SHSAT Math Quiz: Checking With Estimation
20 questions · exam conditions
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Checking With EstimationQuestion 1 of 20

A landscaper must spread 48 cubic yards of mulch. Each wheelbarrow holds 5.2 cubic feet. He estimates he will need about 250 wheelbarrow loads. Which of the following mental approximations best checks his estimate?

Convert 48 cubic yards to about 1,300 cubic feet and divide by 5, giving roughly 260 loads—close to the estimate.
Convert 48 cubic yards to about 1,000 cubic feet and divide by 10, giving roughly 100 loads—far fewer than 250.
Convert 48 cubic yards to about 162 cubic feet and divide by 5, giving roughly 32 loads—much less than 250.
Skip unit conversion and divide 48 by 5 to get about 10 loads—showing 250 is too high.
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SHSAT Math Quiz

SHSAT Math Quiz: Checking With Estimation

Practice Checking With Estimation in SHSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Checking With Estimation, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A landscaper must spread 48 cubic yards of mulch. Each wheelbarrow holds 5.2 cubic feet. He estimates he will need about 250 wheelbarrow loads. Which of the following mental approximations best checks his estimate?

  1. Convert 48 cubic yards to about 1,300 cubic feet and divide by 5, giving roughly 260 loads—close to the estimate. (correct answer)
  2. Convert 48 cubic yards to about 1,000 cubic feet and divide by 10, giving roughly 100 loads—far fewer than 250.
  3. Convert 48 cubic yards to about 162 cubic feet and divide by 5, giving roughly 32 loads—much less than 250.
  4. Skip unit conversion and divide 48 by 5 to get about 10 loads—showing 250 is too high.
Explanation: When you encounter word problems involving different units of measurement, the key is converting to compatible units before doing any calculations. This problem tests your ability to mentally check whether an estimate makes sense by using reasonable approximations. To verify the landscaper's estimate of 250 wheelbarrow loads, you need to convert 48 cubic yards to cubic feet, then divide by the wheelbarrow capacity. Since there are 27 cubic feet in a cubic yard, 48 cubic yards equals 48×27=1,29648 \times 27 = 1,296 cubic feet. For mental math, this rounds to about 1,300 cubic feet. Dividing by the wheelbarrow capacity of 5.2 cubic feet (approximately 5) gives 1,300÷5=2601,300 \div 5 = 260 loads. This confirms the estimate of 250 is reasonable. Choice A correctly follows this logic and arrives at roughly 260 loads, validating the estimate. Choice B uses a reasonable conversion to 1,000 cubic feet but then incorrectly divides by 10 instead of 5, yielding only 100 loads. Choice C makes a major conversion error, calculating 48 cubic yards as only 162 cubic feet—this completely ignores that cubic yards are much larger than cubic feet. Choice D skips the essential unit conversion entirely, comparing cubic yards directly to cubic feet, which is meaningless. Remember: when checking estimates in multi-step problems, each step must use compatible units. Always convert first, then calculate. The SHSAT often includes answer choices that reflect common mistakes like skipping conversions or using wrong conversion factors.

Question 2

A social media post was shared 238 times the first day, then daily shares dropped by about one-third each day. A reporter projects about 500 total shares after 3 days. Which estimation is the best quick check?

  1. Day 1 ≈ 240; Day 2 about two-thirds, 160; Day 3 two-thirds of 160, ~110; total ≈ 510, close to 500. (correct answer)
  2. Add 238 three times (714) and divide by 3, yielding 238 shares, less than 500.
  3. Multiply 238 by 3 to get 714, showing 500 is too low.
  4. Assume zero shares after day 1, so 500 is impossible.
Explanation: When you encounter exponential decay problems, you need to track how a quantity decreases by a consistent percentage over time. Here, shares drop by "about one-third" each day, meaning each day retains about two-thirds of the previous day's shares. Let's work through the calculation systematically. Day 1 starts with 238 shares. For quick estimation, round this to 240. On Day 2, shares drop by one-third, so you keep two-thirds: 240×23=160240 \times \frac{2}{3} = 160. On Day 3, apply the same reduction: 160×23≈107160 \times \frac{2}{3} \approx 107, which rounds to about 110. Adding these up: 240+160+110=510240 + 160 + 110 = 510, which is very close to the projected 500 total shares. Choice A correctly models this exponential decay process and arrives at a reasonable estimate of 510, confirming the reporter's projection. Choice B incorrectly assumes the same number of shares each day (238), then divides the total by 3 for some unclear reason. This completely ignores the decreasing pattern. Choice C simply multiplies the first day by 3 (714), assuming no decrease at all. This fundamentally misunderstands the problem's exponential decay. Choice D unrealistically assumes zero shares after day 1, which contradicts the given information about gradual decline. Study tip: For exponential decay problems on the SHSAT, always track the pattern day-by-day rather than looking for shortcuts. Round numbers early to make mental math easier, but preserve the multiplicative relationship between consecutive terms.

