Single Step SAT Prep

Digital SAT Math · Problem-Solving & Data Analysis

Units & conversions

A conversion item is one multiplication chain. You write the starting quantity with its units, then multiply by fractions equal to 1 until the units you do not want cancel and the unit the question asked for is the only one left. The arithmetic is last; the setup is the whole problem. The test hides those chains inside density and rate stories so the conversion factors never look like the point.

On the test

DomainProblem-Solving and Data Analysis (score report)
CB skillRatios, rates, proportional relationships, and units
What it looks likeA quantity in one unit, a short story (rate, density, capacity), and a request for a different unit — conversion factors given in parentheses unless they are basic time or metric
Often asked“What is the … in [new unit]?”, “What is the density in …?”, “At this rate, how many …?”
FormatMultiple choice and student-produced response
CalculatorAllowed throughout; useful for a long product of decimals, useless if the conversion fraction is upside down

Recognition cues: per, density, in terms of, parenthetical equivalencies (1 mi = 1.609 km), square / cubic, how many … in …, mixed unit systems in one stem.

Pattern recognition

Almost every item is one of these:

  1. Straight chain — start with a quantity, convert through one or more factors until the asked unit remains.
  2. Rate or density hide — a rate (L/min, mi/h) or a density (g/cm³, kg/m³) must be rewritten in different units; the conversion sits inside a quotient, not alone.
  3. Squared or cubed units — area or volume, so the linear conversion factor is applied twice (area) or three times (volume).

Method

  1. Write the starting quantity with its units. Include the full unit string — 36 L/min, not 36.
  2. List the conversion factors you need. The stem supplies any non-basic factor in parentheses. Basic time (60 s = 1 min, 60 min = 1 h) and metric prefixes (kilo-, centi-, milli-) you are expected to know.
  3. Chain fractions equal to 1 so unwanted units cancel. Put each factor so the unit you are removing sits opposite its current position. For a squared unit, write the linear factor twice (or square it once). For a cubed unit, three times.
  4. Do the arithmetic last. Multiply numerators, multiply denominators, divide once. Keep full precision until the final step; round only as the stem instructs.
  5. Confirm the leftover unit is the one asked for. If the stem asked for m³ and you still have L, you stopped early or inverted a factor.
Words in the stemWhat they force
per / ratea quotient unit — convert numerator and denominator separately
density / mass per unit volumemass over volume; convert mass, volume, or both
parenthetical 1 A = k Ba free conversion factor — choose the orientation that cancels
square / area / cm² / m²linear factor applied twice
cubic / volume / cm³ / m³linear factor applied three times
rounded to the nearest…finish the full product first, then round once

Worked example 1 — multi-step rate chain

Stem. A pipeline pumps water at a constant rate of 36 liters per minute. How many cubic meters of water does it pump in 2.5 hours? (1 m³ = 1 000 L; 1 h = 60 min)

Step 1 — write the starting quantity with units.

36 Lmin36\,\frac{\mathrm{L}}{\mathrm{min}}

and a time of 2.5 h2.5\,\mathrm{h}. The asked unit is m³.

Step 2 — chain factors so L and min and h cancel.

36 Lmin⋅2.5 h⋅60 min1 h⋅1 m31000 L36\,\frac{\mathrm{L}}{\mathrm{min}}\cdot 2.5\,\mathrm{h}\cdot\frac{60\,\mathrm{min}}{1\,\mathrm{h}}\cdot\frac{1\,\mathrm{m}^3}{1000\,\mathrm{L}}

Step 3 — arithmetic last.

36⋅2.5⋅601000=54001000=5.4\frac{36\cdot 2.5\cdot 60}{1000} = \frac{5400}{1000} = 5.4

Check. In 2.5 hours there are 2.5×60=1502.5 \times 60 = 150 minutes, so the pipeline delivers 36×150=5 40036 \times 150 = 5\,400 liters, which is 5 400/1 000=5.45\,400 / 1\,000 = 5.4 cubic meters. Leftover unit: m³. Answer: 5.4 m³.

Trap watch. Skipping the hour-to-minute factor leaves 36×2.5/1000=0.0936 \times 2.5 / 1000 = 0.09. Inverting the m³ factor multiplies by 1 000 instead of dividing and produces 5 400 000. Reporting 5 400 answers in liters — the given volume unit — when the stem asked for cubic meters.

