Decimals Exercise – 8.1
Question 1. Which is greater?
- 0.3 or 0.4
- 0.07 or 0.02
- 3 or 0.8
- 0.5 or 0.05
- 1.23 or 1.2
- 0.099 or 0.1 9
- 1.5 or 1.50
- 1.431 or 1.490
- 3.3 or 3.300
- 5.64 or 5.603
Solution: Before comparing, we write both terms in like decimals:
- 0.3 <0.4
- 0.07 >0.02
- 3.0 or 0.8⇒ 3.0 > 0.8
- 0.50 or 0.05 ⇒ 0.50 > 0.05
- 1.23 or 1.20 ⇒ 1.23 >1.20
- 0.099 or 0.190 ⇒ 0.099 < 0.190
- 1.50 or 1.50 ⇒ 1.50 = 1.50
- 1.431 < 1.490
- 3.300 or 3.300 ⇒ 3.300 = 3.300
- 5.640 or 5.603 ⇒ 5.640 >5.603
Question 2. Make five more examples and find the greater number from them.
Solution: Five examples are :
- 1.8 or 1.82
- 1.0009 or 1.09
- 10.01 or 100.1
- 5.1 or 5.01
- 4.213 or 421.3
Before comparing, we write both terms in decimals
- 1.80 or 1.82 ⇒ 1.82 is greater than 1.8
- 1.0009 or 1.0900 ⇒ 1.09 is greater than 1.0009
- 10.01 or 100.10 ⇒ 100.1 is greater than 10.01
- 5.10 or 5.01 ⇒ 5.1 is greater than 5.01
- 4.213 or 421.300 ⇒ 421.3 is greater than 4.213
Decimals Exercise – 8.2
1. Express as rupees using decimals.
- 5 paise
- 75 paise
- 20 paise
- 50 rupees 90 paise
- 725 paise
Solution:
1 paisa \(=₹ \frac{1}{100}\)
5 paise \(=₹\left(\frac{1}{100} \times 5\right)=₹ 0.05\)
1 paisa \(=₹ \frac{1}{100}\)
75 paise \(=₹\left(\frac{1}{100} \times 75\right)=₹ 0.75\)
1 paisa \(=₹ \frac{1}{100}\)
20 paise \(=₹\left(\frac{1}{100} \times 20\right)=₹ 0.20\)
1 paisa \(=₹ \frac{1}{100}\)
₹ 50 + 90 paise \(=₹\left(50+\frac{1}{100} \times 90\right)\)
= ₹ 50.90
1 paisa \(=₹ \frac{1}{100}\)
725 paise \(=₹\left(\frac{1}{100} \times 725\right)=\frac{725}{100}=₹ 7.25\)
Question 2. Express as metres using decimals.
- 15 cm
- 6 cm
- 2 m 45 cm
- 9 m 7 cm
- 419 cm
Solution:
1 cm \(=\frac{1}{100} \mathrm{~m}\)
15 cm \(=\frac{1}{100} \times 15 \mathrm{~m}=0.15 \mathrm{~m}\)
1 cm \(=\frac{1}{100} \mathrm{~m}\)
6 cm \(=\frac{1}{100} \times 6 \mathrm{~m}=0.06 \mathrm{~m}\)
1 cm \(=\frac{1}{100} \mathrm{~m}\)
2 m 45 cm \(2 \mathrm{~m} 45 \mathrm{~cm}=\left(2+\frac{1}{100} \times 45\right) \mathrm{m}=2.45 \mathrm{~m}\)
1 cm \(=\frac{1}{100} \mathrm{~m}\)
9 m 7 cm \(=\left(9+\frac{1}{100} \times 7\right) \mathrm{m}=9.07 \mathrm{~m}\)
1 cm \(=\frac{1}{100} \mathrm{~m}\)
419 cm \(=\frac{1}{100} \times 419 \mathrm{~m}=\frac{419}{100}=4.19 \mathrm{~m}\)
Question 3. Express as cm using decimals.
- 5 mm
- 60 mm
- 164 mm
- 9 cm 8 mm
- 93 mm
- Solution:
1 mm \(=\frac{1}{10} \mathrm{~cm}\)
5 mm \(=\frac{1}{10} \times 5 \mathrm{~cm}=0.5 \mathrm{~cm}\)
1 mm \(=\frac{1}{10} \mathrm{~cm}\)
60 mm \(=\frac{1}{10} \times 60 \mathrm{~cm}=6 \mathrm{~cm}\)
1 mm \(=\frac{1}{10} \mathrm{~cm}\)
164 mm \(=\frac{1}{10} \times 164 \mathrm{~cm}=16.4 \mathrm{~cm}\)
1 mm \(=\frac{1}{10} \mathrm{~cm}\)
9 cm 8 mm \(=\left(9+\frac{1}{10} \times 8\right) \mathrm{cm}=9.8 \mathrm{~cm}\)
1 mm \(=\frac{1}{10} \mathrm{~cm}\)
93 mm \(=\frac{1}{10} \times 93 \mathrm{~cm}=9.3 \mathrm{~cm}\)
Question 4. Express as km using decimals.