Question 3

A student calculated that a car traveling at 68 mph will cover 952 miles in exactly 14 hours. Which estimation best checks whether this calculation is reasonable?

  1. 70×15=1,05070 \times 15 = 1,050 miles, so the answer seems too high
  2. 70×14=98070 \times 14 = 980 miles, so the answer seems reasonable (correct answer)
  3. 65×15=97565 \times 15 = 975 miles, so the answer seems reasonable
  4. 60×14=84060 \times 14 = 840 miles, so the answer seems too low
Explanation: Round 68 mph to 70 mph and keep 14 hours exact. 70 × 14 = 980 miles, which is very close to the calculated 952 miles, confirming the calculation is reasonable. Choice A incorrectly changes 14 hours to 15 hours. Choice C changes both values unnecessarily. Choice D rounds 68 mph too far down to 60 mph, making the estimation less accurate.

Question 4

A cylindrical water tank has a radius of 4.8 meters and height of 7.2 meters. Using the formula V=πr2hV = \pi r^2 h, a student calculated the volume as approximately 521 cubic meters (using π≈3.14\pi \approx 3.14). Which estimation best checks this result?

  1. Using π≈3\pi \approx 3 and dimensions 5×5×75 \times 5 \times 7, estimate is 525525 cubic meters, confirming the calculation (correct answer)
  2. Using π≈3\pi \approx 3 and dimensions 5×5×85 \times 5 \times 8, estimate is 600600 cubic meters, suggesting an error
  3. Using π≈3.14\pi \approx 3.14 exactly gives 539539 cubic meters, so the calculation has a small error
  4. Using π≈4\pi \approx 4 and dimensions 5×5×75 \times 5 \times 7, estimate is 700700 cubic meters, suggesting an error
Explanation: For estimation, use π ≈ 3, round 4.8 to 5, and 7.2 to 7. Volume ≈ 3 × 5² × 7 = 3 × 25 × 7 = 525 cubic meters. This is very close to 521, confirming the calculation is reasonable. Choice B rounds 7.2 up to 8, creating unnecessary error. Choice C uses exact values rather than estimation. Choice D uses π ≈ 4, which is too far from the actual value of π.

Question 5

A store sells items with a 15% discount. A customer calculated that an item originally priced at $78.40 will cost $66.64 after the discount. To verify this calculation using estimation, what approach should be used?

  1. Estimate: 10%10\% of $80=$8\$80 = \$8, so final price ≈$72\approx \$72, which differs significantly
  2. Estimate: 20%20\% of $78=$15.60\$78 = \$15.60, so final price ≈$62.40\approx \$62.40, which differs significantly
  3. Estimate: 15%15\% of $80=$12\$80 = \$12, so final price ≈$68\approx \$68, which is close to $66.64 (correct answer)
  4. Calculate exactly: 15%15\% of $78.40=$11.76\$78.40 = \$11.76, so final price is $66.64, which matches
Explanation: When you encounter a question asking you to verify a calculation using estimation, you're being tested on your ability to use mental math and rounding to quickly check if an answer is reasonable. The correct approach is option C because it uses smart rounding and applies the discount properly. By rounding $78.40 up to $80, you get a nice round number that's easy to work with. Then, $15%15\% ofof \80 = 0.15 \times 80 = $12 . Subtracting this discount: $80 - $12 = $68 . This estimate of $68 is very close to the calculated $66.64, confirming the calculation is reasonable. Option A fails because it uses $10\%$$ instead of the actual $$15\%$$ discount, leading to an underestimated discount and an inflated final price of $72. Option B makes two errors: it uses $20%$$ (too high) and doesn't round $78.40 to make the calculation easier, yet still gets a final price that's too low. Option D misses the point entirely—it's doing exact calculation, not estimation, which defeats the purpose of quickly verifying an answer. For estimation problems on the SHSAT, always round to numbers that make mental math easy (like multiples of 10), but keep your rounding reasonable—don't round $78.40 to $100. Use the correct percentages given in the problem, and remember that estimation should get you close to the exact answer, not exactly match it.

Question 6

A bakery produces 1,847 cupcakes per day. The owner calculates that in 23 days, they will produce 42,481 cupcakes total. To verify this calculation using estimation, what should the owner conclude?

