Yes we can. After Maria remove the cookies the remaining number of cookies is \(1-\frac{5}{6}=\frac{1}{6}\) .
Then Andrea removes \(\frac{3}{5}\)remaining cookies. So cookies left is \(1-\frac{3}{5}=\frac{2}{5}\).


Original Number of cookies ---------> After Maria removed ---------------------> After Andrea removed cookies

x ---------------------------------------------> \(\frac{1}{6}x\) ----------------------------------------> \(\frac{1}{6}\times\frac{2}{5}x=\frac{2}{30}x\).

Now \(\frac{2}{30}x\) and \(\frac{1}{6}x\) must be whole numbers. Hence number of cookies must be a multiple of 30.

You can identify this from the denominator of fractions \(\frac{5}{6}\) and \(\frac{2}{5}\).

Why didn't u simplify 2/30 into 1/15? Because at 1/15, then we say the number of cookies must be a multiple of 15 and therefore, our answers would be different.

Thanks for pointing out. You are correct!

Both \(\frac{2}{30}x\) and \(\frac{1}{6}x\) must be whole numbers. At an intermediate steo the number of cookies cannot be a fraction.

Note:

First of all someone removed 6/5 of the cookies. It means that total number is cookie must be the multiple of 6. So, eliminate the options that are not the multiple of 6.

15 ....200.......340. Cancelled.

Second point is that remaining cookies must be the multiple of 5.

Option A)

6 * (5/6) = 1..............out

Option C)

30 * (5/6) = 25 that is a multiple of 5.

Option D)

60 *(5/6) = 50 , multiple of 5.

Option F)

210 * (5/6) = 35, a multiple of 5.



So, A, B E, G are the options.

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