Question

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 44. WHAT IS THE ORIGINAL NUMBER?

Answer 1

Let the digits are x and y
number will be 10x +y
A/q
x = 3y
now interchanging it, and adding
(10y +x) + (10x +y ) = 44
⇒ 11x + 11y = 44
⇒x +y = 4
⇒3y +y = 4
⇒4y = 4
⇒y = 1
so, x = 3
now number = 10x +y = 10×3 +1 = 31
hence the original number will be 31
also if we take y = 3x
x= 1 and y = 3
then number = 10x +y = 10×1 +3 = 13
then the number will be 13

Answer 2

Answer:

$\huge\Large\underline\mathtt\pink{Step-by-step explanation:-}$

$\huge\Large\underline\mathtt\pink{Given :-}$

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.

$\huge\Large\underline\mathtt\pink{Find :-}$

The Number

$\huge\Large\underline\mathtt\pink{Solution:-}$

Let the digits at tens place be x.

And ones place be 3x .

Original Number = 10x + 3x = 13x

Number after interchanging = 10 × 3x + x = 30x + x = 31x

$\huge\Large\underline\mathtt\pink{According to the Question,:-}$

⇒ Original number + New number = 88

⇒ 13x + 31x = 88

⇒ 44x = 88

⇒ x = 88/44

⇒ x = 2

Original Number = 13x = 13 × 2 = 26

Therefore, by considering the tens place and ones place as 3x and x.

Therefore, by considering the tens place and ones place as 3x and x.The two digit number obtained is 62.

Therefore, by considering the tens place and ones place as 3x and x.The two digit number obtained is 62.Hence, the two-digit number may be 26 or 62.

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