A) 15.6 |
B) 73 |
C) 84 |
D) None of the above |
A) 15.6 |
Let ten's digit be x and unit's digit be x2.
Original number = \( \Large 10x + x^{2} \)
New number = \( \Large 10x^{2} + x \)
According to the question,
=\( \Large 10x^{2} + x - 10x - x^{2} = 54 \)
= \( \Large 9x^{2} - 9x = 54 \)
= \( \Large 9 \left(x^{2} - x\right) = 54 \)
= \( \Large x^{2} - x - 6 = 0 \)
= \( \Large x^{2} - 3x + 2x - 6 = 0 \)
= \( \Large x \left(x - 3\right) + 2 \left(x - 3\right) = 0 \)
= \( \Large \left(x - 3\right) \left(x + 2\right) = 0 \)
Therefore, x = 3, -2
Therefore, Ten's digit = x = 3
Unit's digit = \( \Large x^{2} = 3^{2} = 9 \)
Original number = 39
Required number = \( \Large 39 \times \frac{40}{100} = 15.6 \)