IPMAT Indore Simple & Compound Interest — PYPs

The Setup: Person A borrows Rs. $4000$ for $4$ years, splitting the principal between a $3\%$ simple interest rate and a $4\%$ simple interest rate. The total interest earned is Rs. $520$, and we must find the amount borrowed at $3\%$.

Step 1: Define the variables for the split principal.
Let $x$ represent the principal amount borrowed at the $3\%$ rate.
Let $4000 - x$ represent the remaining principal borrowed at the $4\%$ rate.

Step 2: Calculate the annualized interest yield.
The total interest accrued over $4$ years is $520$. Because it is simple interest, the annual interest is constant:
$$\text{Annual Interest} = \frac{520}{4} = 130$$

Step 3: Construct the linear equation for the annual interest.
$$0.03x + 0.04(4000 - x) = 130$$

Step 4: Solve for $x$.
Distribute the terms and isolate the variable:
$$0.03x + 160 - 0.04x = 130$$
$$-0.01x = -30$$
$$x = 3000$$

Final Answer: 3000

The Setup: We evaluate sequential deposits placed at the end of each year with growing principals and scaling simple interest rates over a bounded 5-year timeline from 'today' (Start of Year 1).

Step 1: Chart the timeline and investment parameters.
Five years from today implies the timeline ends exactly at the conclusion of Year 5 (T=5).
Simple Interest Formula: $\text{Interest} = P \times R \times T$.
* Deposit 1 (End of Yr 1, T=1): Earns $10,000$. Rate = $0.5\% = 0.005$. Time invested = $4$ years (T=1 to T=5).
* Deposit 2 (End of Yr 2, T=2): Income rises by 10k, so earns $20,000$. Rate rises by 0.5%, so = $1.0\% = 0.01$. Time invested = $3$ years.
* Deposit 3 (End of Yr 3, T=3): Earns $30,000$. Rate = $1.5\% = 0.015$. Time invested = $2$ years.
* Deposit 4 (End of Yr 4, T=4): Earns $40,000$. Rate = $2.0\% = 0.02$. Time invested = $1$ year.
* Deposit 5 (End of Yr 5, T=5): Earns $50,000$. Rate = $2.5\% = 0.025$. Time invested = $0$ years (cashed exactly as deposited).

Step 2: Calculate interest for each independent deposit.
* Interest 1: $10000 \times 0.005 \times 4 = 200$
* Interest 2: $20000 \times 0.010 \times 3 = 600$
* Interest 3: $30000 \times 0.015 \times 2 = 900$
* Interest 4: $40000 \times 0.020 \times 1 = 800$
* Interest 5: $50000 \times 0.025 \times 0 = 0$

Step 3: Sum the total interest.

$$\text{Total Interest} = 200 + 600 + 900 + 800 + 0 = 2500$$

Final Answer: 2500

The Setup: Let's speedrun this finance glitch. Compound interest beats simple interest only because of interest earned *on* the first year's interest - so demanding that CI be exactly twice SI is a very aggressive ask, and the rate it forces is correspondingly extreme.
Step 1: Define the base formulas. Let $P$ be the principal and $r$ the annual rate as a decimal.

$$SI=P\cdot r\cdot 2=2Pr, \qquad CI=P(1+r)^2-P$$

Step 2: Equate and expand. The condition is $CI=2\times SI$:

$$P(1+r)^2-P=2(2Pr)$$

Divide through by $P$ (a principal of zero is meaningless) and expand:

$$1+2r+r^2-1=4r \implies r^2+2r=4r$$

Step 3: Solve for the rate.

$$r^2-2r=0 \implies r(r-2)=0$$

The root $r=0$ would mean no interest at all, making $CI=SI=0$ - technically satisfying the equation but describing no loan. Discard it, leaving $r=2$ as a decimal, i.e. $200\%$.
Step 4: Sanity-check, because 200% looks absurd. Take $P=100$ at $r=200\%$:
* Simple interest: $100\times 2\times 2=400$
* Compound: $100(1+2)^2-100=900-100=800$

And $800=2\times 400$ exactly. The answer really is 200% - the reason it feels wrong is that doubling CI relative to SI is a far harsher demand than it sounds, and only a triple-per-year growth factor achieves it. Option 2 is the trap: solve correctly, get $r=2$, and then read that 2 as a percentage instead of as the decimal it is.

Final Answer: 200%

The Setup: The difference between Compound Interest (CI) and Simple Interest (SI) on a principal amount invested for $3$ years at a $10\%$ annual rate is INR $527$. We need to determine the original principal amount.

Step 1: State the 3-year CI and SI difference formula.
For a principal $P$ invested for exactly $3$ years at an annual interest rate $R$ (expressed as a percentage), the difference $D$ between CI and SI is given by the standard derived formula:

$$D = P \left(\frac{R}{100}\right)^2 \left(\frac{R}{100} + 3\right)$$

Step 2: Substitute the known values into the formula.
We are given $D = 527$ and $R = 10$.

$$527 = P \left(\frac{10}{100}\right)^2 \left(\frac{10}{100} + 3\right)$$

$$527 = P (0.1)^2 (0.1 + 3)$$

$$527 = P (0.01)(3.1)$$

$$527 = 0.031 P$$

Step 3: Solve for the Principal $P$.

$$P = \frac{527}{0.031} = \frac{527,000}{31}$$

Divide $527$ by $31$ to simplify the fraction:

$$527 \div 31 = 17$$

$$P = 17,000$$

Final Answer: 17000

The Setup: This is a compound interest min-maxing problem. Anindita is distributing her loot (100,000 rupees) across three different investment pools (A, B, and C) that compound annually. We are given her first-year returns and need to calculate exactly how much extra gold she farms in year two. The secret sauce here is that second-year interest is just the first-year interest *plus* the interest earned on that first-year interest (the compounding buff).

