Math 8

7. Determine whether the following straight lines have positive slopes, negative
slopes, zero slopes or not defined.

(a) Y (b) Y (c) (d) YC

A B X' OX
X X' O
X' O X
Y'
θ X Y' D

O

Answer 

1. (a) 1 (b) -7 (c) -1 (d) 3 (e) 5 (f) 2 2. (a) 6 (b) 8
2 8 2 2 3
2 5 -2 5
3. (a) 2 and 9 (b) 3 and 3 (c) -2 and 7 (d) 3 and 3 (e) -2 and 3

4. (ii) 1, 2 (iii) 2 5. Consult your teacher. 6. Consult your teacher.

7. (a) Positive (b) 0 (c) Negative (d) ∞

23.5 Intercepts of a Straight Line

In the given figure, the straight line AB intersects the x–axis at A(a, 0) and y–axis at
(0, b). The length OA and OB are called the x–intercept (a) and y–intercept (b) of the
straight line AB.

To find the intercepts of a straight line from its equation,
we can find the x–intercept by putting y = 0 and y intercept by putting x = 0.

For example:

Find the x–intercept and y–intercept of the line 3x+2y = 12.

Putting x = 0, we get, Y

3x + 2y = 12 B(o, b)

or 3× 0 + 2y = 12 y-intercept
b
or 2y = 12

∴ y = 122 = 6 X' O a A(a, o) X

and putting y = 0, we get,

3x + 2 × 0 = 12 Y' x-intercept

or 3x = 12

Note:
At X–axis, y = 0
∴ x = 132 = 4 At Y–axis, x = 0

Hence, x–intercept = 4 and y–intercept = 6.

Oasis School Mathematics-8 229955

Worked Out Examples

Example: 1

Find the x–intercept and y–intercept from the following graphs.

(i) Y (ii) Y

X' O X

X' O X

Y' Y'

Solution:
(i) From the graph, we get x–intercept (OA) = 2 and y–intercept (OB) = 4.
(ii) From the graph, we get x–intercept (OA) = 3, y–intercept (OB) = –5.

Example: 2

Draw the graph of the line y = 2x – 4 and find the x–intercept and y–intercept made by
the line.

Solution: Y
Here, y = 2x – 4

x0 2 3
y -4 0 2

We see that the line cuts the x–axis at 2 and y–axis at –4. X' (3,2)
(2,0)
Hence, x–intercept = 2 and y–intercept = –4. X

(0,-4)
Y'

296 Oasis School Mathematics-8

Exercise 23.4

1. Find x–intercept and y–intercept in each of the following.

(a) Y (b) Y

X' X X' X
X
Y' (d) Y'
(c) Y Y

X X'

X'

Y' Y'

2. Find the x–intercept and y–intercept of the following equations:

(a) 2x + 5y = 10 (b) 3x + 2y = 12 (c) 4x + 5y = 20

(d) 4x – y = 12 (e) y = 2x + 4

3. Draw the graphs of the following equations and find x–intercept and y-intercept.

(a) y = 2x – 8 (b) 2x + y = 4 (c) 2x + y = 6

(d) y = 2x + 4 (e) y = –32x – 6

4. (a) Find at which point the line 3x – 2y = 6 cuts the x and y axis and draw the graph of
the line.

(b) Find the points where 2x+3y = 12 cuts X-axis and Y-axis.



Answer
1. (a) 2, 4 (b) 3, -5 (c) -3, 2 (d) -4, -3
2. (a) 5, 2 (b) 4, 6 (c) 5, 4 (d) 3, -12 (e) -2, 4 3. Consult your teacher
4. (a) (2, 0), (0, -3) (b) (6, 0) (0, 4)

Oasis School Mathematics-8 229977

23.6 Simultaneous Equations

Let us consider the equations x + y = 8 and 2x–y = 7. These two equations contain two
variables 'x' and 'y' and both are linear equations. These equations are simultaneous
equations.

Hence, a pair of linear equations in two variables are called a system of simultane-
ous linear equations or simply simultaneous equations.

Method of solving simultaneous equations
There are many methods of solving simultaneous equations. Here we will discuss

only three methods.
(i) Graphical method
(ii) Elimination method
(iii) Substitution method

Graphical method of solving simultaneous equation

The steps for solving of simultaneous equations graphically are as follows:

• Draw the graph of the first equation.

