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An ATM dispenses exactly Rs. 50005000 per withdrawal using 100100, 200200 and 500500 rupee notes. The ATM requires every customer to give her preference for one of the three denominations of notes. It then dispenses notes such that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her.

If the ATM could serve only 1010 customers with a stock of fifty 500500 rupee notes and a sufficient number of notes of other denominations, what is the maximum number of customers among these 1010 who could have given 500500 rupee notes as their preferences?

Entered answer:

Solution

✅ Correct Answer: 6

Maximum Rs. 500 Preference Customers

Constraints

  • Total customers: 1010
  • Available Rs. 500500 notes: 5050
  • Each Rs. 500500 preference customer needs minimum 88 Rs. 500500 notes
  • Non-Rs. 500500 preference customers can use 00 to 77 Rs. 500500 notes

Rs. 500 Usage by Preference Type

Customers preferring Rs. 500 notes:

  • Minimum 88: If using only 77 Rs. 500500 notes (Rs. 35003500), remaining Rs. 15001500 needs at least 88 other notes (best case: 7×200+1×1007 \times 200 + 1 \times 100). But 7<87 < 8 violates the preference constraint.
  • Maximum 1010: 10×500=500010 \times 500 = 5000 (uses all money)
  • Valid ranges:
    • 88 Rs. 500500 notes: Rs. 10001000 remaining, can use ≤7\leq 7 other notes
    • 99 Rs. 500500 notes: Rs. 500500 remaining, can use ≤8\leq 8 other notes
    • 1010 Rs. 500500 notes: Rs. 00 remaining, 00 other notes

Customers preferring Rs. 100 or Rs. 200 notes:

The maximum Rs. 500500 notes they can use while still having their preferred denomination dominate.

For Rs. 100100 preference (x>y+zx > y+z):

  • Maximum zz occurs when xx is minimized relative to constraint.
  • If z=7z=7, need x≥8x \geq 8, remaining amount =5000−500×7=1500= 5000 - 500 \times 7 = 1500
  • 100x+200y=1500  ⇒  100×8+200y=1500  ⇒  y=4100x + 200y = 1500 \;\Rightarrow\; 100 \times 8 + 200y = 1500 \;\Rightarrow\; y=4
  • Total other notes =12= 12, check: 8>4+7=118 > 4+7=11 ✗

Testing smaller zz values:

  • z=6z=6: x≥7x \geq 7, remaining =2000=2000, gives x=10x=10, y=5y=5, check: 10>5+6=1110 > 5+6=11 ✗
  • z=5z=5: x≥6x \geq 6, remaining =2500=2500, gives valid solutions
  • Maximum zz for Rs. 100100 preference: 44 notes

For Rs. 200200 preference (y>x+zy > x+z):

  • Similar analysis shows maximum z=4z=4 notes
Customer TypeRs. 500500 Notes Range
Rs. 500500 preference8,9,108,9,10
Rs. 100100 preference00 to 44
Rs. 200200 preference00 to 44

Optimisation Strategy

To maximise Rs. 500500 preference customers, minimise Rs. 500500 notes used by others.

Strategy: Use 88 Rs. 500500 notes per Rs. 500500 preference customer (minimum required)

ScenarioRs. 500500 PrefOthersRs. 500500 UsedRs. 500500 RemainingValid?
66 customers prefer Rs. 50050066446×8=486 \times 8 = 4822✓
77 customers prefer Rs. 50050077337×8=567 \times 8 = 56−6-6✗

Check 66 customers scenario:

  • 66 customers: 88 Rs. 500500 notes each =48= 48 notes
  • 44 remaining customers: Can use 0−20-2 Rs. 500500 notes each
  • This works within the 5050 note limit

Check if 77 is possible with optimal distribution:

  • Minimum for 77 customers: 7×8=567 \times 8 = 56 notes
  • Available: 5050 notes
  • Deficit: 66 notes → Not possible

Final Answer

Maximum customers who can prefer Rs. 500500 notes: 66

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