Chapter-8 Decimals | Class 6th | NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 6 Maths: One of the most important and interesting things to keep in mind is that these solutions available are totally free of cost. This also includes all the solutions to the exercises that are given in the textbook. NCERT solutions for class 6 maths gives you chapter-wise solutions to each and every question. This can help you solve even the tougher with ease.

It is always good to have a strong foundation in order to build a good building. This also is true in class when you are still learning the basics of maths. NCERT solutions for class 6 maths provides you the exact opportunity to build a strong foundation in this subject. Below is an overview of each and every chapter covered in the NCERT textbook.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

Ex 8.1 Class 6 Maths Question 1.
Write the following as numbers in the given table:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 1
Solution:
On filling the given table, we have
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 2

Ex 8.1 Class 6 Maths Question 2.
Write the following decimals in the place value table:
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205
Solution:
Let us make a common place value table, assigning appropriate place value to the digits in the given numbers. We have,
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 3

Ex 8.1 Class 6 Maths 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 htihdred point eight
Solution:
(a) Seven-tenths = 710 = 0.7
(b) Two tens, 9 – tenths = 20 + 910 = 20.9
(c) Fourteen point six = 14.6
(d) One Hundred and 2-ones = 1 x 100 + 0 x 10 + 2 x 1 + 0 x 110
= 100 + 0 + 2 = 102.0
(e) Six hundred point eight = 600.8

Ex 8.1 Class 6 Maths Question 4.
Write each of the following as decimals:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 4
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 5
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 6

Ex 8.1 Class 6 Maths Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form:
(a) 0.6
(b) 2.5
(c) 1.0
(d) 3.8
(e) 13.7
(f) 21.2
(g) 6.4
Solution:
tiwari academy class 6 maths Chapter 8 Decimals 7
tiwari academy class 6 maths Chapter 8 Decimals 8

Ex 8.1 Class 6 Maths Question 6.
Express the following as cm vising decimals:
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4 cm 2mm
(e) 162 mm
(f) 83 mm
Solution:
tiwari academy class 6 maths Chapter 8 Decimals 9
tiwari academy class 6 maths Chapter 8 Decimals 10

Ex 8.1 Class 6 Maths Question 7.
Between which two whole numbers on the number line are the given numbers lie?
Which of these whole numberes is nearer the number?
tiwari academy class 6 maths Chapter 8 Decimals 11
(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.1
(f) 4.9
Solution:
tiwari academy class 6 maths Chapter 8 Decimals 12

Ex 8.1 Class 6 Maths Question 8.
Show the following numbers on the number line.
(a) 0.2
(b) 1.9
(c) 1.1
(d) 2.5
Solution:
(a) Since 0.2 > 0 but <1. There are 2 tenth in it. Divide the unit length between 0 and 1 into 10 equal parts and take 2 parts as shown below. Thus, A represents 0.2.
tiwari academy class 6 maths Chapter 8 Decimals 13
(b)
 Since 1 < 1.9 < 2. Divide the unit length between 1 and 2 into 10 equal parts and take 9 parts as shown below. Thus, A represents 1.9.
tiwari academy class 6 maths Chapter 8 Decimals 14
(c)
 Since 1 < 1.1 < 2. Divide the unit length between 1 and 2 into 10 equal parts and take 1 part as shown below. Thus, A represents 1.1.
tiwari academy class 6 maths Chapter 8 Decimals 15
(d)
 Since 2 < 2.5 < 3. Divide the unit length between 2 and 3 into 10 equal parts and take 5 parts as shown below. Thus, A represents 2.5.
tiwari academy class 6 maths Chapter 8 Decimals 16

Ex 8.1 Class 6 Maths Question 9.
Write the decimal number represented by the points A, B, C, D on the given number line
tiwari academy class 6 maths Chapter 8 Decimals 17
Solution:
Since A lies between 0 and 1 and the unit length between 0 and 1 stands divided in 10 equal parts. Also A is at the 8th point.
Thus, A represents 0.8.
Since, B lies between 1 and 2 and the unit length between 1 and 2 stands divided in 10 equal parts. Also, B lies 3 points ahead of 1. Therefore, B represents 1.3.
Since C and D lies between 2 and 3 and the unit length between 2 and 3 stands divided between 2 and 3 in 10 equal parts. Also C and D respectively lies 2 and 9 points ahead of 2.
∴ C represents 2.2 and D represents 2.9.

Ex 8.1 Class 6 Maths Question 10.
(a) The length of Ramesh’s notebook is 9 cm 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.
Solution:
(a) Length of Ramesh’s notebook
tiwari academy class 6 maths Chapter 8 Decimals 18
(b) Length of gram plant
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 19

Ex 8.2 Class 6 Maths Question 1.
Complete the table with the help of these boxes and use decimals to write the number
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 20
Solution:
Completing the table, we have
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 21

Ex 8.2 Class 6 Maths Question 2.
Write the numbers given in the following place value table in decimal form:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 22
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 23
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 24

Ex 8.2 Class 6 Maths 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
Solution:
Let us make a common place value table, assigning appropriate place value to the digits in the given numbers. We have,
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 25

Ex 8.2 Class 6 Maths Question 4.
Write each of the following as decimals:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 26
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 27

Ex 8.2 Class 6 Maths 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
Solution:
(a) 0.03 = Zero point zero three or three hundredths.
(b) 1.20 = One point two zero.
(c) 108.56 = One hundred eight point five six.
(d) 10.7 = Ten point zero seven.
(e) 0. 032 = Zero point zero three two.
(f) 5.008 = Five point zero zero eight.

Ex 8.2 Class 6 Maths 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
Solution:
All these points lies between 0 and 1.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 28

Ex 8.2 Class 6 Maths Question 7.
Write as fractions in lowest terms:
(a) 0.60
(b) 0.05
(c) 0.75
(d) 0.25
(f) 0.125
(g) 0.066
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 29
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 30

 

Ex 8.3 Class 6 Maths Question 1.
Which is greater?
(a) 0.3 or 0.4
(b) 0.07 or 0.02
(c) 3 or 0.8
(d) 0.5 or 0.05
(e) 1.23 or 1.2
(f) 0.099 or 0.19
(g) 1.5 or 1.50
(h) 1.431 or 1.490
(i) 3.3 or 3.300
(j) 5.64 or 5.603
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 31
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 32
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 33
In this case, the two numbers have the same parts upto tenths. The hundredth part of 5.64 is greater than that of 5.603.
∴ 5.64 > 5.603

Ex 8.3 Class 6 Maths Question 2.
Make five more examples and find the greater number from them.
(i) 0.5 and 0.8
(ii) 2.0 or 0.9
(iii) 0.042 or 0.22
(iv) 3.012 or 2.99
(v) 0.055 or 0.15
Solution:
Their solutions are:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 34
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 35

Ex 8.4 Class 6 Maths Question 1.
Express as rupees using decimals:
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90 paise
(e) 725 paise
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 36

Ex 8.4 Class 6 Maths Question 2.
Express as metres using decimals:
(a) 15 cm
(b) 6 cm
(c) 2 m 45 m
(d) 9 m 7 cm
(e) 419 cm
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 37

Ex 8.4 Class 6 Maths Question 3.
Express as cm using decimals:
(a) 5 nun
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93 mm
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 38

Ex 8.4 Class 6 Maths Question 4.
Express as km using decimals:
(a) 8 m
(b) 88 m
(c) 8888 m
(d) 70 km 5 m
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 39

Ex 8.4 Class 6 Maths Question 5.
Express as kg using decimals:
(a) 2 g
(b) 100 g
(c) 3750 g
(d) 5 kg 8 g
(e) 26 kg 50 g
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 40
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 41

Ex 8.5 Class 6 Maths 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
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 42

Ex 8.5 Class 6 Maths Question 2.
Rashid spent ₹ 35.75 for Maths book and ₹ 32.60 for Science book. Find the total amount spent by Rashid.
Solution:
Money spent on Maths book = ₹ 35.75
Money spent on Science book=₹ 32.60
∴ Total money spent is =₹ 35.75 +₹ 32.60 = ₹ 68.35
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 43

Ex 8.5 Class 6 Maths Question 3.
Radhika’s mother gave her ₹ 10.50 and her father gave her ₹ 15.80, find the total amount given to Radhika by the parents.
Solution:
Money given by Radhika’s mother = ₹ 10.50
Money given by Radhika’s father = ₹ 15.80
∴ Total amount given to Radhika = ₹ 10.50 + ₹ 15.80 = ₹ 26.30

Ex 8.5 Class 6 Maths 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.
Solution:
Cloth bought by Nasreen for her shirt = 3 m 20 cm
Cloth bought by Nasreen for her trouser = 2 m 5 cm
∴ Total cloth bought by Nasreen = 3 m 20 cm + 2 m 5 cm
= 5 m 25 cm
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 45

Ex 8.5 Class 6 Maths 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 walked in the morning = 2 km 35 m = 2.035 km
Distance walked in the evening = 1 km 7 m = 1.007 km
∴ Total distance walked by Naresh = 2.035 km +1.007 km = 3.042 km
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 46

Ex 8.5 Class 6 Maths Question 6.
Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order 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 7 m = 7.007 km
Distance travelled on foot = 500 m = 0.500 km
Total distance of school from her residence
= 15.268 km + 7.007 km + 0.500 km = 22.775km
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 47

Ex 8.5 Class 6 Maths Question 7.
Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.
Solution:
Weight of rice purchased = 5 kg 400 g = 5.400 kg
Weight of sugar purchased = 2 kg 20 gm = 2.020 kg
Weight of flour purchased = 10 kg 850 g = 10.850 kg
Total weight of purchases = 5.400 kg +2.020 kg +10.850 kg = 18.270 kg
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 48

Ex 8.6 Class 6 Maths Question 1.
Subtract:
(a) ₹ 18.25 from ₹ 20.75
(b) ₹ 202.54 m from 250 m
(c) ₹ 5.36 from ₹ 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 49

Ex 8.6 Class 6 Maths Question 2.
Find the value of:
(a) 9.756 – 6.28
(b) 21.05 – 15.27
(c) 18.5 – 6.79
(d) 11.6 – 9.
Solution:
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 50
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 51

Ex 8.6 Class 6 Maths 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:
Cost of the book = ₹ 35.65
Money given to the shopkeeper by Raju = ₹ 50
Money he got back = ₹ 50 – ₹ 35.65 = ₹ 14.35
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 52

Ex 8.6 Class 6 Maths Question 4.
Rani had ₹ 18.50. She bought one ice-cream for ₹ 11.75. How much money does she have now?
Solution:
Money with Rani = ₹ 18.50
Money spent on ice-cream=₹ 11.75
Money left with Rani = ₹ 18.50 – ₹ 11.75 = ₹ 6.75
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 53

Ex 8.6 Class 6 Maths 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?
Solution:
Total length of cloth = 20 m 5 cm = 20.05 m
Length cut out for curtain = 4 m 50 cm = 4.50 m
Cloth left over = 20.05 m – 4.50 m = 15.55 m
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 54

Ex 8.6 Class 6 Maths 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 by Namita
20 km 50 m = 20.050 km
Distance travelled by bus = 10 km 200 m = 10.200 km
∴ Distance travelled by auto = 20.050 km – 10.200 km
= 9.850 km
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 55

Ex 8.6 Class 6 Maths 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:
Total weight of vegetables bought = 10 kg
Weight of onions = 3 kg 500 g = 3.500 kg
Weight of tomato= 2 kg 75 g = 2.075 kg
∴ Total weight of these vegetables =3.500 kg + 2.075 kg = 5.575 kg
As per question
Weight of potatoes = 10 kg – 5.575 kg = 4.425 kg
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 56

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Chapter-7 Fractions | Class 6th | NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 6 Maths: One of the most important and interesting things to keep in mind is that these solutions available are totally free of cost. This also includes all the solutions to the exercises that are given in the textbook. NCERT solutions for class 6 maths gives you chapter-wise solutions to each and every question. This can help you solve even the tougher with ease.

It is always good to have a strong foundation in order to build a good building. This also is true in class when you are still learning the basics of maths. NCERT solutions for class 6 maths provides you the exact opportunity to build a strong foundation in this subject. Below is an overview of each and every chapter covered in the NCERT textbook.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Ex 7.1 Class 6 Maths Question 1.
Write the fraction representing the shaded portion.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 1
Solution:
The fraction representing the shaded portion in the given diagram are as under:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 2

Ex 7.1 Class 6 Maths Question 2.
Colour the part according to the given fraction
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 3
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 4
Solution:
The figures are shaded as per indicated fraction shown against them.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 5

Ex 7.1 Class 6 Maths Question 3.
Identify the error, if any?
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 6
Solution:
In all the figures, the shaded portions do not represent the given fractions.

Ex 7.1 Class 6 Maths Question 4.
What fraction of a day is 8 hours?
Solution:
Since, there are 24 hours in a day, therefore, the required fraction
=824=13.

Ex 7.1 Class 6 Maths Question 5.
What fraction of an hour is 40 minutes?
Solution:
Since, there are 60 minutes in an hour, therefore, the required fraction =4060=23.

Ex 7.1 Class 6 Maths Question 6.
Arya, Abhimanyu and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share?
(b) What part of a sandwich will each boy receive?
Solution:
(a) To divide 2 sandwiches brought by Arya equally among Arya, Abhimanyu and Vivek (i.e., 3 persons), we have to divide each sandwich into 3 equal parts. Then two parts thus obtained be given to each of them.
(b) 13 of a sandwich will be received by each boy but in total each will get 23.

Ex 7.1 Class 6 Maths Question 7.
Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Solution:
Total number of dresses = 30
Work finished = 20
∴ The fraction of dresses she has finished ==2030=23.

Ex 7.1 Class 6 Maths Question 8.
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Solution:
Natural numbers from 2 to 12 are:
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 i.e., 11 in number.
Prime numbers out of these numbers are:
2, 3, 5, 7, 11 i.e., 5 in number.
∴ Required fraction = =412=13

 

Ex 7.1 Class 6 Maths Question 9.
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Solution:
Natural numbers from 102 to 113 are:
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113 i.e., 12 in number.
Prime numbers out of the above numbers are:
103, 107, 109, 113 i.e., 4 in number.
∴ Required fraction ==412=13

Ex 7.1 Class 6 Maths Question 10.
What fraction of these circles have X’s in them?
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 7
Solution:
∴Required fraction = =48=12

Ex 7.1 Class 6 Maths Question 11.
Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Solution:
Total number of CDs = 3 + 5 = 8
Number of CDs purchased = 3
Fraction of CDs purchased = 38
Fraction of CDs received as gift = 58 .

Ex 7.2 Class 6 Maths Question 1.
Draw number lines and locate the points on them:
(a) 12,14,34,44
(b) 18,28,38,78
(c) 25,35,85,45
Solution:
(a) We know that 12 is greater than zero and less than 1, so it should lie 2 between 0 and 1.
To show 12 on the number line, we divide the gap between 0 and 1 into two equal parts and the point P represents the fraction 12 .
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 8
To represent 14,34,44 on the number line, we divide the gap between 0 and 1 into 4 equal parts, then the points P, R and A respectively represent the fraction 14,34 and 44 .
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 9
(b)
 To rfepresent 18,28,38 and 78 on the number line, we divide the gap between 0 and 1 into 8 equal parts, then the points P, Q, R and S respectively represent the fraction 18,28,38 and 78 .
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 10
(c)
 To represent 25,35 and 45 on the number line, we divide the gap between 0 and 1 into 5 equal parts, then the points P, Q and R respectively represent the fraction 25,35 and 45 .
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 11
For representing 85
Draw a line. Take a point 0 on it. Let it represent 0.
Now, 85=135=1+35
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 12

From O, set off unit distances to the right. Let these segments be OA, AB. Then, clearly the points A and B represent 1 and 2 respectively. Take 1 full unit length to the right of O. Divide the 2nd unit AB into 5 equal parts. Take 3 parts out of these 5 parts to reach P. Then, P represents 85.

Ex 7.2 Class 6 Maths Question 2.
Express the following as mixed fractions:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 13
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 14
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 15

Ex 7.2 Class 6 Maths Question 3.
Express the following as improper fractions:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 16
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 17

Ex 7.3 Class 6 Maths Question 1.

