NIMCET 2009 — Reasoning PYQ

1. Identify the numbers containing the digit 6:

In the range $1$ to $100$, the digit $6$ appears in two places: the units place and the tens place.

• 6 in the units place:

  $6,16,26,36,46,56,66,76,86,96$ (Total: 10 times)

• 6 in the tens place:

  $60,61,62,63,64,65,66,67,68,69$ (Total: 10 times)

2. Calculate the variation (increase):

Each time a $6$ is replaced by a $9$, the value increases. The amount of increase depends on the position of the digit.

• Case 1: Units Place

  When $6$ in the units place becomes $9$, the increase per number is:

  $$9-6=3$$

  Since there are $10$ such numbers (including $66$), the total increase from the units place is:

  $$10\times 3=30$$

• Case 2: Tens Place

  When $6$ in the tens place (value $60$) becomes $9$ (value $90$), the increase per number is:

  $$90-60=30$$

  Since there are $10$ such numbers (including $66$), the total increase from the tens place is:

  $$10\times 30=300$$

3. Total Variation:

Add the increases from both positions:

Final Answer

The algebraic sum varies by 330.

Correct Option: (a)