The ten’s digit of a 3 digit number is 3. If the number is reversed, the number thus formed will be 396 more than the original number. If the sum of the digits at unit’s place and hundred’s place is 14, then the original number is?

953

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539

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935

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339

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None of the above

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Solution

The correct option is B 539
Let the original number be x3y = 100x + 30 +y
When reversed y3x = 100y + 30 + x
We know, y3x - x3y = 99 (y - x) = 396 $\Rightarrow$ y - x = 4
Also, we know y + x = 14
solving we get y = 9 and x = 5
so, the original number x3y = 539

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The ten’s digit of a 3 digit number is 3. If the number is reversed, the number thus formed will be 396 more than the original number. If the sum of the digits at unit’s place and hundred’s place is 14, then the original number is?

The correct option is B 539Let the original number be x3y = 100x + 30 +y When reversed y3x = 100y + 30 + x We know, y3x - x3y = 99 (y - x) = 396 ⇒ y - x = 4 Also, we know y + x = 14 solving we get y = 9 and x = 5 so, the original number x3y = 539

The correct option is B 539
Let the original number be x3y = 100x + 30 +y
When reversed y3x = 100y + 30 + x
We know, y3x - x3y = 99 (y - x) = 396 $\Rightarrow$ y - x = 4
Also, we know y + x = 14
solving we get y = 9 and x = 5
so, the original number x3y = 539