Write the fractions. Are all these fractions equivalent?
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 18
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 19
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 20

Ex 7.3 Class 6 Maths Question 2.
Write the fractions and pair up the equivalent fractions from each row.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 21
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 22

Ex 7.3 Class 6 Maths Question 3.
Replace □ in each of the following by the correct number:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 23
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 24
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 25
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 26

Ex 7.3 Class 6 Maths Question 4.
Find the equivalent fraction of 35 having
(a) denominator 20
(b) numerator 9
(c) denominator 30
(d) numerator 27
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 27
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 28

Ex 7.3 Class 6 Maths Question 5.
Find the equivalent fraction of 3648 with
(a) numerator 9
(b) denominator 4
Solution:
(a) ∵ 36 ÷ 9 = 4
∴ We divide the numerator and denominator by 4
3648=36÷448÷4=912
(b) ∵ 48 ÷ 4 = 12
∴ We divide the numerator and denominator by 12
3648=36÷1248÷12=34

Ex 7.3 Class 6 Maths Question 6.
Check whether the given fractions are equivalent:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 29
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 30

Ex 7.3 Class 6 Maths Question 7.
Reduce the following fractions to simplest form:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 31
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 32
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 33
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 34
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 35

Ex 7.3 Class 6 Maths Question 8.
Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of their of her/his pencils?
Solution:
Fractions of pencils used by Ramesh, Sheelu and Jamaal are 1020,2550 and 4080 respectively.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 36
Thus, these fractions are equal.

Ex 7.3 Class 6 Maths Question 9.
Match the equivalent fractions and write two more for each:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 37
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 38
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 39
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 40
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 41

Ex 7.4 Class 6 Maths Question 1.
Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘>’, ‘=’, ‘>’ between the fractions:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 42
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 43
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 44
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 45

Ex 7.4 Class 6 Maths Question 2.
Compare the fractions and put an appropriate sign.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 46
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 47
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 48

Ex 7.4 Class 6 Maths Question 3.
Make five more such pairs and put appropriate signs.
Solution:
Five more such pairs may be taken as under:
Compare the fractions and put an appropriate sign.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 49

Ex 7.4 Class 6 Maths Question 4.
Look at the figures and write ‘<’ or ‘>’, ‘=’ between the given pairs of fractions.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 50
Make five more such problems and solve them with your friends.
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 51
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 52

Ex 7.4 Class 6 Maths Question 5.
Five more such problems are as under:
Look at the figures given in question 5 and write ‘<’ or ‘>’ or ‘=’ between the pairs of fractions.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 53
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 54

Ex 7.4 Class 6 Maths Question 6.
How quickly can you do this? Fill appropriate sign. (<, =, >)
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 55
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 56
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 57

Ex 7.4 Class 6 Maths Question 7.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 58
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 59
On separating them into three groups of equivalent fractions, we have
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 60

Ex 7.4 Class 6 Maths Question 8.
Find answers to the following. Write and indicate how you solved them.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 61
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 62

Ex 7.4 Class 6 Maths Question 9.
Ila read 25 pages of a book containing 100 pages. Lalita read 25 of the same book. Who read less?
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 63

Ex 7.4 Class 6 Maths Question 10.
Rafiq exercised for 36 of an hour, while Rohit exercised for 34 of
an hour. Who exercised for a longer time?
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 64

Ex 7.4 Class 6 Maths Question 11.
In a class A of 25 students, 20 passed in first class; in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?
Solution:
Fraction of students getting first class in class A
=2025=20÷525÷5=45
Fraction of students getting first class in class B
=2430=24÷630÷6=45
Clearly, an equal number of students got 1st class in both the classes.

Ex 7.5 Class 6 Maths Question 1.
Write these fractions appropriately as additions or subtractions:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 65
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 66

Ex 7.5 Class 6 Maths Question 2.
Solve:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 67
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 68
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 69

Ex 7.5 Class 6 Maths Question 3.
Shubham painted 23 of the wall space in his room. His sister Madhavi helped and painted 13 of the wall space. How much did they paint together?
Solution:
Portion of wall painted by Shubham = 23
Portion of wall painted by Madhavi = 13
Wall painted by both of them = 23+13=2+13=33=1
i.e., full wall was painted by them.

Ex 7.5 Class 6 Maths Question 4.
Fill in the missing fractions.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 70
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 71
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 72

 

Ex 7.5 Class 6 Maths Question 5.
Javed was given 57 of a basket of oranges. What fraction of oranges was left in the basket? ‘
Solution:
Portion of oranges received by Javed = 57
Portion of oranges left in the basket
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 73

Ex 7.6 Class 6 Maths Question 1.
Solve:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 74
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 75
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 76
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 77
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 78

Ex 7.6 Class 6 Maths Question 2.
Sarita bought 25 metre of ribbon and Lalita 34 metre of ribbon.
What is the total length of the ribbon they bought?
Solution:
Ribbon bought of Sarita = 25 m
Ribbon bought of Lalita = 34 m
Total length of the ribbon bought = (25+34) m
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 79

Ex 7.6 Class 6 Maths Question 3.
Naina was given 112 piece of cake and Najma was given 113piece of cake. Find the total amount of cake was given to both of them.
Solution:
Number of pieces of cake consumed by Naina = 112
Number of pieces of cake consumed by Najma = 113
Total amount of cake consumed = 112+113
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 80

Ex 7.6 Class 6 Maths Question 4.
Fill in the boxes:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 81
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 82
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 83
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 84

Ex 7.6 Class 6 Maths Question 5.
Complete the addition-subtraction box.
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 85
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 86
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 87

Ex 7.6 Class 6 Maths Question 6.
A piece of wire 78 metre long broke into two pieces. One piece was 14 metre long. How long is the other piece?
Solution:
Total length of wire = 78 metre
Length of a piece of wire = 14 metre
∴ Other piece of wire =(7814) metre
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 88

Ex 7.6 Class 6 Maths Question 7.
Nandini’s house is 910 km from her school. She walked some distance and then took a bus for 12 km to reach the school. How
Solution:
Total distance of Nandini’s house from her school = 910 km
Distance for which bus was taken = 12 km
∴ Distance walked by Nandini for home
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 89

Ex 7.6 Class 6 Maths Question 8.
Asha and Samuel have bookshelves of the same size partly filled with hooks. Asha’s shelf is 25 th full and Samuel’s shelf is 56 th full. Whose bookshelf is more full? By what fraction?
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 90
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 91

Ex 7.6 Class 6 Maths Question 9.
Jaidev takes 215 minutes to walk across the school ground. Rahul takes 74 minutes to do the same. Who takes less time and by what fraction?
Solution:
NCERT Solutions for Class 6 Maths Chapter 7 Fractions 92

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Chapter-6 Integers | Class 6th | NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 6 Maths: One of the most important and interesting things to keep in mind is that these solutions available are totally free of cost. This also includes all the solutions to the exercises that are given in the textbook. NCERT solutions for class 6 maths gives you chapter-wise solutions to each and every question. This can help you solve even the tougher with ease.

It is always good to have a strong foundation in order to build a good building. This also is true in class when you are still learning the basics of maths. NCERT solutions for class 6 maths provides you the exact opportunity to build a strong foundation in this subject. Below is an overview of each and every chapter covered in the NCERT textbook.

NCERT Solutions for Class 6 Maths Chapter 6 Integers

Ex 6.1 Class 6 Maths Question 1.
Write opposites of the following:
(a) Increase in weight
(b) 30 km north
(c) 326 BC
(d) Loss of ₹ 700
(e) 100 m above sea level
Solution:
(a) Decrease in weight
(b) 30 km South
(c) 326 AD
(d) Gain of ₹ 700
(e) 100 m below sea level

Ex 6.1 Class 6 Maths Question 2.
Represent the following numbers as integers with appropriate signs.
(a) An aeroplane is flying at a height two thousand metre above the ground.
(b) A submarine is moving at a depth, eight hundred metre below the sea level.
(c) A deposit of rupees two hundred.
(d) Withdrawal of rupees seven hundred.
Solution:
(a) +2000
(b) -800
(c) +200
(d) -700.

Ex 6.1 Class 6 Maths Question 3.
Represent the following numbers on a number line:
(a) +5
(b) -10
(c) +8
(d) -1
(e) -6
Solution:
In order to represent integers on a number line, we draw a line and mark a point O almost in the middle of it as shown. Now, set off equal distances on the right hand side as well as on the left hand side of O. On the right side of O, label the points of subdivisions as 1,2, 3 etc. and the point O as 0. Since negative integers are opposite of positive integers (i.e., natural numbers), so we represent negative integers in the opposite direction Le., on the left side of O on the number line. On the left side of point O, label the points of subdivision by -1, -2, -3 etc. as shown in figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 1
Clearly (a) +5 (b) -10 (c) +8 (d) -1 and (e) -6 are represented by the points A, B, C, D and E respectively on the number line.

Ex 6.1 Class 6 Maths Question 4.
Adjacent figure is a vertical number line, representing integers. Observe it and locate the following points:
NCERT Solutions for Class 6 Maths Chapter 6 Integers 2
(a) If point D is+ 8, then which point is -8?
(b) Is point G a negative integer or a positive integer?
(c) Write integers for points B and E.
(d) Which point marked on this number line has the least value?
(e) Arrange all the points in decreasing order of value.
Solution:
From the given vertical number line, clearly
(a) The point F represents -8.
(b) Since G lies below O, so it is a negative integer.
(c) Point B represents +4 and E represents -10.
(d) E has the least value.
(e) Points in decreasing order are as under: D, C, B, A, O, H, G, F, E

 

Ex 6.1 Class 6 Maths Question 5.
Following is the list of temperatures of five places in India, on a particular day of the year.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 3
(a) Write the temperatures of these places in the form of integers in the blank column.

(b) Following is the number line representing the temperature in degree Celsius.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 4
Plot the name of the city against its temperature.
(c) Which is the coolest place?
(d) Write the names of the places where temperatures are above 10°C.
Solution:
(a) Siachin: -10 °C Shimla: -8°C
Ahmedabad: +30°C Delhi: +35°C
Srinagar: -2°C
(b) Plotting the name of the city against its temperature on the number line representing the temperature in degree Celsius, we have
NCERT Solutions for Class 6 Maths Chapter 6 Integers 5
(c)
 The coolest place is Siachin.
(d) The places having temperatures, above 10°C are Ahmedabad anti Delhi.

Ex 6.1 Class 6 Maths Question 6.
In each of the following pairs, which number is to the right of the other on the number line?
(a) 2, 9
(b) -3, -8
(c) 0, -1
(d) -11, 10
(e) -6, 6
(f) 1, -100
Solution:
(a) 9 is on the right of 2 on the number line.
(b) -3 is on the right of -8 on the number line.
(c) 0 is on the right of -1 on the number line.
(d) 10 is on the right of -11 on the number line.
(e) 6 is on the right of-6 on the number line.
(f) 1 is on the right of -100 on the number line.

Ex 6.1 Class 6 Maths Question 7.
Write all the integers between the given pairs (write them in the increasing order).
(a) 0 and -7
(b) -4 and 4
(c) -8 and -15
(d) -30 and -23
Solution:
(a) All the integers between 0 and -7 in the increasing order are -6, -5, -4, -3, -2,-1.
(b) All the integers between -4 and 4 in the increasing order are -3, -2, -1, 0, 1, 2, 3.
(c) All the integers between -8 and -15 in the increasing order are -14, -13,-12,-11,-10,-9.
(d) All the integers between -30 and -23 in the increasing order are -29, -28, -27, -26, -25, -24.

Ex 6.1 Class 6 Maths Question 8.
(a) Write four negative integers greater than 20.
(b) Write four negative integers less than 10.
Solution:
(a) Four negative integers greater than -20 are -20 +1, -20 + 2, -20 + 3 and -20 + 4
i. e., -19 , -18, -17 and -16.
(b) Four negative integers less than-10 are -10 -1, -10 -2, -10 -3 and -10 -4
i. e., -11, -12, -13, -14.

Ex 6.1 Class 6 Maths Question 9.
For the following statements, write true (T) or false (F). If the statement is false, correct the statement.
(a) -8 is to the right of -10 on a number line.
(b) -100 is to the right of -50 on a number line.
(c) Smallest negative integer is -1.
(d) -26 is larger than -25.
Solution:
(a) True.
(b) False. The correct statement should be “ -100 is to the left of -50 on a number line”.
(c) False. The correct statement should be “ There is no last or smallest negative number”.
(d) False. The correct statement should be “ -26 is smaller than -25’.

Ex 6.1 Class 6 Maths Question 10.
Draw a number line and answer the following:
(a) Which number will we reach if we move 4 numbers to the right of -2?
(b) Which number will we reach if we move 5 numbers to the left of 1?
(c) If we are at -8 on the number line, in which direction should we move to reach -13?
(d) If we are at -6 on the number line, in which direction should we move to reach -1?
Solution:
(a) We want to obtain integer on moving 4 numbers to the right of -2. So, we start from -2 and proceed 4 units to the right of -2 to obtain 2 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 6
Hence, we will reach the number 2.
(b) We want to obtain integer on moving 5 numbers to the left of 1. So, we start from 1 and proceed 5 units to the left of 1 to obtain -4 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 7
Hence, we will reach the number -4. ,
(c) Since -13 < -8. So, to reach -13 from -8 on the number line we should move to the left of -8.
(d) Since -1 > -6. So,, to reach -1 from -6 on the number line, we should move to the right of -6.

 

Ex 6.2 Class 6 Maths Question 1.
Using the number line write the integer which is:
(a) 3 more than 5
(b) 5 more than -5
(c) 6 less than 2
(d) 3 less than -2
Solution:
(a) We want to obtain integer 3 more than 5. So, we start from 5 and proceed 3 units to the right to obtain 8 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 8
Hence, 3 more than 5 is 8.
(b) We want to obtain integer 5 more than -5. So, we start from -5 and proceed 5 units to the right to obtain 0 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 9
Hence, 5 more than -5 is 0.
(c) We want to obtain an integer 6 less than 2. So, we start from 2 and proceed 6 units to the left of it to obtain -4 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 10
Hence, 6 less than 2 is -4.
(d) We want to obtain an integer 3 less than -2. So, we start from -2 and proceed 3 units to the left of it to obtain -5 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 11
Hence, 3 less than -2 is -5.

Ex 6.2 Class 6 Maths Question 2.
Use number line and add the following integers:
(a) 9 + (-6)
(b) 5 + (-11)
(c) (-1) + (-7)
(d) (-5) + 10
(e) (-1) + (-2) + (-3)
(f) (-2) + 8 + (-4)
Solution:
(a) First we move to the right of 0 by 9 steps reaching 9. Then we move 6 steps to the left of 9 reaching 3 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 12
Thus, 9 + (-6) = 3.
(b) First we move to the right of 0 by 5 steps reaching 5. Then, we move 11 steps to the left of 5 reaching -6 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 13
Thus, 5 + (-11) = -6.
(c) First we move to the left of 0 by 1 step reaching -1. Then we move 7 steps to* the left of -1 reaching -8 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 14
Thus, (-1) + (-7) = -8.
(d) First we move to the left of 0 by 5 steps reaching -5. Then we move 10 steps to the right of -5 reaching 5 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 15
Thus, (-5) + 10 = 5.
(e) First we move to the left of 0 by 1 step reaching -1. Then we move 2 steps to the left of -1 reaching -3. Again, we move 3 steps to the left of -3 reaching -6 as shown in the figure.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 16
Thus, (-1) + (-2) + (-3) = -6.
(f) First we move to the left of 0 by 2 steps reaching -2. Then we move 8 steps to the right of -2 reaching 6. Again, we move 4 steps to the left of 6 reaching -2 as shown in the figure.
Thus, (-2) + 8 + (-4) = 2.
NCERT Solutions for Class 6 Maths Chapter 6 Integers 17

Ex 6.2 Class 6 Maths Question 3.
Add without using number line:
(a) 11 + (-7)
(b) (-13) + (+18)
(c) (-10) + (+19)
(d) (-250) + (+150)
(e) (-380)+ (-270)
(f) (-217) + (-100)
Solution:
(a) 11 + (-7) = 4 + 7 + (-7)
= 4 + 0 = 4
(b) (-13) + (+18) = (-13) + (+13) + (+ 5)
= 0 + (+5) = 5
(c) (-10) + (+19) = (-10) + (+10) + (+9)
= 0 + (+9) = +9
(d) (-250) + (+150) = (-100) + (-150) + (+150)
= (-100) + 0 = -100
(e) (-380) + (-270) = -650
(f) (-217) + (-100) = -317

Ex 6.2 Class 6 Maths Question 4.
Find the sum of:
(a) 137 and -354
(b) -52 and 52
(c) -312, 39 and 192
(d) -50, – 200 and 300
Solution:
(a) (+137) + (-354) = (+137) + (-137) + (-217)
= 0 + (-217) =-217
(b) (-52) + (+52) = 0
(c) (-312) + (+39) + (+192) = (-312) + (+231)
= (-81) + (-231) + (+231)
= (-81) + 0 = -81
(d) (-50) + (-200) + (+300) = (-250) + (+300)
= (-250) + (+250) + (+50)
= 0 + (+50) = +50

