Question

1. aiden’s yard is a 5m by 12m rectangle and his lawnmower can cut the grass in 15 minutes. the field behind beth’s house measures 30m by 50m and her ride-on lawnmower can cut the grass in 1 hour. there is a 400-m² area between the two properties. how much faster would it take beth and her ride-on mower to mow this space compared to aiden. show the calculations that lead to your final answer.

Explanation:

Step1: Calculate Aiden's lawn area

Aiden's yard is a rectangle with length \( l = 12\,\text{m} \) and width \( w = 5\,\text{m} \). The area of a rectangle is \( A = l \times w \), so \( A_{\text{Aiden}} = 12 \times 5 = 60\,\text{m}^2 \).

Step2: Find Aiden's mowing rate

Aiden mows \( 60\,\text{m}^2 \) in \( 15\,\text{minutes} \). To find the rate per hour, we convert minutes to hours: \( 15\,\text{minutes} = \frac{15}{60} = 0.25\,\text{hours} \). Rate \( r_{\text{Aiden}} = \frac{\text{Area}}{\text{Time}} = \frac{60}{0.25} = 240\,\text{m}^2/\text{hour} \).

Step3: Determine Beth's mowing time

Beth's lawnmower rate is \( 400\,\text{m}^2/\text{hour} \) (wait, no—wait, the problem says "her ride - on lawnmower can cut the grass in 1 hour" for the 400 - m² area? Wait, re - reading: "the field behind Beth’s house measures 30m by 50m and her ride - on lawnmower can cut the grass in 1 hour. There is a 400 - m² area between the two properties. How much faster would it take Beth and her ride - on mower to mow this space compared to Aiden."

Wait, correction: The 400 - m² is the area to mow. So for Aiden: time to mow 400 m². His rate is \( 60\,\text{m}^2 \) in 15 minutes, so per minute: \( \frac{60}{15}=4\,\text{m}^2/\text{minute} \). Time for Aiden to mow 400 m²: \( t_{\text{Aiden}}=\frac{400}{4} = 100\,\text{minutes} \).

For Beth: Her lawnmower can mow (wait, the 30m×50m is \( 30\times50 = 1500\,\text{m}^2 \) in 1 hour (60 minutes). So her rate is \( \frac{1500}{60}=25\,\text{m}^2/\text{minute} \). But we need to mow 400 m². Time for Beth: \( t_{\text{Beth}}=\frac{400}{25}=16\,\text{minutes} \).

Wait, maybe simpler: Aiden's rate: 60 m² in 15 min, so rate \( r_A=\frac{60}{15}=4\,\text{m}^2/\text{min} \). Time for 400 m²: \( t_A=\frac{400}{4}=100\,\text{min} \).

Beth's rate: The field behind her is 30×50 = 1500 m², mowed in 1 hour (60 min), so rate \( r_B=\frac{1500}{60}=25\,\text{m}^2/\text{min} \). Time for 400 m²: \( t_B=\frac{400}{25}=16\,\text{min} \).

Difference in time: \( 100 - 16 = 84\,\text{minutes} \). Wait, but maybe the 400 m² is the area to mow, and we compare the time to mow 400 m².

Alternative approach:

Aiden: Area = 5×12 = 60 m², time = 15 min. So time per m²: \( \frac{15}{60}=\frac{1}{4}\,\text{min/m}^2 \). Time for 400 m²: \( 400\times\frac{1}{4}=100\,\text{min} \).

Beth: Her lawn (30×50 = 1500 m²) takes 1 hour (60 min). So time per m²: \( \frac{60}{1500}=\frac{1}{25}\,\text{min/m}^2 \). Time for 400 m²: \( 400\times\frac{1}{25}=16\,\text{min} \).

Time difference: \( 100 - 16 = 84\,\text{minutes} \), which is \( 1\) hour and \( 24\) minutes, or \( \frac{84}{60}=1.4\) hours. But the question is "how much faster", so the time saved is 84 minutes.

