# Question 8994e

Jan 13, 2018

$= K 68.25$
$= K 3.25$

#### Explanation:

This can be solved using the formula:
$I = \text{Pi} n$
where:
$I = \text{Interest}$
$P = \text{Principal}$
$i = \text{interest rate}$
$n = \text{no. of years}$

Plug in data provided to find the Interest $\left(I\right)$ after after 3 years .
$I = \text{Pin}$
$I = K 650 \times \frac{0.035}{\cancel{y r}} \times 3 \cancel{y r}$
$I = K 68.25$

Thus, the interest $\left(I\right)$ per year is $K 22.75$

If the interest rate is increased by 3.75% after the first year; the Interest $\left(I\right)$ earned for the remaining 2 years can be computed as;
$I = \text{Pin}$
$I = K 650 \times \frac{0.0375}{\cancel{y r}} \times 2 \cancel{y r}$
$I = K 48.75$

Thus, the total Interest earned before and after the increase can be computed as shown;
I_(t)=I_(fy)+I_(s&ty)
${I}_{t} = K 22.75 + K 48.75$
${I}_{t} = K 71.50$

Therefore, an increase of the Interest earned is
=I_(3.50&3.75%)-I_(3.50%)
$= K 71.50 - L 68.25$
$= K 3.25$

Jan 13, 2018

(a) Interest earned after 3 years @ 3.5% Interest = $\textcolor{p u r p \le}{K 68.25}$

(b) Increase in Interest due to rise in interest to 3.75% after one year = $\textcolor{g r e e n}{71.5 - 68.25 = K 3.25}$

#### Explanation:

$I = \frac{P N R}{100}$

Where I = Simple Interest earned, P = Principal, N = No. of years & R = Rate of Interest in percentage.

(a) Given P = K 560, N = 3 years, R = 3.5%

$I = \frac{650 \cdot 3 \cdot 3.5}{100} = \textcolor{red}{K 68.25}$

(b) New Rate of Interest R_1 = 3.75% & ${N}_{1} = 2 y e a r s$

Interest earned ${I}_{1} = \frac{650 \cdot 2 \cdot 3.75}{100} = \textcolor{b l u e}{K 48.75}$

Old rate of interest R = 3.5%, Period N_1 = 1 year

${I}_{o} = \frac{650 \cdot 3.5 \cdot 1}{100} = \textcolor{b l u e}{22.75}$

Total Interest = 48.75 + 22.75 = K 71.5

Increase in interest with the revised interest rate after 1 year $\textcolor{g r e e n}{{I}_{i} = 71.5 - 68.25 = K 3.25}$

Jan 13, 2018

$\text{Part a: } K 68.25$

$\text{Part b: } K 3.25$

#### Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

The symbol % cab be thought of as representing a fraction but one that is worth $\frac{1}{100}$. So we have:

3.5% ->3.5xx% ->3.5xx1/100=3.5/100

A direct comparison is the the metric measure of centimetres. The part word centi signifies $\frac{1}{100}$. So we have $\frac{1}{100} \times 1 \text{ metre}$

It is usual to translate the word of (as in $\frac{1}{100}$ of 1 metre) to mean multiply. So 3.5% of 'something' is stating 3.5%xx"'something'"

The wording 'per anum'$\to$'per' means for each of 1 and 'anum' means 'year'. So 'peranum' is stating 'for 1 year'
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$

color(brown)(ul("For EACH year we have:"))" simple interest"

3.5%" of "K650 color(white)("d") -> color(white)("d") 3.5/100xxK650 = color(cadetblue)(K22.75) color(red)(larr"Checked")

3.75%" of "K650 color(white)("d") ->color(white)("d") 3.75/100xxK650= color(sienna)(K24.375) color(red)(larr"Checked")

..............................................................
color(brown)("Question part a")" 3 years at 3.5%"

$\text{3 years at 3.5% is: "3xxcolor(cadetblue)(K22.75) = K68.25 color(red)(larr" Corrected}$

.............................................................
$\textcolor{b r o w n}{\text{Question part b}}$
Interest starts at 3.5% but at the end of the first year it is changed to 3.75%

1 year at color(white)(.)3.5% color(white)("d") = color(white)("d") 1xx color(cadetblue)(K22.750)color(white)(d)=color(cadetblue)(Kcolor(white)(.)22.75)
2 years at 3.75%=color(white)("d")2xxcolor(sienna)(K24.375)=ul(Kcolor(white)(".")48.75larr" Add")
$\textcolor{w h i t e}{\text{ddddddddddddddddddddddddddddd")Kcolor(white)(".}} 71.50$

So the increase in interest is:

$K \textcolor{w h i t e}{\text{.}} 71.50$
ul(Kcolor(white)(".")68.25 larr" Subtract")
Kcolor(white)(".")color(white)("d")3.25color(red)(larr" Corrected"