Total cost of 30 TV sets = ? 300000
Total cost of 15 TV sets = 9000 x 15 = 135000
Total cost of remaining 15 TV sets = 300000 - 135000 = 165000

? Average cost of remaining TV sets = 165000/15 = ? 11000

Total age of 30 girls = 30 x 13 = 390 yr
Total age of 18 girls = 18 x 15 = 270 yr.
? Age of remaining 12 girls = 390 - 270 = 120 yr
? Required average = 120/12 = 10 yr

Let the numbers A, B, C, and D are N , N+1, N + 2 and N + 3, respectively.
According to the question,
[N + (N+1) + (N+2) + (N+3)] / 4 = 49.5
? 4N + 6 = 4 x 49.5
? 4N = 198 - 6 ? N = 192/4
? N = 48

The required product = B x D
= (N + 1) x (N + 3)
= (48 +1) (48 +3 )
= 49 x 51
= 2499

Overall height = (n₁a₁ + n₂a₂) / (n₁ + n₂)
n₁= 20,
a₁ = 5 ft 11 inches = 5 x 12 + 11 = 71 inches
n₂= 18,
a₂ = 6 ft 2 inches = 6 x 12 + 2 = 74 inches

? Overall height = [(20) (71) + (18) (74)] / (20+18)
= (1420 + 1332) / 38 = 72.42 inches

According to the fundamental to the fundamental formula
Average(A)= Sum (S) / Number(N)

From the question 60 = S / 13
? S = 60 x 13 = 780

Sum of first seven result = 59 x 7 = 413
Sum of last seven result = 61 x 7 = 427

? 7th result = (Sum of first seven result) + (Sum of last seven results) - (Sum of all the results)
= (413 + 427 - 780 )
= (840 - 780)
= 60

NoteSum of first seven when added to the sum of last seven result, then repetition of 7th result takes place.

Here n₁ = 19, a₁ = 74, n₂ = 38, a₁ = 63
Total average weight = (n₁a₁ + n₂ a₂) / (n₁ + n₂)
= [(19) (74) + (38) (63)] / (19 + 38)
= 3800/57 ? 67 kg

Here, n = 10, a = 60, b = 62
Number of runs scored in 11th inning
= n (b - a) + b
= 10 (62 - 60) + 62
= 20 + 62 = 82

Here n= 34, a = 42 and b = 42.4
Weight of the teacher.
= n(b -a) + b
= 34 ( 42.4 - 42) + 42.4
= 13.6 + 42.4 = 56.0 kg

Here, n = 21, a = 64, b = 65
Weight of teacher = n(b - a) + b
= 21( 65 - 64) + 65
= 21 + 65
= 86 kg

Sum of 14 girls and teacher age = 15 x 15 = 225
Without teacher's age sum of 14 girls age = 14 x 14 = 196
? Teacher's age = 225 -196 = 29 year