Unit 6 - Practice Quiz

In the context of simple and compound interest, 'P' always represents the Principal, which is the initial amount of money borrowed or invested.

Using the formula for Simple Interest, . Here, P=SI = \frac{500 \times 10 \times 2}{100} = $100.

The Amount is the final sum after a certain period. It is calculated as Amount = Principal + Interest.

Compound interest is calculated on the principal plus the accumulated interest from previous periods. This 'interest on interest' effect makes it grow faster than simple interest, which is always calculated only on the original principal.

The key feature of compound interest is that interest is earned on previously earned interest. So, for the second year, the new principal is the original principal plus the interest from the first year.

The standard formula for the final amount (A) when interest is compounded annually is , where P is the principal, R is the rate, and T is the time in years.

In simple interest, the interest earned each year is constant. Therefore, if the interest for one year is 50 \times 3 = $150.

For the first year, compound interest is the same as simple interest. $CI = SI = \frac{1000 \times 5 \times 1}{100} = $50.

'T' represents the Time period for which the money is borrowed or invested, usually measured in years.

First, calculate the Simple Interest: 800 + $120 = $920.

In the first year, compound interest is calculated only on the principal, just like simple interest. The 'compounding' effect only begins from the second year onwards.

'Per annum' is a Latin phrase that means 'for each year' or 'annually'. It specifies that the interest rate is for a period of one year.

The time period must be in years. 6 months is 0.5 years. So, $SI = \frac{1000 \times 10 \times 0.5}{100} = $50.

Simple interest is always calculated on the original principal amount, which does not change over the investment or loan period.

'R' stands for the Rate of Interest, which is the percentage at which interest is charged on the principal, usually on a per-year basis.

To double your money, the interest earned must equal the principal (SI = P). Using , we get . This simplifies to , so R = 10%.

'Compounded annually' means the interest earned during the year is added to the principal at the end of the year. This new, larger principal is then used to calculate the interest for the next year.

Let P be the principal. The simple interest for 1 year would be . For 2 years, it would be . The compound interest for 2 years is . If , then . The simple interest is $0.2P = 0.2 \times 1000 = $200.

Simple interest is calculated only on the original principal. The frequency of calculation (e.g., annually, semi-annually) does not affect the total simple interest earned. This frequency is a concept relevant to compound interest.

First, calculate the interest for one year: 2000 + $100 = $2100.

The interest for the period between the 5th and 8th year (3 years) is the difference in the amounts. \nSI for 3 years = ₹12,005 - ₹9,800 = ₹2,205. \nSI for 1 year = ₹2,205 / 3 = ₹735. \nSI for 5 years = ₹735 × 5 = ₹3,675. \nPrincipal (P) = Amount after 5 years - SI for 5 years = ₹9,800 - ₹3,675 = ₹6,125. \nRate (R) = () = () = 12%.

Let the two parts be and . We have . \nGiven, SI on = SI on . \n() = (). \n, which simplifies to . \nThe sum is divided in the ratio 3:4. \nFirst part () = . \nInterest on this part = .

First, calculate the amount after 2 full years. \nAmount after 2 years = . \nNow, calculate the simple interest on this amount for the remaining 6 months (0.5 years). \nSI for next 6 months = . \nTotal Amount = ₹9,680 + ₹484 = ₹10,164. \nTotal Compound Interest = Total Amount - Principal = ₹10,164 - ₹8,000 = ₹2,164.

For 2 years, the formula for the difference between CI and SI is: Difference = . \nGiven, Difference = ₹128 and R = 8%. \n. \n. \n. \n.

Let the principal be P and the rate be R. \nWe have . \nGiven, . \n. \nTaking the cube root, . \nNow, we need to find T when the sum becomes 16 times. \n. \n. \nSubstitute the value of from the first part: \n. \nSince , we get T = 4 years.

Simple Interest for 2 years = ₹1,600. So, SI for 1 year = ₹1,600 / 2 = ₹800. \nFor the first year, CI is the same as SI, so CI for 1st year = ₹800. \nCompound Interest for 2 years = ₹1,680. \nCI for 2nd year = Total CI - CI for 1st year = ₹1,680 - ₹800 = ₹880. \nThe difference between the CI of the 2nd year and the 1st year is the interest earned on the first year's interest. \nDifference = ₹880 - ₹800 = ₹80. \nThis ₹80 is the interest on ₹800 for 1 year. \nRate = () = () = 10%.

Let the value two years ago be P. The rate of depreciation is R = 20%. \nThe formula for depreciation is: Present Value = . \nHere, Present Value = ₹1,60,000, R = 20%, T = 2 years. \n. \n. \n.

Let the shares of the younger (Y) and older (O) son be ₹Y and ₹O respectively. \nThe younger son's money is invested for 18 - 13 = 5 years. \nThe older son's money is invested for 18 - 15 = 3 years. \nThe amounts they receive at age 18 are equal. \n. \n. \n. \nThe ratio of their shares Y:O is 400:441. \nYounger son's share = .

Case 1: Compounded Annually. \nInterest for 1st year = . Amount = ₹5200. \nInterest for next 1/2 year = . \nTotal CI (Annually) = ₹200 + ₹104 = ₹304. \nCase 2: Compounded Half-yearly. \nRate = 4%/2 = 2% per half-year. Time = half-years. \nAmount = . \nCI (Half-yearly) = ₹5306.04 - ₹5000 = ₹306.04. \nDifference = ₹306.04 - ₹304 = ₹2.04.

The sum borrowed (Principal) is the sum of the present worths of the two installments. \nPresent Worth of 1st installment = . \nPresent Worth of 2nd installment = . \nTotal Principal borrowed = ₹4200 + ₹4000 = ₹8,200. \nTotal Amount paid = 2 × ₹4,410 = ₹8,820. \nTotal Interest paid = Total Amount Paid - Principal = ₹8,820 - ₹8,200 = ₹620.

Let the sum lent (Principal) be P. \nRate R = 4%, Time T = 5 years. \nSimple Interest (SI) = . \nAccording to the question, SI = P - 520. \nSo, . \nP = \frac{520 \times 5}{4} = 130 \times 5 = ₹650$$.

Amount after 3 years = ₹926.10. Amount after 2 years = ₹882. \nThe interest for the 3rd year is the difference between these amounts. \nInterest for 3rd year = ₹926.10 - ₹882 = ₹44.10. \nThis interest is calculated on the amount at the end of the 2nd year (₹882). \nRate (R) = . \nNow, let the principal be P. Amount after 2 years = . \n. \n$.

When interest rates are different for different years, the amount is calculated as: \n. \n. \n. \n. \n$.

Let the sum lent at 8% be ₹x. Then the sum lent at 10% is ₹(4000 - x). \nTotal annual interest = Interest from part 1 + Interest from part 2. \n. \nMultiply by 100: . \n. \n. \n. \n.

First, convert percentages to fractions for easier calculation: \n. . . \nPopulation after 1st year = . \nPopulation after 2nd year = . \nPopulation after 3rd year = .

Let the Principal be P. For the sum to double, the Amount must be 2P. \nThis means the Simple Interest (SI) earned must be equal to the Principal. \nSI = Amount - Principal = 2P - P = P. \nUsing the SI formula, . \n. \nDivide by P on both sides: . \n years.

We use the compound interest formula: . \nGiven, A = 13,310, P = 10,000, T = 3 years. \n. \n. \n. \nRecognizing that and , we get: \n. \nTaking the cube root of both sides: . \n. \n. \n.

Step-by-step calculation: \nEnd of Year 1: \nInterest = 10% of 20,000 = ₹2,000. \nAmount due = 20,000 + 2,000 = ₹22,000. \nHe pays ₹5,000. Remaining principal for Year 2 = 22,000 - 5,000 = ₹17,000. \nEnd of Year 2: \nInterest = 10% of 17,000 = ₹1,700. \nAmount due = 17,000 + 1,700 = ₹18,700. \nHe pays ₹5,000. Remaining principal for Year 3 = 18,700 - 5,000 = ₹13,700. \nEnd of Year 3: \nInterest = 10% of 13,700 = ₹1,370. \nTotal amount to be paid = 13,700 + 1,370 = ₹15,070.

Compounded Annually: \n. \n. \nCompounded Half-Yearly: \nRate = 10%/2 = 5% per half year. Time = 3 years × 2 = 6 half-years. \n. \n. \nDifference: \nDifference = . The closest option is ₹227.

In compound interest, the interest for a year is calculated on the principal plus the accumulated interest of the previous years. \nThe interest for the first year is ₹2,250. The interest for the second year is ₹2,385. \nThe extra interest in the second year is due to the interest earned on the first year's interest. \nExtra Interest = ₹2,385 - ₹2,250 = ₹135. \nThis ₹135 is the interest on ₹2,250 for one year. \nRate = () × 100 = () × 100 = 6%.

The difference in compound interest between two consecutive years is the interest on the previous year's interest.
Interest on the 2nd year's interest = CI for 3rd year - CI for 2nd year = $1440 - $1200 = $240.
The rate of interest, R, can be calculated as:
.
Now, let P be the principal. The interest for the first year (CI₁) is .
The interest for the second year (CI₂) is the interest on the amount after the first year, i.e., on .
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We know .
Since , we have:
.

The present value of the loan must equal the sum of the present values of the installments. Let each installment be . The rate .
The present value (PV) of the loan is given by:







Since , we have:
.

Maybe the question should be: The difference between CI and SI for 3 years is times the difference for 2 years. Then .
Let's re-write the current question to be solvable. Maybe the value is wrong.
If P=8000, R=10%, then .
.
.
Let's change the question value to 168. Then the answer is 8000.
New question text: The difference between the compound interest and simple interest on a certain sum for 3 years is $168 more than the difference for 2 years. If the rate of interest is 10% per annum in both cases, find the sum.
Let's re-calculate: . Using the derived relation: .
. This works. My logic was correct, the initial value was just not clean.

. . .
Numbers are not clean again. Let me check the effective rate. Effective CI rate = .
Effective gain rate = . Let's assume the gain is $1350 for clean numbers.
New Question: ...At the end of one year, he gains $1,350...
Gain = .
Given gain is $1,350.
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So the sum borrowed is $62,500.

First, calculate the nominal amount after 3 years.
. (This is option C, a common mistake).
Next, calculate the real rate of return.

So, the real rate of return is approximately 4.854%.
The real value of the investment is the initial principal grown at the real rate of return.
.
Option B is the result of using the approximation , which gives . Option D is the deflated value of the principal itself, not the grown investment.

Let the time be n years.
Simple Interest (SI) after n years: .
Compound Interest (CI) after n years: Rate is per half-year, and time is half-years.
.
We need to find the minimum n such that .


.
This inequality is not solved algebraically. We must test the options.
Maybe I should make it "at least double", i.e. 100% greater.
Let's try that. .
.
Let's test options for this new condition:

• n=10: . . We need . False.
• n=11: . . We need . False.
• n=12: . . We need . False.

• n=13: . . We need . False.

  • n=14: . . We need . True.
    Okay, the "double" question works. New Question: ...minimum number of full years (n) required for the total compound interest earned to be at least double the total simple interest earned.
    New options: ["12 years", "13 years", "14 years", "15 years"]
    Correct option: "14 years".
    Explanation: ... we need .

    Test n=13: . And . Here .
    Test n=14: . And . Here .
    So, n=14 is the first full year where the condition is met.

Let the principal be P and the rate be R. The amount A after time t is .
We are given:
(Eq. 1)
(Eq. 2)
We need to find .
Divide Eq. 2 by Eq. 1:


. This is the growth factor for a 4-year period.
Now, we know that .

.

Let the principal be P, rate be R, and .
The difference for 2 years is .
The difference for 3 years is .
We are given and .
We can rewrite in terms of :
.
Now, substitute the given values:



.
Since , the rate .

Let's check option C: . Then . This means they are equal, ratio is 1:1, which is not 11:10.
Let's check option D: . . Ratio $1.178$.
Let's re-calculate . Ah! . is not integer. Let's assume the question meant time periods are 1 and 2 years.
. . Still the same ratio.
Okay, there must be a typo in the sum. Let's make the sum $23,100 as I tested before.
..`
Then . .
.
So the second part is $11,000$. Let's provide this as an option.
New Options: ["$11,000", "$12,100", "$10,000", "$11,550"]. Correct: "$11,000".

Let the principal be P and the annual rate be R%. Let .
Interest for 1 year (annual compounding) = .
Interest for 1 year (semi-annual compounding) = .
The difference is .
Given (Eq. 1).
Interest for 2 years (annual compounding) = .
Given (Eq. 2).
Substitute Eq. 1 into Eq. 2:


.
Now we have two equations: and .
Divide the second by the first: . This is very ugly. Let me check my numbers.
Let's assume a rate, say R=20%, r=0.2.
.
.
. This is close to 3362. The rate must be slightly lower.
Let's try R=18%, r=0.18.
.
.
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There is an error in my reasoning or the numbers are flawed. Let's start with a nice answer, P=20000.
If P=20000 and the difference is 81: . Not nice.
Let's rethink the problem. . This is the difference in interest for ONE YEAR. It's also the difference between CI and SI for 1 year, where CI is semi-annual.
Let's make the numbers cleaner. Let , . .
Diff = .
Interest for 2 years = .
Let's use these values. Question: ...difference... would be $50 more... interest for two years would be $4,200...
With these new values:
.
.
.
Now, . So .
And . This works perfectly.
I will use these new numbers in the question.

Maybe the numbers were intended to be simpler. Let's say FV for B was $3000. PW = 3000/1.2 = 2500. Then no settlement.
What if FV for B was $3060? PW = 3060/1.2 = 2550. Settlement B pays A $50.
This is a better question. Let's modify it.
B owes A $3,060 due in 2 years at 10% SI...`
Then and . B pays A $50. This is clean and tests the same concept.

Let P be the principal and R be the rate of interest.
Statement I: Gives Simple Interest (SI) for 3 years.
.
This is one equation with two variables (P and R), so it is not sufficient to find P.
Statement II: Gives the difference between CI and SI for 2 years.
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This is also one equation with two variables, so it is not sufficient by itself.
Combining both statements:
We have two distinct equations:
1)
2)
Divide equation (2) by equation (1):
.
Now substitute R=10 back into equation (1):
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Since we can find a unique value for P, both statements together are sufficient.

Okay, the option $8,988 is likely based on an intermediate rounding or a slight number change. Let's find values that lead to it.
If , . . . . Then SI/yr = . Amount 2y = 7000+1862=8862. Not 8800.
There must be a typo in the original question's values or options. Let me create clean values.
P=10000, R=10%. SI 1yr=1000. Amt 2y=12000. Amt 3y=13000.
..amounts to $12,000 in 2 years and $13,000 in 3 years...`
Then P=10000, R=10%. CI Amount = .

Let the principal be P and rate be R. The formula for compound amount is .
Given that the sum becomes 8 times in 3 years:
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Taking the cube root of both sides:
. This means the sum doubles every year, so the rate R is 100%.
Now we need to find the time (t) for the sum to become 32 times itself:
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Substitute the value of we found:
.
We know that .
Therefore, years.

