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The cost of 2 tables and 3 chairs is ₹540, while that of 2 tables and 1 chair is ₹470. What is the cost of 2 tables and 2 chairs combinedly?
₹505
₹525
₹545
₹485
₹505
Subtract the two equations to find the price of 2 chairs: (2T+3C)−(2T+1C)=540−470=70. Since 2 chairs cost ₹70, the cost of 2 tables and 2 chairs is (2T+1C)+1C=470+35=505. Alternatively, find 2T=470−35=435 and add 2C=70, total =435+70=505.
Cost of 2 tables + 3 chairs = ₹540; Cost of 2 tables + 1 chair = ₹470.
2T+3C=540, 2T+1C=470
Subtract the two equations to find the price of 2 chairs: (2T+3C)−(2T+1C)=540−470=70. Since 2 chairs cost ₹70, the cost of 2 tables and 2 chairs is (2T+1C)+1C=470+35=505. Alternatively, find 2T=470−35=435 and add 2C=70, total =435+70=505.
Many students try to solve for the cost of 1 table and 1 chair individually, which leads to unnecessary calculations, or they miscalculate the difference as 1 table instead of 2 chairs.
Define Equations
Let T be the cost of a table and C be the cost of a chair. From the problem: (1) 2T+3C=540 and (2) 2T+1C=470.
2T+3C=540
Find cost of 2 chairs
Subtract equation (2) from equation (1) to eliminate T: (2T+3C)−(2T+1C)=540−470. This gives 2C=70.
2C=70
Calculate target sum
We need the cost of 2 tables and 2 chairs (2T+2C). Since 2T+1C=470 and C=35 (as 2C=70), we calculate 470+35=505.
2T+2C=505
A is correct because the difference between the two costs is the price of 2 chairs (540−470=70), making 1 chair cost ₹35, and thus 2 tables and 2 chairs cost 470+35=505.
This type of simultaneous equation problem is common in RRB exams; always look for the difference between total quantities first to see if it directly maps to the quantity you need to find.
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