Ten chairs and six tables together cost ₹5,140; three chairs and two tables together cost ₹1,635. The cost of 1 chair and 1 table is:

Correct — B. Let c be a chair and t a table:
10c + 6t = 5140
3c + 2t = 1635

Scale the second by 3 so the table terms match:
9c + 6t = 4905

Subtract it from the first — the 6t cancels:
c = 5140 − 4905 = 235

Put c back into 3c + 2t = 1635:
705 + 2t = 1635 → 2t = 930 → t = 465
c + t = 235 + 465 = ₹700 → option (b)

Why the others are wrong

• (a)
  ₹800 — ₹800 forces c = 35 and t = 765 through 3c + 2t = 1635. Those prices make 10 chairs and 6 tables cost ₹4,940, not the ₹5,140 given.
• (c)
  ₹900 — ₹900 makes the chair cost negative: 3c + 2(900 − c) = 1635 gives c = 1635 − 1800 = −₹165. A price cannot be below zero.
• (d)
  ₹600 — ₹600 forces c = 435 and t = 165, which price 10 chairs and 6 tables at ₹5,340 — ₹200 more than the stem allows.

Two equations, two unknowns, solved by elimination: scale one equation until a variable carries the same coefficient in both, then subtract.

Here 3 × (3c + 2t) gives 9c + 6t, matching the 6t of the first equation, so one subtraction leaves c alone. Nothing cleverer is required.

But notice what is actually asked: c + t, not c and t. You can build that combination straight out of the two equations — 2 × (3c + 2t) − ½ × (10c + 6t) = c + t — and read off 3,270 − 2,570 = ₹700 without ever finding either price.

Key facts

• Three times 3c + 2t = 1,635 is 9c + 6t = 4,905, which shares the 6t of the first equation.
• Subtracting leaves c = ₹235, and back-substitution gives t = ₹465.
• 2 × (3c + 2t) − ½ × (10c + 6t) equals c + t, so 3,270 − 2,570 = ₹700 without solving for either price.
• A chair and a table together cost ₹700.

Common traps

• Adding the two equations instead of scaling one first, which eliminates nothing.
• Solving correctly for c = 235 and then answering with the chair price alone.
• Reading 'the cost of 1 chair and 1 table' as the cost of either one rather than their sum.

No directly related past PYQ was found.