If x = 5t and $y = \frac{1}{3}( t +1)$, then the value of t for x = 3y is :

$\frac{1}{3}$ $\frac{1}{2}$ $\frac{1}{4}$ $\frac{1}{5}$

$\frac{1}{4}$

X = 5t $y = \frac{1}{3}( t +1)$ X = 3Y  We have, X = 3Y      ... (A)  So, put the value of X from equation (A) in X  = 5t  = 3Y = 5t      ... (B) Now, put the value of $y = \frac{1}{3}( t +1)$ in equation (B) we have, = 3 × \(\frac{1}{3}\) × ( t+1) = 5t = t +1 = 5t  = 4t = 1  = t = \(\frac{1}{4}\)