## AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers InText Questions and Answers.

### 10th Class Maths 1st Lesson Real Numbers InText Questions and Answers

Do this

Question 1.

Find q and r for the following pairs of positive integers a and b, satisfying a = bq + r. (Page No. 3)

i) a = 13, b = 3

Answer:

13 = 3 × 4 + 1

here q = 4 ; r = 1

ii) a = 8, b = 80

Answer:

Take a = 80, b = 8

80 = 8 × 10 + 0 here q = 10 ; r = 0

iii) a = 125, b = 5

Answer:

125 = 5 × 25 + 0

here q = 25 ; r = 0

iv) a = 132, b = 11

Answer:

132 = 11 × 12 + 0

here q = 12 ; r = 0

Question 2.

Find the HCF of the following by using Euclid division lemma,

i) 50 and 70 (Page No. 4)

Answer:

For given two positive integers a > b;

there exists unique pair of integers q and r satisfying a = bq + r; 0≤r<b.

∴ 70 = 50 × 1 + 20

Here a = 70, b = 50, q = 1, r = 20.

Now consider 50, 20

50 = 20 × 2 + 10

Here a = 50, b = 20, q = 2, r = 10.

Now taking 20 and 10.

20 = 10 × 2 + 0

Here the remainder is zero.

∴ 10 is the HCF of 70 and 50.

ii) 96 and 72

Answer:

96 = 72 × 1 + 24

72 = 24 × 3 + 0

∴ HCF = 24

iii) 300 and 550

Answer:

550 = 300 × 1 + 250

300 = 250 × 1 + 50

250 = 50 × 5 + 0

∴ HCF = 50

iv) 1860 and 2015

Answer:

2015 = 1860 × 1 + 155

1860 = 155 × 12 + 0

∴ HCF = 155

Think & Discuss

Question 1.

From the above questions in ‘DO THIS’, what is the nature of q and r? (Page No. 3)

Answer:

Given: a = bq + r

q > 0 and r lies in between 0 and b

i.e. q > 0 and 0 ≤ r < b

Question 2.

Can you find the HCF of 1.2 and 0.12? Justify your answer. (Page No. 4)

Answer:

Given: 1.2 and 0.12

we have 1.2 = \(\frac{12}{10}\) = \(\frac{120}{100}\)

0.12 = \(\frac{12}{100}\)

Now considering the numerators 12 and 120, their HCF is 12.

∴ HCF of 1.2 and 0.12 is \(\frac{12}{100}\) = 0.12

i.e., if x is a factor of y then x is the HCF of x and y.

Question 3.

If r = 0, then what is the relationship between a, b and q in a = bq + r of Euclid divison lemma? (Page No. 6)

Answer:

Given: r = 0 in a = bq + r then a = bq

i.e., b divides a completely.

i.e., b is a factor of a.

Do this

Question 1.

Express 2310 as a product of prime factors. Also see how your friends have factorized the number. Have they done it as you ? Verify your final product with your friend’s result. Try this for 3 or 4 more numbers. What do you conclude? (Page No. 7)

Answer:

Given: 2310

2310 = 2 × 1155

= 2 × 3 × 385

= 2 × 3 × 5 × 77

2310 = 2 × 3 × 5 × 7 × 11

We notice that this prime factorization is unique.

And also notice that prime factorization of any number is unique i.e., every composite number can be expressed as a product of primes and this factorization is unique.

E.g: 144 = 2 × 72

= 2 × 2 × 36

= 2 × 2 × 2 × 18

= 2 × 2 × 2 × 2 × 9

= 2 × 2 × 2 × 2 × 3 × 3

= 24 × 32

320 = 2 × 160

= 2 × 2 × 80

= 2 × 2 × 2 × 40

= 2 × 2 × 2 × 2 × 20

= 2 × 2 × 2 × 2 × 2 × 10

= 2 × 2 × 2 × 2 × 2 × 2 × 5

= 26 × 5

125 = 5 × 25

= 5 × 5 × 5

= 5^{3}

Question 2.

