Exercise 2.5
Q 1, Ex 2.5 -Class 7th - NCERT Fractions and Decimals - Chapter 2 - Maths
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
Q 2, Ex 2.5 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
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.
Q 3, Ex 2.5 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
(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
Q 4, Ex 2.5 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
Express in kg:
(i) 200 g
Sol :
(i) 200 g $=\dfrac{200}{1000}$ kg = 0.2 kg
ALTERNATE METHOD
(i) 200 g1 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 g1 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 gFirstly , 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
Q 5, Ex 2.5 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
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 :
$=2\times 1 + 0\times \dfrac{1}{10}+3\times \dfrac{1}{100}+4\times \dfrac{1}{1000}$
Q 6, Ex 2.5 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
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
Q 7, Ex 2.5 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
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 :
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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.
Q 8, Ex 2.5 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
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
$\therefore$Sarala bought more fruits.
Q 9, Ex 2.5 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
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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