Datasets:

euclaise

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mathqa_programs

like 1

a ) 176 , b ) 253 , c ) 342 , d ) 458 , e ) 560

b

divide(factorial(23), multiply(factorial(subtract(23, const_2)), factorial(const_2)))

if 23 lines are drawn in a plane such that no two of them are parallel and no three are concurrent , then in how many points do they intersect ?

"if two lines are not parallel , then they will intersect in exactly one point . lines can be extended infinitely on both ends so somewhere they will intersect with each other if they are not parallel . we are also given that no three lines are concurrent . this means that no three lines intersect at the same point . so every pair of two lines we select will have a unique point of intersection which they will not share with any third line . the number of ways to select 2 lines from 23 lines is 23 c 2 = 253 the answer is b ."

a = math.factorial(23) b = 23 - 2 c = math.factorial(b) d = math.factorial(2) e = c * d f = a / e

a ) 369.42 , b ) 762.42 , c ) 162.42 , d ) 362.42 , e ) 333.42

d

divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(1, divide(5.5, multiply(5.5, const_3)))), const_100)

what is the compound interest on rs . 6500 at 5.5 % p . a . compounded half - yearly for 1 year .

"compound interest : a = p ( 1 + r / n ) nt a = 6 , 862.42 c . i . > > 6 , 862.42 - 6500 > > rs . 362.42 answer : d"

a = 3 * 100 b = a * 100 c = 5 * 5 d = 5 / 5 e = 1 * d f = b * e g = f / 100

a ) 140 , b ) 84 , c ) 88 , d ) 72 , e ) 12

b

multiply(6, multiply(const_2, divide(sqrt(1225), divide(6, const_2))))

the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 6 units ?

"given that the area of the square = 1225 sq . units = > side of square = √ 1225 = 35 units the radius of the circle = side of the square = 35 units length of the rectangle = 2 / 5 * 35 = 14 units given that breadth = 6 units area of the rectangle = lb = 14 * 6 = 84 sq . units answer : b"

a = math.sqrt(1225) b = 6 / 2 c = a / b d = 2 * c e = 6 * d

a ) 173 , b ) 185 , c ) 197 , d ) 209 , e ) 221

a

add(add(add(add(add(add(add(20, const_1), add(add(20, const_1), const_1)), add(add(add(20, const_1), const_1), const_2)), add(add(add(add(20, const_1), const_1), const_2), const_1)), add(add(add(add(add(20, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(20, const_1), const_1), const_2), const_1), const_1), const_1)), add(add(add(add(add(add(add(20, const_1), const_1), const_2), const_1), const_1), const_1), const_1))

the sum of the non - prime numbers between 20 and 30 , non - inclusive , is

"sum of consecutive integers from 21 to 29 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 21 + 29 ) / 2 * 9 = 25 * 9 = 225 sum of non - prime numbers b / w 20 and 30 , non inclusive = = = > 225 - 52 ( i . e . , 23 + 29 , being the prime # s in the range ) = 173 answer : a"

a = 20 + 1 b = 20 + 1 c = b + 1 d = a + c e = 20 + 1 f = e + 1 g = f + 2 h = d + g i = 20 + 1 j = i + 1 k = j + 2 l = k + 1 m = h + l n = 20 + 1 o = n + 1 p = o + 2 q = p + 1 r = q + 1 s = m + r t = 20 + 1 u = t + 1 v = u + 2 w = v + 1 x = w + 1 y = x + 1 z = s + y A = 20 + 1 B = A + 1 C = B + 2 D = C + 1 E = D + 1 F = E + 1 G = F + 1 H = z + G

a ) 20 , b ) 31 , c ) 41 , d ) 53 , e ) 64

c

divide(add(60, 22), const_2)

there are 60 supermarkets in the fgh chain . all of them are either in the us or canada . if there are 22 more fgh supermarkets in the us than in canada , how many fgh supermarkets are there in the us ?