Find the HCF and LCM of the following given pairs of numbers by prime factorization, (Page No. 8)

i) 120, 90

Answer:

We have 120 = 2 × 2 × 2 × 3 × 5

= 2^{3} × 3 × 5

90 = 2 × 3 × 3 × 5

= 2 × 3^{2} × 5

∴ HCF = 2 × 3 × 5 = 30

LCM = 2<sup>3</sup> × 3<sup>2</sup> × 5 = 360

ii) 50, 60

Answer:

We have

50 = 2 × 5 × 5 = 2 × 5^{2}

60 = 2 × 2 × 3 × 5 = 2^{2} × 3 × 5

∴ HCF = 2 × 5 = 10

LCM = 2^{2} × 3 × 5^{2} = 300

iii) 37, 49

Answer:

We have

37 = 1 × 37

49 = 7 × 7 = 7^{2}

∴ HCF = 1

LCM = 37 × 7^{2}

Note: H.C.F. of two relatively prime numbers is 1 and LCM is equal to product of the numbers.

Try this

Question 1.

Show that 3^{n} × 4^{m} cannot end with the digit 0 or 5 for any natural numbers ‘n’ and’m’. (Page No. 8)

Answer:

Given number is 3^{4} × 4^{m}.

So the prime factors to it are 3 and 2 only.

I: but if a number want to be end with zero it should have 2 and 5 as its prime factors, but the given hasn’t ‘5’ as its prime factor.

So it cannot be end with zero.

II : now if a number went to be end with 5 it should have ‘5’ as its one of prime factors. But given 3^{n} × 4^{m} do not have 5 as a factor.

So it cannot be end with 5.

Hence proved.

Do this

Question 1.

Write the following terminating decimals in the form of p/q, q ≠ 0 and p, q are co-primes.

i) 15.265

ii) 0.1255

iii) 0.4

iv) 23.34

v) 1215.8

What can you conclude about the denominators through this process? (Page No. 10)

Answer:

i) 15.265

ii) 0.1255

iii) 0.4

0.4 = \(\frac{4}{10}\) = \(\frac{2}{5}\)

iv) 23.34

v) 1215.8

Two and five are the factors for the denominator.

Question 2.

Write the following rational numbers in the form of p/q, where q is of the form 2^{n}.5^{m} where n, m are non-negative integers and then write the numbers in their decimal form. (Page No. 11)

i) \(\frac{3}{4}\)

ii) \(\frac{7}{25}\)

iii) \(\frac{51}{64}\)

iv) \(\frac{14}{25}\)

v) \(\frac{80}{100}\)

Answer:

i) \(\frac{3}{4}\)

\(\frac{3}{4}\) = \(\frac{3}{2 \times 2}\) = \(\frac{3}{2^{2}}\)

Decimal form of \(\frac{3}{4}\) = 0.75

ii) \(\frac{7}{25}\)

\(\frac{7}{25}\) = \(\frac{7}{5 \times 5}\) = \(\frac{7}{5^{2}}\)

Decimal form of \(\frac{7}{25}\) = 0.28

iii) \(\frac{51}{64}\)

\(\frac{51}{64}\) = \(\frac{51}{2^{6}}\)

[∵ 64 = 2 × 32

= 2^{2} × 16

= 2^{3} × 8

= 2^{4} × 4 = 2^{5} × 2 = 2^{6}]

Decimal form of \(\frac{51}{64}\) = 0.796875

iv) \(\frac{14}{25}\)

v) \(\frac{80}{100}\)

\(\frac{80}{100}\) = \(\frac{80}{2^{2} \times 5^{2}}\) = \(\frac{80}{10^{2}}\) = 0.80

Question 3.

Write the following rational numbers as decimal form and find out the block of repeating digits in the quotient. (Page No. 11)

i) \(\frac{1}{3}\)

ii) \(\frac{2}{7}\)

iii) \(\frac{5}{11}\)

iv) \(\frac{10}{13}\)

Answer:

i) \(\frac{1}{3}\)

\(\frac{1}{3}\) = 0.3333…. = \(0 . \overline{3}\)

Block of digits, repeating in the quotient = period = 3.

ii) \(\frac{2}{7}\)

Decimal form of \(\frac{2}{7}\) = 0.285714….

