The average life of an Indian is 56 years. Find the number of times the human heart beats in the life of an Indian, if the heart beats once in 0.8 s

The correct option is 2.2x 10⁹ times.

To find the total number of times the human heart beats in the average lifespan of an Indian, we need to calculate the total number of seconds in 56 years and then divide this total time by the time interval of a single heartbeat.

Step 1: Calculate the total time in seconds
First, we find the number of days in 56 years. Assuming a standard year has 365 days:
Total days = 56 years * 365 days/year = 20,440 days

Next, convert the total days into seconds. We know that:
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
So, 1 day = 24 * 60 * 60 = 86,400 seconds

Therefore, the total time in seconds is:
Total seconds = 20,440 days * 86,400 seconds/day

$$\text{Total seconds}=\mathrm{1,765,992,000}\text{s}$$

This can be written in scientific notation as:
Total seconds ≈ 1.766 * 10⁹ s

Step 2: Calculate the number of heartbeats
The heart beats once every 0.8 seconds. Thus, the total number of heartbeats is given by dividing the total seconds by 0.8 s/beat:

$$\text{Number of heartbeats}=\frac{\mathrm{1,765,992,000}}{0.8}$$

$$\text{Number of heartbeats}=\mathrm{2,207,490,000}$$

Converting this result into scientific notation gives:
Number of heartbeats ≈ 2.2 * 10⁹ times

Hence, the human heart beats approximately 2.2x 10⁹ times in the average lifespan of 56 years.