  1. The calculation appears correct since 1,800×25=45,0001,800 \times 25 = 45,000 is reasonably close (correct answer)
  2. The calculation appears incorrect since 1,900×23=43,7001,900 \times 23 = 43,700 is too different
  3. The calculation appears incorrect since 2,000×20=40,0002,000 \times 20 = 40,000 is too different
  4. The calculation appears correct since 1,850×20=37,0001,850 \times 20 = 37,000 is reasonably close
Explanation: Round 1,847 to 1,800 and 23 to 25 (or keep 23). 1,800 × 25 = 45,000, which is close enough to 42,481 to confirm the calculation is reasonable. The actual calculation: 1,847 × 23 = 42,481 is correct. Choice B uses 1,900 (too high) but keeps 23. Choice C changes both numbers too drastically. Choice D uses 20 instead of 23, which creates too much error.

Question 7

A recipe calls for 3.75 cups of flour per batch. Diego mistakenly thinks 5 batches need 28 cups of flour. Which quick estimation best checks if his total is reasonable?

  1. Round 3.75 up to 4 cups, multiply by 5 to get 20 cups—Diego's 28 cups is far too high. (correct answer)
  2. Round 3.75 to 3 cups, multiply by 5 to get 15 cups—Diego's 28 cups is close.
  3. Double 3.75 to estimate two batches, getting about 7.5 cups, then add 3 more cups to reach 5 batches—near 10.5 cups, so 28 cups is reasonable.
  4. Use the exact value 3.75 and multiply by 8 instead of 5—giving 30 cups, so 28 cups makes sense.
Explanation: When you encounter estimation problems, you're being tested on your ability to use rounding and mental math to quickly check whether a calculation is reasonable. The goal isn't precision—it's catching major errors through smart approximation. Let's find what 5 batches actually need. Since each batch requires 3.75 cups, we need 3.75×53.75 \times 5. For quick estimation, round 3.75 to the nearest whole number: 4 cups. Then 4×5=204 \times 5 = 20 cups. Diego's answer of 28 cups is significantly higher than our estimate of 20 cups, suggesting his calculation contains a substantial error. Looking at the wrong answers: Choice B rounds 3.75 down to 3 cups, giving 3×5=153 \times 5 = 15 cups, then incorrectly concludes that 28 is "close" to 15—but 28 is nearly double 15, which isn't close at all. Choice C uses an unnecessarily complicated approach, doubling 3.75 to get about 7.5 cups for 2 batches, then somehow concludes that adding 3 more cups reaches 5 batches (it doesn't—that would only be about 3 batches), arriving at an unreasonably low estimate. Choice D abandons the estimation concept entirely by using the exact value 3.75, then inexplicably multiplies by 8 instead of 5, which doesn't test Diego's work at all. Study tip: For estimation problems, round to the nearest convenient number (usually whole numbers), do the mental math quickly, then compare your estimate to the given answer. If they're far apart, the original calculation likely has an error.

Question 8

A factory produces 468 bolts per hour. During an 8-hour shift the manager predicts 3,600 bolts will be produced. Which mental calculation best evaluates whether this prediction is reasonable?

  1. Approximate 468 as 500; 500 × 8 = 4,000, so 3,600 is a bit low but still close.
  2. Approximate 468 as 400; 400 × 8 = 3,200, so 3,600 is too high by a wide margin.
  3. Approximate 468 as 450; 450 × 8 = 3,600, matching the prediction closely. (correct answer)
  4. Approximate 468 as 300; 300 × 8 = 2,400, so the prediction is greatly overstated.
Explanation: When you encounter estimation problems, you're being tested on your ability to choose approximations that make mental math easier while staying reasonably close to the original numbers. The key is finding the best approximation for 468 that makes multiplying by 8 simple. Let's evaluate each approach: Choice C approximates 468 as 450, giving us 450×8=3,600450 \times 8 = 3,600. This matches the manager's prediction exactly and uses a reasonable approximation since 450 is only 18 away from 468. Choice A rounds 468 to 500, yielding 500×8=4,000500 \times 8 = 4,000. While this makes the multiplication easy, 500 is 32 away from 468 – a larger approximation error that leads to concluding 3,600 is "a bit low" when it's actually quite accurate. Choice B approximates 468 as 400, giving 400×8=3,200400 \times 8 = 3,200. This 68-unit difference from the original creates a significant underestimate, making 3,600 appear "too high by a wide margin" when the prediction is actually reasonable. Choice D uses 300, which is 168 away from 468 – far too low an approximation. This gives 300×8=2,400300 \times 8 = 2,400, making the prediction seem "greatly overstated" when it's not. Strategy tip: For estimation problems, choose approximations that balance computational ease with accuracy. Numbers ending in 50 or 00 are usually easiest to work with, but pick the one closest to your original value. Don't automatically round to the nearest hundred – sometimes rounding to a "nice" number like 450 gives you both accuracy and simple arithmetic.

