RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 Textbook Exercise Questions and Answers.

RBSE Class 7 Maths Solutions Chapter 2 Fractions and Decimals Ex 2.5

Question 1.
Which is greater?
(i) 0.5 or 0.05
Answer:
0.5 or 0.05
0.5 is greater as its tenths part is greater.

(ii) 0.7 or 0.5
Answer:
0.7 or 0.5
0.7 is greater as its tenths part is greater.

(iii) 7 or 0.7
Answer:
7 or 0.7
7 is greater as its ones part is greater.

(iv) 1.37 or 1.49
Answer:
1.37 or 1.49
1.49 is greater as its tenths part is greater.

(v) 2.03 or 2.30
Answer:
2.03 or 2.30
2.30 is greater as its tenths part is greater.

(vi) 0.8 or 0.88
Answer:
0.8 or 0.88
0.88 is greater as its hundredths part is greater.

Question 2.
Express as rupees using decimals :
(i) 7 paise
Answer:
7 paise
∵ 1 paisa = \(\frac{1}{100}\) rupees
∴ 7 paise = \(\frac{7}{100}\) = 0.07 rupees

(ii) 7 rupees 7 paise
Answer:
7 rupees 7 paise
∵ 1 paisa = \(\frac{1}{100}\) rupees
∴ 7 paise = \(\frac{7}{100}\) = 0.07 rupees
∴ 7 rupees 7 paise = 7 + 0.07 = 7.07 rupees

(iii) 77 rupees 77 paise
Answer:
77 rupees 77 paise
∵ 1 paisa = \(\frac{1}{100}\) rupees
∴ 77 paise = \(\frac{77}{100}\) = 0.77 rupees
∴ 77 rupees 77 paise = 77 + 0.77 = 77.77 rupees

(iv) 50 paise
Answer:
50 paise
∵ 1 paisa = \(\frac{1}{100}\) rupees
∴ 50 paise = \(\frac{50}{100}\) = 0.5 rupees

(v) 235 paise
Answer:
235 paise
∵ 1 paisa = \(\frac{1}{100}\) rupees
∴ 235 paise = \(\frac{235}{100}\) = 2.35 rupees

Question 3.
(i) Express 5 cm in metre and kilometre.
Answer:
Express 5 cm in metre and kilometre
∵ 1 cm = \(\frac{1}{100}\) m
∴ 5 cm = \(\frac{5}{100}\) = 0.05 m
∵ 1 cm = \(\frac{1}{100000}\) km
∴ 5 cm = \(\frac{5}{100000}\) = 0.00005 km

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

Question 4.
Express in kg:
(i) 200 g
Answer:
200 g
∵ 1 g = \(\frac{1}{1000}\) kg
∴ 200 g = \(\frac{200}{1000}\) = 0.2 kg

(ii) 3470 g
Answer:
3470 g
∵ 1 g = \(\frac{1}{1000}\)
∴ 3470 g = \(\frac{3470}{1000}\) = 3.47 kg

(iii) 4 kg 8 g
Answer:
4 kg 8 g
∵ 1 g = \(\frac{1}{1000}\)
∴ 8 g = \(\frac{8}{1000}\) kg = 0.008 kg
∴ 4 kg 8 g = 4 + 0.008 = 4.008 kg

Question 5.
Write the following decimal numbers in the expanded form :
(i) 20.03
Answer:
20.03
= 2 × 10 + 0 × 1 + 0 × \(\frac{1}{10} \)+ 3 × \(\frac{1}{100}\)

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

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

(iv) 2.034
Answer:
2.034
= 2 × 1 + 0 × \(\frac{1}{10}\) + 3 × \(\frac{1}{100}\) + 4 × \(\frac{1}{1000}\)

Question 6.
Write the place value of 2 in the following decimal numbers:
(i) 2.56
Answer:
2.56
Place value of 2 = 2 × 1 = 2

(ii) 21.37
Answer:
21.37
Place value of 2 = 2 × 10 = 20

(iii) 10.25
Answer:
10.25
Place value of 2 = 2 × \(\frac{1}{10}\) = 0.2

(iv) 9.42
Answer:
9.42
Place value of 2 = 2 × \(\frac{1}{100}\) = 0.02

(v) 63.352
Answer:
63.352
Place value of 2 = 2 × \(\frac{1}{1000}\) = 0.002

Question 7.
Dinesh went 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?

Answer:
Distance travelled by Dinesh
= Distance AB + Distance BC
= 7.5km + 12.7km
= 20.2 km
Distance travelled by Ayub
= Distance AD + Distance DC
= 9.3 km + 11.8 km
= 21.1 km
∵ 21.1 > 20.2
∴ Ayub travelled more and by 21.1 - 20.2 = 0.90 km

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?
Answer:
Fruits bought by Shyama
= 5 kg 300 g + 3 kg 250 g
= 5.300 kg + 3.250 kg = 8.550 kg
Fruits bought by Sarala
= 4 kg 800 g + 4 kg 150 g
= 4.800 kg + 4.150 kg = 8.950 kg
∵ 8.950 > 8.550
∴ Sarala bought more fruits.

Question 9.
How much less is 28 km than 42.6 km?
Answer:
∵ 42.6 - 28.0 = 14.6 km
∴ 28 km is less than 42.6 km by 14.6 km.