Why did they had "Monetary Pang of Guilt"?

Let the original number of mints be $n$ . Then, the number of mints left after:

• Sita took and returned, $n_s = n - \frac 13 n + 4 = \frac 23 n + 4$
• Fatima took and returned, $n_f = n_s - \frac 14 n_s + 3 = \frac 34 n_s + 3$
• Eswari took and returned, $n_e = n_f - \frac 12 n_f + 3 = \frac 12 n_f + 2$

And after Eswari took and returned there were 17 mints remaining. Therefore, we have:

$\begin{aligned} n_e & = 17 \\ \frac 12 n_f + 2 & = 17 \\ \implies n_f & = 30 \\ \frac 34 n_s + 3 & = 30 \\ \implies n_s & = 36 \\ \frac 23 n + 4 & = 36 \\ \implies n & = \boxed{48} \end{aligned}$

Let us assume that the total number of mints present in the bowl be x . No. of mints taken by Sita (in terms ofx) =1/3 x-4 =x/3-4 =(x-12)/3 Balance of mints in the bowl after the first raid from Sita =x-((x-12)/3) =(3x-(x-12))/3 =(2x+12)/3 No. of mints taken by Fatima (in terms ofx) =1/4 [(2x+12)/3]-3 =1/4×(2(x+6))/3-3 =(x+6)/6-3 =((x+6)-18)/6 =(x-12)/6 Balance of mints in the bowl after the second raid from Fatima =[(2x+12)/3]-[(x-12)/6]
=(4x+24-x+12)/6
=(3x+36)/6 =(3(x+12))/(3(2))
=(x+12)/2
No. of mints taken by Eshwari (in terms ofx) =1/2 [(x+12)/2]-2 =(x+12)/4-2 =(x+4)/4 Balance of mints in the bowl after the last raid from Eshwari =(x+12)/2-(x+4)/4 =(2x+24-x-4)/4 =(x+20)/4 According to the question, 17 mints were left in the bowl after the raid was over. ∴(x+20)/4=17 x+20=68 x=48

The initial number of mints must be a multiple of 3, so 48 is the only possible answer from all the options given.