SAT math question #2

[BehindEnemyLines]

I'VE MADE A MISTAKE AGAIN, TOO MUCH READING LATELY!!! CORRECTIONS ARE MADE TO #3. ALL ANSWERS ARE FROM THE BOOK AND HAD BEEN CHECKED.

Hope AT'ers here can help again.

1. (Comparison Question - I had to change the symbols. It should not change anything)

For all positive numbers n and k, let n @ k be defined by n @ k = (n - k)k.

0 < r < s

Column A: r @ s
Column B: s @ r

A. if the quanity in Column A is greater;
B. if the quanity in Column B is greater;
C. if the two quantities are equal;
D. if the relationship cannot be determined from the information given.

2. The first term of a sequence of numbers is 2. If each term after the first is 3 more than
twice the preceding term, what is the fourth term of this sequence? (Student-Response)

CORRECTED::
3. For a concert, tickets that were purchased in advance of the day of the of the conert cost
$5.00 each, and tickets purchased the day of the concert cost $8.00 each. The total amount collected in ticket sales was the same as if every ticket purchased had cost $5.50. If 100 tickets were purchased in advance, what was the total number of tickets purchased? (Student-Response)

4. Chairs ready for shipment at the Northern Chair factory come down a ramp in single file.
Inspector A checks every third chair, beginning with the third. Inspector B checks every fifth
chair, beginning with the fifth. If 98 chairs came down the ramp while both inspectors were
working on Monday, how many of these chairs were NOT checked by either of these two inspectors?
(Student-Response)

Answers:
1. B
2. 37
3. 120
4. 53

Explanations how you arrived at the answers will really help.
Thanks.

 

[Noriaki]

n @ k = (n-k)k = nk - k^2

assuming no negatives:
if n > k then n @ k > 0
if n = k then n @ k = 0
if n < k then n @ k < 0

So then with 0 < r < s, since it's strictly less than and not less than or equal, we can ignore the middle line above.
Thus:
r @ s < 0 (this is Column A)
s @ r > 0 (this is Column B)

Edit: Sorry I mixed up my columns.

A < 0 < B

Column B is always bigger.

2.

2, 7, 17, 37

 

[MSantiago]

#2?

2. The first term of a sequence of numbers is 2. If each term after the first is 3 more than
twice the preceding term, what is the fourth term of this sequence? (Student-Response)

1st in sequence: 2 (given)
2nd in sequence: 2 * 2 + 3 = 7
3rd in sequence: 7 * 2 + 3 = 17
4th in sequence: 17 * 2 + 3 = 37

So, the sequence looks like: 2, 7, 17, 37

 

[MSantiago]

Originally posted by: Noriaki
Column A is always bigger.

Actually, column B is always bigger.

(r - s)s < (s-r)r

You can verify it by plugging in sample numbers. Since 0 < r < s, let's use r = 1 and s = 2

A: (r - s)s = (1 - 2)2 = -2
B: (s - r)r = (2 - 1)1 = 1

B > A

 

[Dudd]

1) Arbitrarily assign real values to r and s. It wouldn't be valid as a proof, but for the SAT it works fine. We will use 1 and 2. r@s= 1@2= (1-2)2=-2 s@r=2@1=(2-1)1=1. 1 is greater than -2, B is the answer.

2) Just complete the sequence. 2, (2x2)+3=7, (7x2)+3=17, (17x2)+3=37

3) Not a clue.

4) I'd just count it out.

Ins #1: 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96
Ins #2: 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95

Ins #2 gets 6 out of every 30. Ins #1 gets 10 out of thirty. Two overlap, meaning 14 out of thirty get checked. 16 do not. 16 x3=48. From 91-98, 3 out of 8 are checked, making the grand total 53.

There are probably are quicker ways to do this, but it's 2:40 in the morning and I'm tired.

 

[Noriaki]

Originally posted by: MSantiago

Originally posted by: Noriaki
Column A is always bigger.

Actually, column B is always bigger.

