CBSE Class 6

These NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Exercise 9.2

Question 1.
Total number of animals in five villages are as follows:
Village A : 80
Village B : 120
Village C : 90
Village D : 40
Village E: 60
Prepare a pictograph of these animals using one symbol to represent 10 animals and answer the following questions:
(a) How many symbols represent animals of village E?
(b) Which village has the maximum number of animals?
(c) Which village has more animals: village A or village C?
Answer:

(a) 6
(b) Village B
(c) Village C has more animals than Village A

Question 2.
Total number of students of a school in different years is shown in the following table

A. Prepare a pictograph of students using one symbol to represent 100 students and answer the following questions:
(a) How many symbols represent total number of students in the year 2002?
(b) How many symbols represent total number of students for the year 1998?

B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative?
Answer:

(a) 6
(b) Five completed and one incomplete.
B.

Pictograph B is more informative than A.

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3

Question 1.
Which is greater:
(a) 0.3 or 0.4
(b) 0.07 or 0.02
(c) 3.0 or 0.8
(d) 0.5 or 0.05
(e) 1.23 or 1.20
(f) 0.099 or 0.19
(g) 1.5 or 1.5
(h) 1.431 or 1.490
(i) 5.64 or 5.603
(j) 5.640 or 5.603
Before comparing, we write both terms in like decimals:
(a) 0.3 < 0.4
(b) 0.07 >0.02
(c) 3.0 or 0.8 ⇒ 3.0 > 0.8
(d) 0.5 or 0.05 ⇒ 0.5 > 0.05
(e) 1.23 or 1.20 ⇒ 1.23 > 1.20
(f) 0.099 or 0.19 ⇒ 0.099 < 0.19
(g) 1.5 or 1.5 ⇒ 1.5 = 1.5
(h) 1.431 < 1.490
(i) 3.300 or 3.300 ⇒ 3.300 = 3.300
(j) 5.640 or 5.603 ⇒ 5.64 > 5.603

Question 2.
Make five more examples and find the greater number from them.
Answer:
(a) 1.8 or 1.82
(b) 1.0009 or 1.09
(c) 10.01 or 100.1
(d) 5.100 or 5.0100
(e) 04.213 or 0421.3
Before comparing, we write both terms in like decimals
(a) 1.80 or 1.82 ⇒ 1.82 is greater than 1.8
(b) 1.0009 or 1.0900 ⇒ 1.09 is greater than 1.0009.
(c) 10.01 or 100.10 ⇒ 100.1 is greater than 10.01.
(d) 5.1000 or 5.0100 ⇒ 5.100 is greater than 5.0100.
(e) 04.213 or 0421.300 ⇒ 0421.3 is greater than 04.213.

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.2

Question 1.
Complete the table with the help of these boxes and use decimals to write the number.

Answer:
Explanation:
(a) In this figure,
If small block out of 100 are shaded
∴ Decimal representation = 0.26
(b) 100 small blocks + 38 small blocks are shaded
∴ \(\frac{100}{100}+\frac{38}{100}=1+\frac{38}{100}\) = 1.28
1.38 is represented by the shaded portion.
(c) 100 parts + 28 parts are shaded
1.28 is represented by the shaded portion. Therefore, the table is compiled as :

Question 2.
Write the numbers given in the following place value table in decimal form.

Answer:
(a) 0 x 100 + 0 x 10 + 3 x 1 + 2 x \(\frac { 1 }{ 10 }\) + 5 x 100 + 0 x \(\frac { 1 }{ 1000 }\)
= 0 + 0 + 3 +0.2 + 0.05 + 0 = 3.25

(b) 1 x 100 + 0 x 10 + 2 x 1 + 6 x \(\frac { 1 }{ 10 }\) + 3 x \(\frac { 1 }{ 100 }\) + 0 x \(\frac { 1 }{ 1000 }\)
= 1 + 0 + 2 + 0.6 + 0.03 + 0 = 102.63

(c) 0 x 100 + 3 x 10 + 0 x 1 + 0 x \(\frac { 1 }{ 10 }\) + 2 x \(\frac { 1 }{ 100 }\) + 5 x \(\frac { 1 }{ 1000 }\)
= 0 + 30 + 0 + 0 + 0.02 + 0.005
= 30.025

