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

x = 88

Explanation:

We know that 36% of the letters are vowels. Of these, one-third are Os

(Since 33 1/3% = 33 1/3 divided by 100 = 1/3.) One-third of 36% is 12%, so 12% of the total letters in Pythagoras' soup are Os. When these are removed from the soup, 100 - 12=88% of the original letters remain.

So, x = 88

Hope this helps!

Answer 2

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.

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 =
`(36)/(100)n=0.36n`

Also Given:

33 1/3% of those lentils are green

Amount of green lentils =
`33(1)/(3)\%\ \ Or \ \ (100)/(3)\%`

Amount of green lentils =
`(100)/(3)* 0.36n*(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)* 100=88n \ \ \ Or \ \ \ 88\%`

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