Chapter 3 solution

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Ganita Manjari — Grade 9

Chapter 3: The World of Numbers — Solutions Manual

Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 End-of-Chapter Exercises

Exercise Set 3.1

Question 1

A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?

Solution:

We are given the ratio of spice bags to copper ingots:

2 bags of spices = 15 copper ingots

To find the number of ingots per bag of spices:

1 bag of spices = 15 / 2 copper ingots

For 12 bags of spices, the number of copper ingots received is:

Number of ingots = 12 × (15 / 2) = 6 × 15 = 90

Answer: The merchant will leave with 90 copper ingots.

Question 2

Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.

Solution:

Common Property: All these numbers—11, 13, 17, and 19—are prime numbers (numbers greater than 1 that have no positive divisors other than 1 and themselves) in increasing order between 10 and 20.

The next three prime numbers following 19 are:

23, 29, 31

Answer: The numbers are consecutive prime numbers. The next three numbers in the pattern are 23, 29, and 31.

Question 3

We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.

Solution:

Answer: No, Natural Numbers (ℕ = {1, 2, 3, 4, ...}) are not closed under subtraction.

A set is closed under an operation if performing that operation on any two elements of the set always results in an element that is also in the set.

Examples:

1. Take natural numbers 3 and 5.
   3 - 5 = -2.
   Here, -2 is a negative integer, not a natural number (-2 ∉ ℕ).
2. Take natural numbers 4 and 4.
   4 - 4 = 0.
   Here, 0 is a whole number/integer, not a natural number (0 ∉ ℕ).

Question 4*

Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?

Solution:

• Each of the 4 non-thumb fingers has 3 finger-joints (phalanxes).
• Using the tip of the thumb to point to each joint, one hand can count: 4 fingers × 3 joints = 12 joints.

Relation to Base-12 (Duodecimal) System:

This counting method directly gave rise to the base-12 system. By using the other hand to keep track of complete cycles of 12 (5 fingers on the second hand), ancient merchants could count up to 12 × 5 = 60 items on two hands, which also forms the basis of the sexagesimal (base-60) system used in measuring hours, minutes, seconds, and angles.

Exercise Set 3.2

Question 1

The temperature in the high-altitude desert of Ladakh is recorded as 4°C at noon. By midnight, it drops by 15°C. What is the midnight temperature?

Solution:

Initial temperature at noon = +4°C

Temperature drop = 15°C (represented as subtraction of 15 or adding -15)

Midnight Temperature = 4°C - 15°C = -11°C

Answer: The midnight temperature is -11°C.

Question 2

A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing.

Solution:

Using Brahmagupta's framework of Dhana (Fortune/Positive) and Rina (Debt/Negative):

• Loan (Debt) = -850
• Profit (Fortune) = +1200
• Loss (Debt) = -450

Equation:

Final Financial Standing = (-850) + 1200 + (-450)

Calculation:

(-850) + (-450) + 1200 = -1300 + 1200 = -100

Answer: His final financial standing is -₹100 (a net debt of ₹100).

Question 3

Calculate the following using Brahmagupta's laws:
(i) (-12) × 5
(ii) (-8) × (-7)
(iii) 0 - (-14)
(iv) (-20) ÷ 4

Solution:

1. (-12) × 5: Product of a debt and a fortune is a debt.
   
   (-12) × 5 = -60
2. (-8) × (-7): Product of two debts is a fortune.
   
   (-8) × (-7) = +56
3. 0 - (-14): Subtracting a debt is equivalent to adding a fortune.
   
   0 - (-14) = 0 + 14 = 14
4. (-20) ÷ 4: Quotient of a debt divided by a fortune is a debt.
   
   (-20) ÷ 4 = -5

Question 4

Explain, 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).

Solution:

Real-World Financial Analogy:

• Imagine you have ₹10 in cash (Fortune = +10).
• You also owe your friend ₹5 (Debt = -5).
• If your friend decides to forgive/remove your debt of ₹5, they are taking away (-) a debt (-5).
• Removing a debt makes you ₹5 richer than you were before, effectively increasing your total net worth.

Net Worth = 10 - (-5) = 10 + 5 = 15

Therefore, subtracting a negative number (removing a debt) has the exact same net effect as adding a positive number (receiving a fortune).