• Draw the graph of the second equation with the same axes and the same scale.

• From the graph, read the point of intersection of the two straight lines drawn.

• The required solutions are the x and y coordinates of the point which is the point of
intersection of two simultaneous equations.

Worked Out Examples

Example: 1 Y (4,3)
O
Solve graphically: 2x – y = 5 and x – y = 1 X' X
Y'
Solution:

Here, 2x – y = 5

or 2x – 5 = y

∴ y = 2x – 5 …(i) And x – y = 1

or x –1 = y

y = x – 1 …(ii)

x0 1 4 x0 2 3
y -5 -3 3 2
y -1 1

298 Oasis School Mathematics-8

Now, plotting these points on the graph, we get two lines.
From the graph,
Point of intersection of equation (i) and (ii) is (4,3)
∴ x = 4, y =3.

Example: 2 Y

Solve graphically: 2x + 3y = 13, y = 3.

Solution:

Given equations, 2x +3y = 13 ………….(i) (2, 3)

or, 3y = 13 – 2x 2x+3y=13y = 3
or,
y = 13–2x X' O X
x 3

2 -1 5 Y'

y = 13–2x 3 51
3

y = 3………… (ii)
Again,

x 1 23
y=3 3 3 3


From the graph,

Point of intersection of (i) and (ii) is (2, 3).

∴ x = 2, y = 3

Exercise 23.6

1. Write down the solution of each pair of equations:

(a) Y X (b) Y X
Y'
X' 4 X'
3
2
1

123

Y'

Oasis School Mathematics-8 229999

(c) Y (d) Y X
O
X' X X' O



Y' Y'

2. Complete the following table and plot them on the graph and find the solution.

( a) x x

y = 2x+1 y = 4x–1

(b) x x
y=2 y = x+1

(c) x x

y = x–6 y=2–x


3. Solve the following equations graphically:

(a) x + y = 6, x – y = 2 (b) 2x – y = 5, x + y = 4

(c) x + 2y = 1, x – y = 4 (d) x – 2y = 5, 2x –y =1

(e) 2x = 4 + y, x + 3 = –2y (f) x + 2y + 16 = 0, 2x + 4 – 3y = 0
(h) x = 3, 2x = 26 – 5y
(g) y= 3-5x , y = –4
3

(i) 3x +y = 2, 3x – 2y = 5

4. Convert the following verbal problems into the equations and solve them
graphically.

(a) Find the two numbers whose sum is 7 and difference is 1.

(b) The sum of two numbers is 6 and their difference is 2. Find the numbers.

(c) The difference of two numbers is 2. One number is one third of the other. Find
the numbers.

Answer

1. Consult your teacher. 2. Consult your teacher.
3. (a) 4, 2 (b) 3, 1 (c) 3, -1 (d) -1, -3 (e) 1, -2 (f) -8, -4 (g) 3, -4 (h) 3, 4 (i) 1, -1
4. (a) 4,3 (b) 4, 2 (c) 1, 3)

300 Oasis School Mathematics-8

Additional Text

Elimination method and substitution method of solving simultane-
ous equations

Elimination Method

In this method we make the coefficient of one of the variables equal and we add or sub-
tract the given equations to eliminate that variable. We substitute the value of that varia-
ble in an equation to get the value of other variable.

For example, solve the equations, x +y = 5 and x –y = 3

Solution:

Here, x +y = 5 …………… (i)

x–y = 3 …………… (ii)

Adding equation (i) and (ii),

x + y = 5

x – y = 3

2x = 8

or, x = 8 = 4
2

Substituting the value of x in equation (i), Note:

x + y = 5 If the co-efficeints of a variable are
same in both equations and their sign

or, 4+y = 5 is opposite add the equations. If their

or, y = 5 – 4 sign is same, subtract the equations.

y = 1

∴ x = 4

y = 1.

Substitution method

In this method, we express one of the variables in terms of other and we substitute its
value in other equation to get the values of the variable.

For example solve the equations x +y = 7 and 2x –y = 5

Solution:

Given equation, x +y = 7 ……………(i)

2x – y = 5 ……………(ii)

From equation (i) y = 7 – x

Oasis School Mathematics-8 330011

Substituting this value in equation (ii)

we get, 2x –y = 5

or, 2x – (7–x) = 5

or, 2x – 7 + x = 5

or, 3x = 5 + 7

or, 3x = 12

or, x = 12 = 4.
3

Again, substituting the value of x in (i)

x + y = 7

or, 4 + y = 7

or, y = 7–4 = 3

∴ x = 4 and y=3.

