Category: Class 6

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry InText Questions

Check Your Progress (Page No. 139)

Question 1.
Measure the lengths of \(\overline{\mathrm{AP}}\) and \(\overline{\mathrm{PB}}\) in both the constructions. What do you observe?
Solution:
In the construction we observe that \(\overline{\mathrm{AP}}=\overline{\mathrm{PB}}\)
That is P can divide AB into two equal parts.

Let’s Think (Page No. 140)

Question 1.
In the construction of perpendicular bisector in step 2, what would happen if we take the length of radius to be smaller than half the length of \(\overline{\mathrm{AB}}\) ?
Solution:
If we take the length of radius to be smaller than half the length of \(\overline{\mathrm{AB}}\)
Arcs cannot intersect each other. So, we can’t construct the perpendicular bisector to the given line \(\overline{\mathrm{AB}}\)

(Page No. 144)

Question 1.
Construct angles of 180°, 240°, 300°.
Solution:

i) ∠AOB = 180°
Steps of construction:

• Draw any ray \(\overline{\mathrm{OA}}\) of any length.
• Place the pointer of the compasses at ‘O’ with ‘O’ as centre any convinient radius draw an arc cutting OA at M.
• With M as centre and without altering radius draw an arc which cuts the previous arc at P.
• Draw an arc with the same radius from P which cuts the previous arc at Q and from Q draw another arc which meets at R.
• Join OR. ( \(\overline{\mathrm{OB}}\) ). Then ∠AOB is the required angle.
  Hence the required angle ∠AOB = 180° is constructed.

ii) ∠PQR = 240°
Steps of construction:

• Draw any ray \(\overline{\mathrm{QP}}\) of any length.
• Place the pointer of the compasses at O with ‘O’ as centre any convinient radius draw an arc cutting QP at A.
• With A as centre and without altering radius draw an arc which cuts the previous arc at B.
• Draw an arc with the same radius from B which cuts the previous arc at C and from C draw another arc which meets at D.
• Draw an arc with the same radius from D draw an arc which cuts the previous arc at E.
• Join \(\overrightarrow{\mathrm{QE}}(\overrightarrow{\mathrm{QR}})\). Then ∠PQR is the required angle.
  Hence the required angle ∠PQR = 240° is constructed.

iii) ∠XYZ – 300°
Steps of construction :

• Draw any ray \(\overline{\mathrm{YZ}}\) of any length.
• Place the pointer of the compasses at Y with Y as centre any convinient radius draw an arc cutting YZ at P.
• With P as centre and without altering radius draw arc which quts the previous arc at Q and from Q draw another arc which cuts the previous arc at R.
• Draw an arc with the same radius from R which cuts the previous arc at S and from S draw another arc which meets at T.
• Draw an arc with the same radius from T which cuts the previous arc at U.
• Join \(\overrightarrow{\mathrm{YU}}(\overrightarrow{\mathrm{YX}})\). Then ∠XYZ is required angle.
  Hence the required angle ∠XYZ = 300° is constructed.
  

(Page No. 145)

Question 1.
Construct an angle of 45° by using compasses.
Solution:
45°
Steps of construction :
i) Draw any ray \(\overline{\mathrm{OP}}\) of any length.
ii) Draw arcs with the same radius from A to B and from B to C which cuts the previous arc at B and C respectively.
iii) Draw arcs from B and from C with same radius which can intersect at X.
iv) Join \(\overrightarrow{\mathrm{OX}}(\overrightarrow{\mathrm{OD}})\) i.e., ∠POD = 90°.
v) Draw the bisector to ZPOD which is \(\overrightarrow{\mathrm{OQ}}\)
vi) Now, ∠POQ = ∠QOD = \(\frac{\angle \mathrm{POD}}{2}=\frac{90}{2}\) = 45°
∴ ∠POQ = 45° (Q.E.D)

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes InText Questions

(Page No. 125)

Question 1.
Draw six different types of rough sketches of polygons in your notebook. In which case, it is not possible to form a polygon ?
Hence, what is the least number of sides needed to form a polygon ? Obviously three.
Solution:

From the above figure we conclude that with one side and two sides cannot form a polygon. So, atleast three sides needed to form a polygon.

Check Your Progress (Page No. 127)

Look at the adjacent figure.

Question 1.
Which points are marked in the interior of A GHI ?
Solution:
The points marked in the interior of AGHI are A, B and 0.

Question 2.
Which points sure marked on the triangle?
Solution:
The points marked on the triangle are G, P, H, I and Y.

Question 3.
Which points are’marked in the exterior of A GHI ?
Solution:
The points marked in the exterior of AGHI are M, R, S, X and Z.

(Page No. 130)

Question 1.
Take a rectangular sheet (like a post-card). Fold it along its length so that one half fits exactly over the other half. Is this fold a line of symmetry ? Why ?
Solution:
Yes. Because the two parts of the rectangular sheet (post-card) coincide exactly with each other.
So, the folded line is the line of symmetry.

Question 2.
Open it up now and again fold along its width in the same way. Is this second fold also a line of symmetry ? Why ?
Solution:
Yes. Because the two parts of the rectangular sheet (post-card) coincide exactly with each other.
So, the second folded line also a line of symmetry.

Question 3.
Do you find that these two lines are the lines of symmetry ?
Solution:
Yes. These two lines are the lines of symmetry. One is line of symmetry along length and the other is line of symmetry along width.

Project (Page No. 130)

Question 1.
Collect symmetrical figures from your surroundings and prepare a scrap book.
Solution:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts InText Questions

(Page No. 111)

i) How many rays are there ?
Solution:
Four

ii) How many coins are close to player Q ?
Solution:
B, C and D coins are close to player Q.

iii) While striking with striker there is a possibility of a coin touches with any other. Draw all such possibilities in the given picture by means of the line segments.
Solution:
\(\overline{\mathrm{CB}}\) and \(\overline{\mathrm{DE}}\) .

iv) How many such line segments can be drawn in the picture ?
Solution:
\(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{ED}}, \overline{\mathrm{AC}}, \overline{\mathrm{AE}}, \overline{\mathrm{CD}}, \overline{\mathrm{CE}}, \overline{\mathrm{AD}}, \overline{\mathrm{BD}} \text { and } \overline{\mathrm{BE}}\)

Let’s Do (Page No. 112)

Question 1.
Observe the table and their notations and fill the gaps.

Solution:

Lets Explore (Page No. 114)

Question 1.
Measure the lengths of all line segments in the given figures by using divider and scale. Then compare the sides of the given figures.

Solution:
i) In ΔABC; \(\overline{\mathrm{AB}}\) = 2.2 cm; \(\overline{\mathrm{BC}}\) = 2 cm and \(\overline{\mathrm{AC}}\) = 2.2 cm ‘
2.2 cm = 2.2 cm > 2 cm i.e., \(\overline{\mathrm{AB}}=\overline{\mathrm{AC}}>\overline{\mathrm{BC}}\)
Two sides are equal and one side is different in length.

ii) In PQRS rectangle \(\overline{\mathrm{PS}}=\overline{\mathrm{QR}}\) = 2.7 cm; \(\overline{\mathrm{P Q}}=\overline{\mathrm{RS}}\) = 1.8 cm and \(\overline{\mathrm{PR}}=\overline{\mathrm{QS}}\) = 3.2 cm.
Opposite sides are equal and diagonals are equal in length,

iii) In KLMN square \(\overline{\mathrm{KL}}=\overline{\mathrm{LM}}=\overline{\mathrm{MN}}=\overline{\mathrm{KN}}\) = 1.8 cm
All sides are equal in length.

Let’s Do (Page No. 115)

Question 1.
(i) Find the parallel lines in the below figure. Name, write and read them.

Solution:
a) l, m are parallel lines.
We denote this by writing l || m and can be read as l is parallel to m.
b) n, o are parallel lines.
We denote this by writing n||o and can be read as n is parallel to o.
c) p, q are parallel lines.
We denote this by writing p||q and can be read as p is parallel to q.

ii) Find the intersecting lines in the above figure. Name, write and read them.
Solution:
Intersecting lines are (l, q); (m,q); (m, r); (n,q); (p, r); (o, r); (o, q); (q, r);

iii) Find the concurrent lines in the above figure. Name, write and read them.
Solution:
Three or more lines passing through the same point are called concurrent lines. Concurrent lines are (l, o, p) & (m, n, p).

iv) Find the perpendicular lines in the above figure. Name, write and read them.
Solution:
a) p, o are perpendicular lines.
We denote this by writing p ⊥ o and can be read as p is perpendicular to o.
b) p, n are perpendicular lines.
We denote this by writing p ⊥ n and can be read as p is perpendicular to n.
c) n, q are perpendicular lines.
We denote this by writing n ⊥ q and can be read as n is perpendicular to q.
d) q, o are perpendicular lines.
We denote this by writing q⊥ o and can be read as q is perpendicular to o.

(Page No. 119)

Question 1.
Measure the angles at the vertices.

Solution:

In triangle ABC,
m∠BAC = 60°
m∠ABC = 60°
m∠ACB = 60°

In triangle XYZ,
m∠YXZ = 40°
m∠XYZ = 70°
m∠XZY = 70°

In triangle PQR,
m∠QPR = 35°
m∠PQR = 38°
m∠PRQ = 107°

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra InText Questions

Let’s Explore (Page No. 102)

Question 1.
Arrange 2 matchsticks to form the shape Continue the same shape for 2 times, 3 times and 4 times. Frame the rule for repeating the pattern.
Solution:
To make the given shape 2 matchsticks are needed.
To make the given 2 shapes 4 matchsticks are needed.
To make the given 3 shapes 6 matchsticks are needed.
To make the given 4 shapes 8 matchsticks are needed.
Continue and arrange the information in the following table.

Number of matchsticks required = 2 × Number of shapes to be formed
= 2 × x = 2x

Question 2.
Rita took matchsticks to form the shape
She repeated the pattern and gave a rule.
Number of matchsticks needed = 6.y, where y is the number of shapes to be formed. Is it correct ? Explain.
What is the number of sticks needed to form 5 such shapes ?
Solution:

To make the given shape 6 matchsticks are needed.
To make the given 2 shapes 12 matchsticks are needed.
To make the given 3 shapes 18 matchsticks are needed.
Continue and arrange the information in the following table.

Yes, it is correct.
Number of matchsticks required = 2 × Number of shapes to be formed
= 2 × y = 2y
Number of matchsticks needed to form 5 such shapes = 6 × 5 = 30

Let’s Explore (Page No. 103)

Question 1.
A line of shapes is constructed using matchsticks.

