9th maths ch 4 extra questions
Answers
Question.13 The cost of a notebook is Rs 5 less than twice the cost of a pen. Write this statement as a linear equation in two variables.
Solution. Let Rs x be the cost of a notebook and Rs y be the cost of a pen, then, we have
x = 2y – 5
⇒ x – 2y + 5 = 0.
Question.14 Total cost of a laptop and a mobile phone is Rs 60000. Write a linear equation in two variables to represent this statement.
Solution. Let the cost price of a laptop be Rs l and cost price of a mobile phone be Rs m.
Since combined cost of a laptop and a mobile phone is Rs 60000 So, l + m = 60000
Question.15 The cost of a table is Rs 100 more than half the cost of a chair. Write this statement as a linear equation in two variables.
Solution.
SHORT ANSWER QUESTIONS TYPE-I
Question.16 In some countries temperature is measured in Fahrenheit, whereas in countries like India it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
F=[9/5] c+32
If the temperature is – 40°C, then what is the temperature in Fahrenheit ?[CBSE-14-GDQNI3 W]
Solution.
Question.17 Write linear equation representing a line which is parallel to y-axis and is at a distance of 2 units on the left side of y-axis. [CBSE-14-GDQNI3W]
Solution.Here, required equation is parallel to y-axis at a distance of 2 units on the left side of y-axis.
x = – 2 or x + 2 = 0
Question.18 Find the value ofk, if (1, -1) is a solution of the equation 3x-ky = 8. Also, find the coordinates of another point lying on its graph. [CBSE-15-NS72LP7]
Solution.
Question.19 If (p, 2p + 1) is the solution of the linear equation 4x + 3y = 23. Find the value of p.
Solution.
Question.20 Find the value of m, if (5,8) is a solution of the equation 11 x-2y = 3m, then find one more solution of this equation. [CBSE March 2013]
Solution.
Question.21 If π x + 3y = 25, write y in terms of x and also, find the two solutions of this equation. [CBSE-14-17DIG1U]
Solution.
Question.22 Find four solutions of 2x-y = 4. [CBSE March 2011 ]
Solution.
Question.23 Give equation of two lines on same plane which are intersecting at the point (2, 3). [CBSE March 2012]
Solution.
Since there are infinite lines passing through the point (2, 3).
Let, first equation is x + y = 5 and second equation is 2x + 3y = 13
Clearly, the lines represented by both equations intersect at the point (2, 3).
24. Let y varies directly as x. Ify = 12 when x = 4, then write a linear equation. What is the value of y,when x = 5 ? [NCERT Exemplar Problem]
Solution.
Question.25 Give the equations of two lines passing through (2, 14). How many more such lines are there and why ?
Solution.
SHORT ANSWER QUESTIONS TYPE-II
Question.26 Find the value of a for which the equation 2x + ay = 5 has (1, -1) as a solution. Find two more solutions for the equation obtained. [CBSE March 2011 ]
Solution.
Question.27 A fraction becomes 1/4 when 2 is subtracted from the numerator and 3 is added to the denominator. Represent this situation as a linear equation in two variables. Also, find two solutions for this. [CBSE-15-17DIG1U]
Solution. Let numerator and denominator of the given fraction be respectively x and y. According to the statement, we obtain
Question.28 Draw the graph of the linear equation y = 2/3 x + 1/3 . Check from the graph that (7, 5)is a solution of the linear equation.
Solution.
Question.29 The cost of two pizzas and 1 burger is Rs 450. Represent this situation algebraically and also, draw the graph. [CBSE-14-ERFKZ8H], [CBSE-15-6DWMW5A]
Solution. Let cost price of one pizza be Rs x and that of one burger be Rs y.
... According to given statement, we have
2x + y = 450
⇒ y = 450 – 2x
When x = 100, y = 250
Question.30 Express y in terms of x for the equation 3x – 4y + 7 = 0. Check whether the points (23, 4) and (0,7/4)lie on the graph of this equation or not.
Solution. Given equation is 3x – 4y + 7 = 0
Question.31 The path of an aeroplane is given by the equation 3x – 4y = 12. Represent the graph graphically. Also, show that the point (- 4, – 6) lies on the graph. [CBSE-15-NS72LP7]
Solution. Given equation is 3x – 4y = 12
Question.32 Rupinder and Deepak two students of a vidyalaya contribute to charity. The Contribution of Rupinder is 2/5 of the contribution of Deepak. Write a linear equation According to the above statement and draw the graph for the linear equation.
Solution.
Question.33 Draw the graph of the equation y = mx + c for m = 3 and c = – 1 (a straight line in Cartesian plane). Read from the graph the value of when x = 2.
