Question

a two digit number is 3 more than 4 times the sum of its digits if 18 is added to the number its digits are reversed find the number

Answer 1

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Let the digit in ones place be x and tens place be y
Original number = 10y + x
Number formed by reversing the digits = 10x + y
Given 10y + x = 4(x + y) + 3
⇒ 10y + x − 4x − 4y = 3
⇒ 6y� − 3x� = 3
⇒ 2y� − x� = 1 → (1)
Also given that when 18 is added to the number the digits gets interchanged.
Hence (10y + x) + 18 = 10x + y
⇒ 9x − 9y = 18
⇒ x − y = 2� → (2)
Add (1) and (2), we get
2y� − x� = 1
x − y = 2
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y = 3
Put y=3 in x − y = 2, we get
x − 3 = 2
∴ x = 5
Hence (10y + x) = 10(3) + 5 = 35

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Answer 2

The answer is given below :

Let, the two digits at tens and ones places are a and b respectively.

Then, the number is (10a + b).

When the digits are reversed, the number becomes (10b + a).

By the given condition,

the original number = 4(sum of the digits) + 3

=> 10a + b = 4(a + b) + 3

=> 10a + b = 4a + 4b + 3

=> 6a = 3b + 3

=> 2a = b + 1 .....(i)

and

the original number + 18 = the reversed digits number

=> 10a + b + 18 = 10b + a

=> 9a + 18 = 9b

=> a + 2 = b .....(ii)

Now, from (ii) b = a + 2 is got and putting this in (i), we get

2a = a + 2 + 1

=> a = 3

Putting a = 3 in (ii), we get

b = 3 + 1

=> b = 5

So, the original number is

= 3×10 + 5

= 30 + 5

= 35

and

the reversed digits number is

= 5×10 + 3

= 50 + 3

= 53

Thank you for your question.