The expenditure of a company on five heads A, B, C, D and E is in proportion to 3:5:7:11:? respectively. The expenditure on E is equal to the arithmetic mean of the expenditures on A and D. If the expenditure on C exceeds that on B by ₹18,000, find the total expenditure (₹) of the company.

Correct — B. Work in parts of the ratio, then convert to rupees once.

E is the mean of A and D: (3 + 11) ÷ 2 = 7 parts
C exceeds B by 7 − 5 = 2 parts, and 2 parts = ₹18,000
One part: 18,000 ÷ 2 = ₹9,000

Add all five heads: 3 + 5 + 7 + 11 + 7 = 33 parts
Total: 33 × 9,000 = ₹2,97,000 → option (b)

Why the others are wrong

• (a)
  ₹2,40,000 — ₹2,40,000 is not a whole number of ₹9,000 parts: 2,40,000 ÷ 9,000 ≈ 26.7. With whole-number ratio terms the total must be a whole number of parts, here 33.
• (c)
  ₹1,80,000 — ₹1,80,000 is only 20 parts. The four given heads A to D alone add to 26 parts, ₹2,34,000, so the total must be larger.
• (d)
  ₹2,52,000 — ₹2,52,000 is 28 parts, which would need E = 2 parts. E is the mean of A and D, (3 + 11) ÷ 2 = 7 parts, making 33 parts in all.

A ratio fixes shares in parts, not rupees. One fact that turns parts into rupees then settles every share and the total.

Here that fact is a difference: C − B = 7 − 5 = 2 parts is ₹18,000, so one part is ₹9,000.

A missing term defined by a condition, such as the mean of A and D, is found in parts first, before any rupee value is known.

Key facts

• Value of one part = rupee difference ÷ difference in parts.
• The arithmetic mean of two quantities is their sum divided by 2.
• With whole-number ratio terms, the total is a whole-number multiple of one part.

Common traps

• Treating ₹18,000 as one part instead of two, which doubles every share.
• Totalling only the four printed terms, 26 parts, and leaving out E.

No directly related past PYQ was found.