Word problem: distance and profit

[NotaMathPerson]

What I have tried for prob 30

Let $k+t=$total time taken by A
$t=$ total time taken by B and C

since $D_{A}=D_{B}=D_{C}$

$(k+t)c=(c+d)t$ solving for $t$

$t=\frac{ck}{d}$ hours

$\frac{c^2dk+cdk}{d}$ miles.

For 18, I don't know how to really start.

Please guide through correct solution for both problems. Thanks a lot!

 

[MarkFL]

30.) I would let $x$ be the number of hours after $B$ started walking that $C$ began walking. Since $A$ and $B$ walked the same distance $D$, we may state:

\(\displaystyle D=ct=(c+d)(t-k)\)

\(\displaystyle ct=ct-ck+dt-dk\)

\(\displaystyle t=\frac{k(c+d)}{d}\)

Since $A$ and $C$ also walked the same distance, we have:

\(\displaystyle D=ct=(c+2d)(t-(k+x))=ct-c(k+x)+2dt-2d(k+x)\)

\(\displaystyle t=\frac{(c+2d)(k+x)}{2d}\)

Equating the two expressions for $t$, we obtain:

\(\displaystyle \frac{k(c+d)}{d}=\frac{(c+2d)(k+x)}{2d}\)

\(\displaystyle 2k(c+d)=(c+2d)(k+x)\)

\(\displaystyle k+x=\frac{2k(c+d)}{c+2d}\)

\(\displaystyle x=\frac{2k(c+d)}{c+2d}-k=k\left(\frac{2(c+d)-(c+2d)}{c+2d)}\right)=\frac{ck}{c+2d}\)

\(\displaystyle D=\frac{ck(c+d)}{d}\)

 

[MarkFL]

18.) Let's let $N$ be the total number of apples bought, where:

\(\displaystyle N_3=\) the number bought at 3 per cent.

\(\displaystyle N_4=\frac{5}{6}N_3=\) the number bought at 4 per cent.

Hence:

\(\displaystyle N=N_3+N_4=\frac{11}{6}N_3\)

So, the total amount $A$ paid (in cents) is:

\(\displaystyle A=\frac{1}{3}N_3+\frac{1}{4}N_4=\frac{1}{3}N_3+\frac{5}{24}N_3=\frac{13}{24}N_3=\frac{13}{44}N\)

The amount received from selling $S$ is then:

\(\displaystyle S=\frac{3}{8}N=\frac{13}{44}N+\frac{7}{2}\)

And so we find:

\(\displaystyle N=44\)