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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?

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

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.

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Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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........

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. _________________

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

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.

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.

Re: Problem Solving Territory: 700+ questions only.............. [#permalink]

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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