I know this question has been discussed, but I can't find it
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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.
The question is: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 !
U+g = 100heshamelaziry wrote:I know this question has been discussed, but I can't find it
U+g = 100heshamelaziry wrote:I know this question has been discussed, but I can't find it
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.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 ?
Correct. From the stem and from (2), we know there are 5/2 as many females as males.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 ?
Testluv wrote:Correct. From the stem and from (2), we know there are 5/2 as many females as males.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 ?
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:heshamelaziry wrote:Testluv wrote:Correct. From the stem and from (2), we know there are 5/2 as many females as males.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 ?
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 ?
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%".maihuna wrote:U+g = 100heshamelaziry wrote:I know this question has been discussed, but I can't find it
(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%
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