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Melvin and Danny had some water guns. Melvin gave$${2 \over 5}{}{}$$of his water guns to Danny. Danny then gave $${1\over 2}{}{}$$ of his water guns to Melvin. As a result, Melvin had 3 times as many water guns as Danny. If Melvin gave Danny 28 more water guns than what Danny gave to Melvin, how many water guns did Danny have at first?

Aug 20, 2021

#1
+115733
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I agree, this is a really confusing question.

Melville give danny 2/5 of his so

afer that                      Melville has 3/5M                  and      Danny has    D+2/5M

Now Danny gives Melville half of what he has

so now        Melville has          3/5M+1/2(D+2/5M)   and      Danny has    1/2(D+2/5M)

Now Melville has 3 times more than Danny

$$\frac{3M}{5}+\frac{1}{2}(D+\frac{2M}{5})=3(\frac{1}{2}(D+\frac{2M}{5})\\ 10*[\frac{3M}{5}+\frac{1}{2}(D+\frac{2M}{5})]=10*[3(\frac{1}{2}(D+\frac{2M}{5}))]\\ 6M+5D+2M=15D+6M\\ 8M+5D=15D+6M\\ M=5D$$

Melville gave Danny    $$\frac{2M}{5}=\frac{2*5D}{5}=2D$$

Danny gave Melville    $$0.5(D+\frac{2M}{5})=\frac{5D+2M}{10}=\frac{5D+2*5D}{10}=\frac{15D}{10}=\frac{3D}{2}$$

We are told that Melvin gave Danny 28 more water guns than what Danny gave to Melvin (though it is worded badly)

$$\frac{3D}{2}+28=2D\\ 3D+56=4D\\ D=56$$

So Danny started with 56 water guns.

I leave it to you to check.

LaTex

\frac{3M}{5}+\frac{1}{2}(D+\frac{2M}{5})=3(\frac{1}{2}(D+\frac{2M}{5})\\

10*[\frac{3M}{5}+\frac{1}{2}(D+\frac{2M}{5})]=10*[3(\frac{1}{2}(D+\frac{2M}{5}))]\\
6M+5D+2M=15D+6M\\
8M+5D=15D+6M\\
M=5D

Aug 22, 2021
#2
0

You spelled Melvin wrong

Aug 26, 2021