A=R[(1+i)^n-1]/i(1+i)^n
Distance travelled by boat in upstream=24km
Time taken = 6 h
Speed of the boat in upstream = 24/6 = 4 km/h
And distance travelled by boat in downstream = 20 km
Time taken = 4h Speed of the boat in downstream = 20/4 km/h = 5 km/h
Now, speed of the boat in still water = 1/2 [ speed of the boat in upstream + speed of the boat in downstream] =1/2 [4 + 5] = 1/2 × 9 = 4.5 km/h
And speed of the current = 1?2 [speed of the boat in downstream ? speed of the boat in upstream] = 1/2 [5 ? 4] = 1/2 × 1 = 0.5 km/h
R=S[i/(1+i)^n-1]
The third proportional of two numbers p and q is defined to be that number r such that
p : q = q : r.
Here, required third proportional of 10 & 20, and let it be 'a'
=> 10 : 20 = 20 : a
10a = 20 x 20
=> a = 40
Hence, third proportional of 10 & 20 is 40.
Downstream speed = 8/X kmph
upstream speed = 4/X kmph
Now 80/(8/X) + 80/(4/X)=20
X=2/3
downstream Speed = 8/(2/3)= 12 kmph
upstream Speed = 4/(2/3)= 6 kmph
Rate of the stream = (12+6)/2= 9 kmph
R=S[i/(1+i)^n-1]
let speed of boat= X, speed of stream= Y
Upstream speed= X-Y
Downstream speed= X+Y
Sum of upstream & downstream= (X-Y) +(X+Y)= 2X
So, 2X= 40
X= 20 km/hr
Speed of boat : speed of stream= 600+100 :100= 7:1
So speed of Stream= 20/7 km/hr
ATQ, D/( X-Y) + D/( X+Y) = 5
D/(120/7) + D/(160/7)= 5
D= 480×5/49= 48.97 km= 50 Km(approx)
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