Mathematics – Class 7 – Chapter 12 – Algebraic Expression – Exercise 12.4 – NCERT Exercise Solution

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.

Mathematics - Class 7 - Chapter 12 - Algebraic Expression - Exercise 12.4 - NCERT Exercise Solution

Solution:

(a) According to the question it is given that the number of segments required to form n digits of the given kind are (5n + 1)

Now,

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

= (25 + 1)

= 26

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

= (50 + 1)

= 51

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

= (500 + 1)

= 501

(b) According to the question it is given that the number of segments required to form n digits of the given kind (3n + 1)

Now,

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

= (15 + 1)

= 16

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

= (30 + 1)

= 31

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

= (300 + 1)

= 301

(c) According to the question it is given that the number of segments required to form n digits of the given kind (5n + 2)

Now,

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

= (25 + 2)

= 27

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

= (50 + 2)

= 52

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

= (500 + 2)

= 502

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

Mathematics - Class 7 - Chapter 12 - Algebraic Expression - Exercise 12.4 - NCERT Exercise Solution

Solution:

(i)According to the table,

 (2n – 1)

We have to find, 100th term =?

Where, n =100

Now,

= (2 × 100) – 1

= 200 – 1

= 199

(ii) According to 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) According to 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)According to 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)According to 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

Hence, the table is completed below.

Mathematics - Class 7 - Chapter 12 - Algebraic Expression - Exercise 12.4 - NCERT Exercise Solution

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