Ex 6.2 Class 6 Maths Question 5.
Find the sum:
(a) (-7) + (-9) + 4 + 16
(b) (37) + (-2) + (-65) + (-8)
Solution:
(a) (-7) + (-9) + 4 + 16 = (-16) + 4 + 16
= (-16) + (+16) + 4 = 0 + 4 = 4
(b) (37) + (-2) + (-65) + (-8) = 37 + (-75)
= 37+ (-37)+ (-38)
= 0 + (-38) = -38

 

Ex 6.3 Class 6 Maths Question 1.
Find:
(a) 35-(20)
(b) 72 (90)
(c) (-15) – (-18)
(d) (-20) – (13)
(e) 23 (-12)
(f) (-32) – (-40)
Solution:
(a) 35 – 20 = (15 + 20) – 20
= 15 + (20 – 20) = 15 + 0 = 15
(b) 72 – 90 = 72 – (72 + 18)
= (72 – 72) + (-18)
= 0 + (-18) = -18 1
(c) (-15) – (-18) = (-15) + (additive inverse -18)
= -15 + 18 = -15 +(15 + 3)
= (-15 + 15) 43 = 0 + 3 = +3
(d) (-20) -13 = -20 -13 = -(20 + 13) = -33
(e) 23 – (-12) = 23 + 12 = 35
(f) (-32) – (-40) = -32 + 40 = (-32 + 32) +8
= 0 + 8 + 8

 

Ex 6.3 Class 6 Maths Question 2.
Fill in the blanks with >, < or = sign:
(a) (-3) + (-6) …… (-3) -(-6)
(b) (-21) – (-10) …… (-31) + (-11)
(e) 45 – (-11) ……. 57 + (-4)
(d) (-25) – (-42) …….. (-42) – (-25)
Solution:
(a) Here, (-3) + (-6) = -3 – 6 = =9
and (-3)-(-6) = -3 + 6 =3
As -9 < 3
∴ (-3) + (-6) < (-3) – (-6) (b)

(b) Here, (-21) – (-10) = -21 +10 = -11
and (-31)+ (-11) = -31 -11 = -42
As -11 > -42
∴ (-21) – (-10) > (-31) + (-11)

(c) Here, 45 – (-11) = 45 + 11 = 56
and 57 + (-4) = 57-4 = 53
As 56 > 53
∴ 45 – (-11) > 57 + (-4)

(d) Here, -25 – (-42) = -25 + 42
= -25 + (25 + 17)
= (-25 + 25) +17 = 0 + 17 = 17
and, (-42) – (-25) = -42 + 25
= (-17 – 25) + 25
= -17 + (-25 + 25) = -17 + 0 = -17
As 17 > -17
∴ (-25) – (-42) > (-42) – (-25)

Ex 6.3 Class 6 Maths Question 3.
Fil in the blanks:
(a) (-8)+ ….. = 0
(b) 13 + …….. = 0
(c) 12 + (-12) = ……
(d) (-4) + ……. = -12
(e) ………. -15 = -10
Solution:
(a) (-8) + (+8) = 0
(b) 13 + (-13) = 0
(c) 12 + (-12) = 0
(d) (-4) +(-8) = -12 [∵ -12 – (-4) = -12 + 4 = -8]
(e) 5-15 = -10 [∵ -10 + 15 = 5]

Ex 6.3 Class 6 Maths Question 4.
Find:
(a) (-7) – 8- (-25)
(b) (-13) + 32 – 8 – 1
(c) (-7) + (-8) + (-90)
(d) 50 – (-40) – (-2)
Solution:
(a) (-7) – 8 – (-25) = -7 – 8 + 25
= -15 + 25 = 10.
(b) (-13) + 32 – 8-1 = -13 – 8 – 1 + 32
= -22 + 32 = 10
(c) (-7) + (-8) + (-90) = -7 – 8 – 90 = -105.
(d) 50 – (-40) – (-2) = 50+ 40 + 2 = 92.

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Chapter-5 Understanding Elementary Shapes | Class 6th | NCERT MAths Solutions | Edugrown

NCERT Solutions for Class 6 Maths: One of the most important and interesting things to keep in mind is that these solutions available are totally free of cost. This also includes all the solutions to the exercises that are given in the textbook. NCERT solutions for class 6 maths gives you chapter-wise solutions to each and every question. This can help you solve even the tougher with ease.

It is always good to have a strong foundation in order to build a good building. This also is true in class when you are still learning the basics of maths. NCERT solutions for class 6 maths provides you the exact opportunity to build a strong foundation in this subject. Below is an overview of each and every chapter covered in the NCERT textbook.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

Ex 5.1 Class 6 Maths Question 1.
What is the disadvantage in comparing line segments by mere observation?
Solution:
The disadvantage in comparing line segments by mere observation is the inaccuracy in judgement.

Ex 5.1 Class 6 Maths Question 2.
Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Solution:
It is better to use a divider with ruler as it given an accurate measurement of the line segment.

Ex 5.1 Class 6 Maths Question 3.
Draw any line segment, say AB. Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
[Note: If A, B, C are any three points on a line, such that AC + CB = AB, then we can be sure that C lies between A and B.]
Solution:
On measuring the lengths of line segments AB, BC and AC using ruler, we find
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 1
AB =6.5 cm,
CB = 4.2 cm,
and AC = 2.3 cm
Now, AC +CB = 2.3 cm +4.2 cm = 6.5 cm = AB
AB = AC+CB.

 

Ex 5.1 Class 6 Maths Question 4.
If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Solution:
Since, AB + BC = 5 cm +3 cm = 8 cm = AC
∴ A, B, C are collinear and B lies between A and C.

Ex 5.1 Class 6 Maths Question 5.
Verify, whether D is the mid point of AG¯¯¯¯¯¯¯¯.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 2
Solution:
Since AD = AB +BC + CD = 3 units
and, DG = DE +EF + FG = 3 units
∵ AD = DG [∵ Each = 3 units]
Thus, D is the mid-point of AG¯¯¯¯¯¯¯¯.

Ex 5.1 Class 6 Maths Question 6.
If B is the mid point of AC¯¯¯¯¯¯¯¯ and C is the mid point of BD¯¯¯¯¯¯¯¯, where A, B, C, D lie on a straight line, say why AB = CD?
Solution:
∵ B is the mid point of AC¯¯¯¯¯¯¯¯. Therefore, AB = BC.
∵ C is the mid-point of BD¯¯¯¯¯¯¯¯. Therefore, BC = CD.
Thus, AB =BC = CD i.e., AB = CD.

Ex 5.1 Class 6 Maths Question 7.
Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.
Solution:
Draw any five triangles, T1, T2, T3, T4 and T5. label each one as ∆ ABC. Measure, in each case, the three sides a = BC, b = CA and c = AB.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 3
Let us tabulate the measurements of sides as under:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 4
From the above table, we find that
(i) each value of b + c – a is positive;
(ii) each value of c + a – b is positive;
(iii) each value of a + b – c is positive.
Now, b + c – a is positive ⇒ b + c – a > 0 ⇒ b + c > a
c + a – b is positive ⇒ c + a- b > 0 ⇒ c + a > b
a+ b – c is positive ⇒ a + b – c > 0 ⇒ a + b > c
Thus, it is verified that the sum of any two sides of a triangle is greater than the third side.
∴ The statement ‘the sum of two sides is ever less than the third side’ is never true.

Ex 5.2 Class 6 Maths Question 1.
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3
Solution:
(a) 12 of a revolution
(b) 14 of a revolution
(c) 14 of a revolution
(d) 34 of a revolution
(e) 34 of a revolution
(f) 34 of a revolution

Ex 5.2 Class 6 Maths Question 2.
Where will the hand of a clock stop if it
(a) starts at 12 and makes 12 of a revolution, clockwise?
(b) starts at 2 and makes 12 of a revolution clockwise?
(c) starts at 5 and makes 14 of a revolution, clockwise?
(d) starts at 5 and makes 34 of a revolution, clockwise?
Solution:
(a) If the hand of clock starts at 12 and makes 12 of a revolution clockwise it reaches at 6.
(b) If the hand of clock starts at 2 and makes 12 of a revolution clockwise it reaches at 8.
(c) If the hand of clock starts at 5 and makes 14 of a revolution clockwise it reaches at 8.
(d) If the hand of clock starts at 5 and makes 34 of a revolution clockwise it reaches at 2.

Ex 5.2 Class 6 Maths Question 3.
Which direction will you face if you start facing
(a) east and make 12 of a revolution clockwise?
(b) east and make 112 of a revolution clockwise?
(c) west and make 34 of a revolution anti-clockwise?
(d) south and make one full revolution?
(Should we specify clockwise or anit-clockwise for this last question? Why nor?)
Solution:
(a) Facing east and on making 12 of a revolution clockwise, we will face west.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 5

 

(b) Facing east and on making 112 of a revolution clockwise, we will face west.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 6
(c)
 Facing west and on making 34 of a revolution anti-clockwise, we will face north.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 7
(d)
 Facing south and on making 1 full revolution we will face south again.
There is no need to specify clockwise or anti-clockwise in the last question as turning by one full revolution (i.e., two straight angles) make us to reach the original position.
tiwari academy class 6 maths Chapter 5 Understanding Elementary Shapes 8

 

Ex 5.2 Class 6 Maths Question 4.
What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
(b) south and turn clockwise to face east?
(c) west and turn clockwise to face east?
Solution:
Figures drawn will dearly show the amount of revolution:
(a) 34 of a revolution
tiwari academy class 6 maths Chapter 5 Understanding Elementary Shapes 9
(b)
 34 of a revolution
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 10

(c) 12 of a revolution
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 11

Ex 5.2 Class 6 Maths Question 5.
Find the number of right angles turned through by the hour hand of clock when it goes from
(a) 3 to 6
(b) 2 to 8
(c) 5 to 11
(d) 10 to 1
(e) 12 to 9
(f) 12 to 6
Solution:
Since turning by two straight angles (or four right angles) in the same direction makes a full turn. Thus for the movement of clock hand from 12 – 1,1 – 2, 2 – 3, … 11 – 12 i.e., 12 in number gives 4 right angles turn. Therefore,
(a) From 3 to 6 : 3 – 4, 4 – 5 and 5 – 6 i.e., 3 in number.
∴ Number of right angles turned through by the hour hand of a clock when it goes from 3 to 6
(412×3) right angles = 1 right angles

(b) From 2 to 8 :2 – 3,3 – 4, 4 – 5, 5 – 6,6 – 7 and 7 – 8 i.e., 6 in number.
∴ Number of right angles turned through by the hour hand of a clock
when it goes from 2 to 8
(412×6) right angles = 2 right angles

(c) From 5 to II : 5 – 6, 6 – 7, 7 – 8, 8 – 9, 9 – 10 and 10 – 11 i.e., 6 in number.
∴ Number of right angles turned through by the hour hand of a clock when it goes from 5 to 11
(412×6) right angles = 2 right angles

(d) From 10 to 1 :10 – 11,11 -12 and 12 – 1 i.e., 3 in number.
∴ Number of right angles turned through by the hour hand of a clock when it goes from 10 to 1
(412×3) right angles = 1 right angles

(e) From 12 to 9 : 12 – 1, 1 – 2, 2 – 3, 3 – 4, 4 – 5, 5 – 6, 6 – 7, 7 – 8 and 8 – 9 i.e., 9 in number
∴ Number of right angles turned through by the hour hand of a clock when it goes from 12 to 9
(412×9) right angles = 1 right angles

(f) From 12 to- 6 : 12 – 1, 1 – 2, 2 – 3, 3 – 4, 4 – 5 and 5 – 6 i.e., 6 in number.
∴ Number of right angles turned through by the hour hand of a clock when it goes from 12 to 6
(412×6) right angles = 2 right angles

Ex 5.2 Class 6 Maths Question 6.
How many right angles do you make if you start facing
(a) south and turn clockwise to west?
(b) north and turn anti-clockwise to east?
(c) west and turn to west?
(d) south and turn to north?
Solution:
Figures drawn will clearly indicate the number of right angles we make if we start facing:
(a) south and turn clockwise to west is 1 right angle.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 12
(b)
 north and turn anti-clockwise to east are 3 right angles.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 13

 

(c) west and turn to west are 4 right angles.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 14
(d)
 south and turn to north are 2 right angles.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 15

Ex 5.2 Class 6 Maths Question 7.
Where will the hour hand qfa clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7 and turns through 2 straight angles?
Solution:
We know that in 12 hours hour hand turns through 4 right angles i.e., in a turn of 1 right angle, the hour hand moves 3 hours.
(a) If the hour hand of a clock starts from 6 and turns through 1 right angle then its hour hand stops at (6 +3) i.e., 9.
(b) If the hour hand of a clock starts from 8 and turns through 2 right
angles then its hour hand stops at (8 + 2 x 3) = 14 i.e., 2. .
(c) If the hour hand of a clock starts from 10 and turns through 3 right angles then its hour hand stops at (10+3×3) = 19 i.e., 7.
(d) If the hour hand of a clock starts from 7 and turns through 2 straight angles i.e., 4 right angles then its hour hand stops at (7 + 4 x 3) = 19 i.e., 7.

Ex 5.3 Class 6 Maths Question 1.
Match the following:
(i) Straight angle                             (a) Less than one-fourth of a revolution
(ii) Right angle                               (b) More than half a revolution
(iii) Acute angle                             (c) Half of a revolution
(iv) Obtuse angle                         (d) One-fourth of a revolution
(v) Reflex angle                         (e) Between 14 and 12 of a revolution
(f) One complete revolution
Solution:
On matching, we find
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (e)
(v) → (b)

Question 2.
Classify each one of the following angles as right, straight, acute, obtuse or reflex:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 16
Solution:
Clearly, we find that
(a) → acute angle
(b) → obtuse angle
(c) → right angle
(d) → reflex angle
(e) → straight angle
(f) → acute angles.

Ex 5.4 Class 6 Maths Question 1.
What is the measure of
(i) a right angle?
(ii) a straight angle?
Solution:
(i) 1 right angle = 90°
(ii) 1 straight angle = 2 right angles = 2 x 90°=180°

 

Ex 5.4 Class 6 Maths Question 2.
Say True or False:
(a) The measure of an acute angle < 90°.
(b) The measure of an obtuse angle < 90°.
(c) The measure of a reflex angle > 180°.
(d) The measure of one complete revolution = 360°.
(e) If m∠A = 53° and m∠B = 35°, then m∠A > m∠B.
Solution:
(a) True
(b) False
(c) True
(d) True
(e) True

 

Ex 5.4 Class 6 Maths Question 3.
Write down the measures of
(a) some acute angles.
(b) some obtuse angles.
(give at least two examples of each).
Solution:
(a) Since the measure of an acute angle < 90°. So, two acute angles may be of 60° and 30°.
(b) Since the measure of an obtuse angle > 90°. So, two obtuse angles may be of 110° and 120°.

Ex 5.4 Class 6 Maths Question 4.
Measure the angles given below using the Protractor and write down the measure.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 17
Solution:
On measuring the given angles using protractor, we find that
(a) 45°
(b) 125°
(c) 90°
(d) ∠1 = 40°, ∠2 = 125° and ∠3 = 95°.

 

Ex 5.4 Class 6 Maths Question 5.
Which angle has a large measure ? First estimate and then measure.
Measure of Angle A =
Measure of Angle B =
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 18
Solution:
Observing the given angles, we find that
∠B > ∠A
On measuring, we find that
Measure of Angle A =40°
Measure of Angle B = 65°
∴ ∠B > ∠A.

 

Ex 5.4 Class 6 Maths Question 6.
From these two angles which has larger measure? Estimate and then confirm by measuring them.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 19
Solution:
On observing, we find that
∠B > ∠A
On measuring, we find that
∠A = 45° and ∠B =60°
Thus, ∠B > ∠A.

Ex 5.4 Class 6 Maths Question 7.
Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is ……. .
(b) An angle whose measure is greater than that of a right angle is ……… .
(c) An angle whose measure is the sum of the measures of two right angles is ……. .
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is …… .
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be …… .
Solution:
(a) acute
(b) obtuse
(c) straight
(d) acute
(e) obtuse

 

Ex 5.4 Class 6 Maths Question 8.
Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor).
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 20
Solution:
On estimating without measuring, the measure of the angle shown in each figure is
(a) 45°
(b) 60°
(c) 125°
(d) 135°

On measuring with the protractor, we find their measures as under:
(a) 40°
(b) 65°
(c) 130°
(d) 135°

Ex 5.4 Class 6 Maths Question 9.
Find the angle measure between the hands of the clock in each figure:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 21
Solution:
The angle measure between the hands of the clock in the time shown is as under:
At 9.00 a.m. : 90°
At 1.00 p.m. : 30°
At 6.00 p.m. : 180°

Ex 5.4 Class 6 Maths Question 10.
Investigate
In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 22
Solution:
Looking through a magnifying glass does not make the angle larger and so the size of the angle does not change.

