Guided example: solution

Step 1 · Place the decimal so one nonzero digit leads

The first nonzero digit is the 1. The decimal moves to sit just after it: M = 1.66.

Guided example: solution

The point passes nine zeros and then the 1.

Guided example: solution

The radius is less than 1 m, so the decimal moved right: n = −10.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer B

B: the point moves left past 4, 9, 6 and five zeros, 8 places. A flipped the sign: a number above 1 takes a positive n. C counted only the five zeros and skipped 4, 9 and 6. D stopped one place early: 14.96 × 10⁷ leaves two digits before the point.

Practice 1: the route on the map

Worked example 1: solution

The small number moves right (negative n); the large number moves left (positive n).

Worked example 1: solution

The small number moves right (negative n); the large number moves left (positive n).

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 2: answer D

D: the point moves right past four zeros and the 7, 5 places, and M keeps all three digits. A dropped the zero between 7 and 6: a zero caught between nonzero digits is a digit of M. B flipped the sign: a number below 1 takes a negative n. C counted the four zeros and skipped the 7.

Practice 2: the route on the map

Practice 3 · answer: C

A keyed the −6 as +6: 2.0 × 10⁶ × 4.5 × 10³ = 9.0 × 10⁹. B keyed × 10 before EE, an extra ten: 2.0 × 10 × 10⁻⁶ × 4.5 × 10³ = 9.0 × 10⁻². D read the display back with the sign dropped: E−3 is 10⁻³, not 10³.

Practice 3: the route on the map

Worked example 3: solution

Multiply · coefficients times, exponents added

30. has two digits before the point. It is not yet scientific notation.

Worked example 3: solution

30. becomes 3.0: the point moves 1 place left.

Worked example 3: solution

Moved left adds the place to n: −2 + 1 = −1.

Worked example 3: solution

Worked example 3: the route on the map

Practice 4: answer A

Practice 4: answer A

Practice 4: answer A

Practice 4: the route on the map

Guided example: solution

Step 1 · Nonzero digits always count

The 2 and the 5 count.

Guided example: solution

The zero between 2 and 5 counts.

Guided example: solution

No zero sits in front of the 2. This rule removes nothing.

Guided example: solution

The final zero follows a decimal point, so it counts.

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: D

A counted only the nonzero digits: 2. B read the zeros of 100 as placeholders: 3. They sit between the 1 and the 5, so they are captive. C dropped the trailing zero after the decimal point: 4.

Practice 1: the route on the map

Worked example 1: solution

Step 1 · Nonzero digits always count

The 4 and the 3 count.

Worked example 1: solution

The zero between 4 and 3 counts.

Worked example 1: solution

The two zeros in front only locate the decimal point.

Worked example 1: solution

The number has a decimal point, so the final zero counts.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Nonzero digits always count

The 1 and the 2 count.

Worked example 2: solution

No zero sits between nonzero digits, and none leads. Neither rule applies here.

Worked example 2: solution

There is no decimal point. The two zeros are placeholders.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 · answer: B

The 2 and 6 count, the captive zero between them counts, and the decimal point makes the final zero count. A dropped the trailing zero: 3. C started counting at the first zero after the decimal point: 5. D counted every written digit: 6.

Practice 2: the route on the map

Worked example 3: solution

Count each factor's sig figs

8.5 carries two; 4.27 and 1.36 carry three each. The fewest is two.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 3 · answer: A

B read the count as a measurement with one sig fig: 0.8636 → 0.9. C multiplied: 43.18 × 50 = 2159, heavier than the whole box. D flipped the division: 50 ÷ 43.18 = 1.158 clips per gram.

Practice 3: the route on the map

Worked example 4: solution

A common first attempt

Rounding to three sig figs threw away the tenths digit the cylinder measured. Addition and subtraction do not count sig figs.

Worked example 4: solution

The cylinder is estimated in the tenths, the burette in the hundredths. The sum ends where the least precise input ends: the tenths.

Worked example 4: solution

Worked example 4: the route on the map

Practice 4 · answer: B

A subtracted backwards, initial minus final: +1.7 °C reads as warming. C used the multiplication rule: 21.4 has three sig figs, so −1.66. A difference keeps places, not counts. D cut off the 6 instead of rounding: a first dropped digit of 6 rounds the tenths up.

Practice 4: the route on the map

Worked example 5: solution

Subtract, and mark the entitlement

Do not round yet: carry 0.362 unrounded into the division.

Worked example 5: solution

Worked example 5: solution

Worked example 5: the route on the map

Practice 5 · answer: C

A rounded the sum to 22.0 before dividing: 22.0 ÷ 150.0 = 0.1467 → 0.147. B used one rule throughout: 1.9 has 2 sig figs, so 0.15. D stopped at the sum: 22.0 g is the salt, not the salt per milliliter. E kept every calculator digit.

Practice 5: the route on the map

Practice 6 · answer: E

A rounded the difference before dividing: 2.9 ÷ 25.0 = 0.116 → 0.12. B stopped at the sample: 2.9 g of sugar, not grams per milliliter. C used the count rule on the subtraction: 52.4 has three sig figs, so 2.86 ÷ 25.0 → 0.114. D flipped the division: 25.0 ÷ 2.86 = 8.7 mL per gram.

