Question

the digit at the tens place of a two digit number the three times of a digit at the one place if the sum of a number and the number formed by reversing the digit is 88 find the number​

Answer 1

Answer:

Let the digit in one's place be x.

the digit in ten's place= 3x

the original number= 10(3x) + x (we multiplied by 10 because we have to find the value in ten's place.)

when the digits are interchanged

the number = 10x + 3x

a/q 10x+3x + (10(3x)+x) = 88

= 13x+30x+x = 88

= 44x = 88

= x = 88/44

∴ x= 2

digit in unit's place= 2

digit in ten's place = 3x= 3*2= 6

the original number= 6*10 + 2

= 60+2

= 62

so the required number is 62.

Step-by-step explanation:

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Answer 2

$\large\bf\underline{Given:-}$

• tens place digit is three times of unit place digit.
• sum of number and the number formed by reversing the digits is 88

$\large\bf\underline {To \: find:-}$

• The Number

$\huge\bf\underline{Solution:-}$

Let the tens place digit be x

and units place digit be y.

Then the number = 10x + y.

Reversing the digits then the number formed :-

• 10y + x

According to question:-

sum of a number and the number formed by reversing the digit is 88.

➝ 10x + y + 10y + x = 88

➝ 11x + 11y = 88

Dividing both side by 11, we get,

➝ x + y = 8 .........(i)

we know that tens place digit is 3 times of units place digit that was given in question.

➝ x = 3y

Substituting value of x in eq. (i)

we get,

➝ x + y = 8

➝ 3y + y = 8

➝ 4y = 8

➝ y = 8/4

➝ y = 2

So, units place digit is 2

Tens place digit x = 3y = 3 × 2 = 6

So, the original number 10x + y

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 10 × 6 + 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 60 + 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 62

Original number = 62