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

[catherine19]

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

 

[tkhunny]

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]

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

 

[catherine19]

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]

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

You need to CLARIFY that.
Can you ask someone to help you post your problem IN FULL?
As example, what does "It" mean: the 1600 ?, the account in which deposited/invested?

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

 

[MarkFL]

I would write:

[MATH]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=\,?[/MATH]

 

[catherine19]

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

1600 *12*.04 = 768 euros

 

[Denis]

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]

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