1. ## 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.

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.

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

$A_0 = 13$

$t = 3$

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

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

$A_0 = 13$

$t = 3$
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!

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

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

$A_0 = 13$

$t = 3$

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