(d) 2 x 100 + 1 x 10 + 1 x 1 + 9 x \(\frac { 1 }{ 10 }\) + O x \(\frac { 1 }{ 100 }\) + 2 x \(\frac { 1 }{ 1000 }\)
= 200 + 10 + 1 + 0.9 + 0 + 0.002
= 211.902

(e) 0 x 100 + 1 x 10 + 2 x 1 + 2 x \(\frac { 1 }{ 10 }\) + 4 x \(\frac { 1 }{ 100 }\) + 1 x \(\frac { 1 }{ 1000 }\)
= 0 + 10 + 2 + 0.2 + 0.04 + 0.001
= 12.241

Question 3.
Write the following decimals in the place value table:
(a) 0.29
(b) 2.08
(c) 19.60
(d) 148.32
(e) 200.812
Answer:

Question 4.
Write each of the following as decimals:
(a) 20 + 9 + \(\frac{4}{10}+\frac{1}{100}\)
(b) 137 + \(\frac{5}{100}\)
(c) \(\frac{7}{10}+\frac{6}{100}+\frac{4}{1000}\)
(d) 23 + \(\frac{2}{10}+\frac{6}{1000}\)
(e) 700 + 20 + 5 + \(\frac{9}{100}\)
Answer:
(a) 20 + 9 + 0.4 + 0.01 = 29.41
(b) 137 + 0.05 = 137.05
(c) 0.7 + 0.06 + 0.004 = 0.764
(d) 23 + 0.2 + 0.006 = 23.206
(e) 700 + 20 + 5 + 0.09 = 725.09

Question 5.
Write each of the following decimals in words:
(a) 0.03
(b) 1.20
(c) 108.56
(d) 10.07
(e) 0.032
(f) 5.008
Answer:
(a) Zero point zero three
(b) One point two zero
(c) One hundred and eight point five six
(d) Ten point zero seven
(e) Zero point zero three two
(f) Five point zero zero eight

Question 6.
Between which two numbers in tenths place on the number line does each of the given number lie?
(a) 0.06
(b) 0.45
(c) 0.19
(d) 0.66
(e) 0.92
(f) 0.57
Answer:
All the numbers lie between 0 and 1
(a) 0.06 is nearer to 0.1.
(b) 0.45 is nearer to 0.5.
(c) 0.19 is nearer to 0.2.
(d) 0.66 is nearer to 0.7.
(e) 0.92 is nearer to 0.9.
(f) 0.57 is nearer to 0.6.

Question 7.
Write as fractions in lowest terms:
(a) 0.60
(b) 0.05
(c) 0.75
(d) 0.18
(e) 0.25
(f) o .125
(g) 0.066
Answer:
(a) 0.60 = \(\frac{60}{100}=\frac{3}{5}\)
(b) 0.05 = \(\frac{5}{100}=\frac{1}{20}\)
(c) 0.75 = \(\frac{75}{100}=\frac{3}{4}\)
(d) 0.18 = \(\frac{18}{100}=\frac{9}{50}\)
(e) 0.25 = \(\frac{25}{100}=\frac{1}{4}\)
(f) o .125 = \(\frac{125}{1000}=\frac{1}{8}\)
(g) 0.066 = \(\frac{66}{1000}=\frac{33}{500}\)

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

Question 1.
Write the following as numbers in the given table.

Answer:

Question 2.
Write the following decimals in the place value table.
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205.9
Answer:

Question 3.
Write each of the following as decimals:
(a) Seven-tenths
(b) Two tens and nine-tenths
(c) Fourteen point six
(d) One hundred and two ones
(e) Six hundred point eight
Answer:
(a) The decimal-form of seven-tenths is 7 = \(\frac{7}{10}\) = 0.7
(b) 2 tens and 9-tenths
= 2 x 10 + \(\frac{9}{10}\) = 20 + 0.9
= 20.9
(c) Fourteen point six = 14.6
(d) One hundred and 2-ones
= 100 + 2 x 1 = 100 + 2 = 102
(e) Six hundred point eight = 600.8

