+318 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?
+118673 I agree, this is a really confusing question.
Let Melville start with M and Danny start with D
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