Worked example 2 — density hides the conversion

Stem. A metal block has a mass of 2.4 kilograms and a volume of 300 cubic centimeters. Density is mass per unit volume. What is the density of the block in grams per cubic centimeter? (1 kg = 1 000 g)

Step 1 — write density with the given units.

2.4 kg300 cm3\frac{2.4\,\mathrm{kg}}{300\,\mathrm{cm}^3}

Step 2 — convert mass so the top unit becomes grams.

2.4 kg300 cm3⋅1000 g1 kg=2.4×1000300 gcm3\frac{2.4\,\mathrm{kg}}{300\,\mathrm{cm}^3}\cdot\frac{1000\,\mathrm{g}}{1\,\mathrm{kg}} = \frac{2.4 \times 1000}{300}\,\frac{\mathrm{g}}{\mathrm{cm}^3}

Step 3 — arithmetic.

2400300=8\frac{2400}{300} = 8

Check. 2.4 kg is 2 400 g; 2 400 g in 300 cm³ is 8 g per cm³. Leftover unit: g/cm³, which is what was asked. Answer: 8 g/cm³.

Trap watch. Dividing without converting the mass gives 2.4/300=0.0082.4 / 300 = 0.008 — correct arithmetic in the wrong unit system. Inverting to volume ÷ mass gives 125. Reporting 2 400 is the mass in grams with the volume forgotten. Rounding is not the issue here; the inverted factor and the skipped conversion are.

Worked example 3 — squared units

Stem. A square tile measures 25 centimeters on each side. What is the area of the tile in square meters? (1 m = 100 cm)

Step 1 — area in the given unit first (optional but clear).

25 cm×25 cm=625 cm225\,\mathrm{cm} \times 25\,\mathrm{cm} = 625\,\mathrm{cm}^2

Step 2 — convert cm² to m². Square the linear factor.

1 m=100 cm⇒1 m2=1002 cm2=10 000 cm21\,\mathrm{m} = 100\,\mathrm{cm} \quad\Rightarrow\quad 1\,\mathrm{m}^2 = 100^2\,\mathrm{cm}^2 = 10\,000\,\mathrm{cm}^2 625 cm2⋅1 m210 000 cm2=0.0625 m2625\,\mathrm{cm}^2\cdot\frac{1\,\mathrm{m}^2}{10\,000\,\mathrm{cm}^2} = 0.0625\,\mathrm{m}^2

Alternative route — convert the side first, then square.

25 cm⋅1 m100 cm=0.25 m,0.25×0.25=0.0625 m225\,\mathrm{cm}\cdot\frac{1\,\mathrm{m}}{100\,\mathrm{cm}} = 0.25\,\mathrm{m}, \quad 0.25 \times 0.25 = 0.0625\,\mathrm{m}^2

Same answer. Prefer whichever keeps the numbers cleaner for you.

Check. A 0.25 m by 0.25 m square is a quarter-metre on each side; its area is one-sixteenth of a square metre, and 1/16=0.06251/16 = 0.0625. Leftover unit: m². Answer: 0.0625 m².

Trap watch. Applying the linear factor only once gives 625/100=6.25625 / 100 = 6.25 — the classic squared-unit miss. Reporting 0.25 answers the side length in metres, not the area. Reporting 625 leaves the answer in cm² when the stem asked for square meters.

Practice

Answer before you open the explanation. Two items will be student-produced response (type the number, no choices), matching the real test, and at least two will use squared or cubed units — the error the free web names and almost never prices into a choice list. Every wrong choice below is a specific named trap.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

A digital camera records video at a constant rate of 30 frames per second. How many frames does the camera record in 2 minutes?

Show the answer Choice B

Why it is right

Write the starting rate with its units: 30 frames/s. The asked time is 2 minutes, so insert the factor that cancels minutes into seconds: 30 frames/s · 2 min · (60 s / 1 min). The minutes and seconds cancel, leaving frames. Arithmetic last: 30 · 2 · 60 = 3600. Check: in 1 minute the camera records 30 · 60 = 1800 frames, so in 2 minutes it records 3600.

Why each other choice fails

Choice A
Skips the minutes-to-seconds link in the chain: 30 · 2 = 60 treats the rate as frames per minute instead of frames per second.
Choice C
Converts only one minute: 30 · 60 = 1800. That is the frame count for 1 minute, not for the 2 minutes the stem asks about.
Choice D
Uses 6 in place of 60 in the time factor: 30 · 2 · 6 = 360. The leftover unit is still frames, but the arithmetic dropped a factor of 10.

