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NCHMCT JEE 2024 Solved Paper

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Question : 26 of 200

Marks: +1, -0

If

(\mathrm{a}+\mathrm{b}):(\mathrm{b}+\mathrm{c}):(\mathrm{a}+\mathrm{c})=6: 7: 8

\mathrm{a}+\mathrm{b}+\mathrm{c}=14

, then the value of a is :

$\frac{2}{7}$
4
$\frac{11}{3}$
$\frac{14}{3}$

Solution:

Concept:

Use a common multiplier to convert the ratio into equations and solve for

a

.

Explanation:

Let the common multiplier be

k

. Then:

a+b=6k

,

\;b+c=7k

,

\;a+c=8k

Adding all three:

2(a+b+c)=6k+7k+8k=21k

Given

a+b+c=14

, so:

2(14)=21k

28=21k

k=\frac{4}{3}

Now,

2a=(a+b)+(a+c)-(b+c)

2a=6k+8k-7k=7k

a=\frac{7k}{2}=\frac{7}{2}\times \frac{4}{3}=\frac{14}{3}

Answer:

\frac{14}{3}

So the correct option is D.

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