These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.5

Question 1.

Which is greater?

(i) 0.5 or 0.05

(ii) 0.7 or 0.5

(iii) 7 or 0.7

(iv) 1.37 or 1.49

(v) 2.03 or 2.30

(vi) 0.8 or 0.88

Answer:

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

5 > 0

∴ 0.5 >0.05

So, 0.5 is greater.

(ii) Comparing the digits at tenths places, we have

7 > 5

∴ 0.7 > 0.5

So, 0.7 is greater.

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

7 > 0

∴ 7 > 0.7

So, 7 is greater.

(iv) Since the digits at ones place are same, comparing the digits at tenths place, we have

3 < 4

∴ 1.37 <1.49

So, 1.49 is greater.

(v) Since the digits at ones place are same, comparing the digits at tenths place, we have

0 < 3

∴ 2. 03 < 2.30

So, 2.30 is greater.

(vi) 0.8 can be written as 0.80. Now, digits at tenths place are same, comparing the digits at hundredths place, we have

0 < 8

∴ 0.80 <0.88

So, 0.88 is greater.

Question 2.

Express as rupees using decimals:

(i) 7 paise

(ii) 7 rupees 7 paise

(iii) 77 rupees 77 paise

(iv) 50 paise

(v) 235 paise.

Answer:

(i) 7 paise = ₹ = 7 × \(\frac{1}{100}=\frac{7}{100}\) = ₹ 0.07

(ii) rupees 7 paise = ₹ = 7 + 7 × \(\frac{1}{100}\)

= ₹ 7 + ₹ 0.07 = ₹ 7.07

(iii) 77 rupees 77 paise = ₹ 77 + ₹ 77 × \(\frac{1}{100}\) = ₹ 77 + ₹ 0.77 = ₹ 77.77

(iv) 50 paise = ₹ 50 × \(\frac{1}{100}\) = ₹ 0.50

(v) 235 paise= 200 paise + 35 paise

= ₹ 2 + ₹ 35 × \(\frac{1}{100}\) = ₹ 2 + ₹ 0.35

= ₹ 2.35

Question 3.

(i) Express 5 cm in metre and kilometre

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

Answer:

We know that 100 cm = 1 m,

1000 m = 1 km

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

5 cm = \(\frac{5}{100 \times 100}\) km

= 0.00005 km.

(ii) We know that 1 cm =10 mm

35 mm = \(\frac{35}{10 \times 100}\) m = \(\frac{35}{1000}\)

35mm = \(\frac{35}{10 \times 100}\) m = \(\frac{35}{1000}\)

= 0.035 m

35 mm \(\frac{35}{10 \times 100 \times 1000}\) km

= \(\frac{35}{1000000}\) = 0.000035 km

Question 4.

Express in kg:

(i) 200 g

(ii) 3470 g

(iii) 4 kg 8 g

Answer:

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 = \(\frac{347}{100}\) kg = 3.47 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

Answer:

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.

Answer:

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

∴ place value of 2 is 2 × 1 = 2.

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

∴ place value of 2 is 2 × 10 = 20.

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

∴ place value of 2 is 2 × \(\frac{1}{10}=\frac{2}{10}\)

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

∴ The place value of 2 is 2 × \(\frac{1}{100}=\frac{2}{100}\)

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

∴ The place value of 2 is 2 × \(\frac{1}{1000}=\frac{2}{100}\)

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 from A to B = 7.5 km

Distance from B to C = 12.7 km

Distance from A to C through

B = (7.5 + 12.7) km

= 20.2 km

∴ Distance travelled by Dinesh = 20.2 km

Again

Distance from A to D = 9.3 km

Distance from D to C = 11.8 km

Distance from A to C through D

= (9.3 + 11.8) km

= 21.1 km

∴ Distance travelled by Ayub = 21.1 km 21. 1 > 20.2

∴ Ayub travelled more distance.

Difference = 21.1 km – 20.2 km

= 0.90 km (or) 900 m

∴ 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?

Answer:

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

8.950 > 8.550

So, Sarala bought more fruits

Difference in weight = 8 kg 950 g – 8 kg 550 g = 0 kg 400 g

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

Question 9.

How much less is 28 km than 42.6 km?

Answer:

Difference = 42.6 km – 28 km = 14.6 km 28 km is less than 42.6 km by 14.6 km