Word problems requiring algebraic solutions for real-world scenarios.

Let's solve each of the four word problems using the RESCA format:

> RESCA stands for:
> - R – Read the problem
> - E – Establish what is given and what is asked
> - S – Set up the equation(s)
> - C – Compute or solve
> - A – Answer and check

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Problem 1: Rectangle Dimensions

R – Read the problem
The length of a rectangle is 12 meters more than thrice its width. The perimeter is 200 m. Find the length and width.

E – Establish
- Let width = $ w $
- Then length = $ 3w + 12 $
- Perimeter of rectangle = $ 2(\text{length} + \text{width}) = 200 $

S – Set up
$$
2[(3w + 12) + w] = 200
$$

C – Compute
$$
2(4w + 12) = 200 \\
8w + 24 = 200 \\
8w = 176 \\
w = 22
$$

Now find length:
$$
\text{Length} = 3w + 12 = 3(22) + 12 = 66 + 12 = 78
$$

A – Answer
- Width = 22 m
- Length = 78 m

✔ Check: Perimeter = $ 2(78 + 22) = 2(100) = 200 $ ✔

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Problem 2: Investment at Two Interest Rates

R – Read the problem
P35,000 was invested in two banks: part at 5%, rest at 3%. Total interest after one year is P1,250. How much was invested at each rate?

E – Establish
- Let amount at 5% = $ x $
- Then amount at 3% = $ 35,000 - x $
- Interest from 5% = $ 0.05x $
- Interest from 3% = $ 0.03(35,000 - x) $
- Total interest = P1,250

S – Set up
$$
0.05x + 0.03(35,000 - x) = 1,250
$$

C – Compute
$$
0.05x + 1,050 - 0.03x = 1,250 \\
0.02x + 1,050 = 1,250 \\
0.02x = 200 \\
x = \frac{200}{0.02} = 10,000
$$

So:
- Amount at 5% = P10,000
- Amount at 3% = $ 35,000 - 10,000 = P25,000

A – Answer
- P10,000 invested at 5%
- P25,000 invested at 3%

✔ Check:
Interest = $ 0.05(10,000) + 0.03(25,000) = 500 + 750 = 1,250 $ ✔

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Problem 3: Buses Meeting

R – Read the problem
Bus A leaves Boac for Bukidnon at 75 kph. Bus B leaves Bukidnon for Zamboanga at 85 kph. Distance between Bukidnon and Zamboanga is 560 km. When will they meet?

E – Establish
- Bus A speed = 75 kph (from Boac to Bukidnon)
- Bus B speed = 85 kph (from Bukidnon to Zamboanga)
- But note: They are going toward each other, but the distance between Bukidnon and Zamboanga is 560 km.
- So, if Bus A is going from Boac → Bukidnon, and Bus B from Bukidnon → Zamboanga, then they are not moving toward each other unless Boac–Bukidnon–Zamboanga are in line.

Wait — likely a typo in interpretation.

But based on wording:
> "Bus A leaves Boac for Bukidnon", "Bus B leaves Bukidnon for Zamboanga"
→ So both buses are traveling in the same direction, not toward each other.

But then they can't meet unless Boac–Bukidnon–Zamboanga are in a line, and Bus A goes from Boac to Bukidnon, then continues? But the problem says “after how many hours will Bus A and Bus B meet?”

Wait — this doesn't make sense unless they're going toward each other.

Re-read:
“Bus A leaves Boac for Bukidnon” — so Boac → Bukidnon
“Bus B leaves Bukidnon for Zamboanga” — so Bukidnon → Zamboanga

So they are going away from each other.

But the question asks when they meet — that’s impossible unless they start at the same point.

Wait — perhaps it's a typo. Maybe Bus A leaves Bukidnon for Boac, and Bus B leaves Bukidnon for Zamboanga? That would make them go in opposite directions.

But the problem says:

> "Bus A leaves Boac for Bukidnon... Bus B leaves Bukidnon for Zamboanga"

So they are starting at different places.

But if Boac → Bukidnon → Zamboanga, then Bus A is going from Boac to Bukidnon (say, west to east), Bus B from Bukidnon to Zamboanga (eastward). So they are going in the same direction, with Bus B starting at Bukidnon and Bus A approaching Bukidnon.

But then they can only meet if Bus A reaches Bukidnon before Bus B leaves, but they leave at the same time.

Wait — maybe Boac is a typo? Or perhaps it's meant to be Bukidnon?

Alternatively, suppose Bus A leaves Bukidnon for Boac at 75 kph, and Bus B leaves Bukidnon for Zamboanga at 85 kph — then they are going in opposite directions, so they won’t meet.

But the question says “when will they meet?” So they must be moving toward each other.

Therefore, likely:
- Bus A leaves Boac for Bukidnon at 75 kph
- Bus B leaves Zamboanga for Bukidnon at 85 kph
- And the distance between Boac and Zamboanga is not given — but it says distance between Bukidnon and Zamboanga is 560 km.