"x + ( x - 22 ) = 60 - - > x = 41 . answer : c ."

a = 60 + 22 b = a / 2

a ) 51 , b ) 53 , c ) 55 , d ) 56 , e ) 57

a

divide(divide(multiply(add(multiply(1000, const_2), multiply(subtract(const_12, const_1), 100)), const_12), const_2), add(add(multiply(const_3, const_100), multiply(multiply(const_2, const_3), const_10)), add(const_2, const_3)))

a salt manufacturing company produced a total of 1000 tonnes of salt in january of a particular year . starting from february its production increased by 100 tonnes every month over the previous months until the end of the year . find its ave 66 rage daily production for that year ?

total production of salt by the company in that year = 1000 + 1100 + 1200 + . . . . + 2100 = 18600 . average monthly production of salt for that year = 18600 / 365 â ‰ ˆ 51 answer : a

a = 1000 * 2 b = 12 - 1 c = b * 100 d = a + c e = d * 12 f = e / 2 g = 3 * 100 h = 2 * 3 i = h * 10 j = g + i k = 2 + 3 l = j + k m = f / l

a ) 25 liters , b ) 27 liters , c ) 30 liters , d ) 32 liters , e ) 35 liters

a

subtract(divide(multiply(const_2, const_100), subtract(20, multiply(divide(75, const_100), 20))), 15)

how much water should be added to 15 liters of a 20 % - solution of alcohol to reduce the concentration of alcohol in the solution by 75 % ?

"let x ltr water to be added 2 ltr alcohol to be represented as ( 20 ( 1 - 3 / 4 ( new soln . = 15 + x ) ) ) 2 = 5 % * ( 15 + x ) - - - - - - - - > x = 25 ans a"

a = 2 * 100 b = 75 / 100 c = b * 20 d = 20 - c e = a / d f = e - 15

a ) 17 , b ) 19 , c ) 16 , d ) 18 , e ) 20

e

divide(add(add(add(add(3, const_4), add(3, const_4)), add(const_4, const_4)), 70), 5)

the sum of ages of 5 children born 3 years different each is 70 yrs . what is the age of the elder child ?

"let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 70 5 x = 40 x = 8 . x + 12 = 8 + 12 = 20 e"

a = 3 + 4 b = 3 + 4 c = a + b d = 4 + 4 e = c + d f = e + 70 g = f / 5

a ) 12 , b ) 18 , c ) 24 , d ) 30 , e ) 36

c

subtract(divide(multiply(subtract(subtract(subtract(const_100, multiply(divide(160, 800), const_100)), 27), subtract(const_100, 75)), 800), const_100), divide(multiply(subtract(const_100, 75), 800), const_100))

in a sample of 800 high school students in which all students are either freshmen , sophomores , juniors , or seniors , 27 percent are juniors and 75 percent are not sophomores . if there are 160 seniors , how many more freshmen than sophomores are there among the sample of students ?

"200 are sophomores . the number of freshmen is 600 - 160 - 0.27 ( 800 ) = 224 the answer is c ."

a = 160 / 800 b = a * 100 c = 100 - b d = c - 27 e = 100 - 75 f = d - e g = f * 800 h = g / 100 i = 100 - 75 j = i * 800 k = j / 100 l = h - k

a ) 18 , b ) 20 , c ) 24 , d ) 30 , e ) 22.5

e

divide(add(add(6, const_4), subtract(36, const_4)), const_2)

find the average of all the numbers between 6 and 36 which are divisible by 5 .

"solution average = ( 10 + 15 + 20 + 25 + 30 + 35 ) / 6 ) = 135 / 6 = 22.5 answer e"

a = 6 + 4 b = 36 - 4 c = a + b d = c / 2

a ) 1 / 3 , b ) 1 / 2 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6

b

divide(const_4, add(multiply(const_4, 2), const_1))

tom , working alone , can paint a room in 10 hours . peter and john , working independently , can paint the same room in 5 hours and 2 hours , respectively . tom starts painting the room and works on his own for two hour . he is then joined by peter and they work together for two hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ?

