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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.

Let cost of 1 pencil be Rs. ,v and cost of 1 pen be Rs. y

Case I. Cost of 5 pencils = 5x

Cost of 7 pens = 7y

<>According to question,

 

5x + 7y = 50

Case II. Cost of 7 pencils = 7x

Cost of 5 pens = 5y

According to question,

7x + 5y = 46

Thus, we have


Let cost of 1 pencil be Rs. ,v and cost of 1 pen be Rs. y
Case I. Cos

Let cost of 1 pencil be Rs. ,v and cost of 1 pen be Rs. y
Case I. Cos
Thus, we have following table :

Let cost of 1 pencil be Rs. ,v and cost of 1 pen be Rs. y
Case I. Cos

Fig. 3.5.

When we plot the graph of the given equation, we find that both the lines intersect at the point (3, 5). So, x = 3,y = 5 is the required solution of the pair of linear equation.

Hence, the cost of 1 pencil be Rs. 3 cost of 1 pen be Rs. 5.

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The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and graphically.

Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y

Case I. Cost of 3 bats = 3x

Cost of 6 balls = 6y

According to question,

3x + 6y = 3900

Case II. Cost of I bat = x

Cost of 3 more balls = 3y

According to question,

x + 3y = 1300

So, algebraically representation be

3x + 6y = 3900

x + 3y = 1300

Graphical representation :

We have,    3x + 6y = 3900

⇒    3(x + 2y) = 3900

⇒    x + 2y = 1300

⇒    a = 1300 - 2y

Thus, we have following table :


Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y
Case I. Cos

We have,    x + 3y = 1300

⇒    x = 1300 - 3y

Thus, we have following table :


Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y
Case I. Cos

When we plot the graph of equations, we find that both the lines intersect at the point (1300. 0). Therefore, a = 1300, y = 0 is the solution of the given system of equations.


Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y
Case I. Cos

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(i) 10 students of Class X took part in 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.


(i) Let the number of boys be x and number of girls be y.

Case I.    x + y = 10    ...(i)

Case II.    y = x + 4

⇒    x - y = -4    ...(ii)

We have, x + y = 10

⇒    x = 10 - y

Thus, we have following table :


(i) Let the number of boys be x and number of girls be y.
Case I.  ?

Fig. 3.4.

We have, x - y = -4

⇒    x = y - 4

Thus we have following table :


(i) Let the number of boys be x and number of girls be y.
Case I.  ?

When we plot the graph of the given equation, we find that both the lines intersect at me point (3, 7). So.r = 3,y = 7 is the required solution of the pair of linear equation.

Hence, the number of boys be 3 and the number of girls be 7, who took part in quiz.

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Aftab tells his daughter, 'Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be'. (Isn’t this interesting?) Represent this situation algebraically and graphically.


Let present age of Aftab be x years and present age of his daughter be y years.

Case I. Seven years ago,

Age of Aftab = (x - 7) years

Age of his daughter = (y - 7) years

According to question :

(x - 7) = 7 (y - 7)

⇒ x - 7 = 7y - 49

⇒ x - 7y = -42

Case II.

Three years later,

Age of Aftab = (x + 3) years

Age of his daughter = (y + 3) years

Accoring to questions,

x + 3 = 3 (y + 3)

⇒ x + 3 = 3y + 9

⇒ x — 3y = 6

So, algebraic expression be

x - 7y = -42    ...(i)

x - 3y = 6    ...(ii)

Graphical representation

For eq. (i), we have

x - 7y = -42

⇒    x — 7y — 42

Thus, we have following table :


Let present age of Aftab be x years and present age of his daughter b

From eqn. (ii), we have

x -3y = 6

⇒    x = 3y + 6

Thus, we have following table


Let present age of Aftab be x years and present age of his daughter b

When we plot the graph of equations. We find that both the lines intersect at the point (42, 12). Therefore, x = 42, y = 12 is the solution of the given system of equations.


Let present age of Aftab be x years and present age of his daughter b

Fig. 3.1.

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The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.

Let the cost of 1 kg of apples be Rs. x and of 1 kg of grapes be Rs. y. So, algebraic representation

<>2x + y = 160

 

4x + 2y = 300

⇒ 2x + y = 150

Graphical representation, we have

2x + y = 160

⇒    y = 160 - 2x


Let the cost of 1 kg of apples be Rs. x and of 1 kg of grapes be Rs.

We have,

2x + y = 150

⇒    y = 150 - 2x


Let the cost of 1 kg of apples be Rs. x and of 1 kg of grapes be Rs.

When we plot the graph of the equation we find that two lines do not intersect i.e. they are parallel.


Let the cost of 1 kg of apples be Rs. x and of 1 kg of grapes be Rs.

Fig. 3.3.

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