Shape-1 Shape-2 Shape-3 Shape-4
i) Find the rule that shows how many sticks are needed to make a line of such shapes ?
ii) How many matchsticks are needed to form shape -12 ?
Solution:

Number of matchsticks 3 5 7 9
i) Let us know the pattern
S₁ = 3 = 2 + 1 = (1 × 2) + 1
S₂ = 5 = 4 + 1 = (2 × 2) + 1
S₃ = 7 = 6 + 1 = (3 × 2) + 1
S₄ – 9 = 8 + 1 = (4 × 2) + 1
Now the rule for this pattern is number of matchsticks.

ii) Used to make ‘n’ number of shapes is Sn = (n × 2) + 1 = 2n + 1
Number of matchsticks needed to form shape – 12 is
S₁₂ = 2(12) + 1 = 24 + 1 = 25 sticks.

Check Your Progress (Page No. 105)

Question 1.
Fill the following table as instructed. One is shown for you.

Solution:

Let’s Explore ? (Page No. 106)

Question 1.
Find the general rule for the perimeter of a rectangle. Use variables T and ‘b’ for length and breadth of the rectangle respectively.
Solution:
Given length of rectangle = l
breadth of rectangle = b
We know that the perimeter of rectangle is twice the sum of its length and breadth.
Sum of length and breadth = l + b
Twice the sum of length and breadth = 2 × (l + b)
Rule for the perimeter of a rectangle = 2(l + b)

Question 2.
Find the general rule for the area of a square by using the variable ‘s’ for the side of a square.
Answer:
Given side of a square = s
We know that the area of a square is the product of side and side.
Area of a square = side × side
Rule for the area of a square = s.s

(Page No. 107)

Question 1.
Find the nth term in the following sequences.
0 3, 6, 9, 12, ii) 2, 5, 8, 11, iii) 1, 4, 9, 16,
Solution:
i) Given number pattern is 3, 6, 9, 12,……………..
To find the nth term in the given pattern, we put the sequence in a table.

First number = 3 × 1
Second number = 3 × 2
nth number = 3 × n = 3n
So, the nth term of the pattern 3, 6, 9, 12, is 3n.

ii) Given number pattern is 2, 5, 8, 11,
To find the nth term in the given pattern, we put the sequence in a table.

First number = 2 = 3 × 1 – 1
Second number = 5 = 3 × 2 – 1
Third number = 8 = 3 × 3 – 1
nth number = 3 × n – 1 = 3n – 1
So, the nth term of the pattern 2, 5, 8, 11 is 3n – 1.

iii) Given number pattern is 1, 4, 9, 16,
To find the nth term in the given pattern, we put the sequence in a table.

First number =1 = 1 × 1
Second number = 4 = 2 × 2
Third number =9 = 3 × 3
nth number = n × n = n²
So, the nth term of the pattern 1, 4, 9, 16 is n².

Check Your Progress (Page No. 108)

Question 1.
Complete the table and find the value of ‘p’ for the equation \(\frac{\mathbf{p}}{\mathbf{3}}\) = 4

Solution:

Question 2.
Write LHS and RHS of following simple equations.
i) 2x + 1 = 10
ii) 9 = y – 2
iii) 3p + 5 = 2p + 10
Solution;
i) 2x+ 1 = 10
Given equation is 2x + 1 = 10
L.H.S = 2x + 1
R.H.S = 10

ii) 9 = y – 2
Given equation is 9 = y – 2
LHS = 9
RHS = y – 2

iii) 3p + 5 = 2p + 10
Given equation is 3p + 5 = 2p + 10
LHS = 3p + 5
RHS = 2p + 10

Question 3.
Write any two simple equations and write their LHS and RHS.
Solution:
i) Consider 8x + 3 = 4 is a simple equation.
L.H.S = 8x + 3
RHS = 4

ii) Consider 5a + 6 = 8a – 3 is a simple equation.
LHS = 5a + 6
RHS = 8a – 3

Let’s Explore (Page No. 109)

Observe for what value of m, the equation 3m = 15 has both LHS and RHS become equal.
Solution:
Given equation is 3m = 15
If m = 1, then the value of 3m = 3(1) = 3≠15 ∴ LHS ≠RHS
If m = 2, then the value of 3m = 3(2) = 6 ≠ 15 ∴ LHS ≠ RHS
If m = 3, then the value of 3m = 3(3) = 9≠15 ∴ LHS ≠ RHS
If m = 4, then the value of 3m = 3(4) = 12 ≠ 15 ∴ LHS ≠ RHS
If m = 5, then the value of 3m = 3(5) = 15 = 15 ∴ LHS = RHS
From the above when m = 5 the both LHS and RHS are equal

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic InText Questions

Check Your Progress (Page No. 84)

Question 1.
Express the terms 45 and 70 by using ratio symbol.
Solution:
Given terms are 45 and 70
Ratio = 45 : 70
It can be read as 45 is to 70.

Question 2.
Write antecedent in the ratio 7:15.
Solution:
Given ratio 7 : 15
In the ratio first term is called antecedent.
In 7 : 15 antecedent is 7.

Question 3.
Write the consequent in the ratio 8 : 13.
Solution:
Given ratio 8 : 13
In the ratio second term is called consequent.
In 8 : 13 consequent is 13.

Question 4.
Express the ratio 35 : 55 in the simplest form.
Solution:
Given ratio 35 : 55 (or)
To write the ratio in the simplest form we have to divide by the common factor of two terms 35 and 55.
Common factor is 5.
Now divide by 5,
\(\frac{35}{55}=\frac{35 \div 5}{55 \div 5}=\frac{7}{11}\)
Simplest form of \(\) is \(\frac{7}{11}\)

Question 5.
In the given figure, find the ratio of
i) Shaded part to unshaded parts.
ii) Shaded part to total parts,
iii) Unshaded parts to total parts.

Solution:
i) In the given figur.e,
Number of shaded parts = 1
Number of unshaded parts = 3
Ratio = shaded parts : unshaded parts = 1:3

ii) Number of shaded parts = 1
Total parts = 4
Ratio = shaded parts : total parts = 1:4

iii) Number of unshaded parts = 3
Total parts = 4
Ratio = unshaded parts : total parts = 3:4

Question 6.
Express the following in the form of ratio.
a) The length of a rectangle is triple its breadth. ‘
b) In a school, the workload of teaching 19 sections has been assigned to 38 teachers.
Solution:
a) Let breadth of rectangle = x or one part = 1 part
length of rectangle = triple the breadth
= 3 x x = 3x = 3 parts
Ratio = l : b = x : 3x =\(\frac{1 x}{3 x}=\frac{1}{3}\) = 1:3

b) Given number of sections = 19
Number of teachers = 38
Ratio = 19 : 38 = \(\frac{19}{38}=\frac{1}{2}\) = 1 : 2

(Page No. 88)

Question 1.
Which ratio is larger in the following pairs ?
(a) 5 : 4 or 9 : 8
(b) 12 : 14 or 16 : 18
(c) 8: 20 or 12: 15
(d)4:7 or 7:11
Solution:
a) 5 : 4 or 9 : 8
Write the given ratios as fractions, we have 5 : 4 = \(\frac{5}{4}\) and 9 : 8 = \(\frac{9}{8}\)
Now find the LCM of the denominators of 4 and 8 is 8.
Make the denominator of the each fraction equal to 8.
We have \(\frac{5}{4} \times \frac{2}{2}=\frac{10}{8}\) and \(\frac{9}{8} \times \frac{1}{1}=\frac{9}{8}\)
Clearly we know that 10 > 9
∴ \(\frac{10}{8}>\frac{9}{8}\) (or) 10 : 8 > 9 : 8
10 : 8 is equal to 5 : 4
Therefore the larger ratio is 5 : 4.

b) 12 : 14 or 16:18
12 : 14 = \(\frac{12}{14}=\frac{6}{7}\) and 16 : 18 = \(\frac{16}{18}=\frac{8}{9}\)
Now find the LCM of the denominators of 7 and 9 is 63.
Make the denominator of the each fraction equal to 63.
we have \(\frac{6}{7} \times \frac{9}{9}=\frac{54}{63}\) and \(\frac{8}{9} \times \frac{7}{7}=\frac{56}{63}\)
Clearly, we know that 54 < 56
∴ \(\frac{54}{63}<\frac{56}{63}\) (or) 54:63 < 56:63
56 : 63 is equal to 16 : 18 (or) 8 : 9
∴ The larger ratio is 16 : 18.

c) 8 : 20 or 12 : 15
Write the given ratios as fractions we have
8:20 = \(\frac{8}{20}=\frac{2}{5}\) and 12:15 = \(\frac{12}{15}=\frac{4}{5}\)
\(\frac{2}{5}\) and \(\frac{4}{5}\)

Clearly \(\frac{2}{5}\) < \(\frac{4}{5}\)
i.e., 2:5 < 4 : 5 (or) 8: 20 < 12: 15
Therefore the larger ratio is 12 : 15.

d) 4: 7 or 7: 11
Write the given ratios as fractions, we have 4 7
4 : 7 = \(\frac{4}{7}\) and 7:11 = \(\frac{7}{11}\) .
Now find the LCM of the denominators of 7 and 11 is 77.
Make the denominators of the each fraction equal to 77.
We have \(\frac{4}{7} \times \frac{11}{11}=\frac{44}{77}\) and \(\frac{7}{11} \times \frac{7}{7}=\frac{49}{77}\)
\(\frac{44}{7}\) and \(\frac{49}{77}\)
Clearly we know that 44 < 49
∴ \(\frac{44}{77}<\frac{49}{77}\) (or) 44 : 77 < 49 : 77
i.e.,4: 7 < 7 : 11
Therefore the larger ratio is 7 : 11

Question 2.
Find three equivalent ratios of 12 : 16.
Solution:
Given ratio is 12 : 16
Write the given ratio as fraction we have 12:16= \(\frac{12}{16}=\frac{3}{4}\)
Now, write equivalent fractions of \(\frac{3}{4}\)

i. e., 6 : 8 = 9 : 12 = 12 : 16 = 15 : 20 = 18 : 24
∴ Equivalent ratios of 12 : 16 are 6 : 8, 9 : 12, 12 : 16, 15 : 20 and 18 : 24.

(Page No. 90)

Question 1.
Check whether the following terms are in proportion ?
1) 5,6,7,8
2) 3,5,6,10
3) 4,8,7,14
4) 2,12,3,18
Solution:
1) Given, 5, 6, 7, 8
If a, b, c, d are in proportion i.e., a : b :: c : d
If 5, 6, 7, 8 are in proportion i.e., 5 : 6 : : 7 : 8
We know that, product of extremes = Product of means [a x d : b x c]
5 x 8 = 6 x 7
40 ≠ 42
So, 5, 6, 7, 8 are not in proportion.

