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Question
Two water taps together can fill a tank in `1(7)/(8)` hours. The tap with longer diameter takes 2 hours less than the tap with a smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
Solution
Let the first tap takes x hours to completely fill tank
⇒ Second tap will take 2 hours less
⇒ According to question
`(1)/("x") + (1)/("x" -2) = (8)/(15)`
`("x"-2+"x")/("x"("x"-2)) = (8)/(15)`
`(2"x"-2)/("x"("x"-2)) = (18)/(15)`
`((2"x"-1))/("x"("x"-2)) = (18)/(15)`
`15 ("x"-1) = 4"x" ("x" -2)`
`15"x" - 15 = 4"x"^2 - 8"x"`
`4"x"^2 - 23"x" + 15 = 0`
`4"x"^2 - 20"x" - 3"x" + 15 = 0`
`4"x"("x" - 5) - 3 ("x" -5) = 0`
`("x" - 5) (4"x" - 3) = 0`
`"x" = 5 or (3)/(4)`
Since `(3)/(4) - 2 = "Negative time" (3)/(4)` is not possible.
Which is not possible
⇒ x = 5
Rate of 1st pipe = 5 hours
Rate of 2nd pipe = 5 - 2 = 3hours
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