SSC MTS 4 Feb 2026 Shift 3 Paper

Concept:The problem uses the concept of mixture and alligation, where the honey concentration changes when two mixtures are combined.Explanation:First, find the honey concentration in each jar. J1 has ratio $3:1$, so honey fraction = $\frac{3}{4}=75\%$. J2 has ratio $1:3$, so honey fraction = $\frac{1}{4}=25\%$. After transfer, J2's honey fraction becomes $\frac{1}{2}=50\%$.Let the amount transferred from J1 be $x$ ml. Initially, J2 contains $80$ ml with $20$ ml honey and $60$ ml water. The transferred $x$ ml from J1 contains $\frac{3}{4}x$ honey and $\frac{1}{4}x$ water.After transfer, total honey in J2 = $20 + \frac{3}{4}x$, total water = $60 + \frac{1}{4}x$. For the final ratio $1:1$, these must be equal:$20 + \frac{3}{4}x = 60 + \frac{1}{4}x$$\Rightarrow \frac{1}{2}x = 40$$\Rightarrow x = 80$ ml.Answer:The volume of mixture transferred from J1 to J2 is $80$ ml (Option C).