Ex 5.4 Class 6 Maths Question 11.
Measure and classify each angle:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 23
Solution:
The measure of angle and its classification is recorded in the given table as under:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 24

Ex 5.5 Class 6 Maths Question 1.
Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming the letter ‘L’.
(d) The letter V.
Solution:
(a) Yes
(b) No
(c) Yes
(d) No

 

Ex 5.5 Class 6 Maths Question 2.
Let PQ¯¯¯¯¯¯¯¯ be the perpendicular to the line segment XY¯¯¯¯¯¯¯¯. Let PQ¯¯¯¯¯¯¯¯ and XY¯¯¯¯¯¯¯¯ intersect in the point A. What is the measure of ∠PAY?
Solution:
The measure of ∠PAY =90° [∵ PQ ⊥ XY, given]
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 25

Ex 5.5 Class 6 Maths Question 3.
There are two “set-squares” in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Solution:
In one, the angles are of 30°, 60°, 90° and in the other the angles are of 90°, 45°, 45°.
Clearly, the common angle = 90°

Ex 5.5 Class 6 Maths Question 4.
Study the diagram. The line l is perpendicular to line m.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 26

(a) Is CE = EG?
(b) Does PE bisect CG?
(c) Identify any two line segments for which PE is the perpendicular bisector. ‘ ’
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC< EH
Solution:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 27

Ex 5.6 Class 6 Maths Question 1.
Name the types of following triangles:
(a) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ∆AJBC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(c) ∆PQR such that PQ = QR = PR = 5 cm.
(d) ∆DEF with m∠D = 90°
(e) ∆XYZ with m∠Y = 90° and XY = YZ.
(f) ∆LMN with m∠L = 30°, m∠M = 70° and m∠N = 80°
Solution:
(a) Since all the sides are of different lengths
∴ It is scalene triangle.
(b) In ∆ABC, AB ≠ BC ≠ CA.
∴ AABC is scalene triangle.
(c) In ∆PQR, PQ = QR = PR
∴ ∆PQR is equilateral triangle.
(d) In ∆DEF, m∠D =90°
∴ ∆DEF is right angled triangle.
(e) In ∆ XYZ, m∠Y = 90° and XY = YZ
∴ ∆ XYZ is an isosceles right angled.
(f) As each angle of ∆ LMN is < 90°
∴ ∆ LMN is an acute angled triangle.

Ex 5.6 Class 6 Maths Question 2.
Match the following:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 28
Solution:
The matching is as shown:
(i) → (e)
(ii)→ (g)
(iii) → (a)
(iv) → (f)
(v) → (d)
(vi) → (c)
(vii) → (b)

Ex 5.6 Class 6 Maths Question 3.
Name each of the following triangles in two different ways: (you may judge the nature of the angle by observation)
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 29
Solution:
(a) Isosceles triangle; Acute angled triangle.
(b) Scalene triangle; Right angled triangle.
(c) Isosceles triangle; Obtuse angled triangle.
(d) Isosceles triangle; Right angled triangle.
(e) Equilateral triangle; Acute angled triangle.
(f) Scalene triangle; Obtuse angled triangle.

Ex 5.6 Class 6 Maths Question 4.
Try to construct triangles match sticks. Some are shown here. Gan you make a triangle with
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 30
(a) 3 matchsticks?

(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case)
Name the type of triangle in each case.
If you cannot make a triangle, think of reasons for it.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 31
Solution:
(a) Yes, an equilateral triangle.
(b) No (as sum of two sides is always > the third side but here 1 + 1 = 2 ≱ ≯ 2).
(c) Yes, an isosceles triangle.
(d) Yes, an equilateral triangle.

 

Ex 5.7 Class 6 Maths Question 1.
Say True or False:
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to ont i^Atber.
(d) All the sides of rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.
Solution:
(a) True
(b) True
(c) True
(d) True
(e) False
(f) False

Ex 5.7 Class 6 Maths Question 2.
Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Squares, rectangles, parallelograms are all quadrilaterals.
(e) Square is also a parallelogram.
Solution:
(a) All the properties of a rectangle are there in a square.
(b) All the properties of a parallelogram are there in a rectangle.
(c) All the properties of a rhombus are there in a square.
(d) All the four sided closed plane figure are known as quadrilateral.
(e) All the properties of a parallelogram are there in a square.

Ex 5.7 Class 6 Maths Question 3.
A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Solution:
A regular quadrilateral is a square, regular pentagon, regular hexagon, regular octagon etc.

Ex 5.8 Class 6 Maths Question 1.
Examine whether the following are polygons. If any one among them is not, say why?
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 32
Solution:
(a) No, because it is not closed,
(b) Yes.
(c) No, because it is not drawn by using line segments.
(d) No, because it is not closed by line segments.

Ex 5.8 Class 6 Maths Question 2.
Name each polygon:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 33
make two more examples of each of these.
Solution:
(a) Quadrilateral
(b) Triangle
(c) Pentagon
(d) Octagon

Two more examples of :
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 34

Ex 5.8 Class 6 Maths Question 3.
Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Solution:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 35
Let ABCDEF be a rough sketch of a regular hexagon.
Join AC
A ∆ ABC is obtained in which AB = BC. Thus,
∆ ABC is an isosceles triangle.

Ex 5.8 Class 6 Maths Question 4.
Draw a rough sketch of a regular octagon.
(Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Solution:
Let ABCDEFGH be a rough sketch of a regular octagon. Join AC, CE, EG and GA to obtain a rectangle ACEG.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 36

 

Ex 5.8 Class 6 Maths Question 5.
A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Solution:
Let ABCDE be a rough sketch of a pentagon. To draw its diagonals, join AC, AD, BD, DE and CE. Then, AC, AD, BD, BE and CE are its diagonals.
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 37

 

Ex 5.9 Class 6 Maths Question 1.
Match the following:
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 38
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes 39
Give two new examples of each shape.
Solution:
Matching is as under:
(a) → (ii)
(b) → (iv)
(c) → (v)
(d) → (iii)
(e) → (i)

Two new examples of

(a) Cone: Conical tent.
(b) Sphere: Tennis ball, a ball of wool.
(c) Cylinder: Measuring jars, gas cylinder.
(d) Cuboid: Brick, Match box.
(e) Pyramid: Pyramids of Egypt.

Ex 5.9 Class 6 Maths Question 2.
What shape is
(a) Your instruments box?
(b) A brick?
(c) A match box?
(d) A road-roller?
(e) A sweet laddu?
Solution:
(a) Cuboid
(b) Cuboid
(c) Cuboid
(d) Cylinder
(e) Sphere

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Chapter-4 Basic Geometrical Ideas | Class 6th | NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 6 Maths: One of the most important and interesting things to keep in mind is that these solutions available are totally free of cost. This also includes all the solutions to the exercises that are given in the textbook. NCERT solutions for class 6 maths gives you chapter-wise solutions to each and every question. This can help you solve even the tougher with ease.

It is always good to have a strong foundation in order to build a good building. This also is true in class when you are still learning the basics of maths. NCERT solutions for class 6 maths provides you the exact opportunity to build a strong foundation in this subject. Below is an overview of each and every chapter covered in the NCERT textbook.

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas

Ex 4.1 Class 6 Maths Question 1.
Use the figure to name:
(a) Five points
(b) A line
(c) Four rays
(d) Five line segments
Solution:
Clearly, from the given figure
(a)
 5 points are points, O, C, B, D and E.
(b) A line is [altex]\overleftrightarrow { BD } [/latex]
(c) 4 rays are OC←→,OB←→,OD←→ and ED←→.
(d) 5 line segments are OE¯¯¯¯¯¯¯¯,OD¯¯¯¯¯¯¯¯,DE,¯¯¯¯¯¯¯¯¯OB¯¯¯¯¯¯¯¯ and BD¯¯¯¯¯¯¯¯.

 

Ex 4.1 Class 6 Maths Question 2.
Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.
Solution:
All the possible ways of naming the given by choosing only two letters at a time out of 4 letters are
AB←→,AC←→,AD←→,BC←→,BD←→,CD←→,DC←→,DB←→,CB←→,DA←→,CA←→ and BA←→.

Ex 4.1 Class 6 Maths Question 3.
Use the figure to name :
(a) Line containing point E.
(b) Line passing through A.
(c) Line on which O lies.
(d) Two pairs of intersecting lines.
Solution:
Clearly, from the given figure
(a) Line containing pqint E is AE←→.
(b) Line passing through A is AE←→.
(c) O lies on the line CB←→.
(d) Two pairs of intersecting lines are CO←→,AE←→ and AE←→,EF←→.

Ex 4.1 Class 6 Maths Question 4.
How many lines can pass through
(a) one given point?
(b) two given points?
Solution:
(a) An unlimited number of lines can be drawn passing through a given point.
(b) Exactly one line can be drawn passing through two different given points in a plane.

Ex 4.1 Class 6 Maths Question 5.
Draw a rough figure and label suitably in each of the following cases:
(a) Point P lies on AB¯¯¯¯¯¯¯¯.
(b) XY←→ and PQ←→ intersect at M.
(c) Line l contains E and F but not D.
(d) OP←→ and OQ←→ meet at O.
Solution:
A rough figure and labelled suitably for the given cases are as under:
(a) Point P lies on AB←→ :
(b)
 XY←→ and PQ←→ intersect at M:
(c)
Line l contains E and F but not D:
(d)
 OP←→ and OQ←→ meet at O.

Ex 4.1 Class 6 Maths Question 6.
Consider the following figure of line MN. Say whether following statements are true or false in context of the given figure.
(a) Q, M, O, N, P are points on the line MN←→− .

(b) M, O, N are points on a line segment MN¯¯¯¯¯¯¯¯¯¯.
(c) M and N are end points of line segment MN¯¯¯¯¯¯¯¯¯¯.
(d) O and N are end points of line segment OP¯¯¯¯¯¯¯¯.
(e) M is one of the end points of line segment QO¯¯¯¯¯¯¯¯.
(f) M is point on ray OP−→−.
(g) Ray OP−→− is different from ray QP−→−.
(h) Ray OP−→− is same as ray OM−→−.
(i) Ray OM−→− is not opposite to ray OP−→−.
(j) O is not an initial point of OP−→−.
(k) N is the initial point of NP−→− and NM−→−.
Solution:
In the context of the given figure, the given statement is:
(a) True
(b) True
(c) True
(d) False
(e) False
(f) False
(g) True
(h) False
(i) True
(j) False
(k) True

Ex 4.2 Class 6 Maths Question 1.
Classify the following curves as
(i) Open or
(ii) Closed.
Solution:
Open curves are (a) and (c) and closed curves are (b), (d) and (e).

 

Ex 4.2 Class 6 Maths Question 2.
Draw rough diagrams to illustrate the following:
(a) Open curve
(b) Closed curve
Solution:
Rough diagram to illustrate
(a) Open curve is

(b) Closed curve is

Ex 4.2 Class 6 Maths Question 3.
Draw any polygon and shade its interior.
Solution:
We know that in a closed curve, the interior is inside of the curve. Thus, shaded portion of the polygon indicate its interior.

Ex 4.2 Class 6 Maths Question 4.
Consider the given figure and answer the questions:
(a) Is it a curve?

(b) Is it closed?
Solution:
(a) Yes, the given figure represents a curve.
(b) Yes, the curve is closed.

Ex 4.2 Class 6 Maths Question 5.
Illustrate, if possible, each one of the following with a rough diagram:
(a) A closed curve that is not a polygon.
(b) An open curve made up entirely of line segments.
(c) A polygon with two sides.
Solution:
(a) Impossible, as a close plane figure bounded by lines is called a polygon.
(b) Yes, it may be as
(c)
 No, impossible, as a polygon of two sides cannot be drawn.

Ex 4.3 Class 6 Maths Question 1.
Name the angles in the given figure.
Solution:
The angles in the given figure are named as ∠ABC, ∠BCD, ∠CDA and ∠DAB.

 

Ex 4.3 Class 6 Maths Question 2.
In the given diagram, name the point(s)
(a) In the interior of ∠DOE
(b) In the exterior of ∠EOF
(c) On ∠EOF
Solution:
Clearly, from the given figure :
(a)
 The point A lies in the interior of ∠DOE. .
(b) The points A and C lie in the exterior of ∠EOF.
(c) The points E, B, O and F lie on ∠EOF.

Ex 4.3 Class 6 Maths Question 3.
Draw rough diagrams of two angles such that they have
(a) One point in common
(b) Two points in common
(c) Three points in common
(d) Four points in common
(e) One ray in common
Solution:
(a) The diagram is shown as below
Here, ∠ROQ and ∠QOP have one point O in common.
(b) The diagram is shown as below
Here, ∠MON and ∠ONR have two points O and N in common.
(c) Drawing a diagram of two angles, such that they have three points in common, is not possible.
(d) Drawing a diagram of two angles, such that they have four points in common, is not possible.
(e) In the figure given below, ∠SOT and ∠POT have one ray OT−→− in common.

Ex 4.4 Class 6 Maths Question 1.
Draw a rough sketch of a triangle ABC. Mark a point P in its interior and a point Q in its exterior. Is the point A in its exterior or in its interior?
Solution:
Rough sketch of a ∠ABC as shown.
Points P and Q are marked in the interior and exterior of ∆ABC.
Point A lies on the ∆ABC.

 

Ex 4.4 Class 6 Maths Question 2.
(a) Identify three triangles in the figure.
(b) Write the names of seven angles.
(c) Write the names of six line segments.
(d) Which two triangles have ZB as common?
Solution:
(a) Three triangles are identified in the figure as ∆ ABC, ∆ ABD and ∆ ACD.
(b) The names of seven angles are ∠BAD, ∠BAC, ∠CAD, ∠ABD, ∠ACD, ∠ADC and ∠ADB.
(c) The names of six line segments are AB, BD, DC, CA, BC and AD.
(d) ∆ ABC and ∆ ABD have ∠B as common.

Ex 4.5 Class 6 Maths Question 1.
Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them. Is the meeting point of the diagonals in the interior or exterior of the quadrilateral?
Solution:
PQRS is a quadrilateral having PR and QS as its diagonals intersecting at the point O, which is in the interior of the quadrilateral PQRS.
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 22
Ex 4.5 Class 6 Maths Question 2.
Draw a rough sketch of a quadrilateral KLMN. State,
(a) two pairs of opposite sides,
(b) two pairs of opposite angles,
(c) two pairs of adjacent sides,
(d) two pairs of adjacent angles.
Solution:
KLMN is a quadrilateral.
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 23
(a)
 KL, MN and LM, KN are two pairs of its opposite sides.
(b) ∠K, ∠M and∠L, ∠N are two pairs of opposite angles.
(c) KL, LM and KN, NM are two pairs of adjacent sides.
(d) ∠K, ∠N and ∠L, ∠M are two pairs of adjacent angles.

Ex 4.5 Class 6 Maths Question 3.
Investigate:
Use strips and fasteners to make a triangle and a quadrilateral.
Try to push inward at any one vertex of the triangle. Do the same to the quadrilateral.
Is the triangle distorted? Is the quadrilateral distorted? Is the triangle rigid?
Why is it that structures like electric towers make use of triangular shapes and not quadrilaterals?
Solution:
On pushing inward at any one vertex of the triangle, we find that the triangle is not distorted. Whereas doing so with the quadrilateral we find that it is distorted. Triangle is rigid. Thus, we make use of triangular shapes in structures like electric towers as triangular shapes are more rigid.

Ex 4.6 Class 6 Maths Question 1.
From the figure, identify:
(a) the centre of circle
(b) three radii
(c) a diameter
(d) a chord
(e) two points in the interior
(f) a point in the exterior
(g) a sector
(h) a segment
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 24
Solution:
(a) O is the centre of the circle.
(b) OA, OB and OC are three radii of circle.
(c) AC is a diameter of circle.
(d) ED is a chord of circle.
(e) Points O and P are in the interior of circle.
(f) Point Q is in the exterior of circle.
(g) OAB is a sector of circle.
(h) Shaded region in the interior of a circle enclosed by a chord ED.

Ex 4.6 Class 6 Maths Question 2.
(a) Is every diameter of a circle also a chord?
(b) Is every chord of a circle also a diameter?
Solution:
(a) Yes, a diameter is the longest chord.
(b) Not always.

