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NCERT · Class 7 · Maths

Chapter 2: Fractions and Decimals

Practice each question, then open the answer or worked solution.

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Exercise 2.5

Textbook page 47 · 9 questions
Q1Written answer

Which is greater?

(i) 0.5 or 0.05

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Worked solution

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

Q2Written answer

Express as rupees as decimals:

(i) 7 paise

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Worked solution

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

Q3Written answer

(i) Express 5 cm in meter and kilometer

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Worked solution

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

Q4Written answer

Express in kg:

(i) 200 g

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Worked solution

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

Q5Written answer

Write the following decimal numbers in the expanded form:

(i) 20.03

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Worked solution

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))

Q6Written answer

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

(i) 2.56

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Worked solution

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.

Q7Written answer

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?

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 100

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Worked solution

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.

Q8Written answer

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?

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Worked solution

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.

Q9Written answer

How much less is 28 km than 42.6 km?

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Worked solution

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.