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Director
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Quote: 3. Fresh Grapes contain 90% water and 10% grape pulp whereas Dry Grapes
contain 20% water and 80% grape pulp. How many kgs of Dry Grapes can
be obtained from 20 kgs of Fresh Grapes?
Amount of grape pulp will remain same.
80Z/100 =10/100*20
z=2.5
Answer is 2.5 kgs
Can/Will someone explain the answer a little less tersley?
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ggarr wrote: Quote: 3. Fresh Grapes contain 90% water and 10% grape pulp whereas Dry Grapes
contain 20% water and 80% grape pulp. How many kgs of Dry Grapes can
be obtained from 20 kgs of Fresh Grapes?
Amount of grape pulp will remain same.
80Z/100 =10/100*20
z=2.5
Answer is 2.5 kgs Can/Will someone explain the answer a little less tersley? The key in this question is to consider that the weight of grape pulp remain constant.
So,
First, we have to calculate the weight of grape pulp in the fresh graps:
10%*20 = 2 kg
Second, we know that 2 kg represents 80% of the weight of dry grapes.
Lastly, to have the weight of dry grapes, we have to know what represents 100%. Hence,
100*2/80 = 2,5 kg
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Director
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thanks. it makes sense now. thanks for putting that in perspective.
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Regarding the circle question, actually I can see it being in GMAT. GMAT likes to test geometry questions that seems complicated but can be solved intuitively and quickly if you find the way to do it.
For this question, we can see that OS=r, OT=r-2, ST=r-1. Then we know (r-1)^2+(r-2)^2=r^2. Also, knowing that a special triangle is 3,4,5, we can immediately know that r=5.
Attachments
tempgraph.GIF [ 3.83 KiB | Viewed 411 times ]
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HongHu wrote: Regarding the circle question, actually I can see it being in GMAT. GMAT likes to test geometry questions that seems complicated but can be solved intuitively and quickly if you find the way to do it.
For this question, we can see that OS=r, OT=r-2, ST=r-1. Then we know (r-1)^2+(r-2)^2=r^2. Also, knowing that a special triangle is 3,4,5, we can immediately know that r=5.
Congrats, HongHu, this is what i want ?
Once u get the logic this is simple. Isn't it?
And someone in the earlier post commented that this will not of the GMAT type because this is too lengthy.
I felt bad, but finally i got one person who is able to judge this question correctly..................
Anyway guys, since i want to contribute some knowledge to the students from this thread, i am not going to worry about the comments here........
Keeping rocking folks..............................
Be ready for the next one
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Definitely. You've been doing a great job. Keep it up! 
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I agree with HongHu  .... (and Yezz before)
U should continue :D ... Nice job
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Hey, Fig, HongHu and Yezz thanx a lot.........
I'll definitely continue ...........
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Here comes our next question........
7. Find the total number of different mixed doubles tennis games that can be conducted among 7 married couples if no husband and wife play in the same game.
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o Couple 1 vs Others = 1C2*1C6*1C2
o Couple 2 vs Others = 1C2*1C5*1C2
o Couple 3 vs Others = 1C2*1C4*1C2
o Couple 4 vs Others = 1C2*1C3*1C2
o Couple 5 vs Others = 1C2*1C2*1C2
o Couple 6 vs Others = 1C2*1C1*1C2
Sum = 4*(6+5+4+3+2+1) = 4*21 = 84
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FIG could you please expalin the above solution in detail..
Thanks!!
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yogeshsheth wrote: FIG could you please expalin the above solution in detail..
Thanks!!
At least, I can try
We are looking for the number of possible matches, thus the order does not matter.
We can consider a "first" couple. The husband or the wife, 1C2, can play against a wife or a husband, 1C2, of 1 of the 6 other couples, 1C6.
Thus, the "first" couple could play : 1C2*1C6*1C2 matches.
Then, we can consider a "second" couple. One more time, The husband or the wife, 1C2, can play against a wife or a husband, 1C2, of 1 of the 5 remaining other couples, 1C5.
Thus, the "second" couple could play : 1C2*1C5*1C2 matches.
The patern indicates that we decrease the remaining number of couple.
And so on... till we arrive at couple 6 that can play only the wife or husband of the latest couple.
