If 1 is added to both sides of x – 1 = 18, then we have:
x – 1 + 1 = 18 – 1
x – 1 + 1 = 18 + 1
x – 1 + 1 = 18 – 2
B.
x – 1 + 1 = 18 + 1
The digits of a two digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, then we get 99. Find the original number.
Let the digit at unit place be 'x'.
∴ The digit at the tens place = (x+5)
∴ original number =10(x+5) + x
With interchange of digits, the new number = 10x + (x + 5)
Now, According to the condition, we have
[original number] + [New number] = 99
or [10(x + 5) + x] + [10x + (x + 5)] = 99
or [10x + 50 +x] + [10x + x + 5] = 99
or 11x + 50 + 11x + 5 = 99
or 22x + 55 = 99
Transposing 55 to RHS, we have
22x = 99 - 55 = 44
Dividing both sides by 22, we have
x = 44 22 = 2
∴ x = 2
i.e. Unit place digit = 2
∴ Tens place digit = 2 + 5 = 7
Thus the original number = 72.
If 1 is subtracted from both sides of x + 1 = 5, then we have:
x = 5 – 1
x + 2 4
x = 1 – 6
A.
x = 5 – 1