Two taps together can fill in 1 and 7/8 hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately . Find the time in which tap can fill the tank separately.
Answers
Step-by-step explanation:
Let the time taken by smaller tap to fill the tank completely = x
hours
The volume of the tank filled by a smaller tap in 1 hour =. 1/ x
The time taken by larger tap to fill the tank completely = X-2
hours
The volume of the tank filled by a larger tap in 1 hour = 1 / X-2
Time taken by the tank to fill = 1 (7/8) hours
= 15 /8 hours
The volume of the tank filled by a smaller tap in 15/8 hours = 1/x × 15/8
The volume of the tank filled by a larger tap in
15 / 8 hours = 1 / x -2 × 15 / 8
1/x × 15/8 + 1/x -2 × 15/8 =1
15(x - 2)+15x / x ( x - 2) 8 = 1
15(x - 2) + 15x = 8x(x-2)
15(x - 2 + x) = 8x(x - 2)
=> 15(2x - 2) = 8x^2 - 16
= >30 x - 30 = 8x^2 - 16
=> 8x2 - 46x + 30 = 0
=>2(4x^2 - 23x + 15) = 0
=>4x^2 - 23x + 15 = 0
=>4x2 - 20x - 3x + 15 = 0
=> 4x(x -5) - 3(x - 5) = 0
=> (4x-3)(x-5) = 0
= 5 and 3/4
At x = 5
Time taken by smaller tap to fill the tank completely = 5 hours
Time taken by larger tap to fill tank completely = 5
- 2 = 3 hours
At x = 3/4
Time taken by smaller tap to fill the tank
completely = 3/4 hours
Time taken by larger tap to fill the tank completely = 3/4 -2 hours
Time is not possible to be negative.
So, x = 3/4 is not possible.
Therefore the time for smaller tap is 5 hrs and for larger tap is 3 hrs.
HOPE IT HELPS