Worked Out Examples

Example: 1

Solve the given equations by elimination method: 3x – 2y = 5 and 2x +y = 8

Solution:

Given equation, 3x –2y = 5 ………………….. (i)

2x +y = 8 ……………………(ii)

Multiplying equation (ii) by 2 and adding these equations,

3x –2y = 5

4x +2y = 16

Adding, 7x = 21 Substituting the value of x in (i)
or,
or, x = 21 3x – 2y = 5
or, 7
or, or, 3×3–2y = 5
or, x = 3.
∴ x = 3 and y = 2
2y = 4 or, 9 –2y = 5

y = 4 or, –2y = 5 – 9
2
or, –2y = –4
y = 2.

Cancel (–) sign from both sides.

302 Oasis School Mathematics-8

Example: 2

Solve the given equations by substitution method: 2x +y =5 and 3x –2y = 4

Solution:

Given equations, 2x +y = 5 ………………(i)

3x – 2y = 4 ……………..(ii)

From equation(i), y = 5 – 2x

Substituting this value in equation (ii)

3x –2y = 4

or, 3x – 2(5–2x) = 4
or,
or, 3x – 10+4x = 4 Substituting the value of x in (i) 2x +y = 5
or, 2×2+y = 5
or, 7x – 10 = 4 or, 4 + y = 5
or, or, y = 5 –4
or, 7x = 4+10 or, y = 1
∴ x = 2 and y = 1.
7x = 14

x = 14
7

x = 2

Exercise 23.7

1. Solve the following simultaneous equations (by elimination method).

(a) x + y = 3, x – y = 1 (b) 2x + y = 7 and 3x – y = 8

(c) 3x – 2y = 4, 3x + y = 7 (d) 2x + y = 8, 2x + 3y = 12

(e) 3x – y = 15, 3x + 2y = 24 (f) 2x + 3y = 15, x –2y = –3

(g) 2x + y = 1, 3x – 2y = 5

2. Solve the following simultaneous equations (by substitution method).

(a) –y = –x + 4, 2x – 3y = 6 (b) x – y = 2, x + y = 6

(c) x + 2y = 7, 2x + 3y = 12 (d) 3x + 2y = 5, 3x – y = 2

(e) 2x – y = 3, 3x + 2y = 8 (f) 2x + y = 6, 2y = 2 + x

(g) y = 2x, 4x + 3y = 20

Answer
1. (a) 2, 1 (b) 3,1 (c) 2,1 (d) 3, 2 (e) 6, 3 (f) 3,3 (g) 1, -1
2. (a) 6, 2 (b) 4, 2 (c) 3, 2 (d) 1,1 (e) 2, 1 (f) 2,2 (g) 2,4

Oasis School Mathematics-8 330033

23.7 Quadratic Equation

Let's take an equation x² + 2x + 5 = 0. It is the second degree equation with one
variable which is in the form of ax² + bx + c = 0. This type of equation is the quadratic
equation

where, a = 1, b = 2 and c = 5
Hence an algebraic equation having degree two is the quadratic equation.
It is in the form of ax2 + bx + c = 0. where a, b, c are constants.
Examples:
x2 – 5x + 6 = 0
x2 – 2x = 0
x2 – 8 = 0, etc.
All these are in the form of ax2 + bx + c = 0.
In the first equation, a = 1, b = –5, c = 6
In the second equation, a = 1, b = –2, c = 0
and in the third equation, a = 1, b = 0, c = –8

Solution of Quadratic Equations
A quadratic equation can be solved by different methods. Here we are discussing

the solution of quadratic equation by factorisation method.

Steps

• Simplify the given equation and convert it in the form of ax2+bx+c = 0
• Factorise the expression on the left hand side.
• Write each factor is equal to zero to get the values of x.

Worked Out Examples

Example: 1

Solve: (x+1) (x–3) = 0

Solution:

Here, (x + 1) (x – 3) = 0

Either, (x + 1) = 0

or, x – 3 = 0

or, x = 3, or, x = –1

Hence, x = –1, 3.

