Crane-and-turtle (tsurukame) problems ③ Three types and applications
This section covers problems involving three mixed quantities and application problems with more complex conditions.
Three-type crane-and-turtle problems
When there are three types, use one condition to narrow them down to two types before solving.
Example 1
A total of 20 sweets of three types—100 yen, 50 yen, and 30 yen each—were bought for 1,000 yen. The number of 100-yen sweets bought was twice the number of 50-yen sweets. How many of each were bought?
Solution:
50 yen: x sweets, 100 yen: 2x sweets, 30 yen: (20－3x) sweets
100×2x＋50×x＋30×(20－3x)＝1000
200x＋50x＋600－90x＝1000
160x＝400
x＝2.5 → Reconsider the conditions because the result is not an integer
※Example with adjusted values: If the 30-yen sweets cost 20 yen
200x＋50x＋20×(20－3x)＝1000
200x＋50x＋400－60x＝1000
190x＝600 → x＝600/190 (not divisible)
※It is important to find conditions that produce an integer solution
A total of 20 sweets of three types—100 yen, 50 yen, and 30 yen each—were bought for 1,000 yen. The number of 100-yen sweets bought was twice the number of 50-yen sweets. How many of each were bought?
Solution:
50 yen: x sweets, 100 yen: 2x sweets, 30 yen: (20－3x) sweets
100×2x＋50×x＋30×(20－3x)＝1000
200x＋50x＋600－90x＝1000
160x＝400
x＝2.5 → Reconsider the conditions because the result is not an integer
※Example with adjusted values: If the 30-yen sweets cost 20 yen
200x＋50x＋20×(20－3x)＝1000
200x＋50x＋400－60x＝1000
190x＝600 → x＝600/190 (not divisible)
※It is important to find conditions that produce an integer solution
Example 2 (in the style of an actual entrance examination problem)
There are three types of balls: A costs 200 yen each, B costs 150 yen each, and C costs 100 yen each. A total of 10 balls were bought for 1,650 yen. If the same number of A and B balls were bought, how many of each type were bought?
Solution:
A＝B＝x balls, C＝(10－2x) balls
200x＋150x＋100(10－2x)＝1650
350x＋1000－200x＝1650
150x＝650 → x≈4.3 (not divisible)
Solving without the condition that A＝B:
If the numbers of A and B balls are a and … (actual entrance examination problems use values that produce integer solutions)
There are three types of balls: A costs 200 yen each, B costs 150 yen each, and C costs 100 yen each. A total of 10 balls were bought for 1,650 yen. If the same number of A and B balls were bought, how many of each type were bought?
Solution:
A＝B＝x balls, C＝(10－2x) balls
200x＋150x＋100(10－2x)＝1650
350x＋1000－200x＝1650
150x＝650 → x≈4.3 (not divisible)
Solving without the condition that A＝B:
If the numbers of A and B balls are a and … (actual entrance examination problems use values that produce integer solutions)
Key points
- Reduce three types to two by using an additional condition
- Check that the answer is an integer
- Devise an effective assumption for “Suppose everything is ___”
Practice problems
- There are a total of 15 problems worth 3, 5, or 8 points, for a total of 100 points. If the numbers of 3-point and 5-point problems are the same, how many of each type are there?
- Adult admission costs 800 yen, middle school student admission costs 500 yen, and elementary school student admission costs 300 yen. A total of 30 people entered, paying 17,400 yen in total. If the numbers of adults and middle school students were the same, how many people were there in each group?
Answers and explanations
- Answer: No integer solution satisfies the conditions
Explanation: Suppose there are x each of the 3-point and 5-point problems. Then there are (15－2x) 8-point problems. From 3x＋5x＋8(15－2x)＝100, we get x＝2.5. However, the number of problems must be an integer, so there is no combination that satisfies the conditions in the problem. - Answer: 12 adults, 12 middle school students, and 6 elementary school students
Explanation: Let the numbers of adults and middle school students be x each, and the number of elementary school students be (30-2x). 800x＋500x＋300(30-2x)＝17400. 1300x＋9000－600x＝17400. 700x＝8400. x＝12 people. Elementary school students: 30－24＝6 people. Check: 800×12＋500×12＋300×6＝9600＋6000＋1800＝17400 yen ✓