CBSE Class 6

These NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes InText Questions

NCERT In-text Question Page No. 88

Question 1.
Take any post card. Use the above technique to measure its two adjacent sides.
Answer:
Do it yourself.

Question 2.
Select any three objects having a flat top. Measure all sides of the top using a divider and a ruler.
Answer:
Do it yourself.

NCERT In-text Question Page No. 91

Question 1.
What is the angle name for half a revolution?
Answer:
∵ 2 straight angles = 1 revolution
∴ \(\frac { 1 }{ 2 }\) [2 straight angles] = \(\frac { 1 }{ 2 }\) [1 revolution]
or A straight angle = \(\frac { 1 }{ 2 }\) revolution
Thus, half a revolution means a straight angle.

Question 2.
What is the angle name for one-fourth revolution?
AnswerL
∵ 4 right angles = 1 revolution
∴ A right angle = \(\frac { 1 }{ 4 }\) revolution 4
Thus, one-fourth a revolution means a right angle.

Question 3.
Draw five other situations of one-fourth, half and three-fourth revolution on a clock.
Answer:
(i) One-fourth revolution [i.e. a turn of a right angle]
For one-fourth revolution, the clock hand moves from:
(a) 12 to 3
(b) 3 to 6
(c) 1 to 4
(d) 9 to 12
(e) 6 to 9

(ii) Half-revolution [i.e. a turn of a straight angle]
For a half-revolution, the clock hand moves from
(a) 12 to 6
(b) 3 to 9
(c) 1 to 7
(d) 2 to 8
(e) 4 to 10

(iii) Three-fourth revolution [i.e. a turn of 3 right angles ]
Note: There is no special name for three- fourth of a revolution.
For a three-fourth revolution, the clock- hand moves from:
(a) 12 to 9
(b) 1 to 10
(c) 3 to 12
(e) 9 to 6

NCERT In-text Question Page No. 93

Question 1.
The hour hand of a clock moves from 12 to 5. Is the revolution of the hour hand more than 1 right angle?

Answer:
Yes, the revolution of the hour hand from 12 to 5 is more than 1 right angle.

Question 2.
What does the angle made by the hour hand of the clock look like when it moves from 5 to 7. Is the angle moved more than 1 right angle?

Answer:
No, in this case the angle is less than a right angle.

Question 3.
Draw the following and check the angle with your RA tester.
(a) going from 12 to 2
(b) from 6 to 7
(c) from 4 to 8
(d) from 2 to 5
Answer:
(a) Going from 12 to 2: The angle formed by the hour hand in going from 12 to 2 is shown in the figure alongside. Checking this angle by RA [right angle] tester, we find that it is less than a right angle.

(b) Going from 6 to 7: Checking the angle formed by RA tester, we find

(c) Going from 4 to 8: Checking the angle so formed by RA tester, we find

(d) Going from 2 to 5: Checking the angle so formed by RA tester, we find

Question 4.
Take five different shapes with corners. Name the corners. Examine them with your tester and tabulate your results for each case:

Answer:
It is an activity. Do it yourself.

NCERT In-text Question Page No. 94

Question 1.
Look around you and identify edges meeting at corners to produce angles. List ten such situations.
Answer:
Do it yourself.

Question 2.
List ten situations where the angles made are acute.
Answer:
Do it yourself.

Question 3.
List ten situations where the angles made are right angles.
Answer:
Do it yourself.

Question 4.
Lind five situations where obtuse angles are made.
Answer:
Do it yourself.

Question 5.
List five other situations where reflex angles may be seen.
Answer:
Do it yourself.

These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.3

Question 1.
Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.
(Hint: Three possible expressions are 5 + (8 – 7), 5 – (8 – 7), (5 × 8) + 7; make the other expressions.)
Answer:
(a) (8 × 5) – 7
(b) (8 + 5) – 7
(c) (8 × 7) – 5
(d) (8 + 7) – 5
(e) 5 × (7 + 8)
(f) 5 + (7 × 8)
(g) 5 + (8 – 7)
(h) 5 – (7 + 8)

Question 2.
Which out of the following are expressions with numbers only?
(a) y + 3
(b) (7 × 20) – 8z
(c) 5 (21 – 7) + 7 × 2
(d) 5
(e) 3x
(f) 5- 5n
(g) (7 × 20)- (5 × 10) – 45 + p
Answer:
(c) and (d)

Question 3.
Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.
(a) z + 1, z – 1, y + 17, y- 17
(b) 17y, \(\frac{\mathrm{y}}{17}\) , 5z
(c) 2y + 17, 2y- 17
(d) 7m,- 7m + 3,- 7m- 3
Answer:
(a) z + 1 Addition z – 1 → Subtraction
y + 17 → Additiony – 17 → Subtraction

