BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Gmat Prep Please Help

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 72
Joined: Sat Aug 23, 2008 3:26 am
Thanked: 1 times

Gmat Prep Please Help

by dubeystuts » Sat Jan 17, 2009 11:31 am
Ann $450,000
Bob $360,000
Cal $190,000
Dot $210,000
Ed $680,000
The table above shows the total sales recorded in July for the five salespeople. It was discovered that one of Cal’s sales was incorrectly recorded as one of Ann’s sales. After this error was corrected, Ann’s total sales were still higher than Cal’s total sales, and the median of 5 sales totals was $330,000. What was the value of the incorrectly recorded sale?

I don't know how to approach this question. Can someone help please? Many Thanks.
Top
Source: — Problem Solving |
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

by DanaJ » Sat Jan 17, 2009 12:04 pm
I hope I remember this one correctly, but the median of a group would be the middle value of this group. Say you have 2, 4, 6, 7 and 10. The median would be 6, since it's got 2 numbers smaller than it and two numbers bigger than it. When your series has an even number of elements, then you calculate the arithmetic mean of the two middle values. Like in 2, 4, 6, 8, 10, 11, the mean would be (6+8)/2 = 7.
Well, now let's get back to the problem you presented. Since this is a series having an even number of elements, that means that one of the elements is the mean of the whole thing. Since Bob, Dot and Ed did not have sales of 330 000, that means that either Ann or Cal had sales of 330 000.
So, first of all, if Ann had sales of 330 000, that means that the sale that was incorrectly recorded was worth 120 000 and Cal's sales were 190 000 + 120 000 = 310 000. Since Cal's sales are still below Ann's sales, we can safely say this is the correct guess.

So the answer would be 120 000. please tell me if I guessed it. If not, I need to brush up on my statistics....
Top
Senior | Next Rank: 100 Posts
Posts: 60
Joined: Mon Jun 04, 2007 7:37 am
Thanked: 6 times

by sonu_thekool » Sat Jan 17, 2009 12:32 pm
DanaJ wrote:I hope I remember this one correctly, but the median of a group would be the middle value of this group. Say you have 2, 4, 6, 7 and 10. The median would be 6, since it's got 2 numbers smaller than it and two numbers bigger than it. When your series has an even number of elements, then you calculate the arithmetic mean of the two middle values. Like in 2, 4, 6, 8, 10, 11, the mean would be (6+8)/2 = 7.
Well, now let's get back to the problem you presented. Since this is a series having an even number of elements, that means that one of the elements is the mean of the whole thing. Since Bob, Dot and Ed did not have sales of 330 000, that means that either Ann or Cal had sales of 330 000.
So, first of all, if Ann had sales of 330 000, that means that the sale that was incorrectly recorded was worth 120 000 and Cal's sales were 190 000 + 120 000 = 310 000. Since Cal's sales are still below Ann's sales, we can safely say this is the correct guess.

So the answer would be 120 000. please tell me if I guessed it. If not, I need to brush up on my statistics....
Hi DanaJ,

You guessed it correctly. I followed similar logic.

Since this is a median question, the first step is to order the elements in the increasing order. For simplicity, I am eliminating the trailing 3 zeros.

C - D - B - A - E = 190 - 210 - 360 - 450 - 680
The median in the original list is 360.

D - C - A - B - E = 210 - 310 - 330 - 360 - 680
The new median in the original list is 330.

When the sale was corrected between Ann and Cal, the median came down to 330, which means that their places have changed in the order as no one else had any changes.

It is given that Ann's sales are more than Cal's even after the correction so, Ann has to have the new median of 330. This means Ann had to give away 120 to Cal and the new total for Cal is still 310 which is less than 330.

So, 120000 is the right answer.

Pretty much similar to what you have proposed but since you said you weren't sure, I wanted to post an explanation. If I am not mistaken, this problem was also discussed sometime ago but this helps us revise the fundamentals once again.
Top
Post new topic Post Reply
• Page 1 of 1