Question

In a two digit number, the digit in the unit's place is equal to the square of the digit in ten's place and the difference between the numbers obtained by interchanging the digits is 54. What is 40% of the original number?

• A. 23.4
• B. 15.6
• C. 37.2
• D. 39

Hint:

For any two-digit number, the difference between the number formed by reversing its digits and the original number is always a multiple of 9: \[ (10y+x)-(10x+y)=9(y-x) \] Use this shortcut to solve digit-reversal problems quickly.

Answer 1

The Correct Option is B

Step 1: Form the equations Given that the unit's digit is the square of the ten's digit: \[ y=x^2 \] Also, the difference between the interchanged number and the original number is 54: \[ (10y+x)-(10x+y)=54 \] \[ 9(y-x)=54 \] \[ y-x=6 \]

Step 2: Solve for the digits Substituting \(y=x^2\): \[ x^2-x=6 \] \[ x^2-x-6=0 \] \[ (x-3)(x+2)=0 \] Since a digit cannot be negative, \[ x=3 \] Therefore, \[ y=3^2=9 \] Hence, the original number is: \[ 39 \]

Step 3: Find 40% of the number \[ 40\% \text{ of } 39 =\frac{40}{100}\times39 \] \[ =15.6 \] Therefore, the required value is: \[ {15.6} \]