NCERT Exemplar Class 10 Maths Solutions Chapter 1 Real Numbers

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# NCERT Exemplar Class 10 Maths Solutions Chapter 1 Real Numbers

Edited By Ravindra Pindel | Updated on Sep 05, 2022 12:09 PM IST | #CBSE Class 10th

NCERT exemplar Class 10 Maths solutions chapter 1 is very important for all students who want to pursue mathematics in higher classes. This chapter provides a detailed understanding of real numbers. NCERT exemplar Class 10 Maths chapter 1 solutions can help the students build a strong base which will provide an extra edge while solving questions of NCERT Class 10 Maths.

These Class 10 Maths NCERT exemplar chapter 1 solutions are complete and descriptive which makes the learning of concepts of real numbers extremely effective. CBSE Syllabus for Class 10 has been kept in vision while preparing the NCERT exemplar Class 10 Maths solutions chapter 1.

Question:1

If the value of m is odd number than option A give odd number. Therefore, option A is not the answer of given question. When value of m is even, then option B always given odd number.
For example let m = 2, then m + 1 = 2 + 1 = 3
Which is odd hence B is not the answer.
When value of m is odd/even then option C gives always even number.
For example let m = 2, then 2 m = 4;
let m = 3, then 2 m = 6
Hence C is correct option.
Option D always gives odd integer for every value of m, hence option D is not the answer.

Question:2

For some integer q, every odd integer is of the form
(A) q
(B) q + 1
(C) 2q
(d) 2q + 1

If the value of q is even number than option A gives even number. Therefore, option A is not the answer of given question.
When value of q is odd, then option B always gives even number.
For example let q = 3, then q + 1 = 3 + 1 = 4
Which is even hence B is not the answer.
When value of q is odd/even then option C gives always even number.
For example let m = 2, then 2 m = 4;
let m = 3, then 2 m = 6
Hence C is incorrect option.
Option D always gives odd integer for every value of q, hence option D is the answer.
For example let q = 2, then 2q + 1 = 2 × 2 + 1 $\Rightarrow$ 5

Question:3

n2 – 1 divisible by 8, if n is
(A) an integer
(B) a natural number
(C) an odd integer
(D) an even integer

let n = 2 then n2 – 1 = 4 – 1 = 3, which is not divisible by 8
When n is a natural number,
let n = 4 then n2 – 1 = 16 – 1 = 15, which is not divisible by 8
When n is an odd integer then n2 – 1 is always divisible by 8,
let n = 3, n2 – 1 = 9 – 1 = 8 which is divisible by 8.
Proof: let n = 2q + 1
n2– 1 = (2q + 1)2 – 1
= 4q2 + 1 + 4q – 1
= 4q (q + 1)
Product of two consecutive number is divisible by 2, hence number n2 – 1 is divisible by 8.
When n is an even integer then n2 – 1 is not divisible by 8.
let n = 2,
n2 – 1 = 4 – 1 = 3

Question:4

If the HCF of 65 and 117 is expressible in the form 65 m – 117, then the value of m is
(A) 4 (B) 2 (C) 1 (D) 3

Solution. First of all let us find the HCF of 65 and 117
65 = 1 × 5 × 13 117 = 1 × 3 × 13 × 3
HCF = 13 ….(1)
According to question HCF = 65 m – 117
Using (1)
13 = 65 m – 117
13 + 117 = 65 m
$\frac{130}{65}= m$
m = 2
Therefore, option (b) is correct.

Question:5

The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is
(A) 13 (B) 65 (C) 875 (D) 1750

Solution. First of all subtract remainder from the given number.
(70 – 5) = 65
(125 – 8) = 117
Require number is HCF of 65 and 117
65 = 5 × 13 × 1
117 = 3 × 3 × 13 × 1
HCF = 13
Hence 13 is the largest number which is divides by 70 and 120 & leaves remainder 5 and 8.
Therefore, option (A) is correct.

Question:6

If two positive integers a and b are written as a = x3 y2 and b = xy3; x, y are prime numbers then HCF (a, b)
(A) xy (B) xy2 (C) x3y3 (D) x2y2

$a= x^{3}y^{2}\Rightarrow x\times x\times x\times y\times y$
$b= xy^{3}\Rightarrow x\times y\times y\times y$
For HCF, we have to take all common prime numbers from both the numbers.
Hence,
$\text{HCF = }x\times y\times y= xy^{2}$
$\text{ HCF =}xy^{2}$
Hence option B is correct.

