Question

16. If the position of the digits of a
two digit number are inter-
changed, the number obtained is
smaller than the original number
by 27. If the digits of the num-
ber are in the ratio of 1:2, what
is the original number?
(1) 36
(2) 63
(3) 48
(4) Cannot be determined
(5) None of these
​

Answer 1

let the no be 10x+y

interchanged no.10y+x

difference : 10x+y-(10y+x)

9x+9y

9(x+y)

a/q,

9(x+y)=27

x+y=3

original no.=10x+y

36

Answer 2

Answer:

Option(2)

Step-by-step explanation:

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

The number thus formed = 10x + y.

Number formed by reversing the digits = 10y + x.

According to the given condition,

⇒ (10x + y) - (10y + x) = 27

⇒ 10x + y - 10y - x = 27

⇒ 9x - 9y = 27

⇒ x - y = 3

Given that, digits of the number are in the ratio of 1:2

⇒ (y/x) = (1/2)

⇒ 2y = x

⇒ x - 2y = 0

On solving both the equations, we get

x - y = 3

x - 2y = 0

----------------

    y = 3

Substitute y = 3 in (i), we get

⇒ x - y = 3

⇒ x - 3 = 3

⇒ x = 6

∴ Original number = 10(6) + 3 = 63.

Hope it helps!