Simple equations STD VI

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

Answer:

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.

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?

Answer:

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.

Check: On interchange of digits the number we get is

58. The sum of 85 and 58 is 143 as given.