Question:7

If two positive integers p and q can be expressed as p = ab2 and q = a3b; a, b being prime numbers then LCM (p, q) is
(A) ab (B) a2b2 (C) a3b2 (D) a3b3

Solution.
$p= ab^{2}= a\times b\times b$
$q= a^{3}b= a\times a\times a\times b$
For LCM, we have to take all the primes from prime factorization of two number.
The common prime in both numbers has to be taken only once.
Therefore,
$\text{LCM =}a\times a\times a\times b\times b= a^{3}b^{2}$
Hence option C is correct.

Question:8

The product of a non-zero rational and an irrational number is
(A) always irrational
(B) always rational
(C) Rational or irrational
(D) one

Rational number is the number, which can be written:
in the form of p/q where q not equal to zero. (Here P and Q has to be an integer)
Irrational numbers cannot be written in the form of p/q.
Consider a rational number = 1/2
$\text{ Irrational number =}$ $\sqrt{3}$
$\text{Product =}\frac{1}{2}\times \sqrt{3}= \frac{\sqrt{3}}{2}$ $\text{(irrational)}$
Therefore, option (A) is correct.

Question:9

The least number that is divisible by all the numbers from 1 to 10 is:
(A) 10 (B) 504 (C) 100 (D) 2520

Option A is not the answer of the question, because 10 is not divisible by 3, 4, 6, 7, 8, 9
Option B is also not the answer of the question, because 504 is not divisible by 5 and 10
Option (C) is also not the answer of the question because 100 is not divisible by 3, 6, 7, 8 and 9
Option (D) is the answer of the question because 2520 is divisible by all the numbers from 1 to 10.
Calculation:
Prime factorization of all the numbers:
1 = 1
2 = 2
3 =3
4 =2$\times$2
5 =5
6 = 3$\times$2
7 = 7
8 = 2$\times$2$\times$2
+9 = 3$\times$3
10 = 2$\times$5
For LCM, we have to take all the primes from prime factorization of two number.
The common prime in both numbers has to be taken only once.
Therefore,
LCM = 2$\times$2$\times$2$\times$3$\times$3$\times$3$\times$5$\times$7 =2520
Therefore, option (D) is correct.

Question:10

The decimal expansion of the rational numbers $\frac{14587}{1250}$ will terminate after:
(A) one decimal place
(B) two decimal places
(C) there decimal places
(D) four decimal places

$\text{Rational number =}\frac{14587}{1250}$
$\Rightarrow$ $\frac{14587}{2^{1}\times \left ( 5 \right )^{4}}$

$= \frac{14587}{2^{1}\times 5^{1}\times 5^{3}}$
$= \frac{14587}{10\times 5^{3}}\times \frac{2^{3}}{2^{3}}$ $\text{{Divide and multiply by 23}}$
$= \frac{116696}{10000}= 11.6696$
Hence the given rational number will terminate after four decimal places. (D) four decimal places

Question:1

Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer

The above statement is not true in all the conditions. For example, if we take the value of q as – 1. Then 4q + 2 = – 4 + 2 = – 2 which is not a positive integer.
And moreover, if q is positive then 4q + 2 will be always even. Therefore, we can never write an odd positive number in this form. Hence given statement is false

Question:2

The product of two consecutive positive integer is divisible by 2”. Is this statement true or false”? Justify your answer.

The above statement is true in all the conditions. Consider a number P = q (q + 1)
If q is odd then q+ 1 will be even, hence P will be divisible by 2.
If q is even then q+ 1 will be odd, but P will be divisible by 2.
For example:
let us take numbers 2 and 3:
product is 2 x 3 = 6 which is divisible by 2
Hence given statement is true

Question:3

The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer

The given statement is true in all the conditions.
Let three consecutive positive integers be, x, x + 1 and x + 2.
Divisibility by 3: Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
$\therefore$n = 3p or 3p + 1 or 3p + 2, where p is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 3.
$\Rightarrow$ x (x + 1) (x + 2) is divisible by 3.
Divisibility by 2: Whenever a number is divided 2, the remainder obtained is 0 or 1.
$\therefore$n = 2q or 2q + 1, where q is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.
$\Rightarrow$x (x + 1) (x + 2) is divisible by 2.
Since, x (x + 1) (x + 2) is divisible by 2 and 3.
$\therefore$ x (x + 1) (x + 2) is divisible by 6.

