A group of 630 children is seated in rows for a group photo session. Each row contains three less children than the row in front of it. Which one of…

Correct Option (D)

Let 'x' be the number of children in the first row. Since each subsequent row has three fewer children than the row in front, the number of children in 'n' rows forms an arithmetic progression with the first term 'x' and a common difference 'd = -3'.

The sum of 'n' terms of an arithmetic progression is given by the formula: S_n = n/2 * [2x + (n-1)d].

Given that the total number of children is 630, we have: n/2 * [2x + (n-1)(-3)] = 630.

Simplifying the equation: n * [2x - 3(n-1)] = 1260.

For 'n = 6':

• 6 * [2x - 3(6-1)] = 1260
• 6 * [2x - 3(5)] = 1260
• 6 * [2x - 15] = 1260
• 2x - 15 = 1260 / 6
• 2x - 15 = 210
• 2x = 225
• x = 112.5

Since the number of children in a row must be an integer, x = 112.5 is not a valid solution. Therefore, 6 rows is not a possible arrangement.

Incorrect Options:

For the number of rows to be possible, 'x' (number of children in the first row) must be a positive integer, and the number of children in the last row (x - 3(n-1)) must also be a positive integer.

For 'n = 3':

• 3 * [2x - 3(3-1)] = 1260
• 3 * [2x - 6] = 1260
• 2x - 6 = 420
• 2x = 426
• x = 213

This is an integer. The number of children in the last row (3rd row) would be 213 - 3(2) = 207, which is positive. Thus, 3 rows is possible.

For 'n = 4':

• 4 * [2x - 3(4-1)] = 1260
• 4 * [2x - 9] = 1260
• 2x - 9 = 315
• 2x = 324
• x = 162

This is an integer. The number of children in the last row (4th row) would be 162 - 3(3) = 153, which is positive. Thus, 4 rows is possible.

For 'n = 5':

• 5 * [2x - 3(5-1)] = 1260
• 5 * [2x - 12] = 1260
• 2x - 12 = 252
• 2x = 264
• x = 132

This is an integer. The number of children in the last row (5th row) would be 132 - 3(4) = 120, which is positive. Thus, 5 rows is possible.