the sum of the digits of a number consisting of threedigits is 12. themiddle digit is equal to half the sum of the other two. if the order of the digits is reversed , then the number is diminished by 198. find the number.
Answers
Step-by-step explanation:
Sum of digits = 12
Let unit digit be x ,
tens digit be y &
hundred digit be z
A.T.Q x+z÷2 = y
original number = reversed number +198
100z + 10y + x = 100x + 10y + z +198
(x-100x) + (10y-10y) +(100z-z) = 198
99z - 99x = 198
99(z-x) = 99(2)
z-x = 2
Now,
x+y+z = 12
x+z = 12- y
x+z÷2 = y (above)
12-y÷2 = y
12 -y = 2y
12 = 2y +y
y = 12÷3
y = 4
x+z÷2=y (above)
(put value of y)
x+z = 2×4
x+z = 8
x = 8-z
z-x = 2 ( solved above)
(put value of x )
z-(8-z) = 2
z-8+z = 2
2z = 2+8
z = 10÷2
z = 5
Therefore, x = 8-z
= 8-5
x = 3
y = 4
z = 5
original number = 543
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