CBSE Class 6

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals InText Questions

NCERT In-text Question Page No. 165

Question 1.
Can you now write the following as decimals?

Answer:
We have,
(i) 5 hundreds + 3 tens + 8 ones + 1 tenth
= 5 x 100 + 3 x 10 + 8 x 1 + 1 x \(\frac { 1 }{ 10 }\)
= 500 x 30 + 8 + \(\frac { 1 }{ 10 }\)
= 538 x \(\frac { 1 }{ 10 }\) = 538.1

(ii) 2 hundreds + 7 tens + 3 ones + 4 tenths
= 2 x 100 + 7 x 10 + 3 x 1 + 4 x \(\frac { 1 }{ 10 }\)
= 200 x 70 + 3 + \(\frac { 4 }{ 10 }\)
= 273 x \(\frac { 4 }{ 10 }\) = 273.4

(iii) 3 hundreds + 3 tens + 4 ones + 1 tenth
= 3 x 100 + 5 x 10 + 4 x 1 + 6 x \(\frac { 1 }{ 10 }\)
= 300 x 50 + 4 + \(\frac { 6 }{ 10 }\)
= 354 x \(\frac { 6 }{ 10 }\) = 354.6

Question 2.
Write the lengths of Ravi’s and Raju’s pencils in ‘cm’ using decimals.
Answer:
10 mm = 1 cm
Therefore, 1 mm = \(\frac { 1 }{ 10 }\) cm
So, 5 mm = 5 x \(\frac { 1 }{ 10 }\) cm = \(\frac { 5 }{ 10 }\) cm
and 3 mm = 3 x \(\frac { 1 }{ 10 }\) cm = \(\frac { 3 }{ 10 }\) cm
Now, 7 cm 5 mm = 7 cm + 5 mm
= 7 cm + \(\frac { 5 }{ 10 }\) cm
= 7.5 cm
Again
8 cm 3 mm = 8 cm + 3 mm
= 8 cm + \(\frac { 3 }{ 10 }\)cm
= 8.3 cm

Question 3.
Make three more examples similar to the one given in question 1 and solve them.
Answer:
Please try yourself.

NCERT In-text Question Page No. 167

Question 1.
Write \(\frac{3}{2}, \frac{4}{5}, \frac{8}{5}\) in decimal notation.
Answer:
(i) \(\frac{3}{2}\)
We have, \(\frac{3}{2}=\frac{3 \times 5}{2 \times 5}=\frac{15}{10}=\frac{15}{10}\) = 1.5
∴ \(\frac{3}{2}\) = 1.5

(ii) \(\frac{4}{5}\)
We have, \(\frac{4}{5}=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}\) = 0.8
= \(\frac{8}{10}\) = 0.8
Thus, \(\frac{4}{5}\) = 0.8

(iii) \(\frac{8}{5}\)
We have, \(\frac{8}{5}=\frac{8 \times 5}{5 \times 5}=\frac{16}{10}\) = 1.6
Thus, \(\frac{8}{5}\) = 1.6

NCERT In-text Question Page No. 175

Question 1.
(i) Write 2 rupees 5 paise and 2 rupees 50 paise in decimals.
(ii) Write 20 rupees 7 paise and 21 rupees 75 paise in decimals.
Answer:
(i) (a) 2 rupees 5 paise :
2 rupees + 5 paise = 2 rupees + \(\frac{5}{10}\) rupees
∵ 100 paise = ₹1
∴ 1 paise = ₹ \(\frac{1}{100}\)
= (2 + 0.05) rupees = ₹2.05

(b) 2 rupees 50 paise =
2 rupees + 50 paise = 2 rupee + \(\frac{50}{100}\) rupees
∵ 100 paise = ₹1
∴ 1 paise = ₹ \(\frac{1}{100}\)
= (2 + 0.50) rupees = ₹ 2.50

(ii) (a) 20 rupees 7 paise :
20 rupees +7 paise
= 20 rupees + 7 x\(\frac{1}{100}\) rupees
[ ∴ \(\frac{1}{100}\) rupee = 1 paise]
= 20 rupees + 0.07 rupee
= ₹ (20 + 0.07)
= ₹ 20.07

(b) 21 rupees 75 paise :
21 rupees + 75 paise
= 21 rupees + 75 x -jy rupees
= 21 rupees + -jy rupees
= ₹ [21 + 0.75] = ₹ 21.75

