Class 6

Students can practice TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.6 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Exercise 3.6

Question 1.
Find the LCM and HCF of the following numbers.
(i) 15, 24
(ii) 8, 25
(iii) 12,48
check their relationship.
Answer:
(i) Factors of 15= 3 × 5
Factors of 24 = 3 × 2 × 2 × 2
LCM of 15 and 24 = 3 × 5 × 2 × 2 × 2= 120
HCF of 15 and 24 = 3
LCM × HCF = 120 × 3 = 360
Product of the two numbers = 15 × 24 = 360
∴ LCM × HCF = Product of the two numbers

(ii) Factors of 8 = 2 × 2 × 2
Factors of 25 = 5 × 5
LCM of 8 and 25 = 2 × 2 × 2 × 5 × 5 = 200
HCF of 8 and 25 = 1
∴ LCM × HCF = 200 × 1 = 200
Product of the two numbers = 8 × 25 = 200
∴ LCM × HCF = Product of the two numbers

(iii)

LCM of 12 and 48 = 2 × 2 × 3 × 2 × 2 = 48
HCF of 12 and 48 = 2 × 2 × 3 = 12
LCM × HCF = 12 × 48576
Product of the two numbers = 12 × 48 = 576
∴ LCM × HCF = Product of the two numbers

Question 2.
If the LCM of two numbers is 216 and their product is 7776, what will be its HCF?
Answer:
Product of the two numbers = 7776
LCM of two numbers = 216
We know, LCM × HCF
= Product of the two numbers
∴216 × HCF = 7776
∴HCF = \(\frac{7776}{216}\) = 36

Question 3.
The product of two numbers is 3276.
If their HCF is 6, find their LCM.
Answer:
Product of the two numbers = 3276
HCF of the two numbers = 6
We know, LCM × HCF = Product of the two numbers
LCM × 6 = 3276
∴ LCM = \(\frac{3276}{6}\) = 3276

Question 4.
The HCF of two numbers is 6 and their LCM is 36. If one of the numbers is 12, find the other.
Answer:
The HCF of two numbers = 6
The LCM of two numbers = 36
One of the numbers = 12
Let the other number be x.
HCF × LCM = 6 × 36
Product of the two numbers = 12 × x
We know, LCM × HCF = Product of the two numbers
6 × 36 = 12 × x
(i.e.) 12 × x = 6 × 36
∴ x = \(\frac{6 \times 36}{12}\) = 18
∴ The other number is 18.

Students can practice TS SCERT Class 6 Maths Solutions Chapter 10 Perimeter and Area Ex 10.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area Exercise 10.2

Question 1.
Find the areas of the following figures by counting squares.
(i) 50 cm and 20 cm
(ii) 65 in and 45 in
(iii) 25 cm and 16 cm
(iv) 7 kin and 19 km
Answer:
(i) Length of the rectangle = 50 cm
Breadth of the rectangle = 20 cm
∴ Area of the rectangle
= length × breadth
= 50 cm × 20 cm
= 1000 sq. cm (cm²)

(ii) Length of the rectangle = 65 m
Breadth of the rectangle = 45 m
∴ Area of the rectangle
= length × breadth
= 65 cm × 45 cm
= 2925 sq. m (m²)

(iii) Length of the rectangle = 25 cm
Breadth of the rectangle = 16 cm
∴ Area of the rectangle
= length × breadth
= 25 cm × 16 cm
= 400 sq. cm (cm²)

(iv) Length of the rectangle = 7 km
Breadth of the rectangle =19 km
∴ Area of the rectangle
= length × breadth
= 7 cm × 19 cm
= 133 sq. km (km²)

Question 2.
Find the area of squares with the given sides:
(i) 26 m
(ii) 17 km
(iii) 52 cm
(iv) 8 cm
Answer:
(i) Side of the square = 26 m
Area of the square = side × side
= 26m × 26m
= 676 m²

(ii) Side of the square =17 km
Area of the square = side × side
= 17 cm × 17cm
= 289 km²

(iii) Side of the square = 52 cm
Area of the square = side × side
= 52 cm × 52cm
= 2704 cm²

iv) Side of the square = 8 cm
Area of the square = side × side
= 8 cm × 8 cm
= 64 cm²

Question 3.
The area of rectangular frame is 1,125 sq. cm. If its width is 25 cm, what is its length ?
Answer:
Area of rectangular frame =1125 sq.cm
Width of the frame = 25 cm
Area of rectangular frame
= length × width

Question 4.
The length of a rectangular field is 60 m and the breadth is half of its length. Find the area of the field.
Answer:
The length of a rectangular field = 60 m By problem,
Area of the rectangular field = \(\frac{60 \times 1}{2}\) = 30 m
= length × breadth
= 60 m × 30 m
= 1800 m²

