Questions & Answers

Question

Answers

A. 15

B. 12

C. 5

D. 14

Answer
Verified

* 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$

Solve this quadratic equation by using a quadratic formula.

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 \[x = 2\].

$\therefore x = 15$

×

Sorry!, This page is not available for now to bookmark.