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