A cricketer whose bowling average is 12.4 runs per wicket, takes 5 wickets for 26 runs and there by decreases his average by 0.4. Find out the number of wickets taken by him till the last match?

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 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

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%

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