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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.
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Answers & Comments
Hope it helps you
Please mark me as the brainliest
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.