NCERT Solution class 7 chapter 2 exercise 2.5

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

Sol :

(i) 0.5 or 0.05

Converting these decimal numbers into equivalent fractions,

0.5=\dfrac{5}{10}=\dfrac{5\times10}{10\times 10}=\dfrac{50}{100} and 0.05=\dfrac{5}{100}

It can be observed that both fractions have the same denominator.

As 50 > 5 ,

Therefore , 0.5 > 0.05


 

(ii) 0.7 or 0.5

Converting these decimal numbers into equivalent fractions,

0.7=\dfrac{7}{10} and 0.5=\dfrac{5}{10}

It can be observed that both fractions have the same denominator.

As 7 > 5 ,

Therefore , 0.7 > 0.5


 

(iii) 7 or 0.7

Converting these decimal numbers into equivalent fractions,

0.7=\dfrac{7}{1}=\dfrac{7\times 10}{1\times 10}=\dfrac{70}{10} and 0.7=\dfrac{7}{10}

It can be observed that both fractions have the same denominator.

As 70 > 7 ,

Therefore , 7 > 0.7

 


(iv) 1.37 or 1.49

Converting these decimal numbers into equivalent fractions,

1.37=\dfrac{137}{100} and 1.49=\dfrac{149}{100}

It can be observed that both fractions have the same denominator.

As 137 < 149 ,

Therefore , 1.37 < 1.49

 


 

(v) 2.03 or 2.30

Converting these decimal numbers into equivalent fractions,

2.03=\dfrac{203}{100} and 2.30=\dfrac{230}{100}

It can be observed that both fractions have the same denominator.

As 203 < 230 ,

Therefore , 2.03 < 2.30

 


 

(vi) 0.8 or 0.88

Converting these decimal numbers into equivalent fractions,

0.8=\dfrac{8}{10}=\dfrac{8\times 10}{10 \times 10}=\dfrac{80}{100} and 0.88=\dfrac{88}{100}

It can be observed that both fractions have the same denominator.

As 80 < 88 ,

Therefore , 0.8 < 0.88

 


Question 2

Express as rupees using decimals:

(i) 7 paise

Sol :

1 rupees = 100 paise 

\dfrac{1}{100} ₹ = 1 paise 

multiplying both side by 235 , we get

\dfrac{7}{100} ₹ = 7 paise 

0.07 ₹ = 7 paise 

or 7 paise = 0.07 ₹

ALTERNATE METHOD

(i) 7 paise = ₹ \dfrac{7}{100} = ₹ 0.07

 


(ii) 7 rupees 7 paise

Sol :

Firstly , we convert 7 paise , then we add it to 7 rupees

1 rupees = 100 paise 

\dfrac{1}{100} ₹ = 1 paise 

multiplying both sides by 7 , we get 

\dfrac{7}{100} ₹ = 7 paise

0.07  ₹ = 7 paise 

or 7 paise = 0.07 ₹

= 7 ₹ + 0.07

= 7.07 ₹

ALTERNATE METHOD

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

 


(iii) 77 rupees 77 pasie 

Sol :

Firstly , we convert 77 paise , then we add it to 77 rupees

1 rupees = 100 paise 

\dfrac{1}{100} ₹ = 1 paise 

multiplying both sides by 77 , we get 

\dfrac{77}{100} ₹ = 77 paise 

0.77 ₹ = 77 paise 

or 77 paise = 0.77 ₹

= 77 ₹ + 0.77 ₹

= 77.77 ₹

ALTERNATE METHOD

(iii) 77 rupees 77 pasie = 77 ₹ + \dfrac{77}{100} ₹ = 77.77 ₹

 


(iv) 50 paise 

Sol :

1 rupees = 100 paise 

\dfrac{1}{100} ₹ = 1 paise 

multiplying both side by 50 , we get

\dfrac{50}{100} ₹ = 50 paise 

0.50 ₹ = 50 paise 

or 50 paise = 0.50 ₹

ALTERNATE METHOD

(iv) 50 paise = \dfrac{50}{100} ₹ = 0.50 ₹

 


(v) 235 paise 

Sol :

1 rupees = 100 paise 

\dfrac{1}{100} ₹ = 1 paise 

multiplying both side by 235 , we get

\dfrac{235}{100} ₹ = 235 paise 

2.35 ₹ = 235 paise 

or 235 paise = 2.35 ₹

ALTERNATE METHOD

(v) 235 paise = \dfrac{235}{100} ₹ = 2.35 ₹

Notes : There are 100 paise in 1 rupee. Therefore, if we want to convert paise into rupees, then we have to divide paise by 100.

