Question

The average of the first 7 numbers in a series is 60. When the 8th number is added, the average of the first 8 numbers becomes 63. The 9th number is 11 more than the 8th number. It is also given that the average of the 2nd to the 9th numbers is 66. Find the value of the 1st number in the series.

• A. 68
• B. 70
• C. 71
• D. 75

Hint:

When two groups of the same size overlap almost completely, like first 8 and 2nd to 9th here, which differ only by one term leaving and one term entering, the difference between their two sums directly gives the difference between the leaving and entering terms, without needing every individual value.

Answer 1

The Correct Option is C

Approach: Convert every average into a sum (sum = average × count). The first number is isolated because the two overlapping 8-term sums (first-8 and 2nd-to-9th) share the same middle block \(N_2\ldots N_8\); subtracting peels \(N_1\) out.

Step 1: Turn averages into sums.
First 7: \(S_7 = 7\times60 = 420\).
First 8: \(S_8 = 8\times63 = 504\).
2nd to 9th: \(S_{2\text{-}9} = 8\times66 = 528\).

Step 2: Find \(N_8\) and \(N_9\).
\(N_8 = S_8 - S_7 = 504 - 420 = 84\).
\(N_9 = N_8 + 11 = 95\).

Step 3: Isolate the shared block \(N_2+\cdots+N_8\).
From \(S_{2\text{-}9}\): \((N_2+\cdots+N_8) + N_9 = 528\), so \(N_2+\cdots+N_8 = 528 - 95 = 433\).

Step 4: Extract \(N_1\).
From \(S_8\): \(N_1 + (N_2+\cdots+N_8) = 504\), so
\[ N_1 = 504 - 433 = 71. \]

Final answer: 71 — option (C).

Answer 2

Approach (deviation from an assumed mean): Measure every number as a deviation from 60 (the first average). Since the first 7 numbers average 60, their deviations from 60 already sum to zero, leaving only the new numbers' deviations to track.

Step 1: Let \( d_i = N_i - 60 \). Since the first 7 average 60, \( d_1+\cdots+d_7 = 0 \).
Step 2: First 8 average 63, so \( d_1+\cdots+d_8 = 8(63-60) = 24 \). Since the first-7 deviations sum to 0, \( d_8 = 24 \), i.e. \( N_8 = 84 \). Then \( N_9 = N_8+11 = 95 \), so \( d_9 = 35 \).
Step 3: The 2nd-to-9th average is 66, so \( d_2+\cdots+d_9 = 8(66-60) = 48 \). Subtracting \( d_9=35 \) gives \( d_2+\cdots+d_8 = 13 \).
Step 4: From Step 2, \( d_1 = 24 - (d_2+\cdots+d_8) = 24-13 = 11 \), so
\[ N_1 = 60+11 = 71. \]

Final answer: 71, option (C).

Answer 3

Concept:

• Convert every average into a sum using sum = average times count.
• Notice that "first 8 numbers" and "2nd to 9th numbers" are both 8 terms long and overlap almost completely, the only difference is that the first group includes $N_1$ but not $N_9$, while the second includes $N_9$ but not $N_1$.
• Because of this near-total overlap, the difference between the two group sums directly equals $N_9-N_1$, without needing to know any of the other seven shared terms individually.

Step 1: Convert the given averages into sums.
Sum of first $7$: $S_7=7\times60=420$. Sum of first $8$: $S_8=8\times63=504$. Sum of 2nd to 9th: $S_{2\text{-}9}=8\times66=528$.

Step 2: Find the 8th and 9th numbers.
$N_8=S_8-S_7=504-420=84$. The 9th number is $11$ more than the 8th: $N_9=84+11=95$.

Step 3: Use the overlap between the two 8-term groups.
The group "2nd to 9th" is obtained from "first 8" by removing $N_1$ and adding $N_9$, so
$S_{2\text{-}9}=S_8-N_1+N_9$, which gives $S_{2\text{-}9}-S_8=N_9-N_1$.
Substituting: $528-504=N_9-N_1 \Rightarrow 24=95-N_1$.

Step 4: Solve for $N_1$.
$N_1=95-24=71$.

Final Answer: $N_1=71$