CG Board Class 9 Mathematics Chapter 1 Solutions: History of Mathematics (गणित का इतिहास) – 2026-27

• 10: 1 + 0 = 1
• 11: 1 + 1 = 2
• 321: 3 + 2 + 1 = 6
• 78: 7 + 8 = 15 → 1 + 5 = 6

### Example 2 / उदाहरण 2

• Method: Sum of digits = 8 + 7 + 5 + 6 + 9 + 0 + 4 = 39 → 3 + 9 = 12 → 1 + 2 = 3

### Example 3 / उदाहरण 3

• Sum = 3469 + 2220 + 1239 = 6928
• Check / जाँच:
• Digit sum of 3469 = 4
• Digit sum of 2220 = 6
• Digit sum of 1239 = 6
• Sum of Digit sums = 4 + 6 + 6 = 16 → 1 + 6 = 7
• Digit sum of answer (6928) = 6 + 9 + 2 + 8 = 25 → 2 + 5 = 7
• Since both digit sums equal 7, the answer is verified to be correct.

### Example 4 / उदाहरण 4

• 7816 – 3054 = 4762
• Check / जाँच:
• Digit sum of lower number (3054) + Digit sum of answer (4762) = 3 + 1 = 4
• Digit sum of upper number (7816) = 4
• Both equal 4, hence verified.

### Example 5 / उदाहरण 5

• Product = 371184
• Check / जाँच:
• Digit sum of 456 = 6; Digit sum of 814 = 4
• Product of digit sums = 6 × 4 = 24 → 2 + 4 = 6
• Digit sum of product (371184) = 3 + 7 + 1 + 1 + 8 + 4 = 24 → 6
• Both equal 6, hence verified.

### Example 6 / उदाहरण 6

• भागफल = 241, शेषफल = 10
• Check / जाँच:
• Digit sum of भाज्य (7481) = 7 + 4 + 8 + 1 = 20 → 2
• (Digit sum of भाजक × Digit sum of भागफल) + Digit sum of शेषफल

= (4 × 7) + 1 = 28 + 1 = 29 → 11 → 2

• Both equal 2, hence verified.

### Example 7 / उदाहरण 7

1558

4592

98175

6237

34516548

42957

34699653

• Reduce multiplicand by 1: 438 – 1 = 437
• Append 99: 43799
• Subtract LHS (437): 43799 – 437 = 43362

### Examples 15 to 18 (Ekaadhikena Poorvena Method) / उदाहरण 15 से 18

• Example 15 (12 × 18): LHS = 1 × 2 = 2, RHS = 2 × 8 = 16 → 216
• Example 16 (21 × 29): LHS = 2 × 3 = 6, RHS = 1 × 9 = 09 → 609
• Example 17 (102 × 108): LHS = 10 × 11 = 110, RHS = 2 × 8 = 16 → 11016
• Example 18 (194 × 196): LHS = 19 × 20 = 380, RHS = 4 × 6 = 24 → 38024

### Examples 19 to 25 (Nikhilam Vidhi) / उदाहरण 19 से 25

• Example 19 (12 × 14, Base 10): Dev +2, +4 → 12+4 | 2 × 4 = 168
• Example 20 (16 × 15, Base 10): Dev +6, +5 → 16+5+3(carry) | 30 = 240
• Example 21 (8 × 13, Base 10): Dev -2, +3 → 8+3 | -6 = 110 – 6 = 104
• Example 22 (104 × 108, Base 100): Dev +04, +08 → 104+8 | 32 = 11232
• Example 23 (103 × 101, Base 100): Dev +03, +01 → 103+1 | 03 = 10403
• Example 24 (92 × 107, Base 100): Dev -08, +07 → 92+7 | -56 = 9900 – 56 = 9844
• Example 25 (1014 × 994, Base 1000): Dev +014, -006 → 1014-6 | -084 = 1008000 – 84 = 1007916

### Examples 26 to 28 (Squares) / उदाहरण 26 से 28

• Example 26 (65²): 6 × 7 | 25 = 4225
• Example 27 (64²): 6² | 2(6)(4) | 4² = 36 | 48 | 16 = 4096
• Example 28 (48²): 4² | 2(4)(8) | 8² = 16 | 64 | 64 = 2304

### Example 29 (Square Root by Vilokanam) / उदाहरण 29

The modern decimal number system uses 10 digits (0-9) where every digit has both a face value and a place value. This place value system makes computations simple, scientific, and globally applicable.

