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The Chandlers are moving across the countryMr. Chandler leaves 5 hours before Mrs. Chandlerhe averages 40 mph and she averages 80 mphhow many hours will it take Mrs. Chandler to catch up to Mr. Chandler?

The Chandlers are moving across the countryMr. Chandler leaves 5 hours before Mrs-example-1
The Chandlers are moving across the countryMr. Chandler leaves 5 hours before Mrs-example-1
The Chandlers are moving across the countryMr. Chandler leaves 5 hours before Mrs-example-2
User Damitha Raveendra
by
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1 Answer

23 votes
23 votes

From the information provided, we know that Mr Chandler travels at a speed of 40 miles per hour (40mph). He also left 5 hours before Mrs Chandler.

Therefore, we can tell that for 5 hours he has travelled


\begin{gathered} Dis\tan ce=speed\cdot time \\ \text{Distance}=40*5 \\ \text{Distance}=200 \end{gathered}

We shall now label the total distance covered as d, and the number of hours taken (time) as t.

We now have for Mr Chandler;


d=40t+200---(1)

For Mrs Chandler, we would have;


d=80t---(2)

We would now solve the system of equations as follows;


\begin{gathered} d=40t+200---(1) \\ d=80t---(2) \\ \text{Substitute for the value of d into equation (1)} \\ We\text{ now have;} \\ 80t=40t+200 \\ \text{Collect all like terms;} \\ 80t-40t=200 \\ 40t=200 \\ \text{Divide both sides by 40} \\ t=5 \end{gathered}

This result shows that, travelling at the speed given for both Mr and Mrs Chandler, it would take her 5 hours to catch up with him.

ANSWER:

5 hours

PART B

If Mr Reeds leaves 3.5 hours ahead of Mrs Reeds and he travels at the speed of 55 miles per hour (55 mph), then his distance would be;


\begin{gathered} \text{Distance}=\text{spee}d*\text{time} \\ \text{Distance}=55*3.5 \\ \text{Distance}=192.5 \end{gathered}

If the total distance covered is given as d, then for Mr Reeds the total distance would be;


d=55t+192.5---(1)

For Mrs Reeds, we would have;


d=90t---(2)

Note that d represents the total distance covered and t represents the time taken (in hours).

We shall now solve the system of equations, as follows;


\begin{gathered} d=55t+192.5---(1) \\ d=90t---(2) \end{gathered}

We shall substitute for the value of d into equation two. Note that d = d, which means equation (1) equals equation (2). We now have;


\begin{gathered} 55t+192.5=90t \\ \text{Collect all like terms;} \\ 192.5=90t-55t \\ 192.5=35t \\ \text{Divide both sides by 35} \\ (192.5)/(35)=(35t)/(35) \\ 5.5=t \end{gathered}

The calculations here shows that travelling at the speeds given for both of them, it would take Mrs Reeds 5.5 hours to catch up with Mr Reeds.

ANSWER:

5.5 hours

User GMalc
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