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gsm8k-0008

1.000
1/1 testsΒ· math
Challenge Β· difficulty 2/5
# GSM8K Problem 9

John drives for 3 hours at a speed of 60 mph and then turns around because he realizes he forgot something very important at home.  He tries to get home in 4 hours but spends the first 2 hours in standstill traffic.  He spends the next half-hour driving at a speed of 30mph, before being able to drive the remaining time of the 4 hours going at 80 mph.  How far is he from home at the end of those 4 hours?

Reason step by step, then give ONLY the final answer on the last line as \boxed{ANSWER}.

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Proposed solution
We are given a scenario where John drives away from home for 3 hours at 60 mph, then turns around and tries to get back home in 4 hours. During those 4 hours, his driving is interrupted:

- First 2 hours: standstill traffic (0 mph)
- Next 0.5 hour: drives at 30 mph
- Remaining time: drives at 80 mph

We are to find **how far he is from home at the end of those 4 hours**.

---

### Step 1: Distance John travels away from home initially

He drives for 3 hours at 60 mph:

$$
\text{Distance}_{\text{out}} = 3 \times 60 = 180 \text{ miles}
$$

So he is **180 miles away from home** when he turns around.

Let’s define the direction *away from home* as positive. So his position at turn-around time is +180 miles.

Now, during the next 4 hours, he is trying to return home β€” but his motion may not bring him all the way back.

We need to compute his **net displacement during the 4-hour return attempt**, and subtract that from 180 to get final distance from home.

---

### Step 2: Motion during the 4-hour return attempt

Break it down:

- **First 2 hours**: standstill β†’ displacement = 0
- **Next 0.5 hour**: drives at 30 mph toward home β†’ displacement = $-30 \times 0.5 = -15$ miles (negative because toward home)
- **Remaining time**: total is 4 hours, so remaining = $4 - 2 - 0.5 = 1.5$ hours  
  He drives at 80 mph toward home β†’ displacement = $-80 \times 1.5 = -120$ miles

Total displacement during return attempt:

$$
\Delta x = 0 - 15 - 120 = -135 \text{ miles}
$$

So he moves **135 miles toward home** during the 4 hours.

---

### Step 3: Final distance from home

He started the return trip at +180 miles. After moving βˆ’135 miles:

$$
\text{Final position} = 180 - 135 = \boxed{45} \text{ miles from home}
$$

---

### βœ… Final Answer:

\boxed{45}