1. Concept of zero is found in ancient Vedas using the word ‘Kham’ (Yajurveda).
2. Mentioned by Sanskrit grammarians Panini (500 B.C.), Pingal (200 B.C.), and Rishi Grutsamad.
3. First physical symbol representing zero was found in the Bakhshali Manuscript (300-400 A.D.).

When letters or alphabets are used to represent numbers, it is called a coded/alphanumeric system. Aryabhata used this system extensively in Aaryabhatiya to express large numerical values concisely.

In a rectangle, the area of the square drawn on the diagonal is equal to the sum of the areas of the squares drawn on its two sides. In a right angled triangle, Hypotenuse² = Base² + Perpendicular². Baudhayan described this theorem centuries before Pythagoras.

Aryabhata calculated that a circle of diameter 20000 has a circumference of approximately 62832. Thus, π = (62832)/(20000) = 3.1416, which is accurate to 4 decimal places.

Created by Swami Bharti Krishna Teertha (1884-1960 A.D.). The book contains 16 main Sutras and 13 Upsutras spread across 40 chapters for rapid mental calculations.

Obtaining a single digit number by repeatedly adding all digits of a number is called its Digit sum (Bijank).

6, 2, 7, 9, 7, 3, 1, 2, 4, 4

1. 15: 1 + 5 = 6
2. 38: 3 + 8 = 11 → 1 + 1 = 2
3. 88: 8 + 8 = 16 → 1 + 6 = 7
4. 99: 9 + 9 = 18 → 1 + 8 = 9
5. 412: 4 + 1 + 2 = 7
6. 867: 8 + 6 + 7 = 21 → 2 + 1 = 3
7. 4852: 4 + 8 + 5 + 2 = 19 → 1 + 9 = 10 → 1 + 0 = 1
8. 9875: 9 + 8 + 7 + 5 = 29 → 2 + 9 = 11 → 1 + 1 = 2
9. 24601: 2 + 4 + 6 + 0 + 1 = 13 → 1 + 3 = 4
10. 48956701: 4 + 8 + 9 + 5 + 6 + 7 + 0 + 1 = 40 → 4 + 0 = 4

2, 3, 5, 7

1. 241: Digit sum = 7. Required addition = 9 – 7 = 2
2. 861: Digit sum = 15 → 6. Required addition = 9 – 6 = 3
3. 4441: Digit sum = 13 → 4. Required addition = 9 – 4 = 5
4. 83504: Digit sum = 20 → 2. Required addition = 9 – 2 = 7

Digit sums are used to quickly verify the correctness of addition, subtraction, multiplication, and division operations without full recalculation.

736

1. 44 × 52

• Step 1: 4 × 2 = 8
• Step 2: (4 × 2) + (4 × 5) = 8 + 20 = 28 (write 8, carry 2)
• Step 3: 4 × 5 = 20 + 2 = 22

2288

1. 92 × 37

• Step 1: 2 × 7 = 14 (write 4, carry 1)
• Step 2: (9 × 7) + (2 × 3) = 63 + 6 = 69 + 1 = 70 (write 0, carry 7)
• Step 3: 9 × 3 = 27 + 7 = 34

3404

1. 55 × 55

• Step 1: 5 × 5 = 25 (write 5, carry 2)
• Step 2: (5 × 5) + (5 × 5) = 25 + 25 = 50 + 2 = 52 (write 2, carry 5)
• Step 3: 5 × 5 = 25 + 5 = 30

3025

1. 123 × 321

• Step 1: 3 × 1 = 3
• Step 2: (2 × 1) + (3 × 2) = 8
• Step 3: (1 × 1) + (3 × 3) + (2 × 2) = 14 (write 4, carry 1)
• Step 4: (1 × 2) + (2 × 3) = 8 + 1 = 9
• Step 5: 1 × 3 = 3

