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03 Jan 2016, 13:48
00:00

Difficulty:

55% (hard)

Question Stats:

64% (01:56) correct 36% (01:39) wrong based on 86 sessions

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Mr. Jones spends $76 on movie tickets for a group of adults and children. How many children’s tickets did he purchase? (1) Adult movie tickets cost$ 11 each and children’s movie tickets cost $7 each. (2) Mr. Jones bought two more adult tickets than children’s tickets. _________________ Senior Manager Joined: 10 Mar 2013 Posts: 461 Location: Germany Concentration: Finance, Entrepreneurship Schools: WHU MBA"20 (A$)
GMAT 1: 580 Q46 V24
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Re: Mr. Jones spends $76 on movie tickets for a group of adults and child [#permalink] ### Show Tags 03 Jan 2016, 16:31 Bunuel wrote: Mr. Jones spends$ 76 on movie tickets for a group of adults and children. How many children’s tickets did he purchase?

(1) Adult movie tickets cost $11 each and children’s movie tickets cost$ 7 each.
(2) Mr. Jones bought two more adult tickets than children’s tickets.

(1) 11a + 7c=76 --> $$c=\frac{76-11a}{7}$$, so we have here a unique combination a=5, c=3 Sufficient
(2) Clearly not sufficient, as we don't know the total amount

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Re: Mr. Jones spends $76 on movie tickets for a group of adults and child [#permalink] ### Show Tags 07 Jan 2016, 06:36 BrainLab wrote: Bunuel wrote: Mr. Jones spends$ 76 on movie tickets for a group of adults and children. How many children’s tickets did he purchase?

(1) Adult movie tickets cost $11 each and children’s movie tickets cost$ 7 each.
(2) Mr. Jones bought two more adult tickets than children’s tickets.

(1) 11a + 7c=76 --> $$c=\frac{76-11a}{7}$$, so we have here a unique combination a=5, c=3 Sufficient
(2) Clearly not sufficient, as we don't know the total amount

We know the total 76$so D 5 & 3 resp Math Expert Joined: 02 Aug 2009 Posts: 8281 Re: Mr. Jones spends$ 76 on movie tickets for a group of adults and child  [#permalink]

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07 Jan 2016, 07:19
paidlukkha wrote:
BrainLab wrote:
Bunuel wrote:
Mr. Jones spends $76 on movie tickets for a group of adults and children. How many children’s tickets did he purchase? (1) Adult movie tickets cost$ 11 each and children’s movie tickets cost $7 each. (2) Mr. Jones bought two more adult tickets than children’s tickets. (1) 11a + 7c=76 --> $$c=\frac{76-11a}{7}$$, so we have here a unique combination a=5, c=3 Sufficient (2) Clearly not sufficient, as we don't know the total amount Answer A We know the total 76$ so D
5 & 3 resp

Hi,
you may know the total amount and the info that there are 2 more adult tickets than children's tickets..
but till we know some relation or values of individual tickets, we cannot say anything about the final number..
you are unconsciously carrying forward the info of statement 1 to statement 2..
so 2 is not sufficient..
A
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Re: Mr. Jones spends $76 on movie tickets for a group of adults and child [#permalink] ### Show Tags 07 Jan 2016, 09:45 Hi, you may know the total amount and the info that there are 2 more adult tickets than children's tickets.. but till we know some relation or values of individual tickets, we cannot say anything about the final number.. you are unconsciously carrying forward the info of statement 1 to statement 2.. so 2 is not sufficient.. A[/quote] No 1$ each
39 children
76$Intern Joined: 02 Jul 2015 Posts: 34 WE: Investment Banking (Investment Banking) Re: Mr. Jones spends$ 76 on movie tickets for a group of adults and child  [#permalink]

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07 Jan 2016, 10:14

With statement B, we don't know the prices of each ticket.
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Re: Mr. Jones spends $76 on movie tickets for a group of adults and child [#permalink] ### Show Tags 22 Mar 2019, 01:44 Bunuel wrote: Mr. Jones spends$ 76 on movie tickets for a group of adults and children. How many children’s tickets did he purchase?

(1) Adult movie tickets cost $11 each and children’s movie tickets cost$ 7 each.
(2) Mr. Jones bought two more adult tickets than children’s tickets.

#1
11x+7y= 76
so
x=5 & y =3 sufficient
#2
x=2y
insufficient
IMO A