Question 5.
A square sheet of paper has a perimeter of 40 cm. What is the length of its side ? Also find the area of the square sheet ?
Answer:
Let the side of the square sheet of paper = x cm.
Perimeter of the sheet of paper = 4 × side
= 4 × x
= 4x cm
By problem,
4x = 40
∴ x = \(\frac{40}{4}\) = 10 cm
Side of the square sheet of paper = 10 cm
Area of the sheet of paper = side × side
= 10 cm × 10 cm
= 100 cm²
Area of sheet of paper = 100 cm²

Question 6.
The area of rectangular plot is 2400 square meters and it’s length is 1\(\frac{1}{2}\) times to it’s breadth.What is the perimeter?
Answer:
Area of rectangular plot = 2400 m²
Length is 1\(\frac{1}{2}\) to its breadth i.e, their ratio of its sides = 3:2
Let the length of the plot be 3x m.
Breadth of the plot = 2x m
Area of the plot = length × breadth
= 3x m × 2x m
= 6x²
By problem, 6x² = 2400
x² = \(\frac{2400}{6}\) = 400
x² = 400 = 20 × 20
∴ x = 20
Length of the plot = 3x = 3 × 20 = 60 m
Breadth of the plot = 2x = 2 × 20 = 40 m
Perimeter of the plot
= 2(length + breadth)
= 2(60 + 40) m
= 2 × 100
= 200 m

Question 7.
The length and breadth of a room are 6 m and 4 m respectively. How many square meters of carpet is required to completely cover the floor of the room ? If the carpet costs ₹ 240 a square meter, what will be the total cost of the carpet for completely covering the floor ?
Answer:
Length of the room = 6 m
Breadth of the room = 4 m
Area of the floor of the room
= length × breadth
= 6m × 4m
= 24m²
The carpet required to completely cover the floor of the room = 24 m²
The cost of 1 sq. m of the carpet
= Rs. 240
∴ The cost of 24 sq. m of the carpet
= Rs. 240 × 24 = Rs. 5760

Question 8.
Two fields have the same perimeter. One is a square of side 72 m and another is a rectangle of length 80 m. Which field has the greater area and by how much ?
Answer:

Area of square = l × b
= 72 m × 72 m = 5184 sq.m
Length of a rectangle l = 80 m
Perimeter of a rectangle 2(l + b) = 288 m
= 2 × 80 m + 2b
= 288 m
= 160 m + 2b
= 288 m
⇒ 2b = 288 – 160 = 128 m
b = \(\frac{128}{2}\) = 64 m
Area of rectangle = 80 m × 64 m
= 5120 sq.m
∴ The area of square field is greater than that of rectangular field.
5184 – 5120
= 64 sq.m

Question 9.
The area of a square is 49 sq. dm. A rectangle has the same perimeter as the square. If the; length of the rectangle is 9.3 cm., What is its breadth ? Also find which has greater area ?
Answer:
Area of the square = 49 sq. cm
side × side = 7 cm × 7 cm
Side of the square = 7 cm
The perimeter of the square = 4 × side = 4 × 7 cm
= 28 cm
By problem,
The perimeter of the rectangle = 28 cm
2(l + b) = 28 cm 28
(l + b) = \(\frac{28}{2}\) = 14 cm
9.3 + b = 14
∴ b= 14 – 9.3 = 4.7 cm
Area of the rectangle = length × breadth
= 9.3 cm × 4.7 cm = 43.71 cm²
∴ The square has greater area when compared to the rectangle.

Question 10.
Rahul owns a rectangular field of length 400 m and breadth 200 m. His friend Ramu owns a square field of length 300 m. Find the cost of fencing the two fields at ₹ 150 per meter. If one tree can be planted in an area of 10 sq. m. who can plant more trees in his field ? How many more trees can he plant ?
Answer:
Length of the rectangular field of Rahul
= 400 m
Breadth of the rectangular field of Rahul
= 200 m
Perimeter of rectangular field of Rahul
= 2 [length + breadth]
= 2 [400 + 200]m = 2 × 600 m = 1200 m
Side of the square field of Ramu
= 300 m
Perimeter of square field of Ramu = 4 × side
= 4 × 300 m = 1200 m
Total length of the wire to be fenced = 1200 + 1200 = 2400 m
Cost of fencing 1 meter = Rs. 150

Cost of fencing 2400 meters
= Rs. 2400 × 150
= Rs. 3,60,000

Cost of fencing Ramu’s field
= Rs. 1200 × 150
= Rs. 1,80,000

Area of Rahul’s field = length × breadth
= 400 m × 200 m
= 80,000 m²

A tree occupies an area of 10 m²
Number of trees that can planted in 80,000
Rahul’s field = \(\frac{80,000}{10}\) = 8000
Area of Ramu’s field = side × side
= 300 × 300
= 90,000 m²

A tree occupies an area of 10 m²
Number of trees that can be planted in
Ramu’s field = \(\frac{90,000}{10}\) = 9000
Ramu can plant 1000 trees
(= 9000 – 8000) more.

