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Re: To install cable television in a home, a certain cable company charges [#permalink]
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Stiv wrote:
To install cable television in a home, a certain cable company charges a basic fee of $30 plus a fee of $20 for each cable outlet installed in the home. How much did the cable company charge the Horace family for installing cable television in their home?

(1) The cable company installed three cable outlets in the Horace family home.
(2) The amount that the cable company charged the Horace family for installing cable television in their home was equivalent to an average ( arithmetic mean) charge of the $30 per cable outlet installed.


A logical solution:

Total charge = 30 + 20 * Number of Cable Outlets

(1) The cable company installed three cable outlets in the Horace family home.
Directly gives you the number of cable outlets. Sufficient

(2) The amount that the cable company charged the Horace family for installing cable television in their home was equivalent to an average ( arithmetic mean) charge of the $30 per cable outlet installed.
On the whole, the charge was equivalent to $30 for each outlet. The extra $10 was charged because $30 basic fee was spread across the cable outlets. To give $10 extra to each cable outlet, there must have been 3 cable outlets.
So total charge must have been 3*30 = 90
Sufficient.

Answer (D)
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Re: To install cable television in a home, a certain cable company charges [#permalink]
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

To install cable television in a home, a certain cable company charges a basic fee of $30 plus a fee of $20 for each cable outlet installed in the home. How much did the cable company charge the Horace family for installing cable television in their home?

(1) The cable company installed three cable outlets in the Horace family home.
(2) The amount that the cable company charged the Horace family for installing cable television in their home was equivalent to an average ( arithmetic mean) charge of the $30 per cable outlet installed.


Transforming the original condition and the question, amount that the cable company charge = c, number of each cable outlets installed in home = n. We have c=30+20n thus 2 variable (c,n) and 1 equation (c=30+20n) therefore we need 1 more equation to match the number of equations and variables. Since there is 1 each in 1) and 2), D is likely the answer.

In case of 1), n=3 therefore it is sufficient
In case of 2), 30+20n = 30n, n=3 is sufficient.
Therefore D is the answer.
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Re: To install cable television in a home, a certain cable company charges [#permalink]
My 2 cents on this:

A is easy to get suff. Let's move on

St2 --> Mean of 30 per outlet can be written as total charge/no of outlets. We know the total charge to be 30 + 20(x) with x being number of outlets. So then the equation stands as (30+20x)/x = 30. Then 30+20x = 30x; x =3

Total charge = 30 + 20(x) = 30 + 20*3 = 90; Ie. Suff.

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Re: To install cable television in a home, a certain cable company charges [#permalink]
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Re: To install cable television in a home, a certain cable company charges [#permalink]
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