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CHAPTER 3 — THE WORLD OF NUMBERS
Class IX Mathematics — Number Systems
Preeti Kushwah Classes — Unit Test (Set 2: Intermediate)
General Instructions:
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 6 marks each.
6. Write neat and step-wise solutions wherever required.
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 6 marks each.
6. Write neat and step-wise solutions wherever required.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
The decimal expansion of 0.333... is:
(a) Terminating (b) Non-terminating recurring (c) Irrational (d) An integer
(a) Terminating (b) Non-terminating recurring (c) Irrational (d) An integer
Q2.1
The statement ℕ ⊂ W ⊂ ℤ ⊂ ℚ ⊂ ℝ is:
(a) True (b) False
(a) True (b) False
Q3.1
(−5) × (−3) = ?
(a) −15 (b) 15 (c) −8 (d) 8
(a) −15 (b) 15 (c) −8 (d) 8
Q4.1
A rational number p/q (in lowest terms) has a terminating decimal expansion when the denominator q has only the prime factors ___ and ___.
Q5.1
22/7 is:
(a) Equal to π (b) A rational number (c) An irrational number (d) Undefined
(a) Equal to π (b) A rational number (c) An irrational number (d) Undefined
Q6.1
How many rational numbers exist between 1/3 and 1/2?
(a) 3 (b) 10 (c) 100 (d) Infinitely many
(a) 3 (b) 10 (c) 100 (d) Infinitely many
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Find the decimal expansion of 3/8 and 1/7. Classify each as terminating or non-terminating recurring.
Q8.2
Convert 0.6̄ (i.e., 0.6666...) into p/q form. Show your working clearly.
Q9.2
Find the values of:
(i) (−4) × (−6) (ii) (−15) ÷ 3
State the sign rule used in each case.
(i) (−4) × (−6) (ii) (−15) ÷ 3
State the sign rule used in each case.
Q10.2
Without performing long division, determine whether 7/80 has a terminating decimal expansion. Justify your answer.
Q11.2
Give two examples of irrational numbers whose sum is a rational number.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Convert 0.̅4̅7̅ (i.e., 0.474747...) into p/q form. Show all steps.
Q13.3
Prove that √3 is an irrational number. Write a complete proof using contradiction.
Q14.3
Simplify the following and express as a single fraction in lowest terms:
2/3 + 4/5 − 1/6
2/3 + 4/5 − 1/6
Q15.3
The temperature at 6:00 AM was −4°C. The temperature rose by 3°C every 2 hours.
(i) What was the temperature at 8:00 AM? [1 mark]
(ii) At 10:00 AM? [1 mark]
(iii) At 12:00 noon? [1 mark]
(i) What was the temperature at 8:00 AM? [1 mark]
(ii) At 10:00 AM? [1 mark]
(iii) At 12:00 noon? [1 mark]
Section D — (6 Marks Each) [2 × 6 = 12]
Q16.6
(a) Find 5 rational numbers between 1/2 and 3/4. Show your method clearly. [3 marks]
(b) Explain the density property of rational numbers with an example. Why does it imply there are infinitely many rationals between any two rationals? [3 marks]
(b) Explain the density property of rational numbers with an example. Why does it imply there are infinitely many rationals between any two rationals? [3 marks]
Q17.6
Classify each of the following as rational or irrational. Give a clear reason for each: [1 mark each]
(i) √7 (ii) √16 (iii) 0.101001000100001...
(iv) π (v) 3 + √5 (vi) 0.3̄ (i.e., 0.333...)
(i) √7 (ii) √16 (iii) 0.101001000100001...
(iv) π (v) 3 + √5 (vi) 0.3̄ (i.e., 0.333...)
Bonus Question (Optional) [2 Marks]
Q18.2
★ If a = √2 + 1 and b = √2 − 1, find the value of (a × b) and (a + b). What can you conclude about these values?
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Answer Key & Detailed Solutions
Q1. [1 Mark]
Answer: (b) Non-terminating recurring
0.333... = 0.3̄ is a non-terminating recurring decimal because the digit 3 repeats endlessly. This is the decimal expansion of 1/3, which is a rational number. Recurring (repeating) decimals are always rational.
