MATHEMATICS

CLASS – IX
Topic: Number systems, Polynomials and Linear equations in two variables

Time Allowed: 1 hour Maximum Marks: 30 Marks

General Instructions:
1. All questions are compulsory.
2. The question paper consists of twelve questions divided into 4 sections A, B, C and D.
Section A comprises of five questions of 1 mark each. Section B comprises of two
questions of 2 marks each. Section C comprises of three questions of 3 marks each.
Section D comprises of two questions of 6 marks each.
3. All questions in section A are to be answered in one word, one sentence or as per the
exact requirement of the question.

SECTION A

1. Find the value of k if x = 2, y = 1 is a solution of the equation 2x – 3y = k.

2. Give an example of two irrational numbers whose quotient is a rational number.

3. For polynomial p( x ) = 3 x 3 - 2 x 2 + 2 , write the value of p(–1).

3
4. What is the rationalizing factor of 121?

5. Express x = 3y in the form ax + by + c = 0 and indicate the value of a, b and c in the given
equation.

SECTION B

6. Express y in terms of x, given that 2x – 5y = 7. Check whether the point (–3, –2) is on the
given line.

7. Without actually calculating the cube, evaluate (0.7)3 + ( -0.8)3 + (0.1)3 .

SECTION C

8. The following values of x and y are supposed to satisfy the equation y = ax + b. Find the
values of a and b.
x 0 1 –1 2 –2
 y 3 5 1 7 –1

9. Using factor theorem, factorize : 3 x 3 - 4 x 2 - 12 x + 16 .

5+ 3
10. Find the value of a and b in the following : = 47a + 3b .
7-4 3

SECTION D

11. (i) Using suitable identity, prove that :

(0.87)3 + (0.13)3
=1
(0.87) 2 - (0.87 ´ 0.13) + (0.13) 2
(ii) Simplify :
3 3
- - -3
æ 81 ö 4 æ 25 ö 2 æ2ö
ç ÷ ´ç ÷ ´ç ÷ .
è 16 ø è 9 ø è5ø

æ9ö
12. Linear equation for converting Fahrenheit to Celsius is F = ç ÷ C + 32
è5ø
(i) Draw the graph of this equation using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature is 30º C, find the temperature in Fahrenheit.
(iii) If the temperature is 95º in Fahrenheit, find the temperature in Celsius.
(iv) If the temperature is 0º C, find it in Fahrenheit.
(v) If the temperature is 0º F, find it in Celsius.

MATHEMATICS
CLASS – IX
Topic: Number systems, Polynomials and Linear equations in two variables

SOLUTIONS

1. Putting x = 2, y = 1 in the given equation,

 2(2) – 3(1) = k [½]
4–3=k
1=k [½]

2. Let 1st irrational number be = 8 3 and 2nd irrational number be = 2 3 [½]

8 3
Quotient = = 4 which is a rational number. [½]
2 3

3. Given polynomial is
p( x ) = 3 x 3 - 2x 2 + 2
p( -1) = 3( -1)3 - 2( -1) 2 + 2 [½]
p( -1) = - 3 - 2 + 2
p( -1) = -5 + 2 [½]

4. Rationalsing factor of 3
121 is 3
11 as 3
121´ 11 = 3 11´ 11´ 11 = 11. [1]

5. Given expression is x = 3y
or x – 3y = 0 [½]
Comparing it with ax + by + c = 0
a = 1, b = –3, c = 0 [½]

6. Given equation is 2x – 5y = 7
or 5y = 2x – 7
2x - 7
or y= is the required form. [1]
5
Put x = -3, y = –2 in the LHS of given equation, we have
= 2(–3) –5(–2)
Þ –6 + 10 = 4 ¹ RHS [1]
\ (–3, –2) is not a point on the given line.

7. We know that
a 3 + b 3 + c 3 = 3abc if a + b + c = 0
Here a = 0.7, b = – 0.8, c = 0.1
\ a + b + c = 0.7 – 0.8 + 0.1 = 0 [1]
\ (0.7)3 + ( -0.8)3 + (0.1)3 = 3(0.7)( -0.8)(0.1)
= – 0.168 [1]

8. Given equations is y = ax + b
Putting x = 1, y = 5 we get
5 = a ´ 1+ b
Þ 5=a+b …(i)
Putting x = -1, y = 1 we get
1 = -a + b …(ii) [1]
Adding (i) and (ii)
5 + 1 = 2b
6 = 2b
 6
=b
2
3=b [1]
Putting the value of b in equation (i) we get
5=a+3
5–3=a
2=a [1]

