4.3 Q-20

Question Statement

The average entry test score of engineering candidates was 592 in the year 1998, while the score was 564 in 2002. Assuming that the relationship between time and score is linear, find the average score for the year 2006.

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Background and Explanation

This problem involves finding the average entry test score for 2006 based on a linear relationship between the year and the score. The data provides scores for two years, 1998 and 2002, and we are asked to predict the score for 2006. We can solve this using the slope-intercept form of a line:

• The slope formula helps determine how the score changes over time.
• The point-slope form of the equation will allow us to write a linear equation relating the time (year) to the score.

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Solution

1. Identify the Given Data:
   We are given two points on the line:

   • $(s_1, y_1) = (592, 1998)$ (score and year for 1998),
   • $(s_2, y_2) = (564, 2002)$ (score and year for 2002).

2. Calculate the Slope $m$:
   The slope formula is:
   $m = \frac{y_2 - y_1}{s_2 - s_1}$
   Substituting the values of the given points:
   $m = \frac{2002 - 1998}{564 - 592} = \frac{4}{-28} = \frac{-1}{7}$
   So, the slope of the line is $\frac{-1}{7}$.

3. Write the Equation of the Line:
   Using the point-slope form of the equation $y - y_1 = m(s - s_1)$, and substituting the values $y_1 = 1998$, $s_1 = 592$, and $m = \frac{-1}{7}$, we get:
   $y - 1998 = \frac{-1}{7}(s - 592)$

4. Find the Score for 2006:
   We are asked to find the score for the year 2006, so set $y = 2006$.
   Substitute $y = 2006$ into the equation:
   $2006 - 1998 = \frac{-1}{7}(s - 592)$
   Simplifying:
   $8 = \frac{-1}{7}(s - 592)$

5. Solve for $s$:
   Multiply both sides by 7 to eliminate the fraction:
   $8 \times 7 = -(s - 592)$
   $56 = -(s - 592)$
   Now, simplify the equation:
   $s - 592 = -56$
   Finally, solve for $s$:
   $s = 592 - 56 = 536$

Thus, the average entry test score in 2006 is 536.

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Key Formulas or Methods Used

• Slope Formula:
  $m = \frac{y_2 - y_1}{s_2 - s_1}$

• Point-Slope Form of a Line:
  $y - y_1 = m(s - s_1)$

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Summary of Steps

1. Identify the two given points: $(592, 1998)$ and $(564, 2002)$.
2. Calculate the slope $m = \frac{-1}{7}$.
3. Use the point-slope form to write the equation of the line:
   $y - 1998 = \frac{-1}{7}(s - 592)$
4. Substitute $y = 2006$ to find the score for 2006.
5. Solve for $s$ to find the score for 2006, which is $536$.