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Tom and Jimmy plan to leave at the same time from home and
walk towards each other. Tom walks 52 meters per minute, Jimmy
walks 70 meters per minute. The two meet at cafe shop on the
way. Let's suppose that Tom leaves 4 minutes earlier than Jimmy
and their speeds remain the same, the two will still meet at the
cafe shop. What is the distance between Tom and Jimmy's home?

Sagot :

To find the distance between the homes of Tom and Jimmy, it is assumed

that the distances from their home to the café are equal.

  • The distance between Tom and Jimmy's home is [tex]\underline{1,617.\overline 7 \ meters}[/tex]

Reasons:

The direction in which Tom and Jimmy walks = Towards each other

The speed at which Tom walks = 52 meters per minute

The speed with which Jimmy walks = 70 meters per minute

The time at which Tom leaves = 4 minutes earlier than Jimmy

The point at which they meet = The café

The rate of their speed = Constant

Required:

The distance between Tom and Jimmy home.

Solution:

Tom and Jimmy had a plan to walk at the same speed and meet up at the café.

We have;

The café is equal distance from Tom and Jimmy's houses.

Which gives the following simultaneous equation.

52 × (4 + t) = The distance of Tom's house from the café

70 × t = The distance of Jimmy's  house from the café

  • 52 × (4 + t) = 70 × t

52 × 4 = 70 × t - 52 × t = 18 × t

[tex]\displaystyle t = \frac{52 \times 4}{18} = \frac{104}{9} = 11.\overline{6}[/tex]

The time it take Jimmy to reach the café, t = [tex]\mathbf{11.\overline6}[/tex] minutes

The distance between their homes, d = 52 × (4 + t) + 70 × t

∴ d = 52 × (4 + [tex]11.\overline6[/tex]) + 70 × [tex]11.\overline6[/tex] = 1,617.[tex]\mathbf{\overline 7}[/tex]

  • The distance between Tom and Jimmy's home = 1,617.[tex]\overline 7[/tex] meters

Learn more about simultaneous equations here:

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