Question 2 Standard

A shipping label lists the mass of a package as 2.8 kilograms. What is the mass of the package in grams? (1 kg = 1,000 g)

Show the answer Choice A

Why it is right

Start with 2.8 kg and multiply by the factor that cancels kilograms into grams: 2.8 kg · (1000 g / 1 kg) = 2800 g. The kilogram labels cancel and grams remain. Check: 1 kg is 1000 g, so 2 kg is 2000 g and 0.8 kg is 800 g, totaling 2800 g.

Why each other choice fails

Choice B
Inverts the conversion factor: 2.8 · (1 / 1000) = 0.0028. That would convert grams into kilograms, the opposite direction of the stem.
Choice C
Treats the kilo- prefix as a factor of 10 instead of 1000: 2.8 · 10 = 28. Metric kilo- always means 1000.
Choice D
Applies a factor of 100 (as if converting with centi- in reverse) rather than 1000: 2.8 · 100 = 280.

Question 3 Standard

An aquarium pump moves water at a constant rate of 18 gallons per minute. At this rate, how many gallons of water does the pump move in 2.5 hours?

Show the answer Choice C

Why it is right

Write the rate with units: 18 gal/min. The time is 2.5 hours, so chain a factor that turns hours into minutes: 18 gal/min · 2.5 h · (60 min / 1 h). Hours and minutes cancel, leaving gallons. Arithmetic last: 18 · 2.5 · 60 = 45 · 60 = 2700. Check: in 1 hour the pump moves 18 · 60 = 1080 gallons, so in 2.5 hours it moves 1080 · 2.5 = 2700 gallons.

Why each other choice fails

Choice A
Skips the hours-to-minutes link: 18 · 2.5 = 45 treats the rate as gallons per hour instead of gallons per minute.
Choice B
Converts only one hour: 18 · 60 = 1080. That is the volume for 1 hour, not for the 2.5 hours the stem asks about.
Choice D
Inverts the time factor, dividing by 60 instead of multiplying: 18 · 2.5 / 60 = 0.75. The leftover unit is gallons per (hour·something), not gallons.

Question 4 Standard

A ceramic glaze sample has a mass of 1.2 kilograms and a volume of 400 cubic centimeters. Density is mass per unit volume. What is the density of the sample in grams per cubic centimeter? (1 kg = 1,000 g)

Show the answer Choice D

Why it is right

Density is mass ÷ volume. Write it with the given units first: 1.2 kg / 400 cm³. Convert mass so the top unit becomes grams: (1.2 kg / 400 cm³) · (1000 g / 1 kg) = 1200 / 400 g/cm³ = 3 g/cm³. Check: 1.2 kg is 1200 g, and 1200 g in 400 cm³ is 3 g per cm³. Leftover unit: g/cm³.

Why each other choice fails

Choice A
Skips the mass conversion and divides the kilogram mass by volume: 1.2 / 400 = 0.003. That is a density in kg/cm³, not the g/cm³ the stem asked for.
Choice B
Converts the mass to grams correctly (1.2 · 1000 = 1200) but then reports that mass alone, forgetting to divide by the 400 cm³ volume.
Choice C
Inverts density to volume ÷ converted mass: 400 / 1200 ≈ 0.333. That is cm³/g, the reciprocal of the density asked for.

Question 5 Standard

A square solar panel measures 80 centimeters on each side. What is the area of the panel in square meters? (1 m = 100 cm)

Show the answer Choice A

Why it is right

Area in the given unit first: 80 cm · 80 cm = 6400 cm². Square the linear conversion: 1 m = 100 cm means 1 m² = 100² cm² = 10,000 cm². Then 6400 cm² · (1 m² / 10,000 cm²) = 0.64 m². Alternative route — convert the side first: 80 cm · (1 m / 100 cm) = 0.8 m, then 0.8 · 0.8 = 0.64 m². Same answer. Leftover unit: m².

Why each other choice fails

Choice B
Applies the linear factor only once to the area: 6400 / 100 = 64. Area is length squared, so the linear factor must be applied twice (divide by 10,000, not by 100).
Choice C
Stops after computing the area in square centimeters and never converts: 6400 is the correct area in cm², but the stem asked for square meters.
Choice D
Raises the linear factor to the third power as if converting a volume: 6400 / 1,000,000 = 0.0064. That is the cm³ → m³ scale, not the cm² → m² scale.

