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प्रश्न
| 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) $$\frac{3}{8}$$
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उत्तर
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: \[\text{Part filled} = \frac{5}{4} \times \frac{1}{3} = \frac{5}{12}\]