 


Question 3

(i) Express 5 cm in metre and kilometre

Sol :

(i) 5 cm 

5 cm =\dfrac{5}{100} m = 0.05 m 

5 cm =\dfrac{5}{100000} km = 0.00005 km

 

ALTERNATE METHOD

(i) Converting 5 cm to m 

1 meter = 100 centimeter

\dfrac{1}{100} m = 1 cm

multiplying both sides by 5 , we get

\dfrac{5}{100} m = 5 cm

0.05 m = 5 cm

or 5 cm = 0.05 m 

then converting 0.05 m to km

1 km = 1000 m

\dfrac{1}{1000} km = 1 m

multiplying both sides by 0.05 , we get

\dfrac{0.05}{1000} km = 0.05 m

0.00005 km = 0.05 m 

or 0.05 m = 0.00005 km

5 cm = 0.05 m = 0.00005 km


 

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

Sol :

(ii) 35 mm 

35 mm =\dfrac{35}{10} cm = 3.5 cm

35 mm =\dfrac{35}{1000} m = 0.035 m

35 mm =\dfrac{35}{1000000} km = 0.000035  km

ALTERNATE METHOD

(i) Converting 35 mm to cm 

1 centimeter = 10 milimeter

\dfrac{1}{10} cm = 1 mm

multiplying both sides by 35 , we get

\dfrac{35}{10} cm = 35 mm

3.5 cm = 35 mm

or 35 mm = 3.5 cm

then converting 3.5 cm to m

1 meter = 100 centimeter

\dfrac{1}{100} m = 1 cm

multiplying both sides by 3.5 , we get

\dfrac{3.5}{100} m = 3.5 cm

0.035 m = 3.5 cm

or 3.5 cm = 0.035 m

then converting 0.035 m to km

1 kilometer = 1000 meter 

\dfrac{1}{1000} km = 1 m 

multiplying both sides by 0.035 , we get

\dfrac{0.035}{1000} km = 0.035 m 

0.000035 km = 0.035 m 

or 0.035 m = 0.000035 km

35 mm = 3.5 cm = 0.035 m = 0.000035 km

 


Question 4

Express in kg:
(i) 200 g

Sol :

(i) 200 g =\dfrac{200}{1000} kg = 0.2 kg

ALTERNATE METHOD

(i) 200 g

1 kg = 1000 g

\dfrac{1}{1000} kg = 1 g

multiplying both sides by 200 , we get

\dfrac{200}{1000} kg = 200 g

0.200 kg = 200 g

or 200 g = 0.200 kg

 


 

(ii) 3470 g

Sol :

(ii) 3470 g =\dfrac{3470}{1000} kg = 3.470 kg

ALTERNATE METHOD

(ii) 3470 g 

1 kg = 1000 g

\dfrac{1}{1000} kg = 1 g

multiplying both sides by 3470 , we get

\dfrac{3470}{1000} kg = 3470 g

3.470 kg = 3470 g

or 3470 g = 3.470 kg


 

(iii) 4 kg 8 g

Sol :

(iii) 4 kg 8 g = 4 kg + \dfrac{8}{1000} kg = 4.008 kg

ALTERNATE METHOD

(iii) 4 kg 8 g

Firstly , we convert 8 g to kg then add it to 4 kg

1 kg = 1000 g 

\dfrac{1}{1000} kg = 1 g 

multiplying both sides by 8 , we get

\dfrac{8}{1000} kg =  8 g 

0.008 kg = 8 g 

or 8 g = 0.008 kg

 = 4 kg + 0.008 kg 

= 4.008 kg

 


 

Question 5

Write the following decimal numbers in the expanded form:
(i) 20.03

Sol :

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

 


 

(ii) 2.03

Sol :

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

 


 

(iii) 200.03

Sol :

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

 


 

(iv) 2.034

Sol :

=2\times 1 + 0\times \dfrac{1}{10}+3\times \dfrac{1}{100}+4\times \dfrac{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

Sol :

(i) Ones (ii) Tens (iii) Tenths (iv) Hundredths (v) Thousandths

 


 

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?

Sol :

{image to be added}

Distance travelled by Dinesh = AB + BC = (7.5 + 12.7) km

\begin{array}{r}{7.5} \\ {+12.7} \\ \hline 20.2\end{array}

Therefore, Dinesh travelled 20.2 km.

Distance travelled by Ayub = AD + DC = (9.3 + 11.8) km

\begin{array}{r}{9.3}\\{+11.8}\\\hline21.1\end{array}

Therefore, Ayub travelled 21.1 km.

Hence, Ayub travelled more distance.

Difference = (21.1 — 20.2) km

\begin{array}{r}{21.1}\\{-20.2}\\\hline 0.9\end{array}

Therefore, Ayub travelled 0.9 km more than Dinesh.

 


 

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?

Sol :

Total fruits bought by Shyama = 5 kg 300 g + 3 kg 250 g

= 8 kg 550 g

=\left(8+\dfrac{550}{1000}\right)\text{kg}

= 8.550 kg

Total fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g

= 8 kg 950 g

=\left(8+\dfrac{950}{1000}\right)\text{kg}

= 8.950 kg

\thereforeSarala bought more fruits.


 

Question 9

How much less is 28 km than 42.6 km?

Sol :

\begin{array}{r}{42.6}\\{-28.0}\\\hline 14.6\end{array}

Therefore, 28 km is 14.6 km less than 42.6 km

 

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