In a three digit number, the digit in the unit's place is four times the digit in the hundred's place. If digit in the unit's place and ten's place are interchanged, the new number so formed is 18 more than the original number. If the digit in the hundreds place is one-third of the digit in the ten's place, what is 25% of the original number?

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Solution

Let the hundreds digit be A, the tens digit be B and units digit be C.
Then the number is 100A + 10B + C

The units place digit is four times the digit in the hundreds place. ie C=4A

The units and tens place digits are interchanged.
Then the hundreds digit is A, the ten's digit is C and the units digit is B
The new number is 100A + 10C + B

Since the new number = original number + 18 (since it's 18 more than the original)

New number - Original number = 18
(100A + 10C + B)- (100A + 10B + C) = 18

=> 9C - 9B= 18
Dividing both sides by 9, we get
=> C - B = 2
Now, substitute C=4A and B=3A, we get
4A-3A=2
=> A=2
Substituting B=3A=> 32=6 => B=6
and C=4A=> 42=8 => C=8
Hence the original number is 268
The 25% of the 268
= (25/100) * 268
=(1/4)*268
= 67
Answer = 67

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