2) Given, 3, 5, 6, 10
If a, b, c, d are in proportion i.e., a : b :: c : d
If 3, 5, 6, 10 are in proportion i.e., 3 : 5 :: 6 : 10
We know that, product of extremes = Product of means a x d = b x c
3 x 10 = 5 x 6
30 = 30
So, 3, 5, 6, 10 are in proportion.

3) Given, 4, 8, 7, 14.
If a, b, c, d are in proportion i.e., a : b : : c : d
If 4, 8, 7, 14 are in proportion i.e., 4 : 8 : : 7 : 14
We know that, product of extremes = Product of means a x d = b x c
4 x 14 = 8 x 7
56 = 56
So, 4, 8, 7, 14 are in proportion.

4) Given, 2, 12, 3, 18
If a, b, c, d are in proportion i.e., a : b :: c : d
If 2, 12, 3, 18 are in proportion i.e., 2 : 12 : : 3 : 18
We know that, product of extremes = Product of means [ a x d = b x c ]
2 x 18 = 12 x 3
36 = 36
So, 2, 12, 3, 18 are in proportion.

Let’s Explore (Page No. 92)

Question 1.
Read the table and fill in the boxes.

Prepare two similar problems and ask your friend to solve them
Solution:

(Page. No. 94)

Question 1.
Represent the following in other forms.

Solution:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals InText Questions

(Page No. 63)

Question 1.
Is it true to say that 3 × \(\frac{1}{5}=\frac{1}{5}\) x 3?
Solution:
3 × \(\frac{1}{5}=\frac{1}{5}\) × 3. Yes, it is true.
By using commutative property over multiplication a × b = b × a
3 × \(\frac{1}{5}=\frac{1}{5}\) × 3 = \(\frac{3}{5}\)

Check Your Progress (Page No. 63)

Find :
i) \(5 \times 3 \frac{2}{7}\)
ii) \(2 \frac{5}{9} \times 3\)
iii) \(2 \frac{4}{7} \times 3\)
iv) \(3 \times 1 \frac{3}{4}\)
Solution:

Let’s Explore (Page No. 64)

Question 1.
Observe the products of fractions.
Have you observed the products of any two fractions is always lesser or greater than each of its fraction, write conclusion.
\(\frac{1}{5} \times \frac{2}{3}=\frac{2}{15}\) (Product of two proper fractions)
Solution:
Product of any two proper fractions is always less than each of its fraction.
i. e., \(\frac{2}{15}<\frac{1}{5} \text { and } \frac{2}{15}<\frac{2}{3}\)

ii) \(\frac{3}{2} \times \frac{5}{4}=\frac{15}{8}\) (Product of two improper fractions) •
Solution:
The product of any two improper fractions is always greater than each of its fraction.
i.e, \(\frac{3}{2}<\frac{15}{8} \text { and } \frac{5}{4}<\frac{15}{8}\)

iii) \(\frac{2}{3} \times \frac{5}{3}=\frac{10}{9}\) (Product of proper and improper fractions)
Solution:
The product of a proper fraction and an improper fraction is always greater than its proper fraction and less than its improper fraction.
i.e., \(\frac{2}{3}<\frac{10}{9} \text { and } \frac{5}{3}>\frac{10}{9}\)

(Pg. No. 66)

Question 1.
i) 4 ÷ \(\frac{1}{8}\)
ii) 9 ÷ \(\frac{3}{4}\)
iii) 7 ÷ \(\frac{2}{3}\)
iv) 35 ÷ \(\frac{7}{3}\)
v) 4 ÷ \(\frac{15}{8}\)
Solution:

(Pg. No. 67)

Question 1.
Observation these products and fill in the blanks.

Solution:

Check Your Progress (Page No. 68)

Question 1.
Write the reciprocal of fractions in the given table.

Solution:

(Reciprocal of a fraction \(\frac{\mathrm{a}}{\mathrm{b}}\) is \(\frac{\mathrm{b}}{\mathrm{a}}\))

(Page No. 69)

Question 1.
Find
i) \(\frac{7}{9}\) ÷ 4
ii) \(\frac{3}{4}\) ÷ 9
iii) 4\(\frac{1}{2}\) ÷ 6
iv) \(\frac{1}{5}\) ÷ 3
Solution:

Check Your Progress (Page No. 73, 74 & 75)

Question 1.
Fill in the blanks.

Solution:

Question 2.
Write the place value of the circled digits.

Solution:

Question 3.
a) 700 + 40 + 2 + \(\frac{1}{10}+\frac{3}{100}+\frac{6}{1000}\)
Solution:
700 + 40 + 2 + 0.1 + 0.03 + 0.006 = 742.136

b) 9000 + 800 + 3 + 0.2 + 0.05 + 0.007
Solution:
9000 + 800 + 3 + 0.2 + 0.05 + 0.007 = 983.257

c) 6000 + 400 + 20 + 1 + \(\frac{2}{10}+\frac{5}{100}+\frac{9}{1000}\)
Solution:
6000 + 400 + 20 + 1 + 0.2 + 0.05 + 0.009 = 6421. 259

d) 400 + 5+ \(\frac{1}{10}+\frac{8}{100}\)
Solution:
400 + 5 + 0.1 +0.08 = 405.18

Question 4.
Expand the following into decimals and fractional forms,
a) 164.238
b) 968.054
Solution:
a) 164.238 = 100 + 60 + 4 + 0.2 + 0.03 + 0.008
= 100 + 60 + 4 + \(\frac{2}{10}+\frac{3}{100}+\frac{8}{1000}\)

b) 968.054
Solution:
= 900 + 60 + 8 + 0.0 + 0.05 + 0.004
= 900 + 60 + 8 + 0 + \(\frac{5}{100}+\frac{4}{1000}\)
= 900 + 60 + 8 + \(\frac{5}{100}+\frac{4}{1000}\)

Question 5.
Write fractions as decimals.
1. \(\frac{23}{10}\) = ………..
2. \(\frac{6}{100}\) = ………..
3. \(\frac{3}{8}\) = ………..
4. \(\frac{2}{25}\) = ………..
Solution:
1. \(\frac{23}{10}\) = 2.3
2. \(\frac{6}{100}\) = 0.06
3. \(\frac{3}{8}\) = 0.375
4. \(\frac{2}{25}\) = 0.08

Question 6.
Write decimals as fractions in simplest form.
1. 0.2 = ……………
2. 0.38 = ……………
3. 1.62 = ……………
4. 8.1 = ……………
Solution:
1. 0.2 = \(\frac{2}{10}\)
2. 0.38 = \(\frac{38}{100}\)
3. 1.62 = \(\frac{162}{100}\)
4. 8.1 = \(\frac{81}{10}\)

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers InText Questions

Check Your Progress?(Page No. 47)

Question 1.
Write any five positive integers.
Solution:
1, 2, 3, 4, 5, 6, 7,

Question 2.
Write any five negative integers.
Solution:
-1, -2, -3, -4, -5, -6,

Question 3.
Which number is neither positive nor negative?
Solution:
0 (zero)

Question 4.
Represent the following situations with integers,
(i) A gain of ₹ 500 ( )
(ii) Temperature is below 5°C ( )
Solution:
i) + 7 500
ii) – 5° C

Question 5.
Represent the following using either positive or negative numbers.
a) A bird is flying at a height of 25 meters above the sea level and a fish at a depth of 2 meters.
b) A helicopter is flying at a height of 60m above the sea level and a submarine is at 400m below sea level.
Solution:
a) Height of the flying bird 25 meters from th$ sea level = + 25 meters
Depth of the fish 2 meters from the sea level = – 2 meters
b) Height of the flying helicopter 60 meters from the sea level = + 60 m
Depth of the submarine 400 m from the sea level = – 400 m

Check Your Progress (Page No. 49)

Question 1.
Draw a vertical number line and represent -5,4,0,-6, 2 and 1 on it.
Solution:

Question 2.
Represent opposite integers of – 200 and + 400 on integer number line.
Solution:
Opposite integers means, additive inverse.
∴ Opposite integer (additive inverse) of – 200 is 200.
Opposite integer (additive inverse) of +400 is – 400.

Let’s Think (Page No. 50)

Question 1.
For any two integers, say 3 and 4, we know that 3 < 4.
Is it true to say -3 < -4? Give reason.
Solution:
On the number line, the value of a number increases as we move to right and decreases as we move to the left. As -3 lies right to -4 on the number line.
So, -3 < -4 is not true.

(Pg. No. 52)

Question 1.
What is additive inverse of 7 ?
Solution:
Additive inverse of 7 is -7.

Question 2.
What is additive inverse of -8 ?
Solution:
Additive inverse of -8 is 8.

Let’s Explore (Page.No. 52)

Question 1.
Find the value of the following using a number line.
i) (-3) + 5 ii) (-5) + 3
Make two questions on your own and solve them using the number line.
Solution:
i) (-3) + 5

On the number line, we first move 3 steps to the left of 0 to reach -3.
Then, we move 5 steps to the right of -3 and reach +2. So, (-3) + 5 = 2

ii) (-5) + 3

On the number line, we first move 5 steps to the left of 0 to reach -5. Then, we move 3 steps to the right of -5 and reach -2. So, (-5) + 3 = – 2

iii)(+6) + (-3)

On the number line, we first move 6 steps to the right of 0 to reach +6.
Then, we move 3 steps to the left of 6 and reach +3. So, (+6) + (-3) = 3

iv) (-4) + (-3)

On the number line, we first move 4 steps to the left of 0 to reach -4. Then, we move 3 steps to the left of -4 and reach -7.
So (-4) + (-3) = -7.

Question 2.
Find the solution of the following:
i) (+5) + (-5) (ii) (+6) + (-7) (iii) (-8) + (+2)
Ask your friend five such questions and solve them.
Solution:
i) (+5) + (-5) = 0

On the number line, we first move 5 steps to the right of 0 to reach +5.
Then, we move 5 steps to the left of +5 and reach 0.

ii) (+6) + (-7) = -1

On the number line, we first move 6 steps to the right of 0 to reach +6.
Then, we move 7 steps to the left of +6 and reach -1.

