How to construct linear equations Q2b) Q11 and Q13?

I've found the formula for a which is $r = \frac{v + \mu}{m + 1}$. But I could construct equation for c without getting a bizarre answer. For questions 11, I got the equation 10x+5(41-2x)=205 but x=0 in this case. And lucky last questions 13 I did x=Apollo y=aph and z=ad Z=x+7200 y=x-4000 So we have x+x+7200-400=303200 And go x=98800, complete different to the answer. Can't someone please correct my mistake? And I will love you forever:)

Feb 6, 2017

2(b) $m = \frac{v}{u}$
(11) We have $10$ pieces of length $12 m$ and $5$ pieces of length $17 m$.
(13) Mr. Aphrodite earn $96000, Mr. Apollo earns $100000 and Mr. Adonis earns $107200 Explanation: 2(b) As $\frac{1}{v} + \frac{1}{u} = \frac{2}{r}$, we have $\frac{1}{v} - \frac{1}{r} = \frac{1}{r} - \frac{1}{u}$or $\frac{r - v}{v r} = \frac{u - r}{r u}$i.e. $\frac{r - v}{u - r} = \frac{v r}{r u} = \frac{v}{u}$or $m = \frac{v - r}{r - u} = \frac{v}{u}$(11) let there be $10$pieces of length $x$and $5$pieces of length $y$. As total length is $205 m$, we have $10 x + 5 y = 205$or dividing by $5$$2 x + y = 41$..................(A) Further, as three pieces of size $x$exceed two pieces of size $y$by $2$, we have $3 x - 2 y = 2$..................(B) Now multiplying (A) by $2$and adding to (B), we get $4 x + 2 y + 3 x - 2 y = 82 + 2$i.e. $7 x = 84$and $x = 12$and putting this in (A) we get $2 \times 12 + y = 41$or 24+y=41 i.e. $y = 41 - 24 = 17$Hence, we have $10$pieces of length $12$and $5$pieces of length $17$. (13) Let Mr. Aphrodite earn $x$, then Mr. Apollo earns x+$4000 and Mr. Adonis earns (x+$4000)+$7200 or x+$11200 As their total earnings are $303200, we have

$x + x + 4000 + x + 11200 = 303200$

or $3 x + 15200 = 303200$

or $3 x = 303200 - 15200 = 28800$

and $x = \frac{288000}{3} = 96000$

Hence, Mr. Aphrodite earn $96000, then Mr. Apollo earns $96000+$4000=$100000 and Mr. Adonis earns $100000+$7200 or \$107200#.