Category: Class 8

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Ex 3.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals Exercise 3.1

Construct the quadrilaterals with the measurements given below:

Question (a).
Quadrilateral ABCD with AB = 5.5 cm, BC = 3.5 cm, CD = 4 cm, AD = 5 cm and ∠A = 45°.
Solution:
In Quadrilateral ABCD with AB = 5.5 cm, BC = 3.5 cm, CD = 4 cm, AD = 5 cm and ∠A = 45°.

Construction Steps:

1. Construct a line segment \(\overline{\mathrm{AB}}\) with radius 5.5 cm
2. With the centre A draw a ray and an arc which are equL1 to 45° and 5 cm.
3. These intersecting point is keep as ‘D’.
4. With centres D, B draw two arcs equal to radius 4 cm, 3.5 cm respectively.
5. The intersecting point of these two arcs is keep as ‘C’.
6. Join DC and BC. A F
7. ∴ The required quadrilateral ABCD is formed.

Question (b).
Quadrilateral BEST with BE = 2.9 cm, ES = 3.2 cm, ST = 2.7 cm, BT = 3.4 cm and ∠B=75°.
Solution:
In Quadrilateral BEST
BE = 2.9 cm, ES = 3.2 cm, ST = 2.7 cm, BT = 3.4 cm and ∠B=75°.

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{BE}}\) with radius 2.9 cm.
2. With the centre B, draw a ray of 75° and draw 2.9
   an arc with radius 3.4 cm, keep the intersecting point of these two as T.
3. With the centres T, E draw arcs with radius 2.7 cm, 3.2 cm respectively. These intersection point is keep as S’.
4. Join T, S and E,S.
5. ∴ The required quadrilateral BEST is formed.

Question (c).
Parallelogram PQRS with PQ = 4.5 cm, QR =3 cm and ∠PQR = 60°.
Solution:

In a parallelogram PQRS
PQ = 4.5 cm, QR = 3 cm, ZPQR = 60°.
=RS4.5cmzPS=3crn
[: Opposite sides of a I)aralielograrn are equal]

Construction Steps:

1. Draw a line segment ¡i with radius 4.5 cm.
2. With the centre Q draw a ray and an arc equal to 60° and 3 cm.
3. The intersecting point of these two keep as R’.
4. With the centres R, P draw arcs with 4.5 cm, 3 cm respectively. Keep ‘S’ as the intersecting point of these two arcs.
5. Join P, S and R, S.
6. ∴ The required parallelogram PQRS is formed.

Question (d).
Rhombus MATH with AT =4 cm, ∠MAT =120°.
Solution:

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{MA}}\) with radius 4 cm.
2. With the centre A draw a ray and an arc equal to 120°, 4 cm. These two intersecting point be keep as T.
3. With the centres M, T draw arcs equal to 4 cms.
   These two arcs intersected at the point ‘H’.
4. Join M, H and T, H.
5. ∴ The required rhombus MATH is formed.

Question (e).
Rectangle FLAT with FL =5 cm, LA= 3 cm.
Solution:


In a rectangle FLAT
FL=AT=5cm, LA = TF = 3cm, ∠F = ∠L = ∠A = ∠T = 90°

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{FL}}\) with radius 5 cm.
2. With the centre F draw a ray and an arc equal to 900, 3 cm.
   These to meet at point T.
3. With the centres T, L draw arcs equal to 5 cm, 3 cm respectively.
4. These two arcs meet at the point ‘A’.
5. Join T, A and L, A.
6. ∴ The required rectangle FLAT is formed.

Question (f).
Square LUDO with LU = 4.5 cm.
Solution:


In a square LUDO
LU = UD = DO = OL = 4.5 cm
∠L = ∠U = ∠D = ∠O = 90°

Construction Steps: 45

1. Draw a line segment \(\overline{\mathrm{LU}}\) with radius 4.5 cm.
2. With the centre ‘L’, draw a ray of 90° and an arc with radius 4.5 cm. These two meet at the point ‘O’.
3. Now with the centre U’, draw another ray of 90° and an arc with radius 4.5 cm. These two meet at the point “D”.
4. Join O, D.
5. ∴ The required square LUDO is formed.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.5

Question 1.
Solve the following equations.
i) \(\frac{n}{5}-\frac{5}{7}=\frac{2}{3}\)
Solution:

ii) \(\frac{x}{3}-\frac{x}{4}=14\)
⇒ \(\frac{4 x-3 x}{12}\) = 14
⇒ \(\frac{x}{12}\) = 14
⇒ x = 12 × 14 = 168
∴ x = 168

iii) \(\frac{z}{2}+\frac{z}{3}-\frac{z}{6}=8\)

iv) \(\frac{2 p}{3}-\frac{p}{5}=11 \frac{2}{3}\)

v) \(9 \frac{1}{4}=y-1 \frac{1}{3}\)

vi) \(\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}\)

vii) \(\frac{x}{2}-\frac{1}{4}=\frac{x}{3}+\frac{1}{2}\)

viii) \(\frac{2 x-3}{3 x+2}=\frac{-2}{3}\)
⇒ 3(2x – 3) = – 2(3x + 2)
⇒ 6x – 9 = -6x – 4
⇒ 6x + 6x = -4 + 9
⇒ 12x = 5
∴ x = \(\frac{5}{12}\)

ix) \(\frac{8 p-5}{7 p+1}=\frac{-2}{4}\)
Solution:
⇒ \(\frac{8 p-5}{7 p+1}=\frac{-2}{4}\)
⇒ 2(8p – 5) = – (7p + 1)
⇒ 16p – 10 = – 7p – 1
⇒ 16p + 7p = – 1 + 10
⇒ 23p = 9
∴ x = \(\frac{9}{23}\)

x) \(\frac{7 y+2}{5}=\frac{6 y-5}{11}\)
⇒ 11 (7y + 2) = 5 (6y-5)
⇒ 77y + 22 = 30y – 25
⇒ 77y – 30y = – 25 – 22
⇒ 47y = – 47
∴ y = \(\frac{-47}{47}\)
∴ y = -1

xi) \(\frac{x+5}{6}-\frac{x+1}{9}=\frac{x+3}{4}\)

⇒ 4(x + 13) = 18 (x + 3)
⇒ 4x + 52 = 18x + 54
⇒ 4x – 18x = 54-52
⇒ – 14x = 2
⇒ x = \(\frac{2}{-1}\) = \(\frac{-1}{7}\)
∴ x = \(\frac{-1}{7}\)

xii) \(\frac{3 t+1}{16}-\frac{2 t-3}{7}=\frac{t+3}{8}+\frac{3 t-1}{14}\)
Solution:

⇒ -11t + 55 = 2(19t + 17) = 38t + 34
⇒ -11t – 38t = 34 – 55
⇒ -49t = – 21
⇒ \(\frac{-21}{-49}\) = \(\frac{3}{7}\)
∴ t = \(\frac{3}{7}\)

Question 2.
What number is that of which the third part exceeds the fifth part by 4?
Solution:
Let the number be ‘x’ say.
\(\frac{1}{3}\) rd of a number = \(\frac{1}{3}\) x x = \(\frac{x}{3}\)
\(\frac{3}{7}\) th of a number = \(\frac{1}{5}\) x x = \(\frac{x}{5}\)
According to the sum

∴ The required number is 30.

