two water taps together can fill a tank in 2 11/12 hrs.the tap of smaller diameter takes 2 hrs more than the large one to fill the tank separately . find the time in which each tap can separately fill the tank.
Answers
Let the tap of larger diameter takes time to fill the tank = x hrs
Then, the tap of smaller diameter takes time to fill the tank = (x + 2) hrs
In 1 hour, tap of larger diameter will fill = 1/x part of tank
In 1 hour, tap of smaller diameter will = (1/x + 2) part of tank
Together, both the taps take time to fill the tank = 2 11/12 hrs
In 1 hour, both the taps will fill = 12/35 part of tank
∴ 1/x + 1/(x + 2) = 12/35
Taking the LCM and then solving it.
⇒ (x + 2 + 2)/(x + 2)(x) = 12/35
⇒ (2x + 2)/(x² + 2x) = 12/35
⇒ 12x² + 24x = 70x + 70
⇒ 12x² + 24x - 70x - 70 = 0
⇒ (12x² - 46x - 70 = 0) ÷ 2
⇒ 6x² - 23x - 35 = 0
⇒ 6x² - 30x + 7x - 35 = 0
⇒ 6x(x - 5) + 7(x - 5) = 0
⇒ (6x + 7) = 0 or (x - 5) = 0
⇒ x = - 7/6 or x = 5
x = - 6/7 is not possible because time cannot be negative. So, the correct value of x is 5.
So, time taken by tap of larger diameter is 5 hours and time taken by tap of smaller diameter is 5 + 2 = 7 hours separately.
Answer.
Answer:
Bigger Tap=5 hours Smaller Tap=7 Hours
Step-by-step explanation:
Let time taken for big tap to fill tank = x hours
Let time taken for small tap to fill = x+ 2 hours
therefore x + x+2 = 35/2 hours
i.e. x+x+2= 1 (1 completely tank )
Now...
Time taken by smaller tap to fill in one hour= 1/x+2
Time taken by bigger tap to fill in one hour=1/x
and time would be 12/35 ( reciprocating by 1)
now the equation would look like
1/x + 1/x+2 = 12/35
this would further become an easy quadratic equation which everyone knows how to solve.......
and the final answers for x would come as 5 and -1346/12 ( discarded)
therefore big tap fills in 5 hours
and small tap fills it in 5+2 = 7 hours
that easy.......
Hope you understood it :)