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1. Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets. A 4 B 6 C. 8 D. 10 E. 12

2. A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45 and $70, respectively. On a given day, the price of one stock increased by 15%, while the price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant? A. 20, 35, 70 B. 20, 45, 70 C. 20, 35, 40 D. 35, 40, 70 E. 35, 40, 45

3. A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?

A) -1 and 9 B) 4 and 4 C) 3 and 5 D) 2 and 6 E) 0 and 8

Please share your way of thinking, not only post the answers.

1. Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets. A 4 B 6 C. 8 D. 10 E. 12

Please share your way of thinking, not only post the answers.

OA and explanations to follow.

for this lets assume that Y takes A days to produce w widgets then X takes (A +2)days to produce w widgets. Given both of them produce 5w/4 widgets in 3days, we can get the value of w=12/5 days----eqn 1

Now when working together X and Y will produce w widgets in 1/A + 1/(A+2) -----eqn2 Solving equation 1 and 2 we get A = 4days So X will take A+2 = 4+2= 6days to produce w widgets. so time taken to produce 2w widgets will be 2*6 =12 (option E)

Edited:changed typo for = to +

Last edited by asterixmatrix on 14 Oct 2009, 22:30, edited 1 time in total.

3. A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?

A) -1 and 9 B) 4 and 4 C) 3 and 5 D) 2 and 6 E) 0 and 8

Please share your way of thinking, not only post the answers.

OA and explanations to follow.

for the above set Mean = 0+2+4+6+8/5 = 20/5 = 4. SD1 = sqrt (∑Xi^2/n - Mean^2) for set {0,2,4,6,8} we get SD1 = sqrt ( 120/5 - 16) = 2 sqrt2 now if we add any of the above pair of numbers Mean remains 4. Let the numbers be a and b which is added to the set so new SD (SD2) will be SD2 = sqrt( 120+a^2+b^2 /7 -16) and this needs to be close to SD1 which is 2sqrt2

if we equate SD2 = SD1 then we will get [120 + a^2+ b^2/7 -16] = 8 or 120+a^2+b^2 = 168 then a^2+b^2 = 48. The pair having the closest value to this will be (2,6) so will go with option D

2. A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45 and $70, respectively. On a given day, the price of one stock increased by 15%, while the price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant? A. 20, 35, 70 B. 20, 45, 70 C. 20, 35, 40 D. 35, 40, 70 E. 35, 40, 45

Please share your way of thinking, not only post the answers.

OA and explanations to follow.

for the 5 stocks avg is $42 (sum of all stocks/5 =210/5) and we know that after the changes in price of 2 stocks the avg price rose by approx 2%. So total value of stocks becomes around $214. So change in value of stocks after the decrease and increase in price of 2 stocks is approx $4

We know that price for one of the shares decreased by 35% and another increased by 15%. Logically 35% should be on the share with lowest value considering the avg price is increasing and 15% increase should be on the share which can give a profit of ( $4 + 35% of $20)

Lets assume the stock to have changed be X and Y then X decreased by 35% to become 0.65X and Y increased by 15% to become 1.15Y. now change in Y which is (1.15Y -Y) = 0.15Y needs to be equal to (approx) $4 + 0.35X ( the price of X changed from X to 0.65X) 0.15Y = $4 + 0.35X which is obtained for Y = $70 and X =$20 so I will go with option E

I calculated avg of (0,2,4,6,8) = 4 now std deviation is a measure of central tendency. any thing close to avg is feasible for our purpose. out of options B is the only option not only near but also equal to avg. adding two 4's will result a data in a close spread of avg.

I got this ans without calculation....plz correct me if i am wrong....
_________________

Bhushan S. If you like my post....Consider it for Kudos

I calculated avg of (0,2,4,6,8) = 4 now std deviation is a measure of central tendency. any thing close to avg is feasible for our purpose. out of options B is the only option not only near but also equal to avg. adding two 4's will result a data in a close spread of avg.

