# Real Numbers

Chapters

- ap-scert-10th-class-maths-lesson-1-Real-Numbers
- ap-scert-10th-class-Maths-Lesson-2-Sets
- ap-scert-10th-class-Maths-Lesson-3-Polynomials
- ap-scert-10th-class-Maths-Lesson-4-Pair-of-Linear-Equations-in-Two-Variables
- ap-scert-10th-class-Maths-Lesson-5-Quadratic-Equations
- ap-scert-10th-class-Maths-Lesson-6-Progressions
- ap-scert-10th-class-Maths-Lesson-7-Coordinate-Geometry
- ap-scert-10th-class-Maths-Lesson-8-Similar-Triangles
- ap-scert-10th-class-Maths-Lesson-9-Tangents-and-Secants-to-a-Circle
- ap-scert-10th-class-Maths-Lesson-10-Mensuration
- ap-scert-10th-class-Maths-Lesson-11-Trigonometry
- ap-scert-10th-class-Maths-Lesson-12-Applications-of-Trigonometry
- ap-scert-10th-class-Maths-Lesson-13-Probability
- ap-scert-10th-class-Maths-Lesson-14-Statistics

- ap-scert-10th-class-maths-lesson-1-Real-Numbers
- ap-scert-10th-class-Maths-Lesson-2-Sets
- ap-scert-10th-class-Maths-Lesson-3-Polynomials
- ap-scert-10th-class-Maths-Lesson-4-Pair-of-Linear-Equations-in-Two-Variables
- ap-scert-10th-class-Maths-Lesson-5-Quadratic-Equations
- ap-scert-10th-class-Maths-Lesson-6-Progressions
- ap-scert-10th-class-Maths-Lesson-7-Coordinate-Geometry
- ap-scert-10th-class-Maths-Lesson-8-Similar-Triangles
- ap-scert-10th-class-Maths-Lesson-9-Tangents-and-Secants-to-a-Circle
- ap-scert-10th-class-Maths-Lesson-10-Mensuration
- ap-scert-10th-class-Maths-Lesson-11-Trigonometry
- ap-scert-10th-class-Maths-Lesson-12-Applications-of-Trigonometry
- ap-scert-10th-class-Maths-Lesson-13-Probability
- ap-scert-10th-class-Maths-Lesson-14-Statistics

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Solutions

Use Euclid’s division algorithm to find the HCF of

i) 900 and 270

Answer:

900 = 270 × 3 + 90

270 = 90 × 3 + 0

∴ HCF = 90

ii) 196 and 38220

Answer:

38220 = 196 × 195 + 0

∴ 196 is the HCF of 196 and 38220.

iii) 1651 and 2032

Answer:

2032 = 1651 × 1 + 381

1651 = 381 × 4 + 127

381 = 127 × 3 + 0

∴ HCF = 127

Use Euclid division lemma to show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integers.

Answer:

Let ‘a’ be an odd positive integer.

Let us now apply division algorithm with a and b = 6.

∵ 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.

i.e., ’a’ can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient.

But ‘a’ is taken as an odd number.

∴ a can’t be 6q or 6q + 2 or 6q + 4.

∴ Any odd integer is of the form 6q + 1, 6q + 3 or 6q + 5.

Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1.

Answer:

Let ‘a’ be the square of an integer.

Applying Euclid’s division lemma with a and b = 3

Since 0 ≤ r < 3, the possible remainders are 0, 1, and 2.

∴ a = 3q (or) 3q + 1 (or) 3q + 2

∴ Any square number is of the form 3q, 3q + 1 or 3q + 2, where q is the quotient.

(or)

Let ‘a’ be a positive integer

So it can be expressed as a = bq + r (from Euclideans lemma)

now consider b = 3 then possible values of ‘r’ are ‘0’ or ‘1’ or 2.

then a = 3q + 0 = 3q (or) 3q + 1 or 3q + 2 now square of given positive integer (a^{2}) will be

Case - I: a^{2}- (3q)^{2}= 9q^{2}=3(3q^{2}) = 3p (p = 3q^{2})

Case-II: a^{2}= (3q + l)^{2}= 9q^{2}+ 6q+ 1

= 3[3q^{2}+ 2q] + 1 = 3p+l (Where p = 3q^{2}+ 2q) or

Case - III: a^{2}= (3q + 2)^{2}= 9q^{2}+ 12q + 4 = 9q^{2}+ 12q + 3 + 1

= 3[3q^{2}+ 4q + 1] + 1

= 3p + 1 (where ‘p’ = 3q^{2}+ 4q + 1)

So from above cases 1, 2, 3 it is clear that square of a positive integer (a) is of the form 3p or 3p + 1

Hence proved.

