Math Class 9 Question Bank

Question 1 of 36

1. Question
1 point(s)
Frame a linear equation in the from $ax + by + c = 0 $
(i) $a = – 2, b = 3, c = 4 $
(ii) $ a = 5, b = 0, c = 7$
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Hint

(i) $-2x+3y+4=0$
(ii) $5x+0y+7=0$
Question 2 of 36

2. Question
1 point(s)
Express each of the following equations in the form $y = mx + c $
(i) $3x – y = 4 $
(ii) $2y – x = 2 $
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Hint

(i) $3x-y=4\\ \\ y=3x-4$
(ii) $2y-x=2\\ \\ 2y=2+x\ \Rightarrow\ y=\dfrac{x+2}{2}$
Question 3 of 36

3. Question
1 point(s)
Draw the graph of the linear equation in two variables : $2x + y = 3$
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Hint

$2x+y=3\\ \\ \Rightarrow y=3-2x$
If $x=1$ ,
$y=3-2=1$
If $x=2$
$y=3-2(2)=-1$
If $x=0$
$y=3-2(0)=3$
Question 4 of 36

4. Question
1 point(s)
Find the co-ordinates of points where the graph of equation $4x + 3y = 12$ intersects x-axis and yaxis.
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Hint

Putting $y=0$ in equation $4x+3y=12$
$4x+0=12\\ \\ x=3$
So point intersect x-axis is (3, 0) putting $x = 0$ in equation $4x + 3y = 12$
$4\times0+3y=12\\ \\ y=4$
So point intersect y-axis is (0, 4)
Question 5 of 36

5. Question
1 point(s)
The auto fares in a city are as follows. For the first kilometre the fare is Rs 12 and for the subsequent distance it is Rs 7 per km. Taking the distance covered as x km and the total fare as Rs y, write a linear equation.
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Hint

Let the distance be $x$ km
Total fare $=y$
A/Q
$12+7(x-1)=y\\ \\ 12+7x-7=y\\ \\ 7x-y=-5\\ \\ 7x-y+5=0$
Question 6 of 36

6. Question
1 point(s)
Give geometric representation of equation $3x + 12 = 0$ in (i) one variable (ii) two variables
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Hint

(i) In one veriable, $3x+12=0$ is a point $(-4,0)$
In two variables
$3x+0y+12=0$
Put $y=0$
$3x+0+12=0\\ \\ x=-4$
Put $y=1$
$3x+0\times1+12=0\\ \\ 3x=-12\\ \\ x=-4$
Points are (-4, 0) and (-4, 1)
Question 7 of 36

7. Question
1 point(s)
After 5 years, the age of father will be two times the age of son. Write a linear equation in two variable to represent this statement
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Hint

Let the age of father be $x$ and son be $y$
$x+5=2(y+5)\\ \\ x+5=2y+10\\ \\ \Rightarrow x-2y-5=0$
Question 8 of 36

8. Question
1 point(s)
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to repre-sent this statement.
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Hint

Let the cost of pen $=x$ and cost of a note book $k=y$
$y=2x\\ \\ -2x+y=0$
Question 9 of 36

9. Question
1 point(s)
At what point does the graph of the linear equation $x + y = 5$ meet a line which is parallel to the y-axis, at a distnace of 2 units from the origin in the positive direction of x-axis ?
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Hint

$x+y=5\\ \\ \dfrac{x}{5}+\dfrac{y}{5}=1\\ \\ x=2\\ \\ 2+y=5\ \Rightarrow y=3$
The required point is (2, 3)
Question 10 of 36

10. Question
1 point(s)
Let $y $ varies directly as $x $. If $y=12 $ when $ x=4$ then write a linear equation. What is the value of $y $ when $x=5 $ ?
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Hint

$4\alpha x\\ \\ y=kx\\ \\ y=12,x=4\\ \\ 12=k\times4\\ \\ k=3\\ \\ x=5,y=?\\ \\ y=kx\\ \\ 3\times5=15$
Question 11 of 36

11. Question
1 point(s)
Determine the point on the graph of the linear equation $2x + 5y = 19$ whose ordinate is $1\dfrac{1}{2} $ times of its abscissa.
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Hint

Let abscissa be $a$
ordinate $=\dfrac{3a}{2}$
$2x+5y=19\\ \\ 2(a)+5\times\dfrac{3a}{2}=19\\ \\ \Rightarrow 19a=38\\ \\ \Rightarrow a=2$
ordinate $=\dfrac{3}{2}a=\dfrac{3}{2}\times2=3$
Point $=(2,3)$
Question 12 of 36