"let the time when all three were working together be t hours . then : tom worked for t + 4 hour and has done 1 / 10 * ( t + 4 ) part of the job ; peter worked for t + 2 hour and has done 1 / 5 * ( t + 2 ) part of the job ; john worked for t hours and has done 1 / 2 * t part of the job : 1 / 10 * ( t + 4 ) + 1 / 5 * ( t + 2 ) + 1 / 2 * t = 1 - - > multiply by 10 - - > ( t + 4 ) + ( 2 t + 2 ) + 5 t = 10 - - > t = 1 / 2 ; hence peter has done 1 / 5 * ( 1 / 2 + 2 ) = 1 / 5 * 5 / 2 = 1 / 2 answer : b"

a = 4 * 2 b = a + 1 c = 4 / b

a ) 80 , b ) 26 , c ) 24 , d ) 54 , e ) 15

c

divide(440, multiply(add(60, 6), const_0_2778))

a train 440 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ?

"speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 440 * 3 / 55 = 24 sec . answer : c"

a = 60 + 6 b = a * const_0_2778 c = 440 / b

a ) 09 2 / 3 , b ) 10 2 / 3 , c ) 11 2 / 3 , d ) 12 2 / 3 , e ) 16 2 / 3

b

add(divide(subtract(const_1, add(multiply(subtract(4, 2), add(inverse(15), inverse(20))), multiply(add(inverse(20), add(inverse(10), inverse(15))), 2))), inverse(15)), 4)

a can do a piece of work in 10 days and b can do it in 15 days and c can do it 20 days . they started the work together and a leaves after 2 days and b leaves after 4 days from the beginning . how long will work lost ?

"b 10 2 / 3 2 / 10 + 4 / 15 + x / 20 = 1 x = 32 / 3 = 10 2 / 3"

a = 4 - 2 b = 1/(15) c = 1/(20) d = b + c e = a * d f = 1/(20) g = 1/(10) h = 1/(15) i = g + h j = f + i k = j * 2 l = e + k m = 1 - l n = 1/(15) o = m / n p = o + 4

a ) 64 and 15,625 , b ) 32 and 3,125 , c ) 64 and 15,620 , d ) 64 and 15,635 , e ) 64 and 16,625

b

divide(multiply(multiply(const_100, const_100), const_100), divide(multiply(multiply(const_100, const_100), const_100), subtract(multiply(const_3, const_12), const_4)))

find two integers , neither of which ends in a zero , and whose product is exactly 1 , 00,000

1 , 00,000 = 10 ^ 5 = 10 x 10 x 10 x 10 x 10 = ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) = ( 2 ^ 5 ) x ( 5 ^ 5 ) = 32 x 3125 so the numbers are 32 and 3,125 answer : b

a = 100 * 100 b = a * 100 c = 100 * 100 d = c * 100 e = 3 * 12 f = e - 4 g = d / f h = b / g

a ) 10 km , b ) 12 km , c ) 14 km , d ) 16 km , e ) 18 km

b

multiply(4, divide(6, const_2))

a man starts walking at 3 pm . ha walks at a speed of 4 km / hr on level ground and at a speed of 3 km / hr on uphill , 6 km / hr downhill and then 4 km / hr on level ground to reach home at 9 pm . what is the distance covered on one way

avg speed = 2 * 3 * 6 / ( 3 + 6 ) = 4 so speed of man = 4 km / hr distance = 4 * 6 = 24 km since we want one way distance it will be 24 / 2 = 12 km answer : b

a = 6 / 2 b = 4 * a

a ) 271 , b ) 266 , c ) 225 , d ) 277 , e ) 232

c

subtract(multiply(40, multiply(36, const_0_2778)), 175)

a train 175 m long running at 36 kmph crosses a platform in 40 sec . what is the length of the platform ?

"length of the platform = 36 * 5 / 18 * 40 = 400 – 175 = 225 answer : c"

a = 36 * const_0_2778 b = 40 * a c = b - 175

a ) a ) 600 , b ) b ) 700 , c ) c ) 800 , d ) d ) 900 , e ) e ) 1000

c

divide(1220, multiply(add(const_1, divide(22, const_100)), add(const_1, divide(25, const_100))))

the population of a town increases 22 % and 25 % respectively in two consecutive years . after the growth the present population of the town is 1220 . then what is the population of the town 2 years ago ?

