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Nancy shopped at four department stores and spent an average of $80 p
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03 Feb 2022, 07:30
03 Feb 2022, 07:30
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Nancy shopped at four department stores and spent an average of $80 per store. If she wants to average no more than $70 per store over a total of six stores, what is the most she can average at the two remaining stores?
A $40
B $50
C $55
D $65
E $70
A $40
B $50
C $55
D $65
E $70
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Re: Nancy shopped at four department stores and spent an average of $80 p
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03 Feb 2022, 11:03
03 Feb 2022, 11:03
1
Carcass wrote:
Nancy shopped at four department stores and spent an average of $80 per store. If she wants to average no more than $70 per store over a total of six stores, what is the most she can average at the two remaining stores?
A $40
B $50
C $55
D $65
E $70
A $40
B $50
C $55
D $65
E $70
Nancy shopped at four department stores and spent an average of $80 per store.
We can make things easy for ourselves by saying that Nancy spent exactly $80 at each of the four stores.
If she wants to average no more than $70 per store over a total of six stores, what is the most she can average at the two remaining stores?
Let x = the amount spent at the 5th store
Let y = the amount spent at the 6th store
If the average money spent per store (at all 6 stores) is no more than $70, we can write: \(\frac{80+80+80+80+x+y}{6}≤70\)
Simplify: \(\frac{320+x+y}{6}≤70\)
Multiply both sides of the inequality by 6 to get: \(320+x+y≤420\)
Subtract 320 from both sides to get: \(x+y≤100\)
We want to determine the average spent at the two remaining stores.
In other words we want to find the value of \(\frac{x + y}{2}\)
So, we can just take the inequality \(x+y≤100\) and divide both sides by 2 to get: \(\frac{x+y}{2}≤50\)
Answer: B
gmatclubot
Re: Nancy shopped at four department stores and spent an average of $80 p [#permalink]
03 Feb 2022, 11:03
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