RS Aggarwal Solutions Class 9 Ex 1B

RS Aggarwal Class 9 Ex 1B Chapter 1

Question 1: SIMPLIFY;

(i) ( 4 + 2 ) ( 4 – 2 )

(ii) (5 + 3 ) (53)

(iii) ( 6 – 6) ( 6 + 6)

(iv) (52) (23)

(v) (53 )2

(vi) (3 – 3)2

Solution:

(i) ( 4 + 2 ) ( 4 – 2 )

= 42 – (2)2 [ ( a + b) ( a – b) = a2 – b2]

= 16 – 2

= 14

(ii) (5 + 3 ) (53)

= (5)2 – (3)2

= 5 – 3

= 2

(iii) ( 6 – 6) ( 6 + 6)

= 62 – (6) 2

= 36 – 6

= 30

(iv) (52) (23)

= 5 x 25 x 32 x 2 + 2 x 3

= 1015 – 2 + 6

(v) (53 )2

= (5 )2 + (3 )2 – 2  x 5 3 [(a – b)2 = a2 + b2 – 2ab ]

= 5 + 3 – 215 = 8 – 215

(vi) (3 – 3)2

= 32  + (3)2 – 2  x 3 x 3 [ ( a – b)2 = a2 + b2 – 2ab ]

= 9 + 3 – 63

= 12- 63

Question 2: Represent 3.2 geometrically on the number line.
Solution:

Draw a line segment AB = 3.2 units and extend it to C such that BC = 1 unit.

Now, find the midpoint O of AC.

Taking O as the center and OA as the radius, draw a semicircle.

Now, draw BD AC, intersecting the semicircle at D.

Here,

BD = 3.2 units

Taking B as the centre and BD as the  radius, draw an arc meeting AC produced at E.

Thus, we have:

BE = BD = 3.2 units

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Question 3: Represent 7.28 geometrically on the number line.
Solution:

Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit.

Now, find the midpoint O of AC.

Taking O as the centre and OA as the radius, draw a semicircle.

Now, draw BD AC, intersecting the semicircle at D.

We have:

BD = 7.28 units

Now, taking B as the centre and BD as the radius, draw an arc meeting AC produced at E.

Thus, we have:

BE = BD = 7.28 units

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Question 4: Mention the closure property, associative law , commutative law , the existence of Identity , existence of inverse of each real numbers for each of operations

  1. addition
  2. multiplication on real numbers

Solution:

Addition Properties of Real Numbers:

(i) Closure property: The sum of two real numbers is always a real number.

(ii) Associative law: (a + b) + c = a + (b + c) for all real numbers a, b and c.

(iii) Commutative law: a + b = b + a for all real numbers a and b.

(iv) Existence of additive identity: 0 is called the additive identity for real numbers.

As, for every real number a, 0 + a = a + 0 = a

(v) Existence of additive inverse: For each real number a, there exists a real number ( -a) such that a + (-a) = 0 = (-a) + a. Here, a and (-a) are the additive inverse of each other.

Multiplication properties of Real Numbers:

(i) Closure Property: The product of two real numbers is always a real number.

(ii) Associative law: (ab)c = a(bc) for all real numbers a, b and c.

(iii) Commutative law: a x b = b x a for all real numbers a and b.

(iv) Existence of multiplicative identity: 1 is called the multiplicative identity for real numbers.

As for, every real number a, 1 x a = a x 1 = a

(v) Existence of multiplicative inverse: For each real number a , there exists a real number 1a such that a 1a = 1 = 1aa. Here, a and 1a are the multiplicative inverse of each other.

RATIONALISE THE DENOMINATOR OF EACH OF THE FOLLOWING.

Question 5: 7

Solution:

On multiplying the numerator and denominator of the given number by 7 , we get:

17 = 17 x 77 =  77

Question 6:

Solution: 523

On Multiplying the numerator and denominator of the given number by 7 , we get:

523 = 523 x 33 = 33

Question 7: 12+3

Solution:

On Multiplying the numerator and denominator of the given number by 2 – 3 , we get:

12+3 = 12+3 x 2323

= 23(2)2(3)2

= 2343

= 231

= 2 – 3

Question 8: 152

Solution:

On Multiplying the numerator and denominator of the given number by 5 + 2 , we get:

152 = 152 x 5+25+2

= 5+2(5)2(2)2

= 5+254

= 5+21

= 5+2

Question 9: 15+32

Solution:

On Multiplying the numerator and denominator of the given number by 5 – 32 , we get:

15+32 = 15+32 x 532532

= 532(5)2(32)2

= 5322518

= 5327

Question 10: 165

Solution:

On Multiplying the numerator and denominator of the given number by 6 + 5 , we get:

165 = 165 x 6+56+5

= 6+5(6)2(5)2

= 6+5

Question 11: 47+3

Solution:

On Multiplying the numerator and denominator of the given number by 73 , we get:

47+3 = 47+3 x 7373

= 4(73)(7)2(3)2

= 4(73)73

= 4(73)4

= 73

Question 12: 313+1

Solution:

On multiplying the numerator and denominator of the given number by 3 – 1, we get:

313+1 = 313+1 x 3131

= (31)2(3)2(1)2

= 3+12331

= 4232

= 2(23)2

= 2 – 3

Question 13: 3223+22

Solution:

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

3223+22 = 3223+22 x 322322

= (322)2(3)2(22)2

= 9+812298

= 17 – 12 2

Question 14: 3+131 = a + b 3

Solution:

We have:

3+131

Now, rationalizing the denominator of the given number by multiplying both the numerator and denominator with 3 + 1 we get:

3+131 = 3+131 x 3+13+1

= (3+1)2(3)2(1)2

= 3+1+2331

= 4+232

= 2(2+3)2

= 2 + 3

Therefore,   2 + 3 = a + b 3

So, on comparing the LHS and RHS, we get

a =2 and b = 1

Question 15: 3+232 = a + b 2

Solution:

We have:

3+232

Now, rationalizing the denominator of the given number by multiplying both the denominator and the numerator with 3 + 2, we get:

3+232 = 3+232 x 3+23+2

= (3+2)2(3)2(2)2

= 9+2+6292

= 11+627

Therefore,  117 + 672 = a + b 2

So, on comparing the LHS and the RHS, we get:

a = 117 and b = 67

Question 16: 565+6 a – b 6

Solution:

We have:

565+6

Now, rationalizing the denominator of the given number by multiplying both the denominator and the numerator with 5 – 6, we get:

565+6 = 565+6 x 5656

= (52)2(5)2(6)2

= 25+6106256

= 3110619

= 31191019 6

Therefore, 311910196 = a – b 6

So, on comparing the LHS and the RHS, we get:

a =  3119 and b =  1019

Question 17: 5+237+43 = a – b 3

Solution:

5+237+43

Now, rationalizing the denominator of the given number by multiplying both the denominator and the numerator with 7 – 43, we get:

5+237+43 = 5+237+43 x 743743

= (3514320324)(7)2(43)2

= 11634948

= 11 – 6 3

Therefore, 11 – 6 3 = a – b 3

So, on comparing the LHS and the RHS, we get:

a = 11 and b = 6


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