Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
(i) Forming the equations
Let \(x\) be the number of girls and \(y\) be the number of boys.
Total students are 10, so
\[ x + y = 10. \]
The number of girls is 4 more than the number of boys, so
\[ x = y + 4 \Rightarrow x - y = 4. \]
Thus the required pair of linear equations is
\[ x + y = 10, \qquad x - y = 4. \]
Solving graphically (conceptual steps)
1. For \(x + y = 10\), choose convenient values:
\(x = 0 \Rightarrow y = 10\); \(x = 10 \Rightarrow y = 0\). Plot \((0,10)\) and \((10,0)\) and draw the line.
2. For \(x - y = 4\), choose convenient values:
\(x = 4 \Rightarrow y = 0\); \(x = 6 \Rightarrow y = 2\). Plot \((4,0)\) and \((6,2)\) and draw the line.
3. The two lines intersect at the point \((7,3)\), obtained either from the graph or by solving the equations algebraically.
So \(x = 7, y = 3\).
Conclusion
Girls = 7 and Boys = 3.
(ii) Forming the equations
Let \(x\) be the cost (in rupees) of one pencil and \(y\) be the cost (in rupees) of one pen.
Cost of 5 pencils and 7 pens is ₹ 50:
\[ 5x + 7y = 50. \]
Cost of 7 pencils and 5 pens is ₹ 46:
\[ 7x + 5y = 46. \]
Thus the required pair of equations is
\[ 5x + 7y = 50, \qquad 7x + 5y = 46. \]
Solving graphically (conceptual steps)
1. For \(5x + 7y = 50\), take convenient pairs: if \(x = 5\), then \(25 + 7y = 50 \Rightarrow y = \dfrac{25}{7}\); if \(y = 5\), then \(5x + 35 = 50 \Rightarrow x = 3\). Plot two such points (for easy graph use integer point \((3,5)\)) and draw the line.
2. For \(7x + 5y = 46\), take convenient pairs: if \(x = 4\), then \(28 + 5y = 46 \Rightarrow y = \dfrac{18}{5}\); if \(y = 4\), then \(7x + 20 = 46 \Rightarrow x = \dfrac{26}{7}\). Plot two points and draw the line.
3. The two lines intersect at \((3,5)\), which can also be verified algebraically by solving the pair:
Multiply \(5x + 7y = 50\) by 7 and \(7x + 5y = 46\) by 5:
\[ 35x + 49y = 350, \quad 35x + 25y = 230. \]
Subtracting, \((35x + 49y) - (35x + 25y) = 350 - 230 \Rightarrow 24y = 120 \Rightarrow y = 5. \]
Substitute in \(5x + 7y = 50\):
\[ 5x + 7 \times 5 = 50 \Rightarrow 5x + 35 = 50 \Rightarrow 5x = 15 \Rightarrow x = 3. \]
Conclusion
Cost of one pencil = ₹ 3 and cost of one pen = ₹ 5.
Form variables, translate the sentences into two linear equations, and plot two points for each line. Where the lines meet gives the solution: \((7,3)\) girls/boys in (i) and \((3,5)\) rupee costs in (ii).