Question

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The sum of the digits of A two digit number is 5.
If the digits are reversed , the number is reduced by 27.
Find the number.........




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Answer 1

Answer:

arre dekhte jao...xd

let the unit pla'e digit is X and the second place digit is y

therefore the number is=10y+X

condition 1

X+y=5....(I)

condition 2

10x+y=10y+x-27

9x-9y=-27

x-y=-3........(ii)

solving equations I and ii

y=4

X=1

therefore the number is ......41

Answer 2

Answer:

Let the unit place digit be x and ten's place digit be y.

Then, Number = 10y + x

Sum of digits, x + y = 5 --> ( i )

According to the question,

Reversing the digits,

10y + x - ( 10x + y) = 27

10y + x - 10x - y = 27

9y - 9x = 27

9( y - x ) = 27

y - x = 3

y = x + 3

Putting value of y in equation (i),

x + x + 3 = 5

2x = 5 - 3

x = 2 /2 = 1

y = 1 + 3 = 4

Now, Number = 10y + x = 10 × 4 + 1

Number = 40 + 1 = 41

ANSWER : 41