Chapter Test Paper — Set 3

Answer: (b) Terminating

17/125: First check HCF(17, 125) = 1 (17 is prime; 125 = 5³).
Denominator: 125 = 5³ = 2⁰ × 5³. Since denominator has only factors 2 and 5, the decimal terminates.
17/125 = 17 × 8 / (125 × 8) = 136/1000 = 0.136

Answer: (d) √11

(a) 3/7 — rational (p/q form)
(b) √49 = 7 — rational (natural number)
(c) 0.3̄7̄ = 0.373737... — recurring decimal, hence rational (= 37/99)
(d) √11 — 11 is not a perfect square, so √11 cannot be expressed as p/q. It is irrational.

Answer: (b) 2

Using the identity (a + b)(a − b) = a² − b²:
(√5 + √3)(√5 − √3) = (√5)² − (√3)² = 5 − 3 = 2

Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

π is indeed irrational. The definition of an irrational number is precisely that it cannot be expressed in p/q form where p, q are integers and q ≠ 0. The Reason correctly explains why π is irrational. Both statements are true, and R explains A.

Answer: (b) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.

Assertion is TRUE: Rational + Irrational = Irrational. (If r + x were rational, then x = (r+x) − r = rational − rational = rational, which contradicts x being irrational.)

Reason is TRUE: Irrational numbers are NOT closed under addition (e.g., √2 + (−√2) = 0, which is rational).

However, the Reason (closure of irrationals under addition) does NOT directly explain WHY rational + irrational = irrational. The explanation for A uses a different argument. So B is the correct choice.

Answer: (a) 1.414

√8 = √(4 × 2) = 2√2 = 2 × 1.414 = 2.828
√8 − √2 = 2.828 − 1.414 = 1.414
(Or: √8 − √2 = 2√2 − √2 = √2 = 1.414)

Proof that √5 is irrational:

Step 1: Assume √5 = p/q where p, q are integers, q ≠ 0, HCF(p, q) = 1. [½ mark]

Step 2: Squaring: 5 = p²/q² ⇒ p² = 5q². So p² is divisible by 5 ⇒ p is divisible by 5. Let p = 5k. [½ mark]

Step 3: (5k)² = 5q² ⇒ 25k² = 5q² ⇒ q² = 5k². So q is also divisible by 5. [½ mark]

Step 4: Both p and q divisible by 5 contradicts HCF(p,q) = 1. Hence √5 is irrational. [½ mark]

Let x = 2.353535...    ...(1) [¼ mark]

Note: The non-repeating part before the decimal is "2" and the repeating block "35" has 2 digits.

Multiply by 100:
100x = 235.3535...    ...(2) [½ mark]

Subtract (1) from (2):
100x − x = 235.3535... − 2.3535...
99x = 233 [½ mark]

x = 233/99

Check: HCF(233, 99): 233 = 2 × 99 + 35; 99 = 2 × 35 + 29; 35 = 1 × 29 + 6; 29 = 4 × 6 + 5; 6 = 1 × 5 + 1; 5 = 5 × 1. HCF = 1, so already in lowest terms.
Therefore 2.3̄5̄ = 233/99 [¾ mark]

Successive Magnification for 4.26:

Step 1 — First level: [½ mark]
Locate 4.26 between the integers 4 and 5 on the number line. Divide [4, 5] into 10 equal parts (each = 0.1). The number lies in the interval [4.2, 4.3] (2nd division after 4).

Step 2 — Second level (Zoom in on [4.2, 4.3]): [½ mark]
Divide [4.2, 4.3] into 10 equal parts (each = 0.01). The number 4.26 lies in the 6th part, i.e., between 4.25 and 4.26. Wait — it lies exactly at the 6th division mark: between 4.25 and 4.35? Re-check: [4.2, 4.3] divided into 10 parts gives 4.20, 4.21, ..., 4.26, ..., 4.30. So 4.26 is the 6th mark from 4.20.

Step 3 — Mark the point: [1 mark]
At the 6th equal division within [4.2, 4.3], mark the point. This represents 4.26 on the number line. (Since 4.26 terminates after 2 decimal places, no further magnification is needed.)

[Diagram should show number line with markings at 4, 4.1, 4.2, 4.3 ... and then zoomed portion from 4.2 to 4.3 with 4.26 marked.]

6/(3 + √3)

Multiply numerator and denominator by the conjugate (3 − √3): [½ mark]

= 6(3 − √3) / [(3 + √3)(3 − √3)]

Denominator: (3)² − (√3)² = 9 − 3 = 6 [½ mark]

= 6(3 − √3) / 6
= (3 − √3) [1 mark]

(Or equivalently: 3 − √3 = √3(√3 − 1))

x = 2 − √3

1/x = 1/(2 − √3)

Rationalize: multiply by (2 + √3)/(2 + √3): [½ mark]
1/x = (2 + √3)/[(2)² − (√3)²] = (2 + √3)/(4 − 3) = (2 + √3)/1 = 2 + √3 [½ mark]

x + 1/x = (2 − √3) + (2 + √3) = 2 − √3 + 2 + √3 = 4 [1 mark]

Construction to Represent √3 on the Number Line:

Step 1: Mark point O at origin (0) on the number line. Mark point A at 1 (so OA = 1 unit). Draw a perpendicular AB at A of length 1 unit (so AB = 1). By Pythagoras theorem, OB = √(OA² + AB²) = √(1 + 1) = √2. With centre O and radius OB = √2, draw an arc to cut the number line at C. So OC = √2. [1 mark]

Step 2: At C, draw a perpendicular CD of length 1 unit (CD ⊥ number line, CD = 1 unit). By Pythagoras: OD = √(OC² + CD²) = √(√2² + 1²) = √(2 + 1) = √3. [1 mark]