Question 9

Patricia read 18 pages on Friday, 37 pages on Saturday, and 46 pages on Sunday. She says she read about 120 pages over the weekend. Which estimation is the quickest check?

  1. Round 18 to 20, 37 to 40, 46 to 50; add 20 + 40 + 50 ≈ 110, slightly less than 120. (correct answer)
  2. Add only the two larger numbers, 37 + 46 ≈ 83, and note 83 is less than 120.
  3. Multiply the smallest number by 3 to estimate the total: 18 × 3 ≈ 54, far from 120.
  4. Double the largest number, 46 × 2 ≈ 92, to show the total is 92.
Explanation: When you encounter estimation problems, you're looking for the method that gives you the most reasonable approximation with the least work. The key is balancing accuracy with efficiency. Let's check Patricia's claim by examining each approach. The actual total is 18+37+46=10118 + 37 + 46 = 101 pages, so her estimate of 120 pages is indeed too high. Choice A uses systematic rounding: 18 becomes 20, 37 becomes 40, and 46 becomes 50. Adding these gives 20+40+50=11020 + 40 + 50 = 110. This is close to Patricia's claim of 120 but noticeably lower, suggesting her estimate might be high. This method is both quick and accurate for checking reasonableness. Choice B only considers the two larger numbers: 37+46=8337 + 46 = 83. While this shows that even without the smallest number you're already at 83, it ignores a significant portion (18 pages) and doesn't give you a complete picture for comparison. Choice C multiplies the smallest number by 3: 18×3=5418 \times 3 = 54. This severely underestimates since it assumes all three days had the minimum reading amount, which clearly isn't the case given the much larger numbers for Saturday and Sunday. Choice D doubles the largest number: 46×2=9246 \times 2 = 92. This also underestimates and doesn't account for the third day of reading at all. For estimation problems on the SHSAT, look for methods that account for all given values while using simple rounding. Systematic rounding to nearby tens or hundreds typically provides the best balance of speed and accuracy.

Question 10

A rectangle measures 9.8 cm by 14.2 cm. A student states that its area is about 139 cm². Which mental calculation best evaluates this claim?

  1. Round 9.8 to 10 and 14.2 to 14, multiply to get 140 cm², nearly matching 139 cm². (correct answer)
  2. Add the sides: 9.8 + 14.2 ≈ 24 cm; since 24 is close to 139, the claim checks out.
  3. Double 9.8 to approximate the area, giving about 20 cm², indicating 139 is far off.
  4. Multiply 9 by 14 to get about 126 cm², far below 139, proving the claim wrong.
Explanation: When you need to quickly check if a calculation is reasonable, mental math using rounding is your best tool. For area problems, you're looking for a quick way to estimate the product of two dimensions. The most effective approach is to round each dimension to nearby "friendly" numbers that are easy to multiply mentally. Here, 9.8 cm rounds to 10 cm, and 14.2 cm rounds to 14 cm. Multiplying 10×14=14010 \times 14 = 140 cm², which is extremely close to the student's claim of 139 cm². This suggests the calculation is reasonable. Let's see why the other approaches fail. Choice B makes a fundamental error by adding the dimensions (9.8 + 14.2 ≈ 24) instead of multiplying them. Adding gives you perimeter, not area, and comparing 24 to 139 makes no sense. Choice C suggests doubling one dimension to estimate area, but 2×9.8≈202 \times 9.8 ≈ 20 has no mathematical relationship to the actual area formula. Choice D rounds too aggressively—rounding 9.8 down to 9 and 14.2 down to 14 gives 9×14=1269 \times 14 = 126, which creates unnecessary error and leads to the wrong conclusion that 139 is "far off." Study tip: For estimation problems, round numbers to make mental math easier, but don't round so much that you introduce significant error. When both original numbers are close to their rounded values (like 9.8→10 and 14.2→14), your estimate will be very reliable.

Question 11

A grocery ad lists 4 pounds of apples for $6.36. Howard plans to buy 10 pounds and budgets $15. Does estimation show his budget is enough?

