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 = *a*² – 2*ab* + *b*², 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} = *a*^{2} + 2*ab *+ *b*^{2}, 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*) = *a*^{2} – *b*^{2},
we have

(2 – 0.1)(2 + 0.1) = 2^{2} – (0.1)^{2}

(1.9) × (2.1) = 4 – 0.01

(9.9) (2.1) = 3.99

** **

**2. Factorise : 4 x ^{2}
**

**Solution:**

4*x*^{2} – 9*y*^{2} = 2^{2} *x*^{2} – 3^{2} *y*^{2} = (2*x*)^{2} – (3*y*)^{2}

Substituting *a* – 2*x* and *b *= 3*y* in

(*a*^{2} – *b*^{2}) = (*a + b*) (*a – b*), we have

(2*x*)^{2} – (3*y*)^{2} = (2*x* + 3y) (2*x* – 3*y*)

∴ Factors of 4*x*^{2}
– 9*y*^{2} are (2*x* + 3*y*)
and (2*x* – 3*y*)

** **

**3.
Simplify using identities (i) (3 p **

**Solution:**

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

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

(*x* + *a*) (*x
+ b*) = *x*^{2} + *x*(*a
+ b*) + *ab*

(3*p* + *q*)(3*p
+ r*) = (3*p*)^{2} + 3*p* (*q
+ r*) + (*q *× *r*)

= 3^{2} *p*^{2}
+ 3*p* (*q + r*) + *qr*

(3*p* + *q*)(3*p
+ r*) = 9*p*^{2} + 3*p* (*q
+ r*) + *qr*

**(ii) (3 p + q)(3p – q)**

Substitute *a* = 3*p*
and *b* = *q* in

(*a + b*) (*a – b*) = *a*^{2} – *b*^{2},
we have

(3*p* + *q*)(3*p
– q*) = (3*p*)^{2} – *q*^{2} = 3^{2}*p*^{2} – *q*^{2}

(3*p + q*)(3*p – q*) = 9*p*^{2} – *q*^{2}

** **

**4.
Show that ( x **

**Solution:**

[∵ (*a + b*)2 = *a*2 + 2*ab *+ *b*2

(*a – b*)2 = *a*2 – 2*ab *+ *b*2]

LHS = (*x* + 2*y*)² – (*x* – 2*y*)²

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

== *x*² + 4*xy* + 4*y*² –[*x*² – 4*xy* + 2² *y*²]

= *x*² + 4*xy* + 4*y*² – *x*² + 4*xy* –4 *y*²

= *x*² – *x*² + 4*xy* + 4*xy* + 4*y*² – 4*y*²

*= x*² (1 − 1) + *xy* (4 + 4)
+ *y*² (4 − 4)

= 0*x*² + 8*xy* + 0y² = 8*xy* = RHS

∴ (*x* + 2*y*)² – (*x* – 2*y*)² = 8*xy*

** **

**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

= 50^{2} – 40^{2}

Substituting *a* = 50
and *b* = 40 in

*a*^{2} – *b*^{2} = (*a + b*)(*a – b*) we have

50^{2} – 40^{2 }= (50 + 40) (50–40)

Area of pathway = 90 × 10

Area of the pathway = 900 m^{2}

** **

**Challenge
Problems**

** **

**6.
If X **

**Solution: **

Given X = *a*^{2}
– 1

Y = 1 – *b*^{2}

X + Y = (*a*^{2}–
1) + (1 – *b*^{2})

= *a*^{2} – 1 +
1 – *b*^{2}

We know the identity that *a*^{2}
– *b*^{2} = (*a + b*) (*a – b*)

∴ X + Y = (*a + b*) (*a – b*)

** **

**7.
Find the value of ( x **

**Solution: **

We know that (*a – b*) (*a + b*) = *a*^{2} – *b*^{2}
... (1)

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

(*x – y*)(*x + y*) = *x*^{2} – *y*^{2}

Now (*x – y*) (*x + y*)(*x*^{2} + *y*^{2})
= (*x*^{2} – *y*^{2}) (*x*^{2} + *y*^{2})

Again put *a* = *x*^{2} and *b *= *y*^{2} in (1)

We have (*x*^{2 }–
*y*^{2}) (*x*^{2} + *y*^{2})
= (*x*^{2})^{2} – (*y*^{2})^{2} = *x*^{4} – *y*^{4}

So (*x – y*)(*x + y*) (*x*^{2} + *y*^{2})
= *x*^{4} – *y*^{4}

** **

**8.
Simplify (5 x **

**Solution: **

We have the identities …….. (1)

