AP 10th Class Maths Important Questions Chapter 1 Real Numbers

These AP 10th Class Maths Chapter Wise Important Questions 1st Lesson Real Numbers will help students prepare well for the exams.

AP State Syllabus 10th Class Maths 1st Lesson Important Questions and Answers Real Numbers

Question 1.
Find the HCF of 60 and 100 by using Euclid division lemma.
Solution:
AP 10th Class Maths Important Questions Chapter 1 Real Numbers 1
∴ HCF of 60 and 100 is 20

AP 10th Class Maths Important Questions Chapter 1 Real Numbers

Question 2.
Find the value of log5 \(\sqrt{625}\) Mar. 19
Soluton:
Given that log5 \(\sqrt{625}\)
= log525 = logs552
= 2 log5 5 { ∵ loga xn = n loga x}
= 2 .1 = 2

Question 3.
Express \(\frac{36}{99}\) as a decimal number.
Solution:
\(\frac{36}{99}=\frac{4 \times 9}{9 \times 11}=\frac{4}{11}=0 . \overline{36}\)

Question 4.
Write the Expand Form of log \(\frac{\mathbf{a}^{\mathrm{m}} \mathbf{b}^{\mathrm{n}}}{\mathrm{c}^{\mathrm{z}}}\)
Solution:
log \(\frac{\mathbf{a}^{\mathrm{m}} \mathbf{b}^{\mathrm{n}}}{\mathrm{c}^{\mathrm{z}}}\)
= log ambn – log cz
[ ∵ log \(\frac{\mathrm{x}}{\mathrm{y}}\) = logx – logy].
= log am + log bn – log cz
[ ∵ logxy = logx + logy]
= mloga + nlogb – zlogc.

Question 5.
If x2 + y2 = 7xy, then prove that \(\log \left(\frac{x+y}{3}\right)=\frac{1}{2}(\log x+\log y)\)
Sol.
Given x2 + y2 = 7xy now add (2xy) on both sides. We get
x2 + y2 + 2xy = 7xy + 2xy = 9xy
Considering square root on both sides we get
AP 10th Class Maths Important Questions Chapter 1 Real Numbers 2

Question 6.
Find the HCF of 1260 and 1440 by us¬ing Euclid’s division lemma.
Solution:
Given numbers 1260 and 1440
1440 – 1260 x 1 + 180
1260 = 180 x 7 + 0
∴ HCF of 1260 and 1440 is 180.

Question 7.
Using the Euclid’s division lemma, find the HCF of 4830 and 759.
Solution:
Given numbers are 4830 and 759
4830 = 759 x 6 + 276
759 = 276 x 2 + 207
276 = 207 x 1 + 69
207 = 69 x 3 + 0
∴ HCF of 4830 and 759 = 69

AP 10th Class Maths Important Questions Chapter 1 Real Numbers

Question 8.
If x2 + y2 = lOxy, then prove that 2log (x – y) = log x + log y + 3 log 2.
Solution:
Given x2 + y2 = 10xy
Now subtract ‘2xy’ on both sides
x2 + y2 – 2xy = 10xy – 2xy = 8xy
(x – y)2 = 8xy
taking “logarithm” on both sides
log (x – y)2 = log 8xy = log 8 + log x + logy
⇒ 2 log (x – y) = 3 log 2 + log x + log y

Question 9.
Is log105 rational number or irrational number ? Justify your answer.
Solution:
Let log105 = x
Then 10x = 5
But 2 can’t be written as 10x for any value of x.
∴ log 5 is irrational.