304 Oasis School Mathematics-8

Example: 2

Solve: x2 = 5x

Solution: x2 = 5x
Here, x2 – 5x = 0
or, x(x – 5) = 0
or,
Either, x = 0
or, x – 5 = 0

x = 5

Hence, x = 0, 5

Example: 3

Solve: x2– 49 = 0

Solution: x2 – 49 = 0
Here,

or (x)2 – (7)2 = 0

or, (x + 7)(x – 7) = 0

Either, x + 7 = 0 ⇒ x = –7

or, x – 7 = 0 ⇒ x = 7

Hence, x = –7, 7

Example: 4

Solve: 2x2 – 5x – 12 = 0

Solution: 2x2 – 5x – 12 = 0
Here, 2x2– (8 – 3)x – 12 = 0
or,

or, 2x2 – 8x +3 x – 12 = 0

or, 2x(x – 4) + 3 (x – 4) = 0

or, (x – 4) (2x + 3) = 0

Either, x–4 = 0, or, 2x + 3 = 0

or, x = 4, or , x = -3
Hence, x = 4, -23 2

Oasis School Mathematics-8 330055

Example: 5

Solve: 3x – 7 = x+1
2x – 5 x–1

Solution:

Here, 3x – 7 = x+1
2x – 5 x–1

or (3x – 7) (x – 1) = (x + 1) (2x – 5)

or 3x2 – 3x – 7x + 7 = 2x2 – 5x + 2x – 5

or 3x2 – 10x + 7 = 2x2 – 3x – 5

or 3x2 – 2x2 – 10x + 3x + 7 + 5 = 0

or x2 – 7x + 12 = 0

or x2 – 4x – 3x + 12 = 0

or x(x – 4) – 3(x – 4) = 0

or (x – 4) (x – 3) = 0

Either, x – 4 = 0 or, x – 3 = 0

∴ x = 4 x=3

Hence, x = 4, 3.

Example: 6

If 5 times of a number is added to the square of the number, the result is 66. Find the
numbers.

Solution:

Let the required number be x. Then by the given condition,

x2 + 5x = 66

or, x2 + 5x – 66 = 0

or, x2 + 11x – 6x – 66 = 0

or, x(x + 11) – 6(x + 11) = 0

or, (x + 11) (x – 6) = 0

Either, x + 11 = 0

or, x – 6 = 0

∴ x = –11 ∴ x = 6

Hence, the required number = 6, –11.

306 Oasis School Mathematics-8

Exercise 23.7

Solve the following:

1. (a) (x–3) (x+2) = 0 (b) (x+6) (x–6) = 0

(c) (x+3) (x+4) = 0 (d) (x+1) (x–5) = 0

2. (a) x2–2x=0 (b) 2x2+5x=0

(c) 3x2 = 6x (d) x2 = 5x

3. (a) x2 – 25 = 0 (b) x2 – 81 = 0

(c) x2 – 16 =0 (d) 4x2–81 = 0

4. (a) x2 + 5x + 6 = 0 (b) x2 + x – 20 = 0

(c) x2 – 3x – 10 = 0 (d) 2x2 – x– 6 = 0

(e) x2–9x = 70 (f) x2–10x = 39

(g) 3x2 = 2x + 8 (h) 2x2 = 3x – 1

5. (a) x + 1 = 5 (b) x+1 = x–1
x 2 3x–7 2x–5

6. Solve the following problems:

(a) If 6 is added to the square of a number, the result is 31. Find the number.

(b) If 2 is subtraced from the half of the square of a number, the result is 30. Find
the numbers.

(c) If 5 times a number is added to its square, the result is 6. Find the numbers.

(d) If 6 times of a number is added to the square of the number, the result is 55.
Find the numbers.

(e) One number is 5 less than another and their product is 36. Find the numbers.

(f) If the sum of two numbers is 6 and the product is 8, find the two numbers.



