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22 Feb 2009, 15:54
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22 Feb 2009, 17:43
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Here is my approach.......
It is given that there are 23 cent costing pencil and 21 cent costing pencil.
From stmt 1, it given that there are 6 pencils in total. From here we cannot say how many 23 cent pencils are there. It can range from 1 to 6. So insufficient.
From stmt 2, it is given the total cost pencils is 130. This means 23x + 21y = 130. Hum one eq and two variables......Is it cannot solve?
: Wait a minute......Can we not figure out some random guesses. We should know that here x and y are integers since they represent the number of pencils of each cost.
So we can have at the max 4 - 23 cents costing pencils. So the ans for x will be such that 1<=x<= 4 and 130-23x is divisible by 21. definitely we can find the value of x. Hence sufficient.
IMO B.
It is given that there are 23 cent costing pencil and 21 cent costing pencil.
From stmt 1, it given that there are 6 pencils in total. From here we cannot say how many 23 cent pencils are there. It can range from 1 to 6. So insufficient.
From stmt 2, it is given the total cost pencils is 130. This means 23x + 21y = 130. Hum one eq and two variables......Is it cannot solve?
: Wait a minute......Can we not figure out some random guesses. We should know that here x and y are integers since they represent the number of pencils of each cost. So we can have at the max 4 - 23 cents costing pencils. So the ans for x will be such that 1<=x<= 4 and 130-23x is divisible by 21. definitely we can find the value of x. Hence sufficient.
IMO B.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
