Comments on: More Fun with Weighted Averages
http://www.knewton.com/blog/test-prep/more-fun-with-weighted-averages-2/
Wed, 04 Mar 2015 20:35:51 +0000hourly1http://wordpress.org/?v=4.0.1By: Visitashi
http://www.knewton.com/blog/test-prep/more-fun-with-weighted-averages-2/#comment-2919
Sat, 30 Jul 2011 12:28:00 +0000http://www.knewton.com/?p=18952#comment-2919Thankyou sir.
]]>By: Rich Zwelling
http://www.knewton.com/blog/test-prep/more-fun-with-weighted-averages-2/#comment-2906
Fri, 29 Jul 2011 14:51:00 +0000http://www.knewton.com/?p=18952#comment-2906Yep, you are correct!! Well done, and I like your method!
]]>By: Rich Zwelling
http://www.knewton.com/blog/test-prep/more-fun-with-weighted-averages-2/#comment-2905
Fri, 29 Jul 2011 14:48:00 +0000http://www.knewton.com/?p=18952#comment-2905Here’s the solution. Don’t look if you haven’t yet attempted the problem!!We know balloons come in boxes of 40 or 100, and the average number of balloons per box is 88. If we were to plot these numbers on a number line, it would look like something this:40—————————————–88———100Notice that 88 is a distance of 12 from 100 and a distance of 48 from 40. The two distances of 48 and 12 are in a 4:1 ratio. You could confirm this by dividing the number line into equal parts: 12 12 12 12 1240———X———X———X———88———100This means that there are 4 times as many larger 100-count boxes as there are smaller 40-count boxes. Since there are 70 boxes total, that would mean that there are 14 smaller boxes and 56 larger boxes. You could find this either by using a system (x+y=70 and y=4x, where x represents small and y represents large) or by realizing that 14 and 56 are one-fifth and four-fifths of 70, respectively.Your job is to make the overall average 70. Well, if we have only 40-count boxes and 100-count boxes, then the number of each box must be the same! This is because 70 is the average of 40 and 100 (i.e. right smack in the middle).We have 14 smaller boxes, and we know that won’t change, since we’re only removing larger boxes. That means we must have 14 larger boxes as well. There are now 56, so we must remove 42 of them to get the amount we want. Ans: E
]]>By: Visitashi
http://www.knewton.com/blog/test-prep/more-fun-with-weighted-averages-2/#comment-2900
Thu, 28 Jul 2011 03:40:00 +0000http://www.knewton.com/?p=18952#comment-2900i can assume that all cases have 88 balloons,since the average is 88.
now i have to make the average 70,so i need to remove 18 from each.this means i need to remove a total of 18*70=1260 ballons.but since the condition says that only larger cases must be removed,so i will have to remove 30 balloons from each larger case in order to make the average 70.
so,the number of larger cases removed=1260/30 =42.
is it correct ,sir??
hi ,this is my first post.
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