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RD Chapter 4- Algebraic Identities Ex-4.1 Interview Questions Answers

Question 1 : Evaluate each of the following using identities:
(i) (2x –
1)2
(ii)  (2x + y) (2x – y)
(iii) (a2b – b2a)2
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x2 – 0.3y2) (1.5x2 +0.3y2)

Answer 1 :


Question 2 :
Evaluate each of the following using identities:
(i) (399)2
(ii) (0.98)2
(iii) 991 x 1009
(iv) 117 x 83

Answer 2 :


Question 3 : Simplify each of the following:

Answer 3 :

(i) 175 x 175 +2 x 175 x 25 + 25 x 25 = (175)2 +2 (175) (25) + (25)2

= (175 + 25)2

[Becausea2+ b2+2ab = (a+b)2 ]

= (200)2

= 40000

So, 175 x 175 +2 x 175x 25 + 25 x 25 = 40000.

(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22

= (322)2 –2 x 322 x 22 + (22)2

= (322 – 22)2

[Becausea2+ b2-2ab = (a-b)2]

= (300)2

= 90000

So, 322 x 322 – 2 x322 x 22 + 22 x 22= 90000.

(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x0.24

= (0.76) 2 +2 x 0.76 x 0.24 + (0.24) 2

= (0.76+0.24) 2

[Because a2+ b2+2ab = (a+b)2]

= (1.00)2

= 1

So, 0.76 x 0.76 + 2 x0.76 x 0.24 + 0.24 x 0.24 = 1.

(iv)

Question 4 :

Answer 4 :


Question 5 :

Answer 5 :


Question 6 :

Answer 6 :


Question 7 : If 9x2 +25y2 = 181 and xy = -6, find thevalue of 3x + 5y.

Answer 7 :

9x2 + 25y2 = 181, and xy = -6
(3x + 5y)2 = (3x)2 + (5y)2 +2 x 3x + 5y
9X2 + 25y2 +30xy
= 181 + 30 x (-6)
= 181 – 180 = 1
= (±1 )2
3x +5y = ±1

Question 8 : If 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2.

Answer 8 :

2x + 3y = 8 and xy = 2
Now, (2x + 3y)2 = (2x)2 + (3y)2 +2 x 2x x 3y
 (8)2 = 4x2 +9y2 + 12xy
64 = 4X2 + 9y2 +12 x 2
64 = 4x2 + 9y2 +24
4x2 + 9y2 =64 – 24 = 40
4x2 + 9y2 =40

Question 9 : If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2

Answer 9 :

3x – 7y = 10, xy = -1
3x -7y= 10
Squaring both sides,
(3x – 7y)2 = (10)2
(3x)2 +(7y)2 – 2 x 3x x 7y = 100
 9X2 + 49y2 – 42xy = 100
 9x2 + 49y2 – 42(-l) = 100
9x2 +49y2 + 42 = 100
9x2 +49y2 = 100 – 42 = 58

Question 10 : Simplify each of the following products:

Answer 10 :



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RD Chapter 4- Algebraic Identities Ex-4.1 Contributors

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