Use Euclid’s division lemma to show that the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8.

(OR)

Show that the cube of any positive integer is of form 9m or 9m + 1 or 9m + 8, where m is an integer.

Answer:

Let ‘a’ be positive integer. Then from Euclidean lemma a = bq + r;

now consider b = 9 then 0 ≤ r < 9, it means remainder will be 0, or 1, 2, 3, 4, 5, 6, 7, or 8

So a = bq + r

⇒ a = 9q + r (for b = 9)

now cube of a = a^{3}+ (9q + r)^{3}

= (9q)^{3}+ 3.(9q)^{3}r + 3. 9q.r + r^{3}

= 9^{3}q^{3}+ 3.9^{2}(q^{2}r) + 3.9(q.r) + r^{3}

= 9[9^{2}.q^{3}+ 3.9.q^{2}r + 3.q.r] + r^{3}

a^{3}= 9m + r^{3}(where ‘m’ = 9^{2}q^{3}+ 3.9.q2r + 3.q.r)

if r = 0 ⇒ r^{3}= 0 then a^{3}= 9m + 0 = 9m

and for r = 1 ⇒ r^{3}= l^{3}then a^{3}= 9m + 1

and for r = 2 ⇒ r^{3}= 2^{3}then a^{3}= 9m + 8

for r = 3 ⇒ r^{3}, = 3^{3}⇒ a^{3}= 9m + 27 = 9(m) where m = (9m +3)

for r = 4 ⇒ r^{3}= 4^{3}⇒ a^{3}= 9m + 64 = (9m + 63) + 1 = 9m + 1

for r = 5 ⇒ r^{3}= 125 ⇒ a^{3}= 9m + 125 = (9m + 117) + 8 = 9m + 8

for r = 6 ⇒ r^{3}- 216 ⇒ a^{3}= 9m + 216 = 9m + 9(24) = 9m

for r = 7 ⇒ r^{3}= 243

⇒ a^{3}= 9m + 9(27) = 9m

for r = 8 ⇒ r^{3}= 512

⇒ a^{3}= 9m + 9(56) + 8 = 9m + 8

So from the above it is clear that a^{3}is either in the form of 9m or 9m + 1 or 9m + 8.

Hence proved.

Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.

(Or)

Show that one and only one out of a, a + 2 and a + 4 is divisible by 3 where ‘a’ is any positive integer.

Answer:

Let ‘n’ be any positive integer.

Then from Euclidean’s lemma n = bq + r (now consider b = 3)

⇒ n = 3q + r (here 0 ≤ r < 3) which means the possible values of ‘r’ = 0 or 1 or 2

Now consider r = 0 then ‘n’ = 3q (divisible by 3)

and n + 2 = 3q + 2 (not divisible by 3)

n + 4 = 3q + 4 (not divisible by 3)

Case - II: For r = 1

n = 3q + 1 (not divisible by 3)

n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + l) divisible by 3

n + 4 = 3q + 1 + 4 = 3q + 5 not divisible by 3

Case – III: For r = 2,

n = 3q + 2 not divisible by 3

n + 2 = 3q + 2 + 2 = 3q + 4, not divisible by 3

n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) divisible by 3

So in all above three cases we observe, only one of either (n) or (n + 1) or (n + 4) is divisible by 3.

Hence proved.

Express each of the following numbers as a product of its prime factors.

- i) 140
- ii) 156
- iii) 3825
- iv) 5005
- v) 7429

Answer:

i) 140

∴ 140 = 2 × 2 × 5 × 7 = 2^{2} × 5 × 7

ii) 156

∴156 = 2 × 2 × 3 × 13 = 2^{2} × 3 × 13

iii) 3825

∴ 3825 = 3 × 3 × 5 × 5 × 17 = 3^{2} × 5^{2} × 17

iv) 5005

∴ 5005 = 5 × 7 × 11 × 13

v) 7429

∴7429 = 17 × 19 × 23

Find the L.C.M and H.C.F of the following integers by the prime factorization method.

- i) 12, 15 and 21
- ii) 17, 23 and 29
- iii) 8, 9 and 25
- iv) 72 and 108
- v) 306 and 657

Answer:

i) 12, 15 and 21

12 = 2 × 2 × 3 = 22 × 3

15 = 3 × 5

21 = 3 × 7

L.C.M = 2^{2}× 3 × 5 × 7 = 420

H.C.F = 3

ii) 17, 23 and 29

The given numbers 17, 23 and 29 are all primes.

