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 of sweets, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose?
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 of sweets, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose?
The correct option is A 10
This can be done by trial and error. But to do it systematically, we find the HCF of 420 and 130. This number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will also be the least.
Using Euclid's algorithm to find the HCF of the numbers, we get:
420=130×3+30
130=30×4+10
30=10×3+0
So, the HCF of 420 and 130 is 10.
Therefore, the sweetseller can make stacks of 10 for both kinds of barfi.
10
This can be done by trial and error. But to do it systematically, we find the HCF of 420 and 130. This number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will also be the least.
Using Euclid's algorithm to find the HCF of the numbers, we get:
420=130×3+30
130=30×4+10
30=10×3+0
So, the HCF of 420 and 130 is 10.
Therefore, the sweetseller can make stacks of 10 for both kinds of barfi.
