Question

A, B and C have a few coins with them. 7 times the number of coins that A has is equal to 5 times the number of coins B has while 6 times the number of coins B has is equal to 11 times the number of coins C has. What is the minimum number of coins with A, B and C put together?
110
174
154
165

Answer 1

Answer:

174

Step-by-step explanation:

here 7A=5B

B=7A/5

6B=11C

C=6B/11

=6/11*7A/5=42A/55

A:B:C=A:7A/5:42A/55

=1:7/5:42/55

A:B:C=55:77:42

MINIMUM NO OF COINS

=55+77+42

=174

Answer 2

The minimum number of coins with A, B, and C put together would be 174.

Step-by-step explanation:

Let the number of coins A, B, and C have $a,b, and c$ respectively.

• Then, Seven times a = Five times b

7a = 5b

$a =\frac {5}{7}\times b$

• Now, Six times b = Eleven times c

6b = 11c

$c = \frac{6}{11}\timesb$

• Now, the total number of coins =  a+b+c

Substituting the values of a and c from above, we obtain

• The total number of coins

$= a+b+c$

$= (\frac {5}{7}\times b) + b + (\frac{6}{11} \times b)$

$=\frac{55b+77b+42b}{77}$

$=\frac{174b}{77}$

• Thus, if b has 77 coins, then the total number of coins would become 174.