2 biscuits and 1 chocolate cost ₹69. 2 chocolates and 3 cups of coffee cost ₹127. 3 biscuits, 4 chocolates and 2 cups of coffee cost ₹229. Find the total cost (in ₹) of 5 biscuits, 5 chocolates and 5 cups of coffee.

Correct — A. Three baskets, three unknowns.

Let b, c and f be one biscuit, one chocolate and one cup of coffee.
2b + c = 69 … (i)
2c + 3f = 127 … (ii)
3b + 4c + 2f = 229 … (iii)

From (i), c = 69 − 2b. Into (ii): 138 − 4b + 3f = 127, so 3f = 4b − 11.
Into (iii): −5b + 2f = −47, and with 3f = 4b − 11 this gives b = 17.

So c = 69 − 34 = 35 and 3f = 68 − 11 = 57, giving f = 19.
b + c + f = 71, and 5 × 71 = 355 → option (a).

Why the others are wrong

• (b)
  375 — 375 needs b + c + f = 75. Adding the three given equations gives 5b + 7c + 5f = 425; you must still remove 2c = 70, and removing 50 instead lands here.
• (c)
  304 — 304 is not a multiple of 5, so it cannot be five of each item at any prices at all. That check kills it before any algebra.
• (d)
  345 — 345 is 5 × 69, the first sentence's total scaled up as though ₹69 already bought one of each. The basket in (i) is 2 biscuits and 1 chocolate, with no coffee.

Three linear equations in three unknowns, but the question asks only for b + c + f — a combination, not the individual prices.

That opens a shortcut. Add the three equations: the b-terms give 5b, the c-terms 7c and the f-terms 5f, so 5b + 7c + 5f = 425.

You are two chocolates over the basket you want. Since 2b + c = 69 and c = 35, subtracting 2c = 70 leaves 5b + 5c + 5f = 355 directly.

Full elimination reaches the same place and is safer if the shortcut is not obvious under time pressure.

Key facts

• Adding the three given equations gives 5b + 7c + 5f = 425.
• Subtracting 2c = 70 from that leaves 5b + 5c + 5f = 355, exactly what is asked.
• The prices are ₹17 a biscuit, ₹35 a chocolate and ₹19 a cup of coffee.
• Any total for five of each item must be divisible by 5.

Common traps

• Solving for all three prices when only their sum is needed
• Scaling the wrong basket, for instance multiplying the 2-biscuit-1-chocolate total by 5
• Slipping in the substitution, where the b-terms must collapse to −7b

No directly related past PYQ was found.