This question tests the calculation for a fractional time period under annual compounding.
Bank A (10% p.a. compounded annually for 1.5 years):
For the first full year, compound interest is applied.
Amount after 1 year = .
For the next 0.5 years, simple interest is applied on this new amount.
Interest for next 0.5 year = .
Total Amount from Bank A = .
Total Interest from Bank A = $1,550.
Bank B (9% p.a. compounded semi-annually for 1.5 years):
Rate per period = .
Number of periods = .
Amount from Bank B = .
.
Amount = .
Total Interest from Bank B = $1,411.66125.
Bank A Interest: $1550$.
Bank B Interest: $1411.66...$
Difference is huge. Let me change Bank A rate to be closer. Say 8% annual. Amount 1y = 10800. Interest for 0.5y = . Total A = 11232. Interest = 1232.
Let's try 9.5% for Bank A. Amount 1y = 10950. Interest 0.5y = . Total Int A = 1470.125.
Difference = .
Let's assume the question meant CI for Bank A should be calculated as . This is non-standard but could be intended.
. Interest = $1536$.
Difference = .
Let's change Bank B's rate to make the numbers closer. Say Bank B is 10% semi-annually.
Rate = 5%. Periods = 3. Amount = . Interest B = $1576.25.
Interest A (from before) = $1550$.
Difference = . This is a good scenario.
Let's use these rates: A: 10% annual, B: 10% semi-annual.
I will use these values for the question.

The interest is simple, so it accrues linearly.
The interest earned in the period between 2 years and 3.5 years is:
Interest for years = $11,760 - $10,080 = $1,680.
Now, we can find the interest for 1 year:
SI for 1 year = $\frac{1680}{1.5} = \frac{1680}{3/2} = 1680 \times \frac{2}{3} = 560 \times 2 = $1,120.
Next, find the principal (P). The amount after 2 years is the principal plus 2 years of SI.


$P + 2240 = 10080 \implies P = 10080 - 2240 = $7,840.
Finally, calculate the rate of interest (R):
.
.
. . So it is exactly . My options are bad.
Let's create clean numbers. Let P=8000, R=12%. SI/yr=960.
Amt 2yr = 8000+1920=9920.
Amt 3.5yr = 8000+3.5*960 = 8000+3360=11360.
Q: ...amounts to $9,920 in 2 years and $11,360 in 3.5 years...
Then .
.
.
. This works perfectly.

Let the shares of the younger son (12 years) and elder son (14 years) be Y and E respectively. So, .
The younger son's money will be invested for years.
The elder son's money will be invested for years.
The amounts they receive at age 18 must be equal.
Amount for Younger son: .
Amount for Elder son: .
Setting the amounts equal:


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Now substitute this into the sum equation:


.
.
So, .
The share of the younger son (Y) is:
.
. Not helpful. The sum must be a multiple of 841.
Let's make the sum $126,150Y = (400/841) 126150 = 400150 = 60000$. $E=441*150=66150$. This works.
Let's try with my original option. Is $57,600 correct? If Y=57600, E=120000-57600=62400.
Ratio Y/E = .
We need the ratio to be . So the option is wrong.
Let's change the rate to 10%.
.
.
Ratio is . Sum of parts = 221.
. Not clean.
Let's go back to the ratio for R=5%. Total parts 841. Let's make the total sum $168,200$.
Then . .
This provides a clean answer. The original values must be wrong. I will correct the sum in the question to $168,200 and the option to $80,000.

Let the total capital be P.
The parts are:
at .
at .
The remainder is at .
The total annual simple interest (I) is the sum of the interests from the three parts.




To eliminate fractions, multiply by the LCM of 3 and 6, which is 6.



.
The capital is $6,600.

Let the principal for Scheme A be and for Scheme B be . The time period is 3 years.
Interest from Scheme A (Simple Interest):
.
Interest from Scheme B (Compound Interest):
.
.
We are given that the interests are identical, so .
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To find the ratio , we rearrange the equation:
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Therefore, the ratio of the principals is 331:360.