Question 11.
The length of a rectangular floor is 20 m., more than its breadth. If the perimeter of the floor is 280 m, what is its length ?
Answer:
Let the breadth of the rectangular field = x m
∴ The length of the rectangular field = (x + 20) m

Perimeter of the rectangular field
= 2(length + breadth)
= 2(x + 20 + x) m
= 2(2x + 20)m
= 4x + 40 m
By problem,
4x + 40 = 280 m
4x = 280 – 40
= 240 m
x = \(\frac{240}{2}\) = 60 m

The breadth of the rectangular field
= 60 m
The length of the rectangular field
= x + 20 m
= 60 + 20 m
= 80 m
Its length is 80 m.

Question 12.
A rectangular plot of land is 240 m. by 200 m. The cost of fencing per meter is ₹ 30. What is the cost of fencing the entire field ?
Answer:
The dimensions of the rectangular plot of land are 240 m and 200 m.
Perimeter of the plot of land
= 2 [240 + 200]
= 2[440]m
= 2 × 440 m
= 880 m
Cost of fencing 1 meter is Rs. 30.
Cost of fencing 880 metres is x
Rs. 880 × 30 = Rs. 26400

Question 13.
The side of a square field is 120 meters. The cost of preparing a grass lawn is ₹ 35 per square meter. How much will it cost, if the entire field is converted into a lawn ?
Answer:
The side of the square field = 120 m
Area of the square field = side × side
= 120 m × 120 m
= 14,400 m²
Cost of preparing a sq. m of grass lawn is Rs. 35.
Cost of converting the entire field (14400 m²) into a lawn
= Rs. 14,400 × 35
= Rs. 5,04,000

Question 14.
What will happen to the area of rectangle, if
(i) its length and breadfii are doubled?
(ii) its length is doubled and breadth is tripled ?
Answer:
(i) Let the length of a rectangle be x m and the breadth be y m.
∴ Area of the rectangle = x × y = xy m²
The length and breadth of it are doubled.
Then the length will be 2x (= x + x) and breadth 2y (= y + y).
The area of the rectangle will be
2x × 2y = 4xy m² = 4 (xy m²)
∴ The area of the rectangle increases by 4 times.

(ii) The length after it is doubled will be
x + x = 2x
The breadth after it is tripled will be y + y + y = 3y
The area of the rectangle will be 2x × 3y
= 6 xy m²
= 6 × (xy m²)
∴ The area of the rectangle increases by 6 times.

Question 15.
What will happen to the area of square if its side is:
(i) doubled
(ii) halved
Answer:
(i) Let the side of a square be x metres.
Area of the square = side × side
= x × x
= x² sq. metres
If the side is doubled, it becomes x + x
= 2x metres

Area of the square = side x side
= 2x m × 2x m
= 4x² sq. metres
= 4 x (x² sq. metres)
The area increases by 4 times,

(ii) If the side of the square is halved, it
Area of the square = side × side
= \(\frac{1}{2}\) x m x \(\frac{1}{2}\) x m
= \(\frac{1}{4}\) x² sq. metres
= \(\frac{1}{4}\) × (x² sq. metres)
∴ The area of the square becomes \(\frac{1}{4}\) of the original area.

Students can practice TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.7 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Exercise 3.7

Question 1.
Which of the following numbers are divisible by 4 ?
(i) 572
Answer:
The given number is 572.
The number formed by its last two digits is 72.
It is divisible by 4. So, the given number is divisible by 4.

(ii) 21,084
Answer:
The given number is 21,084.
The number formed by its last two digits is 84.
It is divisible by 4. So, the given number is divisible by 4.

(iii) 14,560
Answer:
The given number is 14,560.
The number formed by its last two digits is 60.
It is divisible by 4. So, the given number is divisible by 4.

(iv) 1,700
Answer:
The given number is 1,700.
1700 = 1000 + 600 + 100
1000, 600 and 100 are multiples of 100, they are completely divisible by 4.
So, the given number is divisible by 4.

(v) 2,150
Answer:
The given number is 2,150.
The number formed by its last two digits is 50.
It is not divisible by 4.
So, the given number is not divisible by 4.

Question 2.
Test whether the following numbers are divisible by 8.
(i) 9774
Answer:
The given number is 9774.
The number formed by its last three digits is 774.
It is not divisible by 8.
So, the given number is not divisible by 8.

(ii) 531048
Answer:
The given number is 531048.
The number formed by its last three digits is 048.
It is divisible by 8.
So, the given number is divisible by 8.

(iii) 5500
Answer:
The given number is 5500.
The number formed by its last three digits is 500.
It is not divisible by 8.
So, the given number is not divisible by 8.

(iv) 6136
Answer:
The given number is 6136.
The number formed by its last three digits is 136.
It is divisible by 8.
So, the given number is divisible by 8.

