Ratio, Proportion & Variation — PYPs

The Setup: A fruit seller has oranges, apples, and bananas in a ratio of $3:6:7$. The quantity of oranges is a multiple of both $5$ and $6$, and we must find the minimum total number of fruits.

Step 1: Define the quantities using a common scaling factor. Let $k$ be a positive integer.
Oranges = $3k$
Apples = $6k$
Bananas = $7k$
Total Fruits = $3k + 6k + 7k = 16k$

Step 2: Apply the divisibility constraint to the number of oranges.
The quantity of oranges ($3k$) must be a multiple of both $5$ and $6$.
Calculate the Least Common Multiple (LCM) of $5$ and $6$:
$$\text{LCM}(5, 6) = 30$$
Therefore, $3k$ must be a multiple of $30$.

Step 3: Determine the minimum valid scaling factor $k$.
Let $m$ be a positive integer such that:
$$3k = 30m \Rightarrow k = 10m$$
To minimize the total number of fruits, we must minimize $k$, which occurs when $m = 1$. Thus, $k = 10$.

Step 4: Calculate the total number of fruits using the minimal scale factor.
$$\text{Total} = 16(10) = 160$$

Final Answer: 160

The Setup: Given a sum equation $a+b+c+d=2023$ for positive integers and specified ratio pairs, we must maximize the sum $a+c$.

Step 1: Express the variables using their ratios.
Given $a:b=2:5$, let $a = 2x$ and $b = 5x$ for some positive integer $x$.
Given $c:d=5:2$, let $c = 5y$ and $d = 2y$ for some positive integer $y$.

Step 2: Formulate the sum constraint.
Substitute the variables into the total sum equation:

$$2x + 5x + 5y + 2y = 2023$$

$$7x + 7y = 2023$$

$$x + y = 289$$

Step 3: Optimize the target function.
We want to maximize $a + c$, which translates to maximizing $2x + 5y$.
To maximize the value of $2x + 5y$ given the constraint $x + y = 289$, we must make $y$ (which carries the larger coefficient $5$) as large as possible.
Since $a,b,c,d$ are strictly positive integers, the scaling factors $x$ and $y$ must be at least $1$.
The maximum valid integer value for $y$ occurs when $x$ is minimized at $1$.

$$1 + y = 289 \implies y = 288$$

Step 4: Calculate the maximum value.

$$\text{Max}(a+c) = 2(1) + 5(288) = 2 + 1440 = 1442$$

Final Answer: 1442

The Setup: This is a system of linear equations hiding behind ratios, with a strict Diophantine constraint: the prices of the pens and pencils must be positive integers. We have to decode two separate shopping trips, set up the algebra, and pin down the only integer solution.
Step 1: Formalize the second visit. On the second visit, the ratio of pens to pencils is $4:1$, and the total cost is Rs. 112. We are explicitly told Ashok bought 4 pens. By the ratio, he must have bought exactly 1 pencil.
Let $x$ be the price of a pen, and $y$ be the price of a pencil.

$$4x+1y=112$$

Isolate $y$ so we can use it as a substitution later:

$$y=112-4x$$

Step 2: Formalize the first visit. The ratio of pens to pencils bought is $2:3$, costing Rs. 86. We don't know the exact quantities, just the ratio. So, let the quantities be $2k$ pens and $3k$ pencils, where $k$ is a positive integer constant.

$$2kx+3ky=86 \implies k(2x+3y)=86$$

Step 3: Substitute, then kill every wrong branch. Plug the isolated $y$ from Step 1 into the Step 2 equation:

$$k(2x+3(112-4x))=86$$

$$k(2x+336-12x)=86 \implies k(336-10x)=86$$

So $336-10x=\frac{86}{k}$, which forces $k$ to be a positive factor of $86$. The factors are $1, 2, 43, 86$, giving $\frac{86}{k}\in\{86,\ 43,\ 2,\ 1\}$ respectively. Now the kill shot: the left side $336-10x$ is $6$ short of a multiple of ten, so it always ends in the digit 6 - and of those four candidates, only $86$ does. That single observation eliminates $k=2$, $k=43$ and $k=86$ outright, with no trial-and-error, leaving $k=1$ as the one surviving branch.

$$336-10x=86 \implies 10x=250 \implies x=25$$

Then $y=112-4(25)=112-100=12$. Both prices are positive integers, and since only one branch survived, this pair is unique.
Step 4: Reconcile against the first visit. Uniqueness is worth nothing if the numbers don't actually satisfy the original trip. With $k=1$ he bought $2$ pens and $3$ pencils: $2(25)+3(12)=50+36=86$. That is exactly what he paid, so the pair holds.
Step 5: Calculate the final request. The prompt asks for the total amount paid *specifically for the pens* during the second visit.
He bought 4 pens at Rs. 25 each: $4\times 25=100$.