NCERT In-text Question Page No. 176

Question 1.
Can you write 4 mm in ’em’ using decimals?
Answer:
Yes,
Since 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm
or 4 mm = 4 x \(\frac { 1 }{ 10 }\) cm = 0.4 cm

Question 2.
How will you write 7 cm 5 mm in ’em’ using decimals.
Answer:
7 cm 5 mm:
Since, 10 mm = 1 cm
∴ 1 mm= \(\frac { 1 }{ 10 }\) cm
Now, 7 cm 5 mm = 7 cm + 5 mm
= 7 cm + 5 x \(\frac { 1 }{ 10 }\) cm =[ 7 + \(\frac { 5 }{ 10 }\) cm
= (7 + 0.5) cm = 7.5 cm

Question 3.
How will you write 2008 m in ‘km’?
Answer:
Yes, we can change the given ‘metres’ into kilometres.
(a) 52 m
∵ 1000m = 1km ∴ 1m = \(\frac{1}{1000}\)
or 52 m = 52 x \(\frac{1}{1000}\) km = 0.052 km

(b) 340m
∵ 1m = \(\frac{1}{1000}\) km
∴ 340m = 340 x \(\frac{1}{1000}\) km
= \(\frac{340}{1000}\) km = 3.40 km

(c) 2008m
∵ 1 m = \(\frac{1}{1000}\) km
∴ 2008m = 2008 x \(\frac{1}{1000}\) km = \(\begin{aligned}
&2008 \\
&\hline 1000
\end{aligned}\) km

= (2 + 0.008) km = 2.008 km

Question 4.
Can you now write 52 m as ‘km’ using decimals? How will you write 340 m as ‘km’ using decimals?
Answer:
1000g = 1kg
∴ 1 g = \(\frac{1}{1000}\)kg
∴ 456g = \(\frac{1}{1000}\) x 456kg
= \(\frac{456}{1000}\) kg = 0.456kg

Question 5.
How will you write 2 kg 9 g in kg’ using decimals?
Answer:
2kg9g = 2kg + 9g
= 2kg + (9 x
= (2 + \(\frac{9}{1000}\))
= (2 + 0.009) kg = 2.009 kg

NCERT In-text Question Page No. 178

Question 1.
Find
(i) 0.29 + 0.36
(ii) 0.7 + 0.08
(iii) 1.54 + 1.80
(iv) 2.66 + 1.85
Answer:
(i) 0.29 + 0.36

∵ (9 + 6) hundredths =15 hundredths = 1
tenths + 5 hundredths
Thus, 0.29 + 0.36 = 0.65

(ii) 0.7 + 0.08

Thus, 0.7 + 0.08 = 0.78

(iii) 1.54 + 1.80

Thus, 1.54 + 1.80 = 3.34
Note: 5 tenths + 8 tenths =13 tenths
and 13 tenths = 10 tenths + 3 tenths = 1 one + 3 tenths

(iv) 2.66 + 1.85

Thus, 2.66 + 1.85 = 4.51

NCERT In-text Question Page No. 180

Question 1.
Subtract 1.85 from 5.46
Answer:
1.85 and 5.46 are ‘like decimals’.

Here, 1 is borrowed from ‘ones’ and given to tenths such that:
4 tenths + 10 tenths =14 tenths
5 ones – 1 one = 4 ones

Question 2.
Subtract 5.25 from 8.28
Answer:
5.25 and 8.28 are ‘like decimals’.

Question 3.
Subtract 0.95 from 2.29
Answer:
0.95 and 2.29 are ‘like decimals’.

Question 4.
Subtract 2.25 from 5.68
Answer:
2.25 and 5.68 are ‘like decimals’.

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.6

Question 1.
Subtract:
(a) ₹ 18.25 from ₹ 20.75
(b) 202.54 m from 250
(c) ₹ 5.36 from ₹ 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg
Answer:
(a) 20.75 – 18.25 = ₹ 2.50
(b) 250 – 202.54 = 47.46 m
(c) 8.40 – 5.36 = ₹ 3.04
(d) 5.206 – 2.051 = 3.155 km
(e) 2.107 – 0.314 = 1.793

Question 2.
Find the value of:
Answer:
(a) 9.756 – 6.28 = 3.476
(b) 21.05 – 15.27 = 5.78
(c) 18.5 – 6.79 = 11.71
(d) 11.6 – 9.847 = 1.753

Question 3.
Raju bought a book of ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Answer:
Total amount given to shopkeeper = ₹ 50
Cost of book = ₹ 35.65
Amount left = ₹ 50.00 – ₹ 35.65
= Rs 14.35
Therefore, Raju got back ₹ 14.35 from the shopkeeper.

