The sum of (p/q)+(q/r)+(r/s)+(s/t) = (239/60). The value of ‘q’ is one less than the value of ‘s’. The average of ‘p’ and ‘s’ is 4. The value of ‘r’ is 3 less than the value of ‘s’. If the value of ‘r’ is the smallest prime number, then find out the value of (p+q+s+t)/4.

Solution

The value of ‘q’ is one less than the value of ‘s’.

q = (s-1)    Eq.(i)

The average of ‘p’ and ‘s’ is 4.

p+s = 4x2 = 8    Eq.(ii)

The value of ‘r’ is 3 less than the value of ‘s’.

r = (s-3)    Eq.(iii)

If the value of ‘r’ is the smallest prime number.

So r = 2 .

Put the value of ‘r’ in Eq.(iii).

2 = (s-3)

s = 3+2

s = 5

Put the value of ‘s’ Eq.(ii).

p+5 = 8

p = 8-5

p = 3

Put the value of ‘s’ Eq.(i).

q = (5-1)

q = 4

The sum of (p/q)+(q/r)+(r/s)+(s/t) = (239/60).

Put the value of ‘p‘, ‘q‘, ‘r‘ and ‘s‘ in the above equation.

(3/4)+(4/2)+(2/5)+(5/t) = (239/60)

(63/20)+(5/t) = (239/60)

(5/t) = (239/60)-(63/20)

(5/t) = (239-189)/60

(5/t) = (50/60)

t = 6

Value of (p+q+s+t)/4 = (3+4+5+6)/4

= 18/4

= 4.5