Question 4.
Write each of the following as decimals:
Answer:
(a) \(\frac{5}{10}\) = 0.5
(b) 3 + \(\frac{7}{10}\) = 3 + 0.7 = 3.7
(c) 200 + 60 + 5 + \(\frac{1}{10}\)
= 200 + 60 + 5 + 0.1 = 265.1
(d) 70 + \(\frac{8}{10}\) = 70 + 0.8 = 70.8
(e) \(\frac{88}{10}=\frac{80+8}{10}=\frac{80}{10}+\frac{8}{10}\)
= 8 + \(\frac{8}{10}\) = 8 + 0.8 = 8.8
(f) 4\(\frac{2}{10}\) = 4 + \(\frac{2}{10}\) = 4 + 0.2 = 4.2
(g) \(\frac{3}{2}=\frac{3 \times 5}{2 \times 5}=\frac{15}{10}=\frac{10+5}{10}\)
= \(\frac{10}{10}+\frac{5}{10}\) = 1 + 0.5 = 1.5
(h) \(\frac{2}{5}=\frac{2 \times 5}{5 \times 2}=\frac{4}{10}\) = 0.4
(i) \(\frac{12}{5}=\frac{12 \times 2}{5 \times 2}=\frac{24}{10}=\frac{20+4}{10}\)
= \(\frac{20}{10}+\frac{4}{10}\) = 2 + 0.4 = 2.4
(j) \(3 \frac{3}{5}=3+\frac{3}{5}=3+\frac{3 \times 2}{5 \times 2}\)
= 3 + \(\frac{6}{10}\) = 3 + 0.6 = 3.6
(k) \(4 \frac{1}{2}=4+\frac{1}{2}=4+\frac{1 \times 5}{2 \times 5}\)
= 4 + \(\frac{5}{10}\) = 4 + 0.5 = 4.5

Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form.
Answer:
(a) 0.6 = \(\frac{6}{10}=\frac{3}{5}\)
(b) 2.5 = \(\frac{25}{10}=\frac{5}{2}\)
(c) 1.0 = \(\frac{10}{10}\) = 1
(d) 3.8 = \(\frac{38}{10}=\frac{19}{5}\)
(e) 13.7 = \(\frac{137}{10}\)
(f) 21.2 = \(\frac{212}{10}=\frac{106}{5}\)
(g) 6.4 = \(\frac{64}{10}=\frac{32}{5}\)

Question 6.
Express the following as cm using decimals.
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4 cm 2 mm
(e) 162 mm
(f) 83 mm
Answer:
(a) ∵ 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 2 mm = \(\frac { 1 }{ 10 }\) x 2 = 0.2 cm

(b) ∵10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 30 mm = \(\frac { 1 }{ 10 }\) x 20 = 3.0 cm

(c) ∵ 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm 10
∴ 116 mm = \(\frac { 1 }{ 10 }\) x 116 = 11.6 cm

(d) ∵ 10 mm = 1 cm
4 cm 2 mm = 4 + \(\frac { 2 }{ 10 }\) cm 10
= 4 + 0.2 = 4.2 cm

(e) ∵ 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm 10
∴ 162 mm = \(\frac { 1 }{ 10 }\) x 162 = 16.2 cm 10

(f) ∵ 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm 10
∴ 83 mm =\(\frac { 1 }{ 10 }\) x 83 = 8.3 cm 10

Question 7.
Between which two whole numbers on the number line are the given numbers lie? Which of these whole numbers is nearer the number?

(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.1
(f) 4.9
Answer:
(a) 0.8 lies between 0 and 1, 0.8 is nearer to 1
(b) 5.1 lies between to 6,5.1 is nearer to 5.
(c) From 2 to 3, 2.6 is nearest to 3.
(d) 2.6 lies between 6 to 7, 6.4 is nearer to 6.
(e) 9.1 lies between 9 to 10,9.1 is nearer to 9.
(f) 4.9 lies between 4 to 5, 4.9 is nearer to 5.

Question 8.
Show the following numbers on the number line:
(a) 0.2
(b) 1.9
(c) 1.1
(d) 2.5
Answer:

Question 9.
Write the decimal number represented by the points A, B, C, D on the given number line.

