Visitors to a show were charged Rs. 15 each on the first day, Rs. 7.50 on the second day, Rs. 2.50 on the third day and total attendance on three days were in the ratio 2:5:13 respectively. The average charge per person for the entire three days is

5358 x 51 = 5358 x (50 +1 )
5358 x 50 + 5358 x 1
=267900 + 5358 = 273258

Total number of cases = ¹⁰c₄
Favorable number of cases = ⁴c₂. ⁶c₂
{Since, we are to select 2 children out of 4 and remaining 2 persons are to be selected from remaining 6 persons ( 2W + 4M)}
? required Probability = ⁴c₂ . ⁶c₂ / ¹⁰c₄
= [{(4 x 3) / (2 x 1)} x {(6 x 5) / (2 x 1)}] / [ (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)]
= [(12/2) x (30/2)] / 210
= 90 / 210
= 9 /12

Let son's age 1 year ago be x years.
? Father's age 1 year ago = 4x years
? 4x + 7 = 2(x + 7) + 9
? 2x = 23 - 7
? x = 8
Ratio of father's and son's present ages = (4x + 1) : (x + 1) = 33 : 9 = 11 : 3

Let A's age be x years, B's age be 2x years and C's age = (x + 17) years

According to the question,
? x + 2x + (x + 17) = 185
? 4x = 185 - 17 = 168
? x = 42
? A's age = 42 years
B's age 84 years
C's age = 42 + 17 = 59 years

Clearly, B covers 18 m in 9 s.
? B's time over the course = 9/18 x 1000 = 500 s
? A' s time over the course = (500 - 9) = 491 s

Let the present age of Ramesh be 3x years and the present age of jayesh = 2x years.

According to question, (3x - 4) = (2x - 4) + 6
? 3x - 4 = 2x - 4 + 6
? 3x - 4 = 2x + 2
? 3x - 2x = 2 + 4
? x = 6
? Present age of Jayesh = 2x = 12 years

Number of possible cases = 100 x 100 for 7^m + 7ⁿ to be divisible by 5,
One of the term has to end with 9 and other with 1 (7^m cannot be divided by 9 )

? m can be 2, 6 , 10, 14, ..... 98 (25 values) and n can be 4, 8, 12 ...... 100(25 values)

Since m and n can interchange
? Required probability = (2 x 25 x 25) / (100 x 100) = 1/8

The probability that head is show in one coin is 1/2.

The probability that the sum of the number on the dice is a prime = the probability that the following pair of number on the dice is a getting on the dice, namely (1, 1), (1, 2), (2, 1), (1, 4 ), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5), (5, 6) = 15/36.

? The required probability = 1/2 x 1/2 x 15/36 = 5/48.

Let the present ages of mother and her daughter are ( x + 25 ) and x years respectively ; then 6( x - 5 ) = x + 25 - 5 ? 6x - 30 = x + 20 ? 5x = 50 ? x = 10
Hence, the age of the mother = 10 + 25 = 35 years

Let us assume x years ago, the ratio of their ages was 3 : 5, then
According to question,
(40 - x)/(60 - x) = 3/5
? (40 - x) 5 = 3 (60 - x)
? 200 - 5x = 180 - 3x
? 2x = 20
? x = 10 years