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