Exercise Set 3.3

Question 1

Prove that the following rational numbers are equal:
(i) 2/3 and 4/6
(ii) 5/4 and 10/8
(iii) -3/5 and -6/10
(iv) 9/3 and 3

Solution:

Two rational numbers a/b and c/d are equal if and only if ad = bc (cross-multiplication law).

1. 2/3 and 4/6:
   2 × 6 = 12 and 3 × 4 = 12. Since 12 = 12, 2/3 = 4/6.
2. 5/4 and 10/8:
   5 × 8 = 40 and 4 × 10 = 40. Since 40 = 40, 5/4 = 10/8.
3. -3/5 and -6/10:
   (-3) × 10 = -30 and 5 × (-6) = -30. Since -30 = -30, -3/5 = -6/10.
4. 9/3 and 3/1:
   9 × 1 = 9 and 3 × 3 = 9. Since 9 = 9, 9/3 = 3.

Question 2

Find the sum:
(i) 2/5 + 3/10
(ii) 7/12 + 5/8
(iii) -4/7 + 3/14

Solution:

1. 2/5 + 3/10:
   LCM of 5 and 10 is 10.
   
   2/5 + 3/10 = 4/10 + 3/10 = (4 + 3)/10 = 7/10
2. 7/12 + 5/8:
   LCM of 12 and 8 is 24.
   
   7/12 + 5/8 = 14/24 + 15/24 = (14 + 15)/24 = 29/24
3. -4/7 + 3/14:
   LCM of 7 and 14 is 14.
   
   -4/7 + 3/14 = -8/14 + 3/14 = (-8 + 3)/14 = -5/14

Question 3

Find the difference:
(i) 5/6 - 1/4
(ii) 11/8 - 3/4
(iii) -7/9 - (-2/3)

Solution:

1. 5/6 - 1/4:
   LCM of 6 and 4 is 12.
   
   5/6 - 1/4 = 10/12 - 3/12 = 7/12
2. 11/8 - 3/4:
   LCM of 8 and 4 is 8.
   
   11/8 - 3/4 = 11/8 - 6/8 = 5/8
3. -7/9 - (-2/3):
   
   -7/9 + 2/3 = -7/9 + 6/9 = -1/9

Question 4

Find the product:
(i) (2/3) × (3/10)
(ii) (7/11) × (5/8)
(iii) (-4/7) × (5/14)

Solution:

1. (2/3) × (3/10) = (2 × 3)/(3 × 10) = 6/30 = 1/5
2. (7/11) × (5/8) = (7 × 5)/(11 × 8) = 35/88
3. (-4/7) × (5/14) = (-4 × 5)/(7 × 14) = -20/98 = -10/49

Question 5

Find the quotient:
(i) (2/3) ÷ (3/10)
(ii) (7/11) ÷ (5/8)
(iii) (-4/7) ÷ (5/14)

Solution:

Rule: (a/b) ÷ (c/d) = (a/b) × (d/c)

1. (2/3) ÷ (3/10) = (2/3) × (10/3) = 20/9
2. (7/11) ÷ (5/8) = (7/11) × (8/5) = 56/55
3. (-4/7) ÷ (5/14) = (-4/7) × (14/5) = (-4 × 2)/5 = -8/5

Question 6

Show that: (1/2 + 3/4) × (8/3) = (1/2) × (8/3) + (3/4) × (8/3).

Solution:

Left Hand Side (LHS):

(1/2 + 3/4) × (8/3) = (2/4 + 3/4) × (8/3) = (5/4) × (8/3) = (5 × 8) / (4 × 3) = 40 / 12 = 10 / 3

Right Hand Side (RHS):

(1/2) × (8/3) + (3/4) × (8/3) = 8/6 + 24/12 = 4/3 + 2 = 4/3 + 6/3 = 10/3

Since LHS = RHS = 10/3, the distributive property is verified.

Question 7

Simplify the following using the distributive property: (7/9) × (6/7 - 3/4).

