Question

Marshall, randy and roger has some coins. Marshalls 5 times coin is equal with randys 8 times coin. Marshalls 8 times coin is equal with rogers 5 times coin. Find the minimum coin the three have in total.

Answer 1

The minimum coin the three have in total would be 129

Step-by-step explanation:

Let A, B and C represents the coins contained by Marshall, Randy and Roger respectively,

∵ Marshall's 5 times coin is equal with Randy's 8 times coin

i.e. $5\times A = 8\times B$

⇒ $\frac{A}{B}=\frac{8}{5}$

Now, Marshall's 8 times coin is equal with Roger's 5 times coin.

i.e $8\times A = 5\times C$

$\frac{A}{C}=\frac{5}{8}$

LCM(8, 5) = 40,

$\frac{A}{B}=\frac{8\times 5}{5\times 5}=\frac{40}{25}$

$\frac{A}{C}=\frac{5\times 8}{8\times 8}=\frac{40}{64}$

Thus, the ratio of coins contained by Marshall, Randy and Roger = 40 : 25 : 64

Since, number of coins can't be a fraction,

Hence, the minimum number of total coins = 40 + 25 + 64 = 129

#Learn more :

In a palace, three different types of coins are there namely gold, silver and bronze. the number of gold, silver and bronze coins is 18000, 9600 and 3600 respectively. find the minimum number of rooms required if in each room should give the same number of coins of the same type.

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