(r - s)s < (s-r)r

You can verify it by plugging in sample numbers. Since 0 < r < s, let's use r = 1 and s = 2

A: (r - s)s = (1 - 2)2 = -2
B: (s - r)r = (2 - 1)1 = 1

B > A

Uh yeah...my last sentence was wrong, but the work wasn't.

r @ s < 0
s @ r > 0

r @ s is smaller than s @ r, thus column B is bigger. I mixed up my columns.

 

[Noriaki]

Originally posted by: BehindEnemyLines3. For a concert, tickets that were purchased in advance of the day of the of the conert cost
$8.00 each. The total amount collected in ticket sales was the same as if every ticket purchased
had cost $5.50. If 100 tickets were purchased in advance, what was the total number of tickets
purchased? (Student-Response)

Is there some missing numbers there?

$8.00 *100 + $x * y = $5.50 * n

$x = the amount of a non-advance ticket
y = the number of non-advance tickets

n = the number of total tickets

We have 3 unkowns and 1 equation....doesn't work...maybe I'm just asleep because it's late and school's not in session so my brain is off...but I don't see how this works...

 

[Noriaki]

Originally posted by: Dudd
4) I'd just count it out.

Ins #1: 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96
Ins #2: 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95

Ins #2 gets 6 out of every 30. Ins #1 gets 10 out of thirty. Two overlap, meaning 14 out of thirty get checked. 16 do not. 16 x3=48. From 91-98, 3 out of 8 are checked, making the grand total 53.

There are probably are quicker ways to do this, but it's 2:40 in the morning and I'm tired.

That works fine.

If you want
A counts every 3, so he counts the floor of 98/3 (floor just means round down), or 32 chairs
B counts every 5, so he counts the floor of 98/5, or 19 chairs

They BOTH count every 15th chair, the floor of 98/15, or 6, so you have to subtract this amount once.

32 + 19 - 6 = 45
45 total chairs ARE counted.

Since it's asking how many are skipped, you need 98 - 45 = 53.

I find it questionable whether this method is actually easier than Dudd's.
It's faster if you know you can remember it...but his is alot easier and less likely to result in a mistake, if you remember part but forget something minor.

 

[Dudd]

Originally posted by: Noriaki

Originally posted by: BehindEnemyLines3. For a concert, tickets that were purchased in advance of the day of the of the conert cost
$8.00 each. The total amount collected in ticket sales was the same as if every ticket purchased
had cost $5.50. If 100 tickets were purchased in advance, what was the total number of tickets
purchased? (Student-Response)

Is there some missing numbers there?

$8.00 *100 + $x * y = $5.50 * n

$x = the amount of a non-advance ticket
y = the number of non-advance tickets

n = the number of total tickets

We have 3 unkowns and 1 equation....doesn't work...maybe I'm just asleep because it's late and school's not in session so my brain is off...but I don't see how this works...

There has to be something missing there. It looks to be impossible without knowing more information.

 

[Noriaki]

Originally posted by: Dudd
There has to be something missing there. It looks to be impossible without knowing more information.

Glad I'm not the only one that thinks so....I was afraid my brain had stopped...

 

[matsuhisa]

3. For a concert, tickets that were purchased in advance of the day of the of the conert cost
$8.00 each. The total amount collected in ticket sales was the same as if every ticket purchased
had cost $5.50. If 100 tickets were purchased in advance, what was the total number of tickets
purchased? (Student-Response)

from the answer you gave us... this question is not possible:
120*$5.50=$660.
however, it says that there were 100 tickets sold at $8.00, which is 100*$8.00=$800. so how can there be a ticket sale of no less than $880 but the answer is $660. you prolly wrote down the problem wrong and didn't leave us enough information to work with.

 

[Noriaki]

Originally posted by: matsuhisa

3. For a concert, tickets that were purchased in advance of the day of the of the conert cost
$8.00 each. The total amount collected in ticket sales was the same as if every ticket purchased
had cost $5.50. If 100 tickets were purchased in advance, what was the total number of tickets
purchased? (Student-Response)

from the answer you gave us... this question is not possible:
120*$5.50=$660.
however, it says that there were 100 tickets sold at $8.00, which is 100*$8.00=$800. so how can there be a ticket sale of no less than $880 but the answer is $660. you prolly wrote down the problem wrong and didn't leave us enough information to work with.