0.333... = 0.3̄ is a non-terminating recurring decimal because the digit 3 repeats endlessly. This is the decimal expansion of 1/3, which is a rational number. Recurring (repeating) decimals are always rational.
Q2. [1 Mark]
Answer: (a) True
The hierarchy of number sets is correct:
ℕ (Natural) ⊂ W (Whole) ⊂ ℤ (Integers) ⊂ ℚ (Rationals) ⊂ ℝ (Reals)
Every natural number is a whole number, every whole number is an integer, every integer is rational, and every rational is a real number.
The hierarchy of number sets is correct:
ℕ (Natural) ⊂ W (Whole) ⊂ ℤ (Integers) ⊂ ℚ (Rationals) ⊂ ℝ (Reals)
Every natural number is a whole number, every whole number is an integer, every integer is rational, and every rational is a real number.
Q3. [1 Mark]
Answer: (b) 15
Sign Rule: Negative × Negative = Positive
(−5) × (−3) = +(5 × 3) = 15
Sign Rule: Negative × Negative = Positive
(−5) × (−3) = +(5 × 3) = 15
Q4. [1 Mark]
Answer: 2 and 5
A rational number p/q (in lowest terms) has a terminating decimal expansion if and only if the denominator q can be written in the form 2m × 5n, where m and n are non-negative integers. Example: 3/8 = 3/2³ terminates as 0.375.
A rational number p/q (in lowest terms) has a terminating decimal expansion if and only if the denominator q can be written in the form 2m × 5n, where m and n are non-negative integers. Example: 3/8 = 3/2³ terminates as 0.375.
Q5. [1 Mark]
Answer: (b) A rational number
22/7 is a rational number (it is in p/q form with integers p = 22 and q = 7).
It is only an approximation of π, not equal to it. The actual value of π = 3.14159265... (irrational), while 22/7 = 3.142857... (recurring rational). They are not equal.
22/7 is a rational number (it is in p/q form with integers p = 22 and q = 7).
It is only an approximation of π, not equal to it. The actual value of π = 3.14159265... (irrational), while 22/7 = 3.142857... (recurring rational). They are not equal.
Q6. [1 Mark]
Answer: (d) Infinitely many
By the density property of rational numbers: between any two distinct rational numbers, there are infinitely many rational numbers. So between 1/3 and 1/2, there are infinitely many rationals.
By the density property of rational numbers: between any two distinct rational numbers, there are infinitely many rational numbers. So between 1/3 and 1/2, there are infinitely many rationals.
Q7. [2 Marks]
3/8:
8 = 2³. Since the denominator has only factor 2, the decimal terminates.
3 ÷ 8 = 0.375 (terminating) [1 mark]
1/7:
7 is a prime number other than 2 or 5. So the decimal is non-terminating recurring.
1 ÷ 7 = 0.142857142857... = 0.̅1̅4̅2̅8̅5̅7̅ (period 6) [1 mark]
8 = 2³. Since the denominator has only factor 2, the decimal terminates.
3 ÷ 8 = 0.375 (terminating) [1 mark]
1/7:
7 is a prime number other than 2 or 5. So the decimal is non-terminating recurring.
1 ÷ 7 = 0.142857142857... = 0.̅1̅4̅2̅8̅5̅7̅ (period 6) [1 mark]
Q8. [2 Marks]
Let x = 0.6666... ...(1) [½ mark]
Multiply both sides by 10:
10x = 6.6666... ...(2) [½ mark]
Subtract (1) from (2):
10x − x = 6.6666... − 0.6666...
9x = 6
x = 6/9 = 2/3 [1 mark]
Therefore 0.6̄ = 2/3.
Multiply both sides by 10:
10x = 6.6666... ...(2) [½ mark]
Subtract (1) from (2):
10x − x = 6.6666... − 0.6666...
9x = 6
x = 6/9 = 2/3 [1 mark]
Therefore 0.6̄ = 2/3.
