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3 years ago

Two investments total $1000. If the combined annual interest is $68, how much was invested at 8% and how much at 5%?

Mathematics

1 answer:

yulyashka [42]3 years ago

5 0

Answer:

600 was invested at 8% and $400 at 5%

Step-by-step explanation:

Let x represent the amount of $ invested at 8% and y the amount at 5%.

Then x + y = $1000 is the total investment. Therefore, y = 10000 - x

Interest earned is then:

0.08x + 0.05(y) = 68, or (after substitution)

0.08x + 0.05(1000) - 0.05x = 68

0.08x + 50 - 0.05x = 68, or

0.03x = 18. Dividing both sides by 0.03, we find that x = 18/0.03 = 600.

If x = $600, then y = $1000 - $600 = $400.

$600 was invested at 8% and $400 at 5%.

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2 years ago

A carpenter wants to put four shelves on a 8-foot wall(=b) so that the five spaces created decrease by 7 inches(=a) as we move u

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⇒ Height of wall = 8 × 12 inches

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