(iii) (-8) + (+2)

On the number line, we first move 8 steps to the left of 0 to reach -8. Then, we move 2 steps to the right of -8 and reach -6.

iv)(-4) + (+8) = +4

On the number line, we first move 4 steps to the left of 0 to reach -4. Then, we move 8 steps to the right of -4 and reach +4.

v) (+3) + (-4) = -1

On the number line, we first move 3 steps to the right of 0 to reach +3. Then, we move 4 steps to the left of +3 and reach -1.

vi) (+5) + (-6) = -1

On the number line, we first move 5 steps to the right of 0 to reach +5. Then, we move 6 steps to the left of +5 and reach -1.

vii) (+4) + (-4) = 0

On the number line, we first move 4 steps to the right of 0 to reach +4. Then, we move 4 steps to the left of +4 and reach 0.

viii) (-6) +(+4) =-2

On the number line, we first move 6 steps to the left of 0 to reach -6. Then, we move 4 steps to the right of -6 and reach -2.
So, (-6) + (+4) = -2

Let’s Explore (Page No. 55)

Question 1.
Take any two integers a and b. Check whether a+b is also an integer.
Case (i) : Consider two integers 3 and -2 (Positive and negative)
Sum = 3 + (-2) = +1 + 2 – 2 = +1 + 0 = +1 is also an integer.
Case (ii) : Consider two ihtegers 5 and 6 (Both are positive)
Sum = 5 +6 = + 11 is also an integer
Case (iii) : Consider two integers -4 and -6 (Both are negative)
Sum = -4 + (-6) = -4 -6 = -10 is also an integer
Case (iv) : Consider two integers -5 and 4 (Negative and positive)
Sum = -5 + 4 = -1 -4 + 4 = -1 + 0 = -1 is also an integer.
So, if a and b are integers, then their sum a + b is also an integer. Integers are closed under addition.

Question 2.
Check the following Properties on integers, a, b, c are any integers.
i) Closure Property under subtraction ‘
ii) Commutative Property under addition and subtraction (a + b = b + a ?, a – b = b – a?)
iii) Associative Property under addition and subtractioji.
(a + b) + c = a + (b + c) ? (a – b).- c = a – (b – c)?
Solution:
i) Closure Property under subtraction :
Case (i) : Consider two integers 4, -5 (positive and negative)
Then, difference a – b = 4 – (-5) = 4 + 5 = + 9is also an integer.
Case (ii) : Consider two integers 3, 8 (Both are positive)
Then, difference a – b = 3 – (+8) = 3-8
= +3 – 3 – 5 = 0 – 5 = -5is also an integer.
Case (iii) : Consider two integers -2, -6 (Both are negative) .
Then, difference a – b = -2 – (-6) = -2 + 6
= -2 + 2 + 4 = 0 + 4 = +4 is also an integer.

Case (iv) : Consider two integers -3, 2 (Negative and positive)
Then, difference a – b = -3 – (+2) = -3 -2 = -5 is also an integer.
So, if a and b are any two integers, then their difference a – b is also an integer. Integers are closed under subtraction.

ii) Commutative Property under addition and subtraction :
(a + b = b + a, a-b = b-a)
(A) Case (Q : Consider two integers-3 and 5 (Negative and positive)
Then, a + b = -3 + (+5)
= -3 + 5 = – 3 + 3 + 2 = 0 + 2 = + 2
b + a = +5 + (-3) = +2 + 3- 3 = +2 + 0 = + 2
∴ a + b = b + a

Case (ii) : Consider two integers +4 and +2 (Both are positive)
Then, a + b = +4 + (+2) = +4 + 2 = + 6 ‘b + a = +2 + (+4) = +2 + 4 = + 6
∴ a + b = b + a

Case (iii) Consider two integers -5 and -3 (Both are negative) Then, a + b = -5 + (-3) = -5 – 3 = -8 b + a = -3 + (-5) = – 3 – 5 = -8
∴ a + b = b + a

Case (iv) : Consider two integers +4 and -1 (Positive and negative)
Then, a + b = +4 + (*1) = +4 -1 = +3 + 1 -1 = +3 + 0 = +3 b + a = -1 + (+4) =-l + 4 = -l + l+ 3 = 0 + 3 = + 3
∴ a + b = b + a
So, integers are commutative under addition.

(B) Consider two integers -4 and +6 (Negative and positive)
Then, a – b = -4 – (+6) = -4 – 6 = -10 b-a = + 6-(-4) = +6 +4 = +10 -10*10
a – b ≠ b – a
So, integers are not commutative under subtraction.

iiO Associative property under addition and subtraction :
(a + b) + c = a + (b + c) ; (a – b) – c = a – (b – c)
(A) Case (i) : Consider any three integers 2, 4, -5
(a + b) + c = (2 + 4) + (-5) = 6 – 5 = +1 + 5- 5 = +1 + 0 = +1
a + (b + c) = 2 + (4 + (-5)) = 2 + (4 – 5) = 2 + (4 – 4 -1)
= 2 + (0-1) = + 2 – 1 = + 1 + 1 – 1 = +1 + 0 = +1
∴ (a + b) + c = a + (b + c)

Case (ii) : Consider any three integers +2, -5, +3
Then, (a + b) + c = [2 + (-5)] + 3 = [+2 -5] + 3 = +2 -2 -3 + 3 = 0 + 0 = 0
a + (b + c) = +2 + [( – 5)+3] = +2 + [-2 – 3 + 3] = +2 + (-2 + 0) = +2 -2 = 0
∴ (a + b) + c = a + (b + c)

Case (iii) : Consider any three integers 3, 4, 6
Then, (a + b) + c = [2 + (-5)] + 3 [+2-5] + 3
a + (b + c) = 3 + (4 + 6) = 3 + 10 = + 13
∴ (a + b) + c = a + (b + c)

Case (iv) : Consider any three integers -4, -2, +5
Then, (a + b) + c = [-4 + (-2)] + 5 = [-4 -2] + 5 = -6 + 5
= -1-5 + 5 = – 1 + 0 = – 1
a + (b + c) = -4 + [(-2) + 5] = -4 + [-2 + 5]
= _4 + [_2 + 2 +3] = -4 + 0 + 3 =-1-3+ 3
= -1 + 0 = -1
∴ (a + b) + c = a + (b + c)

Case (v) : Consider any three integers-3, 4, 1
Then, (a + b) + c = (-3 + 4) + 1 = (-3 + 3 + 1) + 1 = 0 + 1 + 1 = + 2
a + (b + c) = -3 + (4 + 1) = -3 + 5 = -3 + 3 + 2 = 0 + 2 = +2
∴ (a + b) + c = a + (b + c)

Case (vi) : Consider any three integers -2, 6, -7
Then, (a + b) + c = (-2 + 6) + (-7) = (-2 + 2 + 4) + (-7) = 0 + 4 – 7
= + 4 – 4 – 3 = 0 – 3 = -3
a + (b + c) = -2 + [6 + (-7)] = -2 + (6 – 7) = -2 + [+6 – 6 -1]
= -2 + (-1) = -2 -1 = -3
∴ (a + b) + c = a + (b + c)

Case (vii) : Consider any three integers +6, -3, -1
Then, (a + b) +c = [+6 + (-3)] + (-1) = (6 – 3) – 1 = (+3 +3 -3) -1
=+3-1 =+2 + 1 – 1 =+2 + 0 = + 2
a + (b + c) = +6 + [-3 + (-1)] = 6 + [-3 – 1] = 6 + (-4) .
= +2 + 4 – 4 = 2 + 0 = + 2
∴ (a + b) + c = a + (b + c)

Case(viii)
Consider any three integers -4, -1, -7
Then, (a + b) + c = [-4 + (-1)] + (-7) = (-4 -1) – 7 = -5 -7 = -12
a + (b + c) = -4 + [(-1) + (-7)] = -4 + [-1 -7] = -4 + (-8)
= -4 – 8 = -12
∴ (a + b) + c = a + (b + c)
From all the above cases we conclude that, integers are associative under addition.

(B) Consider any three integers +5, -4, 1
Then, (a – b) – c = (+5 – (-4)) – (+1) = (5 + 4) – 1 .
= + 9 – 1 = + 8 + 1 – 1 = + 8 + 0 = + 8
a – (b – c) = +5 – [- 4 – (+1)] = + 5 – [-4 – 1]
= + 5 – [ -5] =+ 5 + 5 = + 10
+ 8 ≠ +10
∴ (a – b) – c ≠ a – (b – c)
So, integers are not associative under subtraction.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM InText Questions

(Page No. 35)

Question 1.
How does the Sieve of Eratosthenes work ?
Solution:
The following example illustrates how the Sieve of Eratosthenes, can be used to find all the prime numbers that are less than 100. .
Step 1: Write the numbers from 1. to 100 in ten rows as shown below.
Step 2: Cross out 1 as 1 is neither a prime nor a composite number.
Step 3: Circle 2 and cross out all the multiples of 2. (2, 4, 6, 8, 10, 12, ………… )
Step 4: Circle 3 and cross out all the multiples of 3. (3, 6, 9, 12, 15, 18,………….)
Step 5: Circle 5 and cross out all the multiples of 5. (5, 10. 15, 20, 25……………… )
Step 6: Circle 7 and cross out all multiples of 7. (7, 14, 21. 28, 35, ………………….. )

Circle all the numbers that are not crossed out and they are the required prime numbers less than 100.

First arrange the numbers ffom 1 to 100 in a table as shown above.
Enter 6 numbers in each row until the last number 100 is reached.
First we select a number and we strike off all the multiples of it.
Start with 2 which is greater than 1.
Round off number 2 and strike off entire column until the end.
Similarly strike off 4th column and 6th column as they are divisible by 2.
Now round off next number 3 and strike off entire column until end.
The number 4 is already gone.

Now round off next number 5 and strike off numbers in inclined fashion as shown in the figure (they are all divisible by 5). When striking off ends in some row, start again striking off with number in another end which is divisible by 5. New striking off line should be parallel to previous strike off line as. shown in the figure.
The number 6 is already gone.
Now round off number 7 and strike off numbers as we did in case of number 5.
8,9,10 are also gone. .
Stop at this point.
Count all remaining numbers. Answer will be 25.

Prime numbers :
There are 25 prime numbers less than 100.
These are:

What if we go above 100 ? Around 400 BC the Greek mathematician. Euclid, proved that there are infinitely many prime numbers.

Co-primes: Two numbers are said to be co-prime if they have no factors in common. Example: (2, 9), (25, 28)
Any two consecutive numbers always form a pair of co-prime numbers.

Example:
Co-prime numbers are also called relatively prime number to one another.
Example: 3, 5, 8, 47 are relatively prime to one another/co-prime to each other.

Twin primes: Two prime numbers are said to be twin primes, if they differ by 2.
Example: (3, 5), (5, 7), (11, 13), …etc.

Prime factorization: The process of expressing the given number as the product of prime numbers is called prime factorization.
Example: Prime factorization of 24 is
24 = 2 x 12
= 2 x 2 x 6
= 2 x 2 x 2 x 3, this way is unique.

