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(a) Solve the following pair of linear equations by graphical method : $2x + y = 9$ and $x - 2y = 2$
OR
(b) Nidhi received simple interest of ₹1,200 when invested $\text{\text{Rs} } x$ at $6\% \text{ p.a.}$ and $\text{\text{Rs} } y$ at $5\% \text{ p.a.}$ for 1 year. Had she invested $\text{\text{Rs} } x$ at $3\% \text{ p.a.}$ and $\text{\text{Rs} } y$ at $8\% \text{ p.a.}$ for that year, she would have received simple interest of ₹1,260. Find the values of $x$ and $y$.
OR
(b) Nidhi received simple interest of ₹1,200 when invested $\text{\text{Rs} } x$ at $6\% \text{ p.a.}$ and $\text{\text{Rs} } y$ at $5\% \text{ p.a.}$ for 1 year. Had she invested $\text{\text{Rs} } x$ at $3\% \text{ p.a.}$ and $\text{\text{Rs} } y$ at $8\% \text{ p.a.}$ for that year, she would have received simple interest of ₹1,260. Find the values of $x$ and $y$.
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(a) Correct graph of each equation ($2 + 2$ marks)
Solution $x = 4, y = 1$ or $(4, 1)$ (1 mark)
OR
(b) $\frac{6}{100}x + \frac{5}{100}y = 1200 \Rightarrow 6x + 5y = 120000 \dots(i)$ ($1\frac{1}{2}$ marks)
$\frac{3}{100}x + \frac{8}{100}y = 1260 \Rightarrow 3x + 8y = 126000 \dots(ii)$ ($1\frac{1}{2}$ marks)
Solving (i) and (ii) we get, $x = 10000$ and $y = 12000$ ($1 + 1$ marks)
Solution $x = 4, y = 1$ or $(4, 1)$ (1 mark)
OR
(b) $\frac{6}{100}x + \frac{5}{100}y = 1200 \Rightarrow 6x + 5y = 120000 \dots(i)$ ($1\frac{1}{2}$ marks)
$\frac{3}{100}x + \frac{8}{100}y = 1260 \Rightarrow 3x + 8y = 126000 \dots(ii)$ ($1\frac{1}{2}$ marks)
Solving (i) and (ii) we get, $x = 10000$ and $y = 12000$ ($1 + 1$ marks)