model 9 average on cricket/exam Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

Runs scored in the next innings = x (let)

According to the question,

10 × 32 + x = 11 × 38

⇒ 320 + x = 418

⇒ x = 418 – 320 = 98

Aliter : Using Rule 18,

If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,

increases by 't’ years

Then, the age of the new person = T + N.t.

Here, T = 32, N= (10 + 1) = 11, t = 6

Required Runs = T + Nt

= 32 + 11 × 6

= 32 + 66 = 98

Answer: (a)

Let the number of runs scored in 11th innings be x.

∴ 10 × 50 + x = 11 × 52

⇒ 500 + x = 572

⇒ x = 572 – 500 = 72 runs

Aliter : Using Rule 18,

If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,

increases by 't’ years

Then, the age of the new person = T + N.t.

Here, T = 50, N = (10 + 1) = 11, t = 2

Required Runs = T + Nt

= 50 + 11 × 2 = 72

Answer: (d)

Let the number of wickets before the last match be x.

According to the question,

12.4x + 22 = (x + 5 ) × 12

⇒ 12.4x + 22 = 12x + 60

⇒ 12.4x – 12x = 60 – 22

⇒ 0.4x = 38

⇒ x = $38/{0.4}$ =$380/4$ = 95

Answer: (c)

Let the batsman make x runs. Total runs in 10 innings = 10 × 32 = 320

∴ ${320+x}/11$ = 32+4

⇒ 320 + x = 36 × 11

⇒ x = 396 – 320 = 76

Aliter : Using Rule 18,

If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,

increases by 't’ years

Then, the age of the new person = T + N.t.

Here, T = 32, N = 11, t = 4

Required Run = T + Nt

[Here N is taken as (n + 1)]

= 32 + 11 × 4

= 32 + 44 = 76

Answer: (d)

If the average in 10 tests be x, then,

x × 10 + 100 = (x + 5) × 11

⇒ 11x – 10x = 100 – 55

⇒ x = 45

∴ Required average = 50

Answer: (b)

Let the highest score be x.

∴ Lowest score = x – 172

∴ x + x – 172 = 40 × 50 – 38 × 48

⇒ 2x – 172 = 2000 – 1824 = 176

⇒ 2x = 176 + 172 = 348

∴ x = $348/2$ = 174