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)
Given that, 30 % of A = 20 % of B
? A/B = 20/30 = 2/3
? A : B = 2 : 3
Let initial quantity be Q, and final quantity be F
F = Q(1 - 8/Q)
=> Q = 20
? log5[(x2 + x ) / x] = 2
? log10(x + 1) = 2
? x + 1 = 25
? x = 24
∴ Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
We know that speed is inversely proportional to time.
Given that, (Speed of A ) : (speed of B ) = 2 : 7
?(Time taken by A ) : (Time taken by B ) = 1/2 : 1/7 = 7 : 2
Let th distance = D
and usual speed = V
According to the question,
D/(3V/4) - D/V = 2
? D/V = 3 x 2 = 6
Time taken to cover the distance with usual speed = 6 h
?2n = 64
? 2n/2 = 26
? n/2 = 6
? n = 12
Distance = 160 km
Relative Speed = 8 + 2 = 10
Time = Distance/Relative speed = 160/10 = 16 h
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.