- 8 m
- 88 m
- 8888 m
- 70 km 5 m
Solution:
1 m \(=\frac{1}{1000} \mathrm{~km}\)
8 m \(=\frac{1}{1000} \times 8 \mathrm{~km}=0.008 \mathrm{~km}\)
1 m \(=\frac{1}{1000} \mathrm{~km}\)
88 m\(=\frac{1}{1000} \times 88 \mathrm{~km}=0.088 \mathrm{~km}\)
1 m \(=\frac{1}{1000} \mathrm{~km}\)
8888 m \(=\frac{1}{1000} \times 8888 \mathrm{~km}=8.888 \mathrm{~km}\)
1 m \(=\frac{1}{1000} \mathrm{~km}\)
70 km 5 m \(=\left(70+\frac{1}{1000} \times 5\right) \mathrm{km}=70.005 \mathrm{~km}\)
Question 5. Express as kg using decimals.
- 2 g
- 100 g
- 3750 g
- 5 kg 8 g
- 26 kg 50 g
Solution:
1 g \(=\frac{1}{1000} \mathrm{~kg}\)
2 g \(=\frac{1}{1000} \times 2 \mathrm{~kg}=0.002 \mathrm{~kg}\)
1 g \(=\frac{1}{1000} \mathrm{~kg}\)
100 g \(=\frac{1}{1000} \times 100 \mathrm{~kg}=0.1 \mathrm{~kg}\)
1 g \(=\frac{1}{1000} \mathrm{~kg}\)
3750 g \(=\frac{1}{1000} \times 3750 \mathrm{~kg}=3.750 \mathrm{~kg}\)
1 g \(=\frac{1}{1000} \mathrm{~kg}\)
5 kg 8 g \(=\left(5+\frac{1}{1000} \times 8\right) \mathrm{kg}=5.008 \mathrm{~kg}\)
1 g \(=\frac{1}{1000} \mathrm{~kg}\)
26 kg 50 g \(=\left(26+\frac{1}{1000} \times 50\right) \mathrm{kg}=26.050 \mathrm{~kg}\)
Decimals Exercise – 8.3
Question 1. Find the sum in each of the following:
- 0.007 + 8.5 + 30.08
- 15 + 0.632 + 13.8
- 27.076 + 0.55 + 0.004
- 25.65 + 9.005 + 3.7
- 0.75 + 10.425 + 2
- 280.69 + 25.2 + 38
Solution:
1. \(\begin{array}{r}
0.007 \\
8.500 \\
+30.080 \\
\hline 38.587 \\
\hline
\end{array}\)
2. \(+\begin{array}{r}
15.000 \\
0.632 \\
13.800 \\
\hline 29.432 \\
\hline
\end{array}\)
3. \(\begin{array}{r}
27.076 \\
0.550 \\
+0.004 \\
\hline \underline{27.630} \\
\hline
\end{array}\)
4. \(\begin{array}{r}
25.650 \\
9.005 \\
+3.700 \\
\hline 38.355 \\
\hline
\end{array}\)
5. \(\begin{array}{r}
0.750 \\
10.425 \\
+2.000 \\
\hline 13.175 \\
\hline
\end{array}\)
6. \(\begin{array}{r}
280.69 \\
25.20 \\
+38.00 \\
\hline 343.89 \\
\hline
\end{array}\)
Question 2. Rashid spent ₹ 35.75 for the Maths book and ₹ 32.60 for the Science book. Find the total amount spent by Rashid.
Solution:
Money spent formats books =₹ 35.75
Money spent on Science book = ₹ 32.60
Total money spent = ₹ 35.75 + ? 32.60 = ₹ 68.35
Therefore, the total money spent by Rashid is ₹ 68.35
Question 3. Radhika’s mother gave her ₹ 10.50 and her father gave her ₹ 1 5.80, find the total amount given to Radhika by the parents.
Solution:
Money given by mother = ₹ 10.50
Money given by father ₹ 15.80
Total money received by Radhika = ₹ 10.50 + ₹ 15.80 = ₹ 26.30
Therefore, the total money received by Radhika is ₹ 26.30
Question 4. Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of clothes bought by her.
