When Professor Sum was asked by Mr. Little how many students in his classes, he answered. “All of them study either languages, or not at all. One-half of them study languages only. One-fourth of them study French, one-seventh study physics only, and 20 do not study at all.” How many students does Professor Sum have?
To solve this problem, we need to use a systematic approach. Let's break down the information given by Professor Sum:
1. "All of them study either languages or not at all." This means that every student in Professor Sum's classes either studies languages or does not study at all.
2. "One-half of them study languages only." This means that half of the students in the classes only study languages and don't study any other subject.
3. "One-fourth of them study French." This means that one-fourth of the students study French.
4. "One-seventh study physics only." This means that one-seventh of the students only study physics and don't study any other subject.
5. "20 do not study at all." This means that 20 students don't study any subject.
Now, let's use this information to solve the problem step by step:
Step 1: Let's assume the total number of students in Professor Sum's classes is "x".
Step 2: We know that one-half of them study languages only, so the number of students studying languages only is (1/2)x.
Step 3: We know that one-fourth of them study French, so the number of students studying French is (1/4)x.
Step 4: We know that one-seventh study physics only, so the number of students studying physics only is (1/7)x.
Step 5: We know that 20 students do not study at all.
Putting it all together, we can write an equation based on the information given:
(1/2)x + (1/4)x + (1/7)x + 20 = x
Now let's solve the equation to find the value of x:
To simplify the equation, let's find a common denominator. The common denominator for 2, 4, and 7 is 28.
(14/28)x + (7/28)x + (4/28)x = x - 20
(25/28)x = x - 20
Multiplying both sides of the equation by 28 to get rid of the fraction:
25x = 28x - 560
Subtracting 28x from both sides of the equation:
25x - 28x = -560
-3x = -560
Dividing both sides of the equation by -3:
x = -560 / -3
x = 186.67
Since the number of students cannot be in decimal form, we need to find the nearest whole number.
Therefore, Professor Sum has approximately 187 students in his classes.
First, let's assign variables to represent the number of students in each category. Let:
- A be the number of students studying only languages
- B be the number of students studying both languages and French
- C be the number of students studying both languages and physics
- D be the number of students studying only French
- E be the number of students studying only physics
- F be the number of students not studying at all
We are given the following information:
- One-half of the students study only languages, so A = 1/2(A + B + C)
- One-fourth of the students study French, so D = 1/4(A + B + D)
- One-seventh of the students study physics only, so E = 1/7(C + E)
We also know that there are 20 students who do not study at all, so F = 20.
Let's solve these equations step-by-step:
1. A = 1/2(A + B + C)
Expand the equation: A = 1/2A + 1/2B + 1/2C
Move 1/2A to the left side: A - 1/2A = 1/2B + 1/2C
Combine like terms: 1/2A = 1/2B + 1/2C
2. D = 1/4(A + B + D)
Expand the equation: D = 1/4A + 1/4B + 1/4D
Move 1/4A to the left side: D - 1/4A = 1/4B + 1/4D
Combine like terms: 3/4D = 1/4A + 1/4B
3. E = 1/7(C + E)
Expand the equation: E = 1/7C + 1/7E
Move 1/7E to the left side: E - 1/7E = 1/7C
Combine like terms: 6/7E = 1/7C
4. F = 20
Now, let's solve these equations simultaneously.
From equation 1, we have:
1/2A = 1/2B + 1/2C
From equation 2, we have:
3/4D = 1/4A + 1/4B
From equation 3, we have:
6/7E = 1/7C
From equation 4, we have:
F = 20
Let's substitute the values one by one to find the values of A, B, C, D, and E:
1. Substitute C = 7E/6 into equation 3:
6/7E = 1/7(7E/6)
Multiply both sides by 7: 6E = 7E/6
Multiply both sides by 6: 36E = 7E
Divide both sides by E: 36 = 7
Since 36 does not equal 7, this equation has no solution. Therefore, we cannot determine the values of A, B, C, D, and E.
Therefore, we cannot determine the total number of students Professor Sum has.