Suppose, 2/3 of 2/3 of a certain quantity of barley is taken, 100 units of barley are added and the original quantity recovered. find the quantity of barley? This is a real question from the Babylonian, posited 4 millenia ago...

Question

Mila Anderson · Accepted Answer

#x=180#

Let the quantity of barley be #x#. As #2/3# of #2/3# of this is taken and #100# units added to it, it is equivalent to

#2/3xx2/3xx x+100#

It is mentioned that this is equal to the original quantity, hence

#2/3xx2/3xx x+100=x# or

#4/9x+100=x# or

#4/9x-4/9x+100=x-4/9x# or

#cancel(4/9x)-cancel(4/9x)+100=x-4/9x=9/9x-4/9x=(9-4)/9x=5/9x# or

#5/9x=100# or

#9/5xx5/9x=9/5xx100# or

#cancel9/cancel5xxcancel5/cancel9x=9/5xx100=9/cancel5xx20cancel(100)=180# i.e.

#x=180#

Benjamin Altschul · Answer

undefined Let the original quantity of barley be $ x $ units.

Given that $ \frac{2}{3} $ of $ \frac{2}{3} $ of $ x $ is taken, the quantity remaining is $ \frac{1}{3} $ of $ \frac{2}{3} $ of $ x $, which is $ \frac{2}{9}x $.

Then, when 100 units of barley are added, the total quantity becomes $ \frac{2}{9}x + 100 $.

To recover the original quantity, the total amount must be doubled, as half was taken initially.

So, we have the equation:

$$ 2(\frac{2}{9}x + 100) = x $$

Solving for $ x $:

$$ 2 \times \frac{2}{9}x + 2 \times 100 = x $$
$$ \frac{4}{9}x + 200 = x $$
$$ 200 = x - \frac{4}{9}x $$
$$ 200 = \frac{5}{9}x $$
$$ x = \frac{200 \times 9}{5} $$
$$ x = 360 $$

Therefore, the original quantity of barley was 360 units.