Question 3.
The difference between two positive integers is 36. The quotient when one integer is
divided by other is 4. Find the integers.
(Hint: If one number is ‘X’, then the other number is ‘x – 36’)
Solution:
Let the two positive numbers be x, (x – 36) say.
If one number is divided by second tten the quotient is 4.
∴ \(\frac{x}{x-36}=4\)
⇒ x = 4(x – 36) = 4x – 144
⇒ 4x – x = 144
3x = 144
x = 48
∴ x – 36 = 48 – 36 = 12
∴ The required two positive intgers are 48, 12.

Question 4.
The numerator of a fraction is 4 less than the denominator. If 1 is added to both its
numerator and denominator, it becomes 1/2 . Find the fraction.
Solution:
Let the denominator of a fractin be x.
The numerator of a fraction is 4 less than the denominator.
∴ The numerator = x – 4
∴ Fraction \(\frac{x-4}{x}\)
If ‘1’ is added to both, its numerator and denominator, it becomes \(\frac{1}{2}\)
∴ \(\frac{1+x-4}{1+x}=\frac{1}{2}\)
2 + 2x – 8 = 1 + x
2x – x = 1 + 6 = 7
x = 7
∴ The denominator = 7
The numerator = 7 – 4 = 3
∴ Fraction = \(\frac{3}{7}\)

Question 5.
Find three consecutive numbers such that if they are divided by 10, 17, and 26 respectively,
the sum of their quotients will be 10.
(Hint: Let the consecutive numbers = x, x+ 1, x+ 2, then \(\frac{x}{10}+\frac{x+1}{17}+\frac{x+2}{26}=10\))
Solution:
Let the three consecutive numbers be assume that x, (x + 1), (x + 2) respectively.
Given that x, (x + 1), (x + 2) are divided by 10, 17, 26 respectively, the sum of the quotients is 10. Then
⇒ \(\frac{x}{10}+\frac{x+1}{17}+\frac{x+2}{26}=10\)
⇒ \(\frac{x \times 221+130(x+1)+85(x+2)}{2210}=10\)
⇒ 221x + 130x + 85x + 130 + 170 = 22,100
⇒ 436x + 300 = 22,100
⇒ 436x = 22,100 – 300
⇒ 436x = 21,800
⇒ \(\frac{21800}{436}\)
∴ x = 50
∴ The required three consecutive num-bers are x = 50
x + 1 =50+ 1 = 51
x + 2 = 50 + 2 = 52

Question 6.
In class of 40 pupils the number of girls is three-fifths of the number of boys. Find the
number of boys in the class.
Solution:
Let the number of boys = x say.
Total number of students = 40
Number of girls = \(\frac{3}{5}\) × x = \(\frac{3x}{5}\)
According to the sum 3x

∴ x = 25
∴ Number of boys in the class room = 25

Question 7.
After 15 years , Mary’s age will be four times of her present age. Find her present age.
Solution:
Let the present age of Mary = x years say.
After 15 years Mary’s age = (x + 15) years
According to the sum
(x + 15) = 4 x x
⇒ x + 15 = 4x
⇒ 4x – x =15
⇒ 3x = 15
⇒ x = 5
∴ The present age of Mary = 5 years.

Question 8.
Aravind has a kiddy bank. It is full of one-rupee and fifty paise coins. It contains 3 times
as many fifty paise coins as one rupee coins. The total amount of the money in the bank is
₹ 35. How many coins of each kind are there in the bank?
Solution:
Number of 1 rupee coins = x say.
Number of 50 – paise coins = 3 x x = 3x
The value of total coins = \(\frac{3x}{2}\) + x
[∵50 paisa coins of 3x = ₹\(\frac{3x}{2}\)
According to the sum
⇒ \(\frac{3x}{2}\) + x = 35
⇒ \(\frac{3 x+2 x}{2}\) = 35
⇒ 5x = 2 × 35
⇒ x = 2 × \(\frac{35}{5}\)
∴ x = 14
∴ Number of 1 rupee coins = 14
Number of 50 paisa coins = 3 × x = 3 × 14 = 42

Question 9.
A and B together can finish a piece of work in 12 days. If ‘A’ alone can finish the same work in 20days , in how many days B alone can finish it?
Solution:
A, B can do a piece of work in 12 days.
(A + B)’s 1 day work = \(\frac{1}{12}\) th part.
A can complete the same work in 20 days.
Then his one day work = \(\frac{1}{20}\)
B’s one day work = (A+B)’s 1 day work – A’s 1 day work
\(=\frac{1}{12}-\frac{1}{20}=\frac{5-3}{60}=\frac{2}{60}=\frac{1}{30}\)
∴ Number of days to take B to com¬plete the whole work = 30.

Question 10.
If a train runs at 40 kmph it reaches its destination late by 11 minutes . But if it runs at 50 kmph it is late by 5 minutes only. Find the distance to be covered by the train.
Solution:
Let the distance to be reached = x km say. Time taken to travel ‘x’ km with speed x
40 km/hr = \(\frac{x}{40}\) hr.
Time taken to travel ‘x’ km with speed 50 km/hr = \(\frac{x}{50}\) hr.
According to the sum the difference between the times

∴ The required distance to be trav¬elled by a train = 20 kms‘.

