Category: Class 6

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.1

Question 1.
Convert the following numerical expressions into Tree diagrams.
(i) 8 + (6 × 2)
(ii) 9 – (2 × 3)
(iii) (3 × 5) – (4 ÷ 2)
(iv) [(2 × 4) + 2] × (8 ÷ 2)
(v) [(6 + 4) × 7] ÷ [2 × (10 – 5)]
(vi) [(4 × 3) ÷ 2] + [8 × (5 – 3)]
Solution:
(i) 8 + (6 × 2)

(ii) 9 – (2 × 3)

(iii) (3 × 5) – (4 ÷ 2)

(iv) [(2 × 4) + 2] × (8 ÷ 2)

(v) [(6 + 4) × 7] ÷ [2 × (10 – 5)]

(vi) [(4 × 3) ÷ 2] + [8 × (5 – 3)]

The first expression goes to the right side branch. So that the value does not change.

Question 2.
Convert the following Tree diagrams into numerical expressions.

Solution:
(i) The numerical Expression is 9 × 8
(ii) The numerical expression is (7 + 6) – 5
(iii) The numerical expression is (8 + 2) – (6 + 1)
(iv) The numerical expression is (5 × 6) – (10 ÷ 2)

Question 3.
Convert the following algebraic expressions into tree diagrams.
(i) 10 V
(ii) 3a – b
(iii) 5x + y
(iv) 20t × p
(v) 2(a + b)
(vi) (x × y) – (y × z)
(vii) 4x + 5y
(viii) (lm – n) ÷ (pq + r)
Solution:
(i) 10 V

(ii) 3a – b

(iii) 5x + y

(iv) 20t × p

(v) 2(a + b)

(vi) (x × y) – (y × z)

(vii) 4x + 5y

(viii) (lm – n) ÷ (pq + r)

Question 4.
Convert tree diagram into Algebraic expression.

Solution:
(i) Algebraic Expression is p + q
(ii) Algebraic Expression is l – m
(iii) Algebraic Expression is (a × b) – c (or) (ab) – c
(iv) Algebraic Expression is (a + b) – (c + d)
(v) Algebraic Expression is (8 ÷ a) + [ (6 ÷ 4) + 3]

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Intext Questions

Try These (Textbook Page No. 61)

Question 1.
Complete the following table. In any triangle.

Solution:

Try These (Textbook Page No. 62)

Question 1.
Complete the table.

Solution:

Try These (Textbook Page No. 64)

Question 1.
Can a triangle be formed with the given sides? If yes, state the type of triangle formed.

Solution:

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions

Question 1.
Name the type of the following triangles.
(a) ∆PQR with m∠Q = 90°
(b) ∆ABC with m∠B = 90° and AB = BC
Solution:
(a) One of the angles is 90°
It is a right-angled triangle
(b) Since two sides are equal.
It is an isosceles triangle. Also m∠B = 90°
It is an Isosceles right-angled triangle

Question 2.
Classify the triangles (scalene, isosceles, equilateral) given below.
(a) ∆ABC, AB = BC
(b) ∆PQR, PQ = QR = RP
(c) ∆ABC, ∠B = 90°
(d) ∆EFG, EF = 3 cm, FG = 4 cm and GE = 3 cm
Solution:
(a) Isosceles triangle
(b) Equilateral triangle
(c) Right angled triangle
(d) Isosceles triangle

Question 3.
In triangle ∆ABC, AB = BC = CA = 5 cm. Then what is the value of ∠A, ∠B and ∠C?
Solution:
Since AB = BC = CA
∠A = ∠B = ∠C
We know that ∠A + ∠B + ∠C = 180°
∴ ∠A = ∠B = ∠C = 60°

Question 4.
In ∆PQR, ∠P = ∠Q = ∠R = 60°, then what can you say about the length of sides of ∆PQR? Also, write the name of the triangle?
Solution:
∠P = ∠Q = ∠R = 60°
So PQ = QR = RP
∆PQR is equilateral triangle

Question 5.
In ∆ABC, AB = BC and ∠A = 50°. Then find the value of ∠C?
Solution:

It is an isosceles triangle.
∠A = ∠C
∠C = 50°

Question 6.
(a) Try to construct triangles using matchsticks.
(b) Can you make a triangle with?
(i) 3 matchsticks?
(ii) 4 matchsticks?
(iii) 5 matchsticks?
(iv) 6 matchsticks?
Name the type of triangle in each case. If you cannot make a triangle think of the reason for it.
Solution:

(b) (i) With the help of 3 matchsticks, we can make an equilateral triangle. Since all three matchsticks are of equal length.

(ii) With the help of 4 matchsticks, we cannot make any triangle because in this case, sum of two sides is equal to the third side and we know that the sum of the lengths of any two sides of a triangle is always greater than the length of the third side.
(iii) With the help of 5 matchsticks, we can make an isosceles triangle. Since we get two sides equal in this case.

(iv) With the help of 6 matchsticks, we can make an equilateral triangle. Since we get three sides equal in length.

Question 7.
A table is bought for ₹ 4500 and sold for ₹ 4800. Find the profit or loss.
Solution:
C.P = ₹ 4500
S.P. = ₹ 4800
Here S.P< C.P
Profit = S.P – C.P = ₹ 4800 – ₹ 4500 = ₹ 300

Question 8.
Draw any line segment \(\overline{\mathbf{P Q}}\). Take any point R not on it. Through R, draw a perpendicular to \(\overline{\mathbf{P Q}}\).
Solution:
Construction:
(i) Drawn a line segment \(\overline{\mathbf{P Q}}\) using scale and taken a point R outside of \(\overline{\mathbf{P Q}}\).
(ii) Placed a set-square on \(\overline{\mathbf{P Q}}\) such that one arm of its right angle aligns along \(\overline{\mathbf{P Q}}\).
(iii) Placed a scale along the other edge of the right angle of the set-square
(iv) Slide the set-square along the line till the point R touches the other arm of its right angle.
(v) Joined RS along the edge through R meeting \(\overline{\mathbf{P Q}}\) at S.
Hence \(\overline{\mathbf{R S}}\) ⊥ \(\overline{\mathbf{P Q}}\).

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.2

Question 1.
Draw a line segment AB = 7 cm and Mark a point P on it. Draw a line perpendicular to the given line segment at P.
Solution:
Construction:
(i) Drawn a segment \(\overline{\mathrm{AB}}\) such that \(\overline{\mathrm{AB}}\) = 7 cm and took a point P anywhere on the line.
(ii) Placing the set square on the line in such a way that the vertex of its right angle coincides with P and one arm of the right angle coincides with the line AB.
(iii) Drawn a line PQ through P along the other arm of the right angle of the set square,
(iv) The line PQ is perpendicular to the line AB at P. ie PQ ⊥ AB and ∠APQ = ∠BPQ = 90°

Question 2.
Draw a line segment LM = 6.5 cm and take a point P not lying on it. Using set square construct a line perpendicular to LM through P.
Solution:
Construction:
(i) Drawn a line segment \(\overline{\mathbf{L M}}\) such that \(\overline{\mathbf{L M}}\) = 6.5 cm and marked a point P anywhere above \(\overline{\mathbf{L M}}\)
(ii) Placing one of the arms of the right angle of the set square along the line segment LM.
(iii) Sliding the set square along the line segment in such a way that the other arm of its right angle touches the point P. Draw a line along this side, passing through point P meeting \(\overline{\mathbf{L M}}\) at Q.
(iv) The line PQ is perpendicular to the line segment LM. ie LM ⊥ PQ.

Question 3.
Find the distance between the given lines using a set square at two different points on each of the pairs of lines and check whether they are parallel.

