Question

A sweet seller has $420$ kaju barfis and $130$ badam barfis. She wants to stack them in such a way that each stack has the same number and they take up the least area of the tray. What is the maximum number of the barfis that can be placed in each stack for this purpose?

Answer 1

Hint: Here, a sweet seller has a fixed number of two different types of sweets and she wants to stack them in such a way that each stack has the same number of sweets. satisfying this condition we have to find the maximum number of sweets in each stack and the maximum number of sweets is the highest common factor of the number of two different types of sweets.

Complete step-by-step solution:
The area of the tray that is used up in stacking the burfis will be least if the sweet seller stacks maximum number of burfis in each stack. Since each stack has the same number of burfis. Therefore, the number of stacks will be least if the number of burfis in each stack is equal to the HCF of $420$ and $130$.
Now, we have to find the HCF by prime factorization methods.
Here, we have to write the prime factors of $420$ and $130$.
$420 = 2 \times 2 \times 3 \times 5 \times 7$
$130 = 2 \times 5 \times 13$
The prime factor $2$ and $5$ is common to the factor of both numbers.
So, HCF $ = 2 \times 5$
Thus, HCF of $420$ and $130$ is $10$.

Hence, the maximum number of sweets arranged in each stack is $10$ burfis of each kind to cover the least area of the tray.

Note: Here, the Highest common factor of $420$and $130$ can also be calculated by long division method. to find HCF by this method, divide the bigger number by smaller number and then divide smaller number by the remainder of the first one and then repeat these steps until the remainder becomes zero.