Solution:
Cloth bought for shirt = 3 m 20 cm = 3.20 m
Cloth bought for trouser = 2 m 5 cm = 2.05 m
Total length of cloth bought by Nasreen = 3.20 m + 2.05 m = 5.25 m
Therefore, the total length of cloth bought by Nasreen is 5.25 m
Question 5. Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Solution:
Distance travelled in the morning = 2 km 35 m = 2.035 km
Distance travelled in the evening =1 km 7m = 1.007 km
Total distance travelled = (2.035 + 1.007) km- 3.042 km
Therefore, the total distance travelled by Naresh is 3.042 km.
Question 6. Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot to reach her school. How far is her school from her residence?
Solution:
Distance travelled by bus = 15 km 268 m = 15.268 km
Distance travelled by car = 7 km 7m = 7.007 km
Distance travelled on foot = 500 m = 0.500 km
Total distance travelled = (15.268 + 7.007 + 0.500) km = 22.775 km
Therefore, the total distance travelled by Sunita is 22.775 km
Question 7. Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850g flour. Find the total weight of his purchases.
Solution:
Weight of Rice = 5 kg 400 g = 5.400 kg
Weight of Sugar = 2 kg 20 g = 2.020 kg
Weight flour = 10 kg 850 g = 10.850 kg
Total weight = (5.400 + 2.020 + 10.850) kg = 18.270 kg
Therefore, the total weight of Ravi’s purchases is 18.270 kg
Decimals Exercise – 8.4
Question 1. Subtract:
- ₹ 18.25 from ₹ 20.75
- 202.54 m from 250 m
- ₹ 5.36 from ₹ 8.40
- 2.051 km from 5.206 km
- 0.314 kg from 2.107 kg
Solution:
1. \(\begin{array}{r}
20.75 \\
-18.25 \\
\hline 02.50 \\
\hline
\end{array}\)
₹ 20.75- ₹ 18.25 2.50
2. \(\begin{array}{r}
250.00 \\
-202.54 \\
\hline 47.46 \\
\hline
\end{array}\)
250 m- 202.54 m = 47.46 m
3. \(\begin{array}{r}
8.40 \\
-5.36 \\
\hline 3.04 \\
\hline
\end{array}\)
₹ 8.40 – ₹ 5.36 = ? 3.04
4. \(\begin{array}{r}
5.206 \\
-2.051 \\
\hline 3.155 \\
\hline
\end{array}\)
5.206 km- 2.051 km = 3.155 km
5. \(\begin{array}{r}
2.107 \\
-0.314 \\
\hline 1.793
\end{array}\)
2.107 kg – 0.314 kg = 1.793 kg
Question 2. Find the value of:
- 9.756-6.28
- 21.05-15.27
- 18.5-6.79
- 11.6-9.847
Solution:
1. \(\begin{array}{r}
9.756 \\
-6.28 \\
\hline 3.476 \\
\hline
\end{array}\)
2. \(\begin{array}{r}
21.05 \\
-15.27 \\
\hline 05.78 \\
\hline
\end{array}\)
3. \(\begin{array}{r}
18.50 \\
-6.79 \\
\hline 11.71 \\
\hline
\end{array}\)
4. \(\begin{array}{r}
11.600 \\
-9.847 \\
\hline 1.753 \\
\hline
\end{array}\)
Question 3. Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Solution:
The total amount is given to shopkeeper=₹ 50
Cost of book = ₹ 35.65
Amount left = ₹ 50.00- ₹ 35.65 = ₹ 14.35
Therefore, Raju got back ₹ 14.35 from the shopkeeper.
Question 4. Rani had ₹ 18.50. She bought one ice cream for ₹ 1 1.75. How much money does she have now?
Solution:
Total money = ₹ 18.50
Cost of ice-cream = ₹ 11.75
Amount left = ₹ 18.50- ₹ 11.75 = ₹ 6.75
Therefore, Rani has ₹ 6.75 now
Question 5. Tina had 20 m 5 cm long cloth. She cuts 4m 50 cm length of cloth from this for making a curtain. How many clothes are left with her?
Solution:
Total length of cloth = 20 m 5 cm = 20.05 m
Length of cloth used = 4 m 50 cm = 4.50 m
Remaining cloth = 20.05 m- 4.50 m = 15.55 m
Therefore, 15.55 m of cloth is left with Tina
Question 6. Namita travels 20 km 50 m every day. Out of this, she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?
Solution:
Total Distance travelled = 20 km 50 m
= 20.050 km
Distance travelled by auto = (20.050- 10.200) km = 9.850 km
Therefore, 9.850 km distance is travelled by auto.
Question 7. Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Solution:
Weight of onions=3kg 500 g = 3.500kg
Weight of tomatoes = 2 kg 75 g = 2.075 kg
Total weight of onions and tomatoes = (3.500 + 2.075) kg = 5.575 kg
Therefore, weight of potatoes = (10.000- 5.575) kg = 4.425 kg
Thus, the weight of potatoes is 4.425 kg.