Problem 24 - Algebra for College Students

Question

A well driller charges \(\$ 9.00\) per foot for the first 10 feet, \(\$ 9.10\) per foot for the next 10 feet, \(\$ 9.20\) per foot for the next 10 feet, and so on, at a price increase of \(\$ 0.10\) per foot for succeeding intervals of 10 feet. How much does it cost to drill a well to a depth of 150 feet? \$1455

Step-by-Step Solution

Answer

The total cost to drill 150 feet is $1455.

Step 1: Calculate Cost for First 10 Feet

The driller charges \( \\(9.00 \) per foot for the first 10 feet. So, the cost for the first 10 feet is \( 9.00 \times 10 = \\)90.00 \).

Step 2: Calculate Cost for Next 10 Feet Intervals

From 10 to 20 feet, the cost is \( \\(9.10 \) per foot. Thus, \( 9.10 \times 10 = \\)91.00 \). Similarly, calculate for the next intervals:- 20 to 30 feet: \( 9.20 \times 10 = \\(92.00 \)- 30 to 40 feet: \( 9.30 \times 10 = \\)93.00 \)- 40 to 50 feet: \( 9.40 \times 10 = \\(94.00 \)- 50 to 60 feet: \( 9.50 \times 10 = \\)95.00 \)- 60 to 70 feet: \( 9.60 \times 10 = \\(96.00 \)- 70 to 80 feet: \( 9.70 \times 10 = \\)97.00 \)- 80 to 90 feet: \( 9.80 \times 10 = \\(98.00 \)- 90 to 100 feet: \( 9.90 \times 10 = \\)99.00 \).This pattern continues for each interval of 10 feet with the price increasing by \( \$0.10 \) per foot.

Step 3: Calculate Cost for Remaining Feet Beyond 100

Continue the pattern for the intervals from 100 to 150 feet:- 100 to 110 feet: \( 10.00 \times 10 = \\(100.00 \)- 110 to 120 feet: \( 10.10 \times 10 = \\)101.00 \)- 120 to 130 feet: \( 10.20 \times 10 = \\(102.00 \)- 130 to 140 feet: \( 10.30 \times 10 = \\)103.00 \)- 140 to 150 feet: \( 10.40 \times 10 = \$104.00 \).These calculations account for each additional 10-foot interval.

Step 4: Sum Total Cost for 150 Feet

Add all the costs for each 10-foot interval:\[ \\(90 + \\)91 + \\(92 + \\)93 + \\(94 + \\)95 + \\(96 + \\)97 + \\(98 + \\)99 + \\(100 + \\)101 + \\(102 + \\)103 + \\(104 = \\)1455 \]. The total cost for drilling to 150 feet is \( \$1455 \).

Key Concepts

Cost Calculation

Calculating the cost of drilling a well might seem complicated at first, mainly due to changing prices at different depths. However, understanding the pattern makes it straightforward. Initially, the well driller charges \( \\(9.00 \) per foot for the first 10 feet. This price then increases by \( \\)0.10 \) every subsequent 10 feet. So, from 10 to 20 feet, the price is \( \\(9.10 \) per foot, then \( \\)9.20 \) from 20 to 30 feet, and so forth.

• Understanding the calculation requires breaking down the total into smaller, manageable parts.
• Each interval of 10 feet contributes incrementally to the total cost based on its specific rate.
• Once you notice this pattern, you can easily compute the cost for each 10-foot segment and keep a running total.

The arithmetic sequence here makes it easier, as the pattern of increasing costs extends systematically all the way to 150 feet. With separate calculations for each 10-foot section, the total is easily summed to obtain the final cost of \( \$1455 \).

Step-by-Step Problem Solving

Breaking down the problem into manageable steps is crucial to solve it efficiently. This involves understanding each part of the question and tackling them one at a time.
First, calculate the cost for the first interval. At the start, for the first 10 feet, multiply \( \\(9.00 \) by 10 to get \( \\)90.00 \). This illustrates how critical accurate multiplication and focus on individual sections are.

• Next, proceed to the 10-foot intervals, where the price increases by \( \$0.10 \) per subsequent 10 feet.
• Apply the same multiplication method for each interval, ensuring to adjust the rate according to the specific depth range you are calculating.
• Continue this method for all intervals, up to the depth you need, which in this case is 150 feet.

Completing these steps methodically ensures no part is overlooked and simplifies checking the calculations for errors. By the end, summation of all individual costs gives the correct total cost for the well drilling.

Real-world Application in Algebra

This problem is an excellent application of arithmetic sequences, a fundamental concept in algebra that deals with sequences of numbers in which the difference of any two successive members is constant. In real-world applications, such as calculating cost of services with incremental price increases, algebra helps break down the problem and organize information efficiently.

• An arithmetic sequence here describes the increasing cost pattern as depth increases - each set of 10 feet is a term in the sequence.
• Understanding the pattern of incremental charges allows one to employ algebraic thinking to calculate larger totals without errors.
• Armed with this sequence, planning, budgeting, or forecasting costs becomes not only possible but also straightforward.

In solving problems like this, algebra doesn't just give an answer; it teaches a process of analytical thinking that extends beyond mathematics, helping to navigate various everyday problems with method and clarity.