Quantity of A in mixture left = | ❨ | 7x - | 7 | x 9 | ❩ | litres = | ❨ | 7x - | 21 | ❩ litres. |
12 | 4 |
Quantity of B in mixture left = | ❨ | 5x - | 5 | x 9 | ❩ | litres = | ❨ | 5x - | 15 | ❩ litres. |
12 | 4 |
∴ |
|
= | 7 | |||||
|
9 |
⟹ | 28x - 21 | = | 7 |
20x + 21 | 9 |
⟹ 252x - 189 = 140x + 147
⟹ 112x = 336
⟹ x = 3.
So, the can contained 21 litres of A.
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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