Averages Quiz Set 020

Let the present ages of the three friends be a, b and c. We are given ${(b - 5) + (c - 5)}/2$ = 10. Which gives b + c = 30. We are also given ${(a - 3) + (b - 3) + (c - 3)}/3$ = 27, which gives a + b + c = 3 * 27 + 9 = 90. Putting b + c here we get a = 90 - (b + c) = 90 - 30 = 60 years.

The runs already scored = scoring rate × overs = $4.3 × 20$ = 86. Required number of runs = 326 - 86 = 240 in 50 - 20 = 30 overs. Required rate = 240/30 = 8.

We have three equations (p + q)/2 = average of pq, (q + r)/2 = average of qr and (r + p)/2 = average of rp. Adding these three we get p + q + r = (average of pq + average of qr + average of rp) = (28 + 48 + 72) = 148. So p = 148 - (q + r) = 148 - (2 × average of q and r) = 148 - 2 × 48 = 52.

Let the total score of the class before grace marks be x. Then, by average formula $52 = x/24$, which gives x = $52 × 24 = 1248$. When grace marks = 8 are added for each of the 24 students, the new total becomes $1248 + 24 × 8 = 1440$, the new average becomes $1440/24 = 60$. TIP: As a shortcut, the new average = old average + grace marks.

Let the required average be x. Then $28x - (24 + 58) = 26 × (x - 1).$ which gives $28x - 82 = 26 × (x - 1).$ Solving for x we get x = 28.