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)
2nd opinion that’s your answer
10 lol that’s the answer.
Answer:
75,361,700
Step-by-step explanation:
find the hundredths place, if the number to the right is higher than or equal to 5 increase the number in the hundredth place by 1, if not keep the number the same (so 6 will turn into 7) and make the numbers after that 0s. Copy down the numbers exactly like they are but with the rounded hundredth place.
