Students can Download Maths Chapter 1 Number System Ex 1.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.1

Question 1.

Write the decimal numbers for the following pictorial representation of numbers.

Solution:

(i) Tens 2 ones 2 tenths = 12.2

(ii) Tens 1 ones 3 tenths = 21.3

Question 2.

Express the following in cm using decimals.

(i) 5 mm

(ii) 9 mm

(iii) 42 mm

(iv) 8 cm 9 mm

(v) 375 mm

Solution:

(i) 5 mm

1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm

5 mm = \(\frac { 5 }{ 10 } \) = 0.5 cm

(ii) 9 mm

1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm

9 mm = \(\frac { 9 }{ 10 } \) cm = 0.9 cm

(iii) 42 mm

1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm

42 mm = \(\frac { 42 }{ 10 } \) cm = 4.2 cm

(iv) 8 cm 9 mm

1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm

8 cm 9 mm = 8 cm + \(\frac { 9 }{ 10 } \) cm = 8.9 cm

(v) 375 mm

1 mm = \(\frac { 1 }{ 10 } \) cm = 0.1 cm

375 mm = \(\frac { 375 }{ 10 } \) cm = 37.5 cm

Question 3.

Express the following in metres using decimals.

(i) 16 cm

(ii) 7 cm

(iii) 43 cm

(iv) 6 m 6 cm

(v) 2 m 54 cm

Solution:

(i) 16 cm

1 cm = \(\frac { 1 }{ 100 } \) cm = 0.01 m

16 cm = \(\frac { 16 }{ 100 } \) m = 0.16 m

(ii) 7 cm

1 cm = \(\frac { 1 }{ 100 } \) cm = 0.01 m

1 cm = \(\frac { 7 }{ 100 } \) m = 0.07 m

(iii) 43 cm

1 cm = \(\frac { 1 }{ 100 } \) cm = 0.01 m

43 cm = \(\frac { 43 }{ 100 } \) m = 0.43 m

(iv) 6 m 6 cm

1 cm = \(\frac { 1 }{ 10 } \) m = 0.01 m

6 m 6 cm = 6 m + \(\frac { 6 }{ 100 } \) m = 6 m + 0.06 m = 6.06 m

(v) 2 mm 54 cm

1 cm = \(\frac { 1 }{ 100 } \) cm = 0.01 m

2 m 54 cm = 2 m + \(\frac { 54 }{ 100 } \) m = 2 m + 0.54 m = 2.54 m

Question 4.

Expand the following decimal numbers.

(i) 37.3

(ii) 658.37

(iii) 237.6

(iv) 5678.358

Solution:

(i) 37.3 = 30 + 7 + \(\frac { 3 }{ 10 } \) = 3 × 10^{1} + 7 × 10^{0} + 3 × 10^{-1}

(ii) 658.37 = 600 + 50 + 8 + \(\frac { 3 }{ 10 } \) + \(\frac { 7 }{ 100 } \)

= 6 × 10^{2} + 5 × 10^{1} + 8 × 100 + 3 × 10^{-1} + 7 × 10^{-2}

(iii) 237.6 = 200 + 30 + 7 + \(\frac { 6 }{ 10 } \)

= 2 × 10^{2} + 3 × 10^{1} + 7 × 10^{0} + 6 × 10^{-1}

(iv) 5678.358 = 5000 + 600 + 70 + 8 + \(\frac { 3 }{ 10 } \) + \(\frac { 5 }{ 100 } \) + \(\frac { 8 }{ 1000 } \)

= 5 × 10^{3} + 6 × 10^{2} + 7 × 10^{1} + 8 × 10^{0} + 3 × 10^{-1} + 5 × 10^{-2} + 8 × 10^{-3}

Question 5.

Express the following decimal numbers in place value grid and write the place value of the underlined digit.

(i) 53.61

(ii) 263.271

(iii) 17.39

(iv) 9.657

(v) 4972.068

Solution:

(i) 53.61

(ii) 263.271

(iii) 17.39

(iv) 9.657

(v) 4972.068

Objective Type Questions

Question 6.

The place value of 3 in 85.073 is _____

(i) tenths

(ii) hundredths

(iii) thousands

(iv) thousandths

Answer:

(iv) thousandths

Hint: 1000 g = 1 kg; 1 g = \(\frac { 1 }{ 1000 } \) kg

Question 7.

To convert grams into kilograms, we have to divide it by

(i) 10000

(ii) 1000

(iii) 100

(iv) 10

Answer:

(ii) 1000

Hint: 85.073 = 8 × 10 + 5 × 1 + 0 × \(\frac { 1 }{ 10 } \) + 7 × \(\frac { 1 }{ 100 } \) + 3 × \(\frac { 1 }{ 1000 } \)

Question 8.

The decimal representation of 30 kg and 43 g is ____ kg.

(i) 30.43

(ii) 30.430

(iii) 30.043

(iv) 30.0043

Answer:

(iii) 30.043

Hint: 30 kg and 43 g = 30 kg + \(\frac { 43 }{ 1000 } \) kg = 30 + 0.043 = 30.043

Question 9.

A cricket pitch is about 264 cm wide. It is equal to _____ m.

(i) 26.4

(ii) 2.64

(iii) 0.264

(iv) 0.0264

Answer:

(ii) 2.64

Hint: 264 cm = \(\frac { 264 }{ 100 } \) m = 2.64 m