Let the mean and variance of four numbers 3, 7, x and y(x > y) be 5 and 10 respectively. Then the mean of four numbers 3 + 2x, 7 + 2y, x + y and x - y is ______________.

Solution

Maths sir · 5 steps

Turn the mean into x + y and the variance into x² + y², then solve for x and y separately.

1. Use the mean. The mean is 5, so 3 + 7 + x + y = 4 × 5 = 20. That gives x + y = 10.

   σ² = (Σ xᵢ²)/n - x̄² ⇒ 10 = (9 + 49 + x² + y²)/4 - 25

2. Use the variance. This gives 9 + 49 + x² + y² = 140, so x² + y² = 82. Then 2xy = (x+y)² - (x²+y²) = 100 - 82 = 18, so xy = 9.
3. Find x and y. Two numbers with sum 10 and product 9 are 9 and 1. Since x > y, x = 9 and y = 1.
4. New numbers. 3 + 2x = 21, 7 + 2y = 9, x + y = 10 and x - y = 8. Their sum is 21 + 9 + 10 + 8 = 48.
5. So. The mean is 48/4 = 12.

The common slip: 8. Swapping the values to x = 1, y = 9 ignores the condition x > y. The new numbers become 5, 25, 10 and -8, which gives a mean of 8.

The idea to keep. Mean gives you the sum and variance gives you the sum of squares. From those two, (x+y)² - (x²+y²) = 2xy pins down both unknowns. Then use any condition like x > y to fix which is which.

Answer from the NTA final answer key. Solution by PrepWiser.