2 minutes 30 seconds of internet download bill is 18 rupees, then how much will be the bill of 3 minutes 20 seconds? (Up to one decimal place) A. 24 B. 24.1 C. 24.2 D. 23.9
A
The problem asks us to calculate the internet download bill for a duration of 3 minutes 20 seconds, given that the bill for 2 minutes 30 seconds is 18 rupees. This is a problem involving proportional relationships, assuming the cost is directly proportional to the download time.
To work with the durations easily, we should convert both given times into seconds. There are 60 seconds in one minute.
We know the cost for 150 seconds is 18 rupees. We can find the cost per second by dividing the total cost by the total time in seconds.
Let's simplify this fraction:
Cost per second = $\frac{18}{150} = \frac{18 \div 6}{150 \div 6} = \frac{3}{25}$ rupees per second.
Alternatively, we can keep it as $\frac{18}{150}$ for the next step.
Now we need to find the bill for the second duration, which is 200 seconds. We multiply the cost per second by the new total time in seconds.
Let's calculate the value:
New Bill = $\frac{18}{150} \times 200 = \frac{18 \times 200}{150}$
We can simplify by dividing both 200 and 150 by their greatest common divisor, which is 50:
New Bill = $\frac{18 \times (200 \div 50)}{(150 \div 50)} = \frac{18 \times 4}{3}$
Now, simplify further by dividing 18 by 3:
New Bill = $(18 \div 3) \times 4 = 6 \times 4$
New Bill = 24 rupees
The question asks for the answer up to one decimal place. The calculated bill is exactly 24.0 rupees.
| Description | Calculation | Result |
|---|---|---|
| First duration in seconds | $2 \times 60 + 30$ | 150 seconds |
| Second duration in seconds | $3 \times 60 + 20$ | 200 seconds |
| Cost per second | $\frac{18 \text{ rupees}}{150 \text{ seconds}}$ | $\frac{18}{150}$ rupees/second |
| New Bill | $\frac{18}{150} \times 200$ | 24 rupees |
| New Bill (one decimal place) | 24.0 | 24.0 rupees |
Let's compare our calculated bill with the given options:
Our calculated bill of 24.0 rupees matches option A when considering the required precision of one decimal place.
| Concept | Explanation |
|---|---|
| Proportionality | Assuming cost increases linearly with time. |
| Unit Conversion | Converting minutes and seconds into a single unit (seconds). |
| Rate Calculation | Finding the cost per unit of time (e.g., per second). |
| Applying Rate | Using the calculated rate to find the cost for a different duration. |
A proportional relationship exists between two quantities when their ratio is constant. In this problem, the ratio of the total cost to the total time is constant. If $C$ is the cost and $T$ is the time, then $\frac{C}{T} = k$, where $k$ is the constant of proportionality (which is the cost per unit time). So, $C = kT$.
Given $C_1$ at $T_1$ and asked to find $C_2$ at $T_2$, we have:
Substituting the value of $k$ from the first equation into the second equation, we get:
$C_2 = \frac{C_1}{T_1} \times T_2$
In this problem:
Using the formula:
$C_2 = \frac{18 \text{ rupees}}{150 \text{ seconds}} \times 200 \text{ seconds}$
$C_2 = \frac{18}{150} \times 200 = \frac{18 \times 200}{150} = \frac{3600}{150} = \frac{360}{15} = 24$ rupees.
This confirms our earlier calculation and the principle of proportional relationships.
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