NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –99Patterns in a Calendar — In-text Questions
Book page 97 Updated on2026-09-05
Q1.
Let us take the marked 2 × 2 square, and consider the numbers lying on the diagonals; 12 and 20; 13 and 19. Find their sums; 12 + 20, 13 + 19. What do you observe?
Answer
Both diagonals add up to the same number.
12 + 20 = 32 13 + 19 = 32 The two diagonal sums are equal.
Try This: Take the square 5, 6 over 12, 13 from the same November 2024 calendar. 5 + 13 = 18 and 6 + 12 = 18. Equal again.
Q2.
Will the diagonal sums be equal in every 2 × 2 square in this endless grid? How can we be sure?
Answer
Yes — but checking examples can never make us sure, because there are unlimited 2 × 2 squares.
Why it happens: To be sure we need an argument that covers all squares at once. Algebra does exactly that: call the top-left number a and describe the other three in terms of a. One calculation then settles every square.
Q3.
Given that we know the top left number, how do we find the other numbers in this 2 × 2 square?
Answer
Describe them in words first, then in symbols.
the number to the right of a is 1 more than it
the number below a is 7 more than it
the number diagonal to a is 8 more than it
a
a + 1
a + 7
a + 8
Diagonal 1: a + (a + 8) = a + a + 8 = 2a + 8 Diagonal 2: (a + 1) + (a + 7) = a + a + 1 + 7 = 2a + 8 Both diagonal sums are equal — for every value of a.
Why it happens: A calendar week has 7 days, so dropping one row adds 7. Moving one column right adds 1. The diagonal cell does both, so it adds 8.
Q4.
Verify this expression for diagonal sums by considering any 2 × 2 square and taking its top left number to be ‘a’.
Answer
Take the square with top-left number 25 (so a = 25): the square is 25, 26 over 32, 33.
Did you know? This is the power of algebraic modelling — one short proof replaces an unlimited number of checks.
Q5.
Consider a set of numbers from the calendar (having endless rows) forming the plus shape 8 / 14, 15, 16 / 22. Find the sum of all the numbers. Compare it with the number in the centre: 15. Repeat this for another set of numbers that forms this shape. What do you observe?
Answer
8 + 14 + 15 + 16 + 22 = (8 + 22) + (14 + 16) + 15 = 30 + 30 + 15 = 75 And 75 = 5 × 15, five times the centre.
Another plus shape, centred at 26: the numbers are 19 / 25, 26, 27 / 33.
19 + 25 + 26 + 27 + 33 = 130 = 5 × 26 ✔
Observation: the total is always 5 times the number in the centre.
Q6.
Will this always happen? How do you show this? [Hint: Consider a general set of numbers that forms this shape. Take the number at the centre to be ‘a’. Express the other numbers in terms of ‘a’.]
Answer
Yes, always. Let the centre be a.
a – 7
a – 1
a
a + 1
a + 7
Sum = (a – 7) + (a – 1) + a + (a + 1) + (a + 7) = a + a + a + a + a + (–7 + 7) + (–1 + 1) = 5a + 0 + 0 = 5a
Why it happens: The shape is symmetric about its centre. The number above cancels the number below (–7 and +7), and the number on the left cancels the one on the right (–1 and +1). Only five copies of the centre survive.
Q7.
Find other shapes for which the sum of the numbers within the figure is always a multiple of one of the numbers.
Answer
Any shape that is symmetric about one cell works. Taking the centre as a:
Shape
Numbers in terms of a
Sum
Row of 3 (horizontal)
a – 1, a, a + 1
3a
Column of 3 (vertical)
a – 7, a, a + 7
3a
Diagonal of 3 (↘)
a – 8, a, a + 8
3a
The X of 5 (four corners + centre)
a – 8, a – 6, a, a + 6, a + 8
5a
Full 3 × 3 block
nine numbers around a
9a
3 × 3 block: (a – 8) + (a – 7) + (a – 6) + (a – 1) + a + (a + 1) + (a + 6) + (a + 7) + (a + 8) = 9a
Why it happens: In every one of these shapes the cells pair up around the centre, and each pair adds to 2a. With k such shapes the total is always a multiple of the centre number.
Check it yourself: Take the 3 × 3 block centred on 15 (that is 7 to 23 in the November calendar rows). Its nine numbers add to 135 = 9 × 15 ✔