Question 145651
1) Let x = the amount invested at 9% simple interest. Then $15,000-x = the amount invested at 10% simple interest.
The interest earned on these two amounts can be expressed as, after changing the percentages to their decimal equivalents:
0.09x is the interest earned at 9%.
0.1($15,000-x) is the interest earned at 10%
The sum of these two amounts is $1,432,so we can set up an equation to find x:
0.09x+0.1($15,000-x) = $1,432 Simplify.
0.09x+1500-0.1x = 1432 Combine like-terms.
(0.09x-0.1x)+1500 = 1432
-0.01x+1500 = 1432 Subtract 1500 from both sides.
-0.01x = -68 Divide both sides by -0.01
x = 6800, so...
$6,800 was invested at 9% simple interest and $15,000-$6,800 = $8,200 was invested at 10% simple interest.
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2) Use the distance formula: d = rt where d = distance traveled, r = rate of speed, and t = time of travel.
Car 1:
{{{d[1] = r[1]*t}}} and...
Car 2:
{{{d[2] = r[2]*t}}}
For car 1, substitute {{{d[1] = 300}}} and for car 2 substitute {{{r[2] = r[1]-5}}} and {{{d[2] = 275}}}
The time of travel,t, is the same for both cars, so...
1) {{{t = 300/r[1]}}}
2) {{{t = 275/(r[1]-5)}}}, so we can write...
{{{300/r[1] = 275/(r[1]-5)}}} Simplify.
{{{300(r[1]-5) = 275(r[1])}}}
{{{300r[1]-1500 = 275r[1]}}} Add 1500 to both sides.
{{{300r[1] = 275r[1]+1500}}} Subtract {{{275r[1]}}} from both sides.
{{{25r[1] = 1500}}} Divide both sides by 25.
{{{r[1] = 60}}}
The first car was traveling at 60mph and the second car was traveling at (60-5) = 55mph.