15 chairs and 3 tables cost ₹7,800 and 14 chairs and 6 tables cost ₹9,200. What is the cost of 17 chairs and 18 tables?
- AOption (a)
₹17,596
- BOption (b)
₹17,600
- COption (c)
₹17,604
- DOption (d)
₹17,598
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Correct answer
₹17,600
Why this answer is correct
Shortcut Trick
Simplify the first cost equation to: 5 chairs + 1 table = ₹2,600.
Multiply this by 6 to match tables in the second case: 30 chairs + 6 tables = ₹15,600.
Difference with second equation: (30 − 14) chairs = ₹15,600 − ₹9,200 ⇒ 16 chairs = ₹6,400.
Cost of 1 chair = ₹400 and cost of 1 table = 2,600 − (5 × 400) = ₹600.
Required total cost = 17 × 400 + 18 × 600 = 6,800 + 10,800 = ₹17,600.
∴ The correct answer is ₹17,600.
Alternate Method
Given:
15 chairs (C) + 3 tables (T) = ₹7,800
14 chairs (C) + 6 tables (T) = ₹9,200
Formula Used:
Linear Equation solving using the elimination method.
⇒ Let chair cost be C and table cost be T.
⇒ 15C + 3T = 7,800 ⇒ 5C + T = 2,600 ... (Equation 1)
⇒ 14C + 6T = 9,200 ⇒ 7C + 3T = 4,600 ... (Equation 2)
⇒ Multiply Equation 1 by 3: 15C + 3T = 7,800
⇒ Subtract Equation 2 from (Equation 1 × 3): (15C + 3T) − (7C + 3T) = 7,800 − 4,600
⇒ 8C = 3,200 ⇒ C = 400
⇒ Substitute C in Eq 1: 5(400) + T = 2,600 ⇒ T = 2,600 − 2,000 = 600
⇒ Cost of 17 chairs and 18 tables = 17 × 400 + 18 × 600
⇒ 6,800 + 10,800 = 17,600
∴ The correct answer is ₹17,600.
Additional Information
Linear Equations in Two Variables
Equations of the form ax + by = c represent a straight line. Solving two such equations finds the intersection point (x, y).
Elimination Method
A process of eliminating one variable by making its coefficients equal in both equations through multiplication, then adding or subtracting.
Substitution Method
Expressing one variable in terms of the other from one equation and substituting it into the second equation to solve for one variable first.