"explanation : formula : ( after = 100 denominator ago = 100 numerator ) 1220 * 100 / 122 * 100 / 125 = 800 answer : option c"

a = 22 / 100 b = 1 + a c = 25 / 100 d = 1 + c e = b * d f = 1220 / e

a ) 720 , b ) 864 , c ) 900 , d ) 936 , e ) 1296

d

subtract(power(add(4, const_2), 4), multiply(divide(factorial(add(4, const_2)), multiply(factorial(const_2), factorial(4))), factorial(4)))

in how many v ways can a 4 - letter password be chosen , using the letters a , b , c , d , e , and / or f , such that at least one letter is repeated within the password ?

total number of four letter passwords = 6 * 6 * 6 * 6 = 1296 - - - - - - ( 1 ) total number of passwords in which no letter repeats = 6 c 4 * 4 ! = 15 * 24 = 360 - - - - - - ( 2 ) therefore required value v = ( 1 ) - ( 2 ) = 1296 - 360 = 936 . d

a = 4 + 2 b = a ** 4 c = 4 + 2 d = math.factorial(c) e = math.factorial(2) f = math.factorial(4) g = e * f h = d / g i = math.factorial(4) j = h * i k = b - j

a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) none of the above

d

divide(add(add(multiply(85, 6), multiply(60, 4)), 30), add(6, 4))

the average expenditure of a labourer for 6 months was 85 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 is

"income of 6 months = ( 6 × 85 ) – debt = 510 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 780 average monthly income = 780 ÷ 10 = 78 answer d"

a = 85 * 6 b = 60 * 4 c = a + b d = c + 30 e = 6 + 4 f = d / e

a ) 5 , b ) 9 , c ) 10 , d ) 20 , e ) 30

e

multiply(subtract(343, 10), 10)

what is the greatest positive integer x such that 7 ^ x is a factor of 343 ^ 10 ?

"343 ^ 10 = ( 7 ^ 3 ) ^ 10 = 7 ^ 30 answer : e"

a = 343 - 10 b = a * 10

a ) 0.1 , b ) 0.3 , c ) 0.5 , d ) 0.7 , e ) 0.9

c

divide(divide(add(0, 2), 2), divide(add(const_4.0, 4), 2))

in the xy - coordinate system , what is the slope of the line that goes through the point ( 4 , 4 ) and is equidistant from the two points p = ( 0 , 2 ) and q = ( 12 , 8 ) ?

"first , get the middle coordinate between ( 0,2 ) and ( 12,8 ) . . . x = 0 + ( 12 - 0 ) / 2 = 6 y = 2 + ( 8 - 2 ) / 2 = 5 second , get the slope of ( 6,5 ) and ( 4,4 ) . m = 5 - 4 / 6 - 4 = 1 / 2 = 0.5 answer : c"

a = 0 + 2 b = a / 2 c = 4 + 0 d = c / 2 e = b / d

a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10

c

inverse(add(divide(const_1, 24), divide(const_1, 12)))

a sum of money is sufficient to pay b ' s wages for 12 days and c ' s wages for 24 days . the same money is sufficient to pay the wages of both for ?

let the total money be $ x b ' s 1 day work = $ x / 12 c ' s 1 day work = $ x / 24 a + b 1 day work = $ x / 8 money is sufficient to pay the wages of both for 8 days answer is c

a = 1 / 24 b = 1 / 12 c = a + b d = 1/(c)

a ) 100 , b ) 200 , c ) 350 , d ) 50 , e ) 250

b

multiply(const_4, negate(subtract(550, add(350, 250))))

jonathan , matthew and zac are picking strawberries from their yard . together they have picked a total of 550 strawberries . jonathan and matthew have picked 350 strawberries together . matthew and zac have picked 250 strawberries together . how many strawberries has zac picked by himself ?

jonathan = matthew + zac = 550 strawberries j + m = 350 m + z = 250 use j + m = 350 to plug into the original formula ( j + m ) + z = 550 350 + z = 550 subtract 350 from each side z = 200 answer : b

a = 350 + 250 b = 550 - a c = 4 * negate

a ) 55 , b ) 80 , c ) 85 , d ) 90 , e ) 95

a

divide(subtract(divide(const_3600, const_10), multiply(70, const_2)), const_4)

in a certain parallelogram the degree measure of one angle exceeds that of the other by 70 what is the degree measure of the smaller angle ?

