Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 1.

Which is greater?

(i) 0 0.5 or 0.05

(ii) 0.7 or 0.5

(iii) 1 or 0.7

(iv) 1.37 or 1.49

(v) 2.03 or 2.30

(vi) 0.8 or 0.88

Solution:

(i) Comparing the digits at tenths place, we have 5 > 0

ā“ 0.5 > 0.05

(ii) Comparing the digits at tenths place, we have 7 > 5

ā“ 0.7 > 0.5

(iii) Comparing the digits at ones place, we have 7 > 0

ā“ 7 > 0.7

(iv) Since, the digits at ones place are same.

ā“ Comparing the digits at tenths place, we have 3 < 4

ā“ 1.37 < 1.49 or 1.49 > 1.37

(v) Since the digits at ones place are same.

ā“ Comparing the digits at tenths place, we have 0 < 3

ā“ 2.03 < 2.30 or 2.30 > 2.03

(vi) 0.8 can be written as 0.80.

Now, digits at tenths place are same.

ā“ Comparing the digits at hundredths place, we have 0 < 8

ā“ 0.80 < 0.88 or 0.88 > 0.8

Question 2.

Express as rupees using decimals:

(i) 7 paise

(ii) 7 rupees 1 paise

(iii) 77 rupees 77 paise

(iv) 50 paise

(v) 235 paise

Solution:

We know that:

100 paise = ā¹ 1 ā 1 paise = ā¹ \(\frac { 1 }{ 100 }\)

(i) 7 paise = ā¹ 7 x = ā¹ \(\frac { 1 }{ 100 }\) = ā¹ 0.07

(ii) 7 rupees 7 paise = ā¹ 7 + ā¹ 0.07 = ā¹ 7.07

(iii) 77 rupees 77 paise= ā¹ 77 + 77 paise

= ā¹ 77 + ā¹ 77 x \(\frac { 1 }{ 100 }\)

= ā¹ 77 + ā¹ 0.77 = ā¹ 77.77

(iv) 50 paise = ā¹ 50 x \(\frac { 1 }{ 100 }\)

(v) 235 paise= 200 paise + 35 paise

= ā¹ 2 + ā¹ 35 x \(\frac { 1 }{ 100 }\) [āµ 200 paise = ā¹ 2]

= ā¹ 2 + ā¹ 0.35 = ā¹ 2.35

Relation between commonly used Measures:

Question 3.

(i) Express 5 cm in metre and kilometre

(ii) Express 35 mm in cm, m and km

Solution:

We know that 1 cm = 10 mm, 100 cm = 1 m, 1000 m = 1 km

(i) 5 cm = \(\frac { 5 }{ 100 }\) m = 0.05 m

5 cm = \(\frac { 5 }{ 100000 }\) km = 0.00005 km

(ii) 35 mm = \(\frac { 35 }{ 10 }\) cm = 3.5 cm

35 mm = \(\frac { 35 }{ 1000 }\)m = 0.035 m

35 mm = \(\frac { 35 }{ 100000 }\) km = 0.000035 km

Question 4.

Express in kg:

(i) 200 g

(ii) 3470 g

(iii) 4 kg 8 g

Solution:

We know that 1000 g = 1 kg

(i) 200 g = \(\frac { 200 }{ 1000 }\) kg

= \(\frac { 2 }{ 10 }\) kg

= 0.2 kg

(ii) 3470 g = \(\frac { 3470 }{ 1000 }\) kg

= 3.470 kg

(iii) 4 kg 8 g = 4 kg + \(\frac { 8 }{ 1000 }\) kg

= 4 kg + 0.008 kg = 4.008 kg

Question 5.

Write the following decimal numbers in the expanded form:

(i) 20.03

(ii) 2.03

(iii) 200.03

(iv) 2.034

Solution:

(i) 20.03 = 2 x 10 + 0 x 1 + 0 x \(\frac { 1 }{ 10 }\) + 3 x \(\frac { 1 }{ 100 }\)

= 2 x 10 + \(\frac { 3 }{ 100 }\)

(ii) 2.03 = 2 x 1 + 0 x \(\frac { 1 }{ 10 }\) + 3 x \(\frac { 1 }{ 100 }\)

= 2 x 1 + \(\frac { 3 }{ 100 }\)

(iii) 200.03 = 2 x 100 + 0 x 10 + 0 x 1 + 0 x \(\frac { 1 }{ 10 }\) + 3 x \(\frac { 1 }{ 100 }\)

= 2 x 100 + \(\frac { 3 }{ 100 }\)

(iv) 2.034 = 2 x 1 + 0 x \(\frac { 1 }{ 10 }\) + 3 x \(\frac { 1 }{ 100 }\) + 4 x \(\frac { 1 }{ 100 }\)

= 2 x 1 + \(\frac { 3 }{ 100 }\) + \(\frac { 4 }{ 1000 }\)

Question 6.

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

(i) 2.56

(ii) 21.37

(iii) 10.25

(iv) 9.42

(v) 63.352

Solution:

(i) In 2.56, the digit 2 is at the ones place.

ā“ Place value of 2 is 2 x 1, i.e. 2.

(ii) In 21.37, the digit 2 is at the tens place.

ā“ Place value of 2 is 2 x 10, i.e. 20.

(iii) In 10.25, the digit 2 is at the tenths place.

ā“ Place value of 2 is 2 x \(\frac { 1 }{ 10 }\), i.e. \(\frac { 2 }{ 10 }\).

(iv) In 9.42, the digit 2 is at the hundredths place.

ā“ The place value of 2 is 2 x \(\frac { 1 }{ 100 }\), i.e. \(\frac { 2 }{ 100 }\).

(v) In 63.352, the digit 2 is at the thousandths place.

ā“ The place value of 2 is 2 x \(\frac { 1 }{ 100 }\), i.e. \(\frac { 2 }{ 1000 }\).

Question 7.

Dinesh went A from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?

Solution:

āµ Distance from A to B = 7.5 km

Distance from B to C = 12.7 km

ā“ Distance from A to C through B

= (7.5 + 12.7) km

= 20.2 km

ā“ Distance travelled by Dinesh = 20.2 km

Again, Distance from A to D = 9.3 km

Distance from D to C = 11.8 km

ā“ Distance from A to C through D

= (9.3 + 11.8) km = 21.1 km

ā“ Distance travelled by Ayub = 21.1 km

Since (21.1 – 20.2) km = 0.9 km or 900 m

or 900 m

ā“ Ayub travelled more distance by 900 m.

Question 8.

Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?

Solution:

Since, Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes,

Total fruits bought by Shyama

= 5 kg 300 g + 3 kg 250 g

= 8 kg 550 g

Since, Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas,

Total fruits bought by Sarala

= 4 kg 800 g + 4 kg 150 g = 8 kg 950 g

Since, 8 kg 950 g – 8 kg 550 g = 0 kg 400 g

or \(\frac { 400 }{ 1000 }\) kg = 0.4 kg.

Thus, Sarala bought more fruits by 0.4 kg.

Question 9.

How much less is 28 km than 42.6 km?

Solution:

Since 42.6 km – 28 km = 14.6 km

ā“ 28 km is less than 42.6 km by 14.6 km.