Solution.
hope it helps
Q.1: Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) x – y/5 – 10 = 0
(ii) -2x+3y = 6
(iii) y – 2 = 0
Solution:
(i) The equation x-y/5-10 = 0 can be written as:
(1)x + (-1/5) y + (-10) = 0
Now compare the above equation with ax + by + c = 0
Thus, we get;
a = 1
b = -⅕
c = -10
(ii) –2x + 3y = 6
Re-arranging the given equation, we get,
–2x + 3y – 6 = 0
The equation –2x + 3y – 6 = 0 can be written as,
(–2)x + 3y +(– 6) = 0
Now comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0
We get, a = –2
b = 3
c = -6
(iii) y – 2 = 0
Solution:
y – 2 = 0
The equation y – 2 = 0 can be written as,
0x + 1y + (–2) = 0
Now comparing 0x + 1y + (–2) = 0with ax + by + c = 0
We get, a = 0
b = 1
c = –2
Q.2. Write four solutions for each of the following equations:
(i) 2x + y = 7
Solution:
To find the four solutions of 2x + y = 7 we substitute different values for x and y
Let x = 0
Then,
2x + y = 7
(2×0)+y = 7
y = 7
(0,7)
Let x = 1
Then,
2x + y = 7
(2×1)+y = 7
2+y = 7
y = 7 – 2
y = 5
(1,5)
Let y = 1
Then,
2x + y = 7
2x+ 1 = 7
2x = 7 – 1
2x = 6
x = 3
(3,1)
Let x = 2
Then,
2x + y = 7
2(2)+y = 7
4+y = 7
y = 7 – 4
y = 3
(2,3)
The solutions are (0, 7), (1,5), (3,1), (2,3)
(ii) πx + y = 9
Solution:
To find the four solutions of πx + y = 9 we substitute different values for x and y
Let x = 0
Then,
πx + y = 9
(π × 0)+y = 9
y = 9
(0,9)
Let x = 1
Then,
πx + y = 9
(π×1)+y = 9
π+y = 9
y = 9-π
(1,9-π)
Let y = 0
Then,
πx + y = 9
πx +0 = 9
πx = 9
x =9/π
(9/π,0)
Let x = -1
Then,
πx + y = 9
(π(-1))+y = 9
-π + y = 9
y = 9+π
(-1,9+π)
The solutions are (0,9), (1,9-π),(9/π,0),(-1,9+π)
Q.3: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
The given equation is
2x + 3y = k
According to the question, x = 2 and y = 1.
Now, Substituting the values of x and y in the equation 2x + 3y = k,
We get,
⇒(2 x 2)+ (3 × 1) = k
⇒4+3 = k
⇒7 = k
⇒k = 7
The value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k, is 7.
Q.4: Draw the graph of each of the following linear equations in two variables:
(i)y = 3x
Solution:
To draw a graph of linear equations in two variables, let us find out the points to plot.
To find out the points, we have to find the values for which x and y satisfies the given equation.
Here,
y=3x
Substituting the values for x,
When x = 0,
y = 3x
y = 3(0)
⇒ y = 0
When x = 1,
y = 3x
y = 3(1)
⇒ y = 3
x y
0 0
1 3
The points to be plotted are (0, 0) and (1, 3)
Class 9 maths Chapter 4 Important Question 4.i
(ii) 3 = 2x + y
Solution:
To draw a graph of linear equations in two variables, let us find out the points to plot.
To find out the points, we have to find the values for which x and y satisfies the given equation.
Here,
3 = 2x + y
Substituting the values for x,
When x = 0,
3 = 2x + y
⇒ 3 = 2(0) + y
⇒ 3 = 0 + y
⇒ y = 3
When x = 1,
3 = 2x + y
⇒ 3 = 2(1) + y
⇒ 3 = 2 + y
⇒ y = 3 – 2
⇒ y = 1
x y
0 3
1 1
The points to be plotted are (0, 3) and (1, 1)
Class 9 Maths Chapter 4 Important Questions 4.ii
Q.5: If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
The given equation is
3y = ax + 7
According to the question, x = 3 and y = 4
Now, Substituting the values of x and y in the equation 3y = ax + 7,
We get,
(3×4) = (ax3) + 7
⇒ 12 = 3a+7
⇒ 3a = 12–7
⇒ 3a = 5
⇒ a = 5/3
The value of a, if the point (3, 4) lies on the graph of the equation 3y = ax + 7 is 5/3.
Q.6: Show that the points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7.
Solution:
We have the equation,
y = 9x – 7
For A (1, 2),
Substituting (x,y) = (1, 2),
We get,
2 = 9(1) – 7
2 = 9 – 7
2 = 2
For B (–1, –16),
Substituting (x,y) = (–1, –16),
We get,
–16 = 9(–1) – 7
-16 = – 9 – 7
-16 = – 16
For C (0, –7),
Substituting (x,y) = (0, –7),
We get,
– 7 = 9(0) – 7
-7 = 0 – 7
-7 = – 7
Hence, the points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.
Thus, A (1, 2), B (–1, –16) and C (0, –7) are solutions of the linear equation y = 9x – 7
Therefore, the points A (1, 2), B (–1, –16), C (0, –7) lie on the graph of linear equation y = 9x – 7.
Q.7: Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts X and Y-axis?
Solution: Given equation,
3x + 4y = 6.
We need at least 2 points on the graph to draw the graph of this equation,
Thus, the points the graph cuts
(i) x-axis
Since the point is on the x-axis, we have y = 0.
Substituting y = 0 in the equation, 3x + 4y = 6,
We get,
3x + 4×0 = 6
⇒ 3x = 6
⇒ x = 2
Hence, the point at which the graph cuts x-axis = (2, 0).
(ii) y-axis
Since the point is on the y-axis, we have, x = 0.
Substituting x = 0 in the equation, 3x + 4y = 6,
We get,
3×0 + 4y = 6
⇒ 4y = 6
⇒ y = 6/4
⇒ y = 3/2
⇒ y = 1.5
Hence, the point at which the graph cuts y-axis = (0, 1.5).
Plotting the points (0, 1.5) and (2, 0) on the graph.
Hope it will help you (≧▽≦)