Q9. [2 Marks]
(i) (−4) × (−6):
Sign Rule: Negative × Negative = Positive
(−4) × (−6) = +(4 × 6) = 24 [1 mark]
(ii) (−15) ÷ 3:
Sign Rule: Negative ÷ Positive = Negative
(−15) ÷ 3 = −(15 ÷ 3) = −5 [1 mark]
Sign Rule: Negative × Negative = Positive
(−4) × (−6) = +(4 × 6) = 24 [1 mark]
(ii) (−15) ÷ 3:
Sign Rule: Negative ÷ Positive = Negative
(−15) ÷ 3 = −(15 ÷ 3) = −5 [1 mark]
Q10. [2 Marks]
7/80:
First, check if 7 and 80 share any common factors: HCF(7, 80) = 1. So the fraction is already in lowest terms. [½ mark]
Now factorize the denominator:
80 = 16 × 5 = 24 × 51 [½ mark]
Since 80 = 24 × 51 has only factors of 2 and 5, the decimal expansion of 7/80 is terminating. [1 mark]
(7/80 = 0.0875)
First, check if 7 and 80 share any common factors: HCF(7, 80) = 1. So the fraction is already in lowest terms. [½ mark]
Now factorize the denominator:
80 = 16 × 5 = 24 × 51 [½ mark]
Since 80 = 24 × 51 has only factors of 2 and 5, the decimal expansion of 7/80 is terminating. [1 mark]
(7/80 = 0.0875)
Q11. [2 Marks]
Example 1: [1 mark]
Let a = √3 and b = −√3
Both are irrational numbers.
a + b = √3 + (−√3) = 0, which is rational.
Example 2: [1 mark]
Let a = (2 + √5) and b = (2 − √5)
Both are irrational numbers.
a + b = (2 + √5) + (2 − √5) = 4, which is rational.
Let a = √3 and b = −√3
Both are irrational numbers.
a + b = √3 + (−√3) = 0, which is rational.
Example 2: [1 mark]
Let a = (2 + √5) and b = (2 − √5)
Both are irrational numbers.
a + b = (2 + √5) + (2 − √5) = 4, which is rational.
Q12. [3 Marks]
Let x = 0.474747... ...(1) [½ mark]
Since the repeating block "47" has 2 digits, multiply by 100:
100x = 47.4747... ...(2) [1 mark]
Subtract (1) from (2):
100x − x = 47.4747... − 0.4747...
99x = 47 [1 mark]
x = 47/99
Check: HCF(47, 99) = 1 (47 is prime; 99 = 9 × 11; neither 9 nor 11 is divisible by 47).
Therefore 0.474747... = 47/99 [½ mark]
Since the repeating block "47" has 2 digits, multiply by 100:
100x = 47.4747... ...(2) [1 mark]
Subtract (1) from (2):
100x − x = 47.4747... − 0.4747...
99x = 47 [1 mark]
x = 47/99
Check: HCF(47, 99) = 1 (47 is prime; 99 = 9 × 11; neither 9 nor 11 is divisible by 47).
Therefore 0.474747... = 47/99 [½ mark]
Q13. [3 Marks]
Theorem: √3 is irrational.
Proof by Contradiction:
Assume √3 is rational. Then √3 = p/q where p, q are integers, q ≠ 0, and HCF(p, q) = 1. [½ mark]
Squaring: 3 = p²/q² ⇒ p² = 3q² [½ mark]
So p² is divisible by 3 ⇒ p is divisible by 3 (since if p² is divisible by prime 3, so is p).
Let p = 3k for some integer k. [1 mark]
Substituting: (3k)² = 3q² ⇒ 9k² = 3q² ⇒ q² = 3k² [½ mark]
So q² is divisible by 3 ⇒ q is divisible by 3.
But then both p and q are divisible by 3, contradicting HCF(p, q) = 1.
Therefore, our assumption was wrong. √3 is irrational. ■ [½ mark]
Proof by Contradiction:
Assume √3 is rational. Then √3 = p/q where p, q are integers, q ≠ 0, and HCF(p, q) = 1. [½ mark]
Squaring: 3 = p²/q² ⇒ p² = 3q² [½ mark]
So p² is divisible by 3 ⇒ p is divisible by 3 (since if p² is divisible by prime 3, so is p).
Let p = 3k for some integer k. [1 mark]
Substituting: (3k)² = 3q² ⇒ 9k² = 3q² ⇒ q² = 3k² [½ mark]
So q² is divisible by 3 ⇒ q is divisible by 3.
