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

NCERT Solutions for Class 7 Maths Exercise 2.5 Chapter 2 Fractions and Decimals, students can download the PDF which is given here. Students are going to learn about decimal numbers and how to express rupees in decimals in this exercise of NCERT Solutions Maths for Class 7 Chapter 2. Expert tutors at BYJU’S have prepared the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals. Students gain more knowledge by referring to these solutions.

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

### Access Answers to NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals Exercise 2.5

1. Which is greater?

(i) 0.5 or 0.05

Solution:-

By comparing whole number, 0 = 0

By comparing the tenths place digit, 5 > 0

∴ 0.5 > 0.05

(ii) 0.7 or 0.5

Solution:-

By comparing whole number, 0 = 0

By comparing the tenths place digit, 7 > 5

∴ 0.7 > 0.5

(iii) 7 or 0.7

Solution:-

By comparing whole number, 7 > 0

∴ 7 > 0.7

(iv) 1.37 or 1.49

Solution:-

By comparing whole number, 1 = 1

By comparing the tenths place digit, 3 < 4

∴ 1.37 < 1.49

(v) 2.03 or 2.30

Solution:-

By comparing whole number, 2 = 2

By comparing the tenths place digit, 0 < 3

∴ 2.03 < 2.30

(vi) 0.8 or 0.88

Solution:-

By comparing whole number, 0 = 0

By comparing the tenths place digit, 8 = 8

By comparing the hundredths place digit, 0 < 8

∴ 0.8 < 0.88

2. Express as rupees as decimals:

(i) 7 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 paise = ₹ (7/100)

= ₹ 0.07

(ii) 7 rupees 7 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 rupees 7 paise = ₹ 7 + ₹ (7/100)

= ₹ 7 + ₹ 0.07

= ₹ 7.07

(iii) 77 rupees 77 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 77 rupees 77 paise = ₹ 77 + ₹ (77/100)

= ₹ 77 + ₹ 0.77

= ₹ 77.77

(iv) 50 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 50 paise = ₹ (50/100)

= ₹ 0.50

(v) 235 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 235 paise = ₹ (235/100)

= ₹ 2.35

3. (i) Express 5 cm in meter and kilometer

Solution:-

We know that,

= 1 meter = 100 cm

Then,

= 1 cm = (1/100) m

= 5 cm = (5/100)

= 0.05 m

Now,

= 1 km = 1000 m

Then,

= 1 m = (1/1000) km

= 0.05 m = (0.05/1000)

= 0. 00005 km

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

Solution:-

We know that,

= 1 cm = 10 mm

Then,

= 1 mm = (1/10) cm

= 35 mm = (35/10) cm

= 3.5 cm

And,

= 1 meter = 100 cm

Then,

= 1 cm = (1/100) m

= 3.5 cm = (3.5/100) m

= (35/1000) m

= 0.035 m

Now,

= 1 km = 1000 m

Then,

= 1 m = (1/1000) km

= 0.035 m = (0.035/1000)

= 0. 000035 km

4. Express in kg:

(i) 200 g

Solution:-

We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 200 g = (200/1000) kg

= (2/10)

= 0.2 kg

(ii) 3470 g

Solution:-

We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 3470 g = (3470/1000) kg

= (3470/100)

= 3.470 kg

(ii) 4 kg 8 g

Solution:-

We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 4 kg 8 g = 4 kg + (8/1000) kg

= 4 kg + 0.008

= 4.008 kg

5. Write the following decimal numbers in the expanded form:

(i) 20.03

Solution:-

We have,

20.03 = (2 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(ii) 2.03

Solution:-

We have,

2.03 = (2 × 1) + (0 × (1/10)) + (3 × (1/100))

(iii) 200.03

Solution:-

We have,

200.03 = (2 × 100) + (0 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(iv) 2.034

Solution:-

We have,

2.034 = (2 × 1) + (0 × (1/10)) + (3 × (1/100)) + (4 × (1/1000))

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

(i) 2.56

Solution:-

From the question, we observe that,

The place value of 2 in 2.56 is ones

(ii) 21.37

Solution:-

From the question, we observe that,

The place value of 2 in 21.37 is tens

(iii) 10.25

Solution:-

From the question, we observe that,

The place value of 2 in 10.25 is tenths.

(iv) 9.42

Solution:-

From the question, we observe that,

The place value of 2 in 9.42 is hundredth.

(v) 63.352

Solution:-

From the question, we observe that,

The place value of 2 in 63.352 is thousandth.

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? Solution:-

From the question, it is given that,

Distance travelled by Dinesh = AB + BC

= 7.5 + 12.7

= 20.2 km

∴Dinesh travelled 20.2 km

Distance travelled by Ayub = AD + DC

= 9.3 + 11.8

= 21.1 km

∴Ayub travelled 21.1km

Clearly, Ayub travelled more distance by = (21.1 – 20.2)

= 0.9 km

∴Ayub travelled 0.9 km more than Dinesh.

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:-

From the question, it is given that,

Fruits bought by Shyama = 5 kg 300 g

= 5 kg + (300/1000) kg

= 5 kg + 0.3 kg

= 5.3 kg

Fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g

= (4 + (800/1000)) + (4 + (150/1000))

= (4 + 0.8) kg + (4 + .150) kg

= 4.8 kg + 4.150kg

= 8.950 kg

So, Sarala bought more fruits.

9. How much less is 28 km than 42.6 km?

Solution:-

Now, we have to find the difference of 42.6 km and 28 km

42.6

-28.0

14.6

∴ 14.6 km less is 28 km than 42.6 km.

### Access Other Exercises of NCERT Solutions For Class 7 Maths Chapter 2 – Fractions and Decimals

Exercise 2.1 Solutions

Exercise 2.2 Solutions

Exercise 2.3 Solutions

Exercise 2.4 Solutions

Exercise 2.6 Solutions

Exercise 2.7 Solutions

Also, explore –

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7

NCERT Solutions