Rational Number Word Problems - Pre Algebra

1. A weather station records temperature changes over four hours. The temperature drops $3.5°F$ in the first hour, rises $2.25°F$ in the second hour, drops $4.75°F$ in the third hour, and rises $1.5°F$ in the fourth hour. If the final temperature is $-2.5°F$, what was the initial temperature?
   Answer: $2.0°F$ Let the initial temperature be $x$. The total change is: $-3.5 + 2.25 - 4.75 + 1.5 = -4.5°F$. So $x + (-4.5) = -2.5$, which gives us $x = -2.5 + 4.5 = 2.0°F$. Choice B adds instead of subtracting the net change. Choice C miscalculates the total change as $-6.5$. Choice D results from calculation errors in the temperature changes.
2. A scientist measures the change in a chemical solution's pH level during an experiment. The pH starts at $7.5$, decreases by $2.25$, then increases by $0.75$, decreases by $1.5$, and finally increases by $3.25$. If a solution is considered acidic when pH < 7, basic when pH > 7, and neutral when pH = 7, what type of solution resulted, and by how much did the pH change from the starting point?
   Answer: Basic solution; increased by $0.25$ This question tests your ability to track sequential changes using positive and negative numbers, then interpret the final result. When you see a series of increases and decreases, work through them step-by-step while keeping track of the cumulative effect. Starting at pH $7.5$, let's trace each change: - Decreases by $2.25$: $7.5 - 2.25 = 5.25$ - Increases by $0.75$: $5.25 + 0.75 = 6.0$ - Decreases by $1.5$: $6.0 - 1.5 = 4.5$ - Increases by $3.25$: $4.5 + 3.25 = 7.75$ The final pH is $7.75$. Since $7.75 > 7$, this is a basic solution. To find the net change, subtract the starting value: $7.75 - 7.5 = +0.25$, meaning an increase of $0.25$. Answer A is wrong because the solution isn't neutral (pH ≠ 7) and there was a net change. Answer B incorrectly classifies the solution as acidic when $7.75 > 7$ makes it basic. Answer C correctly identifies that the pH changed by $0.25$ but wrongly calls it a decrease and misclassifies the solution as acidic. Answer D correctly identifies both the basic nature (pH > 7) and the net increase. Study tip: For sequential change problems, work step-by-step rather than trying shortcuts. Always distinguish between the direction of individual changes and the overall net change from start to finish.
3. A football team's field position changes during a series of plays. Starting at their own 25-yard line, they gain $12.5$ yards, lose $8.25$ yards, gain $15.75$ yards, lose $4.5$ yards, and gain $7.25$ yards. Field position is measured as yards from their own goal line. If they need to reach the 50-yard line to be in the opponent's territory, how many more yards do they need?
   Answer: $2.25$ yards This problem tests your ability to work with positive and negative numbers in a real-world context. When a team gains yards, you add to their position; when they lose yards, you subtract from their position. Starting at the 25-yard line, let's track each play: $25 + 12.5 - 8.25 + 15.75 - 4.5 + 7.25$. Work left to right: $25 + 12.5 = 37.5$, then $37.5 - 8.25 = 29.25$, then $29.25 + 15.75 = 45$, then $45 - 4.5 = 40.5$, and finally $40.5 + 7.25 = 47.75$. The team ends up at the 47.75-yard line. Since they need to reach the 50-yard line, they need $50 - 47.75 = 2.25$ more yards, which is answer C. Answer A ($22.75$ yards) likely comes from incorrectly subtracting their final position from their starting position: $47.75 - 25 = 22.75$. This tells you how far they've moved total, not how far they need to go. Answer B ($7.75$ yards) might result from calculation errors in tracking the plays, perhaps by mixing up signs or decimal places. Answer D ($27.25$ yards) could come from subtracting their starting position from the target: $50 - 25 = 25$, then making a small arithmetic error. Remember: in multi-step problems involving gains and losses, carefully track each change in sequence, then compare your final result to the target. Don't confuse "distance traveled" with "distance remaining."
