Question

$\huge{\fbox{\fbox{\orange{\mathfrak{Explanation\:Required}}}}}$




$<font color=red>$
Answer please




Answer will be ☑ verified by brainly experts ​

Answer 1

ANSWER:

• The original number is 54

GIVEN:

• A two digit number is six times the sum of its digit.
• when the sum of the digits is added to the original number It results = 63.

SOLUTION:

• Let the digit at tens place be x.
• Let the digit at ones place be y.
• Original number = 10x+y

Now according to question:

$\implies \: 10x + y = 6(x + y) \\ \implies \: 10x + y = 6x + 6y \\ \implies10x - 6x = 6y - y \\ \implies4x = 5y \\ \implies \: x = \frac{5y}{4} \: \: \: \: .....(i)$

Now ;

$\implies \: x + y + 10x + y = 63 \\ \implies11x + 2y = 63 \\ putting \: x = \frac{5y}{4} \: from \: eq(i) \\ \implies \: 11 \times \frac{5y}{4} + 2y = 63 \\ \implies \: \frac{55y + 8y}{4} = 63 \\ \implies \: 63y = 252 \\ \implies \: y = 4 \\ \\ \implies \: x = \frac{5y}{4} \\ \implies \: x = \frac{5 \times 4}{4} \\ \implies \: x = 5$

So the original number = 10*5+4

=50+4

=54

Answer 2

$\huge{\fbox{\fbox{\orange{\mathfrak{Hey\:learner!}}}}}$

$<font color=red>$

........Given.......

⭕A two digit number is six times the sum of its digit.

⭕ If we add the two digits we get 63 as their sum.

⭕Let the first digit be X and second be Y.

⭕Then we get 10x+y.

______________________________

Now.

=> 10x+y =6(x+y)

=> 10x+y = 6x+6y

=> 10x-6x = 6x-y

=> 4x = 5y

=> x = 5y/4. (let it be eqn. 1)

______________________________

Now,

x + y + 10x + y = 63

=> 11x + 2y = 63

Putting x = 5y/4

=> 11×5y/4+2y = 63

=> 5y + 8y/4 = 63

=> 63y = 252

So now we get y as 4.

__________________________________

x = 5y/4 ( as we know)

=> x = 5×4/4

=> x = 5

__________________________________

So,

$\huge{\fbox{\fbox{\blue{\mathfrak{The\:required\:number:-}}}}}$

$<font color=red>$

10×5+4

=50+4

=54