Class 9 · MathsNCERT 2026 Revised Syllabus

Chapter 3: The World of Numbers (0904CH03)

Ganita Manjari Part I · 43 questions · 6 exercises

Chapter code: 0904ch03 · Book: Ganita Manjari Part I · Class 9 Maths

Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part I Chapter 3. Every exercise has a clear, step-by-step answer for chapter code 0904CH03.

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Exercise 3.1(4 questions)

1A merchant in Lothal exchanges bags of spices for copper ingots — 15 ingots for every 2 bags. If he brings 12 bags, how many ingots does he leave with?

School Level Solution

No. of ingots received by the merchant for every 2 bags = 15 No. of ingots received by the merchant for every bag = 152 No. of ingots received by the merchant for 12 bags = 15 × 12⁶2 = 90 Therefore, the merchant will leave with 90 ingots.

2The Ishango-bone column has 11, 13, 17, 19. What do these numbers have in common? List the next three numbers in the pattern.

School Level Solution

The nos. 11, 13, 17, 19 have in common that they are the prime nos. between 11 to 20. The next 3 nos. in the series are: 23, 29, 31

3Natural numbers are closed under addition. Are they closed under subtraction? Justify with examples.

School Level Solution

No, natural nos. are not closed under subtraction. This is because closure means, when we subtract two nos. of natural nos., we at get a natural no., which is not true. e.g. 2 − 3 = −1 ∉ N 5 − 26 = −21 ∉ N

4★ Ancient Indians counted finger joints (3 per finger, thumb used to count). How many can you count on one hand? How does this relate to base-12?

School Level Solution

Except thumb, there are 4 fingers in each hand. Each finger is having 3 joints each. Therefore, no. of counts on one hand = 4 × 3 = 12 Ans. Base 12 is preferred over base 10 because 12 is divisible by 2, 3, 4, & 6. Making it easier for trade, sharing and measurement.

Exercise 3.2(4 questions)

1Temperature in Ladakh is 4°C at noon. By midnight it drops by 15°C. What is the midnight temperature?

School Level Solution

Initial temp. = 4°C The temperature drops by 15°C Temperature at midnight = 4°C − 15°C = −11°C Ans.

2A spice trader takes a loan of ₹850, makes a profit of ₹1,200 the next day, then a loss of ₹450 the following week. Write this as an equation using integers and find his final financial standing.

School Level Solution

Loan (debt) taken by the trader = ₹850 ∵ debt ∴ negative Profit made by him the next day = ₹1,200 ∵ fortune ∴ positive Loss incurred in the following week = ₹450 ∵ debt ∴ negative Writing the sequence as an equation, we get = ₹(−850 + 1200 − 450) = ₹(1200 − 1300) = ₹ −100 His final financial standing = ₹ −100 Ans.

3Calculate the following using Brahmagupta's laws:

School Level Solution

i) A debt multiplied by a fortune ⇒ the product of two debts is a debt. The product of a fortune and a debt is a debt. ∴ (−12) × 5 = −60 Ans.

ii) The product of two debts is a fortune. ∴ (−8) × (−7) = 56 Ans.

iii) Zero minus a debt is the same no. but positive. A debt subtracted from zero, it be a fortune. ∴ 0 − (−14) = 0 + 14 = 14

iv) (−20) ÷ 4 The quotient of a debt and a fortune is a debt. ∴ −204 = −5

4Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 − (−5) = 15).

School Level Solution

Real World eg. Suppose you owe ₹500 to a friend (debt = − ₹500) If my friend says "I remove ₹200 from my debt", that means: −500 − (−200) = −500 + 200 = −300 Your debt reduces from 500 to 300 — you are ₹200 better off. Insight: Subtracting a debt (−200) is same as gaining ₹200. Subtracting a negative = Adding a positive a − (−b) = a + b

Exercise 3.3(8 questions)

1Prove that the following rational numbers are equal:

School Level Solution

i) 23 and 46 LHS = 23 RHS = 4²6₃ = 23 LHS = RHS Hence proved.

ii) 54 and 108 LHS = 54 RHS = 10⁵8₄ = 54 LHS = RHS Hence proved.

iii) −35 and −610 LHS = −35 RHS = −6³10₅ = −35 LHS = RHS Hence proved.

iv) 93 and 3 LHS = 9³3 = 3 RHS = 3 LHS = RHS Hence proved.