in a parallelogram opposite angles are equal and the angles at each side are supplementary to each other ( supplementary angles are two angles that add up to 180 ° ) . given : x + ( x + 70 ) = 180 - - > x = 55 . answer : a .

a = 3600 / 10 b = 70 * 2 c = a - b d = c / 4

a ) 4000 , b ) 4500 , c ) 5000 , d ) 5500 , e ) 6000

b

divide(1350, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100)))

a candidate got 35 % of the votes and lost to the rival by 1350 votes . how many votes were cast ?

"35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % = 1350 10 % = 450 100 % of the votes = 4500 answer : b"

a = 35 / 100 b = 1 - a c = 35 / 100 d = b - c e = 1350 / d

a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5

b

sqrt(add(power(sqrt(subtract(13, multiply(const_2, 2028))), const_2), multiply(const_4, 2028)))

the product of two numbers is 2028 and their h . c . f . is 13 . the number of such pairs is :

"explanation : let the numbers be 13 a and 13 b . then , 13 a * 13 b = 2028 = > ab = 12 . now , co - primes with product 12 are ( 1 , 12 ) and ( 3 , 4 ) . so , the required numbers are ( 13 * 1 , 13 * 12 ) and ( 13 * 3 , 13 * 4 ) . clearly , there are 2 such pairs . answer is b"

a = 2 * 2028 b = 13 - a c = math.sqrt(b) d = c ** 2 e = 4 * 2028 f = d + e g = math.sqrt(f)

a ) 87 kmph , b ) 65 kmph , c ) 68 kmph , d ) 16 kmph , e ) 18 kmph

c

subtract(multiply(16, multiply(160, const_0_2778)), 140)

a train 140 m long crosses a platform 160 m long in 16 sec ; find the speed of the train ?

"d = 140 + 160 = 300 t = 16 s = 300 / 16 * 18 / 5 = 67.5 kmph answer : c"

a = 160 * const_0_2778 b = 16 * a c = b - 140

a ) 35 kmph , b ) 32 kmph , c ) 30 kmph , d ) 31 kmph , e ) 20 kmph

c

divide(multiply(55, 6), add(5, 6))

a zeebra chases a tiger 5 hours after the tiger runs . zeebra takes 6 hours to reach the tiger . if the average speed of the zeebra is 55 kmph , what s the average speed of the tiger ?

tiger take 11 hours and zeebra take 6 hours . . . then distance chased by them is 55 * 6 . so speed of tiger is ( 55 * 6 ) / 11 = 30 kmph . answer is c

a = 55 * 6 b = 5 + 6 c = a / b

a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 8 , d ) 1 / 16 , e ) 1 / 32

e

power(divide(1, 2), 5)

if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first 4 flips but not on the last flip ?

p ( hhhht ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 the answer is e .

a = 1 / 2 b = a ** 5

a ) 2 : 5 , b ) 2 : 3 , c ) 2 : 4 , d ) 1 : 1 , e ) 2 : 9

d

divide(subtract(15.8, 15.4), subtract(16.2, 15.8))

the average age of students of a class is 15.8 years . the average age of boys in the class is 16.2 years and that of the girls is 15.4 years . the ration of the number of boys to the number of girls in the class is :

"let the ratio be k : 1 . then , k * 16.2 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.2 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 / 0.4 = 1 / 1 required ratio = 1 / 1 : 1 = 1 : 1 . answer : d"

a = 15 - 8 b = 16 - 2 c = a / b

a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 15

a

divide(subtract(75, multiply(const_3, 9)), multiply(const_3, const_2))

a number is doubled and 9 is added . if resultant is trebled , it becomes 75 . what is that number

"explanation : = > 3 ( 2 x + 9 ) = 75 = > 2 x + 9 = 25 = > x = 8 answer : option a"

a = 3 * 9 b = 75 - a c = 3 * 2 d = b / c

a ) 40 % , b ) 50 % , c ) 65 % , d ) 71.4 % , e ) 75 %

d

multiply(divide(subtract(210, add(multiply(3, 8), multiply(8, 3))), 210), const_100)

a batsman scored 210 runs which included 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets

