RS Aggarwal Solutions: Number System- 4 | Extra Documents & Tests for Class 9 PDF Download

Q.1. Add:
(i) (2√3 − 5√2) and (√3+2√2)
(ii) (2√2 + 5√3 − 7√5)and(3√3 − √2 + √5)
(iii)
Ans.
(i) 2√3 − 5√2 + √ 3 + 2√2
= (2√3 + √3) + (2√2 − 5√2)
= 3√3− 3√2
(ii) 2√2+ 5√3 − 7√5 + 3√3 − √2 + √5
= 2√2 - √2 + 5√3 + 3√3 + √5  − 7√5
= √2 + 8√3 −6√5
(iii)


Q.2. Multiply:
(i) 3√5 by 2√5
(ii) 6√15 by 4√3
(iii) 2√6 by 3√3
(iv) 3√8 by 3√2
(v) √10 by √40
(vi) 3√28 by 2√7

Ans.
(i) 3√5 × 2√5 = 3 × 2 × √5 x √5 = 6 × 5 = 30
(ii) 6√15 × 4√3 = 6 × 4 × √5 × √3 × √3 = 24 × 3 × √5 = 72√5
(iii) 2√6 × 3√3 = 2 × 3 × √2 × √3 × √3 = 6 × 3 × √2 = 18√2
(iv) 3√8 × 3√2 = 3 × 3 × √2 × √2 × √2 × √2 = 9 × 4 = 36

(v) √10 × √40 = √2 × √5 × √2 × √2 × √2 × √5 = √2 × √2 × √2 × √2 × √5 × √5 = 2 × 2 × 5 = 20

(vi) 3√28 × 2√7 =  × √7 = 6 × 7 × √4 = 42 × 2 = 84

Q.3. Divide:
(i) 16√6 by 4√2

(ii) 12√5 by 4√3
(iii) 18√21 by 6√7
Ans. 
(i)  
(ii)
(iii)

Q.4. Simplify
(i) (3 − √11) (3 + √11)
(ii) (−3 + √5) (−3 − √5)
(iii) (3 − √3)²
(iv) (√5 − √3)²

(v) (5 + √7) (2 + √5)
(vi) (√5 – √2) (√2 - √3)
Ans. (i) (3 − √11) (3 + √11)
=3² − (√11)²    [(a − b)( a + b) =a² − b²]
=9 − 11
= −2
(ii) (−3 + √5) (−3 − √5)
=(−3)²−(√5)²   [(a + b)(a − b) = a² − b²]
=9 − 5
=4
(iii) (3 − √3)²
=3²+(√3)² − 2 × 3 × √3   [(a−b)²=a²+b²−2ab]

=9 + 3 − 6√3
=12 − 6√3
(iv) (√5 − √3)²
=(√5)²+ (√3)² − 2 × √5√3   [(a−b)² = a² + b² − 2ab]
=5 + 3 − 2√15 = 8 − 2√15

= √5 × √2 − √5 × √3 - √2 × √2 + √2 × √3
= √10 − √15 − 2 + √6
(v) (5 + √7) (2 + √5)
(5 + √7) (2 + √5)
= 5 × 2 + 5 × √5+ √7 × 2 +√7 × √5
=10 + 5√5 + 2√7 + 3√5
(vi) (√5 − √2) (√2 − √3)
(√5 - √2)(√2 − √3)
= √5 × √2 – √5 × √3 − √2 × √2 +  √2  × √3

= √10 − √15 − 2 + √6

Q.5. Simplify (3 + √3) (2 + √2)².
Ans. (3 + √3) (2+√2)²
= (3 + √3) [2²+(√2)² + 2 × 2√2]
=(3 + √3) [4 + 2 + 4√2]
= (3 + √3) [6 + 4√2]
=3 × 6 + 3 × 4√2 + √3 × 6 + √3 × 4√2
=18 + 12√2 + 6√3 + 4√6

Q.6. Examine whether the following numbers are rational or irrational:
(i) (5 − √5) (5 + √5)
(ii) (√3 + 2)²
(iii)
(iv) √8 + 4√32 − 6√2
Ans. (i) (5 − √5) (5 + √5)
=5² − (√5)²   [(a − b)(a + b) = a² − b²]
= 25 − 5
= 20, which is an integer
Hence, (5 − √5) (5 + √5) is rational.
(ii) (√3 + 2)²
= (√3)²+2²+2×√3×2    [(a + b)²=a² + b² + 2ab]
= 3 + 4 + 4√3
= 7 + 4√3
Since, the sum and product of rational numbers and an irrational number is always an irrational.
⇒7 + 4√3 is irrational.
Hence, (√3 + 2)² is irrational.
(iii)


= −1, which is an integer
Hence, is rational.

(iv) √8 + 4√32 −6√2
=2√2 + 4 × 4√2 − 6√2
=2√2 + 16√2 − 6√2
=12√2
Since, the product of a rational number and an irrational number is always an irrational.
Hence, √8 + 4√32 − 6√2 is rational.

Q.7. On her birthday Reema distributed chocolates in an orphanage. The total number of chocolates she distributed is given by (5+ √11) (5− √11).
(i) Find the number of chocolates distributed by her.
(ii) Write the moral values depicted here by Reema.
Ans.

(i) As, (5 + √11) (5 − √11)
=5²−(√11)² [(a+b)(a−b)=a²−b²]
=25 − 11
=14
Hence, the number of chocolates distributed by Reema is 14.
(ii) The moral values depicted here by Reema is helpfulness and caring.
Disclaimer: The moral values may vary from person to person.

Q.8. Simplify
(i)
(ii)
(iii)
Ans.
(i) 3√45 − √125 + √200 − √50

= 3×3√5 – 5√5 + 10√2 − 5√2
= 9√5 − 5√5 + 5√2
=4√5 + 5√2
(ii)

(iii)

= 6√2 + 20√2 − 3√2
= 23√2