# Two brands of chocolates are available in packs of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy?

A. 2 of first kind, 8 of second kind

B. 5 of first kind, 8 of second kind

C. 5 of first kind, 7 of second kind

D. 2 of first kind, 7 of second kind

Last updated date: 28th Mar 2023

•

Total views: 307.2k

•

Views today: 4.84k

Answer

Verified

307.2k+ 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.

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

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE