Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum

Monthly Update sign up
Mailing List signup
Learning Math Home
Patterns, Functions, and Algebra
Session 6 Part A Part B Part C Part D Homework
Algebra Site Map
Session 6 Materials:



Solutions for Session 6, Homework

See solutions for Problems: H1 | H2 | H3 | H4 | H5

Problem H1

The circles must each be 9, because the mobile splits twice. The clue tells us the cubes cannot weigh more than 5 each. Because the hourglass and 2 cubes weigh 18 and the hourglass weighs less than 10, each cube is 5 and the hourglass is 8.

<< back to Problem H1


Problem H2

The upside-down cone is 10 (2 splits), and the circle is 5 (3 splits). Because the cube and the diamond make 5 together, the clue tells us the diamond is 2, and the cube is 3. The others: hourglass = 7, cylinder = 6, horizontal bar = 1, pyramid = 8, flat disk = 9.

<< back to Problem H2


Problem H3

The flat disk must be 9 (2 splits). The cube must be an even number, and larger than the circle -- this makes it 8. The others: circle = 5, pyramid = 3, cylinder = 2, diamond = 1.

<< back to Problem H3


Problem H4

One possible course is to find the value of the diamond first. After the diamond in the third set, there are three consecutive splits (to the cylinders), so the total weight after the diamond must be a multiple of 8. This limits the possible values of the diamond shape to 4 or 12, because the diamond on the left, when paired with a circle, weighs a total of 15. Investigating the circles in the left branch shows that the circle must be 11, and the diamond must be 4. The others: cylinder = 7, flat disk = 8, pyramid = 14, hourglass = 6, cube = 3, upside-down cone = 10.

<< back to Problem H4


Problem H5

Backtrack or work from false position.

Backtracking: Each brother got 4 cookies, so the brothers shared 12. This is half of the original number of cookies, so the original number is 24 (multiply 12 by 2).

False position: Pick a convenient number of cookies for Marcus, then see what would have happened. Say that Marcus had 60 cookies (a good number, since 60 is very divisible). Half go to David, so there are 30 to split among 3 brothers. If Marcus had 60, each brother gets 10 cookies.

So, to turn 10 into 4 we need to multiply by 0.4, which shows that Marcus had (60)(0.4) = 24 cookies to start.

As with Problem B10, finding the solution by covering up is possible, but difficult. A solution found by doing the same thing to both sides is only possible after creating an equation for the entire situation.

<< back to Problem H5


Learning Math Home | Algebra Home | Glossary | Map | ©

Session 6: Index | Notes | Solutions | Video


© Annenberg Foundation 2016. All rights reserved. Legal Policy