Question

The average number of copies of a book sold per day by a shopkeeper is 60 in the initial seven days and 63 in the initial eight days, after the book launch. On the ninth day, she sells 11 copies less than the eighth day, and the average number of copies sold per day from the second day to the ninth day becomes 66. The number of copies sold on the first day of the book launch is:

Hint:

When dealing with averages over overlapping time intervals, convert each average to a total sum. Then, use differences of these sums to find individual day values and set up equations to solve for the unknowns.

Answer 1

Correct Answer: 59

Approach: Every "average" here is really a hidden total. Convert each average into a sum of copies, then the single overlapping day (or the leftover day) pops out by simple subtraction.

Step 1: First 7 days average \(60\), so the 7-day total is \[ S_7 = 7 \times 60 = 420. \]

Step 2: First 8 days average \(63\), so the 8-day total is \[ S_8 = 8 \times 63 = 504. \] The 8th-day sale is the difference: \[ x_8 = S_8 - S_7 = 504 - 420 = 84. \]

Step 3: The 9th day is \(11\) less than the 8th day: \[ x_9 = 84 - 11 = 73. \]

Step 4: Days 2 to 9 are 8 days with average \(66\), so their total is \[ S_{2\text{ to }9} = 8 \times 66 = 528. \]

Step 5: Total for days 1 to 9 is \[ S_9 = S_8 + x_9 = 504 + 73 = 577. \] But the same 9-day total is also day 1 plus days 2 to 9: \[ S_9 = x_1 + 528. \] So \[ x_1 = 577 - 528 = 49. \]

Final answer: \( x_1 = 49 \) copies on the first day.

Answer 2

Method: Let the sales on days \(1\) to \(9\) be \(d_1,\dots,d_9\). The sum of the first 7 days \(=60\times7=420\), and the sum of the first 8 days \(=63\times8=504\), so \(d_8 = 504-420=84\).

Day 9 sales are \(11\) less than day 8, so \(d_9 = 84-11=73\).

The sum of days 2 to 9 (8 days) \(=66\times8=528\). But days 2 to 9 equal (sum of days 1 to 9) minus \(d_1\), and the sum of days 1 to 9 \(= 504 + 73 = 577\).

So \(577 - d_1 = 528\), giving \(d_1 = 577-528\).

\[ \boxed{d_1 = 49 \text{ copies}} \]