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📝 How Use a Problem Solving Strategy (13 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 13 questions available

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What is How Use a Problem Solving Strategy?

Definition:
A problem-solving strategy is a systematic step-by-step approach used to tackle word problems and real-world scenarios in algebra. It typically involves: 1) Read and understand the problem, 2) Assign a variable to the unknown, 3) Translate the situation into an equation, 4) Solve the equation, and 5) Check if the answer is reasonable in the original context.

Working:
This strategy works by breaking down complex sentences into manageable parts, identifying relationships between quantities, and converting those relationships into algebraic expressions. For example, if a problem states 'the sum of a number and 5 is 12', we assign $x$ as the number, write $x + 5 = 12$ , solve to get $x = 7$ , and verify by substitution.

Example:
Problem: 'A number increased by 9 gives 20.' Use the strategy: read, let $x$ be the number, translate to $x + 9 = 20$ , subtract 9: $x = 11$ . Check: $11 + 9 = 20$ , correct.

Reason:
This structured approach reduces confusion, minimizes errors, and ensures that all aspects of the problem are considered, making it essential for solving application problems.

Easy

Medium

Hard

📝 All How Use a Problem Solving Strategy MCQs

Q1. A school plans to buy notebooks for $240$ students. Each student needs $3$ notebooks, but the school already has $185$ usable notebooks. If notebooks are sold in packs of $20$ , which strategy correctly determines the minimum number of packs needed?

A.Multiply

240

3

, subtract

185

, then divide by

20

and round down

B.Multiply

240

3

, subtract

185

, then divide by

20

and round up ✅

C.Add

240

3

, and

185

, then divide by

20

D.Subtract

240

from

185

, then divide the result by

20

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A sound problem-solving strategy first identifies the total requirement, subtracts the available supply, and then accounts for whole packages. Since a partial package cannot be purchased, the quotient must be rounded upward. This sequence preserves the meaning of each quantity.

Q2. A student immediately writes an equation after reading a word problem but obtains an answer that does not fit the situation. Which step was most likely skipped?

A.Choosing a calculator before reading the problem

B.Checking whether the answer is reasonable in context ✅

C.Rounding every number before forming the equation

D.Replacing all words with numerical values

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Problem solving is not complete when an equation produces a numerical result. Interpreting the result and checking whether it is reasonable can reveal incorrect assumptions, units, or arithmetic. This verification step is especially important in modeling situations.

Q3. A delivery company charges a fixed fee of $500$ rupees plus $75$ rupees per package. A customer has a budget of $2,000$ rupees. Which approach best models the maximum number of packages the customer can send?

A.Solve

500+75p=2000

and accept any real-valued solution

B.Solve

500-75p=2000

and round the result

C.Solve

500+75p\le2000

and choose the greatest whole-number value of

p

✅

D.Solve

75p-500\le2000

and choose the smallest value of

p

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The fixed fee is incurred regardless of the number of packages, while the variable charge depends on $p$ . Because the customer cannot exceed the budget, an inequality is appropriate. Since packages are counted discretely, the greatest whole-number solution represents the practical maximum.

Q4. A farmer wants to fence a rectangular plot using $120$ meters of fencing. He decides that the length should be twice the width. Which sequence is the most effective modeling strategy?

A.Set

w=2l

, substitute into

l+w=120

, and solve

B.Represent the perimeter correctly, express one dimension using the other, substitute, and verify the resulting dimensions ✅

C.Divide

120

2

immediately and use the result as both dimensions

D.Assume the area is

120

, then solve for the dimensions

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: An effective strategy identifies the relevant geometric relationship, translates the condition between dimensions, solves the resulting equation, and checks the answer. The perimeter is $120$ , not the area, and the relationship between length and width must be preserved.

Q5. A taxi fare is modeled by $F=150+40d$ , where $d$ is distance in kilometers. A passenger has $1,350$ rupees available. Which reasoning correctly determines whether a $30$ -km trip is affordable?

A.Calculate

40(30)

only because the fixed fee is unrelated to distance

B.Calculate

150+40(30)

, then compare the total with

1,350

✅

C.Calculate

1,350-150

, then conclude the trip costs that amount

D.Divide

1,350

30

and compare the result with

40

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The model contains two cost components: a fixed charge and a distance-dependent charge. Substituting $d=30$ gives the complete fare, $150+40(30)=1,350$ . Comparing that total with the available budget directly answers the affordability question.

Q6. A student solves $4x+18=70$ by writing $4x=70+18$ , followed by $x=22$ . What is the best diagnosis of the error?

A.The student should multiply

70

18

first

B.The student should divide

18

4

before subtracting

C.The student changed the side of the equation without changing the sign of

18

✅

D.The student should have added

4

to both sides

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: To isolate $4x$ , the $18$ must be removed by subtracting $18$ from both sides, giving $4x=52$ . The student's addition increases the constant instead of canceling it, so the resulting value of $x$ is incorrect.

Q7. A theater's seating chart shows the number of seats sold increasing approximately linearly as showtime approaches. At $6:00$ p.m., $120$ seats are sold; at $7:00$ p.m., $180$ are sold. If the same trend continues, which estimate is most reasonable for $7:30$ p.m.?

A.150 seats

B.180 seats

C.210 seats ✅

D.240 seats

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The increase is $60$ seats per hour, or $30$ seats per half hour. Starting from $180$ seats at $7:00$ , the estimate at $7:30$ is therefore $180+30=210$ . This uses the observed rate as a simple model.

Q8. A water tank contains $900$ liters and loses $35$ liters every minute. A student models the amount after $t$ minutes as $900+35t$ . Which correction best explains the modeling error?

A.The initial amount should be

35

liters

B.Time should be measured in hours

C.Because water is being lost, the rate should be subtracted:

900-35t

✅

D.The model should multiply

900

35t

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The initial quantity is $900$ liters, and the phrase 'loses $35$ liters every minute' represents a negative rate. Therefore the amount decreases with time, producing $A=900-35t$ . The student's positive rate incorrectly represents the tank filling.

Q9. A graph of a quantity versus time starts at $80$ , rises steadily to $140$ by $t=3$ , and then remains horizontal until $t=5$ . Which interpretation is most consistent with the graph?

A.The quantity decreases at a constant rate throughout

B.The quantity increases for

5

units of time at the same rate

C.The quantity increases during the first

3

units, then stays constant ✅

D.The quantity is constant first and then increases

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The upward segment indicates that the quantity is increasing as time passes. The horizontal segment from $t=3$ to $t=5$ indicates no further change. A graph therefore helps identify different stages of a model rather than merely providing numerical output.

Q10. Two methods are proposed for estimating the cost of producing $500$ items. Method A uses a fixed cost of $8,000$ rupees plus $25$ rupees per item. Method B uses $12,000$ rupees plus $15$ rupees per item. Which conclusion follows from comparing the models?

A.Method A is always more expensive

B.Method B is always more expensive

C.The methods have equal cost at

400

items, so the cheaper method depends on production level ✅

D.The methods have equal cost at

500

items

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Set the costs equal: $8000+25x=12000+15x$ . This gives $10x=4000$ , so $x=400$ . Below $400$ items, Method A has the lower fixed cost and is cheaper; above $400$ , Method B's lower variable cost becomes advantageous.

Q11. A student estimates that a rectangular garden with perimeter $50$ meters and width $10$ meters has length $20$ meters. Another student claims the length is $15$ meters because $50-10-10=30$ and then $30/2=15$ . Which evaluation is correct?

A.The first student is correct because

20+10=30

B.The second student is correct because the remaining perimeter is

30

meters, giving

15

meters for each of the two lengths ✅

C.Both are correct because either length can represent the longer side

D.Neither is correct because perimeter must be divided by

4

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A rectangle has two lengths and two widths. After removing the two $10$ -meter widths from the $50$ -meter perimeter, $30$ meters remain for the two equal lengths. Thus $2l=30$ , giving $l=15$ . The first answer fails the perimeter check.

Q12. A charity needs to raise at least $50,000$ rupees. It already has $12,500$ rupees and expects each fundraiser to contribute $2,500$ rupees. Which reasoning correctly finds the minimum number of fundraisers required?

A.Solve

12,500+2,500n=50,000

, obtaining

n=15

B.Solve

12,500+2,500n\ge50,000

, obtaining

n\ge15

, so

15

fundraisers are needed ✅

C.Solve

2,500n-12,500\ge50,000

, obtaining

n\ge25

D.Divide

50,000

2,500

and ignore the existing funds, requiring

20

fundraisers

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The phrase 'at least' requires an inequality because reaching or exceeding the target is acceptable. Subtracting the existing $12,500$ gives $37,500$ still needed. Dividing by $2,500$ gives $15$ , so $15$ fundraisers are sufficient and minimal.

Q13. A machine's output is modeled by $Q=120t-2t^2$ , where $t$ is the number of hours operated. An engineer wants the first time after startup when output reaches $160$ units. Which strategy is most appropriate?

A.Assume a constant rate and solve

120t=160

B.Set

120t-2t^2=160

, solve for all possible

t

-values, then select the smallest positive value and verify it in the model ✅

C.Set

120t-2t^2=0

and choose the larger root

D.Add

120

and

2

, then divide by

160

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The model is quadratic, so the target output must be represented by $120t-2t^2=160$ . This may produce more than one time value because output can rise and later fall. Since the question asks for the first occurrence, the smallest positive valid solution must be selected and checked.

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