Question

Andile deposits R900 into a savings account paying 6.5% interest per year, compounded quarterly.After three and a half years he withdraws R1000 from the account and deposit it into a second account paying 11% simple interest per year.

1.How much is the total amount accrued in the first account two years after withdrawing R1000 the correct answer rounded to the nearest Rand is?

*R145

*R1105

*R605

*R128

2. The number of years after Andile deposited the 1000 ino the second account,accrued amount in the second account is 1605.The time (correct to two decimal places) the money was is invested for, is

*4.36 years

*5.50 years

*9.31 years

*4.53 Years

Answer 1

Solution 1

Here P = R900

Interest rate $r=6.5$ percent $r=\frac{6.5}{100}$

The interest is compounded quarterly, so $n=4$

And the number of years $t = 3.5$    (after withdrawing R 1000) 

Then using the formula $A = P (1 + \frac{r}{100n})^{nt}$

$A = 900 (1 + \frac{6.5}{100(4)})^{(4)(3.5t)}$

$A = R 1127.85$

 

So, the amount after 3.5 years is $A = R 1127.85$

And when he withdraws R 1000, the amount left is

$R 1127.85 – R 1000 = R 127.85$

 

Now after two years,

Investment Amount          $P = R 127.85$

Number of years                $t=2$

Therefore, using $A = P (1 + \frac{r}{100n})^{nt}$

$A = 127.85 (1 + \frac{6.5}{100\times4})^{4\times2}$

$A = R 145.44$

The correct answer is first option R 145 

Solution 2

$I = \frac {PRT} {100}$

Here,

$I$ is Interest, which is calculated as

$I = R1605 - R1000$

$I = R605$

Using the values in $I = \frac {PRT} {100}$ , we get

$605 = \frac {PRT} {100}$

$605\times 100=PRT\\$

$605\times 100=1000\times 11\times T$

$T=\frac{605\times 100}{(1000)(11)}\\$

$T=5.5$

Hence $T=5.5$  years

Therefore, the second option is correct.