L.C.M = their product

= 17 × 23 × 29 = 11339

∴ H.C.F = 1

iii) 8, 9 and 25

8 = 2 × 2 × 2 = 2^{3}

9 = 3 × 3 = 3^{2}

25 = 5 × 5 = 5^{2}

L.C.M = 2^{3}× 3^{2}× 5^{2}= 1800

(or)

8, 9 and 25 are relatively prime, therefore L.C.M is equal to their product,

(i.e.,) L.C.M = 8 × 9 × 25 = 1800

H.C.F = 1

iv) 72 and 108

72 = 2^{3}× 3^{2}

108 = 2^{2}× 3^{3}

L.C.M = 2^{3}× 3^{3}= 8 × 27 = 216

H.C.F = 2^{2}× 3^{2}= 4 × 9 = 36

v) 306 and 657

306 = 2 × 3^{2}× 17

657 = 32 × 73

L.C.M = 2 × 3^{2}× 17 × 73 = 22338

H.C.F = 3^{2}= 9

Check whether 6

Answer:

Given number = 6^{n}= (2 × 3)^{n}

The prime factors here are 2 and 3 only.

To be end with 0; 6^{n}should have a prime factor 5 and also 2.

So, 6^{n}can’t end with zero.

Question 4.

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer:

Given numbers are 7 × 11 × 13

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

⇒ 13(7 × 11 + 1) and

5(7 × 6 × 4 × 3 × 2 × 1 + 1)

⇒ 13 K and 5 L, where K = 78 and L = 7 × 6 × 4 × 3 × 2 × 1 + 1 = 1009

As the given numbers can be written as product of two numbers, they are composite.

Question 5.

How will you show that (17 × 11 × 2) + (17 × 11 × 5) is a composite number? Explain.

Answer:

(17 × 11 × 2) + (17 × 11 × 5)

= (17 × 11) (2 + 5)

= (17 × 11) (7)

= 187 × 7

Now the given expression is written as a product of two integers and hence it is a composite number.

What is the last digit of 6100?

Answer:

We know that

6^{1}= 6

6^{2}= 36

6^{3}= 216

6^{4}= 1296

6^{5}= 7776

We see that 6^{n}for any positive integer n ends is 6.

i.e., unit digit is always 6.

∴ Unit digit of 6^{100}is 6.

Question 1.

Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating repeating decimals.

i)3/8

ii)229/400

iii)4 1/5

iv)2/11

v)81/25

i)3/8

[!! Denominator 8 = 2^{3}, consists of only 2’s. Hence a terminating decimal.]

∴3/8= 0.375 is a terminating decimal.

ii) 229/400

[!! Denominator 400 = 2^{4} × 5^{2} = 2^{n} × 5^{m}. Hence a terminating decimal.]

∴ 229400 = 0.5725 is a terminating decimal.

iii) 4 1/5

4 1/5 = 4 + 1/5

[!! Denominator is 5. Hence a terminating decimal.]

∴ 415 = 4.2 is a terminating decimal.

iv) 2/11

Without performing division, state whether the following rational numbers will have a terminating decimal form or a non-terminating, repeating decimal form.

- i)13/3125
- ii)11/12
- iii)64/455
- iv)15/1600
- v)29/343
- vi)23/2
^{3}5^{2} - vii)129/2
^{2}.5^{7}.7^{5} - viii)9/15
- iX)36/100
- X)77/210

Answer:

i)13/3125

Note: We check whether the denominator is of the form 2^{n}. 5^{m}or not? If yes, the rational number can be expressed as a terminating decimal. If not, it can’t be expressed as a terminating decimal.

[!! Denominator is of the form 2^{m}× 5^{n}. Hence a terminating decimal.]

3125 = 5^{5}

∴13/3125is a terminating decimal.

ii)11/12

The denominator 12 is not a factor of 11. Moreover 12 = 2^{2}× 3.

[!! Denominator is not of the form 2^{m}× 5^{n}.]

∴11/12 is a non terminating, repeating decimal.

iii) 64/455

[!! Denominator is not of the form 2^{m}× 5^{n}. Hence a non terminating decimal.]

∴ 455 = 5 × 7 × 13

Hence 64/455 is a non terminating, repeating decimal.

iv) 15/1600

∴ 1600 = 2^{6}× 5^{2}[∵ The denominator is of the form 2^{n}. 5^{m}]

Hence 15/1600 is a terminating decimal.

343 = 7^{3}[Not of the form 2^{n}. 5^{m}]

∴29343is a non terminating, repeating decimal.

vi)23/2^{3}.5^{2}

23/2^{3}.5^{2}is a terminating decimal.

