IPMAT Indore Profit & Loss — PYPs

The Setup: We have a system of profit/loss constraints across two different scenarios for an unknown quantity of apples and mangoes with fixed unit costs. We must solve the linear equations to find the exact number of apples.

Step 1: Define variables and calculate per-item cost variations.
Let $A$ be the number of apples (Cost = $20$ each).
Let $M$ be the number of mangoes (Cost = $25$ each).
* $+10\%$ on apples = $+2$ profit per apple.
* $-20\%$ on mangoes = $-5$ loss per mango.
* $-20\%$ on apples = $-4$ loss per apple.
* $+10\%$ on mangoes = $+2.5$ profit per mango.

Step 2: Formulate Scenario 1 to find the quantity ratio.
In the first scenario, the net profit is exactly $0$:

$$2A - 5M = 0 \implies 2A = 5M \implies M = 0.4A$$

Step 3: Formulate Scenario 2 to solve for $A$.
In the second scenario, the net loss is $150$ (meaning profit is $-150$):

$$-4A + 2.5M = -150$$

Multiply by $-1$ for clarity:

$$4A - 2.5M = 150$$

Step 4: Substitute the ratio and calculate $A$.
Substitute $M = 0.4A$ into the second equation:

$$4A - 2.5(0.4A) = 150$$

$$4A - 1A = 150$$

$$3A = 150 \implies A = 50$$

Final Answer: 50

The Setup: Revenue is $\text{Price}\times\text{Quantity}$, so a percentage change in each multiplies rather than adds. Anyone who subtracts $25\%-20\%=5\%$ has fallen for the trap this question exists to set.
Step 1: Define the baseline. Let the original price be $P$ and quantity $Q$, so the original revenue is $R_1=PQ$.
Step 2: Apply the percentage shifts as multipliers.
* Price falls 25%: new price $=(1-0.25)P=0.75P$
* Quantity rises 20%: new quantity $=(1+0.20)Q=1.20Q$

Step 3: Calculate the new revenue.

$$R_2=(0.75P)(1.20Q)=(0.75\times 1.20)PQ=0.90\,PQ$$

Step 4: Convert to a percentage drop. Revenue went from $1.00PQ$ to $0.90PQ$, a fall of $0.10PQ$:

$$\frac{R_1-R_2}{R_1}\times 100=\frac{0.10PQ}{PQ}\times 100=10\%$$

So $x=10$. Sanity check with real numbers: at $P=100$, $Q=100$, revenue goes from 10,000 to $75\times 120=9{,}000$ - down 1,000, which is 10%.

Final Answer: 10

The Setup: A chocolate's price undergoes a sequential $x\%$ increase followed by an $x\%$ decrease. The resulting price is $96.76\%$ of the original price, and we must determine the value of $x$.

Step 1: Construct the sequential price multiplier function.
Let $P$ represent the baseline price.
An $x\%$ increase applies a multiplier of $(1 + \frac{x}{100})$.
An $x\%$ decrease applies a subsequent multiplier of $(1 - \frac{x}{100})$.
$$\text{New Price} = P \times \left(1 + \frac{x}{100}\right) \times \left(1 - \frac{x}{100}\right)$$

Step 2: Simplify via the difference of squares identity.
$$\text{New Price} = P \left(1 - \frac{x^2}{10000}\right)$$

Step 3: Equate to the provided net proportional change.
The final state is $96.76\%$ (or $0.9676$) of $P$.
$$P \left(1 - \frac{x^2}{10000}\right) = 0.9676 P$$
Divide out $P$ from both sides since initial price is arbitrary:
$$1 - \frac{x^2}{10000} = 0.9676$$

Step 4: Solve for the absolute rate $x$.
$$\frac{x^2}{10000} = 1 - 0.9676 = 0.0324$$
$$x^2 = 324$$
Because a percentage rate scaling magnitude must be positive, take the principal square root:
$$x = 18$$

Final Answer: 18

The Setup: A solid golden ball costing INR $1,000,000$ is melted into smaller spherical beads. The radius of each bead is one-fifth of the original ball's radius. The beads are sold at a $20\%$ discount for a total overall profit of $20\%$. We must find the listed price per bead.

Step 1: Calculate the number of beads produced.
Let the radius of the original ball be $R$ and the radius of a bead be $r$. We are given $r = \frac{R}{5}$.
The volume of a sphere is proportional to the cube of its radius.

$$\text{Number of beads} = \frac{\text{Volume of original ball}}{\text{Volume of one bead}} = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = \left(\frac{R}{r}\right)^3 = (5)^3 = 125$$

So, $125$ identical beads are produced.

Step 2: Calculate the required revenue per bead.
The total cost is INR $1,000,000$.
The total profit made is $20\%$, so the total revenue generated is $120\%$ of the cost:

$$\text{Total Revenue} = 1,000,000 \times 1.20 = 1,200,000$$

Since this revenue comes from selling $125$ beads, the selling price ($SP$) per bead is:

$$SP = \frac{1,200,000}{125} = 9600$$

Step 3: Calculate the listed price.
The selling price is derived after a $20\%$ discount is applied to the listed price ($L$).

$$0.80 \times L = 9600$$

$$L = \frac{9600}{0.80} = 12000$$

Final Answer: 12000

The Setup: This is a classic Profit & Loss side quest. Ram is basically running a whole fruit cartel. We are given the final profit margins and need to reverse-engineer the original marked price (the bait) of the mangoes. We'll break this down by isolating the cost and revenue for each fruit type.

Step 1: Calculate the base stats (Total Cost Price).
Ram drops some cash upfront to acquire the inventory. We calculate the total cost price by multiplying the quantity by the price per unit.
$$Total\ CP = 3 \times 20 + 5 \times 38 = 60 + 190 = 250\ rupees$$

Step 2: Calculate the total loot (Total Selling Price).
The problem states he secures a massive 32% overall profit on the entire stash. We apply a $1.32$ multiplier to the total cost.
$$Total\ SP = 250 \times 1.32 = 330\ rupees$$

Step 3: Isolate the orange revenue.
The oranges alone gave a cracked 50% profit buff. The cost price for just the oranges was 60 rupees.
$$Oranges\ SP = 60 \times 1.5 = 90\ rupees$$

Step 4: Find the mango selling price.
By subtracting the orange revenue from the total revenue, we find exactly how much the mango bag sold for.
$$Mangoes\ SP\ (for\ 5) = 330 - 90 = 240\ rupees$$
To find the selling price of just one mango, divide by the squad size of 5.
$$SP\ of\ one\ mango = \frac{240}{5} = 48\ rupees$$

Step 5: Reverse-engineer the Marked Price.
Ram offered a 20% discount on mangoes to pass the vibe check for his customers. This means the final selling price is exactly 80% of the marked price ($SP = 0.80 \times MP$). We divide the single mango SP by $0.80$ to find the original inflated tag.
$$MP = \frac{48}{0.80} = 60\ rupees$$

Final Answer: 60