$36$ pens and $42$ pencils cost $\$460$. How much for $18$ pens and $21$ pencils? Different approaches give different answers.

Question

Cost of $36$ pens and $42$ pencils is $\$460$. What is the cost of $18$ pens and $21$ pencils?

Solution:

Method 1: Unitary method

36 pens and 42 pencils costs $460
36 pens and 1 pencil will cost $460/42
1 pen   and 1 pencil will cost $460/(42*36)
1 pen and 21 pencils will cost ($460/(42*36))*21
18 pens and 21 pencils will cost ($460/(42*36))*21 *18

Answer is $115$.

Method 2: Direct proportion

Cost is ∝ Total number of items
Therefore,
*(Cost/Total number of items)=k(constant)*
 (460/78)=(x/39)
 x=230

Answer is $230$.

Correct answer is given as $230$.

Answer 1

You are doing unitary wrong.

If you have $A + B = C$ and you divide $B$ and $C$ by $42$ you must divide both terms $A, B$ by $42$.

$36a + 42b = 460$ So

$\frac {36a + 42b}{42} = \frac {460}{42}$ is the next step.

$36a + \frac {42}b= \frac {460}{42}$ is completely wrong.

So $\frac {36}{42}a + b = \frac {460}{42}$

So $\frac 67 a+b = \frac {230}{21}$.

Now then $21(\frac 67 a + b) = 21*\frac {230}{21}$ will give us

$\frac {21*6}7a + 21b =230$

$18a + 21b = 230$ and we are done.

But notice we got lucky. We were asked to solve something that was proportional.

If we wear asked to find out how much $20$ pens and $15$ pencils cost we wouldn't get anywhere.

$\frac {36}{42}a + b = \frac {460}{42}$ so

$15(\frac {36}{42}a + b) = 15\frac {460}{42}$ so

$\frac {15*6}{7}a + 15b = \frac {230*15}{21}$

$\frac {90}7a + 15 b = \frac {1150}7$

But that doesn't help us because $\frac {90}7\ne 30$

Or we could do it

$\frac {20}{36}(36a + 42b) = \frac {20}{36}460$

so $20a + \frac{70}3 b = \frac {1840}{9}$

But that doesn't help us because $\frac {70}3 \ne 15$.

We can not solve it unless pens and pencils are bought in the same proportions.

.....

Method 2) makes the unjustified assumptions that pens and pencils cost the same. In this case that didn't matter. Was we are buying the pens and pencils in the same proportions, they could for all we know, cost the same. But in general this wouldn't work.

If pens and pencils both cost $\frac {460}{36+42} = \frac {460}{78} = \frac {230}{39}$ each, then yes buying $39$ pens and pencils will by $230$.

But what if pens cost $10$ and pencils cost $\frac{460 - 36*10}{42} = \frac{100}{42} =\frac {50}{21}$.

Well, we would still have $18$ pens and $21$ pencils cost $18*10 + \frac {50}{21}*21 = 180 + 50 = 230$.

But that's because we bought them in proportion.

If I asked how much $20$ pens and $15$ pencils cost, this method we figure $(20+15)*\frac {230}{39}= \frac {35*230}{39}=206\frac {16}{39}$.

But the real answer would be $20*10 + 15*\frac {50}{21}= 200+\frac{250}7= 235\frac 57$.

Answer 2

This ridiculous question is now alive for many hours, and has not been deleted by the CRUDE militia.

Given such data Occam's razor tells us to make the simplest assumptions compatible with the data. The simplest assumptions are that there is a prize $a$ for a pen and a prize $b$ for a pencil. More complicated tariffs dealing with the prices of various numbers of pens and of pencils should have been announced in the statement of the problem. Given that $$36 a+42 b=460$$ it is then obvious that $$18 a+21 b={1\over2}(36a+42 b)=230\ .$$