Question 11.
One fourth of a herd of deer has gone to the forest. One third of the total number is
grazing in a field and remaining 15 are drinking water on the bank of a river. Find the total
number of deer.
Solution:
Number of deer = x say.
Number of deer has gone to the forest
= \(\frac{1}{4}\) × x = \(\frac{x}{4}\)
Number of deer grazing in the field
= \(\frac{1}{3}\) × x =\(\frac{x}{3}\)
Number of remaining deer =15
According to the sum

∴ x = 36
∴ The total number of deer = 36

Question 12.
By selling a radio for ₹903, a shop keeper gains 5%. Find the cost price of the radio.
Solution:
The selling price of a radio = ₹ 903
Profit % = 5%
C.P = ?
C.P = \(\frac{\mathrm{S.P} \times 100}{(100+\mathrm{g})}\)
= \(\frac{903 \times 100}{(100+5)}\)
= \(\frac{903\times 100}{105}\)
C.P. = 8.6 × 100 = 860
∴ The cost price of the radio = ₹ 860

Question 13.
Sekhar gives a quarter of his sweets to Renu and then gives 5 sweets to Raji. He has 7 sweets left. How many did he have to start with?
Solution:
Number of sweets with Sekhar = x say.
Number of sweets given to Renu
= \(\frac{1}{4}\) × x = \(\frac{x}{4}\)
Number of sweets given to Raji = 5
Till he has 7 sweets left.
x – ( \(\frac{x}{4}\) + 5) = 7
⇒ x – \(\frac{x}{4}\) – 5 = 7
⇒ x – \(\frac{x}{4}\) = 7 + 5 = 12
⇒ \(\frac{4 x-x}{4}\) = 12
⇒ \(\frac{3x}{4}\) = 12
⇒ x = 12 × \(\frac{4}{3}\) = 16
∴ x = 4 × 4 = 16
∴ Number of sweets with Sekhar at the beginning = 16

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.4

Question 1.
Find the value of ’x’ so that l || m

Solution:
Given l|| m. Then 3x – 10° = 2x + 15°
[Vertically opposite angles and corresponding angles are equal.]
⇒ 3x – 10 = 2x + 15
⇒ 3x – 2x = 15 + 10
∴ x = 25°

Question 2.
Eight times of a number reduced by 10 is equal to the sum of six times the number and 4. Find the number.
Solution:
Let the number be ‘x’ say.
8 times of a number = 8 × x = 8x
¡f10 is reduced from 8x then 8x – O
6 times of a number = 6 × x = 6x
If 4 is added to 6x then 6x + 4
According to the sum,
8x – 10 = 6x + 4
⇒ 8x – 6x = 4 + 10
⇒ 2x = 14
⇒ x = 7
∴ The required number = 7

Question 3.
A number consists of two digits whose sum is 9. If 27 is subtracted from the number its digits are reversed. Find the number.
Solution:
Let a digit of two digit number be x.
The sum of two digits = 9
∴ Another digit = 9 – x
The number = 10 (9 – x) + x
= 90 – 10x + x
= 90 – 9x
If 27 is subtraçted from the number its digits are reversed.
∴ (90 – 9x) – 27 = 10x + (9 – x)
63 – 9x = 9x + 9
9x + 9x = 63 – 9
18x = 54
∴ x = \(\frac { 54 }{ 18 }\) = 3
∴ Units digit = 3
Tens digit = 9 – 3 = 6
∴ The number = 63

Question 4.
A number is divided into two parts such that one part is 10 more than the other. If the two parts are in the ratio 5:3, find the number and the two parts.
Solution:
If a number is divided into two parts in he ratio of 5 : 3, let the parts be 5x, 3x say.
According to the sum,
5x = 3x + 10
⇒ 5x – 3x = 10
⇒ 2x = 10
∴ x = \(\frac { 10 }{ 2 }\)
∴ x = 5
∴ The required number be
x + 3x = 8x
= 8 × 5 = 40
And the parts of number are
5 = 5 × 5 = 25
3 = 3 × 5 = 15

Question 5.
When I triple a certain number and add 2, I get the same answer as I do when I subtract the number from 50. Find the number.
Solution:
Let the number be x’ say.
3 times of a number = 3 × x = 3x
If 2 is added to 3x then 3x + 2
If ‘xis subtracted from 50 then it becomes 50 – x.
According to the sum,
3x + 2 = 50 – x
3x + x = 50 – 2
4x = 48 .
x = 12
∴ The required number 12

Question 6.
Mary is twice older than her sister. In 5 years time, she will be 2 years older than her sister. Find how old are they both now.
Solution:
Let the age of Marys sister = x say.
Mary’s age = 2 × x = 2x
After 5 years her sister’s age
= (x + 5) years
After 5 years Mary’s age
= (2x + 5) years
According to the sum,
2x + 5 = (x + 5) + 2
= 2x – x = 5 + 2 – 5
∴ The age of Mary’s sister = x = 2 years
Mary’s age = 2x = 2 x 2 = 4 years

Question 7.
In 5 years time, Reshma will be three times old as she was 9 years ago. How old is she now?
Solution:
Reshma’s present age = ‘x’ years say.
After 5 years Reshmats age
= (x + 5) years
Before 9 years Reshma’s age
=(x – 9) years
According to the sum
= x+ 5 = 3(x – 9) = 3x – 27
x – 3x = -27-5
-2x = -32
x = \(\frac{-32}{-2}\) = 16
∴ x = 16
∴ Reshma’s present age = 16 years.

Question 8.
A town’s population increased by 1200 people, and then this new population decreased 11%. The town now had 32 less people than it did before the 1200 increase. Find the original population.
Solution:
Let th population of a town after the increase of 1200 is x say.
11% of present population

The present population of town
= 11,200 – 1200 = 10,000

Question 9.
A man on his way to dinner shortly after 6.00 p.m. observes that the hands of his watch form an angle of 110°. Returning before 7.00 p.m. he notices that again the hands of his watch form an angle of 1100. Find the number of minutes that he has been away.
Solution:
Let the number be ‘x ray.
\(\frac { 1 }{ 3 }\) rd of a number = \(\frac { 1 }{ 3 }\) x x = \(\frac { x }{ 3 }\)
\(\frac { 1 }{ 5 }\) th of a number = \(\frac { 1 }{ 5 }\) x x = \(\frac { x }{ 5 }\)
According to the sum

∴ x = 30
∴ The required number is 30.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.3

