MAK LABS — Std 9 Maths Part 1 Chapter 2 Real Numbers — Practice Sets 2.1 to 2.5 and Problem Set 2 Solutions

1Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.

Rule: if the denominator (in lowest terms) has no prime factors other than 2 and 5, the decimal form is terminating; otherwise it is non-terminating recurring.

(i) 135 — denominator 5 has only the prime factor 5. Terminating (= 2.6)

(ii) 211 — denominator 11 is a prime other than 2 or 5. Non-terminating recurring (= 0.18)

(iii) 2916 — 16 = 2⁴. Terminating (= 1.8125)

(iv) 17125 — 125 = 5³. Terminating (= 0.136)

(v) 116 — 6 = 2 × 3, and 3 is not 2 or 5. Non-terminating recurring (= 1.83)

2Write the following rational numbers in decimal form.

(i) 127200 = 0.635

(ii) 2599 = 0.252525… = 0.25

(iii) 237 = 3.285714285714… = 3.285714

(iv) 45 = 0.8

(v) 178 = 2.125

3Write the following rational numbers in pq form.

(i) 0.6

Let x = 0.666… ∴ 10x = 6.666…

10x − x = 6.666… − 0.666… ⇒ 9x = 6 ⇒ x = 69 = 23

0.6 = 23

(ii) 0.37

Let x = 0.3737… ∴ 100x = 37.3737…

100x − x = 37 ⇒ 99x = 37 ⇒ x = 3799

0.37 = 3799

(iii) 3.17

Let x = 3.1717… ∴ 100x = 317.1717…

100x − x = 317.1717… − 3.1717… ⇒ 99x = 314 ⇒ x = 31499

3.17 = 31499

(iv) 15.89

Let x = 15.8989… ∴ 100x = 1589.8989…

100x − x = 1589.8989… − 15.8989… ⇒ 99x = 1574 ⇒ x = 157499

15.89 = 157499

(v) 2.514

Let x = 2.514514… ∴ 1000x = 2514.514514…

1000x − x = 2514.514… − 2.514… ⇒ 999x = 2512 ⇒ x = 2512999

2.514 = 2512999

1Show that 4√2 is an irrational number.

Assume that 4√2 is rational. Then 4√2 = pq, where p, q are integers and q ≠ 0.

∴ √2 = p4q

p and 4q are integers and 4q ≠ 0, so p4q is a rational number. This would make √2 rational.

But √2 is irrational. This is a contradiction, so our assumption is wrong.

4√2 is an irrational number.

2Prove that 3 + √5 is an irrational number.

Assume that 3 + √5 is rational. Let 3 + √5 = pq, where p, q are integers and q ≠ 0.

∴ √5 = pq − 3 = p − 3qq

p − 3q and q are integers, so the right side is rational, while the left side √5 is irrational.

This is a contradiction, so our assumption is wrong.

3 + √5 is an irrational number.

3Represent the numbers √5 and √10 on a number line.

√5: 5 = 2² + 1², so √5 is the hypotenuse of a right triangle with sides 2 and 1.

Draw a number line. Mark O at 0 and A at 2, so OA = 2 units.

At A draw a perpendicular and mark B on it with AB = 1 unit. Join OB.

By Pythagoras, OB² = 2² + 1² = 5, so OB = √5.

With O as centre and radius OB, draw an arc cutting the number line at P. The co-ordinate of P is √5.

OP = OB = √5 ≈ 2.24

√10: 10 = 3² + 1², so take OC = 3 units and CD = 1 unit perpendicular to the line at C. Then OD² = 9 + 1 = 10, so OD = √10. An arc with centre O and radius OD cuts the number line at Q, whose co-ordinate is √10.

OQ = OD = √10 ≈ 3.16

4Write any three rational numbers between the two numbers given below.

Sample answers (many others are possible):

(i) 0.3 and −0.5 — 0, 0.1, 0.2 (also −0.4, −0.1, …)

(ii) −2.3 and −2.33 — −2.31, −2.32, −2.325

(iii) 5.2 and 5.3 — 5.21, 5.25, 5.29

(iv) −4.5 and −4.6 — −4.51, −4.55, −4.59

Writing both numbers with more decimal places helps: −2.3 = −2.300 and −2.33 = −2.330, so any number from −2.301 to −2.329 lies between them.