Question 10.
Express 1.2333333 In \(\frac{\mathbf{p}}{\mathbf{q}}\) form where p and q are co-primes.
Solution:
Let x = 1.2333333
= \(1.2 \overline{3}\)
AP 10th Class Maths Important Questions Chapter 1 Real Numbers 3

Question 11.
Prove that \(\sqrt{5}+\sqrt{7}\) is irrational.
Solution:
To show that \(\sqrt{5}+\sqrt{7}\) is an irrational. Let us prove this by contradiction.
So let us assume \(\sqrt{5}+\sqrt{7}\) = a
Where ‘a’ is a rational number.
⇒ \(\sqrt{7}\) = a – \(\sqrt{5}\) now squaring on both sides we get
(\(\sqrt{7}\))2 = (a-\(\sqrt{5}\))2
⇒ 7 = a2 + 5 – 2a\(\sqrt{5}\)
2a\(\sqrt{5}\) = a2 + 5 – 7 = a2 – 2
⇒ \(\sqrt{5}=\frac{\mathrm{a}^{2}-2}{2 \mathrm{a}}\) in which L.H.S part
\(\sqrt{5}\) is an irrational and RHS part \(\left(\frac{a^{2}-2}{2 a}\right)\)
(∵ ‘a’ is a rational)
So from the above a rational number and an irrational numbers are equal, which is impossible, which is wrong, So our assumption is wrong.
Hence given \(\sqrt{5}+\sqrt{7}\) won’t be a rational.
Therefore \(\sqrt{5}+\sqrt{7}\) is an irrational.

Question 12.
Show that \(\sqrt{3}\) is irrational.
Solution :
Let \(\sqrt{3}\) be rational then \(\sqrt{3}\) = a/b
(where a, b are coprimes and b ≠ 0) ⇒ b\(\sqrt{3}\) = a
Squaring on both sides, we get
3b2 = a2
3 divides a2 => 3 divides a (1)
There exists an integer c, such that
a = 3c
⇒ a2 = 9c2
⇒ 3b2 = 9c2 (∵ a2 = 3b2)
⇒ b2 = 3c2
3 divides b2 => 3 divides b (2)
From (1) & (2)
3 is a common factor for a and b This contradicts the fact that a and b are coprimes
∴ Our assumption that \(\sqrt{3}\) is rational is wrong.
∴ \(\sqrt{3}\) is irrational.

AP 10th Class Maths Important Questions Chapter 1 Real Numbers

Question 13.
Show that \(2+5 \sqrt{3}\) is irrational.
Solution:
Given that \(2+5 \sqrt{3}\) is an irrational. On the contrary, let us assume that \(2+5 \sqrt{3}\) is a rational number.
AP 10th Class Maths Important Questions Chapter 1 Real Numbers 4
Here p, q being integers we can say that \(\frac{p-2 q}{5 q}\) is a rational number. This contradicts the fact that \(\sqrt{3}\) is an irra-tional number. This is due to our as-sumption “2 + 5\(\sqrt{3}\) is a rational num-ber”.
Hence our assumption is wrong.
∴ 2 + 5\(\sqrt{3}\) is an irrational number.

Question 14.
Prove that \(\sqrt{5}+\sqrt{11}\) is irrational.
Solution:
Let us assume that \(\sqrt{5}+\sqrt{11}\) is ratio¬nal number.
Let us \(\sqrt{5}+\sqrt{11}=\frac{\mathrm{a}}{\mathrm{b}}\) where a, b are b
co – primes b ≠ 0
∴ \(\sqrt{5}=\frac{a}{b}-\sqrt{11}\)
Squaring on both sides, we get
AP 10th Class Maths Important Questions Chapter 1 Real Numbers 5
Since, a, b are integers \(\frac{a^{2}+6 b^{2}}{2 a b}\) is rational number.
So \(\sqrt{11}\) is rational number.
This contradicts the fact that \(\sqrt{11}\) is irrational number.
Hence \(\sqrt{5}+\sqrt{11}\) is irrational number.

