Question

Guys, I need not the answer but how to do this problem, a.k.a the steps. Also, i need to know why. Alright.

Susan begins counting backward from 1298 by 4, saying one number every 5 seconds. At the same time, Jim begins counting forward from 171 by 3, saying 1 number every 45 seconds. What number will they both say at the same time?

Please explain thoroughly as this is one of the question for the hunter college exam and going to hunter is my life dream....

Answer 1

It's not an essay

Answer 2

lol thinking, chillax :D

Answer 3

this is arithmetic sequence

Answer 4

Can anybody help me?

Answer 5

Please?

Answer 6

So like... Susan is counting down from 1298.
She's subtracting 4 from that number every 5 seconds.

So if x represents time, in seconds, then every x/5 she'll subtract 4 from the number.
\[\large\rm 1298-4\left(\frac x5\right)\]

Answer 7

Ok...

Answer 8

So ummm

Answer 9

Wait I know one thing

Answer 10

If we want this problem to make sense, then she shouldn't be subtracting pieces of 4 at say, x=1, 2, 3, 4 and all these values that are not multiples of 5.

So maybe we need to take the floor of this x/5.
Ehhh maybe that just makes things more confusing tho :U

Answer 11

The equation is "supposedly" 1298-4x=171+3x

Answer 12

But that's where I got completely confused

Answer 13

hmm that doesn't look right :O
Especially if Jimbo is adding every 45 seconds.

That wasn't a typo, right?

Answer 14

Um no

Answer 15

Wait I think the time does not matter because they are counting at same intervals

Answer 16

That's just what I think

Answer 17

They are counting at same intervals ... hmm interesting :D sec thinking lol

Answer 18

Ok

Answer 19

Hmm ya that equation does work.
I'm having trouble wrapping my head around the fact that we can ignore the 5 second and 45 second counting markers tho, so weird :U hmm

Answer 20

45?

Answer 21

Oh shoot

Answer 22

Srry that was a typo

Answer 23

45 is 5

Answer 24

LOL you bozo XD dang that's what I was asking about earlier lol

Answer 25

Oops XD

Answer 26

Sorry

Answer 27

So let's assume t is measuring time in seconds.
And we'll let x be multiples of 5: \(\large \rm x=5t\)

Then every new x (every 5 seconds that passes), we'll subtract 4 from Susan's value.
\(\large\rm 1298-4x\)

So for example, if x=2 (which means 2 groups of 5 seconds have passed), then we've subtracted 2 fours from our starting value.
\(\large \rm 1298 - 4(2)\)

So we have our equation for Susan's counting.

What do you think so far, does any of this make sense? :U

Answer 28

Oh maybe Hero can help XD

Answer 29

I think I know what the answer is but I can't say the answer out loud.

Answer 30

Hero?

Answer 31

Uh I know the answer from other sites that I checked but the other sites wouldn't tell me how they got it

Answer 32

The answer is 654

Answer 33

But I dont know how you get it

Answer 34

The same can be done for Jim's counting,
every 5 seconds (every x), we'll add 3.
\(\large\rm 171+3x\)

Answer 35

This is the equation for Jim's counting.

Answer 36

expression*

Answer 37

And we want to know what value of x gives us the same number for each of them. So we'll equate these to one another.
\(\large\rm 1298-4x=171+3x\)

Answer 38

Alright

Answer 39

Then simply solve for x.
To isolate x, we'll get the x-stuff on one side,
and everything else on the other side.

Subtract 171 from each side,
\(\large\rm 1298-171-4x=3x\)
Add 4x to each side,
\(\large\rm 1298-171 = 3x+4x\)

Answer 40

\(\large\rm 1127=7x\)

Answer 41

Divide by 7,
\(\large\rm x=161\)

Answer 42

Wait

Answer 43

Why do you do the first part

Answer 44

Where you apparently isolate x

Answer 45

Susan is counting down.
After x amount of time, she arrives at some value.

Jim is counting up.
After the same x amount of time, he arrives at the same value as Susan.

So we set the expressions equal to one another because they represent the same resulting value that they ended up with.

And x is the time that passed for them to each arrive at this number.

Answer 46

Mmm I dunno, I'm struggling to explain this well :p hmm

Answer 47

Hmmmmmmm.......

Answer 48

We're trying to solve for x, the time that passes in order for these "pieces" to be equal.

