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Part of $10,000 was invested at 6% interest and the rest at 7.5%.?
If the annual income was $712.50, how much was invested at each rate.
Could someone show me how to solve?
Thank you.
2 Answers
- KevinMLv 71 decade agoFavorite Answer
The answer above is correct. But my favorite way to do problems like this is to say that you earned 6% on the entire $10,000, and an extra 1.5% on only that part invested at 7.5%.
You earn 6% on $10,000, making $600 per year. That leaves $112.50 left.
How much do you have to invest at 1.5% to make $112.50? $112.50 / 1.5% = $7500
So it was $2500 at 6% and $7500 at 7.5%.
- RaymondLv 71 decade ago
Sum invested at 6% = x
income from this portion = 6% of x = (6/100)x
The rest (10,000 - x) was invested at 7.5%
Income from this portion: 7.5% of (10,000 - x) = (7.5/100)(10,000-x)
Total income = (6/100)x + (7.5/100)(10,000-x) = 712.50
Solve for x