Every number can be expressed as product of primes in a unique manner. We can factorize a given number in to product of primes in two methods. They are
a) Division method
b) Factor tree method

Common factors: The set of all factors which divides all the given numbers are called their common factors.
Example: Common factors to 24, 36 & 48 are 1, 2, 3, 4, 6 & 12
Factors of 24 = 1, 2, 3, 4, 6, 8, 12 & 24
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 & 36
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24 & 48
Common factors to 24, 36 & 48 are 1, 2, 3, 4, 6 & 12
We can see that among their common factors 12 is the highest common factor. It is called H.C.F. of the given numbers. So H.C.F. of 24, 36 & 48 is 12.

H.C.F./G.C.D : The highest common factor or the greatest common divisor of given numbers is the greatest of their common factors.
H.C.F. of given two or more numbers can be found in two ways.
a) By prime factorization
b) By continued division
H.C.F. of any two consecutive numbers is always 1.
H.C.F. of relatively prime/co-prime numbers is always 1.
H.C.F. of any two consecutive even numbers is always 2.
H.C.F. of any two consecutive odd numbers is always 1.

Common multiples:
Multiples of 8: 8, 16, 24, 32, 40, 48,
Multiples of 12: 12, 24, 36, 48, … .
Multiples.common to 8 & 12: 24, 48; 72, 96, …
Least among the common multiple is 24. This is called L.C.M. of 8 & 12. The number of common multiples of given two or more numbers is infinite, as such greatest common multiple cannot be determined.

L.C.M.: The least common multiple of two or more numbers is the smallest natural number among their common multiples.
L.C.M. of given numbers can be found by the
a) Method of prime factorization.
b) Division method.
L.C.M. of any two consecutive numbers is always equal to their product.
L.C.M. of 8 &9 is 8 x 9 = 72
L.C.M. of co-prime numbers is always equal to their product.
L.C.M. of 8 & 15 is 8 x 15 = 120

Relation between the L.C.M. & H.C.F:
For a given two numbers Nj & N2 , the product of the numbers is equal to the product of their L.C.M.(L) & H.C.F.(H)
N₁ x N₂ = L x H

Check Your Progress (Page No. 29)

Question 1.
Are the numbers 900, 452, 9534, 788 divisible by 2? Why?
Solution:
Yes. Because these numbers have 0, 2, 4 and 8 in their ones place. The numbers having 0, 2, 4, 6 and 8 in their ones place are divisible by 2.

Question 2.
Are the numbers 953, 457, 781, 325, 269 divisible by 2? Why?
Solution:
No. Because, these numbers have 3, 7, 1, 5 and 9 in their ones place. The numbers having 0, 2,
4, 6 and 8 in their ones place are only divisible by 2. So, these are not divisible by 2.

Question 3.
Are the numbers 452, 673, 259, 356 divisible by 2? Verify.
Solution:
452 and 356 have 2 and 6 in their ones place respectively.
So, they are divisible by 2.
673 and 259 have 3 and 9 in their ones place respectively.
So, they are not divisible by 2.

Question 4.
Check whether the following numbers are divisible by 3 (using rule). Verify by actual division.
(i) 12345 (ii) 61392 (iii)8747
Solution:
i)12345
1 + 2 + 3 + 4 + 5=15 is a multiple of 3.
If the sum of the digits of a number is the multiple of 3, then the number is divisible by 3.
So, 12345 is divisible by 3.

ii) 61392
6 + 1 + 3 + 9 + 2 = 21is a multiple,of 3.
If the sum of the digits of a number is a multiple of 3, then the number is divisible
by 3.

So, 61392 is divisible by 3.

iii) 8747
8 + 7 + 4 + 7 = 26is not a multiple of 3.
If the sum of the digits of a number is a multiple of 3, then the numbei is divisible by 3. So, 8747 is not divisible by 3.

So, 8747 is not divisible by 3.

Let’s Explore (Page No. 29)

Question 1.
Is 8430 divisible by 6? Why?
Given number is 8430.
The given number has zero in the ones place.
So, 8430 is divisible by 2. –
And the unit sum is8 + 4 + 3 + 0 = 15 is a multiple of 3.
So, 8430 is divisible by 3.
If a number is divisible by both 2 and 3, then only it is divisible by 6.
8430 is divisible by both 2 and 3.
Therefore 8430 is divisible by 6.

Question 2.
Take any three 4 digit numbers and check whether they are divisible by 6.
Solution:
Consider: i) 5632, ii) 6855, iii) 9600 are three 4 digit numbers.
i) 5632 has 2 in its ones place. So, 5632 is divisible by 2.
The unit sum is
5 + 6 + 3 + 2 = 16 is not a multiple of 3. So, 5632 is not divisible by 3.
If a number is divisible by both 2 and 3, then only it is divisible by 6.
5632 is qnly divisible by 2, but not divisible by 3.
So, 5632 is not divisible by 6.

ii) 6855 has 5 in the ones place. So, 6855 is not divisible by 2.
6 + 8 + 5 + 5 = 24 is a multiple of 3.
So, 6855 is divisible by 3.
If a number is divisible by both 2 and 3, then only it is divisible by 6.
6855 is not divisible by 2, but it is divisible by 3.
So, 6855 is not divisible by 6.

iii) 9600 has ‘0’ in the ones place. So, 9600 is divisible by 2.
9 + 6 + 0 + 0 = 15 is a multiple of 3.
So, 9600 is divisible by 3. .
If a number is divisible by both 2 and 3, then only it is divisible by 6, 9600 is divisible by 1 both 2 and 3.
So, 9600 is divisible by 6.

Question 3.
Can you give an example of a number which is divisible by 6 but not by 2 and 3? Why?
Solution:
No. We can’t give any example, because if any number is divisible by both 2 and 3, then only it is divisible by 6. Otherwise it is not possible.

Check Your Progress (Page No. 30 & 31))

Question 1.
Test whether 6669 is divisible by 9. ,
Solution:
Given number is 6669.
Sum of the digits = 6 + 6 + 6 + 9 = 27 is divisible by 9
If the sum of the digits of a number is divisible by 9 then, it is divisible by 9. ” 27 is divisible by 9. So, 6669 is divisible by 9.

Question 2.
Without actual division, find whether 8989794 is divisible by 9.
Solution:
Given number is 8989794.
Sum of the digits =8+9+8+9+7+9+4=54 – 1
If the sum of the digits of a number is divisible by 9.
Then, it is divisible by 9.
54 is divisible by 9. So, 8989794 is divisible by 9.

Question 3.
Are the numbers 28570, 90875 divisible by 5? Verify by actual division also.
Solution:
a) Given number be 28570.
The numbers with zero or five at ones place are divisible by 5
28570 has zero in its ones place. So, 28570 is divisible by 5.
Actual division :

So, 28570 is completely divisible by 5.

b) Given number is 90875
In 90875, ones place digit is 5. So, 90875 is divisible by 5.
Actual division :

In the given number 90875 the digit in the units place is 5,
‘ So, it is divisible by 5.
So, 90875 is divisible by 5.

Question 4.
Check whether the number 598, 864, 4782 and 8976 are divisible by 4. Use divisibility rule and verify by actual division.
Solution:
a) Given number is 598.
The number formed by the digits in tens and ones places of 598 is 98.
If the number formed by last two digits (Ones and Tens) of the number is divisible
by 4, then the number is divisible by 4.
98 is not divisible by 4. So, 598 is not divisible by 4.

b) Given number is 864.
The number formed by tens and ones places of 864 is 64.
If the number formed by last two digits (ones and tens)
of the number is divisible by 4, then the number is divisible by 4.
64 is divisible by 4. So, 864 is divisible by 4.

c) Given number is 4782.
The number formed by ones and tens places of 4782 is 82.
If the number formed by last two digits (ones and tens) of the number is divisible by 4, then the number is divisible by 4. 82 is not divisible by 4.

d) Given number is 8976.
The number formed by the digits in tens and ones places of 8976 is 76.
If the number formed by last two digits
(Tens and ones) of the number is divisible by 4. Then the number is divisible by 4.
76 is divisible by 4. So, 8976 is divisible by 4.

Question 1.
Fill the blanks and complete the table. (Page No. 32)
Solution:

Lets Explore (Page No.33)

Question 1.
1221 is a polindrome number, which on reversing its digits gives the same number. Thus, every polindrome number with even number of digits is always divisible by 11. Write polindrome number of 6 – digits.
Solution:

There are some polindrome number of 6 – difits.

Check Your Progress (Page No.34)

Question 1.
Find the factors of 60.
Solution:
60 = 1 x 60
60 = 2 x 30
60 = 3 x 20
60 = 4 x 15
60 = 5 x 12
60 = 6 x 10
∴ The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Question 2.
Do all the factors of a given number divide the number exactly? Find the factors of 30 and verify by division.
30 = 1 x 30
30 = 3 x 10
30 = 2 x 15
30 = 5 x 6
The factors of 30 are 1, 2, 3, 5, 6,10, 15 and 30.

Yes, the factors of a given numbers are divide the number exactly.

Question 3.
3 is a factor of 15 and 24. Is 3 a factor of their difference also?
Solution:
Difference = 24 – 15 = 9 is the multiple of 3.
Yes, 3 is a factor of difference of 15 and 24.

Let’s Explore (Page No. 35)

Question 1.
What is the smallest prime number?
Solution:
2

Question 2.
What is the smallest composite number?
Solution:
4

Question 3.
What is the smallest odd prime number?
Solution:
3

Question 4.
What is the smallest odd composite number?
Solution:
9

Question 5.
Write 10 odd and 10 even composite numbers.
Solution:
Odd composite numbers are 9, 15, 21, 25, 27, 33, 35, 39, 45, 49.
Even composite numbers are 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.
Except 2, every even number is a composite number.

Let’s Explore (Page No. 36)

Question 1.
Can you guess a prime number which when on reversing its digits, gives another prime number? (Hint a 2 digit prime number)
Solution:
13 and 31; 17 and 71, 37 and 73 79 and 97.

Question 2.
311 is a prime number. Can you find the other two prime numbers just by rearrang-ing the digits?
Solution:
113, 131

Check Your Progress (Page No. 36)

Question 1.
From the following numbers identify different pairs of co-primes. 2, 3,4, 5,6,7, 8, 9 and 10.
Solution:
The numbers which have only 1 as the common factor are called co-primes, (or) Numbers having no common factors, other than 1 are called co-primes.
2,3; 2, 5; 2, 7; 2, 9; 3,4; 3, 5; 3, 7; 3, 8; 3,10; 4,5; 4, 7; 4, 9; 5, 6; 5, 7; 5,8; 5, 9; 6, 7; 8, 9 and 9,10. These are the different pairs of co-primes with 2, 3, 4, 5, 6, 7, 8, 9 and 10.
1) Any two primes always forms a pair of co-primes.
2) Any two consecutive numbers always form a pair of co-primes.
3) Any two primes cilways form a pair of co-primes. .

