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If $a, b, c$ are three positive integers such that $a$ and $b$ are in the ratio $3:4$ while $b$ and $c$ are in the ratio $2:1$, then which one of the following is a possible value of $(a + b+c)$?

1. $201$
2. $205$
3. $207$
4. $210$

Given that,

• $a:b=3:4$
• $b:c=2:1$

We can combine the ratios,

• $a:b=(3:4)\times 2 = 6:8$
• $b:c=(2:1)\times 4 = 8:4$

$\Rightarrow a:b:c =6:8:4=3:4:2$

Let   $a=3k,b=4k,c=2k$

$\therefore$ The value of $a+b+c = 3k+4k+2k= 9k$

The number $207$ is divisible by $9$. We can write $207 = 9 \cdot 23.$

$\textbf{PS:}\; \textsf{Divisibility by 9:-}$ The sum of digits of the number must be divisible by $9.$

Correct Answer $:\text{C}$

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