(*a + b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}

(*a – b*)^{2} = *a*^{2} – 2*ab *+ *b*^{2}

So (5*x* – 3*y*)^{2} – (5*x* + 3*y*)^{2} = (5*x*)^{2} – (2 × 5*x* × 3*y*)
+ (3*y*)^{2} – [(5*x*)^{2} + 2(5*x*) (3*y*) + (3*y*)^{2}]

= 5^{2}*x*^{2}
– 30*xy* + 3^{2}*y*^{2} – [5^{2}*x*^{2} – 30*xy* + 3^{2}*y*^{2}]

= 25*x*^{2} – 30*xy* + 9*y*^{2} – [25*x*^{2}
+ 30*xy* + 9*y*^{2}]

= 25*x*^{2} – 30*xy* + 9*y*^{2} – 25*x*^{2}
– 30*xy* – 9*y*^{2}

= *x*^{2} (25–25)
– *xy* (30 + 30) + *y*^{2} (9 – 9)

= 0*x*^{2} – 60*xy* + 0*y*^{2} = – 60 *xy*

∴ (5*x *– 3*y*)^{2} – (5*x* + 3*y*)^{2} = –60*xy*

** **

**9.
Simplify: (i) (a+b) ^{2} – (a–b)^{2 }(ii) (a+b)^{2} + (a–b)^{2}**

**Solution: **

Applying the identities

(*a + b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}

(*a –b*)^{2} = *a*^{2} – 2*ab* + *b*^{2}

(i) (*a + b*)^{2}
– (*a –b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}– [*a*^{2} – 2*ab* + *b*^{2}]

= *a*^{2} + 2*ab* + *b*^{2}–
*a*^{2} + 2*ab* – *b*^{2}

=* a*^{2 }(1 – 1
) + *ab* (2 + 2 ) + *b*^{2 }( 1 – 1)

= 0*a*^{2} + 4*ab* + 0*b*^{2} = 4*ab*

(*a + b*)^{2} –
(*a –b*)^{2} = 4* ab*

(ii) (*a + b*)^{2}
+ (*a –b*)^{2 }= *a*^{2} + 2*ab* + *b*^{2 }+ (*a*^{2} – 2*ab* + *b*^{2 })

= *a*^{2} + 2*ab* + *b*^{2
}+ *a*^{2} – 2*ab* + *b*^{2
}

= *a*^{2 }(1 + 1
) + *ab* (2 – 2 ) + *b*^{2 }( 1 + 1)

= 2*a*^{2} + 0*ab* + 2*b*^{2} = 2*a*^{2} + 2*b*^{2} = 2 (*a*^{2}
+ *b*^{2})

∴ (*a + b*)^{2} –
(*a – b*)^{2} = 2(*a*^{2} + *b*^{2})

** **

**10.
A square lawn has a 2 m wide path surrounding it. If the area of the path
is 136 m^{2} , 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)^{2}
– *a*^{2}

136 = *a*^{2} +
(2 × *a* × 4) + 4^{2} – *a*^{2}

136 = *a*^{2} +
8*a* + 16 – *a*^{2}

136 = 8*a* + 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)m^{2} = 225 m^{2}

∴ Area of the lawn = 225 m^{2}

** **

**11. Solve the following inequalities.**

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

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

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

**Solution: **

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

Subtracting 3*n* both
sides

4*n* + 7 – 3*n* ≥ 3*n* + 10 – 3*n*

*n*(4 – 3) + 7 ≥ 3*n* + 10 – 3*n*

*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.**

6*x* + 36 ≥ 5*x* – 15

Subtracting 5*x* on both
sides

6*x* + 36 – 5*x* ≥ 5*x*
– 15 – 5*x*

*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 ≤ 5*x* + 2 – 2 ≤ 32
– 2

–15 ≤ 5*x* ≤ 30

Dividing throughout by 5

–15/5 ≤ 5*x*/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.

** **

__ANSWERS:__

**Exercise 3.3 **

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

2. (2*x* + 3*y*)
(2*x* – 3*y*)

3. (i) 9*p*^{2}
+ 3*p* (*q* + *r*) + *qr* (ii) 9*p*^{2}
– *q*^{2}

4. 900 *sq.m*

**Challenge problems**

6. (a – b) (a + b)

7. x^{4} – y^{4}

8. –60*xy*

9. (i) 4ab (ii) 2(a^{2}
+ b^{2})

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

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7th Maths : Term 3 Unit 3 : Algebra : Exercise 3.3 | Questions with Answers, Solution | Algebra | Term 3 Chapter 3 | 7th Maths

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