  1. Approximate $6.36 as $6.40; 4 lb for $6.40 means 8 lb for $12.80, so 10 lb is about $16—over his budget. (correct answer)
  2. Assume 4 pounds cost $5; then 10 pounds cost $12.50, so the budget is safe.
  3. Double 4 lb to get 8 lb for $6.36, showing 10 lb costs only $6.36.
  4. Assume each pound costs roughly $1, so 10 lb cost $10, which fits the budget.
Explanation: When you encounter estimation problems on the SHSAT, you need to make reasonable approximations that simplify calculations while staying close to the actual values. The goal is to get a ballpark figure that helps you make decisions. Let's work through this systematically. Howard needs to estimate whether $15 is enough for 10 pounds of apples when 4 pounds cost $6.36. Choice A uses sound estimation by rounding $6.36 to $6.40 (close to the original). If 4 pounds cost $6.40, then 8 pounds cost $12.80 (doubling both quantity and price). To get from 8 pounds to 10 pounds, you need 2 more pounds. Since 4 pounds cost $6.40, each pound costs $1.60, so 2 pounds cost $3.20. Adding: $12.80 + $3.20 = $16.00, which exceeds Howard's $15 budget. Choice B drastically underestimates by assuming $5 for 4 pounds instead of $6.36—that's off by over $1, making the estimation unreliable. Choice C makes a critical error by claiming that doubling the weight (4 to 8 pounds) doesn't change the cost, which violates basic proportional reasoning. Choice D assumes $1 per pound, but $6.364=1.59\frac{6.36}{4} = 1.59 $ per pound, so this significantly underestimates the true cost. For estimation problems, make approximations that are reasonably close to the original numbers and maintain proportional relationships. Avoid rounding that changes values too dramatically, as this defeats the purpose of getting a reliable estimate.

Question 12

A company bought 128 computers at $739 each and 15 printers at $129 each. An intern estimated the equipment cost at about $100,000. Which mental math check makes the most sense?

  1. Round each computer to $750: 128 × 750 ≈ 96,000; add printers rounded to $125: 15 × 125 ≈ 1,900; total ≈ 98,000—close to $100,000. (correct answer)
  2. Round computers to $1,000 each for 128,000, ignore printers, showing the estimate is low.
  3. Assume printers and computers each cost about $100, so 143 items × 100 ≈ 14,300—far below $100,000.
  4. Use $500 per computer: 128 × 500 ≈ 64,000; add printers at $100 each to get 65,500—well under $100,000.
Explanation: When you encounter estimation problems, you're being tested on your ability to use reasonable rounding that maintains accuracy while simplifying calculations. The key is finding a balance between making the math manageable and staying close to the actual values. Choice A demonstrates proper estimation technique. Rounding $739 to $750 is reasonable (only $11 difference) and makes multiplication easier: $128×750=96,000128 \times 750 = 96,000 .Addingtheprintersat$125eachgives$. Adding the printers at $125 each gives $15 \times 125 = 1,875$$, so the total is approximately $97,875—very close to the $100,000 estimate. Choice B rounds computers to $1,000 each, which is $261 higher than the actual price—a 35% increase that's far too aggressive. This creates a massive overestimate of $128,000 just for computers, making it useless for checking the intern's work. Choice C makes the opposite error by rounding everything down to 100.Thisunderestimatescomputersbyover85100. This underestimates computers by over 85% (739 vs $100), producing a wildly inaccurate total of around $14,300 that tells you nothing useful about whether $100,000 is reasonable. Choice D rounds computers down to $500—a $239 underestimate per computer. With 128 computers, this creates a significant shortfall that makes the total unrealistically low at about $65,500. Strategy tip: For estimation problems, round to numbers that make multiplication easy (like multiples of 25, 50, or 100) while staying within 10-15% of the original values. Dramatic rounding in either direction defeats the purpose of checking your work.

Question 13

During a fundraiser, tickets cost $3.75 for children and $5.25 for adults. If 112 children and 86 adults attended, is an announced revenue of about $900 reasonable?

  1. Round prices to $4 and $5; revenue ≈ 112 × 4 + 86 × 5 = 448 + 430 = 878, close to $900. (correct answer)
  2. Assume every ticket cost $10; 198 tickets × 10 = 1,980, so revenue is far higher.
  3. Multiply total attendees by $4 to get 792, slightly under $900, so correct.
  4. Use $5 for all tickets: 198 × 5 = 990, proving the revenue should be almost $1,000.
Explanation: When you encounter estimation problems involving different prices and quantities, your goal is to find reasonable approximations that make mental math easier while staying close to the actual values. To check if $900 is reasonable, you need to calculate the estimated revenue using rounded ticket prices. The actual prices are $3.75 for children and $5.25 for adults. Choice A correctly rounds these to $4 and $5 respectively, which are close approximations that simplify calculations. With 112 children and 86 adults, the estimated revenue becomes: $112×4+86×5=448+430=878112 \times 4 + 86 \times 5 = 448 + 430 = 878 $. This gives $878, which is very close to the announced $900, confirming it's reasonable. Choice B uses $10 per ticket, which is far too high compared to the actual prices of $3.75 and $5.25. This leads to a grossly inflated estimate of $1,980. Choice C makes a fundamental error by multiplying the total number of attendees (198) by $4, treating all tickets as if they cost the same. This ignores the fact that adult tickets cost more than children's tickets. Choice D assumes all tickets cost $5, which overestimates since children's tickets are significantly cheaper at $3.75. For SHSAT estimation problems, round to convenient numbers that are close to the original values, then calculate each component separately when dealing with different prices or rates. Don't oversimplify by using the same value for clearly different quantities.

