Question

If (a + b) : (b + c) : (c + a) = 6 : 7 : 8 and \( a + b + c = 14 \), then the value of \( a \) is:

• A. 7
• B. 4
• C. \( \frac{11}{3} \)
• D. \( \frac{14}{3} \)

Hint:

When dealing with ratios involving combinations of variables, convert each part to a multiple of a common variable (like \( x \)), then use the given total to find individual values.

Answer 1

The Correct Option is D

Let us assume: \[ a + b = 6x,\quad b + c = 7x,\quad c + a = 8x \] Now, add all three equations: \[ (a + b) + (b + c) + (c + a) = 6x + 7x + 8x = 21x \Rightarrow 2(a + b + c) = 21x \Rightarrow a + b + c = \frac{21x}{2} \] But it is given that: \[ a + b + c = 14 \Rightarrow \frac{21x}{2} = 14 \Rightarrow x = \frac{28}{21} = \frac{4}{3} \] Now use \( x = \frac{4}{3} \) to find \( a \): We know: \[ a + b = 6x = 6 \times \frac{4}{3} = 8 b + c = 7x = \frac{28}{3} \Rightarrow c = \frac{28}{3} - b c + a = 8x = \frac{32}{3} \Rightarrow a = \frac{32}{3} - c \] Substitute \( c = \frac{28}{3} - b \) into equation for \( a \): \[ a = \frac{32}{3} - \left( \frac{28}{3} - b \right) = \frac{32}{3} - \frac{28}{3} + b = \frac{4}{3} + b \] Now, from \( a + b = 8 \), substitute \( a = \frac{4}{3} + b \): \[ \frac{4}{3} + b + b = 8 \Rightarrow \frac{4}{3} + 2b = 8 \Rightarrow 2b = 8 - \frac{4}{3} = \frac{20}{3} \Rightarrow b = \frac{10}{3} \Rightarrow a = \frac{4}{3} + \frac{10}{3} = \frac{14}{3} \]