Buying two different kinds of toys.

My sister in grade 4 came to me for some help on this question, and I was kind of embarrassed that I didn't know how to do it.

"Kimikos mother and father own a toy store. Today Kimiko and her mother are buying some tiny clay animals for the store. Her mother likes the seals that are 22 cents each. Kimiko likes the bears that are 29 cents each. Her mother gets some of both and pays 5.39. How many of each kind of animal did Kimikos mother buy?"

Could someone please help us out here? I'm genuinely curious as to how one would go about solving this now.

(Also I'm sorry if this is the wrong place to ask, I wasn't quite sure where to post this)

re: Buying two different kinds of toys.

This is an example of solving a system of equations of two variables. We let s be the number of seals purchased and b be the number of bears purchased. Then we know that .22s+.29b = 5.39 since you pay .22 per seal, .29 per beat, and 5.39 total.

Normally one would expect some more info such as the total number of items purchased was ____ ..... in this case, we know that $\displaystyle s\geq 0$ and $\displaystyle b\geq 0$ since some of each item were bought. You can play around with different values for s and b now to find s=10 and b=11

(normally you wouldn't have to play around with values because normally they would have told you a total of 21 items were purchased)

Re: Buying two different kinds of toys.

Quote:

Originally Posted by

**MonroeYoder** This is an example of solving a system of equations of two variables. We let s be the number of seals purchased and b be the number of bears purchased. Then we know that .22s+.29b = 5.39 since you pay .22 per seal, .29 per beat, and 5.39 total.

Normally one would expect some more info such as the total number of items purchased was ____ ..... in this case, we know that $\displaystyle s\geq 0$ and $\displaystyle b\geq 0$ since some of each item were bought. You can play around with different values for s and b now to find s=10 and b=11

(normally you wouldn't have to play around with values because normally they would have told you a total of 21 items were purchased)

I just subtracted 29 from 539 until I got 220, which was divisible by 22. Yeah, I didn't like this question in the first place. Thanks!

Re: Buying two different kinds of toys.

Hello, Childhood!

Grade 4?

If she doesn't know Algebra, your method should be acceptable.

Quote:

Kimiko's mother and father own a toy store.

Today Kimiko and her mother are buying some tiny clay animals for the store.

Her mother likes the seals that are 22 cents each.

Kimiko likes the bears that are 29 cents each.

Her mother gets some of both and pays $5.39.

How many of each kind of animal did they buy?

$\displaystyle S$ seals were bought at $\displaystyle 22\cent$ each.

. . Their cost was $\displaystyle 22S$ cents.

$\displaystyle B$ bears were bought at $\displaystyle 29\cent$ each.

. . Their cost was $\displaystyle 29B$ cents.

The total cost was $\displaystyle 539\cent:\;22S + 29B \:=\:539$

Solve for $\displaystyle S:\;\;S \:=\:\dfrac{539-29B}{22} \quad\Rightarrow\quad S\;=\;24 - B + \dfrac{11-7B}{22}$ .[1]

Since $\displaystyle B$ is an integer, $\displaystyle 11-7B$ must be a multiple of 22.

Hence: .$\displaystyle 11-7B \:=\:22m\;\text{ for some integer }m.$

Solve for $\displaystyle B:\;\;B \:=\:\dfrac{11-22m}{7} \quad\Rightarrow\quad B \:=\:1 - 3m + \frac{4-m}{7}$

Since $\displaystyle B$ is an integer, $\displaystyle 4-m$ must be a multiple of 7.

The first time this happens is when $\displaystyle m = \text{-}3.$

. . $\displaystyle B \;=\;1 - 3(\text{-}3) + \dfrac{4 - (\text{-}3)}{7} \quad\Rightarrow\quad \boxed{B \:=\:11}$

Sustitute into [1]: .$\displaystyle S \:=\:24 - 11 + \dfrac{11-7(11)}{22} \quad\Rightarrow\quad \boxed{S \:=\:10}$

Therefore, they bought: .10 seals and 11 bears.