fractions

[ricnik]

need help

There were 20 dogs at the animal shelter. One half of the dogs were retrievers. One tenth of the dogs were pointers. One forth of the dogs were setters. The rest of the dogs were other breeds.
1. How many dogs were retrievers? ------
2. How many dogs were pointers? -----
3. How many dogs were setters? ------
4. How many dogs were other breeds? ----

Thanks so much for helping.

 

[winw]

Hi,

fist write out what you have

There were 20 dogs at the animal shelter. One half of the dogs were retrievers. One tenth of the dogs were pointers. One forth of the dogs were setters. The rest of the dogs were other breeds.
1. How many dogs were retrievers? ------
2. How many dogs were pointers? -----
3. How many dogs were setters? ------
4. How many dogs were other breeds? ----

Let r represent the number of retrievers
let p represent the number of pointers
let s represent the number of setters
let 0 represent the number of other

there are 20 dogs in total, if one half of them are retrievers then that means (1/2)(20)=r

and you continue like that for the rest of the questions. All the dogs have to be accounted for so you will have r+p+s-120=o

hope that helps!

 

[soroban]

Hello, ricnik!

This is not an algebra problem!

There were 20 dogs at the animal shelter.

One half of the dogs were retrievers. . . \(\displaystyle \frac{1}{2} \times 20 \:=\:10\text{ retrievers}\)

One tenth of the dogs were pointers. . . \(\displaystyle \frac{1}{10} \times 20 \:=\:2\text{ pointers}\)

One fourth of the dogs were setters. . . \(\displaystyle \frac{1}{4} \times 20 \:=\:5\text{ setters}\)

The rest of the dogs were other breeds. . . \(\displaystyle 20 - 10 - 2 - 5 \:=\:3\text{ other breeds}\)

Hard hard was that?