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NCERT Solutions for Class 7 Maths Exercise 2.5

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NCERT Solutions for Class 7 Maths Exercise 2.5 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 7 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Fractions and Decimals Download as PDF

NCERT Solutions for Class 7 Maths Exercise 2.5

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:

{tex}\because {/tex} 100 paise = Re. 1

{tex}\therefore {/tex} 1 paisa = Re. {tex}\frac{1}{{100}}{/tex}

7 paise = Re. {tex}\frac{7}{{100}}{/tex} = Re. 0.07

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

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

50 paise = Re. {tex}\frac{{50}}{{100}}{/tex} = Re. 0.50

235 paise = Re. {tex}\frac{{235}}{{100}}{/tex} = 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.

{tex}\because {/tex} 100 cm = 1 meter

{tex}\therefore {/tex} 1 cm = {tex}\frac{1}{{100}}{/tex} meter {tex} \Rightarrow {/tex} 5 cm = {tex}\frac{5}{{100}}{/tex} = 0.05 meter.

Now, {tex}\because {/tex} 1000 meters = 1 kilometers

{tex}\therefore {/tex} 1 meter = {tex}\frac{1}{{1000}}{/tex} kilometer

{tex} \Rightarrow {/tex} 0.05 meter = {tex}\frac{{0.05}}{{1000}}{/tex} = 0.00005 kilometer

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

{tex}\because {/tex} 10 mm = 1 cm

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

Now, {tex}\because {/tex} 100 cm = 1 meter

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

Again, {tex}\because {/tex} 1000 meters = 1 kilometers

{tex}\therefore {/tex} 1 meter = {tex}\frac{1}{{1000}}{/tex} kilometer

{tex} \Rightarrow {/tex} 0.035 meter = {tex}\frac{{0.035}}{{1000}}{/tex} = 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 {tex} \Rightarrow {/tex} 1 g = {tex}\frac{1}{{1000}}{/tex} kg

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

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

4 kg 8 g = 4 kg + {tex}\left( {8 \times \frac{1}{{1000}}} \right){/tex} 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 = {tex}2 \times 10 + 0 \times 1 + 0 \times \frac{1}{{10}} + 3 \times \frac{1}{{100}}{/tex}

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

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

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


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 = {tex}2 \times \frac{1}{{10}}{/tex} = 2 tenths

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

(v) Place value of 2 in 63.352 = {tex}2 \times \frac{1}{{1000}}{/tex} = 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.

NCERT Solutions for Class 7 Maths Exercise 2.5

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