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A certain college has a student-to-teacher ratio of 11 to 1. [#permalink]
27 Jan 2008, 09:39

00:00

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Difficulty:

5% (low)

Question Stats:

87% (02:10) correct
13% (00:59) wrong based on 88 sessions

A certain college has a student-to-teacher ratio of 11 to 1. The average (arithmetic mean) annual salary for teachers is $26,000. If the college pays a total of $3,380,000 in annual salaries to its teachers, how many students does the college have ?

Hi, this is a pretty easy question, I know. Still, It took me a bit more than 3 min to perform all the calculation.. Are there any tips on how to calculate everything quicker? In general, I will greatly appreciate if anyone can share any ideas on how to do calculations quickly.. Calculations of fractions-decimals or just large numbers take my time. moreover I just tend to make careless mistakes here being pressed on time.......Please, advise..

Re: PS: Please, advise on how to calculate quicker!!! [#permalink]
27 Jan 2008, 18:58

chica wrote:

A certain college has a student-to-teacher ratio of 11 to 1. The average (arithmetic mean) annual salary for teachers is $26,000. If the college pays a total of $3,380,000 in annual salaries to its teachers, how many students does the college have ? (A) 130 (B) 169 (C) 1,300 (D) 1,430 (E) 1,560

Hi, this is a pretty easy question, I know. Still, It took me a bit more than 3 min to perform all the calculation.. Are there any tips on how to calculate everything quicker? In general, I will greatly appreciate if anyone can share any ideas on how to do calculations quickly.. Calculations of fractions-decimals or just large numbers take my time. moreover I just tend to make careless mistakes here being pressed on time.......Please, advise..

Re: PS: Please, advise on how to calculate quicker!!! [#permalink]
12 Sep 2013, 04:47

Could someone show me a purely algebraic approach. I can't connect those elements (two equations). I solve fast via the common sense, but I it looks like I am doing something silly on the proportions side.

I have this ready, after taking out three 0's, \(26t = 3800\) (after 3800/t=26), and then since we have \(11t=s\) it must be that \(\frac{s}{11}=\frac{t}{1}\)? But then, if it is true, I am confused with my common sense...please help me untangle myself. How do you express ratios as stated in the problem? _________________

There are times when I do not mind kudos...I do enjoy giving some for help

Re: PS: Please, advise on how to calculate quicker!!! [#permalink]
12 Sep 2013, 05:19

obs23 wrote:

Could someone show me a purely algebraic approach. I can't connect those elements (two equations). I solve fast via the common sense, but I it looks like I am doing something silly on the proportions side.

I have this ready, after taking out three 0's, \(26t = 3800\) (after 3800/t=26), and then since we have \(11t=s\) it must be that \(\frac{s}{11}=\frac{t}{1}\)? But then, if it is true, I am confused with my common sense...please help me untangle myself. How do you express ratios as stated in the problem?

chica,

This calculation is v easy if you know 13^2 = 169.

Re: PS: Please, advise on how to calculate quicker!!! [#permalink]
12 Sep 2013, 06:33

obs23 wrote:

Could someone show me a purely algebraic approach. I can't connect those elements (two equations). I solve fast via the common sense, but I it looks like I am doing something silly on the proportions side.

I have this ready, after taking out three 0's, \(26t = 3800\) (after 3800/t=26), and then since we have \(11t=s\) it must be that \(\frac{s}{11}=\frac{t}{1}\)? But then, if it is true, I am confused with my common sense...please help me untangle myself. How do you express ratios as stated in the problem?

This is an arithmetic question, using algebra and adding variables only makes this problem complicated. Additionally, you shouldn't be concerned about expressing the solution in ratios if that doesn't appear as an answer form. But regarding your equation. Your math is correct s/11 = t/1 (This is confirmed through backsolving with the arithmetic... 1430/11 = 130/1) But this proportion doesn't actually give you the answer to the problem since there are two unsolved variables, so I'm not sure what use it is.

