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Darden2010
Joined: 30 Mar 2007
Last visit: 22 Jul 2020
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Schools: Darden '15
GMAT Date: 12-07-2011
Updated on: 24 Jun 2008, 10:12
Originally posted by Darden2010 on 28 Nov 2007, 04:11.
Last edited by Darden2010 on 24 Jun 2008, 10:12, edited 1 time in total.
Last edited by Darden2010 on 24 Jun 2008, 10:12, edited 1 time in total.
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28 Nov 2007, 04:17
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I guess it's easy to assume that you need 1 - but if you think about it, 2 already tells you that there are 6 pencils. 130/21 = 6+ 130/23 = 5+ So there must be more than 5 pencils in total and there can't be more than 6 (you can only buy whole pencils).
It's easy to work out that the only way to get 130 from 21s and 23s is 4*21 + 2*23.
Answer B
It's easy to work out that the only way to get 130 from 21s and 23s is 4*21 + 2*23.
Answer B
28 Nov 2007, 04:33
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To my mind the answer here is B.
You can get the info given in first statment (that there was 6 pencils) from the info in statement 2:
If the total sum of money spent is 130 cents then minimum number of pencils bought is 130/23 = 5.6 or 6 because the number of pencils can't be a fractional number, while the maximum number of pencils bought is 130/21=6.19 or once again 6. So we have figured out that the number of pencils bought was 6.
How to understand the number of 21-cent pencils and 23-cent pencils? My suggestion is to try out different options. The thing is that there is only one possible solution for this problem.
1) Let's try the option in the middle: 3*21+3*23=132 >130 - it's too big so we should decrease the number of 23-cent pencils.
2) 4*21+2*23= 130 BINGO!
Total value of all other variants will be either bigger or smaller.
You can get the info given in first statment (that there was 6 pencils) from the info in statement 2:
If the total sum of money spent is 130 cents then minimum number of pencils bought is 130/23 = 5.6 or 6 because the number of pencils can't be a fractional number, while the maximum number of pencils bought is 130/21=6.19 or once again 6. So we have figured out that the number of pencils bought was 6.
How to understand the number of 21-cent pencils and 23-cent pencils? My suggestion is to try out different options. The thing is that there is only one possible solution for this problem.
1) Let's try the option in the middle: 3*21+3*23=132 >130 - it's too big so we should decrease the number of 23-cent pencils.
2) 4*21+2*23= 130 BINGO!
Total value of all other variants will be either bigger or smaller.
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!
