Hi, I tired to use the average chart to solve these problem, but I not sure if I have done then correctly.
1-
25% of the children at the daycare center are 4 years old or older while 35% of the children at this daycare center are 2 years old or younger. What is the median age of the children at this daycare center?
A) 1
B) 2
C) 3
D) 4
E) It cannot be determined from the information given
2-
Each of the 25 boxes in a particular storage room weighs either 30 pounds or 50 pounds. The average (arithmetic mean) weight of the boxes in this storage room is 38 pounds. If the average weight of the boxes in the storage room is increased to 40 pounds by removing some of the 30-pound boxes, how many 30-pound boxes must be removed?
A) 5
B) 7
C) 9
D) 10
E) 12
Average problem
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We don't know the exact ages of the children. Hence we cannot determine the median age of the children.phoenix9801 wrote:25% of the children at the daycare center are 4 years old or older while 35% of the children at this daycare center are 2 years old or younger. What is the median age of the children at this daycare center?
The correct answer is E.
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Say, the number of 30 pounds box in the room = x and later n number of 30 pounds boxes were removed.phoenix9801 wrote:Each of the 25 boxes in a particular storage room weighs either 30 pounds or 50 pounds. The average (arithmetic mean) weight of the boxes in this storage room is 38 pounds. If the average weight of the boxes in the storage room is increased to 40 pounds by removing some of the 30-pound boxes, how many 30-pound boxes must be removed?
Hence, 50*(25 - x) + 30x = 25*38 ................................ (1)
Later, 50*(25 - x) + 30(x - n) = (25 - n)*40
-----> 50*(25 - x) + 30x - 30n = 25*40 - 40n ..........................(2)
Subtracting (1) form (2),
--> -30n = (25*40 - 25*38) - 40n
--> 10n = 25*(40 - 38) = 25*2 = 50
--> n = 5
The correct answer is A.
Last edited by Anurag@Gurome on Fri Jun 01, 2012 6:33 am, edited 1 time in total.
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- Ashujain
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@Anurag...I guess it can be determined.Anurag@Gurome wrote:We don't know the exact ages of the children. Hence we cannot determine the median age of the children.phoenix9801 wrote:25% of the children at the daycare center are 4 years old or older while 35% of the children at this daycare center are 2 years old or younger. What is the median age of the children at this daycare center?
The correct answer is E.
Lets say there are 100 children. 25 are 4 years old or older, 35 are 2 years old or younger and hence, 40 are 3 years old.(I don't think GMAT wants us to think of ages other than integers)
So if we arrange these students in the ascending order of their age then 3 will be the median.
It will always be 3 whatever be the no. of children.
IMO: C
What is the OA?
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That's a pretty strong assumption you are making.Ashujain wrote:...I don't think GMAT wants us to think of ages other than integers
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- Ashujain
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well the reason behind it is that i have done a similar question somewhere (I have forgotten the source though) and answer to that was 3.Anurag@Gurome wrote:That's a pretty strong assumption you are making.Ashujain wrote:...I don't think GMAT wants us to think of ages other than integers
@phoenix...can you please share the source of this question and what is the OA?
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for forming equation 2 you are assuming that number of boxes is 25, is this correct?
it is mentioned that some boxes are removed but no where it is mentioned that boxes are replaced or added
it is mentioned that some boxes are removed but no where it is mentioned that boxes are replaced or added
Anurag@Gurome wrote:Say, the number of 30 pounds box in the room = x and later n number of 30 pounds boxes were replaced 40 pound boxes.phoenix9801 wrote:Each of the 25 boxes in a particular storage room weighs either 30 pounds or 50 pounds. The average (arithmetic mean) weight of the boxes in this storage room is 38 pounds. If the average weight of the boxes in the storage room is increased to 40 pounds by removing some of the 30-pound boxes, how many 30-pound boxes must be removed?
Hence, 50*(25 - x) + 30x = 25*38 ................................ (1)
Later, 50*(25 - x) + 30(x - n) + 40n = 25*40
-----> 50*(25 - x) + 30x + 10n = 25*40 ..........................(2)
Subtracting (1) form (2),
--> 10n = (25*40 - 25*38) = 25*(40 - 38) = 25*2 = 50
--> n = 5
The correct answer is A.
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- eagleeye
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Hi outreach:outreach wrote:for forming equation 2 you are assuming that number of boxes is 25, is this correct?
it is mentioned that some boxes are removed but no where it is mentioned that boxes are replaced or addedAnurag@Gurome wrote:Say, the number of 30 pounds box in the room = x and later n number of 30 pounds boxes were replaced 40 pound boxes.phoenix9801 wrote:Each of the 25 boxes in a particular storage room weighs either 30 pounds or 50 pounds. The average (arithmetic mean) weight of the boxes in this storage room is 38 pounds. If the average weight of the boxes in the storage room is increased to 40 pounds by removing some of the 30-pound boxes, how many 30-pound boxes must be removed?
Hence, 50*(25 - x) + 30x = 25*38 ................................ (1)
Later, 50*(25 - x) + 30(x - n) + 40n = 25*40
-----> 50*(25 - x) + 30x + 10n = 25*40 ..........................(2)
Subtracting (1) form (2),
--> 10n = (25*40 - 25*38) = 25*(40 - 38) = 25*2 = 50
--> n = 5
The correct answer is A.
Anurag's setup of the second equation is correct. He is not assuming that there are 25 boxes.
In fact,
50*(25 - x) + 30(x - n) + 40n = 25*40
is the same as 50(25-x) + 30(x-n) = 40(25-n). (He is setting it up using 25-n as the final no.)
Let me know if this helps
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Thanks for pointing it out.outreach wrote:for forming equation 2 you are assuming that number of boxes is 25, is this correct?
it is mentioned that some boxes are removed but no where it is mentioned that boxes are replaced or added
I misread the problem and hence made a wrong assumption.
Edited the reply.
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