Correct option is (A) 81
Let shares of A, B, C and D are a, b, c & d respectively.
According to given conditions,
3b = a+c+d __________(1)
4c = a+b+d __________(2)
5d = a+b+c __________(3)
Subtract equation (1) from (2), we get
4c - 3b = (a+b+d) - (a+c+d)
4c - 3b = b - c
4c + c = 3b + b
\(\Rightarrow\) 5c = 4b
\(\Rightarrow\) c = \(\frac45b\) __________(4)
Subtract equation (1) from (3), we get
5d - 3b = b - d
\(\Rightarrow\) 5d + d = b + 3b
\(\Rightarrow\) 6d = 4b
\(\Rightarrow\) d = \(\frac46b=\frac23b\) __________(5)
Subtract equation (2) from (3), we get
5d - 4c = c - d
\(\Rightarrow\) 5d + d = c + 4c
\(\Rightarrow\) 6d = 5c
\(\Rightarrow\) d = \(\frac56c\) __________(6)
From (1), (4) & (5), we get
\(3b=a+\frac45b+\frac23b\)
\(\Rightarrow\) \(a=3b-\frac23b-\frac45b\)
\(=\frac{45b-10b-12b}{15}\)
\(=\frac{23}{15}b\) __________(7)
From (4), (5), (6) & (7), we conclude that
c < b, d < b, d < c, a > b
\(\Rightarrow\) d < c < b < a
\(\therefore\) largest share = a
smallest share = d
\(\therefore\) a+d = 99 __________(8) \((\because\) Sum of largest and smallest share is 99 (given))
\(\Rightarrow\frac{23}{15}b+\frac23b=99\) (From (7) & (5))
\(\Rightarrow\frac{23b+10b}{15}=99\)
\(\Rightarrow\) 33b = 99 \(\times\) 15
\(\Rightarrow b=\frac{99\times15}{33}=3\times15=45\)
\(\therefore c=\frac45b\) (From (4))
\(=\frac45\times45\)
= 4 \(\times\) 9 = 36
\(\therefore\) b+c = 45+36 = 81
\(\therefore\) Sum of the other two shares = 81
