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Example 8: The sum of three consecutive multiples of 4 is 444. Find the multiples.

Solution: Let x be a multiple of 4. Therefore, the next two consecutive multiples of 4 will be x + 4,
and x + 8.

According to the question,
x + (x + 4) + (x + 8) = 444
⇒ 3 x + 12 = 444 ⇒ 3x = 444 – 12 = 432 (Transposing 12 to RHS)
432
⇒ x = = 144
3
Therefore, the second multiple = 144 + 4 = 148
And, the third multiple = 144 + 8 = 152

Thus, the three consecutive multiples of 4 whose sum is 444 are 144, 148 and 152.
Example 9: Half of a number exceeds one-fifth of its succeeding number by 4. Find the number.
Solution: Let the required number be x. Its succeeding number = x + 1
According to the question,

1 1 5x − 2 x + 1)
(
x − ( x + 1) = 4 ⇒ = 4 (Taking LCM of 2 and 5)
2 5 10
⇒ 5x – 2x – 2 = 40 (Transposing 10 to RHS)
⇒ 3 x = 40 + 2 ⇒ 3x = 42
⇒ x = 14 (Dividing both sides by 3)
Thus, the required number is 14.
Example 10: The present age of Rahul’s father is two times the present age of Rahul. After 5 years,
their ages will add to 55 years. Find their present ages.
Solution: Let the present age of Rahul be y years and his father’s age is 2y years.

After 5 years, Rahul’s age = (y + 5) years. Rahul’s father age = (2y + 5) years.
Since, the sum of their ages after 5 years = 55
∴ ( y + 5) + (2y + 5) = 55 ⇒ 3y + 10 = 55

⇒ 3 y = 55 – 10 (Transposing 10 to RHS)
⇒ 3 y = 45 ⇒ y = 15 years
Thus, Rahul’s present age is 15 years and his father’s present age is 2 × 15 = 30 years.
Example 11: Somya has 20 orange bars more than four times the number of orange bars Raju has.
If both of them together have 160 orange bars, how many orange bars does Raju have?
Solution: Let Raju’s collection of orange bar be x.

Therefore, the Somya’s collection of orange bars = 4x + 20
According to the question,
Total orange bars = x + 4x + 20 = 5x + 20

But they together have 160 orange bars.
So, 5x + 20 = 160 ⇒ 5x = 160 – 20 = 140
140
⇒ x = = 28
5
Thus, Raju has 28 orange bars and Somya has 4 × 28 + 20 = 132 orange bars.

41 Linear Equations in One Variable