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

Redeem

Jackfruits and pumpkins

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

Jackfruits and pumpkins

by sanju09 » Fri Apr 23, 2010 4:25 am
Jackfruits and pumpkins are the only items on sale in a veggie store. Average (arithmetic mean) weight of all pumpkins on sale is 6 lbs. What percentage of the total are pumpkins?

(1) The average (arithmetic mean) weight of all the jackfruits was 7 ½ lbs.

(2) The average (arithmetic mean) weight of the total was twice as close to the average (arithmetic mean) weight of the jackfruits as it was to the average (arithmetic mean) weight of the pumpkins.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Source: — Data Sufficiency |
Master | Next Rank: 500 Posts
Posts: 114
Joined: Mon Sep 22, 2008 3:51 am
Thanked: 8 times
GMAT Score:680

by Fiver » Fri Apr 23, 2010 6:56 am
Choose B.

St1] by itself is irrelavant as it no idea of the overall avg or the ratio.
St2] The average (arithmetic mean) weight of the total was twice as close to the average (arithmetic mean) weight of the jackfruits as it was to the average (arithmetic mean) weight of the pumpkins.

There are 2 ways to interpret this statement.
A] logical Interpretation: The statement says that the overall avg is twice closer to the avg. of J than to the avg. of P.
This means that the avg of Js is contributing 2 time more than the avg. of Ps in arriving at the overall avg.
This in turn means that the number of Js is twice the number of Ps.
This means that the P ~ 33.33% of the overall number.

B] Algebraic interpretaion:

Assume J to be the avg. weight of all jackfruits.
Assume 'x' to be the overall avg.

Given avg. weight of Pumkins is 6.
To find (J-x)/(x-6) = ?
According to st2] 2(J-x) = (x-6)
Hence the ratio becomes (J-x)/2(J-x) = 1/2 and hence the ratio of Pumkins to total is 1/3 ~ 33.33%
Top
User avatar
Legendary Member
Posts: 537
Joined: Sun Jul 19, 2009 7:15 am
Location: Nagpur , India
Thanked: 41 times
Followed by:1 members

by rockeyb » Fri Apr 23, 2010 8:00 am
I dont quite agree with the explanation above , I may be wrong though but this is my view .
A] logical Interpretation: The statement says that the overall avg is twice closer to the avg. of J than to the avg. of P.
This means that the avg of Js is contributing 2 time more than the avg. of Ps in arriving at the overall avg.
This in turn means that the number of Js is twice the number of Ps.
This means that the P ~ 33.33% of the overall number.
Look at the statement in bold are we saying that each Jackfruit weighs equal to each Pumpkin ?

If we are assuming weight of each J = weight of each P then we can say number of Js is twice the number of Ps .

No where in the question is the relation between the weights of J and P is given and hence we can not say how many J = number of P's .

I my opinion the answer should be E .
"Know thyself" and "Nothing in excess"
Top
Master | Next Rank: 500 Posts
Posts: 114
Joined: Mon Sep 22, 2008 3:51 am
Thanked: 8 times
GMAT Score:680

by Fiver » Fri Apr 23, 2010 8:24 am
rockeyb wrote:I dont quite agree with the explanation above , I may be wrong though but this is my view .
A] logical Interpretation: The statement says that the overall avg is twice closer to the avg. of J than to the avg. of P.
This means that the avg of Js is contributing 2 time more than the avg. of Ps in arriving at the overall avg.
This in turn means that the number of Js is twice the number of Ps.
This means that the P ~ 33.33% of the overall number.
Look at the statement in bold are we saying that each Jackfruit weighs equal to each Pumpkin ?

If we are assuming weight of each J = weight of each P then we can say number of Js is twice the number of Ps .
No where in the question is the relation between the weights of J and P is given and hence we can not say how many J = number of P's .

I my opinion the answer should be E .
If the weight of each J were to be equal to the weight of each P, then the the overall avg would be = weight of each J = weight of each P.
In other words the avg. is the same as the individual weights and thus invalidates st2. and makes it impossible to find an exact ratio between the 2 categories.

