Class 6

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.1

Question 1.
Construct a line segment of length 6.9 cms using ruler and compass.

Solution:
Steps of construction:

1. Draw a line l. Mark a point A on the line l.
2. Place the metal pointer of the compass on the zero mark of the ruler. Open the compass, so that pencil point touches the 6.9 cm mark on the ruler.
3. Place the pointer at A on the line l and draw an arc to cut the line. Mark the point where the arc cuts the line as B
4. On the line l, we got the line segment AB of required length.
   

Question 2.
Construct a line segment of length 4.3 cms using ruler.
Solution:

Steps of construction:

1. Place the ruler on paper and hold it firmly.
2. Mark a point with a sharp edged pencil against 0 cm mark on the ruler. Name the point as P.
3. Mark another point against 3 small divisions just after the 4 cm mark. Name this point as Q.
4. Join points P and Q along the edge of the ruler.
   Therefore, PQ is the required line segment of length 4.3 cm.

Question 3.
Construct a circle with centre M and radius 4 cm.
Solution:

Steps of construction:

1. Mark a point M on the paper.
2. By using compasses take 4 cm as the radius on with the scale.
3. Place the metal point on M and draw a circle from M.
   Hence required circle is constructed with center M and radius 4 cm.

Question 4.
Draw any circle and mark three points A, B and C such that
(i) A is on the circle
(ii) B is in the interior of the circle
(iii) C is in
Solution:

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

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Unit Exercise

Question 1.
Write the integers for the following situations.
i) A kite is flying at a height of 225 m in the sky. ( )
ii) A whale is at a depth of 1250 m in the ocean. ( )
iii) The temperature in Sahara desert is 12°C below freezing temperature. ( )
iv) Ravi withdrawn Rs. 3800 from ATM using his debit card. ( )
Answer:
i) 225 m
ii) – 1250 m
iii) – 12°C
iv) -3800

Question 2.
Justify the following statements with an example.
i) A positive integer is always greater than a negative integer.
ii) All positive integers are natural numbers.
iii) Zero is greater than a negative integer.
iv) There exist infinite integers in the number system.
v) All whole numbers are integers.
Answer:
i) A positive integer is always greater than a negative integer:
Consider
a) 3, -5 are two integers
3 > -5
b) – 10, 1 are two integers
1 > -10
∴ A positive integer is always greater than a negative integer.

ii) All positive integers are natural numbers:
Positive integers = 1, 2, 3, …..
Natural numbers = 1, 2, 3, …..
So, all positive integers are natural numbers.

iii) Zero is greater than a negative integer:
On a number line, for a given pair of numbers the number on R.H.S is always greater than the number on L.H.S.
All negative integers lie on the L.H.S. of zero, on a number line.
As such all negative numbers are less than zero or zero is greater than all negative integers.

iv) There exist infinite integers in the number system:
If we write integer on a number line, as the line extends on both sides endlessly so as the integers.
(Or)
For every integer there exists another integer which is 1 more than the given integer. Hence the integers are infinite.

v) All whole numbers are integers:
Whole numbers : 0, 1, 2, 3,……
Integers: ……, -3, -2, -1, 0, 1, 2, 3,…..
So, whole number are part of integers.
∴ All whole numbers are integers.

Question 3.
Represent
i) 3+4
ii) 8+(-3)
iii) -7-2
iv) 6-(5)
v)-5-(-4)
on number line.
Answer:
i) 3 + 4 = +7.

On the number line we first move 3 steps to the right of 0 to reach +3.
Then, we move 4 steps to the right of +3 and reach + 7.

ii) 8 + (-3) = +5.

On the number line we first move 8 steps to the right of 0 to reach +8.
Then, we move 3 steps to the left of +8 and reach +5.

iii) -7 -2 = -9

On the number line we first move 7 steps to the left of 0 to reach -7.
Then we move 2 steps to the left of -7 and reach -9.

iv) 6 – (5) = + 1

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

v) -5 – (-4) = -1

-5 – (-4) = -5 + 4 = -1 (∵ -(-a) = a)
On the number line, we first move 5 steps to the left of 0 to reach -5.
Then, we move 4 steps to the right of -5 and reach -1.