The pieces being: 1298-4x and 171+3x

Answer 49

Yeah, ok

Answer 50

Go on

Answer 51

So after we apply those basic Algebra steps,
we find that x=161.

161 of these 5 second intervals must pass in order for Susan's number to shrink down to be equal with Jim's rising number.

Answer 52

From there, you can plug 161 into either counting expression to see which number they ended up at.

Answer 53

Susan: \(\large\rm 1298-4(161)=654\)

Answer 54

Or you could plug the value into Jim's expression:
\(\large\rm 171+3(161)=654\)

Answer 55

Ok..... only part i dont get is the thing where

Answer 56

Subtract 171 and add 4x

Answer 57

Could you please explain but simply?

Answer 58

Because I dont understand...

Answer 59

When you arrive at an equation like this:\[\large\rm 1298-4x=171+3x\]Your goal is to try and get it to look like this:\[\large\rm x=stuff\]
We're trying to figure out this special value of x which makes the equation hold true. That requires us to get x alone, get all of the "other stuff" away from the x.

Answer 60

We have two separate terms containing x, (4x) and (3x) so it's a lil trickier.
We'll try to get both x terms onto one side of the equation
We do so because then we can combine like-terms afterwards.

Answer 61

Why does 1298-171=7x equate to 1298=7x? Shouldn't it be 1127=7x

Answer 62

Correct, it equates to 1127=7x.
I think that's what I wrote above as well.

Answer 63

No but why isn't it1127? Why did the 117 just dissapear

Answer 64

Lemme do the Algebra steps again real quick,
I'll color the steps for clarity.

Answer 65

Ok

Answer 66

\[\large\rm 1298-4x=171+3x\]

Add 4x to each side,\[\large\rm 1298-4x\color{orangered}{+4x}=171+3x\color{orangered}{+4x}\]We do this so the -4x and +4x will cancel out on the left.\[\large\rm 1298\cancel{-4x\color{orangered}{+4x}}=171+3x\color{orangered}{+4x}\]\[\large\rm 1298=171+3x\color{orangered}{+4x}\]

Answer 67

Then we'll subtract 171 from each side,
so that only x-stuff remains on the right side of the equation.

Answer 68

Give me a sec to figure this out..

Answer 69

\[\large\rm 1298\color{royalblue}{-171}=171\color{royalblue}{-171}+3x\color{orangered}{+4x}\]

Answer 70

We're doing this so the 171-171 will disappear from the right side of the equation,\[\large\rm 1298\color{royalblue}{-171}=\cancel{171\color{royalblue}{-171}}+3x\color{orangered}{+4x}\]\[\large\rm 1298\color{royalblue}{-171}=3x\color{orangered}{+4x}\]

Answer 71

Uh but If on the other side of the equation there is 1298-171 then shouldn't the result be 1127=7x?

Answer 72

Yes, our next step would be to apply the subtraction on the left side,\[\large\rm 1127=3x\color{orangered}{+4x}\]And add together the like-terms on the right,\[\large\rm 1127=7x\]

Answer 73

Give me a sec

Answer 74

Notice we don't quite have it the form of x=stuff yet.
There is still some junk attached to our x.

Answer 75

Ohhhhh I see now

Answer 76

Dude

Answer 77

\[\large\rm 1127=7x\]The 7 is multiplying the x,
so to get it off the x, we apply the inverse of multiplying. We divide 7.\[\large\rm \frac{1127}{7}=\frac{7x}{7}\]And apply cancellation,\[\large\rm \frac{1127}{7}=\frac{\cancel7x}{\cancel7}\]

Answer 78

\[\large\rm 161=x\]

Answer 79

One of the weird things to keep in mind, for this problem, ...

Answer 80

?

Answer 81

x represents multiples of 5.
So if they had asked the question a little differently, say,
"How much time must pass for them to get the same value?"

Then we would say 161*5 seconds must pass.

Answer 82

They didn't ask for that though, so no big deal.

Answer 83

Ok, so then we would do one of the two either 171+161×4 or 1298-161×4 right?

Answer 84

To get 654

Answer 85

Correct.
Since we expect them to end at the same value,
either expression should work.

Answer 86

Yayyyy we got it! :D

Answer 87

Thanks man

Answer 88

For spending your time on me

Answer 89

np XD

Answer 90

Ok I just became your fan

Answer 91

yes because to become fans you need to be helped