Question 14

A smartphone battery drains 3.8% per hour while streaming. If the phone starts at 92% charge, a student predicts it will be empty after 20 hours. Which estimation best tests that prediction?

  1. Round 3.8% to 4%; 4% × 20 = 80%, so charge drops from 92% to about 12%, not empty. (correct answer)
  2. Assume 5% per hour: 5% × 20 = 100%, exactly empty, confirming the student.
  3. Use 2% per hour: 2% × 20 = 40%, leaving 52%, so prediction fails.
  4. Double 3.8 to 7.6% for safety: 7.6 × 10 = 76%, leaving 16%, showing prediction wrong.
Explanation: When you encounter estimation problems, you're being asked to check whether a rough calculation supports or contradicts a given prediction. The key is finding a reasonable approximation that's easy to calculate while staying close to the original values. Let's test the student's prediction that the phone will be empty after 20 hours. Starting at 92% charge with a 3.8% drain per hour, we need to estimate how much battery will remain. Choice A provides the best estimation approach: rounding 3.8% to 4% makes the math simple while staying very close to the actual rate. After 20 hours: 4%×20=80%4\% \times 20 = 80\% total drain. Starting at 92%, the phone would drop to approximately 92%−80%=12%92\% - 80\% = 12\%, meaning it wouldn't be empty. This directly contradicts the student's prediction. Choice B uses 5% per hour, which is too far from 3.8% to be a good estimation—it inflates the drain rate by over 30%. Choice C uses 2% per hour, which severely underestimates the actual drain rate. Choice D unnecessarily complicates the estimation by doubling the rate and halving the time, creating a "safety margin" that isn't asked for and makes the estimation less accurate. The best estimations for testing predictions stay close to the original values while simplifying calculations. Here, rounding 3.8% to 4% achieves both goals and clearly shows the student's prediction is wrong. Strategy tip: When estimating, round to the nearest "friendly" number that keeps calculations simple but doesn't stray too far from the original—typically within 10-15% of the actual value.

Question 15

A delivery route is 246 miles long. A truck averages 52 miles per hour. The driver claims the trip will take about 4 hours. Which estimation checks this quickly?

  1. Round 52 mph to 50 mph; 250 miles ÷ 50 mph ≈ 5 hours, longer than claimed. (correct answer)
  2. Double the speed to 100 mph: 246 ÷ 100 ≈ 2.5 hours, confirming 4 hours.
  3. Add 52 and 246 to verify the answer, yielding 298.
  4. Divide 246 by 25 mph, getting about 10 hours, so 4 hours is too short.
Explanation: When you encounter a word problem asking you to check whether an estimate is reasonable, you need to use quick mental math to verify the calculation. This question tests your ability to use rounding and estimation to check if the driver's 4-hour claim makes sense. The correct approach is option A: Round 52 mph to 50 mph, then calculate 250 miles ÷ 50 mph = 5 hours. Since the actual trip should take about 5 hours, the driver's claim of 4 hours is too optimistic. This estimation method uses simple rounding (246 → 250, 52 → 50) to create numbers that are easy to divide mentally. Option B contains flawed logic. Doubling the speed to 100 mph and getting 2.5 hours doesn't confirm that 4 hours is correct at the actual speed of 52 mph. The calculation is mathematically sound but irrelevant to checking the driver's estimate. Option C makes no sense mathematically. Adding speed (52 mph) and distance (246 miles) gives you nothing meaningful since you can't add different units. This violates basic principles of unit analysis. Option D uses an arbitrary speed of 25 mph instead of the given 52 mph. While it correctly identifies that 10 hours would make 4 hours too short, it's not checking the driver's claim using the actual conditions of the problem. Study tip: For estimation problems, always round to numbers that make mental math easy (like 50 or 100), keep your units consistent, and remember that time=distancespeed\text{time} = \frac{\text{distance}}{\text{speed}}.

Question 16

A box contains 298 light bulbs packed equally into 8 cartons. A worker says each carton holds about 40 bulbs. Which estimation best evaluates that statement?