Question:4

Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer

Any positive integer can be written in the form of 3q or 3q+1 or 3q + 2.
Since square of 3q = 9q² = 3(3q²) = 3m (where 3q² = m)
Square of (3q+1) = (3q+1) ² = 9q²+1+6q = 3(3q²+2q) +1 = 3m + 1
(where, m = 3q² + 2q)
Square of (3q+2) = (3q+2) ² = 9q²+4+12q = 3(3q²+4q) +4 = 3m + 4 (where, m = 3q²+2q)

Thus, there is not any square of a positive integer, which can be written in the form of 3m + 2
Hence the given statement is false.

Question:5

A positive integer is of the form 3q + 1, being a natural number. Can you write its square in any form other than 3m + 1, i.e. 3m or 3m + 2 for some integer m? Justify your answer

Let us take the square of (3q + 1)
(3q + 1)2 = (3q) 2 + (1)2 + 2 × 3q × 1
= 9q2 + 1 + 6q
= 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3m + 1
{where m = 3q2 + 2q}
Hence, the square of 3q + 1 (where q is a positive integer) cannot be
written in any form other than 3m + 1
Hence the given statement is false.

Question:6

The number 525 and 3000 are both divisible only by 3. 5, 15, 25 and 75 what is HCF (525, 3000)? Justify your answer.

The highest common divisor of any two numbers is called HCF.
Here 75 is the highest common divisor.
We can also find HCF:
Applying Euclid’s division algorithm on 525 and 3000 we get
Hence 75 is the HCF of numbers 525 and 3000

Question:7

Explain why 3 $\times$ 5 $\times$ 7 + 7 is a composite number

Composite number – the number which is divisible by a number other than one and itself. Here, 3 x 5 x 7 + 7 = 112 and 112 is divisible by 2, 4, 7, 8, 14, 16, 28 and 56. Hence it is a composite number.

Question:8

Can two numbers have 18 as their HCF and 380 as then LCM? Give reasons.

As we know, LCM is a multiple of HCF
$\frac{LCM}{HCF}$has to be an integer.
If the given statement is true:
380 = 18n where n is any integer.
But there is no such type of n exist as we can say that 380 is not divisible by 18.
Hence two numbers, with 18 as their HCF and 380 as their LCM, do not exist.

Question:9

Without actually performing the long division, find if 987/10500 will have terminating or non-terminating decimal expansion give reason for your answer.

We can also write this number as
:$\frac{987}{10500}= \frac{3\times 7\times 47}{3\times 7\times 2^{2}\times 5^{3}}= \frac{47}{2^{2}\times 5^{3}}$

= $\frac{47}{2^{2}\times 5^{3}}\times \frac{2}{2}= \frac{47\times 2}{5^{3}\times 2^{3}}$
= $\frac{94}{\left ( 10 \right )^{3}}= \frac{94}{1000}= 0\cdot 094$
Here, we have a finite number of digits after the decimal point.
Hence it is terminating decimal expansion.

Question:10

A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the formp/q? Give reason.

The given decimal number is 327.7881.
$327\cdot 7081= \frac{3277081}{10000}= \frac{p}{q}$
The q is 10000, therefore its prime factorization, can be written as:
q = 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
= 24 × 54 = (2 × 5)4

Question:1

Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q

Any positive integer can be written in the form of 4m or 4m + 1 or 4m + 2 or 4m + 3.
Case 1:
A = 4m
(4m)2 = 16 m2
=4(4m2)
=4q (Here q = 4m2)
Case 2:
A = 4m + 1
(4m + 1)2= (4m)2 + 1 + 8m
=16m2 + 8m + 1
=4(4m2 + 2m) + 1
=4q + 1 (here q = 4m2 + 2m)
Case 3:
A = 4m + 2
(4m + 2) 2
= 16 m2 + 4 + 16 m
=4 × (4m2 + 1) + 4 m
= 4q
Here q = (4m2 + 1 + 4m)
Case 4:
A = 4m + 3
(4m + 3)2 = 16 m2 + 9 + 24 m
4(4m2 + 6m + 2) + 1
4q + 1
here q = (4m2 + 6m +2)
Hence square of any positive integer is either of the form 4q or 4q + 1.