1State the order of the surds given below.

The order is the index of the root (no index written means square root, order 2).

(i) ³√7 — 3 (ii) 5√12 — 2 (iii) ⁴√10 — 4 (iv) √39 — 2 (v) ³√18 — 3

2State which of the following are surds. Justify.

(i) ³√51 — 51 is not the cube of any rational number, so ³√51 is irrational. Surd

(ii) ⁴√16 — 2⁴ = 16, so ⁴√16 = 2, a rational number. Not a surd

(iii) ⁵√81 — 81 = 3⁴ is not the fifth power of any rational number, so ⁵√81 is irrational. Surd

(iv) √256 — 16² = 256, so √256 = 16. Not a surd

(v) ³√64 — 4³ = 64, so ³√64 = 4. Not a surd

(vi) √227 — 227 is a positive rational number, but it is not the square of any rational number (√227 = √1547 and 154 is not a perfect square). So √227 is irrational. Surd

3Classify the given pair of surds into like surds and unlike surds.

Write each surd in its simplest form and compare the radicands.

(i) √52, 5√13 — √52 = √4 × 13 = 2√13. Both have radicand 13. Like surds

(ii) √68, 5√3 — √68 = 2√17; radicands 17 and 3 differ. Unlike surds

(iii) 4√18, 7√2 — 4√18 = 4 × 3√2 = 12√2. Like surds

(iv) 19√12, 6√3 — 19√12 = 19 × 2√3 = 38√3. Like surds

(v) 5√22, 7√33 — both are already in simplest form; radicands 22 and 33 differ. Unlike surds

(vi) 5√5, √75 — √75 = 5√3; radicands 5 and 3 differ. Unlike surds

4Simplify the following surds.

(i) √27 = √9 × 3 = 3√3

(ii) √50 = √25 × 2 = 5√2

(iii) √250 = √25 × 10 = 5√10

(iv) √112 = √16 × 7 = 4√7

(v) √168 = √4 × 42 = 2√42

5Compare the following pair of surds.

For positive numbers, the one with the larger square is larger. So compare the squares.

(i) 7√2, 5√3 — (7√2)² = 98, (5√3)² = 75. 98 > 75 ∴ 7√2 > 5√3

(ii) √247, √274 — 247 < 274 ∴ √247 < √274

(iii) 2√7, √28 — (2√7)² = 28, (√28)² = 28 ∴ 2√7 = √28

(iv) 5√5, 7√2 — 125 and 98. 125 > 98 ∴ 5√5 > 7√2

(v) 4√42, 9√2 — 16 × 42 = 672 and 81 × 2 = 162. 672 > 162 ∴ 4√42 > 9√2

(vi) 5√3, 9 — 75 and 81. 75 < 81 ∴ 5√3 < 9

(vii) 7, 2√5 — 49 and 20. 49 > 20 ∴ 7 > 2√5

6Simplify.

(i) 5√3 + 8√3
= (5 + 8)√3 = 13√3

(ii) 9√5 − 4√5 + √125
= 9√5 − 4√5 + 5√5 (√125 = √25 × 5 = 5√5)
= (9 − 4 + 5)√5 = 10√5

(iii) 7√48 − √27 − √3
= 7 × 4√3 − 3√3 − √3
= (28 − 3 − 1)√3 = 24√3

(iv) √7 − 35√7 + 2√7
= (1 − 35 + 2)√7 = 5 − 3 + 105√7 = 125√7

7Multiply and write the answer in the simplest form.

(i) 3√12 × √18
= 3√12 × 18 = 3√216 = 3√36 × 6 = 3 × 6√6 = 18√6

(ii) 3√12 × 7√15
= 21√12 × 15 = 21√180 = 21√36 × 5 = 21 × 6√5 = 126√5

(iii) 3√8 × √5
= 3√40 = 3√4 × 10 = 3 × 2√10 = 6√10

(iv) 5√8 × 2√8
= 10 × (√8)² = 10 × 8 = 80

8Divide, and write the answer in simplest form.

(i) √98 ÷ √2
= √982 = √49 = 7

(ii) √125 ÷ √50
= √12550 = √52 = √5√2 × √2√2 = √102

(iii) √54 ÷ √27
= √5427 = √2

(iv) √310 ÷ √5
= √3105 = √62

9Rationalize the denominator.