Question 6 Harder

A delivery drone flies at a constant speed of 48 miles per hour. What is the drone's speed in meters per second, rounded to the nearest tenth? (1 mi = 1.609 km; 1 km = 1,000 m; 1 h = 3,600 s)

Show the answer Choice B

Why it is right

Chain factors so miles, kilometers, and hours cancel into meters per second: 48 mi/h · (1.609 km / 1 mi) · (1,000 m / 1 km) · (1 h / 3,600 s) = 48 · 1.609 · 1000 / 3600. Because 1000/3600 = 1/3.6, the product is 48 · 1.609 / 3.6 = 77.232 / 3.6 = 21.4533… m/s. Rounded to the nearest tenth: 21.5. Carry full precision through the chain; round only once at the end.

Why each other choice fails

Choice A
Mid-chain rounding: replaces 1.609 with 1.6 before multiplying, so 48 · 1.6 / 3.6 = 76.8 / 3.6 = 21.333…, which rounds to 21.3. The full-precision product 21.4533… rounds to 21.5, not 21.3.
Choice C
Stops after the miles-to-kilometers step and reports 48 · 1.609 = 77.232 ≈ 77.2. That is the speed in kilometers per hour; the stem asked for meters per second, so the 1000 m and 3600 s factors were never applied.
Choice D
Skips the miles-to-kilometers link and treats the given 48 as if it were already in km/h: 48 · 1000 / 3600 = 13.333… ≈ 13.3. The leftover unit is m/s only if the starting unit had been kilometers, which it was not.

Question 7 Harder Student-produced response

A rooftop rain barrel collects water at a steady rate of 15 liters every 4 minutes. How many cubic meters of water does the barrel collect in 8 hours? (1 m³ = 1,000 L)

Show the answer 1.8

Why it is right

Write the rate as a fraction with units: 15 L / 4 min. Chain factors that turn hours into minutes and liters into cubic meters: (15 L / 4 min) · 8 h · (60 min / 1 h) · (1 m³ / 1,000 L). Minutes, hours, and liters cancel, leaving m³. Arithmetic last: (15 · 8 · 60) / (4 · 1,000) = 7,200 / 4,000 = 1.8. Check: in 8 hours there are 8 · 60 = 480 minutes, so there are 480 / 4 = 120 collection intervals of 15 L each, totaling 120 · 15 = 1,800 L = 1.8 m³.

Answers students type instead

1800
Completes the volume chain correctly in liters — (15 · 8 · 60) / 4 = 1,800 — but never divides by 1,000, so the answer is left in the given volume unit (liters) when the stem asked for cubic meters.
0.03
Skips the hours-to-minutes link: (15 · 8) / (4 · 1,000) = 120 / 4,000 = 0.03. That treats the rate as liters per 4 hours instead of liters per 4 minutes.
7.2
Drops the 4-minute interval from the denominator: (15 · 8 · 60) / 1,000 = 7.2. That treats the rate as 15 L per minute rather than 15 L every 4 minutes.

Question 8 Harder

A rectangular storage tank has interior dimensions 120 centimeters by 80 centimeters by 50 centimeters. What is the volume of the tank in cubic meters? (1 m = 100 cm)

Show the answer Choice C

Why it is right

Volume in the given unit first: 120 cm · 80 cm · 50 cm = 480,000 cm³. Cube the linear conversion: 1 m = 100 cm means 1 m³ = 100³ cm³ = 1,000,000 cm³. Then 480,000 cm³ · (1 m³ / 1,000,000 cm³) = 0.48 m³. Alternative route — convert each side first: 1.2 m · 0.8 m · 0.5 m = 0.48 m³. Same answer. Leftover unit: m³.

Why each other choice fails

Choice A
Applies the linear factor only once to the volume: 480,000 / 100 = 4,800. Volume is length cubed, so the linear factor must be applied three times (divide by 1,000,000, not by 100).
Choice B
Applies the linear factor only twice, as if converting an area: 480,000 / 10,000 = 48. That is the cm² → m² scale (100²), not the cm³ → m³ scale (100³).
Choice D
Stops after computing the volume in cubic centimeters and never converts: 480,000 is the correct volume in cm³, but the stem asked for cubic meters.

