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More # 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 kindB. 5 of first kind, 8 of second kindC. 5 of first kind, 7 of second kindD. 2 of first kind, 7 of second kind

Last updated date: 28th Mar 2023
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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.

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.