Fractions and Decimals Ex-2.5

NCERT solutions for Maths Fractions and Decimals 

NCERT Solutions for Class 7 Maths Fractions and Decimals

Class –VII Mathematics (Ex. 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) 0.5 > 0.05

(ii) 0.7 > 0.5

(iii) 7 > 0.7

(iv) 1.37 < 1.49

(v) 2.03 < 2.30

(vi) 0.8 < 0.88

NCERT Solutions for Class 7 Maths Exercise 2.5

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. $\frac{1}{{100}}$

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

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

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

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

235 paise = Re. $\frac{{235}}{{100}}$ = 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 = $\frac{1}{{100}}$ meter $\Rightarrow$ 5 cm = $\frac{5}{{100}}$ = 0.05 meter.

Now, $\because$ 1000 meters = 1 kilometers

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

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

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

$\because$ 10 mm = 1 cm

$\therefore$ 1 mm = $\frac{1}{{10}}$ cm $\Rightarrow$ 35 mm = $\frac{{35}}{{10}}$ = 3.5 cm

Now, $\because$ 100 cm = 1 meter

$\therefore$ 1 cm = $\frac{1}{{100}}$ meter $\Rightarrow$3.5 cm = $\frac{{3.5}}{{100}}$ = 0.035 meter

Again, $\because$ 1000 meters = 1 kilometers

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

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

Question 4.Express in kg.:

(i) 200 g

(ii) 3470 g

(iii)4 kg 8 g

Answer:

Les us consider , 1000 g = 1 kg $\Rightarrow$ 1 g = $\frac{1}{{1000}}$ kg

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

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

4 kg 8 g = 4 kg + $\left( {8 \times \frac{1}{{1000}}} \right)$ kg = 4 kg + 0.008 kg = 4.008 kg

NCERT Solutions for Class 7 Maths Exercise 2.5

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 = $2 \times 10 + 0 \times 1 + 0 \times \frac{1}{{10}} + 3 \times \frac{1}{{100}}$

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

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

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

NCERT Solutions for Class 7 Maths Exercise 2.5

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 = $2 \times \frac{1}{{10}}$ = 2 tenths

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

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

NCERT Solutions for Class 7 Maths Exercise 2.5

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

NCERT Solutions for Class 7 Maths Exercise 2.5

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.

NCERT Solutions for Class 7 Maths Exercise 2.5

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.