Ex 4.6 Class 6 Maths Question 3.
Draw any circle and mark
(a) its centre
(b) a radius
(c) a diameter
(d) a Sector
(e) a segment
(f) a point in its interior
(g) a point in its exterior
(h) an arc
Solution:
Draw any circle and its various parts are as under :
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 25
(a) Its centre is O.
(b) Its radii are OA, OB and OC.
(c) Its diameter is AOC.
(d) Its sector is OAB.
(e) Shaded region in the interior of a circle enclosed by a chord PQ.
(f) M is a point in its interior.
(g) N is a point in its exterior.
(h) BC is an arc.

Ex 4.6 Class 6 Maths Question 4.
Say true or false:
(a) Two diameters of a circle will necessarily intersect.
(b) The centre of a circle is always in its interior.
Solution:
(a) True, because each diameter passes through the centre of a circle.
(b) True.

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Chapter-3 Playing With Numbers | Class 6th | NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 6 Maths: One of the most important and interesting things to keep in mind is that these solutions available are totally free of cost. This also includes all the solutions to the exercises that are given in the textbook. NCERT solutions for class 6 maths gives you chapter-wise solutions to each and every question. This can help you solve even the tougher with ease.

It is always good to have a strong foundation in order to build a good building. This also is true in class when you are still learning the basics of maths. NCERT solutions for class 6 maths provides you the exact opportunity to build a strong foundation in this subject. Below is an overview of each and every chapter covered in the NCERT textbook.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Ex 3.1 Class 6 Maths Question 1.
Write all the factors of the following numbers :
(a) 24
(b) 15
(c) 21
(d) 27
(e) 12
(f) 20
(g) 18
(h) 23
(i) 36
Solution:
(a) We have,
24 = 1 x 24
24 = 2 x 12
24 = 3 x 8
24 = 4 x 6
24 = 6 x 4
Stop here, because 4 and 6 have occurred earlier.
Thus, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

(b) We have,
15 = 1 x 15
15 = 3 x 5
15 = 5 x 3
Stop here, because 3 and 5 have occurred earlier.
Thus, all the factors of 15 are 1, 3, 5 and 15.

(c) We have,
21 = 1 x 21
21 = 3 x 7
21 = 7 x 3
Stop here, because 3 and 7 have occurred earlier.
Thus, all the factors of 21 are 1, 3, 7 and 21.

(d) We have,
27 = 1 x 27
27 = 3 x 9
27 = 9 x 3
Stop here, because 3 and 9 have occurred earlier.
Thus, all the factors of 27 are 1, 3, 9 and 27.

(e) We have,
12 = 1 x 12
12 = 2 x 6
12 = 3 x 4
12 = 4 x 3
Stop here, because 3 and 4 have occurred earlier.
Thus, all the factors of 12 are 1, 2, 3, 4, 6 and 12.

(f) We have,
20 = 1 x 20
20 = 2 x10
20 = 4 x 5
20 = 5 x 4
Stop here, because 4 and 5 have occurred earlier.
Thus, all the factors of 20 are 1, 2, 4, 5, 10 and 20.

(g) We have,
18 = 1 x 18
18 = 2 x 9
18 = 3 x 6
18 = 6 x 3
Stop here, because 3 and 6 have occurred earlier.
Thus, all the factors of 18 are 1, 2, 3, 6, 9 and 18.

(h) We have,
23 = 1 x 23
23 = 23 x 1
Stop here, because 1 and 23 have occurred earlier.
Thus, all the factors of 23 are 1 and 23.

(i) We have,
36 = 1 x 36
36 = 2 x 18
36 = 3 x 12
36 = 4 x 9
36 = 1 x 6
Stop here, because both the factors (6) are same.
Thus, all the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Ex 3.1 Class 6 Maths Question 2.
Write first five multiples of:
(a) 5
(b) 8
(c) 9
Solution:
(a) In order to obtain first five multiples of 5, we multiply it by 1, 2, 3, 4 and 5 respectively. ,r
We have,
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4=20
5 x 5 = 25
Hence, the first five multiples of 5 are 5, 10, 15, 20 and 25 respectively.

(b) In order to obtain first five multiples of 8, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
8 x 1=8
8 x 2=16
8 x 3=24
8 x 4 = 32
8 x 5 = 40
Hence, the first five multiples of 8 are 8, 16, 24, 32 and 40 respectively,

(c) In order to obtain first five multiples of 9, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
Hence, the first five multiples of 9 are 9, 18, 27, 36 and 45 respectively.

Ex 3.1 Class 6 Maths Question 3.
Match the items in column 1 with the items in column 2.
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 1
Solution:
Matching is as under :
(i) → (c) ∵ 35 x 2 = 70
(ii) →(d) ∵ 30 = 15 = 2
(iii) → (a) ∵ 8 x 2 = 16
(iv)→ (f) ∵ 20 = 20 = 1
(v) → (e) ∵ 50 = 25 = 2

Ex 3.1 Class 6 Maths Question 4.
Find all the multiples of 9 upto 100.
Solution:
All the multiples of 9 upto 100 are
9 x 1, 9 x 2, 9 x 3, 9 x 4, 9 x 5, 9 x 6, 9 x 7, 9 x 8, 9 x 9, 9 x 10 and 9 x 11.
i. e., 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

Ex 3.2 Class 6 Maths Question 1.
What is the sum of any two
(a) Odd numbers?
(b) Even numbers?
Solution:
(a) Sum of two odd numbers is even.
(b) Sum of two even numbers is even.

Ex 3.2 Class 6 Maths Question 2.
State whether the following statements are True or False :
(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.
Solution:
(a) False
(b) True
(c) True
(d) False
(e) False
(f) False
(g) False
(h) True
(i) False
(j) True

Ex 3.2 Class 6 Maths Question 3.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.
Solution:
By the Sieve of Eratosthenes method find the prime numbers between 1 and 100. We find that these are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.
Out of these, a pair of prime numbers having same digits are 13, 31; 17, 71; 37, 73, 79, 97.

Ex 3.2 Class 6 Maths Question 4.
Write down separately the prime and composite numbers less than 20.
Solution:
Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19.
Composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18.

Ex 3.2 Class 6 Maths Question 5.
What is the greatest prime number between 1 and 10?
Solution:
Prime numbers between 1 and 10 are 2, 5 and 7.
∴ Greatest prime number between 1 and 10 = 7

Ex 3.2 Class 6 Maths Question 6.
Express the following as the sum of two odd primes
(a) 44
(b) 36
(c) 24
(d) 18
Solution:
(a) 44 = 13 + 31
(b) 36 = 5 + 31
(c) 24 = 11 +13
(d) 18 = 7+11
Note : In 1742, mathematician Goldbach had a conjecture (guess) for which he could not provide a proof. It may be stated as “Every even number greater than 4 can be expressed as the sum of two odd prime numbers”.

Ex 3.2 Class 6 Maths Question 7.
Give three pairs of prime numbers whose difference is 2.
Solution:
Three pairs of prime number whose difference is 2 are 3, 5; 5, 7 and 11, 13.
Note : Two prime numbers are known as twin-primes if there is one composite number between them. In other words, two prime numbers whose difference is 2 are called twin-primes.

Ex 3.2 Class 6 Maths Question 8.
Which of the following numbers are prime? ,
(a) 23
(b) 51
(c) 37
(d) 26
Solution:
(a) We find that 23 is not exactly divisible by any of the prime numbers 2, 3, 5, 7 and 11 (i.e., upto half of 23). So, it is a prime number.
(b) We find that 51 is divisible by 3. So, it is not a prime number.
(c) We find that 37 is not exactly divisible by any of the prime numbers 2, 3, 5, 7, 11, 13 and 17 (i.e., upto half of 37). So, it is a prime number.
(d) We find that 26 is exactly divisible by 2 and 13. So, it is not a prime number.

Ex 3.2 Class 6 Maths Question 9.
Write seven consecutive composite numbers less-than 100 so that there is no prime number between them.
Solution:
Seven consecutive composite numbers less than 100 so that there is no prime number between them are 90, 91, 92, 93, 94, 95 and 96.

Ex 3.2 Class 6 Maths Question 10.
Express each of the following numbers as the sum of three odd primes:
(a) 21
(b) 31
(c) 53
(d) 61
Solution:
Expressing the given numbers as the sum of three odd primes, we have
(a) 21 = 3 + 5 + 13
(b) 31 = 3 + 5 + 23
(c) 53 = 3 + 19 + 31
(d) 61 = 3 + 11 + 47

Ex 3.2 Class 6 Maths Question 11.
Write five pairs of prime numbers less than 20 whose sum is divisible by 5.
Solution:
Prime numbers below 20 are 2, 3, 5, 7, 13, 17 and 19.
Possible sum of pairs of these numbers :
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 2
Clearly, five pairs of prime numbers whose sum is divisible by 5 are 2, 3; 2, 13; 3, 7; 3,17 and 7, 13.

Ex 3.2 Class 6 Maths Question 12.
Fill in the blanks :
(a) A number which has only two factors is called a
(b) A number which has more than two factors is called a
(c) 1 is neither nor
(d) The smallest prime number is
(e) The smallest composite number is
(f) The smallest even number is
Solution:
(a) Prime
(b) composite
(c) prime, composite
(d) 2
(e) 4
(f) 2

Ex 3.3 Class 6 Maths Question 1.
Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 4; by 11 (say yes or no) :
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 3
Solution:
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 4

Ex 3.3 Class 6 Maths Question 2.
Using divisibility tests, determine which of the following numbers are divisible by 4; by 8 :
(a) 572
(b) 726352
(c) 5500
(d) 6000
(e) 12159
(f) 14560
(g) 21084
(h) 31795072
(i) 1700
(j) 2150
Solution:
We know that a number is divisible by 4, if the number formed by its digits in ten’s and unit’s place is divisible by 4.
(a) In 572, 72 is divisible by 4. So, 572 is divisible by 4.
(b) In 726352, 52 is divisible by 4. So, it is divisible by 4.
(c) In 5500, 00 is divisible by 4. So, it is divisible by 4.
(d) In 6000, 00 is divisible by 4. So, it is divisible by 4.
(e) In 12159, 59 is not divisible by 4. So, it is not divisible by 4.
(f) In 14560, 60 is divisible by 4. So, it is divisible by 4.
(g) In 21084, 84 is divisible by 4. So, it is divisible by 4.
(h) In 31795072, 72 is divisible by 4. So, it is divisible by 4.
(i) In 1700,00 is divisible by 4. So, it is divisible by 4.
(j) In 2150, 50 is not divisible by 4. So, it is not divisible by 4.

Also, we know that a number is divisible by 8, if the number formed by its hundred’s, ten’s and unit’s places is divisible by 8.

(a) 572 is not divisible by 8.
(b) In 726352, 352 is divisible by 8. So, it divisible by 8.
(c) In 5500, 500 is not divisible by 8. So, it is not divisible by 8.
(d) In 6000, 000 is divisible by 8. So, it is divisible by 8.
(e) In 12159, 159 is not divisible by 8. So, it is not divisible by 8.
(f) In 14560, 560 is divisible by 8. So, it is divisible by 8.
(g) In 21084, 084 is not divisible by 8. So, it is not divisible by 8.
(h) In 31795072, 072 is divisible by 8. So, it is divisible by 8.
(i) In 1700, 700 is not divisible by 8. So, it is not divisible by 8.
(j) In 2150, 150 is not divisible by 8. So, it is not divisible by 8.

Ex 3.3 Class 6 Maths Question 3.
Using divisibility tests, determine which of the following
numbers are divisible by 6 : .
(a) 297144
(b) 1258
(c) 4335
(d) 61233
(e) 901352
(f) 438750
(g) 1790184
(h) 12583
(i) 639210
(j) 17852
Solution:
We know that a number is divisible by 6, if it is divisible by 2 and 3 both.
(a) Given number = 297144
Its unit’s digit is 4. So, it is divisible by 2.
Sum of its digits = 2+ 9+ 7 + 1 + 4 + 4 = 27, which is divisible by 3.
∴ 297144 is divisible by 6.

(b) Given number =1258
Its unit’s digit is 8. So, it is divisible by 2.
Sum of its digits = 1+ 2 + 5 + 8 = 16, which is not divisible by 3.
∴ 1258 is not divisible by 6.

(c) Given number = 4335 .
Its unit’s digit is 5. So, it is not divisible by 2.
∴ 4335 is also not divisible by 6.

(d) Given number = 61233
Its unit’s digit is 3. So, it is not divisible by 2.
∴ 61233 is also not divisible by 6.

(e) Given number = 901352
Its unit’s digit is 2. So, it is divisible by 2.
Sum of its digits = 9+ 0 + 1 + 345 + 2 = 20, which is not divisible by 3.
∴ 901352 is not divisible by 6.

(f) Given number = 438750
Its unit’s digit is 0. So, it is divisible by 2.
Sum of its digits = 4+ 3 + 8 + 7 + 5 + 0=27, which is divisible by 3.
∴ 438750 is divisible by 6.

(g) Given number = 1790184
Its unit’s digit is 4. So, it is divisible by 2.
Sum of its digits = 1+ 7 + 9+ 0+ 1+ 8 + 4 = 30, which is divisible by 3.
∴ 1790184 is divisible by 6.

(h) Given number = 12583 .
Its unit’s digit is 3. So, it is not divisible by 2.
∴ 12583 is not divisible by 6.

(i) Given number = 639210
Its unit’s digit is 0. So, it is divisible by 2.
Sum of its digits = 6+ 3+ 9+ 2 + 1 + 0 = 21, which is divisible by 3.
∴ 639210 is divisible by 6.

(j) Given number = 17852
Its unit’s digit is 2. So, it is divisible by 2.
Sum of its digits = 1 + 7 + 8 + 5 + 2 = 23, which is not divisible by 3.
∴ 17852 is not divisible by 6.

Ex 3.3 Class 6 Maths Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 11:
(a) 5445
(b) 10824
(c) 7138965
(d) 70169308
(e) 10000001
(f) 901153
Solution:
We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
(a) Given number = 5445
Sum of its digits at odd places = 5 + 4 = 9
Sum of its digit at even places = 4 + 5 = 9
Difference of these two sums = 9 – 9 = 0
∴ 5445 is divisible by 11.

(b) Given number = 10824
Sum of its digits at odd places = 4+ 8 + 1 = 13
Sum of its digits at even places =2 + 0 =2
Difference of these two sums =13 – 2 = 11, which is a multiple of 11.
∴ 10824 is divisible by 11.

(c) Given number = 7138965
Sum of its digits at odd places = 5+ 9+ 3 + 7= 24
Sum of its digits at even places = 6+ 8 + 1 = 15
Difference of these two sums = 24 – 15 = 9, which is not a multiple of 11.
∴ 7138965 is not divisible by 11.

(d) Given number = 70169308
Sum of its digits at odd places = 8 + 3 + 6 + 0=17
Sum of its digits at even places = 0 + 9 + 1 + 7 = 17
Difference of these two sums =17 – 17 = 0
∴ 70169308 is divisible by 11.

(e) Given number = 10000001
Sum of its digits at odd places = 1 + 0 + 0 + 0 = 1
Sum of its digits at even places = 0 + 0 + 0 + 1 = 1
Difference of these two sums = 1 – 1 = 0
∴ 10000001 is divisible by 11.

(f) Given number = 901153
Sum of its digits at odd places = 3 + 1 + 0 = 4
Sum of its digits at even places = 5 + 1 + 9=15
Difference of these two sums =15 – 4 = 11,
which is a multiple of 11.
∴ 901153 is divisible by 11.

Ex 3.3 Class 6 Maths Question 5.
Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3 :
(a) … 6724
(b) 4765 … 2
Solution:
We know that a number divisible by 3, if the sum of its digits is divisible by 3.
(a) … 6724
For … 6724, we have 6 + 7 + 2 + 4 =19, we add 2 to 19, the resulting number 21 will be divisible by 3.
∴ The required smallest digit is 2.
Again, if we add 8 to 19, the resulting number 27 will be divisible by 3. .’. The required largest digit is 8.

(b) 4765 … 2
For 4765 … 2, we have 4 + 7 + 6 + 5 + 2 = 24, it is divisible by 3.
Hence the required smallest digit is 0.
Again, if we add 9 to 24, the resulting number 33 will be divisible by 3.
∴ The required largest digit is 9.

Ex 3.3 Class 6 Maths Question 6.
Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11:
(a) 92 … 389
(b) 8 … 9484
Solution:
We know that a number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or divisible by 11.
(a) 92 … 389
For 92 … 389, sum of the digits at odd places and sum of digits at even places
= 9 + 3 + 2 = 14
= 8 + required digit + 9
= required digit+ 17
Difference between these sums
= required digit + 17 – 14
= required digit + 3
For (required digit + 3) to become 11, we must have the required digit as 8 (∵ 3+ 8 gives 11).
Hence, the required smallest digit = 8 .

