The sum of the digits of a 2-digit number is 8. The number obtained by
interchanging the digits exceeds the given number by 18. Find the given numbers.
The ones digit of a 2-digit number is twice the tens Jigit. When the number
formed by reversing the digits is added to the original number, the sum is 99. Find
the original number.
Answers
Answer:
Let the number be in the form of 10x+y, where x and y are the tens digit and the units digit respectively. Applying the first conditions, we get
x+y=8 (1)
Applying the second condition, given in the problem,
10x+y+18=10y+x
⇒9x−9y=−18
⇒x−y=−2 (2)
By solving Equation. (1) and (2), we get
x=3 and y=5 and the number is 35
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Answer:
the number is 35
Step-by-step explanation:
The sum of the digits of a 2 digit number is 8.
Let the digits at 1s place be x. Then, the digits at 10s place is (8 - x)
Therefore original number = 10 (8 - x) + x
80 - 10x + x
80 - 9x
On interchanging the digits from 1s to 10s and vice versa
New number obtained = 10x + 8 - x
= 9x + 8
According to the statement, the number obtained by interchanging the digits exceed the original number by 18. The number to find out.
New number - Original number = 18
9x + 8 - (80 - 9x) = 18
9x + 8 - 80 + 9x = 18
18x - 72 = 18
18x = 18 + 72
18x = 90
x = 90 / 18
x = 5
Hence, the digit at 1s place is 5.
Now, digits at 10s place = (8 - 5) = 3.
Thus, the original number is 35 and the new number after interchanging the places of digits is 53.
Answer the original number is 35 and the new number is 53