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Question

If the digits of a three digit number are reserved, then the number so obtained is less than the original number by 297. If the sum of the digits of the number is 8 and its hundred's digit has the largest possible value, then the ten's digit of the number is

A
3
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B
2
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C
1
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D
0
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Solution

The correct option is C 1
Let the number be x+10y+100z
According to the question,
Upon reversing, the number obtained =z+10y+100x
x+10y+100zz10y100x=297
99(zx)=297
z=x+3

Also,
x+y+z=8
Possible cases :
(I)x=1,z=4,y=3
(II)x=2,z=5,y=1

Given that Hundredth's place is maximum possible, therefore z=5
Ten's digit =1

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