The cost of one pencil, two pens and four erasers is ₹ 22 while the cost of five pencils, four pens and two erasers is ₹ 32. How much will three pencils, three pens and three erasers cost?

### Concept & Logic When asked to find the combined cost of equal quantities of multiple items (like 3 of each), look for a way to add or subtract the given equations to produce symmetric coefficients. You often don't need to find the individual cost of each item. ### Step-by-Step Solution **Given:** Let the cost of a pencil be $c$, a pen be $p$, and an eraser be $e$. 1. $1c + 2p + 4e = 22$ 2. $5c + 4p + 2e = 32$ **Calculation:** * Notice that adding the coefficients of the two equations yields a uniform sum for each variable. Let's add Equation 1 and Equation 2 together: $(1c + 5c) + (2p + 4p) + (4e + 2e) = 22 + 32$ $6c + 6p + 6e = 54$ * We need the cost of 3 pencils, 3 pens, and 3 erasers ($3c + 3p + 3e$). To get this, simply divide the entire new equation by 2: $\frac{6c + 6p + 6e}{2} = \frac{54}{2}$ $3c + 3p + 3e = 27$ * The total cost is ₹ 27. ### Exam Strategy & Shortcut Scan the target question first. Seeing "three pencils, three pens and three erasers" suggests a symmetrical target ($3x+3y+3z$). Look at the given quantities: $(1,2,4)$ and $(5,4,2)$. Adding them vertically gives $(6,6,6)$, which is a clean multiple of the target $(3,3,3)$. This bypasses the impossible task of solving a 3-variable system with only 2 equations. ### Common Pitfall The biggest trap is trying to solve for the individual prices of a pencil, pen, and eraser. With 3 variables and only 2 equations, finding unique individual prices is mathematically impossible, leading to wasted time and frustration. ### Final Answer **Therefore, the correct answer is ₹ 27.**