Students can Download Maths Chapter 1 Number System Additional 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 Additional Questions

Additional Questions and Answers

Exercise 1.1

Question 1.

Express as rupees using decimals.

(i) 4 paise

(ii) 4 rupees 4 paise

(iii) 44 rupees 44 paise

(iv) 50 paise

(v) 625 paise

Solution:

We know that 100 paise = ₹ 1

1 paise = ₹ \(\frac { 1 }{ 100 } \)

(i) 4 paise = ₹ 4 × \(\frac { 1 }{ 100 } \) = ₹ \(\frac { 4 }{ 100 } \) = ₹ 0.04

(ii) 4 rupees 4 paise = ₹ 4 + ₹ 0.04 = ₹ 4.04

(iii) 44 rupees 4 paise = ₹ 44 + 44 paise = ₹ 44 + ₹ \(\frac { 44 }{ 100 } \) = ₹ 44 + ₹ 0.44 = ₹ 44.44

(iv) 625 paise = 600 paise + 25 paise = ₹ 6 + ₹ \(\frac { 25 }{ 100 } \) = ₹ 6 + ₹ 0.25 = ₹ 6.25

Question 2.

Express 7 cm in metre and kilometer.

Solution:

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

7 cm = \(\frac { 7 }{ 10000 } \) km = 0.00007 km

Question 3.

Write the following decimal numbers in the expanded form.

(i) 30.04

(ii) 3.04

(iii) 300.04

Solution:

(i) 30.04 = 3 × 10 + 0 × 1 + 0 × \(\frac { 1 }{ 10 } \) + 4 × \(\frac { 1 }{ 100 } \) = 3 × 10 + \(\frac { 4 }{ 100 } \)

(ii) 3.04 = 3 × 1 + 0 × \(\frac { 1 }{ 10 } \) + 4 × \(\frac { 1 }{ 100 } \) = 3 × 1 + \(\frac { 4 }{ 100 } \)

(iii) 300.04 = 3 × 100 + 0 × 10 + 0 × 1 + 0 × \(\frac { 1 }{ 10 } \) + 4 × \(\frac { 1 }{ 100 } \) = 3 × 100 + \(\frac { 4 }{ 100 } \) = 3 × 100 + \(\frac { 4 }{ 100 } \)

Question 4.

Write the place value of 2 in the following decimal numbers.

(i) 2.47

(ii) 26.89

(iii) 36.28

Solution:

(i) 2.47 Place value of 2 in 2.47 is ones.

(ii) 26.89 Place value of 2 in 26.89 is Tens.

(iii) 36.28 Place value of 2 in 36.28 is tenths

Exercise 1.2

Question 1.

Explain the following as fractions.

(i) Ajar containing 3.6 litres of milk.

(ii) A cup containing 9.63 mg of medicine.

Solution:

(i) 3.6 = 3 + \(\frac { 6 }{ 10 } \) = 3 + \(\frac { 3 }{ 5 } \) = 3 \(\frac { 3 }{ 5 } \) litre of milk

(ii) 9.63 = 9 + \(\frac { 6 }{ 10 } \) + \(\frac { 3 }{ 100 } \) = \(\frac { 900+60+3 }{ 100 } \) = \(\frac { 963 }{ 100 } \) mg of medicine

Question 2.

Convert into decimal.

(i) Three hundred three and nine hundredths.

(ii) Six and fifty five thousands

Solution:

(i) Three hundred three and nine hundredths

= 303 + \(\frac { 9 }{ 100 } \) = 303 + 0 × \(\frac { 1 }{ 10 } \) + 9 × \(\frac { 1 }{ 100 } \) = 303.09

(ii) Six and fifty five thousands

6 + \(\frac { 55 }{ 100 } \) = 6 + \(\frac { 5 }{ 100 } \) + \(\frac { 5 }{ 1000 } \) = 6 + \(\frac { 0 }{ 10 } \) + \(\frac { 5 }{ 100 } \) + \(\frac { 5 }{ 1000 } \) = 6.055

Question 3.

Find the decimal form of (i) 194 + 20 + 3 + \(\frac { 7 }{ 10 } \) + \(\frac { 2 }{ 100 } \)

(ii) 111 + 11 + 1 + \(\frac { 1 }{ 10 } \) + \(\frac { 1 }{ 1000 } \)

Solution:

(i) 194 + 20 + 3 + \(\frac { 7 }{ 10 } \) + \(\frac { 2 }{ 100 } \) = 217 + 7 × \(\frac { 1 }{ 10 } \) + 2 × \(\frac { 1 }{ 100 } \) = 217.72

(ii) 111 + 11 + 1 + \(\frac { 1 }{ 10 } \) + \(\frac { 1 }{ 1000 } \) = 123 + 1 × \(\frac { 1 }{ 10 } \) + 0 × \(\frac { 1 }{ 100 } \) + 1 × \(\frac { 1 }{ 1000 } \) = 123.101

Exercise 1.3

Question 1.

Maya bought 5 kg 300 g bananas and 3 kg 250 kg oranges. Diya bought 4 kg 800 g apples and 4 kg 150 g of mangoes. Who bought more fruits.

Solution:

Total fruits bought by Maya = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g = 8.550 kg

Total fruits Diya bought = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g = 8.950 kg

Comparing the whole number parts, they are equal.

Comparing thet tenths place we get 9 > 5.

∴ 8.950 kg > 8.550 kg

∴ Diya bought more fruits.

Question 2.

Which is greater 28 km or 42.6 km.

Solution:

Comparing the whole number part 42 > 28.

42.6 km is greater than 28 km.

Exercise 1.4

Question 1.

Show that the following numbers in a number line.

(i) 0.2

(ii) 1.8

(iii) 1.1

(iv) 2.6

Solution:

Question 2.

Write the decimal numbers represented by the points A, B, C, D, E and F.

Solution:

A (0.5); B (1.2); C (2.3); D (2.8); E (3.4); F (3.9)