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#1 2014-10-23 04:13:00

Georgeyyyyyyyyyyyyy
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Hard algebra q

1.    You are sitting in Costa Coffee in Armada Way when you meet an elderly American tourist who enters into a conversation with you.  The last time she was in the UK was 1961.  Her long term memory is good and she remembers that in 1961 she needed $3 more for each £1 received in exchange.  She knows that she is now receiving £40 more for each $100 exchanged but her short term memory is not that good and she cannot remember the current exchange rate.  You don’t know the exchange rate and neither does anyone in the coffee shop.  There is a signal blackout so that you cannot consult your phone and are forced to consider using your mathematical knowledge to help the old lady.  Assuming that the correct exchange rate is $d = £1:

    •    you write down expressions for the number of £s she received for $100
    •    now                               
    •    in 1961.                           

    •    You form an equation and show by algebra that it reduces to the quadratic equation

4d2 + 12d – 30 = 0                       

    •    You then solve this equation and tell the lady what the current exchange rate is, correct to the nearest penny. [You will need to use the quadratic equation formula]

#2 2014-10-23 04:58:24

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Hard algebra q

Hi;

If all you need is the quadratic equation solved then:

d = 1.622498999199199 round to the nearest penny we get 1.62 which is close to the current conversion.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2014-10-23 07:04:12

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Hard algebra q

hi Georgeyyyyyyyyyyyyy

I'm having a lot of difficulty making sense of the question.

she remembers that in 1961 she needed $3 more for each £1

More than what?

she is now receiving £40 more for each $100 exchanged

Again, more than what?

Here's my attempt to make an equation. 

If the exchange rate is $d for £1 then $100 will become £(100/d)

In 1961 it took $3 more to get a £ so the exchange rate would be $(d+3) to make one £1.  So $100 got £ 100/(d+3) in 1961.

So if she is now receiving £40 more that means  100/d  = 100/(d+3) + 40

=> 100(d+3) = 100d + 40d(d+3) 

This will simplify to the required equation.

Hope that helps.



Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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