Question:2

Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 4
r is remainder when we divide A by 4, therefore:
$0\leq\ r$ $<4$ $r = 0, 1, 2, 3$
A = 4q + r …(1)
Case 1:
A = 4q
A3 = (4q)3 = 64q3 = 4(16q3)
A3 = 4m
(m = 16 q3)
Case 2:
A = 4q + 1
(4q + 1)3 = (4q)3 + (1)3 + 3(4q)2 (1) + 3(4q) (1)2
(Using (a + b)3 = a3+ b3 + 3a2b + 3ab2)
= 64q3 + 1 + 48q2 + 12q
= 4(16q3 + 12q2 + 3q) + 1
A3 = 4m + 1
(m = 16q3 + 12q2 + 3q)
Case 3:
A = 4q + 2
= (4q + 2)3 = (4q)3 + (2)3 + 3(4q)2 (2) + 3(4q) (2)2
(using (a + b)3 = a3 + b3 + 3a2b + 3ab2)
= 64q3 + 8 + 96q2 + 48q
= 4(16q3 + 2 + 24q2+ 12q)
= 4m
(m = 16q3 + 2 + 24q2 + 12 q)
Case 4:
A = 4q + 3
A3 = (4q + 3)2 = (4q)3 + (3)3 + 3(4q)2 (3) + 3(4q) (3)2
(Using (a + b)3 = a3 + b3 + 3a2b + 3ab2)
= 64q3 + 27 + 144q2 + 108q
= 64q3 + 24 + 144 q2 + 108q + 3
= 4(16q3 + 6 + 36q2 + 27q) + 3
= 4m + 3
(m = 16q3 + 6 + 36q2 + 27q)
Hence any positive integer’s cube can be written in the form of
4m, 4m + 1, 4m + 3.

Question:3

Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

By Euclid’s division –
Any positive integer can be written as:
A = bm + r
Here b = 5
r is remainder when we divide A by 5, therefore:
0 $\leq$ r $<$ 5, r = 0, 1, 2, 3.4
A = 5m + r …(1)
Case 1:
A = 5m
(5m)2 = 25 m2
= 5(5m2)
= 5q
(Here q = 5 m2)
Case 2:
A = 5m + 1
(5m + 1)2 = 25 m2 + 1 + 10 m
25m2 + 10 m + 1
5(5m2 + 2m) + 1
5q + 1
(Here q = 5m2 + 2m)
Similarly, we can verify it for 5m + 2, 5m + 3, 5m + 4
Here, square of any positive integer is in the form of 5q, 5q + 1
Hence, square of any positive integer cannot be of the form 5q + 2 or 5q + 3.

Question:4

Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 6
r is remainder when we divide A by 5, therefore:
0 $\leq$ r $<$ 6, r = 0, 1, 2, 3,4,5
A = 6q + r …(1)
Case 1:
A = 6q
A2 = (6q)2 = 36q2 = 6(6q2)
A2 = 6m
(m = 6q2)
Case 2:
A = 6q + 1
A2 = (6q + 1)2
A2 = 36q2 + 1 + 2(6q)
(Using (a + b)2 = a2 + b2 + 2ab)
A2 = 36q2 + 1 + 12q
A2 = 6(6q2 + 2q) + 1
A2 = 6m + 1
(m = 6q2 + 2q)
Similarly, we can verify it for 6q + 2, 6q + 3, 6q + 4, 6q + 5
Now, it is clear that the square of any positive integer cannot be of the form 6m + 2, 6m + 5.

Question:5

Show that the square of any odd integer is of the form 4q + 1 for some integer q.

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 4
r is remainder when we divide A by 4, therefore:
0 $\leq$ r $<$ 4, r = 0, 1, 2, 3
Even: A = 4q and 4q + 2
Odd: A = 4q + 1 and 4q + 3
Case 1:
Now squaring the odd terms, we get
(4q + 1)2 = (4q) 2 + (1)2 + 8q
= 16q2 + 8q + 1
= 4(4q2 + 2q) + 1
= 4m + 1
Here m = 4q2 + 2q
Case 2:
(4q + 3)2 = (4q)2 + (3)2 +24q
= 16q2 + 8 + 1 + 24q
=16q2 + 24q + 8 + 1
=4(4q2 + 6q + 2) + 1
=4m + 1
Here m = 4q2 + 6q + 2
Hence we can say that square of any odd integer is of the form 4q + 1 for some integer q.

Question:6

If n is an odd integer, then show that n2 – 1 is divisible by 8.