Answer
1. (a) 3, -2 (b) -6, 6 (c) -3, -4 (d) -1, 5 2. (a) 0, 2 (b) 0, -5/2 (c) 0, 2 (d) 0, 5
3. (a) 5, -5 (b) 9, -9 (c) 4, -4 (d) 9/2, -9/2 4. (a) -3, -2 (b) -5, 4 (c) 5, -2 (d) 2, -3/2
(e) -5, 14 (f) -3, 13 (g) 2, -4/3 (h) 1, 1/2
5. (a) 2, 1/2 (b) 3, 4 6. (a) 5, -5 (b) 8, -8 (c) -6, 1 (d) -11, 5 (e) 4, 9 and -4,-9 (f) 4, 2

Oasis School Mathematics-8 330077

23.8 Inequality (Inequation)

Inequality: Let us consider the mathematical statement x + 3 = 7. Value of x deter-
mines whether the given statement is true or false.

If x = 2, 2+3 = 7 (false statement)
If x = 4, 3+4 = 7 (true statement)
Such open statement containing (=) sign is the equation. Open statement containing

trichotomy signs such as <, ≤,>,≥ are called inequality.
For example: x < 5 : x is less than 5
x ≤ 4 : x is less than or equal to 4
x > 6 : x is greater than 6
x ≥ 7 : x is greater than or equal to 7.

S.N. Some mathematical notations: Meanings

1. a > b → a is greater than b.
2. a < b → a is smaller than b.
3. a ≥ b → a is greater than or equal to b.
4. a ≤ b → a is smaller than or equal to b.

Note: <means less than, > means greater than, ≤ means less than or equal to ≥ means greater than or equal to

Properties of inequalities

1. If same number is added to both sides or subtracted from the both sides, the sign of
inequality remains the same.

i.e. a < b ⇒ a + c < b + c
and a – c < b – c.

2. If both sides of an inequality are multiplied or divided by a positive number, the

inequality sign remains the same.

i.e. a < b ⇒ ac < bc and a < bc if c > 0.
c

3. If both sides of an inequality are multiplied or divided by a negative number, the

inequality sign alters.

i.e. a < b ⇒ ac > bc and a < bc if c < 0. and -a > -b
c

Solution sets on a number line

Suppose we have to show all the numbers between any two given numbers say –2
and 5. See the following number line.

-5 -4 -3 -2 -1 0 12 34 5

Here the line represents all the real numbers. The right arrow shows the direction for
increasing order of numbers while the left arrow shows the direction for decreasing

308 Oasis School Mathematics-8

order of numbers. Hence the line represents a real number line.

Inequality in the number line:

Let's observe the following number line.

(i) x≥2

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

(ii) x ≥ 1

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

(iii) -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 –1< x ≤ 6

(iv) -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 -2 ≤ x ≤ 3

Worked Out Examples

Example: 1

Show the following inequalities in the number line.
(a) x ≥ –3 (b) x < 2 (c) 1 ≤ x < 5
Solution:

(a)
x ≥ -3

-4 -3 -2 -1 0 1 2 3 4

(b) x<2

-4 -3 -2 -1 0 1 2 3 4 5

(c) 1≤x<5

-4 -3 -2 -1 0 1 2 3 4 56

Example: 2

Write the inequality represented by the given graph.

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

Solution:
This number line represents all the numbers less than or equal to 4.
∴ Required inequality is x ≤ 4.

Oasis School Mathematics-8 330099

Example: 3

Find the solution set for the given inequality in number line.
2x + 5 ≥ x + 8
Solution:
2x+5 ≥ x+8
or, 2x+5-5 ≥ x+8–5 [Subtracting 5 from both sides]
or, 2x ≥ x+3
or, 2x–x ≥ x–x+3 [Subtracting x from both sides]
or, x ≥ 3

-3 -2 -1 0 1 2 3 4 5

Example: 4

Solve the following inequality and find its solution set.

2x – 4 ≤ 6 ∈ N

Solution:

2x – 4 ≤ 6

or, 2x – 4 + 4 ≤ 6 + 4

or, 2x ≤ 10

or, 2xx ≤ 10
x

or, x ≤ 5.

Since x is less than or equal to 5, and x ∈ N.

