### CAT preparation - Ratio and Proportion (Tough)

**Question**

John
has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates
of types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the
ratio 3 : 5. If there are more chocolates of type C than of type B, and more of
type B than of type A, what is the minimum possible number of chocolates
overall?

**Explanation**

Again,
big thanks to Mukund Sukumar for excellent solution.

Let
john have 3x chocolates of type A and 7x of type B

Let
Mike have 5y chocolates of type B and 4y of type C

Let
john have 3z chocolates of type C and 5z of type A

So
in total A=3x+5z ; B=7x+5y ; C=4y+3z

Since
C>B we get solving y<3z-7x --->(1)

Since
B>A we get solving 5y>5z-4x ---> (2)

What
gets inferred from above 2 statements is z>=3. so when x=1,z=3, we get only
y=1 as choice, for which second condition doesnt satisfy.

So,
when x=1,z=4, we get y<5 from first condition and when y > 3.2 from
second condition. so which gives choice the only y=4.

Hence
x=1,y=4 and z=4 works and is the best possible answer.

For
these values, we get A=23,B=27,C=28.

Minimum
possible number of chocolates overall is 76.

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