NIMCET 2008 — Reasoning PYQ

Step 1: Set up the AP Formula

Let:

• $n$ = number of rows.

• $a$ = number of children in the first (front) row.

• $d$ = common difference = $-3$ (since each subsequent row has 3 fewer children).

• $S_n$ = total number of children = $630$.

The formula for the sum of an AP is:

Step 2: Create the Equation

Substituting the known values:

For a configuration to be possible, $a$ must be a positive integer, and the number of children in the last row must also be greater than zero.

Step 3: Test the Options

• Case (a): $n=3$

  $$2a=\frac{1260}{3}+3(3)-3=420+9-3=426$$

  $a=213$ (Integer). Possible.

• Case (b): $n=4$

  $$2a=\frac{1260}{4}+3(4)-3=315+12-3=324$$

  $a=162$ (Integer). Possible.

• Case (c): $n=5$

  $$2a=\frac{1260}{5}+3(5)-3=252+15-3=264$$

  $a=132$ (Integer). Possible.

• Case (d): $n=6$

  $$2a=\frac{1260}{6}+3(6)-3=210+18-3=225$$

  $a=112.5$

  Since the number of children ($a$) cannot be a fraction ($112.5$), 6 rows are not possible.

Final Answer:

The number of rows that is not possible is 6.

Correct Option: (d)