Please Can You Teach Me How To Answer This
- Thread starter JoeG
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pka
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As a beginning algebra learner you need to learn to use variables.
So let \(\displaystyle N\) be the number of apples Ned has, \(\displaystyle P\) be the number of apples Peter has and \(\displaystyle M\) be the number of apples Morris has.
Now the given tells us that \(\displaystyle N+M+P=~?^{***}\). Can you answer that?
Also we are told that \(\displaystyle N=\frac{7}{3}P\). Why is that? You need to explain!
Moreover \(\displaystyle M=\frac{1}{2}N\) WHY is that?
Substitute in *** & solve.
Thank you very much. But I’m afraid I am still stuck. Maybe if you could show me how you did it i can see how it’s done and work a process backwards for myself?
My initial reaction was to use decimals. So P * 2.33 = N
And N / 2 = M
But can’t work out how to solve it from there
My initial reaction was to use decimals. So P * 2.33 = N
And N / 2 = M
But can’t work out how to solve it from there
pka
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Decimals are not precise. \(\displaystyle 2.33\ne\frac{7}{3}\)
If you want to learn to do mathematics then put the calculator into a drawer and let it stay there.
\(\displaystyle \frac{7}{3}P+\frac{7}{6}P+P=~?\)
It is an odd problem because qqq is not standard algebraic notation.
But pka is absolutely right. Before trying to do anything much, you need to assign in writing a letter to represent each numeric quantity that is unknown. In math speak, this is naming the variables.
[MATH]p = \text {number of Peter's apples.}[/MATH]
[MATH]n = \text {number of Ned's apples.}[/MATH]
[MATH]m = \text {number of Morris's apples.}[/MATH]
[MATH]t = \text {total number of apples.}[/MATH]
Then you write down in mathematical form the mathematical facts that you have.
[MATH]n = \dfrac{7}{3} * p \implies 3n = 7p.[/MATH]
[MATH]m = \dfrac{1}{2} * n \implies 2m = n.[/MATH]
[MATH]m + n + p = t.[/MATH]
In purely mathematical terms, what is shown above does not have a unique solution because there are four mysteries (the variables) and only three clues (the mathematical facts).
But in this find of puzzle, there are implied clues. One such clue is that m, n, and p are whole numbers rather than fractions. (To be a true math problem rather than a puzzle, you should not have to make a guess, but it is the kind of common sense sort of clue often needed in puzzles and applied math.) Another clue is that q is a non-zero digit from 1 through 9 and qqq means
[MATH]t = 100q + 10q + q.[/MATH]
These extra clues should let you solve the puzzle (if my guesses are right). It is sort of a combination logic and math puzzle.
But pka is absolutely right. Before trying to do anything much, you need to assign in writing a letter to represent each numeric quantity that is unknown. In math speak, this is naming the variables.
[MATH]p = \text {number of Peter's apples.}[/MATH]
[MATH]n = \text {number of Ned's apples.}[/MATH]
[MATH]m = \text {number of Morris's apples.}[/MATH]
[MATH]t = \text {total number of apples.}[/MATH]
Then you write down in mathematical form the mathematical facts that you have.
[MATH]n = \dfrac{7}{3} * p \implies 3n = 7p.[/MATH]
[MATH]m = \dfrac{1}{2} * n \implies 2m = n.[/MATH]
[MATH]m + n + p = t.[/MATH]
In purely mathematical terms, what is shown above does not have a unique solution because there are four mysteries (the variables) and only three clues (the mathematical facts).
But in this find of puzzle, there are implied clues. One such clue is that m, n, and p are whole numbers rather than fractions. (To be a true math problem rather than a puzzle, you should not have to make a guess, but it is the kind of common sense sort of clue often needed in puzzles and applied math.) Another clue is that q is a non-zero digit from 1 through 9 and qqq means
[MATH]t = 100q + 10q + q.[/MATH]
These extra clues should let you solve the puzzle (if my guesses are right). It is sort of a combination logic and math puzzle.
Last edited:
Oh. Blame my eyesight. Then we have enough clues and don't need any guesses.
[MATH]n = \dfrac{7}{3} * p.[/MATH]
[MATH]m = \dfrac{1}{2} * n.[/MATH]
[MATH]m + n + p = t.[/MATH]
[MATH]t = 999.[/MATH]
Four variables, four facts. A general way to solve such problems is called substitution.
We can substitute from the fourth equation into the third giving us
[MATH]m + n + p = 999.[/MATH]
We can substitute from the first equation into the one immediately above.
[MATH]m + n + \dfrac{7}{3} * n = 999 \implies m + \dfrac{3}{3} * n + \dfrac{7}{3} * n = 999 = m + \dfrac{10}{3} * n = 999.[/MATH]
And we can substitute from the second equation into the one immediately above.
[MATH]\dfrac{1}{2} * n + \dfrac{10}{3} * n = 999.[/MATH]
Let's eliminate the fractions by multiplying both sides of the equation by 6.
[MATH]3n + 20n = 5994 \implies 23n = 5994.[/MATH]
Now divide both sides of that equation by 23 to get
[MATH]n = 260 + \dfrac{14}{23}.[/MATH]
[MATH]\therefore m = \dfrac{1}{2} * n = 130 + \dfrac{7}{23}.[/MATH]
[MATH]\text {And }p = \dfrac{7}{3} * n = \dfrac{7}{3} * (260 - 2) + \dfrac{7}{3} * 2 + \dfrac{7}{3} * \dfrac{14}{23} =[/MATH]
[MATH]7 * 86 + \dfrac{7}{3} * \dfrac{46}{23} + \dfrac{7}{3} * \dfrac{14}{3} = 602 + \dfrac{7}{3} * {60}{23} =[/MATH]
[MATH]602 + 7 * \dfrac{20}{ 23} = 602 + \dfrac{140}{23} = 602 + 6 + \dfrac{2}{23} = 608 + \dfrac{2}{23}.[/MATH]
That is a very silly answer in practical terms so let's check.
[MATH]130 + \dfrac{7}{23} + 260 + \dfrac{14}{23} + 608 + \dfrac{2}{23} = 998 + \dfrac{7 + 14 + 2}{23} = 999 \ \checkmark.[/MATH]
Name your variables. Write down your numerical facts in mathematical terms. Use substitution and simplify where you can. Check your work. That is algebra.
Dr.Peterson
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Don't blame your eyesight; it's an error in printing (or else a really bad font).
But you made an error in your work, replacing p with 7/3 n rather than n with 7/3 p. Doing it correctly, the answers come out to integers.
I used pka's equation from post #4.