# 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 Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

For representing the situation graphically and algebraically, we need to form linear equations

(i) Let number of girls = x

Let number of boys = y

According to the question, Total no of students is equal to 10,  ................................eq(i)

Now we will find different points to plot the equation. We can take any value of y and put in eq (i) to obtain the value of x at that point

Putting y = 4, 5 and 6. we get,

at x = 4

X = 10 – 4 = 6

at x = 5

X = 10 – 5 = 5

at x = 6

X = 10 – 6 = 4 Number of girls is 4 more than number of boys …………Given

So,

x = y + 4
⇒ y = x - 4          ................................(ii)

Now for plotting the points on graph, take any values of x and put them in eq (ii) to obtain values of y

Putting x = 3 ,5 and 7 we get
at x = 3
y = 3 - 4 = -1
at x = 4
y = 5 - 4 = 1
at x = 7
y = 7 - 4 = 3 Graphical representation :
Plotting the points obtain on graph we get, As, both lines intersect each other at (7, 3)
Solution of this pair of equation is (7, 3)
i.e.
No of girls, x = 7
No of boys, y = 3

(ii)
Let cost of one pencil = Rs. X

Let cost of one pen = Rs. Y

According to the question,  5 pencils and 7 pens together cost Rs 50

5x + 7y = 50

⇒5x = 50 - 7y

⇒ Putting value of y = 0 , 5 , 10 we get,

x = 10 - 0 = 10   Now,

7 pencils and 5 pens together cost Rs. 46

7x + 5y = 46

⇒ 5y = 46 – 7x Putting x = -2 , 3 , 8 we get,    Graphical Representation:

Plotting the points obtain on graph we get, As, both lines intersect each other at (3, 5)
Solution of this pair of equation is (3, 5)
i.e.
cost of pencil, x = 3
cost of pen, y = 5

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