Question 170949
1. lets call the number of nickels and dimes, n and d , respectively.
:
.05n+.1d=5.65............eq 1
.05(n+8)+.1(2d)=10.45....eq 2
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lets distribute and collect terms from eq 2
:
.05n+.4+.2d=10.45
.05n+.2d=10.05....revised eq 2
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now lets subtract eq 1 from revised eq 2 and eliminate the n terms .05n-.05n=0
we end up with .2d-.1d and 10.05-5.65
:
.2d-.1d=10.05-5.65
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.1d=4.4
:
{{{highlight(d=44)}}}dimes
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now plug in d's found value into either equation. I choose eq 1
:
.05n+.1(44)=5.65---->.05n=1.25
:
{{{highlight(n=25)}}}nickels
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2) d=rt.   we know the distance in both cases is 50. Lets call the local trains rate r and its time t.  So the express's rate would be 2r and its time t-1
:
50=rt..........eq 1
50=2r(t-1).....eq 2
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lets rewrite eq 1 as t=50/r and plug that value into eq 2
:
{{{50=2r((50/r)-1)}}}----->{{{50=2r((50-r)/r)}}}----
{{{50=2cross(r)(50-r)/cross(r)}}}
{{{50=100-2r}}}---->{{{-2r=-50}}}
{{{highlight(r=25)}}}MPH-speed of the local train
{{{highlight(2(25)=50)}}}MPH-speed of the express
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3)amount invested times rate of interest will equal interest earned. In this case we have two different amounts. amount at 9% we will call x so the amount invested at 8% will be 9000-x. Multiply each of those by there respective interest rates and that will equal 770
:
.09x+.08(9000-x)=770
:
.09x+720-.08x=770
:
.01x=50
:
{{{highlight(x=5000)}}}dollars invested at 9%
{{{highlight(9000-5000=4000)}}}dollars invested at 8%