A sweet seller has 420 Kaju burfis and 130 Ba | vedclass

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Question

(A) Let $d$ be the number of items in each stack.Since each stack must contain the same number of items,$d$ must be a common divisor of $420$ and $130

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) Let $d$ be the number of items in each stack.Since each stack must contain the same number of items,$d$ must be a common divisor of $420$ and $130$.To occupy the least area of the tray,the number of stacks must be minimized,which implies that the number of items in each stack $(d)$ must be the greatest common divisor $(GCD)$ of $420$ and $130$.Using Euclid's division algorithm:$420 = 130 	imes 3 + 30$$130 = 30 	imes 4 + 10$$30 = 10 	imes 3 + 0$The last non-zero divisor is $10$.Therefore,$GCD(420, 130) = 10$.Thus,the maximum number of burfis that can be placed in each stack is $10$.

$A$ sweet seller has $420$ Kaju burfis and $130$ Badam burfis. He wants to stack them in such a way that each stack carries the same number of items and they take up the least area of the tray. What is the maximum number of burfis that can be placed in each stack for this purpose?

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