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# Joe sells twice as many \$10 tickets as Sue and Sue sells

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Senior Manager
Joined: 02 Oct 2005
Posts: 297

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Joe sells twice as many \$10 tickets as Sue and Sue sells [#permalink]

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09 Nov 2005, 10:42
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Joe sells twice as many \$10 tickets as Sue and Sue sells three times as many \$5 tickets as Joe. How many tickets did Joe sell? (tickets only come in \$10 or \$5)
(1) Sue sold a total of 35 tickets.
(2) Together Joe and Sue sold 70 tickets for \$500.

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VP
Joined: 22 Aug 2005
Posts: 1111

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Location: CA

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09 Nov 2005, 10:49
B.

let:
x = number of \$10 tickets Sue sold
y = number of \$5 tickets Joe sold

given:
joe sold: 2x + y tickets
Sue sold: x + 3y tickets

we need to find 2x+y

(1) x + 3y = 35
cannot get anything. two variables one equation.

(2) gives two equations:
3x + 4y = 70
30x + 20y = 500

solve for x,y and get 2x+y

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Current Student
Joined: 23 Oct 2005
Posts: 243

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Schools: Cranfield SOM

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09 Nov 2005, 12:26
duttsit wrote:
B.

let:
x = number of \$10 tickets Sue sold
y = number of \$5 tickets Joe sold

given:
joe sold: 2x + y tickets
Sue sold: x + 3y tickets

we need to find 2x+y

(1) x + 3y = 35
cannot get anything. two variables one equation.

(2) gives two equations:
3x + 4y = 70
30x + 20y = 500

solve for x,y and get 2x+y

I was thinking C. But you are correct. Answer has to be (b).

I did not realize that you can form an equation out of the total sales.

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Senior Manager
Joined: 02 Oct 2005
Posts: 297

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09 Nov 2005, 21:03
Thats correct, OA is B. I was about to choose C, but then thought about the available option again and found the answer to be B.

Just wanted to post it so that forum members does not fall for the wrong answer.

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Senior Manager
Joined: 07 Jul 2005
Posts: 402

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09 Nov 2005, 21:20
(1) Insufficient.

We get only 3 equations, but we have 4 unknowns

(2) Sufficient.

We have 4 equations and 4 unknowns.

I pick B.

Kudos [?]: 61 [0], given: 0

09 Nov 2005, 21:20
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