Question

Three truck drivers, Amar, Akbar and Anthony stop at a road side eating joint. Amar orders 10 rotis, 4 plates of tadka, and a cup of tea. Akbar orders 7 rotis, 3 plates of tadka, and a cup of tea. Amar pays 80 for the meal and Akbar pays 60. Meanwhile, Anthony orders 5 rotis, 5 plates of tadka and 5 cups of tea. How much (in ₹) will Anthony pay?

• A. 75
• B. 80
• C. 95
• D. 100
• E. None of the above

Hint:

- Always assign variables to each item’s price.
- Often, specific linear combinations (like \(r+t+c\)) are uniquely determined even if individual prices are not.
- Check consistency by substituting into multiple equations.

Answer 1

The Correct Option is D

Step 1: Assign variables.
Let the cost of one roti = \(r\), one plate tadka = \(t\), and one cup tea = \(c\).
Step 2: Form equations from Amar and Akbar’s orders.
Amar’s bill: \[ 10r + 4t + c = 80 \] Akbar’s bill: \[ 7r + 3t + c = 60 \] Step 3: Eliminate variables.
Subtract the second equation from the first: \[ (10r+4t+c)-(7r+3t+c)=80-60 \] \[ 3r+t=20. \] Step 4: Express Anthony’s order.
Anthony orders 5 rotis, 5 tadkas, 5 teas: \[ \text{Cost} = 5r+5t+5c=5(r+t+c). \] Step 5: Find \(r+t+c\).
From Akbar’s equation: \[ 7r+3t+c=60. \] Rewrite as: \[ (3r+t)+(4r+2t+c)=60. \] But \(3r+t=20\), so \[ 20+(4r+2t+c)=60 \;\Rightarrow\;4r+2t+c=40. \] Divide through by 2: \[ 2r+t+\tfrac{c}{2}=20. \] Better approach: Use direct combination. From Amar’s bill: \(10r+4t+c=80\). Factor as \[ (3r+t)+(7r+3t+c)=20+60=80, \] consistent. Now try to find \(r+t+c\). Subtract \((3r+t=20)\) from \((7r+3t+c=60)\): \[ 4r+2t+c=40 \;\Rightarrow\;2r+t+\tfrac{c}{2}=20. \] Still underdetermined, but notice symmetry: If we want \(r+t+c\), take (Amar eqn) – (Akbar eqn): \[ (10r+4t+c)-(7r+3t+c)=20 \;\Rightarrow\;3r+t=20. \] Now add (Akbar eqn): \((7r+3t+c)=60\). Total: \[ (3r+t)+(7r+3t+c)=20+60=80 \;\Rightarrow\;10r+4t+c=80, \] already known. To compute \(r+t+c\), observe Anthony’s bill: \(5(r+t+c)\). We must extract \(r+t+c\) from given relations. From Amar eqn: \(10r+4t+c=80\). Divide by 2: \[ 5r+2t+\tfrac{c}{2}=40. \] Not enough. But since problem gives neat option, assume unique. Solve system systematically: Two equations in three unknowns. But Anthony’s combination is symmetric, so it may be determinable. Let’s compute: From (3r+t=20) $\Rightarrow$ \(t=20-3r\). Substitute in Akbar eqn: \(7r+3(20-3r)+c=60\Rightarrow7r+60-9r+c=60\Rightarrow-2r+c=0\Rightarrow c=2r.\) Then Amar eqn: \(10r+4(20-3r)+2r=80\Rightarrow10r+80-12r+2r=80\Rightarrow0r+80=80\). Okay, consistent. So relation holds: \(t=20-3r,\;c=2r\). Now Anthony’s total: \(5(r+t+c)=5[r+(20-3r)+2r]=5[20]=100.\) \[ \boxed{100} \]