Practice 6: the route on the map

Practice 1 · answer: A

B is the working unit of the bench, not the base unit; 1 kg = 1000 g ties every gram reading to the kilogram. C measures volume, the space a sample fills. D measures length.

Practice 2 · answer: D

A has the right unit and an absurd size: 1 kg is a bag of sugar, and a paper clip is about a gram. B misnames the kelvin; it is one of the five base units. C reports a length; a volume takes mL or L.

Guided example: solution

Step 1 · Name the quantity

Liters and milliliters both measure volume. One equality links them.

Guided example: solution

The prefix milli- means one thousandth, so the equality is 1 L = 1000 mL.

Guided example: solution

Both orientations equal 1. Only one cancels L:

Guided example: solution

Guided example: the route on the map

Practice 3 · answer: C

A flipped the factor: 18 × 1000 = 1.8 × 10⁴, a count that grew while the unit got larger. B used the centi power, 1 g = 100 mg: 18 ÷ 100 = 0.18. D relabeled the unit without converting: 18 mg is not 18 g.

Practice 3: the route on the map

Practice 4 · answer: A

Two moves: build the volume, then apply the exact identity.

B squared the edge: 1.6 × 1.6 = 2.6 cm², the area of one face. C multiplied the edge by 3: 1.6 × 3 = 4.8 cm, still a length. D converted to liters: 4.1 cm³ × (1 L / 1000 cm³) = 4.1 × 10⁻³ L, not mL.

Practice 4: the route on the map

Practice 5 · answer: D

Three moves: build the volume, rename cm³ as mL, then the prefix equality.

Practice 5 · answer: D

Practice 5: the route on the map

Practice 6 · answer: E

Three moves: build the volume, convert the mass, build the density.

Practice 6 · answer: E

Practice 6: the route on the map

Guided example: solution

One conversion factor is needed.

Step 2 · Pick the factor that cancels its unit

The equality gives two factors. The old unit, lb, goes on the bottom:

Guided example: solution

kg survives after one factor, so the chain stops.

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: D

A flipped the factor: 16.0 / 2.54 = 6.30. B used 1 in = 25.4 cm, the millimeter number: 16.0 × 25.4 = 406. C relabeled the unit without converting: 16.0. The 2.54 is exact by definition, so the three sig figs of 16.0 carry through.

Practice 1: the route on the map

Worked example 1: solution

One conversion factor is needed.

A common first attempt

No unit cancels, and the answer is not in meters.

Worked example 1: solution

Both factors come from the same equality, and both equal 1. Only one cancels the given unit:

Worked example 1: solution

One factor completes the plan: cm cancels, m survives.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Two conversion factors are needed.

Step 2 · Pick the factor that cancels its unit

The prefix equality gives the first factor, with µm in the denominator:

Worked example 2: solution

The result is in meters, not the wanted millimeters. The second factor cancels m and leaves mm:

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 · answer: A

B stopped mid-plan at grams: 21 × 10⁻³ = 0.021 g. C flipped the kilogram factor: 21 × 10⁻³ × 10³ = 21, the starting number again. D flipped both factors: 21 × 10³ × 10³ = 2.1 × 10⁷.

Practice 2: the route on the map

Worked example 3: solution

One conversion factor is needed, cubed.

A common first attempt

m³ is m × m × m. One linear factor cancels one m; two survive, and the answer is not in cm³.

Worked example 3: solution

Cancelling all three copies of m takes the whole equality cubed, number and unit:

Worked example 3: solution

One cubed factor completes the plan: m³ cancels, cm³ survives.

Worked example 3: solution

Worked example 3: the route on the map

Practice 3 · answer: C

A cubed the unit without cubing the number, dividing by only 100: 7.5 × 10⁵ / 100 = 7500. B pointed the cubed factor the wrong way, so nothing cancels: 7.5 × 10⁵ × 10⁶ = 7.5 × 10¹¹. D used the linear factor, inverted: 7.5 × 10⁵ × 100 = 7.5 × 10⁷.

Practice 3: the route on the map

Practice 4 · answer: B

Four sig figs survive: 2500. carries four, and the English–metric factor 3.785 counts too, so nothing limits below four. A stopped mid-plan: 2500. mL is 2.500 L, and liters are not gallons. C flipped the gallon factor: 2500./1000 × 3.785 = 9.463. D treated milliliters as liters: 2500./3.785 = 660.5.

Practice 4: the route on the map

Worked example 4: solution

Four conversion factors are needed, one per arrow of the plan.

Step 2 · Pick the factor that cancels its unit

The first arrow removes mL. The prefix equality gives the factor, with mL in the denominator:

Worked example 4: solution

Each factor cancels the unit left by the one before. The chain ends when km survives:

Worked example 4: solution

Worked example 4: the route on the map

Practice 5 · answer: D

The flow rate is a measured equality, 12.0 L = 1 min, and it converts liters to minutes. A flipped the gallon factor: 5.00 / 3.785 / 12.0 × 60 = 6.60. B skipped the gallon hop and treated gallons as liters: 5.00 / 12.0 × 60 = 25.0. C stopped at minutes: 5.00 × 3.785 / 12.0 = 1.58, a count of min, not s.

