Answer
452.7k+ views
Hint: By prime factorization method take the LCM of the number of chocolates of both brands. To get the least number of boxes, find LCM (24, 15) to the number of chocolate per box.
“Complete step-by-step answer:”
Given the number of chocolate of ${{1}^{st}}$ brand in one pack = 24
The number of chocolate of ${{2}^{nd}}$ brand in one pack = 15
To find the least number of boxes of each kind first we have to find the LCM of 24 and 15.
LCM by prime factorization method.
The method of prime factorization is used to “break down” or express a given number as a product of prime numbers.
The prime number will occur more than once in the prime factorization.
Where the prime number is a whole number which is greater than 1, it is only divisible by 1 and itself.
LCM of 24 $=2\times 2\times 2\times 3={{2}^{3}}\times 3$
LCM of 15 $=3\times 5$
LCM of two numbers = product of the greater power of each prime factor involved in the numbers, with highest power.
LCM of 24 \[=2\times 2\times 2\times \times 5\]
LCM of 15 $=\times 5$
$\therefore $ LCM of 24 and 15 $=2\times 2\times 2\times 3\times 5$
LCM (24, 15) = 120
Hence, the number of packet of ${{1}^{st}}$ brand $=\dfrac{\text{LCM }\left( 24,15 \right)}{\text{Number of chocolate in }{{\text{1}}^{st}}\text{ brand}}$
Number of packets in ${{2}^{nd}}$ brand $=\dfrac{\text{LCM }\left( 24,15 \right)}{\text{Number of chocolate in }{{\text{2}}^{nd}}\text{ brand}}$
$=\dfrac{120}{15}=8$
$\therefore $ 5 of first kind, 8 of second kind
Note: Another way of finding the least number of boxes of both kind,
Number of packet of ${{1}^{st}}$ brand = 24 = 3 x 8
Number of packets in ${{2}^{nd}}$ brand = 15 = 3 x 5
24 = 3 x 8
15 = 3 x 5
$\therefore $ 8 of ${{1}^{st}}$kind and 5 of ${{2}^{nd}}$kind.
“Complete step-by-step answer:”
Given the number of chocolate of ${{1}^{st}}$ brand in one pack = 24
The number of chocolate of ${{2}^{nd}}$ brand in one pack = 15
To find the least number of boxes of each kind first we have to find the LCM of 24 and 15.
LCM by prime factorization method.
The method of prime factorization is used to “break down” or express a given number as a product of prime numbers.
The prime number will occur more than once in the prime factorization.
Where the prime number is a whole number which is greater than 1, it is only divisible by 1 and itself.
![seo images](https://www.vedantu.com/question-sets/40ab519e-4ea8-4f6e-a41e-488b38e5aa7e3077923194837488636.png)
LCM of 24 $=2\times 2\times 2\times 3={{2}^{3}}\times 3$
LCM of 15 $=3\times 5$
LCM of two numbers = product of the greater power of each prime factor involved in the numbers, with highest power.
LCM of 24 \[=2\times 2\times 2\times \times 5\]
LCM of 15 $=\times 5$
$\therefore $ LCM of 24 and 15 $=2\times 2\times 2\times 3\times 5$
LCM (24, 15) = 120
Hence, the number of packet of ${{1}^{st}}$ brand $=\dfrac{\text{LCM }\left( 24,15 \right)}{\text{Number of chocolate in }{{\text{1}}^{st}}\text{ brand}}$
Number of packets in ${{2}^{nd}}$ brand $=\dfrac{\text{LCM }\left( 24,15 \right)}{\text{Number of chocolate in }{{\text{2}}^{nd}}\text{ brand}}$
$=\dfrac{120}{15}=8$
$\therefore $ 5 of first kind, 8 of second kind
Note: Another way of finding the least number of boxes of both kind,
Number of packet of ${{1}^{st}}$ brand = 24 = 3 x 8
Number of packets in ${{2}^{nd}}$ brand = 15 = 3 x 5
24 = 3 x 8
15 = 3 x 5
$\therefore $ 8 of ${{1}^{st}}$kind and 5 of ${{2}^{nd}}$kind.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)