(v) 4152
Answer:
The given number is 4152.
The number formed by its last three digits is 152.
It is divisible by 8.
So, the given number is divisible by 8.

Question 3.
Check whether the following numbers are divisible by 11.
(i) 859484
Answer:
The given number is 859484.
Sum of the digits at odd places = 4 + 4 + 5 = 13
Sum of the digits at even places = 8 + 9 + 8 = 25
Their difference = 25 – 13 = 12
This difference is not either 0 or divisible by 11.
So, the given number is not divisible by 11.

(ii) 10824
Answer:
The given number is 10824.
Sum of the digits at odd places. = 4 + 8 + 1 = 13
Sum of the digits at even places = 2 + 0 = 2
Their difference =13 – 2 = 11
This difference 11 is divisible by 11.
∴ The given number is divisible by 11.

(iii) 20801
Answer:
The given number is 20801.
Sum of the digits at odd places = 1 + 8 + 2 = 11
Sum of the digits at even places = 0 + 0 = 0
Their difference = 11 – 0 = 11
This difference 11 is divisible by 11.
∴ The given number is divisible by 11.

Question 4.
Verify whether the following numbers are divisible by 4 and by 8 ?
(i) 2104
Answer:
The given number is 2104.
The number formed by its last two digits is 04.
It is divisible by 4. So, the given number is divisible by 4,
The number formed by its last three digits is 104. It is divisible by 8.
So, the given number is divisible by 8.

(ii) 726352
Answer:
The given number is 726352.
The number formed by its last two ‘ digits is 52.
It is divisible by 4. So, the given number is divisible by 4.
The number formed by its last three digits is 352. It is divisible by 8.
So, the given number is divisible by 8.

(iii) 1800
Answer:
The given number is 1800.
1800 = 1000 + 800
1000 and 800 are multiples of 100.
We know that 100 is divisible by 4. So, the given number is divisible by 4.
The number formed by its last three digits is 800. It is divisible by 8.
So, the given number is divisible by 8.

Question 5.
Find the smallest number that must be added to 289279, so that it is divisible by 8 ?
Answer:
The given number is 289279.
The number formed by its last three digits is 279.
If 279 is to be exactly divisible by 8, we have to add 1 to it.
(i.e.,) 279 + 1 = 280; it is divisible by 8.
So, 1 must be added to the given number, so that it is divisible by 8.

Question 6.
Find the smallest number that can be subtracted from 1965, so that it becomes divisible by 4 ?
Answer:
The given number is 1965.
The number formed by its last two digits is 65.
The smallest number that can be subtracted from 65 is 1, so that it becomes divisible by 4.
(i.e.) 65 – 1 = 64
We know that 64 is divisible by 4.

Question 7.
Write all the possible numbers between 1000 and 1100, that are divisible by 11 ?
Answer:
We know that 990 is divisible by 11 (∵ 90 × 11 = 990)
990 is a multiple of 11.
The possible numbers divisible by 11 are 1001, 1012, 1023, 1034, 1045, 1056, 1067, 1078, 1089.

Question 8.
Write the nearest number to 1240 which is divisible by 11 ?
Answer:
11 × 112 = 1232
11 × 113 = 1243
∴ The nearest number to 1240 is 1243 but not 1232.
∴ 1243 is the nearest number to 1240 which is divisible by 11.

Question 9.
Write the nearest number to 105 which is divisible by 4?
Answer:
We know that 4 × 25 = 100
4 × 26 = 104
4 × 27 = 108
∴ 104 is the nearest number to 105 which is divisible by 4.

Students can practice TS SCERT Class 6 Maths Solutions Chapter 10 Perimeter and Area InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area InText Questions

Try This

Question 1.
Give five examples of situations where you need to know the perimeter.
Answer:

• To construct the house.
• To pave the tiles of a room.
• To find the perimeter of a flat.
• To determine the perimeter of a field.
• To construct a well around the school ground. For the above cases the knowledge of concept of perimeter is required.

Do This

Question 1.
What would be the perimeter of these shapes ? Fill in the blanks given and in each case start from the point A.

Answer:
(i) Perimeter = AB + BC + CD + DA
= 10m + 40m + 10m + 40m
= 100 m

= 100m + 120m + 90m + 45m + 60m + 80m
= 495 m

Try This

Find the perimeter of the following :
Question 1.
A table with sides equal to 30 cm, 15 cm, 30 cm, 15 cm respectively.
Answer:
Perimeter of table = 2 (l + b)
= 2 (30cm + 15cm) = 2 (45)cm = 90 cms.

Question 2.
Measure the length of the sides of your text book cover. What is the perimeter ?
Answer:
The length of the sides of the text book (Maths) is 30 cm, 20 cm.
Its perimeter = 2 (l + b)
= 2 (30 + 20)cm
= 2 × 50cm
= 100 cms.