(b) 17y → Multiplication y/17 → Division
5z → Multiplication

(c) 2y + 17 → Multiplication and Addition
2y – 17 → Multiplication and Subtraction

(d) 7m → Multiplication
-7m + 3 → Multiplication and Addition
7m – 3 —> Multiplication and Subtraction

Question 4.
Give expressions for the following cases.
(a) 7 added to p
(b) 7 subtracted from p
(c) p multiplied by 7
(d) p divided by 7
(e) 7 subtracted from – m
(f) – p multiplied by 5
(g) – p divided by 5
(h) p multiplied by – 5
Answer:
(a) p + 7
(b) p – 7
(c) 7p
(d) p/7
(e) -m – 7
(f) -5 p
(g) -p/5
(h) -5p

Question 5.
Give expressions in the following cases.
(a) 11 added to 2m
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by – 8
(f) y is multiplied by – 8 and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by – 5 and the result is added to 16
Answer:
(a) 2m + 11
(b) 2m – 11
(c) 5y + 3
(d) 5y – 3
(e) -8y
(f) -8y + 5
(g) 16 – 5y
(h) -5y + 16

Question 6.
(a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Answer:
(a) t + 4, t – 4, 4 – t, \(\frac{t}{4}, \frac{4}{t}\)
(b) 2y + 7, 2y – 7, 7y + 2, 7y – 2 and so on.

These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.2

Question 1.
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Answer:
Side of equilateral triangle = l
Therefore, Perimeter of equilateral triangle = 3 × side = 3l

Question 2.
The Side of a regular hexagon (see the given figure) is denoted by l. Express the perimeter of the hexagon using l.
(Hint: A regular hexagon has all its six sides equal in length.)

Answer:
Side of hexagon = l
Therefore,
Perimeter of Hexagon = 6 × side = 6l

Question 3.
A cube is a three-dimensional figure as shown in the given figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.

Answer:
Length of one edge of cube = l
Number of edges in a cube = 12
Therefore, total length = 12 × 1 = 12l

Question 4.
The diameter of a circle is a line which joins two points on the circle and also passed through the centre of the circle. (In the adjoining figure AB is a diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius(r).

Answer:
Since, length of diameter is double the length of radius.
Therefore, d = 2r

Question 5.
To find sum of three numbers 14,27 and 13, we can have two ways:
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus, (14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on whole numbers, in a general way, by using variables a, b and c.
Answer:
(a + b) + c = a + (b + c )

These NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.5

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.
Answer:
(a) The adjacent edges of a table top.

.’. They form perpendicular lines,

(b) The lines of a railway track

Here, the two adjacent sides don’t meet.
χ They do not form perpendicular lines

(c) The line segments forming the letter ‘L’.

Here, the line segments form a right angle
∴ They form perpendicular lines.

(d) Here, the angle between the lines is not a right angle.

∴ They do not form perpendicular lines.

Question 2.
Let \(\overline{\mathrm{PQ}}\) be the perpendicular to the line segment \(\overline{\mathrm{XY}}\). Let \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{XY}}\) intersect in the point A. What is the measure of ∠PAY?
Answer:

Since, \(\overline{\mathrm{XY}}\) ⊥ PQ
Angle between them is a right angle.
.’. ∠PAY = 90°

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?
Answer:

The angle are 90°, 45°, 45°
The angles are 90°, 60°, 30°
So, Angle 90° is common between them

Question 4.
Study the diagram. The line l is perpendicular to line m.

(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
Answer:
(a) CE = CD + DE = 1 + 1=2 EG = EF + FG = 1 + 1 = 2

(b) PE & CG Intersect at point E & CE = EG
.’. PE is the bisector of CG

(c) PE is perpendicular bisector for
\(\overline{\mathrm{CG}}\)
As CE = EG = 2
& \(\overline{\mathrm{CE}}\) = EG = 2
\(\overline{\mathrm{XY}}\) ⊥ \(\overline{\mathrm{CG}}\)
\(\overline{\mathrm{BH}}\)
As \(\overline{\mathrm{BE}}=\overlline{\mathrm{BE}}\) = 3
\(\overline{\mathrm{PE}}=\overlline{\mathrm{BH}}\)

(d) (i) True
AC = AB + BC = 1 + 1 = 2
FG = 1
∴ AC > FG

(ii) True
CD = 1
GH = 1
CD > GH

(iii) True
BC = 1
EH = EF + FG + GH = 1 + 1 + 1 = 3
∴ CD > GH

These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.1

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.

