A fruit seller has a stock of mangoes, bananas and apples, with at least one fruit of each type. At the beginning of a day, mangoes constitute 40% of his total stock. During the day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he finds that he has sold exactly 50% of the total fruits. The smallest possible total number of fruits in the stock at the beginning of the day is:
Answer and solution
Correct answer: 340
Solution
Answer: 340
1. Let the total number of fruits at the beginning be N.
2. Mangoes make up 40% of the stock, so the seller has 0.4N mangoes. He sells half of them, which means he sells 0.2N mangoes.
3. He also sells 96 bananas and 40% of the apples. Let the number of apples be A.
4. The total number of fruits sold is 50% of the stock:
5. 0.2N + 96 + 0.4A = 0.5N
6. So,
7. 0.4A = 0.3N − 96
A = 0.75N − 240
8. For the number of mangoes and apples, the required percentages must give whole numbers. Since 40% of the stock is an integer, N must be a multiple of 5. Since 40% of the apples must also be a whole number, A must be a multiple of 5. This makes N a multiple of 20.
9. Also, there must be at least one apple, so:
10. 0.75N − 240 > 0
N > 320
11. The smallest multiple of 20 greater than 320 is 340.
12. At N = 340, the number of apples is 15, mangoes are 136 and bananas are 189. All conditions are satisfied.
13. Answer: 340 (TITA).
Why this method works:
Integer fruit counts constrain the solution beyond the sales equation. From 4A=3N−960, N must be divisible by 4; the mango fraction requires N divisible by 5. Together these force N divisible by 20.
Independent check and common trap:
At the boundary N=320 the apple count is zero, contradicting at least one of each fruit. The next allowed total is 340. Stocks 136,189,15 allow sales 68,96,6, totaling 170, exactly half the original stock.
Apply: A=3N/4−240>0 gives N>320. Whole-fruit sales require N multiple 20. Smallest is 340; check M 136, A 15, B 189 and sales 68+6+96=170.
Check: At the boundary N=320 the apple count is zero, contradicting at least one of each fruit. The next allowed total is 340. Stocks 136,189,15 allow sales 68,96,6, totaling 170, exactly half the original stock.
1. Let the total number of fruits at the beginning be N.
2. Mangoes make up 40% of the stock, so the seller has 0.4N mangoes. He sells half of them, which means he sells 0.2N mangoes.
3. He also sells 96 bananas and 40% of the apples. Let the number of apples be A.
4. The total number of fruits sold is 50% of the stock:
5. 0.2N + 96 + 0.4A = 0.5N
6. So,
7. 0.4A = 0.3N − 96
A = 0.75N − 240
8. For the number of mangoes and apples, the required percentages must give whole numbers. Since 40% of the stock is an integer, N must be a multiple of 5. Since 40% of the apples must also be a whole number, A must be a multiple of 5. This makes N a multiple of 20.
9. Also, there must be at least one apple, so:
10. 0.75N − 240 > 0
N > 320
11. The smallest multiple of 20 greater than 320 is 340.
12. At N = 340, the number of apples is 15, mangoes are 136 and bananas are 189. All conditions are satisfied.
13. Answer: 340 (TITA).
Why this method works:
Integer fruit counts constrain the solution beyond the sales equation. From 4A=3N−960, N must be divisible by 4; the mango fraction requires N divisible by 5. Together these force N divisible by 20.
Independent check and common trap:
At the boundary N=320 the apple count is zero, contradicting at least one of each fruit. The next allowed total is 340. Stocks 136,189,15 allow sales 68,96,6, totaling 170, exactly half the original stock.
G Strategy shortcut
Spot: Arithmetic – Percentages.Apply: A=3N/4−240>0 gives N>320. Whole-fruit sales require N multiple 20. Smallest is 340; check M 136, A 15, B 189 and sales 68+6+96=170.
Check: At the boundary N=320 the apple count is zero, contradicting at least one of each fruit. The next allowed total is 340. Stocks 136,189,15 allow sales 68,96,6, totaling 170, exactly half the original stock.
G Strategy: Spot: Arithmetic – Percentages. Apply: A=3N/4−240>0 gives N>320. Whole-fruit sales require N multiple 20. Smallest is 340; check M 136, A 15, B 189 and sales 68+6+96=170. Check: At the boundary N=320 the apple count is zero, contradicting at least one of each fruit. The next allowed total is 340. Stocks 136,189,15 allow sales 68,96,6, totaling 170, exactly half the original stock.