(b) 8…9484
For 8 … 9484, sum of the digits at odd places
= 4 + 4 + required digit
= 8 + required digit
and sum of digits at even places
= 8 + 9 + 8 = 25
Difference between these sums
= 25 – (8 + required digit)
= 17- required digit
For (17 — required digit) to become 11 we must have the required digit as 6 (∵ 17-6 =11).
Hence the required smallest digit = 6

Ex 3.4 Class 6 Maths Question 1.
Find the common factors of:
(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120
Solution:
(a) We have, 20 = 1 x 20
= 2 x 10
= 4 x 5
∴ All the factors of 20 are 1, 2, 4, 5, 10 and 20
Again, 28 = 1 x 28
28 = 2 x 14
28 = 4 x 7
∴ All the factors of 28 are 1, 2, 4, 7, 14 and 28.
Out of these 1, 2 and 4 occur in both the lists.
∴ 1, 2 and 4 are common factors of 20 and 28.

(b) We have, 15 = 1 x 15
15 = 3 x 5
∴ All the factors of 15 are 1, 3, 5 and 15.
Again, 25 = 1 x 25
25 = 5 x 5
∴ All the factors of 25 are 1, 5 and 25.
Out of these 1 and 5 occur in both the lists.
∴1 and 5 are common factors of 15 and 25.

(c) We have, 35 = 1 x 35
35 = 5 x 7
∴ All the factors of 35 are 1, 5, 7 and 35.
Again, 50 = 1 x 50
50 = 2 x 25
50 = 5 x 10
∴ All the factors of 50 are 1, 2, 5, 10, 25 and 50. ,
Out of these 1 and 5 occur in both the lists.
∴ 1 and 5 are common factors of 35 and 50.

(d) We have, 56 = 1 x 56
56 = 2 x 28
56 = 4 x 14
56 = 7 x 8
∴ All the factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.
Again, 120 = 1 x 120
120 = 2 x 60
120 = 3 x 40
120 = 4 x 30
120 = 5 x 24
120 = 6 x 20
120 = 8 x 15
120 = 10 x 12
∴ All the factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
Out of these 1, 2, 4 and 8 occur in both the lists.
∴ 1, 2, 4 and 8 are common factors of 56 and 120.

Ex 3.4 Class 6 Maths Question 2.
Find the common factors of:
(a) 4, 8 and 12
(b) 5, 15 and 25
Solution:
(a) We have, 4=1 x 4
4 = 2 x 2
∴ All the factors of 4 are 1, 2 and 4.
Again, 8 = 1 x 8
8 = 2 x 4
∴All the factors of 8 are 1, 2, 4 and 8.
Again, 12 = 1 x 12
12 = 2 x 6
12 = 3 x 4
∴ All the factors of 12 are 1, 2, 3, 4, 6 and 12.
Out of these 1, 2 and 4 occur in all the three lists.
∴ 1, 2 and 4 are common factors of 4, 8 and 12.

(b) We have, 5 = 1 x 5
∴All the factors of 5 are 1 and 5.
15 = 1 x 15
15 = 3 x 5
∴ All the factors of 15 are 1, 3, 5 and 15.
25 = 1 x 25
25 = 5 x 5
∴ All the factors of 25 are 1, 5 and 25.
Out of these 1 and 5 occur in all the three lists.
∴ 1 and 5 are common factors of 5, 15 and 25.

Ex 3.4 Class 6 Maths Question 3.
Find first three common multiples of:
(a) 6 and 8
(b) 12 and 18
Solution:
(a) Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,…
Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72,…
Out of these 24, 48, 72, … occur in both thfe lists.
∴ The first three common multiples of 6 and’8 are 24, 48 and 72.
(b) Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, …
Multiples of 18 are 18, 36, 54, 72, 90, 108, …
Out of these 36, 72, 108, … occur in both the lists.
∴ The first three common multiples of 12 and 18 are 36, 72 and 108.

Ex 3.4 Class 6 Maths Question 4.
Write all the numbers less than 100 which are corrimon multiples of 3 and 4.
Solution:
Common multiples of 3 and 4 are multiples of 3 x 4 i. e., 12.
∴ Common multiples of 3 and 4 less than 100 are 12, 24, 36,48, 60, 72, 84 and 96.

Ex 3.4 Class 6 Maths Question 5.
Which of the following numbers are co-prime?
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16
Solution:
(a) Factors of 18 are 1, 2, 3, 6, 9 and 18 and, that of 35 are 1, 5, 7 and 35.
∴ Common factor of 18 and 35 is 1.
Thus, 18 and 35 are co-prime.

(b) Factors of 15 are 1, 3, 5 and 15 and, that of 37 are 1 and 37.
∴ Common factor of 15 and 37 is 1.
Thus, 15 and 37 are co-prime.

(c) Since 5 is a common factor of 30 and 415.
∴ 30 and 415 are not co-prime.

(d) ∴ 68 + 17 = 4 i.e., 17 is a common factor of 17 and 68.
∴ 17 and 68 are not co-prime.

(e) Since 1 is the only common factor of 216 and 215.
∴ 216 and 215 are co-prime.

(f) Since 1 is the only common factor of 81 and 16.
∴ 81 and 16 are co-prime.

Ex 3.4 Class 6 Maths Question 6.
A number is divisible by both 5 and 12. By which other number will that number be always divisible?
Solution:
Since a number is divisible by both 5 and 12.
So, it is also divisible by 5 x 12 i. e., 60.

Ex 3.4 Class 6 Maths Question 7.
A number is divisible by 12. By what other numbers will that number be divisible?
Solution:
Factors of 12 are 1, 2, 3, 4 and 12.
Since a number is divisible by 12. So, it is also divisible by the factors of 12.
Thus, the number is also divisible by 2, 3 and 4.

Ex 3.5 Class 6 Maths Question 1.
Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a number is divisible by 9, it must be divisible by 3.
(c) A number is divisible by 18, if it is divisible by both 3 and 6.
(d) If a number is divisible by 9 and 10 both, then it must be , divisible by 90.
(e) If two numbers are co-primes’, at least one of them must be prime.
(f) All numbers which are divisible by 4 must also be divisible by 8.
(g) All numbers which are divisible by 8 must also be divisible by 4.
(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Solution:
Statements (b), (d), (g) and (i) are true.

Ex 3.5 Class 6 Maths Question 2.
Here are two different factor trees for 60. Write the missing numbers.
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 5
Solution:
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 6

Ex 3.5 Class 6 Maths Question 3.
Which factors are not included in the prime factorisation of a
composite number?
Solution:
1 and composite numbers are not included in the prime factorisation of a composite number.

Ex 3.5 Class 6 Maths Question 4.
Write the greatest 4-digit number and express it in terms of its prime factors.
Solution:
The greatest 4-digit number = 9999
We have,
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 7
∴ 9999 = 3 x 3 x 11 x 101

Ex 3.5 Class 6 Maths Question 5.
Write the smallest 5-digit number and express it in the form of its prime factors.
Solution:
Smallest 5-digit number = 10000
We have,
tiwari academy class 6 maths Chapter 3 Playing with Numbers 8
∴ 10000=2 x 2 x 2 x 2 x 5 x 5 x 5 x 5

Ex 3.5 Class 6 Maths Question 6.
Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.
Solution:
We have,
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 9
∴ 1729 = 7 x 13 x 19
Relation between two consecutive prime factors is stated as “difference between two consecutive prime factors is 6”.

Ex 3.5 Class 6 Maths Question 7.
The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Solution:
Let us consider the product of three consecutive numbers as under :
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 10
In all the products, we have 6 or 4 or 0 in the unit’s place, so each product is divisible by 2.
Also, sum of digits in these products are divisible by 3. So, each of the product is divisible by 3.
Since 2 and 3 are co-prime, so the product 2 x 3 = 6 divides each of the above products.
Thus, the product of three consecutive numbers is always divisible by 6.

Ex 3.5 Class 6 Maths Question 8.
The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of same examples.
Solution:
3 + 5 = 8 and 8 is divisible by 4.
5 + 7 = 12 and 12 is divisible by 4.
7 + 9=16 and 16 is divisible by 4.
9 +11 = 20 and 20 is divisible by 4.

Ex 3.5 Class 6 Maths Question 9.
In which of the following expressions, prime factorisation has been done ? ‘
(a) 24 = 2 x 3 x 4
(b) 56 = 7 x 2 x 2 x 2
(c) 70 = 2 x 5 x 7
(d) 54 = 2 x 3 x 9
Solution:
In (b) and (c) prime factorisation has been done.

Ex 3.5 Class 6 Maths Question 10.
Determine if 25110 is divisible by 45.
Solution:
Since 45 = 5 x 9, where 5 and 9 are co-primes.
So to check the divisibility of 25110 by 45, test it for 5 and 9.
In 25110, unit’s digit = 0
So, 25110 is divisible by 5.
Sum of its digits = 2 + 5 + 1 + 1+ 0 = 9, which is divisible by 9.
So, 25110 is also divisible by 9.
Hence, 25110 is divisible by 45.

Ex 3.5 Class 6 Maths Question 11.
18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 x 6 = 24 ? If not, give an example to justify your answer.
Solution:
Not necessarily, as the numbers 12,36,60 etc. are each divisible by both 4 and 6. But these numbers are not divisible by 4 x 6 = 24.

Ex 3.5 Class 6 Maths Question 12.
I am the smallest number, having four different prime factors. Can you find me?
Solution:
Smallest four different prime numbers are 2, 3, 5 and 7.
∴ Required number = 2 x 3 x 5 x 7 = 210.

Ex 3.6 Class 6 Maths Question 1.
Find the HCF of the following numbers :
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75
Solution:
(a) First, we write the prime factorisation of each of the given numbers. We have,
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 11
∴ 18 = 2 x 3 x 3
and 48 = 2 x 2 x 2 x 2 x 3
We find that 2 and 3 each occurs as a common factor in the given numbers at least once.
∴ Required HCF = 2 x 3 = 6

(b) First, we write the prime factorisation of each of the given numbers.
We have,
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 12
∴ 30 = 2 x 3 x 5
and 42 = 2 x 3 x 7
We find that 2 and 3 each occurs as a common factor in the given numbers at least once.
Required HCF = 2 x 3 = 6

(c) First, we write the prime factorisation of each of the given numbers.
We have,
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 13
∴ 18 = 2 x 3 x 3
and 60 = 2 x 2 x 3 x 5
We find that 2 and 3 each occurs as a common factor in

(d) 27,63
Factors of 27 are 1, 3, 9 and 27.
Factors of 63 are 1, 3, 7, 9, 21 and 63.
Common factors of 27 and 63 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 27 and 63 is 9.

(e) 36,84
Factors of 36 are 1, 2, 3,4, 6, 9,12, 18 and 36.
Factors of 84 are 1,2, 3,4,6,7, 12, 14,21, 28, 42 and 84.
Common factors of 36 and 84 are 1,2, 3,4, 6 and 12.
Highest of these common factors is 12.
∴ H.C.F. of 36 and 84 is 12. if) 34,102 • ‘
Factors of 34 are 1,2, 17 and 34.
Factors of 102 are 1, 2, 3,6,17, 34. 51 and 102.
∴ Common factors of 34 and 102 are 1, 2, 17 and 34.
Highest of these common factors is 34.
∴ H.C.F. of 34 and 102 is 34.

(g) 70,105,175
Factors of 70 are 1. 2, 5. 7, 10. 14, 35 and 70.
Factors of 105 are 1, 3, 5. 7. 15. 21. 35 and 105.
Factors of 175 are 1. 5, 7. 25. 35 and 175. .’. Common factors of 70, 105 and 175 are 1, 5 and 35.
Highest of these common factors is 35.
∴ H.C.F. of 70. 105 and 175 are 35.

(h)
 91,112,49
Factors of 91 are 1,7, 13 and 91.
Factors of 112 are 1,2. 4. 7, 8, 14. 16. 28, 56 and 112.
Factors of 49 are 1.7 and 49.
Common factors of 91,112 and 49 are 1 and 7.
Highest of these common factors is 7.
∴ H.C.F. of 91, 112 and 49 is 7.

(i) 18,54,81
Factors of 18 are 1. 2, 3. 6, 9 and 18. Factors of 54 are 1, 2. 3, 6. 9, 18. 27 and 54.
Factors of 81 are 1. 3, 9, 27 and 81.
∴ Common factors of 18,54 and 81 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 18, 54 and 81 is 9.
(j) 12, 45, 75
Factors of 12 are 1, 2, 3, 4, 6 and 12. Factors of 45 are 1, 3, 5, 9, 15 and 45.
: Factors of 75 are 1, 3, 5, 15, 25 and 75.
∴ Common factors of 12,45 and 75 are 1 and 3.
Highest of these common factors is 3.
H.C.F. of 12. 45 and 75 is 3.

Ex 3.6 Class 6 Maths Question 2.
What is the H.C.F. of two consecutive :
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution :
(a) The H.C.F. of two consecutive numbers is 1.
(b) The H.C.F. of two consecutive even numbers is 2.
(c) The H.C.F. of two consecutive odd numbers is 1.

Ex 3.6 Class 6 Maths Question 3.
H. C.F. of co-prime numbers 4 and 15 was found as follows factorization: 4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F?
Solution :
No, the answer is not correct. The correct answer is as follows :
H.C.F. of 4 and 15 is 1.

Ex 3.7 Class 6 Maths Question 1.
Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer e×act number of times.
Solution :
Factors of 75 are 1, 3, 5, 15, 25 and 75.
Factors of 69 are 1, 3, 23 and 69.
∴ Common factors of 75 and 69 are 1 and 3.
Highest of these common factors is 3.
∴ H.C.F. of 75 and 69 is 3.
Hence, the maximum value of weight which can measure the weight of the fertilizer e×act number of times is 3 kg.

Ex 3.7 Class 6 Maths Question 2.
Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm, and 77 cm, respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
Solution :
NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers 27
∴ L.C.M. of 63, 70 and 77
= 2 × 3 × 3 × 5 × 7 × 11 = 6930.
Hence, the minimum distance each should cover so that all can cover the distance in complete steps is 6930 cm.

Ex 3.7 Class 6 Maths Question 3.
The length, breadth, and height of a room are 825 cm, 675 cm, and 450 cm, respectively. Find the longest tape which can measure the three dimensions of the room exactly.
Solution :
Factors of 825 are 1, 3, 5, 11, 15, 25, 33, 55,75,165, 275 and 825.
Factors of 675 are 1, 3, 5,9,15, 25, 27,45,75, 135, 225 and 675.
Factors of 450 are 1,2,3,5,6,9,10,15,18,25, 30,45, 50, 75, 90, 150, 225 and 450.
∴ Common factors of 825, 675 and 450 are 1,3,5,15, 25 and 75.
Highest of these common factors is 75.
Hence, the length of the longest tape which can measure the three dimensions of the room exactly is 75 cm.

Ex 3.7 Class 6 Maths Question 4.
Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Solution :
tiwari academy class 6 maths Chapter 3 Playing With Numbers 28
∴ L.C.M. of 6, 8 and 12 = 2 × 2 × 2 × 3 = 24. Multiples of 24 are 24,48,72,96,120,144,
Hence, the smallest 3-digit number which is exactly divisible by 6, 8 and 12 is 120.

Ex 3.7 Class 6 Maths Question 5.
Determine the largest 3-digit number exactly divisible by 8, 10 and 12.
Solution :
NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers 29
∴ L.C.M. of 8,10 and 12 = 2×2×2×3×5
= 120.
Multiples of 120 are :
120 × 1 = 120,120 × 2 = 240,120 × 3 = 360,120 × 4 = 480,120 × 5 = 600,120 × 6 = 720,120 × 7 = 840,
120 × 8 = 960,120 × 9 = 1080,
Hence, the largest 3-digit number exactly divisible by 8, 10 and 12 is 960.

Ex 3.7 Class 6 Maths Question 6.
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 am at what time will they change simultaneously again?
Solution :
NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers 30
∴ L.C.M. of 48,72 and 108 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432.
432 seconds = 7 min 12 seconds.
Hence, they will change simultaneously again after 7 min 12 seconds from 7 a.m.

Ex 3.7 Class 6 Maths Question 7.
Three tankers contain 403 liters, 434 liters and 465 liters of diesel respectively. Find the ma×imum capacity of a container that can measure the diesel of the three containers e×act a number of times.
Solution :
Factors of 403 are 1, 13, 31 and 403. Factors of 434 are 1, 2, 7, 14, 31, 62, 217 and 434.
Factors of 465 are 1, 3, 5, 15, 31, 93, 155 and 465.
Common factors of 403,434 and 465 are 1 and 31.
Highest of these common factors is 31.
∴ H.C.F. of 403. 434 and 465 is 31.
Hence, the maximum capacity of the container that can measure the diesel of the three containers an e×act number of times is 31 litres.

