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 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. Fig. 3.2 29 (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.
Step-by-Step Solution
Key Concept: Translate the word statements into two linear equations in two variables, then locate the point of intersection of the two straight lines (graphical solution). The coordinates of the intersection give the required numbers or costs.
### Part (i)
1. Let \(b\) = number of boys and \(g\) = number of girls.
2. Total students: \(b+g = 10\) (Equation 1)
3. Girls are 4 more than boys: \(g = b + 4\) (Equation 2)
4. Graphical method – Plot the two lines on the \(b\)-\(g\) plane:
- For Eq. 1, take \(b=0\) ⇒ \(g=10\) and \(g=0\) ⇒ \(b=10\). Draw the line joining (0,10) and (10,0).
- For Eq. 2, take \(b=0\) ⇒ \(g=4\) and \(b=6\) ⇒ \(g=10\). Draw the line joining (0,4) and (6,10).
5. The two lines intersect at the point where both equations are satisfied. Solving algebraically gives:
\[g = b+4 \Rightarrow b+(b+4)=10 \Rightarrow 2b=6 \Rightarrow b=3\]
Substituting back: \(g = 3+4 = 7\).
6. Hence the intersection point is \((b,g) = (3,7)\).
Answer: 3 boys and 7 girls.
### Part (ii)
1. Let \(p\) = cost of one pencil (in rupees) and \(q\) = cost of one pen.
2. From the data:
- 5 pencils + 7 pens cost \(\text{Rs }50\): \(5p + 7q = 50\) (Equation 1)
- 7 pencils + 5 pens cost \(\text{Rs }46\): \(7p + 5q = 46\) (Equation 2)
3. Graphical method – Plot the two lines on the \(p\)-\(q\) plane:
- For Eq. 1, choose \(p=0\) ⇒ \(q=\frac{50}{7}\approx7.14\) and \(q=0\) ⇒ \(p=10\). Plot points (0,7.14) and (10,0).
- For Eq. 2, choose \(p=0\) ⇒ \(q=\frac{46}{5}=9.2\) and \(q=0\) ⇒ \(p=\frac{46}{7}\approx6.57\). Plot points (0,9.2) and (6.57,0).
4. The intersection of the two lines gives the simultaneous solution. Solving algebraically (as done for checking):
\[\begin{aligned}
5p+7q &= 50 \quad\text{(1)}\\
7p+5q &= 46 \quad\text{(2)}
\end{aligned}\]
Multiply (1) by 5 and (2) by 7:
\[25p+35q = 250\]
\[49p+35q = 322\]
Subtract: \(24p = 72 \Rightarrow p = 3\).
Substitute \(p=3\) into (1): \(5(3)+7q = 50 \Rightarrow 15+7q = 50 \Rightarrow 7q = 35 \Rightarrow q = 5\).
5. Hence the intersection point is \((p,q) = (3,5)\).
Answer: Cost of one pencil = Rs 3, cost of one pen = Rs 5.
*Graphical interpretation*: In both parts the straight lines representing the two equations intersect at a unique point; the coordinates of that point give the required numbers (boys‑girls) or costs (pencil‑pen).
Correct Answer: (i) Boys = 3, Girls = 7. (ii) Cost of one pencil = Rs 3, Cost of one pen = Rs 5.