Oh...I missed the 120...I thought those were possible answers for #4...

yeah....I think BehindEnemyLines need to look at that problem again...unless they were giving out money with non-advance ticket "sales" that doesn't work. (Edit..sorry needed to clarify who I was talking to in this sentence).

ie
$8.00*100 + $x*y = $5.50*120

$x*y must be negative to satisfy this...since y is a quantity, y being negative makes no sense, so x must be negative. Seems a little weird....

Actually I can update this equation. Since n is the total number of ticket, and 100 is a portion of those ticket sales, so y = n-100:
$8.00*100 + $x*(n-100) = $5.50*n
or
$8.00*100 + $x*(20) = $5.50*120

n-100 shouldn't be the negative part, like I said for y above..


If the box office gave out $7 with each purcase:
$8.00*100 + (-$7)*20 = $5.50*120
$800 - $140 = $660

Then it works...
But we aren't trying to solve for $x given n, we are trying to solve for n...
Without knowing n, we have 2 unknowns and 1 equation. Still doens't work.
We need $x to solve for it...and having $x be negative seems insane....but it would have to be for this to work (-$7 to be exact).

Are you sure there is no typo in that question?

 

[BehindEnemyLines]

Sorry again, I fixed the question for #3...I left out some info. The answers are correct thou.

I still have questions on the following. How did you find the highlighted part without using Dudd's listing method? I find Dudd's method is time consuming but accuracy is better. Is there a better time/accuracy ratio method?

Quote: Noriaki
--------------------------------------------------------------------------------
Originally posted by: Dudd
4) I'd just count it out.

Ins #1: 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96
Ins #2: 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95

Ins #2 gets 6 out of every 30. Ins #1 gets 10 out of thirty. Two overlap, meaning 14 out of thirty get checked. 16 do not. 16 x3=48. From 91-98, 3 out of 8 are checked, making the grand total 53.

There are probably are quicker ways to do this, but it's 2:40 in the morning and I'm tired.
--------------------------------------------------------------------------------

That works fine.

If you want
A counts every 3, so he counts the floor of 98/3 (floor just means round down), or 32 chairs
B counts every 5, so he counts the floor of 98/5, or 19 chairs

They BOTH count every 15th chair, the floor of 98/15, or 6, so you have to subtract this amount once.

32 + 19 - 6 = 45
45 total chairs ARE counted.

Since it's asking how many are skipped, you need 98 - 45 = 53.

I find it questionable whether this method is actually easier than Dudd's.
It's faster if you know you can remember it...but his is alot easier and less likely to result in a mistake, if you remember part but forget something minor.

 

[dafatha00]

$5.00y + $8.00x = $5.50 (y+x)

y=100 tickets sold in advance

$500 + $8.00x = $5.50 (100+x)

$500 + $8.00x = $550 + 5.5x

2.5x = $50

x= 20 tickets sold on day of concert

y+x = total number of tickets sold

y+x = 120 total tickets sold

 

[soccerbud34]

Here is the method that i used to approach # 3:

-The most important aspect of this problem is to find the ratio of tickets purchased in advanced to tickets purchased on the day of concert

- will use 'x"as the unkown variable. It should show the ratio of tickets sold in advanced to tickets sold on day of concert

**calculation for the ratio**

(5x + 8.00)/(x + 1) = 5.50.
- it says to that the average price of both kind tickets sold was $5.50; therefore, we used this equation to see how many advanced tickets must be sold, plus 1 ticket sold on the day of concert will equal out to $5.50.

- if we solve for x, x should equal to 5. Therefore, the ratio of advanced ticket to ticket sold on the day of concert is 5:1, which means that for every 5 advanced ticket sold, 1 ticket is sold on the day of concert.

- Therefore, according to logic, if 100 tickets were sold in advance, 20 tickets would be sold on the day of concert or 100 / 5 = 20.

- Thus, the total amount of tickets sold was 100 + 20 = 120

**edit: dafa beat me to it **

 

[rgwalt]

In order to figure out that the inspectors are both inspecting every 15th chair you have to think about sequences in the problem.

