#### catherine19

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- Jun 16, 2019

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KATIE invests £1600 a month. It earns 4% interest in one year.

How much interest does she get?

How much interest does she get?

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- Thread starter catherine19
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- Jun 16, 2019

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KATIE invests £1600 a month. It earns 4% interest in one year.

How much interest does she get?

How much interest does she get?

Last edited by a moderator:

- Joined
- Jun 16, 2019

- Messages
- 5

Would it be

I = 1600 * .04*1 = 64 euros interest for 1 year

Or

5.33 EUROS per month

Plz suggest

I = 1600 * .04*1 = 64 euros interest for 1 year

Or

5.33 EUROS per month

Plz suggest

- Joined
- Jun 16, 2019

- Messages
- 5

But for 1 mark what would be the easiest approach plzYou 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.

You need to CLARIFY that.KATIE invests £1600 a month. It earns 4% interest in one year.

How much interest does she get?

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?

- Joined
- Jun 16, 2019

- Messages
- 5

my calculation i get 64 euros

(1600*1*4)/100 = 64

or

1600 *12*.04 = 768 euros

- Joined
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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