Question

There are three boxes of three different colors -- Green, Blue and Red -- and 6 toys of which 2 are of Green colour, 2 are of Blue colour and 2 are of Red colour. The toys are packed in the three boxes such that each box has 2 toys of different colours in it and also the colour of the box is different from the colour of the toys packed in it. Now, 10 chocolates are kept in these boxes in such a way that the Green box has the maximum possible chocolates in it whereas, the Red box has the least possible chocolates in it. Each box should have at least one chocolate and no two boxes have the same number of chocolates. Determine which of the following is definitely true ?

• A. The box which has the toys of Blue and Green colors has 3 chocolates in it
• B. (B) The box which has the toys of Green and Red colors has 2 chocolates in it
• C. (C) The Green box, the Blue box and Red box have 6, 3 and 1 chocolate/s in them respectively
• D. (D) Green Box has not more than one chocolate in it
• E. (E) The box which has the toys of Red and Blue colors has 8 chocolates in it

Hint:

In box-arrangement logic, first determine which toys go in which box from the colour constraints, then test the chocolate distribution by maximizing/minimizing as required.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Each box has 2 toys of different colours, not matching the box colour. Maximize Green box chocolates; minimize Red box chocolates. Total = 10, all different, min 1 each.

Step 2: Detailed Explanation:
Box colours: Green, Blue, Red. Toy colours: 2G, 2B, 2R. Each box gets 2 toys, different colours, different from box colour. Green box: toys not green $\to$ Blue+Red toys. Blue box: toys not blue $\to$ Green+Red toys. Red box: toys not red $\to$ Green+Blue toys. Chocolates: total=10, all distinct, $\geq1$ each. Green max, Red min. Max distribution: Green=7, Blue=2, Red=1 but $7+2+1=10$ $\checkmark$ distinct. Or Green=6, Blue=3, Red=1 ($10$) $\checkmark$. Or Green=5, Blue=4, Red=1. Green box has Blue+Red toys; Blue box has Green+Red toys; Red box has Green+Blue toys. Option (B): ``box with Green and Red toys'' = Blue box; has 2 chocolates. In distribution 7,2,1: Blue=2 $\checkmark$. In 6,3,1: Blue=3 $\times$. In 5,4,1: Blue=4 $\times$. Only definitely true for max configuration $\to$ official answer is (B).

Step 3: Final Answer:
The box which has the toys of Green and Red colors has 2 chocolates in it.