Analyze Solar-System Unit Rates and Scale - 6th Grade Math

1. If the rover runs under this pattern for exactly 5 hours (300 minutes), how far does it travel in total?
   Answer: $180 \text{ m}$ This question tests rate problems with a twist: the rover doesn't drive the entire time, so you must separate active driving time from paused time. When a rate problem includes breaks or interruptions, always find the true amount of "working" time before you multiply by the rate. Start by counting the cycles. Each full cycle is $60$ minutes (48 driving + 12 pausing), and $300 \div 60 = 5$ complete cycles. In each cycle the rover only drives for 48 minutes, so total driving time is $5 \times 48 = 240$ minutes, which equals $240 \div 60 = 4$ hours. At $45$ meters per hour, the distance is $45 \times 4 = 180$ meters. The choice of $225 \text{ m}$ comes from ignoring the pauses entirely and using all $5$ hours of driving: $45 \times 5 = 225$. That's the trap for students who forget the rover stops. The choice of $144 \text{ m}$ mistakenly treats the 12-minute pauses as the driving time: $5 \times 12 = 60$ minutes $= 1$ hour is wrong reasoning, or it mixes up minutes and rates carelessly. The choice of $36 \text{ m}$ likely comes from using only a single cycle's driving time or dividing incorrectly, capturing far too little travel. Study tip: In any rate-with-breaks problem, ask yourself, "How much time is the object actually moving?" Subtract the idle time first, convert minutes to hours so your units match the rate, and only then multiply distance $=$ rate $\times$ time.
2. Using unit rates, how many more degrees per hour does Jupiter rotate compared to Earth?
   Answer: $21°\text{/hr}$ Whenever a problem asks about a unit rate, remember that "unit" means "per one" — so a rate like degrees per hour tells you how much rotation happens in exactly one hour. To find it, divide the total amount by the total time. Start with each planet separately. Earth rotates $360°$ in $24$ hours, so its unit rate is $360 \div 24 = 15°\text{/hr}$. Jupiter rotates $360°$ in only $10$ hours, so its unit rate is $360 \div 10 = 36°\text{/hr}$. The question asks how many more degrees per hour Jupiter rotates, which is a difference: $36 - 15 = 21°\text{/hr}$. That confirms $21°\text{/hr}$. Now look at the traps. The choice $2.4°\text{/hr}$ comes from dividing $24 \div 10$ — comparing the times instead of finding rotation rates. The choice $9°\text{/hr}$ mistakenly divides $360$ by the difference in hours ($24-10=14$)... actually it comes from careless arithmetic like $24-15$; either way it skips computing Jupiter's true rate. The choice $14°\text{/hr}$ is just the difference in hours ($24-10$), not a rate in degrees per hour at all — a classic "wrong units" trap. The takeaway: when comparing two rates, always compute each unit rate first, then subtract at the very end. Watch your units — if the answer should be in degrees per hour, any number that's really just a difference in hours (like $14$) is a red flag that you compared the wrong quantities.
3. Mission control needs the probe to send a 960-kilobit image. How much real time (including the interference gaps) will it take to fully download the image?
   Answer: $50 \text{ seconds}$ Whenever a question involves a "steady rate" interrupted by regular gaps, the trick is to separate two different questions: how much actual sending time do I need? and how much extra real time do the gaps add? Rushing to multiply or divide without accounting for the pauses is exactly the trap here. Start with the pure transmission time. The probe sends $32$ kilobits every second it has a signal, so to send $960$ kilobits it needs $960 \div 32 = 30$ seconds of actual data-sending time. But the probe only sends data $3$ out of every $5$ seconds. So those $30$ working seconds don't happen back-to-back — they're spread out across cycles. In each $5$-second cycle, only $3$ seconds do work. To get $30$ working seconds, you need $30 \div 3 = 10$ full cycles, and each cycle is $5$ seconds long: $10 \times 5 = 50$ seconds of real time. Now the wrong answers: **$30$ seconds is the pure sending time — it ignores the interference gaps entirely. $18$ seconds** comes from mistakenly multiplying $30$ by $\frac{3}{5}$, shrinking the time instead of stretching it. **$33$ seconds** is a guess from adding a small gap ($30 + 3$) rather than scaling by the whole cycle ratio. The key takeaway: when work only happens part of the time, real time gets longer, not shorter. Find the working time first, then multiply by the ratio of total-time to working-time ($\frac{5}{3}$ here) to stretch it out.
