# Some More Applications on the Basis of Solving Equations Having the Variable on Both Sides

#### Example

The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number?
Let us take the two-digit number such that the digit in the units place is b.
The digit in the tens place differs from b by 3.
Let us take it as b + 3.
So, the two-digit number is 10(b + 3) + b = 10b + 30 + b = 11b + 30.
With interchange of digits, the resulting two-digit number will be 10b + (b + 3) = 11b + 3.
If we add these two two-digit numbers, their sum is (11b + 30) + (11b + 3) = 11b + 11b + 30 + 3 = 22b + 33
It is given that the sum is 143.
Therefore, 22b + 33 = 143.
or 22b = 143 - 33
or 22b = 110
or b = 110/22
or b = 5
The units digit is 5 and therefore the tens digit is 5 + 3 which is 8. The number is 85.

#### Example

Arjun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.

Let us take Shriya’s present age to be x years.
Then Arjun’s present age would be 2x years.
Shriya’s age five years ago was (x – 5) years.
Arjun’s age five years ago was (2x – 5) years.
It is given that Arjun’s age five years ago was three times Shriya’s age.
Thus,           2x – 5 = 3(x – 5)
or                2x – 5 = 3x – 15
or                15 – 5 = 3x – 2x
or                  10 = x
So, Shriya’s present age = x = 10 years.
Therefore, Arjun’s present age = 2x = 2 × 10 = 20 years.

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Problems on Linear Equations in One Variable [00:12:52]
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