## Question 3

Is 3a + 2b + 5c even if 0<a<b<c and a, b and c are integers?

- 9a+7c is not even
- a
^{3}*(c-1)^{2}is odd

### Correct Answer

D

### Solution

**Steps 1 & 2: Understand Question and Draw Inferences**

In expression (3a + 2b + 5c), 2b is an Even term, and so will not impact the even-odd nature of the expression.

So, the expression will be even if (3a + 5c) is even.

Now, 3 and 5 are odd numbers. So in the product 3*a and 5*c respectively, 3 and 5 play no role in the even-odd nature of the product.

So, (3a + 5c) will be even if (a + c) is even.

And, (3a + 5c) is odd if (a + c) is odd.

With this understanding, let’s analyse the given statements.

**Step 3: Analyze Statement 1**

9a+7c is not even

–> 9a+7c is odd

–> a + c is odd

Therefore, (3a + 5c) is odd.

Sufficient.

**Step 4: Analyze Statement 2**

a^{3}*(c-1)^{2} is odd

–> a*(c-1) is odd

–> a is odd and c – 1 is odd

–> a + c – 1 is even

–> a + c is odd

Therefore, (3a + 5c) is odd

Sufficient.

**Step 5: Analyze Both Statements Together (if needed)**

We get a unique answer in step 3 and step 4, so this step is not required

**Answer: Option (D)**

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