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Three bags of apples and two bags of oranges weigh $32$ pounds.

Four bags of apples and three bags of oranges weigh $44$ pounds.

All bags of apples weigh the same. All bags of oranges weigh the same.

What is the weight of two bags of apples and one bag of oranges?

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    $\begingroup$ What grade is your daughter in? As far as we know, her instructor could be expecting her to invert the relevant square matrix. $\endgroup$ Jan 9, 2013 at 11:08
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    $\begingroup$ @peoplepower: It might be different in various countries but usually at point when you invert matrices you tend to not work on apples and oranges... $\endgroup$ Jan 9, 2013 at 14:57
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    $\begingroup$ @MaciejPiechotka We're not just talking about apples and oranges, we are talking about bags of them. $\endgroup$ Jan 12, 2013 at 3:13

6 Answers 6

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$x$: weight of a bag of apples (in pounds)

$y$: weight of a bag of oranges (in pounds)

First we "translate" the givens into algebraic equations:

  • $(1)$ "Three bags of apples and two bags of oranges weigh $32$ pounds." $\implies 3x + 2y = 32$.
  • $(2)$ "Four bags of apples and three bags of oranges weigh $44$ pounds." $\implies 4x + 3y = 44$

This gives us the system of two equations in two unknowns: $$3x + 2y = 32\tag{1}$$ $$4x + 3y = 44\tag{2}$$

Ask your daughter to solve the system of two equations in two unknowns to determine the values of $x$ and $y$.

Hints for your daughter:

  • multiply equation $(1)$ by $3$, and multiply equation $(2)$ by $2$:

$$9x + 6y = 96\tag{1.1}$$ $$8x + 6y = 88\tag{2.1}$$

  • subtract equation $(2.1)$ from equation $(1.1)$, which will give the value of $x$.

  • Solve for $y$ using either equation $(1)$ or $(2)$ and your value for $x$.

  • Then determine what $2x + y$ equals. That will be your (her) solution.

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    $\begingroup$ It is interesting to see that how all the four solutions have used the same variables $x,y$. Whenever, there is a system of linear equation involving two variables, all of us seem to agree that $x$ and $y$ are the best variables to use. Somethings are drilled hard in our head during our early mathematical training. :) $\endgroup$
    – user17762
    Jan 9, 2013 at 3:02
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    $\begingroup$ @Marvis The conventions go stronger than that though - x, y & z are for real random variables, a, b & c are for constants and i, j & k are for intiger random variables $\endgroup$
    – Dale M
    Jan 9, 2013 at 3:17
  • $\begingroup$ AFAIK the i,j,k convention is a holdover from punch-card Fortran, where the 2nd column of values (which started with 'i') were always integers. I have no idea how it leaked back to math academia. $\endgroup$ Jan 9, 2013 at 7:17
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    $\begingroup$ @Marvis, well, you know it is a convention since Middle ages at very least: en.wikipedia.org/wiki/X#Metalinguistic_usage. So you might as well be surprised that answer used Arabic numbers. $\endgroup$ Jan 9, 2013 at 9:10
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    $\begingroup$ @scraimer Variables i and j have been used as indices for hundreds of years, long before FORTRAN. Their use as indices there is either a coincidence or a result of their use in mathematics, not the other way around. $\endgroup$ Jan 9, 2013 at 13:40
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Let one bag of apple weight $x$ pounds and one bag of orange weight $y$ pounds. We then have \begin{align} 3x+2y & = 32\\ 4x+3y & = 44 \end{align} We need the weight of $2$ bags of apples and $1$ bag of orange i.e. we need $2x+y$. Note that $$2x+y = 2(3x+2y) - (4x+3y) = 2 \times 32 - 44 = 20$$

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Let $x$ be the weight of a bag of apples and $y$ the weight of a bag of oranges. We’re told that $3x+2y=32$ and $4x+3y=44$:

$$\left\{\begin{align*} &3x+2y=32\\ &4x+3y=44\;. \end{align*}\right.\tag{1}$$

We want to know what $2x+y$ is.

One way to answer the question is to solve $(1)$ for $x$ and $y$ and substitute into $2x+y$. Multiply the top equation of $(1)$ by $3$ and the bottom by $2$, so as to get equations with the same coefficient on $y$:

$$\left\{\begin{align*} &9x+6y=96\\ &8x+6y=88\;. \end{align*}\right.\tag{2}$$

If you now subtract the bottom equation in $(2)$ from the top you find that $x=8$. Substitute that value of $x$ into any of the equations in $(1)$ or $(2)$ to find $y$; I’ll use the top equation in $(1)$, since it has the smallest coefficients. From it I find that $3\cdot8+2y=32$, $24+2y=32$, $2y=8$, and $y=4$. Thus, $2x+y=2\cdot8+4=20$.

If you happen to notice that $2(3x+2y)-(4x+3y)=2x+y$, you can take advantage of a shortcut (which I see Marvis has already pointed out), but if not, solving the system is guaranteed to work, and fairly mechanically, too.

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Let the weight of $1$ bag of apple be $x$, and the weight of $1$ bag of orange be $y$. Then you have

\begin{equation} 3x + 2y = 32 \end{equation}

and

\begin{equation} 4x + 3y = 44 \end{equation}

Now, solving these equations gives you $x = 8$ and $y = 4$. Thus, the weight of $2$ bags of apple and $1$ bag of orange is $2x+ y = 16 + 4 = 20$.

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Here's another approach that shows how one might be able to intuit the answer for this particular problem without being able to solve general systems of linear equations:

"We know that 3 bags of apples and 2 bags of oranges weigh a total of 32 pounds, and we know that 4 bags of apples and 3 bags of oranges weigh 44 pounds. Then by subtracting the first from the second, we find out that one bag of apples and one bag of oranges weighs 44-32=12 pounds.

But now that we have this, we can take a bag of apples and a bag of oranges away from the 3-and-2-bags pile to see that 2 bags of apples and 1 bag of oranges weighs 32-12=20 pounds."

(This is, of course, roughly equivalent to Marvis's observation, but involves no multiplication whatsoever, just a couple of successive subtractions. And you could go on to note that you can take another bag-of-apples-and-bag-of-oranges away from the 2-and-1 pile to find that a bag of apples weighs 20-12=8 pounds all by itself and then that a bag of oranges weighs 12-8=4 pounds, but that's tangential and of course mostly a function of this special set of parameters.)

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Let one bag of apple weight $x$ pounds and one bag of orange weight $y$ pounds. We then have \begin{align} 3x&+2y=32\qquad\\ 4x&+3y=44\qquad \end{align}

To find the value of $x$ and $y$: \begin{align} (3x+2y=32) \cdot 3\\ (4x+3y=44) \cdot 2 \end{align} yields \begin{align} 9x+6y=96 \qquad(1)\\ 8x+6y=88 \qquad(2) \end{align} Subtracting $(2)$ from $(1)$, we get $x=8$ and $y = 4$. So the weight of $2$ bags of apples and $1$ bag of orange is $$ 2\cdot8 + 4 =16+4 =20\text{ pounds}. $$

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