  1. Round 298 to 320; 320 ÷ 8 = 40, so the worker's number is reasonable. (correct answer)
  2. Multiply 8 by 50 to get 400, proving each carton must hold 50 bulbs.
  3. Add 298 and 8 to get 306, divide by 10 to get 30.6 bulbs.
  4. Round 298 to 240; 240 ÷ 8 = 30, so the worker's number is too high.
Explanation: When you encounter estimation problems, you're testing whether someone's approximation is reasonable by using your own rounded numbers to check their work. The key is choosing rounding that makes the division simple while staying close to the original number. To evaluate the worker's claim that each carton holds about 40 bulbs, you need to estimate how many bulbs per carton there actually are. Since 298÷8298 \div 8 involves awkward numbers, round 298 to something that divides evenly by 8. The number 320 is close to 298 and divides cleanly: 320÷8=40320 \div 8 = 40. This matches the worker's estimate exactly, confirming it's reasonable. Choice A uses this correct approach. Choice B makes a fundamental error by working backwards—multiplying 8 by an assumed 50 bulbs per carton. This doesn't evaluate the worker's statement; it creates a different scenario entirely and then incorrectly claims this "proves" each carton must hold 50 bulbs. Choice C adds two unrelated quantities (298 bulbs plus 8 cartons) and then divides by 10. This mathematical operation has no logical meaning—you can't add a count of objects to a count of containers and expect a meaningful result. Choice D rounds 298 down to 240, which gives 240÷8=30240 \div 8 = 30. While the math is correct, rounding from 298 to 240 is unnecessarily extreme when 320 is much closer and gives cleaner division. For estimation problems, choose the rounding that's closest to the original number while making the arithmetic manageable. This gives you the most reliable check of someone's approximation.

Question 17

Each week a bookstore's profit grew by 12% over the previous week. After 5 weeks a clerk estimates the total growth to be about 60%. Which estimation best checks the clerk's figure?

  1. Multiply 12% by 5 to get 60%, which provides a reasonable approximation for moderate growth rates over short periods. (correct answer)
  2. Square 12% to get 144%, proving growth is over 100%.
  3. Add 12 and 5 to get 17%; growth is just 17%.
  4. Raise 1.12 to the 5th power, noting it is about 1.8, or 80% growth, so 60% is too high.
Explanation: When you encounter compound growth problems, you need to distinguish between simple approximations and precise calculations, especially when checking reasonableness of estimates. The clerk's 60% estimate can be checked using a simple linear approximation. For moderate growth rates over short periods, multiplying the weekly growth rate by the number of weeks gives a reasonable ballpark figure: 12%×5=60%12\% \times 5 = 60\%. This method works well when growth rates are relatively small (under 15-20%) and the time period is short, making choice A correct. Let's examine why the other options fail. Choice B misunderstands the problem entirely—squaring 12% gives 1.44%, not 144%, and has no relevance to 5-week growth. Choice C makes a nonsensical calculation by adding the percentage rate to the number of weeks, treating completely different units as if they were the same. Choice D actually performs the precise compound growth calculation: (1.12)5≈1.76(1.12)^5 \approx 1.76, which means 76% growth, not 80%. More importantly, this shows the clerk's 60% estimate is too low, not too high. The key insight is that choice A represents exactly the type of quick estimation technique that's useful for checking reasonableness, while choice D performs an exact calculation but misinterprets the result. Strategy tip: On compound growth problems, remember that linear approximations (rate × time periods) work well for quick checks with moderate rates, but actual compound growth will always be slightly higher than the linear estimate due to compounding effects.

Question 18

A cylindrical water pipe has a length of 2.3 m and an inner diameter of 8 cm. An apprentice calculates its volume as 115 liters. Which estimation quickly tests this volume?

  1. Diameter 0.08 m, radius 0.04 m; volume ≈ 3.14 × (0.04)² × 2.3 ≈ 0.012 m³ = 12 L, far below 115 L. (correct answer)
  2. Use diameter 0.1 m and length 2 m: volume ≈ 0.015 m³ = 15 L, confirming 115 L is reasonable.
  3. Multiply length 2.3 by diameter 8 to get roughly 18 L, close enough to validate 115 L.
  4. Double the length to 4.6 and divide by diameter 8, yielding about 0.6 L as expected.
Explanation: When you encounter volume problems involving unrealistic answers, estimation is your best friend for quickly checking whether a calculation makes sense. To estimate a cylinder's volume, use V=πr2hV = \pi r^2 h. First, convert all units consistently: the 8 cm diameter becomes 0.08 m, so the radius is 0.04 m. Using π≈3.14\pi \approx 3.14, you get: V≈3.14×(0.04)2×2.3≈3.14×0.0016×2.3≈0.012 m3V \approx 3.14 \times (0.04)^2 \times 2.3 \approx 3.14 \times 0.0016 \times 2.3 \approx 0.012 \text{ m}^3. Since 1 m³ = 1000 L, this equals about 12 liters. The apprentice's answer of 115 liters is nearly 10 times too large, clearly indicating an error. Looking at the wrong answers: Choice B uses rounded values (0.1 m diameter, 2 m length) that are significantly larger than the actual dimensions, making the estimate artificially high and misleading you into thinking 115 L is reasonable. Choice C commits a fundamental error by multiplying length times diameter (2.3 × 8 = 18), which gives a meaningless result since volume requires area times length, not diameter times length. Choice D uses bizarre operations (doubling length, then dividing by diameter) that have no mathematical relationship to volume calculation. The key strategy here is dimensional analysis and order-of-magnitude checking. Always convert units first, use reasonable approximations for π and squared terms, and ask yourself: "Does this result make physical sense?" A thin pipe less than a meter wide and about 2 meters long shouldn't hold over 100 liters of water.

