The digits of a two-digit number differ by 3. If digits are interchanged and the resulting number is added to the original number, we get 121. Find the original number.
Answers
Answer:
74
Step-by-step explanation:
Let the ten's place of the number be x and unit's place of the number be y and x>y.
According to the question,
x - y = 3 ....(1)
Original number = x(10) + y(1) = 10x + y
Reversed number = y(10) + x(1) = 10y + x
According to the second part of the question,
10x + y + 10y + x = 121
11x + 11y = 121
11[x + y] = 121
x + y = 11 .....(2)
If we solve the (1) and (2) eqns, we will get,
x = 7 and y = 4
The original number is 7(10) + 4(1) = 70 + 4 = 74 [answer]
Hope this helps you!!!
Let us take the two digit number such that the digit in the units place is x. The digit
in the tens place differs from x by 3. Let us take it as x + 3. So the two-digit number is
10 (x + 3) + x = 10x + 30 + x = 11x + 30.
With interchange of digits, the resulting two-digit number will be 10x + (x + 3) = 11x + 3
If we add these two two-digit numbers, their sum is
(11x + 30) + (11x + 3) = 11x + 11x + 30 + 3 = 22x + 33
It is given that the sum is 121.
Therefore, 22x + 33 = 121
22x = 121 – 33
22x = 88
x=4
The units digit is 4 and therefore the tens digit is 4 + 3 = 7.
Hence, the number is 74 or 47.