Homework problem: All the students have the same age(in years) apart from 3...

[e=mc²]

The problem:
There is a class in an elementary school. The total age of the students therein is 208(the age of every student added up together). All the students have the same age(in years) apart from 3; 2 are a year older than the rest, and 1 is a year younger than the rest.

-Find the number of the students
-Find the age of each student.

Now the thing is, we've only been taught to solve first degree equations(linear) in one variable, or equations that can be solved by being written in that form. It feels like this homework exercise is completely out of my level. I tried everything but I had no results. And yet, the instructor deliberately chose this specific exercise, and even told us he will give +2 points(not much but when he does that it's usually easier to get a perfect score in the exam itself than solve the problem, no joke there.)

This is the first time I find myself in need of help from others, in a math problem. I'm pretty sure there's some sort of fooling, or trick that I didn't see.

I tried to solve it myself but I could only just name 3 variables: y being the number of students in total, x the age of students that are equal in age, and z, the number of said students. I didn't get anything coherent past that.

P.S I'm not very sure if my post belongs in pre-algebra or beginning algebra, so forgive me if I'm in the wrong section.

 

[Deleted member 4993]

The problem:
There is a class in an elementary school. The total age of the students therein is 208(the age of every student added up together). All the students have the same age(in years) apart from 3; 2 are a year older than the rest, and 1 is a year younger than the rest.

-Find the number of the students
-Find the age of each student.

Now the thing is, we've only been taught to solve first degree equations(linear) in one variable, or equations that can be solved by being written in that form. It feels like this homework exercise is completely out of my level. I tried everything but I had no results. And yet, the instructor deliberately chose this specific exercise, and even told us he will give +2 points(not much but when he does that it's usually easier to get a perfect score in the exam itself than solve the problem, no joke there.)

This is the first time I find myself in need of help from others, in a math problem. I'm pretty sure there's some sort of fooling, or trick that I didn't see.

I tried to solve it myself but I could only just name 3 variables: y being the number of students in total, x the age of students that are equal in age, and z, the number of said students. I didn't get anything coherent past that.

P.S I'm not very sure if my post belongs in pre-algebra or beginning algebra, so forgive me if I'm in the wrong section.

Assume:

# of students = n

(n-3) students are of age = A

2 students are of age = A+1

1 student age = A-1

We are going to assume 'n' and 'A' are integers.

Then

(n-3)*A +2*(A+1) + A - 1 = 208

n *A = 207

Can you factorize 207 and take a guess at "n" and "A"?

 

[e=mc²]

The factors of 207 are 207, 69, 23, 9, 3 and 1. Given that it's a single class, in an elementary school, the most likely outcome is that n = 23 while A = 9. No other outcome is possible given that 1 and 3 are too small numbers to make a class and aren't ages suitable for elementary school, which automatically crosses 207 and 69 out of the list. Am I right?

Thanks for the help by the way, never occurred to me to factorize the number of students.
So this means that 20 students are age 9, 2 are age 10 and a single student at the age of 8.

 

[JeffM]

The problem:
There is a class in an elementary school. The total age of the students therein is 208(the age of every student added up together). All the students have the same age(in years) apart from 3; 2 are a year older than the rest, and 1 is a year younger than the rest.

-Find the number of the students
-Find the age of each student.

Now the thing is, we've only been taught to solve first degree equations(linear) in one variable, or equations that can be solved by being written in that form. It feels like this homework exercise is completely out of my level. I tried everything but I had no results. And yet, the instructor deliberately chose this specific exercise, and even told us he will give +2 points(not much but when he does that it's usually easier to get a perfect score in the exam itself than solve the problem, no joke there.)

This is the first time I find myself in need of help from others, in a math problem. I'm pretty sure there's some sort of fooling, or trick that I didn't see.

I tried to solve it myself but I could only just name 3 variables: y being the number of students in total, x the age of students that are equal in age, and z, the number of said students. I didn't get anything coherent past that.

P.S I'm not very sure if my post belongs in pre-algebra or beginning algebra, so forgive me if I'm in the wrong section.

I know it is common practice to start by teaching how to solve linear equations in one variable, but it is a very bad practice.

In addition there are indeed two "tricks."

I do not generally give solutions, but I shall make an exception in this case.

\(\displaystyle n = number\ of\ students\ of\ same\ age.\)

\(\displaystyle t = total\ students.\)

\(\displaystyle a = age\ of\ all\ but\ three\ students.\)

\(\displaystyle b = age\ of\ two\ older\ students.\)

\(\displaystyle c = age\ of\ one\ younger\ student.\)

See how easy the first step is when you provide a symbol (letter) for each unknown.

Now in general to solve for q unknowns, you need q equations. We seem to need 5 equations. Three of them are obvious, namely

\(\displaystyle t = n + 3.\)

\(\displaystyle b = a + 1.\)

\(\displaystyle c = a - 1.\)

That too was easy. But now comes the first trick. We are dealing with a sum of t ages, but we do not know what t is except
n + 3. But we don't need to know the individual ages. There are n students at age a so they contribute n * a = an to the total age. There are 2 students at age b so they contribute 2 * b = 2b. And the one student of age c contributes c. That gives the equation

\(\displaystyle an + 2b + c = 208.\)

The most general method for solving systems of equations is called substitution. We use the information in one equation to get rid of an unknown in the other equations. Here goes.

\(\displaystyle 208 = an + 2b + c = an + 2(a + 1) + c = an + 2a + 2 + c \implies\)

\(\displaystyle 206 = a(n + 2) + c = a(n + 2) + a - 1 \implies\)

\(\displaystyle 207 = a(n + 3) = at.\)

We are left with a single equation in two unknowns. Such equations have an infinite number of possible solutions, which effectively means they can't be solved. But now comes the second trick. We require a solution in positive integers. This extra requirement sometimes lets us solve what is otherwise insoluble.

\(\displaystyle 3^2 * 23 = 207 \implies\)

a = 3 and t = 69 or a = 9 and t = 23. The latter looks more realistic. Let's check.

\(\displaystyle 20 * 9 + 2 * 10 + 8 = 180 + 20 + 8 = 208.\)