4. Sunlight takes about 8 minutes to travel the $150{,}000{,}000$ km from the Sun to Earth. Using this rate, about how many minutes does it take sunlight to reach Saturn, which is about $1{,}500{,}000{,}000$ km from the Sun?
   Answer: 80 minutes Whenever a problem gives you a rate ("8 minutes per 150,000,000 km") and asks about a new distance, think proportional reasoning: if the distance grows, the time grows by the same factor. The smart first move is to compare the two distances. Look at the numbers carefully. Saturn is $1{,}500{,}000{,}000$ km from the Sun, while Earth is $150{,}000{,}000$ km. Dividing, $\frac{1{,}500{,}000{,}000}{150{,}000{,}000} = 10$, so Saturn is 10 times farther. Since light travels at a constant speed, it takes 10 times as long: $8 \times 10 = 80$ minutes. The choice of $800$ minutes comes from multiplying by 100 instead of 10 — a place-value slip from miscounting the zeros. Both numbers have the same leading digits, so the only difference is one extra zero, meaning a factor of 10, not 100. The choice of $40$ minutes multiplies by 5, which would only be right if Saturn were 5 times farther, but it's 10 times. The choice of $8$ minutes is the original Earth time unchanged — a trap for anyone who forgets that a farther distance needs more time, not the same amount. Your strategy: when comparing large numbers with the same leading digits, just count the extra zeros. Each extra zero means the number is 10 times bigger. Here, Saturn's distance has exactly one more zero than Earth's, so multiply the time by 10. Setting up the ratio before calculating keeps you from mismatching the zeros.
5. A class builds a model using a scale of $1$ cm $= 10,000$ km for planet diameters. If that same scale were used for distances, the model Earth should sit $150$ m from the model Sun, since Earth's actual distance from the Sun is about $150,000,000$ km. However, because of limited hallway space, the class instead placed the model Earth only $1.5$ m from the model Sun. About how many times smaller is the distance scale actually used compared to the diameter scale?
   Answer: 100 times smaller Whenever you compare two scales, you're really asking "how much does 1 cm represent in each case?" The scale that squeezes more real distance into the same model length is the smaller (more shrunken) scale. Set up both as ratios of real distance to model distance and compare. Start with what the diameter scale actually is: $1 \text{ cm} = 10{,}000 \text{ km}$. Now find the distance scale the class really used. Earth's real distance is $150{,}000{,}000$ km, mapped onto a model distance of $1.5$ m $= 150$ cm. That means each centimeter stands for $150{,}000{,}000 \div 150 = 1{,}000{,}000$ km. Comparing the two: $1{,}000{,}000 \div 10{,}000 = 100$. Each centimeter of the distance model covers 100 times more real space, so the distance scale is 100 times smaller than the diameter scale. The "$10$ times smaller" choice comes from only shrinking once (comparing $150$ m to $1.5$ m gives a factor of 100, not 10). The "$1{,}000$ times smaller" answer overshoots — that's roughly the raw ratio before dividing out the original scale. The "$15$ times smaller" choice grabs the loose number $15$ from the distances without doing any real scaling calculation; it's a trap for pattern-matching digits instead of computing ratios. A reliable tip: convert every scale into "how much real distance does 1 cm represent," keep all units consistent (turn meters into centimeters!), then divide the two representations. The bigger the km-per-cm, the smaller the scale.
6. On a scale model laid out across a football field, the model Earth is placed $10$ m from the model Sun. Earth's actual average distance from the Sun is $150,000,000$ km, and Mars's actual average distance from the Sun is $228,000,000$ km. Using the same scale, about how far (to the nearest meter) should the model Mars be placed from the model Sun?
   Answer: 15 m Whenever you see a scale model problem, think proportions: the ratio between the model and the real thing stays the same for every object. Your job is to find that scale, then apply it consistently. Start by finding the scale from Earth's numbers. The model puts Earth $10$ m from the Sun, while the real distance is $150,000,000$ km. So every $15,000,000$ km of real distance equals $1$ m on the model (since $150,000,000 \div 10 = 15,000,000$). Now apply that same scale to Mars, which is $228,000,000$ km away: $228,000,000 \div 15,000,000 = 15.2 \text{ m}$ Rounded to the nearest meter, that's 15 m. The choice of $23$ m comes from treating $228$ million as if it were $228$ meters — dividing incorrectly or dropping too many zeros so the scale gets distorted. The answer of $20$ m looks like a rough guess that Mars should be "about double" Earth's distance, but $228$ million isn't double $150$ million — it's only about $1.5$ times as far. The choice of $7$ m flips the ratio the wrong way, making Mars closer than Earth, when Mars is actually farther out. Your takeaway: in scale problems, always find the "unit rate" first — how many real units equal one model unit — then multiply or divide the same way for every object. Setting up a clean proportion like $\frac{10}{150{,}000{,}000} = \frac{x}{228{,}000{,}000}$ keeps you from misplacing zeros.
