The problem asks for the number of ways a lady can choose a quantity of apples and bananas, given their individual costs and the total amount spent. This can be modeled using a linear Diophantine equation.

Mathematical Formulation

Let A be the number of apples and B be the number of bananas purchased.

The cost of apples is Rs. 25 each, and the cost of bananas is Rs. 6 each.

The total cost is Rs. 378.

The equation representing the total cost is:

We need to find the number of pairs of non-negative integers $(A, B)$ that satisfy this equation. Often, "some" implies a positive quantity, so we will look for solutions where $A > 0$ and $B > 0$.

Finding Integer Solutions

We need to find integer solutions for $A$ and $B$. First, let's find one particular solution.

Rearranging the equation:

This shows that $378 - 6B$ must be divisible by 25.

We can test values for $B$ or use modular arithmetic. Let's find the remainder when 378 is divided by 6:

So, $378$ is a multiple of 6. This means $25A$ must also be a multiple of 6.

Since 25 and 6 are coprime ($gcd(25, 6) = 1$), $A$ must be a multiple of 6.

Let's test multiples of 6 for $A$ starting from a value that makes $25A$ less than 378.

• If $A = 6$: $25(6) + 6B = 378 \implies 150 + 6B = 378 \implies 6B = 228 \implies B = 38$. This gives the solution $(A, B) = (6, 38)$.
• If $A = 12$: $25(12) + 6B = 378 \implies 300 + 6B = 378 \implies 6B = 78 \implies B = 13$. This gives the solution $(A, B) = (12, 13)$.
• If $A = 18$: $25(18) = 450$, which is greater than 378. So, $A$ cannot be 18 or larger.

We can also find the general solution. From $25A + 6B = 378$, a particular solution is $(A_0, B_0) = (12, 13)$.

The general solution is:

where $t$ is an integer.

Applying Constraints

We assume the lady bought a positive number of apples and bananas, so $A > 0$ and $B > 0$.

• Constraint $A > 0$:
$ 12 + 6t > 0 \implies 6t > -12 \implies t > -2 $
• Constraint $B > 0$:
$ 13 - 25t > 0 \implies 13 > 25t \implies t < \frac{13}{25} $

Combining the constraints, we need integers $t$ such that $-2 < t < \frac{13}{25}$.

The integers that satisfy this condition are $t = -1$ and $t = 0$.

Counting the Ways

The possible integer values for $t$ are -1 and 0. Each value corresponds to a valid way of choosing apples and bananas.

• For $t = -1$: $A = 12 + 6(-1) = 6$, $B = 13 - 25(-1) = 38$. Pair: $(6, 38)$.
• For $t = 0$: $A = 12 + 6(0) = 12$, $B = 13 - 25(0) = 13$. Pair: $(12, 13)$.

There are 2 pairs of $(A, B)$ that satisfy the condition $A > 0$ and $B > 0$. Therefore, there are 2 ways she could have chosen the numbers of apples and bananas.