In a two-digit positive 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 number and the number obtained by interchanging the digits is 54. What is 40% of the original number ?
A. 15.6
B. 24
C. 37.2
D. 39
E. None of these
Answer: Option A
Solution(By Examveda Team)
Let ten's digit = xThen, unit's digit = x2
Then, number = 10x + x2
Clearly, since x2 > x,
So, the number formed by interchanging the digits is greater than the original number.
$$\eqalign{ & \therefore \left( {10{x^2} + x} \right) - \left( {10x + {x^2}} \right) = 54 \cr & \Leftrightarrow 9{x^2} - 9x = 54 \cr & \Leftrightarrow {x^2} - x = 6 \cr & \Leftrightarrow {x^2} - x - 6 = 0 \cr & \Leftrightarrow {x^2} - 3x + 2x - 6 = 0 \cr & \Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0 \cr & \Leftrightarrow x = 3 \cr} $$
So. ten's digit = 3, unit's digit = 32 = 9
∴ Original number = 39
Required result :
= 40% of 39
= 15.6
Related Questions on Problems on Numbers
If one-third of one-fourth of a number is 15, then three-tenth of that number is:
A. 35
B. 36
C. 45
D. 54
E. None of these
A. 9
B. 11
C. 13
D. 15
E. None of these
A. 3
B. 4
C. 9
D. Cannot be determined
E. None of these
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