SIMPLE INTEREST: "KATIE invests £1600 a month. It earns 4% interest in one year. How much interest does she get?"

catherine19

New member
KATIE invests £1600 a month. It earns 4% interest in one year.
How much interest does she get?

Last edited by a moderator:

tkhunny

Moderator
Staff member
You should have I = Prt. Is this familiar?

Month 1: I = Prt ==> I = 1600*.04*12 months = 1600 * .04 * 1 year
Month 2: I = Prt ==> I = 1600*.04*11 months = 1600 * .04 * (11/12) year

Continue in this way.

catherine19

New member
Would it be
I = 1600 * .04*1 = 64 euros interest for 1 year
Or
5.33 EUROS per month
Plz suggest

catherine19

New member
You should have I = Prt. Is this familiar?

Month 1: I = Prt ==> I = 1600*.04*12 months = 1600 * .04 * 1 year
Month 2: I = Prt ==> I = 1600*.04*11 months = 1600 * .04 * (11/12) year

Continue in this way.
But for 1 mark what would be the easiest approach plz

Denis

Senior Member
KATIE invests £1600 a month. It earns 4% interest in one year.
How much interest does she get?
You need to CLARIFY that.
As example, what does "It" mean: the 1600 ?, the account in which deposited/invested?

Also, what is intended with "Simple Interest"? No compounding?

MarkFL

Super Moderator
Staff member
I would write:

$$\displaystyle I=\sum_{k=1}^{12}\left(1600\cdot0.04\cdot\frac{k}{12}\right)=\frac{16}{3}\sum_{k=1}^{12}(k)=\frac{8\cdot12\cdot13}{3}=4\cdot8\cdot13=\,?$$

catherine19

New member
this is the exact question.
my calculation i get 64 euros
(1600*1*4)/100 = 64
or

1600 *12*.04 = 768 euros

Denis

Senior Member
So you're starting with Pounds and ending with Euros....

Is this problem translated from another language?
Are you a student: if so, what grade? Thank you.

Paul Belino

New member
Assuming that the interest is compounded per month then 4% per year then per month interest 4/12% per month.
As this is (1/3)% but converted to a fraction is (1/3)/100 = 1/300
At the end of month 1 Katie has saved £1600 + plus her interest = £ 1600 + 1600 * (1/300)
To make this easier to deal with
Amount at end of month 1 is £1600*(1 + 1/300)
but let I = (1+1/300)
So at the end of
Month 1 Katie has £1600 * I
Month 2 Katie has £[(1600*I) + 1600]*I = 1600*I^2 + 1600*I = 1600*(I^2 + I)
Month 3 Katie has £[(1600*I) + 1600]*I + 1600]*I = 1600*(I^3 + I^2 +1)
Month 4 Katie has £[(1600*I) + 1600]*I + 1600]*I + 1600]*I = 1600*(I^4+I^3+I^2+1)
and so on to
Month 12 = 1600*(I^12+I^11+I^10+I^9+I^9+I^8+I^7+I^6+I^5+I^4+I^3+I^2+I^1+I^0)
remember I^1 = I and I^0 = 1

Can you see the pattern?

The I^12 etc part is called a geometric series.

You don't have to work out of that!
There is a shortcut formula that you may have to learn by heart.

GOOGLE Sum of a geometrical series to find it.

As a clue your number of terms is 12 and your common ratio is I