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Yuki888 [10]
3 years ago
9

A church building fund has invested some money in two ways: part of the money at 7% interest and four times as much at 11%. Find

the amount invested at each rate if the total annual income from interest is 7650.
Mathematics
1 answer:
aliya0001 [1]3 years ago
4 0

Answer: he invested $15000 in part 1 and $60000 in part 2

Step-by-step explanation:

Let x represent the amount of money invested in part one.

Let 4x represent the amount of money invested in part 2 .

Total amount of money invested in part 1 and part 2 is

x + 4x = 5x

The formula for simple interest is expressed as

I = PRT/100

Where

P is the principal or initial amount.

T is the duration in years

R is the number rate.

For part 1

R = 7%

T = 1 year

P = x

I = (x × 7 × 1)/100 = 0.07x

For part 2,

R = 11%

T = 1 year

P = 4x

I = (4x × 11 × 1)/100 = 0.44x

if the total annual income from interest is 7650. This means that

0.44x + 0.07x = 7650 - - - - - - - -2

0.51x = 7650

x = 7650/0.51

x = 15000

Amount invested in part 1 is $15000

Amount invested in part 2 is 4×15000 = $60000

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If the length of a rectangle is 3m longer than its width, then:
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To find out let's substitute the numbers:
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154=154
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