9. Let p( x ) = 3 x 3 - 4 x 2 - 12 x + 16
Factors of constant term [16] are ± 1, ± 2, ± 4, ± 8, ± 16
We find that
p(–2) = 3( -2)3 - 4( -2)2 - 12( -2) + 16
= –24 – 16 + 24 + 16 = 0
\ (x + 2) is a factor [1]

3x2 – 10x + 8
x+2 3x3 – 4x2 – 12x + 16
3x3 + 6x2
– –
–10x2 – 12x + 16
–10x2 – 20x
+ +
8x + 16
8x + 16
– –
0 [1]
Thus we have
p(x) = ( x + 2)(3 x 2 - 10 x + 8)
= ( x + 2)(3 x 2 - 6 x - 4 x + 8)
= ( x + 2)[3 x ( x - 2) - 4( x - 2)]
= ( x + 2)( x - 2)(3 x - 4) [1]

5+ 3
10. = 47a + 3b
7-4 3
Multiplying Nr. and Dr. by 7 + 4 3 , we have

5+ 3 5+ 3 7+4 3 (5 + 3 )(7 + 4 3 )
= ´ = [1]
7-4 3 7-4 3 7+4 3 (7 - 4 3 )(7 + 4 3 )

5+ 3 35 + 5 ´ 4 3 + 7 ´ 3 + 3 ´ 4 3
Þ =
7-4 3 ( )
72 - 4 3
2

5+ 3 35 + 20 3 + 7 3 + 4 ´ 3
Þ = [1]
7-4 3 49 - 48

5+ 3 35 + 27 3 + 12
Þ = = 47 + 27 3
7-4 3 1
 5+ 3
\ = a + 3b
7-4 3
Þ 47 + 27 3 = a + 3b [1]
Þ a = 47 and b = 27.

11. (i) Take a = 0.87 and b = 0.13

We have a 3 + b 3 = (a + b )(a 2 - ab + b 2 )

(0.87) 3 + (0.13) 3
LHS = [1]
(.87) 2 - (0.87 ´ 0.13) + (0.13) 2
(0.87 + 0.13)[(0.87) 2 - (0.87 ´ 0.13) + (0.13) 2 ]
=
(0.87) 2 - (0.87 ´ 0.13) + (0.13) 2
= (0.87 + 0.13) = 1 = RHS [1]
3 3
- - -3
æ 81 ö 4 æ 25 ö 2 æ2ö
(ii) ç ÷ ´ç ÷ ´ç ÷
è 16 ø è 9 ø è5ø
3 3
3
æ 16 ö 4 æ 9 ö 2 æ 5 ö
= ç ÷ ´ç ÷ ´ç ÷ [1]
è 81 ø è 25 ø è2ø
3 3
æ 24 ö 4 æ 32 ö 2 æ 5 ö3
= çç 4 ÷ ´ç ÷ ´ç ÷
÷ ç 52 ÷
è3 ø è ø è2ø
3 3
éæ 2 ö 4 ù 4 éæ 3 ö 2 ù 2 æ 5 ö 3
= êç ÷ ú ´ êç ÷ ú ´ ç ÷ [1]
ëêè 3 ø ûú êëè 5 ø ûú è2ø
3 3
4´ 2´ 3
é2ù 4 é3 ù 2 æ5ö
=ê ú ´ê ú ´ç ÷
ë3 û ë5 û è2ø
3 3 3
æ2ö æ3ö æ5ö
= ç ÷ ´ç ÷ ´ç ÷ [1]
è3ø è5ø è2ø
3
é2 3 5ù
=ê ´ ´ ú
ë3 5 2û
= [1]
3

=1 [1]
12. (i) The graph of the line y
9 100 (35, 95)
F = C + 32
5
is shown in figure 90 (30, 86)
86
Required table is: 80 9
F= C + 32
5
C 5 -5 10 70
F 41 23 50
60

Fahrenheit (ºF)
(ii) From the graph: 50 (10, 50)
C = 30°
Þ F = 86° 40 (5, 41)
[1]
(iii) If F = 95° Þ C = 35° 30
on reading the graph. (– 5, 23)
[1] 20
(iv) If C = 0° (0, –17.8)
Þ F = 32° 10
on reading the graph
[1] -40 -30 -20 -10 0 10 20 30 x
(v) If F = 0° Celsius (ºC)
-10
Þ C = -17.8°
[1]
-20
(approximately)
-30

(-40, -40) -40

[2]