Question

The ones place digit of a two digit number is twice the tens place digit. The number obtained by reversing the order of digits increases by 18 than the original number. Find the number.​

Answer 1

Answer:

Let the one's digit be y and tens digit be x,

Number = 10x + y

Then,x=3y⋯(i)  

Reversed number = 10y + x  

A.t.Q :- (10x+y)−(10y+x)=36 Put x = 3y in eq. (i)

⇒9x−9y=36

⇒x−y=4⋯(ii)

⇒3y−y=4

∴2y=4  x=3y  ∴x=6

y=2

∴ Number = 62

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Step-by-step explanation:

Answer 2

Answer:

24

Step-by-step explanation:

Let the ones place digit be: y

& the tens place digit be: x

∴ The number= 10x + y

& the number obtained by reversing the digits= 10y + x

Now,

(10x + y) = (10y + x) + 18 ....(given)

10x - x = 10y - y + 18

y = 2x ...(given).... (eqn 1)

∴ 10x + 2x is the number

∴ The number is 12x

Now, the reverse number= 10y + x = 20x + x = 21x

Now,

12x + 18 = 21x ...(given)

21x - 12x = 18

9x = 18

x =  $\frac{18}{9}$

∴ x = 2

Now,

The number = 12x = 12(2) = 24

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