Set of three PS questions

Question

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.

OA and explanations to follow.

asterixmatrix · Answer

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 +

asterixmatrix · Answer

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   = 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

asterixmatrix · Answer

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

bhushan252 · Answer

My approach for last Q

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.... :wink:

asterixmatrix · Answer

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

Bunuel · Answer

Answers:
1. E. (As asterixmatrix did it) 
2. E. (As asterixmatrix did it)
3. D. (Explanation in previous post)

+1 to asterixmatrix

PussInBoots · Answer

asterixmatrix your solutions is all sweet and nice... with a calculator, which is not allowed on GMAT.

Question #3 will not be asked on GMAT.

Bunuel · Answer

You can check new set of: PS problems: good-set-of-ps-85414.html DS problems: good-set-of-ds-85413.html

amit2k9 · Answer

12.

interesting one.

2,6

doe007 · Answer

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

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.

doe007 · Answer

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.

doe007 · Answer

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

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

Answer is E.

mau5 · Answer

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

rx*(t+2) = w

ry*(t) = w.

Also, w/(t+2) + w/t = (5w/4)*(1/3)

or 1/t + 1/(t+2) = 5/12 --> (3+2)/12 = 1/4+1/6

Thus, t = 4. Replacing, rx = w/6,

w/6*T = 2w

or T = 12 days.

E.

bumpbot · Answer

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