Question 2.
Write the pairs of twin primes less than 50.
Solution:
Two prime numbers are said to be twin primes, if they differ each other by 2.
Twin primes less than 50 are (3, 5); (5, 7); (11,13); (17, 19); (29, 31) and (41, 43).

(Page No. 36)

Question 1.
Find the HCF of 12, 16 and 28
Solution:

Thus 12 = 2 x 2 x 3
16 = 2 x 2 x 2 x 2
28 = 2 x 2 x 7
The common factor of 12, 16 and 28 ¡s 2 x 2 = 4.
Hence, H.C.F of 12, 16 and 28 is 4.

Let’s Explore (Page No. 40)

What is the HCF of any two
i) Consecutive numbers ?
ii) Consecutive even numbers ?
iii) Consecutive odd numbers? What do you observe? Discuss with your Mends.
Solution:
Consider the two consecutive number are 5, 6 and 9, 10.

We observed that HCF of any two consecutive numbers is always 1.

ii) Consider two consecutive even numbers are 8, 10 and 20, 22.

We observed that HCF of any two consecutive even numbers is always 2.

iii)Consider the consecutive odd numbers are 7, 9 and 13, 15.

We observed that HCF of any two consecutive odd numbers is always 1.

(Page No. 42)

Question 1.
Find LCM of (i) 3, 4 (ii) 10, 11 (iii) 10, 30 (iv) 12, 24 (v) 3, 12 by prime factorization method.
Solution:
i) Given numbers are 3, 4
Factors of 3 = 1 x 3
Factors of 4 = 2 x 2
LCM of 3, 4 = 1 x 3 x 2 x 2 = 12

ii) Given numbers are 10, 11
Factors of 10 = 2 x 5
Factors of 11 = 1 x 11
LCM of 10, 11 = 2 x 5 x 11 = 110

iii) Given numbers are 10, 30
Factors of 10 = 2×5
Factors of 30 = 2x3x5
LCM of 10, 30 = 2 x 3 x 5 = 30

iv) Given numbers are 12, 24
Factors of 12 = 2x2x3 ,
Factors of 24 =. 2 x 2 x 2 x 3
LCM of 12, 24 = 2 x 2 x 2 x 3 = 24

v) Given numbers are 3,12 Factors of 3 = 1×3
Factors of 12 =2 x 2 x 3,
LCM of 3, 12 = 3 x 2 x 2 = 12

(Page No. 43)

Question 1.
What is the LCM and HCF of twin prime numbers ?
Solution:
LCM = Product of the taken twin primes and HCF = 1

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers InText Questions

Question 1.
Fill the following table with the successor and predecessor of the numbers provided. (Page No. 15)

Solution:

Discuss

Question 1.
Which number has no successor ? (Page No. 15)
Answer:
Each and every number has a successor.

Question 2.
Which number has no predecessor? (Page No. 15)
Answer:
Zero (0) has no predecessor in the set of whole numbers.

Check Your Progress (Page No. 16)

Question 1.
Which is the smallest whole number?
Answer:
Zero(0) is the smallest whole number.

Let’s Think (Page No. 16)

Question 1.
Are all natural numbers whole numbers? .
Solution:
Yes. All the natural numbers are whole numbers.

Question 2.
Are all whole numbers natural numbers?
Solution:
No. All the whole numbers are not natural numbers.

Let’s Do (Page No. 17)

Show these on number line:

i) 5 + 3
Solution:

Draw the number line starting with ‘O’.
Start from 5, make 3 jumps to the right of 5 on the number line. We reach 8.
So, 5 + 3 = 8

ii) 5 – 3
Solution:

Draw the number line starting with zero (0).
Start from 5, make 3 jumps to the left of 5 on the number line, we reach 2.
So, 5 – 3 = 2

iii) 3 + 5
Solution:

Draw the number line starting with zero (0).
Start from 3, we make 5 jumps to the right of 3 on the number line. We reach 8. So, 3 + 5 = 8

iv) 10 + 1
Solution:

Draw the number line which starts with zero (0).
Start from 10, make 1 jump to the right of 10 on the number line. We reach’ll. So, 10 + 1 = 11

v) 8 – 5
Solution:

Draw the number line which starts with zero (0).
Start from 8, we make 5 jumps to the left of 8 on the number line. We reach 3. So, 8-5 = 3

Let’s Explore (Page No. 17)

Find the following by using number line:

Question 1.
Which number should be deducted from 8 to get 5?
Solution:

Draw the number line, which starts with zero (0).
To get 5 from 8. We have to start from 8. We make 3 jumps to the left of 8 on the number line, we reach 5. As we are moving on left side we have minus sign.
So, 8 – 3 = 5
Therefore to get 5 we deduct 3 from 8.

Question 2.
Which number should be deducted from 6 to get 1?
Solution:

Draw the number line starting with zero (0).
To get 1 from 6 we have to start from 6.
We make 5 jumps to the left of 6 on the number line. We reach 1. As we have moved to left side, we have 6-5 = 1 Threfore to get 1 we deduct 5 from 6.

Question 3.
Which number should be added to 6 to get 8?
Solution:

Draw the number line starts with zero (0).
To get 8 from 6, we have to start from 6.
We make 2 jumps to the right of 6 on the number line to reach 8; So, 6 + 2 = 8 ‘
Therefore to get 8 we add 2 to 6.

Question 4.
How many 6 are needed to get 30?
Solution:

Draw the number line starts with zero (0).
Start from 0 and make 6 jumps to the right of the zero as the number line. Now, treat 6 jumps as one step. So, to reach 30, we make 5 steps.
So, 5 × 6 = 30

Question
Raju and Gayatri together made a number line and played a game on it.

Raju asked “Gayatri, where will you reach if you jump thrice, taking leaps of 3, 8 and 5”? Gayatri said the first leap will take me to 3 and then from there I will reach 11 in the second step and another five steps from there to 16′.
Draw Gayatri’s steps and verify her answers.
Play this game using addition and subtraction on this number line with your friend.
Solution:

Lets Think (Page No. 19)

Question 1.
Are the whole numbers closed under subtraction?
Solution:
8 – 5 = 3, a whole number
5 – 8 = -3 is not a whole number
Therefore whole numbers are not closed under subtraction.

Question 2.
Are the whole numbers closed under division?
Solution:
6 ÷ 3 = 2, a whole number
3 ÷ 6 = \(\frac{3}{6}\), not a whole number 6
Therefore, whole numbers are not closed under division.

Check Your Progress (Page No. 19)

Question 1.
Find out 12 ÷ 3 and 4 ÷ 7.
Solution:
12 ÷ 3
12 is divided by 3 means, we subtract 3 from 12 repeatedly till, we get zero i.e., we subtract 3 from 12 again and again till, we get zero.
12 – 3 = 9 once
9-3 = 6 twice
6-3 = 3 thrice
3-3 = 0 four times
So, 12-3 = 4

42 ÷ 7
42 is divided by 7 means, we subtract 7 from 42 repeatedly, i.e., we subtract 7 from 42 again and again till, we get zero a number less than 7.
42 – 7 = 35 once
35 – 7 = 28 twice
28 – 7 = 21 thrice
21-7 = 14 four times
14-7 = 7 five times
7 – 7 = 0 six times
i.e., we can subtract 7 from 42 for 6 times successively. ,
So, 42 ÷ 7 = 6.

Question 2.
What would 6 4-0 and 9 4- 0 be equal to?
Solution:
6 ÷ 0
6 is divided by 0 means, we subtract 0 from 6 repeatedly i.e., we subtract 0 from 6 again and agian from 6.
6 – 0 = 6 once
6 – 0 = 6 twice
6 – 0 = 6 thrice and ……………….
If we subtract zero from 6 successively we can’t get zero at any end.
So, 6 ÷ 0 is not a number that we can reach.
So, division of a whole number by 0 does not give a known number as answer, so it is not defined.
Similarly 9 ÷ 0 is not defined.
So, we can’t say whether they are equal or not.

Let’s Explore (Page No. 20)

Take few examples and check whether

a) Subtraction is commutative over whole numbers or not?
Solution:
Let’s take two whole numbers 4 and 6
6 – 4 = 2 and (4 – 6) = – 2 is not a whole number.
So, 6 – 4 ≠ 4 – 6.
Therefore we say that subtraction is not commutative over the whole numbers.

b) Division is commutative over whole numbers or not ?
Solution:
Let’s take two whole numbers 8 and 2
8 ÷ 2 = 4 and (2 ÷ 8) = \(\frac{1}{4}\) is not a whole number.
So, 8 ÷ 2 ≠ 2 ÷ 8
Therefore, we say that division is not commutative over the whole numbers.

Check Your Progress (Page No. 22)

Verify the following:
i) (5 × 6) × 2 = 5 × (6 × 2)
Solution:
L.H.S : (5 × 6) × 2 = 30 × 2 = 60
R.H.S : 5 × (6 × 2) = 5 × 12 = 60
∴ L.H.S = R.H.S
So (5 × 6) × 2 = 5 × (6 × 2)
∴ Multiplication of whole numbers is associative.

ii) (3 × 7) × 5 = 3 × ( 7 × 5 )
Solution:
L.H.S : (3 × 7) × 5 = 21 × 5 = 105
R.H.S : 3 × (7 × 5) = 3 × 35 = 105
∴ (3 × 7) × 5 = 3 × (7 × 5)
∴ Multiplication of whole numbers is associative.

Check Your Progress (Page No. 22)

Use the commutative and associative properties to simplify the following:

i) 319 + 69 + 81
Solution:
319 + 69 + 81 = 319 +(81 + 69) (Commutative property)
= (319 + 81) + 69 (Associative property)
= 400 + 69 = 469

ii) 431 + 37 + 69 + 63
Solution:
431 + 37 + 69 + 63
= 431 + (37 + 69) + 63
= 431 + (69 + 37) + 63 (Commutative property)
= (431 + 69) + (37 + 63) (Associative property)
=(431 + 69) + 100
= 500+ 100 = 600

iii) 2 × (71 × 5)
Solution:
2 × (71 × 5) = 2 × (5 × 71) (Commutative property)
= (2 × 5) × 71 (Associative property)
= 10 × 71
= 710

iv) 50 × 17 × 2
Solution:
50 × (17 × 2) = 50 × (2 × 17) (Commutative property)
= (50 × 2) × 17 (Associative property)
= 100 × 17 = 1700

Let’s Think (Page No. 22)

a) Is(8 ÷ 2) ÷ 4 = 8 ÷ (2 ÷ 4)?
Is there any associative property for division ?
Check if this property holds for subtraction of whole numbers too.
Solution:
a) (8 ÷ 2) ÷ 4 = (8 ÷ 2) ÷ 4
= 4 ÷ 4 = 1
8 ÷ (2 ÷4) = 8 ÷ \(\left(\frac{2}{4}\right)\)
= 8 – \(\frac{4}{2}\) = 8 × 2 = 16
So, (8 ÷ 2) ÷ 4 ≠ 8 ÷ (2 ÷ 4) .
Therefore, associative property does not holds in division.

b) Is (8 – 2) – 4 = 8 – (2 – 4) ?
Solution:
(8 – 2) – 4 = 6 – 4 .
= 2
8 – ( 2 – 4 ) = 8 – ( – 2)
= 8 + 2 = 10
So, (8 – 2) – 4 ≠ 8 – (2 – 4)
Therefore, associative property does not holds in subtraction,
i. e., whole numbers are not associative w.r.t. subtraction.
They are not equal.
So whole numbers do not satisfy Associative property w.r.t. Subtraction & Division.