To explain further, the overall avg. = (number of Ps * 6 + number of Js * J)/ Number of (Ps + Js)
Here if the overall avg. is closer to the avg. weight of Js, it means that the number of Js is greater by the same degree.
Top
Master | Next Rank: 500 Posts
Posts: 135
Joined: Tue Oct 13, 2009 10:27 am
Thanked: 3 times

by boazkhan » Fri Apr 23, 2010 9:28 am
IMO B -- What is the OA?
Top
Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 am
Thanked: 173 times
Followed by:2 members
GMAT Score:710

by gmatmachoman » Fri Apr 23, 2010 10:29 am
@sanju bhai,

I am not able to get the wordings of st 2.
But in anycase the DS asks for ratio of pumpkins in the basket of(Pumpkin + JF)

Pretty much a weighted average concept asking to find n1 &n2.

IMO E!!

Plz do explain st2.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 435
Joined: Mon Mar 15, 2010 6:15 am
Thanked: 32 times
Followed by:1 members

by eaakbari » Thu Apr 29, 2010 11:04 pm
Skim through the question once more, we are just talking about averages. Nowhere in the question is any quantity mentioned, but for the final answer we do require quantities.

Answer E

Do confirm Sanju
Whether you think you can or can't, you're right.
- Henry Ford
Top
Master | Next Rank: 500 Posts
Posts: 114
Joined: Mon Sep 22, 2008 3:51 am
Thanked: 8 times
GMAT Score:680

by Fiver » Wed Jun 16, 2010 9:56 am
OA ?
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Wed Jun 16, 2010 12:35 pm
Fiver's analysis is spot on.

We can picture weighted averages a number of different ways.

One very useful visualization is to think of the number line:

Average of Item A --------------- Total Average -------------- Average of Item B

If the total average is directly in the middle of the two groups, we have an equal number of each item. If the total average is skewed in one direction, that group must have a greater number of items.

Further, the amount of skew is inversely proportional to the relative numbers of items.

For example, if group A has twice as many items as group B, the Total average will be twice as far from B as from A. We can generalize the relationship to:

Avg of A -------x------- Total Avg ------------------y----------------- Avg of B

(# in A)/(# in B) = y/x

Since statement (2) gives us the exact skew (i.e. the values of x and y), we can determine the percent weight of each group; accordingly, (2) is sufficient alone.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Wed Jun 16, 2010 12:37 pm
eaakbari wrote:Skim through the question once more, we are just talking about averages. Nowhere in the question is any quantity mentioned, but for the final answer we do require quantities.

Answer E

Do confirm Sanju
Hi,

the final answer requires percentages, not quantities. To solve for a relationship, you don't need actual numbers.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Master | Next Rank: 500 Posts
Posts: 186
Joined: Fri May 28, 2010 1:05 am
Thanked: 11 times

by jube » Thu Jun 17, 2010 9:48 am
Stuart Kovinsky wrote:Fiver's analysis is spot on.

We can picture weighted averages a number of different ways.

One very useful visualization is to think of the number line:

Average of Item A --------------- Total Average -------------- Average of Item B

If the total average is directly in the middle of the two groups, we have an equal number of each item. If the total average is skewed in one direction, that group must have a greater number of items.
Stuart, won't the numerator (the sum of whatever we're talking about) also play a role? I'm interpreting the above to mean only the denominator i.e. the no. of items in discussion. Could you clarify this, please?

Thanks.
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Thu Jun 17, 2010 11:02 am
jube wrote:
Stuart Kovinsky wrote:Fiver's analysis is spot on.

We can picture weighted averages a number of different ways.

One very useful visualization is to think of the number line:

Average of Item A --------------- Total Average -------------- Average of Item B

If the total average is directly in the middle of the two groups, we have an equal number of each item. If the total average is skewed in one direction, that group must have a greater number of items.
Stuart, won't the numerator (the sum of whatever we're talking about) also play a role? I'm interpreting the above to mean only the denominator i.e. the no. of items in discussion. Could you clarify this, please?