Question 4.
Write all the integers lying between the given numbers.
i) 7 and 12
ii) -5 and -1
iii) -3 and 3
iv) -6 and 0
Answer:
i) 7 and 12
Integers lying between 7 and 12 are 8, 9, 10, ll.

ii) -5 and -1
Integers lying between -5 and -1 are -2, -3 and -4.

iii) -3 and 3
Integers lying between -3 and 3 are -2, -1, 0, 1, 2.

iv) -6 and 0
Integers lying between -6 and 0 are -5, -4, -3, -2, -1.

Question 5.
Arrange the following integers in ascending order and descending order.
-1000, 10, -1, -100, 0, 1000, 1, -10
Answer:
Given numbers -1000, 10, -1, -100, 0, 1000, 1, -10
Ascending order: – 1000, -100, -10, -1, 0, 1, 10, 1000
Descending order: 1000, 10, 1, 0, -1, -10, -100, -1000

Question 6.
Write a real life situation for each of the following integers.
i) -200 m
ii) +42°C
iii) -4800(cr)
iv) -3.0 kg
Answer:
i) -200 m
In the Singareni coal mines workers will go 200 m below the ground level.
ii) +42°C
In summer average temperature of May month is 42°C.
iii) Rs. 4800 (crores)
Central Government sanctioned Rs. 4800(crores) for education in the Annual Budget.
iv) – 3.0 kg
In a Primary School the ground balance of rice = -3 kg

Question 7.
Find:
i) (-603) + (603)
ii) (-5281) +(1825)
iii) (-32) + (-2) + (-20) + (-6)
Answer:
i) (-603) + (603)
Sum of a number and its additive number is 0.
Additive inverse of – 603 is 603.
So, -603 + 603 = 0

ii) (-5281) + (1825)
= – 5281 + 1825 = – 3456

iii) (-32) + (-2) + (-20) + (-6)
= -32 – 2 – 20 – 6 = -60

Question 8.
Find:
i)(-2) – (+1)
ii) (-270) – (-270)
iii) (1000) – (-1000)
Answer:
i) -2 – (+1)
= -2 -1 = -3

ii) – 270 – (-270)
= -270 + 270 [∵ -(-a) = a]
= 0 [-a + a = 0]

iii) 1000 – (-1000)
= 1000 + 1000 [∵ -(-a) = a]
= 2000

Question 9.
In a quiz competition, where negative score for wrong answer is taken,Team A scored +10, -10, 0, -10, 10, -10 and Team B scored 10,10, -10,0,0,10 in 6 rounds successively .Which team wins the competition? How?
Answer:
Score of Team A = (+10), (-10), 0, (-10), 10, (-10)
Total score of Team A = (+10) + (-10) + (0) + (-10) + (10) + (-10)
= +10 – 10 + 0 – 10 + 10 – 10
= +20 – 30 = -10
Score of Team B = 10, 10, -10, 0, 0, 10
Total Score of Team B = (+10) + (+10) + (-10) + 0 + 0 + (+10)
= +10 + 10 – 10 + 10
= + 30 – 10
= + 20
-10 < + 20
Team A < Team B
So, Team B is the winner. Because, Team B got more score than Team A.

Question 10.
An apartment has 10 floors and two cellars for car parking under the basement. A lift is now, at the ground floor. Ravi goes 5 floors up and then 3 floors up, 2 floors down and then 6 floors down and come to lower cellar for taking his car. Count how many floors does Ravi travel all together? Represent the result on a vertical number line.
Answer:

First move:
Ground to 5th floor up = 5
(Can reach 5th floor)
Second move:
5th floor to 3 floors up = 3
(Can reach 8)
Third move:
8th floor to 2 floors down = 2
(Can reach 6th floor)
Fourth move:
6th floor to 6 floors down = 6
(Can reach ground floor)
Final move:
Ground to lower cellar = 2
(Can reach cellar -2)
No. of floors Ravi travelled = 18

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic Ex 6.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Ex 6.4

Question 1.
Express each of the following percents as fractions in the simplest form,
a) 15%
b) 35%
c) 50%
d) 75%
Answer:
a) Given percent is 15%
= \(\frac{15}{100}\) = \(\frac{3}{20}\) (fraction form) [∵ To express a percentage as a fraction first drop the symbol % and then divide it by 100.]

b) Given percent is 35%
= \(\frac{35}{100}\) = \(\frac{7}{20}\) (fraction form)

c) Given percent is 50%
= \(\frac{50}{100}\) = \(\frac{1}{2}\) (fraction form)

d) Given percent is 75%
= \(\frac{75}{100}\) = \(\frac{3}{4}\) (fraction form)

