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# Each of the dinners served at a banquet was either chicken or beef

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Each of the dinners served at a banquet was either chicken or beef  [#permalink]

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19 Jul 2016, 04:42
5
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Difficulty:

75% (hard)

Question Stats:

58% (02:06) correct 42% (02:25) wrong based on 713 sessions

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Each of the dinners served at a banquet was either chicken or beef or fish. The ratio of the number of chicken dinners to the number of beef dinners to the number of fish dinners served at the banquet was 7:5:2, respectively. If there were more than 5 fish dinners served at the banquet, what was the total number of dinners served at the banquet ?

(1) The total number of beef dinners and fish dinners served at the banquet was less than 30.
(2) The number of chicken dinners served at the banquet was less than 25.
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Re: Each of the dinners served at a banquet was either chicken or beef  [#permalink]

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19 Jul 2016, 05:26
14
1
The number of chicken dinners = 7x
Number of beef dinners = 5x
Number of fish dinners = 2x
Total dinners = 14x and x>2 (as fish dinners are greater than 5) Need to find x.

Statement 1:
Given total no of beef dinners and fish dinners were less than 30.
7x < 30 implies that x is either 3, 4 (As x has to be greater than 2)

This is not sufficient

Statement 2
Chicken dinners, i.e. 7x<25 or x<4
From the question stem we have x>2. With this we have a unique answer as x = 3.

So this statement is sufficient and answer is B.

Hope this helps!
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Each of the dinners served at a banquet was either chicken or beef  [#permalink]

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Updated on: 19 Jul 2016, 05:27
babuvgmat wrote:
Each of the dinners served at a banquet was either chicken or beef or fish. The ratio of the number of chicken dinners to the number of beef dinners to the number of fish dinners served at the banquet was 7:5:2, respectively. If there were more than 5 fish dinners served at the banquet, what was the total number of dinners served at the banquet ?

(1) The total number of beef dinners and fish dinners served at the banquet was less than 30.
(2) The number of chicken dinners served at the banquet was less than 25.

I would like to see the actual solution for this question.

Given ration of C:B:F is 7X:5X:2X , if there were more than 5 fish dinners and asked us to find the total number of dishes served .

Stat 1: The total number of B + F < 30.

9X < 30....then in order to get some that is divisible less than 30 is 27 or 18 or 9.. then X = 3 or 2 or 1...Insufficient.

Stat 2: The number of C < 25.. 7X < 25... then number can be 21 or 14 or 7...then X can be 3 or 2 or 1..Insufficient..

Stat 1 + Stat 2 : Not sufficient...

Waiting for some more responses...not sure if OA B is correct answer..it has to be C or E.

Originally posted by msk0657 on 19 Jul 2016, 05:22.
Last edited by msk0657 on 19 Jul 2016, 05:27, edited 1 time in total.
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Re: Each of the dinners served at a banquet was either chicken or beef  [#permalink]

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21 Jul 2016, 10:38
1
msk0657 wrote:
babuvgmat wrote:
Each of the dinners served at a banquet was either chicken or beef or fish. The ratio of the number of chicken dinners to the number of beef dinners to the number of fish dinners served at the banquet was 7:5:2, respectively. If there were more than 5 fish dinners served at the banquet, what was the total number of dinners served at the banquet ?

(1) The total number of beef dinners and fish dinners served at the banquet was less than 30.
(2) The number of chicken dinners served at the banquet was less than 25.

I would like to see the actual solution for this question.

Given ration of C:B:F is 7X:5X:2X , if there were more than 5 fish dinners and asked us to find the total number of dishes served .

Stat 1: The total number of B + F < 30.

9X < 30....then in order to get some that is divisible less than 30 is 27 or 18 or 9.. then X = 3 or 2 or 1...Insufficient.

Stat 2: The number of C < 25.. 7X < 25... then number can be 21 or 14 or 7...then X can be 3 or 2 or 1..Insufficient..

Stat 1 + Stat 2 : Not sufficient...

Waiting for some more responses...not sure if OA B is correct answer..it has to be C or E.

You have missed a case which states "If there were more than 5 fish dinners served at the banquet" => 2X > 5 or X>2.5. Hence X could be either 3 or 4 for statement 1 and X = 3 for statement 2. Hence answer is B.
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Each of the dinners served at a banquet was either chicken or beef  [#permalink]

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27 Jan 2017, 17:13
2
Quote:
Each of the dinners served at a banquet was either chicken or beef or fish. The ratio of the number of chicken dinners to the number of beef dinners to the number of fish dinners served at the banquet was 7:5:2, respectively. If there were more than 5 fish dinners served at the banquet, what was the total number of dinners served at the banquet ?

(1) The total number of beef dinners and fish dinners served at the banquet was less than 30.
(2) The number of chicken dinners served at the banquet was less than 25.

I felt this was a pretty easy question.

So C:B:F ratio is 7:5:2.
W/ a restriction that states that fish is GREATER than 5. So F > 5.

Now moving on to S1

S1: The total number of beef dinners and fish dinners served at the banquet was less than 30.

