a 84 = 2 * 2 * 2 * 7 b 105 is odd c 54 = 2 * 3 * 3 d 42 = 2 * 3 * 7 e 70 = 2 * 5 * 7 the answer must meet three criteria : odd , have repeating prime factors , and 7 is a prime factor . a is the only answer that meets all criteria | a = 2 ** 2
b = 7 * 3
c = a * b
|
a ) 65 , b ) 69 , c ) 72 , d ) 75 , e ) none | d | divide(add(add(add(add(76, 65), 82), 67), 85), add(const_1, const_4)) | kamal obtained 76 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "sol . average = 76 + 65 + 82 + 67 + 85 / 5 ) = ( 375 / 5 ) = 75 . answer d" | a = 76 + 65
b = a + 82
c = b + 67
d = c + 85
e = 1 + 4
f = d / e
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 6 | e | divide(divide(divide(180, const_3), 2), add(const_3, const_2)) | find the number of factors of 180 that are in the form ( 4 * k + 2 ) , where k is a non - negative integer ? | make sure to follow posting guidelines ( link in my signatures ) . title of the thread must mention the first few words of the question itself . as for the question , the most straightforward way is to list out the factors ( this is fine for this question as 180 is a relatively small number ) . number of factors of 180 ( = 2 ^ 2 * 3 ^ 2 * 5 ) = 18 . so listing the factors gives us : 1 , 180 2 , 90 3 , 60 4 , 45 5 , 36 6 , 30 9 , 20 10 , 18 12 , 15 as the question asks us about the factors of the form 4 k + 2 - - - > eliminate all odd factors . you are now left with 24 , 610 , 1218 , 2030 , 3660 , 90180 for the form 4 k + 2 , the first few values are 2 , 6 , 10 , 14 , 18 , 22 , 26 , 30 . . . already giving you 5 factors ( = 2 , 610 , 1830 ) . you can stop here and mark 6 ( 6 th one = 90 ) as the correct answer as 5 is not given . for checking any particular factor whether they conform to the form 4 k + 2 - - - > do the following 4 k + 2 = 36 - - - > 4 k = 34 - - > this will not give you an integer value for k - - - > reject . proceed with others until you find all the values asked . e is thus the correct answer . | a = 180 / 3
b = a / 2
c = 3 + 2
d = b / c
|
a ) 7 / 13 , b ) 4 / 5 , c ) 9 / 13 , d ) 11 / 13 , e ) 7 / 10 | d | divide(add(subtract(subtract(1, divide(const_1, const_2)), divide(divide(const_1, const_2), 2)), subtract(divide(const_2, 5), divide(divide(const_2, 5), 4))), subtract(subtract(1, subtract(subtract(1, divide(const_1, const_2)), divide(divide(const_1, const_2), 2))), subtract(subtract(1, divide(const_1, const_2)), divide(const_2, 5)))) | every student of a certain school must play an instrument . in last year , 1 / 2 of the students picked a woodwind instrument , 2 / 5 of the students picked a brass instrument , and all of the other students took percussion . in this year , 1 / 2 of the students who play woodwind and 1 / 4 of the students who play brass instruments left school , other students did not leave , and no fresh student come in . what fraction of all students play either a woodwind or brass instrument ? | lets pick smart numbers . total number of students : 20 woodwind ( 1 / 2 ) : 10 brass ( 2 / 5 ) : 8 percussion ( 1 / 10 ) : 2 after leaving school woodwind : 5 brass : 6 percussion : 2 new total number of students : 13 woodwind and brass : 11 answer 11 / 13 or d | a = 1 / 2
b = 1 - a
c = 1 / 2
d = c / 2
e = b - d
f = 2 / 5
g = 2 / 5
h = g / 4
i = f - h
j = e + i
k = 1 / 2
l = 1 - k
m = 1 / 2
n = m / 2
o = l - n
p = 1 - o
q = 1 / 2
r = 1 - q
s = 2 / 5
t = r - s
u = p - t
v = j / u
|
['a ) 250 pi kg', 'b ) 500 pi kg', 'c ) 500 pi ^ 2 kg', 'd ) 1000 pi kg', 'e ) 1000 pi ^ 2 kg'] | a | multiply(200, volume_cylinder(divide(1, const_2), 5)) | a cylinder is being filled with sand weighing 200 kg per cubic foot . the cylinder has a diameter of 1 foot and is 5 feet tall . how much sand is being used ? ? | answer is a . volume of cylinder = pi * r ^ 2 * h - - - - - - - - > pi * ( 1 / 2 ) ^ 2 * 5 - - - - - - - - > pi * ( 1 / 4 ) * 5 - - - - - - - - > ( 5 / 4 ) pi cubic feet . 1 cubic feet = 200 kg . so ( 5 / 4 ) pi cubic feet = ( ( 200 * 5 ) / 4 ) pi - - - - - - - - - > 250 pi regards , abhijit answer is a | a = 1 / 2
b = 200 * volume_cylinder
|
a ) 1 / 7 , b ) 3 / 10 , c ) 1 / 2 , d ) 5 / 6 , e ) 8 / 9 | a | divide(subtract(subtract(4, 1), 3), multiply(4, 3)) | a can complete the job in 4 hours and b can complete the same job in 3 hours . a works for 1 hour and then b joins and both complete the job . what fraction of the job did b complete | a = 1 / 7 | a = 4 - 1
b = a - 3
c = 4 * 3
d = b / c
|
a ) 122 , b ) 140 , c ) 199 , d ) 188 , e ) 190 | e | subtract(divide(multiply(760, subtract(22, const_2)), 16), 760) | there is food for 760 men for 22 days . how many more men should join after two days so that the same food may last for 16 days more ? | "760 - - - - 22 760 - - - - 20 x - - - - - 16 x * 16 = 760 * 20 x = 950 760 - - - - - - - 190 answer : e" | a = 22 - 2
b = 760 * a
c = b / 16
d = c - 760
|
a ) 0 , b ) 1 , c ) 8 , d ) 2 , e ) 4 | c | divide(log(8), log(power(8, 8))) | if n = 8 ^ 8 β 8 , what is the units digit of n ? | "8 ^ 8 - 8 = 8 ( 8 ^ 7 - 1 ) = = > 8 ( 2 ^ 21 - 1 ) last digit of 2 ^ 21 is 2 based on what explanation livestronger is saying . 2 ^ 24 - 1 yields 2 - 1 = 1 as the unit digit . now on multiply this with 8 , we get unit digit as 8 answer : c" | a = math.log(8)
b = 8 ** 8
c = math.log(b)
d = a / c
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.