I got this ans without calculation....plz correct me if i am wrong....

my logic goes to this extent and pardon me if I am incorrect in my understanding Initially I had thought that we should pick 4,4 and since Mean is 4 so SD will remain unchanged. For above set we get SD = sqrt40/5 = 2sqrt2 approx 2.83. Now when we take 4,4 we get 7 terms so SD = sqrt40/7 which is approx 2.4. Now if we take 2 and 6 we get SD = sqrt 48/7 which is approx 2.6 and this is pair of (2,6) will give us SD nearest to the original SD

I calculated avg of (0,2,4,6,8) = 4 now std deviation is a measure of central tendency. any thing close to avg is feasible for our purpose. out of options B is the only option not only near but also equal to avg. adding two 4's will result a data in a close spread of avg.

I got this ans without calculation....plz correct me if i am wrong....

my logic goes to this extent and pardon me if I am incorrect in my understanding Initially I had thought that we should pick 4,4 and since Mean is 4 so SD will remain unchanged. For above set we get SD = sqrt40/5 = 2sqrt2 approx 2.83. Now when we take 4,4 we get 7 terms so SD = sqrt40/7 which is approx 2.4. Now if we take 2 and 6 we get SD = sqrt 48/7 which is approx 2.6 and this is pair of (2,6) will give us SD nearest to the original SD

Asterix Fingers "X"d

OA IS D (2, 6)

Let's eliminate wrong answers:

Mean is 4 and so are the means of all 5 pairs from answers choices.

A. (-1, 9) These two numbers are farthest from the mean and they will stretch the set making SD bigger

B. (4, 4) These two numbers are closest to the mean and the will shrink the set making SD smaller

C. (3, 5) Suitable option so far

D. (2, 6) Suitable option so far

E. (0, 8) These two numbers are also far from mean and they will also stretch the set making SD bigger.

So, when I looked at the options C and D I assumed that C is also too close to the mean and it will affect it more than D. So I ended with D and was correct. But still my logic eliminating C was not sure thing, without the calculations.

And here comes my question:

I was told that GMAT almost never asks to calculate SD. Meaning that you should know what SD means and how it's calculated but no question asks calculations itself.

Have you EVER seen official GMAT question asking calculation of SD?
_________________

3. For the original set 0, 2, 4, 6, 8, standard deviation = 3.16227766

A) For the set -1, 0, 2, 4, 6, 8, 9, standard deviation = 3.872983346 B) For the set 0, 2, 4, 4, 4, 6, 8, standard deviation = 2.581988897 C) For the set 0, 2, 3, 4, 5, 6, 8, standard deviation = 2.645751311 D) For the set 0, 2, 2, 4, 6, 6, 8, standard deviation = 2.828427125 E) For the set 0, 0, 2, 4, 6, 8, 8, standard deviation = 3.464101615

[As the sample size is very small, SD's are calculated using formula for sample.]

Difference between original stdev and stdev of option D is 0.333850535 Difference between original stdev and stdev of option E is 0.301823955

Though option D has a very close call, closest to the original standard deviation is found in option E.

I believe real GMAT would not ask for such lengthy calculations.

2. Total price of 5 stocks before changes = $20+$35+$40+$45+$70 = $210 Total price of 5 stocks after changes = $210 * 1.02 = $214 --> increase by $4 approximately

As the % of increase on price of one stock is quite lower than the % of decrease on price of another stock and the total is increased, it is logical to guess that 35% decrease would be on the smallest price and 15% increase would be on the highest price.

To crosscheck: 15% of $70 - 15% of $20 = $10.5 - $7 = $3.5 which is closer to $4

Note: $3.5 is the highest change in the total stock price that we can obtain after the changes in stock prices; if we consider any other stock in place of those two, the difference will be lower than $3.5.

Answer is E.

Last edited by doe007 on 07 Apr 2013, 00:55, edited 1 time in total.

1. Assuming X takes n days to produce w widgets. So, Y takes n-2 days to produce w widgets. In 3 days, X produces 3w/n widgets and Y produces 3w/(n-2) widgets

3w/n + 3w/(n-2) = 5w/4 --> 24n - 24 = 5*n^2 - 10n --> 5*n^2 - 34n + 24 = 0 --> (5n-4)(n-6) = 0 --> n = 6 [n cannot be 4/5 because in that case Y would need -6/5 days to produce w widgets and that is impossible.]

So, to produce 2w widgets, machine X alone would need 12 days to produce 2w widgets.

1. Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets. A 4 B 6 C. 8 D. 10 E. 12

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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