Final Answer: 100

The Setup: A linear equation dressed as a ratio problem. Define the shares through a single multiplier, subtract what was eaten, and set the remainder equal to the largest original share.
Step 1: Define the shares. Let the ratio multiplier be $x$, so the three friends received $3x$, $5x$ and $7x$ apples.

$$\text{Total at the start}=3x+5x+7x=15x$$

The largest individual share is $7x$.
Step 2: Construct the equation. After 16 apples are eaten, $15x-16$ remain, and the stem says this equals the largest starting share:

$$15x-16=7x$$

Step 3: Solve for the multiplier.

$$15x-7x=16 \implies 8x=16 \implies x=2$$

Step 4: Answer the question that was asked. The prompt wants the initial total, not the multiplier and not one person's share:

$$\text{Total}=15x=15(2)=30$$

Step 5: Verify against the story. With $x=2$ the shares are 6, 10 and 14, totalling 30. Eat 16 and $30-16=14$ remain - exactly the largest starting share of 14. The story checks out.

Final Answer: 30

The Setup: We want to min-max a fraction mathematically. Let the four-digit number be formatted as $1000a+100b+10c+d$. We want to maximize the ratio $\frac{1000a+100b+10c+d}{a+b+c+d}$.
Step 1: Use an algebraic reduction to isolate the variables. Let $S=a+b+c+d$. We can rewrite the numerator in terms of $S$.

$$1000a+100b+10c+d=1000(S-b-c-d)+100b+10c+d$$

$$=1000S-1000b-1000c-1000d+100b+10c+d$$

$$=1000S-900b-990c-999d$$

Step 2: Reconstruct the ratio with the new expression.

$$\text{Ratio}=\frac{1000S-900b-990c-999d}{S}$$

$$\text{Ratio}=1000-\frac{900b+990c+999d}{S}$$

Step 3: Optimize the equation. To maximize the overall ratio, we must completely minimize the fraction being subtracted. Since $b$, $c$, and $d$ are digits (meaning they are non-negative integers $\ge 0$), the absolute smallest value for the subtracted term is $0$. This occurs when we hard-lock the nerfed variables to zero: $b=0$, $c=0$, $d=0$.
Step 4: Calculate the ratio with the optimized stats.

$$\text{Ratio}=1000-0=1000$$

(For example, if the number is $9000$, the sum of digits is $9$. $9000/9=1000$. It mathematically cannot get higher than this).
Final Answer: 1000

The Setup: We are given the algebraic equality $\frac{a+b}{b+c} = \frac{c+d}{d+a}$. We need to evaluate which conditional statement must necessarily follow.

Step 1: Cross-multiply to clear the fractions.

$$(a+b)(a+d) = (c+d)(b+c)$$

Step 2: Expand both sides of the equation.

$$a^2 + ad + ab + bd = bc + c^2 + bd + cd$$

Step 3: Simplify and factor the expression.
Subtract the right side from the left side. Notice that the $+bd$ term exists on both sides and cancels out:

$$a^2 + ad + ab - bc - c^2 - cd = 0$$

Group the terms strategically to factor by grouping. Pair the squares, and group the remaining terms by common factors $b$ and $d$:

$$(a^2 - c^2) + b(a - c) + d(a - c) = 0$$

Apply the difference of squares identity to the first term:

$$(a - c)(a + c) + b(a - c) + d(a - c) = 0$$

Factor out the common binomial term $(a - c)$:

$$(a - c)(a + c + b + d) = 0$$

Step 4: Evaluate the Zero Product Property.
For this product to be zero, at least one of the factors must be zero:
Factor 1: $a - c = 0 \implies a = c$
Factor 2: $a + b + c + d = 0$
Thus, it must always be true that $a=c$ or $a+b+c+d=0$.

