GATE 2024 IN Q.28: solution and answer

GATE 2024 · INQ.28 · 1 mark · NAT

The resistance of a 20 k $Ω$ resistor is measured six consecutive times using an LCR meter. The first five readings are 19 k $Ω$ , 18 k $Ω$ , 23 k $Ω$ , 21 k $Ω$ and 17 k $Ω$ . If the mean of the measurements and the true value are equal, the last reading is ________ k $Ω$ (rounded off to nearest integer).

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Answer and solution

ANSWER

The last reading is 22 kΩ.

Official answer: 22Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

The true value is $R_{t} = 20 k Ω$ . The question says the mean of all six readings equals it: $\overset{ˉ}{R} = \frac{1}{6} \sum_{i = 1}^{6} R_{i} = 20 k Ω$ .
So the six readings must add up to $\sum_{i = 1}^{6} R_{i} = 6 \times 20 = 120 k Ω$ .
Add the first five readings: $19 + 18 + 23 + 21 + 17 = 98 k Ω$ .
The sixth reading is what remains: $R_{6} = 120 - 98 = 22 k Ω$ This is already a whole number, so rounding does not change it.
Check: the deviations of the six readings from $20 k Ω$ are $- 1, - 2, + 3, + 1, - 3, + 2$ . They add up to zero, so the mean of the six readings is exactly $20 k Ω$ , as the question requires.

KEY IDEA

If the arithmetic mean equals the true value, the deviations of the readings from the true value must add up to zero. Equivalently, the sum of all n readings must be n times the true value, and the missing reading is whatever is left over.

WATCH OUT FOR

Averaging only the five known readings (98/5 = 19.6 kΩ) and reporting that, or dividing by 5 instead of 6 when setting up the mean. The condition covers all six readings, so the total must be 6 × 20 = 120 kΩ.

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GATE 2024 IN Q.28: solution and answer

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