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24. The diagram on the right shows the distance
between
and Brother walked
back and forth between and
once
for
km.
If Sister walked from
to
SCHOOL
later. She then came back to
F
and went to
passing
thru . In total, she walked for 20 km for the whole journey. The
distance between
and
was
km.

Sagot :

Answer:

- The distance from (F) to {SCHOOL} (d_{FS} is 8 km

- The distance between the two points Brother walked between (which we assumed to be (F) and (G) is 4 km

Step-by-step explanation:

Given Information:

[tex]{ - \text{ (Total distance Brother walked} \(d_{BB}\): \(8 \, \text{km}}[/tex]

[tex]- {\text{(Total distance Sister walked} \(d_{SS}\): \(20 \, \text{km}}[/tex]

[tex]{- Distance from \: \(F\) \: to \ \: (\text{SCHOOL}\) \: and \: back \: (\(2 \times d_{FS}\)) \: is \: a \: part \: of \: Sister's \: total \: distance.} [/tex]

Step-by-Step Solution:

1. Brother's total distance (back and forth) is twice the distance between two points:

[tex]{2 \times d_{BB} = 8 \, \text{km} \implies d_{BB} = \frac{8}{2} = 4 \, \text{km}}[/tex]

The distance between the two points

[tex] \text{(we will assume these are} (F) and (G) \: is \: (4 , \text{km}\).[/tex]

2. Sister's Journey:

For Sister, the total distance walked is:

[tex] {d_{FS} + d_{FS} + d_{FG} = 20 \, \text{km} \implies 2d_{FS} + d_{FG} = 20 \, \text{km}}[/tex]

[tex] {\text{We already know from Brother's journey that} \(d_{FG} = 4 \, \text{km}}[/tex]

3. Substituting (d_{FG} into the equation:

[tex]2d_{FS} + 4 \, \text{km} = 20 \, \text{km}[/tex]

[tex]2d_{FS} = 20 \, \text{km} - 4 \, \text{km} = 16 \, \text{km}[/tex]

[tex]d_{FS} = \frac{16}{2} = 8 \, \text{km}[/tex]