Question 3.
Around a rectangular park of sides 100 meters and 70 meters a wire has to be put. The cost of the wire is ₹ 20 per meter. What is the total cost of the wire ?
Answer:
Given that length of a park = 100 m
breadth of a park = 70 m
∴ Perimeter of a rectangular park = 2 (l + b)m
= 2 (100 + 70)m = 2 × 170m = 340 m.
The total cost of the wire at the rate of ₹ 20 perimeter = 340m × 20 = ₹ 6,800
Perimeter = AB + BC + £D + DA = 10m + 40m + 10m + 40m = 100 m

Try This

Question 1.
Find the perimeter of the following rectangles.

Answer:

Do This

Question 1.
A square picture frame has sides of 0. 75 mts. If the cost of a coloured paper is ₹ 20 per meter, what is the cost of putting coloured paper around the frame ?
Answer:
The side of sqaured frame = 0.75 m
Its perimeter =4 × s = 4 × 0.75 = 3m
∴ The cost of a coloured paper around the square frame at the rate of ₹ 20 per meter = 3 × 20 = ₹ 60.

Question 2.
There is a string of length 44 cm. How many different rectangles with positive integers as length and breadth can be made with this string ?
Answer:
The length and breadth of rectangle as follows whose perimeter is 44 cm.

Question 3.
If I have a string 41 cm long can I make a rectangle using the string completely ? Give reasons.
Answer:

Try These

Question 1.
Find the perimeter of the following squares. Figures are drawn on 1 cm grids.

Answer:
i) Perimeter of square ABCD = 4 × s = 4 × 4 = 16 cm.
ii) Perimeter of square PQRS = 4 × s = 4 × 6 = 24 cm.
iii) Perimeter of square EFGH = 4 × s = 4 × 5 = 20 cm.
iv) Perimeter of square WXYZ = 4 × s = 4 × 2 = 8 cm.

Question 2.
Find various objects from your surroundings which have regular shapes and their perimeters.
Answer:
The objects which are having the regular shapes are
1) A ‘4’ size squared paper.
2) The pentagon building in U.S.A.
3) A square shaped cake.
4) A square shaped field.

Do This

Question 1.
Find the perimeter of a regular pentagon of side 8 cm.
Answer:
Perimeter of a regular polygon = n × length of its side
∴ Perimeter of a pentagon = 5 × 8 cm = 40 cm

Try This

Question 1.
Find the areas of the following figures by counting squares.
Area of each square is 1 sq.cm

Answer:

Do This

Question 1.
Trace shapes of leaves, flower petals and other such objects on the graph paper and find their area approximately.
Answer:
Student Activity

Question 2.
Draw any line diagram on a graph sheet. Count the squares and use them to estimate the area of the region.
Answer:
Student Activity

Try These

Question 1.
Draw two different rectangles having the same perimeter. Compare their areas. Are they same ? Can you draw two different squares having the same perimeter ?
Answer:
i) If the perimeters of two rectangles are equal then the areas of the rectangles need , not to be equal.
ii) No. We can’t draw two different squares having same perimeter. ;

Do This

Question 1.
Find the area of:
(i) The floor of your classroom.
(ii) A door in your house
(iii) The black board in your classroom.
Answer:
(i) Length of our class room floor = 15m
Breadth of the class room floor = 10m
Area of the floor = lb
= (15m) (10m)
= 150 sq.m

(ii) Length of the door in our house = 5m
Breadth of the door = 1.5 m
Area of the door = lb
= (5m) (1.5m)
= 7.5 sq.m

(iii) Length of the black board = 2.5m
Breadth of the black board = 2m
Area of the Black board = lb = 5sq.m

Try These

Question 1.
The length of one side of few squares are given. Find their areas using graph papers also find side × side. What do you notice from the result obtained ?
i) 4 cm
ii) 6 cm
iii) 2 cm
iv) 8 cm
Answer:

We notice that two results are equal.

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 6 Integers Ex 6.4 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 6 Integers Exercise 6.4

Question 1.
Find:
(i) 40 – (22)
Answer:
40 – (22)
= 40 + (additive inverse of 22)
= 40 – 22
= 18

(ii) 84 – (98)
Answer:
= 84 + (additive inverse of 98)
= 84 – 98
= -14

(iii) (- 16) + (- 17)
Answer:
(- 16) + (- 17)
= -16 – 17
= -33

(iv) (- 20) – (13)
Answer:
(- 20) – (13)
= – 20 + (additive inverse of 13)
= -20 -13
= -33

(v) 38 – (- 6)
Answer:
38 – (- 6)
= 38 + (additive inverse of – 6)
= 38 + 6 = 44

(vi) (-17)-(-36)
Answer:
(-17)-(-36)
= -17 + (additive inverse of – 36)
= -17 + 36
= 19

