B.D. for (3/2) years = Rs. 60
B.D. for 2 years = Rs. (60 x 2/3 x 2) = Rs. 80
Now, B.D. = Rs. 80; T.D. = Rs. 75
and Time = 2 years
? Sum = Rs. (80 x 75 / 5) = Rs. 1200
? Rs. 80 is S.I. on Rs. 1200 for 2 years.
So, rate = (100 x 80/1200 x 2)% = 31/3%B.D. for (3 / 2) years = Rs. 60
B.D. for 2 years = Rs. (60 x 2 / 3 x 2) = Rs. 80
Now, B.D. = Rs. 80; T.D. = Rs. 75 and Time = 2 years
? Sum = Rs. (80 x 75 / 5) = Rs. 1200
? Rs. 80 is S.I. on Rs. 1200 for 2 years.
So, rate = (100 x 80 / 1200 x 2)% = 31/3%
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
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