Question 2.
Express each of the following fractions as percents.
a) \(\frac{15}{2}\)
b) 8\(\frac{1}{4}\)
c) 5\(\frac{3}{4}\)
d) 3\(\frac{1}{3}\)
Answer:
a) Given number \(\frac{15}{2}\) is in fraction form.
To convert the fraction into percent we have to multiply the fraction by 100%
\(\frac{15}{2}\) = \(\frac{15}{2}\) × 100 = 15 × 50 = 750% (percent form)

b) Given number 8\(\frac{1}{4}\) is in fraction form.
To convert the fraction into percent we have to multiply the fraction by 100%
8\(\frac{1}{4}\) = \(\frac{33}{4}\) = \(\frac{33}{4}\) × 100 = 33 × 25 = 825%,(percent form).

c) Given number 5\(\frac{3}{4}\) is in fraction form.
To convert the fraction into percent we have to multiply the fraction by 100 and assign the percentage symbol %
5\(\frac{3}{4}\) = \(\frac{23}{4}\) × 100 = 23 × 25 = 575% (percent form)

d) Given number is 3\(\frac{1}{3}\) in fraction form.
To convert the fraction into percent we have to multiply the fraction by 100%
3\(\frac{1}{3}\) = \(\frac{10}{3}\) × 100 = \(\frac{1000}{3}\) = 333\(\frac{1}{3}\) % (percent form)

Question 3.
Express each of the following ratios as percents.
(a) 3 : 5 (b) 5 : 8 (c) 2.5 : 55 (d) 4 : 36
Answer:
a) 3 : 5
Given ratio is 3 : 5.
3 : 5 = \(\frac{3}{5}\) (fraction form)
To convert the fraction into percent we have to multiply the fraction by 100%
\(\frac{3}{5}\) = \(\frac{3}{5}\) × 100 = 60% (percent form)

b) 5 : 8
Given ratio is 5 : 8
5 : 8 = \(\frac{5}{8}\) (fraction form)
To convert the fraction into percent we have to multiply the fraction by 100%
\(\frac{3}{5}\) = \(\frac{3}{5}\) × 100 = 62 \(\frac{1}{2}\) % (percent form)

c) 2.5 : 55
Given ratio is 2.5 : 55
2.5 : 55 = \(\frac{2.5}{55}\)
To convert the fraction into percent we have to multiply the fraction by 100%

d) 4 : 3 6
Given ratio is 4 : 36
4 : 36 = \(\frac{4}{36}\) (fraction form)
To convert the fraction into percent we have to multiply the fraction by 100%
\(\frac{4}{36}\) = \(\frac{4}{36}\) × 100 = \(\frac{100}{9}\) = 11 \(\frac{1}{9}\) % (percent form)

Question 4.
Express each of the following percents as ratios in the simplest form.
(a) 12% (b) 25% (c) 45% (d) 84%
Answer:
a) Given percent is 12%
= \(\frac{12}{100}\) = \(\frac{3}{25}\) (fraction form)
= 3 : 25 (ratio form)

b) Given percent is 25%
= \(\frac{25}{100}\) = \(\frac{1}{4}\) (fraction form)
= 1 : 4 (ratio form)

c) Given percent is 45%
= \(\frac{45}{100}\) = \(\frac{9}{20}\) (fraction form)
= 9 : 20 (ratio form)

d) Given percent is 84%
= \(\frac{84}{100}\) = \(\frac{21}{25}\) (fraction form)
= 21 : 25 (ratio form)

Question 5.
Express each of the following percents as decimals,
(a) 1% (b) 6% (c) 19% (d) 67%
Answer:
a) Given percent is 1%
= 1% = \(\frac{1}{100}\) (fraction form)
= 0.01 (decimal form)

b) Given percent is 6%
= 6% = \(\frac{6}{100}\) = \(\frac{3}{50}\) (fraction form)
= 0.06 (decimal form)

c) Given percent is 19%
= 19% = \(\frac{19}{100}\) (fraction form)
= 0.19 (decimal form)

d) Given percent is 67%
= 67% = \(\frac{67}{100}\)
= 0.67 (decimal form)

Question 6.
Express each of the following decimals as percent.
(a) 0.04 (b) 0.52 (c) 0.125 (d) 0.0006
Answer:
a) Given decimal number is 0.04.
Convert this into fraction

b) Given decimal number is 0.52.
Convert this into fraction

c) Given decimal number is 0.125.
Convert this into fraction

d) Given decimal number is 0.0006.
Convert this into fraction

Question 7.
Find the number, which is 12\(\frac{1}{2}\)% of 75.
Answer:
Given 12\(\frac{1}{2}\)% of 75
We know that x% of y = \(\frac{x}{100}\) × y