Question 4.
Rani had ₹ 18.50. She bought one ice¬cream for ₹ 11.75. How much money does she have now?
Answer:
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 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?
Answer:
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?
Answer:
Total distance travel = 20 km 50 m = 20.050 km
Distance travelled by bus
= 10 km 200 m = 10.200 km
Distance travelled by auto
= 20.050 – 10.200 = 9.850 km
Therefore, Namita travels 9.850 km by auto.

Question 7.
Aakash bought vegetables weighing 10 kg. Out of this 3 kg 500 g in onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Answer:
Weight of onions = 3 kg 500 g = 3.500 kg
Weight of tomatoes = 2 kg 75 g = 2.075 kg
Total weight of onions and tomatoes = 3.500 + 2.075 = 5.575 kg
Therefore, weight of potatoes = 10.000 – 5.575 = 4.425 kg
Thus, the weight of potatoes is 4.425 kg.

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5

Question 1.
Find the sum in each of the following:
(a) 0.007 + 8.5 + 30.08
(b) 15 + 0.632+ 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38
Answer:

Question 2.
Rashid spent ₹ 35.75 for Maths book and ₹ 32.60 for Science book. Find the total amount spent by Rashid.
Answer:
Money spent for Maths book = ₹ 35.75 Money spent for Science book = ₹ 32.60
Total money spent = ₹ 35.75 + ₹ 32.60
.’. Total amount of money spent by Rashid is = ₹ 68.35
Therefore, total money spent by Rashid is ₹ 68.35.

Question 3.
Radhika’s mother have her ₹ 10.50 and her father gave her ₹ 15.80. Find the total amount given to Radhika by the parents.
Ans. 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 cloth bought by her.
Answer:
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 + 2.05 = 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?
Answer:
Distance travelled in morning = 2 km 35 m = 2.035 km
Distance travelled in evening = 1 km 7 m = 1.007 km
Total distance travelled = 2.035 + 1.007 = 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 by foot in order to reach her school. How far is her school from her residence?
Answer:
Distance travelled by bus =15 km 268 m = 15.268 km
Distance travelled by car = 7 km 7 m = 7.007 km
Distance travelled on foot = 500 m = 0.500 km
Total distance travelled = 15.268 + 7.007 + 0.500 = 22.775 km Therefore, total distance travelled by Sunita is 22.775 km.

Question 7.
Ravi purchases 5 kg 400g rice, 2 kg 20g sugar and 10 kg 850g flour. Find the total weight of his purchases.
Answer:
Weight of Rice = 5 kg 400 g = 5.400 kg
Weight of Sugar = 2 kg 20 g = 2.020 kg
Weight of Flour = 10 kg 850 g = 10.850 kg
Total weight = 5.400 + 2.020 + 10.850
= 18.270 kg
Therefore, the total weight of Ravi’s purchase = 18.270 kg

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4

Question 1.
Express as rupees using decimals:
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90 paise
(e) 725 paise
Answer:
(a) ∵ 1 paisa = ₹ \(\frac { 1 }{ 100 }\)
∴ 5 paise = \(\frac { 1 }{ 100 }\) × 5 = ₹ 0.05

(b) ∵ 1 paisa = \(\frac { 1 }{ 100 }\)
∴ 75 paise = \(\frac { 1 }{ 100 }\) × 75 = ₹ 0.75

(c) ∵ 1 paisa = X \(\frac { 1 }{ 100 }\)
∴ 20 paise = \(\frac { 1 }{ 100 }\) × 20 = ₹ 0.2

(d) ∵ 1 paisa = X \(\frac { 1 }{ 100 }\)
∴ ₹ 50 + 90 paise
= 50 + \(\frac { 1 }{ 100 }\) × 90 = ₹ 50.90

(e) ∵ 1 paisa = ₹ \(\frac { 1 }{ 100 }\)
∴ 725 paise = \(\frac { 1 }{ 100 }\) × 725
= ₹ 7.25