7. A spacecraft travels $450{,}000{,}000$ km to Jupiter in 900 days at a constant speed. At that same speed, how many days would it take the spacecraft to travel $750{,}000{,}000$ km to Saturn?
   Answer: 1,500 days Whenever you see a problem with a constant speed (or constant rate), think proportions: the distance and time increase together at the same ratio. Your first move is to find the unit rate—how far the spacecraft travels in one day. Divide the distance by the time: $450{,}000{,}000 \div 900 = 500{,}000$ km per day. Now use that speed to find the time for the longer trip by dividing distance by rate: $750{,}000{,}000 \div 500{,}000 = 1{,}500$ days. You can also check with a proportion: $\frac{450{,}000{,}000}{900} = \frac{750{,}000{,}000}{x}$, which solves to $x = 1{,}500$ days. The answer of 540 days comes from setting up the proportion upside down—multiplying when you should divide, so you get a smaller time for a longer trip, which makes no sense. The choice of 1,200 days looks like a guess from adding a fixed amount ($900 + 300$) instead of scaling by the speed. The value 1,350 days comes from multiplying 900 by $1.5$ using the wrong ratio ($\frac{750}{500}$ paired incorrectly) or estimating loosely; the true multiplier is $\frac{750}{450} \approx 1.67$, giving 1,500, not 1,350. The key trap here is that a longer distance must take more time at the same speed—so immediately eliminate any answer smaller than 900. On rate problems, always find the unit rate first, then double-check that your answer moves in the logical direction.
8. A classroom scale model represents Earth's actual diameter of $12,000$ km with a marble that is $1$ cm across. Using this same scale, how far apart should the model Earth and model Sun be placed if the real average Earth–Sun distance is $150,000,000$ km?
   Answer: 125 m Whenever you see a scale-model problem, the key idea is that the scale factor is a single ratio that applies to every measurement. Once you find how many real kilometers one model centimeter represents, you use that same ratio for all other distances. Start by finding the scale. The marble is $1$ cm across and represents $12{,}000$ km. So every $1$ cm of model equals $12{,}000$ km of reality. To find the model Earth–Sun distance, divide the real distance by that number: $\frac{150{,}000{,}000 \text{ km}}{12{,}000 \text{ km per cm}} = 12{,}500 \text{ cm}.$ Now convert centimeters to meters (since the answers are in meters): $12{,}500 \text{ cm} \div 100 = 125 \text{ m}.$ That matches 125 m. The trap answer 12,500 m comes from forgetting to convert — $12{,}500$ is the correct value in centimeters, not meters. Choosing 12.5 m means you divided by $100$ twice, over-converting the units. Picking 1,250 m comes from a place-value slip, dividing by only $10$ instead of $100$ when converting cm to m. Your takeaway: in scale problems, always do the division first to stay in the original units, then convert to the units the answer choices use as a separate final step. Remembering that $100$ cm $=1$ m protects you from the most common trap — landing on a number that's right but off by a factor of $100$.
9. Probe Alpha travels $2{,}400{,}000$ km every 2 hours. Probe Beta travels $36{,}000{,}000$ km every 2 days. Which probe has the greater unit rate of speed, and what is that rate in km per hour?
   Answer: Probe Alpha, at 1,200,000 km per hour, dividing 2,400,000 by 2 hours and comparing to Beta's rate To compare fairly, find each probe's speed in km per hour. Probe Alpha travels 2,400,000 km in 2 hours, so its rate is 2,400,000 ÷ 2 = 1,200,000 km per hour. Probe Beta travels 36,000,000 km in 2 days; since 2 days = 48 hours, its rate is 36,000,000 ÷ 48 = 750,000 km per hour. Because 1,200,000 > 750,000, Probe Alpha is faster, so 'Probe Alpha, at 1,200,000 km per hour' is correct. The choice giving Alpha 2,400,000 km per hour never divided by the 2 hours — that number is the total distance, not a per-hour rate. The choice giving Beta 1,500,000 km per hour used only 24 hours (one day) instead of the full 48 hours, and then wrongly named the slower probe as faster. The choice giving Beta 750,000 km per hour computes Beta's speed correctly but picks the slower probe as the winner. Remember: convert all times to the same unit (1 day = 24 hours) before dividing distance by time, then compare.
