Solution:-

Let us assume the point of intersection of CD and line $5x + 3y = 25$be the point Q(a, b)

From the question it is given that, line $5x + 3y = 25$ divides the line segment CD are in the ratio $1: 3$,

So, $CQ: QD= 1: 3$          

So, Point C become $3(b, 4) = (3b, 12)$

D become $1(5, 8) = (5, 8)$

Then, the coordinates of Q are,

$Q(a, b) = Q[((5 + 3b)/4), ((8 + 12)/4)]$

$= Q[((5 + 3b)/4), 5]$                           

Since Q(a, b) lies on the line $5x + 3y = 25$,

Where, $x = (5 + 3b)/4, y = 5$

$5((5 + 3b)/4) + 3(5) = 25$

$(25 + 15b)/4 + 15 = 25$

$25 + 15b + 60 = 100$

$15b + 85 = 100$

$15b = 100 – 85$

$15b = 15$

$b = 15/15$

$b = 1$

Therefore, value of b is $1$.