Question

The digit at ten place of a two digit number is 4 times the digit at unit units place if the sum of this number and the number formed by receiving the digit is 55 find the number

Answer 1

Solution :-

Let the units place and tens place of a two digit number be x and y respectively.

Case I : The digit at tens place of a two digit number is 4 times the digit at units place.

=> y = 4x _____(i)

Case II : The sum of this number and the number formed by reversing the digit is 55.

=> 10y + x + 10x + y = 55

=> 11y + 11x = 55

=> 11(y + x) = 55

=> y + x = 5

=> 4x + x = 5 [from equation (i)]

=> 5x = 5

=> x = 1

Substituting the value of x in equation (i) we get,

=> y = 4x

=> y = 4 × 1 = 4

Therefore,

∴ Number = 10y + x = 10 × 4 + 1 = 41

Answer 2

$\bold{Correct\:Question}$

The digit at ten place of a two digit number is 4 times the digit at unit units place if the sum of this number and the number formed by reversing the digit is 55. find the number.

______________________________

$\huge\mathfrak{Answer\:=\:41}$

_______________________________

$\huge\bigstar\underline\mathfrak\red{Explanation}$

Let the digit at one's place = x

The digit at ten's place = 4x ( Given ).

Number = 10×[4x]+ x

= 40x + x

=41 x

________________________

After reversing the digit,

The digit at one's place = 4x

The digit at ten's place = x

Number = 10[x]+4x

= 10x+4x

=14x

________________________

According to question,

→41x+14x=55

→ 55x=55

→x=55/55

→x=1

______________________

Therefore,

The two digit number = 41×1 = 41 ( required answer )