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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Fri Jan 21, 2011 12:24 am

Zach picked some pumpkins to give to Ann, Beth and Chuck. He gave the 2 heaviest pumpkins to
Ann, and the 5 lightest pumpkins to Beth. The combined weight of the two heaviest pumpkins is
25% of the total weight of all the pumpkins Zach picked. The combined weight of the five lightest
pumpkins is 45% of the total weight of all the pumpkins Zach picked. How many pumpkins did Zach
give Chuck?

Heaviest =25%
Lightest= 45%
Both= 70% and remaining is 30%
45%/5= 9%
Since Z's is greater than 9%, if 10% then 3.

I'm I doing this right? I made big assumptions here.

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Source: Beat The GMAT — Problem Solving | Fri Jan 21, 2011 12:24 am

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Fri Jan 21, 2011 2:12 am

yellowho wrote:Zach picked some pumpkins to give to Ann, Beth and Chuck. He gave the 2 heaviest pumpkins to Ann, and the 5 lightest pumpkins to Beth. The combined weight of the two heaviest pumpkins is 25% of the total weight of all the pumpkins Zach picked. The combined weight of the five lightest pumpkins is 45% of the total weight of all the pumpkins Zach picked. How many pumpkins did Zach give Chuck?

...

I'm I doing this right? I made big assumptions here.

Yes, you're doing right.
But as you're not looking confident, I assume that you're not doing it with a concrete reason...

Say, the required number = z
As you have said, the combined weight of these z pumpkins must be 30% of the total weight of all the pumpkins. This means the average weight of these z pumpkins must be (30/z)% of the total weight of all the pumpkins.

Now, as these z pumpkins are of medium weight, then their average must be greater than the average weight of the 5 lightest pumpkins but less than the average weight of the 2 heaviest pumpkins.

Hence, 9 < (30/z) < 12.5
=> (30/12.5) < z < (30/9)
=> 2.(something) < z < 3.(something)

As z must be an integer, only possible value of z is 3.

Rahul Lakhani
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aleph777: Master | Next Rank: 500 Posts; Posts: 131; Joined: Fri Jun 18, 2010 10:19 am; Location: New York, NY; Thanked: 10 times

by aleph777 » Fri Jan 21, 2011 6:22 am

Yellowho

Another way to deal with this is in terms of fractions, and you can cut down on your total computation.

25% = 1/4
45% = 45/100 or 9/20

So let's add these two together and see what fraction of the total amount we have:

1/4 + 9/20 = 5/20 + 9/20 = 14/20
Then reduce 14/20 to 7/10.

So you know that Ann and Beth have 7 out of 10 pumpkins, which means Zach received the three left over.

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samrendra4u: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Wed Aug 25, 2010 8:09 pm

by samrendra4u » Fri Jan 21, 2011 7:51 am

aleph777 wrote:Yellowho

Another way to deal with this is in terms of fractions, and you can cut down on your total computation.

25% = 1/4
45% = 45/100 or 9/20

So let's add these two together and see what fraction of the total amount we have:

1/4 + 9/20 = 5/20 + 9/20 = 14/20
Then reduce 14/20 to 7/10.

So you know that Ann and Beth have 7 out of 10 pumpkins, which means Zach received the three left over.

Great Approach!

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Fri Jan 21, 2011 9:52 am

aleph777 wrote:Another way to deal with this is in terms of fractions, and you can cut down on your total computation.

25% = 1/4
45% = 45/100 or 9/20

So let's add these two together and see what fraction of the total amount we have:

1/4 + 9/20 = 5/20 + 9/20 = 14/20
Then reduce 14/20 to 7/10.

So you know that Ann and Beth have 7 out of 10 pumpkins, which means Zach received the three left over.

I think you got lucky for this problem with this method.
Frankly, only from ratio, we can't determine the numbers.

Also here the ratios are of weight of some pumpkins. What is your logic behind treating them as the ratio of numbers of pumpkin?

Now why you got lucky...
If I change the problem from 2 heaviest pumpkins to a single heavy pumpkin, i.e. only a single pumpkin whose weight is 25% of the total weight of all the pumpkins is given to Ann, then according to your method the result should be 4. Right? But actually that's not true. If that is the case, then there is two possible results, either 2 or 3.

A simple intuitive explanation behind it is when a single pumpkin weighs 25% of the whole, then number of medium weighted pumpkin cannot increase considering the number and weight percentage of light weighted pumpkins remain the same. But according to your method the number is increasing which is not possible.

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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Fri Jan 21, 2011 8:45 pm

How did you go from: 9 < (30/z) < 12.5
To: (30/12.5) < z < (30/9)

[quote="Rahul@gurome"][quote="yellowho"]Zach picked some pumpkins to give to Ann, Beth and Chuck. He gave the 2 heaviest pumpkins to Ann, and the 5 lightest pumpkins to Beth. The combined weight of the two heaviest pumpkins is 25% of the total weight of all the pumpkins Zach picked. The combined weight of the five lightest pumpkins is 45% of the total weight of all the pumpkins Zach picked. How many pumpkins did Zach give Chuck?

...

I'm I doing this right? I made big assumptions here.[/quote]

Yes, you're doing right.
But as you're not looking confident, I assume that you're not doing it with a concrete reason...

Say, the required number = z
As you have said, the combined weight of these z pumpkins must be 30% of the total weight of all the pumpkins. This means the average weight of these z pumpkins must be (30/z)% of the total weight of all the pumpkins.

Now, as these z pumpkins are of medium weight, then their average must be greater than the average weight of the 5 lightest pumpkins but less than the average weight of the 2 heaviest pumpkins.

Hence, 9 < (30/z) < 12.5
=> (30/12.5) < z < (30/9)
=> 2.(something) < z < 3.(something)

As z must be an integer, only possible value of z is 3.[/quote]

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Fri Jan 21, 2011 10:26 pm

yellowho wrote:How did you go from: 9 < (30/z) < 12.5
To: (30/12.5) < z < (30/9)

You can do that in any of the following way.

Method - 1
Note that, 9 < (30/z) < 12.5 is actually two inequalities,

1. 9 < (30/z) => z < (30/9)
AND
2. (30/z) < 12.5 => z > (30/12.5)

Combine the results => (30/12.5) < z < (30/9)

Method - 2
Apply the concept that if you reciprocate all the terms in a inequality, all the inequality signs also reciprocates, i.e. x > 1 => (1/x) < 1.

Thus, 9 < (30/z) < 12.5
=> (1/9) > (z/30) > (1/12.5)
=> (30/9) > z > (z/12.5)
=> (30/12.5) < z < (30/9)

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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