Question 9 Harder

A office printer uses ink at a constant rate of 2.4 milliliters per page. Ink is sold in cartridges that each hold 0.12 liters. How many pages can be printed with 5 full cartridges? (1 L = 1,000 mL)

Show the answer Choice D

Why it is right

Total ink in liters: 5 · 0.12 = 0.6 L. Convert to milliliters so the rate unit matches: 0.6 L · (1,000 mL / 1 L) = 600 mL. Pages = total ink ÷ ink per page: 600 mL / (2.4 mL/page) = 250 pages. As one chain: 5 · 0.12 L · (1,000 mL / 1 L) · (1 page / 2.4 mL) = (5 · 0.12 · 1,000) / 2.4 = 600 / 2.4 = 250. Leftover unit: pages.

Why each other choice fails

Choice A
Converts and divides for a single cartridge only: 0.12 · 1,000 / 2.4 = 50. That skips the factor of 5 full cartridges the stem specifies.
Choice B
Skips the liters-to-milliliters link and divides the liter total by the milliliter rate: 5 · 0.12 / 2.4 = 0.25. The units do not match, so the quotient is not a page count.
Choice C
Inverts the rate — multiplies total ink by milliliters per page instead of dividing: 5 · 0.12 · 1,000 · 2.4 = 1,440. That treats the rate as pages per milliliter rather than milliliters per page.

Question 10 Hardest

A landscaping crew fills rectangular planter boxes that each measure 50 centimeters by 40 centimeters by 30 centimeters. Soil is delivered at a constant rate of 0.15 cubic meters per hour. How many minutes does it take to fill 4 identical planter boxes completely? (1 m = 100 cm)

Show the answer Choice C

Why it is right

Volume of one box in the given unit: 50 cm · 40 cm · 30 cm = 60,000 cm³. Cube the linear conversion: 1 m = 100 cm means 1 m³ = 100³ cm³ = 1,000,000 cm³, so one box is 60,000 / 1,000,000 = 0.06 m³ and four boxes are 0.24 m³. Time in hours: 0.24 m³ ÷ 0.15 m³/h = 1.6 h. Convert to minutes: 1.6 · 60 = 96. As one chain: 4 · 50 · 40 · 30 cm³ · (1 m³ / 1,000,000 cm³) · (1 h / 0.15 m³) · (60 min / 1 h) = (4 · 60,000 · 60) / (1,000,000 · 0.15) = 14,400,000 / 150,000 = 96. Leftover unit: minutes.

Why each other choice fails

Choice A
Completes the volume and rate chain correctly but leaves the answer in hours: 0.24 / 0.15 = 1.6. The stem asked for minutes, so the final factor of 60 is missing.
Choice B
Fills only one planter box: 0.06 / 0.15 · 60 = 24. That skips the factor of 4 identical boxes the stem specifies.
Choice D
Applies the linear conversion only twice, as if converting an area (divide by 10,000 instead of 1,000,000): 4 · 60,000 / 10,000 = 24, then 24 / 0.15 · 60 = 9,600. Volume is length cubed, so the linear factor must be applied three times.

Question 11 Hardest Student-produced response

A chemical stock solution contains 4 grams of solute per liter of solution. A lab needs 0.48 kilograms of solute. The solution is dispensed at a constant rate of 80 milliliters per second. How many minutes does it take to dispense enough solution to supply the required amount of solute? (1 kg = 1,000 g; 1 L = 1,000 mL)

Show the answer 25

Why it is right

Convert the required solute to grams: 0.48 kg · (1,000 g / 1 kg) = 480 g. Volume of solution needed at 4 g/L: 480 g · (1 L / 4 g) = 120 L. Convert to milliliters to match the rate: 120 L · (1,000 mL / 1 L) = 120,000 mL. Time in seconds: 120,000 mL ÷ 80 mL/s = 1,500 s. Convert to minutes: 1,500 · (1 min / 60 s) = 25. As one chain: 0.48 kg · (1,000 g / 1 kg) · (1 L / 4 g) · (1,000 mL / 1 L) · (1 s / 80 mL) · (1 min / 60 s) = (0.48 · 1,000 · 1,000) / (4 · 80 · 60) = 480,000 / 19,200 = 25. Leftover unit: minutes.

Answers students type instead

1500
Completes the mass, concentration, and volume-rate chain correctly but leaves the answer in seconds: 120,000 / 80 = 1,500. The stem asked for minutes, so the final factor of 1/60 is missing.
0.025
Skips the liters-to-milliliters link and divides the liter volume by the milliliter-per-second rate: (0.48 · 1,000) / (4 · 80 · 60) = 480 / 19,200 = 0.025. The units in the rate quotient do not match, so the result is not a time in minutes.
1.5
Converts mass and concentration to 120 L correctly, then divides by 80 treating the rate as liters per second instead of milliliters per second: 120 / 80 = 1.5. That skips both the 1,000 mL factor and the seconds-to-minutes factor.