12. Question
1 point(s)
Find three solutions of $5x – y + 6 = 0$ after reducing it to $y = mx + c$ form.
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Hint

$5x-y+6=0\\ \\ y=5x-6\\ \\ x=1,\\ \\ y=5-6=-1\\ \\ x=0\\ \\ y=-6\\ \\ y=0\\ \\ 0=5x-6\\ \\ 6=5x\\ \\ x=\dfrac{6}{5}$
The 3 solutions are $\left(\dfrac{6}{5},0\right),(i-1),(0,-6)$
Question 13 of 36

13. Question
1 point(s)
Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units.
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Hint

$x+y=5+5=10\ \Rightarrow \ y=10-x$
For $x=5,y=5$ , lies on the graph of $x+y=10$
For $x=3,y=7$ lies on the graph of $x+y=10$
Coordinates are $(4,6),(5,5),(3,7)$
Question 14 of 36

14. Question
1 point(s)
For what value of $p $, the linear equation $2x + py = 8$ has equal values of $x $ and $y $ for its solution?
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Hint

$2x+Py=8\\ \\ P=\dfrac{8-2x}{y}\\ \\ \text{or}\ \dfrac{8-2x}{x}\\ \\ \text{Put}\ x=1\\ \\ P=\dfrac{8-2\times1}{1}=8-2=6$
Question 15 of 36

15. Question
1 point(s)
Find any three solutions for the equation $15x – 2y = 7$.
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Hint

$15x-2y=7\\ \\ 15x=7+2y\\ \\ x=\dfrac{7+2y}{15}$
$x\ \dfrac{7}{15}\ \dfrac{9}{15}\ \dfrac{11}{15}\\ \\ y\ \ 0\ \ 1\ \ 2$
Question 16 of 36

16. Question
1 point(s)
Draw the graph 2x + y = 4 and find the area of the triangle formed by the line with x-axis and y-axis.
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Hint

$2x+y=4\\ \\ x=0\\ \\ y=4\\ \\ y=0\\ \\ 2x+0=4\\ \\ x=2$
Base of the triangle $=2$
Height $=4$
Area $=\dfrac{1}{2}\times2\times4=4\ \text{Sq units}$
Question 17 of 36

17. Question
1 point(s)
The cost of a table exceeds the cost of the chair by Rs 150. Write a linear equation in two variables to represent this statement. Also, find two solutions for the same equation.
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Hint

Let the cost of table be $x$ and chain be $y$
$x-y=150\\ \\ x=y+150$
$x\ 150\ 151\ 152\\ y\ \ 0\ \ 1\ \ 2$
Question 18 of 36

18. Question
1 point(s)
Solve for $ x:x-1=\dfrac{3}{4}\ (x+1)+\dfrac{1}{2}$
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Hint

$x-1=\dfrac{3}{4}(x+1)+\dfrac{1}{2}\\ \\ \Rightarrow x-1-\dfrac{1}{2}=\dfrac{3x+3}{4}\\ \\ \Rightarrow \dfrac{2x-2-1}{2}=\dfrac{3x+3}{4}\\ \\ \Rightarrow \dfrac{2x-3}{2}=\dfrac{3x+3}{4}\\ \\ \Rightarrow 2(2x-3)=3x+3\\ \\ \Rightarrow 4x-6=3x+3\\ \\ \Rightarrow x=9$
Question 19 of 36

19. Question
1 point(s)
If the number of hours for which a labourer works is x and y are his wages (in rupees) and y = 2x – 1, draw the graph of the work-wages equation. From the graph, find the wages of the labourer if he works for 6 hours.
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Hint

The time for which a labour works is $x$ hours and his wages is $₹\ y$
$y=2x-1$
when $x=1$ we have $y=2-1=1$
when $x=2$ we have $y=4-1=3$
when $x=3$ we have $y=6-1=5$
$y=2x-1\\ \\ 2\times6-1\\ \\ =12-1\\ \\ y=11$
So wages of labour if the works for 6 hour is ₹ 11
Question 20 of 36

20. Question
1 point(s)
Ashish and Deepak contribute to charity. The contribution of Ashish is $\dfrac{1}{2} $ of contribution of Deepak. Write a linear equation to represent the above and draw the graph. From the graph, find the contribution of Ashish, if Deepak contributes Rs 50.
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Hint

Let ashish’s contribution be $x$ and deepak’s contribution be $y$
$x=\dfrac{2}{5}y\\ \\ 5x=2y\\ \\ 5x-2y=0$
If deepak’s contribution is 50 then ashish’s contribution is $x=20$
Question 21 of 36

