The value of k for which the system of equation has no solution :
3x - 4y + 7 = 0, kx + 3y - 5 = 0
- k = 2.25
- k = -2.25
- k = 225
- k = 225
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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 :
1113 Views
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
We have,
2x + y = 150
⇒ y = 150 - 2x
When we plot the graph of the equation we find that two lines do not intersect i.e. they are parallel.
Fig. 3.3.
1699 Views
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
Thus, we have following table :
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.
995 Views
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 :
From eqn. [ii], we have
x -3y = 6
⇒ x = 3y + 6
Thus, we have following table
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.
Fig. 3.1.
1935 Views
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 :
We have, x + 3y = 1300
⇒ x = 1300 - 3y
Thus, we have following table :
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.
2043 Views