Directions (95-98): Read the following information carefully and answer the given questions. A shopkeeper sold five articles: P, Q, R, S and T. Further partial information in given in table below. | Article | Cost price | % Profit | Marked price | |---|---|---|---| | P | - | 40% | $(10.8b + 3m)$ | | Q | - | - | $(12a - n)$ | | R | - | 20% | - | | S | $(5b - m - n - 48)$ | - | - | | T | $(4a + \frac{m}{2})$ | - | $(13a - \frac{n}{2} + 6)$ | Note: (i) Value of '$n$' is 4 times of larger root of given equation. $K^2 - 32K + 252 = 0$ (ii) Article T is marked up ₹1280 more than its cost price and the difference between marked price and cost price of article T is 28% more than profit earned on article S. (iii) Discount given on article Q is 16.66%, and selling price of S is ₹$(5m + 7n + 4a + 296)$. (iv) Value of '$m$' is twice of missing value in the given sequence. 140, 136, 161, (?), 329, -200 If marked price of T is ₹80 less than cost price of R, which is sold after discount of ₹400 or $(\frac{5L}{14})\%$. Find the value of L.

### Concept & Profit Logic Strategy This problem utilizes the decoded variables from the central puzzle to connect the Marked Price of one article to the Cost Price of another. The core formula utilized here connects Discounts to Percentages: $$Discount\ \% = \frac{Discount\ Amount}{Marked\ Price} \times 100$$ ### Step-by-Step Solution **1. Recall Decoded Base Variables:** * From decoding the initial notes (see previous explanations): $a = 150$, $b = 200$, $m = 80$, $n = 72$. **2. Calculate Parameters for T:** * $MP_T = 13a - \frac{n}{2} + 6$ * $MP_T = 13(150) - \frac{72}{2} + 6 = 1950 - 36 + 6 = 1920$. **3. Relate to Article R:** * The problem states $MP_T$ is ₹80 less than $CP_R$. * Therefore, $CP_R = MP_T + 80 = 1920 + 80 = 2000$. * From the main table, Article R has a $20\%$ profit. * $SP_R = CP_R \times 1.20 = 2000 \times 1.2 = 2400$. **4. Apply Discount Data for R:** * Given discount amount = ₹400. * $MP_R = SP_R + Discount = 2400 + 400 = 2800$. * Actual Discount $\%$ = $(\frac{400}{2800}) \times 100 = \frac{100}{7}\%$. * We are given that the discount $\%$ is $(\frac{5L}{14})\%$. * Equating the two: $\frac{5L}{14} = \frac{100}{7} \implies 5L = 200 \implies L = 40$. ### Exam Strategy & Shortcut Instead of recalculating variable $a$ from scratch, store derived variables immediately in a table on your rough sheet during Data Interpretation blocks. From there, tracing the path $MP_T \rightarrow CP_R \rightarrow SP_R \rightarrow MP_R$ is a highly linear arithmetic chain that takes less than 30 seconds to execute. ### Common Pitfall A very common trap is applying the discount directly on the Selling Price rather than the Marked Price. Always remember that $MP = SP + Discount$, and the discount percentage is universally calculated with respect to the $MP$ as the base denominator. ### Final Answer Therefore, the correct answer is **40**.