Answer:

Question 2.
We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.l (given above) give us the same rule as that given by L. Which are these? Why does this happen?
Answer:
The rules to find the number of matchsticks for:
The letter L is 2n The letter C is 3n
The letter V is 2n The letter U is 3n
The letter T is 2n The letter F is 4n
∴ It is same for L, V and T. The number of matchsticks required in each of them is 2.

Question 3.
Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Answer:
Number of rows = n
Cadets in each row = 5
Therefore, total number of cadets = 5n.

Question 4.
If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Answer:
Number of boxes = b
Number of mangoes in each box = 50
Therefore, total number of mangoes = 50b.

Question 5.
The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)
Ans. Number of students = s
Number of pencils to each student = 5
Therefore, total number of pencils needed are = 5s.

Question 6.
A bird flies i kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)
Answer:
Time taken by bird = t minutes
Speed of bird = 1 km per minute
Therefore, Distance covered by bird = speed × time = 1 × t = t km.

Question 7.
Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rowrs? How many dots are there if there are 8 rows? If there are 10 rows?
Answer:

1 row = 9 dots
r rows = (9*r) dots
= 9r dots
8rows = 9*8 = 72dots
10rows = 10*9 = 90dots

Question 8.
Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be × years.
Answer:
Radhas age = x years
Therefore, Leela’s age = (x – 4) years

Question 9.
Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Answer:
Number of laddus gave away = l
Number of laddus remaining = 5
Total number of laddus = (l + 5)

Question 10.
Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?
Answer:
Number of oranges in one box = x
Number of boxes = 2
Therefore, total number of oranges in boxes = 2x
Remaining oranges =10
Thus, number of oranges = 2x + 10

Question 11.
(a) Look at the following matchstick pattern of squares. The squares are not separate. Two neighbouring squares have a common match- stick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares. (Hint: if you remove the vertical stick at the end, you will get a pattern of Cs.)

(b) The given figure gives a matchstick pattern of triangles. Find the general rule that gives the number of matchsticks in terms of the number of triangles.

Answer:

13 matchsticks
If we remove 1 from each then they makes table of 3, i.e., 3, 6, 9, 12…
So the required equation = 3x + 1, where x is number of squares.

If we remove 1 from each then they makes table of 2, i.e., 2, 4, 6, 8…
So the required equation = 2x +1, where x is number of triangles.

These NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion InText Questions

NCERT In-text Question Page No. 245
Question 1.
In a class, there are 20 boys and 40 girls. What is the ratio of the number of boys to the number of girls?
Answer:
Number of boys = 20
Number of girls = 40
Ratio of boys to girls = \(\frac{20}{40}=\frac{1}{2}=\frac{1}{2}\)
hence the required ratio is \(\frac{1}{2}\)

Question 2.
Ravi walks 6 km in an hour while Roshan walks 4 km in an hour. What is the ratio of the distance covered by Ravi to the distance covered by Roshan?
Answer:
Distance covered by Ravi in 1 hr = 6 km
Distance covered by Roshan in 1 hr = 4 km
∴ Ratio of distance covered Ravi in 1 hr to distance covered by Roshan in 1 hr = 6 km : 4
= \(\frac{6}{3}=\frac{4}{2}\)
Note:
(i) We donate ratio by the symbol
(ii) Ratio has no unit.
(iii) Two quantities can be compared only if they are in the same unit.
(iv) In a ratio the order of terms is very important. 2 : 3 and 3 : 2 are different ratios.

NCERT In-text Question Page No. 246
Question 1.
Saurabh takes 15 minutes to reach school from his house and Sachin takes one hour to reach school from his house. Find the ratio of the time taken by Saurabh to the time taken by Sachin.
Answer:
Time taken by Saurab = 15 minutes
Time taken by Sachin = 1 hour = 60 minutes
∴ Ratio of time taken by Saurabh to the time taken by Sachin
= \(\frac{\text { Time taken by Saurabh }}{\text { Time taken by Sachin }}\)
= \(\frac{15}{10}=\frac{15 \div 15}{60 \div 15}=\frac{1}{4}\) 1 : 4
[∵ HCF of 15 and 60 is 15]

Question 2.
Cost of a toffee is 50 paise and cost of a chocolate is ₹ 10. Find the ratio of the cost of a toffee to the cost of a chocolate.
Answer:
Cost of Toffee = 50 paise
Cost of chocolate = ₹ 10 rupees
[∵ ₹ 1 = 100 paise]
∴ 10 ₹ = 1000 paise
So, the cost of chocolate = 1000 paise
The ratio of the cost of toffee to the cost of chocolate = \(\frac{50}{1000}\) = 1 : 20

Question 3.
In a school, there were 73 holidays in one year. What is the ratio of the number of holidays to the number of days in one year?
Answer:
Number of holidays = 73 days
Number of days in one year = 365 days

[∵ HCF of 73 and 365 is 73]

NCERT In-text Question Page No. 248
Question 1.
Find the ratio of number of notebooks to the number of books in your bag.
Answer:
It is activity, please do it yourself.