Question 2.
Fill in the blanks with appropriate >, < or = sign.
(i) (-4) +(-5) ………… (-5)-(-4)
(ii) (-16) – (-23) ……….. (-6) + (-12)
(iii) 44 – (-10) ……….. 47 + (-3)
(iv) (-21) + (-22) ………….. (-22) + (-21)
Answer:
i) (-4) +(-5) =-9 and
(- 5) – (- 4) = – 5 + 4 = – 1
We know that – 9 is less than – 1
∴ (- 4) + (- 5) < (- 5) – (- 4) (ii) (- 16) – (- 23) = – 16 + 23 = 7 and (- 6) + (- 12) = – 18 We know that 7 > – 18
∴ (- 16) – (- 23) > (- 6) + (- 12)

(iii) 44 – (- 10) = 44 + 10 = 54 and
47 + (- 3) = 44
We know that 54 > 44
∴ 44 – (- 10) > 47 + (- 3)

(iv) (- 21) + (-22) = -21 – 22 = -43 and
(-22) +(-21) =-22 – 21 =-43
We know that – 43 is equal to – 43
∴ (-21)+ (-22) = (-22)+ (-21)

Question 3.
Fill In the blanks.
(i) (-13) + ……… = 0
(ii) (- 16) + (16) = ……..
(iii) (- 5) + ………… = – 14
(iv) ………… – 16 = – 22
Answer:
(i) (- 13) + (+13) =0 .
(ii) (- 16) + (16) = 0
(iii) (-5) + (-9) = – 14
(iv) (-6) – 16 = -22

Question 4.
Simplify:
(i) (-6)-(5)-(+2)
(ii) (- 12) + 42 – 7 – 2
(iii) (- 3) + (- 6) + (- 24)
(iv) 40 – (- 50) – (2)
Answer:
(i) (- 6) – (+5) – (+ 2)
= – 6 – 5 – 2 = -13

(ii) ( – 12) + 42 – 7 – 2
= (- 12 – 7 – 2) + 42
= -21 + 42 = 21

(iii) (-3) +(-6)+ (-24)
= – 3 – 6 – 24
= – 33

(iv) 40 – (- 50) – (2)
= 40 + 50 – 2
= 90 – 2
= 88

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 6 Integers Ex 6.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 6 Integers Exercise 6.3

Question 1.
Add the following integers using number line.
(i) 7 + (- 6)
Answer:

∴ (+7) + (-6) = +1

(ii) (- 8) + (- 2)
Answer:

∴ (-8) + (-2) = -10

(iii) (- 6) + (- 5) + (+ 2)
Answer:

∴ (-6) + (-5) + (2) = -9

(iv) (- 8) + (- 9) + (+ 17)
Answer:

∴ (-8) + (-9) + (-17) = 0

(v) (-3) + (-8) + (-5)
Answer:

∴ (-3) + (-8) + (-5) = -16

(vi) (- 1) + 7 + (- 3)
Answer:

∴ (-1) + (7) + (-3) = 3

Question 2.
Add without using number line,
(i) 10 + (-3)
(ii) – 10 + (+ 16)
(iii) (- 8) + (+ 8)
iv) – 215 + (+100)
(v) (- 110) + (- 22)
(vi) 17 + (- 11)
Answer:
(i) 10 + (-3) = 7 + 3 + (- 3)
= 7 + [(3) + (- 3)]
= 7 + 0
= 7

(ii) – 10 + (+ 16) = – 10 + (10 + 6)
= [(- 10) + (+ 10)] +(6)
= 0 + 6
= 6

(iii) (- 8) + (+ 8) = [(- 8) + (+ 8)] = 0

(iv) – 215 + (+100)
= -115 – 100 + (+100)
= -115+ [(-100)+ (+100)]
= -115 + 0
= -115

(v) (-110) + (- 22)
= – 110 – 22
= – 132

(vi) 17 + (- 11)
= 6 + [(11) + (-11)]
= 6 + 0 = 6

Question 3.
Find the sum of:
(i) 120 and – 274
(ii) – 68 and 28
(iii) – 29, 38 and 190
(iv) – 60, – 100 and 300
Answer:
(i) (120) + (-274) .
= (120) + (- 120 – 154)
= (120) + (- 120) + (- 154)
= [(120) + (- 120)] + (- 154)
= 0-154 = -154

(ii) (- 68) + 28
= (- 40 – 28) + 28
= (-40)+ (-28)+ (28)
= -40 + [(-28) + (28)]
= -40 + 0
= -40

(iii) (- 29) + (38) + (190)
= – 29 + [38 + 190]
= – 29 + 228 = 199

(iv) (- 60) + (- 100) + (300)
= (- 60 – 100) + (300)
= – 160 + 300
= 140

Question 4.
Simplify:
(i) (- 6) + (-10) + 5+17
(ii) 30 + (- 30) + (- 60) + (- 18)
(iii) (- 80) + (+ 40) + (- 30) + (+ 6)
(iv) 70 + (- 18) + (- 10) + (- 17)
Answer:
(i) (- 6) + (-10) + 5 + 17
= -6 – 10 + 5 + 17
= -16 + 22
= 6