[∵ The denominator is of the form 2^{n}. 5^{m}]

vii)129/2^{2}.5^{7}.7^{5}

129/2^{2}.5^{7}.7^{5}is a non terminating, repeating decimal.

viii)9/15

9/15=3/5

Denominator is of the form 2^{n}. 5^{m}.

∴915=35is a terminating decimal.

ix)36100

100 = 2^{2}× 5^{2}is of the form 2^{n}. 5^{m}

Hence36/100is a terminating decimal.

x) 77/210

210 = 2 × 3 × 5 × 7 is not of the form 2^{n}. 5^{m}

Given fraction has a non-terminating, repeating decimal expansion.

Write the following rationals in decimal form using Theorem 1.4.

- i)1325
- ii)1516
- iii)23/2
^{3}.5^{2} - iv)7218/3
^{2}.5^{2} - v)143/110

Answer:

i)13/25

ii) 15/16

15/16 = 152×2×2×2

iii) 23/2^{3}?5^{2}

iv) 7218/3^{2}?5^{2}

v) 143/110

The decimal form of some real numbers are given below. In each case, decide whether the number is rational or not. If it is rational, and expressed in form p/q, what can you say about the prime factors of q?

i) 43.12345678?

ii) 0.120120012000120000 ……….

Prove that the following are irrational,

i)1/√2

ii) √3 + √5

iii) 6 + √2

iv) √5

v) 3 + 2√5

Answer:

i)1/√2

On the contrary, suppose that is a 1/√2 rational number;

then 1/√2 is of the form wherepqand q are integers.

∴ 1/√2 = p/q

⇒ √2/1 =q/p

(i.e.,) √2 is a rational number and it is a contradiction. This contradiction arised due to our supposition that 1/√2 is a rational number.

Hence 1/√2 is an irrational number.

ii) Suppose √3 + √5 is not an irrational number.

Then √3 + √5 must be a rational number.

√3 + √5 = p/q, q ≠ 0 and p, q ∈ Z

Squaring on both sides

but √15 is an irrational number.

p^{2}?8q^{2}/2q^{2} is a rational number

(p^{2} - 8q^{2}, 2q^{2} ∈ Z, 2q^{2} ≠ 0)

but an irrational number cant be equal to a rational number, so our supposition that √3 + √5 is not an irrational number is false.

∴ √3 + √5 is an irrational number.

iii) 6 + √2

To prove: 6 + √2 is an irrational number.

Let us suppose that 6 + √2 is a rational number.

∴ 6 + √2 = p/q, q ≠ 0

⇒ √2 = p/q - 6

⇒ √2 = Difference of two rational numbers

⇒ √2 = rational number But this contradicts the fact that √2 is an irrational number.

∴ Our supposition is wrong.

Hence the given statement is true.

6 + √2 is an irrational number.

iv) √5

To prove: √5 is an irrational number.

On the contrary, let us assume that √5 is a rational number.

∴ √5 = p/q, q ≠ 0

If p, q have a common factor, on cancelling the common factor let it be

reduces to a/b where a, b are co-primes.

Now √5 =a/b, where HCF (a, b) = 1

Squaring on both sides we get

⇒ (√5)^{2}=(a/b)2

⇒ 5 =a^{2}/b^{2}

⇒ 5b^{2}= a^{2}

⇒ 5 divides a^{2}and thereby 5 divides 5

Now, take a = 5c

then, a^{2}= 25c^{2}

i.e., 5b^{2}= 25c^{2}

⇒ b^{2}= 5c^{2}

⇒ 5 divides b^{2}and thereby b.

⇒ 5 divides both b and a.

This contradicts that a and b are co-primes.

This contradiction arised due to our assumption that √5 is a rational number.

Hence our assumption is wrong and the given statement is true, i.e., √5 is an irrational number,

v) 3 + 2√5

To Prove: 3 + 2√5 is an irrational.

On the contrary, let us assume that 3 + 2√5 is a rational number.

Here p, q being integers we can say that p-3q/2q is a rational number.

This contradicts the fact that √5 is an irrational number. This is due to our assumption “3 + 2√5 is a rational number”.

Hence our assumption is wrong.

∴ 3 + 2√5 is an irrational number.

Prove that √p + √q is an irrational, where p, q are primes.

Answer:

Given that p, q are primes.

Hence fp and fq are irrationals.

[∵ p, q have no factors other than 1 ∵ they are primes.]

Now √p + √q = sum of two irrational numbers = an irrational number

Hence proved.

