20774_Problem_solving_Year_6_Number_and_place_value_Patterns_and_algebra_1

SOLUTIONS
Note: Many solutions are written statements rather than just numbers. This is to encourage teachers and students to solve
problems in this way.
PROFIT AND LOSS ...................................................... page 9
1.
Items
sold per
week
50 75 100 250 500 1000 1500 3000
Income 2500 3750 5000 12 500 25 000 50 000 75 000 150 000
Total
costs
4000 +
750
4000 +
1125
2. Loss
3. Profit
4. Between 100 and 250
4000 +
1500
4000 +
3750
4000 +
7500
110 113 114 115
Income 5500 5650 5700 5750
Costs 4000 4000 4000 4000
115 items
1650 1645 1710 1725
5560 5695 5710 5725
4000 +
15 000
4 000 +
22 500
4000 +
45 000
CALCULATOR PATTERNS ........................................ page 10
1. (a) 4899
(b) 4900
(c) difference of 1
(d) Numbers will vary but difference will always be 1
(e) Numbers will vary but difference will always be 1
(f) For consecutive numbers, the product of the
number before and the number after is 1 less than
the number squared, number 2 .
The number before is (number – 1),
the number after is (number + 1)
(number – 1) x (number +1)
= number x number + number x 1
– 1 x number – 1 x 1
= number 2 + number – number – 1
= number 2 – 1
No matter what number is chosen, there will
always be a difference of 1
2. (a) 279
(b) 93
(c) 3
(d) e.g. 47
difference between 4 3 and 7 3 is 279
4 2 + (4 x 7) + 7 2 = 93
279 ÷ 93 = 3
Try 83
difference between 8 3 and 3 3 is 485
8 2 + (8 x 3) + 3 2 = 97
485 ÷ 97 = 5
(e) answer is always the difference between the tens
and ones digits
(f) e.g. 346 is 34 and 6 or 3 and 46
Difference is 28 or 43
(g) Pattern
when a 3 digit is split into two parts, the difference
of the cubes of the parts divided by the sum of the
squares and the product of the two parts is the
same as the difference between the two parts.
This result is a consequence of the algebraic
relationship
a 3 – b 3 = (a – b) x (a 2 + ab + b 2 ). Showing students
and asking them to substitute numbers for a and
b will show students it always works – this is
the what the questions have asked them to do in
words.
Some students may be able to multiply out the
brackets part by part in a similar way to the
example above:
(a – b) x (a 2 + ab + b 2 )
= (a x a 2 ) + (a x ab) + (a x b 2 ) – (b x a 2 )
– (b x ab) – (b x b 2 )
= a3 + a 2 b + ab 2 – a 2 b –ab 2 – b 3
= a 3 – b 3
PUZZLE SCROLLS ...................................................... page 11
1. $6.50
2. 169
3. 20
4. 18
5. 25
6. $268 and $232
R.I.C. Publications ® www.ricpublications.com.au Problem-solving in mathematics
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