Solution:

Using p(q - r) = pq - pr:

(7/9) × (6/7 - 3/4) = (7/9 × 6/7) - (7/9 × 3/4)

= (6/9) - (21/36) = 2/3 - 7/12

LCM of 3 and 12 is 12:

= 8/12 - 7/12 = 1/12

Answer: 1/12

Question 8

Find the rational number x such that: (5/6)x + (3/5) = (5/6)x + (1/2).

Solution:

Subtract (5/6)x from both sides of the equation:

3/5 = 1/2

This statement (3/5 = 1/2 or 0.6 = 0.5) is a contradiction and mathematically false for all values of x.

Answer: There is no rational number x that satisfies this equation (No Solution).

Exercise Set 3.4

Question 1

Represent the rational numbers 2/3, -5/4, and 1(1/2) on a single number line.

Solution:

Convert all fractions to a common denominator (LCM of 3, 4, 2 = 12) to place them precisely:

• 2/3 = 8/12 = 0.67 (Lies between 0 and 1)
• -5/4 = -1(1/4) = -15/12 = -1.25 (Lies between -2 and -1)
• 1(1/2) = 3/2 = 18/12 = 1.50 (Lies exactly midway between 1 and 2)

<---|-------|-------|-------|-------|-------|-------|-------|--->
-2 -5/4 -1 0 2/3 1 1(1/2) 2

Question 2

Find three distinct rational numbers that lie strictly between -1/2 and 1/4.

Solution:

Express both fractions with a common larger denominator, e.g., 16:

-1/2 = -8/16   and   1/4 = 4/16

Any integer numerators between -8 and 4 can be chosen:

Chosen numerators: -4, 0, 2

The corresponding rational numbers are:

-4/16 = -1/4,    0/16 = 0,    2/16 = 1/8

Answer: Three such rational numbers are -1/4, 0, and 1/8.

Question 3

Simplify the expression: (-1/4) + (5/12).

Solution:

LCM of 4 and 12 is 12.

(-1/4) + (5/12) = -3/12 + 5/12 = (-3 + 5)/12 = 2/12 = 1/6

Answer: 1/6

Question 4

A tailor has 15(3/4) metres of fine silk. If making one kurta requires 2(1/4) metres of silk, exactly how many kurtas can he make?

Solution:

Convert mixed fractions to improper fractions:

• Total silk available = 15(3/4) = 63/4 metres
• Silk required per kurta = 2(1/4) = 9/4 metres

Number of kurtas = (63/4) ÷ (9/4) = (63/4) × (4/9) = 63 / 9 = 7

Answer: The tailor can make exactly 7 kurtas.

Question 5

Find three rational numbers between 3.1415 and 3.1416.

Solution:

Rewrite the decimal numbers with additional decimal places:

3.14150   and   3.14160

We can pick any numbers strictly between 3.14150 and 3.14160, such as:

3.14151,   3.14152,   3.14155

Expressed as fractions in p/q form:

314151 / 100000,    314152 / 100000,    314155 / 100000

Answer: 3.14151, 3.14152, and 3.14155.

Question 6*

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

Solution:

Yes, there are several methods:

1. Method of Averages (Midpoint Method):
   For any two rational numbers a and b, their average m = (a + b)/2 always lies strictly between a and b.
2. Equivalent Fraction / Common Denominator Method:
   Convert both rational numbers to have a large common denominator N. Then pick any integer numerator strictly between the two resulting numerators.
3. Mediant Method (Farey Fraction Method):
   For two positive rational numbers a/b and c/d, the mediant (a + c)/(b + d) always lies strictly between them.

Exercise Set 3.5

Question 1

Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15, and 13/250. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

Solution:

Rule: A rational number in lowest terms p/q has a terminating decimal expansion if and only if the prime factorization of denominator q contains only powers of 2, 5, or both (q = 2^m × 5ⁿ).

Fraction	Denominator Factorization	Predicted Type	Long Division Check
7/20	20 = 22 × 5	Terminating	7 ÷ 20 = 0.35
4/15	15 = 3 × 5 (contains 3)	Repeating	4 ÷ 15 = 0.2666... = 0.26̄
13/250	250 = 2 × 53	Terminating	13 ÷ 250 = 0.052

Question 2

Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13, etc. What do you notice?

