At a certain fast food restaurant, Brian can buy 3 burgers, 7 shakes and one order of fries for ₹ 120 exactly. At the same place, it would cost ₹ 164.50 for 4 burgers, 10 shakes and one order of fries. How much would it cost for an ordinary meal of one burger, one shake and one fries?

### Concept & Strategy When asked for a specific combination of items (1 burger, 1 shake, 1 fries) from a system that has more variables (3) than equations (2), you cannot find the individual price of each item. Instead, you must find a multiplier for each equation so that when added together, they perfectly produce the target expression. ### Step-by-Step Solution **Given:** Let Burger = $b$, Shake = $s$, Fries = $f$. 1. $3b + 7s + 1f = 120$ 2. $4b + 10s + 1f = 164.5$ Target: Find $1b + 1s + 1f$. **Calculation:** * We need to find two multipliers, let's call them $x$ and $y$, such that: $x(\text{Equation 1}) + y(\text{Equation 2}) = \text{Target}$ $x(3b + 7s + 1f) + y(4b + 10s + 1f) = 1b + 1s + 1f$ * This gives us a mini-system of coefficients to solve: For $f$: $1x + 1y = 1 \implies x = 1 - y$ For $b$: $3x + 4y = 1$ * Substitute $x$: $3(1 - y) + 4y = 1$ $3 - 3y + 4y = 1 \implies y = -2$ * If $y = -2$, then $x = 1 - (-2) = 3$. * Let's verify these multipliers work for $s$ (the shakes): $7x + 10y = 7(3) + 10(-2) = 21 - 20 = 1$. It matches perfectly! * Apply these multipliers to the actual costs: Target Cost = $3(120) - 2(164.5)$ Target Cost = $360 - 329 = 31$ ### Exam Strategy & Shortcut An intuitive shortcut for finding combinations is to eliminate the most "awkward" variable first. Subtracting Eq 1 from Eq 2 gives: $1b + 3s = 44.5$ Now, how do we get $1b + 1s + 1f$? Notice that if we subtract our new equation ($1b + 3s$) from a multiple of the original equations, we can shape it. Actually, a faster visual trick: Eq 1 has $1f$. We need $1f$. If we take 3 times Eq 1, we get $3f$. If we subtract 2 times Eq 2, we get $-2f$, leaving $1f$. Let's try it: $3(3b, 7s, 1f) - 2(4b, 10s, 1f) = (9-8)b + (21-20)s + (3-2)f = 1b + 1s + 1f$. This visual matching is much faster than setting up the $x,y$ multiplier system. ### Common Pitfall The most common mistake is choosing "Cannot be determined". Students learn in basic algebra that you need 3 equations to solve for 3 variables. While true for individual variables, it is a dangerous trap here because the question only asks for a *linear combination* of those variables, which is perfectly solvable. ### Final Answer **Therefore, the correct answer is ₹ 31.**