Repeating part/period = 285714

∴ \(\frac{2}{7}\) = \(0 . \overline{285714}\)

iii) \(\frac{5}{11}\)

Period = 45

Decimal form of \(\frac{5}{11}\) = 0.454545.

= \(0 . \overline{45}\)

iv) \(\frac{10}{13}\)

Decimal form of \(\frac{10}{13}\) = 0.769230.

= \(0 . \overline{769230}\)

Period = 769230

Do this

Question 1.

Verify the statement proved above for p = 2, p = 5 and for a^{2} = 1, 4, 9, 25, 36, 49, 64 and 81. (Page No. 14)

Answer:

From the above we can conclude that if a prime number ‘p’ divides a^{2}, then it also divides a.

Think and Discuss

Question 1.

Write the nature of y, a and x in y = a^{x}. Can you determine the value of x for a given y? Justify your answer. (Page No. 17)

Answer:

y = a^{x} here a ≠ 0

We can determine the value of ‘x’ for a given y.

for example y = 5, a = 2

We cannot express y = a^{x} for y = 5, a = 2 and for y = 7, a = 3, we cannot express seven (7) as a power of 3.

Question 2.

You know that 2^{1} = 2, 4^{1} = 4, 8^{1} = 8 and 10^{1} = 10. What do you notice about the values of log_{2} 2, log_{4} 4, log_{8} 8 and log_{10} 10? What can you generalise from this? (Page No. 18)

Answer:

From the graph log_{2} 2 = log_{4} 4 = log_{8} 8 = log_{10} 10 = 1

We conclude that log_{a} a = 1 where a is a natural number.

Question 3.

Does log_{10} 0 exist? (Page No. 18)

Answer:

No, log_{10} 0 doesn’t exist, i.e a^{x} ≠ 0 ∀ a, x ∈ N.

Question 4.

We know that, if 7 = 2^{x} then x = log_{2} 7. Then what is the value of \(2^{\log _{2} 7}\)? Justify your answer. Generalise the above by taking some more examples for \(\mathbf{a}^{\log _{\mathrm{a}} \mathbf{N}}\). (Page No. 21)

Answer:

We know that if 7 = 2^{x} then x = log_{2} 7

We want to find the value of \(2^{\log _{2} 7}\);

Now put log_{2} 7 = x in the given

∴ \(2^{\log _{2} 7}\) = 2^{x} = 7 (given)

∴ \(2^{\log _{2} 7}\) = 7

Thus \(\mathbf{a}^{\log _{\mathrm{a}} \mathbf{N}}\) = N

a) \(3^{\log _{3} 8}\)

Answer:

If x = \(3^{\log _{3} 8}\) then

log_{3} x = log_{3} 8

⇒ x = 8

b) \(5^{\log _{5} 10}\)

Answer:

If y = \(5^{\log _{5} 10}\)

then log_{5} y = log_{5} 10

⇒ y = 10

Do this

Question 1.

Write the powers to which the bases to be raised in the following. (Page No. 18)

i) 64 = 2^{x}

Answer:

64 = 2^{x}

We know that

64 = 2 × 32

= 2 × 2 × 16

= 2 × 2 × 2 × 8

= 2 × 2 × 2 × 2 × 4

= 2 × 2 × 2 × 2 × 2 × 2

64 = 2^{6}

⇒ x = 6

ii) 100 = 5^{b}

Answer:

Here also 100 cannot be written as any power of 5.

i.e., there exists no integer for b such that 5^{b} = 100

iii) \(\frac{1}{81}\) = 3^{c}

Answer:

We know that 81 = 3 x 27

= 3 × 3 × 9

= 3 × 3 × 3 × 3

= 34

∴ \(\frac{1}{81}\) = 3^{-4 } [∵ a^{-m} = \(\frac{1}{\mathrm{a}^{\mathrm{m}}}\)]

∴ c = – 4

iv) 100 = 10^{z}

Answer:

100 = 10^{2}

z = 2

v) \(\frac{1}{256}\) = 4^{a}

Answer:

We know that 256 = 4 × 64

= 4 × 4 × 16

= 4 × 4 × 4 × 4

∴ \(\frac{1}{256}\) = 4^{-4}

∴ a = – 4

Question 2.

