A cricketer has a certain average for 10 inning. In the eleventh innings. he scored 216 runs, there by increasing his average by 12 runs. Find out his new average?

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 temperature on sunday be x° C.
(20° + 18.6° + 17° + 21.4° + 19° + 19.8° + x) / 7 = 19.5°
? 115.8 + x = 19.5 x 7
? x = 136.5 - 11.58
? x = 20.7°C

Here, P = 9 km/h, Q = 25 km/h, R = 30 km
x = 3 km/h, y = 5 km/h and z = 10 km/h
? Required average speed
= [P + Q + R] / [(P/x) + (Q/y) + (R/z)] = [9+25 + 30] / [(9/3) + (25/5) + (30/10)]
= 64/(3 + 5 + 3)
= 64/11
= 5⁹/_{11 km/h}

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 the ages of mother, father and boys be M, F, B₁ and B₂, respectively.
The total age of four member
= 19 x 4
= 76 yr.

Given (B₁ + B₁)/2 = 11/2
? B₁ + B₂ = 11
M + F + B₁ + B₂ = 76 - 11
? M + F = 65 .......(i)

According to the question,
(B₁ + B₂ + F) / 3 = (B₁ + B₂ + M) / 3 + 3
? B₁ + B₂ + F) = B₁ + B₂ + M + 9
? F = M + 9
? F - M = 9 ............. (ii)

From Eqs (i) and (ii), we get
F = 37 yr and M = 28 yr

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%