Question
Solve the following equations:
1. 7x – 5 = 2x
2. 5x – 12 = 2x – 6
3. 7p- 3 = 3p + 8
4. 8m + 9 = 7m + 8
5. 7z + 13 = 2z + 4
6. 9y + 5 = 15y – 1
7. 3x + 4 = 5(x – 2)
8. 3(t – 3) = 5(2t – 1)
9. 5(p – 3) = 3(p – 2)
10. 5(z + 3) = 4(2z + 1)
11. 15(x – 1) + 4(x + 3) = 2 (7 + x)
12. 3 (5z – 7) +2 (9z – 11) = 4 (8z – 7) – 111
13. 8(x – 3) – (6 – 2x) = 2(x + 2) – 5 (5 – x)
14. 3(n – 4)+2(4n – 5) = 5(n + 2) + 16
Solution:
1. 7x – 5 = 2x
⇒ 7x – 2x = 5
⇒ 5x = 5
⇒ x = \(\frac { 5 }{ 5 }\) = 1
∴ x = 1

2. 5x – 12 = 2x – 6
⇒ 5x – 2x = – 6 + 12
⇒ 3x = 6
⇒ x = \(\frac { 6 }{ 3 }\) = 2
∴ x = 2

3. 7p – 3 = 3p + 8
⇒ 7p – 3p = 8 + 3
⇒ 4p = 11
⇒ p = \(\frac { 11 }{ 4 }\)

4. 8m + 9 = 7m + 8
⇒ 8m – 7m = 8 – 9
∴ m = – 1

5. 7z + 13 = 2z + 4
⇒ 7z – 2z = 4 – 13
⇒ 5z = – 9
∴ z = \(\frac { -9 }{ 5 }\)

6. 9y + 5 = 15y – 1
⇒ 9y – 15y = – 1 – 5
⇒ -6y = -6
⇒ y = \(\frac { -6 }{ -6 }\)
∴ y = 1

7. 3x + 4 = 5(x – 2)
⇒ 3x + 4 =5x – 10
⇒ 3x – 5x= – 10 – 4
⇒ – 2x = – 14
:. x = 71

8. 3(t – 3) = 5(2t – 1)
⇒ 3t – 9 = 10t – 5
⇒ 3t – 10t = – 5+ 9
⇒ – 7t = 4
∴ t = \(\frac { -4 }{ 7 }\)

9. 5 (p – 3) = 3 (p – 2)
⇒ 5p – 15 = 3p – 6
⇒ 5p – 3p = -6 + 15
⇒ 2p = 9
∴ p = \(\frac { 9 }{ 2 }\)

10. 5(z + 3) = 4(2z + 1)
⇒ 5z + 15 = 8z + 4
⇒ 5z – 8z = 4 – 15
⇒ – 3z = – 11
⇒ z = \(\frac { -11 }{ 3 }\)
∴ z = \(\frac { -11 }{ 3 }\)

11. 15(x – 1) + 4(x + 3) = 2(7 + x)
⇒ 15x – 15 + 4x + 12= 14 + 2x
⇒ 19x – 3 = 14 + 2x
⇒ 19x – 2x = 14 + 3
⇒ 17x = 17 ,
x = \(\frac { 17 }{ 17 }\) = 1
∴ x = 1

12. 3(5z – 7)+2(9z – 11) = 4(8z – 7) – 111
⇒ 15z – 21 + 18z – 22 = 32z – 28 – 111
⇒ 33z – 43 = 32z – 139
⇒ 33z – 32z = – 139 + 43
∴ z = – 96

13. 8(x – 3) – (6 – 2x)=2(x+2)-.5(5 – x)
⇒ 8x – 24 – 6 + 2x = 2x + 4 – 25 +5x
⇒ 8x – 30 = 5x – 21
⇒ 8x – 5x= – 21 +30
⇒ 3x = 9
∴ x = 3

14. 3(n – 4) + 2(4n – 5) = 5(n + 2) + 16
⇒ 3n – 12 + 8n – 10 = 5n + 10 + 16
⇒ 11n – 22 = 5n + 26
⇒ 11n – 5n = 26 + 22
⇒ 6n =48
⇒ n = \(\frac { 48 }{ 6 }\) = 8
∴ n = 8

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.2

Question 1.
Find ‘x’ in the following figures?


Solution:
1) In a triangle the exterior angle is equal to the sum of its opposite interior angles.
∴ ∠ACD = ∠B + ∠A
⇒ 123°= x + 56°
⇒ x = 123°- 56° = 67°
∴ x = 67°

ii) Sum of three angles of a triangle = 180°
∠P + ∠Q +∠R = 180°
⇒ 45° + 3x + 16°+ 68° = 180°
⇒ 3x + 129° = 180°
3x = 180 – 129 = 51
∴ x = \(\frac{51}{3}\)
∴ x = 17°

iii) ∠A + ∠B + ∠C = 180°
⇒ 25° + x + 30° = 180°
x + 55°= 180°
x = 180 – 55 = 125°
∴ x = 125°

iv) In ΔXYZ, \(\overline{\mathrm{XY}}=\overline{\mathrm{XZ}}\) then ∠Y = ∠Z
∴ 2x + 7° = 45°
⇒ 2x = 45 – 7
⇒ 2x = 38
⇒ x = \(\frac{38}{2}\)
∴ x = 19°

v) From ΔBOA
\(\overline{\mathrm{AB}}=\overline{\mathrm{OA}} \) ⇒ ∠B = ∠O = 3x + 10° ………(1)
From ΔCOD
\(\overline{\mathrm{OC}}=\overline{\mathrm{CD}} \) ⇒ ∠O = ∠D …………………(2)
[∵ The angles which are opposite to the equal sides are equal].
from (1) & (2)
∠BOA = ∠COD
[∵ Vertically opposite angles are equal.]
But ∠COD = 90 – x
(∵ 2x + ∠O + ∠D = 180
⇒ 2x + ∠O + ∠O = 180 (∵∠O = ∠D)
⇒ 2∠O = 180 – 2x
∠O = \(\frac{180-2 x}{2}\) = 90 – x]
∴ From ∠BOA = ∠COD
⇒ 3x + 10 = 90 – x
⇒ 3x + x = 90 – 10
⇒ 4x = 80
∴ x = 20°

Question 2.
The difference between two numbers is 8. if 2 is added to the bigger number the result will be three times the smaller number. Find the numbers.
Solution:
Let the bigger number be x.
The difference between two numbers 8
∴ Smaller number = x – 8
If 2 is added to the bigger number the result will be three times the smaller
number.
So x + 2 = 3(x – 8)
x + 2 = 3x – 24
x – 3x = -24 – 2
– 2x = -26
∴ x = \(\frac{26}{2}\) = 13
∴ Bigger number = 13
Smaller number = 13 – 8 = 5

