Question

the digit at tens place of two digit number is two times the digit at one places. When 27 is subtracted from the number,its digit are reversed. Find the number.​

Answer 1

Answer:

36

Step-by-step explanation:

Correct option is

A

36

Let unit's digit =x and ten's digit =y

∴ Number =10y+x

Reverse number =10x+y

According to the question

2y=x...(i)

10y+x+27=10x+y ...(ii)

⇒10y+2y+27=20y+y

12y−21y=−27

−9y=−27⇒y=3

x=6

∴Number=36

Answer 2

Step-by-step explanation:

Let the number at tens place be x and the number at ones place be y. The number is 10x+y. and interchanging to the one place of x and ten place of y

i. e x+10y

According to the question,

Condition I,

The digit at tens place of a two digit number is two times the digit at ones place.

or,x=2y - (i)

Condition II,

When 27 is subtracted from the number, its digit are reversed.

or,10x+y−27=10y+x

or,10x−x−27=10y−y

or,9x−27=9y - (ii)

Put the value of x from equation (i) in equation (ii), we get,

or,9(2y)−27=9y

or,18y−9y=27

or,9y=27

or,9×y=9×3

∴y=3

Put value do y in equation (i), we get,

or,x=2y

or,x=2×3

∴x=6

So, (x,y) = (6,3)

10x = 10×6 = 60

10x +y = 60+3 = 63

Therefore, the required two digit number is 63.