Sidewinder, Inc., has sales of $658,000, costs of $334,000, depreciation expense of $79,000, interest expense of $44,000, and a
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Answer:
1276.5
Explanation:
Today, Eric deposited an amount(PO) that is = 1000
The annual nominal Interest rate for the first 7 years = i
This is convertible semiannually.
As such,
The semi-annual interest rate for first 7 years =
The number of semi-annual periods in 7 years (k) = 2*7 = 14
The magnitude of the deposit at end of year 7( P7),
= P7 == P0 * (1 + i/2)14 ------ equation 1
On the other hand,
The nominal interest rate after 7 years = 2i which is convertible quarterly
The effective quarterly interest rate = =
The number of quarters between the end of year 7 and year 10.5 = 4* (10.5 - 7) = 14
The amount of deposit at the end of year 10.5 year = P10.5
=
* (1 +
---equation 2
We are given the following:
=1980
We shall thus substitute this value into the equation as
1980 = *( 1 + i/2)14
Further, we have worked out that
P7 == P0 * (1- i/2)14 *(1
We shall therefore substitute equation 1 into equation 2 as follows
1980 = P0 * (1+ i/2)14 *( 1 + i/2)14
1980 = 1000* (1+ i/2)14 *( 1 + i/2)14
This can also be rewritten as
1000* (1+ i/2)14 *( 1 + i/2)14 =1980
(1+ i/2)14 *( 1 + i/2)14 =1980/1000
(1+ i/2)28 =1.98
(1+ i/2) = 1.981/28
(1+ i/2) = 1.0247
i/2 = 1.0247 -1
i/2 = 0.0247
1 = 0.0247 *2 =0.0494
At the end of year 5, the amount has accumulated to an amount = X
X = Po * ( 1 + i/2 )10
= 1000 * ( 1 + 0.0494/2)10
= 1000 * ( 1 + 0.0247)10
= 1000 + ( 1.0247)10
= 1000* 1.276467
= 1276.5