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  • Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 10

Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 10

Answers (1)

Answer:

\begin{aligned} &x=2000, y=3000, z=5000 \end{aligned}

Given:

Total investment=10000

Let x, y, z be the investments,

\begin{aligned} &x+y+z=10000 \\ &0.1 x+0.12 y+0.15 z=1310 \\ &-0.1 x-0.12 y-0.15 z=190 \end{aligned}

Hint:

X=A-1B is used for the problem

Solution:

Let x, y, z be the investments at the rates of interest of 10%,12%,13% per annum respectively

Total investment=10000

\begin{aligned} &x+y+z=10000 \\ \end{aligned}

Income from the 1st investment of Rs x

=\operatorname{Rs} \frac{10 x}{100}=0.1 x

Income from 2nd investment of Rs y

=\operatorname{Rs} \frac{12y}{100}=0.12 y

Income from 3rd investment of Rs z

=\operatorname{Rs} \frac{15z}{100}=0.15 z

\begin{aligned} &\therefore \text { Total annual income }=\operatorname{Rs}(0.1 x+0.12 y+0.15 z)\\ &0.1 x+0.12 y+0.15 z=1310 \end{aligned}

Its given that the combined income from the first two income is Rs190 short of the income from the third.

\begin{aligned} &0.1 x+0.12 y+0.15 z=190 \\ &-0.1 x-0.12 y+0.15 z=190 \\ &\text { Thus, } x+y+z=10000 \\ &0.1 x+0.12 y+0.15 z=1310 \\ &-0.1 x-0.12 y+0.15 z=190 \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 1 & 1 & 1 \\ 0.1 & 0.12 & 0.15 \\ -0.1 & -0.12 & 0.15 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 10000 \\ 1310 \\ 190 \end{array}\right]} \\ &A X=B \\ &A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 0.1 & 0.12 & 0.15 \\ -0.1 & -0.12 & 0.15 \end{array}\right] \end{aligned}

\begin{aligned} &|A|=1(0.15 \times 0.12+0.15 \times 0.12)-1(0.15 \times 0.1+0.15 \times 0.1)+1(-0.1 \times 0.12+0.12 \times 0.1) \\ &|A|=0.036-0.0370 \\ &|A|=0.006 \end{aligned}

Let i,j be the cofactors of the elements

\begin{aligned} &a_{i j} \text { in } A=\left[a_{i j}\right] \\ &c_{11}=(-1)^{1+1}\left|\begin{array}{cc} 0.12 & 0.15 \\ -0.12 & 0.15 \end{array}\right|=0.036, c_{12}=(-1)^{1+2}\left|\begin{array}{cc} 0.1 & 0.15 \\ -0.1 & 0.15 \end{array}\right|=-0.036 \\ &c_{21}=(-1)^{2+1}\left|\begin{array}{cc} 1 & 1 \\ -0.12 & 0.15 \end{array}\right|=-0.27, c_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 1 \\ -0.1 & 0.15 \end{array}\right|=0.25 \end{aligned}

\begin{aligned} &c_{31}=(-1)^{3+1}\left|\begin{array}{cc} 1 & 1 \\ 0.12 & 0.15 \end{array}\right|=0.3, c_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 1 \\ 0.1 & 0.15 \end{array}\right|=-0.05 \\ &c_{13}=(-1)^{1+3}\left|\begin{array}{cc} 1 & 0.12 \\ -0.1 & -0.12 \end{array}\right|=0, c_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & 1 \\ -0.1 & -0.12 \end{array}\right|=0.02 \\ &c_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ 0.1 & 0.12 \end{array}\right|=0.02 \end{aligned}

\begin{aligned} &a d j A=\left[\begin{array}{ccc} 0.036 & -0.03 & 0 \\ -0.27 & 0.25 & 0.02 \\ 0.03 & -0.05 & 0.02 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 0.036 & -0.27 & 0.03 \\ -0.03 & 0.25 & -0.05 \\ 0 & 0.02 & 0.02 \end{array}\right] \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{|A|} a d j A \\ &=\frac{1}{0.006}\left[\begin{array}{ccc} 0.036 & -0.27 & 0.03 \\ -0.03 & 0.25 & -0.05 \\ 0 & 0.02 & 0.02 \end{array}\right] \\ &X=A^{-1} B \end{aligned}

\begin{aligned} &X=\left[\begin{array}{ccc} 0.036 & -0.27 & 0.03 \\ -0.03 & 0.25 & -0.05 \\ 0 & 0.02 & 0.02 \end{array}\right]\left[\begin{array}{c} 10000 \\ 1310 \\ 190 \end{array}\right] \\ &=\frac{1}{0.006}\left[\begin{array}{c} 360-353.7+5.7 \\ -300+327.5-9.5 \\ 0+26.2+3.8 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1000}{6}\left[\begin{array}{l} 12 \\ 18 \\ 30 \end{array}\right]=\left[\begin{array}{c} 2000 \\ 3000 \\ 5000 \end{array}\right]} \\ &x=2000, y={{3000}, z=5000} \end{aligned}

Thus the three investments are of Rs 2000, Rs.3000, Rs 5000.

Posted by

Gurleen Kaur

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