RS Aggarwal Solutions: Exercise 1F - Number System | Extra Documents & Tests for Class 9 PDF Download

Q.1. Write the rationalising factor of the denominator in
Ans.


Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in

Q.2. Rationalise the denominator of each of the following.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
Ans.
(i)
On multiplying the numerator and denominator of the given number by √7, we get:

(ii)
On multiplying the numerator and denominator of the given number by √3, we get:

(iii)
On multiplying the numerator and denominator of the given number by 2 - √3, we get:

(iv)
On multiplying the numerator and denominator of the given number by √5 +2, we get:

(v)

On multiplying the numerator and denominator of the given number by 5-3√2, we get:

(vi)
Multiplying the numerator and denominator by √7+√6, we get

(vii)
Multiplying the numerator and denominator by √11+√7, we get


(viii)
Multiplying the numerator and denominator by 2+√2, we get
 

(ix)
Multiplying the numerator and denominator by 3-2√2, we get


Q.3. It being given that √2 = 1.414, √3 = 1.732, √5 = 2.236 and √10 = 3.162, find the value of three places of decimals, of each of the following.
(i)
(ii)
(iii)

Ans.
(i)

(ii)

(iii)


Q.4. Find rational numbers a and b such that

(i)
(ii)
(iii)
(iv)

Ans.

(i)


(ii)



(iii)


(iv)




Q.5. It being given  that √3 = 1.732, √5 = 2.236, √6 = 2.449  and √10 = 3.162, find to three places of decimal, the value of each of the following.

(i)
(ii)
(iii)
(iv)
(v)
(vi)
Ans.
(i)


= 0.213
(ii)


=3 × (2.236 − 1.732)
= 1.512
(iii)

(iv)


(v)


= 16.660
(vi)


= 4.441

Q.6. Simplify by rationalising the denominator.
(i)
(ii)
Ans.
(i)



(ii)



Q.7. Simplify
(i)
(ii)
(iii)
(iv)
Ans.
(i)


(ii)


= 0
(iii)


= 16 − √3
(iv)


= 0

Q.8. Prove that
(i)
(ii)
Ans.
(i)


= 2/2
= 1
(ii)




Q.9. Find the values of a and b if

Ans.




Comparing with the given expression, we get
a = 0 and b = 1
Thus, the values of a and b are 0 and 1, respectively.

Q.10. Simplify

Ans.


Q.11. If x = 3 + 2√2, check whether is rational or irrational.
Ans.
x = 3 + 2√2    .....(1)

Adding (1) and (2), we get
which is a rational number
Thus, is rational.

Q.12.
If x = 2 − √3, find value of
Ans.
x = 2 − √3    .....(1)

Subtracting (2) from (1), we get

Thus, the value of

Q.13. If x = 9 − 4√5, find the value of
Ans.
x = 9 − 4√5    .....(1)

Adding (1) and (2), we get

Squaring on both sides, we get

Thus, the value of x² +  is 322.

Q.14. If x =  find the value of
Ans.


Adding (1) and (2), we get

Thus, the value of x +is 5.

Q.15. If a = 3 − 2√2, find the value of a² -
Ans.
a = 3−2√2
⇒ a² = (3−2√2)²
⇒ a² = 9 + 8 − 12√2
⇒ a² = 17 − 12√2    .....(1)
 

Subtracting (2) from (1), we get

Thus, the value of a² -

Q.16. If x = √13 + 2√3, find the value of x −
Ans.

Subtracting (2) from (1), we get

Thus, the value of x

Q.17.

If x = 2 + √3, find the value of
Ans.
 
Adding (1) and (2), we get

Cubing both sides, we get

Thus, the value of

Q.18. If andshow that
Ans.
Disclaimer: The question is incorrect.



The question is incorrect. Kindly check the question.
The question should have been to show that x − y =


Q.19. If a = and b = show that 3a² + 4ab − 3b² =
Ans.
According to question,



Now,
3a² + 4ab − 3b²
= 3(a² − b²) + 4ab
= 3 (a + b)(a − b) + 4ab


Hence, 3a² + 4ab − 3b² =

Q.20.
If a = and b =find the value of a² + b² – 5ab.
Ans.
According to question,


Now,

Hence, the value of a² + b² – 5ab is 93.

Q.21.
If p = and q = find the value of p² + q².
Ans.
According to question,



Now,
p² + q² = (p+q)² − 2pq

Hence, the value of p² + q² is 47.

Q.22. Rationalise the denominator of each of the following.
(i) 
(ii)
(iii)
Ans.
(i)

Hence, the rationalised form is
(ii)

Hence, the rationalised form is
(iii)


Hence, the rationalised form is

Q.23. Given, √2 = 1.414 and √6 = 2.449, find the value of correct to 3 places of decimal.
Ans.


Hence, the value of correct to 3 places of decimal is −1.465.

Q.24. If x = find the value of x³ – 2x² – 7x + 5.
Ans.

Now,

Also,

Now,

Hence, the value of x³ – 2x² – 7x + 5 is 3.

Q.25. Evaluate  it being given that √5 = 2.236 and √10 = 3.162.
Hint

Ans.


Hence,