Average

Hisham, here is the link for the explaination;


https://www.beatthegmat.com/weighted-average-t28451.html

I would appreciate if you can explain why the answer is C.

Pedros wrote:Hisham, here is the link for the explaination;


https://www.beatthegmat.com/weighted-average-t28451.html

I would appreciate if you can explain why the answer is C.

I didn't understand what he wrote. I need some simple explanation to this question !

heshamelaziry wrote:

Pedros wrote:Hisham, here is the link for the explaination;


https://www.beatthegmat.com/weighted-average-t28451.html

I would appreciate if you can explain why the answer is C.

I didn't understand what he wrote. I need some simple explanation to this question !

The question is:

Each student in a certain sociology class is either an undergraduate student or a graduate student. What percentage of the students in the class are graduate students?

(1) The average score on the final exam for undergraduate students was 20 points lower than the average for all students in the class.

(2) The average score on the final exam for graduate students was 40 points higher than the average for all students in the class.

(1) is clearly insufficient by itself because there is no info about graduate students; (2) is insufficient by itself because there is no info about undergraduate students.

(1) + (2)

Okay, so the concept of weighted average. Let the average for all students be x. Let's call the undergradate students UG; let's call the graduate students G. UG is 20 points below X while G is 40 points above. And, of course, 20 is half of 40.

Then:

UG...X......G

Because UG is twice as close to the "center" as G, there must be twice as many UGs as Gs. (Or, because G is twice as far away from the center than UG, there must be half as many Gs as UGs). So the ratio of UG to G is 2:1. This is a part-to-part ratio. 2 + 1 = 3 is the whole. So 2/3 (or approx 67%) of all students are UGs.

Choose C.

And, of course, if you know the tactic well, and you are really confident, you can pick C without even setting up the UG...X......G diagram.

_______________________

This concept shows up quite a bit on the GMAT, and can be applied to other topics that also involve weighted average. For example, average speed of a round trip is a weighted average, and time is the weight. Let's say that from point A to point B there are 100 miles. Let's say a car travels from A to B at 100mph and then returns from B to A at 50 mph. Going from A to B would therefore take the car 1 hour while returning from B to A will take 2 hours. So, the car is travelling at 100 mph for 1 hour and at 50 mph for 2 hours. So, the car's average speed will be the number that is twice as close to 50 than it is to 100, in other words a 2:1 ratio. From 50 to 100, there are 50 units. Dividing 50 into thirds is approximately 17. Therefore, the car's average speed is approx 50 + 17 = 67mph:

50------|-----|-----100
50......67....83....100

heshamelaziry wrote:I know this question has been discussed, but I can't find it

U+g = 100
(x-20)U+(x+40)(100-U)=xu+x(100-u)
Ux -20u + 100x+4000-ux-40u = ux+100x-ux = 100x
4000 = 60u
U = 4000/60 = 200/3 = 67%
G = 33%

heshamelaziry wrote:I know this question has been discussed, but I can't find it

U+g = 100
(x-20)U+(x+40)(100-U)=xu+x(100-u)
Ux -20u + 100x+4000-ux-40u = ux+100x-ux = 100x
4000 = 60u
U = 4000/60 = 200/3 = 67%
G = 33%

Thanks Testluv,

Could statement 1 be interpreted to say that there were less undergraduate students than graduates but the undergraduates scored more on the final test than graduate students ?

heshamelaziry wrote:Thanks Testluv,

Could statement 1 be interpreted to say that there were less undergraduate students than graduates but the undergraduates scored more on the final test than graduate students ?

No. There may be some undergrads who scored better than some graduate students but, from (1), on average, the undergrads must have scored worse than the graduate students. Because there are only undergraduate and graduate students, (1) basically tells us that the undergrads pulled the grand average to the left or down; to compensate for this pull-down, on average, the grads must have scored better than the undergrads.

At a certain company, the average (arithmetic mean) number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company’s male employees to the number of the company’s female employees?

(1) There are 52 male employees at the company.
(2) The average number of years of experience for the company’s male and female employees combined is 9.3 years.



It seems to me that this question is similar to the original question in this thread. Is the requested ratio 5/2 ?

heshamelaziry wrote:At a certain company, the average (arithmetic mean) number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company’s male employees to the number of the company’s female employees?

(1) There are 52 male employees at the company.
(2) The average number of years of experience for the company’s male and female employees combined is 9.3 years.



It seems to me that this question is similar to the original question in this thread. Is the requested ratio 5/2 ?

Correct. From the stem and from (2), we know there are 5/2 as many females as males.

Testluv wrote:

heshamelaziry wrote:At a certain company, the average (arithmetic mean) number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company’s male employees to the number of the company’s female employees?

(1) There are 52 male employees at the company.
(2) The average number of years of experience for the company’s male and female employees combined is 9.3 years.



It seems to me that this question is similar to the original question in this thread. Is the requested ratio 5/2 ?

Correct. From the stem and from (2), we know there are 5/2 as many females as males.

In one DS average question from OG12, statement 1 gave the average for a group and statement B gave the average for another group, and the question asked for the average of the 2 groups. The OA is E, since we need the number of elements in each goup to find the average for the two groups.

The above two questions ask for a ratio, can we find the number of UG and G or female employees and male employees from the information in these two questions ?

heshamelaziry wrote:

Testluv wrote:

heshamelaziry wrote:At a certain company, the average (arithmetic mean) number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company’s male employees to the number of the company’s female employees?

(1) There are 52 male employees at the company.
(2) The average number of years of experience for the company’s male and female employees combined is 9.3 years.



It seems to me that this question is similar to the original question in this thread. Is the requested ratio 5/2 ?

Correct. From the stem and from (2), we know there are 5/2 as many females as males.

In one DS average question from OG12, statement 1 gave the average for a group and statement B gave the average for another group, and the question asked for the average of the 2 groups. The OA is E, since we need the number of elements in each goup to find the average for the two groups.

The above two questions ask for a ratio, can we find the number of UG and G or female employees and male employees from the information in these two questions ?

Nope; just the proportions. We can sum all of this up in a formula. Let's say that there are two groups whose averages are X and Y. Let's say the grand average is just MEAN. Let the distance of X from the mean be x and the distance of Y from the mean be y:

X...MEAN...Y

Then, the proportion of X/Y is just y/x.

maihuna wrote:

heshamelaziry wrote:I know this question has been discussed, but I can't find it

U+g = 100
(x-20)U+(x+40)(100-U)=xu+x(100-u)
Ux -20u + 100x+4000-ux-40u = ux+100x-ux = 100x
4000 = 60u
U = 4000/60 = 200/3 = 67%
G = 33%

Maiuna, while your algebra may be correct, it is far easier (and less error-prone) to say: "because the undergraduates' average was twice as close to the grand average, there must be twice as many undergrads as grads: 2:1, and that, therefore, ratio of undergrads to whole is just 2:3, or 2/3 = 67%".

And if you are confident in applying the above reasoning, it is even easier to say: "because in combination the statements tell me how far each group is from the grand average, the answer is C". (You don't even need to figure out the ratio; b/c it is DS, you just need to know that you CAN compute it.)

Hi Testluv,

This approach is really simple and great !!!

Thanks,
Shobha