4/6 for green and 2/5 for red since first wasn't replaced.
Partition the set A into 4 subsets:
{1, 8}, {2, 7}, {3, 6}, and {4, 5},
each consisting of two integers whose sum is 9. If 5 integers are selected from A,
then by the Pigeonhole Principle at least two must be from the same subset. But
then the sum of these two integers is 9.
It would be D 2 if you take each placement of dots and find the differ by multiply or divide that should get you the other hard to explain but ya
Division x=81/3 that’s right
a = amount invested at 7%
b = amount invested at 9%
we know the amount invested was ₹36000, thus we know that whatever "a" and "b" are, a + b = 36000. We can also say that
since we know the interest earned from the invested was ₹2920, then we say that 0.07a + 0.09b = 2920.
![\begin{cases} a + b = 36000\\\\ 0.07a+0.09b=2920 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{a + b = 36000\implies \underline{b = 36000-a}}~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.07a~~ + ~~0.09(\underline{36000-a})~~ = ~~2920} \\\\\\ 0.07a+3240-0.09a=2920\implies 3240-0.02a=2920\implies -0.02a=-320 \\\\\\ a=\cfrac{-320}{-0.02}\implies \boxed{a=16000}~\hfill \boxed{\stackrel{36000~~ - ~~16000}{20000=b}}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20a%20%2B%20b%20%3D%2036000%5C%5C%5C%5C%200.07a%2B0.09b%3D2920%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20the%201st%20equation%7D%7D%7Ba%20%2B%20b%20%3D%2036000%5Cimplies%20%5Cunderline%7Bb%20%3D%2036000-a%7D%7D~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20on%20the%202nd%20equation%7D%7D%7B0.07a~~%20%2B%20~~0.09%28%5Cunderline%7B36000-a%7D%29~~%20%3D%20~~2920%7D%20%5C%5C%5C%5C%5C%5C%200.07a%2B3240-0.09a%3D2920%5Cimplies%203240-0.02a%3D2920%5Cimplies%20-0.02a%3D-320%20%5C%5C%5C%5C%5C%5C%20a%3D%5Ccfrac%7B-320%7D%7B-0.02%7D%5Cimplies%20%5Cboxed%7Ba%3D16000%7D~%5Chfill%20%5Cboxed%7B%5Cstackrel%7B36000~~%20-%20~~16000%7D%7B20000%3Db%7D%7D)