Solution:
Making the points P, Q, R, S and A, B, C, D on the given lines
PQ = RS = 0.9 cm
AB = CD = 1 cm
Distance between the two lines are equal.
They are parallel lines.

Question 4.
Draw a line segment measuring 7,8cm. Mark a point B above it at a distance of 5 cm. Through B draw a line parallel to the given line segment.
Solution:
Construction:
(i) Using a scale drawn a line segment \(\overline{\mathrm{PQ}}\) = 7.8 cm. Marked a point A on the line.
(ii) Placing the set square in such a way that the vertex of the right angle coincides with A and one of the edges of right angle lies along the line segment PQ. Mark a point B. Such that AB = 5 cm above the line PQ.
(iii) Placed the scale and the set square in such a way the set square is below PQ and one edge that form right angle with PQ. Placed the scale along the other edge of the right angle.
(iv) Holding the scale firmly and sliding the set square along the edge of the scale until the edge touches the point B. Drawn the line BC through B.
(v) Now the line BC is parallel to PQ i.e, BC || PQ.

Question 5.
Draw a line. Mark a point R below it at a distance of 5.4 cm. Through R draw a line parallel to the given line.
Solution:
(i) Using a scale drawn a line AB and marked a point Q on the line.
(ii) Placing the set square in such a way that the vertex of the right angle coincides with Q and one of the edges of the right angle lies along AB. Marked the Point R such that QR = 5.4 cm.
(iii) Placing the set square above AB in such a way that one of the edges that form a right angle with AB. Placed the scale along the other edge of the right angle.
(iv) Holding the scale firmly and slide the set square along the edge of the scale until the edge touches the point R. Drawn the line RS through R. The line RS is parallel to AB i.e RS || AB.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Intext Questions

Try These (Textbook Page No.49)

Question 1.
Arrange in ascending order
(i) C.P, M.P, Discount
(ii) M.P., S.P., Discount
Solution:
(i) Discount < C.P < M.P.
(ii) Discount < S.P < M.P

Question 2.
Which is greater S.P or M.P?
Solution:
M.P is greater than S.P

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Additional Questions

Question 1.
______ is the difference between S.P and C.P., when S.P > C.P.
Solution:
Profit

Question 2.

(a) What is the total bill amount?
(b) Which item is the costliest one?
(c) Write the name of the shop?
Solution:
(a) ₹ 4,050
(b) Tables
(c) ABC Furnitures, Madurai

Question 3.
A vegetable vendor bought a dozen drum sticks for ₹ 42. Two of them become dry. If he has to get a profit of ₹ 8. Find the S.P of each drumstick.
Solution:
12 drumsticks cost price = ₹ 42
Profit = ₹ 8
Profit = S.P – C.P
⇒ 8 = S.P – 42
⇒ S.P = 8 + 42 = ₹ 50
S.P of 10 drumsticks = ₹ 50
S.P of each drumstick = \(\frac { 50 }{ 10 }\) = ₹ 5

Question 4.
Rice is being sold at ₹ 1800 per bag of 25kg at profit of ₹ 250. Find the Cost price of the rice bag.
Solution:
S.P = ₹ 1800
Profit = ₹ 250
Profit = S.P – C.P
⇒ 250 = 1800 – C.P
⇒ C.P = 1800 – 250 = ₹ 1550

Question 5.
Mangai buys a book for ₹ 300. She wants to sell it at a profit of ₹ 50 after making a discount of ₹ 30. What is the M.P of a book?
Solution:
Profit = S.P – C.P
⇒ 50 = S.P – 300
⇒ S.P = 50 + 300 = 350
S.P = M.P – Discount
⇒ 350 = M.P – 30
⇒ M.P = 350 + 30 = ₹ 380

