# ratio prob

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• Oct 10th 2009, 07:33 AM
GAdams
ratio prob
Due to illness a man's weight increases each year in the ratio 10:11
He starts at 13 stones
How heavy will he be at the end of the third year?

I'm not sure how to tackle this, I thought...

10: 11 as a percentage is 91% and after 3 years, 91 stones....? Got to be wrong.
• Oct 10th 2009, 07:35 AM
e^(i*pi)
Quote:

Originally Posted by GAdams
Due to illness a man's weight increases each year in the ratio 10:11
He starts at 13 stones
How heavy will he be at the end of the third year?

I'm not sure how to tackle this, I thought...

10: 11 as a percentage is 91% and after 3 years, 91 stones....? Got to be wrong.

Use the compound interest formula.

$\displaystyle A(t) = A_0\left(1+\frac{11}{10}\right)^t$

$\displaystyle A_0 = 13$

$\displaystyle t = 3$
• Oct 10th 2009, 07:40 AM
GAdams
Quote:

Originally Posted by e^(i*pi)
Use the compound interest formula.

$\displaystyle A(t) = A_0\left(1+\frac{11}{10}\right)^t$

$\displaystyle A_0 = 13$

$\displaystyle t = 3$

That's basically what i did:

13 x (1.91)^3 = 90.6 stones (1 dp)

I just thought I had gone wrong....90 stones is over 500 kg!
• Oct 10th 2009, 07:43 AM
GAdams
Quote:

Originally Posted by e^(i*pi)
Use the compound interest formula.

$\displaystyle A(t) = A_0\left(1+\frac{11}{10}\right)^t$

$\displaystyle A_0 = 13$

$\displaystyle t = 3$

Hmm....why 11/10 and not 10/11?