10. Using the same scale, how far from the Sun should Mars be placed in the model if Mars's actual distance from the Sun is about 225 million km?
    Answer: $22.5 \text{ m}$ Whenever you see a scale model problem, the key idea is that every real distance is shrunk by the same factor. Your first job is to find that scale factor, then apply it consistently to every object. Here, Earth's actual distance of 150 million km is represented by 15 meters. To find how many million km each model meter represents, divide: $150 \div 15 = 10$ million km per meter. In other words, you shrink real distances by dividing by 10 (in these units). So for Mars, take its actual distance and divide the same way: $225 \div 10 = 22.5$ meters. This keeps the model proportional, which is why 22.5 m is correct. The choice 2.25 m comes from dividing by 100 instead of 10 — a decimal-place slip. Watch your place value: since $15$ (not $1.5$) matched $150$, the factor is $10$, not $100$. The choice 15 m is just Earth's model distance repeated. Mars is farther from the Sun than Earth, so its model distance must be larger than 15 m, not equal to it. The choice 225 m ignores the scale entirely — it's just Mars's actual number in millions written as meters. You must apply the shrinking factor, not copy the real distance. A reliable strategy for scale problems: set up a proportion, $\frac{15}{150} = \frac{x}{225}$, and solve. Cross-multiplying gives $150x = 15 \times 225$, so $x = 22.5$. Setting up the ratio the same way on both sides protects you from decimal and copying errors.
11. In a scale model, a foam ball with a diameter of $14$ cm represents the Sun, whose actual diameter is about $1,400,000$ km. Using this same scale, what should be the diameter of the model Earth, given that Earth's actual diameter is about $12,600$ km?
    Answer: 0.126 cm Whenever you see a scale model problem, the key idea is that everything shrinks by the same ratio. Your first job is to find that scale factor by comparing the model size to the real size for the object you already know—here, the Sun. The Sun's real diameter is $1{,}400{,}000$ km, and its model is $14$ cm. So the scale factor is: $\frac{14 \text{ cm}}{1{,}400{,}000 \text{ km}} = \frac{1}{100{,}000}$ Every real measurement gets divided by $100{,}000$ to become a model measurement. Apply this to Earth: $\frac{12{,}600 \text{ km}}{100{,}000} = 0.126 \text{ cm}$ That's why 0.126 cm is correct—you divided Earth's real diameter by the same $100{,}000$ you found from the Sun. The choice 12.6 cm comes from dividing by only $1{,}000$—a common slip when you miscount the zeros in $1{,}400{,}000$. The choice 1.26 cm results from dividing by $10{,}000$, again losing a zero in the scale factor. The choice 0.0126 cm goes the other way, dividing by $1{,}000{,}000$—one too many zeros. All three wrong answers are decimal-point errors from getting the scale factor slightly off. The safe strategy: always write your scale as a clean fraction and simplify it fully before using it. Count the zeros carefully—$1{,}400{,}000 \div 14 = 100{,}000$ has exactly five zeros. Then apply that exact same divisor to every other object. Keeping the ratio consistent is what these scale problems are really testing.
12. How many more kilometers does Mercury travel than Venus in one hour?
    Answer: $46{,}800 \text{ km}$ The speeds are given in kilometers per second, but the question asks about one hour, so convert first: one hour has $60 \times 60 = 3{,}600$ seconds. The phrase "how many more" means you compare, so find the difference in speeds, $48 - 35 = 13$ km per second, then multiply by the seconds in an hour: $13 \times 3{,}600 = 46{,}800$ km. So Mercury travels $46{,}800 \text{ km}$ more each hour. The value $126{,}000 \text{ km}$ is Venus's full hourly distance ($35 \times 3{,}600$) — that's how far Venus goes, not how much farther Mercury goes. The value $780 \text{ km}$ comes from multiplying the $13$ km per-second difference by only $60$, converting for one minute instead of one full hour. The value $172{,}800 \text{ km}$ is Mercury's full hourly distance ($48 \times 3{,}600$) — that's Mercury's total travel, not the difference. Always match your rate's time unit to the question, convert fully, and remember to subtract when a question asks "how many more."
13. Venus's actual distance from the Sun is about 108,000,000 km. Is the student's reasoning and calculation for Venus's model placement valid?
