A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is:
Question:
A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is:
A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is:
Options
Answer: 12
Explanation:
Step 1: Define variables Let: x = number of ₹100 cheques y = number of ₹250 cheques z = number of ₹500 cheques We are given: Total cheques = 100 → x + y + z = 100 Total amount = 15250 → 100x + 250y + 500z = 15250 We are asked to maximize z. Step 2: Simplify the total amount equation Divide the total amount equation by 50: 100x + 250y + 500z = 15250 → 2x + 5y + 10z = 305 So we have the system: x + y + z = 100 2x + 5y + 10z = 305 Step 3: Express x in terms of y and z From x + y + z = 100 → x = 100 - y - z Plug into 2nd equation: 2(100 - y - z) + 5y + 10z = 305 200 - 2y - 2z + 5y + 10z = 305 Combine like terms: (-2y + 5y) + (-2z + 10z) + 200 = 305 → 3y + 8z + 200 = 305 3y + 8z = 105 → Equation (★) Step 4: Express y in terms of z 3y + 8z = 105 → y = (105 - 8z)/3 y must be non-negative integer → 105 - 8z divisible by 3 → 105 - 8z ≡ 0 (mod 3) 105 ≡ 0 (mod 3), since 1+0+5=6 divisible by 3 → 105 ≡ 0 (mod 3) So -8z ≡ 0 (mod 3) → 8z ≡ 0 (mod 3) → 8 ≡ 2 (mod 3) → 2 z ≡ 0 (mod 3) → z ≡ 0 (mod 3) So z = 0, 3, 6, … Step 5: Find maximum z y = (105 - 8z)/3 ≥ 0 → 105 - 8z ≥ 0 → 8z ≤ 105 → z ≤ 13.125 → integer → z ≤ 13 Also z ≡ 0 (mod 3) → possible z = 0, 3, 6, 9, 12 Maximum z = 12 Check corresponding y: y = (105 - 8×12)/3 = (105 - 96)/3 = 9/3 = 3 Then x = 100 - y - z = 100 - 3 -12 = 85 ✅ All non-negative integers → valid. ✅ Step 6: Answer Maximum possible number of ₹500 cheques = 12
Explanation:
Step 1: Define variables Let: x = number of ₹100 cheques y = number of ₹250 cheques z = number of ₹500 cheques We are given: Total cheques = 100 → x + y + z = 100 Total amount = 15250 → 100x + 250y + 500z = 15250 We are asked to maximize z. Step 2: Simplify the total amount equation Divide the total amount equation by 50: 100x + 250y + 500z = 15250 → 2x + 5y + 10z = 305 So we have the system: x + y + z = 100 2x + 5y + 10z = 305 Step 3: Express x in terms of y and z From x + y + z = 100 → x = 100 - y - z Plug into 2nd equation: 2(100 - y - z) + 5y + 10z = 305 200 - 2y - 2z + 5y + 10z = 305 Combine like terms: (-2y + 5y) + (-2z + 10z) + 200 = 305 → 3y + 8z + 200 = 305 3y + 8z = 105 → Equation (★) Step 4: Express y in terms of z 3y + 8z = 105 → y = (105 - 8z)/3 y must be non-negative integer → 105 - 8z divisible by 3 → 105 - 8z ≡ 0 (mod 3) 105 ≡ 0 (mod 3), since 1+0+5=6 divisible by 3 → 105 ≡ 0 (mod 3) So -8z ≡ 0 (mod 3) → 8z ≡ 0 (mod 3) → 8 ≡ 2 (mod 3) → 2 z ≡ 0 (mod 3) → z ≡ 0 (mod 3) So z = 0, 3, 6, … Step 5: Find maximum z y = (105 - 8z)/3 ≥ 0 → 105 - 8z ≥ 0 → 8z ≤ 105 → z ≤ 13.125 → integer → z ≤ 13 Also z ≡ 0 (mod 3) → possible z = 0, 3, 6, 9, 12 Maximum z = 12 Check corresponding y: y = (105 - 8×12)/3 = (105 - 96)/3 = 9/3 = 3 Then x = 100 - y - z = 100 - 3 -12 = 85 ✅ All non-negative integers → valid. ✅ Step 6: Answer Maximum possible number of ₹500 cheques = 12
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