Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question:

If $$(a+b):(b+c):(c+a)=6:7:8$$ and $a+b+c=14$, then find the value of $c$.


My solution:

  1. $$\frac{(a+b)(c+a)}{(b+c)}=\frac{(6)(8)}{7}$$$$\Rightarrow \frac{ac + a^2 + bc + ba}{b+c} = \frac{48}{7}$$$$\Rightarrow \frac{a(b+c)+a^2+bc}{b+c}=\frac{48}{7}$$$$\Rightarrow ????$$

  2. $$\Rightarrow a+b=6x \space\space \text{and} \space \space b+c=7x$$$$\Rightarrow b=6x-a\space\space\space\text{and}\space\space\space b=7x-c$$$$\Rightarrow \text{solving we get}\space\space x = c-a$$$$\Rightarrow ????$$


My query:

I am totally stuck on this problem. Please help.

Thanks a lot!

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Let $$x=b+c,$$ $$y=a+c,$$ $$z=a+b.$$ Then, $x+y+z=2(a+b+c)=28$ and $z:x:y=6:7:8$.

By substituting $z=6k$, $x=7k$, $y=8k$ into $x+y+z=28$, we have $21 k=28$, $k=\frac{4}{3}$, from where we get: $$z=6k=8,$$ $$x=7k=\frac{28}{3},$$ $$y=8k=\frac{32}{3}.$$ Finally, $$a=a+b+c-x=\frac{14}{3},$$ $$b=a+b+c-y=\frac{10}{3},$$ $$c=a+b+c-z=6.$$

share|improve this answer
    
Thanks a lot. Clear and simple solution. –  Gaurang Tandon Feb 15 at 16:56

$$\frac{a+b}6=\frac{b+c}7=\frac{c+a}8=\frac{-(a+b)+(b+c)+(c+a)}{-6+7+8}=\frac{2c}9$$

Similarly, each ratio is equal to $$\frac{a+b+(b+c)+(c+a)}{6+7+8}=\frac{2(a+b+c)}{21}$$

$$\implies \frac{2c}9=\frac{2(a+b+c)}{21}$$

Now, we have $a+b+c=14$

share|improve this answer
    
@GaurangTandon, how about this? –  lab bhattacharjee Feb 16 at 15:33
    
I was wondering how did you get $$\frac{-(a+b)+(b+c)+(c+a)}{-6+7+8}=\frac{2c}9$$. Please explain. Thanks. –  Gaurang Tandon Feb 16 at 16:46
    
@GaurangTandon, Ratio & Proportion formula says : $$\frac aA=\frac bB=\frac cC=\frac{pa+qb+rc}{pA+qB+rC}$$ etc. –  lab bhattacharjee Feb 17 at 5:52
    
Oh, I see. Thank you :) Got to know a new formula ! –  Gaurang Tandon Feb 17 at 5:59
    
@GaurangTandon, please derive this –  lab bhattacharjee Feb 17 at 6:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.