Re: PS: Please, advise on how to calculate quicker!!! [#permalink]
12 Sep 2013, 06:39

2

This post received KUDOS

Expert's post

obs23 wrote:

Could someone show me a purely algebraic approach. I can't connect those elements (two equations). I solve fast via the common sense, but I it looks like I am doing something silly on the proportions side.

I have this ready, after taking out three 0's, \(26t = 3800\) (after 3800/t=26), and then since we have \(11t=s\) it must be that \(\frac{s}{11}=\frac{t}{1}\)? But then, if it is true, I am confused with my common sense...please help me untangle myself. How do you express ratios as stated in the problem?

Hi obs, if you're intent on solving the question through ratios, then you can, but it's not necessarily the best approach. That being said, let's attack it as a ratio problem.

Firstly, there's a slight typo in how you copied the math, it's actually \(26t = 3,380\). You can then figure out that whatever t is, if you multiply it by 11 you'll get the number of students. So yes \(\frac{s}{11}=\frac{t}{1}\), and you already have \(26t = 3,380\) so \(t= \frac{3,380}{26}\). Now solve t for 130, and then \(11t=s\). s is 1430.

A sneakier way to solve this:

The GMAT will always try and trap you, so if it asks for the number of students (s), the number of teachers (t) will also be a trap answer somewhere on the list. If you recognize this, then all you have to do is look for the two answer choices that are 11 away from one another. The bigger one will be the answer and the smaller will be the trap. This kind of "fourth wall" shortcut isn't for everyone, but it can be very helpful when crunched for time or unsure about the math.

Re: PS: Please, advise on how to calculate quicker!!! [#permalink]
14 Sep 2013, 01:19

VeritasPrepRon wrote:

obs23 wrote:

Could someone show me a purely algebraic approach. I can't connect those elements (two equations). I solve fast via the common sense, but I it looks like I am doing something silly on the proportions side.

I have this ready, after taking out three 0's, \(26t = 3800\) (after 3800/t=26), and then since we have \(11t=s\) it must be that \(\frac{s}{11}=\frac{t}{1}\)? But then, if it is true, I am confused with my common sense...please help me untangle myself. How do you express ratios as stated in the problem?

Hi obs, if you're intent on solving the question through ratios, then you can, but it's not necessarily the best approach. That being said, let's attack it as a ratio problem.

Firstly, there's a slight typo in how you copied the math, it's actually \(26t = 3,380\). You can then figure out that whatever t is, if you multiply it by 11 you'll get the number of students. So yes \(\frac{s}{11}=\frac{t}{1}\), and you already have \(26t = 3,380\) so \(t= \frac{3,380}{26}\). Now solve t for 130, and then \(11t=s\). s is 1430.

A sneakier way to solve this:

The GMAT will always try and trap you, so if it asks for the number of students (s), the number of teachers (t) will also be a trap answer somewhere on the list. If you recognize this, then all you have to do is look for the two answer choices that are 11 away from one another. The bigger one will be the answer and the smaller will be the trap. This kind of "fourth wall" shortcut isn't for everyone, but it can be very helpful when crunched for time or unsure about the math.

Hope this helps! -Ron

Thanks Ron, all makes sense. As for the use, I was just a bit rusty on ratios so I wanted to make sure I do the simple stuff correctly in algebraic terms. _________________

There are times when I do not mind kudos...I do enjoy giving some for help

Re: A certain college has a student-to-teacher ratio of 11 to 1. [#permalink]
14 Sep 2013, 06:58

Expert's post

No problem obs23, as mentioned it may not be the most obvious way to solve this problem, but I like to remind students that pretty much any GMAT math problem can be solved in a variety of ways. If you have one way to solve the problem, you're probably okay, but if you have three different ways, then you're definitely in great shape to solve any variation of the problem on test day. Hope this makes sense!

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