Therefore, we can write B + F < 30.
Okay, so it can't be 5 + 2 (the original ratio of beef and fish because there was more than 5 fish dinners.)
So we move on to 15 + 6 < 30. Good
Try 20 + 8 < 30. Also good, therefore S1 is Not sufficient.

Statement 2:
The number of chicken dinner was less than 25.
Okay, so Chicken must be in multiples of 7.

It can be 7, 14, 21. However, we must remember the number of fish dinners too! The original restriction is that fish must be greater than 5!

So it can be 2, 4, 6, 8. However, you'll notice that chicken dinners can really be either 7, 14, or 21. However, the first two iterations (7, and 14) correspond to fish dinner of 2, and 4 which are both less than 5. Therefore, there must be 21 chicken dinners, 6 fish dinners, and 15 beef dinners.

SUFFICIENT (B)
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Re: Each of the dinners served at a banquet was either chicken or beef  [#permalink]

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10 Apr 2017, 10:29
1
1
babuvgmat wrote:
Each of the dinners served at a banquet was either chicken or beef or fish. The ratio of the number of chicken dinners to the number of beef dinners to the number of fish dinners served at the banquet was 7:5:2, respectively. If there were more than 5 fish dinners served at the banquet, what was the total number of dinners served at the banquet ?

(1) The total number of beef dinners and fish dinners served at the banquet was less than 30.
(2) The number of chicken dinners served at the banquet was less than 25.

F>5

1)
7:5:2
C:B:F
21:15:6 (B+F=21 - allowed)
28:20:8 (B+F=28 - allowed)

Not suff

2)
7:5:2
C:B:F
21:15:6 (C=21 - allowed)
28:20:8 (C=28 - not allowed)

Suff. B

kudos if you agree, comment if you have a better method.
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Re: Each of the dinners served at a banquet was either chicken or beef  [#permalink]

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10 Sep 2017, 19:14
Here's how I thought about it. This question is testing whether or not you are paying attention to endpoints.

Chicken : Beef : Fish = 7m : 5m : 2m --> total is 14m where m is the multiplier.

There are more than 5 fish dinners, so 2m > 5 --> m can be 3 or anything greater than 3. It's reasonable to assume we are talking about integers for the multiplier as the question gives no indication that partial dinners are possible.

Statement 1: Beef + Fish = 5m + 2m = 7m and statement says 7m < 30.

This means that m can be as large as 4. That means (given the constraint in the question) that m can be 3 or 4, which gives us two different numbers for the total.

Statement 2: Chicken = 7m and statement says 7m < 25.

In this case, m can be 3 (the restriction in the question makes it impossible for it to be less than 3). This statement is therefore sufficient as there is only one value m can take, which means that total can only take one value (14*3 = 42).
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Re: Each of the dinners served at a banquet was either chicken or beef  [#permalink]

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13 Jul 2018, 14:23
1
babuvgmat wrote:
Each of the dinners served at a banquet was either chicken or beef or fish. The ratio of the number of chicken dinners to the number of beef dinners to the number of fish dinners served at the banquet was 7:5:2, respectively. If there were more than 5 fish dinners served at the banquet, what was the total number of dinners served at the banquet ?

(1) The total number of beef dinners and fish dinners served at the banquet was less than 30.
(2) The number of chicken dinners served at the banquet was less than 25.

We can let the number of chicken, beef and fish dinners be 7x, 5x, 2x, respectively, where x is a positive integer. We are given that more than 5 fish dinners served at the banquet; therefore, x must be at least 3. We need to determine the total number of dinners served at the banquet. That is, we need to determine the value of 7x + 5x + 2x = 14x.

Statement One Alone:

The total number of beef dinners and fish dinners served at the banquet was less than 30.

That is, 5x + 2x < 30, or 7x < 30. So x < 30/7 = 4 2/8. Since x must be an integer, x ≤ 4.

Also, since x ≥ 3, x could be either 3 or 4. Statement one alone is not sufficient.

Statement Two Alone:

The number of chicken dinners served at the banquet was less than 25.

That is, 7x < 25. So x < 25/7 = 3 4/7. Since x must be an integer, x ≤ 3.

Also, since x ≥ 3, x must be 3 and the number of dinners served is 17 x 3 = 51. Statement two alone is sufficient.

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Each of the dinners served at a banquet was either chicken or beef  [#permalink]

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28 Aug 2018, 23:01
Hi,

Let Chicken Dinners be 7x, beef Dinners 5x, Fish Dinners 2x, then total dinners will be 14x. We are given that 2x>5. And we are asked what is the total no of diners which is 14x. So we need to find x.

Stmt 1: 7x<30 and 2x>5

So if we add the II inequality and I inequality we have
7x<30
-2x<-5
5x<25, so x< 5.
We get range of x as
2<x<5 , So x can be 3,4 So insufficient

Stmt 2: 7x<25 and 2x>5
again if we add the II in inequality and I inequality we have
7x<25
-2x<-5
5x<20
So we have x<4.
2<x<4 this means x=3.

Hence B

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Each of the dinners served at a banquet was either chicken or beef   [#permalink] 28 Aug 2018, 23:01
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