Answer:
A = 0 + \(\frac{8}{10}\) = 0.8
B = 1 + \(\frac{3}{10}\) = 1.3
C = 2 + \(\frac{2}{10}\) = 2.2
D = 2 + \(\frac{9}{10}\) = 2.9

Question 10.
(a) The length of Ramesh’s notebook is 9 cm and 5 mm. What will be its length in cm?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
Answer:
(a) 9 cm 5 mm = 9 cm + 5 mm
= 9 + \(\frac{5}{10}\) = 9.5 cm
(b) 65 mm = \(\frac{65}{10}\) cm = 6.5 cm

These NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Exercise 9.1

Question 1.
In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

(a) Find how many students obtained marks equal to or more than 7.
(b) How many students obtained marks below 4?
Answer:
Marks Tally Marks No. of students


(a) Twelve students
(b) Eight students

Question 2.
Following is the choice of sweets of 30 students of Class VI.
Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgul- la, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Laddo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.
(a) Arrange the names of sweets in a ta¬ble using tally marks.
(b) Which sweet is preferred by most of the students?
Answer:
(a) Table using tally marks:

(b) Ladoo, because 11 students prefer to eat.

Question 3.
Catherine threw a dice 40 times and noted the number appearing each time as shown below:

Make a table and enter the data using tally marks. Find the number that appeared.
(a) The minimum number of times
(b) The maximum number of times
(c) Find those numbers that appear an equal number of times.
Answer:

(a) The minimum number of times = 4
(b) The maximum number of times = 5
(c) 1 and 6 are the numbers that

Question 4.
Following pictograph shows the number of tractors in five villages.

Observe the pictograph and answer the following questions:
(i) Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) How many more tractors village C has as compared to village B.
(iv) What is the total number of tractors in all the five villages?
Answer:
(i) Village D
(ii) Village C
(iii) 3 tractor
(iv) 28 tractor

Question 5.
The number of girl students in each class of a co-educational middle school is depicted by the pictograph:

Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in Class VI less than the number of girls in Class V?
(c) How many girls are there in Class VII?
Answer:
(a) Class VIII
(b) No
(c) 3 × 4 = 12 girls

Question 6.
The sale of electric bulbs on different days of a week is shown below:


Observe the pictograph and answer the following questions:
(a) How many bulbs were sold on Friday?
(b) On which day were the maximum number of bulbs sold?
(c) On which of the days same number of bulbs were sold?
(d) On which of the days minimum number of bulbs were sold?
(e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?
Answer:
(a) Number of bulbs sold on Friday are 14.
(b) Maximum number of bulbs (18) were sold on Sunday.
(c) Same number of bulbs (8) were sold on Wednesday and Saturday.
(d) Then minimum number of bulbs (8) were sold on Wednesday and Saturday.
(e) The total number of bulbs sold in the given week were 86.
So the number of carton required to hold 86 bulb = \(\frac{86}{9}\) ≈ 10. Hence, 10 cartons required to hold the bulbs.

Question 7.
In a village, six fruit merchants sold the following number of fruit baskets in a particular season:


Observe this pictograph and answer the following question:
(a) Which merchant sold the maximum number of baskets?
(b) How many fruit baskets were sold by Anwar?
(c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them?
Answer:
(a) Martin
(b) 7 × 100 = 700 fruit basket
(c) Anwar, Martin, Ranjit Singh

These NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.5

Question 1.
Write these fractions appropriately as additions or subtractions:

Answer:

Question 2.
Solve:
Answer:
If the denominator is same, we can add and subtract the numerator

(g) Since 1 and \(\frac { 3 }{ 3 }\) are equivalent fractions.
(h) \(\frac{1}{4}+\frac{0}{4}=\frac{1+0}{4}=\frac{1}{4}\)
∴ \(1-\frac{2}{3}=\frac{2}{3}-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}\)

(i) Since 3 = \(\frac { 3 }{ 1 }\), and equivalent fraction of \(\frac { 3 }{ 1 }\) having denominator as 5 is given by

Question 3.
Shubham painted \(\frac { 2 }{ 3 }\) of the wall space in his room. His sister Madhavi helped and painted \(\frac { 1 }{ 3 }\) of the wall space. How much did they paint together?
Answre:
Space painted by them together = Space painted by Shubham + Space painted by Madhavi
= \(\frac{2}{3}+\frac{1}{3}=\frac{3}{3}=1\)
∴ Shubham and Madhavi together painted one complete wall in a room.

Question 4.
Fill in the missing fractions.
Answer:

Question 5.
Javed was given \(\frac { 5 }{ 7 }\) of a basket of oranges. What fraction of oranges was left in the basket?
Answer:
Fraction of oranges given to Javed = \(\frac { 5 }{ 7 }\)
Fraction of oranges left in the basket = \(1-\frac{5}{7}=\frac{2}{7}\)