(OR)

Question 8.
Pavani secured 85% marks in mathematics paper. If maximum marks in the paper are 80, find the marks secured by her in that paper.
Answer:
Given maximum marks of the paper = 80
Percentage of marks secured by Pavani = 85%
Marks secured by Pavani

∴ Marks secured by Pavani = 68

Question 9.
Siva spends 78% of his monthly income. If he saves Rs. 7,700/- per month, what is his monthly income?
Answer:
Let Siva’s monthly income = Rs. x
Percentage of income spent = 78%
Percentage of income saved = 100% – percentage of spent income
= 100% – 78% = 22%
Saved income = 22% of monthly income = 7700
= 22% of x = 7700
⇒ \(\frac{1}{2}\) × x = 7700
⇒ x = \(\frac{7700×100}{2}\)
⇒ x = Rs. 35000
∴ Siva’s monthly income = Rs. 35000

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)

S.No.	Natural number	Predecessor	Successor
1.	135
2.	237
3.	999

Solution:

S.No.	Natural number	Predecessor	Successor
1.	135	134	136
2.	237	236	238
3.	999	998	1000

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 4 Integers Ex 4.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.4

Question 1.
Find:
i) 40 – (22)
ii) 84 – (98)
iii) (-16) + (-17)
iv) (-20) – (13)
v) (38) – (-6)
vi) (-17) – (-36)
Answer:
i) 40 – (22) = +40 – 22
= +(18 + 22)-22
= 18 +(22-22)
= 18 + 0
∴ 40-(22) = + 18

ii) 84 – (98) = 84 – 98
= + 84 – 84 – 14
= (84 – 84) – 14
= 0 – 14
∴ 84 – (98) = – 14

iii) (-16) + (-17) = -16 – 17 = – 33
∴ (-16) + (-17) = -33

iv) (-20) – (13) = – 20 – 13
= -33
∴ (-20) – (13) = – 33

v) 38 – (-6) = 38 + 6
We know – (-a) = a = + 44
∴ 38 – (-6) = + 44

vi) (-17) – (-36) = -17 + 36
We know – (- a) = a
= -17 + 17 + 19
= (-17+17) + 19
= + 19
∴ (-17) – (-36) = + 19

Question 2.
Fill in the boxes with >, < or = sign:
(i) (-4) + (-5) (-5) – (-4)
(ii) (-16) – (-23) (-6) + (-12)
(iii) 44 – (-10) 47 +(-3)
Answer:
(i) (-4) + (-5) (-5)- (-4)
-4-5  -5 + 4
-9 -1

(ii) (-16) – (-23) (-6) + (-12)
-16 + 23 -6-12  [∵ -(-a) = a]
-16 + 16 + 7 -18
+7 -18

(iii) 44 – (-10) 47 + (-3)
44+10 47-3.
54 44 + 3 – 3
54 44

Question 3.
Fill in the blanks:
i) (-13) + ———– = 0
ii) (-16) + 16 = ———–
iii) (-5) + ———– = -14
iv) ———– + (2 – 16) = – 22
Answer:
i) (-13) + ———– = 0
We know, additive inverse of a is -a (or) – a is a.
Additive inverse of -13 is 13. So, -13 + 13 = 0

ii) (-16) + 16 = ———–
We know, sum of a number and its additive inverse is 0.
i.e., (-a) + a = 0
So, (-16) + (16) = 0

iii) (-5) + ———– = -14
-5 + (-9) = -14

iv) ———– + (2 – 16) = – 22
———– + (-14) = -22
– 8 + (-14) = – 22

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Ex 4.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.3

Question 1.
Add the following integers using number line.
i) 7 + (-6)
ii) (-8) + (-2)
iii) (-6) + (-5) + (+2)
Answer:
i) 7 + (-6)

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

ii) (-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 left of -8 and reach -10.
So,(-8) + (-2) = -10

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

On the number line, we first move 6 steps to the left of 0 to reach -6.
We move 5 steps to the left of -6 and reach -11.
Then, we move 2 steps to the right of -11 and reach -9.
So, (-6) + (-5) + (+2) = -9