(i) 3√5 = 3√5 × √5√5 = 3√55

(ii) 1√14 = 1√14 × √14√14 = √1414

(iii) 5√7 = 5√7 × √7√7 = 5√77

(iv) 69√3 = 69√3 × √3√3 = 6√327 = 2√39

(v) 11√3 = 11√3 × √3√3 = 11√33

1Multiply.

(i) √3(√7 − √3)
= √3 × √7 − √3 × √3 = √21 − 3

(ii) (√5 − √7)√2
= √5 × √2 − √7 × √2 = √10 − √14

(iii) (3√2 − √3)(4√3 − √2)
= 3√2(4√3 − √2) − √3(4√3 − √2)
= 12√6 − 3 × 2 − 4 × 3 + √6
= 12√6 − 6 − 12 + √6 = 13√6 − 18

2Rationalize the denominator.

(i) 1√7 + √2 — multiply by the conjugate √7 − √2.
= √7 − √2(√7)² − (√2)² = √7 − √25

(ii) 32√5 − 3√2 — multiply by the conjugate 2√5 + 3√2.
= 3(2√5 + 3√2)(2√5)² − (3√2)² = 3(2√5 + 3√2)20 − 18 = 6√5 + 9√22

(iii) 47 + 4√3 — multiply by the conjugate 7 − 4√3.
= 4(7 − 4√3)7² − (4√3)² = 4(7 − 4√3)49 − 48 = 28 − 16√3

(iv) √5 − √3√5 + √3 — multiply by the conjugate √5 − √3.
= (√5 − √3)²5 − 3 = 5 + 3 − 2√152 = 8 − 2√152 = 4 − √15

1Find the value.

(i) |15 − 2| = |13| = 13

(ii) |4 − 9| = |−5| = 5

(iii) |7| × |−4| = 7 × 4 = 28

2Solve.

If |a| = b (b > 0), then a = b or a = −b.

(i) |3x − 5| = 1
3x − 5 = 1 ⇒ 3x = 6 ⇒ x = 2 or 3x − 5 = −1 ⇒ 3x = 4 ⇒ x = 43
x = 2 or x = 43

(ii) |7 − 2x| = 5
7 − 2x = 5 ⇒ 2x = 2 ⇒ x = 1 or 7 − 2x = −5 ⇒ 2x = 12 ⇒ x = 6
x = 1 or x = 6

(iii) |8 − x2| = 5
8 − x2 = 5 ⇒ 8 − x = 10 ⇒ x = −2 or 8 − x2 = −5 ⇒ 8 − x = −10 ⇒ x = 18
x = −2 or x = 18

(iv) |5 + x4| = 5
5 + x4 = 5 ⇒ x4 = 0 ⇒ x = 0 or 5 + x4 = −5 ⇒ x4 = −10 ⇒ x = −40
x = 0 or x = −40

1Choose the correct alternative answer for the questions given below.

(i) Which one of the following is an irrational number? — (B) √5
√1625 = 45, 39 = 13 and √196 = 14 are all rational.
(ii) Which of the following is an irrational number? — (D) 0.101001000…
Its decimal expansion never ends and never repeats. The other three are terminating or recurring.
(iii) Decimal expansion of which of the following is non-terminating recurring? — (C) 311
The denominator 11 has a prime factor other than 2 and 5. 311 = 0.27
(iv) Every point on the number line represents which of the following numbers? — (D) Real numbers
(v) The number 0.4 in pq form is — (A) 49
x = 0.444…, 10x − x = 4, so x = 49.
(vi) What is √n, if n is not a perfect square number? — (C) Irrational number
(vii) Which of the following is not a surd? — (C) ³√64
³√64 = 4, a rational number.
(viii) What is the order of the surd ³√√5? — (C) 6
³√√5 = (5^1/2)^1/3 = 5^1/6 = ⁶√5.
(ix) Which one is the conjugate pair of 2√5 + √3? — (A) −2√5 + √3
(2√5 + √3)(−2√5 + √3) = 3 − 20 = −17, a rational number.
(x) The value of |12 − (13 + 7) × 4| is — (B) 68
|12 − 20 × 4| = |12 − 80| = |−68| = 68

2Write the following numbers in pq form.

