CA Foundation 2026 · Paper 3 · Quantitative AptitudeQ34 · 1 mark↻ Appears in 2 of 6 yearsOfficial key verified
If the mean 40, kk, 6k6k, 4k24k^2, 8k−4k28k - 4k^2 is 20, then the value of kk is ______.
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The k2k^2 terms cancel; solving 40+15k5=20\dfrac{40+15k}{5}=20 gives k=4k=4.

Step 1 — Write the mean equation

For the five observations 40, k, 6k, 4k2, 8k−4k240,\ k,\ 6k,\ 4k^2,\ 8k-4k^2:

Mean=40+k+6k+4k2+(8k−4k2)5=20\text{Mean} = \frac{40 + k + 6k + 4k^2 + (8k - 4k^2)}{5} = 20

Step 2 — Simplify the numerator

The k2k^2 terms cancel and the kk terms combine:

4k2−4k2=0,k+6k+8k=15k4k^2 - 4k^2 = 0,\qquad k + 6k + 8k = 15k

⇒sum=40+15k\Rightarrow \text{sum} = 40 + 15k

Step 3 — Solve for k

40+15k5=20  ⇒  40+15k=100  ⇒  15k=60  ⇒  k=4\frac{40 + 15k}{5} = 20 \;\Rightarrow\; 40 + 15k = 100 \;\Rightarrow\; 15k = 60 \;\Rightarrow\; k = 4

Why the other options are wrong: k=15,10,8k=15,10,8 give sums of 265,190,160265,190,160 and means of 53,38,3253,38,32 — none equal 2020. Only k=4k=4 satisfies the equation. …

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