(ii) 30 + (-30)+ (-60) + (-18)
= 30 – 30 – 60 – 18
= 0 – 78
= – 78

(iii) (- 80) + (+ 40) + (- 30) + (+ 6)
= (- 80) + (- 30) + (+ 40) + (+ 6)
= – 80 – 30 + 40 + 6
= -110 + 46
= -64

(iv) 70 + (- 18) + (- 10) + (- 17)
= 70 – 18 – 10 – 17
= 70 – 45
= 35

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 6 Integers Ex 6.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 6 Integers Exercise 6.2

Question 1.
Put appropriate symbol > or < in the space given between the two integers.
(i) -1 ……….. 0
(ii) – 3 ……… – 7
(iii) -10 ………. + 10
(iv) 0 ……… – 5
(v) – 100 ………. 99
(vi) 0 ….. 100
Answer:
(i) – 1 < 0
(ii) – 3 > – 7
(iii) – 10 < 10
(iv) 0 > -5
(v) – 100 < 99
(vi) 0 < 100

Question 2.
Write the following integers in increasing or decreasing order.
(i) – 7, 5, – 3
(ii) -1, 3, 0
(iii) 1, 3, -6
iv) – 5, – 3, – 1
Answer:
(i) Increasing order : – 7, – 3, 5
Decreasing order : 5, – 3, – 7

(ii) Increasing order : – 1, 0, 3
Decreasing order : 3, 0, – 1

(iii) Increasing order :-6, 1, 3
Decreasing order : 3, 1, -6

(iv) Increasing order : – 5, – 3, – 1
Decreasing order : – 1, – 3, – 5

Question 3.
Write True or False, correct those that are false.
(i) Zero is the right of – 3. ( )
(ii) -12 and+12 represent on the number line the same integer. ( )
(iii) Every positive integer is greater than zero. ( )
(iv) – 5 < 8 ( )
(v) (- 100) > (+100) ( )
(vi) – 1 < – 8 ( )
Answer:
(i) True
(ii) False
– 12 and + 12 represent on the number line the different integers.
(iii) True
(iv) True
(v) False ‘(-100) <(+100)
(vi) False (-1)>(-8)

Question 4.
Find all the integers which lie between the given two integers. Also represent them on the number line.
(i) – 1 and 1
Answer:
0 lies between – 1 and 1.

(ii) – 5 and 0
Answer:
– 4, – 3, – 2, – 1 lie between – 5 and 0.

(iii) – 6 and – 8
Answer:
– 7 lies between – 6 and – 8.

(iv) 0 and – 3
Answer:
-2, -1 lie between 0 and – 3.

Question 5.
The temperature recorded in Shimla is – 4°C and in Kufri is – 6°C on the same day. Which place is colder on that day ? How ?
Answer:
The lowest temperature recorded at Kufri is – 6°C. Kufri is colder than Shimla. (∵ – 6°C < – 4°C)

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 6 Integers Ex 6.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 6 Integers Exercise 6.1

Question 1.
Represent the following statements using signs of integers.
(i) An aeroplane is flying at a height of 3000 meters. ( )
Answer:
+ 3000 m

(ii) The fish is 10 meters below the water surface. ( )
Answer:
– 10 m

(iii) The temperature in Hyderabad is 35°C above 0°C. ( )
Answer:
+ 35°C

(iv) Water freezes at 0°C temperature. ( )
Answer:
0°C

(v) The average temperature at the Mount Everest in January is 36°C below zero degree. ( )
Answer:
– 36°C

(vi) The submarine is 500 meters below the surface of the sea. ( )
Answer:
– 500 m

(vii) The average temperature at Dargeeling in July is 19°C below zero degree. ( )
Answer:
– 19°C

(viii) The average low temperature in Visakhapatnam during January is 18°C. ( )
Answer:
+ 18°C

Question 2.
Write any five negative integers.
Answer:
– 320, – 270, – 165, – 89, – 7

Question 3.
Write any five positive integers.
Answer:
94, 175, 236, 348, 6

Question 4.
Mark the integers on the number line given below – 4, 3, 2, 0, – 1, 5

Answer:

Question 5.
Write True or False. If the statement is false, correct the statement,
(i) – 7 is on the right side of – 6 on the number line. ( )
Answer:
False
– 7 is on the left side of – 6 on the number line. (True)

(ii) Zero is a positive number. ( )
Answer:
False
Zero is neither positive nor negative (True)

(iii) 9 is.on the right side of z6ro on the number line. ( )
Answer:
True

(iv) -1 is an integer which lies between – 2 and 0. ( )
Answer:
True

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 8 Data Handling Ex 8.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 8 Data Handling Exercise 8.3

Question 1.
The life span of some animals is given as follows :
Bein’ – 40 years, Bull – 28 years, Camel – 50 years, Dog – 22 years,
Cat – 25 years, Donkey – 45 years, Goat – 15 years, Horse – 10 years,
Cow – 22 years, Elephant – 70 years.
Draw a horizontal bar graph to represent the data.
Answer:

Question 2.
The following table shows the monthly expenditure of Imran’s family on various items.