21. Question
1 point(s)
Give geometric representation of $2y + 7 = 0 $ as an equation :
(i) in one variable
(ii) in two variable
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Hint

$2y+7=0$
In one variable
$2y+7=0\\ \\ y=-\dfrac{7}{2}$
So we can say that in one variable the equation is a point
In two variable,
$0x+2y+7=0\\ \\ 2y=\dfrac{-7}{2}\\ \\ x=1\\ \\ y=\dfrac{-7}{2}$
Points are $\left(0,\dfrac{-7}{2}\right),\left(1,\dfrac{-7}{2}\right)$
Question 22 of 36

22. Question
1 point(s)
Find $a $, if $\dfrac{a+3}{a-2}=\dfrac{a+2}{a+7}a\ne2,a\ne7 $
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Hint

$\dfrac{a+3}{a-2}=\dfrac{a+2}{a+7}\\ \\ \Rightarrow (a+3)\ (a+7)=(a+2)\ (a-2)\\ \\ \Rightarrow a^2+7a+3a+21=a^2-4\\ \\ \Rightarrow 10a=-4-21\\ \\ \Rightarrow 10a=-25\\ \\ \Rightarrow a=\dfrac{-25}{10}=\dfrac{-5}{2}$
Question 23 of 36

23. Question
1 point(s)
Sum of the digits of a two digit number is 14. If we add 18 to the original number, the digits interchange their places. Write two equations for these two statements.
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Hint

Let the unit digit be $y$ and ten’s digit be $x$
original number $=10x+y$
A/Q
$10x+y+18=10y+x\\ \\ \Rightarrow 10x+y-10y-x=-18\\ \\ \Rightarrow 9x-9y=-18\\ \\ \Rightarrow x-y=-2\ …(1)$
Also $x+y=14\ ...(2)$
Question 24 of 36

24. Question
1 point(s)
The food charges in a hostel are as follows : For the first day, the charges are Rs 100 and for the subsequent days it is Rs 50 per day. Taking the number of days as x and total charges as Rs. y, write a linear equation for this information and draw its graph.
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Hint

For $x-1=50(x-1)$
$50x-50\\ \\ y=50x-50+100\\ \\ y=50x+50\\ \\ x\ 0\ 10\ 20\\ y\ 50\ \ 60\ \ 70$
Question 25 of 36

25. Question
1 point(s)
Solve for $\dfrac{7x-1}{4}-\dfrac{1}{3}\left(2x+\dfrac{x-1}{2}\right)=6\dfrac{1}{3} $
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Hint

$\dfrac{7x-1}{4}-\dfrac{1}{3}\left(2x+\dfrac{x-1}{2}\right)=\dfrac{19}{3}\\ \\ \Rightarrow \dfrac{7x-1}{4}-\dfrac{1}{3}\left(\dfrac{3x-2}{2}\right)=\dfrac{19}{3}\\ \\ \Rightarrow \dfrac{21x-3-6x+4}{12}=\dfrac{19}{3}\\ \\ \Rightarrow \dfrac{15x+1}{4}=19\\ \\ \Rightarrow 15x=75\\ \\ \Rightarrow x=5$
Question 26 of 36

26. Question
1 point(s)
Draw the graph of two lines, whose equations are 3x – 2y = 4 and x + y – 3 = 0 on the same graph paper and find the co-ordinates of the point where two lines intersect.
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Hint

$3x-2y=4\qquad x+y-3=0\\ \\ x=0\qquad x=0,y=3\\ \\ 3(0)-2y=4\qquad y=0\\ \\ y=2\qquad x+0-3=0$
Let $y=0\qquad x=3$
$3x-2\times0=4\\ \\ x=1.3$
Intersection is at $(2,1)$
Question 27 of 36

27. Question
1 point(s)
Rs 27 is in the form of 50 paise, 25 paise and 20 paise coins. The number of 25 paise coins is double the number of 20 paise coins but half the number of 50 paise coins. Find the number of coins of each type.
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Hint

Let the number of 50 paise coins $=x,25$ paise coins $=y$ and 20 paise coins $=z$
$50x+25y+20z=2700\ ….(1)\\ \\ y=2z\ ….(2)\\ \\ y=\dfrac{x}{2}$
so, $\dfrac{x}{2}=2z\Rightarrow x=4z\ .....(3)$
$50(4z)+25(2z)+20z=2700\\ \\ 270z=2700\\ \\ z=10\\ \\ z=10\\ \\ y=2z=20\\ \\ x=4z=40$
Question 28 of 36