Thanks.
Hi,

the weighted average formula is:

Weighted average = (% weight group 1)*(average of group 1) + (% weight group 2)*(average of group 2) + (% weight group 3)*(average of group 3) + ...

The sum is only relevant in calculating the average of each individual group; it doesn't play a role when using the weighted average formula (or the number line/ratio method).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Master | Next Rank: 500 Posts
Posts: 265
Joined: Mon Dec 28, 2009 9:45 pm
Thanked: 26 times
Followed by:2 members
GMAT Score:760

by mj78ind » Thu Jun 17, 2010 8:11 pm
rockeyb wrote:I dont quite agree with the explanation above , I may be wrong though but this is my view .
A] logical Interpretation: The statement says that the overall avg is twice closer to the avg. of J than to the avg. of P.
This means that the avg of Js is contributing 2 time more than the avg. of Ps in arriving at the overall avg.
This in turn means that the number of Js is twice the number of Ps.
This means that the P ~ 33.33% of the overall number.
Look at the statement in bold are we saying that each Jackfruit weighs equal to each Pumpkin ?

If we are assuming weight of each J = weight of each P then we can say number of Js is twice the number of Ps .

No where in the question is the relation between the weights of J and P is given and hence we can not say how many J = number of P's .

I my opinion the answer should be E .
@ rockeyb
i agree with your logic but think we can get to an answer using the 2 stmts as follows:

Let us assume, Sum of wts of Jackfruits = J, same for pumpkins = P; number of jackfruits = j and pumps = p

Stmt 1 J/j = 7.5

Stmt 2 (P + J)/(p+j) divides the line joining the ratios in 2:1 ratio. We do not know if Jackfruits average is to the left or right of the combined average.

Stmt 1 and 2:
(P+J)/(p+j) = 7 (1) ; J/j = 7.5 (2) ; P/p = 6 (3), note we come up with the combined average of 7.0 in (1) since it is given that the ratio of division is 1:2 in favor of jackfruit.

We have to find, j/(p+j),
Taking values from (2) and (3) and inputting in (1), we get
(6p + 7.5j) = 7 (p+j) thus we have 0.5j = p, or p = 2j

Thus j/(p+j) = j/(0.5j+j) = 2/3

However if the avg of jackfruits is not given we can not find that the combined average is 7.0 and not solve the question.
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Fri Jun 18, 2010 12:59 am
mj78ind wrote:
However if the avg of jackfruits is not given we can not find that the combined average is 7.0 and not solve the question.
Hi,

you're correct that if we aren't given the average of J we can't find the combined average.

However, the combined average is irrelevant to the question, which is:

What percentage of the total are pumpkins?

To find the percent weight of the pumpkins, we don't need any actual measurements. For the reasons outlined in the posts above, (2) is sufficient alone to answer the question posed.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Master | Next Rank: 500 Posts
Posts: 265
Joined: Mon Dec 28, 2009 9:45 pm
Thanked: 26 times
Followed by:2 members
GMAT Score:760

by mj78ind » Fri Jun 18, 2010 6:14 am
Stuart Kovinsky wrote:
mj78ind wrote:
However if the avg of jackfruits is not given we can not find that the combined average is 7.0 and not solve the question.
Hi,

you're correct that if we aren't given the average of J we can't find the combined average.

However, the combined average is irrelevant to the question, which is:

What percentage of the total are pumpkins?

To find the percent weight of the pumpkins, we don't need any actual measurements. For the reasons outlined in the posts above, (2) is sufficient alone to answer the question posed.
@Stuart I have to agree to that, just relooked at the way you proposed:

Assuming a number line, using the concepts of centre of gravity in physics (12th grade)

(sum of pumpks wts) ---------------2x-------- Total average ------x------ (sum of jacks wts)

We know, (sum of pumpks wt) = 6*p, hence (sum of jackfs wts ) can be calculated:
6p*2x = (sum of jack wts)*x thus sum of jack wts = 12p
Thus B sufficient.

Thanks!
Top
Post new topic Post Reply