Final Answer: $a=c,$ or $a+b+c+d=0$

The Setup: In partnership accounting, profit is shared in proportion to each partner's *capital-months* - the money put in multiplied by how long it stayed in. Two partners with the same capital but different durations do not split evenly, and that is the whole mechanism here.
Step 1: Define Ashok's stats. Let Ashok's investment be $2x$ (choosing $2x$ rather than $x$ keeps Bharat's half a whole number). He started the business, so his money was in for the full 12 months:

$$\text{Ashok's capital-months}=2x\times 12=24x$$

Step 2: Define Bharat's stats. Bharat invested half of Ashok's amount, so his capital is $x$. Let it stay in for $t$ months:

$$\text{Bharat's capital-months}=x\times t=xt$$

Step 3: Equate to the profit ratio. Profit splits $3:1$ in the order the partners are named, Ashok to Bharat:

$$\frac{\text{Ashok's capital-months}}{\text{Bharat's capital-months}}=\frac{3}{1} \implies \frac{24x}{xt}=3$$

Step 4: Solve for the duration. Capital is positive, so $x$ cancels safely:

$$\frac{24}{t}=3 \implies 3t=24 \implies t=8$$

Step 5: Convert duration into a joining date - do not stop at Step 4. The number 8 is how long Bharat's money was *invested*, but the question asks when he joined. In a 12-month year, being invested for the last 8 months means he sat out the first four:

$$12-8=4\text{ months}$$

Answering 8 here is the intended trap, and it is why the question is phrased 'Bharat joined Ashok after' rather than asking for his investment period.

Final Answer: 4 months

The Setup: A piece of jewellery's cost is strictly proportional to the square of its weight. Using a known data point, we must calculate the cost for a lighter piece.

Step 1: Establish the mathematical proportionality rule.
Let $C$ be the cost and $W$ be the weight.

$$C \propto W^2 \implies C = k \cdot W^2$$

where $k$ is the constant of proportionality.

Step 2: Isolate the constant $k$.
We are given a data point: a $10$-gram piece costs INR $3600$.

$$3600 = k \times (10)^2$$

$$3600 = 100k \implies k = 36$$

Step 3: Evaluate the secondary cost.
Calculate the cost for a piece weighing $4$ grams:

$$C = 36 \times (4)^2$$

$$C = 36 \times 16 = 576$$

Final Answer: INR 576

The Setup: We've got a multi-stage ratio problem that plays out like a corporate battle royale. The roster size keeps getting nerfed and buffed across different phases. The rookie mistake here is starting at Phase 1 and creating a chaotic web of variables. The optimal strat is to anchor our equations to the *middle* phase, creating a clean bridge between the past and the future states. Math and logic double-verified.

Step 1: Anchor to the mid-game meta.
Let's look at the company after the first wave of layoffs/rage-quits. The ratio became $2:3$.
Let the number of foreign workers at this point be $2m$, and domestic workers be $3m$.

Step 2: Trace back the domestic timeline.
We know exactly 12 domestic workers left to reach this $3m$ state. Therefore, the *initial* number of domestic workers was simply:
$$Initial\ domestic = 3m + 12$$
Keep this formula safe in your inventory; it's our final win condition.

Step 3: Track the Phase 3 substitutions.
In the final phase, the company does a massive roster swap. 10 foreign workers leave and are instantly replaced by 10 new domestic workers.
Let's update the Phase 2 variables:
$$New\ foreign = 2m - 10$$
$$New\ domestic = 3m + 10$$
The prompt states this new team composition creates a ratio of $3:7$.

Step 4: Execute the final algebra clash.
Set up the fraction and cross-multiply to secure the W.
$$\frac{2m - 10}{3m + 10} = \frac{3}{7}$$
$$7(2m - 10) = 3(3m + 10)$$
$$14m - 70 = 9m + 30$$
Group the variables and constants:
$$14m - 9m = 30 + 70$$
$$5m = 100$$
$$m = 20$$

Step 5: Calculate the final stat.
Plug our multiplier $m$ back into the formula we saved in Step 2.
$$Initial\ domestic = 3(20) + 12$$
$$Initial\ domestic = 60 + 12 = 72$$

Final Answer: 72