Question 15.
Show that \(\sqrt{5}+\sqrt{3}\) is an irrational number.
Solution:
Suppose \(\sqrt{5}+\sqrt{3}\) is not an irrational number. \(\sqrt{5}+\sqrt{3}\) is a rational number.
Let \(\sqrt{5}+\sqrt{3}\) = p/q where q ≠ 0 and
p, q ∈ Z squaring on both sides
5 + 3 – 2\(\sqrt{15}\) = \(\frac{\mathrm{p}^{2}}{\mathrm{q}^{2}}\)
\(\sqrt{15}=\frac{8 q^{2}-p^{2}}{2 q^{2}}\)
p, q ∈ Z & q ≠ 0
8q2 – p2 & 2q2 ∈ Z and also 2q2 ≠ 0.
So \(\frac{8 q^{2}-p^{2}}{2 q^{2}}\) is a rational number.
but \(\sqrt{15}\) is an irrational number.
An irrational number never be equal to a rational number.
So our supposition that \(\sqrt{5}+\sqrt{3}\) is not an irrational number is false.
∴ \(\sqrt{5}+\sqrt{3}\) is an irrational number.

Question 16.
Prove that \(\sqrt{7}+\sqrt{11}\) is irrational.
Jane 2019
Solution:
Let us assume that \(\sqrt{7}+\sqrt{11}\) is ratio¬nal number.
Let us \(\sqrt{7}+\sqrt{11}\) = a/b
co – primes b ≠ 0.
∴ \(\sqrt{7}=\frac{a}{b}-\sqrt{11}\)
Squaring on both sides, we get
\(7=\frac{a^{2}}{b^{2}}+11-2 \frac{a}{b} \sqrt{11}\)
Rearranging
AP 10th Class Maths Important Questions Chapter 1 Real Numbers 6
Since, a, b are integers \(\frac{a^{2}+4 b^{2}}{2 a b}\) is ra- tional number.
So \(\sqrt{11}\) is rational number.
This contradicts the fact that \(\sqrt{11}\) is irrational number.
Hence \(\sqrt{7}+\sqrt{11}\) is irrational number.

AP 10th Class Maths Important Questions Chapter 1 Real Numbers

Question 17.
If (2)x + 1 = (3)1 -x, find the value of x.
Solution:
(2)x + 1 = (3) 1 – x
(x+ l)log102 = (1-x)log103
x log102 + 1 log102 = 1 log10 3 – x log10 3
x log10 2 + x log10 3 = log10 3 – log10 2
x (log10 2 + log10 3) – log103 – log10 2
x = \(\frac{\log _{10} 3-\log _{10} 2}{\log _{10} 2+\log _{10} 3}\)

Fundamental Theorem of Arithmetic :

Question 18.
Support the Fundamental theorem of Arithmetic by considering some examples.
Solution:
Every composite number can be ex-pressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
Ex: 1) Now let us consider a com¬posite number 186.
∴ 186 can be factorised 2 x 3 x 31 where 2, 3 and 31 are primes.
Ex: 2) Now consider another com¬posite number 384.
∴ 384 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 = 27 x 3 where these 2 and 3 are primes.
Thus we can express any composite number as a product of some primes. Thus the theorem is valid.

Question 19.
Write 210 as product of prime factors.
Solution:
210 = 3 x 70
= 3 x 7 x 10
= 3 x 7 x 5 x 2
∴ way of 2 x 3 x 5 x 7.
So the composite number 210 can be written as product of primes 2, 3, 5 and 7 in the way of 2 x 3 x 5 x 7.

Question 20.
Is the units digit in the value of 6n where ‘n’ is a natural number is zero? Justify your answer.
Solution:
No, the units number in the value of 6n where ‘n’ is a natural number will not be zero.
Why because, if a (composite) number is written as product of prime factors having 2 and 5 then the units digit becomes zero.
In this case
6 = 2 x 3 that means 6 has only 2 in its factors but not 5.
Hence in the value of 6n, the units digit will not be equal to zero.

Question 21.
If the HCF of 306 and 657 is 9, then find their LCM.
Solution:
The given two numbers are 306, and 657.
HCF of given numbers = 9
From the formula
LCM x HCF = Product of given two numbers
(LCM) (9) = 657 x 306
∴ LCM = \(\frac{657 \times 306}{9}\) = 22, 338
∴ LCM of 306 and 657 is 22,338.

AP 10th Class Maths Important Questions Chapter 1 Real Numbers

Question 22.
Find the HCF of 45,75 from Euclid’s division lemma.
Solution:
Let us write 45 and 75 in the form of a = bq + r where a = 75, b = 45
∴ 75 = 45 (1) + 30
Again using lemma a = bq + r, where
a = 45, and b = 30,
we get, 45 = 30 (1) + 15
then 30 = 15 (2) + 0
Here we get remainder ‘0’ when we
divide it by 15.
So 15 is HCF of 45 and 75.

Question 23.
How do you say 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are complex numbers ? Explain it.
Solution:
The given
7 x 11 x 13 + 13= 13 [(7 x 11) + 1] = 13 [77 + 1] =13 x 78
So we can express 7 x 11 x 13 + 13 as product of 13 and 78 that means 13 and 78 are factors of given 7 x 11 x 13 + 13. Hence it is a composite number.
Similarly
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5[(7 x 6 x 4 x 3 x 2 x 1) + 1]
= 5 [1008 + 1] = 5 x 1009
That means 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 can be written as product of two factors 5 and 1009. Hence (7 x 6 x 5 x 4 x 3 x 2 x 1) + 5 is a complex number because it has more than 2 factors.

Logarithms :

Question 24.
Express the following in log form,
i) 25 = 32
ii) 33 = 27
iii) 53 = 125
Solution:
If ax = N then logaN = x(by definition)
i) if 25 = 32 then log232 =5
ii) if 33 = 27 then log327 = 3 j log forms
iii) if 53 = 125 then log5125 = 3

Question 25.
Calculate the value of log2512.
Solution:
We can write 512 as 29.
log2512 = loga29 = 9 log 22 = 9 (1)
= 9 (∵ logaa = 1)
log2512 = 9

Question 26.
Calculate the value of log10 0.01.
Solution:
We can write 0.01 = \(\frac{1}{100}=\frac{1}{10^{2}}\) = 10-2
Then log100.01= log10 10-2
= -2log1010
= – 2
∴ log100.01 = -2( ∵ log1010= 1)

Question 27.
Express log5125 = 3 in its exponential form.
Solution:
We know that
if logax = N then aN = x
So, log5125 = 3 => 53 = 125
[∵ x = 125, a = 5, N = 3]
∵ 53 = 125 is the exponential form of log5125 = 3

AP 10th Class Maths Important Questions Chapter 1 Real Numbers

Question 28.
Prove that logaxy = logax + logay.
Solution:
Let logaxy = p => ap = xy …………(1)
Let logax = q => aq = x ……….. (2)
Similarly, let logay = r then ar = y …………… (3)
Now put the values of ‘x’ and ‘y’ in the equation (1)
ap = xy
= aq. ar [∵ x = aq, y = ar]
ap = aq + r [ ∵ am . an = am + n]
⇒ p = q + r
Now reproduce the values of p, q and r.
We get, logaxy = logax + logay Hence proved.

Question 29.
[H+] ion concentration in a soap used by Sankar is 9.2 x 10-22. Then find its pH using logarithm.
Solution:
The [H+] ion concentration in the soap is 9.2 x 10-22.
Then the pH = – log10 [H+]
= -log10 (9.2 x 10-22)
= -[log109.2 + log1010-22] = -log109.2-(-22).
logio10
= -0.96 + 22 = 21.04
Hence pH = 21.04 (Actually it is not possible, pH is always less than 14.)

Irrational Numbers :

Question 30.
Choose the irrational numbers from the following given
\(\begin{aligned}
&\sqrt{2}, \sqrt{3}, 5,5.75,1.735, \ldots \ldots \ldots \ldots, \\
&\sqrt{8}, \sqrt{4}, \sqrt{16}
\end{aligned}\) ……………
Solution:
\(\sqrt{2}, \sqrt{3}, \sqrt{8}\), ……………… are irrational numbers.

