Two pipes running together can fill a tank in 100/9 minutes. If one pipe
takes 5 minutes more than the other to fill the tank separately, find the
Answers
Step-by-step explanation:
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Let there be two pipes, A and B.
A takes more time i.e., 5 mins more than B to fill tank separately.
Let time taken by B be x mins. Then time taken by A will be x + 5 mins.
→ 100/(9x) + 100/[9(x + 5)] = 1
→ 100/9 × (1/x) + 100/9 × 1/(x + 5) = 1
→ 1/x + 1/(x + 5) = 9/100
→ (x + 5 + x)/(x² + 5x) = 9/100
→ 100(2x + 5) = 9(x² + 5x)
→ 200x + 500 = 9x² + 45x
→ 9x² - 155x - 500 = 0
D = b² - 4ac
D = (-155)² - 4(9)(-500)
D = 42025
Now, x = (- b ± √D)/2a
→ x = (155 ± √42025)/2(9)
→ x = (155 ± 205)/18
Case 1: When it's + 205,
→ x = (155 + 205)/18
→ x = 20
Case 2: When it's - 205,
→ x = (155 - 205)/18
→ x = - 50/18
“Since time can't be negative hence, value of x is 20 mins.”
Time taken by B = 20 mins
Time taken by A = 20 + 5 = 25 mins
To fill tank separately