Ncert Solutions For Class 7 Maths Ex 12.4

NCERT Solutions For Class 7 Maths Ex 12.4 PDF Free Download

NCERT Solutions for Class 7 Maths Exercise 12.4 Chapter 12 Algebraic Expressions are given here in a simple and detailed way. Using algebraic expressions – formulas and rules is the only topic covered in this exercise of NCERT Solutions for Class 7 Chapter 12. These solutions can be beneficial for the students to clear all their doubts quickly and understand the concepts of this chapter. Students can use them as worksheets to prepare for their exams.

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 12 Perimeter and Area – Exercise 12.4

 

ncert sol class 7 math ch 12 ex 4
ncert sol class 7 math ch 12 ex 4
ncert sol class 7 math ch 12 ex 4
ncert sol class 7 math ch 12 ex 4
ncert sol class 7 math ch 12 ex 4

 

Access other exercises of NCERT Solutions For Class 7 Chapter 12 – Algebraic Expressions

Exercise 12.1 Solutions

Exercise 12.2 Solutions

Exercise 12.3 Solutions

Access answers to Maths NCERT Solutions for Class 7 Chapter 12 – Algebraic Expressions Exercise 12.4

1. Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 6

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 7

Solution:-

(a) From the question it is given that the numbers of segments required to form n digits of the kind
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 8is (5n + 1)

Then,

The number of segments required to form 5 digits = ((5 × 5) + 1)

= (25 + 1)

= 26

The number of segments required to form 10 digits = ((5 × 10) + 1)

= (50 + 1)

= 51

The number of segments required to form 100 digits = ((5 × 100) + 1)

= (500 + 1)

= 501

(b) From the question it is given that the numbers of segments required to form n digits of the kind
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 9is (3n + 1)

Then,

The number of segments required to form 5 digits = ((3 × 5) + 1)

= (15 + 1)

= 16

The number of segments required to form 10 digits = ((3 × 10) + 1)

= (30 + 1)

= 31

The number of segments required to form 100 digits = ((3 × 100) + 1)

= (300 + 1)

= 301

(c) From the question it is given that the numbers of segments required to form n digits of the kind
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 10is (5n + 2)

Then,

The number of segments required to form 5 digits = ((5 × 5) + 2)

= (25 + 2)

= 27

The number of segments required to form 10 digits = ((5 × 10) + 2)

= (50 + 2)

= 52

The number of segments required to form 100 digits = ((5 × 100) + 1)

= (500 + 2)

= 502

2. Use the given algebraic expression to complete the table of number patterns.

S. No.

Expression

Terms

1st

2nd

3rd

4th

5th

10th

100th

(i)

2n – 1

1

3

5

7

9

19

(ii)

3n + 2

5

8

11

14

(iii)

4n + 1

5

9

13

17

(iv)

7n + 20

27

34

41

48

(v)

n2 + 1

2

5

10

17

10001

Solution:-

(i) From the table (2n – 1)

Then, 100th term =?

Where n = 100

= (2 × 100) – 1

= 200 – 1

= 199

(ii) From the table (3n + 2)

5th term =?

Where n = 5

= (3 × 5) + 2

= 15 + 2

= 17

Then, 10th term =?

Where n = 10

= (3 × 10) + 2

= 30 + 2

= 32

Then, 100th term =?

Where n = 100

= (3 × 100) + 2

= 300 + 2

= 302

(iii) From the table (4n + 1)

5th term =?

Where n = 5

= (4 × 5) + 1

= 20 + 1

= 21

Then, 10th term =?

Where n = 10

= (4 × 10) + 1

= 40 + 1

= 41

Then, 100th term =?

Where n = 100

= (4 × 100) + 1

= 400 + 1

= 401

(iv) From the table (7n + 20)

5th term =?

Where n = 5

= (7 × 5) + 20

= 35 + 20

= 55

Then, 10th term =?

Where n = 10

= (7 × 10) + 20

= 70 + 20

= 90

Then, 100th term =?

Where n = 100

= (7 × 100) + 20

= 700 + 20

= 720

(v) From the table (n2 + 1)

5th term =?

Where n = 5

= (52) + 1

= 25+ 1

= 26

Then, 10th term =?

Where n = 10

= (102) + 1

= 100 + 1

= 101

So the table is completed below.

S. No.

Expression

Terms

1st

2nd

3rd

4th

5th

10th

100th

(i)

2n – 1

1

3

5

7

9

19

199

(ii)

3n + 2

5

8

11

14

17

32

302

(iii)

4n + 1

5

9

13

17

21

41

401

(iv)

7n + 20

27

34

41

48

55

90

720

(v)

n2 + 1

2

5

10

17

26

101

10001


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