Find using distributive property : (Pg. No. 22)
i) 2 × (5 + 6)
ii) 5 × (7 + 8)
Solution:
i) 2 × (5 + 6)
Given, 2 × (5 + 6) = (2 × 5) + (2 × 6)
By using distributive property of multiplication over addition.
2 × 11 = 10 + 12
22 = 22
L.H.S. = R.H.S

ii) 5 × (7 + 8)
Given, 5 × (7 + 8) = (5 × 7) + (5 × 8)
By using distributive property of multiplication over addition.
5 × 15 = 35 + 40
75 = 75
L.H.S = R.HS

iii) 19 × 7 + 19 × 3
Given, (19 × 7) + (19 × 3) = 19 × (7 + 3)
By using distributive property of multiplication over addition.
133 + 57 = 19 × 10
190 = 190
L.H.S = R.H.S

Find : i) 25 × 78 ii) 17 × 26 iii) 49 × 68 + 32 × 49 using distributive property. (Page. No. 22)
Solution:
i) 25 × 78
Given, 25 × 78 = 25 × (80 – 2)
By using distributive property of multiplication over subtraction.
= (25 × 80) – (25 × 2)
= 2000 – 50 = 1950
∴ 25 × 78 = 1950

ii) 17 × 26
Given, 17 × 26 = (10 + 7) × 26
By using distributive property of multiplication over addition.
= (10 × 26) + (7 × 26)
= 260 + 182 = 442
7.17 × 26 = 442

(OR)

= 17 × (30 – 4)
By using distributive property of multiplication over subtraction.
= (17 × 30) – (17 × 4)
= 510 – 68 = 442

iii) 49 × 68 + 32 × 49
Given, 49 × 68 + 32 × 49 = (49 × 68) + (49 × 32)
By using distributive property of multiplication over addition.
= 49 × (68 + 32) .
= 49 × 100
∴ (49 × 68) + (32 × 49) = 4900

Let’s Explore (Page. No. 25)

Question 1.
Which numbers can be shown as a line only?
Solution:
Two or more than two numbers can be shown as a line.
i.e., 2, 3, 4, 5, 6,

Question 2.
Which numbers can be shown as rectangles?
Solution:
6, 8, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27,.

Question 3.
Which numbers can be shown as squares?
Solution:
4, 9, 16, 25.

Question 4.
Which numbers can be shown as triangles?
Solution:
3, 6, 10, 15, 21, :
Note : Starting from 3; +3, +4, +5, +6, …………………. are all triangular numbers.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us InText Questions

Write the numbers in expanded form. (Page No. 5)

Question 1.
96,08,54,039
Solution:
96,08,54,039 = 9 × 10,00,00,000 + 6 × 1,00,00,000 + 8 × 10,00,000 + 5 × 10,000 + 4 × 1000 + 3 × 10 + 9 × 1
Ninety six crores eight lakhs fifty four thousand and thirty nine.

Question 2.
857,90,00,756
Solution:
857,90,00,756 = 8 × 100,00,00,000 + 5 × 10,00,00,000 + 7 × 1,00,00,000 + 9 × 10,00,000 + 7 × 100 + 5 × 10 + 6 × 1
Eight hundred fifty seven crores ninety lakhs seven hundred and fifty six.

1 Crore = 10 Ten Lakhs
= 100 Lakhs
= 1000 Ten Thousands
= 10,000 Thousands
= 1,00,000 Hundreds
= 10,00,000 Tens
= 1,00,00,000 Unit’s

Check Your Progress (Page No. 6)

Question 1.
Write 10 crores and 100 crores as in the above table.
Solution:
Ten crores = 10 One crores
= 100 Ten lakhs
= 1000 Lakhs
= ,10,000 Ten thousands
= 1,00,000 Thousands
= 10,00,000 Hundreds
= 1,00,00,000 Tens
= 10,00,00,000 Units

Hundred crores = 100 One crores
= 10 Ten crores
= 10,000 Lakhs
= 1.0. 000 Ten thousands
= 10.0. 000 Thousands
= 1.0. 00.000 Hundreds
= 10.0. 00.000 Tens
= 100.0. 00.000 Units

Check Your Progress (Page No. 8)

Question 1.
Write remaining numbers of the above table in the International System.
Solution:

Question 2.
Fill the boxes in the table with your own numbers and write in words in the International system.
Solution:
a) 896800705

Put comma for each period 896,800,705 in International System.
In expanded form :
= 8 ×x 1,000,000,000 + 9 × 10,000,000 + 6 × 1,000,000 + 8 × 100,000 + 7 × 100 + 5 × 1

In word form :
Eight hundred ninety six millions eight hundred thousand seven hundred and five.

b) 239176507857
Put comma for each period 239,176,507,857 in International System.
In expanded form :
= 2 × 100,000,000,000 + 3 × 10,000,000,000 + 9 × 1,000,000,000 + 1 × 100,000,000 + 7 × 10,000,000 + 6 × 1,000,000 + 5 × 100,000 + 7 × 1,000 + 8 × 100 + 5 × 10 + 7 × 1
In word form :
Two hundred thirty nine billion one seventy six million five hundred seven thousand eight hundred and fifty seven.

c) 452069258932
Put comma for each period 452,069,258,932
In expanded form :
= 4 × 100,000,000,000 + 5 × 10,000,000,000 + 2 × 1,000,000,000 + 6 × 10,000,000 + 9 × 1,000,000 + 2 × 100,000 + 5 × 10,000 + 8 × 1,000 + 9 × 100 + 3 × 10 + 2 × 1
In word form :
Four hundred fifty two billion sixty nine million two hundred fifty eight thousand nine hundred and thirty two.

d) 839241367054
Put comma for each period 839,241,367,054
In expanded form :
8 × 100,000,000,000 + 3 × 10,000,000,000 + 9 × 1,000,000,000 + 2 × 100,000,000 + 4 × 10,000,000 + 1 × 1,000,000 + 3 × 100,000 + 6 × 10,000 + 7 × 1.000 + 5 × 10 + 4 × 1
In word form :
Eight hundred thirty nine billion two hundred forty one million three hundred sixty seven thousand and fifty four.

e) 342056743298
Put comma for each period 342,056,743,298
In expanded form :
3 × 100,000,000,000 + 4 × 10,000,000,000 + 2 × 1,000,000,000 + 5 × 10,000,000 + 6 × 1,000,000 + 7 × 100,000 + 4 × 10,000 + 3 × 1,000 + 2 × 100 + 9 × 10 + 8 × 1
In word form :
Three hundred forty two billion fifty six million seven hundred forty three thousand two hundred and ninety eight.

Check Your Progress (Page No.12)

Question 1.
Round off each to the nearest ten, hundred and thousands.
(1) 56,789 (2) 86,289 (3) 4,56,726 (4) 5,62,724
Solution:

Let’s Explore (Page No.12)

Question 1.
Discuss with your friends about rounding off numbers. Consider the population of A.P., Telangana and India in 2011. Round off the numbers to the nearest lakhs.
Solution:

Estimate the sum by rounding and verify the result. (Page No.12)

Question 1. 8756 + 723
Solution:
Given 8756 + 723
First estimate by rounding = 8800 + 700 = 9500

Thus sum is 9,479.
Think
9479 is close to the estimate of 9500.

Question 2.
56723 + 4567 + 72 + 5
Solution:
Given 56723 + 4567 + 72 + 5
First estimate by rounding = 56720 + 4570 + 70 + 10 = 61370

The sum is 61,367.
Think
61367 is close to the estimate of 61370.

Question 3.
656724 + 8567
Solution:
Given 656724 + 8567
First estimate by rounding = 657000 + 9000 = 666000

The sum is 6,65,291.

Think
665291 is close to the estimate of 666000.

Question 4.
60756 + 2562 + 72
Solution:
Given 60756 + 2562 + 72
First estimate by rounding = 60760 + 2560 + 70 = 63390

The sum is 63,390.
Think
63390 is equal to the estimate of 63390.

Estimate the difference by rounding and verify the result.Pg. No. 13)

Question 1.
7023 – 856
Solution:
Given, 7023 – 856
First estimate by rounding = 7000 – 900 = 6100

Think
6167 is close to the estimate of 6100

Question 2.
9563 – 2847
Solution:
Given, 9563 – 2847
First estimate by rounding = 10000 – 3000 = 7000

Think
6716 is close to the estimate of 7000

Question 3.
52007 – 6756
Solution:
Given, 52007 – 6756
First estimate by rounding = 52000 – 7000 = 45000

Think
45251 is close to the estimate of 45000

Question 4.
95625 – 4235
Solution:
Given, 95625 – 4235
First estimate by rounding = 95600 – 4200 = 91400 .

Think
91390 is close to the estimate of 91400.

Estimate the product by rounding and verify the result.

Question 1.
63 × 85
Solution:
Given, 63 × 85
First estimate by rounding = 60 × 90 = 5400,
Rounding the result to hundreds = 5400

Think
5355 is close to the estimate of 5400.

Question 2.
636 × 78
Solution:
Given, 636 × 78
First estimate by rounding = 640 × 80 = 51200
Rounding the result to hundreds = 51200

Think
49608 is close to the estimate of 51200.

Question 3.
506 × 85
Solution:
Given, 506 × 85
First estimate by rounding = 500 × 90 = 45000

Think
43010 is close to the estimate of 45000.

Question 4.
709 × 98
Solution:
Given, 709 × 98
First estimate by rounding = 700 × 100 = 70000

Think
69482 is close to the estimate of 70000.

Estimate the quotient by rounding and verify the result.

Question 1.
936 ÷ 7
Solution:
Given, 936 ÷ 7
Divide 936 ÷ 7
First estimate by rounding 1000 ÷ 10 = 100

Think
133 is close to the estimate of 100.

