The correct option is C 375
Let the numbers in the 100's place be x, 10's place be y and1's place be z.
H T O
x y z
So,
100x+10y+z=25(x+y+z)
As, the number is 25 times the sum of the digits of the number.
100x+10y+z=25x+25y+25z
75x−15y−24z=0
25x−5y−8z=0....(1)
Again,
100x + 10y + z + 198 = 100z + 10y + x
As, the number will reverse if 198 is added to it
99x−997=−198
x−z=−2.............(2)
Now,
x+z=y+1
As, the sum of extreme two digits is one more than the middle term
x−y+z=1.......(3)
⇒eqn(1)−5×eqn(3)
25x−5y−8z=0
−5x+5y−5z=−5
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20x−13z=−5.....................(4)
⇒eqn(2)×13−eqn(4)
13x−13z=−26
−20x+13z=+5
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−7x=−21
x=3.....................(5)
Now, substituting eqn(5) in eqn(2), we get,
x−z=−2
3−z=−2
z=5.....................(6)
Again, substituting eqn(5) and eqn(6) and z in eqn(3), we get
x−y+z=1
3−y+5=1
−y=−7
y=7
Hence the number is 375.
Check: 375 + 198 = 573 (digits are reversed)