Question 12 Hardest

An industrial coolant has a density of 0.8 grams per cubic centimeter. A technician draws a sample with volume 0.005 cubic meters. What is the mass of this sample in kilograms? (1 m = 100 cm; 1 kg = 1,000 g)

Show the answer Choice A

Why it is right

Convert the sample volume from cubic meters to cubic centimeters by cubing the linear factor: 1 m = 100 cm means 1 m³ = 100³ cm³ = 1,000,000 cm³, so 0.005 m³ = 0.005 · 1,000,000 = 5,000 cm³. Mass in grams: density · volume = 0.8 g/cm³ · 5,000 cm³ = 4,000 g. Convert to kilograms: 4,000 g · (1 kg / 1,000 g) = 4 kg. As one chain: 0.8 g/cm³ · 0.005 m³ · (100 cm / 1 m)³ · (1 kg / 1,000 g) = 0.8 · 0.005 · 1,000,000 / 1,000 = 0.8 · 5 = 4. Leftover unit: kg.

Why each other choice fails

Choice B
Converts volume and multiplies by density correctly to get 4,000 g, but never applies the grams-to-kilograms factor. That answers in grams when the stem asked for kilograms.
Choice C
Applies the linear conversion only twice, as if converting an area (multiply by 10,000 instead of 1,000,000): 0.8 · 0.005 · 10,000 / 1,000 = 0.04. Volume is length cubed, so the linear factor must be applied three times.
Choice D
Applies the linear conversion only once (multiply by 100 instead of 1,000,000): 0.8 · 0.005 · 100 / 1,000 = 0.0004. That is the classic linear-on-cube miss for a volume conversion.

Common mistakes

  1. Inverted conversion factor — multiplying by 1000 L / 1 m³ when the quantity is already in liters and needs cubic meters, or the reverse. The leftover unit is the tell: if it is not the asked unit, a factor is upside down.
  2. Linear factor on a squared or cubed unit — converting cm² → m² by dividing by 100 instead of 10 000, or cm³ → m³ by dividing by 100 instead of 1 000 000. Raise the linear factor to the matching power.
  3. Rounding mid-chain — rounding 1.609 to 1.6 before multiplying through a multi-step speed conversion, then missing the choice that used full precision. Carry extra digits; round only at the end as instructed.
  4. Answering in the given unit — computing 5 400 liters correctly and gridding 5400 when the stem asked for cubic meters. Reread the last line before you commit.
  5. Converting only one side of a rate — changing minutes to hours in the denominator of a speed but leaving the numerator in kilometers when the stem wanted meters. Numerator and denominator each need their own factor.
  6. Skipping a link in the chain — going from hours to an answer in per-minute rates without the 60, or from kg to a g/cm³ density without the 1 000.
  7. Treating density as mass alone (or volume alone) — reporting the converted mass and forgetting to divide by volume, or the reverse.
  8. Confusing which unit the stem asked for — solving a perfectly good chain that ends in m/s when the choices are in km/h. Match the target unit before arithmetic.

FAQ

Do I need to memorize every conversion? No. The test gives any non-basic factor in the stem — miles to kilometers, pounds to kilograms, and so on. You are expected to know time (seconds, minutes, hours, days) and metric prefixes for mass, length and volume (kilo-, centi-, milli-). If a factor is not in the stem and is not one of those, you have misread the problem.

Is this the same as ratios and proportions? College Board files them under the same skill string. Operationally they differ: a pure proportion keeps the unit system fixed and scales a comparison; a conversion multiplies by fractions equal to 1 until the unit label changes. When both appear in one stem — a rate and a unit change — do the conversion inside the rate before you scale.

What about currency conversions? Same method, rare on the Digital SAT. If a stem says 1 USD = 0.92 EUR, that is just another factor equal to a constant, oriented so the currency you hold cancels.

How do I enter a long decimal on an SPR? Type the number the last line asks for — no unit letters. If the stem says “rounded to the nearest tenth,” round once at the end. Equivalent fractions are accepted when the value is exact (for example 1/16 for 0.0625), but a rounded decimal must match the requested place value.

Desmos? Yes for the final product of an already-correct chain. No for deciding which way a factor points — see the box above.

Related skills

Editorial team · last reviewed 2026-08-12