Folks, I will be posting questions from Problem Solving section regularly in this thread............. Here comes our first question 1. A shopkeeper increases the price of the article by X% and then decreases the resulting price by X%. As a result the price of the article is reduced by 180 $. Ater one more such cycle the price is further reduced by 153$. Find the oriiginal price of the article.

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Problem Solving Territory: 700+ questions only..............

1 2 3

ggarr

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11 Oct 2006, 14:43

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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?

Fig

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11 Oct 2006, 14:52

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ggarr

Quote:

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

ggarr

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11 Oct 2006, 18:16

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thanks. it makes sense now. thanks for putting that in perspective.

HongHu

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12 Oct 2006, 20:57

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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.

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cicerone

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12 Oct 2006, 22:54

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HongHu

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

HongHu

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13 Oct 2006, 05:27

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Definitely. You've been doing a great job. Keep it up! :good

Fig

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13 Oct 2006, 05:49

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I agree with HongHu

.... (and Yezz before)

U should continue :D

... Nice job

cicerone

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13 Oct 2006, 09:12

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Hey, Fig, HongHu and Yezz thanx a lot.........
I'll definitely continue ...........

cicerone

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16 Oct 2006, 23:37

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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.

Fig

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17 Oct 2006, 00:57

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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

yogeshsheth

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17 Oct 2006, 11:54

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FIG could you please expalin the above solution in detail..
Thanks!!

Fig

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Updated on: 17 Oct 2006, 13:50

Originally posted by Fig on 17 Oct 2006, 13:43.
Last edited by Fig on 17 Oct 2006, 13:50, edited 1 time in total.

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yogeshsheth

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.

yogeshsheth

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17 Oct 2006, 13:48

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Fig

yogeshsheth

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 :-D

Fig

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17 Oct 2006, 13:49

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U are welcomed

ugo_castelo

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17 Oct 2006, 14:17

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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.

navdeepbajwa

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02 Sep 2009, 12:27

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I think i should be 42

M1W2+.............................M1W7
.
.
.
M7W1+.............................M7W6

6*7=42

srivas

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13 Sep 2009, 12:05

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I got 1200 as the answer for the first question too .. But my working out was way too long.. Urs is really apt cicerone.. :-D

srivas

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13 Sep 2009, 12:21

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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)

srivas

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13 Sep 2009, 12:46

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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?

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

srivas

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13 Sep 2009, 12:55

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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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