Elimination Method ③ How to Think When There Are Three Types
The elimination method works in the same way as the “simultaneous equations” learned in middle school. If you become familiar with the method of “eliminating one side and thinking,” you will be able to solve even difficult problems on your own from elementary school onward.
Elimination Problems with Three Types
When there are three types of items, three conditions are also needed. The basic method is to use two conditions to eliminate one type first, narrow the items down to two types, and then eliminate one more time.
Example Problem
There are three types of items: A, B, and C.
・1 A + 2 B = 500 yen
・1 B + 3 C = 450 yen
・1 A + 1 C = 300 yen
Find the price of one each of A, B, and C.
Solution:
① From condition ③, A can be written as A = 300 − C.
② Substitute this into condition ①: (300 − C) + 2B = 500 → 2B = 200 + C
③ From condition ②, B = 450 − 3C.
④ Combine ② and ③: 2 × (450 − 3C) = 200 + C
900 − 6C = 200 + C
700 = 7C
C = 100
⑤ B = 450 − 3 × 100 = 150 yen
⑥ A = 300 − 100 = 200 yen
Answer: A = 200 yen, B = 150 yen, C = 100 yen
There are three types of items: A, B, and C.
・1 A + 2 B = 500 yen
・1 B + 3 C = 450 yen
・1 A + 1 C = 300 yen
Find the price of one each of A, B, and C.
Solution:
① From condition ③, A can be written as A = 300 − C.
② Substitute this into condition ①: (300 − C) + 2B = 500 → 2B = 200 + C
③ From condition ②, B = 450 − 3C.
④ Combine ② and ③: 2 × (450 − 3C) = 200 + C
900 − 6C = 200 + C
700 = 7C
C = 100
⑤ B = 450 − 3 × 100 = 150 yen
⑥ A = 300 − 100 = 200 yen
Answer: A = 200 yen, B = 150 yen, C = 100 yen
There Is Also a Method of “Eliminating by Addition”
Sometimes a variable can be eliminated by addition rather than subtraction. Depending on the problem, using subtraction and addition effectively can make the calculations much easier.
Key Points
- The number of conditions and the number of types of items to find are usually the same (two conditions for two types, and three conditions for three types).
- Look carefully at the problem and decide whether to “eliminate by subtraction” or “eliminate by addition.”
- The more difficult the problem, the more important it is to write the intermediate calculations carefully in your notebook.
- In an actual entrance examination, the key is how quickly you can recognize “which variable is easiest to eliminate.”
Practice Problems
- One hamburger and two orders of fries cost 620 yen, and two hamburgers and one order of fries cost 790 yen. Find the price of one hamburger.
- There are two types of balls, A and B. 3 A + 1 B = 500 g, and 1 A + 3 B = 700 g. Find the weight of one B.
- The transportation cost for one adult and two children is 1,300 yen, and for two adults and one child it is 1,700 yen. Find the transportation cost for one adult and one child.
Answers and Explanations
- Answer: One hamburger = 320 yen
Explanation: Condition ①: one hamburger + two orders of fries = 620 yen; condition ②: two hamburgers + one order of fries = 790 yen. Eliminate hamburgers: ① × 2: two hamburgers + four orders of fries = 1,240 yen. ① × 2 − ②: three orders of fries = 1,240 − 790 = 450 yen. One order of fries = 150 yen. Hamburger = 620 − 150 × 2 = 620 − 300 = 320 yen. Check: 320 × 2 + 150 = 640 + 150 = 790✓ - Answer: One B = 175 g
Explanation: Condition ①: 3 A + 1 B = 500 g; condition ②: 1 A + 3 B = 700 g. Eliminate A: ② × 3: 3 A + 9 B = 2,100 g. ② × 3 − ①: 8 B = 2,100 − 500 = 1,600 g. One B = 1,600 ÷ 8 = 200 g. One A = (500 − 200) ÷ 3 = 300 ÷ 3 = 100 g. Check: 100 × 3 + 200 = 500✓, 100 + 200 × 3 = 700✓ - Answer: One adult = 700 yen, one child = 300 yen
Explanation: Condition ①: one adult + two children = 1,300 yen; condition ②: two adults + one child = 1,700 yen. Eliminate adults: ① × 2: two adults + four children = 2,600 yen. ① × 2 − ②: three children = 2,600 − 1,700 = 900 yen. One child = 300 yen. One adult = 1,300 − 300 × 2 = 1,300 − 600 = 700 yen. Check: 700 + 300 × 2 = 1,300✓, 700 × 2 + 300 = 1,700✓