Question 2.
956 ÷ 17
Solution:
Given, 956 ÷ 17
Divide 956 ÷ 17
First estimate by rounding 1000 – 20 = 50

Think
56 is close to the estimate of 50.

Question 3.
859 ÷ 23
Given, 859 ÷ 23
Divide 859 ÷ 23
First estimate by rounding 860 ÷ 20 = 43

Think
37 is close to the estimate of 43.

Question 4.
708 ÷ 32
Given, 708 ÷ 32
Divide 708 ÷ 32
First estimate by rounding 710 ÷ 30 = 23

Think
22 is close to the estimate of 23.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra Unit Exercise

Question 1.
The cost of one fan is Rs. 1500. Then what is the cost of ‘n’ fans?
Answer:

Given cost of one fan = Rs. 1500
Number of fans = n
Cost of n fans = cost of one fan × no. of fans = 1500 × n
∴ Cost of n fans = 1500n

Question 2.
Srinu has number of pencils. Raheem has 4 times the pencils as of Srinu. How many pencils does Rahim has? Write an expression.
Answer:
Let number of pencils Srinu has = x
Number of pencils Raheem has = 4 times of Srinu
= 4 × x
∴ Number of pencils Raheem has = 4x

Question 3.
Parvathi has 5 more books than Sofia. How many books are with Parvathi? Write an expression choosing any variable for number of books.
Answer:
Let number of books Sofia has = y
Given Parvathi has 5 more books than Sofia
Number of books Parvathi has = 5 books more than Sofia
= y + 5
∴ Number of books Parvathi has = y + 5

Question 4.
Which of the following are equations?
i) 10 – 4p = 2
ii) 10 + 8x < – 22
iii) x + 5 = 8
iv) m + 6 = 2
v) 22x – 5 = 8
vi) 4k + 5 > – 100
vii) 4p + 7 = 23
viii) y < – 4
Answer:
i) 10 – 4p = 2
We know that, a mathematical statement involving equality symbol is called an equation.
10 – 4p = 2 has equality symbol.
So, it is an equation.

ii) 10 + 8x < – 22
We know that, a mathematical statement involving equality symbol is called an equation.
10 + 8x < – 22 has no equality symbol.
So, it is not an equation [so it is an inequation]

iii) x + 5 = 8 We know that, a mathematical statement involving equality symbol is called an equation.
x + 5 = 8 has equality symbol.
So, it is an equation.

iv) m + 6 = 2 We know that, a mathematical statement involving equality symbol is called an equation.
m + 6 = 2 has equality symbol.
So, it is an equation.

v) 22x – 5 = 8 We know that, a mathematical statement involving equality symbol is called an equation.
22x – 5 = 8 has equality symbol.
So, it is an equation.

vi) 4k + 5 > – 100
We know that, a mathematical statement involving equality symbol is called an equation.
4k + 5 > -100 has no equality symbol.
So, it is not an equation.
It is an inequation.

vii) 4p + 7 = 23
We know that, a mathematical statement involving equality symbol is ailed an equation.
4p + 7 = 23 has equality symbol.
So, it is an equation.

viii) y < – 4
We know that, a mathematical statement involving equality symbol is called an equation.
y < – 4 has no equality symbol.
So, it is not an equation.
It is an inequation.

Question 5.
Write L.H.S and R.H.S of the following equations:
i) 7x + 8 = 22
ii) 9y – 3 = 6
iii) 3k – 10 = 2
iv) 3p – 4q = -19
Answer:
i) 7x + 8 = 22
Given equation is 7x + 8 = 22
LHS = 7x + 8
RHS = 22

ii) 9y – 3 = 6
Given equation is 9y – 3 = 6
LHS = 9y – 3
RHS = 6

iii) 3k – 10 = 2
Given equation is 3k – 10 = 2
LHS = 3k – 10
RHS = 2

iv) 3p – 4q = -19
Given equation is 3p – 4q = -19
LHS = 3p – 4q
RHS = -19

Question 6.
Solve the following equations by trial and error method.
i) x – 3 = 5
ii) y + 6 = 15
iii) y = -1
iv) 2k – 1 = 3
Answer:
i) x – 3 = 5
Given equation is x – 3 = 5
If x = 1, then the value of x – 3 = 1 – 3 = -2 ≠ 5
If x = 2, then the value of x – 3 = 2 – 3 = -l ≠ 5
If x = 3, then the value of x – 3 = 3 – 3 = 0 ≠ 5
If x = 4, then the value of x – 3 = 4 – 3 = l ≠ 5
If x = 5, then the value of x – 3 = 5 – 3 = 2 ≠ 5
If x = 6, then the value of x – 3 = 6 – 3 = 3 ≠ 5
If x = 7, then the value of x – 3 = 7 – 3 = 4 ≠ 5
If x = 8, then the value of x – 3 = 8 – 3 = 5 = 5
From the above when x = 8, the both LHS and RHS are equal.
∴ Solution of the equation x – 3 = 5 is x = 8

ii) y + 6 = 15
Given equation is y + 6 = 15
If y = 1, then the value of y + 6 = 1 + 6 = 7 ≠ 15
If y = 2, then the value of y + 6 = 2 + 6 = 8 ≠ 15
If y = 3, then the value of y +6 = 3 + 6 = 9 ≠ 15
If y = 4, then the value of y + 6 = 4 + 6 = 10 ≠ 15
If y = 5, then the value of y + 6 = 5 + 6 = 11 ≠ 15
If y = 6, then the value of y + 6 = 6 + 6 = 12 ≠ 15
If y = 7, then the value hf y + 6 = 7 + 6 = 13 ≠ 15
If y = 8, then the value of y + 6 = 8 + 6 = 14 ≠ 15
If y = 9, then the value of y + 6 = 9 + 6 = 15 = 15
From the above when y = 9, the both LHS and RHS are equal.
∴ Solution of the equation y + 6 = 15 is y = 9

iii) \(\frac{m}{2}\) = -1
Given equation is \(\frac{m}{2}\) = -1
If m = 1, then the value of \(\frac{m}{2}\) = \(\frac{1}{2}\) ≠ -1
If m = 2, then the value of \(\frac{m}{2}\) = \(\frac{2}{2}\) = 1 ≠ -1
If m = 3, then the value of \(\frac{m}{2}\) = \(\frac{3}{2}\) ≠ -1
Here, we are not getting negative values.
If we take (substitute) m as a negative number we will get negative value.
If m = -1, then the value of \(\frac{m}{2}\) = \(\frac{-1}{2}\) ≠ -1
If m = -2, then the value of \(\frac{m}{2}\) = \(\frac{-2}{2}\) = -1 = -1
From the above when m = -2, the both LHS and RHS are equal.
∴ Solution of the equation \(\frac{m}{2}\) = -1 is m = -2.

iv) 2k – 1 = 3
Given equation is 2k – 1 = 3
If k = 1, then the value of 2k – 1 = 2(1) – 1 = 2 – 1 = 1 ≠ 3
If k = 2, then the value of 2k – 1 = 2(2) – 1 = 4 – 1 = 3 = 3
From the above when k – 2, the both LHS and RHS are equal.
Solution of the equation 2k – 1 = 3 is k = 2.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra Ex 7.3

Question 1.
Identify which of the following are equations.
i) x – 3 = 7
ii) l + 5 > 9
iii) p – 4 < 10
iv) 5 + m = -6
v) 2s – 2 = 12
vi) 3x + 5
Answer:
i) x – 3 = 7
We know that, a mathematical statement involving equality symbol is called an equation.
x – 3 = 7 has equality symbol. So, it is an equation.

ii) l + 5 > 9
We know that, a mathematical statement involving equality symbol is called an equation.
l + 5 > 9 has no equality symbol.
So, it is not an equation. [It is an inequation]

iii) p – 4 < 10
We know that, a mathematical statement involving equality symbol is called an equation.
p – 4 < 10 has no equality symbol.
So, it is not an equation. [It is an inequation]

iv) 5 + m = – 6
We know that, a mathematical statement involving equality symbol is called an equation.
5 + m = – 6 has equality symbol.
So, it is an equation.

v) 2s – 2 = 12
We know that, a mathematical statement involving equality symbol is called an equation.
2s – 2 = 12 has equality symbol.
So, it is an equation.

vi) 3x + 5
It is only an expression. It’s not an equation.

Question 2.
Write LHS and RHS of the following equations.
i) x – 5 = 6
ii) 4y = 12
iii) 2z + 3 = 7
Answer:
i) x – 5 = 6
Given equation is x – 5 = 6
LHS = x – 5
RHS = 6

ii) 4y = 12
Given equation is 4y = 12
LHS = 4y
RHS = 12

iii) 2z + 3 = 7
Given equation is 2z + 3 = 7
LHS = 2z + 3
RHS = 7

Question 3.
Solve the following equation by Trial & Error Method.
i) x + 3 = 5
ii) y – 2 = 7
iii) a + 4 = 9
Answer:
i) x + 3 = 5
Given equation is x + 3 = 5
If x = 1, then the value of x + 3 = 1 + 3 = 4 ≠ 5
If x = 2, then the value of x + 3 = 2 + 3 = 5 = 5
From the above when x = 2, then both LHS and RHS are equal.
∴ Solution of the equation x + 3 = 5 is x = 2

ii) y – 2 = 7
Given equation is y – 2 = 7
If y = 1, then the value of y – 2 = 1 – 2 = -1 ≠ 7
If y = 2, then the value of y – 2 = 2 – 2 = 0 ≠ 7
If y = 3, then the value of y – 2 = 3 – 2 = 1 ≠ 7
If y = 4, then the value of y – 2 = 4 – 2 = 2 ≠ 7
If y = 5, then the value of y – 2 = 5 – 2 = 3 ≠ 7
If y = 6, then the value of y – 2 = 6 – 2 = 4 ≠ 7
If y = 7, then the value of y – 2 = 7 – 2 = 5 ≠ 7
If y = 8, then the value of y – 2 = 8 – 2 = 6 ≠ 7
If y = 9, then the value of y – 2 = 9 – 2 = 7 = 7
From the above when y = 9, the both LHS and RHS are equal.
Solution of the equation y – 2 = 7 is y = 9

ii) a + 4 = 9
Given equation is a + 4 = 9
If a = 1, then the value of a + 4 = 1 + 4 = 5 ≠ 9
If a = 2, then the value of a + 4 = 2 + 4 = 6 ≠ 9
If a = 3, then the value of a + 4 = 3 + 4 = 7 ≠ 9
If a = 4, then the value of a + 4 = 4 + 4 = 8 ≠ 9
If a = 5, then the value of a + 4 = 5 + 4 = 9 = 9
From the above when a = 5, the both LHS and RHS are equal.
∴ Solution of the equation a + 4 = 9 is a = 5