Question- Two water taps together can fill a tank in $9\dfrac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time (in hours) in which a tap of smaller diameter can separately fill the tank.
(A) 25
(B) $\dfrac{{ - 25}}{4}$
(C) 24
(D) $\dfrac{{25}}{4}$
Answer
366.9k+ views
Hint- Here, we will be using the concept of how much part of the tank is filled in one hour.
Let us assume that the tap with smaller diameter can separately fill the tank in $x$ hours.
According to the problem statement it is given that the tap with larger diameter can separately fill the tank in $\left( {x - 10} \right)$ hours.
Also, given that both the taps can together fill the tank in $9\dfrac{3}{8} = \dfrac{{\left( {8 \times 9} \right) + 3}}{8} = \dfrac{{75}}{8}$ hours.
So, the tap with a smaller diameter can fill $\dfrac{1}{x}$ part of the tank in 1 hour. Similarly, the tap with a larger diameter can fill $\dfrac{1}{{\left( {x - 10} \right)}}$ part of the tank in 1 hour. Also, both the taps fill $\dfrac{8}{{75}}$ part of the tank in 1 hour.
Then, \[
\dfrac{1}{x} + \dfrac{1}{{\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{x - 10 + x}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{2x - 10}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow 75\left( {2x - 10} \right) = 8x\left( {x - 10} \right) \\
\Rightarrow 150x - 750 = 8{x^2} - 80x \Rightarrow 8{x^2} - 230x + 750 = 0 \Rightarrow 4{x^2} - 115x + 375 = 0 \\
\Rightarrow 4{x^2} - 100x - 15x + 375 = 0 \Rightarrow 4x\left( {x - 25} \right) - 15\left( {x - 25} \right) = 0 \Rightarrow \left( {x - 25} \right)\left( {4x - 15} \right) = 0 \\
\]
Either $x = 25$ or $4x = 15 \Rightarrow x = \dfrac{{15}}{4}$
When $x = 25$, $\left( {x - 10} \right) = 25 - 10 = 15$ and when $x = \dfrac{{15}}{4}$, $\left( {x - 10} \right) = \dfrac{{15}}{4} - 10 = \dfrac{{15 - 40}}{4} = - \dfrac{{25}}{4}$
Since, the time can never be negative so $x = \dfrac{{15}}{4}$ is rejected.
Hence, $x = 25$ and $\left( {x - 10} \right) = 15$
The tap with smaller diameter can separately fill the tank in 25 hours and the tap with larger diameter can separately fill the tank in 15 hours.
Therefore, option A is correct.
Note- In these types of problems, the concept of part of the work done (here it is filling the tank) in one hour is utilized to obtain an equation in one variable so that we can solve for it.
Let us assume that the tap with smaller diameter can separately fill the tank in $x$ hours.
According to the problem statement it is given that the tap with larger diameter can separately fill the tank in $\left( {x - 10} \right)$ hours.
Also, given that both the taps can together fill the tank in $9\dfrac{3}{8} = \dfrac{{\left( {8 \times 9} \right) + 3}}{8} = \dfrac{{75}}{8}$ hours.
So, the tap with a smaller diameter can fill $\dfrac{1}{x}$ part of the tank in 1 hour. Similarly, the tap with a larger diameter can fill $\dfrac{1}{{\left( {x - 10} \right)}}$ part of the tank in 1 hour. Also, both the taps fill $\dfrac{8}{{75}}$ part of the tank in 1 hour.
Then, \[
\dfrac{1}{x} + \dfrac{1}{{\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{x - 10 + x}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{2x - 10}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow 75\left( {2x - 10} \right) = 8x\left( {x - 10} \right) \\
\Rightarrow 150x - 750 = 8{x^2} - 80x \Rightarrow 8{x^2} - 230x + 750 = 0 \Rightarrow 4{x^2} - 115x + 375 = 0 \\
\Rightarrow 4{x^2} - 100x - 15x + 375 = 0 \Rightarrow 4x\left( {x - 25} \right) - 15\left( {x - 25} \right) = 0 \Rightarrow \left( {x - 25} \right)\left( {4x - 15} \right) = 0 \\
\]
Either $x = 25$ or $4x = 15 \Rightarrow x = \dfrac{{15}}{4}$
When $x = 25$, $\left( {x - 10} \right) = 25 - 10 = 15$ and when $x = \dfrac{{15}}{4}$, $\left( {x - 10} \right) = \dfrac{{15}}{4} - 10 = \dfrac{{15 - 40}}{4} = - \dfrac{{25}}{4}$
Since, the time can never be negative so $x = \dfrac{{15}}{4}$ is rejected.
Hence, $x = 25$ and $\left( {x - 10} \right) = 15$
The tap with smaller diameter can separately fill the tank in 25 hours and the tap with larger diameter can separately fill the tank in 15 hours.
Therefore, option A is correct.
Note- In these types of problems, the concept of part of the work done (here it is filling the tank) in one hour is utilized to obtain an equation in one variable so that we can solve for it.
Last updated date: 03rd Oct 2023
•
Total views: 366.9k
•
Views today: 3.66k
Recently Updated Pages
What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

An alternating current can be produced by A a transformer class 12 physics CBSE

What is the value of 01+23+45+67++1617+1819+20 class 11 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers
