Question

Two water taps together fill a tank in $$1\frac{7}{8}$$ hours. The tape 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 :

(a) $$2(x + 2)$$

(b) $$2(x - 2)$$

(c) $$\frac{2}{x + 2}$$

(d) $$\frac{2}{x - 2}$$

2. The quadratic equation representing the given information is :

(a) $$4x^2 - 23x + 15 = 0$$

(b) $$2x^2 - 23x + 15 = 0$$

(c) $$4x^2 + 23x - 15 = 0$$

(d) $$2x^2 + 23x - 15 = 0$$

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

(a) 5 hours

(b) 3 hours

(c) 7 hours

(d) none of these

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

(a) $$\frac{1}{3}$$

(b) $$\frac{1}{5}$$

(c) $$\frac{3}{5}$$

(d) $$\frac{3}{7}$$

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

(a) $$\frac{5}{12}$$

(b) $$\frac{3}{5}$$

(c) $$\frac{5}{8}$$

(d) $$\frac{3}{8}$$

Answer 1

1. The correct option is (d).

Reasoning:

Time taken by smaller tap alone = x hours

Time taken by larger tap alone = (x – 2) hours

Part of tank filled by larger tap in 1 hour = \[\frac{1}{x - 2}\]

Part of tank filled by larger tap in 2 hours = \[2 \times \frac{1}{x - 2} = \frac{2}{x - 2}\]

2. The correct option is (a).

Reasoning:

Combined time taken = \[1\frac{7}{8} = \frac{15}{8}\text{ hours}\]

Portion of the tank filled by both taps in 1 hour = \[\frac{1}{\frac{15}{8}} = \frac{8}{15}\]

Setting up the equation: \[\frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15}\] 

\[\frac{(x - 2) + x}{x(x - 2)} = \frac{8}{15}\] 

\[\frac{2x - 2}{x^2 - 2x} = \frac{8}{15}\] 

\[\frac{2(x - 1)}{x^2 - 2x} = \frac{8}{15}\] 

\[\frac{x - 1}{x^2 - 2x} = \frac{4}{15}\] 

\[15(x - 1) = 4(x^2 - 2x)\] 

\[15x - 15 = 4x^2 - 8x\] 

\[4x^2 - 23x + 15 = 0\]

3. The correct option is (b).

Reasoning: Solving \[4x^2 - 23x + 15 = 0\]:

\[4x^2 - 20x - 3x + 15 = 0\] 

\[4x(x - 5) - 3(x - 5) = 0\] 

\[(4x - 3)(x - 5) = 0\] 

\[x = 5 \quad \text{or} \quad x = \frac{3}{4}\]

If \[x = \frac{3}{4}\] hours, then time taken by the larger tap would be \[\frac{3}{4} - 2 = -\frac{5}{4}\] hours, which is impossible since time cannot be negative. Thus, x = 5.

Time taken by the smaller tap alone = 5 hours

Time taken by the larger tap alone = \[x - 2 = 5 - 2 = 3\text{ hours}\]

4. The correct option is (c).

Reasoning:

Time taken by smaller tap to fill the tank alone = 5 hours

Part of the tank filled by smaller tap in 1 hour = \[\frac{1}{5}\]

Part of the tank filled by smaller tap in 3 hours = \[3 \times \frac{1}{5} = \frac{3}{5}\]

5. The correct option is (a).

Reasoning:

Time taken by larger tap to fill the tank alone = 3 hours

Part of the tank filled by larger tap in 1 hour = \[\frac{1}{3}\]

Given time = \[1\frac{1}{4}\text{ hours} = \frac{5}{4}\text{ hours}\]

Part of the tank filled in \[\frac{5}{4}\] hours: \[\text{Part filled} = \frac{5}{4} \times \frac{1}{3} = \frac{5}{12}\]