A Sweet seller
has 420 kaju burfis and 130
badam burfis he wants to
stack them in such a way
that each stack has the
same number of burfis and
they take up the least area
of the tray. What is the
number of burfis that can
be placed in each stack for
this purpose?
Answers
Answer:
ANSWER
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 must have the same number of burfis. Therefore, the number of stacks will be least if the number of burfis in each stacks in equal to the
HCF of 420 and 130.
In order to find the HCF of 420 and 130, let us apply Euclids division lemma to 420 and 130 to get
420=130×3+30
Let us now consider the divisor 130 and the remainder 30 and apply division lemma to get
130=30×4+10
Consider now divisor 30 and the remainder 10 and apply division lemma, we get
30=3×10+0
Since the remainder at this stage is zero. Therefore, last divisor 10 is the HCF of 420 and 130.
Hence , the sweet seller can make stacks of 10 burfis of each kind to cover the least area of the tray.
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Answer:
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 must have the same number of burfis. Therefore, the number of stacks will be least if the number of burfis in each stacks in equal to the
HCF of 420 and 130.
In order to find the HCF of 420 and 130, let us apply Euclids division lemma to 420 and 130 to get
420=130×3+30
Let us now consider the divisor 130 and the remainder 30 and apply division lemma to get
130=30×4+10
Consider now divisor 30 and the remainder 10 and apply division lemma, we get
30=3×10+0
Since the remainder at this stage is zero. Therefore, last divisor 10 is the HCF of 420 and 130.
Hence , the sweet seller can make stacks of 10 burfis of each kind to cover the least area of the tray.
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