Question 3.
What are those two numbers whose sum is 58 and difference is 28’?
Solution:
Let the bigger number be ‘x’.
The sum of two numbers = 58
∴ Smaller number = 58 – x
The difference of two numbers = 28
∴ x – (58 – x) = 28
x – 58 + x = 28
2x = 28 + 58 = 86
∴ x = \(\frac{86}{2}\) = 43
∴ Bigger number or one number = 43
Smaller number or second number = 58 – 43 = 15

Question 4.
The sum of two consecutive odd numbers is 56. Find the numbers.
Solution:
Let the two consecutive odd numbers
be 2x 1, 2x + 3 say.
Sum of the odd numbers
=(2x + 1) + (2x + 3) = 56
= 4x + 4 = 56
⇒ 4x = 56 – 4 = 52
x = \(\frac{52}{4}\) = 13
∴ x = 13
∴ 2x + 1 = 2 × 13 + 1
= 26 + 1 = 27
2x + 3 = 2 × 13 + 3
= 26 + 3 = 29
∴ The required two consecutive odd numbers be 27, 29.

Question 5.
The sum of three consecutive multiples of 7 is 777. Find these multiples.
(Hint: Three consecutive multiples of 7 are ‘x’, ‘x+ 7’, ‘x+ 14’)
Solution:
Let the three consecutive multiples of 7 be x, x + 7, x + 14 say.
According to the sum,
The sum of three consecutive multiples of 7 is 777.
⇒ x + (x + 7) + (x + 14)= 777
⇒ 3x + 21 = 777
⇒ 3x = 777 – 21 = 756
x = \(\frac { 756 }{ 3 }\) = 252
x+ 7 = 252 + 7 = 259
x + 14 252 + 14 = 266
∴ The required three consecutive multiples of 7 are 252, 259, 266

Question 6.
A man walks 10 km, then travels a certain distance by train and then by bus as far as twice by the train. 1f the whole journey is of 70km, how far did he travel by train?
Solution:
The distance travelled by walk = 10 km
Let the distance travelled by train = x km say.
The distance travelled by bus
= 2 × x = 2x km
∴ 10 + x + 2x = 70
⇒ 3x = 70 – 10
⇒ 3x = 60
⇒ x = \(\frac { 60 }{ 3 }\) = 20
⇒ x = 20
∴ The distance travelled by train = 20 km.

Question 7.
Vinay bought a pizza and cut it into three pieces. When the weighed the first piece he found that it was 7g lighter than the second piece and 4g.heavier than the third piece. If the whole pizza weighed 300g. How much did each of the three pieces weigh?

(Hint: weight of normal piece be ‘x’ then weight of largest piece is ‘x+ 7’, weight of the smallest piece is ‘x-4’)
Solution:
If pizza is cut into three pieces.
Let the weight of first piece he ‘x’ gm say.
Weight of the second piece = (x + 7) gm
Weight of the third piece = (x – 4) gm
According to the sum
∴ x + (x + 7) + (x – 4) = 300
⇒ 3x + 3 = 300
⇒ 3x = 300 – 3 = 297
⇒ x = \(\frac { 297 }{ 3 }\) = 99
∴ x= 99
x + 7= 99 + 7 =106
x – 4 = 99 – 4 = 95
∴ The required 3 pieces of pizza weighs 95 gm. 99 gm, 106 gm.

Question 8.
The distance around a rectangular field is 400 meters.The length of the field is 26 meters more than the breadth. Calculate the length and breadth of the field’?
Solution:
Let the breadth of a rectangular field = x m
Length = (x + 26) m.
Perimeter of a rectangular field
= 2(l + b) = 400
l + b = 200
x + 26 + x = 200
2x = 200 – 26 = 174
x = \(\frac { 174 }{ 2 }\)
∴ x = 87
∴ The length = x +26
= 87 + 26
= 113 m
Breadth = x = 87 m.

Question 9.
The length of a rectangular field is 8 meters less than twice its breadth. If the perimeter of the rectangular field is 56 meters, find its length and breadth’?
Solution:
Let the breadth of a rectangular field = xm.
Length = 2 × x – 8 = (2x – 8) m.
Perimeter of a field = 56 m.
∴ 2(l + b) = 56
l + b = 28
2x – 8 + x = 28
3x = 28 + 8 = 36
x = \(\frac { 36 }{ 3 }\)
∴ x= 12
∴ Breadth = 12 m
Length = 2x – 8
= 2 × 12 – 8
= 24 – 8 = 16m.

Question 10.
Two equal sides of a triangle are each 5 meters less than twice the third side. if the perimeter of the triangle is 55 meters, find the length of its sides’?
Solution:
A triangle in which the length of the third side = x m. say.
The length of remaining two equal sides = 2 × x – 5 = (2x – 5) m.
Perimeter of a triangle = 55 m.
∴ (2x – 5) + (2x -5) + x = 55
⇒ 5x – 10 = 55
⇒ 5x = 65
⇒ x = \(\frac { 65 }{ 5 }\)
∴ x = 13m
2x – 5 = 2 × 13 – 5 = 26 – 5 = 21m.
∴ The lengths of three sides of a triangle are 13, 21, 21. (in m.)

Question 11.
Two complementary angles differ by 12°, find the angles’?
Solution:
Let one angle in two complementary angles be x.
Sum of the two complementary angles = 90°
∴ Second angle = 90° – x
Here two complementary angles differ by 12°.
∴ x – (90°- x) = 12°
x – 90° + x = 12°
2x = 12° + 90° = 102°
∴ \(\frac{102^{\circ}}{2}\) = 51°
one angle = 51°
Second angle 90° – 51° = 39°

Question 12.
The ages of Rahul and Laxmi arc in the ratio 5:7. Four years later, the sum of their ages will
be 56 years. What are their present ages’?
Solution:
The ratio of ages of Rahul and Lakshmi = 5:7
Let their ages be 5x, 7x say.
After 4 years Rahuls agt. = 5x + 4
After 4 years Lakshmis age = 7x + 4
According to the sum,
After 4 years the sum of their ages = 56
⇒ (5x + 4) + (7x + 4) = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 – 8 = 48
⇒ x= \(\frac { 48 }{ 12 }\) = 4
∴ x = 4
∴ Rahuls present age
= 5x = 5 × 4 = 20 years
∴ Lakshmi’s present age
= 7x = 7 × 4 = 28 years