Express the logarithms of the following into sum of the logarithms. (Page No. 19)

i) 35 × 46

Answer:

log xy = log x + log y

log_{10}35 × 46 = log_{10}35 + log_{10}46

ii) 235 × 437

Answer:

log_{10}235 × 437 = log_{10}235 + log_{10}437 [∵ log xy = log x + log y]

iii) 2437 × 3568

Answer:

log_{10} 2437 × 3568 = log_{10}2437 + log_{10}3568 [∵ log xy = log x + log y]

Question 3.

Express the logarithms of the follow¬ing into difference of the logarithms. (Page No. 20)

i) \(\frac{23}{34}\)

Answer:

log_{10} = \(\frac{23}{34}\) = log_{10} 23 – log_{10} 34

[∵ log \(\frac{x}{y}\) = log x – log y]

ii) \(\frac{373}{275}\)

Answer:

log_{10} = \(\frac{373}{275}\) = log_{10} 373 – log_{10} 275

[∵ log \(\frac{x}{y}\) = log x – log y]

iii) \(\frac{4525}{3734}\)

Answer:

log_{10} = \(\frac{4525}{3734}\) = log_{10} 4525 – log_{10} 3734

[∵ log \(\frac{x}{y}\) = log x – log y]

iv) \(\frac{5055}{3303}\)

Answer:

log_{10} = \(\frac{5055}{3303}\) = log_{10} 5055 – log_{10} 3303

[∵ log \(\frac{x}{y}\) = log x – log y]

Question 4.

By using the formula log_{a}x^{n} = n log_{a} x, convert the following. (Page No. 21)

i) log_{2} 7^{25}

Answer:

log_{2} 7^{25} = 25 log_{2} 7

ii) log_{5} 8^{50}

Answer:

log_{5} 8^{50} = 50 log_{5} 8 = 50 log_{5} 23

= 3 × 50 log_{5}2 = 150 log_{5}2

iii) log 5^{23}

Answer:

log 5^{23} = 23 log 5

iv) log 1024

Answer:

log 1024 = log 2^{10} [∵ 1024 = 2^{10}]

= 10 log 2

Try this

Question 1.

Write the following relation in exponential form and find the values of respective variables. (Page No. 18)

i) log_{2}32 = x

Answer:

log_{2}32 = x

⇒ log_{2}2^{5} = x [∵ 32 = 2^{5}]

⇒ 5 log_{2}2 = x [∵ log a^{m} = m log a]

⇒ 5 × 1 = x [∵ log_{a} a = 1]

∴ x = 5

ii) log_{5}625 = y

Answer:

log_{5}625 = y

⇒ log_{5}4 = y [∵ 625 = 5^{4}]

⇒ 4 log_{5} 5 = y [∵ log a^{m} = m log a]

⇒ 4 × 1 = y [∵ log_{a} a = 1]

∴ y = 4

iii) log_{10}10000 = z

Answer:

log_{10}10000 = z

=> log_{10}10^{4} = z [∵ 10000 = 10 × 10 × 10 × 10 = 10^{4}]

=> 4 log_{10}10 = z [∵ log a^{m} = m log a]

=> 4 × 1 = z [∵ log_{a} a = 1]

∴ z = 4

iv) \(\log _{7} \frac{1}{343}\) = -a

Answer:

Question 2.

i) Find the value of log_{2}32. (Page No. 21)

Answer:

log_{2} 32 = log_{2} 2^{5}

[∵ 32 = 2 × 2 × 2 × 2 × 2 = 2^{5}]

= 5 log_{2} 2 [∵ log a^{m} = m log a]

= 5 × 1 [∵ log_{a} a = 1]

= 5

ii) Find the value of log_{c} √c.

Answer:

= \(\frac{1}{2}\) × 1 [∵ log_{a} a = 1]

= \(\frac{1}{2}\)

iii) Find the value of log_{10}0.001

Answer:

iv) Find the value of \(\log _{\frac{2}{3}} \frac{8}{27}\)

Answer: