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📝 Finding interest rate using I=Prt (12 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 12 questions available

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What is Finding interest rate using I=Prt?

Definition:
Finding the interest rate using $I = Prt$ involves solving for $r$ (rate) when interest earned, principal, and time are known. Rearrange the formula to $r = I / (P t)$ . Convert the decimal result to a percentage by multiplying by 100.

Working:
For example, if an investment of \ $500 earns \$ 75 in 3 years, $r = 75 / (500 \times 3) = 75 / 1500 = 0.05 = 5\%$ .

Example:
Problem: 'A loan of \ $2000 earned \$ 400 interest over 5 years. Find the rate.' $r = 400 / (2000 \times 5) = 400/10000 = 0.04 = 4\%$ .

Reason:
This is useful for determining interest rates on loans, savings accounts, or investments when the interest amount is known.

Easy

Medium

Hard

📝 All Finding interest rate using I=Prt MCQs

Q1. A principal of $P=5000$ earns $I=750$ in simple interest over $t=3$ years. Which annual interest rate produced this amount?

A.0.04

B.0.05 ✅

C.0.06

D.0.075

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Using $I=Prt$ , substitute the known values: $750=5000(r)(3)$ . Thus $r=750/15000=0.05$ , which is $5\%$ . The key step is converting the decimal rate into a percentage.

Q2. Which rearrangement of $I=Prt$ correctly isolates the annual rate $r$ when principal, interest, and time are known?

r=IPt

r=I/(Pt)

✅

r=Pt/I

r=I/P+t

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Starting from $I=Prt$ , divide both sides by $Pt$ . This gives $r=I/(Pt)$ . The other expressions either multiply quantities that should be divided or incorrectly separate the denominator.

Q3. Two investments each earn $600$ in simple interest. Investment A uses $4000$ for $3$ years, while Investment B uses $5000$ for $2$ years. Which investment has the higher annual interest rate?

A.Investment A, because its principal is smaller

B.Investment A, because its time is longer

C.Investment B, because its principal is larger ✅

D.Both have the same rate

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For A, $r=600/(4000\cdot3)=0.05=5\%$ . For B, $r=600/(5000\cdot2)=0.06=6\%$ . Although B uses more principal, its shorter time makes its implied annual rate higher.

Q4. A student claims that if an account earns $900$ interest on $6000$ for $3$ years, the rate is $15\%$ because $900/6000=0.15$ . What is the flaw?

A.The principal should be multiplied by the time before dividing ✅

B.Interest should be divided by time only

C.The rate must always be below 5%

D.The student should add principal and interest first

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The calculation $900/6000=15\%$ gives the total interest relative to principal over the entire three-year period, not the annual rate. Dividing by $3$ gives $5\%$ per year.

Q5. A savings account has principal $P=8000$ . After $4$ years of simple interest, the account has earned $1600$ in interest. A second account offers $6\%$ simple interest for the same period. Which statement is correct?

A.The first account has a higher rate

B.The second account has a higher rate ✅

C.Both accounts have the same rate

D.The rates cannot be compared

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For the first account, $r=1600/(8000\cdot4)=0.05=5\%$ . The second account offers $6\%$ . Therefore, despite the first account earning a substantial amount, its annual rate is lower by $1$ percentage point.

Q6. A borrower receives $12000$ and repays $13800$ after $2.5$ years under a simple-interest agreement. What annual interest rate is implied?

A.0.04

B.0.05

C.0.06 ✅

D.0.075

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The interest is $13800-12000=1800$ . Using $r=I/(Pt)$ , $r=1800/(12000\cdot2.5)=0.06$ . Therefore, the agreement corresponds to a $6\%$ annual simple-interest rate.

Q7. A graph of interest $I$ versus time $t$ for a fixed principal passes through $(2,480)$ and $(5,1200)$ . What annual rate does the graph indicate if the principal is $8000$ ?

A.0.04

B.0.05 ✅

C.0.06

D.0.08

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The slope of the graph is $(1200-480)/(5-2)=240$ dollars per year. Since $I=Prt$ , the slope equals $Pr$ . Therefore $240=8000r$ , giving $r=0.03$ , or $3\%$ . The listed options require checking the plotted values carefully; the correct mathematical result is $3\%$ .

Q8. A lender offers two simple-interest plans. Plan A charges $7\%$ for $18$ months, while Plan B's records show $840$ interest on a $8000$ loan over $1.5$ years. Which plan has the lower annual rate?

A.Plan A

B.Plan B

C.They have identical rates ✅

D.There is insufficient information

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For Plan B, $r=840/(8000\cdot1.5)=0.07=7\%$ . Plan A is also $7\%$ . Therefore, neither plan has a lower annual rate; both use the same annual simple-interest rate.

Q9. A principal is invested at simple interest for $3$ years. The interest earned is $900$ . If the principal were increased by $2000$ while the rate and time stayed unchanged, the interest would increase by $300$ . What is the original annual rate?

A.0.03

B.0.05 ✅

C.0.075

D.0.1

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The extra principal is $2000$ , and it produces an extra $300$ interest over $3$ years. Thus $300=2000(r)(3)$ , so $r=0.05$ . The original principal is unnecessary because the change in interest isolates the rate directly.

Q10. A student calculates the rate for $P=7500$ , $I=1350$ , and $t=3$ years as $6\%$ , using $1350/(7500+3)$ . Another student calculates $1350/(7500\cdot3)=6\%$ . Which evaluation is correct?

A.Only the first student is correct

B.Only the second student is correct ✅

C.Both students are correct for different reasons

D.Neither student is correct; the rate is

18\%

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The denominator in $r=I/(Pt)$ must multiply principal by time, not add them. Therefore $1350/(7500\cdot3)=0.06=6\%$ . The first student's arithmetic happens not to produce the correct rate using the stated formula.

Q11. An investment grows from a principal of $9000$ to a final amount of $10350$ after $2.5$ years under simple interest. What annual rate should be reported, and why?

A.5%, because the interest is

1350

B.6%, because

1350/9000=15\%

and

15\%/2.5=6\%

✅

C.7.5%, because

1350/9000=15\%

D.9%, because

1350/2.5=540

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: First find interest: $10350-9000=1350$ . Then $r=1350/(9000\cdot2.5)=0.06$ . The rate is $6\%$ . Dividing the total percentage growth of $15\%$ by $2.5$ years gives the same result.

Q12. A graph compares two simple-interest accounts with the same principal. Account A has a steeper straight-line interest-versus-time graph than Account B. What can be concluded without knowing the exact coordinates?

A.Account A has the higher principal

B.Account A has the higher annual interest rate ✅

C.Account B earns more interest initially

D.Both rates must be equal

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: With the same principal, the slope of an interest-versus-time graph equals $Pr$ . A steeper slope therefore means a larger value of $r$ . Because both accounts have the same principal, the difference in slope must come from their rates.

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