Question

The
solve the following:
sum of the digits of a number consisting of
three digits is 12 The middle digit is equal to half of
the sum of the other two. If the order of digits is
reversed the number is diminished by 198. Find the number​

Answer 1

Answer:

Let the last and the first digit be x and y respectively

Therefore, the middle term will be 1/2(x+y)

A/Q

x+1/2(x+y)+y=12

x+x/2+y/2+y=12

3x/2+3y/2=12

3(x+y)/2=12

3(x+y)=24

x+y=8 ......(i)

Again,

A/Q

100y+10{1/2(x+y)}+x=100x+ 10{1/2(x+y)}+y-198

100y+10{1/2(8)}+x=100x+ 10{1/2(8)}+y-198

100y+40+x=100x+40+y-198

99y+198=99x

99(y+2)=99x

y+2=x

y-x= -2

x-y=2 .......(ii)

Now,

(i)-(ii)

(x+y) - (x-y) =8-2

x+y-x+y=6

2y=6

y=3 ......(iii)

Putting (iii) in (i),we have

x+3=8

x=8-3

x=5

Therefore, the number is

100x+10{1/2(x+y)}+y

=(5*100)+10{1/2(8)}+3 from (i)

=500+40+3

=543

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