Question 2.
Add without using number line.
(i) 10 + (-3)
(ii) 10 + (+16)
(iii) (-8) + (+8)
Answer:
i) 10 + (-3)
10 + (-3) = 7 + 3 + (-3) = 7 + (3 + (-3))
= 7 + 3 – 3 = 7 + 0
∴ 10 + (-3) = +7

ii) -10 + (+16)
-10 + (+16)
= -10 + 10 + 6
= (-10 + 10) + 6
= 0 + 6
∴ -10 + (+16) = + 6

iii) (-8) + (+ 8)
-8 + (+8) = -8 + 8 = 0
∴ -8 + (+8) = 0

Question 3.
Find the sum of i) 120 and -274 ii) -68 and 28
Answer:
i) 120 and-274
Method -1: Sum = + 120 + (-274)
= + 120 + (- 120 – 154)
= + 120 – 120 – 154
∴ 120 + (-274) = -154

Method – II:
As the given numbers have opposite sign, we subtract one from other

As 274 is having negative (-) sign, the answer is – 154.

ii) -68 and 28
Sum = – 68 + (28)
= – 40 – 28 + 28
∴ – 68 + 28 = – 40

Question 4.
Simplify:
i) (-6) + (-10) + 5 + 17
ii) 30 + (-30) + (-60) + (-18)
Answer:
i) (-6) + (-10) + 5 + 17
– 6 – 10 + 5 + 17 = -16 + 22
= -16 + 16 + 6
= + 6
∴ (-6) + (-10) + 5 + 17 = + 6

ii) 30 + (-30) + (-60) + (-18)
= 30 + (-30 -60 -18)
= 30 + (-108)
= 30 – 108 = – 78
∴ 30 + (-30) + (-60) + (-18) = -78

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic Ex 6.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Ex 6.3

Question 1.
If the cost of 3 apples is Rs. 60/-, then find the cost of 7 apples.
Answer:
Cost of 3 apples = Rs. 60
Cost of 1 apple = \(\frac{60}{3}\) = Rs. 20
∴ Cost of 7 apples = 7 × Rs. 20 = Rs. 140

Question 2.
Uma bought 8 books for a total of ? 120. How much would she pay for just 5 books?
Answer:
Cost of 8 books = Rs. 120
Cost of 1 book = \(\frac{120}{8}\) = Rs. 15
∴ Cost of 5 books = 5 × 15 = Rs. 75

Question 3.
The cost of 5 fans is Rs. 11,000/-. Find the number of funs that can be purchased for Rs. 4,400/-.
Answer:
Cost of 5 fans = Rs. 11000
Cost of 1 fan = \(\frac{11000}{5}\) = Rs. 2200
Number of fans can be purchased for Rs. 4400 = 4400 ÷ cost of 1 fan = \(\frac{4400}{2200}\) = 2

Question 4.
A car is moving at a constant speed covers a distance of 180 km in 3 hours. Find the time taken by the car to cover a distance of 420 km at the same speed.
Answer:
Distance covered = 180 km
Time taken = 3 Hrs.
Distance to be covered = 420 km
Time required = ? = x (say)
Since the car is moving at a constant speed, we have
180 : 3 is as to 420 : x
180 : 3 :: 420 : x
Its a proportion.
So, product of extremes = Product of means
180 × x = 3 × 420

∴ Time required = 7 Hrs.
(OR)
Distance covered by a car in 3 hours = 180 km
Distance covered by a car in 1 hour = 180 ÷ 3
Time taken to cover 60 km distance = \(\frac{180}{3}\) = 60 km
Time taken to cover 420 km distance = 420 ÷ 60 = \(\frac{420}{60}\) = 7 hours

Question 5.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Answer:
108 : 594 :: x : 1650
So, 108 × 1650 = x × 594
∴ x = \(\frac{108 \times 1650}{\hline 594}\) = 300 litres
(OR)
Distance covered by truck with 108 litres diesel = 594 km
Distance covered by truck with 1 litre of diesel = \(\frac{594}{108}\) km
Diesel required to cover 1650 km distance = 1650 ÷ \(\frac{594}{108}\) = 1650 × \(\frac{108}{594}\)
Diesel required to cover 1650 km distance = 300 litres.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.2

Question 1.
Put appropriate symbol > or < in the boxes given.
i) -1 0
ii) -3 -7
iii) -10  +10
Answer:
i) -1 < 0
ii) -3 > -7
iii) -10 < +10