But then both p and q are divisible by 3, contradicting HCF(p, q) = 1.
Therefore, our assumption was wrong. √3 is irrational. ■ [½ mark]
Q14. [3 Marks]
2/3 + 4/5 − 1/6
Step 1: Find LCM of denominators 3, 5, 6.
LCM(3, 5, 6) = 30 [1 mark]
Step 2: Convert each fraction:
2/3 = 20/30
4/5 = 24/30
1/6 = 5/30 [1 mark]
Step 3: Add and subtract:
20/30 + 24/30 − 5/30 = (20 + 24 − 5)/30 = 39/30 = 13/10 [1 mark]
(Or written as 1.3 or 1 3/10)
Step 1: Find LCM of denominators 3, 5, 6.
LCM(3, 5, 6) = 30 [1 mark]
Step 2: Convert each fraction:
2/3 = 20/30
4/5 = 24/30
1/6 = 5/30 [1 mark]
Step 3: Add and subtract:
20/30 + 24/30 − 5/30 = (20 + 24 − 5)/30 = 39/30 = 13/10 [1 mark]
(Or written as 1.3 or 1 3/10)
Q15. [3 Marks]
Starting temperature at 6:00 AM = −4°C. Rise = 3°C every 2 hours.
(i) Temperature at 8:00 AM:
2 hours after 6 AM → 1 rise of 3°C
−4 + 3 = −1°C [1 mark]
(ii) Temperature at 10:00 AM:
4 hours after 6 AM → 2 rises of 3°C
−4 + (2 × 3) = −4 + 6 = 2°C [1 mark]
(iii) Temperature at 12:00 noon:
6 hours after 6 AM → 3 rises of 3°C
−4 + (3 × 3) = −4 + 9 = 5°C [1 mark]
(i) Temperature at 8:00 AM:
2 hours after 6 AM → 1 rise of 3°C
−4 + 3 = −1°C [1 mark]
(ii) Temperature at 10:00 AM:
4 hours after 6 AM → 2 rises of 3°C
−4 + (2 × 3) = −4 + 6 = 2°C [1 mark]
(iii) Temperature at 12:00 noon:
6 hours after 6 AM → 3 rises of 3°C
−4 + (3 × 3) = −4 + 9 = 5°C [1 mark]
Q16. [6 Marks]
(a) Five rational numbers between 1/2 and 3/4: [3 marks]
Method: Convert to equivalent fractions with a larger denominator.
1/2 = 10/20 and 3/4 = 15/20
Rationals between 10/20 and 15/20: 11/20, 12/20, 13/20, 14/20
That gives only 4. Use denominator 40:
1/2 = 20/40 and 3/4 = 30/40
Five rational numbers: 21/40, 22/40 (=11/20), 24/40 (=3/5), 26/40 (=13/20), 28/40 (=7/10)
(Any 5 valid rationals strictly between 1/2 and 3/4 are acceptable.) [3 marks]
(b) Density Property: [3 marks]
Statement: Between any two distinct rational numbers, there is always at least one rational number.
Formal version: If a and b are rational numbers with a < b, then the number m = (a + b)/2 is also a rational number, and a < m < b. [1 mark]
Example: Between 1/2 and 3/4:
m = (1/2 + 3/4)/2 = (2/4 + 3/4)/2 = (5/4)/2 = 5/8
We have 1/2 < 5/8 < 3/4. ✓ [1 mark]
Why infinitely many: We can always find a new rational number between any two rationals by this process. Starting with a and b, we find m1 between them; then we find m2 between a and m1; then m3 between a and m2; and so on forever. This process never ends, producing infinitely many distinct rationals. [1 mark]
Method: Convert to equivalent fractions with a larger denominator.
1/2 = 10/20 and 3/4 = 15/20
Rationals between 10/20 and 15/20: 11/20, 12/20, 13/20, 14/20
That gives only 4. Use denominator 40:
1/2 = 20/40 and 3/4 = 30/40
Five rational numbers: 21/40, 22/40 (=11/20), 24/40 (=3/5), 26/40 (=13/20), 28/40 (=7/10)
(Any 5 valid rationals strictly between 1/2 and 3/4 are acceptable.) [3 marks]
(b) Density Property: [3 marks]
Statement: Between any two distinct rational numbers, there is always at least one rational number.
