Loading a ₹10 Lakh ATM Cassette Against Four Constraints
A currency chest has to load one ATM cassette with exactly ₹10,00,000 using ₹2,000, ₹500 and ₹100 notes. The cassette physically holds a maximum of 2,000 notes. The branch's currency officer imposes four rules: at least 40% of the total value must be in ₹2,000 notes; the number of ₹500 notes must be exactly twice the number of ₹100 notes; no more than 400 notes of ₹2,000 denomination may be used; and no note of any denomination may be short by even one, because the cassette must tally to the rupee.
Two load plans are on the table. Plan P: 200 notes of ₹2,000, 1,080 notes of ₹500 and 600 notes of ₹100. Plan Q: 225 notes of ₹2,000, 1,000 notes of ₹500 and 500 notes of ₹100. The cash officer wants a single plan cleared for loading today, and separately asks how lean the cassette can be made - the smallest total number of notes that can still satisfy every rule.
1. Test Plan P and Plan Q against every constraint and say which, if either, can be loaded.
Read the worked answer
Plan P - value checks out, one rule fails. Value: (200 × 2,000) + (1,080 × 500) + (600 × 100) = 4,00,000 + 5,40,000 + 60,000 = ₹10,00,000. Note count = 200 + 1,080 + 600 = 1,880, within 2,000. ₹2,000 share = 4,00,000/10,00,000 = exactly 40%, which satisfies 'at least 40%'. The ₹2,000 count, 200, is under 400. But the ₹500-to-₹100 ratio is 1,080 : 600 = 1.8 : 1, not 2 : 1. Plan P fails.
Plan Q - passes on all four rules. Value: (225 × 2,000) + (1,000 × 500) + (500 × 100) = 4,50,000 + 5,00,000 + 50,000 = ₹10,00,000. Note count = 225 + 1,000 + 500 = 1,725, within 2,000. ₹2,000 share = 4,50,000/10,00,000 = 45%, which is at least 40%. Ratio 1,000 : 500 = 2 : 1 exactly. The ₹2,000 count of 225 is under 400. Plan Q can be loaded.
Note how easy it is to clear Plan P on a careless reading: it hits the value target and the percentage target, and the ratio rule is the one that quietly breaks. The discipline is to test each rule separately and record a pass or fail for each, rather than stopping at the first few passes.
2. Derive the general condition on the number of ₹2,000 notes and use it to find the smallest total note count that satisfies every rule.
Read the worked answer
Let x, y and z be the counts of ₹2,000, ₹500 and ₹100 notes.
Value: 2000x + 500y + 100z = 10,00,000. Divide by 100: 20x + 5y + z = 10,000. The ratio rule gives y = 2z, so 20x + 10z + z = 10,000, i.e. 11z = 10,000 − 20x.
For z to be a whole number, 10,000 − 20x must be divisible by 11. Since 10,000 ≡ 1 and 20 ≡ 9 (mod 11), the condition is 9x ≡ 1 (mod 11), which gives x ≡ 5 (mod 11).
The other rules bound x: the 40% floor means 2000x ≥ 4,00,000, so x ≥ 200; the denomination cap means x ≤ 400. Total notes N = x + y + z = x + 3z. Substituting z = (10,000 − 20x)/11 gives N = (30,000 − 49x)/11, which falls as x rises - so make x as large as possible.
The largest x ≤ 400 with x ≡ 5 (mod 11) is 390 (11 × 35 = 385, plus 5). Then z = (10,000 − 7,800)/11 = 200 and y = 400.
Check: 390 × 2,000 = ₹7,80,000; 400 × 500 = ₹2,00,000; 200 × 100 = ₹20,000; total ₹10,00,000. Count = 990 notes, ₹2,000 share = 78%, ratio 400 : 200 = 2 : 1, and x = 390 ≤ 400. The leanest valid cassette holds 990 notes.