Fraction help: "Mrs. Carlyle bought a bag of peanuts for her children ,Philip, Joy, Brent, & Preston took peanuts from the bag"

[summerlearnermom]

Mrs. Carlyle bought a bag of peanuts for her children ,Philip, Joy, Brent, & Preston took peanuts from the bag
Philip took 1/3 of the peanuts from the bag
Joy took 1/4 of the REMAINING peanuts from the bag
Brent took 1/2 of the REMAINING Peanuts from the bag
Preston took 10 peanuts
There were 71 remaining peanuts in the bag.
How many were there initially in the bag??
How many peanuts did each child take??

 

[lev888]

Where is the difficulty? Please read posting guidelines.
I would suggest setting up an equation or finding the unknowns in reverse - if 71 remains and Preston took 10, then - ? etc.

 

[summerlearnermom]

Where is the difficulty? Please read posting guidelines.
I would suggest setting up an equation or finding the unknowns in reverse - if 71 remains and Preston took 10, then - ? etc.

Trying to get this figured out. I am not good at this an need to explain it to my son (12 yrs old.) Summer Learning.

 

[MarkFL]

Hello, and welcome to FMH!

We need to define some variables:

\(N\) = the number of nuts initially in the bag.

\(P\) = the number Philip took

\(J\) = the number Joy took

\(B\) = the number Brent took

\(R\) = the number Preston took

Now, let's look at statements in the problem, that we can turn into equations:

"Philip took 1/3 of the peanuts from the bag"

[MATH]P=\frac{1}{3}N[/MATH]
At this point, the number of nuts left is [MATH]\frac{2}{3}N[/MATH] and so:

"Joy took 1/4 of the REMAINING peanuts from the bag"

[MATH]J=\frac{1}{4}\cdot\frac{2}{3}N=\frac{1}{6}N[/MATH]
At this point, the number of nuts left is [MATH]\frac{1}{2}N[/MATH] and so:

"Brent took 1/2 of the REMAINING Peanuts from the bag"

[MATH]B=\frac{1}{2}\cdot\frac{1}{2}N=\frac{1}{4}N[/MATH]
"Preston took 10 peanuts"

[MATH]R=10[/MATH]
Now, we are told "There were 71 remaining peanuts in the bag." And so we may write:

[MATH]N-\left(P+J+B+R\right)=71[/MATH]
Substitute for all the values representing the amount each child took, which will leave only \(N\) as a variable, and we can then solve the equation to find \(N\). Can you proceed?

 

[Otis]

Hi mom! When you have time, please read the forum's submission guidelines. Basically, the tutors here would like to see what you've tried, so we can see where you're stuck.

In this exercise, we can form an equation to solve, and we can form that equation step-by-step (as each child takes some peanuts). I'll get you started.

The first step in a word problem like this is to assign a symbol to represent the unknown number we're trying to find:

Let x = the beginning number of peanuts.

In other words, we start with x peanuts, and then we start subtracting fractional amounts of x, according to the given steps. Philip took 1/3 of the beginning peanuts. That's subtraction.

x - (1/3)(x)

So, now there are x - x/3 peanuts remaining.

Next step: Joy took 1/4 of those remaining peanuts. She took 1/4 of (x - x/3).

Now there are x - x/3 - (1/4)(x - x/3) peanuts remaining. That expression simplifies to x/2, so there are x/2 peanuts remaining after Joy took hers.

Can you continue like this? The last subtraction will be -10, and then we set the final expression equal to 71 and solve for x.

Please show us how far you get. Cheers

?

 

[summerlearnermom]

Hello, and welcome to FMH!

We need to define some variables:

\(N\) = the number of nuts initially in the bag.

\(P\) = the number Philip took

\(J\) = the number Joy took

\(B\) = the number Brent took

\(R\) = the number Preston took

Now, let's look at statements in the problem, that we can turn into equations:

"Philip took 1/3 of the peanuts from the bag"

[MATH]P=\frac{1}{3}N[/MATH]
At this point, the number of nuts left is [MATH]\frac{2}{3}N[/MATH] and so:

"Joy took 1/4 of the REMAINING peanuts from the bag"

[MATH]J=\frac{1}{4}\cdot\frac{2}{3}N=\frac{1}{6}N[/MATH]
At this point, the number of nuts left is [MATH]\frac{1}{2}N[/MATH] and so:

"Brent took 1/2 of the REMAINING Peanuts from the bag"

[MATH]B=\frac{1}{2}\cdot\frac{1}{2}N=\frac{1}{4}N[/MATH]
"Preston took 10 peanuts"

[MATH]R=10[/MATH]
Now, we are told "There were 71 remaining peanuts in the bag." And so we may write:

[MATH]N-\left(P+J+B+R\right)=71[/MATH]
Substitute for all the values representing the amount each child took, which will leave only \(N\) as a variable, and we can then solve the equation to find \(N\). Can you proceed?

Thank you but I am not good at this :-(

 

[Dr.Peterson]

Mrs. Carlyle bought a bag of peanuts for her children ,Philip, Joy, Brent, & Preston took peanuts from the bag
Philip took 1/3 of the peanuts from the bag
Joy took 1/4 of the REMAINING peanuts from the bag
Brent took 1/2 of the REMAINING Peanuts from the bag
Preston took 10 peanuts
There were 71 remaining peanuts in the bag.
How many were there initially in the bag??
How many peanuts did each child take??

You've seen two suggested ways to use algebra -- one using lots of variables, which may be one to hold on to for later, and one using one variable, which is probably easier. Let's look at the other suggested way: working backward, which is a nice introduction to some of the ideas of algebra.

At the end, there are 71 peanuts.

How many were there before Preston took 10? That would be 71 + 10 = 81, right? (We're adding 10, to undo his subtraction of 10. Check: 81 - 10 = 71.)

How many were there before Brent took half of what was there? Well, if he took half, then he left half. So the 81 he left were half of what was there before. Undoing the halving, there must have been 2 times the 81: 162. (Check: 162 - 1/2(162) = 162 - 81 = 81.)

We have two more steps; these are just a little harder than what I've done for you, but I've given you the necessary ideas. If this looks like a way you'd be expected to solve the problem (that depends on whether it said to use algebra), give it a try, and show us your results at each step.

If you think algebra is needed, try working through Otis's suggestion, post #5, and ask any questions you have about it. In order to help you most effectively, we need to know where you and your son each have difficulty, and how much algebra you each understand. (If I were helping face to face, I'd be looking through his book or notes to see how to tie the problem back to what he has seen.)