## Andhra Pradesh Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Textbooks Solutions PDF

Andhra Pradesh State Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Textbooks. These Andhra Pradesh State Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.## Andhra Pradesh State Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Books Solutions

Board | AP Board |

Materials | Textbook Solutions/Guide |

Format | DOC/PDF |

Class | 8th |

Subject | Maths |

Chapters | Maths Chapter 2 Linear Equations in One Variable Ex 2.5 |

Provider | Hsslive |

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## AP Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:Question 1.

Solve the following equations.

i) 𝑛5−57=23

Solution:

ii) 𝑥3−𝑥4=14

⇒ 4𝑥−3𝑥12 = 14

⇒ 𝑥12 = 14

⇒ x = 12 × 14 = 168

∴ x = 168

iii) 𝑧2+𝑧3−𝑧6=8

iv) 2𝑝3−𝑝5=1123

v) 914=𝑦−113

vi) 𝑥2−45+𝑥5+3𝑥10=15

vii) 𝑥2−14=𝑥3+12

viii) 2𝑥−33𝑥+2=−23

⇒ 3(2x – 3) = – 2(3x + 2)

⇒ 6x – 9 = -6x – 4

⇒ 6x + 6x = -4 + 9

⇒ 12x = 5

∴ x = 512

ix) 8𝑝−57𝑝+1=−24

Solution:

⇒ 8𝑝−57𝑝+1=−24

⇒ 2(8p – 5) = – (7p + 1)

⇒ 16p – 10 = – 7p – 1

⇒ 16p + 7p = – 1 + 10

⇒ 23p = 9

∴ x = 923

x) 7𝑦+25=6𝑦−511

⇒ 11 (7y + 2) = 5 (6y-5)

⇒ 77y + 22 = 30y – 25

⇒ 77y – 30y = – 25 – 22

⇒ 47y = – 47

∴ y = −4747

∴ y = -1

xi) 𝑥+56−𝑥+19=𝑥+34

⇒ 4(x + 13) = 18 (x + 3)

⇒ 4x + 52 = 18x + 54

⇒ 4x – 18x = 54-52

⇒ – 14x = 2

⇒ x = 2−1 = −17

∴ x = −17

xii) 3𝑡+116−2𝑡−37=𝑡+38+3𝑡−114

Solution:

⇒ -11t + 55 = 2(19t + 17) = 38t + 34

⇒ -11t – 38t = 34 – 55

⇒ -49t = – 21

⇒ −21−49 = 37

∴ t = 37

Question 2.

What number is that of which the third part exceeds the fifth part by 4?

Solution:

Let the number be ‘x’ say.

13 rd of a number = 13 x x = 𝑥3

37 th of a number = 15 x x = 𝑥5

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.

∴ 𝑥𝑥−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 𝑥−4𝑥

If ‘1’ is added to both, its numerator and denominator, it becomes 12

∴ 1+𝑥−41+𝑥=12

2 + 2x – 8 = 1 + x

2x – x = 1 + 6 = 7

x = 7

∴ The denominator = 7

The numerator = 7 – 4 = 3

∴ Fraction = 37

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 𝑥10+𝑥+117+𝑥+226=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

⇒ 𝑥10+𝑥+117+𝑥+226=10

⇒ 𝑥×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

⇒ 21800436

∴ 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 = 35 × x = 3𝑥5

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 = 3𝑥2 + x

[∵50 paisa coins of 3x = ₹3𝑥2

According to the sum

⇒ 3𝑥2 + x = 35

⇒ 3𝑥+2𝑥2 = 35

⇒ 5x = 2 × 35

⇒ x = 2 × 355

∴ 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 = 112 th part.

A can complete the same work in 20 days.

Then his one day work = 120

B’s one day work = (A+B)’s 1 day work – A’s 1 day work

=112−120=5−360=260=130

∴ 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 = 𝑥40 hr.

Time taken to travel ‘x’ km with speed 50 km/hr = 𝑥50 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

= 14 × x = 𝑥4

Number of deer grazing in the field

= 13 × x =𝑥3

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 = S.P×100(100+g)

= 903×100(100+5)

= 903×100105

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

= 14 × x = 𝑥4

Number of sweets given to Raji = 5

Till he has 7 sweets left.

x – ( 𝑥4 + 5) = 7

⇒ x – 𝑥4 – 5 = 7

⇒ x – 𝑥4 = 7 + 5 = 12

⇒ 4𝑥−𝑥4 = 12

⇒ 3𝑥4 = 12

⇒ x = 12 × 43 = 16

∴ x = 4 × 4 = 16

∴ Number of sweets with Sekhar at the beginning = 16

## AP Board Textbook Solutions PDF for Class 8th Maths

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