P, Q and R are three consecutive odd numbers in ascending order. If the value of three times P is $3$ less than two times R. Find the value of R.

5

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7

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9

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11

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Solution

The correct option is A $9$
Let the three consecutive odd numbers be $P = n - 2$ , $Q = n$ and $R = n + 2$ .

It is given that the value of three times $P$ is $3$ less than two times $R$ , therefore, we have:

$3 P = 2 R - 3 \Rightarrow 3 (n - 2) = 2 (n + 2) - 3 \Rightarrow 3 n - 6 = 2 n + 4 - 3 \Rightarrow 3 n - 6 = 2 n + 1 \Rightarrow 3 n - 2 n = 1 + 6 \Rightarrow n = 7$

Therefore, $n = 7$ and the other two numbers are:

$P = n - 2 = 7 - 2 = 5$ and

$R = n + 2 = 7 + 2 = 9$

Hence, $R = 9$ .

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