Answer:
The time taken by smaller pipe to fill the tank alone is 74.2 min
Step-by-step explanation:
Given as :
The time taken by two pipes to fill tank = 19 min
Let the time taken by larger pipe to fill tank = y
And the time taken by smaller pipe to fill tank = x
According to question
The smaller pipe takes 49 minutes longer than larger pipe
I.e x = y + 49
Now,
+
= ![\dfrac{1}{49}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B49%7D)
Or,
+
= ![\dfrac{1}{49}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B49%7D)
Or,
= ![\dfrac{1}{49}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B49%7D)
Or, 2 y + 49 = ![\dfrac{y^{2}+49y }{19}](https://tex.z-dn.net/?f=%5Cdfrac%7By%5E%7B2%7D%2B49y%20%7D%7B19%7D)
Or, 19 × ( 2 y + 49 ) = y² + 49 y
Or, 38 y + 931 = y² + 49 y
Or, y² + 49 y - 38 y - 931 = 0
or, y² + 11 y - 931 = 0
Or, Applying quadratic equation method to calculate the value of y
so, y = ![\frac{- b \pm \sqrt{b^{2}-4\times a\times c}}{2\times a}](https://tex.z-dn.net/?f=%5Cfrac%7B-%20b%20%5Cpm%20%5Csqrt%7Bb%5E%7B2%7D-4%5Ctimes%20a%5Ctimes%20c%7D%7D%7B2%5Ctimes%20a%7D)
Or, y = ![\frac{- 11 \pm \sqrt{11^{2}-4\times 1\times (-931)}}{2\times 1}](https://tex.z-dn.net/?f=%5Cfrac%7B-%2011%20%5Cpm%20%5Csqrt%7B11%5E%7B2%7D-4%5Ctimes%201%5Ctimes%20%28-931%29%7D%7D%7B2%5Ctimes%201%7D)
Or, y = ![\frac{- 1 \pm \sqrt{121+3724}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%201%20%5Cpm%20%5Csqrt%7B121%2B3724%7D%7D%7B2%7D)
or, y = ![\frac{- 1 \pm \sqrt{3845}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%201%20%5Cpm%20%5Csqrt%7B3845%7D%7D%7B2%7D)
or, y = ![\frac{- 1 \pm 62.008}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%201%20%5Cpm%2062.008%7D%7B2%7D)
∴ y = 25.2 , - 36.5
So, The time taken by larger pipe = y = 25.2 min
and the time taken by smaller pipe to fill the tank = y + 49 = 25.2 + 49 = 74.2 min
Hence The time taken by smaller pipe to fill the tank alone is 74.2 min answer