Grade 10 — Mathematics 10-3 › T3U02 Imperial Measurement

Grade 10 › T3U02 Imperial Measurement › L3
GRADE 10 · ALBERTAOUTCOME MAT10-3-M2

L3 · Converting within imperial units

Learning goal

“I can convert up and down inside the imperial system by chaining the provided relationships - multiplying toward smaller units, dividing toward larger ones - and write results in remainder form (like 2 ft 6 in) when the job calls for it.”

Chain conversion — converting through an in-between unit: yards -> feet -> inches, using one provided relationship per link.
Remainder form — a measurement written as a big unit plus what's left over: 30 in = 2 ft 6 in. Job-site standard.
Relationship factor — the provided count joining two units: 12 (in per ft), 3 (ft per yd), 16 (oz per lb), 8 (fl oz per cup).
Worked example

Convert 5 ft to inches. The chain link is 12 in per ft, and feet are the BIGGER unit, so the count grows: 5 x 12 = 60 in. Now climb back up: convert 100 in to feet. Divide by the link: 100 / 12 = 8.33... - but job sites don't write 8.33 ft. They write the remainder form: 8 full feet use 96 in, leaving 4 in -> 8 ft 4 in. Two-link chain: convert 2 mi to yards directly from the strip: 2 x 1760 = 3520 yd. The imperial routine is identical every time: name the link, decide direction (smaller unit -> bigger count -> multiply; bigger unit -> smaller count -> divide), and use remainder form when a fractional foot would be useless.

Check yourself — explain it

Converting 100 in to feet uses division because...

Paired problem — example still visible

Convert 7 ft to inches -> 7 x 12 = 84 in

Faded completion — last step is yours

"Convert 45 in to remainder form: 45 / 12 = 3 full feet using ___ in, leaving ___ in -> 3 ft ___ in." -> 36; 9; 9

Ready to practise?

10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

Start practice →Take the unit quiz