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.