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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

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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

\thereforeIn 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.

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