Try This — Find the Colours! — The Baudhāyana–Pythagoras Theorem

1. 13 marksGanita Prakash Cl-8 Part 2, Try This, page 54

   There are 3 closed boxes — one containing only red balls, the second only blue balls, and the third only green balls. The boxes are labelled RED, BLUE and GREEN in such a way that no box has the correct label. Find which label goes with which box, if you are allowed to open only one box.

   Hint. List every way the labels can all be wrong. There are fewer possibilities than you might expect.

   Step 1 — list every possible arrangement. The condition is that no box carries its correct label. Write the box by its label and the contents beside it. With three colours there are six ways to place the contents, but only two of them get every label wrong:

   Arrangement A box labelled RED → contains blue box labelled BLUE → contains green box labelled GREEN → contains red

   Arrangement B box labelled RED → contains green box labelled BLUE → contains red box labelled GREEN → contains blue

   Any other placement leaves at least one box correctly labelled, so it is ruled out by the condition.

   Step 2 — notice how the two arrangements differ. They disagree in every single position — the RED box holds blue in one and green in the other, and so on. So learning the contents of any one box tells you which arrangement you are in, and that fixes the other two.

   Step 3 — open one box and finish. Open the box labelled RED and look at one ball. · If it is blue, you are in Arrangement A: the BLUE box holds green and the GREEN box holds red. · If it is green, you are in Arrangement B: the BLUE box holds red and the GREEN box holds blue. It cannot be red, because the RED label is known to be wrong.

   Either way all three boxes are identified after opening just one. Any of the three boxes could have been chosen — the argument is the same.

   Why "no box is correct" is essential. If only some labels were wrong there would be more arrangements to separate, and one look would not be enough. The strength of the puzzle comes from the condition being so strict that just two possibilities survive.

   ✦ Open any one box — say the one labelled RED — and take out a ball. If it is blue then BLUE holds green and GREEN holds red; if it is green then BLUE holds red and GREEN holds blue. One box is always enough.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 2, Reprint 2026-27 (hegp202.pdf), where this is Chapter 2 (pages 33-54) — the ninth chapter of the Class 8 course. Like the rest of Part 2, this PDF carries NO printed answer key, so every answer here was derived from first principles and then independently recomputed in Python before being written — including all the √2-style one-decimal bounds, the six figure triangles, the rhombus side, the odd-square triple generator, and the complete list of Baudhāyana triples with all numbers at most 20. TWO POINTS WHERE THE BOOK'S OWN TEXT NEEDS CARE ARE FLAGGED IN PLACE: (1) on page 48 the book lists four triples with numbers at most 20 and says the list 'contains' them — the complete list has six, since (5, 12, 13) and (8, 15, 17) also qualify and are not multiples of (3, 4, 5); (2) the six right triangles in Figure it Out Q2 on pages 52-53 are labelled only in the printed figure, so each one's right-angle position was read directly off the rendered PDF page before solving, and the reading is stated in the solution so a student can check it against the book.. Questions are referenced from the NCERT textbook for identification.