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

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

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.5
Ex 2.5 Class 7 Maths 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
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 or 1.49 > 1.37

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

(vi) 0.8 or 0.88 ⇒ 0.80 or 0.88
Since tenths place is same.
Comparing the hundredth place, we get 0 < 8
∴ 0.80 < 0.88 or 0.88 > 0.80

Ex 2.5 Class 7 Maths 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
Solution:
(i) Since 1 rupee = 100 paise and 1 paise = \frac{1}{100} rupees
7 paise =\frac{7}{100} rupees = 0.07 rupees

(ii) 7 rupees 7 paise = 7 rupees + \frac{7}{100} rupees
= 7.07 rupees

(iii) 77 rupees 77 paise = 77 rupees + \frac{77}{100} rupees
= 77.77 rupees

(iv) 50 paise = \frac{50}{100} rupees = 0.50 rupees

(v) 235 paise = \frac{235}{100} rupees = 2.35 rupees

Ex 2.5 Class 7 Maths Question 3.
(i) Express 5 cm in metre and kilometre
(ii) Express 35 mm in cm, m and km.
Solution:
(i) 1 metre = 100 cm
1 kilometre = 1000 metre = 100 × 1000 cm
= 100000 cm
∴ 5 cm = \frac{5}{100} metre = 0.05 metre
5 cm = \frac{5}{100000} km = 0.00005 km
Hence, 5 cm = 0.05 m and 0.00005 km

(ii) 1 cm = 10 mm and 1 km = 100000 cm
∴ 35 mm = \frac{35}{10} cm = 3.5 cm,
35 mm = \frac{35}{1000} m = 0.035 m
35 mm = \frac{35}{1000000} km = 0.000035 km
Hence, 35 mm = 3.5 cm, 0.035 m and 0.000035 km.

Ex 2.5 Class 7 Maths Question 4.
Express in kg:
(i) 200 g
(ii) 3470 g
(iii) 4 kg 8 g
Solution:
(i) 200g = \frac{200}{1000} kg [∵ 1 kg = 1000g]
= 0.2 kg
(ii) 3470 g = \frac{3470}{10000} = 3.47 kg [∵ 1 kg = 1000 g]
(iii) 4 kg 8 g = 4 kg + \frac{8}{10000}kg [∵ 1 kg = 1000 g]
= 4 kg + 0.008 kg = 4.008 kg

Ex 2.5 Class 7 Maths Question 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 = 2 × 10 + 0 × 1 + 0 × \frac{1}{10} + 3 × \frac{1}{100}
(ii) 2.03 = 2 × 1 + 0 × \frac{1}{10} + 3 × \frac{1}{100}
(iii) 200.03 = 2 × 100 + 0 × 10 + 0 × 1 + 0 × \frac{1}{10} +3 × \frac{1}{100}
(iv) 2.034 = 2 × 1 + 0 × \frac{1}{10} + 3 × \frac{1}{10} + 4 × \frac{1}{1000}

Ex 2.5 Class 7 Maths 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
Solution:
(i) Place value of 2 in 2.56 = 2 × 1 = 2 i.e. ones
(ii) Place value of 2 in 21.37 = 2 × 10 = 20 i.e. tens
(iii) Place value of 2 in 10.25 = \frac{2}{10} = 0.2 i.e. tenths
(iv) Place value of 2 in 9.42 = \frac{2}{100} = 0.02 i.e. hundredths
(v) Place value of 2 in 63.352 = \frac{2}{1000} = 0.002 i.e. thousandths.

Ex 2.5 Class 7 Maths 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?
Solution:
Distance travelled by Dinesh from A to C
= AB + BC
= 7.5 km +- 12.7 km
= 20.2 km
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 1
Distance travelled by Ayub from A to C
= AD + DC
= 9.3 km + 11.8 km = 21.1 km
Since 21.1 km > 20.2 km.
Hence, Ayub travelled more distance.

Ex 2.5 Class 7 Maths 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?
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.

Ex 2.5 Class 7 Maths Question 9.
How much less is 28 km than 42.6 km?
Solution:
Since 28 km < 42.6 km
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 2
Hence, 28 km is less than 42.6 km by 14.6 km.

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