NIMCET 2025 — Mathematics PYQ

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Question

NIMCET 2025 — Mathematics PYQ

NIMCET | Mathematics | 2025

The number of all even integers between $99$ and $999$ which are not multiples of $3$ and $5$ is:

Choose the correct answer:

Explanation

Solution

Step 1: Define the Range

Integers between $99$ and $999$ are $\{100, 101, \dots, 998\}$ .

We only need even integers:

$S = \{100, 102, 104, \dots, 998\}$

Total number of even integers ( $n$ ):

998 = 100 + (n-1)2

898 = (n-1)2 \implies 449 = n-1 \implies n = 450

Step 2: Multiples of 3 (within even set)

An even multiple of $3$ is a multiple of $6$ ( $2 \times 3$ ).

$A = \{102, 108, \dots, 996\}$

996 = 102 + (n_A - 1)6

894 = (n_A - 1)6 \implies 149 = n_A - 1 \implies n_A = 150

Step 3: Multiples of 5 (within even set)

An even multiple of $5$ is a multiple of $10$ ( $2 \times 5$ ).

$B = \{100, 110, \dots, 990\}$

990 = 100 + (n_B - 1)10

890 = (n_B - 1)10 \implies 89 = n_B - 1 \implies n_B = 90

Step 4: Multiples of both 3 and 5 (within even set)

An even multiple of $15$ ( $3 \times 5$ ) is a multiple of $30$ ( $2 \times 15$ ).

$A \cap B = \{120, 150, \dots, 990\}$

990 = 120 + (n_{A \cap B} - 1)30

870 = (n_{A \cap B} - 1)30 \implies 29 = n_{A \cap B} - 1 \implies n_{A \cap B} = 30

Step 5: Apply Inclusion-Exclusion Principle

Number of even integers that are multiples of $3$ or $5$ :

n(A \cup B) = n_A + n_B - n_{A \cap B}

n(A \cup B) = 150 + 90 - 30 = 210

Step 6: Required Count

Subtract those that are multiples of $3$ or $5$ from the total even integers:

\text{Required} = \text{Total even} - n(A \cup B)

\text{Required} = 450 - 210 = 240

Final Answer:

The number of such integers is 240. (Option A)