2Find the sum:

School Level Solution

i) 25 + 310 = 4 + 310 = 710

ii) 712 + 58 = 14 + 1524 = 2924 = 1524

3Find the difference:

School Level Solution

i) 56 − 14 = 10 − 312 = 712 Ans.

ii) 118 − 34 = 11 − 68 = 58 Ans.

iii) −79 − −23 = −79 + 23 = −7 + 279 = 209 = 229

4Find the product:

School Level Solution

i) 23 × 310 = 15 Ans.

ii) 711 × 58 = 3588 Ans.

iii) −47 × 514 = −1049 Ans.

5Find the quotient:

School Level Solution

i) 23 ÷ 310 = 23 × 103 = 209 = 229 Ans.

ii) 711 ÷ 58 = 711 × 85 = 5655 = 1155 Ans.

iii) −47 ÷ 514 = −47 × 145 = −85 = −135

6Show that: (12 + 34) × 83 = 12 × 83 + 34 × 83.

School Level Solution

LHS RHS (12 + 34) × 83 12×83 + 34×83 = (2 + 34) × 83 = 43 + 63 = 54 × 83 = 103 = 103 = 103 LHS = RHS Hence proved.

7Simplify the following using the distributive property: 79 (67 − 34).

School Level Solution

79 ( 67 − 34 ) = (79 × 6²7) − (79 × 34) = 23 − 712 = 8 − 712 = 112

8Find the rational number x such that: 56 (x + 35) = 56 x + 12.

School Level Solution

56 ( x + 35 ) = 56 x + 12 56 x + 1530 = 56 x + 12 56 x + 12 = 56 x + 12 56 x = 56 x x = 56 × 65 x x = x Any value of x can satisfy the given equation.

Exercise 3.4(6 questions)

1Represent the rational numbers 23, −54 and 1½ on a single number line.

School Level Solution

Required nos. to be represented on the no. line are → 23, −54 and 112

-54 = -1.25 R −54 P: 23 = 0.667 P 23 Q: 32 = 1.5 Q 1½

P represents 23 Q represents 112 = 32 R represents −54 = −114

2Find three distinct rational numbers that lie strictly between −12 and 14.

School Level Solution

−12 = −1×22×2 = −24 × 1010 = −2040 −24 = −24 × 1010 = −2040 14 = 14 × 1010 = 1040 Three distinct rational nos. btwⁿ them are: −2040 < −2140 < 140 < 340 < 1040 −12 < −2140 < 140 < 340 < 14 The required nos. are: −2140, 140, 340 Ans.

3Simplify the expression: (−14) + (512).

School Level Solution

= −3 + 512 = 212₆ = 16 Ans.

4A tailor has 15¾ metres of fine silk. If making one kurta requires 2¼ metres of silk, exactly how many kurtas can he make?

School Level Solution

Length of fine silk a tailor has = 1534 m = 634 m Length of fine silk required to make one kurta = 214 m = 94 m No. of kurtas he can make = 634 ÷ 94 = 634 × 49 = 7 Therefore, the tailor can make 7 kurtas from the length of cloth he has.

5Find three rational numbers between 3.1415 and 3.1416.

School Level Solution

3.1415 = 3141510000 = 31415×1010000×10 = 314150100000 3.1416 = 3141610000 = 31416×1010000×10 = 314160100000 Three rational nos. between them are: 314150100000 < 314151100000 < 314152100000 < 314153100000 < 314160100000 3.1415 < 3.14151 < 3.14152 < 3.14153 < 3.1416 The req. nos. are: 3.14151, 3.14152, 3.14153 Ans.

6★ Can you think of other way(s) to find a rational number between any two rational numbers?

School Level Solution

There are multiple ways we can find rational nos. between any two rational nos. Some of them are: 1) Taking the LCM and multiplying the numerator and denominator with 10, 100 or 10ⁿ and writing the new rational nos. between the new rational nos. formed [As they are equal to the earlier ones]. 2) Dividing the numerator by the denominator, converting them into decimal and writing the rational nos. between them. 3) Converting the denominator to 10ⁿ and by multiplying both the numerator and denominator with a same no. and then converting them to decimal and writing the rational nos. btwⁿ them, etc.

Exercise 3.5(5 questions)

1Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 720, 415 and 13250. Then check your answers by explicitly performing the long divisions.

School Level Solution

720 Here, the denominator is 20. 20 = 2 × 2 × 5 = 2² × 5¹ It can be expressed as 2^m × 5ⁿ and has no other factor than 2 and 5, it will have terminating decimal digits Verify:

Q = 0.35 R = 0 720 = 0.35 which has 2 digits of decimal. Hence, verified.

415 Here, the denominator is 15. 15 = 3 × 5 = 3¹ × 5¹ It cannot be expressed as 2^m × 5ⁿ only and has factors other than 2 and 5, it will have non-terminating repeating decimal digits Verify: 415 = 0.26 Q = 0.26, R = 10 which has non-terminating repeating decimal digits. Hence, verified.

13250 Here denominator is 250. 250 = 2 × 5 × 5 × 5 = 2¹ × 5³ It can be expressed as 2^m × 5ⁿ and has no factor other than 2 and 5, it will have terminating decimal digits Verify: Q = 0.052 R = 0 13250 = 0.052 which has 3 digits of decimal. Hence, verified.

2Find the value of the following by using identities:(i) 102 × 106(ii) 34 × 36(iii) 98 × 102(iv) 3.1²

School Level Solution

(i) 102 × 106 = (100 + 2)(100 + 6) = 100 × 100 + 100 × 6 + 2 × 100 + 2 × 6 = 10000 + 600 + 200 + 12 = 10812 Ans.

(ii) 34 × 36 = (35 – 1)(35 + 1) = 35² – 1² = 1225 – 1 = 1224 Ans.

(iii) 98 × 102 = (100 – 2)(100 + 2) = 100² – 2² = 10000 – 4 = 9996 Ans.

(iv) 3.1² = (3 + 0.1)² = 3² + 2 × 3 × 0.1 + 0.1² = 9 + 0.6 + 0.01 = 9.61 Ans.

3Find the value of the following by using identities:(i) 4.5 × 3.5(ii) 198 × 202(iii) 0.98 × 1.02(iv) 20.1²

School Level Solution

(i) 4.5 × 3.5 = (4 + 0.5)(4 – 0.5) = 4² – 0.5² = 16 – 0.25 = 15.75 Ans.

(ii) 198 × 202 = (200 – 2)(200 + 2) = 200² – 2² = 40000 – 4 = 39996 Ans.

(iii) 0.98 × 1.02 = (1 – 0.02)(1 + 0.02) = 1² – 0.02² = 1 – 0.0004 = 0.9996 Ans.

(iv) 20.1² = (20 + 0.1)² = 20² + 2 × 20 × 0.1 + 0.1² = 400 + 4 + 0.01 = 404.01 Ans.

4If x + 1/x = 5, find the value of:(i) x² + 1/x²(ii) x⁴ + 1/x⁴

School Level Solution

(i) Given: x + 1/x = 5 Squaring both sides: (x + 1/x)² = 5² x² + 2 × x × 1/x + 1/x² = 25 x² + 2 + 1/x² = 25 x² + 1/x² = 25 – 2 = 23 Ans.

(ii) x² + 1/x² = 23 Squaring both sides: (x² + 1/x²)² = 23² x⁴ + 2 × x² × 1/x² + 1/x⁴ = 529 x⁴ + 2 + 1/x⁴ = 529 x⁴ + 1/x⁴ = 529 – 2 = 527 Ans.

5If x – 1/x = 4, find the value of:(i) x² + 1/x²(ii) x⁴ + 1/x⁴

School Level Solution

(i) Given: x – 1/x = 4 Squaring both sides: (x – 1/x)² = 4² x² – 2 × x × 1/x + 1/x² = 16 x² – 2 + 1/x² = 16 x² + 1/x² = 16 + 2 = 18 Ans.

(ii) x² + 1/x² = 18 Squaring both sides: (x² + 1/x²)² = 18² x⁴ + 2 × x² × 1/x² + 1/x⁴ = 324 x⁴ + 2 + 1/x⁴ = 324 x⁴ + 1/x⁴ = 324 – 2 = 322 Ans.

End-of-Chapter Exercises(16 questions)

Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division: (i) 3/50 (ii) 2/9

School Level Solution

3/50 = 0.06, so it is terminating. 2/9 = 0.222… = 0.2, so it is non-terminating and repeating.

Prove that √5 is an irrational number.

School Level Solution

Assume √5 = p/q in lowest terms, where p and q are non-zero integers and q > 0. Then p² = 5q². Hence 5 divides p², so 5 divides p; write p = 5k. Substitution gives 25k² = 5q², so q² = 5k², and therefore 5 divides q. This contradicts gcd(p,q) = 1. Hence √5 is irrational.

Convert the following decimal numbers in the form of p/q: (i) 12.6 (ii) 0.0120 (iii) 3.052 (iv) 1.235 (v) 0.23 (vi) 2.05 (vii) 2.125 (viii) 3.125 (ix) 2.1625

School Level Solution

(i) 12.6 = 126/10 = 63/5. (ii) 0.0120 = 120/10000 = 3/250. (iii) Let x = 3.052. Then 1000x = 3052.052 and x = 3.052; subtracting gives 999x = 3049, so x = 3049/999. (iv) 1.235 = (1235 − 1)/999 = 1234/999. (v) 0.23 = 23/99. (vi) 2.05 = 2 + 5/99 = 203/99. (vii) 2.125 = (2125 − 21)/990 = 1052/495. (viii) 3.125 = (3125 − 312)/900 = 2813/900. (ix) 2.1625 = (21625 − 2)/9999 = 21623/9999.

Locate the following rational numbers on the number line: (i) 0.532 (ii) 1.15

School Level Solution

(i) 0.532 is the point 532/1000 = 133/250, lying between 0.5 and 0.6 on the number line. (ii) 1.15 = 52/45 = 1.1555…, so locate it between 1.15 and 1.16, closer to 1.16.

Find 6 rational numbers between 3 and 4.

School Level Solution

One possible set is 31/10, 32/10, 33/10, 34/10, 35/10, 36/10.

Find 5 rational numbers between 2/5 and 3/5.

School Level Solution

Using denominator 50: 21/50, 22/50, 23/50, 24/50, 25/50 all lie strictly between 20/50 and 30/50.

Find 5 rational numbers between 1/6 and 2/5.

School Level Solution

Using denominator 30: 6/30 < 7/30 < 8/30 < 9/30 < 10/30 < 11/30 < 12/30, so 7/30, 8/30, 9/30, 10/30, 11/30 are five rational numbers between 1/6 = 5/30 and 2/5 = 12/30.

If x/3 + x/5 = 16/15, find the rational number x.

School Level Solution

x(1/3 + 1/5) = 16/15, so x(8/15) = 16/15 and x = 2.

Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.

School Level Solution

a + 1/b = 0 gives a = −1/b. Hence ab = −1, which is negative.

A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form p/10^4, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 2^4 or 5^4? Give reasons.

School Level Solution

A terminating decimal with its last non-zero digit in the 4th decimal place can be written p/10^4 with p not divisible by 10. In lowest terms the denominator contains only factors 2 and 5, and because the decimal terminates exactly at the 4th place, the larger exponent of 2 or 5 in the lowest denominator is 4. Therefore the denominator is necessarily divisible by 2^4 or by 5^4, though not necessarily by both.

Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.

School Level Solution

125 = 5^3, so the denominator has only the prime factor 5. Therefore 18/125 has a terminating decimal with 3 decimal places: 18/125 = 0.144.

A rational number in its lowest form has denominator 2^3 × 5. How many decimal places will its decimal expansion have? Explain your answer.

School Level Solution

2^3 × 5 = 40. In lowest terms such a denominator can be converted to 1000 by multiplying numerator and denominator by 25, so the decimal expansion has 3 places.

★13

Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ − k₁ > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition k₂ − k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.

School Level Solution

Take m = 48. Then a = 28/48 and b = 40/48, so k₂ − k₁ = 12 > 6. Five distinct rational numbers between them are 29/48, 30/48, 31/48, 32/48 and 33/48. To place n distinct integer numerators strictly between k₁ and k₂, there must be at least n+1 integer steps available; equivalently k₂ − k₁ > n+1 is a sufficient condition for this construction.

★14

Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.

School Level Solution

From x+y+z=0, squaring gives x²+y²+z²+2(xy+yz+zx)=0. Since xy+yz+zx=0, we obtain x²+y²+z²=0. For rational (hence real) numbers, each square is non-negative, so x=y=z=0.

★15

Show that the rational number (a+b)/2 lies between the rational numbers a and b.

School Level Solution

If a < b, then a < (a+b)/2 < b because adding a

16
Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.

School Level Solution

The nine right triangles in the square-root spiral have successive hypotenuses √2, √3, √4, √5, √6, √7, √8, √9 and √10; equivalently √2, √3, 2, √5, √6, √7, 2√2, 3 and √10.

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