The average age of two boys and their father is greater then average age of those two boys and their mother by 3 yr. The average age of the four is 19 yr. If the average age of the two boys be 5 ¹/ ₂, then find the age of the father and mother.

Let x be the number of wickets taken till the last match
According to the question,
(12.4 x + 26) = 12(x + 5)
? 12.4 x + 26 = 12x + 60
? 0.4 x = 34
? x = 34/0.4 = 85

Let average after 10th innings = x.
According to the question.
10x + 216 = 11(x +12)
? x = 216 - 132 = 84
? New average = 84 + 12 = 96

Distance between Raipur and Somgarh
= Average speed x Time
= 69 x 35/60= 161/4 km
New speed = (69 + 36) km/h = 105 km/h
? Time = Distance/ Speed = 161/4 x 105 h
= 161 x 60/4 x 105min = 23 min

Let the minimum score = x
? Maximum score = x + 100
? x + (x + 100) = 30 x 42 - 40 x 28
? 2x + 100 = 1260 - 1120 = 140
? 2x = 140 - 100 = 40
? x = 20
Hence, the maximum score = x + 100
= 20 + 100 = 120

According to the question
(P + T + R) / 3 = 54¹/₃
? P + T + R = (163/3) x 3
? P + T + R = 163 ............(i)
Also (T + F + G)/3 = 53
? T + G + F = 159 ..........(ii)
Clearly, from the two Eqs. (i) and (ii), the problem cannot be solved.

Let money of C be ? N.

According to the question
Total money of B = N + N x 50%
= N + 50N/100 = 3N/2

Total money of A = 2 x 3N/2 = 3N

Average money of three persons = 12000
? Total money of three = 12000 x 3
? 3N + (3N/2) + N = 12000 x 3
? (6N + 3N + 2N)/2 = 36000
? N = (36000 x 2)/11 = 72000/11

Now, money of A = 3N = (3 x 72000)/11
= ? 216000/11

Let the required runs be N.

According to the question,
(80 x 99 + N) / 80 = 100
? 7920 + N = 8000
? N = 80

Let the total number of students be 100.
Also, let the mean score = N

According to the question,
20 x 80 + 25 x 31 + (100 - 20 - 25) x N = 100 x 52
? 1600 + 775 + (55) x N = 5200
? 55N = 5200 - 1600 - 775
? N = 2825/55 = 51.36 = 51.4%

So, mean score of remaining 55% is 51.4%

Sum of the weight of A, B, C and D = 67 x 4 = 268 kg
and average weight of A, B , C , D and E = 67 - 2 = 65kg

? Sum of the weight of A, B, C, D and E = 65 x 5 = 325 kg
? Weight of E = 325 - 268 = 57 kg
? Weight of F = 57 + 4 = 61 kg

Now, average weight of F, B, C, D and E = 64 kg
? Sum of the weight of F, B, C, D and E = 64 x 5 = 320 Kg

? Sum of the weights of B, C and D = 320 - 57 - 61 = 202 kg
? Weight of A = 268 - 202 = 66 kg

Total age of Rashmi and Surbhi (5 yr ago) 20 x 2 = 40 yr.
Total age of Rashmi and Surbhi (at present) = 40 + 5 + 5 = 50 yr.
Total age of Rashmi, Surbhi and Meeta (at present) = 30 x 3 = 90 yr.
? Present age of Meeta = 90 - 50 = 40 yr.
? Age of meeta after 15yr. = 40 + 15 = 55 yr.