← Sensim Lab Sensim Math

AMC 8 · 2018 · #12

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

rateratio-proportionfraction-arithmetic dimensional-analysisidentify-subproblems ↑ Prerequisites: fraction-arithmeticratio-proportion

📏 Medium solution 💡 3 insights

Problem

The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping, he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual time?

Pick an answer.

(A)

5:50

(B)

6:00

(C)

6:30

(D)

6:55

(E)

8:10

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Sri's car clock runs fast at a steady rate. At noon, both his accurate wristwatch and the car clock read $12{:}00$. By the time his watch reads $12{:}30$, the car clock already reads $12{:}35$. Later that day his watch is gone, and the car clock reads $7{:}00$. What is the true time?

Givens: At the start, both clocks read $12{:}00$ noon; When the accurate watch advances $30$ minutes, the car clock advances $35$ minutes; The car clock gains time at a constant rate; Later the car clock reads $7{:}00$; Answer choices: (A) $5{:}50$, (B) $6{:}00$, (C) $6{:}30$, (D) $6{:}55$, (E) $8{:}10$

Unknowns: The actual (true) time when the car clock displays $7{:}00$

Understand

Restated: Sri's car clock runs fast at a steady rate. At noon, both his accurate wristwatch and the car clock read $12{:}00$. By the time his watch reads $12{:}30$, the car clock already reads $12{:}35$. Later that day his watch is gone, and the car clock reads $7{:}00$. What is the true time?

Givens: At the start, both clocks read $12{:}00$ noon; When the accurate watch advances $30$ minutes, the car clock advances $35$ minutes; The car clock gains time at a constant rate; Later the car clock reads $7{:}00$; Answer choices: (A) $5{:}50$, (B) $6{:}00$, (C) $6{:}30$, (D) $6{:}55$, (E) $8{:}10$

Plan

Primary tool: #8 Analyze the Units

Secondary: #5 Look for a Pattern, #6 Guess and Check

This is a rate problem with two different time "units" — car-clock minutes and real minutes. Tool #8 (Analyze the Units) keeps them straight: the data point ($35$ car-min per $30$ real-min) defines an exchange rate between the two. Tool #5 (Look for a Pattern) turns that data point into the cleaner repeating block "$7$ car-min $= 6$ real-min," which is easy to scale by counting how many blocks fit in $420$ car-minutes. Tool #6 (Guess and Check) gives a quick sanity test against the multiple-choice answers at the end.

Execute — Answer: B

#8 Analyze the Units 6.RP.A.1 Step 1

Translate the synchronization data into a clean ratio.
The car clock gains $5$ extra minutes for every $30$ real minutes, so the ratio of car-clock time to real time is $35{:}30$.
Dividing both numbers by $5$ gives the simplest form $7{:}6$.

$$\dfrac{\text{car min}}{\text{real min}} = \dfrac{35}{30} = \dfrac{7}{6}$$

💡 Forming a ratio between two matched quantities (car-time to real-time) is exactly the Grade 6 definition of a ratio.

#5 Look for a Pattern 6.RP.A.1 Step 2

Restate the ratio as a repeating pattern: every $6$ real minutes that pass, the car clock advances $7$ minutes.
This pattern is the same forever because the car clock gains at a constant rate, so it can be scaled to any chunk of time.

$$\underbrace{6 \text{ real min}}_{\text{one block}} \;\longleftrightarrow\; \underbrace{7 \text{ car min}}_{\text{one block}}$$

💡 Spotting a repeatable "$6$-and-$7$" block turns a one-time data point into a pattern you can scale.

#8 Analyze the Units 5.MD.A.1 Step 3

Figure out how much car-clock time has passed since noon.
The car clock now reads $7{:}00$ and started at $12{:}00$, so $7$ hours have passed on the car clock.
Convert to minutes so the units match the ratio: $7 \times 60 = 420$ car-minutes.

$$7 \text{ hr} \times 60 \tfrac{\text{min}}{\text{hr}} = 420 \text{ car-min}$$

💡 Converting hours into minutes within the same time system is a Grade 5 standard-unit conversion.