28. Question
1 point(s)
Solve : $\dfrac{y-3}{5}+\dfrac{y-4}{7}=6-\left[\dfrac{2y-1}{35}\right] $
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Hint

$y-\dfrac{3}{5}+\dfrac{y-4}{7}=6-\left(\dfrac{2y-1}{35}\right)\\ \\ \dfrac{7y-21+3y-20}{35}=\dfrac{210-2y+1}{35}\\ \\ \Rightarrow 10y-41=211-2y\\ \\ \Rightarrow 12y=252\\ \\ \Rightarrow y=21$
Question 29 of 36

29. Question
1 point(s)
A number consists of 2 digits. The digit at tens place is 2 times the digit in units place. The number formed by reversing the digit is 27 less than the original number. Find the number.
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Hint

Let the digit at tens place $=y$
unit place $=x$
Number $=10y+x$
A/Q
$(10y+x)-(10x+y)=27\\ \\ 10y+x-10x-y=27\\ \\ 9y-9x=27\\ \\ y-x=3$
Also $y=2x$
$2x-x=3\\ \\ x=3\\ \\ y=6$
Number $=63$
Question 30 of 36

30. Question
1 point(s)
A man leaves half his property to his wife, one third of the remaining to his son and the rest to his daughter. If daughter’s share is Rs 15000, how much money did the man leave? How much money did his wife and son each get?
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Hint

Let the property be $x$
wife $=\dfrac{x}{2}$
Son $=\dfrac{x}{2}-\dfrac{x}{3}=\dfrac{x}{6}$
Daughter $=x-\left(\dfrac{x}{2}+\dfrac{x}{6}\right)=\dfrac{x}{3}$
A/Q
$15000=\dfrac{x}{3}$
$x=45000$
wife $=\dfrac{45000}{2}=22500$
Son $=\dfrac{45000}{6}=7500$
Question 31 of 36

31. Question
1 point(s)
You know that the force applied on a body is directly proportional to the acceleration produced in the body. If constant of proportionality is 2, write an equation to express this situation and plot the graph of equation.
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Hint

Let the force be $y$ and $x$ is acceleration
$y\alpha x\\ \\ y=kx\\ \\ y=2x\\ \\ x=0,y=0\\ \\ x=1,y=2$
Question 32 of 36

32. Question
1 point(s)
I am three times as old as my son. Five years later, I shall be two and a half times as old as my son. How old am I and how old is my son?
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Hint

Let $y$ age be $x$ years
Son’s age $=y$ years
$x=3y\\ \\ x+5=\dfrac{5}{2}(y+5)\\ \\ \Rightarrow 3y+5=\dfrac{5y+25}{2}\\ \\ \Rightarrow 6y+10=5y+25\\ \\ \Rightarrow y=15\\ \\ x=45$
My age is 45 years and my son’s age is 15 years
Question 33 of 36

33. Question
1 point(s)
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is 2 units.
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Hint

Let the distance be $x$
work done $=y$
$y=kx\\ \\ y=5x\\ \\ x\ 0\ 1\\ y\ 0\ 5$
Question 34 of 36

34. Question
1 point(s)
The following values of $x $ and $y $ are thought to satisfy a linear equation.
Draw the graph, using the values of $x $ and $y $. At what points the graph cuts the x-axis and y-axis.
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Hint

The graph cuts the x-axis at $\left(\dfrac{1}{2},0\right)$ and y-axis at $(0,-1)$
Question 35 of 36

35. Question
1 point(s)
A and B are friends. A is elder to B by 5 years. B’s sister C is half the age of B while A’s father D is 8 years older than twice the age of B. If the present age of D is 48 years, find the present ages of A, B and C.
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Hint

Let the present age of B be $x$
$A=5+x\\ \\ C=\dfrac{x}{2}\\ \\ D=8+2x\\ \\ \text{A/Q}\ 8+2x=48\\ \\ 2x=40\Rightarrow x=20\ years$
The present age of A = 20 + 5 = 25 years
The present age of B = 20 years
The present age of C = 10 years
Question 36 of 36

36. Question
1 point(s)
Solve for $x:\dfrac{3x-5}{3}+\dfrac{4x+2}{5}=\dfrac{25x+7}{15} $
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Hint

$\dfrac{3x+5}{3}+\dfrac{4x+2}{5}=\dfrac{25x+7}{15}\\ \\ \dfrac{15x-25+12x+6}{15}=\dfrac{25x+7}{15}\\ \\ \Rightarrow 27x-19=25x+7\\ \\ \Rightarrow 2x=26\\ \\ \Rightarrow x=13$