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, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is
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Official Correct Answer: 340. First, the problem says the seller has mangoes, bananas, and apples, with at least one of each. At the beginning of the day, mangoes make up 40% of his stock. So, if the total number of fruits is, say, T, then mangoes are 0.4T. That means bananas and apples together make up 60% of T, right? So bananas + apples = 0.6T.
Then, during the day, he sells half of the mangoes. So he sells 0.5 * 0.4T = 0.2T mangoes. He also sells 96 bananas and 40% of the apples. So apples sold are 0.4 * (number of apples). Let the number of apples be $2. So apples sold are 0.4A.
At the end of the day, he has sold 50% of the fruits. So total fruits sold that day are 0.5T.
We have the total fruits sold:
Mangoes sold: 0.2T
Bananas sold: 96
Apples sold: 0.4A
So total sold = 0.2T + 96 + 0.4A = 0.5T
So, 0.2T + 96 + 0.4A = 0.5T
Let me rearrange this equation:
0.4A = 0.5T - 0.2T - 96
0.4A = 0.3T - 96
So, 0.4A = 0.3T - 96
I can write this as:
A = (0.3T - 96) / 0.4
A = (3T/10 - 96) / (2/5)
A = (3T/10 - 96) * (5/2)
A = (15T/20 - 480/2) / 2
Wait, maybe I should compute it differently.
Wait, 0.3T is 3T/10, so 3T/10 - 96, divided by 0.4 which is 2/5. So dividing by 2/5 is multiplying by 5/2.
So A = (3T/10 - 96) * (5/2)
A = (3T/10 * 5/2) - (96 * 5/2)
A = (15T/20) - 240
Simplify 15T/20 to 3T/4.
So A = (3T/4) - 240
But A must be an integer because it's the number of apples. So (3T/4) - 240 must be an integer. Therefore, 3T/4 must be an integer plus 240, which is also an integer. So 3T must be divisible by 4, meaning T must be divisible by 4/ gcd(3,4) = 4. So T must be a multiple of 4.
Also, since A is the number of apples, it must be positive. So:
(3T/4) - 240 > 0
3T/4 > 240
T > (240 * 4)/3
T > 320
So T must be greater than 320. But we need the smallest possible T, so T should be just above 320. But T must also satisfy that A is an integer.
Also, let's remember that at the beginning, the number of mangoes is 0.4T, which must be an integer because you can't have a fraction of a mango. So 0.4T must be integer. 0.4 is 2/5, so 2T/5 must be integer. Therefore, T must be a multiple of 5/ gcd(2,5) = 5. So T must be a multiple of 5.
So T must be a multiple of both 4 and 5, so LCM(4,5) is 20. So T must be a multiple of 20.
We have T > 320, so the smallest multiple of 20 greater than 320 is 340. Let me check if T=340 works.
Compute A:
A = (3*340)/4 - 240
A = 1020/4 - 240
A = 255 - 240
A = 15
So apples are 15. Let's check if that's valid.
Total fruits initially: 340
Mangoes: 0.4*340 = 136
Bananas: 340 - 136 - 15 = 189
Wait, bananas are 189? Let me check.
Wait, total fruits: mangoes + bananas + apples = 136 + bananas + 15 = 340
So bananas = 340 - 136 -15 = 189
So bananas are 189.
Now, during the day, he sells half the mangoes: 136 / 2 = 68 mangoes sold.
He sells 96 bananas.
He sells 40% of apples: 40% of 15 is 6 apples.
Total sold: 68 + 96 + 6 = 170
Total fruits initially: 340, so 50% is 170. That matches.
So yes, T=340 works.
Is there a smaller T? . The next lower multiple of 20 is 320, but T must be greater than 320, so 340 is the next one.
Wait, but let me check if T=340 is indeed the smallest. Suppose T=340 is the answer, but if T=320 is possible.
Wait, T must be greater than 320, so 320 is not allowed. So 340 is the next.
But let me check if T=340 is indeed the minimal. Suppose T=340 is the minimal, but if there's a T between 320 and 340 that is a multiple of 20. The next multiple after 320 is 340, so yes, 340 is the minimal.
Therefore, the smallest possible total number of fruits is 340