Practice 5: the route on the map

Practice 6 · answer: E

A stopped at milligrams: 46.0 / 2.205 × 25.0 = 522 mg, a mass, not a volume. B treated pounds as kilograms: 46.0 × 25.0 × 5.00 / 250. = 23.0. C flipped the liquid's factor: 522 × 250. / 5.00 = 2.61 × 10⁴. D flipped the pound factor: 46.0 × 2.205 × 25.0 × 5.00 / 250. = 50.7.

Practice 6: the route on the map

Guided example: solution

Step 2 · Pick the linking equation

The arrow from Kelvin to Celsius takes the shift back off: subtract 273.15. Nothing stretches, because a kelvin and a Celsius degree are the same size.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: B

A added the shift instead of taking it off: 358 + 273.15 = 631.15. C answered in Fahrenheit, the wrong scale: 1.8(84.85) + 32 = 184.73. D stretched by 1.8 on the way to Celsius: 1.8 × 84.85 = 152.73; the 1.8 belongs to Fahrenheit.

Practice 1: the route on the map

Worked example 1: solution

Step 2 · Pick the linking equation

Celsius to Kelvin is one shift. No stretching, because a kelvin and a Celsius degree are already the same size.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

The 32 is already in Fahrenheit-sized degrees. Only the reading gets stretched.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 · answer: D

A dropped the minus sign: 78 + 273.15 = 351.15. C subtracted the shift instead of adding it: −78 − 273.15 = −351.15, and negative kelvins do not exist. B built Fahrenheit instead of Kelvin: 1.8(−78) + 32 = −108.4.

Practice 2: the route on the map

Practice 3 · answer: A

B stopped at Celsius: −50.15, never stretched or shifted into °F. C divided by 1.8, the coming-back move: −50.15 / 1.8 + 32 = 4.14. D dropped the minus sign: 1.8(50.15) + 32 = 122.27. E fed kelvins straight into the Fahrenheit equation: 1.8(223) + 32 = 433.4.

Practice 3: the route on the map

Practice 4 · answer: C

Two moves are needed: Fahrenheit to Celsius, then Celsius to Kelvin.

A never divided by 1.8: 250. − 32 + 273.15 = 491.15. B stopped at Celsius: (250. − 32) / 1.8 = 121.11. D added 273.15 straight to the Fahrenheit reading: 250. + 273.15 = 523.15. E divided before subtracting: 250. / 1.8 − 32 + 273.15 = 380.04.

Practice 4: the route on the map

Guided example: solution

One conversion factor is needed. As a rate: each of the 500. milliliters carries 1.26 g, so the tank receives 500 times 1.26 g.

Guided example: solution

The given is 500. mL, so the fraction must cancel milliliters.

Guided example: solution

The equality gives two orientations. Only one cancels mL:

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: C

A flipped the fraction: 4.20 ÷ 7.85 = 0.535, and nothing cancels. B slipped a decimal: 4.20 × 7.85 = 33.0, not 3.30. D divided the density by the volume: 7.85 ÷ 4.20 = 1.87, not a mass.

Practice 1: the route on the map

Worked example 1: solution

A common first attempt

Nothing cancels, and no quantity carries g²/mL. The fraction is upside down.

Worked example 1: solution

The given unit is grams, so grams belong in the denominator. Flip the fraction: 1 mL over 19.3 g.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Practice 2 · answer: B

A divided the density by the mass: 10.5 ÷ 170. = 0.0618, and its units are not milliliters. C slipped a decimal: 170. ÷ 10.5 = 16.2, not 1.62. D flipped the fraction: 170. × 10.5 = 1.79 × 10³, with units of g²/mL.

Practice 2: the route on the map

Worked example 2: solution

Build the fraction from the data

Mass on top, volume underneath, units in place. The surviving unit, g/mL, is a density.

Worked example 2: solution

Of the candidates, only iron matches 7.87 g/mL. The pellets are iron.

Worked example 2: solution

Worked example 2: the route on the map

Practice 3 · answer: A

B divided by the after reading: 109.3 ÷ 33.6 = 3.25, but that volume counts the water too. C flipped the fraction: 12.2 ÷ 109.3 = 0.112 mL/g, not a density. D divided by the before reading: 109.3 ÷ 21.4 = 5.11, the water alone.

Practice 3: the route on the map

Worked example 3: solution

Three conversion factors are needed.

Step 2 · Start with the given Step 3 · Write the density fraction

The metric factor converts kilograms to grams. The density fraction, written 1 mL over 0.789 g, then cancels grams.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 4 · answer: D

Practice 4 · answer: D

Practice 4: the route on the map

Practice 5 · answer: E

Three conversion factors are needed. The aluminum's mass is also the gasoline's mass.

Practice 5 · answer: E

Practice 5: the route on the map