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

S.No.	Numbers	Increasing order	Decreasing order
i)	-7, 5, -3	-7 < 5 < -3	5 > -3 > -7
ii)	-1, 3, 0	-1 < 0 < 3	3 > 0 > -1
iii)	1, 3, -6	-6 < 1 < 3	3 > 1 > -6
iv)	-5, -3, -1	-5 < -3 < -1	-1 > -3 > -5

Question 3.
Write True or False.
i) Zero is on 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) (-100) > (+100) ( )
Answer:
i) True
ii) False
iii) True
iv) False

Question 4.
Find all integers which lie between the given two integers. Represent them on number line:
i) -1 and 1
ii) -5 and 0
iii) -6 and -8
iv) 0 and -3
Answer:
i) Given integers -1 and +1

Integers between -1 and 1 is 0.

ii) Given integers -5 and 0

Integers between -5 and 0 are -4, -3, -2, -1.

iii) Given, integers -6 and -8

Integers between -6 and -8 is -7.

iv) Given integers 0 and -3

Integers between 0 and -3 are -1, -2.

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? Why?
Answer:
Given temperatures of Shimla and Kufri are -4°C and -6°C.
We know that -6°C < – 4°C
Therefore, temperature in Kufri is colder than Shimla.

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:

S.No.	Number	Nearest ten	Nearest hundred | Nearest thousand
1.	56,789	56,790	56,800	57,000
2.	86,289	86,290	86,300	86,000
3.	4,56,726	4,56,730	4,56,700         ’	4,57,000
4.	5,62,724	5,62,720	5,62,700	5,63,000

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:

State	Population in 2011	Round off the nearest lakhs
Andhra Pradesh	4,92,94,020	4,93,00,000
Telangana	3,52,86,757	3,53,00,000
India	1,21,08,54,977	1,21,09,00,000

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 6 Basic Arithmetic Ex 6.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Ex 6.1

Question 1.
Express the following in the terms of ratios.
i) The length of a rectangle is 5 times to its breadth.
ii) For preparing coffee, 2 cups of water require to 1 cup of milk.
Answer:
i) Let the breadth of rectangle = x = 1 part
length of rectangle = 5x = 5 parts
Ratio = length : breadth = 5x : x = \(\frac{5x}{x}\) = \(\frac{5}{1}\) = 5 : 1

ii) To prepare coffee,
Required cups of water = 2
Required cups of milk = 1
Ratio = water : milk = 2 : 1

Question 2.
Express the following in the simplest form.
i) 24 : 9
ii) 144 : 12
iii) 961 : 31
iv) 1575 : 1190
Answer:
i) 24 : 9
Given ratio is 24 : 9
To write the given ratio in the simplest form.
First find the HCF of two terms.

HCF of 24 and 9 is 3.
Then, divide each term by their HCF.
Simplest form of the ratio = 24 ÷ 3 : 9 ÷ 3
Required ratio = 8 : 3

ii) 144 : 12
Given ratio is 144 : 12
To write the given ratio in the simplest form.
First find the HCF of two terms.

HCF of 144 and 12 is 12.
Then, divide each term by their HCF.
Simplest form of the ratio = 144 ÷ 12 : 12 ÷ 12
Required ratio = 12 : 1

iii) 961 : 31
Given ratio is 961 : 31
To write the given ratio in the simplest form.
First find the HCF of two terms.

HCF of 961 and 31 is 31.
Then, divide each term by their HCF.
Simplest form of the ratio = 961 ÷ 31 : 31 ÷ 31
Required ratio = 31 : 1

iv) 1575 : 1190
Given ratio is 1575 : 1190
To write the given ratio in the simplest form.
First find the HCF of two terms.

HCF of 1575 and 1190 is 35.
Then, divide each term by their HCF.
Simplest form of the ratio = 1575 ÷ 35 : 1190 ÷ 35
∴ Required ratio = 45 : 34

Question 3.
Write the antecedents and consequents of the following ratios.
(i) 36 : 73
(ii) 65 : 84
(iii) 58 : 97
(iv) 69 : 137
Answer:

S.No.	Ratio	Antecedents	Consequents
(i)	36: 73	36	73
(ii)	65 : 84	65	84
(iii)	58:97	58	97
(iv)	69 : 137	69	137

