Word problem leading to simple equation

[NotaMathPerson]

Hello! I need on these problems

My attempt in prob 38

$x=$ number of candidates failed
$4x=$ number of candidates passed
$5x=$ total number of candidates

From the given scenerio we add 14 more cadidates
$5x+14$

Six less failed
$x-6=$ new number of candidtes failed
$4x+6=$ new number of cadidates passed

Now

$5x+14-(x-6)=$ new number candidates passed

Then
$5x+14-(x-6)=5(x-6)$
Solving for x

$x=50$

Therefore $5x=5*50=250$ candidates.

But when I did the reverese

$5x+14-(4x+6)=$ new number of candidates failed

Then
$4x+6=5(5x+14-(4x+6))$
$4x+6=5(x+8)$

Solving for x gives a different answer.

The first answer agrees with the answer from the book.

Can you tell why I get different answers from the different methods I presented? Thanks much!

 

[Deveno]

Yes, I can tell you that.

When you write:

$5x + 14 - (x - 6) = 5(x - 6)$

To get:

$4x + 20 = 5x - 30$

and thus:

$50 = x$

you are perfectly correct.

However, if there are $14$ additional candidates, and $6$ fewer failed, then that means that not only do we add that $6$ to the number of candidates that pass, but we must also add the $14$ additional candidates, so that the new number of candidates that passed is:

$4x + 6 + 14 = 4x + 20$.

So now you should have had, in your second method:

$5x + 14 - (4x + 20)$ = new number of failed candidates

$4x + 20 = 5 (5x + 14 - (4x + 20))$
$4x + 20 = 25x + 70 - 20x - 100$
$4x + 20 = 5x - 30$
$4x + 50 = 5x$
$50 = x$

which agrees with method 1.

 

[NotaMathPerson]

Yes, I can tell you that.

When you write:

$5x + 14 - (x - 6) = 5(x - 6)$

To get:

$4x + 20 = 5x - 30$

and thus:

$50 = x$

you are perfectly correct.

However, if there are $14$ additional candidates, and $6$ fewer failed, then that means that not only do we add that $6$ to the number of candidates that pass, but we must also add the $14$ additional candidates, so that the new number of candidates that passed is:

$4x + 6 + 14 = 4x + 20$.

So now you should have had, in your second method:

$5x + 14 - (4x + 20)$ = new number of failed candidates

$4x + 20 = 5 (5x + 14 - (4x + 20))$
$4x + 20 = 25x + 70 - 20x - 100$
$4x + 20 = 5x - 30$
$4x + 50 = 5x$
$50 = x$

which agrees with method 1.

Does it mean the 14 new candidates passed?. The problem did not say that they passed. Why is that? I am still confused.

 

[NotaMathPerson]

Hello everyone! This is my solution for prob 34. Kindly check if I got it right.

Let $x=$ B's rate for the first part of the walk
$2x=$ B's rate when A has traveled 15 miles.

Now, A has a rate of $5$mph for the entire walk. Assuming they started walking at the same time, it woukd take A $3$ hrs to travel $15$ miles.
That time B has traveled $3x$ miles.
So after $6$ hrs, A has traveled $30$ more miles while B has travled $12x$ miles. The total distance traveled by A and B is $45$ miles

So $12x+3x=45$ the sum of distances traveled by B in first $3$ hrs and the next $6$ hrs where he has doubled his rate

Solving for $x=3$ mph.

Am I correct? And if possible can you show me another way of solving it? Thanks!

 

[MarkFL]

Hello everyone! This is my solution for prob 34. Kindly check if I got it right.

Let $x=$ B's rate for the first part of the walk
$2x=$ B's rate when A has traveled 15 miles.

Now, A has a rate of $5$mph for the entire walk. Assuming they started walking at the same time, it woukd take A $3$ hrs to travel $15$ miles.
That time B has traveled $3x$ miles.
So after $6$ hrs, A has traveled $30$ more miles while B has travled $12x$ miles. The total distance traveled by A and B is $45$ miles

So $12x+3x=45$ the sum of distances traveled by B in first $3$ hrs and the next $6$ hrs where he has doubled his rate

Solving for $x=3$ mph.

Am I correct? And if possible can you show me another way of solving it? Thanks!

I would solve the problem in the same way...the distance walked by both is 45 miles...and B walks for 3 hours with velocity $v$ (in mph) and 6 hours with velocity $2v$, hence:

$$45=3v+6(2v)=15v\implies v=3$$

 

[NotaMathPerson]

I would solve the problem in the same way...the distance walked by both is 45 miles...and B walks for 3 hours with velocity $v$ (in mph) and 6 hours with velocity $2v$, hence:

$$45=3v+6(2v)=15v\implies v=3$$

Hello MarkFl!

Could you take a look at prob 38. Because I don't understand why the 14 additional candidates is added to the number of candidates who passed successfuly. Because the problem did not state that these 14 additional passed. Thank you!

 

[MarkFL]

Hello MarkFl!

Could you take a look at prob 38. Because I don't understand why the 14 additional candidates is added to the number of candidates who passed successfuly. Because the problem did not state that these 14 additional passed. Thank you!

Think about it...if 14 more candidates are added, and at the same time 6 fewer fail, then it can only be that 20 more passed than initially. :)

 

[Deveno]

Does it mean the 14 new candidates passed?. The problem did not say that they passed. Why is that? I am still confused.

We don't know that all 14 "extra" candidates passed-but what we do know is that 6 fewer failed OVERALL.

We have 6 fewer failures, and 14 more people. Suppose our original number of candidates is $n$. Of these, we have $x$ successes, and $y$ failures, so:

$x + y = n$.

Now we add 14 people, so we have $n + 14$, and the new number of failures is $y - 6$. What is the new number of successes?

$? + (y - 6) = n + 14$
$? + y - 6 = x + y + 14$
$? - 6 = x + 14$
$? - 6 + 6 = x + 14 + 6$
$? + 0 = x + 20$
$? = x + 20$

So, 20 more successes than before. Note that whatever $n$ is/was, does not affect our determination of the number of new successes.