Question

A typical person has an average heart rate of 72.0 beats/min. Calculate the following: (a) How many beats does she have in 6.0 years? (b) How many beats in 6.00 years? (c) And finally, how many beats in 6.000 years? Pay close attention to significant figures in this question.

Answer 1

Rule: After multiplication or division final result should have same number of significant figures as the quantity which has lowest number of significant figures.

Avg heart rate = 72.0 beats/min

1 year = 365 day = 365*24 day = 365*24*60 min

6 yr = 6*365*24*60 = 3153600 min

Now in 6 years number of beats will be:

Total number of beats = time*Avg heart rate = 3153600 min*(72.0 beats/min)

Total number of beats = 227059200 Beats

Part A.

Now when year is denoted as 6.0 yrs, then

(6.0 and 72.0 between these number 6.0 have 2 significant figures, while 72.0 have 3 significant figures, So final should have lowest (2) significant figures)

Total number of beats = 227059200 Beats = 2.3*10^8 Beats

Part B.

Now when year is denoted as 6.00 yrs, then

(6.00 and 72.0 between these number 6.00 have 3 significant figures, while 72.0 also have 3 significant figures, So final should have lowest (3) significant figures)

Total number of beats = 227059200 Beats = 2.27*10^8 Beats

Part C.

Now when year is denoted as 6.000 yrs, then

(6.000 and 72.0 between these number 6.000 have 4 significant figures, while 72.0 also have 3 significant figures, So final should have lowest (3) significant figures)

Total number of beats = 227059200 Beats = 2.27*10^8 Beats

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