4. A stock's price changes over five days are: Monday $+\$2.75$, Tuesday $-\$4.50$, Wednesday $+\$1.25$, Thursday $-\$3.75$, Friday $+\$2.50$. If the stock's price on Friday evening was $38.75, and an investor bought $150$ shares on Monday morning, what was the total value of their investment when they purchased it?
   Answer: $6,037.50 Total change over the week: $2.75 - 4.50 + 1.25 - 3.75 + 2.50 = -1.75$. If Friday's price was $38.75 and the total change was $-\$1.75$, then Monday's opening price was $38.75 - (-1.75) = 38.75 + 1.75 = \$40.25$. Investment value: $150 × 40.25 = \$6,037.50$. Choice A uses Friday's price instead of Monday's. Choice C miscalculates the weekly change. Choice D adds the change incorrectly.
5. An elevator starts at the ground floor (floor 0) and makes the following moves: up $12$ floors, down $5.5$ floors, up $8.25$ floors, down $11.75$ floors, and up $4.5$ floors. If the building charges $0.75 per floor traveled (regardless of direction) and gives a $2.50 discount for trips over $35$ floors total, what is the cost of this elevator trip?
   Answer: $28.75 When you see a problem involving distance traveled and cost calculations, you need to track two separate things: the total distance moved and any special conditions that affect pricing. First, calculate the total floors traveled by adding up all movements regardless of direction: $12 + 5.5 + 8.25 + 11.75 + 4.5 = 42$ floors total. Notice that we count both up and down movements since the building charges per floor traveled, not net displacement. Next, apply the pricing structure. At $0.75 per floor, the base cost would be $42 \times \$0.75 = \$31.50$. However, since the total distance (42 floors) exceeds 35 floors, you qualify for the $2.50 discount. The final cost is $\$31.50 - \$2.50 = \$28.75$. Answer A ($26.25) represents a calculation error where someone might have incorrectly applied the discount or miscalculated the total distance. Answer B ($31.25) comes from calculating $\$31.50 - \$0.25$, suggesting confusion about the discount amount. Answer D ($29.00) results from applying a $2.50 discount incorrectly to a wrong base calculation. The key strategy here is to carefully read multi-step problems and identify each component: distance calculation (including all movements), base pricing, and conditional discounts. Always check whether you qualify for special conditions before applying them, and double-check your arithmetic on problems involving decimals.
6. A delivery truck's weight changes throughout the day. It starts with $2,450$ pounds of cargo. At the first stop, it delivers $375.5$ pounds and picks up $128.25$ pounds. At the second stop, it delivers $892.75$ pounds and picks up $445.5$ pounds. At the third stop, it delivers $520.25$ pounds. If the empty truck weighs $4,200$ pounds, what is the truck's total weight after all stops?
   Answer: $5,435.25$ pounds Starting cargo: $2,450$ pounds. After first stop: $2,450 - 375.5 + 128.25 = 2,202.75$ pounds. After second stop: $2,202.75 - 892.75 + 445.5 = 1,755.5$ pounds. After third stop: $1,755.5 - 520.25 = 1,235.25$ pounds of cargo. Total truck weight: $4,200 + 1,235.25 = 5,435.25$ pounds. Choice B adds an extra $200$ pounds. Choice C gives only the cargo weight, forgetting the truck's weight. Choice D adds an extra $400$ pounds.
7. Maria's checking account has a balance of $-\$47.25$. She makes three deposits of $28.50, $15.75, and $22.00, then writes checks for $18.25 and $35.50. If her bank charges a $25.00 overdraft fee for each day her account is negative, and she started the day negative, what is her balance after all transactions and fees?
   Answer: $-\$59.75$ Starting balance: $-\$47.25$. After deposits: $-47.25 + 28.50 + 15.75 + 22.00 = \$19.00$. After checks: $19.00 - 18.25 - 35.50 = -\$34.75$. Since the account started negative and ended negative, she incurs one overdraft fee of $25.00. Final balance: $-34.75 - 25.00 = -\$59.75$. Choice A forgets the overdraft fee. Choice C includes only transactions without the fee. Choice D incorrectly applies two overdraft fees.

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