Question

sum of the digits of a two-digit number is 5 when the digits are reversed the resulting number is greater than the original number by 27 find the number​

Answer 1

S O L U T I O N :

Let the ten's place digit be x & one's place digit be y respectively.

$\boxed{\bf{Original\:number = 10x + y}}$

$\boxed{\bf{Reversed\:number = 10y + x}}$

A/q

$\underbrace{\sf{1^{st}\:Case\::}}$

$\longrightarrow\tt{x + y = 5}$

$\longrightarrow\tt{ y = 5 - x..............(1)}$

$\underbrace{\sf{2^{nd}\:Case\::}}$

$\longrightarrow\tt{10y + x = 10x + y + 27}$

$\longrightarrow\tt{10y - y + x - 10x = 27}$

$\longrightarrow\tt{9y-9x = 27}$

$\longrightarrow\tt{9(y-x) = 27}$

$\longrightarrow\tt{(y-x) = \cancel{27/9}}$

$\longrightarrow\tt{y-x = 3}$

$\longrightarrow\tt{5-x-x = -3\:\:[from(1)]}$

$\longrightarrow\tt{5-2x = 3}$

$\longrightarrow\tt{-2x = 3 - 5}$

$\longrightarrow\tt{-2x = -2}$

$\longrightarrow\tt{x = \cancel{-2/-2}}$

$\longrightarrow\bf{x = 1}$

∴ Putting the value of x in equation (1),we get;

$\longrightarrow\tt{y = 5-1}$

$\longrightarrow\bf{y = 4}$

Thus,

→ The number = 10x + y

→ The number = 10(1) + 4

→ The number = 10 + 4

→ The number = 14 .

Answer 2

Answer:-

Let the digit in the unit place be 'y' and 'x' in the tens digit place .

Hence,The original number is 10x+y......(i)

Case 1 :-

x+y=5

y=5-x....(ii)

Case 2 :-

$\blue{10y + x = 10x + y + 27}$

$\blue{10y - y = 10x - x + 27}$

$\blue{9y = 9x + 27}$

Dividing throughout by 9 we get :-

$\red{y = x + 27}$

$\red{5 - x = x + 27.....(from \: it)}$

$\red{5 - 27 = x + x}$

$\red{2x = - 22}$

$\red{x = \frac{ - 22}{2}}$

$\red{x = - 11}$

Substituting x=-11 in.... (ii)

$\green{y = 5 - ( - 11)}$

$\green{y = 5 + 11}$

$\green{y = 16}$

$\purple{now \: subsituting \: x = - 11......(1)}$

$\purple{and \: y = 16......(1)}$

Original number⤵️⤵️⤵️⤵️⤵️

$\pink{10( - 11) + 16 \: }$

$\pink{ - 110 + 16}$

$\pink{ = - 94}$

Hence,The number is -94

Hope this helps you ✌✌