"number of runs made by running = 210 - ( 3 x 4 + 8 x 6 ) = 210 - ( 60 ) = 150 now , we need to calculate 60 is what percent of 120 . = > 150 / 210 * 100 = 71.4 % d"

a = 3 * 8 b = 8 * 3 c = a + b d = 210 - c e = d / 210 f = e * 100

a ) 70000 , b ) 80000 , c ) 90000 , d ) 90090 , e ) 100000

a

multiply(112, power(add(const_4, const_1), const_4))

( 112 x 5 ^ 4 ) = ?

"( 112 x 54 ) = 112 x 10 4 = 112 x 104 = 1120000 = 70000 2 24 16 a )"

a = 4 + 1 b = a ** 4 c = 112 * b

a ) 2 , b ) 3.25 , c ) 4.25 , d ) 5.25 , e ) 6.25

a

subtract(power(2, 2), 2)

x + ( 1 / x ) = 2 find x ^ 2 + ( 1 / x ^ 2 )

"squaring on both sides ( x + 1 / x ) ^ 2 = 2 ^ 2 x ^ 2 + 1 / x ^ 2 = 4 - 2 x ^ 2 + 1 / x ^ 2 = 2 answer : a"

a = 2 ** 2 b = a - 2

a ) 344545629 , b ) 723437481 , c ) 354595321 , d ) 964564944 , e ) 458449909

b

subtract(427398, multiply(floor(divide(427398, 15)), 15))

what least no . must be subtracted from 427398 so that remaining no . is divisible by 15 ?

"explanation : on dividing 427398 by 15 we get the remainder 3 , so 3 should be subtracted option b"

a = 427398 / 15 b = math.floor(a) c = b * 15 d = 427398 - c

a ) 1 / 6 , b ) 2 / 9 , c ) 6 / 11 , d ) 9 / 16 , e ) 3 / 4

c

multiply(divide(subtract(12, 3), 12), divide(subtract(subtract(12, 3), const_1), subtract(12, const_1)))

in a box of 12 pens , a total of 3 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ?

"# defective pens = 3 # good pens = 9 probability of the 1 st pen being good = # of favorable outcomes / # of total outcomes = 9 / 12 probability of the 2 nd pen being good = # of remaining favorable outcomes / # of total remaining outcomes = 8 / 11 total probability = 9 / 12 * 8 / 11 = 6 / 11 answer will be c"

a = 12 - 3 b = a / 12 c = 12 - 3 d = c - 1 e = 12 - 1 f = d / e g = b * f

a ) 12 , b ) 12.5 , c ) 6.67 , d ) 13.5 , e ) 14

c

divide(divide(const_2, divide(15, const_100)), const_2)

if a sum of money doubles itself in 15 years at simple interest , the ratepercent per annum is

"explanation : let sum = x then simple interest = x rate = ( 100 * x ) / ( x * 15 ) = 6.67 option c"

a = 15 / 100 b = 2 / a c = b / 2

a ) 25 , b ) 76 , c ) 29 , d ) 12 , e ) 20

d

divide(subtract(subtract(subtract(883, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6)

alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 883 . find the cost of each ice - cream cup ?

"explanation : let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 883 96 + 225 + 490 + 6 x = 883 6 x = 72 = > x = 12 . answer : d"

a = 16 * 6 b = 883 - a c = 5 * 45 d = b - c e = 7 * 70 f = d - e g = f / 6

a ) 50 km , b ) 58 km , c ) 60 km , d ) 70 km , e ) 40 km

e

multiply(8, divide(20, subtract(16, 8)))

if a person walks at 16 km / hr instead of 8 km / hr , he would have walked 20 km more . the distance traveled by him if he walked at 16 km / hr ?