∴ Required solution set is {5, 4, 3, 2, 1}

Example: 5

Solve the given inequalities and represent in a number line
–3 < x –2 ≤ 4
Solution:
Here, –3 < x – 2 ≤ 4
Now, –3 < x – 2 ≤ 4
or, –3+2<x–2+2 ≤ 4 + 2 [adding 2 in all sides]
or –1 < x ≤ 6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

310 Oasis School Mathematics-8

Exercise 23.7

1. Show the following inequalities in the number line.

(a) x < 2 (b) x ≥ -5 (c) x ≤ -3

(d) x > 4 (e) -1 ≤ x < 3 (f) 3 < x < 7

2. Write down the inequalities represented by each of the following graphs.

(a)
-4 -3 -2 -1 0 1 2 3 4 5

(b)

-4 -3 -2 -1 0 1 2 3

(c)

-4 -3 -2 -1 0 1 2

(d)

-4 -3 -2 -1 0 1 2 3

(e)

-4 -3 -2 -1 0 1 2 3

(f)
-4 -3 -2 -1 0 1 2 3 4 5

3. If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, find the solution set of the
following inequations.

(a) x ≥ 6 (b) x – 5 > 2 (c) x – 3 < 5 (d) 3x – 1 ≤ 5

4. Solve the following:

(a) x + 3 > 6 (b) 2x – 5 ≥ 5 (c) 3y + 2 ≤ 8

(d) 3x – 10 > 5x + 4 (e) 2x + 8 ≤ 3x + 4 (f) 2(x + 3) ≤ x + 2

(g) 2x+3 (x+1) > 2x + 9 (h) x+2 ≤ x+3 (i) 2x3-1 > 3x+1
4 5 2

5. Solve the following inequalities and find the solution set.

(a) 2x – 3 < 5; x ∈ N (b) 2x + 16 ≤ x + 18, x ∈ W

(c) 6 – 3x > 9; x ∈ I (d) – 1 ≤ x + 1 < 3, x ∈ I

6. Solve the following inequalities and show them in the number line.

(a) 2(x+5) < 4 + 5x (b) x –2 > 3x–1 (c) 2x+8 ≥ x + 5
3 4

(d) 2x+1 ≤ x+2 (e) -2 ≤ x – 4 ≤ 2 (f) – 1 ≤ 2x– 5 ≤ 3
4 3 2 2 2

(g) –6 ≤ 2x + 4 < 10

 Oasis School Mathematics-8 331111

Answer
1. Consult your teacher.
2. (a) x > 2 (b) x ≤ 2 (c) x<1 (d) x>-2 (e) -3 < x ≤ 2 (f) -1<x<4
3. (a) {6, 7, 8, 9} (b) {8, 9} (c) {0, 1, 2, 3, 4, 5, 6, 7} (d) {0, 1, 2}
4. (a) x > 3 (b) x ≥ 5 (c) y ≤ 2 (d) x < -7 (e) x ≤ 4 (f) x ≤ -4 (g) x>2 (h) x ≤ 2 (i) x<-1
5. (a) {1,2,3} (b) {0, 1, 2} (c) {..., -4, -3, -2} (d) {-2,-1, 0,1}
6. (a) x > 2 (b) x<-1 (c) x ≥ -3 (d) x ≤ 5/2 (e) 2 ≤ x ≤ 6 (f) 1≤ x ≤ 2 (g) -5 ≤ x < 3

Assessment Test Paper

Full marks : 30

Attempt all the questions.

Group 'A' [4 × 1 = 4]

1. (a) Factorise : x3 – 8

(b) Solve : 52x = 1

(c) Find the slope of the straight line 3y = 6x + 5

(d) Solve : 2x ≤ –4

Group 'B' [5 × 2 = 10]

2. (a) Simplify : a2 + b2
a–b b–a

(b) Simplify : 128x6y-3
64x-3y3

(c) Simplify : 5x+1–5x
4×5x

(d) Draw a graph of the line 2x + y = 5.

(e) Solve the given inequality and show it in the number line.

7 – 3x > 10.

Group 'C' [4 × 4= 16]

3. (a) If a + 1 = 5, find the value of a2+ 1 .
a a2

(b) Find the H.C.F. of 8a3–27, 4a2–12a+9.

(c) Simplify: 1 + 1 – 1
a+b a–b a2–b2

(d) Solve graphically: y = x+2 and y = 3x



312 Oasis School Mathematics-8

BLE New Model Question

Time : 3 hrs. F.M. 100

(Compulsory Mathematics)

(Based on New Curriculum and Specification Grid)

A C
(Group 'A'] [10×1=10]
H [Ans: ∠DHF]
1. a. What is the corresponding angle of E F
∠BGH in the given figure? G [Ans: 77cm²]

B D
7cm
A O B

b. Find the area of the given semi-circle.

