A class eats 2/5 of chocolates on 1st day. On the 2nd day they eat 3/4 of the remainder. How many chocolates were there initially if still 75 chocolates are left?

Options

A. a. 250
B. b. 750
C. c. 500
D. d. 1000

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Define the total number of chocolates
Let the total number of chocolates initially be \( x \).

### Step 2: Calculate chocolates eaten on the 1st day
On the 1st day, the class eats \( \frac{2}{5} \) of the total chocolates. Therefore, the amount eaten on the 1st day is:
\[ \text{Chocolates eaten on 1st day} = \frac{2}{5} x \]

### Step 3: Calculate remaining chocolates after the 1st day
After the 1st day, the remaining chocolates are:
\[ \text{Remaining chocolates} = x - \frac{2}{5} x = \frac{3}{5} x \]

### Step 4: Calculate chocolates eaten on the 2nd day
On the 2nd day, the class eats \( \frac{3}{4} \) of the remaining chocolates. Therefore, the amount eaten on the 2nd day is:
\[ \text{Chocolates eaten on 2nd day} = \frac{3}{4} \left(\frac{3}{5} x\right) = \frac{9}{20} x \]

### Step 5: Calculate remaining chocolates after the 2nd day
After the 2nd day, the remaining chocolates are:
\[ \text{Remaining chocolates} = \frac{3}{5} x - \frac{9}{20} x \]

To perform this subtraction, we need a common denominator. The common denominator for 5 and 20 is 20:
\[ \frac{3}{5} x = \frac{12}{20} x \]
Thus,
\[ \text{Remaining chocolates} = \frac{12}{20} x - \frac{9}{20} x = \frac{3}{20} x \]

### Step 6: Set up the equation based on remaining chocolates
According to the problem, after the 2nd day, there are 75 chocolates left:
\[ \frac{3}{20} x = 75 \]

### Step 7: Solve for \( x \)
To find \( x \), we can multiply both sides by 20:
\[ 3x = 75 \times 20 \]
\[ 3x = 1500 \]
Now, divide both sides by 3:
\[ x = \frac{1500}{3} = 500 \]

### Conclusion
The total number of chocolates initially was \( 500 \).

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