Students and Teachers

Let the initial number of teachers be $x$ , and the initial number of students be $y$ .

From the ratio above, $x:y = 7:8$ . Convert it into fractions and you get

$\dfrac{x}{y} = \dfrac{7}{8}$

$y=\dfrac{8}{7}x\quad\implies$ Equation 1

Next, the final number of teachers is $x+11$ , and the final number of students is $y+14$ . The ratio is given to be $6:7$ . Writing this in a fraction gives

$\dfrac{x+11}{y+14} = \dfrac{6}{7}$

$7(x+11)=6(y+14)\quad\implies$ Equation 2

Substitute Equation 1 into Equation 2 and simplify (do it yourself!). The value of $x$ calculated will be the initial number of teachers in the room, which is $x = \boxed{49}$

Alternate method:

The ratio of teachers to students is 7:8. When converted to fractions, we get:

$\dfrac{\text{Number of teachers}}{\text{Number of students}} = \dfrac{7}{8}$

Note that this is just the simplest ratio of teachers to students. The actual value could be any of these:

$\dfrac{\text{Number of teachers}}{\text{Number of students}} = \dfrac{7}{8} = \dfrac{14}{16} = \dfrac{21}{24} = \dfrac{28}{32} = \dfrac{35}{40} = \dots$

In other words, we can say that

$\dfrac{\text{Initial number of teachers}}{\text{Initial number of students}} = \dfrac{7x}{8x}$

for some integer $x$ . (Note: $x$ cannot be a decimal, because that will make $7x$ into a decimal number, and we can't have a decimal number of people!)

After students and teachers enter the room, we have

$\dfrac{\text{Final number of teachers}}{\text{Final number of students}} = \dfrac{7x + 11}{8x + 14} = \dfrac{6}{7}$

From here, we now have an equation with a single variable:

$\dfrac{7x + 11}{8x + 14} = \dfrac{6}{7}$

Solving this will give you $x = 7$ . Note that the initial number of teachers is $7x$ , therefore

$\text{Initial number of teachers} = 7(7) = \boxed{49}$

Initially, T/S=7/8 .....................(1) writing this in linear form: 8T=7S Later on, T+11/S+14=6/7 imply that: 7T-6S=7 .......................(2)

solving (1) & (2) T=49