Ex 3.7 Class 6 Maths Question 8.
Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Solution :
NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers 31
∴ L.C.M. of 6, 15 and 18 = 2 × 3 × 3 × 5 = 90.
Hence, the required number is 90 + 5 i.e., 95.

Ex 3.7 Class 6 Maths Question 9.
Find the smallest four digit number which is divisible by 18, 24 and 32.
Solution :
NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers 32
∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288.
Multiples of 288 are :
288 × 1 = 288, 288 × 2 = 576, 288 × 3 = 864, 288×4= 1152,
Hence, the smallest four digit number which is divisible by 18, 24 and 32 is 1152.

Ex 3.7 Class 6 Maths Question 10.
Find the L.C.M. of the following numbers:
(a) 9 and 4
(b) 12 and 5
(c) 6 and 5
(d) 15 and 4.
Observe a common property in the obtained L.C.M.s. Is L.C.M. the product of two numbers in each case?
Solution :
(a) 9 and 4
NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers 33
∴ L.C.M. of 9 and 4 = 2 × 2 × 3 × 3 = 36 (= 9 × 4).

(b) 
12 and 5
tiwari academy class 6 maths Chapter 3 Playing With Numbers 34
∴ L.C.M. of 12 and 5 = 2×2×3×5 = 60 (= 12 × 5).

(c)
 6 and 5
NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers 35
∴ L.C.M. of 6 and 5 = 2×3×5 = 30 (= 6 × 5).

(d)
 15 and 4
NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers 36
∴ L.C.M. of 15 and 4 = 2 × 2 × 3 × 5 = 60 (=15×4).
We observe a common property in the obtained L.C.M.’s that L.C.M. is the product of two numbers in each case.

Ex 3.7 Class 6 Maths Question 11.
Find the L.C.M. of the following numbers in which one number is the factor of the other.
(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45.
What do you observe in the results obtained?
Solution :
(a) 5, 20
Prime factorisations of 5 and 20 are as follows: 5 = 5
20 = 2 × 2 × 5 ∴ L.C.M. of 5 and 20
=2×2×5 = 20.

(b)
 6, 18
Prime factorisations of 6 and 18 are as follows:
6 = 2×3 18 = 2×3×3
∴ L.C.M. of 6 and 18 = 2×3×3 = 18.

(c)
 12, 48
Prime factorisations of 12 and 48 are as follows:
12 = 2 × 2 × 3
48 = 2×2×2×2×3
∴ L.C.M. of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48.

(d)
 9, 45
Prime factorisations of 9 and 45 are as follows:
9 = 3×3
45 = 3 × 3 × 5
∴ L.C.M. of 9 and 45 = 3 × 3 × 5 = 45.
In the results obtained, we observe that L.C.M. of the two numbers in which one number is the factor of the other is the greater number.

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Chapter-2 Whole Numbers | Class 6th | NCERT Solutions | Edugrown

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers

Exercise 2.1

 

Question 1.
Write the next three natural numbers after 10999.
Solution:
The next three natural numbers after 10999 are 11000, 11001 and 11002. 

Question 2.
Write three whole numbers occurring just before 10001.
Solution:
10001 – 1 = 10000
10000 – 1 = 9999
9999 – 1 = 9998
Hence, three whole numbers just before 10001 are 10000, 9999 and 9998.

Question 3.
Which is the smallest whole number?
Solution:
0 is the smallest whole number.

Question 4.
How many whole numbers are there between 32 and 53?
Solution:
The whole numbers between 32 and 53 are 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52.

Question 5.
Write the successor of:
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
Solution:
(a) Successor of 244070 is 244070 + 1 = 244071
Hence, successor of 244070 is 244071.
(b) Successor of 100199 is 100199 + 1 = 100200
Hence, successor of 100199 is 100200.
(c) Successor of 1099999 is 1099999 + 1 = 1100000
Hence, successor of 1099999 is 1100000.
(d) Successor of 2345670 is 2345670 + 1 = 2345671
Hence, successor of 2345670 is 2345671

 Question 6.
Write the predecessor of:
(a) 94
(b) 10000
(c) 208090
(d) 7654321
Solution:
(a) Predecessor of 94 is 94 – 1 = 93
Hence, predecessor of 94 is 93.
(b) Predecessor of 1000 is 10000 – 1 = 9999
Hence, predecessor of 10000 is 9999.
(c) Predecessor of 208090 is 208090 -1 = 208089
Hence, predecessor of 208090 is 208089.
(d) Predecessor of 7654321 is 7654321 – 1 = 7654320
Hence, predecessor of 7654321 is 7654320.

 Question 7.
In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them,
(a) 530, 503
(b) 370, 307
(c) 98765, 56789
(d) 9830415,10023001
Solution:
We know that the smaller number is always on the left side of the greater number on number line.
(a) 530, 503
Clearly 503 is smaller than 530.
Hence, 503 will be on left side of 530 on number line.
Expression: 503 < 530 or 530 > 503

 

(b) 307 < 370
Clearly 307 is smaller than 370.
Hence, 307 will be on the left side of 370 on number line.
Expression: 307 < 370 or 370 > 307.

(c) 98765, 56789
Clearly 56789 is smaller than 98765.
Hence, 56789 will be on left side of 98765 on number line.
Expression: 56789 < 98765 or 98765 > 56789.

(d) 9830415, 10023001
Clearly, 9830415 is smaller than 10023001
Hence, 9830415 will be on the left side of 10023001 on the number line.
Expression: 9830415 < 10023001 or 10023001 > 9830415.

Question 8.
Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
(b) 400 is the predecessor of 399.
(c) Zero is the smallest whole number.
(d) 600 is the successor of 599.
(e) All natural numbers are whole numbers.
(f) All whole numbers are natural numbers.
(g) The predecessor of a two-digit number is never a single-digit number.
(h) 1 is the smallest whole number.
(i) The natural number 1 has no predecessor.
(j) The whole number 1 has no predecessor.
(k) The whole number 13 lies between 11 and 12.
(l) The whole number 0 has no predecessor.
(m) The successor of a two-digit number is always a two-digit number.
Solution:
(a) This statement is false (F)
(b) This statement is false (F)
(c) This statement is true (T)
(d) This statement is true (T)
(e) This statement is true (T)
(f) This statement is false (F)
(g) This statement is false (F)
(h) This statement is false (F)
(i) This statement is true (T)
(J) This statement is false (F)
(k) This statement is false (F)
(l) This statement is true (T)
(m) This statement is false (F).

Exercise 2.2Question 1.
Find the sum by suitable arrangement:
(a) 837 + 208 + 363
(b) 1962 + 453,+ 1538 + 647
Solution:
(a) 837 + 208 + 363 = (837 + 363) + 208
= 1200 + 208 [Using associative property]
= 1408

(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100 = 4600

Question 2.
Find the product by suitable arrangement:
(а) 2 x 1768 x 50
(b) 4 x 166 x 25
(c) 8 x 291 x 125
(d) 625 x 279 x 16
(e) 285 x 5 x 60
(f) 125 x 40 x 8 x 25
Solution:
(a) 2 x 1768 x 50 = (2 x 50) x 1768 = 176800
(b) 4 x 166 x 25 = 166 x (25 x 4) = 166 x 100 = 16600
(c) 8 x 291 x 125 = (8 x 125) x 291 = 1000 x 291 = 291000
(d) 625 x 279 x 16 = (625 x 16) x 279 = 10000 x 279 = 2790000
(e) 285 x 5 x 60 = 285 x (5 x 60) = 285 x 300 = (300 – 15)x 300 = 300 x 300 – 15 x 300 = 90000 – 4500 = 85500
(f) 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25) = 1000 X 1000 = 1000000

Question 3.
Find the value of the following:
(а) 297 x 17 + 297 x 3
(б) 54279 x 92 + 8 x 54279
(c) 81265 x 169 – 81265 x 69
(d) 3845 x 5 x 782 + 769 x 25 x 218
Solution:
(a) 297 x 17 x 297 x 3 = 297 x (17 + 3)
= 297 x 20 = 297 x 2 x 10
= 594 x 10 = 5940

(b) 54279 x 92 + 8 x 54279 = 54279 x (92 + 8)
= 54279 x 100 = 5427900

(c) 81265 x 169 – 81265 x 69
= 81265 x (169 – 69)
= 81265 x 100 = 8126500

(d) 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + 769 x 5 x 5 x 218
= 3845 x 5 x 782 + (769 x 5) x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x (782 + 218)
= 3845 x 5 x 1000
= 19225 x 1000
= 19225000

 Question 4.
Find the product using suitable properties.
(a) 738 x 103
(b) 854 x 102
(c) 258 x 1008
(d) 1005 x 168
Solution:
(a) 738 x 103 = 738 x (100 + 3)
= 738 x 100 + 738 x 3 [Using distributive property]
= 73800 + 2214 = 76014

(b) 854 x 102 = 854 x (100 + 2)
= 854 x 100 + 854 x 2 [Using distributive property]
= 85400 + 1708 = 87108

(c) 258 x 1008 = 258 x (1000 + 8)
= 258 x 1000 + 258 x 8 [Using distributive property]
= 258000 + 2064 = 260064

(d) 1005 x 168 = (1000 + 5) x 168
= 1000 x 168 + 5 x 168 [Using distributive property]
= 168000 + 840 = 168840

Question 5.
A taxidriver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litre of petrol. If the petrol cost ₹44 per litre, how much did he spend in all on petrol?
Solution:
Petrol filled on Monday = 40 litres
Cost of petrol = ₹44 per litre
Petrol filled on Tuesday = 50 litre
Cost of petrol = ₹44 pet litre
∴ Total money spent in all
= ₹(40 x 44 + 50 x 44)
= ₹(40 + 50) x 44 = ₹90 x 44 = ₹3960

Question 6.
A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ₹15 per litre, how much money is due to the vendor per day?
Solution:
Milk supplied in the morning = 32 litres
Cost of milk = ₹15 per litre
Milk supplied in the evening = 68 litres
Cost of milk = ₹15 per litre

 Question 7.
Match the following:
(i) 425 x 136 = 425 x (6 + 30 + 100)
(ii) 2 x 49 x 50 = 2 x 50 x 49
(iii) 80 + 2005 + 20 = 80 + 20 + 2005
Hence (i) ↔ (c), (ii) ↔ (a) and (iii) ↔ (b)
∴ Money paid to the vendor
= ₹ (32 x 15 + 68 x 15)
= ₹(32 + 68) x 15
= ₹100 x 15
= ₹1500
(a) Commutativity under multiplication
(b) Commutativity under addition
(c) Distributivity of multiplication over addition

xercise 2.3

 Question 1.
Which of the following will not represent zero:
(a) 1 + 0
(b) 0 x 0
(c) 02
(d) 10102
Solution:
(a) 1 + 0 = 1 ≠ 0, does not represent zero.
(b) 0 x 0 = 0, represents zero
(c) 02 = 0, represents zero.
( d) 10102 = 02 = 0 represents zero.

Question 2.
If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Solution:
Yes, Examples:
5 x 0 = 0
0 x 8 = 0
0 x 0 = 0

 Question 3.
If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
Solution:
This is only true, when each of the number are 1.
1 x 1 = 1

 Question 4.
Find using distributive property:
(а) 728 x 101
(b) 5437 x 1001
(c) 824 x 25
(d) 4275 x 125
(e) 504 x 35
Solution:
(a) 728 x 101 = 728 x (100 + 1)
= 728 x 100 + 728 x 1
= 72800 + 728
= 73528

(b) 5437 x 1001 = 5437 x (1000 + 1)
= 5437 x 1000 + 5437 x 1
= 5437000 + 5437
= 5442437

(c) 824 x 25 = 824 x (20 + 5)
= 824 x 20 + 824 x 5
= 16480 + 4120
= 20600

(d) 4275 x 125 = 4275 x (100 + 20 + 5)
= 4275 x 100 + 4275 x 20 + 4275 x 5
= 427500 + 85500 + 21375
= 534375

(e) 504 x 35 = (500 + 4) x 35
= 500 x 35 + 4 x 35
= 17500 + 140
= 17640

 Question 5.
Study the pattern:
1 x 8 + 1= 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?
Solution:
Step I: 123456 x 8 + 6 = 987654
Step II: 1234567 x 8 + 7 = 9876543

Working pattern:
(1) x 8 + 1 = 9
(12) x 8 + 2 = (11 + 1) x 8 + 2 = 98
(123) x 8 + 3 = (111 + 11 + 1) x 8 + 3 = 987
(1234) x 8 + 4 = (1111 + 111 + 11 + 1) x 8 + 4 = 9876
(12345) x 8 + 5 = (11111 + 1111 + 111 + 11 + 1) x 8 + 5 = 98765

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CLASS 12TH CHAPTER -16 Environmental Issues |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

All chapter class 12 Biology NCERT Solutions are prescribed by the subject teachers based on the NCERT textbook questions and explained all solutions in a better way for easy understanding. In-depth knowledge of the biology subject can be attained by the students with the help of detailed NCERT Solutions of Class 12 Biology

NCERT Solutions for Class 12 Biology Chapter :16 Environmental Issues

Question 1.
What are the various constituents of domestic sewage ? Discuss the effects of sewage discharge on a river.
Solution:
The domestic sewage contains every-thing that goes down the drain into the sewer of the house. The various constituents of domestic sewage are suspended solids, colloidal particles, pathogenic contaminants and dissolved materials. Suspended solids are sand and silt. Colloidal particles include clay, faecal matter, fine fibres of paper and cloth. Pathogenic contaminants are eggs of coliforms and enterococci. Dissolved materials includes inorganic nutrients such as nitrates, phosphates, ammonia, sodium and calcium. Effects of sewage discharge on a river :

  • Water becomes unfit for bathing and drinking and also for domestic or industrial use as it becomes colored, turbid with a lot of particulate matter floating on water.
  • The domestic sewage adds nitrates and phosphates into the river. These nitrates and phosphates encourage a thick bloom of blue green algae, which depletes the oxygen content of the water during night. This suffocates the fish and other aquatic life. Consequently river become highly polluted.

Question 2.
List all the wastes that you generate, at home, school, or during your trips to other places. Could you very easily reduce the generation of these wastes? Which would be difficult or rather impossible to reduce?
Solution:
Wastes generated at home include plastic bags, paper napkins, toiletries, kitchen wastes (such as peelings of vegetables and fruits, tea leaves), domestic sewage, glass, etc.

Wastes generated at school include waste paper, plastics, vegetable and fruit peels, food wrapping, sewage, etc. Wastes generated at trips or picnics include plastic, paper, vegetable and fruit peels, disposable cups, plates, spoons etc.

Yes, wastes can be easily reduced by the judicious use of the above materials. Wastage of paper can be minimized by writing on both sides of the paper and by using recycled paper.

Plastic and glass waste can also be reduced by recycling and re-using. Also, substituting plastic bags with biodegradable jute bags can reduce wastes generated at home, school or during trips. Domestic sewage can be reduced by optimizing the use of water while bathing, cooking, and other household activities.

Non-biodegradable wastes such as plastic, metal, broken glass, etc. are difficult to decompose because microorganisms do not have the ability to decompose them.

Question 3.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Solution:

  • Global warming is a rise in the mean temperature of the lower atmosphere and the earth’s surface. Causes – increase in the quantity of radioactively active greenhouse gases CO2, CH4, N2O, CFCs. They allow heat waves to reach the surface and prevent their escape.
  • They are produced by combustion of fossil fuels, biomass [CO2]; burning of nitrogen-rich fuels [N2O]; paddy fields, fermentation in cattle and wetlands [CH4]; refrigerators, aerosols, drying, cleaning [CFCs].
  • Effects: Heating of earth surface [mean temperature is increased] Climatic changes e.g.: El Nino effect.
  • Increased melting of polar ice caps and Himalayan snowcaps. Increased sea levels and coastal areas will submerge.
  • Measures – Decreased use of fossil fuels, improve the efficiency of energy usage, Reduce deforestation, plant trees Control of man-made sources of greenhouse gases like vehicles, aerosol sprays.

Question 4.
Match the items given in Column A and B
column A                                       Column B
(a) Catalytic converter               (i) Particulate matter
(b) Electrostatic precipitator    (ii) Carbon monoxide and nitrogen oxides
(c) Earmuffs                               (iii) High noise level
(d) Landfills                               (iv) Solid wastes
Solution:
(a) – (ii); (b) – (i); (c) – (iii); (d) – (iv).

