Data Sufficiency

At a certain company, a test was given to a group of men and women seeking for promotions. If the average (artihmetic mean) score for the group was 80, was the average score for the women more than 85?

(1) The average score for the men was less than 75.
(2) The group consisted of more men than women.

[spoiler]OA:C[/spoiler]

Can anyone please explain this problem using the tug of war method?Also,what range does this problem follow?700-800?

Thanks

From Statement 1, we know that the average score for the women must be greater than 80 (since the men's average is below 75 and the overall average is 80). However, if we have a lot more women than men, maybe an average of 82 for the women would be enough to balance out an average of 74 for the men (this would require a 3:1 ratio of women:men). If we have an equal number of men and women, then we could get an average of 80 if the men's average was 74 and the women's average was 86.

Statement 2 does not do much for us on its own. Either group could have the higher average (men = 82, women = 74, men:women=3:1 OR women = 86, men = 77, men:women = 2:1).

When we put them together, we know that men's average is less than 75, and that there are more men than women. By the tug of war method, this would mean that the women's average must be farther away from 80 than the men's average is. Since the men's average must be more than 5 less than 80, the women's average must be more than 5 more than 80, meaning it will always be above 85.

I find DS questions based on similar concepts very hard to understand.Is there a way out?Any material that you could lead me to #link.There was nothing much about these in the Veritas guide.

Bill@VeritasPrep wrote: Also,can you please explain this for me---How can statement 1 tell us that B is more than 80?There's oo much doubt right?

From Statement 1, we know that the average score for the women must be greater than 80 (since the men's average is below 75 and the overall average is 80). However, if we have a lot more women than men, maybe an average of 82 for the women would be enough to balance out an average of 74 for the men (this would require a 3:1 ratio of women:men). If we have an equal number of men and women, then we could get an average of 80 if the men's average was 74 and the women's average was 86.

We are mixing two groups (men and women) and coming up with an overall average of 80. If the men's average is below 80 (which Statement 1 guarantees), then the women's average must be above 80 to bring the overall average up. If the men's average is 70 and the women's average is 79, there's no way for the weighted average to be higher than either of the individual group averages.

I find the best way to approach abstract weighted average DS questions is to try numbers intelligently. When it says the men's average is less than 75, try 74 using the tug of war/number line method, but also try something lower to see if there's a trick.

Where could I find more problems like these?One doubt here too.

If we have a equation in a DS question like this,

3x+5y<40(It derived from a statement)

Can we modify the equation to 3x+5y=35 or anything less than 40?It tends to give us the wrong answer.

This was the problem

Material A costs $3 per kilogram, and material B costs $5 per kilogram. If 10 kilograms of material K consists of x kilograms of material A and y kilograms of material B, is x > y?

(1) y > 4
(2) The cost of the 10 kilograms of material K is less than $40.

Thanks Bill.

Bill@VeritasPrep wrote:From Statement 1, we know that the average score for the women must be greater than 80 (since the men's average is below 75 and the overall average is 80). However, if we have a lot more women than men, maybe an average of 82 for the women would be enough to balance out an average of 74 for the men (this would require a 3:1 ratio of women:men). If we have an equal number of men and women, then we could get an average of 80 if the men's average was 74 and the women's average was 86.

Statement 2 does not do much for us on its own. Either group could have the higher average (men = 82, women = 74, men:women=3:1 OR women = 86, men = 77, men:women = 2:1).

When we put them together, we know that men's average is less than 75, and that there are more men than women. By the tug of war method, this would mean that the women's average must be farther away from 80 than the men's average is. Since the men's average must be more than 5 less than 80, the women's average must be more than 5 more than 80, meaning it will always be above 85.

Where could I find more problems like these?

This is a weighted average and inequality problem. You can use the GMATFix App to filter questions and create drills targeting the specific topic you want to master; you can also filter by difficulty level if you wish.





-Patrick
GMATFix

Oh ok.Cool.Thats an excellent app.I got a few right,a few wrong.I cant comprehend how the weighed equations work

A(x)+B(y)=Weighed Average

I find it hard to know what A and B stand for in the problem,also x and y.I find it difficult to recognise and isolate the variables in this problem

Any advice is appreciated.

Patrick_GMATFix wrote:

Where could I find more problems like these?

This is a weighted average and inequality problem. You can use the GMATFix App to filter questions and create drills targeting the specific topic you want to master; you can also filter by difficulty level if you wish.





-Patrick
GMATFix

dddanny2006 wrote:Oh ok.Cool.Thats an excellent app

Thanks that's kind of you. I'm actually writing code for it right now (well I stopped to type this reply).

dddanny2006 wrote:A(x)+B(y)=Weighed Average

I find it hard to know what A and B stand for in the problem,also x and y.

Weighted Avg means that the average is the result of a mix adjusted for the weights of the components. For example, if a room contains 60% women and 40% men, the weighted avg of any measure (age, test scores, salaries) will be 60% of the women's measure + 40% of the men's measure: W.A.=0.6x + 0.4y where x and y are the measures that actually get averaged.

In all such questions you need to be clear about two things:
(1) what measure gets averaged?
(2) what are the groups' weights? (percentage distribution which always adds up to 1 or 100%)

Example 1: A store sells two types of widgets. Type A sells for $32 and type B sells for $20. If 25% of widgets sold were of Type A, what is the average selling price?
(1) what measure gets averaged? Prices.
(2) what are the weights (% distribution - must add up to 1 or 100%). 25% A, 75% B
Equation: $32*(0.25) + $20*(0.75) = Average

Example 2: The average salary of all employees is $32,500. If the ratio of male to female employees is 3:5, and each female earns $34,000, how much does each male earn?
(1) What measure gets averaged? earnings.
(2) what are the weights (% distribution - must add up to 1 or 100%). 3/8 Males, 5/8 Females (ratio 3:5 means 8 total parts)
Equation: x*(3/8) + $34,000*(5/8) = $32,500

Example 3: I have two investments. Investment A grew by 15% and investment B grew by 8%. If the average growth of both investments is 10%, what fraction of my capital was invested in A?
(1) What measure gets averaged? percent growth!
(2) what are the weights (% distribution - must add up to 1 or 100%). Unknown. Call them a for investment A and 1-a (for investment B) so the weights add up to 1
Equation: (15%)*a + (8%)*(1-a) = 10%

Example 4: 10 kilograms of material K is made up of x kilograms of A costing $3 per kilo, and y kilograms of B costing $5 per kilo. Build a formula for average price of material K.
(1) What measure gets averaged? price per kilo
(2) what are the weights (% distribution - must add up to 1 or 100%). x/10 for A, and y/10 for B. Since they add up to 1, it makes sense to use x/10 and (10-x)/10
Equation: $3*(x/10) + $5(y/10) = Price of K
even better: $3*(x/10) + $5*(10-x)/10 = Price of K

Does that help?
-Patrick

Thanks for that Brent.Well I saw that video prior to starting this entire thread.Its easy to get ratio's,but I cant use the same method on DS Weighted Average problems.I dont know what the 2 end of the tug rope should be.

Brent@GMATPrepNow wrote:If anyone is interested, we have a free and comprehensive video on the topic of weighted averages: https://www.gmatprepnow.com/module/gmat- ... ics?id=805

Cheers,
Brent

GMAT fix app is awesome,Im saying this again because it so easily filters problems according to a variety of parameters.

Patrick_GMATFix wrote: