Home | | Maths 7th Std | Exercise 3.3

# Exercise 3.3

7th Maths : Term 3 Unit 3 : Algebra : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

Exercise 3.3

Miscellaneous Practice problems

1. Using identity, find the value of

( i) (4.9)2

(ii) (100.1)2

(iii) (1.9)×(2.1)

Solution:

(i) (4.9)2

(4.9)² = (5 – 0.1)2

Substituting a = 5 and b = 0.1 in

(a – b)² = a² – 2ab + b², we have

(5 – 0.1)² = 5² – 2 (5) (0.1) + (0.1)2

(4.9)2 = 25 – 1 + 0.01 = 24 + 0.01

(4.9)2 = 24.01

(ii) (100.1)2

(100.1)2 = (100 + 0.1)2

Substituting a = 100 and b = 0.1 in

(a + b)2 = a2 + 2ab + b2, we have

(100 + 0.1)2 = (100)2 + 2 (100) (0.1) + (0.1)2

(100.1)2 = 10000 + 20 + 0.01

(100.1)2 = 10020.01

(ii) (1.9) × (2.1)

(1.9) × (2.1) = (2 – 0.1) × (2 + 0.1)

Substituting a = 100 and b = 0.1 in

(a – b)(a + b) = a2b2, we have

(2 – 0.1)(2 + 0.1) = 22 – (0.1)2

(1.9) × (2.1) = 4 – 0.01

(9.9) (2.1) = 3.99

2. Factorise : 4x 2 9 y2

Solution:

4x2 – 9y2 = 22 x2 – 32 y2 = (2x)2 – (3y)2

Substituting a – 2x and b = 3y in

(a2b2) = (a + b) (a – b), we have

(2x)2 – (3y)2 = (2x + 3y) (2x – 3y)

Factors of 4x2 – 9y2 are (2x + 3y) and (2x – 3y)

3. Simplify using identities (i) (3 p + q )(3 p +r) (ii) (3 p + q )(3p q)

Solution:

(i) (3p + q)(3p + r)

Substitute x = 3p, a = q and b = r in

(x + a) (x + b) = x2 + x(a + b) + ab

(3p + q)(3p + r) = (3p)2 + 3p (q + r) + (q × r)

= 32 p2 + 3p (q + r) + qr

(3p + q)(3p + r) = 9p2 + 3p (q + r) + qr

(ii) (3p + q)(3p – q)

Substitute a  = 3p and b = q in

(a + b) (a – b) = a2b2, we have

(3p + q)(3p – q) = (3p)2q2 = 32p2q2

(3p + q)(3p – q) = 9p2q2

4. Show that (x + 2 y)2 (x 2y)2 =8xy

Solution:

[ (a + b)2 = a2 + 2ab b2

(a – b)2 = a2 – 2ab b2]

LHS = (x + 2y)² – (x – 2y

= x² + (2 × x × 2y) + (2y)² – [x² – (2 × x × 2y) + (2y)²]

== x² + 4xy + 4y² –[x² – 4xy + 2² y²]

= x² + 4xy + 4y² – x² + 4xy –4 y²

= x² – x² + 4xy + 4xy + 4y² – 4y²

= x² (1 − 1) + xy (4 + 4) + y² (4 − 4)

= 0x² + 8xy + 0y² = 8xy = RHS

(x + 2y)² – (x – 2y)² = 8xy

5. The pathway of a square paddy field has 5 m width and length of its side is 40 m. Find the total area of its pathway. (Note: Use suitable identity)

Solution:

Given side of the square = 40 m

Also width of the pathway = 5 m Side of the larger square =. 40m + 2(5)m = 40m + 10m = 50m

Area of the path way = area of large square – area of smaller square

= 502 – 402

Substituting a = 50 and b = 40 in

a2b2 = (a + b)(a – b) we have

502 – 402 = (50 + 40) (50–40)

Area of pathway = 90 × 10

Area of the pathway = 900 m2

Challenge Problems

6. If X = a2 1 and Y = 1 b2 , then find X +Y and factorize the same.

Solution:

Given X = a2 – 1

Y = 1 – b2

X + Y = (a2– 1) + (1 – b2)

= a2 – 1 + 1 – b2

We know the identity that a2b2 = (a + b) (a – b)

X + Y = (a + b) (a – b)

7. Find the value of (x y )(x + y)(x2 + y2 )

Solution:

We know that (a – b) (a + b) = a2b2 ... (1)

Put a = x and b = y in the identity (1) then

(x – y)(x + y) = x2y2

Now (x – y) (x + y)(x2 + y2) = (x2y2) (x2 + y2)

Again put a = x2 and b = y2 in (1)

We have (x2 y2) (x2 + y2) = (x2)2 – (y2)2 = x4y4

So (x – y)(x + y) (x2 + y2) = x4y4

8. Simplify (5x 3 y)2 (5x +3y)2

Solution:

We have the identities …….. (1)

(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

So (5x – 3y)2 – (5x + 3y)2 = (5x)2 – (2 × 5x × 3y) + (3y)2 – [(5x)2 + 2(5x) (3y) + (3y)2]