Formal version: If a and b are rational numbers with a < b, then the number m = (a + b)/2 is also a rational number, and a < m < b. [1 mark]
Example: Between 1/2 and 3/4:
m = (1/2 + 3/4)/2 = (2/4 + 3/4)/2 = (5/4)/2 = 5/8
We have 1/2 < 5/8 < 3/4. ✓ [1 mark]
Why infinitely many: We can always find a new rational number between any two rationals by this process. Starting with a and b, we find m1 between them; then we find m2 between a and m1; then m3 between a and m2; and so on forever. This process never ends, producing infinitely many distinct rationals. [1 mark]
Q17. [6 Marks]
(i) √7 — Irrational
7 is not a perfect square. √7 cannot be expressed as p/q. Its decimal is non-terminating non-recurring (2.6457...). [1 mark]
(ii) √16 — Rational
√16 = 4 (since 4 × 4 = 16). 4 = 4/1, which is in p/q form. It is a rational number (in fact, a natural number). [1 mark]
(iii) 0.101001000100001... — Irrational
The pattern of zeros between the 1s keeps increasing (1, 2, 3, 4, ...). The decimal is non-terminating and non-recurring (the pattern never exactly repeats). Hence irrational. [1 mark]
(iv) π — Irrational
π = 3.14159265... is a well-known irrational number. It cannot be expressed as p/q for any integers p and q. [1 mark]
(v) 3 + √5 — Irrational
√5 is irrational. Adding the rational 3 to an irrational number gives an irrational number. (If 3 + √5 were rational, then √5 = (3 + √5) − 3 would also be rational — contradiction.) [1 mark]
(vi) 0.3̄ = 0.333... — Rational
This is a non-terminating recurring decimal. Recurring decimals are always rational. 0.3̄ = 3/9 = 1/3. [1 mark]
7 is not a perfect square. √7 cannot be expressed as p/q. Its decimal is non-terminating non-recurring (2.6457...). [1 mark]
(ii) √16 — Rational
√16 = 4 (since 4 × 4 = 16). 4 = 4/1, which is in p/q form. It is a rational number (in fact, a natural number). [1 mark]
(iii) 0.101001000100001... — Irrational
The pattern of zeros between the 1s keeps increasing (1, 2, 3, 4, ...). The decimal is non-terminating and non-recurring (the pattern never exactly repeats). Hence irrational. [1 mark]
(iv) π — Irrational
π = 3.14159265... is a well-known irrational number. It cannot be expressed as p/q for any integers p and q. [1 mark]
(v) 3 + √5 — Irrational
√5 is irrational. Adding the rational 3 to an irrational number gives an irrational number. (If 3 + √5 were rational, then √5 = (3 + √5) − 3 would also be rational — contradiction.) [1 mark]
(vi) 0.3̄ = 0.333... — Rational
This is a non-terminating recurring decimal. Recurring decimals are always rational. 0.3̄ = 3/9 = 1/3. [1 mark]
Q18. Bonus [2 Marks]
Given: a = √2 + 1, b = √2 − 1
a × b: [1 mark]
(√2 + 1)(√2 − 1) = (√2)² − 1² [using (x+y)(x−y) = x²−y²]
= 2 − 1 = 1 (rational)
a + b: [1 mark]
(√2 + 1) + (√2 − 1) = 2√2 = 2√2 (irrational)
Conclusion: The product of two irrational numbers (a and b) can be rational (= 1), while their sum can be irrational (= 2√2). This shows that the product/sum of two irrationals need not always be irrational.
a × b: [1 mark]
(√2 + 1)(√2 − 1) = (√2)² − 1² [using (x+y)(x−y) = x²−y²]
= 2 − 1 = 1 (rational)
a + b: [1 mark]
(√2 + 1) + (√2 − 1) = 2√2 = 2√2 (irrational)
Conclusion: The product of two irrational numbers (a and b) can be rational (= 1), while their sum can be irrational (= 2√2). This shows that the product/sum of two irrationals need not always be irrational.