Question 5.
Write critical notes on the following :
(a) Eutrophication
(b) Biological magnification
(c) Groundwater depletion and ways for its replenishment
Solution:
(a) Eutrophication: The natural aging process of lakes by nutrient enrichment of their water. In young lake water is cold and clear and supports only little life. With time, streams introduce nutrients into lake which increases lakes’ fertility and encourages aquatic growth. Over centuries silts and organic debris pile up, and lake becomes shallow and warmer. It supports plants and later gets converted into land. Lakes span depends on the climate, size of lake.

(b) Biological magnification: Industrial wastes released into water contain toxic substances, such as arsenic, cadmium, lead, zinc, copper, mercury, and cyanides, besides some salts, acids, and alkalies. All these materials can prove harmful for our health.
NCERT Solutions for Class 12 Biology Chapter 16 Environmental Issues 5.1
They may reach the human body directly with contaminated food or indirectly by way of plants and other animals. The concentration of the toxic materials increases at each trophic level of a food chain. This is called biological magnification. River water may have a very low concentration of DDT, but the carnivorous fish in that river may contain a high concentration of DDT and become unfit for eating by man. Mercury discharged into rivers and lakes is changed by bacteria to the neurotoxic form called methyl mercury. The latter is highly poisonous and may be directly absorbed by fish.

(c) Groundwater depletion and ways for its replenishment: Groundwater depletion is defined as long-term water-level decline caused by sustained groundwater pumping. The volume of ground water in storage is decreasing in many areas of the world in response to pumping. Some of the negative effects of groundwater depletion include increased pumping costs, deterioration of water quality, reduction of water in streams and lakes.
Some ways for water replenishment are:

  • Reduction in consumption: Sprinkler and subsurface irrigation techniques reduce the amount of water used in irrigation.
  • Rain water harvesting: Rain water collected over roofs is allowed to pass into the ground through deep water pipes.

Question 6.
Why does the ozone hole form over Antarctica? How will enhanced ultraviolet radiation affect us?
Solution:
Ozone hole forms over Antarctica where no one lives and no pollution is present but not over Newyork, Bangalore etc., (polluted cities). It is because CFCs and ozone-depleting substances (ODS) released worldwide accumulates in the stratosphere and drifts towards, Antarctica in winters (July – August) when temperatures is -’85° C in Antarctica.

In winters polar ice clouds are formed over Antarctica. It provides a catalytic surface for (CFCs and other ODS to release CL and other free radicals that breakdown ozone layer forming an ozone hole during spring in presence of sunlight. In summer, the ozone hole disappears due to mixing of air worldwide.

Ozone holes allow UV radiations (UVA & UVB) to reach earth’s surface. Which was earlier reflected by the ozone layer. UVB damages DNA, skin cells and causes mutations and skin cancers respectively. UVB even causes corneal damage (Snow Blindness).

Question 7.
Discuss the role of women and communities in protection and conservation of forests.
Solution:
Forest Conservation and Management:
It is time to think deeply and act seriously in order to protect this vital natural resource. Some of the measures of conservation are

  1. Social forestry programme: It was started in 1976 and involves the affor­estation on public and common lands for fuel, fodder, timber for agricul­tural equipment and fruits. These are mainly meant for rural people.
  2. Agroforestry programme: It involves the multiple use of same land for agriculture, forestry and animal husbandary. Taungya System and Jhum are examples.
    • Taungya System: It involves growing agricultural crops between planted trees.
    • Jhum (Slash and burn agriculture): It involves felling and burning of forests, followed by the cultivation of crops for a few years. Later the cultivation is abandoned for the growth of forests. It is a traditional agroforestry system.
  3. Urban forestry programme: It involves afforestation in urban land ar­eas e.g. along the roads, big parks, big compounds etc. with ornamental and fruit trees.
  4. Commercial forestry: It involves planting of fast-growing trees on avail­able land to fulfill commercial demand.
  5. Conservation forestry: It involves protection of degraded forest to allow recoupment of their flora and fauna.

Reforestation: It is the process of restoring a forest that once existed, but was removed at some point of time in the past. Reforestation may occur naturally in a deforested area. The above-said methods speed up the refor­estation programme.

Question 8.
What measures, as an individual, you would take to reduce environmental pollution?
Solution:
To reduce environmental pollution, we should change our habits and lifestyle so as to reduce the use of disposable materials. We should use preferably those items which can easily be recycled and also minimise the use of fossil fuels. We should also take measures to improve the quality of air by using CNG gases wherever possible instead of using diesel or petrol. We should also use the catalytic converter in our vehicles.

Question 9.
Discuss briefly the following:
(a) Radioactive wastes
(b) Defunct ships and e-wastes
(c) Municipal solid wastes
Solution:
a. Radioactive waste materials are released from thermonuclear explosions. Radioactive isotopes, such as radium-226, thorium- 232, potassium-40, uranium-235, carbon-14, etc. are spread all over the world and contaminate air, soil, water, vegetation and animals.

b. Irrepairble electronic goods and computers are called electronic wastes (e-waste).
Ships that are no longer in use or that are to be dismantled are called defunct ships. Asbestos, Polychlorinated Biphenyl (PCB) produced during dismantling defunct ship cause serious health hazards especially cancer.

c. Municipal solid wastes are wastes from homes, offices, stores, schools, hospitals, etc., that are collected and disposed of by the municipality.

Question 10.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi?
Solution:
Under the direction of Supreme Court of India, the State Government of Delhi took the following measures to improve the quality of air:

  • Switching over the entire fleet of public transport buses from diesel to CNG (Compressed Natural Gas) by the end of 2002.
  • Phasing out of old vehicles.
  • Use of unleaded petrol.
  • Use of low sulphur petrol and diesel.
  • Use of catalytic converters in vehicles.
  • Application of Euro II norms for vehicles.

Because of above mentioned measures adopted by the Government the air quality of Delhi has improved with a substantial fall in S02, CO, Nox level between 1997-2005.

Question 11.
Discuss in brief the following:
(a) Green house gases
(b) Catalytic converter
(c) Ultraviolet B
Solution:
(a) Green house gases: The gases which are transparent to solar radiation but retain and partially reflect back long wave heat radiations are called greenhouse gases. Green house gases are essential for keeping the earth warm and hospitable. They are also called radiatively active gases. They prevent a substantial part of long wave radiations emitted by earth to escape into space. Rather green house gases radiate a part of this energy back to the earth. The phenomenon is called greenhouse flux. Because of greenhouse flux, the mean annual temperature of the earth is 15°C. In its absence, it will fall to – 18°C.

However, recently the concentration of greenhouse gases has started rising to result in an enhanced greenhouse effect that is resulting in increasing the mean global temperature. It is called global warming. A regular assessment of the abundance of greenhouse gases and their impact on the global environment is being made by IPCC (Intergovernmental Panel on Climate Change). The various green house gases are CO2 (warming effect 60%), CH4 (effect 20%) , chlorofluorocarbons or CFCs (14%) andT nitrous oxide (N2O, 6%). Others of minor significance are water vapors and ozone.

(b) Catalytic converter: Catalytic converters are devices that are fitted into automobiles for reducing the emission of gases. These have expensive metals (platinum – palladium, and rhodium) as catalysts. As the exhaust passes through the catalytic converters, unburnt hydrocarbons are converted into CO2 and H2O and carbon monoxide and nitric oxide are changed to CO2 and N2 respectively. Vehicles fitted with catalytic converters should be run on unleaded petrol as leaded petrol would inactivate the catalyst in the converters.

(c) Ultraviolet B – UV-B having 280-320nm wavelength. Their harmful radiations penetrate through the ozone hole to strike the earth. On earth, these can affect human beings and other animals by causing :

  • Skin cancer
  • Blindness and increased incidence of cataract in eyes, and
  • Malfunctioning of the immune system.
  • Higher number of mutations.
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CLASS 12TH CHAPTER -15 Biodiversity and Conservation |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

All chapter class 12 Biology NCERT Solutions are prescribed by the subject teachers based on the NCERT textbook questions and explained all solutions in a better way for easy understanding. In-depth knowledge of the biology subject can be attained by the students with the help of detailed NCERT Solutions of Class 12 Biology

NCERT Solutions for Class 12 Biology Chapter :15 Biodiversity and Conservation

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CLASS 12TH CHAPTER -14 Ecosystem |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

All chapter class 12 Biology NCERT Solutions are prescribed by the subject teachers based on the NCERT textbook questions and explained all solutions in a better way for easy understanding. In-depth knowledge of the biology subject can be attained by the students with the help of detailed NCERT Solutions of Class 12 Biology

NCERT Solutions for Class 12 Biology Chapter :14 Ecosystem

Question 1.
Fill in the blanks
(a) Plants are called as ___ because they fix carbon dioxide.
(b) In an ecosystem dominated by trees, the pyramid (of numbers) is ___ type.
(c) In aquatic ecosystem, the limiting factor for the productivity is ___
(d) Common detritivores in our ecosystem are ____
(e) The major reservoir of carbon on earth is ___
Solution:
(a) producers
(b) inverted or spindle
(c) light
(d) saprotrophs
(e) oceans

Question 2.
Which one of the following has the largest population in a food chain?
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposer’s
Solution:
(d) decomposer’s

Question 3.
The second trophic level in a lake is
(a) phytoplankton
(b) zooplankton
(c) benthos
(d) fishes.
Solution:
(b) zooplankton

Question 4.
Secondary producers are
(a) herbivores
(b) producers
(c) carnivores
(d) none of these
Solution:
(a) herbivores

Question 5.
What is the percentage of photosynthetically active radiation (PAR), in the incident solar radiation.
(a) 100%
(b) 50%
(c) 1 – 5%
(d) 2 – 10%
Solution:
(b) 50%

Question 6.
Distinguish between
(a) Grazing food chain and detritus food chain
(b) Upright and inverted pyramid
(c) Litter and detritus
(d) Production and decomposition
(e) Food chain and food web
(f) Primary and secondary productivity
Solution:
(a) Differences between grazing food chain and detritus food chain are as follows
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.1

(b) Differences between upright and inverted pyramids are as follows :
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.2

(c) Differences between litter and detritus are as follows :
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.3

(d) Differences between production and decomposition are as follows :
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.4

(e) Differences between food chain and food web are as follows:
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.5

(f) Differences between primary productivity and secondary productivity are as follows :
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 6.6

Question 7.
Describe the components of an ecosystem.
Solution:
Ecosystem: The system resulting from the interaction between organisms and their environment is called an ecosystem.
(a) Producers: Organisms, which can synthesize their own food are included under producers, e.g., Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

(b) Consumers:

  • Primary consumer: Animals, which feed on producers are included in this category e.g., Daphnia, Cyclops, Paramoecium, Amoeba, and small fishes.
  • Secondary consumers: Primary consumers also serve as food for water snakes, a few tortoises, few types of fish, etc. hence, these are carnivores.
  • Tertiary consumers: Secondary consumers also serve as food for aquatic birds like kingfishers, cranes, big fish and these together form a top-class carnivorous group and called tertiary consumers.

(c) Decomposers: All producers and consumers die and accumulate on the floor of the pond. Even the waste material and feces of these animals get accumulated on the floor of the pond. Similarly, the floor of the pond is also occupied by decomposers, which include bacteria and fungi. These decomposers decompose complex organic compounds of then- bodies into simpler forms which are finally mixed with the soil of the floor of ponds. These are again absorbed by the roots of producer plants and thus matter is recycled.

Question 8.
Define ecological pyramids and describe with examples, pyramids of number and biomass.
Solution:
An ecological pyramid is a graphic representation of an ecological parameter, as a number of individuals present in various trophic levels of a food chain with producers forming the base and top carnivores the tip. Ecological pyramids were developed by Charles Elton (1927) and are, therefore, also called Eltonian pyramids.

There are three types of ecological pyramids, namely,

  • Pyramid of numbers
  • Pyramid of biomass
  • Pyramid of energy

Pyramid of numbers: It is a graphic representation of the number of individuals per unit area of various trophic levels stepwise with producers at the base and top carnivores at the tip. In a grassland, the producers, which are mainly grasses, are always maximum in number. This number then shows a decrease towards the apex, as the primary consumers (herbivores) like rabbits, mice, etc. are lesser in number than the grasses; the secondary consumers, snakes, and lizards are lesser in number than the rabbits and mice. Finally, the top (tertiary) consumers hawks or other birds, are the least in number. Thus, the pyramid becomes upright.
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 8.1

Pyramid of biomass: The amount of living organic matter (fresh and dry weight) is called biomass. Here, different trophic level of the ecosystem are arranged according to the biomass of the organisms. In grassland and forest, there is generally a gradual decrease in biomass of organisms at successive levels from the producers to the top carnivores. Thus these pyramids are upright. But in pond ecosystem, it is inverted because the biomass gradually increases from the producers to carnivores.
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 8.2

Question 9.
What is primary productivity? Give a brief description of factors that affect primary productivity.
Solution:
The rate of biomass production is called productivity.
It is expressed in terms of g-2yr-1  or(Kcal-m-2) yr-1 to compare the productivity of ecosystems.
It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NFP).

Gross Primary Productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilized by plants in respiration.

Gross primary productivity minus respiration losses (R), is the Net Primary Productivity (NPP). GPP – R=NPP.
Primary productivity depends on:

  • The plant species inhabiting a particular area.
  • The environmental factors.
  • Availability of nutrients.
  • Photosynthetic capacity of plants.

Question 10.
Define decomposition and describe the processes and products of decomposition.
Solution:
Decomposition is the breakdown of dead or wastes organic matter by micro-organisms. Decomposition is both physical and chemical in nature. Processes involved in decomposition are – fragmentation, catabolism & leaching.

  • Fragmentation – The process primarily due to the action of detritus feeding invertebrate (detritivores) causes it to break into smaller particles. The detritus gets pulverized when passing through the digestive tracts of animals. Due to fragmentation, the surface area of detritus particles is greatly increased.
  • Catabolism – Enzyme degradation of detritus into simpler organic substances by bacteria and fungi.
  • Leaching – The process by which nutrients, chemicals, or contaminants are dissolved & carried away by water, or are moved into a lower layer of soil.

Various inorganic and organic substances are obtained by decomposition. Inorganic substances are obtained in the process of mineralization while organic substances are obtained in humification. A dark coloured amorphous substance called humus is formed by decomposition. Humus is highly resistant to microbial action & undergoes extremely slow decomposition. It serves as a reservoir of nutrients.

Question 11.
Give an account of energy flow in an ecosystem.
Solution:
Ecosystems require a constant input of energy as every component of an ecosystem is regularly dissipating energy.

Two laws of thermodynamics govern this flow of energy. According to the first law of thermodynamics, energy can be transferred as well as transformed but is neither created nor destroyed. According to the second law of thermodynamics, every activity involving energy transformation is accompanied by the dissipation of energy. Except for deep hydrothermal ecosystems, the source of energy in all ecosystems is solar energy. 50% of the solar energy incident over the earth is present in PAR (photosynthetically active radiation).

Energy flow in an ecosystem is always unidirectional or one way, i.e., solar radiation → producers → herbivores → carnivores. It cannot pass in the reverse direction. There is a decrease in the content and flow of energy with the rise in trophic level. Only 10% of energy is transferred from one trophic level to the next.
Producer biomass (1000 K cal) → Herbivore biomass (100 K cal) → Carnivore I biomass (10 Kcal) Carnivore II biomass (1 Kcal)
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 11.1

Question 12.
Write important features of a sedimentary cycle in an ecosystem
Solution:
Sedimentary Biogeochemical cycle:- It is the circulation of a biogeochemical between the biotic and abiotic compound of an ecosystem is a nongaseous being lithosphere or sediments of the earth. Sedimentary cycles occur in the case of phosphorus, calcium, magnesium, zinc, copper, etc.

Question 13.
Outline salient features of carbon cycling in an ecosystem.
Solution:
Carbon constitutes 49 percent of the dry weight of organisms and is next only to water. 71 percent of carbon is found dissolved in oceans. This ocean reservoir regulates the amount of carbon dioxide in the atmosphere. Fossil fuels also represent a reservoir of carbon. Carbon cycling occurs through the atmosphere, ocean, and living and dead organisms. 4 x 1013 kg of carbon is fixed in the biosphere through photosynthesis annually.

A considerable amount of carbon returns to the atmosphere as Co2 through respiratory activities of the producers and consumers. Decomposers also contribute substantially to the CO2 pool by their processing of waste materials and dead organic matter of land or oceans. Some amount of fixed carbon is lost to sediments and removed from circulation. Burning of wood, forest fire and combustion of organic matter, fossil fuels, volcanic activity are additional sources for releasing Co2 into the atmosphere.

Human activities have significantly influenced the carbon cycle. Rapid deforestation and the massive burning of fossil fuels for energy and transport have significantly increased the rate of release of carbon dioxide into the atmosphere.
NCERT Solutions for Class 12 Biology Chapter 14 Ecosystem 13.1

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