If the 3-year moving averages of five values a, b, 20, c, d are 12, 14 & 16. Where a, b, c, d > 0, then match the following:
| List-I | List-II | ||
| (A) | a + d | (I) | 6 |
| (B) | c + d | (II) | 16 |
| (C) | c − a | (III) | 22 |
| (D) | b + a | (IV) | 28 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
The three 3-year moving averages of a, b, 20, c, d are:
\( \displaystyle \frac{a+b+20}{3}=12\ \Rightarrow\ a+b=16 \)
\( \displaystyle \frac{b+20+c}{3}=14\ \Rightarrow\ b+c=22 \)
\( \displaystyle \frac{20+c+d}{3}=16\ \Rightarrow\ c+d=28 \)
The four values cannot be found individually, but every item in List-I is a combination of these three sums.
(A) a + d
This matches (III) 22.
\( \displaystyle a+d=(a+b)+(c+d)-(b+c) \)
\( \displaystyle =16+28-22=22 \)
(B) c + d
This matches (IV) 28, directly from the third average.
(C) c − a
This matches (I) 6.
\( \displaystyle c-a=(b+c)-(a+b)=22-16=6 \)
(D) b + a
This matches (II) 16, directly from the first average.
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).