Question 707181: Caroline Case won $60,000 on a slot machine in Las Vegas. She invested part of the money at 2% simple interest and the rest at 3%. In one year, she earned a total of $1600 in interest. How much was invested at the 3% rate?
Please explain step by step how to solve thiS!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! let x = amount invested at 2%, y = amount invested at 3%
She invested $60,000 total, so
x+y = 60000
solve for y to get: y = 60000 - x
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If she invests x dollars at 2% interest, then she will earn 0.02x dollars in interest alone after one year. Similarly, she will earn 0.03y dollars if she invests y dollars at 3%
In total, she will earn 0.02x + 0.03y dollars in interest alone from both accounts combined.
We're told that "In one year, she earned a total of $1600 in interest", so
Total Interest Earned = 1600
0.02x + 0.03y = 1600
Now plug in y = 60000 - x and solve for x
0.02x + 0.03y = 1600
0.02x + 0.03(60000 - x) = 1600
0.02x + 0.03(60000) - 0.03x = 1600
0.02x + 1800 - 0.03x = 1600
-0.01x + 1800 = 1600
-0.01x = 1600 - 1800
-0.01x = -200
x = -200/(-0.01)
x = 20000
So $20,000 was invested at 2% simple interest
y = 60000 - x
y = 60000 - 20000
y = 40000
and $40,000 was invested at 3% simple interest
Check:
we can see that the two amounts $20,000 and $40,000 add to $60,000, so that checks out
$20,000 invested at 2% gives you 20000*0.02*1 = 400 dollars in interest alone
$40,000 invested at 3% gives you 40000*0.03*1 = 1200 dollars in interest alone
In total, you get 400+1200 = 1600 dollars in interest alone
So the answers are verified.
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