NCERT Solutions
Class 10 - Mathematics - Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES - Exercise 3.1

Question 1

(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.