1) Dick takes a loan of $8,000 to buy a used truck at the rate of 9 % simple Interest.Calculate the annual interest to be paid for the loan amount.

From the details given in the problem Principle = P = $8,000 and R = 9% or 0.09 expressed as a decimal.

As the annual Interest is to be calculated, the time period T =1.

Plugging these values in the simple Interest formula,

I = P x T x R

= 8,000 x 1 x 0.09

= 720.00

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2) Steve invested $ 10,000 in a savings bank account that earned 2% simple interest. Find the interest earned if the amount was kept in the bank for 4 years.

Principle P = $ 10,000 Time Period T =4 years and Rate of Interest = 2% = 0.02

Plugging these values in the simple Interest formula,

I = PX T X R

= 10,000 X 4 x 0.02

= $ 800

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3) Ryan bought $ 15,000 from a bank to buy a car at 10% simple Interest. If he paid $ 9,000 as interest while clearing the loan, find the time for which the loan was given.

T = I/(PR)

= 9000/(15,000 x 0.10)

= 6 years.

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4) In how much time will the simple interest on $3,500 at the rate of 9% p.a be the same as simple interest on $4,000 at 10.5% p.a for 4 years?

S.I on $4,000 at rate 10.5% = 10.5/100 = 0.105 for 4 years

S.I = ( P x R x T) /100

= 4000 x 0.105 x 4

S.I = $ 1,680

The interest of $1,680 is the same as that on $3,500 at 9% p.a for suppose 't' years.

S.I x 100

Time = t = ------------

P x R

1680 x 100

Time = t = ------------

3500 x 9

168,000

Time = t = ------------

31,500

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