Question 6.
Manila buys a dress for ₹ 750. She wants to sell it at a profit of ₹ 75 on sales and she marks ₹ 900. What is the discount that she will give to her customers?
Solution:
Profit = S.P – C.P
⇒ 75 = S.P – 750
⇒ S.P = 75 + 750 = ₹ 825
S.P = M.P – Discount
⇒ 825 = 900 – Discount
⇒ Discount = 900 – 825 = ₹ 75

Question 7.
Complete the following table:

Solution:
(i) Profit = S.P – C.P = 60 – 50 where CP < SP = ₹ 10
(ii) S.P = C.P – loss = 80 – 15 = ₹ 65

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Intext Questions

Activity (Text book Page No. 39)

Question 1.
Observe the following shapes and answer the questions given below.

i) Mark the closed shapes as ‘✓’ and open shapes as ‘✗’
ii) Find the measure of the boundary of closed shapes by using a ruler.
iii) Which closed shape has the shortest boundary?
iv) Which closed shape has the longest boundary?
Activity to be done by the students themselves

Try this (Text book Page No. 40)

Question 1.

Is the perimeter of the given shape possible ? why ?
Solution:
Not possible. Because perimeter is the length of the boundaries of any closed shape. It is not a closed shape.

Try These (Text book Page No.41)

Question 1.
Draw a shape with perimeter 16 cm in a dot sheet.
Solution:

Question 2.
What is the perimeter of a rectangle if the length is twice its breadth?
Solution:
Perimeter of a rectangle = 2(l + b) units
= 2 × (2b + b) units
= 2 × 3b units = 6b units

Perimeter of a rectangle = 6b units.

Question 3.
What would be the perimeter of a square if its side is reduced to half?
Solution:
Let a side of a square = s units.

Question 4.
What is the perimeter of a triangle if all sides are equal in length?
Solution:
Perimeter of a triangle P = sum of its 3 sides.
If all sides are equal, it is an equilateral triangle with side say a.
Perimeter = a + a + a units = 3a units

Activity (Text book Page No. 41)

Question 1.
Choose any five items like Table, A4 sheet, Note-book, etc in the classroom. Guess the approximate length of each side by observation and write down the estimated perimeter of the item. Then, measure by using ruler and record the actual perimeter and find the difference in the following table (to the nearest cm)

Solution:
Activity to be done by the students themselves

Try this (Text book Page No. 42)

Question 1.

Can different shapes have the same perimeter
Solution:
Yes, different shapes can have the same perimeter.

Try These (Text book Page No. 43)

Question 1.
Find the breadth of the rectangle with perimeter 14 m and length 4 m.
Solution:
Given perimeter of the rectangle P = 14 m
length l = 4 m

Question 2.
The perimeter of an isosceles triangle is 21 cm. Find the measure of equal sides given that the third side is 5 cm.
Solution:
Perimeter of the triangle P = (a + b + c) units.
21 = [(a + b) + 5] cm
21 – 5 = a + b
16 = a + b
Here a and b are equal sides and let a = b
16 = a + a
2a = 16
\(a=\frac{16}{2}\)
a = 8 cm.
∴ Equal sides measure 8 cm.

Try these (Text book Page No. 44)

Question 1.
Find the number of tiles required to fill the area of following figures

Solution:
i) Total number of tiles required = 16
Number of tiles already filled = 7
Remaining required tiles = 16 – 7 = 9
ii) Total number of tiles required = 12
Number of tiles already filled = 6
Remaining required tiles = 12 – 6 = 6
iii) Total number of tiles required = 12
Number of tiles already filled = 6
Remaining required tiles = 12 – 6 = 6
iv) Total number of tiles required = 16
Number of tiles already filled = 8
Remaining required tiles = 16 – 8 = 8

Activity (Text book Page No. 45)

Question 1.
Mark the base and height of the following right angled triangle.