Question 2.
Express as metres using decimals:
(a) 15 cm
(b) 6 cm
(c) 2 m 45 cm
(d) 9 m 7 cm
(e) 419 cm
Answer:
(a) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 15 cm = \(\frac { 1 }{ 100 }\) × 15 = 0.15 m 100
(b) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 6 cm = \(\frac { 1 }{ 100 }\) × 6 = 0.06 m

(c) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 2 m 45 cm = 2 + \(\frac { 1 }{ 100 }\) × 45
= 2.45 m

(d) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 9 m 7 cm = 9 x \(\frac { 1 }{ 100 }\) × 9 = 9.07 m

(e) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 419 = \(\frac { 1 }{ 100 }\) × 419 = 4.19 m

Question 3.
Express as cm using decimals:
(a) 5 mm
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93 mm
Answer:
(a) ∵ 1mm = \(\frac { 1 }{ 10 }\) cm
5mm = \(\frac { 1 }{ 10 }\) × 5 = 0.5 cm

(b) ∵ 1 mm = \(\frac { 1 }{ 10 }\) cm 10
∴ 60 mm = \(\frac { 1 }{ 10 }\) × 60 = 6 cm

(c) ∵ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 164 mm = \(\frac { 1 }{ 10 }\) × 164 = 16.4 cm

(d) ∵ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 9 cm 8 mm = 9 + \(\frac { 1 }{ 10 }\) × 8
= 9 + 0.8 cm = 9.8 cm

(e) ∵ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 93 mm = \(\frac { 1 }{ 10 }\) × 93 cm

Question 4.
Express as km using decimals:
(a) 8m
(b) 88 m
(c) 8888 m
(d) 70 km 5 m
Answer:
(a) ∵ 1 m = \(\frac { 1 }{ 1000 }\) km
∴ 8m = \(\frac { 1 }{ 1000 }\) × 8 = 0.008 km

(b) ∵ 1 m = \(\frac { 1 }{ 1000 }\) km
∴ 88 m: 1 \(\frac { 1 }{ 1000 }\) × 88 = 0.088 km

(c) ∵ 1 m = \(\frac { 1 }{ 1000 }\) km
∴ 8888m = \(\frac { 1 }{ 1000 }\) × 88 = 8.888km

(d) ∵ 1m = \(\frac{1}{100}\)km
∴ 70km 5m = 70 + \(\frac{1}{100}\) × 5
= 70.005 km

Question 5.
Express as kg using decimals:
(a) 2g
(b) 100g
(c) 3750 g
(d) 5 kg 8 g
(e) 26kg 50g
Answer:
(a) 1 g = \(\frac{1}{1000}\) kg
∴ 2g = \(\frac{1}{1000}\) × 2 = 0.002kg

(b) ∵ 1g = \(\frac{1}{1000}\) kg
∴ 100g= \(\frac{1}{1000}\) × 100 = 0.1 kg

(c) ∵ 1g = \(\frac{1}{1000}\) kg
∴ 3750g = \(\frac{1}{1000}\) × 3750 = 3.750kg

(d) ∵ 1g = \(\frac{1}{1000}\) kg
∴ 5kg50g = 5 + \(\frac{1}{1000}\) × 8
= 5.008kg

(e) ∵ 1g = \(\frac{1}{1000}\) kg
∴ 26kg 50g = 26 + \(\frac{1}{1000}\) × 50
= 26.050kg

These NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Exercise 9.4

Question 1.
A survey of 120 school students was done to find which activity they prefer to do in their free time.

Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students.
Which activity is preferred by most of the students other than playing?
Answer:
Steps :
(i) Draw two perpendicular lines OX and OY.
(ii) Draw vertical bars (rectangles) of same width keeping equal spacing between them.
(iii) ∵ 1 unit length = 5 students,
∴ Heights of bars will be
For playing = 45 ÷ 5 = 9 units
For reading story books = 30 ÷ 5 = 6 units
For watching TV = 20 ÷ 5 = 4 units
For listening to music = 10 ÷ 5 = 2 units
For painting = 15 ÷ 5 = 3 units
The required graph is as under :

Reading story books is preferred by most of the students other than playing.