"let the actual distance traveled be x km . then , x / 8 = ( x + 20 ) / 16 2 x - x = 20 = > x = 20 km . the required distance is = ( 20 + 20 ) km = 40 km answer : e"

a = 16 - 8 b = 20 / a c = 8 * b

a ) 2 / 3 , b ) 5 / 6 , c ) 7 / 12 , d ) 11 / 12 , e ) 17 / 24

c

divide(add(multiply(divide(1, 3), const_12), multiply(divide(const_12, 2), divide(1, 2))), const_12)

drum x is 1 / 2 full of oil and drum y , which has twice the capacity of drum x , is 1 / 3 full of oil . if all of the oil in drum x is poured into drum y , then drum y will be filled to what capacity ?

"( 1 / 2 ) x = ( 1 / 4 ) y ( 1 / 4 ) y + ( 1 / 3 ) y = ( 7 / 12 ) y the answer is c ."

a = 1 / 3 b = a * 12 c = 12 / 2 d = 1 / 2 e = c * d f = b + e g = f / 12

a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 90 kmph

e

divide(divide(subtract(160, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778)

a train 160 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is

"speed of train relative to man : 160 / 6 * 18 / 5 km / hr = 96 km / hr let speed of train = x therefore x + 6 = 96 x = 96 - 6 x = 90 km / hr answer : e"

a = 6 * const_0_2778 b = a * 6 c = 160 - b d = c / 6 e = d / const_0_2778

a ) 520 , b ) 620 , c ) 820 , d ) 740 , e ) 864

e

divide(divide(multiply(240, 6), add(const_1, divide(const_2, const_3))), const_2)

an aeroplane covers a certain distance at a speed of 240 kmph in 6 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of :

"distance = ( 240 x 6 ) = 1440 km . speed = distance / time speed = 1440 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = ( 1440 x 3 / 5 ) km / hr = 864 km / hr answer e ) 864 km / hr"

a = 240 * 6 b = 2 / 3 c = 1 + b d = a / c e = d / 2

a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) none of them

c

multiply(13, multiply(divide(multiply(3, 24), multiply(9, 8)), const_2))

if 9 engines consume 24 metric tonnes of coal , when each is working 8 hoursday , bow much coal will be required for 8 engines , each running 13 hours a day , it being given that 3 engines of former type consume as much as 4 engines of latter type ?

let 3 engines of former type consume 1 unit in 1 hour . then , 4 engines of latter type consume 1 unit in 1 hour . therefore 1 engine of former type consumes ( 1 / 3 ) unit in 1 hour . 1 engine of latter type consumes ( 1 / 4 ) unit in 1 hour . let the required consumption of coal be x units . less engines , less coal consumed ( direct proportion ) more working hours , more coal consumed ( direct proportion ) less rate of consumption , less coal consumed ( direct prportion ) number of engines 9 : 8 working hours 8 : 13 } : : 24 : x rate of consumption ( 1 / 3 ) : ( 1 / 4 ) [ 9 x 8 x ( 1 / 3 ) x x ) = ( 8 x 13 x ( 1 / 4 ) x 24 ) = 24 x = 624 = x = 26 . hence , the required consumption of coal = 26 metric tonnes . answer is c .

a = 3 * 24 b = 9 * 8 c = a / b d = c * 2 e = 13 * d

a ) 2 , b ) 5 , c ) 10 , d ) 15 , e ) 20

a

log(divide(log(subtract(5625, multiply(add(const_4, const_1), const_1000))), log(add(const_4, const_1))))

the value of log 2 ( log 5625 ) is

solution let log 5625 = x . then , 5 x = 625 = 54 ‹ = › x = 4 . let log 2 ( log 5625 ) = y . then , log 24 = y ‹ = › 2 y = 4 y ; 2 . answer a

a = 4 + 1 b = a * 1000 c = 5625 - b d = math.log(c) e = 4 + 1 f = math.log(e) g = d / f h = math.log(g)

a ) 95 , b ) 90 , c ) 85 , d ) 80 , e ) 64

e

divide(16, divide(subtract(100, 75), 100))

if 16 percent of the students at a certain school went to a camping trip and took more than $ 100 , and 75 percent of the students who went to the camping trip did not take more than $ 100 , what percentage of the students at the school went to the camping trip ?