Solution:
Activity to be done by the students themselves

Try These (Text book Page No. 45)

Question 1.
Draw the following in a graph sheet?
i) Two different rectangles whose areas are 16cm²
Solution:

Activity (Text book Page No. 46)

Question 1.
Find the area of the given ‘L’ shaped rectangular figure by dividing it into squares of equal sizes
Solution:
Activity to be done by the students themselves

Try this (Text book Page No.46)

Question 2.
Can you find the area of ‘L’ shaped figure as the difference between two areas.
Solution:
Here Area of L shape = (Area of the rectangle ABCD)
(Area of the rectangle GFHB) = (2 × 3) – (2 × 1) cm²
= 6 – 2 cm² = 4 cm²

Try these (Text book Page No. 46)

Question 1.
Measure using ruler and find the perimeter of each of the following diagram

Solution:
i) Perimeter = 2.5 + 0.5 + 1 + 1 + 0.5 + 1 + 0.5 + 1 + 0.5 + 1.5 = 10 cm
ii) Perimeter = 2.5 + 0.5 + 2.5 + 0.5 + 0.5 cm = 6.5 cm
iii) Perimeter = 0.5 + 1 + 0.5 + 0.5 + 1 + 1 + 0.5 + 1 + 1.5 + 0.5 + 1 + 0.5 cm = 9 cm
iv) Perimeter = 2 + 0.5 + 1.5 + 1.5 + 0.5 + 0.5 + 2 + 1 cm = 9.5 cm

Activity (Text book Page No 46)

Question 1.
Form all possible shapes of perimeter 80 cm with 9 identical squares, each of side 4 cm.
Solution:
Activity to be done by the students themselves

Activity

Question 2.
Cut a rectangular sheet along one of its diagonals. Two identical scalene right angled triangles are obtained. Join them along their sides of identical length in all possible ways. Six different shapes can be obtained. Four of them are given. Find the remaining two shapes. Find the perimeter of all the six shapes and fill in the table.



Based on the above activity answer the following questions:
i) Are the perimeters same for all the shapes?
ii) Which shape has the longest perimeter?
iii) Which shape has the shortest perimeter?
iv) Are the areas of all the shapes same? why?
Solution:
Activity to be done by the students themselves

Try These (Text book Page No. 49)

Question 1.
Find the approximate area of the following figures:

Solution:

Try these (Text book Page No. 50)

Question 2.
Fill in the blanks.
(i) 7 cm² = ____ mm²
Hint: 1 cm² = 100 m
Solution:
700

(ii) 10 m² = ___ cm²
Hint: 1 m² = 10000 cm²
Solution:
1,00,000

(iii) 3 km² = ____ m²
Hint: 1 km² = 1000000
Solution:
30,00,000

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Intext Questions

Try These (Textbook Page No. 26)

Question 1.
Complete the following table:

Solution:

Question 2.
Determine which metric unit you would use to express the following:

1. The length of your middle finger.
2. The weight of an elephant.
3. The weight of the ring.
4. The weight of the tablet.
5. The length of the safety pin.
6. The height of the building.
7. The length of the seashore in Tamilnadu.
8. The volume of a cup of coffee.
9. The capacity of water in the tank.

Solution:

1. Cm
2. Kilogram
3. gram
4. milligram
5. millimetre
6. metre
7. Kilometre
8. Millilitre
9. Litre

Try These (Textbook Page No. 28)

Question 1.
Convert the following.
(i) 23 km into m
(ii) 1.78 m into cm
(iii) 7814 m into km
(iv) 8.67 mm into cm
(v) 16 l into ml
(vi) 1500 ml into l
(vii) 2360 l into kl
(viii) 873 l into ml
(ix) 40 mg into g
(x) 1550 g into kg
(xi) 6.5 kg into g
(xii) 723 g into mg
Solution:

Try These (Textbook Page No. 30)

Question 1.
Five kilograms of compost is needed for a coconut tree for every six months. How many kilograms of compost is needed for 50 such coconut trees for one and a half years?
Solution:
Compost needed for a tree for 6 months = 5 kg
Compost needed for 50 trees for half a year = 5 × 50 = 250 kg.
Compost needed for 1\(\frac { 1 }{ 2 }\) years = 250 × 3 = 750 kg. [∴ 1\(\frac { 1 }{ 2 }\) years = 3 half years]

