{"id":41354,"date":"2022-09-20T20:44:37","date_gmt":"2022-09-20T20:44:37","guid":{"rendered":"https:\/\/wpapp.kaptest.com\/study\/?p=41354"},"modified":"2022-09-20T20:44:38","modified_gmt":"2022-09-20T20:44:38","slug":"psat-math-sample-test","status":"publish","type":"post","link":"https:\/\/wpapp.kaptest.com\/study\/psat\/psat-math-sample-test\/","title":{"rendered":"PSAT Math Sample Test"},"content":{"rendered":"\n
<\/section>\n

Systems of Equations<\/h2>\n

The linear equations detailed in the previous chapter are well suited for modeling a variety of scenarios and for solving for a single variable in terms of another that is clearly defined (e.g., what is the cost of a data plan if you consume 4 GB of data in a month). However, sometimes you will be given a set of multiple equations with multiple variables that are interdependent. For example, suppose a $50\/month cell phone plan includes $0.05 text messages and $0.40 voice calls, with a cap of 1,000 combined text messages and voice calls.<\/p>\n

This scenario can be represented by the following system of equations:<\/p>\n

\n \n \n \n \n $<\/mi>\n 0.05<\/mn>\n t<\/mi>\n +<\/mo>\n $<\/mi>\n 0.40<\/mn>\n v<\/mi>\n <\/mtd>\n \n =<\/mo>\n <\/mtd>\n \n $<\/mi>\n 50<\/mn>\n <\/mtd>\n <\/mtr>\n \n \n t<\/mi>\n +<\/mo>\n v<\/mi>\n <\/mtd>\n \n =<\/mo>\n <\/mtd>\n \n 1<\/mn>\n ,<\/mo>\n 000<\/mn>\n <\/mtd>\n <\/mtr>\n <\/mtable>\n <\/math>\n <\/div>\n

Solving such a system would enable you to determine the maximum number of text messages and voice calls you could make under this plan, while optimizing total usage. To solve systems of equations, you\u2019ll need to rely on a different set of tools that builds on the algebra you\u2019re already familiar with. The following question shows an example of such a system in the context of a test-like question.<\/p>\n

\n
\n PSAT<\/span>\n Math<\/span>\n assessment<\/span>\n quiz<\/span>\n <\/section>\n
    \n
  1. \n container<\/span>\n <\/li>\n
  2. \n
    \n single-select<\/span>\n <\/section>\n
    \n

    If 3r<\/i> + 2s<\/i> = 24 and r<\/i> + s<\/i> = 12, what is the value of r<\/i> + 6 ?<\/span><\/p>\n <\/section>\n

      \n
    1.   0<\/li>\n
    2.   4<\/li>\n
    3.   6<\/li>\n
    4. 12<\/li>\n <\/ol>\n
      \n

      Difficulty:<\/b> Medium<\/p>\n

      Category:<\/b> Heart of Algebra \/ Linear Equations<\/p>\n

      <\/p>\n

      Getting to the Answer:<\/b> Although substitution will certainly work here, there is a quicker way. Rewrite the first equation so that (r + s<\/i>)<\/span> is a component. First, express 3r<\/i> as r<\/i> + 2r<\/i>.<\/span> Then factor a 2 out of the left-hand side to get r<\/i> + 2(r<\/i> + s<\/i>) = 24.<\/span> Now you can substitute 12 in for (r + s<\/i>),<\/span> simplify, and solve for r.<\/i> Be careful. The question is asking for r +<\/i> 6,<\/span> so A is a trap corresponding to the value of r<\/i> alone. Instead, (C) is correct.<\/p>\n <\/section>\n <\/li>\n <\/ol>\n <\/section>\n

      You might be tempted to switch on math autopilot at this point and employ substitution, solving the second equation<\/span> for s<\/i><\/span> in terms of r<\/i>:<\/span><\/p>\n

      s<\/i> = 12 \u2013 r<\/i><\/span><\/p>\n

      You could plug the resulting expression back into the other equation and eventually solve for r<\/i><\/span>, but remember, the PSAT<\/span> tests your ability to solve math problems in the most efficient way. The following table contains some strategic thinking designed to help you find the most efficient way to solve this problem on Test Day, along with some suggested scratchwork.<\/p>\n

      \n \n \n \n \n \n \n \n
      Strategic Thinking<\/th>\n Math Scratchwork<\/th>\n <\/tr>\n <\/thead>\n
      \n

      Step 1: Read the question, identifying and organizing important information as you go<\/b><\/p>\n

      In this case, you\u2019re looking for the value of r<\/i>. There are two equations that involve r<\/i> and s<\/i>.
      \n <\/p>\n <\/td>\n

      		\n


      \n <\/p>\n

      3r<\/i> + 2s<\/i> = 24<\/p>\n

      r<\/i> + s<\/i> = 12<\/p>\n <\/td>\n <\/tr>\n

      \n

      Step 2: Choose the best strategy to answer the question<\/b><\/p>\n

      Is there any way you can make the first equation look like the second one? Does the quantity r + s exist in the first equation in some form?<\/i><\/p>\n

      <\/p>\n

      How can you effectively use both equations?<\/i><\/p>\n

      Once you\u2019ve written the first equation in terms of r<\/i> + s<\/i>, substitute the value of r<\/i> + s<\/i> (which is 12) into the second equation and solve for r<\/i>.<\/p>\n <\/td>\n

      		\n

      3r<\/i> + 2s<\/i> = 24<\/p>\n

      r<\/i> + 2r<\/i> + 2s<\/i> = 24<\/p>\n

      r<\/i> + 2(r<\/i> + s<\/i>) = 24<\/p>\n

      r<\/i> + 2(12) = 24<\/p>\n

      r<\/i> = 0<\/p>\n <\/td>\n <\/tr>\n

      \n

      Step 3: Check that you answered the<\/b> right<\/i><\/b> question<\/b><\/p>\n

      Be careful! The question isn\u2019t asking for the value of r<\/i>. Add 6 to your result and you should see that (C) is the correct answer.<\/p>\n <\/td>\n

      		\n


      \n <\/i><\/p>\n

      r<\/i> + 6 = 0 + 6<\/p>\n

      r<\/i> + 6 = 6<\/p>\n <\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n