Question 2.
Find the ratio of number of desks and chairs in your classroom.
Answer:
It is activity, please do it yourself.

Question 3.
Find the number of students above twelve years of age in your class. Then, find the ratio of number of students with age above twelve years and the remaining students.
Answer:
It is activity, please do it yourself.

Question 4.
Find the ratio of number of doors and the number of windows in your classroom.
Answer:
It is activity, please do it yourself.

Question 5.
Draw any rectangle and find the ratio of its length to its breadth.
Answer:
It is activity, please do it yourself.

NCERT In-text Question Page No. 254
Question 1.
Determine if the following are in proportion.
If yes, then write them in the proper form.
1. 1 : 5 and 3 : 15
2. 2 : 9 and 18 : 81
3. 15 : 45 and 5 : 25
4. 4 : 12 and 9 : 27
5. ₹ 10 to ₹15 and 4 to 6
Answer:
1. 1 : 5 and 3 : 15
We have, 3 : 15 = \(\frac{3}{15}=\frac{3 \div 3}{15 \div 3}=\frac{1}{5}\) = 1 : 5
[ ∵ HFC of 3 and 15 is 3]
i.e,. 1 : 5 and 3 : 15 are in proportion.
Thus, the proper form is 1 : 5 :: 3 : 15

2. 2 : 9 and 18 : 81
We have, 18 : 81= \(\frac{18}{81}=\frac{18 \div 9}{81 \div 9}=\frac{2}{9}\) = 2 : 9
[∵ HFC of 18 and 81 is 9]
i. e,. 2 : 9 and 18:81 are in proportion.
Thus, the proper form is 2 : 9 :: 18:81

3. 15 : 45 and 5 : 25
We have, 15 : 45 = \(\frac{15}{45}=\frac{15 \div 15}{45 \div 15}=\frac{1}{3}\) = 1 : 3
[ ∵ HFC of 15 and 45 is 15]
We have, 5 : 25 = \(\frac{5}{25}=\frac{5 \div 5}{25 \div 5}=\frac{1}{5}\) = 1 : 5
[∵HCF of 5 and 25 is 5]
∵ 1 : 3 ≠ 1 : 5 ∴ 15 : 45 ≠ 5 : 25
i. e,. 15 : 45 and 5 : 25 are not in proportion.

4. 4 : 12 and 9: 27
We have, 4 : 12 = \(\frac{4}{12}=\frac{4 \div 4}{12 \div 4}=\frac{1}{3}\) = 1 : 3
[∵ HFC of 4 and 12 is 4]
Again we have, 9 : 27 = \(\frac{9}{27}=\frac{9 \div 9}{27 \div 9}=\frac{1}{3}\) =1:3
[∵ HFC of 9 and 27 is 9]
4 : 12 = 9 : 27
or 4 : 12 and 9 : 27 are in proportion,
i.e., 4 : 12 : : 9 : 27 is the correct form.

5. ₹ 10 to ₹ 15 and 4 to 6
We have, ₹10 : ₹15 = \(=\frac{10}{15}=\frac{10 \div 5}{15 \div 5}=\frac{2}{3}\) = 2 : 3
[∵ HCF of 10 and 15 is 5]
Again we have, 4 : 6 = \(\frac{4}{6}=\frac{4 \div 2}{6 \div 2}=\frac{2}{3}\) = 1 : 3
[∵ HFC of 4 and 6 is 2]
Thus ₹ 10 : ₹ 15 = 4 : 6.
or ₹ 10, ₹ 15, 4 and 6 are in proportion.
so, the correct from is ₹ 10 : ₹ 15 : : 4 : 6

NCERT In-text Question Page No. 257
Question 1.
Prepare five similar problems (on unitary method) and ask your friends to solve them.
Answer:
It is a project work. Please do it yourself.

Question 2.
Read the table and fill in the boxes

Answer:
∵ Distance travelled by Kriti in 2 hours = 6 km
∴ Distance travelled by Kriti in 1 hour = \(\frac { 6 }{ 2 }\) = 3 km
∴ Distance travelled by Karan in 1 hour = 5 km
∴ Distance travelled by Karan in 4 hours = 4 x 4 km = 16 km
∴ Distance travelled by Kriti in 1 hour = 3 km
∴ Distance travelled by Kriti in 4 hour = 3 x 4 km = 12 km
Thus, the table is written as given below :