Math, asked by Anonymous, 2 months ago

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

Answers

Answered by Anonymous
16

Answer:

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ANSWER

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Given :-

  • A sweetseller has 420 kaju barfis and 130 badam barfis

To Find :-

As She wants to stack them in such a way that each stack has the same number of barfis and they take me number of barfis and they take up the least area of the tray:

  • We have to find the maximum number of barfis that can be placed in each stack for this purpose?

  • Hence , we have to find the HCF of 420 and 130.

Solution :-

Here , I have found the HCF by using Euclid's Division Algorithm :-

Let a = 420 (as it is bigger number)

and let b = 130 ( as it is smaller number)

Now , by using Euclid's Division Algorithm , we have:

a = bq + r

where , a is dividend , b is divisor , q is quotient and r is remainder.

a = b × q + r

420 = 130 × 3 + 30 (case 1)

130 = 30 × 4 + 10 ( case 2)

30 = 10 × 3 + 0 ( case 3)

Now , the remainder has come 0 (zero) in case 3 , so we stop the process

So , The HCF of 420 and 130 came as 10.

(in place of b in case 3)

Hence , the required answer is 10 .

Which means the maximum number of barfis that can be placed in each stack for this purpose is 10.

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Additional Information

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  • What is Euclid's Division Algorithm or Euclid's Division Lemma ?

=> A lemma is already a proved statement which is used to prove others statements.

According to "Euclid's Division Algorithm or Euclid's Division Lemma" for any two given positive integers a and b there exist a unique whole numbers q and r

such that ,

a = bq + r , where 0 < r < b

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Understanding the Formula

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  • Divide a and b :-

Case 1 :-

b ) a ( q

r

=> dividend = (divisor × quotient) +

remainder

=> a = b×q+r

Hence , the formula is proved.

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