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## AP Board Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Book Answers

 AP Board Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Book Answers

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

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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

## Andhra Pradesh Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Textbooks for Exam Preparations

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