Question
The sum of (p/q)+(q/r)+(r/s)+(s/t) = (239/60). The value
of βqβ is one less than the value of βsβ. The average of βpβ and βsβ is 4. The value of βrβ is 3 less than the value of βsβ. If the value of βrβ is the smallest prime number, then find out the value of (p+q+s+t)/4.Solution
The value of βqβ is one less than the value of βsβ.
q = (s-1)Β Β Eq.(i)
The average of βpβ and βsβ is 4.
p+s = 4x2 = 8Β Β Eq.(ii)
The value of βrβ is 3 less than the value of βsβ.
r = (s-3)Β Β Eq.(iii)
If the value of βrβ is the smallest prime number.
So r = 2 .
Put the value of βrβ in Eq.(iii).
2 = (s-3)
s = 3+2
s = 5
Put the value of βsβ Eq.(ii).
p+5 = 8
p = 8-5
p = 3
Put the value of βsβ Eq.(i).
q = (5-1)
q = 4
The sum of (p/q)+(q/r)+(r/s)+(s/t) = (239/60).
Put the value of βpβ, βqβ, βrβ and βsβ 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
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