Unit Test (Solutions): Number Systems | Mathematics (Maths) Class 9 PDF Download

Time: 1 hour
M.M.: 30 
Attempt all questions. 
Question numbers 1 to 5 carry 1 mark each. 
Question numbers 6 to 8 carry 2 marks each. 
Question numbers 9 to 11 carry 3 marks each. 
Question numbers 12 & 13 carry 5 marks each.

Q1:  What is the product of a rational and an irrational number? (1 Mark)
(i) Always an integer
(ii)Always a rational number
(iii) Always an irrational number
(iv)Sometimes rational and sometimes irrational

Ans:(iii)
The product of a rational and an irrational number is always an irrational number. For example, 2 is a rational number and √3 is irrational. Thus, 2√3 is always an irrational number.

Q2: What is the value of (256)^0.16 X (256)^0.09? (1 Mark)
(i)4
(ii)16
(iii)64
(iv)256.25

Ans: (i)

(256)^0.16 x (256)^0.09 = (256)^{(0.16 + 0.09)}

= (256)^0.25

= (256)^(25/100)

= (256)^(1/4)

= (4⁴)^(1/4)

= 4^4(1/4)

= 4

Q3: Add 2√2+ 5√3 and √2 – 3√3. (1 Mark)

Ans:

(2√2 + 5√3) + (√2 – 3√3)

= 2√2 + 5√3 + √2 – 3√3

= (2 + 1)√2 + (5 – 3)√3

= 3√2 + 2√3

Q4: Simplify: (√3+√7) (√3-√7). (1 Mark)

Solution:

(√3 + √7)(√3 – √7)

Using the identity (a + b)(a – b) = a² – b²,

(√3 + √7)(√3 – √7) = (√3)² – (√7)²

= 3 – 7

= -4

Q5: Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number. (1 Mark)

Ans: No, since the square root of a positive integer 16 is equal to 4. Here, 4 is a rational number.

Q6: Find five rational numbers between 3/5 and 4/5. (2 Marks)

Ans:

We have to find five rational numbers between 3/5 and 4/5.

So, let us write the given numbers by multiplying with 6/6, (here 6 = 5 + 1)

Now,

3/5 = (3/5) × (6/6) = 18/30

4/5 = (4/5) × (6/6) = 24/30

Thus, the required five rational numbers will be: 19/30, 20/30, 21/30, 22/30, 23/30

Q7: Rationalise the denominator of 1/[7+3√3]. (2 Marks)

Ans:

1/(7 + 3√3)

By rationalizing the denominator,

= [1/(7 + 3√3)] [(7 – 3√3)/(7 – 3√3)]

= (7 – 3√3)/[(7)² – (3√3)²]

= (7 – 3√3)/(49 – 27)

= (7 – 3√3)/22

Q.8: Simplify: (2 Marks)

(i) 7^2/3.7^1/5
(ii) 10^1/2/10^1/4

Ans:

(i) 7^2/3.7^1/5

Bases are equal, so add the powers.

7^{(2/3 + 1/5)}

= 7^{(10 + 3)/15}

= 7^13/15

(ii) 10^1/2/10^1/4

Bases are equal, so subtract the powers.

= 10 ^{(1/2 – 1/4)}

= 10^1/4

Q9: Find three different irrational numbers between the rational numbers 5/7 and 9/11. (3 Marks)

Ans:

The given two rational numbers are 5/7 and 9/11.

5/7 = 0.714285714…..

9/11 = 0.81818181……

Hence, the three irrational numbers between 5/7 and 9/11 can be:

0.720720072000…

0.730730073000…

0.808008000…

Q.10: Represent √(9.3) on the number line. (3 Marks)

Ans:

Representation of √9.3 on the number line is given below:

Q11: Locate √3 on the number line. (3 Marks)

Ans:

Construct BD of unit length perpendicular to OB (here, OA = AB = 1 unit) as shown in the figure.

By Pythagoras theorem,

OD = √(2 + 1) = √3

Taking O as the centre and OD as radius, draw an arc which intersects the number line at the point Q using a compass.

Therefore, Q corresponds to the value of √3 on the number line.

Q12: Find the decimal expansions of 10/3, 7/8 and 1/7. (5 Marks)

Ans:

Therefore, 10/3 = 3.3333…

7/8 = 0.875

1/7 = 0.1428571…

Let x = 0.3333…. 

Multiply with 10,

10x = 3.3333…

Now, 3.3333… = 3 + x (as we assumed x = 0.3333…)

Thus, 10x = 3 + x

10x – x = 3

9x = 3

x = 1/3

Therefore, 0.3333… = 1/3. Here, 1/3 is in the form of p/q and q ≠ 0.

Q13: What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer. (5 Marks)

Ans:

Thus, 1/17 = 0.0588235294117647….

Therefore, 1/17 has 16 digits in the repeating block of digits in the decimal expansion.