Q1.
What is the 100th even number?
Answer
The 100th even number is 200.
nth even number = 2n
100th even number = 2 × 100 = 200
100th even number = 2 × 100 = 200
Tip: knowing this immediately gives the 100th odd number as 200 – 1 = 199.
NCERT Solutions Ganita Prakash (Part 1) Chapter 6 In-text Questions — Some Explorations in Grids
Book page 133 Updated on2026-09-05
Q2.
Write a formula to find the nth odd number.
Answer
The nth odd number is 2n – 1.
Step 1: find the even number at that position → 2n
Step 2: subtract 1 from it → 2n – 1
Step 2: subtract 1 from it → 2n – 1
| n | 2n | 2n – 1 |
|---|---|---|
| 1 | 2 | 1 |
| 2 | 4 | 3 |
| 7 | 14 | 13 |
| 50 | 100 | 99 |
Why it happens: 2n is always even, so 2n – 1 is always odd, and as n runs through 1, 2, 3, … the values 1, 3, 5, 7, … appear in order without a gap.
Q3.
Are you able to see what the circled numbers represent?
Answer
Yes — the number in each yellow circle is the sum of the row or column it sits beside.
For the grid shown in the book:
| C1 | C2 | C3 | Row sum | |
|---|---|---|---|---|
| R1 | 4 | 7 | 5 | 16 |
| R2 | 6 | 1 | 2 | 9 |
| R3 | 3 | 9 | 8 | 20 |
| Column sum | 13 | 17 | 15 | — |
4 + 7 + 5 = 16 • 6 + 1 + 2 = 9 • 3 + 9 + 8 = 20
4 + 6 + 3 = 13 • 7 + 1 + 9 = 17 • 5 + 2 + 8 = 15
4 + 6 + 3 = 13 • 7 + 1 + 9 = 17 • 5 + 2 + 8 = 15
Check it yourself: 16 + 9 + 20 = 45 and 13 + 17 + 15 = 45 — both equal 1 + 2 + … + 9.
Q4.
Fill the grids below based on the rule mentioned above:
Answer
Use the numbers 1 – 9, each exactly once, so that every row and every column adds to the circled number.
First grid — row sums 13, 14, 18 and column sums 24, 9, 12, with 9 in the top-left corner and 5 in the bottom-right corner.
| C1 | C2 | C3 | Row sum | |
|---|---|---|---|---|
| R1 | 9 | 1 | 3 | 13 |
| R2 | 8 | 2 | 4 | 14 |
| R3 | 7 | 6 | 5 | 18 |
| Column sum | 24 | 9 | 12 | 45 |
How it was found:
Row 1: 9 + ? + ? = 13 → the other two add to 4 → they are 1 and 3
Column 1: 9 + ? + ? = 24 → the other two add to 15 → they are 7 and 8
Row 3: ? + ? + 5 = 18 → the first two add to 13 → so 7 goes in R3C1 and 6 in R3C2
Column 2: 1 + ? + 6 = 9 → the middle entry is 2
Row 2: 8 + 2 + ? = 14 → the last entry is 4 ✓
Column 1: 9 + ? + ? = 24 → the other two add to 15 → they are 7 and 8
Row 3: ? + ? + 5 = 18 → the first two add to 13 → so 7 goes in R3C1 and 6 in R3C2
Column 2: 1 + ? + 6 = 9 → the middle entry is 2
Row 2: 8 + 2 + ? = 14 → the last entry is 4 ✓
Second grid — row sums 24, 15, 6 and column sums 12, 16, 17, with 4 in the middle-left cell and 3 in the bottom-right cell.
| C1 | C2 | C3 | Row sum | |
|---|---|---|---|---|
| R1 | 7 | 8 | 9 | 24 |
| R2 | 4 | 6 | 5 | 15 |
| R3 | 1 | 2 | 3 | 6 |
| Column sum | 12 | 16 | 17 | 45 |
Row 1 must be 24, and the largest possible sum of three different numbers is 9 + 8 + 7 = 24 → row 1 is 7, 8, 9 in some order
Row 3 must be 6, the smallest possible sum, and 3 is already there → the other two are 1 and 2
Column 1: ? + 4 + ? = 12 → 7 + 4 + 1 = 12 ✓
Column 2: 8 + 6 + 2 = 16 ✓ Column 3: 9 + 5 + 3 = 17 ✓
Row 3 must be 6, the smallest possible sum, and 3 is already there → the other two are 1 and 2
Column 1: ? + 4 + ? = 12 → 7 + 4 + 1 = 12 ✓
Column 2: 8 + 6 + 2 = 16 ✓ Column 3: 9 + 5 + 3 = 17 ✓
Did you know? the second grid has one more answer: 7, 9, 8 / 4, 5, 6 / 1, 2, 3 also fits every circle. The first grid, however, has exactly one answer.
Why it happens: the three row sums always add to 1 + 2 + … + 9 = 45, and so do the three column sums. Checking this first (13 + 14 + 18 = 45, 24 + 9 + 12 = 45) tells you the puzzle is worth attempting.