Determine the values of the following,

i) log_{25}5

Answer:

ii) log_{81}3

Answer:

iii) log_{2}(1 1/6)

Answer:

iv) log_{7}1

Answer:

log_{7}1 = log_{7}7^{0}= 0 log_{7}7 = 0

v) log_{x}√x

Answer:

vi) log_{2}512

Answer:

log_{2}512 = log_{2}2^{9} [∵ 512 = 2^{9}]

= 9log_{2}2[∵ log x^{m}= m log x]

= 9 × 1 [∵ log_{a}a = 1]

= 9

vii) log_{10}0.01

Answer:

viii)log 2/3(8/27)

Answer:

ix)2^{2}+log_{2}3

Answer:

2^{2}+log_{2}3= 2^{2}.2^{log}_{2} 3[∵ a^{m}. a^{n}= a^{m+n}]

= 4 × 3 [∵log_{a}N= N]

= 12

Write the following expressions as log N and find their values.

i) log 2 + log 5

Answer:

log 2 + log 5

= log 2 × 5 [∵ log m + log n = log mn]

= log 10

= 1

ii) log_{2}16 - log_{2}2

Answer:

iii) 3 log_{64}4

Answer:

iv) 2 log 3 - 3 log 2

Answer:

2 log 3 - 3 log 2

= log 3^{2} log 2^{3}

= log 9 - log 8

= log 9/8

v) log 10 + 2 log 3 - log 2

Answer:

log 10 + 2 log 3 - log 2

= log 10 + log 3^{2} log 2

= log 10 + log 9 - log 2 [∵ m log a = log a^{m}]

= log 10×9/2 [∵ log a + log b = log ab; log a - log b = log a/b]

= log 45

Evaluate each of the following in terms of x and y, if it is given x = log

i) log_{2}15

Answer:

log_{2}15 = log_{2}3 × 5

= log_{2}3 + log_{2}5[∵ log mn = log m + log n]

= x + y

ii) log_{2}7.5

Answer:

iii) log_{2}60

Answer:

log_{2}60 = log_{2}2^{2}× 3 × 5

= log_{2}2^{2}+ log_{2}3 + log_{2}5

= 2 log_{2}2 + x + y

= 2 + x + y

iv) log_{2}6750

Answer:

log_{2}6750

= log_{2}2 × 3^{3}× 5^{3}

= log_{2}2 + log_{2}33 + log_{2}5^{3}

= 1 + 3 log_{2}3 + 3 log_{2}5

= 1 + 3x + 3y

Expand the following,

i) log 1000

Answer:

log 1000 = log 10^{3}

= 3 log 10

= 3 × 1

= 3

ii)log[128/625]

Answer:

iii) log x^{2}y^{3}z^{4}

Answer:

log x^{2}y^{3}z^{4}= logx^{2}+ logy^{3}+ logz^{4}[∵ log ab = log a + log b]

= 2 log x + 3 log y + 4 log z

[∵ log a^{m}= m log a]

iv)log p^{2}q^{3}/r

Answer:

If x

Answer:

Given: x^{2}+ y^{2}= 25xy

We know that (x + y)^{2}= x^{2}+ y^{2}+ 2xy

= 25xy + 2xy [∵ x^{2}+ y^{2}= 25xy given]

(x + y)^{2}= 27xy

Taking ‘log’ on both sides

log (x + y)^{2}= log 27xy

2 log (x + y) = log 27 + log x + log y

= log 3^{3}+ log x + log y

⇒ 2 log (x + y) = 3log3 + log x + log y

If log (x+y/3)=1/2(log x + log y), then find the value of x/y + y/x.

Answer:

(squaring on both sides)

⇒ (x + y)^{2}= (3√xy)^{2}

⇒ x^{2}+ y^{2}+ 2xy = 9xy

⇒x^{2}+ y^{2}= 9xy - 2xy = 7xy

If (2.3)

Answer:

Given (2.3)^{x} = (0.23)^{y}= 1000 = 10^{3}

If 2

Answer:

Given: 2^{x+1}= 3^{1-x}

log 2^{x+1}= log 3^{1-x}

(x + 1) log 2 = (1 - x) log 3

x log 2 + log 2 = log 3 - x log 3

x log 2 + x log 3 = log 3 - log 2

x (log 3 + log 2) = log 3 - log 2

Is i) log 2 is rational or irrational? Justify your answer.

Answer:

Let log_{10}2 = x

Then 10^{x}= 2

But 2 can’t be written as 10^{x}for any value of x

∴ log 2 is irrational.

ii) log 100 is rational or irrational? Justify your answer.

Answer:

Let log_{10}100 = x

⇒ log_{10}10^{2}= x

⇒ 2 log_{10}10 = x = 2

∴ log 100 is rational.

∴ log 100 = 2

Hence rational.

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