Answer :
Firsly, let's work out an equation for the situation where both send e-mails in 15 minutes.
If x is how long it takes for the older computer to send one e-mail and y is how long it takes for the new computer to send the e-mail, then:
In 1 minute, the older comoputer can send 1/x fraction of the e-mail and the new computer 1/y fractio of the e-mail.
In 15 minutes, the older computer sent 15/x fraction of the e-mail and the new computer sent 15/y fractio of it. In total, they sent the whole e-mail, so, the sum of these fractions has to be 1:
[tex]\frac{15}{x}+\frac{15}{y}=1[/tex]Since the older computer take twice as long as the new computer, then x is equal to two times y:
[tex]x=2y[/tex]Substituteing this, we have:
[tex]\begin{gathered} \frac{15}{2y}+\frac{15}{y}=1 \\ \frac{15}{2y}+\frac{30}{2y}=1 \\ \frac{45}{2y}=1 \\ 45=2y \\ y=\frac{45}{2} \\ y=22.5 \end{gathered}[/tex]And now, for x, we have:
[tex]\begin{gathered} x=2y \\ x=2\cdot22.5 \\ x=45 \end{gathered}[/tex]Since y is the time for the new computer, it will take 22.5 minutes for the newer computer to send the e-mail on its own.