We can represent any odd positive integer in the form of (4q + 1) or (4q + 3) for some integer q.
Case 1:
If n = 4q + 1
n2 – 1 = (4q + 1)2 –1
n2– 1 = (4q)2 + (1)2 + 2(4q) (1) – 1
Using (a + b)2 = a2 + b2 + 2ab
n2 – 1 = 16q2 + 1 + 8q – 1
n2 – 1 = 16q2 + 8q = 8q (2q + 1)
Which is divisible by 8.
Case 2:
If n = 4q + 3
n2 – 1 = (4q + 3)2 – 1
n2 – 1 = (4q)2 + (3)2 + 2(4q) (3) – 1
(Using (a + b)2 = a2 + b2 + 2ab)
n2 – 1 =16q2 + 9 + 24q –1
n2 – 1 = 16q2 + 8 + 24q = 8(2q2 + 1+3q)
Which is divisible by 8
Hence any odd positive integer n when written in n2 – 1 from is divisible by 8.

Question:7

Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.

We can represent any positive integer in the form of (2q +1) for any integer q.
Let x = 2q + 1; y = 2p + 1
x2 + y2 = (2q + 1)2 + (2p + 1)2
x2 + y2 = (2q)2 + (1)2 + 2(2q) (1) + (2p)2 + (1)2 + 2(2p) (1)
(Using (a + b) 2 = a2 + b + 2ab)
x2 + y2 = 4q2 + 1 + 4q + 4p2 + 1 + 4p
x2 + y2 = 2(2q2 + 2q + 2p2 + 2p + 1)
(2q2 + 2q + 2p2 + 2p + 1) is odd
Hence, x2 + y2 is divisible by 2 but not divisible by 4.

Question:8

Use Euclid’s division algorithm to find the HCF of 441, 567, 693

First, we find the HCF of 441 and 567 by using Euclid division algorithm.
Here HCF of (567, 41) = 63 Now, we will find the HCF of 63 and 693

693 = 63 $\times$11 + 0
Therefore HCF of (63, 693) = 63
Thus HCF of (441, 467, 693) = 63.

Question:9

Using Euclid is division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.

As we know that 1, 2 and 3 are the remainders of given number 1251, 9377 and 15628
After subtracting remainders, we get
1251 – 1 = 1250
9377 – 2 = 9375
15628 – 3 = 15625
Now HCF of 1250, 9375, 15625 can be calculated as:
First, we will calculate the HCF of 9375 and 15625;
HCF (15625, 9375) = 3125 Now we will calculate the HCF of 3125 and 1250.
HCF of 3125 and 1250 is 625. Here HCF of (1250, 9375, 15625) = 625
Hence largest number is 625.

Question:10

Prove that $\sqrt{3}+\sqrt{5}$ is irrational.

We will do it by the method of contradiction: We will assume $\sqrt{3}+\sqrt{5}$ is a rational number. If it leads to some absurd outcome then it is a wrong assumption.
Let $\sqrt{3}+\sqrt{5}$ is a rational number
$\therefore \sqrt{3}+\sqrt{5}= \frac{a}{b},b\neq 0,a,b\, \epsilon z$
$\sqrt{5}= \frac{a}{b}-\sqrt{3}$
Squaring both side.
$\left ( \sqrt{5} \right )^{2}= \left ( \frac{a}{b} -\sqrt{3}\right )^{2}$
$5= \left ( \frac{a}{b} \right )^{2}+\left ( \sqrt{3} \right )^{2}-2\left ( \frac{a}{b} \right )\left ( \sqrt{3} \right )$
(Using (a – b)2 = a2 + b2 = 2ab)
$\left ( \frac{a}{b} \right )^{2}+3-2\left ( \frac{a}{b} \right )\left ( \sqrt{3} \right )= 5$
$\frac{a^{2}}{b^{2}}+3-5= 2\sqrt{3}\left ( \frac{a}{b} \right )$
$\frac{a^{2}-2b^{2}}{b^{2}}= 2\sqrt{3}\left ( \frac{a}{b} \right )$
$\frac{b\left ( a^{2} -2b^{2}\right )}{2ab^{2}}= \sqrt{3}$
$\frac{a^{2}-2b^{2}}{2ab}= \sqrt{3}$
LHS is a rational number but RHS is an irrational number.
This is not possible, hence our assumption was wrong.
Hence $\sqrt{3}+\sqrt{5}$ is a irrational number.

Question:11

Show that 12n cannot end with the digit 0 or 5 for any natural number n.