2. a. Find the length of AB in the given A B
right angled triangle ABC. 5cm C
15cm
20cm [Ans: 3cm]
4cm
b. Draw the compass bearing of 060°.

3. a. If A = {a, e, i, o, u} and B = {e, a, r}, find A∩B. [Ans: {e, a}]

b. Convert 6.201 × 104 into normal form. [Ans: 62010]

4. a. Find the range: (Temperature): 25°C, 28°C, 30°C, 32°C, 33°C. [Ans: 8°C]

b. Factorise: 4x² – 1 [Ans: (2x + 1) (2x – 1)]

5. a. If mx=ax, find the value of x. [Ans: 0]

b. Find the slope of the line: y – 2x = 6 [Ans: 2]

E
(Group 'B') [17×2=34]

6. a. Find the value of x in the AB
given figure. 3x+5
C 1250 HD [Ans: 40°]

A

b. Show that ∠A = 90° in the 3a

given figure. 2a
a

B B 25cm C
10cm
c. In the figure, if ∆ABC ~ ∆ADE,
DE
find the length of AB. [Ans: 37.5cm]
A

A

7. a. Prove that

∆ADB ≅ ADC in the figure.

B DC

b. In the given semi-circle if the length of C

ACB is 11cm, find the length of AB. A B [Ans: 7 cm]
O

Oasis School Mathematics-8 331133

c. Draw a net of a cube.

Y

8. a. Find the length of AB in A X [Ans: 5 units]
the given figure. X' O

B

Y'

b. If the length and area of a rectangle are 6m and 30m² respectively, find its perimeter. [Ans: 22m]

c. If U = {1, 2, 3, 4, 5} and A = {2, 4}, find A. [Ans: {1, 3, 5}]

9. a. If a, m and b are in proportion show that m² = ab.

b. Convert 1002 into decimal. [Ans: 4]
c Find the median: 12, 13, 17, 10, 18, 15, 25 [Ans: 15]

10. a. Factorise: x² – 5x + 6 [Ans: (x – 2) (x – 3)]

b. If a = 2 and b = 2, find the value of ab × ba. [Ans: 16]
c. [Ans: 1]
11. a. Simplify: 3x+1–3x
3x × 2

Show the equation y = x in graph.

b. Solve and show in number line: -5x ≤ 10. [Ans: x ≥ -2]


(Group'C') [14×4=56]

12. If U = {1, 2, 3, ....10}, A = {prime numbers less than 10}, B = {even numbers up to 10} and

C = {1, 2, 3, 4, 5,}, find A∩B∩C by using Venn-diagram. [Ans: {2}]

13. If x – y = 5, find the value of x3–y3 – 15xy. [Ans: 125]

14. Find H.C.F. : x² – 4, x² + 3x + 2 [Ans: x + 2]
[Ans: 7x + 3 ]
15. Simplify: 3 + 4 – 24
x+3 x–3 x2–9 [Ans: 3, 2]

16. Solve graphically: x + y = 5 and x – y = 1

17. Find the mean from the given data: [Ans: 40]

(Marks obtained) 10 20 30 40 50 60
(No. of student) 2 3 54 5 6

18. The length of a hall is 20m and breadth is one fourth of length. If the volume of the hall is 400m³, find

the height of the room. [Ans: 4m]

19. Simplify: 110112 + 111112 – 100012 [Ans: 1010012]

20. A shopkeeper bought a watch for Rs.1,000 and fixed its price 25% above it. If he sold it by allowing

10% discount, find the selling price. [Ans: Rs.1125]

21. 20 men can complete a piece of work in 36 days. How many men should be added to complete the

work in 24 days? [Ans: 10]

22. Find the rate of interest if Rs. 4,000amounts to Rs.6,000 in 2 years. [Ans: 25%]

23. Draw a triangle ABC with vertices A(-1, 3), B(3, 6) and C(5, -2) on a graph paper. Reflect it in X-axis

and show the image A'B'C' on the same graph paper. [Ans: (-1, -3), B'(3, -6), C' (5, 2)]

24. Verify experimentally that diagonals of a parallelogram bisect each other. [Two figures of different
measures are necessary]

25. Construct a rectangle ABCD where diagonal AC =BD = 10 cm and angle between them is 60°.

314 Oasis School Mathematics-8