One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
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Verified answer
Answer:
Let the digit in units place =x
The digit in tens place =3x
The number is 10(3x)+x---(1)
Digits are interchanged
Digit in units place =3x
Digit in tens place =x
The new number =10(x)+3x ---(2)
It is given that, sum of these two numbers =88
⇒10(3x)+x+10(x)+3x=88
⇒30x+x+10x+3x=88
⇒44x=88
⇒x=8844
⇒x=2
from (1)
The number =10(3x)+x=10(3×2)+2=60+2=62
Therefore, the required numbers are 62 or 26
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Here is your answer hope it helps
Let the digit in units place be x.
Then, digit in ten's place =3x
So, original number =10(3x)+x=31x
On interchanging the digits, new number =10x+3x=13x
According to the given condition,
31x+13x=88
⇒44x=88
⇒x=2
So, the original number is 31x or 31×2=62.
On the other hand, if we consider the digit in ten's place as x, then the digit in the unit's place will be 3x.
So, original number =10(x)+3x=13x
On interchanging the digits, new number =10(3x)+x=31x
According to the given condition,
13x+31x=88
⇒44x=88
⇒x=2
So, the original number is 13x or 13×2=26.
Here, both answers are correct, as 26+62=88.
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