Question

$36\%$ of the beans in Pythagoras's soup are lentils, and $33\frac13\%$ of those lentils are green. If Pythagoras removes all the green lentils from his soup, then $x\%$ of the original beans remains. What is $x$?

Answer 1

Question is not proper, Proper question is given below;

36% of the beans in Pythagoras's soup are lentils, and 33 1/3% of those lentils are green.

If Pythagoras removes all green lentils from his soup, then x% of the original beans remain. What is x?

Answer:

88% of the original beans remains $(x)$ after removing green lentils.

Step-by-step explanation:

Given:

Let the Original beans be 'n'.

Now given:

36% of the beans in Pythagoras's soup are lentils.

So we can say that;

Amount of lentils = $\frac{36}{100}n=0.36n$

Also Given:

33 1/3% of those lentils are green

Amount of green lentils = $33\frac{1}{3}\%\ \ Or \ \ \frac{100}{3}\%$

Amount of green lentils = $\frac{100}{3}\times 0.36n\times\frac{1}{100}= 0.12n$

Now we need find the percentage of original beans remain after removing green lentils.

Solution:

percentage of original beans remain  ⇒ $x\%$

To percentage of original beans remain after removing green lentils we will subtract Amount of green lentils from Original beans  and then multiplied by 100.

framing in equation form we get;

$x\%$ = $(n-0.12n)\times 100=88n \ \ \ Or \ \ \ 88\%$

Hence 88% of the original beans remains $(x)$ after removing green lentils.