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#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 74 sub question (ii) math

Answer: Rs 5000 invested in the first bond and Rs 25000 invested in the second bond

Hint:

$S I=\frac{P R T}{100}$ where P is principal, R is rate, T is time.

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given: Total amount to invest in 2 different types of bonds=Rs 30000

First bond pays 5% interest per year

Second bond pays 7% interest per year

Let Rs x can be invested in the first bond then, the sum of money invested in the second bond will be $Rs \ \ (30000 -x )$

First bond pays 5% interest per year second bond pays 7% interest per year

$\therefore$In order to obtain an annual total interest of Rs 2000, we have

$\Rightarrow\left[\begin{array}{ll} x & (30000-x) \end{array}\right]\left[\begin{array}{c} \frac{5}{100} \\ \frac{7}{100} \end{array}\right]=2000\\\\$

$\therefore$ simple interest for 1 year

$=\frac{\text { principal } \times \text { rate }}{100}\\$

$\\ \Rightarrow \frac{5 x}{100}+\frac{7(30000-x)}{100}=2000\\\\ \Rightarrow 5 x+210000-7 x=200000\\\\ \Rightarrow 210000-2 x=200000\\\\ \Rightarrow 2 x=210000-200000\\\\ \Rightarrow 2 x=10000\\\\ \Rightarrow x=\frac{10000}{2}\\\\ \Rightarrow x=5000$

Thus, in order to obtain an annual total interest of Rs 2000, the trust should invest Rs 5000 in the first bond and the remaining Rs25000 in the second bond.