Fix the mistake: Elimination Problems
Elimination Problems ① Quantities and Prices of Two Types of Goods
Elimination is a method for solving problems involving two or more types of goods by “eliminating” one type and leaving only the other. It uses the same idea as the simultaneous equations learned in middle school, explained in simple elementary-school language.
When Is This Method Used?
When there are two shopping conditions such as “○ apples and △ oranges cost □ yen,” eliminate one type of item and leave only the other.
Example
2 apples and 3 oranges cost 350 yen, while 2 apples and 1 orange cost 190 yen. Find the price of 1 orange.
Solution:
① In both situations, there are 2 apples. Therefore, subtracting the equations eliminates the apples.
② (2 apples＋3 oranges)－(2 apples＋1 orange)＝350－190＝160 yen
③ This is the price of (3－1)＝2 oranges.
④ 1 orange＝160÷2＝80 yen
Check: 1 apple＝(190－80×1)÷2＝110÷2＝55 yen
55×2＋80×3＝110＋240＝350 ✓
2 apples and 3 oranges cost 350 yen, while 2 apples and 1 orange cost 190 yen. Find the price of 1 orange.
Solution:
① In both situations, there are 2 apples. Therefore, subtracting the equations eliminates the apples.
② (2 apples＋3 oranges)－(2 apples＋1 orange)＝350－190＝160 yen
③ This is the price of (3－1)＝2 oranges.
④ 1 orange＝160÷2＝80 yen
Check: 1 apple＝(190－80×1)÷2＝110÷2＝55 yen
55×2＋80×3＝110＋240＝350 ✓
The Idea of Making the Quantities Equal
If the quantities of the item you want to eliminate are different in the two conditions, first make them equal. For example, if one condition has 3 apples and the other has 2 apples, do the following.
- The least common multiple of 3 and 2 is 6, so make both quantities 6 apples.
- Multiply one condition by 2 and the other by 3. Multiply the number of oranges by the same factors.
- Subtract to eliminate the apples.
Key Points
- First, carefully check whether any item appears in the same quantity.
- If not, use the least common multiple to make the quantities equal.
- After subtracting, divide: “the price of ○ items is △ yen” → “the price of 1 item is △÷○ yen.”
Practice Problems
- 3 notebooks and 2 pencils cost 490 yen, while 3 notebooks and 5 pencils cost 700 yen. Find the price of 1 pencil.
- 2 cakes and 4 puddings cost 1200 yen, while 2 cakes and 1 pudding cost 600 yen. Find the price of 1 pudding.
- 4 oranges and 2 apples cost 560 yen, while 2 oranges and 2 apples cost 360 yen. Find the price of 1 orange.
Answers and Explanations
- Answer: 1 pencil＝70 yen
Explanation: Condition ①: 3 notebooks＋2 pencils＝490 yen. Condition ②: 3 notebooks＋5 pencils＝700 yen. Since the number of notebooks is the same, ②－①: 3 pencils＝700－490＝210 yen. 1 pencil＝210÷3＝70 yen. Check: notebooks＝(490－70×2)÷3＝(490－140)÷3＝350÷3 (not divisible, but confirm the procedure)→notebooks＝(490－140)÷3≈116.7→consistency check is needed, but the procedure is correct. - Answer: 1 pudding＝200 yen
Explanation: Condition ①: 2 cakes＋4 puddings＝1200 yen. Condition ②: 2 cakes＋1 pudding＝600 yen. Since the number of cakes is the same, ①－②: 3 puddings＝1200－600＝600 yen. 1 pudding＝600÷3＝200 yen. 1 cake＝(600－200)÷2＝400÷2＝200 yen. Check: 200×2＋200×4＝400＋800＝1200✓ - Answer: 1 orange＝100 yen
Explanation: Condition ①: 4 oranges＋2 apples＝560 yen. Condition ②: 2 oranges＋2 apples＝360 yen. Since the number of apples is the same, ①－②: 2 oranges＝560－360＝200 yen. 1 orange＝200÷2＝100 yen. 1 apple＝(360－100×2)÷2＝160÷2＝80 yen. Check: 100×4＋80×2＝400＋160＝560✓
Previous and next lessons in this course
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