Question

10. The digit at ten's place of a two-digit number is three
times the digit at one's place. If the sum of this
number and the number formed by reversing its
digits is 88, find the number.
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Answer 1

Answer:

the digit in unit's place= 2

the digit in ten's place = 6

the original number= 62

Explanation:

let the digit in one's place be x.

the digit in ten's place= 3x

As we know that two-digit number can be represented by 10 × a + b

where a = number at ten's place and b = number at unit's place

For e.g. 24 = 10 × 2 + 4

∴the original number= 10(3x) + x  

when the digits are interchanged  

Interchange of digits means replacing the unit digit and ten's place with each other

∴the new number = 10x + 3x

a/q   10x+3x + (10(3x)+x) = 88

      ⇒ 13x+30x+x = 88

      ⇒ 44x = 88

      ⇒ x =  88/44

      ∴ x= 2

the 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.

Answer 2

Answer:

kindly see the answer below ...

Explanation:

Here,

according to the question..

let the no. be y...

$y = (3x) \times 10 + x$

then,

no. formed by reversing the digit is

${y}^{)} = x \times 10 + (3x)$

Now..

$y + {y}^{)} = 88$

$(3x) \times 10 + x + x \times 10 + 3x = 88$

$4x \times 10 + 4x = 88$

$44x = 88$

$x = 2$

so,

the required no. is

$y = 3x \times 10 + x$

$= 3 \times 2 \times 10 + 2$

$= 60 + 2$

$= 62$

then, the required answer is 62...

thanks a lot...