= 52x2 – 30xy + 32y2 – [52x2 – 30xy + 32y2]

= 25x2 – 30xy + 9y2 – [25x2 + 30xy + 9y2]

= 25x2 – 30xy + 9y2 – 25x2 – 30xy – 9y2

= x2 (25–25) – xy (30 + 30) + y2 (9 – 9)

= 0x2 – 60xy + 0y2 = – 60 xy

(5x – 3y)2 – (5x + 3y)2 = –60xy

9. Simplify: (i) (a+b)2 – (a–b)2  (ii) (a+b)2 + (a–b)2

Solution:

Applying the identities

(a + b)2 = a2 + 2ab + b2

(a –b)2 = a2 – 2ab + b2

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

= a2 + 2ab + b2a2 + 2abb2

= a2 (1 – 1 ) + ab (2 + 2 ) + b2 ( 1 – 1)

= 0a2 + 4ab + 0b2 = 4ab

(a + b)2 – (a –b)2 = 4 ab

(ii) (a + b)2 + (a –b)2 = a2 + 2ab + b2 + (a2 – 2ab + b2 )

= a2 + 2ab + b2 + a2 – 2ab + b2

= a2 (1 + 1 ) + ab (2 – 2 ) + b2 ( 1 + 1)

= 2a2 + 0ab + 2b2  = 2a2 + 2b2 = 2 (a2 + b2)

(a + b)2 – (a – b)2 = 2(a2 + b2)

10. A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m2 , find the area of lawn.

Solution:

Let the side of the lawn = a m

then side of big square = (a + 2(2))m

= (a + 4)m Area of the path = Area of large square – Area of smaller square

136 = (a + 4)2a2

136 = a2 + (2 × a × 4) + 42a2

136 = a2 + 8a + 16 – a2

136 = 8a + 16

136 = 8 (a + 2)

Dividing by 8

17 = a + 2

Subtracting 2 on both sides

17 – 2 = a + 2 – 2

15 = a

Side of small square = 15m

Area of Square = (Side × Side) Sq. units.

Area of the lawn = (15 × 15)m2 = 225 m2

Area of the lawn = 225 m2

11. Solve the following inequalities.

(i) 4n + 7 ≥ 3n + 10, n is an integer.

(ii) 6(x + 6) ≥ 5(x − 3), x is a whole number.

(iii) −13 ≤ 5x + 2 ≤ 32, x is an integer.

Solution:

(i) 4n + 7 > 3n + 10, n is an integer.

Subtracting 3n both sides

4n + 7 – 3n   ≥  3n + 10 – 3n

n(4 – 3) + 7   ≥  3n + 10 – 3n

n (4 – 3) + 7   ≥  n (3 – 3) + 10

n + 7 ≥ 10

Subtracting 7 on both sides

n + 7 – 7 ≥ 10 – 7

n ≥ 3

Since the solution is an integer and is greater than or equal to 3, the solution will be 3, 4, 5, 6, 7,...

n = 3, 4, 5, 6, 7,...

(ii) 6(x +6) ≥ 5(x – 3), x is a whole number.

6x + 36 ≥ 5x – 15

Subtracting 5x on both sides

6x + 36 – 5x ≥ 5x – 15 – 5x

x(6 – 5) + 36 ≥ x (5 – 5) – 15

x + 36 ≥ – 15

Subtracting 36 on both sides

x + 36 – 36 ≥ –15 –36

x ≥ –51

The solution is a whole number and which is greater than or equal to –51

The solution is 0, 1, 2, 3, 4,...

x = 0, 1,2, 3, 4,...

(iii) –13 ≤ 5x + 2 ≤ 32, x is an integer.

Subtracting throughout by 2

–13 –2 ≤ 5x + 2 – 2 ≤ 32 – 2

–15 ≤ 5x ≤ 30

Dividing throughout by 5

–15/5   ≤   5x/5   ≤   30/5

– 3 ≤ x ≤ 6

Since the solution is an integer between –3 and 6 both inclusive, we have the solution as –3, –2, –1,0, 1,2, 3, 4, 5, 6.

i.e. x = –3, –2, 0, 1, 2, 3, 4, 5 and 6.

Exercise 3.3

1. (i) 24.01 (ii) 10020.01 (iii) 3.99

2. (2x + 3y) (2x – 3y)

3. (i) 9p2 + 3p (q + r) + qr (ii) 9p2q2

4. 900 sq.m

Challenge problems

6. (a – b) (a + b)

7. x4 – y4

8. –60xy

9. (i) 4ab (ii) 2(a2 + b2)

10. 225 sq.m

11. (i) n = 3, 4, 5, 6, ....  (ii) x = 0, 1, 2, 3, ...  (iii) x = –3, –2, –1, 0, 1, 2, 3, 4, 5 and 6

Tags : Questions with Answers, Solution | Algebra | Term 3 Chapter 3 | 7th Maths , 7th Maths : Term 3 Unit 3 : Algebra
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
7th Maths : Term 3 Unit 3 : Algebra : Exercise 3.3 | Questions with Answers, Solution | Algebra | Term 3 Chapter 3 | 7th Maths