Question 13.
There are 180 multiple choice questions in a test. A candidate gets 4 marks for every correct answer, and for every un-attempted or wrongly answered questions one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the
test how many questions did he answer correctly ?
Solution:
Number of questions attempted for correct answers = x say
umber of questions attempted for wrong answers = 180 – x
4 marks are awarded for every correct answer.
Then numl)er of marks ohtained for correct answers 4 × x = 4x
1 mark is deducted for every wrong answer.
∴ Number of marks deducted for wrong answers
=(180 – x) × 1 = 180 – x
According to the sum.
4x – (180 – x) = 450
⇒ 4x – 180 + x = 450
⇒ 5x = 450 + 180
⇒ 5x = 630
x = \(\frac { 630 }{ 5 }\)
∴ x = 126
∴ Number of questions attempted for correct answers = 126

Question 14.
A sum of ₹ 500 is in the form of denominations of ₹ 5 and ₹ 10. If the total number of notes is 90 find the number of notes of each denomination.
(Hint: let the number of 5 rupee notes be ‘x’, then number of 10 rupee notes = 90 – x)
Solution:
Number of ₹ 5 notes = x say.
Number of ₹ 10 notes = 90 – x
5x + 10(90 – x) = 500
5x + 900 – 10x = 500
– 5x = 500 – 900 = – 400
x = \(\frac { -400 }{ -5 }\)
∴ x = 80
∴ Number of ₹ 5 notes = 80
Number of ₹ 10 notes
= 90 – x = 90 – 80 = 10

Question 15.
A person spent ₹ 564 in buying geese and ducks,if each goose cost ₹ 7 and each duck ₹ 3 and if the total number of birds bought was 108, how many of each type did he buy?

Solution:
Let the number of pens be x.
The total number of things = 108
∴ The number of pencils = 108 – x
Thecostofpens of x = ₹7 × x = ₹ 7x
The cost of pencils of (108- x)
= ₹3(108 – x) = ₹ (324 – 3x)
Total amount to t)uy eflS and Pencils = ₹564
∴ 7x + (324 – 3x) 5M
7x + 324 – 3x = 564
4x = 564 – 324 = 240
∴ x = \(\frac { 240 }{ 4 }\) = 60
The number of pens = 60
The number of pencils = 108 – 60 = 48

Question 16.
The perimeter ofa school volleyball court is 177 ft and the length is twice the width. What are the dimensions of the volleyball court’?

Solution:
Breadth of a volleyball court = x feet say.
∴ Its length = 2 × x = 2 x feet.
The perimeter of a court = 177 feet.
⇒ 2(l + b) = 177
⇒ 2(2x + x) = 177
⇒ 2 x 3x = 177
⇒ 6x = 177
⇒ x =\(\frac { 177 }{ 6 }\)
∴ x = 29.5
∴ The breadth of a volleyball court = 29.5 ft
The length of a volleyball court = 2x = 2 × 29.5 = 59ft

Question 17.
The sum of the page numbers on the facing pages of a book is 373. What are the page numbers?
(Hint :Let the page numbers of open pages are x and (x + 1)

Solution:
Number of first page of a opened book x
Number ol second page = x + 1
∴ The surï of the numbers of two pages = 373
⇒ x + (x + 1) = 373
⇒ 2x + 1 = 373
2x = 372
⇒ x = 186
∴x + 1 = 186 + 1 = 187
∴ Numbers of two c(nlsecutive pages = 186, 187

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.1

Question
Solve the following Simple Equations:

(i) 6m = 12
(ii) 14p -42
(iii) -5y = 30
(iv) – 2x = – 12
(v) 34x = – 51
(vi) \(\frac{n}{7}\) = -3
(vn) \(\frac{2x}{3}\) = 8
(vui) 3x+1 = 16
(ix) 3p – 7 = 0
(x) 13 – 6n = 7
(xi) 200y – 51 = 49
(xii) 11n + 1 = 1
(xiii) 7x – 9 = 16
(xiv) 8x + \(\frac{5}{2}\) =13
(xv) 4x – \(\frac{5}{3}\) = 9
(xvi) x – \(\frac{4}{3}\) = 3\(\frac{1}{2}\)
Solution:
i) 6m = 12 ⇒ m = \(\frac{12}{6}\) ⇒ m = 2

ii) 14p = – 42p ⇒ P = \(\frac{-42}{14}\)
∴ p = -3

iii) -5y = 30 ⇒ y = \(\frac{30}{-5}\) = -6
∴ y = -6

iv) -2x = -12
⇒ 2x = 12
x = \(\frac{30}{-5}\)
= 6
∴ x = 6

v) 34x = -51
⇒ \(\frac{-3}{2}\) = \(\frac{-3}{2}\)
∴ x = \(\frac{-3}{2}\)

vi) \(\frac{n}{7}\) = -3
⇒ n = -3 x 7 = -21
∴ n = -21

vii) \(\frac{2x}{3}\) = 18 ⇒ 18 x \(\frac{3}{2}\) = 27
∴ x = 27

viii) 3x + 1 = 16
3x = 16 – 1 = 15
3x = 15
x = \(\frac{15}{3}\)
∴ x = 5

ix) 3p – 7 = 0
⇒ 3p = 7
∴ p = \(\frac{7}{3}\)

x) 13 – 6n = 7 ⇒ -6n = 7 – 13
⇒ -6n = -6 ⇒ n= \(\frac{-6}{-6}\)
∴ n = 1

xi) 200y – 51 = 49
⇒ 200y = 49 + 51
⇒ 200y = 100
⇒ y = \(\frac{100}{200}\)
∴ y = \(\frac{1}{2}\)

xii) 11n + 1 = 1
⇒ 11n = 1 – 1
= 11n = 0
⇒ n = \(\frac{0}{11}\) = 0
∴ n = 0

xiii) 7x – 9 = 16
⇒ 7x = 16 + 9
⇒ 7x = 25
∴ x = \(\frac{25}{7}\)

xiv) 8x + \(\frac{5}{2}\) = 13

xv) 4x – \(\frac{5}{3}\)

xvi) x + \(\frac{4}{3}\) = 3\(\frac{1}{2}\)
⇒ \(x+\frac{4}{3}=\frac{7}{2}\)
⇒\(\frac{7}{2}-\frac{4}{3}=\frac{21-8}{6}\)
∴ x = \(\frac{13}{6}\)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers Exercise 1.3

