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tough PS n DS

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tough PS n DS [#permalink] New post 28 Jan 2006, 09:18
can someone pls solve n explain these problems?
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 [#permalink] New post 28 Jan 2006, 09:25
The area is 18

AD = 10

AE =5
CD=4
CE=3
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 [#permalink] New post 28 Jan 2006, 09:30
Before calculate you should establish why those triangles are equal.

In Ds it may hurt you doing otherwise.
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 [#permalink] New post 28 Jan 2006, 09:55
wait! first lemme get it straight, what is the area of that quad? l*b right

celix if u tellin area = 18 ==> AB*BC =6*3 , but since as per ur solution if cd=4 then cb =4 and then area will be 6*4 ...

where am i going wrong, please explain
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 [#permalink] New post 28 Jan 2006, 10:12
1. 18

Triangle ABD and triangle ECD are similar triangles (share the same angle)
Hence EC = 3 (3,4,5 triangle)

Also triangle AEF and ECD are similar triangles, since they are formed by parallel lines cut by a line (transversal as math geeks would call it)

Infact these traingles are equal since AE = ED = 5

Area of the quadrilateral = 1/2 (4) (3+6) = 18

2. C (It will not pass), assuming that each of the L's vote YES on the bill, Otherwise it is E.

From #1 we get total L's who vote = (55/100)*400*(85/100) = 187 (But we don't know about Rs and Ds.
From #2 we know that none of the Rs or Ds vote.

Hence total votes = 187 < 189. Hence C. (or E if we don't know each one will vote YES or NO)

3. 11 (div by 78) + 9 (div by 91) = 20.

78*10 = 780 (including 78 there are 10 numbers. Excluding 78 there are 9 of them)
78+78*2 = 936 < 1000. So 2 more available.

Hence No divisible by 78 =9+2 = 11
Similarly for 91 there are 10-1 = 9

Total = 20.

OA's please?
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 [#permalink] New post 28 Jan 2006, 10:24
Approximation is very fast for the third question

xxx is a multiple of either 78 or 91

You see 10*78 + 2*78 = 936

So there are 11 multiples of 78

Similiary for 91:

10*91=91

So there are 9 multples of 91


Next step are there multiples counted twice?

91= 7*13
78= 2*3*13

Both are equal if 2*3 and 7 have a common multiple, wich is 7 and 6 respectively. Notice that above you needn't to look out for common multiples above 999.

So 7*2*3*13=546, or 6*7*13 is counted twice


So 19 is the answer
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 [#permalink] New post 28 Jan 2006, 10:25
giddi77 wrote:
1. 18

Triangle ABD and triangle ECD are similar triangles (share the same angle)
Hence EC = 3 (3,4,5 triangle)

Also triangle AEF and ECD are similar triangles, since they are formed by parallel lines cut by a line (transversal as math geeks would call it)

Infact these traingles are equal since AE = ED = 5

Area of the quadrilateral = 1/2 (4) (3+6) = 18

2. C (It will not pass), assuming that each of the L's vote YES on the bill, Otherwise it is E.

From #1 we get total L's who vote = (55/100)*400*(85/100) = 187 (But we don't know about Rs and Ds.
From #2 we know that none of the Rs or Ds vote.

Hence total votes = 187 < 189. Hence C. (or E if we don't know each one will vote YES or NO)

3. 11 (div by 78) + 9 (div by 91) = 20.

78*10 = 780 (including 78 there are 10 numbers. Excluding 78 there are 9 of them)
78+78*2 = 936 < 1000. So 2 more available.

Hence No divisible by 78 =9+2 = 11
Similarly for 91 there are 10-1 = 9

Total = 20.

OA's please?


:cool u nailed them all , can u please explain the 3rd

how did u come to this conclusion ( i know its very silly) i am not able to understand it.
:dunnow
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 [#permalink] New post 28 Jan 2006, 10:27
Are you sure it's 20?
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 [#permalink] New post 28 Jan 2006, 10:28
allabout wrote:


Next step are there multiples counted twice?

91= 7*13
78= 2*3*13

Both are equal if 2*3 and 7 have a common multiple, wich is 7 and 6 respectively. Notice that above you needn't to look out for common multiples above 999.

So 7*2*3*13=546, or 6*7*13 is counted twice


So 19 is the answer


Tooo good allabout! Thanks for explaining it very clearly! I'll remember this.
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 [#permalink] New post 28 Jan 2006, 10:30
razrulz wrote:
wait! first lemme get it straight, what is the area of that quad? l*b right

celix if u tellin area = 18 ==> AB*BC =6*3 , but since as per ur solution if cd=4 then cb =4 and then area will be 6*4 ...

where am i going wrong, please explain


The area = (AB+CE)*BC/2
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 [#permalink] New post 28 Jan 2006, 10:33
hey all about, i got this question from a grp! someone had posted the questions and answers and asked for explanations. by far , her answer for the 3rd question was 20 , so i am not sure !

so i dont have a OA. I am sorry!
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 [#permalink] New post 28 Jan 2006, 10:36
razrulz wrote:
hey all about, i got this question from a grp! someone had posted the questions and answers and asked for explanations. by far , her answer for the 3rd question was 20 , so i am not sure !

so i dont have a OA. I am sorry!


Ok, the question may be ambiguous, since "different numbers" is omitted.

Generally 546 is one number, so the answer is 19 not 20.
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 [#permalink] New post 28 Jan 2006, 10:40
allabout wrote:
razrulz wrote:
hey all about, i got this question from a grp! someone had posted the questions and answers and asked for explanations. by far , her answer for the 3rd question was 20 , so i am not sure !

so i dont have a OA. I am sorry!


Ok, the question may be ambiguous, since "different numbers" is omitted.

Generally 546 is one number, so the answer is 19 not 20.


AYE AYE SIR ! :) thnks again
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 [#permalink] New post 28 Jan 2006, 18:53
According to me the two possible answers are 18 and 22.5. Since 18 is the only one given it should be it.
  [#permalink] 28 Jan 2006, 18:53
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