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Ezekiel went fishing on each day of last week starting on

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Ezekiel went fishing on each day of last week starting on [#permalink]

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New post 02 Jun 2009, 23:05
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Ezekiel went fishing on each day of last week starting on Monday and caught a total of 76 fish. Each day he either caught more or the same number of fish as the day before, and he never caught more than 12 fish in one day. Did Ezekiel catch more than 9 fish on Tuesday?

(1) Ezekiel caught 8 fish on Monday.

(2) Ezekiel caught 4 more fish on Wednesday than he did on Tuesday.
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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 00:13
Answer C


1, insufficient..

76-8 = 68..
for maximum fishes to be caught on tuesday.. 6x = 68 ==> x>11..
also, there is a possibility taht he caught only 8 fishes.. 5x = 60 ==> x =10..

2, insufficient..

for maximum fishes to be caught on tuesday .. 0 + x + 5x+20 = 76
=> 6x = 56 => x > 9
if the number of fishes caught on monday is more than 2 then X < 9..

Combined...

monday fishes caught = 8.. total remaining is 68

x + 5 ( X+4 ) = 68
6x+20 = 68
6x = 48
x = 6.. maximum caught on tuesday is 6.. hence less than 9..

Whats the OA ?

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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 00:16
Not C


Hrm I'm not following your logic.

Try drawing it out on a piece of paper and labelling the days and penciling in the conditions. Try to get arrangements with all the rules... maybe that will help
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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 00:30
D

1, Mon=8, Tue cannot be 9, because 9+4=13 (but no more than 12 fish per day), Tue must be 8, suff
2. The same logic
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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 00:36
sondenso wrote:
D

1, Mon=8, Tue cannot be 9, because 9+4=13 (but no more than 12 fish per day), Tue must be 9, suff
2. The same logic


Hrm.... What's your logic for (2)?
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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 00:52
Hades wrote:
sondenso wrote:
D

1, Mon=8, Tue cannot be 9, because 9+4=13 (but no more than 12 fish per day), Tue must be 9, suff
2. The same logic


Hrm.... What's your logic for (2)?



Logic1: Each day he either caught more or the same number of fish as the day before
Logic2: Ezekiel caught 4 more fish on Wednesday than he did on Tuesday
Tuesday=x, Wednesday=x+4 and
Logic3: he never caught more than 12 fish in one day

So x cannot be more than 8.
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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 03:50
If he caught 8 on Monday from (1) he could have gotten 8 or 10 on Tuesday.
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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 04:06
Hades wrote:
Ezekiel went fishing on each day of last week starting on Monday and caught a total of 76 fish. Each day he either caught more or the same number of fish as the day before, and he never caught more than 12 fish in one day. Did Ezekiel catch more than 9 fish on Tuesday?

(1) Ezekiel caught 8 fish on Monday.

(2) Ezekiel caught 4 more fish on Wednesday than he did on Tuesday.



(1)
8 , 8 ,12,12,12,12,12 --> Tue More than 9 fish ? --> No
8 , 10,10,12,12,12,12 --> Tue More than 9 fish ? --> Yes

multiple solutions
not sufficient

(2)
Max fish caught on any day =12
assume that Wed Ez caught 12 fish .. Tues day Ez caught 8 fish
So <9
B is the answer
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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 09:43
Conditions as stated in the question:

\(\frac{\frac{}{Monday} \leq \frac{}{Tuesday} \leq \frac{}{Wednesday} \leq \frac{}{Thursday} \leq \frac{}{Friday} \leq \frac{}{Saturday} \leq \frac{}{Sunday}}{\Sigma{76 Fish}}\)

(1)

\(\frac{\frac{8}{Monday} \leq \frac{8}{Tuesday} \leq \frac{12}{Wednesday} \leq \frac{12}{Thursday} \leq \frac{12}{Friday} \leq \frac{12}{Saturday} \leq \frac{12}{Sunday}}{\Sigma{76 Fish}}\)
\(\longrightarrow NO\)

\(\frac{\frac{8}{Monday} \leq \frac{10}{Tuesday} \leq \frac{10}{Wednesday} \leq \frac{12}{Thursday} \leq \frac{12}{Friday} \leq \frac{12}{Saturday} \leq \frac{12}{Sunday}}{\Sigma{76 Fish}}\)
\(\longrightarrow YES\)

\(\longrightarrow INSUFFICIENT\).