39483

1. 414 × 232

• Step 1: 4 × 2 = 8
• Step 2: (1 × 2) + (4 × 3) = 14 (write 4, carry 1)
• Step 3: (4 × 2) + (4 × 2) + (1 × 3) = 19 + 1 = 20 (write 0, carry 2)
• Step 4: (4 × 3) + (1 × 2) = 14 + 2 = 16 (write 6, carry 1)
• Step 5: 4 × 2 = 8 + 1 = 9

96048

1. 504 × 618

• Step 1: 4 × 8 = 32 (write 2, carry 3)
• Step 2: (0 × 8) + (4 × 1) = 4 + 3 = 7
• Step 3: (5 × 8) + (4 × 6) + (0 × 1) = 64 + 0 = 64 (write 4, carry 6)
• Step 4: (5 × 1) + (0 × 6) = 5 + 6 = 11 (write 1, carry 1)
• Step 5: 5 × 6 = 30 + 1 = 31

311472

1. 812 × 453

• Step 1: 2 × 3 = 6
• Step 2: (1 × 3) + (2 × 5) = 13 (write 3, carry 1)
• Step 3: (8 × 3) + (2 × 4) + (1 × 5) = 37 + 1 = 38 (write 8, carry 3)
• Step 4: (8 × 5) + (1 × 4) = 44 + 3 = 47 (write 7, carry 4)
• Step 5: 8 × 4 = 32 + 4 = 36

367836

5643

1. 4378 × 9999

• LHS = 4378 – 1 = 4377; RHS = 9999 – 4377 = 5622

43775622

1. 87 × 999

• Multiplicand = 087; LHS = 086; RHS = 999 – 086 = 913

86913

1. 345 × 99999

• Multiplicand = 00345; LHS = 00344; RHS = 99999 – 00344 = 99655

34499655

1. 48 × 9

• Condition 3: LHS = 48 – 1 = 47; Combine = 479
• Subtract LHS: 479 – 47 = 432

1. 9457 × 999

• Condition 3: LHS = 9457 – 1 = 9456; Combine = 9456999
• Subtract LHS: 9456999 – 9456 = 9447543

4x² + 21x + 5

1. (4x + 2y)(3x + 3y):

• Step 1: 2y × 3y = 6y²
• Step 2: (4x × 3y) + (2y × 3x) = 12xy + 6xy = 18xy
• Step 3: 4x × 3x = 12x²

12x² + 18xy + 6y²

1. (x – 3y)(x + 3y):

• Step 1: (-3y)(3y) = -9y²
• Step 2: (x)(3y) + (-3y)(x) = 0
• Step 3: x × x = x²

x² – 9y²

1. (x + 4)(x + 4):

• Step 1: 4 × 4 = 16
• Step 2: 4x + 4x = 8x
• Step 3: x × x = x²

x² + 8x + 16 (Note: Official back-of-book key misprints the middle coefficient as 4x, giving x² + 4x + 16, but the complete algebraic derivation gives x² + 8x + 16).

1. (x² + 2x + 1)(x² + 3x + 4):

• Vertical right: 1 × 4 = 4
• Cross 2 terms: (2x × 4) + (1 × 3x) = 11x
• Cross 3 terms: (x² × 4) + (1 × x²) + (2x × 3x) = 11x²
• Cross 2 left terms: (x² × 3x) + (2x × x²) = 5x³
• Vertical left: x² × x² = x⁴

x⁴ + 5x³ + 11x² + 11x + 4

1. (2x² + 3y – 4)(3x² + 4y + 5):

• Term 1: 2x² × 3x² = 6x⁴
• Term 2: (2x² × 4y) + (3y × 3x²) = 17x²y
• Term 3: (2x² × 5) + (-4 × 3x²) + (3y × 4y) = 10x² – 12x² + 12y² = -2x² + 12y²
• Term 4: (3y × 5) + (-4 × 4y) = 15y – 16y = -y
• Term 5: -4 × 5 = -20

6x⁴ + 17x²y – 2x² + 12y² – y – 20