A fuel tank leaks 2% of its volume per day. After 5 days, what fraction of the tank remains?

A fuel tank leaks 2% of its volume per day. After 5 days, what fraction of the tank remains?

• A. 0.90
• B. 0.905
• C. 0.91
• D. 0.92

Answer: (B)

To determine the fraction of the fuel tank that remains after 5 days of leaking 2% of its volume per day, we can use the concept of exponential decay. Each day, the tank retains 98% of its volume since it loses 2%. This can be mathematically expressed as multiplying the remaining volume by 0.98 each day. After one day, the remaining volume can be calculated as: \[ V_1 = V \times 0.98 \] After two days, it will be: \[ V_2 = V_1 \times 0.98 = V \times 0.98^2 \] Following this pattern, after 5 days, the remaining volume will be: \[ V_5 = V \times 0.98^5 \] To find the fraction of the tank that remains, we calculate \( 0.98^5 \): \[ 0.98^5 \approx 0.9039 \] This indicates that approximately 90.39% of the tank remains after 5 days. When considering the provided answer choices, 0.905 is closest to this calculated value. Thus, the fraction that remains after 5 days of leakage is

To determine the fraction of the fuel tank that remains after 5 days of leaking 2% of its volume per day, we can use the concept of exponential decay.

Each day, the tank retains 98% of its volume since it loses 2%. This can be mathematically expressed as multiplying the remaining volume by 0.98 each day.

After one day, the remaining volume can be calculated as:

[ V_1 = V \times 0.98 ]

After two days, it will be:

[ V_2 = V_1 \times 0.98 = V \times 0.98^2 ]

Following this pattern, after 5 days, the remaining volume will be:

[ V_5 = V \times 0.98^5 ]

To find the fraction of the tank that remains, we calculate ( 0.98^5 ):

[ 0.98^5 \approx 0.9039 ]

This indicates that approximately 90.39% of the tank remains after 5 days. When considering the provided answer choices, 0.905 is closest to this calculated value.

Thus, the fraction that remains after 5 days of leakage is