₹ 1290 is divided between A, B, and C such that A’s share is 1½ times B’s share and B’s share is 1¾ times C’s share. What is C’s share?

Introduction / Context:
This chained multiplicative relationship is most easily solved by expressing all shares in terms of C. Then enforce the total to find C numerically and back out other shares if desired.

Given Data / Assumptions:

• A = 1.5 B = (3/2)B
• B = 1.75 C = (7/4)C
• Total A + B + C = ₹ 1290

Concept / Approach:
Substitute B in terms of C into A = (3/2)B to obtain A in terms of C. Sum A + B + C in terms of C to match the total, then solve for C.

Step-by-Step Solution:
B = (7/4)CA = (3/2)B = (3/2)*(7/4)C = (21/8)CSum = (21/8 + 7/4 + 1)C = (21/8 + 14/8 + 8/8)C = (43/8)C(43/8)C = 1290 ⇒ C = 1290 * 8 / 43 = 240

Verification / Alternative check:
B = (7/4)*240 = 420; A = (21/8)*240 = 630; total 630 + 420 + 240 = 1290.

Why Other Options Are Wrong:
200, 300, 400, 420 do not satisfy the given multiplicative relationships when summed to 1290.

Common Pitfalls:
Adding 1.5 and 1.75 to get a direct ratio; always multiply through the chain correctly.

Final Answer:
₹ 240