Step 3: With centre O and radius OD = √3, draw an arc to cut the number line at E. The point E represents √3 on the number line. [1 mark]

(i) (3 + √2)(3 − √2):
= (3)² − (√2)² = 9 − 2 = 7 [1 mark]

(ii) (1 + √5)²:
= 1² + 2(1)(√5) + (√5)² = 1 + 2√5 + 5 = 6 + 2√5 [1 mark]

(iii) √75 − √48 + √27:
√75 = √(25×3) = 5√3
√48 = √(16×3) = 4√3
√27 = √(9×3) = 3√3
= 5√3 − 4√3 + 3√3 = (5 − 4 + 3)√3 = 4√3 [1 mark]

Method of Means between 2/7 and 3/7:

m₁ = (2/7 + 3/7)/2 = (5/7)/2 = 5/14   [Check: 2/7 = 4/14 < 5/14 < 6/14 = 3/7 ✓] [½ mark]

m₂ = (2/7 + 5/14)/2 = (4/14 + 5/14)/2 = (9/14)/2 = 9/28   [4/14 < 9/28 < 5/14 ✓] [½ mark]

m₃ = (5/14 + 3/7)/2 = (5/14 + 6/14)/2 = (11/14)/2 = 11/28   [5/14 < 11/28 < 6/14 ✓] [½ mark]

m₄ = (2/7 + 9/28)/2 = (8/28 + 9/28)/2 = (17/28)/2 = 17/56   [8/28 < 17/56 < 9/28 ✓] [½ mark]

m₅ = (3/7 + 11/28)/2 = (12/28 + 11/28)/2 = (23/28)/2 = 23/56   [11/28 < 23/56 < 12/28 ✓] [1 mark]

Five rationals between 2/7 and 3/7: 5/14, 9/28, 11/28, 17/56, 23/56

(i) Product of two irrationals that IS irrational: [1 mark]
Let a = √2 and b = √3 (both irrational).
a × b = √2 × √3 = √6
√6 is irrational (6 is not a perfect square). So the product is irrational.

(ii) Product of two irrationals that IS rational: [1 mark]
Let a = √5 and b = √5 (both irrational).
a × b = √5 × √5 = 5
5 is rational.
(Another example: √2 × √8 = √16 = 4, which is rational.)

(iii) Conclusion: [1 mark]
The product of two irrational numbers may be either rational or irrational. There is no definite rule; it depends on the specific numbers. Irrationals are NOT closed under multiplication.

(a) Prove (3 − √5) is irrational: [3 marks]

Assume, for contradiction, that (3 − √5) is rational. [½ mark]

Then 3 − √5 = r for some rational number r. [½ mark]

⇒ √5 = 3 − r

Since 3 is rational and r is rational, the difference (3 − r) is rational. [1 mark]

Therefore √5 is rational — but this contradicts the given fact that √5 is irrational. [½ mark]

Our assumption was wrong. Hence (3 − √5) is irrational. ■ [½ mark]

(b) Simplify (2√3 + 3√2)² − (2√3 − 3√2)²: [3 marks]

Using the identity a² − b² = (a+b)(a−b):
Let a = (2√3 + 3√2) and b = (2√3 − 3√2). [½ mark]

a + b = (2√3 + 3√2) + (2√3 − 3√2) = 4√3 [½ mark]
a − b = (2√3 + 3√2) − (2√3 − 3√2) = 6√2 [½ mark]

a² − b² = (a+b)(a−b) = (4√3)(6√2) = 24√6 [1 mark]

Answer: 24√6 [½ mark]

Alternative (expanding directly): (2√3 + 3√2)² = 12 + 12√6 + 18 = 30 + 12√6; (2√3 − 3√2)² = 12 − 12√6 + 18 = 30 − 12√6; Difference = 24√6.

(a) Convert to p/q form: [3 marks — 1 mark each]

(i) 0.2̄ = 0.222...:
x = 0.222...; 10x = 2.222...; 10x − x = 2; 9x = 2; x = 2/9 (Rational) ✓

(ii) 0.1̄2̄ = 0.121212...:
x = 0.121212...; 100x = 12.1212...; 99x = 12; x = 12/99 = 4/33 (Rational) ✓

(iii) 0.9̄ = 0.999...:
x = 0.999...; 10x = 9.999...; 9x = 9; x = 1 (or 9/9 = 1) (Rational) ✓
Interesting result: 0.9̄ = 1 exactly. This shows that 0.999... and 1 are the same number.

(b) Why every terminating decimal is rational: [3 marks]

Definition: A terminating decimal is one that has a finite number of decimal places, e.g., 0.375, 2.6, 0.05. [½ mark]

General Argument: Any terminating decimal with n decimal places can be written as a fraction with denominator 10ⁿ. [1 mark]
For example, a decimal with 3 places: x.abc = (integer)/1000. Since integers are rational and 1000 is a non-zero integer, the result is in p/q form. [1 mark]

Example: 0.136 = 136/1000 = 17/125. This is in p/q form with p = 17, q = 125, q ≠ 0. So it is rational. [½ mark]

Conclusion: Every terminating decimal equals some fraction of the form p/10ⁿ, which is always a rational number. Hence, all terminating decimals are rational.

2^1/2 × 2^3/4 ÷ 2^1/4

Using law of exponents: a^m × aⁿ ÷ a^p = a^m+n−p [½ mark]

= 2^{(1/2 + 3/4 − 1/4)}
= 2^{(2/4 + 3/4 − 1/4)}
= 2^4/4
= 2¹ [1 mark]

= 2 (which is 2¹ = 2, a rational integer) [½ mark]

As a surd: 2 = √4 or simply 2 (it is already a rational number).