We know that, if any number ends up with digit 0 or 5 then it must be divisible by 5.
Here 12n = (2 × 2 × 3)n
12n = (22 × 3)n
12n = (22n × 3n)
5 is not there, in the prime factorization form.
Hence 12n can’t end with the digit 0 or 5.

Question:12

On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can covers the same distance in complete steps?

Step-length of first person = 40 cm
Step-length of second person = 42 cm
Step-length of third person = 45 cm
For finding minimum distance which will be multiple of all step-lengths. we have to find the LCM of (40, 42, 45)
Prime factorization forms: 40 = 2 × 2 × 2 × 5 × 1
42 = 2 × 3 × 7 × 1
45 = 5 × 3 × 3 × 1
LCM = 2 × 3 × 5 × 2 × 2 × 7 × 3 = 2520
Hence minimum distance is 2520 cm.

Question:13

Write the denominator of the rational number 257/5000 in the form 2m × 5n, where m, n are non-negative integers. Hence, write its decimal expansion, without actual division.

Here, denominator = 5000
It can be written as:
5000 = 2 × 2 × 2 × 5 × 5 × 5 × 5
Hence the given fraction $\frac{257}{2^{3}\times 5^{4}}$ can be written as:
$\Rightarrow \frac{257}{\left ( 2^{3}\times 5^{4} \right )5^{1}}$
$\Rightarrow \frac{257}{1000\times 5^{1}}\times \frac{2^{1}}{2^{1}}$ (On dividing & multiply by 21)
$\Rightarrow \frac{514}{10000}$
$\Rightarrow 0\cdot 0514$
Hence, it is a terminating decimal.

Question:14

We will do it by method of contradiction: We will assume $\sqrt{p}+\sqrt{q}$ is a rational number. If it leads to some absurd outcome then it is wrong assumption.
Let $\sqrt{p}+\sqrt{q}$ is a rational number
$\sqrt{p}+\sqrt{q}= \frac{a}{b},b\neq a,a,b\epsilon z$
$\sqrt{q}= \frac{a}{b}-\sqrt{p}$
Squaring both sides
$\left ( \sqrt{q} \right )^{2}=\left ( \frac{a}{b}-\sqrt{p} \right )^{2}$
$q= \left ( \frac{a}{b} \right )^{2}+\left ( \sqrt{p} \right )^{2}-2\left ( \frac{a}{b} \right )\left ( \sqrt{p} \right )$
(Using (a – b)2 = a2 + b2 – 2ab)
$\frac{a^{2}}{b^{2}}+p-q= 2\sqrt{p}\frac{a}{b}$
$2\sqrt{p}\frac{a}{b}= \frac{a^{2}}{b^{2}}+p-q$
$\sqrt{p}= \frac{b\left ( a^{2}+b^{2}\left ( p-q \right ) \right )}{2ab^{2}}$
$\sqrt{p}= \frac{a^{2}+b^{2}\left ( p-q \right )}{2ab}$
RHS is a rational number but LHS is an irrational number.
This is not possible, hence our assumption was wrong.
Here $\sqrt{p}+\sqrt{q}$ has to be irrational.

Question:1

Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 6
r is remainder when we divide A by 5, therefore:
0 $\leq$ r $<$ 6, r = 0, 1, 2, 3,4,5
A = 6q + r
Here A = 6q + r
A3 = (6q + r)3 = 216q3 + r3 + 3.6q.r (6q + r)
[$\because$ (a + b)3 = a3 + b3 + 3ab (a + b)]
A3 = (216q3 + 108q2r + 18qr2) + r3 …(1)
Put r = 0, 1, 2, 3, 4, 5
Case 1:
A = 6q
A2 = 216q3
A3 =6 (36q3)
=6 (m )
Case 2:
A = 6q + 1
A3 = (216q3 + 108q2 + 18q) + 1
=6(36q3 + 18q2 + 3q) + 1
a3 = 6m + 1
{m = 36q3 + 18q2 + 3q}
Case 3:
A = 6q + 2
A3 = (216q3 + 216q3 + 72q) + 8
= (216q3 + 216q2 + 72q + 6) + 2
= 6m + 2
Case 4:
A = 6q + 3
A3=216q3 + 324q2 + 162q + 24 + 3
=6(36q3 + 54q2 + 27q + 4) + 3
= 6m + 3
{m = 36 q3 + 54q2 + 27q + 4}
Case 5:
A = 6q + 4
A3 = (216 q3 + 432 q2 + 288q) + 64
=6(36q3 + 72q2 + 48 q) + 60 + 4
=6(36q3 + 72q2 + 48q + 10) + 4
=6m + 4
{m = 36q3 + 72q2 + 48q + 10}
Case 6:
A = 6q + 5
A3 = (216 q3 + 540 q2 + 450 q) + 125
= 216 q3 + 540 q2 + 450 q + 120 + 5
= 6 (36q3 + 90q2 + 75 q + 20) + 5
= 6m + 5
(where m = 36 q3 + 90q2 + 75q + 20)
Hence, cube of a positive integer of the form 6q + r, q is a integer can be express in the form 6m + r where r = 0, 1, 2, 3, 4 and 5.