Question 31.
Define the irrational numbers. Give examples.
Solution:
Definition of irrational numbers :
A number which can be expressed nei-ther as a terminating decimal nor as a non-terminating decimal is called an irrational number.
These are denoted with ‘S’ or Q’
(OR)
A number which cannot be repre¬sented in the form of p/q is called an irrational.
Ex : \(\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{11}, \sqrt{12}, \ldots \ldots \ldots\)

AP 10th Class Maths Important Questions Chapter 1 Real Numbers

Question 32.
Write the differences between rational and irrational numbers.
Solution:

Irrational Numbers Rational Numbers
1. These are neither in the form of terminating decimal nor as a non-terminating, recurring decimals. 1. These are either in the form of termina
ting or as a non-terminating, recurring decimal.
2. We cannot express them in the form of p/q (q ≠ 0). 2. We can express them in the form of p/q (q ≠ 0)
3. \(\sqrt{\mathrm{p}}\) is an irrational when p is a prime. 3. All natural numbers/integers are rational numbers.
4. We denote them with ‘Q’’ or ‘S’. 4. We denote them with ‘Q’.
5. We cannot find their square root value exact.
Ex : \(\sqrt{2}, \sqrt{3}, \sqrt{5}, \ldots \ldots \ldots\) …………….
5. Ex :  \(\frac{5}{8}, \frac{6}{7}\)1, – 3,…………..

Question 33.
Plot \(\sqrt{2}, \sqrt{3}, \sqrt{5}\) on number line.
Solution:
Now consider a right angle triangle with 1 cm each leg as shown in figure. AB = AC = 1 cm, then
hypotenuse = BC = \(\sqrt{1^{2}+1^{2}}=\sqrt{2}\)
Now plot a point ‘D’ on the line AB such that AD = BC = \(\sqrt{2}\)
AP 10th Class Maths Important Questions Chapter 1 Real Numbers 7
Now plot a point ‘E’ on the line AB
such that AE = CD = \(\sqrt{3}\)
Now plot a point ‘F’ on the line AB
such that AF = CE = 2
Now plot another point ‘G’ on the line
AB such that
AG = CG = \(\sqrt{1^{2}+2^{2}}=\sqrt{1+4}=\sqrt{5}\)
The point ‘D’ is \(\sqrt{2}\) , ‘E’ is \(\sqrt{3}\) , ‘F’ is \(\sqrt{4}\) , and the point ‘G’ is \(\sqrt{5}\) on num¬ber line.

Question 34.
Find the value of \(\sqrt{2}\) upto 4 decimals.
Solution:
AP 10th Class Maths Important Questions Chapter 1 Real Numbers 8

Question 35.
Where do we use irrational numbers in our day to day life ?
Solution:
1. We use the value of π in finding perimeter and area of a circle (n is an irrational).
2. To find the length of the diagonal of a square we use \(\sqrt{2}\) which is an irrational one.
3. To find the T.S.A., curved surface area and volumes of solids viz cyl-inder, sphere, and cone etc. We use K in above case which is an irratio¬nal one.
4. For calculation of compound inter-est, E.M.I. (equal monthly install-ments) in Banks they use ‘e’ value which is an irrational one.
5. We use these numbers in space and economics related matters thor-oughly.

Question 36.
Establish the relation between diago-nal and side of a square.
Solution:
Let the side of square ABCD = a
Then its diagonal AC = Hypotenuse of ΔABC.
AP 10th Class Maths Important Questions Chapter 1 Real Numbers 9
So the diagonal of a square is a\(\sqrt{2}\) when its side is ‘a’.

Question 37.
Is π a rational or irrational ? Give reasons.
Solution:
π is an irrational number. The value of
π is approximately \(\frac{22}{7}\)
But not exactly,
\(\frac{22}{7}\) = 3.14285714286. eventhough 22
\(\frac{22}{7}\) is in the form of p/q.
π is an irrational.

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