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A children’s theater sells tickets to a show. Tickets for children...

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A children’s theater sells tickets to a show. Tickets for children...  [#permalink]

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New post 19 Nov 2018, 18:04
6
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

38% (02:24) correct 62% (01:33) wrong based on 133 sessions

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A children’s theater sells tickets to a show. Tickets for children cost $10 and tickets for adults cost $35. If ticket revenues from the last performance were $390, and everyone at the performance had a ticket, how many people were at the performance?

(1) The number of children was more than 3 times the number of adults.

(2) The maximum capacity of the theater is 32 seats.


Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Looking for a time efficient way to solve this one. Thank you!
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A children’s theater sells tickets to a show. Tickets for children...  [#permalink]

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New post Updated on: 20 Nov 2018, 08:50
1
1
Dorsano505 wrote:
A children’s theater sells tickets to a show. Tickets for children cost $10 and tickets for adults cost $35. If ticket revenues from the last performance were $390, and everyone at the performance had a ticket, how many people were at the performance?

(1) The number of children was more than 3 times the number of adults.

(2) The maximum capacity of the theater is 32 seats.


Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Looking for a time efficient way to solve this one. Thank you!


The answer is C.
This question can be solved precisely - actually solving.

From the question stem, we can build an equation with two variables (10x+35y=390).
(1) gives us an inequality, not an equation (x+y<32), so we can't know if it's solvable without actually trying.
The data works for 39 children and 0 adults (39*10+0=390, or two adults and 32 children (35*2+32*10=70+320=390). Two different possible solution - insufficient!
(2) only gives us an inequality, not an equation (x+y<32), so we can't know if it's solvable without actually trying.
This data works for four children and 10 adults (4*10+35*10=390). It also works for 8 adults and 11 children (11*10+8*35=110+280=390). Two different possible solution - insufficient!

Combined - this works for four adults and 25 children (4*35+25*10=140+250=390). Any less adults and there will be more than 32 people (0 adults+ 39 children, 2 adults+35 children). Any more adults and there won't be more than three times as many children (6 adults, 18 children - 6*34+18*10=210+180=390). This is the only solution - sufficient!
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Originally posted by DavidTutorexamPAL on 20 Nov 2018, 00:42.
Last edited by DavidTutorexamPAL on 20 Nov 2018, 08:50, edited 2 times in total.
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Re: A children’s theater sells tickets to a show. Tickets for children...  [#permalink]

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New post 20 Nov 2018, 06:33
1
DavidTutorexamPAL wrote:
Dorsano505 wrote:
A children’s theater sells tickets to a show. Tickets for children cost $10 and tickets for adults cost $35. If ticket revenues from the last performance were $390, and everyone at the performance had a ticket, how many people were at the performance?

(1) The number of children was more than 3 times the number of adults.

(2) The maximum capacity of the theater is 32 seats.


Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Looking for a time efficient way to solve this one. Thank you!


The answer is A.
This question can be solved part logically, and part precisely - actually solving.

From the question stem, we can build an equation with two variables (10x+35y=390).
(1) gives us another equation (3x=y), giving us two equations with two variables - this is definitely solvable! B, C and E are eliminated.
(2) only gives us an inequality, not an equation (x+y<32), so we can't know if it's solvable with actually trying.
This data works for four children and 10 adults (4*10+35*10=390). It also works for 8 adults and 11 children (11*10+8*35=110+280=390). Two different possible solution - insufficient!


(1) Doesn't say it's equal, it says it is >=
Then how to solve?
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Re: A children’s theater sells tickets to a show. Tickets for children...  [#permalink]

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New post 20 Nov 2018, 07:28
1
Correct answer should be C.
Both statements alone are insufficient. When combined, note the following:

When adult tickets sold = 4, children tickets sold = 25. Note that 25/4 > 3

This is the only possible combination.

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Re: A children’s theater sells tickets to a show. Tickets for children...  [#permalink]

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New post 20 Nov 2018, 08:51
1
RK007 wrote:
DavidTutorexamPAL wrote:
Dorsano505 wrote:
A children’s theater sells tickets to a show. Tickets for children cost $10 and tickets for adults cost $35. If ticket revenues from the last performance were $390, and everyone at the performance had a ticket, how many people were at the performance?

(1) The number of children was more than 3 times the number of adults.

(2) The maximum capacity of the theater is 32 seats.


Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Looking for a time efficient way to solve this one. Thank you!


The answer is A.
This question can be solved part logically, and part precisely - actually solving.

From the question stem, we can build an equation with two variables (10x+35y=390).
(1) gives us another equation (3x=y), giving us two equations with two variables - this is definitely solvable! B, C and E are eliminated.
(2) only gives us an inequality, not an equation (x+y<32), so we can't know if it's solvable with actually trying.
This data works for four children and 10 adults (4*10+35*10=390). It also works for 8 adults and 11 children (11*10+8*35=110+280=390). Two different possible solution - insufficient!


(1) Doesn't say it's equal, it says it is >=
Then how to solve?


you are right - good catch! thanks. Modified my answer above - look now.
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Re: A children’s theater sells tickets to a show. Tickets for children...  [#permalink]

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New post 29 Nov 2018, 01:31
C is the answer

(1) gives us an inequality x+y<32 so we can't know if it's solvable without actually trying.

(2) an inequality x+y<32 so we can't know answer

Combined four adults and 25 children (4*35+25*10=140+250=390).
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Re: A children’s theater sells tickets to a show. Tickets for children...  [#permalink]

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New post 29 Nov 2018, 12:23
Dorsano505 wrote:
A children’s theater sells tickets to a show. Tickets for children cost $10 and tickets for adults cost $35. If ticket revenues from the last performance were $390, and everyone at the performance had a ticket, how many people were at the performance?

(1) The number of children was more than 3 times the number of adults.

(2) The maximum capacity of the theater is 32 seats.


Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Looking for a time efficient way to solve this one. Thank you!


Let c = number of children and a = number of adults.

We know 10c + 35a = 390, and c and a must be integers because these are human beings.

Therefore we know the following are possible solutions
C A
39 0
32 2
25 4
18 6
11 8
4 10

S1) c > 3a. We have multiple solutions satisfying this in our list. INSUFFICIENT
S2) c+a < 32. Again we have multiple solutions satisfying this in our list. INSUFFICIENT
Together) Only 25 and 4 satisfy both S1 and S2. Total = 25+4 = 29. SUFFICIENT

Answer C
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Re: A children’s theater sells tickets to a show. Tickets for children... &nbs [#permalink] 29 Nov 2018, 12:23
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