1
$\begingroup$

There are $42$ dogs signed up to compete in the dog show. There are $36$ more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?

I have tried to subtract $42$ from $36$ and then divide by two. I don't know the way forward from here again, can I get help with this?

$\endgroup$
3
  • $\begingroup$ $S + L = 42$ and $S - L = 36$. $\endgroup$ Jul 27 at 23:24
  • 2
    $\begingroup$ the "proper" way to answer this question is perhaps assign variables to the number of small dogs and number of large dogs and come up with some equations and try to solve them. E.g. x small dogs and y large, with x+y=42 and x-y=36. (Try solve this.) Alternatively, start with 36 small dogs and no large dogs, and keep increasing the number in each group (to maintain difference of 36) until you get the correct sum 42. E.g. next try is 37 small dogs and 1 large, total is 38, too small. Next 38 small, 2 large, total 40, not enough. Finally 39 small dogs and 3 large, total is 42, just right. $\endgroup$
    – Mirko
    Jul 27 at 23:27
  • 1
    $\begingroup$ you probably want to subtract 36 from 42, not the other way around. So, 42-36=6, and, inspired by the answer by Theo Bendit, the interpretation of this number 6 is that in this group of 6 remaining dogs there are an equal number of small and large dogs. So there are 3 small dogs and 3 large dogs in this group of 6 dogs. These 3 small dogs together with the 36 dogs that were temporarily subtracted makes 39 small dogs total. Welcome to MSE! $\endgroup$
    – Mirko
    Jul 27 at 23:36
4
$\begingroup$

First of all, we know there are a total of $42$ dogs.

Let $x$ be the number of small dogs and $y$ be the number of big dogs.

We know that $x = y + 36$ and $x + y = 42$

Substituting $y + 36$ in the second equation, we get: $$y + y + 36 = 42$$ $$2y + 36 = 42$$

Simplifying this, we get: $$2y = 42-36$$ $$2y = 6$$ $$y = 3$$

There are a total of 3 big dogs.
Subtracting 3 from the original total ($42$). We get $39$.

There are $39$ small dogs in the show.

$\endgroup$
2
$\begingroup$

Let's draw a picture here to help us picture what's happening. There are $42$ dogs in show, of various sizes, which I shall represent with the following $7 \times 6$ array:

$$\begin{matrix} ? & ? & ? & ? & ? & ? & ? \\ ? & ? & ? & ? & ? & ? & ? \\ ? & ? & ? & ? & ? & ? & ? \\ ? & ? & ? & ? & ? & ? & ? \\ ? & ? & ? & ? & ? & ? & ? \\ ? & ? & ? & ? & ? & ? & ? \end{matrix}$$

I put $?$ into it to show that we don't know the size of the dogs. Now, we know that there are $36$ more small dogs than big Dogs. So, at least, we know $36$ of these are small dogs:

$$\begin{matrix} d & d & d & d & d & d & ? \\ d & d & d & d & d & d & ? \\ d & d & d & d & d & d & ? \\ d & d & d & d & d & d & ? \\ d & d & d & d & d & d & ? \\ d & d & d & d & d & d & ? \end{matrix}$$

If we remove these dogs from the show, we get just $6$ dogs:

$$\begin{matrix} ? \\ ? \\ ? \\ ? \\ ? \\ ? \end{matrix}$$ Importantly, now there are the same number of small dogs left as big Dogs, so half of these remaining dogs are small, and half are big, making our diagram look like this:

$$\begin{matrix} d \\ d \\ d \\ D \\ D \\ D \end{matrix}$$

If we add the other $36$ dogs back, our show looks like this in total:

$$\begin{matrix} d & d & d & d & d & d & d \\ d & d & d & d & d & d & d \\ d & d & d & d & d & d & d \\ d & d & d & d & d & d & D \\ d & d & d & d & d & d & D \\ d & d & d & d & d & d & D \end{matrix}$$

So, as you can see, there are $3$ big Dogs, and the other $39 = 36 + 3$ are small dogs.

It's all well and good to do some operations, but you should try to keep track of what the numbers you get actually mean in terms of the real world!

$\endgroup$
2
  • $\begingroup$ :D :D :D ++++++++++ $\endgroup$
    – Mirko
    Jul 27 at 23:30
  • $\begingroup$ @Mirko I'm glad you liked it. :-) $\endgroup$ Jul 27 at 23:31
2
$\begingroup$

if you want to see how equations could be used, and also to relate to your starting point of
$42-36=6$, and the equations
$x+y=42$ and
$x-y=36$ (where $x$ is the number of small dogs and $y$ is the number of large dogs). You could obtain a new equation by subtracting the left hand side (LHS) of the second equation from the LHS of the first equations, and also subtracting the RHS (right hand sides) in a similar way, and making a new equation like this:
$x+y-(x-y)=42-36$. Then we have
$x+y-x+y=2y=42-36=6$. So $2y=6$ and $y=3$ the number of large dogs. Finally
$x=y+36=3+36=39$, the number of small dogs.

Note that if we added the original two equations we would have found the number of small dogs first (rather than the number of large dogs first). That is, start with
$x+y=42$ and
$x-y=36$ and add the corresponding sides of the two equations to obtain a new equation
$x+y+x-y=42+36$, that is $2x=78$ and dividing both sides by $2$ we obtain $x=39,$ the number of small dogs.

Alternatively, as I wrote in a comment, you could start with a group of $36$ small dogs and no large dogs (i.e. a group of no large dogs) and increase the number in each group (by one dog at a time, in each group) till we get the correct total number of dogs. E.g. next try is $37$ small dogs and $1$ large, total is $38$, too small. Next $38$ small, $2$ large, total $40$, not enough. Finally $39$ small dogs and $3$ large, total is $42$, just right. Notice also that instead of adding one dog at a time in each group, we could have figured how many dogs to add in each group (adding the same number in each group) to get the correct total. So, start with $36$ small dogs and no large dogs. We need to add $6$ dogs (that is $42-36=6$ dogs) to get a total of $42$ dogs. Half of these $6$ extra dogs must be small and the other half large, so we need to add $3$ small dogs and $3$ large dogs. So in the end we have $36+3=39$ small dogs, and $3$ large dogs. 🐶 🐶 🐶

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.