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The sum of two - digits number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the numbers.

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42 and 24

41 and 24

43 and 24

40 and 24

answer is A.

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Detailed Solution

It is given that,
The sum of two - digits number and the number obtained by reversing the digits is 66.
Let the unit’s digit be y and the ten’s digit be x.
Then, Number

= 10 x + y

According to question,
The sum of a two-digit number and the result of reversing the digits is 66 so,

10 x + y + x + 10 y = 66 11 x + 11 y = 66 11 x + y = 66 x + y = 6611 x + y = 6 … … 1

Also,
Digits of the number differ by 2 so,

x - y = 2

… (2)
Standard form should be used to write equations (1) and (2) above.

a 1 x + b 1 y + c 1 = 0

and

a 2 x + b 2 y + c 2 = 0.

In standard form, the following equations are written:

x + y - 6 = 0

… (3)

x - y - 2 = 0

… (4)
Equations (3) and (4) in comparison to the Standard form we get,

a 1 = 1, b 1 = 1 a n d c 1 = − 6 a 2 = 1, b 2 = − 1 a n d c 2 = − 2

We have,

a 1 a 2 = 11, b 1 b 2 = 1 − 1, c 1 c 2 = − 6 − 2 = 31 ∵ a 1 a 2 ≠ b 1 b 2 ≠ c 1 c 2

Using cross - multiplication method, we get:
Question Image

Then,

x 1 × − 2 − − 1 × − 6 = y − 6 × 1 − − 2 × 1 = 1 1 × − 1 − 1 × 1

x -2 - 6 = y − 6 + 2 = 1 − 1 − 1

x -8 = y − 4 = 1 -2

I        II          III

On taking I and III terms:

x -8 = 1 -2 x = − 8 − 2 x = 4

On taking II and III terms:

y − 4 = 1 -2 y = − 4 − 2 y = 2

So,
Original number

= 10 x + y

= 10 (4) + 2

= 40 + 2

= 42

On reversing number

= x + 10 y

= 4 + 10 (2)

= 4 + 20

= 24

Therefore, the two-digit number is 42 and 24. There are two such numbers.
Hence, option (1) is correct.

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