Question

5. The sum of the digits of a two digit number is 9. The digit in the ones place is 1 more than
3 times the digit in the tens place. Find the numbers.

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Answer 1

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Q - The sum of the digits of a two digit number is 9 .

The digit in the ones place is 1 more than 3 times the digit in the tens place . Find the number .

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Let the digit at ones place be “x”

&

the digit at tens place be “y”

According to the Question ,

$\rm \: ones \: place \: digit \: + tens \: place \: digit = 9$

$\mapsto \rm \: x + y = 9$

$\mapsto \rm \: x = 9 - y.....(i)$

&

$\rm \: ones \: place \: digit = (3 \times tens \: place \: digit) + 1$

$\mapsto \rm \: x = 3y + 1$

put the value of x from equation (i)

$\mapsto \rm \: 9 - y = 3y + 1$

$\mapsto \rm \: 9 - 1 = 3y + y$

$\mapsto \rm \: \cancel4y = \cancel8$

$\mapsto \boxed{ \rm \: y = 2}$

Now , put the value of y in eq(i)

$\mapsto \rm \: x = 9 - y$

$\mapsto \rm \: x = 9 - 2$

$\mapsto \boxed{ \rm \: x = 7}$

So , the final number would be

$\underline{ \: \: } \: \underline{ \: \: }$

in these blanks we have to fill the tens digit and ones digit

$\underline{ \rm \: y} \: \underline{ \rm \: x}$

$\implies \boxed{ \rm number \rightarrow\: 27}$

So ,

The number would be 27 .

Answer 2

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