Construct a vertical bar diagram to represent the above data.
Answer:

Question 3.
TravellIng time from Hyderabad to Tirupathi by different means of transport are:
Car -8 hours, Bus – 15 hours, Train – 12 hours, Aeroplane – 1 hour. Represent the information using a bar diagram.
Answer:

Question 4.
A survey of 120 school students was conducted to find which activity they prefer to do in their free time.

Draw a bar graph to illustrate the above data.
Answer:

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 8 Data Handling Ex 8.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 8 Data Handling Exercise 8.2

Question 1.
The number of wrist watches manufactured by a factory in a week are as follows.

Represent the data using a pictograph. Choose a suitable scale.
Answer:

Question 2.
Details of fruits sold in a week by Ahmed, a fruit vendor are given here under. Prepare a pictograph for the data : [Scale : Represent 5 fruits with a symbol]

Answer the following questions :
i) How many symbols represent the fruits sold on Tuesday ?
ii) How many symbols represent the fruits sold on Friday ?
Answer:

Question 3.
Votes polled for various candidates in a sarpanch election are shown below, against their symbols in the following table.

Represent the data using a pictograph. Choose a suitable scale. Answer the following questions :
i) Which symbol got least votes ?
ii) Which symbol’s candidate won in the election ?
Answer:

i) Watch symbol got least votes.
ii) The candidate having pot symbol won in the election.

Question 4.
The following pictograph shows the number of students have cycles, in five classes of a school.

Answer the following questions based on the pictograph given above :
i) Which class students have the maximum number of cycles ?
ii) Which class students have the minimum number of cycles ?
iii) Which class students have 9 cycles ?
iv) What is the total number of cycles in all the five classes ?
Answer:
i) IX class students have the maximum number of cycles. They are 12 in number.
ii) VI class students have the minimum number of cycles. They are 5 in number.
iii) VIII class students have 9 cycles.
iv) The total number of cycles in all the five classes is 43.

Question 5.
The sale of television sets of different companies on a day is shown in the pictograph given below.

Answer the following questions :
(i) How many TVs of company A were sold ?
(ii) Which company’s TVs did people like more ?
(iii) Which company sold 15 TV sets ?
(iv) Which company had the least sale ?
Answer:
i) TVs of company A sold were 25.
ii) The people more like the TVs of company C.
iii) Company E sold 15 TV sets.
iv) Company D had the least sale. This company sold only 5 TV sets.

Question 6.
Monthly salaries of 5 workers are shown in the pictograph given below :

Answer the following questions :
(i) What is the scale used in the pictograph ?
(ii) How much is Sachin’s salary ?
(iii) Who earns more salary ?
(iv) How much is Ramesh’s salary more than Vilas’s ?
Answer:
(i) The scale used in the pictograph is 1000 rupees.
(ii) The salary of Sachin is Rs. 10,000.
(iii) Dinesh and Sachin earn more salary (i.e.,) Rs. 10,000 each.
(iv) Ramesh earns Rs. 8,000 whereas Vilas. Rs. 7000.
Ramesh’s salary is Rs. 1000 more than Vilas’s

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 8 Data Handling Ex 8.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 8 Data Handling Exercise 8.1

Question 1.
A child’s kiddy bank is opened and the coins collected are in the following denomination.

Represent the data in a frequency distribution table using tally marks.
Answer:

Question 2.
The favourite colours of 25 students in a class are given below :
Blue, Red, Green, White, Blue, Green, White, Red, Orange, Green, Blue, White, Blue, Orange, Blue, Blue, White, Red, White, White, Red, Green, Blue, Blue, White.
Write a frequency distribution table using tally marks for the data. Which is the least favourite colour for the students ?
Answer:

The least favourite colour for the students is Orange.

Question 3.
A TV channel invited a SMS poll on ‘Ban of Liquor’ giving options.
A – Complete ban B – Partial ban C – Continue sales
They received the following SMS; in the first hour.

Represent the data in a frequency distribution table using tally marks.
Answer:
SMS poll on ‘Ban of Liquor’

Question 4.
Vehicles that crossed a checkpost between 10AM and 11AM are as follows :
Car, lorry, bus, lorry, auto, lorry, lorry, bus, auto, bike, bus, lorry, lorry, zeep, lorry, bus, zeep, car, bike, bus, car, lorry, bus, lorry, bus, bike, car, zeep, bus, lorry, lorry, bus, car, car, bike, auto.
Represent the data in a frequency distribution table using tally marks.
Answer:
Vehicles that crossed a checkpost.