Fractions and Decimals - Solutions 5

CBSE Class –VII Mathematics
NCERT Solutions
Chapter 2 Fractions and Decimals (Ex. 2.5)

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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) 0.50 > 0.05

(ii) 0.7 > 0.5

(iii) 7.0 > 0.7

(iv) 1.37 < 1.49

(v) 2.03 < 2.30

(vi) 0.80 < 0.88

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: ∵$\because$ 100 paise = Re. 1

∴$\therefore$ 1 paisa = Re. 1100$\frac{1}{100}$

7 paise = Re. 7100$\frac{7}{100}$ = Re. 0.07

7 rupees 7 paise = Rs. 7 + Re. 7100$\frac{7}{100}$ = Rs. 7 + Re. 0.07 = Rs. 7.07

77 rupees 77 paise = Rs. 77 + Re. 77100$\frac{77}{100}$ = Rs. 77 + Re. 0.77 = Rs. 77.77

50 paise = Re. 50100$\frac{50}{100}$ = Re. 0.50

235 paise = Re. 235100$\frac{235}{100}$ = Rs. 2.35

Question 3. (i) Express 5 cm in metre and kilometer.

(ii) Express 35 mm in cm, m and km.

Answer: (i) Express 5 cm in meter and kilometer.

∵$\because$ 100 cm = 1 meter

∴$\therefore$ 1 cm = 1100$\frac{1}{100}$ meter ⇒$\Rightarrow$ 5 cm = 5100$\frac{5}{100}$ = 0.05 meter.

Now, ∵$\because$ 1000 meters = 1 kilometers

∴$\therefore$ 1 meter = 11000$\frac{1}{1000}$ kilometer

⇒$\Rightarrow$ 0.05 meter = 0.051000$\frac{0.05}{1000}$ = 0.00005 kilometer

(ii) Express 35 mm in cm, m and km.

∵$\because$ 10 mm = 1 cm

∴$\therefore$ 1 mm = 110$\frac{1}{10}$ cm ⇒$\Rightarrow$ 35 mm = 3510$\frac{35}{10}$ = 3.5 cm

Now, ∵$\because$ 100 cm = 1 meter

∴$\therefore$ 1 cm = 1100$\frac{1}{100}$ meter ⇒$\Rightarrow$3.5 cm = 3.5100$\frac{3.5}{100}$ = 0.035 meter

Again, ∵$\because$ 1000 meters = 1 kilometers

∴$\therefore$ 1 meter = 11000$\frac{1}{1000}$ kilometer

⇒$\Rightarrow$ 0.035 meter = 0.0351000$\frac{0.035}{1000}$ = 0.000035 kilometer

Question 4. Express in kg.:

(i) 200 g

(ii) 3470 g

(iii)4 kg 8 g

Answer: 1000 g = 1 kg ⇒$\Rightarrow$ 1 g = 11000$\frac{1}{1000}$ kg

200 g = (200×11000)$\left(200\times \frac{1}{1000}\right)$ kg = 0.2 kg

3470 g = (3470×11000)$\left(3470\times \frac{1}{1000}\right)$ kg = 3.470 kg

4 kg 8 g = 4 kg + (8×11000)$\left(8\times \frac{1}{1000}\right)$ 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: (i) 20.03 = 2×10+0×1+0×110+3×1100$2\times 10+0\times 1+0\times \frac{1}{10}+3\times \frac{1}{100}$

(ii) 2.03 = 2×1+0×110+3×1100$2\times 1+0\times \frac{1}{10}+3\times \frac{1}{100}$

(iii) 200.03 = 2×100+0×10+0×1+0×110+3×1100$2\times 100+0\times 10+0\times 1+0\times \frac{1}{10}+3\times \frac{1}{100}$

(iv) 2.034 = 2×1+0×110+3×1100+4×11000$2\times 1+0\times \frac{1}{10}+3\times \frac{1}{100}+4\times \frac{1}{1000}$

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) Place value of 2 in 2.56 = 2 x 1 = 2 ones

(ii) Place value of 2 in 21.37 = 2 x 10 = 2 tens

(iii) Place value of 2 in 10.25 = 2×110$2\times \frac{1}{10}$ = 2 tenths

(iv) Place value of 2 in 9.42 = 2×1100$2\times \frac{1}{100}$ = 2 hundredth

(v) Place value of 2 in 63.352 = 2×11000$2\times \frac{1}{1000}$ = 2 thousandth

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 travelled by Dinesh when he went from place A to place B = 7.5 km and from place B to C = 12.7 km.

Total distance covered by Dinesh = AB + BC

= 7.5 + 12.7 = 20.2 km

Total distance covered by Ayub = AD + DC

= 9.3 + 11.8 = 21.1 km

On comparing the total distance of Ayub and Dinesh,

21.1 km > 20.2 km

Therefore, Ayub covered more distance by 21.1 – 20.2 = 0.9 km = 900 m

Question 8. Shyam 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: Total weight of fruits bought by Shyam = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g

Total weight of fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g

On comparing the quantity of fruits,

8 kg 550 g < 8 kg 950 g

Therefore, Sarala bought more fruits.

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

Answer: We have to find the difference of 42.6 km and 28 km.

42.6 – 28.0 = 14.6 km

Therefore 14.6 km less is 28 km than 42.6 km.