# In one senior class, 1/3 of the students are honor students. Of these honor students, 2/7 are varsity athletes. If they are four athletes in the class, how many students are there in the class? a. 84 b. 42 c. 52 d. 60

May 16, 2018

B. 42

#### Explanation:

With the given information, we know that 2/7 of 1/3 of the class is equal to 4. This can be written as:

$\frac{2}{7} x \cdot \frac{1}{3} = 4$, where $x$ is the number of total students. All we need to do now is find $x$.

$\frac{2}{7} x = 12$

$2 x = 84$

$x = 42$

May 16, 2018

There are (b) 42 students in the class.

#### Explanation:

From the given information, we know that $\frac{1}{3}$ of the total number of students is the number of honour students:

$\frac{1}{3} \times \text{(total students)" = "(honour students)}$

We also know that $\frac{2}{7}$ of the honour students are athletes:

$\frac{2}{7} \times \text{(honour students)" = "(athletes)}$

But we also know that the number of athletes in the class is $4$.
And since we also know that the number of honour students is "$\frac{1}{3}$ the total number of students in the class", we substitute these values in like this:

$\frac{2}{7} \times \text{(honour students)" = "(athletes)}$

$\frac{2}{7} \times \left(\frac{1}{3} \times \text{total students}\right) = 4$

Let's allow $x$ to represent the total number of students. Then our equation to solve is:

$\frac{2}{7} \times \frac{1}{3} \times x = 4$

We can combine the two fractions on the left into one through multiplication:

$\frac{2}{21} \times x = 4$

We can also isolate $x$ by dividing both sides by $\frac{2}{21}$ (which is the same a multiplying both sides by $\frac{21}{2}$):

$\frac{21}{2} \left(\frac{2}{21} \times x\right) = \frac{21}{2} \times 4$

$x = \frac{21}{2} \times 4$

The 4 and the 2 reduce to "2 over 1":

$x = \frac{21}{1} \times 2$

And now we multiply $21 \times 2$:

$x = 42$