Hence, we sum up all these matches:
Sum = 4*(6+5+4+3+2+1) = 4*21 = 84.
Last edited by Fig on 17 Oct 2006, 13:50, edited 1 time in total.
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Fig wrote: yogeshsheth wrote: FIG could you please expalin the above solution in detail..
Thanks!! At least, I can try We are looking for the number of possible matches, thus the order does not matter. We can consider a "first" couple. The husband or the wife, 1C2, can play against a wife or a husband, 1C2, of 1 of the 6 other couples, 1C6. Thus, the "first" couple could play : 1C2*1C6*1C2 matches. Then, we can consider a "second" couple. One more time, The husband or the wife, 1C2, can play against a wife or a husband, 1C2, of 1 of the 5 remaining other couples, 1C5. Thus, the "second" couple could play : 1C2*1C5*1C2 matches. The patern indicates that we decrease the ramining number of couple. And so on... till we arrive at couple 6 that can play only the wife or husband of the latest couple. Hence, we sum up all these matches: Sum = 4*(6+5+4+3+2+1) = 4*21 = 84. Thanks for the nice expalnation FIG 
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U are welcomed 
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Well I get 42
7 * 6 ( 7C1* 6C1 , 7 * each couple time 6 different couples)
It'd be 84 if the matches were played twice.
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Re: Problem Solving Territory: 700+ questions only.............. [#permalink]
 02 Sep 2009, 12:27
I think i should be 42
M1W2+.............................M1W7
.
.
.
M7W1+.............................M7W6
6*7=42
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Re: Problem Solving Territory: 700+ questions only.............. [#permalink]
13 Sep 2009, 12:05
I got 1200 as the answer for the first question too .. But my working out was way too long.. Urs is really apt cicerone..
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Re: Problem Solving Territory: 700+ questions only.............. [#permalink]
13 Sep 2009, 12:21
Now for question 2, My Soln is:
In Tri(ABC) and Tri(AEB)
Angle ABC= Angle AEB and
Angle A is common to both triangles.
Hence third angle will also be equal.
Therefore by AAA property
=> Tri(ABC) ~ Tri(AEB)
Since for Similar triangles , the ratio for corresponding sides are equal, Therefore
=> AB/AE = BC/EB = AC/AB
taking AB/AE = AC/AB and substituting values for AB = 6 and AC = 5
we get => AE = 36/5
Now CE = AE - AC = (36/5) - 5 = 11/5 (Ans CE = 11/5)
To find DC, lets consider tri(BAD) and tri(BCA).
Given that Angle ACD= Angle BAD
and since Angle B is common to both
Again by AAA property we get
=> Tri(BAD) ~ Tri(BCA)
Since for Similar triangles , the ratio for corresponding sides are equal, Therefore
=> BA/BC = AD/CA = BD/BA
taking BA/BC = BD/BA and substituting values for AB = 6 and BD = 4
we get => BC = 36/4 = 9
Now CD = BC - BD = 9 - 4 = 5(Ans CD = 5)
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Re: Problem Solving Territory: 700+ questions only.............. [#permalink]
13 Sep 2009, 12:46
3. Fresh Grapes contain 90% water and 10% grape pulp whereas Dry Grapes
contain 20% water and 80% grape pulp. How many kgs of Dry Grapes can
be obtained from 20 kgs of Fresh Grapes?
Soln: 20 kgs of Fresh grapes has 18 kg water and 2 kg grape pulp.
The quantity of pulp does not change. Hence
1 kg of Dry grapes contains .8 kg of pulp
X kgs of Dry grapes will contain 2 kg of pulp
X = 2/.8 = 2.5 kgs
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Re: Problem Solving Territory: 700+ questions only.............. [#permalink]
13 Sep 2009, 12:55
5. When a clock indicates completion of 24 hours it actually loses one hour. How much time does it loose in 5 days?
Soln: When it shows completion of 24 hours it had lost 1 hr
Hence it shows only 24 hours for every 25 hours
therefore it loses 1 hour every 25 hours
it would have lost x hours in 120 hours (5 days)
x = 120/25 = 4.8 hours
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Re: Problem Solving Territory: 700+ questions only..............
[#permalink]
13 Sep 2009, 12:55
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