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Question

What is the value of \((2x + 4y)\), if mean of the following distribution is 42?

Class0-2020-4040-6060-8080-100Total
Frequency820x2y70

This question was previously asked in
RRB Group D 2025 Question Paper (18-Aug-2026) (Shift 1)
The correct answer is

86

Class midpoints: 10, 30, 50, 70, 90.

Total frequency: \(8 + 20 + x + 2 + y = 70\), so \(x + y = 40\).

\(\sum f_i x_i = 8(10) + 20(30) + 50x + 2(70) + 90y = 820 + 50x + 90y\).

Mean = \(\dfrac{\sum f_i x_i}{\sum f_i} = 42\), so \(820 + 50x + 90y = 42 \times 70 = 2940\), giving \(50x + 90y = 2120\).

Using \(x = 40 - y\): \(50(40 - y) + 90y = 2000 + 40y = 2120\), so \(y = 3\) and \(x = 37\).

Therefore \(2x + 4y = 74 + 12 = 86\).

Hence, the value of \((2x + 4y)\) is 86.

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