Two purchases of the same two items

Primary 4 Four Operations of Decimals study guide: the whole chapter, step by step

About this subtopic

The idea in brief. You cannot compare two purchases that differ in both items. So multiply one whole purchase, price and all, until one of its counts matches the other purchase's. Now the two purchases differ in one item only, and the difference between the two prices buys exactly that difference.

Lesson video, with the working

Comparing two purchases

3 min 42 s

The film teaches 2 questions in turn. Each question shows first; its working is written as the film explains it.

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Today's question

The total cost of 1 cap and 3 badges is $17.40. The total cost of 2 such caps and 9 such badges is $48.30. Find the cost of 1 badge.

  1. 1 cap + 3 badges = $17.40
  2. 2 caps + 6 badges = 2 × 17.40 = $34.80
  3. 2 caps + 9 badges = $48.30
  4. 3 badges = 48.30 − 34.80 = $13.50
  5. One badge costs $13.50 ÷ 3 = $4.50.

Ans: $4.50

Example 1

The total cost of 1 pencil case and 2 markers is $9.70. The total cost of 4 such pencil cases and 2 such markers is $22.90. Find the cost of 1 pencil case.

  1. 1 pencil case + 2 markers = $9.70
  2. 4 pencil cases + 2 markers = $22.90
  3. 3 pencil cases = 22.90 − 9.70 = $13.20
  4. One pencil case costs $13.20 ÷ 3 = $4.40.

Ans: $4.40

How to solve it
  1. Find which item to match. Double the first purchase so both purchases contain 2 caps.
  2. Multiply the whole purchase, including its price: 2 caps and 6 badges cost 2 times $17.40, which is $34.80.
  3. Compare with 2 caps and 9 badges costing $48.30. The difference pays for 3 extra badges.
  4. Subtract: $48.30 minus $34.80 is $13.50. Divide by 3 to find one badge: $4.50.
  5. If one item already matches, subtract without multiplying. The purchases with 2 markers differ by 3 pencil cases.
  6. The difference is $22.90 minus $9.70, which is $13.20. One pencil case costs $13.20 divided by 3, which is $4.40.

When one item matches, the difference in total cost belongs to the extra items. Divide by their count to find the cost of one.

Practice questions with worked solutions

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Intermediate

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Question 1 of 7 · Intermediate · free to tryRavi bought 2 books and 3 files for $23.10. Mei Ling bought 2 such books and…Ravi bought 2 books and 3 files…

Ravi bought 2 books and 3 files for $23.10. Mei Ling bought 2 such books and 7 such files for $35.50. Find the cost of 1 book.

2books+3files$23.102books+7files$35.50
$

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Video solution · 1 min 54 s

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Working

  1. 2 books + 3 files = $23.10
  2. 2 books + 7 files = $35.50
  3. 4 files = 35.50 − 23.10 = $12.40
  4. 1 file = 12.40 ÷ 4 = $3.10
  5. 3 files = 3 × 3.10 = $9.30
  6. 2 books = 23.10 − 9.30 = $13.80
  7. 1 book = 13.80 ÷ 2 = $6.90
  8. 1 book costs $6.90.

Answer: $6.90

For parents

How can my child solve two purchases problems without using algebra?

Your child uses two purchases containing the same kinds of item to work out an individual price.

What your child learns here

Your child uses two purchases containing the same kinds of item to work out an individual price. This is an application of Primary 4 decimal operations and mathematical problem solving. The key is to compare purchases that contain the same quantity of one item, so the remaining difference has a clear meaning.

  • Identify the quantity of each item in each purchase and its total cost, keeping item counts distinct from prices written in dollars and cents.
  • Recognise that the same item has the same unit price throughout the problem unless the question explicitly states a different condition.
  • Repeat a whole purchase when needed to match one item's count, multiplying both item counts and its total by the same whole number.
  • Subtract the matched purchase totals to find the cost of the extra items, then divide by their count to obtain one item's price.
  • Use the recovered price in an original purchase to find the other price, and read the final question again before choosing which amount to report.
  • Apply comparisons to masses as well as prices, including three pair totals where each of the three objects is counted twice overall.
Does the answer make sense?

After finding both unit prices, your child can calculate the cost of each original purchase using its own item counts. Both totals must match the amounts given; fitting just one total is not enough to confirm that the two prices are correct.

How to practise these questions

This practice set has Intermediate and Challenge questions. Intermediate items include purchases whose item counts already match, a purchase that must be repeated before comparison, three pair masses, and a new combination of containers after individual masses are found. Challenge items add change after finding a price or compare what the same money can buy using a relationship between item prices. Ask your child to identify what is equal before subtracting any totals. For pair-mass questions, discuss how often each object is counted when the three pairs are combined. Once an individual value is known, check the final request carefully: the answer may be another bundle's total or the money remaining.

Questions parents ask

Does my P4 child need simultaneous equations for this page? No formal algebra is needed for the method described here. Your child works with item counts and purchase totals, repeating a purchase when necessary before comparing. These are applications of multiplication, subtraction and division, with the reason for each calculation stated in words.

Why must both item counts change when a purchase is repeated? Repeating a purchase means buying its entire collection again. Every item count and the total cost therefore change by the same multiplier. Doubling only the item whose count you want to match would create a different purchase whose total is not supplied by the question.

What should I ask if my child subtracts the totals immediately? Ask which items are present in equal numbers in the two purchases. If one item's counts already match, the difference can be useful straight away. If neither matches, your child needs to establish a suitable comparison first so that the difference represents a known number of just one item.