Exercise 2.4

Find the value of ‘x’ so that l || m.

Solution:

The angles (3x - 10°) and (2x + 15°) are corresponding angles.

For lines l and m to be parallel, corresponding angles must be .

Therefore: 3x - 10 = 2x + 15

x - =

x =

Therefore, the value of x is 25.

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 'x' represent the unknown number.

- = 6x +

8x - 10 - 6x = 6x + 4 - 6x

- 10 =

2x - 10 + 10 = 4 + 10

2x =

2x2 = 142

x =

Therefore, the number is 7.

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 the tens digit be 'x'.

Since the sum of the digits is 9, the units digit will be .

The number can be represented as + (9 - x).

When the digits are reversed, the tens digit becomes and the units digit becomes .

The reversed number can be represented as 10(9-x)10(x-9) + .

The problem states that when 27 is subtracted from the original number, the digits are reversed. So:

10x + (9 - x) + = 10(9 - x) + x +

+ 9 - = - 10x9x + x

9x - 18 = 90 - 9x

9x + 9x = 90 + 18

x =

x = 10818

x =

Units digit = 9 - x = 9 - 6 =

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: Let the two parts be 5x and 3x.

= 3x + (or) 3x = 5x +

5x - 3x = 10

= 10

x =

First part: 5x = 5 × 5 =

Second part: 3x = 3 × 5 =

Original number = First part + Second part = 25 + 15 =

Verification:

The parts are 25 and 15. 25 is indeed more than 15.

The ratio of 25:15 simplifies to .

Therefore, the number is , and the two parts are and .

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 'x' represent the unknown number.

3x + 2 = 50 - x

3x + 2 + x = 50 - x + x

4x + 2 =

4x + 2 - 2 = 50 - 2

4x =

4x4 = 484

x =

Therefore, the number is 12.

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 'm' represent Mary's current age and 's' represent her sister's current age.

Mary is twice as old as her sister: m =

In 5 years, Mary will be 2 years older than her sister: m + = s + + 2

m + 5 = s + 7

Substituting: + 5 = s + 7

2s - s = 7 -

s =

m = 2s

m = 2 × 2

m =

Therefore, Mary is currently years old, and her sister is currently years old.

7. In 5 years time, Reshma will be three times old as she was 9 years ago. How old is she now?

Solution:

Let 'r' represent Reshma's current age.

Reshma's age in 5 years will be .

Reshma's age 9 years ago was .

Thus, r + 5 = (r - 9)

r + 5 = -

r - 3r = -27 - 5

-2r = -32

r =

Therefore, Reshma is currently 16 years old.

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 d> id before the 1200 increase. Find the original population.

Solution:

Let 'p' represent the original population.

Population after increase = p +

Population after decrease = (p + 1200) + × (p + 1200)

The population after the decrease is 32 less than the population before the decrease: (p + 1200) = (p + 1200) -

0.89p + 0.89 × 1200 = p + 1200 - 32

0.89p + = p +

1068 - 1168 = p - 0.89p

=

p = −1000.11

p = (fraction form)

p = (decimal form)

Since population cannot be negative, there appears to be an error in the problem statement. The population decreased after the increase, which is unusual. If the problem meant that the population after the decrease was 32 less than the original population, the equation would be:

(p + 1200) = p +

0.89p + = p - 32

1068 + 32 = p - 0.89p

1100 =

p = 11000.11

p =

Therefore, if the population after the decrease was 32 less than the original population, then the original population was 10000.