Q. 13.8( 262 Votes )

# Form the pair of

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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