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. 

Few more questions - Mixed bag

Have given below four questions across topics. Will discuss the answers shortly.

1. Average of 6 distinct positive integers is 33. The median of the three largest numbers is 43. What is the difference between the highest and lowest possible median of the 6 numbers?

2. In class A, the ratio of boys to girls is 2:3. In class B the ratio of boys to girls is 4 : 5. If the ratio of boys to girls in both classes put together is 3 : 4, what is the ratio of number of girls in class A to number of girls in class B?

3. N is an 80-digit positive integer (in the decimal scale). All digits except the 44th digit (from the left) are 2. If N is divisible by 13, find the 26th digit.

4. A page is torn from a novel. The sum of the remaining digits is 10000. What is the sum of the two page-numbers on the torn page of this novel?