Tenth Grade Mathematics: Applied Word Problems: Motion, Work, and Economy

The true power of mathematics lies in its ability to translate real-world descriptions, constraints, and relationships into precise algebraic equations. Applied word problems require you to dissect narrative descriptions, identify two unknown quantities, assign algebraic variables to them, formulate a system of two independent linear equations, and interpret the resulting mathematical solution within the practical context of the problem. This chapter focuses on the three major classic categories of applied problems: relative motion in rivers and transit, collaborative work and rate of completion, and financial economics involving cost and investment.

General Strategy for Formulating Word Problems

Every word problem can be conquered by following a structured four-stage translation process:

The Four-Stage Modeling Protocol:
1. Identify Unknowns: Read the question carefully to pinpoint the two quantities
   being sought. Assign variables x and y with precise units.
2. Translate Relationships: Re-read the problem sentence by sentence to write down
   two independent equations connecting x and y.
3. Solve Algebraically: Choose the most efficient algebraic method (substitution,
   elimination, or cross-multiplication) to find the numerical values.
4. Verify Context: Check that the answers make practical sense (e.g., speeds,
   lengths, and times must be positive; item counts must be integers).

Category 1: Relative Motion (Upstream and Downstream)

Problems involving boats in rivers, swimmers in currents, or airplanes in wind currents rely on the principle of relative velocity.

Definitions:
Let the speed of the boat in still water = x km/h
Let the speed of the water current       = y km/h   (where x > y)

1. Downstream Motion (Boat moves in the SAME direction as the water):
   The current aids the boat:
   Downstream Speed = (x + y) km/h

2. Upstream Motion (Boat moves AGAINST the direction of the water):
   The current opposes the boat:
   Upstream Speed = (x - y) km/h

Fundamental Motion Identity:
Time = Distance / Speed
  Time Downstream = Distance / (x + y)
  Time Upstream   = Distance / (x - y)

Motion Worked Example

A boat covers 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can cover 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.

Formulation:
Let speed in still water = x km/h, and speed of stream = y km/h.
Upstream speed = (x - y) km/h, Downstream speed = (x + y) km/h.

Equation 1:  30 / (x - y) + 44 / (x + y) = 10
Equation 2:  40 / (x - y) + 55 / (x + y) = 13

Let u = 1 / (x - y)  and  v = 1 / (x + y):
  30*u + 44*v = 10    --- (1)
  40*u + 55*v = 13    --- (2)

Multiply (1) by 4 and (2) by 3 to eliminate u:
  120*u + 176*v = 40
  120*u + 165*v = 39
Subtracting gives: 11*v = 1  ===>  v = 1 / 11.
Substitute v = 1/11 into (1):
  30*u + 44*(1/11) = 10  ===>  30*u + 4 = 10  ===>  30*u = 6  ===>  u = 1 / 5.

Reverting substitutions:
  x - y = 5     --- (A)
  x + y = 11    --- (B)

Adding (A) and (B): 2*x = 16 ===> x = 8 km/h.
Subtracting (A) from (B): 2*y = 6 ===> y = 3 km/h.
Conclusion: Speed of boat in still water = 8 km/h; speed of stream = 3 km/h.

Category 2: Collaborative Work and Rates

Problems where multiple workers or machines cooperate to complete a task are solved by considering the work done per unit time (work rate).

Rate Principle:
If worker A completes the entire job in x days, work done by A in 1 day = 1 / x.
If worker B completes the entire job in y days, work done by B in 1 day = 1 / y.
If both work together for D days to finish the job:
                        D * [1/x + 1/y] = 1 (Total Job)

Category 3: Economic Transactions and Mixture Problems

Economic problems involve unit costs, ticket pricing, or fixed and variable charges (such as taxi fares or library rentals).

Two-Part Cost Model:
Total Cost = Fixed Base Charge + (Rate per Unit) * (Units Consumed)
            y = F + r * x

Chapter Exercises

Exercise 1: 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Exercise 2: A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 5/6. Find the original fraction.

Exercise 3: The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is 105 dollars, and for a journey of 15 km, the charge paid is 155 dollars. What are the fixed charges and the charge per kilometer? How much does a person have to pay for travelling a distance of 25 km?

Solutions and Explanations

Solution 1: Let the time taken by 1 woman alone to finish the work be x days, and the time taken by 1 man alone be y days. In 1 day, 1 woman completes 1/x of the work, and 1 man completes 1/y of the work. For the first condition: 2 women and 5 men work for 4 days to finish the job: 4(2/x + 5/y) = 1 ==> 2/x + 5/y = 1/4. For the second condition: 3 women and 6 men work for 3 days: 3(3/x + 6/y) = 1 ==> 3/x + 6/y = 1/3 ==> 1/x + 2/y = 1/9. Let u = 1/x and v = 1/y: 2u + 5v = 1/4 and u + 2v = 1/9. From the second equation, u = 1/9 - 2v. Substitute into the first: 2(1/9 - 2v) + 5v = 1/4 ==> 2/9 - 4v + 5v = 1/4 ==> v = 1/4 - 2/9 = (9 - 8)/36 = 1/36. Then u = 1/9 - 2(1/36) = 1/9 - 1/18 = 1/18. Inverting: x = 1/u = 18 days, and y = 1/v = 36 days. Thus, 1 woman alone takes 18 days, and 1 man alone takes 36 days.

Solution 2: Let the fraction be x/y, where x is the numerator and y is the denominator. Condition 1: (x + 2)/(y + 2) = 9/11 ==> 11(x + 2) = 9(y + 2) ==> 11x + 22 = 9y + 18 ==> 11x - 9y = -4. Condition 2: (x + 3)/(y + 3) = 5/6 ==> 6(x + 3) = 5(y + 3) ==> 6x + 18 = 5y + 15 ==> 6x - 5y = -3. Solve using substitution: from Equation 2, x = (5y - 3)/6. Substitute into Equation 1: 11((5y - 3)/6) - 9y = -4 ==> 55y - 33 - 54y = -24 ==> y - 33 = -24 ==> y = 9. Substitute y = 9 into x: x = (5*(9) - 3)/6 = (45 - 3)/6 = 42/6 = 7. Thus, the original fraction is 7/9.

Solution 3: Let the fixed charge be x dollars, and the running charge per kilometer be y dollars. For a 10 km trip: x + 10y = 105 (Equation 1). For a 15 km trip: x + 15y = 155 (Equation 2). Subtract Equation 1 from Equation 2: (x + 15y) - (x + 10y) = 155 - 105 ==> 5y = 50 ==> y = 10 dollars per kilometer. Substitute y = 10 into Equation 1: x + 10(10) = 105 ==> x + 100 = 105 ==> x = 5 dollars. Thus, the fixed charge is 5 dollars, and the rate per kilometer is 10 dollars. For a distance of 25 km, the total charge is: x + 25y = 5 + 25(10) = 5 + 250 = 255 dollars.