Question:2

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

Any positive integer can be written in form of 3m, 3m + 1 or 3m + 2
Case-1:
n = 3m
3m is divisible by 3
n + 2 = 3m + 2
Here when n + 2 is divided by 3 it leaves remainder 2
n + 4 = 3m + 4
= 3m + 3 + 1
= 3(m + 1) + 1
Here on dividing by 3, it leaves remainder 1
In this case n is divisible by 3 but (n + 2) and (n + 4) are not divisible by 3
Case-2 :
n = 3m + 1
on dividing by 3, it leaves remainder 1
n + 2 = 3m + 1 + 2
n + 2 = 3m + 3
= 3(m + 1)
on dividing by 3 it leaves remainder 0
Hence it is divisible by 3
x + 4 = 3m + 1 + 4
x + 4 = 3m + 3 + 2
= 3(m + 1) + 2
on dividing by 3, it leaves remainder 2
In this (n + 2) is divisible by 3 but n and (n + 4) are not divisible by 3
Hence through both the cases we conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.
Similarly, we can show for n = 3m + 2 form.

Question:3

Prove that one of any three consecutive positive integers must be divisible by 3.

Let three consecutive integers are n, n + 1, n + 2 when n is a natural number
i.e., n = 1, 2, 3, 4, ….
for n = 1, chosen numbers :(1, 2, 3) and 3 is divisible by 3
for n = 2, chosen numbers :(2, 3,4) and 3 is divisible by 3
for n = 3, chosen numbers :( 3, 4, 5) and 3 is divisible by 3
for n = 4, chosen numbers :(4, 5, 6) and 6 is divisible by 3
for n = 5, (chosen numbers :(5, 6, 7) and 6 is divisible by 3
Proof:
Case 1 : Let n is divisible by 3, it means n can be written as :
n = 3m,
Now, n +1 = 3m + 1 ; it gives remainder 1, when divided by 3
Now, n +2 = 3m + 2; it gives remainder 2, when divided by 3

Case 2: Let n + 1 is divisible by 3, it means n +1 can be written as :
n +1 = 3m,
Now, n = 3m – 1= 3(m – 1) + 2 ; it gives remainder 2, when divided by 3
Now, n +2 = 3m + 1; it gives remainder 1, when divided by 3
Case 3: Let n + 2 is divisible by 3, it means n +2 can be written as :
n +2 = 3m,
Now, n = 3m – 2= 3(m – 1) + 1 ; it gives remainder 1, when divided by 3
Now, n +1 = 3m – 1=3(m – 1) + 2 ; it gives remainder 2, when divided by 3

Question:4

For any positive integer n, prove that n3 – n is divisible by 6.

Let three consecutive positive integers be, x, x + 1 and x + 2.
Divisibility by 3: Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
.$\therefore$n = 3p or 3p + 1 or 3p + 2, where p is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 3.
$\Rightarrow$ x (x + 1) (x + 2) is divisible by 3.

Divisibility by 2: Whenever a number is divided 2, the remainder obtained is 0 or 1.
$\therefore$ n = 2q or 2q + 1, where q is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.$\Rightarrow$
x (x + 1) (x + 2) is divisible by 2.
Since, x (x + 1) (x + 2) is divisible by 2 and 3.
$\therefore$x (x + 1) (x + 2) is divisible by 6.
Now the given number is:
P = n3 – n
P = n (n2 – 1)
P = n (n – 1) (n + 1)
P = (n – 1) n (n + 1)
Therefore, P is the product of three consecutive numbers.
Now P can be written as:
P = x (x + 1) (x + 2) which is is divisible by 6.
Hence P is divisible by 6.
Hence proved

