Excess and Shortage Problems ③ Slightly Difficult Application Problems
Let us use the way of thinking for excess and shortage problems to tackle slightly more advanced questions. We will carefully practice understanding the conditions and setting up equations.
A Problem Involving Two Ways of Distributing
Example Problem
We distribute notebooks to students. If we give 5 notebooks to each student, 4 notebooks are left over; if we give 6 notebooks to each student, we are 8 notebooks short. How many students are there, and how many notebooks are there?
Solution:
① Number of students = (4＋8)÷(6－5)＝12÷1＝12 students
② Number of notebooks＝5×12＋4＝64 notebooks
Check: 6×12－8＝72－8＝64 ✓
We distribute notebooks to students. If we give 5 notebooks to each student, 4 notebooks are left over; if we give 6 notebooks to each student, we are 8 notebooks short. How many students are there, and how many notebooks are there?
Solution:
① Number of students = (4＋8)÷(6－5)＝12÷1＝12 students
② Number of notebooks＝5×12＋4＝64 notebooks
Check: 6×12－8＝72－8＝64 ✓
The Type “If the Number of People Increases, There Will Not Be Enough”
Example Problem 2
We were supposed to distribute apples to children, 6 apples per child, until all the apples were distributed exactly. However, 4 more children came, so we could distribute only 4 apples per child, with 8 apples still left over. Find the original number of children.
Solution: Let the original number of children be N.
① Number of apples＝6×N (original plan)
② In fact, 4 apples were given to each of (N＋4) children, with 8 apples left over: 4×(N＋4)＋8
③ The number of apples is the same, so
6N＝4×(N＋4)＋8
6N＝4N＋16＋8
6N＝4N＋24
2N＝24
N＝12 children
④ Number of apples＝6×12＝72 apples
We were supposed to distribute apples to children, 6 apples per child, until all the apples were distributed exactly. However, 4 more children came, so we could distribute only 4 apples per child, with 8 apples still left over. Find the original number of children.
Solution: Let the original number of children be N.
① Number of apples＝6×N (original plan)
② In fact, 4 apples were given to each of (N＋4) children, with 8 apples left over: 4×(N＋4)＋8
③ The number of apples is the same, so
6N＝4×(N＋4)＋8
6N＝4N＋16＋8
6N＝4N＋24
2N＝24
N＝12 children
④ Number of apples＝6×12＝72 apples
Key Points
- Read the problem carefully and make sure that “the number of items does not change.”
- For “problems in which the number of people changes,” carefully set up a new equation for each way of distributing.
- If the calculation does not work, read the problem again.
- Excess and shortage problems and difference-collection problems have the same basic way of thinking. Remember to set up the equations consistently.
Practice Problems
- If ribbon is cut into 4cm pieces for children, 20cm is left over; if it is cut into 5cm pieces, it is 10cm short. How many children are there?
- If 3 sweets are given to each child, 15 sweets are left over. When 3 more children come and the same number of sweets is given to each child, there are none left. How many sweets are given to each child?
- If 7 pencils are given to each person, 3 pencils are left over; if 9 pencils are given to each person, 13 pencils are short. How many pencils are there altogether?
Answers and Explanations
- Answer: There are 6 children
Explanation: With 4cm pieces, 20cm is left over; with 5cm pieces, there is a shortage of 10cm. Number of children＝(surplus＋shortage)÷(difference in amount distributed)＝(20＋10)÷(5－4)＝30÷1＝30 children. Ribbon＝4×30＋20＝140cm. Check: 5×30－10＝140✓. - Answer: 5 sweets were given to each child
Explanation: Giving 3 sweets each to N children leaves 15 sweets, so sweets＝3N＋15. When 3 more children receive x sweets each and none are left, sweets＝x×(N＋3). 3N＋15＝x(N＋3). x is an integer. N＋3 must divide 3N＋15. 3N＋15＝3(N＋5). x(N＋3)＝3(N＋5). x＝3(N＋5)÷(N＋3)＝3×(1＋2÷(N＋3)). For N＋3 to divide 2: N＋3＝1 (N＝－2, impossible) or N＋3＝2 (N＝－1, impossible). It does not divide evenly → Check the problem setting. When N＝2: sweets＝6＋15＝21. 21÷5＝4.2 for 5 children (not divisible). When N＝3: sweets＝9＋15＝24. 24÷6＝4 sweets for 6 children✓. When N＝5: sweets＝15＋15＝30. 30÷8＝3.75 for 8 children (not divisible). When N＝6: sweets＝33. 33÷9＝3.67 for 9 children (not divisible). When N＝12: sweets＝51. 51÷15＝3.4 for 15 children (not divisible). One possible answer: When N＝3 children, 4 sweets were given to each child (24 sweets total÷6 children). - Answer: There are 59 pencils
Explanation: With 7 pencils per person, 3 are left over; with 9 pencils per person, there is a shortage of 13. Number of people＝(3＋13)÷(9－7)＝16÷2＝8 people. Number of pencils＝7×8＋3＝56＋3＝59 pencils. Check: 9×8－13＝72－13＝59✓