Ryan

 

[soccerbud34]

I still have questions on the following. How did you find the highlighted part without using Dudd's listing method? I find Dudd's method is time consuming but accuracy is better. Is there a better time/accuracy ratio method?

Will, first you have to realize that there is the possiblity of counting the same chair twice.

Thus, as you investigate further, you will realize that there are multiples of 3's and multiples of 5 which are of the same figures
i.e. 15, 30, 45, 60, 75, 90 (forgot what the exact technical term is). And if you number each chair from 1 to 98, you will realize that the chairs numbered 15, 30, 45, 60, 75, 90, which is a grand total of 6, all have been counted twice.

hope that clears it up a bit for you

 

[Skyclad1uhm1]

Or to write it down so even a republican will understand: ()

5.5 - 5 = 0.5 <- You will need 0.5 extra per ticket to compensate for these costing less.

Other tickets:
8 - 5.5 = 2.5 <- the amount you have extra for each of the other tickets.

2.5 / 0.5 = 5 <- Each of those tickets can compensate for 5 of the cheap ones.

So you need:
100 / 5 = 20 extra tickets. 100 + 20 = 120 total.

 

[Noriaki]

Originally posted by: BehindEnemyLinesI still have questions on the following. How did you find the highlighted part without using Dudd's listing method?

15 is the Lowest Common Multiple of 3 and 5.

So each of them counts every 15th chair.

Basically you have a sequence of all multiples of 3, and a sequence of all multiples of 5.
You make a 3rd sequence which is multiples of the LCM (15), and that will give you the set of numbers that appear in both of the first two sequences.

If you just use floor(98/3) and floor(98/5) every 15th chair (again the LCM of 3 and 5) is counted twice. So you need to subtract that amount once, to account for the duplication.

(Just so we're clear, this is counting the chairs that ARE inspected, you will need to subtract this number from the total to get the amount that aren't inspected)



Here's my version of #3 hehehe.

$5*100 + $8*y = $5.50*n

y is the number of tickets sold the day of, n is the total. Since n is the total, and 100 were sold in advance, y = (n-100).

So $5*100 + $8*(n-100) = $5.50*n
$500 + $8n - $800 = $5.5*n
-$300 = -$2.5n
n = $300/$2.5
n = 120

 

[BehindEnemyLines]

Thank you all. I understand them now, although some methods are easier to understand. I may have more questions coming so, hope you all can help again

Thanks.

 

[matsuhisa]

CORRECTED::
3. For a concert, tickets that were purchased in advance of the day of the of the conert cost
$5.00 each, and tickets purchased the day of the concert cost $8.00 each. The total amount collected in ticket sales was the same as if every ticket purchased had cost $5.50. If 100 tickets were purchased in advance, what was the total number of tickets purchased? (Student-Response)

3. 120

i guess being a applied major at ucla was helpful... hehe...

well you have
100*$5.00=$500.00 (amount sold for advanced tickets)

the total amount made from tickets sales for advanced and day off tickets is:
x*$5.50

therefore, to find out the total number of tickets sold is:
total amount made from ticket sales=amount sold for advanced tickets+price of day off tickets*(total number of tickets sold-tickets sold in advanced)
in equation form:
($5.5)*x=(x-100)*$8+$500

so you get
5.5x=8x-800+500
5.5x=8x-300
300=2.5x
120=x

x is the total number of tickets sold, so here is the answer. if you don't understand the explanation... let me know. or maybe someone else can make it clearer.

 

[BehindEnemyLines]

matsuhisa & and all of you,

yes I understand them now. The actual hard part is putting words and numbers into an equation(s) that is workable. I need more practice (know more technique/methods) setting up equations.

thanks again.

 

[Noriaki]

Originally posted by: BehindEnemyLines
matsuhisa & and all of you,

yes I understand them now. The actual hard part is putting words and numbers into an equation(s) that is workable. I need more practice (know more technique/methods) setting up equations.

thanks again.

It just takes practice. Eventually you can just see the equations from reading the words. You just need to practice a bunch, and eventually it'll just come to you.