Case Study III

Two water taps together fill a tank in 1 $\dfrac{7}{8}$ hours. The tap with larger diameter takes 2 hours less than the tap with smaller one to fill the tank completely. Based on the above information, answer the following questions:

1. If time taken by the tap with smaller diameter to fill the tank alone be x hours, then part of the tank filled by the tap with larger diameter alone in 2 hours is:

1. 2(x + 2)

2. 2(x - 2)

3. $\dfrac{2}{x + 2}$

4. $\dfrac{2}{x - 2}$

2. The quadratic equation representing the given information is:

1. 4x² − 23x + 15 = 0
2. 2x² − 23x + 15 = 0
3. 4x² + 23x − 15 = 0
4. 2x² + 23x − 15 = 0

3. Time taken by the larger tap to fill the tank alone is:

1. 5 hours
2. 3 hours
3. 7 hours
4. none of these

4. The part of the tank which can be filled by the smaller tap in 3 hours is:

1. $\dfrac{1}{3}$

2. $\dfrac{1}{5}$

3. $\dfrac{3}{5}$

4. $\dfrac{3}{7}$

5. The part of the tank which can be filled by the larger tap in $1\dfrac{1}{4}$ hours is:

1. $\dfrac{5}{12}$

2. $\dfrac{3}{5}$

3. $\dfrac{5}{8}$

4. $\dfrac{3}{8}$

1. Let time taken by smaller tap to fill tank = x hours

Part of tank filled by smaller tap in one hour = $\dfrac{1}{x}$

Time taken by larger tap to fill tank = (x - 2) hours

Part of tank filled by larger tap in one hour $\dfrac{1}{x - 2}$

Part of tank filled by larger tap in two hour $\dfrac{2}{x - 2}$

Hence, option (4) is the correct option.

2. Given,

Together smaller and larger tap take $1\dfrac{7}{8} = \dfrac{15}{8}$ hours to fill tank.

Rate at which the taps fill tank in one hour = $\dfrac{1}{\dfrac{15}{8}} = \dfrac{8}{15}$ tank/hr

From question 1 we have,

Part of tank filled by smaller tap in one hour = $\dfrac{1}{x}$

Part of tank filled by larger tap in one hour = $\dfrac{1}{x - 2}$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{x - 2} = \dfrac{8}{15} \\[1em] \Rightarrow \dfrac{x - 2 + x}{x^2 - 2x}= \dfrac{8}{15} \\[1em] \Rightarrow 15(2x - 2) = 8(x^2 - 2x) \\[1em] \Rightarrow 30x - 30 = 8x^2 - 16x \\[1em] \Rightarrow 0 = 8x^2 - 16x - 30x + 30 \\[1em] \Rightarrow 8x^2 - 46x + 30 = 0 \\[1em] \Rightarrow 2(4x^2 - 23x + 15) = 0 \\[1em] \Rightarrow 4x^2 - 23x + 15 = 0$

Hence, option (1) is the correct option.

3. Solving equation from question 2,

⇒ 4x² - 23x + 15 = 0

⇒ 4x² - 20x - 3x + 15 = 0

⇒ 4x(x - 5) - 3(x - 5) = 0

⇒ (4x - 3)(x - 5) = 0

⇒ (4x - 3) = 0 or (x - 5) = 0     [Using zero-product rule]

⇒ x = $\dfrac{3}{4}$ or x = 5

Case 1 : x = $\dfrac{3}{4}$

Larger tap take (x - 2) hours = $\dfrac{3}{4} - 2 = \dfrac{3 - 8}{4} = \dfrac{-5}{4}$

= -1.25 hours, which is not possible.

Case 2 : x = 5

The time taken by the smaller tap is x = 5 hours.

The time taken by the larger tap is x − 2 = 5 - 2 = 3 hours.

Hence, option (2) is the correct option.

4. The smaller tap fills the tank in x = 5 hours.

Part of tank filled by smaller tap in one hour = $\dfrac{1}{5}$

Part of tank filled by smaller tap in 3 hours = $\dfrac{3}{5}$

Hence, option (3) is the correct option.

5. The larger tap fills the tank in 3 hours.

Part of tank filled by larger tap in one hour $\dfrac{1}{3}$

Part of tank filled by larger tap in $1\dfrac{1}{4} = \dfrac{5}{4}$ hours

= $\dfrac{\dfrac{5}{4}}{3}$

= $\dfrac{5}{12}$

Hence, option (1) is the correct option.