    Answer: No, because the correct scale distance for Venus is 10.8 cm, since Venus's real distance is not exactly double Mercury's real distance. Whenever you work with scale models, remember that each measurement must be converted using the scale — you can't rely on shortcuts based on "about" relationships between the real values. The scale here is $1 \text{ cm} = 10{,}000{,}000 \text{ km}$, so every distance gets divided by 10,000,000 to find its model length. Apply the scale to Venus directly: $108{,}000{,}000 \div 10{,}000{,}000 = 10.8 \text{ cm}$. That's the correct placement. The student's error was assuming Venus is exactly twice Mercury's distance. But $58{,}000{,}000 \times 2 = 116{,}000{,}000$, which is not the same as Venus's real distance of 108,000,000 km. Because "about twice" isn't exactly twice, doubling Mercury's model distance overshoots the true answer. This is why the choice stating the correct distance is 10.8 cm, since Venus's real distance is not exactly double Mercury's is right. The choice claiming doubling Mercury's model distance correctly reflects the real distances is wrong because it treats an approximation as an exact relationship. The choice saying Mercury should be at 0.58 cm miscalculates the scale — dividing 58,000,000 by 10,000,000 gives 5.8, not 0.58 (that would divide by 100,000,000). The choice claiming the ratio doesn't matter, only total distance is confused: total distance from the Sun is what you scale, and the student did use it wrongly. Study tip: In scale problems, always convert the actual number directly with the scale factor. Never trust "about" or "roughly" statements to give exact model measurements — approximations create errors.
14. A group of students builds a solar-system distance model using the scale $1$ cm of model distance $= 20,000,000$ km of actual distance. They calculate model distances as follows: Mercury (actual $58,000,000$ km) at $2.9$ cm, Venus (actual $108,000,000$ km) at $5.4$ cm, Earth (actual $150,000,000$ km) at $6.5$ cm, and Mars (actual $228,000,000$ km) at $11.4$ cm. Which planet's model distance does NOT match the stated scale?
    Answer: Earth Whenever you work with a scale like "$1$ cm $= 20,000,000$ km," the key idea is that dividing the actual distance by the scale factor should always give the model distance. So to check each planet, take its real distance and divide by $20,000,000$ km per cm. For Earth, the actual distance is $150,000,000$ km. Dividing gives $150,000,000 \div 20,000,000 = 7.5$ cm. But the students listed Earth at $6.5$ cm, so Earth doesn't match the scale — it should be $7.5$ cm, not $6.5$ cm. Check the others to confirm they're correct. Mercury: $58,000,000 \div 20,000,000 = 2.9$ cm, which matches exactly. Venus: $108,000,000 \div 20,000,000 = 5.4$ cm, also a perfect match. Mars: $228,000,000 \div 20,000,000 = 11.4$ cm, correct as well. Since these three follow the scale precisely, they cannot be the odd one out — only Earth's value is off. A quick shortcut: since $20,000,000$ km equals $1$ cm, you can just drop seven zeros and divide by $2$. For Earth, $150 \div 2 = 75$, meaning $7.5$ cm. When a scale question asks which value is "wrong," don't assume — test every choice with the same division. The one whose numbers don't produce the listed answer is your culprit.
15. A science museum builds a scale model of the solar system in which the real distance from the Sun to Earth, $150{,}000{,}000$ km, is represented by a model distance of $3$ cm. Using this same scale, what model distance should represent the real distance from the Sun to Neptune, which is about $4{,}500{,}000{,}000$ km?
    Answer: 90 cm Whenever you see a scale model or map problem, you're really working with a proportion — the ratio of real distance to model distance stays the same everywhere in the model. Your job is to find the scale factor once, then apply it consistently. Start by figuring out how many real kilometers each centimeter represents. Since $150{,}000{,}000$ km maps to $3$ cm, each centimeter stands for $150{,}000{,}000 \div 3 = 50{,}000{,}000$ km. Now use that same rate for Neptune: $4{,}500{,}000{,}000 \div 50{,}000{,}000 = 90$ cm. Another way to see it: Neptune is $4{,}500{,}000{,}000 \div 150{,}000{,}000 = 30$ times farther than Earth, so its model distance is $30 \times 3 = 90$ cm. The choice of $9$ cm comes from just multiplying $3 \times 3$, as if Neptune were only 3 times farther — a misread of the zeros. The $900$ cm answer multiplies by an extra factor of 10, likely from a place-value slip while dividing the large numbers. The $45$ cm answer comes from taking $4{,}500{,}000{,}000 \div 100{,}000{,}000$ or halving figures incorrectly — it doesn't match the true scale factor of $50{,}000{,}000$ km per cm. The takeaway: in scale problems, always find the unit rate first (how much one unit represents), then multiply. Counting zeros carefully with these huge astronomical numbers is where most mistakes happen — line them up and cancel matching zeros before dividing.

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