Sam used to walk to school every morning, and it takes him 20 min. Once on his way, he realized that

Q: Why is the fraction of distance equal to the fraction of time?

Because Sam walks at a constant speed. Distance is directly proportional to time ($d = v \times t$). Therefore, the ratio of the distance to the turning point over the total distance is $(v \times t_1) / (v \times 20)$, which simplifies perfectly to $t_1 / 20$.

Q: Why do we add 20 minutes for the final trip? Didn't he already cover some distance?

When Sam turns back, he goes all the way back to his home to get the notebook. His final trip starts from home all over again, so he has to cover the entire distance to school from scratch, which takes the full 20 minutes.

Q: How did we get the bell time as 28 minutes?

The problem states that if he took his usual 20 minutes, he would arrive 8 minutes *before* the bell. This means the bell rings 8 minutes *after* his usual arrival time. So, $20 + 8 = 28$ minutes from when he left home.

Q: What if his speed going back was different?

If his speed going back was, say, twice as fast, the time to return would be $t_1 / 2$. The total time equation would then be $t_1 + t_1/2 + 20 = 38$. But since the problem says 'same speed', the return time is exactly $t_1$.

Q: Does the actual distance to the school matter?

Not at all! Since we are asked for a fraction, the actual distance $D$ and speed $v$ cancel out in the final step. We only need to work with the time ratios, which makes the math much cleaner.

Question

Sam used to walk to school every morning, and it takes him 20 min. Once on his way, he realized that he had forgotten his homework notebook at home. He knew that if he continued walking to school at the same speed, he would be there 8 min before the bell, so he went back home for the notebook and arrived the school 10 min after the bell. If he had walked all the way with his usual speed, what fraction of the way to school had he covered till the moment he turned back?

Options:

(A) 8/20

(B) 9/20

(C) 10/20

(D) 12/20

Sigma Percentile · Accepted Answer

### The Case of the Forgotten Notebook: A Kinematics Mystery

We've all been there. You are walking to school, lost in thought, and suddenly it hits you—you left your homework on the desk. This classic student dilemma is actually a beautiful exercise in relative time and constant velocity kinematics. Let's break down Sam's chaotic morning step by step.

### Setting the Baseline: The Perfect Morning

Before we analyze the chaos, we must understand the baseline. On a normal day, Sam walks from his home to school at a constant speed, $v$. We are told this journey takes exactly $20$ minutes.

We are also given a crucial piece of information to anchor our timeline: if he had continued at his usual speed, he would have arrived $8$ minutes *before* the school bell rings.

Let's define the time the bell rings as $T_{	ext{bell}}$. Since his usual arrival time is $20$ minutes after leaving home, we can write:

$$20 = T_{	ext{bell}} - 8$$

Solving this simple equation tells us exactly when the bell rings relative to his departure:

$$T_{	ext{bell}} = 28 	ext{ minutes}$$

### The Chaotic Journey

Now, let's look at what actually happened. Sam walks for a certain amount of time before realizing his mistake. Let's call this time $t_1$.

Because his speed $v$ is constant, the time it takes him to walk back home is exactly the same: $t_1$.

Once he is back home, he grabs his notebook and has to walk the entire distance to school again. We already know this full trip takes $20$ minutes.

So, what is the total time Sam spent walking that morning? It is the sum of these three segments:

$$T_{	ext{total}} = t_1 	ext{ (out)} + t_1 	ext{ (back)} + 20 	ext{ (final trip)}$$

$$T_{	ext{total}} = 2t_1 + 20$$

### The Grand Equation

The problem states that on this chaotic day, Sam arrives at school $10$ minutes *after* the bell rings.

We already calculated that the bell rings at the $28$-minute mark. Therefore, his actual arrival time is:

$$T_{	ext{total}} = 28 + 10 = 38 	ext{ minutes}$$

Now, we simply equate our two expressions for the total time:

$$2t_1 + 20 = 38$$

Subtracting $20$ from both sides gives:

$$2t_1 = 18$$

$$t_1 = 9 	ext{ minutes}$$

### The Final Reveal

The question asks for the fraction of the way to school he had covered before turning back.

Since Sam walks at a constant speed, the distance he covers is directly proportional to the time he walks ($d = v 	imes t$). Therefore, the fraction of the total distance is exactly equal to the fraction of the total usual time.

$$	ext{Fraction} = \frac{	ext{Distance to turn}}{	ext{Total Distance}} = \frac{v \cdot t_1}{v \cdot 20} = \frac{t_1}{20}$$

Substituting our value of $t_1 = 9$, we get our final answer:

**Fraction = 9/20**

By carefully tracking the timeline and trusting the constancy of his speed, a seemingly complex word problem unravels into a beautiful, logical sequence.

Select Answer:

Visualized Solution

$T_{usual} = 20 min$

$T_{arrive} = T_{bell} - 8$

$T_{bell} = 28 min$

$Time to turn = t_{1}$

$Time to return = t_{1}$

$Final trip = 20 min$

$T_{total} = 2 t_{1} + 20$

$2 t_{1} + 20 = 28 + 10$

$t_{1} = 9 min$

$Fraction = \frac{9}{20}$

$What if v_{return} = 2 v ?$

The Sigma Insight: Motion in a Straight Line

The Case of the Forgotten Notebook

Setting the Baseline

The Chaotic Journey

The Grand Equation

The Final Reveal

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tusual​=20 min

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tarrive​=Tbell​−8

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tbell​=28 min

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time to turn=t1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time to return=t1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final trip=20 min

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ttotal​=2t1​+20

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2t1​+20=28+10

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​=9 min

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fraction=209​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if vreturn​=2v?