Source: Princeton Review
Kevin buys beer in bottles and cans. He pays $1.00 for each can of beer and $1.50 for each bottle of beer. If he buys a total of 15 bottles and cans of beer, how many bottles of beer did he buy?
1) Kevin spent a total of $18.00 on beer.
2) Kevin bought 3 more cans of beer than bottles of beer.
The OA is D.
Kevin buys beer in bottles and cans. He pays $1.00 for each
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Target question: How many bottles of beer did Kevin buy?BTGmoderatorLU wrote:Source: Princeton Review
Kevin buys beer in bottles and cans. He pays $1.00 for each can of beer and $1.50 for each bottle of beer. If he buys a total of 15 bottles and cans of beer, how many bottles of beer did he buy?
1) Kevin spent a total of $18.00 on beer.
2) Kevin bought 3 more cans of beer than bottles of beer.
The OA is D.
Given: Kevin pays $1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer
Let C = the NUMBER of Cans that Kevin bought
Let B = the NUMBER of Bottles that Kevin bought
So, we can write: C + B = 15
Statement 1: Kevin spent a total of $18.00 on beer
The COST of C cans = ($1.00)C = 1C
The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer: D
Cheers,
Brent