Compare relative performance: Shantanu's marks are 40% less than Kamal's. By what percent are Kamal's marks greater than Shantanu's?

Introduction / Context:
This problem flips a percentage comparison. If one value is 40% less than another, the reverse comparison produces a larger-than-40% increase due to the smaller base.

Given Data / Assumptions:

• Let Kamal's marks = K.
• Shantanu's marks = 60% of K = 0.6K.
• We seek the percent by which K exceeds 0.6K.

Concept / Approach:
Percent increase from Shantanu to Kamal = (K − 0.6K) / 0.6K * 100%.

Step-by-Step Solution:

Verification / Alternative check:
Pick K = 100: then Shantanu = 60. Increase from 60 to 100 is 40 on a base of 60, i.e., 40/60 = 66.666…%.

Why Other Options Are Wrong:

• 33 1/3 % and 44 3/5 % come from misplacing the base.
• 55 2/3 % underestimates the true ratio.
• 60% equates to the original difference relative to the larger base, not the smaller.

Common Pitfalls:
Using 40% directly for the reverse comparison; always compute relative to the smaller base when asked “how much more than Shantanu.”

Final Answer:
66 2/3 %