Question 4.
Find the ratios of the following in their simplest form.
i) 25 minutes to 55 minutes
ii) 45 seconds to 30 minutes
iii) 4 m 20 cm to 8 m 40 cm
iv) 5 litres to 0.75 litres
v) 4 weeks to 4 days
vi) 5 dozen to 2 scores (1 score = 20 items)
Answer:
i) 25 minutes to 55 minutes
Given 25 minutes to 55 minutes = 25 : 55

HCF of 25 and 55 is 5 = 25 ÷ 5 : 55 ÷ 5 (Divide each term by 5)
∴ Required ratio = 5:11

ii) 45 seconds to 30 minutes
Given 45 seconds to 30 minutes.
To write the given ratio in the simplest form.
First convert the two terms into same units.
1 minute = 60 seconds
30 minutes = 30 × 60 = 1800 seconds
Then, find the HCF of two terms and divide them by their HCF.

HCF of 45 and 1800 is 45.
= 45 ÷ 45 : 1800 ÷ 45 (divide each term by 45)
∴ Required ratio = 1 : 40

iii) 4 m 20 cm to 8 m 40 cm
Given 4 m 20 cm to 8 m 40 cm
To write the given ratio in the simplest form.
First we convert them into cm.
1 m = 100 cm
4 m = 400 cm
8 m = 800 cm
4 m 20 cm = 400 + 20 = 420 cm
8 m 40 cm = 800 + 40 = 840 cm
Then, find the HCF of 420 and 840 and divide them by HCF.

HCF of 420 and 840 is 420.
= 420 ÷ 420 : 840 ÷ 420 (divide each term by 420)
Required ratio = 1 : 2.

iv) 5 litres to 0.75 litres
Given 5 litres to 0.75 litres
To write the given ratio in the simplest form, first we convert litres into millilitres.
1 litre = 1000 ml
5 litres = 5000 ml
0.75 litres = 750 ml
Then, find the HCF of 5000; 750 and divide them by HCF.

HCF of 5000 and 750 is 250,
= 5000 ÷ 250 : 750 ÷ 250 (divide each term by 250)
Required ratio = 20 : 3

v) 4 weeks to 4 days
Given 4 weeks to 4 days.
To write the given ratio in the simplest form, first we convert them into same units (days).
1 week = 7 days
4 weeks = 4 × 7 = 28 days
Then, find the HCF of 28 and 4, then divide them by HCF.

HCF of 28 and 4 is 4 = 28 ÷ 4 : 4 ÷ 4 (divide each term by 4)
Required ratio = 7 : 1

vi) 5 dozen to 2 scores (1 score = 20 items)
Given 5 dozen to 2 scores.
To write the given ratio in the simplest form, first we convert them into same units.
1 dozen = 12 items
5 dozens = 5 × 12 = 60 items
1 score = 20 items
2 scores = 2 × 20 = 40 items
Then, find the HCF of 60 and 40, divide them by HCF.

HCF of 60 and 40 is 20 = 60 ÷ 20 : 40 ÷ 20 (divide each term by 20)
Required ratio = 3 : 2

Question 5.
Rahim works in a software company and earns Rs. 75,000/- per month. He saves Rs. 28,000/- per month from his earnings. Find the ratio of
i) His savings to his income
ii) His income to his expenditure
iii) His savings to his expenditure
Answer:
Given Rahim’s monthly income = Rs. 75000
monthly savings = Rs. 28000
Monthly expenditure = income – savings
= 75000 – 28000 = 47,000/-
i) Ratio of savings to income = 28000 : 75000
To convert the ratio into the simplest form divide each term by their HCF is 1000
= 28000 ÷ 1000 : 75000 ÷ 1000 (divide by 1000)
= 28 : 75
Required ratio of savings to income = 28 : 75

ii) Ratio of income to expenditure = 75000 : 47000
To convert the ratio into the simplest form divide each term by their HCF is 1000 = 75000 ÷ 1000 : 47000 ÷ 1000 (divide by 1000)
= 75 : 47
Required ratio of income to expenditure = 75 : 47

iii) Ratio of savings to expenditure = 28000 : 47000
To convert the ratio into the simplest form divide each term by their HCF is 1000 = 28000 ÷ 1000 : 47000 ÷ 1000 (divide by 1000)
= 28 : 47
Required ratio of savings to expenditure = 28 : 47

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Ex 4.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.1

Question 1.
Write True or False against each of the following statements.
i) -7 is on the right side of -6 on the number line.
ii) Zero is a positive number.
iii) 29 is on the right side of zero on the number line.
iv) -1 lies between the integers -2 and 1.
v) There are nine integers between -5 and +5.
Answer:
i) False
ii) False
iii) True
iv) True
v) True