Wait, maybe the problem is: Aiden's lawnmower cuts 60 m² in 15 min, so rate is \( \frac{60}{15}=4\,\text{m}^2/\text{min} \). Time to mow 400 m²: \( \frac{400}{4}=100\,\text{min} \).

Beth's lawnmower: the area behind her is 30×50 = 1500 m², mowed in 1 hour (60 min), so rate is \( \frac{1500}{60}=25\,\text{m}^2/\text{min} \). Time to mow 400 m²: \( \frac{400}{25}=16\,\text{min} \).

Difference: \( 100 - 16 = 84\,\text{minutes} \), which is \( 1\) hour and \( 24\) minutes, or \( 1.4\) hours. But let's check the initial calculation again.

Wait, the problem says "the field behind Beth’s house measures 30m by 50m and her ride - on lawnmower can cut the grass in 1 hour. There is a 400 - m² area between the two properties. How much faster would it take Beth and her ride - on mower to mow this space compared to Aiden."

So Aiden's rate: 5×12 = 60 m² in 15 min. So rate \( r_A=\frac{60}{15}=4\,\text{m}^2/\text{min} \). Time for 400 m²: \( t_A=\frac{400}{4}=100\,\text{min} \).

Beth's rate: 30×50 = 1500 m² in 60 min, so \( r_B=\frac{1500}{60}=25\,\text{m}^2/\text{min} \). Time for 400 m²: \( t_B=\frac{400}{25}=16\,\text{min} \).

Time saved: \( 100 - 16 = 84\,\text{minutes} \), which is \( 1\) hour and \( 24\) minutes, or \( \frac{84}{60}=1.4\) hours. But maybe the question is in hours? Let's convert 84 minutes to hours: \( 84\div60 = 1.4\) hours, or 1 hour 24 minutes.

But let's do it step by step correctly:

Step1: Calculate Aiden's mowing rate

Aiden's lawn area: \( A_A=5\times12 = 60\,\text{m}^2 \). Time taken: \( t_A = 15\,\text{minutes}=\frac{15}{60}=0.25\,\text{hours} \). Rate \( r_A=\frac{A_A}{t_A}=\frac{60}{0.25}=240\,\text{m}^2/\text{hour} \).

Step2: Calculate Beth's mowing rate

Beth's lawn area: \( A_B = 30\times50=1500\,\text{m}^2 \). Time taken: \( t_B = 1\,\text{hour} \). Rate \( r_B=\frac{A_B}{t_B}=\frac{1500}{1}=1500\,\text{m}^2/\text{hour} \). Wait, no! Wait, the 400 - m² is the area to mow, not her lawn. So we need to find the time for each to mow 400 m².

Step3: Time for Aiden to mow 400 m²

Using rate \( r_A = 240\,\text{m}^2/\text{hour} \), time \( t_{A,400}=\frac{400}{r_A}=\frac{400}{240}=\frac{5}{3}\approx1.6667\,\text{hours} \), which is \( \frac{5}{3}\times60 = 100\,\text{minutes} \).

Step4: Time for Beth to mow 400 m²

Beth's rate for her lawn is 1500 m² per hour, so her rate is \( 1500\,\text{m}^2/\text{hour} \). Time \( t_{B,400}=\frac{400}{1500}=\frac{4}{15}\approx0.2667\,\text{hours} \), which is \( \frac{4}{15}\times60 = 16\,\text{minutes} \).

Step5: Calculate the time difference

Time difference \( \Delta t=t_{A,400}-t_{B,400}=\frac{5}{3}-\frac{4}{15}=\frac{25 - 4}{15}=\frac{21}{15}=\frac{7}{5}=1.4\,\text{hours} \), or \( 1.4\times60 = 84\,\text{minutes} \).

Answer:

Beth's ride - on mower takes 84 minutes (or 1.4 hours) faster than Aiden's lawnmower to mow the 400 - m² area.