Math Problem Statement

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 makes 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:

Solution

The problem in the image reads:

Q24. 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 makes 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:

Let's solve this problem step by step.

Step 1: Define Variables

Let TT be the total number of fruits at the beginning of the day. Let MM be the number of mangoes, BB be the number of bananas, and AA be the number of apples at the beginning of the day.

Given:

  • M=0.4TM = 0.4T (since mangoes make up 40% of the total stock)
  • The seller sells M2\frac{M}{2} mangoes.
  • The seller sells 96 bananas.
  • The seller sells 0.4A0.4A apples.

Step 2: Calculate the Fruits Sold

The total number of fruits sold: Total fruits sold=M2+96+0.4A\text{Total fruits sold} = \frac{M}{2} + 96 + 0.4A

We know that at the end of the day, the seller has sold 50% of his fruits, so: T2=M2+96+0.4A\frac{T}{2} = \frac{M}{2} + 96 + 0.4A

Substitute M=0.4TM = 0.4T into the equation: T2=0.4T2+96+0.4A\frac{T}{2} = \frac{0.4T}{2} + 96 + 0.4A T2=0.2T+96+0.4A\frac{T}{2} = 0.2T + 96 + 0.4A

Step 3: Simplify the Equation

Subtract 0.2T0.2T from both sides: T20.2T=96+0.4A\frac{T}{2} - 0.2T = 96 + 0.4A 0.3T=96+0.4A0.3T = 96 + 0.4A

Step 4: Solve for AA and TT

Solve for AA: 0.4A=0.3T960.4A = 0.3T - 96 A=0.3T960.4A = \frac{0.3T - 96}{0.4} A=0.75T240A = 0.75T - 240

Since AA must be a positive integer, 0.75T2400.75T - 240 must also be positive. Therefore, TT must satisfy: 0.75T>2400.75T > 240 T>320T > 320

The smallest integer value for TT that satisfies this inequality is T=321T = 321.

Step 5: Check Consistency

For T=321T = 321:

  • M=0.4×321=128.4M = 0.4 \times 321 = 128.4, but since the number of fruits must be an integer, T=321T = 321 is not valid.

Let's try T=400T = 400 (next logical step to check for consistency):

  • M=0.4×400=160M = 0.4 \times 400 = 160
  • A=0.75×400240=60A = 0.75 \times 400 - 240 = 60

Now check: 0.3×400=96+0.4×600.3 \times 400 = 96 + 0.4 \times 60 120=96+24120 = 96 + 24 This satisfies the equation, so the minimum possible total number of fruits at the beginning of the day is 400.

Final Answer

The smallest possible total number of fruits in the stock at the beginning of the day is 400.

Would you like more details or have any questions?

Here are five related questions:

  1. How would the problem change if the percentage of mangoes was different?
  2. What if the seller sold a different percentage of apples?
  3. How can the problem be modified to involve more types of fruits?
  4. What if the total fruits sold were 60% instead of 50%?
  5. Can you solve a similar problem where the fruit percentages are given for different fruits?

Tip: When solving percentage problems, always ensure that the final quantities are reasonable and consistent with the conditions of the problem.

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Math Problem Analysis

Mathematical Concepts

Algebra
Percentage
Equation Solving

Formulas

M = 0.4T (Mangoes make up 40% of total stock)
Total fruits sold = M/2 + 96 + 0.4A
T/2 = M/2 + 96 + 0.4A
0.3T = 96 + 0.4A

Theorems

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Suitable Grade Level

Grades 9-10