(i) 0.555
As printed, 0.555 is a terminating decimal: 0.555 = 5551000 = 111200

If the question means the recurring decimal 0.555… = 0.5, then x = 0.555…, 10x − x = 5, 9x = 5, so x = 59.

(ii) 29.568
Let x = 29.568568… ∴ 1000x = 29568.568568…
999x = 29568.568… − 29.568… = 29539 ∴ x = 29539999

(iii) 9.315315…
Let x = 9.315315… ∴ 1000x = 9315.315315…
999x = 9306 ∴ x = 9306999 = 1034111

(iv) 357.417417…
Let x = 357.417417… ∴ 1000x = 357417.417417…
999x = 357060 ∴ x = 357060999 = 119020333

(v) 30.219
Let x = 30.219219… ∴ 1000x = 30219.219219…
999x = 30189 ∴ x = 30189999 = 10063333

3Write the following numbers in its decimal form.

(i) −57 = −0.714285714285… = −0.714285

(ii) 911 = 0.8181… = 0.81

(iii) √5 = 2.2360679… (non-terminating, non-recurring, since √5 is irrational)

(iv) 12113 = 9.307692307692… = 9.307692

(v) 298 = 3.625

4Show that 5 + √7 is an irrational number.

Assume that 5 + √7 is rational. Let 5 + √7 = pq, where p, q are integers and q ≠ 0.

∴ √7 = pq − 5 = p − 5qq

The right side is rational, but √7 is irrational. This is a contradiction.

5 + √7 is an irrational number.

5Write the following surds in simplest form.

(i) 34√8 = 34 × 2√2 = 32√2

(ii) −59√45 = −59 × 3√5 = −53√5

6Write the simplest form of rationalising factor for the given surds.

Simplify the surd first; the simplest rationalising factor is the square root of the remaining radicand.

(i) √32 = 4√2 ⇒ √2 (4√2 × √2 = 8)

(ii) √50 = 5√2 ⇒ √2 (5√2 × √2 = 10)

(iii) √27 = 3√3 ⇒ √3 (3√3 × √3 = 9)

(iv) 35√10 ⇒ √10 (35√10 × √10 = 6)

(v) 3√72 = 3 × 6√2 = 18√2 ⇒ √2 (18√2 × √2 = 36)

(vi) 4√11 ⇒ √11 (4√11 × √11 = 44)

7Simplify.

(i) 47√147 + 38√192 − 15√75
√147 = 7√3, √192 = 8√3, √75 = 5√3
= 47 × 7√3 + 38 × 8√3 − 15 × 5√3
= 4√3 + 3√3 − √3 = 6√3

(ii) 5√3 + 2√27 + 1√3
2√27 = 6√3 and 1√3 = √33
= 5√3 + 6√3 + 13√3 = (11 + 13)√3 = 343√3

(iii) √216 − 5√6 + √294 − 3√6
√216 = 6√6, √294 = 7√6, 3√6 = 3√66 = 12√6
= (6 − 5 + 7 − 12)√6 = 152√6

(iv) 4√12 − √75 − 7√48
= 4 × 2√3 − 5√3 − 7 × 4√3
= (8 − 5 − 28)√3 = −25√3

(v*) 2√48 − √75 − 1√3
= 2 × 4√3 − 5√3 − 13√3
= (8 − 5 − 13)√3 = 83√3

8Rationalize the denominator.

(i) 1√5 = 1√5 × √5√5 = √55

(ii) 23√7 = 23√7 × √7√7 = 2√721

(iii) 1√3 − √2 = √3 + √2(√3)² − (√2)² = √3 + √23 − 2 = √3 + √2

(iv) 13√5 + 2√2 = 3√5 − 2√2(3√5)² − (2√2)² = 3√5 − 2√245 − 8 = 3√5 − 2√237

(v) 124√3 − √2 = 12(4√3 + √2)(4√3)² − (√2)² = 12(4√3 + √2)48 − 2 = 12(4√3 + √2)46 = 24√3 + 6√223

2. Real Numbers

Practice Set 2.1

Practice Set 2.2

Practice Set 2.3

Practice Set 2.4

Practice Set 2.5

Problem Set 2