Question 2.
Say True or False and Justify.
(a) Is it correct: 4 m + 3 cm = 7 m
(b) Can we add the following?
(i) 6 litre + 7 kg
(ii) 3 m + 5 l
(iii) 400 ml + 300 g
Solution:
(a) False.
4 m + 3 cm = 400 cm + 3 cm = 403 cm.
(b) (i) No, we can’t add.
(ii) No. We can’t add.
(iii) No. We can’t add.

Recap (Textbook Page No. 33)

Question 1.
Read and write the time in the appropriate place.

Solution:

Try These (Textbook Page No. 33)

Question 1.
Say the following time in two ways.
(a) 9.20
(b) 4.50
(c) 5.15
(d) 6.45
(e) 11.30
Solution:
(a) 9.20
20 minutes past 9
40 minutes to 10
(b) 4.50
50 minutes past 4
10 minutes to 5
(c) 5.15
15 minutes past 5
45 minutes to 6
(d) 6.45
45 minutes past 6
15 minutes to 7
(e) 11.30
30 minutes past 11
30 minutes to 12

Try These (Textbook Page No. 34)

Question 1.
Convert the following:
(i) 4 hours = _____ minutes
(ii) 240 minutes = ____ hours
(iii) 30 minutes = _____ seconds
(iv) 3600 seconds = _____ hours
(v) 2 hours = _____ seconds
Solution:
(i) 4 hours = 240 minutes
4 hours = 4 × 60 minutes = 240 minutes
(ii) 240 minutes = 4 hrs
240 minutes = \(\frac{240}{60}\) = 4 hrs
(iii) 30 minutes = 1800 seconds
30 minutes = 30 × 60 seconds = 1800 seconds
(iv) 3600 seconds = 1 hrs
3600 seconds = \(\frac{3600}{60}\) minutes = 60 minutes = \(\frac{60}{60}\) hours = 1 hour
(v) 2 hrs = 7200 seconds
2 hrs = 2 × 60 minutes = 120 minutes = 120 × 60 seconds = 7200 seconds

Try These (Textbook Page No. 36)

Question 1.
Convert the 12-hour format into the 24-hour format and vice versa.

Solution:
10 : 40 a.m. = 10 : 40 hours
11 a.m. = 11 : 00 hours
1 : 15 a.m. = 01 : 15 hours
5 a.m. = 05 : 00 hours
16 : 20 hours = 4 : 20 p.m.
00 : 40 hours = 12 :40 a.m.
1 p.m. = 13 : 00 hours
11 : 15 p.m. = 23 : 15 hours
3 p.m. = 15 :00 hours
12 midnight = 00 : 00 hours
12 : 25 hrs = 00 : 25 p.m.
4 : 10hrs = 4 : 10 a.m.

Try These (Textbook Page No. 39)

Question 1.
Check whether the following years as Ordinary or Leap Year?
1994; 1985; 2000; 2007; 2010; 2100
Solution:
We know that a leap year is divisible by 4.
In the case of a century which is divisible by 400 is a leap year.
1994 – Not a leap year (Not divisible by 4)
1985 – Not a leap year (Not divisible by 4)
2000 – It is a leap year (divisible by 400)
2007 – Not a leap year (Not divisible by 4)
2010 – Not a leap year (Not divisible by 4)
2100 – Not a leap year (Not divisible by 400)

Question 2.
How many days are there from 1st April to 30th June?
Solution:
Months from April to June are:

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 2 Integers Intext Questions

(Try These Text Book Page No. 26)