Question 19

A baker needs 5.6 kilograms of sugar for a recipe scaled up by a factor of 6. The original recipe calls for 0.92 kilograms. The baker's figure is questioned. Which estimation best checks it?

  1. Round 0.92 to 1 kg; 1 × 6 = 6 kg, so 5.6 kg is close. (correct answer)
  2. Add 0.92 and 6 to get 6.92 kg, proving 5.6 kg is off.
  3. Multiply 0.92 by 10 to get 9.2 kg, much higher than 5.6 kg.
  4. Multiply 0.5 kg by 6 to get 3 kg, showing 5.6 kg is too much.
Explanation: When you encounter a problem asking you to check whether a calculation is reasonable, you need to use estimation to quickly verify if the answer makes sense. The baker claims that scaling up a recipe by a factor of 6 means multiplying the original amount by 6. Let's check: if the original recipe needs 0.92 kg of sugar, then scaling up by 6 should give us 0.92×6=5.520.92 \times 6 = 5.52 kg. The baker's figure of 5.6 kg is indeed very close to this exact calculation. Choice A correctly demonstrates good estimation technique. By rounding 0.92 to 1 kg (a reasonable approximation), then multiplying by 6 to get 6 kg, we can see that 5.6 kg is reasonably close to our estimate. This confirms the baker's calculation is likely correct. Choice B makes a fundamental error by adding instead of multiplying. When scaling a recipe, you multiply each ingredient by the scale factor, not add the scale factor to the original amount. Choice C arbitrarily multiplies by 10 instead of 6, which has nothing to do with the scaling factor mentioned in the problem. Choice D uses 0.5 kg instead of 0.92 kg, making the estimation too far from the original amount to be useful for checking the calculation. Strategy tip: For estimation problems on the SHSAT, round to numbers that make mental math easy (like 1 instead of 0.92), but make sure you're using the correct operation. Always ask yourself: "Does my estimated answer land in the same ballpark as the given answer?"

Question 20

To tile a floor, 1 box covers 18.2 square feet. A room is 11.5 ft by 14 ft. A contractor orders 8 boxes. Which estimation best checks if 8 boxes will likely be enough?

  1. Estimate the room as 12 ft by 14 ft for about 168 ft²; 8 boxes cover roughly 8 × 18 ≈ 144 ft², probably not enough. (correct answer)
  2. Round the room to 10 ft by 15 ft ≈ 150 ft²; 8 boxes cover about 200 ft², easily enough.
  3. Double the longer side to 28 ft and compare to 8 boxes, showing coverage is sufficient.
  4. Ignore room size and assume each box covers about 10 ft²; 8 boxes give 80 ft², so coverage is short.
Explanation: When you encounter estimation problems involving area and coverage, you need to make reasonable approximations that simplify calculations while staying close to the actual values. Let's find the room's area first: 11.5×14=16111.5 \times 14 = 161 square feet. With 8 boxes covering 8×18.2=145.68 \times 18.2 = 145.6 square feet, there's clearly not enough material since 145.6 < 161. Now let's examine how well each estimation approach works. Choice A rounds 11.5 ft to 12 ft, giving 12×14=16812 \times 14 = 168 ft². It estimates box coverage as 8×18≈1448 \times 18 \approx 144 ft². Since 144 < 168, this correctly predicts insufficient coverage and closely matches the actual calculation. Choice B rounds to 10×15=15010 \times 15 = 150 ft² and estimates 8 boxes cover "about 200 ft²." This box coverage estimate is wildly inaccurate—8×18.2=145.68 \times 18.2 = 145.6, not 200. This leads to the wrong conclusion that coverage is sufficient. Choice C suggests doubling the longer side to 28 ft, which creates an entirely different room (11.5×28=32211.5 \times 28 = 322 ft²) that bears no resemblance to the original problem. Choice D assumes each box covers only 10 ft² instead of 18.2 ft², drastically underestimating box coverage and ignoring the room size entirely. Study tip: Good estimation maintains the essential relationships in the original problem. Round numbers to make calculations easier, but keep your estimates reasonably close to actual values. Dramatic changes to given numbers usually lead to meaningless results.