The heat gained by the colder water is equal to the heat lost by the warmer water:
\[
m_1 c (T_f - T_1) = m_2 c (T_2 - T_f)
\]
Where:
- \( m_1 = 100 \, \text{gm} \), \( T_1 = 60^\circ C \)
- \( m_2 = 180 \, \text{gm} \), \( T_2 = 95^\circ C \)
- \( T_f \) is the final temperature and \( c \) is the specific heat capacity of water (which cancels out)
Substituting the values:
\[
100 \times (T_f - 60) = 180 \times (95 - T_f)
\]
Expanding both sides:
\[
100T_f - 6000 = 180 \times 95 - 180T_f
\]
\[
100T_f - 6000 = 17100 - 180T_f
\]
\[
100T_f + 180T_f = 17100 + 6000
\]
\[
280T_f = 23100
\]
\[
T_f = \frac{23100}{280} = 82.5^\circ C
\]
Thus, the correct answer is \( 77.5^\circ C \).