Question

Question #8994e

Answer 1

`=K68.25`
`=K3.25`

Explanation:

This can be solved using the formula:
`I="Pi"n`
where:
`I="Interest"`
`P="Principal"`
`i="interest rate"`
`n="no. of years"`

Plug in data provided to find the Interest `(I)` after after 3 years .
`I="Pin"`
`I=K650xx(0.035)/(cancel(yr))xx3cancel(yr)`
`I=K68.25`

Thus, the interest `(I)` per year is `K22.75`

If the interest rate is increased by 3.75% after the first year; the Interest `(I)` earned for the remaining 2 years can be computed as;
`I="Pin"`
`I=K650xx(0.0375)/(cancel(yr))xx2cancel(yr)`
`I=K48.75`

Thus, the total Interest earned before and after the increase can be computed as shown;
`I_(t)=I_(fy)+I_(s&ty)`
`I_(t)=K22.75+K48.75`
`I_(t)=K71.50`

Therefore, an increase of the Interest earned is
`=I_(3.50&3.75%)-I_(3.50%)`
`=K71.50-L68.25`
`=K3.25`

Answer 2

(a) Interest earned after 3 years @ 3.5% Interest = `color(purple)(K 68.25)`

(b) Increase in Interest due to rise in interest to 3.75% after one year = `color(green)(71.5 - 68.25 = K 3.25)`

Explanation:

`I = (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 = ( 650 * 3 * 3.5) / 100 =color(red)( K 68.25)`

(b) New Rate of Interest `R_1 = 3.75%` & `N_1 = 2 years`

Interest earned `I_1 = (650 * 2 * 3.75) / 100 = color(blue)(K 48.75)`

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

`I_o = (650 * 3.5 * 1) / 100 = color (blue)(22.75)`

Total Interest = 48.75 + 22.75 = K 71.5#

Increase in interest with the revised interest rate after 1 year `color(green)(I_i = 71.5 - 68.25 = K 3.25)`

Answer 3

`"Part a: "K68.25`

`"Part b: "K3.25`

Explanation:

`color(blue)("Preamble")`

The symbol % cab be thought of as representing a fraction but one that is worth `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 `1/100`. So we have `1/100xx1" metre"`

It is usual to translate the word of (as in `1/100` of 1 metre) to mean multiply. So `3.5%` of 'something' is stating `3.5%xx"'something'"`

The wording 'per anum'`->`'per' means for each of 1 and 'anum' means 'year'. So 'peranum' is stating 'for 1 year'
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("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%"`

`"3 years at 3.5% is: "3xxcolor(cadetblue)(K22.75) = K68.25 color(red)(larr" Corrected"`

.............................................................
`color(brown)("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")`
`color(white)("ddddddddddddddddddddddddddddd")Kcolor(white)(".")71.50`

So the increase in interest is:

`Kcolor(white)(".")71.50`
`ul(Kcolor(white)(".")68.25 larr" Subtract")`
`Kcolor(white)(".")color(white)("d")3.25color(red)(larr" Corrected"`