Step 1: Define the initial loadout.
Total investment is 100,000.
We already know Scheme A gets 30,000. That means the remaining 70,000 is split between B and C.
$$I_B + I_C = 70000$$

Step 2: Analyze the first-year interest.
First-year compound interest behaves exactly like simple interest. Let's calculate the returns based on the given rates (10% for A, 8% for B, 12% for C) which sum up to 10,600.
$$0.10(30000) + 0.08(I_B) + 0.12(I_C) = 10600$$
$$3000 + 0.08(I_B) + 0.12(I_C) = 10600$$
Subtract the guaranteed 3,000 from Scheme A:
$$0.08(I_B) + 0.12(I_C) = 7600$$

Step 3: Solve the system of equations.
We have a basic system here. Let's multiply the equation from Step 1 by 0.08 to set up an elimination:
$$0.08(I_B) + 0.08(I_C) = 5600$$
Subtract this from the interest equation in Step 2:
$$(0.08 I_B + 0.12 I_C) - (0.08 I_B + 0.08 I_C) = 7600 - 5600$$
$$0.04(I_C) = 2000$$
$$I_C = \frac{2000}{0.04} = 50000$$
If $I_C = 50000$, then $I_B$ must be 20,000 to complete the 70,000 remainder.

Step 4: Calculate the first-year interest breakdown.
Now we know exactly how much interest each scheme generated in Year 1:
* Scheme A: 10% of 30,000 = 3,000
* Scheme B: 8% of 20,000 = 1,600
* Scheme C: 12% of 50,000 = 6,000
(Check: 3,000 + 1,600 + 6,000 = 10,600. The math checks out perfectly.)

Step 5: Calculate the second-year interest.
Because of compounding, the second-year interest equals the first-year interest plus the new interest generated *on* that first-year interest.
* Scheme A Year 2 Interest: 3,000 + (10% of 3,000) = 3,000 + 300 = 3,300
* Scheme B Year 2 Interest: 1,600 + (8% of 1,600) = 1,600 + 128 = 1,728
* Scheme C Year 2 Interest: 6,000 + (12% of 6,000) = 6,000 + 720 = 6,720

Sum it all up for the final score:
Total Year 2 Interest = 3,300 + 1,728 + 6,720 = 11,748

Final Answer: 11748

The Setup: A principal sum of $10000$ rupees is split into two investment schemes with different interest protocols (Compound vs Simple). Setting up a linear equation evaluating their final amounts will isolate the initial split quantities.

Step 1: Define the variables and growth formulas for 1 year.
Let the investment in Scheme A be $x$.
Let the investment in Scheme B be $10000 - x$.
Scheme A ($10\%$ p.a. compounded half-yearly): The rate per half-year period is $5\% = 0.05$, and there are $2$ compounding periods.
$$\text{Amount}_A = x(1 + 0.05)^2 = x(1.1025) = 1.1025x$$
Scheme B ($12\%$ p.a. simple interest):
$$\text{Amount}_B = (10000 - x)(1 + 0.12) = 1.12(10000 - x)$$

Step 2: Construct the equation based on the given constraint.
The value of Scheme B exceeds Scheme A by $2310$ at the end of the year.
$$\text{Amount}_B - \text{Amount}_A = 2310$$
$$1.12(10000 - x) - 1.1025x = 2310$$

Step 3: Solve for $x$.
$$11200 - 1.12x - 1.1025x = 2310$$
$$11200 - 2310 = 2.2225x$$
$$8890 = 2.2225x$$
Recognize that $2.2225 = \frac{22225}{10000} = \frac{889}{400}$.
$$x = 8890 \times \frac{400}{889} = 10 \times 400 = 4000$$
So, $4000$ was invested in A, and $6000$ was invested in B.

Step 4: Calculate the total interest earned.
Interest from A $= 1.1025(4000) - 4000 = 410$
Interest from B $= 6000 \times 0.12 = 720$
$$\text{Total Interest} = 410 + 720 = 1130$$

Final Answer: 1130

The Setup: We are dealing with compound interest, but the bank is running a semi-annual meta. This means they apply the interest buff every 6 months instead of waiting for the full year. Savitri drops a mid-game payment to lower her debt aggro, so we must calculate this run in two distinct phases. Math, logic, and syntax have been double-verified.

Step 1: Calibrate the interest stats.
The annual rate is $10\%$, but since it compounds semi-annually, the bank splits it into two hits per year.
Half-year rate = $5\%$.
This translates to a growth multiplier of $1.05$ every 6 months.

Step 2: Calculate the Year 1 damage.
The initial principal is 10000. In one year, there are two compounding cycles (two half-years).
$$Amount\ after\ Year\ 1 = 10000 \times (1.05)^2$$
$$Amount\ after\ Year\ 1 = 10000 \times 1.1025 = 11025\ rupees$$

Step 3: Process the mid-game transaction.
Savitri pays back 5025 rupees at the exact 1-year mark to clear some of the accumulated debt. We subtract this from the total to find our new baseline for Phase 2.
$$New\ Principal = 11025 - 5025 = 6000\ rupees$$

Step 4: Calculate the final Year 2 boss phase.
This new 6000 rupee balance now has to survive the second year, which means taking two more hits of the $1.05$ multiplier.
$$Final\ Amount = 6000 \times (1.05)^2$$
$$Final\ Amount = 6000 \times 1.1025$$
To do the math cleanly: $6000 \times 1.1 = 6600$, and $6000 \times 0.0025 = 15$.
$$Final\ Amount = 6600 + 15 = 6615\ rupees$$

Final Answer: 6615