"let x be the number of students in the school . 0.16 x students went to the trip and took more than 100 $ . they compose ( 100 - 75 ) = 25 % of all students who went to the trip . therefore the toal of 0.16 x / 0.25 = 0.64 x students went to the camping which is 64 % . the answer is e"

a = 100 - 75 b = a / 100 c = 16 / b

a ) 52 kmph , b ) 56 kmph , c ) 60 kmph , d ) 62 kmph , e ) 72 kmph

c

subtract(multiply(divide(320, 10), const_3_6), 55)

a man sitting in a train which is traveling at 55 kmph observes that a goods train , traveling in opposite direction , takes 10 seconds to pass him . if the goods train is 320 m long , find its speed

"explanation : relative speed = 320 / 10 m / sec = ( ( 320 / 10 ) × ( 18 / 5 ) ) kmph = 115 kmph . speed of goods train = ( 115 - 55 ) kmph = 60 kmph answer : option c"

a = 320 / 10 b = a * const_3_6 c = b - 55

a ) 1 : 2 , b ) 3 : 5 , c ) 4 : 5 , d ) 16 : 15 , e ) none

a

divide(multiply(90, const_2), multiply(90, const_3))

a man invests some money partly in 10 % stock at 90 and partly in 5 % stock at 90 . to obtain equal dividends from both , he must invest the money in the ratio :

"solution for an income of rs . 1 in 10 % stock at 90 , investment = rs . ( 90 / 10 ) = rs . 9 . for an income of rs . 1 in 5 % stock at 90 , investment = rs . ( 90 / 5 ) = rs . 18 . ∴ ratio of investments = 9 : 18 = 1 : 2 answer a"

a = 90 * 2 b = 90 * 3 c = a / b

a ) a ) 70 , b ) b ) 76 , c ) c ) 78 , d ) d ) 98 , e ) e ) 88

d

subtract(multiply(add(32, 6), add(10, const_1)), multiply(10, 32))

average of 10 matches is 32 , how many runs one should should score to increase his average by 6 runs .

"explanation : average after 11 innings should be 38 so , required score = ( 11 * 38 ) - ( 10 * 32 ) = 418 - 320 = 98 answer : option d"

a = 32 + 6 b = 10 + 1 c = a * b d = 10 * 32 e = c - d

a ) - 8 , b ) - 9 , c ) - 12 , d ) - 4 , e ) 12

c

divide(negate(add(23, 1)), 2)

solve below question 2 x + 1 = - 23

1 . subtract 1 from both sides : 2 x + 1 - 1 = - 23 - 1 2 . simplify both sides : 2 x = - 24 3 . divide both sides by 2 : 4 . simplify both sides : x = - 12 c

a = 23 + 1 b = negate / (

a ) 241 , b ) 242 , c ) 244 , d ) 247 , e ) 240

b

multiply(85184, 352)

calculate 85184 ÷ ? = 352

"answer let 85184 ÷ x = 242 then x = 85184 / 242 = 352 . option : b"

a = 85184 * 352

a ) four , b ) six , c ) nine , d ) ten , e ) twelve

e

add(8.1, subtract(8.1, multiply(1.5, 0.4)))

7.51 8.22 7.86 8.36 8.09 7.83 8.30 8.01 7.73 8.25 7.96 8.53 a vending machine is designed to dispense 8 ounces of coffee into a cup . after a test that recorded the number of ounces of coffee in each of 1000 cups dispensed by the vending machine , the 12 listed amounts , in ounces , were selected from the data above . if the 1000 recorded amounts have a mean of 8.1 ounces and a standard deviation of 0.4 ounces , how many of the 12 listed amounts are within 1.5 standard deviation of the mean ?

mean = 8.1 standard deviation = 0.4 within 1.5 standard deviation of the mean i . e . 1.5 standard deviation above the mean = 8.1 + 1.5 * 0.4 = 8.7 and 1.5 standard deviation below the mean = 8.1 - 1.5 * 0.4 = 7.5 all values are within the range hence , 12 values fall within 1.5 sd from mean answer : option e

a = 1 * 5 b = 8 - 1 c = 8 + 1