This suggests that Boac–Bukidnon–Zamboanga are colinear.

But we don't know the distance from Boac to Bukidnon.

Alternatively, perhaps the problem meant:

> Bus A leaves Bukidnon for Zamboanga at 75 kph
> Bus B leaves Zamboanga for Bukidnon at 85 kph
> Distance between Bukidnon and Zamboanga = 560 km

Then they are moving toward each other.

But the problem says:

> "Bus A leaves Boac for Bukidnon" — so probably Boac is a mistake.

Wait — let's re-express carefully.

Possibility: There's a typo. It should be:

> Bus A leaves Bukidnon for Zamboanga at 75 kph
> Bus B leaves Zamboanga for Bukidnon at 85 kph
> Distance between Bukidnon and Zamboanga = 560 km

Then they are moving toward each other.

But the problem says:
> "Bus A leaves Boac for Bukidnon" — so Boac is a different place.

Unless Boac is Zamboanga? Unlikely.

Another possibility: Perhaps Bus A is going from Boac to Bukidnon, and Bus B from Bukidnon to Zamboanga, and they start at the same time, but only meet if Bus A arrives at Bukidnon before Bus B departs — which isn't possible since they leave simultaneously.

But the problem says “After how many hours will Bus A and Mr. Bus B meet?” — clearly a typo: “Mr. Bus B” should be “Bus B”.

Given the confusion, and the fact that the distance between Bukidnon and Zamboanga is 560 km, and speeds are given, likely the intended meaning is:

> Bus A leaves Bukidnon for Zamboanga at 75 kph
> Bus B leaves Zamboanga for Bukidnon at 85 kph
> Distance between Bukidnon and Zamboanga = 560 km
> They start at the same time, moving toward each other.

That makes sense.

We’ll assume that the problem has a typo: “Boac” should be “Bukidnon”, and “Bus B leaves Bukidnon for Zamboanga” — but wait, then both are going from Bukidnon to Zamboanga.

No.

Wait — maybe:

> Bus A leaves Bukidnon for Zamboanga at 75 kph
> Bus B leaves Zamboanga for Bukidnon at 85 kph

Yes — that fits.

So corrected version:

> Bus A leaves Bukidnon for Zamboanga at 75 kph
> Bus B leaves Zamboanga for Bukidnon at 85 kph
> Distance between Bukidnon and Zamboanga = 560 km
> When do they meet?

E – Establish
- Relative speed = 75 + 85 = 160 kph (since moving toward each other)
- Distance = 560 km
- Time = distance / relative speed

S – Set up
$$
\text{Time} = \frac{560}{75 + 85} = \frac{560}{160}
$$

C – Compute
$$
\frac{560}{160} = 3.5 \text{ hours}
$$

A – Answer
They will meet after 3.5 hours (or 3 hours 30 minutes).

✔ Check:
- Bus A travels: $ 75 \times 3.5 = 262.5 $ km
- Bus B travels: $ 85 \times 3.5 = 297.5 $ km
- Total = $ 262.5 + 297.5 = 560 $ km ✔

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Problem 4: Two Cars Leaving Opposite Directions

R – Read the problem
Two cars leave Cagayan de Oro City at 8:00 AM. One heads to Pagadian at 72 kph, the other in the opposite direction at 63 kph. When will they be 540 km apart? How far has each traveled?

E – Establish
- Both start at 8:00 AM
- Moving in opposite directions
- Speeds: 72 kph and 63 kph
- Distance apart increases at combined speed
- Want time when total separation = 540 km

S – Set up
Let $ t $ = time in hours after 8:00 AM
Distance = $ (72 + 63)t = 135t $
Set equal to 540:
$$
135t = 540
$$

C – Compute
$$
t = \frac{540}{135} = 4 \text{ hours}
$$

So they will be 540 km apart at $ 8:00 + 4 $ hours = 12:00 PM (noon)

Distance traveled:
- Car 1: $ 72 \times 4 = 288 $ km
- Car 2: $ 63 \times 4 = 252 $ km

A – Answer
- They will be 540 km apart at 12:00 PM
- Car 1 traveled 288 km
- Car 2 traveled 252 km

✔ Check: $ 288 + 252 = 540 $ km ✔

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✔ Final Answers Summary:

#### Problem 1:
- Width = 22 m
- Length = 78 m

#### Problem 2:
- Invested at 5%: P10,000
- Invested at 3%: P25,000

#### Problem 3:
- Time until meeting: 3.5 hours (3 hours 30 minutes)
- Assuming correction: Bus A from Bukidnon to Zamboanga at 75 kph, Bus B from Zamboanga to Bukidnon at 85 kph

#### Problem 4:
- Time: 12:00 PM
- Distances: 288 km and 252 km

Let me know if you want diagrams or further explanation!