Question 2.
Observe the following number line and answer the following questions.

i) Which is the nearest positive integer to -1?
ii) How many negative numbers you will find on the left side of Zero?
iii) How many integers are there in between -3 and 7?
iv) Write 3 integers lesser than -2.
v) Write 3 integers more than -2.
Answer:
i) +1 is the nearest positive integer to -1.
ii) On the given number line negative numbers to left of zero are -1, -2, -3, -4, -5.
So, there are 5 in number.
iii) On the number line integers between -3 and 7 are -2, -1, 0, 1, 2, 3, 4, 5, 6.
So, there are 9 in number.
iv) Integers less than -2 means numbers left side of -2.
They are -3, -4, -5, -6, -7, ………
So, 3 integers lesser than -2 are -3, -4, -5.
v) Integers more than -2 means numbers right side of -2.
They are -1, 0, 1, 2, 3, 4, ……..
So, 3 integers more than -2 are -1, 0, 1.

Question 3.
Represent the integers on a number line as given below.
i) Integers lies between -7 and -2.
ii) Integers lies Between -2 and 5.
Answer:

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

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Unit Exercise

Question 1.
Classify the given numbers according to their divisibility.

Answer:

(OR)

Question 2.
Write the divisibility rule by 11 with one example.
Answer:
Divisibility by 11:
A given number is divisible by 11, if the difference between the sum of the digits at odd places and the sum of the digits at even places (from the right) is either ‘O’ or a multiple of 11.
Ex: 123321
Sum of the digits at odd places = 1 + 3 + 2 = 6.
Sum of the digits at even places = 2 + 3 + 1 = 6
Their difference = 6 – 6 = 0
So, 123321 is divisible by 11.

Question 3.
Fill the table with correct answer.

Answer:

Question 4.
Find HCF of 70, 105 and 175 by prime factorization method.
Answer:
Given numbers are 70, 105 and 175.
The HCF of 70, 105 and 175 can be formed by prime factorization as follows:

Thus, 70 = 2 × 5 × 7
105 = 3 × 5 × 7
175 = 5 × 5 × 7
Common factors of 70, 105 and 175 are: 5, 7
Their product: 5 × 7 = 35
Hence, HCF of 70, 105 and 175 is 35.

Question 5.
Find HCF of 18, 54, 81 by continued division method.
Answer:
Given numbers are 18, 54 and 81 To find the HCF of 18, 54 and 81
First find the HCF of 18 and 54

HCF of 18 and 54 is 18.
Then find the HCF of the third number and the HCF of first two numbers.
i.e., let us find the HCF of 81 and 18

Last divisor is 9 when remainder is zero.
∴ HCF of 18, 54 and 81 is 9.

Question 6.
Find LCM of 4, 12, 24 by two methods.
Answer:
Given numbers are 4, 12 and 24
a) LCM by prime factors method:
Factors of 4 = 2 × 2
Factors of 12 = 2 × 2 × 3
Factors of 24 = 2 × 2 × 2 × 3
LCM of 4, 12 and 24= 2 × 2 × 2 × 3 = 24

b) Division method:

Thus, the LCM of 4, 12 and 24 is 2 × 2 × 2 × 3 = 24

Question 7.
What is the capacity of the largest vessel which can empty the oil from three vessels containing 32 liters, 24 liters and 48 liters respectively, an exact number of times?
Answer:
Given capacity of three vessels are 32 liters, 24 liters and 48 liters.
To find the capacity of the largest vessel.
We have to find the HCF of 32, 24 and 48.
First find the HCF of 32 and 24.
HCF of 32 and 24 is 8.

Then, find the HCF of third number and the HCF of first two numbers.
i.e., let us find the HCF of 48 and 8.

HCF of 32, 24 and 48 is 8.
Hence, the capacity of the largest vessel which can empty the oil, then the three vessels containing 32 l, 24 l and 48 l of exact number of times is 8 liters.

Question 8.
HCF, LCM of two numbers are 9 and 54 respectively. If one of those two numbers is 18, find the other number.
Answer:
Given, HCF of two numbers = 9
LCM of two numbers = 54
One of the two numbers a = 18
Then, the other number b = ?
We know, that the product of two numbers = LCM × HCF
a × b = LCM × HCF
18 × b = 54 × 9
b = \(\frac{54 × 9}{18}\) = 27
∴ The other number = 27