#8 Analyze the Units 6.RP.A.3 Step 4

Count how many "$7$ car-min" blocks fit in $420$ car-minutes.
Since $420 \div 7 = 60$, exactly $60$ blocks have passed.
Each block represents $6$ real minutes, so the total real time is $60 \times 6 = 360$ real-minutes.

$$\dfrac{420 \text{ car-min}}{7 \tfrac{\text{car-min}}{\text{block}}} = 60 \text{ blocks} \;\Rightarrow\; 60 \times 6 = 360 \text{ real-min}$$

💡 Using the unit rate "$6$ real-min per block" to scale the count is exactly Grade 6 rate reasoning.

#8 Analyze the Units 4.MD.A.2 Step 5

Convert $360$ real-minutes back into hours, then add to the noon starting time.
$360 \div 60 = 6$ hours, and $12{:}00 + 6{:}00 = 6{:}00$.
The actual time is $6{:}00$, which matches choice (B).

$$360 \text{ min} \div 60 = 6 \text{ hr}, \quad 12{:}00 + 6{:}00 = 6{:}00 \;\Rightarrow\; \textbf{(B)}$$

💡 Adding elapsed hours to a starting clock time is a Grade 4 distance-and-time word-problem move.

[1] #8 6.RP.A.1 Translate the synchronization data into a clean ratio. The car clock gains $5$ e

[2] #5 6.RP.A.1 Restate the ratio as a repeating pattern: every $6$ real minutes that pass, the

[3] #8 5.MD.A.1 Figure out how much car-clock time has passed since noon. The car clock now read

[4] #8 6.RP.A.3 Count how many "$7$ car-min" blocks fit in $420$ car-minutes. Since $420 \div 7

[5] #8 4.MD.A.2 Convert $360$ real-minutes back into hours, then add to the noon starting time.

Review

Reasonableness: The car clock runs $\tfrac{7}{6}$ as fast as real time, so it should always read a bigger number than the actual elapsed time. The car shows $7$ hours elapsed; the real elapsed time should be smaller, and $6$ hours is indeed smaller. Also, $7 \times \tfrac{6}{7} = 6$ exactly, so the answer comes out as a whole number of hours with no leftover minutes — a clean match to choice (B) $6{:}00$. Choices (A) $5{:}50$ and (C) $6{:}30$ would require the ratio to be different from $7{:}6$, and (D) and (E) are clearly too far off.

Alternative: Tool #6 (Guess and Check) on the choices: for each candidate real time $T_A$, compute the car-clock reading $T_C = \tfrac{7}{6} T_A$. (A) $5{:}50$ ($350$ min) gives $T_C = 408.3$ min, not $420$. (B) $6{:}00$ ($360$ min) gives $T_C = 420$ min, exactly $7{:}00$ on the car clock. ✓. The other choices overshoot or undershoot, confirming (B).

CCSS standards used (min grade 6)

4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money (Adding $6$ elapsed hours to the noon start time to land on the actual clock reading of $6{:}00$.)
5.MD.A.1 Convert among different-sized standard measurement units within a given system (Converting $7$ hours on the car clock into $420$ minutes so the units match the $7{:}6$ ratio.)
6.RP.A.1 Understand the concept of a ratio and use ratio language (Reading the $35$-car-min vs. $30$-real-min data as the ratio $7{:}6$ and as a repeating $6$-real-to-$7$-car block.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Scaling the $7{:}6$ ratio across $420$ car-minutes to recover the $360$ real-minutes that have actually elapsed.)

⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — turning "$35$ car-min : $30$ real-min" into the clean $7{:}6$ ratio — that you already know!

⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — turning "$35$ car-min : $30$ real-min" into the clean $7{:}6$ ratio — that you already know!