Question 1.
Express each of the following decimal in the \(\frac{p}{q}\) form.
(i) 0.57 (ii) 0.176 (iii) 1.00001 (iv) 25.125
Solution:
(i) 0.57 = \(\frac{57}{100}\) (∵ two digits are there after the decimal poing)
(ii) 0.176 = \(\frac{176}{1000}\)
(iii) 1.00001 = \(\frac{100001}{100000}\)
(iv) 25.125 = \(\frac{25125}{1000}\)

Question 2.
Express each of the following decimals in the rational form \(\frac{p}{q}\)
(1) \(0 . \overline{9}\)
(ii) \(0 . \overline{57}\)
(iii) \(0 .7 \overline{29}\)
(iv) \(12.2 \overline{8}\)
Solution:
(i) \(0 . \overline{9}\)
Let x = \(0 . \overline{9}\)
⇒ x = 0.999 ………………. (1)
Here periodicity is 1. So, equation (1) should be multiplied both sides with
= 10 × x = 10 × 0.999
10 x = 9.999 ………….. (2)

\(0 . \overline{9}\) = 1

Second Method:
\(0 . \overline{9}=0+\overline{9}=0+\frac{9}{9}\)
= 0 + 1 = 1

(ii) \(0 . \overline{57}\)
x = \(0 . \overline{57}\) ⇒ x = 0.5757…………(1)
Here periodicity is 2. So, we should multiply with 100
⇒ 100 × x = 100 x 0.5757 …………..
100 × =57.57 ……………………. (2)

(iii) \(0 .7 \overline{29}\)
x = \(0 .7 \overline{29}\)
x = \(0 .7 \overline{29}\) ⇒ x = 0.7979…………(1)
Here periodicity is 2. So, equation (1) should multiply with 100
⇒ 100 × x = 100 × 0.72929 …………..
100 × = 72.929 …………………… (2)

(iv) \(12.2 \overline{8}\)
x = (iv) \(12.2 \overline{8}\)
⇒ x = 12.288 ………..(1)
Here periodicity is 1. So, equation (1) should multiply with 10
⇒ 100 × x = 100 × 12.288 …………..
10 x = 122.888 …………………… (2)

Question 3.
Find(x + y) ÷ (x – y) if
(i) x = \(\frac{5}{2}\), y = \(-\frac{3}{4}\)
(ii) x = \(\frac{1}{4}\), y = \(\frac{3}{2}\)
Solution:
If x = \(\frac{5}{2}\), y = \(-\frac{3}{4}\) then

ii) x = \(\frac{1}{4}\), y = \(\frac{3}{2}\)

Question 4.
Divide the sum of \(-\frac{13}{5}\) and \(\frac{12}{7}\) by the product of \(-\frac{13}{7}\) and \(-\frac{1}{2}\)
Solution:
Sum of \(-\frac{13}{5}\) and \(\frac{12}{7}\)

the product of \(-\frac{13}{7}\) and \(-\frac{1}{2}\)

Question 5.
If \(\frac{2}{5}\) of a number exceeds \(\frac{1}{7}\) of the same number by 36. Find the number.
Solution:
Let the number be ‘x’ say.
\(\frac{2}{5}\) part of x = \(\frac{2}{5}\) × x = \(\frac{2x}{5}\)
\(\frac{1}{7}\) part of x = \(\frac{1}{7}\) × x = \(\frac{x}{7}\)
∴ According to the sum,

Question 6.
Two pieces of lengths 2\(\frac{2}{5}\) m and 3\(\frac{3}{10}\) mare cut off from a rope 11 m long. What is the length of the remaining rope?
Soltuion:
The length of the remaining rope

∴ The length of remaining rope
= 5\(\frac{1}{10}\) mts.

Question 7.
The cost of 7\(\frac{2}{3}\) meters of cloth is ₹12\(\frac{3}{4}\) . Find the cost per metre.
Solution:
The cost of 7\(\frac{2}{3}\) mts (\(\frac{23}{3}\) mts ) of cloth
= ₹ \(12 \frac{3}{4}\) = ₹ \(\frac{51}{4}\)
∴ The cost of 1m cloth
= \(\frac{51}{4} \div \frac{23}{3}=\frac{51}{4} \times \frac{3}{23}=\frac{153}{92}\) = ₹ 1.66

Question 8.
Find the area of a rectangular park which is 18\(\frac{3}{5}\)m long and 8\(\frac{2}{3}\) in broad.
Solution:
The length of the rectangular park
= 18\(\frac{3}{5}\)m = \(\frac{93}{5}\)
Its width / breath = 8\(\frac{2}{3}\) m = \(\frac{26}{3}\) m
∴ Area of the rectangular park
(A) = l × b

Question 9.
What number should \(-\frac{33}{16}\) be divided by to get \(-\frac{11}{4}\)
Solution:
Let the dividing number be ‘x’ say.

Question 10.
If 36 trousers of equal sizes can be stitched with 64 meters of cloth. What is the length of the cloth required for each trouser?
Solution:
36 trousers of equal sizes can he stitched with 64 mts of cloth, then the length of the cloth ¡s required for each trouser
= 64 ÷ 36
= \(\frac{64}{36}=\frac{16}{9}\) = 1 \(\frac{7}{9}\)

Question 11.
When the repeating decimal 0.363636 …. is written in simplest fractional form\(\frac{p}{q}\) , find the sum p+ q.
Solution:
x = 0.363636………………………….. (1)
Here periodicity is ‘2’. So, equation (1) should be multiplied both sides with 100.
⇒ 100 × x = 100 × 0.363636 …………..
100 x = 36.3636 ………..(2)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers Exercise 1.2

Question 1.
Represent these numbers on the number line.
(i) \(\frac{9}{7}\)
(ii) \(-\frac{7}{5}\)
Solution:
(i) \(\frac{9}{7}\)

(ii) \(-\frac{7}{5}\)

Question 2.
Represent \(-\frac{2}{13}, \frac{5}{13}, \frac{-9}{13}\) on the number line.
Solution:

