1. Let the specific heats of liquids $A$ and $B$ be $c_A$ and $c_B$ respectively.

2. Use the first mixing condition:

• Mass of $A = 100\,\text{g}$, initial temperature $=100^\circ\text{C}$
• Mass of $B = 50\,\text{g}$, initial temperature $=75^\circ\text{C}$
• Final temperature $=90^\circ\text{C}$

Since heat lost by $A$ = heat gained by $B$,

$100c_A(100-90)=50c_B(90-75)$

$100c_A\cdot 10 = 50c_B\cdot 15$

$1000c_A = 750c_B$

$\frac{c_A}{c_B}=\frac{3}{4}$

3. Now use the second condition:

• Mass of $A = 100\,\text{g}$ at $100^\circ\text{C}$
• Mass of $B = 50\,\text{g}$ at $50^\circ\text{C}$
• Let final temperature be $T$

Again,

$100c_A(100-T)=50c_B(T-50)$

Substitute $c_A=\frac{3}{4}c_B$:

$100\left(\frac{3}{4}c_B\right)(100-T)=50c_B(T-50)$

Cancel $c_B$:

$75(100-T)=50(T-50)$

$7500-75T=50T-2500$

$10000=125T$

$T=80^\circ\text{C}$

4. Check options:

• A: $60^\circ\text{C}$ ❌
• B: $70^\circ\text{C}$ ❌
• C: $85^\circ\text{C}$ ❌
• D: $80^\circ\text{C}$ ✅

Therefore, the correct answer is Option D.