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

NCERT Solutions Class 7 Mathematics 
Chapter – 2 (Fractions and Decimals)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 2 Fractions and Decimals Exercise 2.5 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 2 Fractions and Decimals

Exercise – 2.5

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

Solution –
(i) 0.5 or 0.05
Comparing the tenths place, we get 5 > 0
∴ 0.5 > 0.05

(ii) 0.7 or 0.5
Comparing the tenths place, we get 7 > 5
∴ 0.7 > 0.5

(iii) 7 or 0.7
Comparing the one’s place, we get 7 > 0
∴ 7 > 0.7

(iv) 1.37 or 1.49
Comparing the tenths place, we get 3 < 4
∴ 1.37 < 1.49

(v) 2.03 or 2.30
Comparing the tenths place, we get 0 < 3
∴ 2.03 < 2.30

(vi) 0.8 or 0.88
Since tenths place is same.
Comparing the hundredth place, we get 0 < 8
∴ 0.8 < 0.88

2. Express as rupees as decimals:
(i) 7 paise                   (ii) 7 rupees 7 paise                  (iii) 77 rupees 77 paise
(iv) 50 paise              (v) 235 paise

Solution –
(i) 7 paise
We know that,
⇒ ₹ 1 = 100 paise
⇒ 1 paise = ₹ \frac{1}{100}

∴ 7 paise = ₹ \frac{7}{100} = ₹ 0.07

(ii) 7 rupees 7 paise

We know that,
⇒ ₹ 1 = 100 paise
⇒ 1 paise = ₹ \frac{1}{100}

∴ 7 rupees 7 paise = ₹ 7 + ₹ \frac{7}{100}
= ₹ 7 + ₹ 0.07
= ₹ 7.07

(iii) 77 rupees 77 paise
We know that,
⇒ ₹ 1 = 100 paise
⇒1 paise = ₹ \frac{1}{100}

∴ 77 rupees 77 paise = ₹ 77 + ₹ \frac{77}{100}

= ₹ 77 + ₹ 0.77
= ₹ 77.77

(iv) 50 paise

We know that,
⇒ ₹ 1 = 100 paise
⇒ 1 paise = ₹ \frac{1}{100}

∴ 50 paise = ₹ \frac{50}{100} = ₹ 0.50

(v) 235 paise
We know that,
⇒ ₹ 1 = 100 paise
⇒ 1 paise = ₹ \frac{1}{100}

∴ 235 paise = ₹ \frac{235}{100} = ₹ 2.35

3.
(i) Express 5 cm in meter and kilometer
(ii) Express 35 mm in cm, m and km.

Solution –
(i) Express 5 cm in meter and kilometer
We know that,
⇒ 1 meter = 100 cm
Then,
= 1 cm = \frac{1}{100} m

= 5 cm = \frac{5}{100} = 0.05 m
Now,
= 1 km = 1000 m
Then,
= 1 m = \frac{1}{1000} km

= 0.05 m = \frac{0.05}{1000} km = 0. 00005 km

(i) Express 35 mm in cm, m and km
We know that,
= 1 cm = 10 mm
Then,
= 1 mm = \frac{1}{10} cm

= 35 mm = \frac{35}{10} cm = 3.5 cm
And,
= 1 meter = 100 cm
Then,
= 1 cm = \frac{1}{100} m

= 3.5 cm = \frac{3.5}{100} m

= \frac{35}{1000} m  = 0.035 m

Now,
= 1 km = 1000 m
Then,
= 1 m = \frac{1}{1000} km

= 0.035 m = \frac{0.035}{1000} km = 0. 000035 km

4. Express in kg:
(i) 200 g                      (ii) 3470 g                  (iii) 4 kg 8 g

Solution –
(i) 200 g
We know that,
= 1 kg = 1000 g
Then,
= 1 g = \frac{1}{1000} kg

= 200 g = \frac{200}{1000} kg = \frac{2}{10} kg
= 0.2 kg

(ii) 3470 g
We know that,
= 1 kg = 1000 g
Then,
= 1 g = \frac{1}{1000} kg

= 3470 g = \frac{3470}{1000} kg
= 3.470 kg

(iii) 4 kg 8 g
We know that,
= 1 kg = 1000 g
Then,
= 1 g = \frac{1}{1000} kg

= 4 kg 8 g = 4 kg + \frac{8}{1000} kg
= 4 kg + 0.008
= 4.008 kg

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
⇒ 20.03 = (2 × 10) + (0 × 1) + (0 × \frac{1}{10} ) + (3 × \frac{1}{100})

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

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

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

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) 2.56
Place value of 2 in 2.56 = 2 × 1 = 2 i.e. ones

(ii) 21.37
Place value of 2 in 21.37 = 2 × 10 = 20 i.e. tens

(iii) 10.25
Place value of 2 in 10.25 = \frac{2}{10} = 0.2 i.e. tenths

(iv) 9.42
Place value of 2 in 9.42 = \frac{2}{100} = 0.02 i.e. hundredths

(v) 63.352
Place value of 2 in 63.352 = \frac{2}{1000} = 0.002 i.e. thousandths.

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?
NCERT Class 7 Maths Solution

Solution –
Distance travelled by Dinesh from A to C.
= AB + BC
= 7.5 km + 12.7 km
= 20.2 km
∴ Dinesh travelled 20.2 km

Distance travelled by Ayub from A to C
= AD + DC
= 9.3 km + 11.8 km
= 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 –
Fruits bought by Shyama
= 5 kg 300 g apples + 3 kg 250 g mangoes
= 5.300 kg apples + 3.250 kg mangoes
= 8.550 kg of fruits
Fruits bought by Sarala
= 4 kg 800 g oranges +- 4 kg 150 g bananas
= 4.800 kg oranges + 4.150 kg bananas
= 8.950 kg of fruits
Since 8.950 kg > 8.550 kg
Hence, Sarala bought more fruits.

9. How much less is 28 km than 42.6 km?
Solution –
Since 28 km < 42.6 km
⇒ 42.6 km – 28.0 km = 14.6 km
∴ 14.6 km less is 28 km than 42.6 km.

 

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