Question 2.
The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:

Draw a bar graph to represent the above information choosing the scale of your choice.
Answer:
Steps:
(i) Draw two perpendicular lines OX and OY.
(ii) Draw bars (rectangles) of equal width on OX having same spacing between them.
(iii) Taking an appropriate scale (here 1 unit length = 10 books) fix the heights of various bars, such as:
For Sunday = 65 ÷ 10 = 6.5 units
For Monday = 40 ÷ 10 = 4 units
For Tuesday = 30 ÷ 10 = 3 units
For Wednesday = 50 ÷ 10 = 5 units
For Thursday = 20 ÷ 10 = 2 units
For Friday = 70 ÷ 10 = 7 units
∴ The required graph is as under :

Question 3.
Following table shows the number of bicycles manufactured in a factory during the years 1998 to 2002. Illustrate this data using a bar graph. Choose a scale of your choice.

(a) In which year was the maximum number of bicycles manufactured?
(b) In which year was the minimum number of bicycles manufactured?
Answer:
Steps :
(i) Draw two perpendicular lines OX and OX
(ii) Draw bars of equal width on OX having same spacing between them.
(iii) For fixing heights of various bars, choose an appropriate scale (here: 1 unit length = 200 bicycles) so that, we have:
∴ For 1998: 800 ÷ 200 = 4 units
For 1999: 600 ÷ 200 = 3 units
For 2000: 900 ÷ 200 = 4.5 units
For 2001: 1100 ÷ 200 = 5.5 units
For 2002: 1200 ÷ 200 = 6 units

(a) The maximum number of bicycles manufactures in 2002.
(b) The minimum number of bicycles manufactured in 1999.

Question 4.
Number of persons in various age groups in a town is given in the following table.

Draw a bar graph to represent the above information and answer the following
questions. (take 1 unit length = 20 thousands)
(a) Which two age groups have same population?
(b) All persons in the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
Answer:
Steps :
(i) Draw two perpendicular lines OX and OY.
(ii) Draw bars of equal width on OX having same spacing between them.
(iii) Since the scale is given:
1 unit length = 20 thousand person
∴ The heights of various bars are given below such as :
For 1 – 14 : 200000 ÷ 20000 = 10 units
For 15 – 29 : 160000 ÷ 20000 = 8 units
For 30 – 44 : 120000 ÷ 20000 = 6 units
For 45 – 59 : 120000 ÷ 20000 = 6 units
For 60 – 74 : 80000 ÷ 20000 = 4 units
For 75 and above
: 40000 ÷ 20000 = 2 units
Now, the required bar graph is as follows

(a) Group 30 – 44 and group 45 – 59 have same population.
(b) 80,000 + 40,000 = 1,20,000 senior citizens are there in the town.

These NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Exercise 9.3

Question 1.
The bar graph given alongside shows the amount of wheat purchased by government during the year 1998 – 2002.

Read the bar graph and write down your observations. In which year was
(a) the wheat production maximum?
(b) the wheat production minimum?
Answer:
(a) In 2002, production of wheat was maximum.
(b) In 1998, production of wheat was minimum.

Question 2.
Observe this bar graph which is showing the sale of shirts in a ready made shop from Monday to Saturday.

Now answer the following questions:
(a) What information does the above bar graph give?
(b) What is the scale chosen on the horizontal line representing number of shirts?
(c) On which day was the maximum number of shirts sold? How many shirts were sold on that day?
(d) On which day was the minimum number shirts sold?
(e) How many shirts were sold on Thursday?
Answer:
(a) This bar graph shows the sale of shirt in a readymade shop from Monday to Saturday.
(b) 1 unit = 5 shirts
(c) On Saturday, maximum number of shirts, 60 shirts were sold.
(d) On Tuesday, minimum number of shirts were sold.
(e) On Tuesday, 35 shirts were sold.

Question 3.
Observe this bar graph which shows the marks obtained by Aziz in half-yearly examination in different subjects.
Answer the given questions.
(a) What information does the bar graph give?
(b) Name the subject in which Aziz scored maximum marks.
(c) Name the subject in which he has scored minimum marks.
(d) State the name of the subjects and marks obtained in each of them.

Answer:
(a) This bar graph shows the marks obtained by Aziz in half yearly examination in different subjects.
(b) Hindi.
(c) Social Studies.
(d) Hindi 80, English 60, Mathematics 70, Science 50, Social Studies 40.