Data Sufficiency

I cannot unnderstand the explanation of this example:

A certain bank has ten branches. What is the total amount of assets under management at the bank?

(1) There is an average of 400 customers per branch. When each branch's average assets under management per customer is computed, these values are added together and this sum is divided by 10. The result is $400,000 per customer.

(2) The bank has a total of 4,000 customers. When the total assets per branch are added up, each branch is found to manage, on average, 160 million dollars in assets.
OAb

Hi!

IMO D

Let X be the number of branches and Y be the total amount of assets under management at the bank.
X = 10
Y = ?

(1)
Let C be the number of customers
C/X = C/10 = 400
C = 4,000
Y/C = $400,000
Y = $400,000*4,000 = $1,600,000,000
The statement (1) ALONE is SUFFICIENT
We should eliminate answer choices B and C.

(2)
Let C be the number of customers
C = 4,000
Y/X = $160,000,000
Y = $160,000,000*10 = $1,600,000,000
The statement (2) ALONE is SUFFICIENT
We should eliminate answer choices A and E.

If you don't agree with my reasoning, please post here. Thanks!

Hope it helps!
Best,
Maciek

Maciek wrote:Hi!

IMO D

Let X be the number of branches and Y be the total amount of assets under management at the bank.
X = 10
Y = ?

(1)
Let C be the number of customers
C/X = C/10 = 400
C = 4,000
Y/C = $400,000
Y = $400,000*4,000 = $1,600,000,000
The statement (1) ALONE is SUFFICIENT
We should eliminate answer choices B and C.

(2)
Let C be the number of customers
C = 4,000
Y/X = $160,000,000
Y = $160,000,000*10 = $1,600,000,000
The statement (2) ALONE is SUFFICIENT
We should eliminate answer choices A and E.

If you don't agree with my reasoning, please post here. Thanks!

Hope it helps!
Best,
Maciek

Well...this is exactly the way I solved this question and reached to the same conclusion. However, the OA is A.

The explanation provided in the doc (papgust's 300 maths qns) is that the 2nd statement is wrong since "we know average for each branch but we don't know how many customers each branch has. The stem gives average, but not weighted average. So, insufficient".

This problem is relate to weighted averages.....there's a catch here...that's what i'm trying to figure out.

Thanks for clarifying Singhsa!

Check this:

Breaking Down GMATPrep Weighted Average Problems (Part 1 of 2)
https://www.beatthegmat.com/mba/2009/12/ ... art-1-of-2

Breaking Down GMATPrep Weighted Average Problems (Part 2 of 2)
https://www.beatthegmat.com/mba/2009/12/ ... art-2-of-2

Hope it helps!
Best,
Maciek

I checked this in the document.Actually OA is B.

In stmt1,we are given average of all branch customers.From this we can't calculate the numver of customer in each branch.

Let Assets of branch1 be B1 and so on to B10

Let number of customers of branch1 be C1 and so on to C10.

1/10[B1/C1+B2/C2.....+B10/C10]=400,000

C1+C2+C3+...+C10/10=400 or C1+C2+...C10=4000

Since we dont know each value of C1,C2 ...and C10,we cannot calculate B1+B2+..B10

Hence B

Maciek wrote:Thanks for clarifying Singhsa!

Check this:

Breaking Down GMATPrep Weighted Average Problems (Part 1 of 2)
https://www.beatthegmat.com/mba/2009/12/ ... art-1-of-2

Breaking Down GMATPrep Weighted Average Problems (Part 2 of 2)
https://www.beatthegmat.com/mba/2009/12/ ... art-2-of-2

Hope it helps!
Best,
Maciek

thanks for the link.........

Maciek wrote:Thanks for clarifying Singhsa!

Check this:

Breaking Down GMATPrep Weighted Average Problems (Part 1 of 2)
https://www.beatthegmat.com/mba/2009/12/ ... art-1-of-2

Breaking Down GMATPrep Weighted Average Problems (Part 2 of 2)
https://www.beatthegmat.com/mba/2009/12/ ... art-2-of-2

Hope it helps!
Best,
Maciek

Thanks a ton bro!!!

selango wrote:I checked this in the document.Actually OA is B.

In stmt1,we are given average of all branch customers.From this we can't calculate the numver of customer in each branch.

Let Assets of branch1 be B1 and so on to B10

Let number of customers of branch1 be C1 and so on to C10.

1/10[B1/C1+B2/C2.....+B10/C10]=400,000

C1+C2+C3+...+C10/10=400 or C1+C2+...C10=4000

Since we dont know each value of C1,C2 ...and C10,we cannot calculate B1+B2+..B10

Hence B

Oh yes!!! thanks for pointing that out. I've made the corrections.