Question 1.
Read the following numbers orally.
(i) + 24
(ii) -13
(iii) -9
(iv) 8
Solution:
(i) + 24 is read as ‘ plus 24’ or ‘positive 24’
(ii) -13 is read as ‘minus 13’ or ‘negative 13’
(iii) -9 is read as ‘minus 9’ or ‘ negative 9’
(iv) 8 is read as ‘plus 8’ or ‘positive 8’ or simply ‘8’

Question 2.
Draw a number line and mark the following integers.
(i) 0
(ii) -6
(iii) 5
(iv) -8
Solution:

Question 3.
Are all natural numbers integers?
Solution:
Yes, all natural numbers are positive integers

Question 4.
Which part of the integers are not whole numbers?
Solution:
The numbers 0, 1, 2, 3, are whole numbers. Therefore negative part of the integers are not whole numbers.

Question 5.
How many units should you move to the left of 3 to reach -4.
Solution:

From 3, we should move 7 units to the left to reach -4.

(Activity Textbook Page No. 27)

Question 1.
Ask your parents / grandparents about the depth at which the various types of vegetables (seeds) should be planted, for their better and efficient growth. For the same, draw a number line indicating the depth of various vegetable seeds. (Draw the planting chart!).
Solution:
Activity to be done by the students themselves

(Try These Text Book Page No. 29)

Question 1.
Find the opposite of the following numbers :
(i) 55
(ii) -300
(iii) +5080
(iv) -2500
(v) 0
Solution:
For every positive integer, there is a corresponding negative integer as its opposite number and vice versa

Question 2.
Represent the following situations as integers.
i) A loss of ₹ 2000
ii) 2018 CE (AD)
iii) Fishes found at 60 m below the sea level
iv) 18°C below 0°C
v) Gaining 13 points
vi) A jet plane at a height of 2500 m
Solution:
(i) Since loss is negative, a loss of t 2000 is denoted as – ₹ 2000
(ii) Here CE (AD) can be considered as a positive number. Hence CE (AD) 2018 can be represented as + 2018.
(iii) Below sea level is represented as -60 m.
(iv) 18°C below 0°C is represented as – 18°C
(v) Gaining 13 points can be represented as + 13
(vi) Height of 2500 m is represented as + 2500 m.

Question 3.
Suppose in a building, there are 2 basement floors. If the ground floor is denoted as zero, how can we represent the basement floors?
Solution:
If the ground floor is denoted as 0, basement floors are represented as -1 and -2.

(Try This Textbook Page No. 31)

Question 1.
For two numbers, say 3 and 5, we know that 5 > 3. Will there be a change in the inequality if both the numbers have negative sign before them?
Solution:
We know that 5 > 3
But when we change the sign of both the numbers we get – 5 and – 3.
Here -5 is 2 units to the left of-3 and so – 5 < 3.
∴ The inequality changes when the signs of both the numbers are changed.

(Activity Textbook Page No. 31)

Question 1.
Take two cards from a deck of playing cards and identify, which is greater between them, assuming that the Joker card represents zero, black cards represent positive integers, red cards represent negative integers and the cards A, J, Q and K represent 1,11,12 and 13 respectively.
Solution:
Activity to be done by the students themselves

(Try These Textbook Page No. 31)

Question 1.
Is -15 < -26? Why?
Solution:
Given statement is wrong.
We will draw the number line and plot the numbers on it.

So -26 < -15

Question 2.
Which is smaller -3 or -5? Why?
Solution:
Comparing the numbers from the number line.

– 5 is 2 units to the left of -3 and so -5 is smaller than -3. i.e. – 5 < – 3

Question 3.
Which is greater 7 or -4? Why?
Solution:
7 is a positive integer and lies to the right of 0.-4 is a negative integer and lies to the left of 0
∴ 7 > -4

Question 4.
Which is the greatest negative integer?
Solution:

– 1 is the greatest negative integer.

Question 5.
Which is the smallest positive integer?
Solution:

+ 1 is the smallest positive integer.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 4 Geometry Intext Questions

Try this (Text book Page No. 56)

Question 1.

