Challenge Problem Showdown – October 31th, 2011

We invite you to test your GMAT knowledge for a chance to win! Each week, we will post a new Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for a free Manhattan GMAT Prep item. Tell your friends to get out their scrap paper and start solving!

Here is this week’s problem:

The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12. The mixture will be changed so that the ratio of A to C is quadrupled and the ratio of A to D is decreased. The ratio of A to B will be held constant. If B will constitute 20% of the weight of the new mixture, by approximately what percent will the ratio of A to D be decreased?

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By: Tom | 2011/10/31 | Challenge Problem | Trackback | Comments [RSS 2.0]

2 Responses to Challenge Problem Showdown – October 31th, 2011 »»


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  1. Nsk
    Comment by Nsk | 2011/11/01 at 08:11:02

    for, 4 part of A , 1 part of D is reduced from mixture

    2. comment_type != "trackback" && $comment->comment_type != "pingback" && !ereg("", $comment->comment_content) && !ereg("", $comment->comment_content)) { ?>
  2. Mahbub
    Comment by Mahbub | 2011/11/10 at 18:40:37

    Say new D is D. So the new ratio becomes 4:7:16:D. Where D < 12.

    Say the total new mixture is 100 unit. So B is 20 unit. The mixture becomes:
    4x + 7x + 16x + Dx = 100
    27x + Dx = 100

    Here 7x = 20, so x = 20/7

    so Dx = 100 – 27x
    D = 100/x – 27
    D = 100 * 7/20 – 27 = 35 – 27 = 8

    Before A/D = 4/12 = 1/3 = 0.33
    Now A/D = 4/8 = 1/2 = 0.50
    % Change = (1/2 – 1/3) / (1/3) = 1/6 / 1/3 = 1/2 = 50% Decrease

    50% Increase in the ratio.


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  1. Nsk
    Comment by Nsk | 2011/11/01 at 08:11:02

    for, 4 part of A , 1 part of D is reduced from mixture

    2. comment_type == "trackback" || $comment->comment_type == "pingback" || ereg("", $comment->comment_content) || ereg("", $comment->comment_content)) { ?>

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    1. Mahbub
      Comment by Mahbub | 2011/11/10 at 18:40:37

      Say new D is D. So the new ratio becomes 4:7:16:D. Where D < 12.

      Say the total new mixture is 100 unit. So B is 20 unit. The mixture becomes:
      4x + 7x + 16x + Dx = 100
      27x + Dx = 100

      Here 7x = 20, so x = 20/7

      so Dx = 100 – 27x
      D = 100/x – 27
      D = 100 * 7/20 – 27 = 35 – 27 = 8

      Before A/D = 4/12 = 1/3 = 0.33
      Now A/D = 4/8 = 1/2 = 0.50
      % Change = (1/2 – 1/3) / (1/3) = 1/6 / 1/3 = 1/2 = 50% Decrease

      50% Increase in the ratio.

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