(2)

Let \(X\) represent the # of fish caught Tuesday. Then:

\(\frac{\frac{}{Monday} \leq \frac{X}{Tuesday} \leq \frac{X+4}{Wednesday} \leq \frac{}{Thursday} \leq \frac{}{Friday} \leq \frac{}{Saturday} \leq \frac{}{Sunday}}{\Sigma{76 Fish in Total}}\)


Let's figure out what the other values can be. Let's assign some variables.

\(\frac{\frac{A}{Monday} \leq \frac{X}{Tuesday} \leq \frac{X+4}{Wednesday} \leq \frac{B}{Thursday} \leq \frac{C}{Friday} \leq \frac{D}{Saturday} \leq \frac{E}{Sunday}}{\Sigma{76 Fish}}\)

We want:

\(A + X + (X + 4) + B + C + D + E = 76\)
\(0 \leq A \leq X \leq (X+4) \leq B \leq C \leq D \leq E \leq 12\)

What are the maximum possible values for each? 12. Substituting in we get:

\(0 \leq A \leq X \leq (X+4) \leq B \leq C \leq D \leq 12 \leq 12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq B \leq C \leq 12 \leq 12 \leq 12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq B \leq 12 \leq 12 \leq 12 \leq 12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(X+4=12 \longrightarrow X=8\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq 8 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(X=8\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq 8 \leq 8 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(A=8\)
\(X=8\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)


\(A + X + (X + 4) + B + C + D + E = 8 + 8 + (8+4) + 12 + 12 + 12 + 12 = 8 + 8 + 60 = 76\)

So there is only one possible arrangement possible, because if we take anything less than the maximum, the days won't sum to 76 as required in the question.

\(\frac{\frac{8}{Monday} \leq \frac{8}{Tuesday} \leq \frac{12}{Wednesday} \leq \frac{12}{Thursday} \leq \frac{12}{Friday} \leq \frac{12}{Saturday} \leq \frac{12}{Sunday}}{\Sigma{76 Fish}}\)

\(\longrightarrow SUFFICIENT\)


Final Answer, \(B\).

\(\surd\)
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Last edited by Hades on 03 Jun 2009, 12:12, edited 2 times in total.

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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 11:58
Hades wrote:
Conditions as stated in the question:

\(\frac{\frac{}{Monday} \leq \frac{}{Tuesday} \leq \frac{}{Wednesday} \leq \frac{}{Thursday} \leq \frac{}{Friday} \leq \frac{}{Saturday} \leq \frac{}{Sunday}}{\Sigma{76 Fish}}\)

(1)

Only 1 possible arrangement:

\(\frac{\frac{8}{Monday} \leq \frac{8}{Tuesday} \leq \frac{12}{Wednesday} \leq \frac{12}{Thursday} \leq \frac{12}{Friday} \leq \frac{12}{Saturday} \leq \frac{12}{Sunday}}{\Sigma{76 Fish}}\)

There is no other possible arrangment (go ahead and play around with it).

\(\longrightarrow SUFFICIENT\).

(2)

Let \(X\) represent the # of fish caught Tuesday. Then:

\(\frac{\frac{}{Monday} \leq \frac{X}{Tuesday} \leq \frac{X+4}{Wednesday} \leq \frac{}{Thursday} \leq \frac{}{Friday} \leq \frac{}{Saturday} \leq \frac{}{Sunday}}{\Sigma{76 Fish in Total}}\)


Let's figure out what the other values can be. Let's assign some variables.

\(\frac{\frac{A}{Monday} \leq \frac{X}{Tuesday} \leq \frac{X+4}{Wednesday} \leq \frac{B}{Thursday} \leq \frac{C}{Friday} \leq \frac{D}{Saturday} \leq \frac{E}{Sunday}}{\Sigma{76 Fish}}\)

We want:

\(A + X + (X + 4) + B + C + D + E = 76\)
\(0 \leq A \leq X \leq (X+4) \leq B \leq C \leq D \leq E \leq 12\)

What are the maximum possible values for each? 12. Substituting in we get:

\(0 \leq A \leq X \leq (X+4) \leq B \leq C \leq D \leq 12 \leq 12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq B \leq C \leq 12 \leq 12 \leq 12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq B \leq 12 \leq 12 \leq 12 \leq 12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(X+4=12 \longrightarrow X=8\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq 8 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(X=8\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq 8 \leq 8 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(A=8\)
\(X=8\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)


\(A + X + (X + 4) + B + C + D + E = 8 + 8 + (8+4) + 12 + 12 + 12 + 12 = 8 + 8 + 60 = 76\)

So there is only one possible arrangement possible, because if we take anything less than the maximum, the days won't sum to 76 as required in the question. This is also the same arrangement as in (1).