The diagonal of a rectangle divides it into two equal halves but it is not a line of symmetry. why?
Solution:
We get two equal halves. But one half of the figure does not coincide exactly with
the other half with the line of symmetry. Therefore it is not a line of symmetry.

Try These (Text book Page No. 57)

Question 1.
Is the dotted line shown in each figure a line of symmetry? If yes put ✓otherwise put ✗ Justify your answer.

Question 2.
Check the following figures for symmetry ? Write YES or NO.

Try these (Text Book Page No. 59)

Question 1.
Draw the following figures in a paper. Cut out each of them and fold so that the two parts of each figure exactly coincide.

i) Which of the above figures have one, two or more lines of symmetry?
ii) Which of the above figures do not have any line of symmetry?
Solution:

2) All have lines of symmetry

Question 2.
Write the numbers from 0 to 9.
(i) Which numbers have a line of symmetry?
(ii) List out the numbers which do not have a line of symmetry.
Solution:
i) Numbers 1 and 3 have one line of symmetry.
ii) Numbers 2, 4, 5, 6, 7 and 9 do not have a line of symmetry7.

Activity (Text Book Page No. 59)

Question 1.
Complete the other half of the following figures such that the dotted line is the line of symmetry.

Activity to be done by the students themselves

Try this (Text Book Page No. 60)

Question 1.
Which other capital letters of English alphabets look like the same in the mirror?
Solution:
A, H, I, M, O, T, U, V, W, X and Z.

Activity (Text book Page No. 62)

Symmetrical figures by ink blots
Step 1 : Take a sheet of paper and fold it into half to make the crease.
Step 2 : Put some ink blots on one side of the crease of the paper.
Step 3 : Fold the paper along the crease and press it.
Step 4 : Open the paper, you will find an imprint of the ink blots on the other part also which is symmetrical about the crease.

Activity to be done by the students themselves

Try These (Text book Page No. 62)

Question 1.
Find the password :
“Kannukkiniyal has a new game app in her laptop protected with a password. She has decided to challenge her friends with this paragraph which contains that password”.
If you follow the steps given below, you will find it.
Steps :
i) Write the above paragraph in capita! letters.
ii) Turn the paper upside down and look at it in the mirror.
iii) The word which remains unchanged in the mirror is the password.
Solution:
i) KANNUKK IN IYAL HAS ANEW GAME APP IN HER LAPTOP PROTECTED WITH A PASSWORD SHE HAS DECIDED TO CHALLENGE HER FRIENDS WITH THIS PARAGRAPH WHICH CONTAINS THAT PASSWORD

iii) “DECIDED” is the password.

Question 2.
Form words using the Setters B, C, D, E,H, I, K, O and X. Write those words in paper in capital letters. Turn it upside down and look at them in the mirror.
Solution:
i) List the letters which have horizontal and vertical line of symmetry
ii) Do the words HIKE, DICE, COOK remain unchanged in the mirror?
iii) The words which you have found that remain unchanged in the mirror are ______, ______, ______
Solution:
B, C, D, E, H, I, K, O, X
i) Vertical lines of symmetry A, H, I, M, O, T, U, V, W, X, Y
Horizontal lines of symmetry B, C, D, E, H, I, K, O, X
ii) Yes
iii) HOOK, KICK, ICE

Try this (Text book Page No. 64)

Question 1.
Can you identify the object which does not have rotational symmetry in the above situations? Why?
Solution:
1) Computer mouse
2) Trapezium
3) Chair
4) School bags etc.

Question 2.
Suppose, the man hole cover of the water sump is in circular shape
i) The number of ways to dose that circular lid is ___
ii) What is its order of rotational symmetry?

Solution:
i) many
ii) Infinite

Activity (Text book Page No. 64)

Question 1.
Find the order of rotational symmetry by fixing the relevant shape in different ways.

Activity to be done by the students themselves