A boy was carrying a basket of cups. He fell down and some of the cups broke. The boy had 10 cups left with him. When asked by his mother how many cups were broken, the boy could not recall. However, here called that when he counted the total number of cups at a time. 1 cup was left and when he counted them 5 at a time, no cup was left. At least how many cups were broken?

Options

A. 15
B. 25
C. 30
D. 35

Answer

To solve the problem, we need to determine the total number of cups the boy originally had before some broke. We know that after the fall, he has 10 cups left.

Let’s denote the total number of cups the boy originally had as \( x \). The number of cups that broke can be represented as \( b \). Thus, we have:

\[ x - b = 10 \]

From the problem, we know two conditions regarding the total number of cups \( x \):

1. When he counted the cups one at a time, he had a remainder of 1.
2. When he counted the cups five at a time, there was no remainder.

These conditions can be expressed mathematically as:

1. \( x \equiv 1 \mod 1 \) (This condition is always true for any integer.)
2. \( x \equiv 0 \mod 5 \)

Additionally, since he had 10 cups left, we can express this as:

\[ x = 10 + b \]

Now, we need to find the smallest \( b \) such that \( x \equiv 1 \mod 2 \) and \( x \equiv 0 \mod 5 \).

### Step 1: Analyze the conditions

From the second condition \( x \equiv 0 \mod 5 \), we know that \( x \) must be a multiple of 5. The possible values of \( x \) that are greater than or equal to 10 (since he has 10 left) are:

- 10
- 15
- 20
- 25
- ...

### Step 2: Check each candidate for the first condition

Now we check each of these values to see if they satisfy the first condition \( x \equiv 1 \mod 2 \):

- For \( x = 10 \):
- \( 10 \mod 2 = 0 \) (does not satisfy)

- For \( x = 15 \):
- \( 15 \mod 2 = 1 \) (satisfies)

- For \( x = 20 \):
- \( 20 \mod 2 = 0 \) (does not satisfy)

- For \( x = 25 \):
- \( 25 \mod 2 = 1 \) (satisfies)

### Step 3: Calculate the number of broken cups

Now we have two valid candidates for \( x \): 15 and 25. We can calculate the number of broken cups \( b \) for each case:

1. If \( x = 15 \):
\[ b = x - 10 = 15 - 10 = 5 \]

2. If \( x = 25 \):
\[ b = x - 10 = 25 - 10 = 15 \]

### Conclusion

The minimum number of cups that could have broken is 5.

Thus, the answer is:

**At least 5 cups were broken.**

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