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A number when reversed becomes 45% greater than original. By how much percentage is the units place greater than ten s place digit place?
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A number when reversed becomes 45% greater than original. By how much ...
Problem Statement: A number when reversed becomes 45% greater than original. By how much percentage is the units place greater than ten's place digit place?

Solution:

Let's assume the number to be AB (where A is the tens place digit and B is the units place digit).

When the number is reversed, it becomes BA, which is 45% greater than the original number AB.

Mathematically, we can represent this as:

BA = AB + 0.45AB

Simplifying this equation, we get:

BA = 1.45AB

Now, we can substitute BA with 10A + B and AB with 10B + A (since A is in the tens place and B is in the units place).

So, we get:

10A + B = 1.45(10B + A)

Simplifying this equation, we get:

5A = 4B

Therefore, the percentage by which the units place digit (B) is greater than the tens place digit (A) can be calculated as:

Percentage increase = (B - A)/A x 100

Substituting 5A/4 for B, we get:

Percentage increase = (5A/4 - A)/A x 100

Simplifying this equation, we get:

Percentage increase = 25%

Therefore, the units place digit is 25% greater than the tens place digit in the original number.

Conclusion:

The units place digit is 25% greater than the tens place digit in the original number.
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A number when reversed becomes 45% greater than original. By how much ...
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A number when reversed becomes 45% greater than original. By how much percentage is the units place greater than ten s place digit place?
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