Question

# One tank can be filled by two taps in $6$ hours. The smaller tap alone takes $5$ hours more than the bigger tap alone. Then the time required (in hours) by the smaller tap to fill the tank separately.A. 15B. 12C. 5D. 14

Hint: Here we assume the time required by smaller tap as a variable and then write time required by larger tap using the variable. Then using the unitary system we find the amount filled by each tap separately and add them together to equate with the amount filled by two taps together.
* If we have a value of multiple units then we can find the value of a single unit by dividing the value of multiple units by the number of units.

Assign a variable to time required for a smaller tap alone to fill the tank.
Let $x$ is time in hours required for a smaller tap alone to fill the tank.
Find the time required for a bigger tap alone to fill the tank.
Since a smaller tap takes 5 hours more than a bigger tap.
$\therefore$ Time required for bigger tap alone to fill the tank $= x - 5$
Find the amount of water filled in $1$ hour by each tap.
Amount of water filled by smaller tap in x hours $= 1$ unit
Therefore, the amount of water filled by smaller tap in 1 hour $= \dfrac{1}{x}$ unit
Amount of water filled by larger tap in $x - 5$ hours $= 1$unit
Therefore, the amount of water filled by larger tap in 1 hour $= \dfrac{1}{{x - 5}}$
Since both taps fill up the tank together in 6 hours.
Amount of tank filled by both taps in 6 hours $= 1$unit
Therefore, amount of tank filled by both taps in 1 hour $= \dfrac{1}{6}$unit
Use the given condition to form an equation.
The amount of water filled by the smaller tap and the amount of water filled by the larger tap equals the amount of water filled by both taps together.
$\therefore \dfrac{1}{x} + \dfrac{1}{{x - 5}} = \dfrac{1}{6}$
Solve this equation for x.
$\therefore \dfrac{1}{x} + \dfrac{1}{{x - 5}} = \dfrac{1}{6}$
Take LCM on the left hand side of the equation.
$\Rightarrow \dfrac{{x - 5 + x}}{{x(x - 5)}} = \dfrac{1}{6} \\ \Rightarrow \dfrac{{2x - 5}}{{{x^2} - 5x}} = \dfrac{1}{6} \\$
Cross multiply the denominator of opposite sides to the numerator of opposite sides.
$\Rightarrow (2x - 5)6 = {x^2} - 5x \\ \Rightarrow 12x - 30 = {x^2} - 5x \\$
Shift all the values to one side of the equation
$\Rightarrow {x^2} - 5x - 12x + 30 = 0$
$\Rightarrow {x^2} - 17x + 30 = 0$
Comparing $\Rightarrow {x^2} - 17x + 30 = 0$ with $a{x^2} + bx + c = 0$ we get:
$a = 1, b = - 17, c = 30$.
Substituting these values of $a, b, c$ in $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ we get:
$x = \dfrac{{ - ( - 17) \pm \sqrt {{{( - 17)}^2} - 4(1)(30)} }}{{2(1)}} \\ x = \dfrac{{17 \pm \sqrt {289 - 120} }}{2} \\$

$x = \dfrac{{17 \pm \sqrt {169} }}{2} \\ x = \dfrac{{17 \pm 13}}{2} \\$
Now, we solve for both values
$x = \dfrac{{17 + 13}}{2},x = \dfrac{{17 - 13}}{2}, \\ x = \dfrac{{30}}{2},x = \dfrac{4}{2} \\ x = 15, x = 2 \\$
If $x = 15$ then $x - 5 = 15 - 5 = 10$
If $x = 2$ then $x - 5 = 2 - 5 = - 3$
where $x$ and $x - 5$ are time values so they cannot be negative, So we reject the value of $x = 2$.
$\therefore x = 15$

So, the correct answer is “Option A”.

Note: Students many times make the mistake of considering both the values of time which is wrong. Also, keep in mind that whenever a value is greater than the other value we always add the value to the smaller value and equate it with the greater value.