Solution:

Performing long division for 1/13:

1/13 = 0.076923076923... = 0.076923̄

Repeating Block: 076923 (length 6).

Evaluating multiples of 1/13:

• 1/13 = 0.076923̄
• 2/13 = 0.153846̄
• 3/13 = 0.230769̄
• 4/13 = 0.307692̄
• 9/13 = 0.692307̄

Observation: The fractions for n/13 split into two distinct cyclic families of 6 digits:

1. Family A (from 1/13, 3/13, 4/13, 9/13, 10/13, 12/13): Uses cyclic shifts of 076923.
2. Family B (from 2/13, 5/13, 6/13, 7/13, 8/13, 11/13): Uses cyclic shifts of 153846.

Question 3

Classify the following numbers as rational or irrational:
(i) √81
(ii) √12
(iii) 0.33333...
(iv) 0.123451234512345...
(v) 1.01001000100001...
(vi) 23.560185612239874790120... (non-terminating non-repeating)
Find the explicit fractions in case they are rational.

Solution:

1. √81 = 9 = 9/1 — Rational.
2. √12 = 2√3 — Irrational (since 12 is not a perfect square).
3. 0.33333... = 0.3̄ — Rational.
   Let x = 0.333... ⇒ 10x = 3.333... ⇒ 9x = 3 ⇒ x = 3/9 = 1/3.
4. 0.1234512345... = 0.12345̄ — Rational (pure repeating block of 5 digits).
   Let x = 0.12345̄ ⇒ 100000x - x = 12345 ⇒ 99999x = 12345 ⇒ x = 12345 / 99999 = 4115 / 33333.
5. 1.010010001... — Irrational (pattern expands with increasing zeros; non-repeating and non-terminating).
6. 23.560185612239874790120... — Irrational (non-terminating, non-repeating).

Question 4

The number 0.9̄ (which means 0.99999...) is a rational number. Using algebra (let x = 0.9̄, multiply by 10, and subtract), explain why 0.9̄ is exactly equal to 1.

Solution:

Let x = 0.99999...   --- (Equation 1)

Multiply both sides by 10:

10x = 9.99999...   --- (Equation 2)

Subtract Equation 1 from Equation 2:

10x - x = (9.99999...) - (0.99999...)

9x = 9

x = 9 / 9 = 1

Conclusion: 0.99999... = 1 exactly. This reflects the non-uniqueness of decimal representations for terminating rationals.

Question 5*

We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers (n) whose reciprocals (1/n) produce decimals with repeating blocks that are cyclic.

Solution:

Primes p whose reciprocal 1/p generates a full period repeating block of length p - 1 are called full-repetition primes or long primes.

Examples of such numbers n include:

• n = 7: 1/7 = 0.142857̄ (period length 6 = 7-1). Cyclic number = 142857.
• n = 17: 1/17 = 0.0588235294117647̄ (period length 16 = 17-1). Cyclic number = 0588235294117647.
• n = 19: 1/19 = 0.052631578947368421̄ (period length 18 = 19-1). Cyclic number = 052631578947368421.
• Other full-period primes: 23, 29, 47, 59, 61.

End-of-Chapter Exercises

Question 1

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

Solution:

1. 3/50:
   Long division of 3 by 50: 3.00 ÷ 50 = 0.06.
   Answer: 0.06 (Terminating decimal).
2. 2/9:
   Long division of 2 by 9: 2.000... ÷ 9 = 0.222....
   Answer: 0.2̄ (Non-terminating repeating decimal).

Question 2

Prove that √5 is an irrational number.

Solution (Proof by Contradiction):

Step 1: Assume on the contrary that √5 is a rational number. Then √5 = p/q, where p, q ∈ ℤ, q ≠ 0, and gcd(p, q) = 1 (coprime).

Step 2: Squaring both sides: 5 = p² / q² ⇒ 5q² = p².   --- (Eq. 1)

Step 3: Since 5 divides p², 5 must also divide p (since 5 is prime). Let p = 5k for some integer k.

Step 4: Substitute p = 5k into Eq. 1:
5q² = (5k)² = 25k² ⇒ q² = 5k².

Step 5: Since 5 divides q², 5 must also divide q.

Step 6 (Contradiction): From Steps 3 and 5, 5 is a common factor of both p and q. This contradicts our initial assumption that p and q are coprime.

Conclusion: Our assumption is false. Therefore, √5 is an irrational number.

Question 3

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

Solution:

1. 12.6: 126/10 = 63/5
2. 0.0120: 120/10000 = 3/250
3. 3.052: 3052/1000 = 763/250
4. 1.235̄:
   Let x = 1.235̄ ⇒ 100x = 123.5̄ and 1000x = 1235.5̄.
   900x = 1235 - 123 = 1112 ⇒ x = 1112 / 900 = 278 / 225.
5. 0.23̄:
   Let x = 0.23̄ ⇒ 10x = 2.3̄ and 100x = 23.3̄.
   90x = 21 ⇒ x = 21 / 90 = 7 / 30.
6. 2.05̄:
   Let x = 2.05̄ ⇒ 10x = 20.5̄ and 100x = 205.5̄.
   90x = 185 ⇒ x = 185 / 90 = 37 / 18.
7. 2.125: 2125/1000 = 17/8
8. 3.125: 3125/1000 = 25/8
9. 2.1625: 21625/10000 = 173/80

Question 4

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

Solution:

• (i) 0.532: Lies between 0.5 and 0.6 (more specifically between 0.53 and 0.54). Divide the unit interval [0, 1] into tenths, expand [0.5, 0.6] into hundredths, and locate 0.532 slightly past 0.53.
• (ii) 1.15: Lies between 1 and 2, exactly halfway between 1.1 and 1.2 (or 23/20 = 1(3/20)).

Question 5

Find 6 rational numbers between 3 and 4.

Solution:

To find n = 6 rational numbers, express 3 and 4 as fractions with denominator n + 1 = 7:

3 = 21 / 7   and   4 = 28 / 7

Six rational numbers between 21/7 and 28/7 are:

22/7,   23/7,   24/7,   25/7,   26/7,   27/7

Question 6

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

Solution:

Multiply numerators and denominators by 5 + 1 = 6:

2/5 = 12 / 30   and   3/5 = 18 / 30

Five rational numbers between 12/30 and 18/30 are:

13/30,   14/30 (7/15),   15/30 (1/2),   16/30 (8/15),   17/30

Question 7

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

Solution:

First convert to a common denominator (LCM of 6 and 5 is 30):

1/6 = 5 / 30   and   2/5 = 12 / 30

Pick any 5 rational numbers between 5/30 and 12/30:

6/30 (1/5),   7/30,   8/30 (4/15),   9/30 (3/10),   10/30 (1/3)

Question 8

If x/35 + x/16 = 15, find the rational number x.

Solution:

Combine terms on LHS by taking LCM of 35 and 16 (35 × 16 = 560):

(16x + 35x) / 560 = 15

51x / 560 = 15

51x = 15 × 560 = 8400

x = 8400 / 51 = 2800 / 17

Answer: x = 2800 / 17 (or 164(12/17)).

Question 9

Let a and b be two non-zero rational numbers such that a/b + b/a = 10. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.

Solution:

Simplify the given equation:

(a² + b²) / ab = 10 ⇒ a² + b² = 10ab

Since a and b are non-zero real rational numbers, a² > 0 and b² > 0. Therefore, the sum of squares a² + b² must be strictly positive (a² + b² > 0).

Thus, 10ab = a² + b² > 0 ⇒ ab > 0.

Answer: ab is positive.

Question 10

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 / 10⁴, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in lowest form, is divisible by 2⁴ or 5⁴? Give reasons.

Solution:

Let the decimal expansion be x = d₀.d₁d₂d₃d₄ where d₄ ≠ 0.

Multiply by 10⁴:

10⁴x = d₀d₁d₂d₃d₄ = p (an integer)

Since d₄ ≠ 0, the last digit of p is non-zero, meaning p is not divisible by 10.

Thus, x = p / 10⁴ = p / (2⁴ × 5⁴).

Is lowest-form denominator divisible by 2⁴ or 5⁴?

Since p is not divisible by 10, p cannot be simultaneously divisible by both 2 and 5. Therefore, canceling common factors with 2⁴ × 5⁴ can at most reduce one prime factor power (2 or 5), but cannot reduce both. Hence, in lowest form a/b, the denominator b MUST retain either 2⁴ or 5⁴ as a factor. So yes, it is necessary.

Question 11

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.

Solution:

The fraction 18 / 125 is in lowest terms (gcd(18, 125) = 1).

Prime factorize denominator: 125 = 5³ = 2⁰ × 5³.

Since denominator factors contain only 5s, the decimal expansion is terminating.

Number of decimal places = max(0, 3) = 3 decimal places.

(Verification: 18 / 125 = (18 × 8) / (125 × 8) = 144 / 1000 = 0.144).

Question 12

A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.

Solution:

Let the rational number be p / (2³ × 5) in lowest form.

To convert to a power of 10, multiply numerator and denominator by 5²:

[p × 5²] / [2³ × 5³] = (25p) / 10³

Since 25p is divided by 10³, the decimal expansion terminates after 3 decimal places (determined by max(3, 1) = 3).

Question 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 two rational numbers using this method.

Solution:

1. Common denominator for 12 and 6 is 12: 7/12 and 10/12. Difference in numerators is 10 - 7 = 3.

To get k₂ - k₁ > 6, multiply numerators and denominator by 3:

a = 21 / 36 (k₁ = 21),    b = 30 / 36 (k₂ = 30),    m = 36

Check condition: k₂ - k₁ = 30 - 21 = 9 > 6 (Satisfied!).

2. Five rational numbers between 21/36 and 30/36:

22/36,   23/36,   24/36,   25/36,   26/36

3. Why k₂ - k₁ > n + 1 is necessary:
The number of strictly intermediate integers between two integers k₁ and k₂ is given by (k₂ - k₁ - 1). To ensure there are at least n integer numerators available, we must have k₂ - k₁ - 1 ≥ n ⇒ k₂ - k₁ ≥ n + 1. Hence k₂ - k₁ > n + 1 ensures strict availability of n distinct intermediate fractions.

Question 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.

Solution:

Consider the identity for the sum of squares of three real variables:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

Substitute the given conditions x + y + z = 0 and xy + yz + zx = 0:

0² = x² + y² + z² + 2(0)

x² + y² + z² = 0

Since x, y, z are rational (and real) numbers, their squares x² ≥ 0, y² ≥ 0, z² ≥ 0 are non-negative.

The sum of non-negative real numbers can equal zero if and only if each individual term is zero:

x² = 0,   y² = 0,   z² = 0 ⇒ x = 0, y = 0, z = 0

Conclusion: x = y = z = 0 simultaneously.

Question 15*

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

Solution:

Without loss of generality, assume a < b.

Part 1: Show a < (a + b)/2

Add a to both sides of a < b:
a + a < a + b ⇒ 2a < a + b ⇒ a < (a + b)/2

Part 2: Show (a + b)/2 < b

Add b to both sides of a < b:
a + b < b + b ⇒ a + b < 2b ⇒ (a + b)/2 < b

Combining Part 1 and Part 2:

a < (a + b)/2 < b

Thus, (a + b)/2 lies strictly between a and b.

Question 16

Find the lengths of the hypotenuses of all the right triangles in the square root spiral.

Solution:

In the square root spiral (Theodorous spiral):

• Triangle 1: Legs are 1 and 1.
  Hypotenuse h₁ = √(1² + 1²) = √2.
• Triangle 2: Legs are √2 and 1.
  Hypotenuse h₂ = √((√2)² + 1²) = √(2 + 1) = √3.
• Triangle 3: Legs are √3 and 1.
  Hypotenuse h₃ = √((√3)² + 1²) = √(3 + 1) = √4 = 2.
• Triangle 4: Legs are √4 and 1.
  Hypotenuse h₄ = √((√4)² + 1²) = √5.
• In general, for the n-th triangle:
  Legs are √n and 1.
  Hypotenuse h_n = √(n + 1).

Answer: The sequence of hypotenuses is √2, √3, √4 (2), √5, √6, √7, √8, √9 (3), ..., √(n+1).

Ganita Manjari Grade 9 Solutions Manual © 2026. All rights reserved.

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