Question 3.
Write five rational numbers which are smaller than \(\frac{5}{6}\)
Solution:
The rational number which are less than
\(\frac{5}{6}=\left\{\frac{4}{6}, \frac{3}{6}, \frac{2}{6}, \frac{1}{6}, \frac{0}{6}, \frac{-1}{6}, \frac{-2}{6} \ldots \ldots .\right\}\)

Question 4.
Find 12 rational numbers between -1 and 2.
Solution:

Question 5.
Find a rational number between \(\frac{2}{3}\) and \(\frac{3}{4}\)
[Hint : First write the rational numbers with equal denominators.]
Solution:
The given rational numbers are \(\frac{2}{3}\) and \(\frac{3}{4}\)
\(\frac{2}{3} \times \frac{4}{4}=\frac{8}{12}, \frac{3}{4} \times \frac{3}{3}=\frac{9}{12}\)
The rational numbers between \(\frac{8}{12}, \frac{9}{12}\) is
\(\frac{\left(\frac{8}{12}+\frac{9}{12}\right)}{2}=\frac{\frac{17}{12}}{2}=\frac{17}{24}\)
(∵ the rational number between a, b is \(\frac{a+b}{2}\) )
∴ the rational number between \(\frac{2}{3}\) and \(\frac{3}{4}\) is \(\frac{17}{24}\)

Question 6.
Find ten rational numbers between \(-\frac{3}{4}\) and \(\frac{5}{6}\)
Solution:

The 10 rational numbers between \(-\frac{9}{12}\) and \(\frac{10}{12}\) are

∴ We can select any 10 rational numbers from the above number line.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers Exercise 1.1

Question 1.
Name the properly Involved in the following examples.

vii) 7a + (-7) = 0
viii) x + \(\frac{1}{x}\) = 1(x ≠ 0)
ix) (2 x x) + (2 x 6) = 2 x (x + 6)
Solution:
i) Additive identity
ii) Distributive law
iii) Multiplicative identity
iv) Multiplicative identity
v) Commutative law of addition
vi) Closure law in multiplication
vii) Additive inverse
viii) Multiplicative inverse
ix) Distributive

Question 2.
Write the additive and the multiplicative inverses of the following.
i) \(\frac{-3}{5}\)
ii) 1
iii) 0
iv) \(\frac{7}{9}\)
v) -1
Solution:

Question 3.
Fill in the blanks



Solution:
i) \(\left(\frac{-12}{5}\right)\)
ii) \(\left(\frac{4}{3}\right)\)
iii) \(\left(\frac{9}{11}\right)\)
iv) \(\left(\frac{6}{7}\right)\)
v) \(\left(\frac{3}{4}, \frac{1}{3}\right)\)
vi) 0

Question 4.
Multiply \(\frac{2}{11}\) by the reciprocal of \(\frac{-5}{14}\)
Solution:
The reciprocal of \(\frac{-5}{14}\) is \(\frac{-14}{5}\)
( ∵ \(\left(\frac{-5}{14}\right) \times\left(\frac{-14}{5}\right)=1\) )
∴ The product of \(\frac{2}{11}\) and \(\frac{-14}{5}\) is
\(\frac{2}{11} \times\left(\frac{-14}{5}\right)=\frac{-28}{55}\)

Question 5.
Which properties can be used computing \(\frac{2}{5} \times\left(5 \times \frac{7}{6}\right)+\frac{1}{3} \times\left(3 \times \frac{4}{11}\right)\)
Solution:
The following properties are involved in the product of
\(\frac{2}{5} \times\left(5 \times \frac{7}{6}\right)+\frac{1}{3} \times\left(3 \times \frac{4}{11}\right)\)
i) Multiplicative associative property.
ii) Multiplicative inverse.
iii) Multiplicative identity.
iv) Closure with addition

Question 6.
Verify the following
\(\left(\frac{5}{4}+\frac{-1}{2}\right)+\frac{-3}{2}=\frac{5}{4}+\left(\frac{-1}{2}+\frac{-3}{2}\right)\)
Solution:

Question 7.
Evaluate \(\frac{3}{5}+\frac{7}{3}+\left(\frac{-2}{5}\right)+\left(\frac{-2}{3}\right)\) after rearrangement.
Solution:

Let x = \(1.2 \overline{4}\)
⇒ x = 1.244……. …………………(1)
Here periodicity of equation (1) is 1. So
it should be multiplied by 10 on both
sides.
⇒ 10 x x = 10 x 1.244
10x = 12.44 …………..(2)

Question 8.
Subtract
(i) \(\frac{3}{4}\) from \(\frac{1}{3}\)
(ii) \(\frac{-32}{13}\) from 2
(iii) -7 from \(\frac{-4}{7}\)
Solution:

Question 9.
What numbers should be added to \(\frac{-5}{8}\) so as to get \(\frac{-3}{2}\) ?
Solution:
Let the number to be add ‘x’ say

∴ \(\frac{-7}{8}\) should be added to \(\frac{-5}{8}\) then we will get \(\frac{-3}{2}\)

Question 10.
The sum of two rational numbers is 8 If one of the numbers is \(\frac{-5}{6}\) find the other.
Let the second number be ‘x’ say
⇒ \(x+\left(\frac{-5}{6}\right)=8\)
⇒\(8+\frac{5}{6}=\frac{48+5}{6}=\frac{53}{6}\)
∴ The other number (x) = \(\frac{53}{6}\)

Question 11.
Is subtraction associative in rational numbers? Explain with an example.
Solution:
Let \(\frac{1}{2}, \frac{3}{4}, \frac{-5}{4}\) are any 3 rational numbers.
Associative property under subtraction
a – (b – c) = (a – b) – c

∴ L.H.S. ≠ R.H.S.
∴ a – (b – c) ≠ (a – b) – c
∴ Subtraction is not an associative in rational numbers.

Question 12.
Verify that – (-x) = x for
(i) x = \(\frac{2}{15}\)
(ii) x = \(\frac{-13}{15}\)
Solution:

Question 13.
Write-
(i) The set of numbers which do not have any additive identity
(ii) The rational number that does not have any reciprocal
(iii) The reciprocal of a negative rational number.
Solution:
i) Set of natural numbers ’N’ doesn’t possesses the number ‘0’.
ii) The rational number ‘0’ has no multiplicative inverse.
[ ∵ 1/0 is not defined]
iii) The reciprocal of a negative rational number is a negative rational number.
Ex : Reciprocal of \(\frac{-2}{5}=\frac{-5}{2}\)