This answer key gives a full worked solution for each of the 10 problems, using prime factorizations and lists of multiples. Each word problem states whether the GCF or the LCM is needed and why.
Problem 1: List the factors
Test pairs that multiply to 36: \(1 \times 36\), \(2 \times 18\), \(3 \times 12\), \(4 \times 9\), \(6 \times 6\).
1, 2, 3, 4, 6, 9, 12, 18, 36
Problem 2: Prime factorization
Split 60 into \(6 \times 10\), then split each: \(6 = 2 \times 3\) and \(10 = 2 \times 5\). So \(60 = 2 \times 2 \times 3 \times 5\).
Check: \(4 \times 3 \times 5 = 60\).
\(2^2 \times 3 \times 5\)
Problem 3: GCF of two small numbers
Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 18: 1, 2, 3, 6, 9, 18. The common factors are 1, 2, 3, and 6.
\(\text{GCF} = 6\)
Problem 4: GCF of larger numbers
\(48 = 2 \times 2 \times 2 \times 2 \times 3\) and \(72 = 2 \times 2 \times 2 \times 3 \times 3\). Shared primes: three 2s and one 3. \(2 \times 2 \times 2 \times 3 = 24\).
Check: \(48 : 24 = 2\) and \(72 : 24 = 3\).
\(\text{GCF} = 24\)
Problem 5: LCM of 6 and 8
Multiples of 8: 8, 16, 24. Is 8 a multiple of 6? No. Is 16? No. Is 24? \(6 \times 4 = 24\), yes.
\(\text{LCM} = 24\)
Problem 6: LCM of 9 and 12
Multiples of 12: 12, 24, 36. 12 is not divisible by 9, 24 is not, but \(9 \times 4 = 36\).
\(\text{LCM} = 36\)
Problem 7: Factor a sum
\(54 = 2 \times 3^3\) and \(36 = 2^2 \times 3^2\). The GCF is \(2 \times 3^2 = 18\). Divide each addend by 18: \(54 : 18 = 3\) and \(36 : 18 = 2\).
So \(54 + 36 = 18(3 + 2)\). Check: \(54 + 36 = 90\) and \(18 \times 5 = 90\).
\(18(3 + 2)\)
Problem 8: Fruit baskets
The number of baskets must divide both 36 and 60, so find the GCF. \(36 = 2^2 \times 3^2\) and \(60 = 2^2 \times 3 \times 5\), so the GCF is \(2^2 \times 3 = 12\).
Apples per basket: \(36 : 12 = 3\). Oranges per basket: \(60 : 12 = 5\).
12 baskets, each with 3 apples and 5 oranges.
Problem 9: Bus schedules
The buses meet again after a number of minutes that is a multiple of both 8 and 12, so find the LCM. Multiples of 12: 12, 24. \(24 = 8 \times 3\), so the LCM is 24.
24 minutes after 7:00 AM is 7:24 AM.
7:24 AM
Problem 10: Sticker sets
\(56 = 2^3 \times 7\) and \(84 = 2^2 \times 3 \times 7\). The GCF is \(2^2 \times 7 = 28\). Sets: 28. Stars per set: \(56 : 28 = 2\). Hearts per set: \(84 : 28 = 3\).
Factored sum: \(56 + 84 = 28(2 + 3)\). Check: \(140 = 28 \times 5\).
28 sets, each with 2 stars and 3 hearts; \(56 + 84 = 28(2 + 3)\).
See every chapter on the Grade 6 math page.