\(\frac{\frac{8}{Monday} \leq \frac{8}{Tuesday} \leq \frac{12}{Wednesday} \leq \frac{12}{Thursday} \leq \frac{12}{Friday} \leq \frac{12}{Saturday} \leq \frac{12}{Sunday}}{\Sigma{76 Fish}}\)

\(\longrightarrow SUFFICIENT\)


Final Answer, \(D\).

\(\surd\)


Statement 1 is not sufficient.
How about below scenario?
8 , 10,10,12,12,12,12 --> Tue More than 9 fish ? --> Yes
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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 12:10
x2suresh wrote:
Hades wrote:
Conditions as stated in the question:

\(\frac{\frac{}{Monday} \leq \frac{}{Tuesday} \leq \frac{}{Wednesday} \leq \frac{}{Thursday} \leq \frac{}{Friday} \leq \frac{}{Saturday} \leq \frac{}{Sunday}}{\Sigma{76 Fish}}\)

(1)

Only 1 possible arrangement:

\(\frac{\frac{8}{Monday} \leq \frac{8}{Tuesday} \leq \frac{12}{Wednesday} \leq \frac{12}{Thursday} \leq \frac{12}{Friday} \leq \frac{12}{Saturday} \leq \frac{12}{Sunday}}{\Sigma{76 Fish}}\)

There is no other possible arrangment (go ahead and play around with it).

\(\longrightarrow SUFFICIENT\).

(2)

Let \(X\) represent the # of fish caught Tuesday. Then:

\(\frac{\frac{}{Monday} \leq \frac{X}{Tuesday} \leq \frac{X+4}{Wednesday} \leq \frac{}{Thursday} \leq \frac{}{Friday} \leq \frac{}{Saturday} \leq \frac{}{Sunday}}{\Sigma{76 Fish in Total}}\)


Let's figure out what the other values can be. Let's assign some variables.

\(\frac{\frac{A}{Monday} \leq \frac{X}{Tuesday} \leq \frac{X+4}{Wednesday} \leq \frac{B}{Thursday} \leq \frac{C}{Friday} \leq \frac{D}{Saturday} \leq \frac{E}{Sunday}}{\Sigma{76 Fish}}\)

We want:

\(A + X + (X + 4) + B + C + D + E = 76\)
\(0 \leq A \leq X \leq (X+4) \leq B \leq C \leq D \leq E \leq 12\)

What are the maximum possible values for each? 12. Substituting in we get:

\(0 \leq A \leq X \leq (X+4) \leq B \leq C \leq D \leq 12 \leq 12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq B \leq C \leq 12 \leq 12 \leq 12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq B \leq 12 \leq 12 \leq 12 \leq 12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq X \leq (X+4) \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(X+4=12 \longrightarrow X=8\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq A \leq 8 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(X=8\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)

\(0 \leq 8 \leq 8 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12 \leq 12\)
\(A=8\)
\(X=8\)
\(B=12\)
\(C=12\)
\(D=12\)
\(E=12\)


\(A + X + (X + 4) + B + C + D + E = 8 + 8 + (8+4) + 12 + 12 + 12 + 12 = 8 + 8 + 60 = 76\)

So there is only one possible arrangement possible, because if we take anything less than the maximum, the days won't sum to 76 as required in the question. This is also the same arrangement as in (1).

\(\frac{\frac{8}{Monday} \leq \frac{8}{Tuesday} \leq \frac{12}{Wednesday} \leq \frac{12}{Thursday} \leq \frac{12}{Friday} \leq \frac{12}{Saturday} \leq \frac{12}{Sunday}}{\Sigma{76 Fish}}\)

\(\longrightarrow SUFFICIENT\)


Final Answer, \(D\).

\(\surd\)


Statement 1 is not sufficient.
How about below scenario?
8 , 10,10,12,12,12,12 --> Tue More than 9 fish ? --> Yes


You're absolutely right, I even came up with that case myself. I'll fix it
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Re: Tough DS 4 [#permalink]

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New post 03 Jun 2009, 23:26
its actually lot simpler than i thought it was..

statement 1 : insufficinet.
statement 2 : max fishes caught in a day is 12, that means max fishes that can be caught on tuesday is 8..
hence answer B..

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Re: Tough DS 4   [#permalink] 03 Jun 2009, 23:26
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