Great question — this is a classic Systems of Equations problem dressed up in a discount/pricing context. The trap most students fall into is trying to work backwards from the answer choices instead of setting up the algebra cleanly. Here's the step-by-step:
Key concept tested: Setting up and solving a system of two linear equations.
Let T = full price of one table, C = full price of one chair.
Step 1: Translate Condition 1.
"Tables at 40% discount, chairs at full price → total $22,000"
20 × (0.6T) + 50 × C = 22,000
→ 12T + 50C = 22,000 ... (Equation 1)
Step 2: Translate Condition 2.
"All items at 10% discount → total $27,000"
20 × (0.9T) + 50 × (0.9C) = 27,000
→ 18T + 45C = 27,000 ... (Equation 2)
Step 3: Eliminate one variable.
Multiply Equation 1 by 9: 108T + 450C = 198,000
Multiply Equation 2 by 10: 180T + 450C = 270,000
Subtract Equation 1 from Equation 2:
72T = 72,000 → T = 1,000
Step 4: Solve for C.
Substitute T = 1,000 into Equation 1:
12(1,000) + 50C = 22,000
50C = 10,000 → C = 200
Step 5: Answer the actual question.
Customer buys 1 table + 6 chairs at full price:
1(1,000) + 6(200) = 1,000 + 1,200 = 2,200
Answer: B
Common trap: Many students misread Condition 1 and apply a 40% discount to chairs as well, or forget that "40% discount" means you pay 60% of the price. Always translate "x% discount" as multiplying by (1 − x/100) before writing the equation.
Takeaway: Whenever a PS question gives you two "total cost/revenue" conditions involving two unknown prices, immediately set up two equations and solve — don't try to back-solve from the choices unless the algebra gets messy.