The given question can be illustrated using the figure below:
AB and AC are two smooth planes inclined to the horizontal at ∠θ1 and ∠θ2 respectively.
The height of both the planes is the same, therefore, both the stones will reach the bottom with same speed.
As P.E. at O = K.E. at A = K.E. at B
mgh = 1/2 mv12 = 1/2 mv22
∴ v1 = v2
As it is clear from fig. above, acceleration of the two blocks are
a1 = g sin θ1 ,
a2 = g sin θ2
As θ2 > θ1
∴ a2 > a1
From v = u + at
= 0 + at
t = v/a
As t ∝ 1/a, and a2 > a1
∴ t2 < t1
That is, the second stone will take lesser time and reach the bottom earlier than the first stone.