Question:5

Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

By Euclid’s division –
Any positive integer can be written as:
n = bm + r
Here b = 5
r is remainder when we divide n by 5, therefore:
0 $\leq$ r $<$ 5, r = 0, 1, 2, 3.4
n = 5m + r, therefore n can have values:
n = 5m, 5m + 1, 5m + 2, 5m + 3, 5m + 4.
Here m is natural number
Case 1 : Let n is divisible by 5, it means n can be written as :
n = 5m,
Now, n +4 = 5m + 4;it gives remainder 4, when divided by 5
Now, n + 8 = 5m + 8= 5(m +1) +3; it gives remainder 3, when divided by 5
Now, n + 12 = 5m + 12= 5(m +2) +2; it gives remainder 2, when divided by 5
Now, n + 16 = 5m + 16= 5(m +3) +1; it gives remainder 1, when divided by 5
Case 2: Let n + 4 is divisible by 5, it means n + 4 can be written as :
n + 4 = 5m,
Now, n = 5m – 4 = 5(m – 1) + 1 ; it gives remainder 1, when divided by 5
Now, n + 8 = 5m + 4; it gives remainder 4, when divided by 5
Now, n + 12 = 5m + 8 = 5(m +1) +3; it gives remainder 3, when divided by 5
Now, n + 16 = 5m + 12 = 5(m +2) +2; it gives remainder 2, when divided by 5
Similarly, we can show for other cases.

## NCERT Exemplar Solutions Class 10 Maths Chapter 1 Important Topics:

• In this chapter, the students will learn to write any specific type of number in terms of variables; such as the representation of even number, odd number et cetera.
• We will discuss a rational number and rational number and their properties.
• In this chapter, we will use the remainder theorem to prove many other theorems
• NCERT exemplar Class 10 Maths solutions chapter 1 explains the fundamental theorem of arithmetic.
• In this chapter, students will learn about Euclid’s division lemma method.

## NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

 Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry & Its Equations Chapter 9 Circles Chapter 10 Constructions Chapter 11 Areas related to Circles Chapter 12 Surface Areas and Volumes Chapter 13 Statistics and Probability

## Features of NCERT Exemplar Class 10 Maths Solutions Chapter 1:

• These Class 10 Maths NCERT exemplar chapter 1 solutions provide a basic knowledge of real numbers which has great importance in higher classes.

• The questions based on Real Numbers can be practiced in a better way along with these solutions.

• The NCERT exemplar Class 10 Maths chapter 1 solution Real Numbers has a good amount of problems for practice and are sufficient for a student to easily sail through other books such as NCERT Class 10 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths.

NCERT exemplar Class 10 Maths solutions chapter 1 pdf download are available through online tools for the students to access this content in an offline version so that no breaks in continuity are faced while practicing NCERT exemplar Class 10 Maths chapter 1.

### Check Chapter-Wise Solutions of Book Questions

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

### Also, Check NCERT Books and NCERT Syllabus here

1. What is the divisibility test by three?

If the sum of digits is divisible by three then the number will be divisible by three.

2. Is the chapter Real Numbers important for competitive examinations like JEE Advanced?

The chapter on Real Numbers is important for competitive examinations like JEE Advanced. A better understanding of the concepts will help the student solve complex problems in higher classes and competitive examinations.

3. What is the mark distribution of Real Numbers in the final boards examination?

The chapter on Real Numbers holds around 7-9% of the total marks of the paper. NCERT exemplar Class 10 Maths solutions chapter 1 can help the students learn the concepts and provide ample amount of practice questions.

4. Is the chapter Real Numbers important for Board examinations?

The chapter of Real Numbers is extremely important for Board examinations as it holds around 7-8% weightage of the whole paper.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

According to CBSE norms, a student must be 14 years old by the end of the year in which the exam will be held in order to sit for the 10th board exam. If your age is greater than 14, however, there are no limits. Therefore, based on your DOB, you will be 12 years and 6 months old by the end of December 2022. After the actual 16 June 2024, you'll be qualified to take your 10th board.

Hello aspirant,

Central Board Of Secondary Education(CBSE) is likely to declare class 10 and 12 terms 2 board result 2022 by July 15. The evaluation process is underway. Students are demanding good results and don't want to lack behind. They are requesting the board to use their best scores in Term 1 and Term 2 exams to prepare for the results.

CBSE concluded board exams 2022 for 10, and 12 on June 15 and May 24. Exams for both classes began on April 26. A total of 35 lakh students including 21 lakh class 10 students and 14 lakh class 12 students appeared in exams and are awaiting their results.

You can look for your results on websites- cbse.gov (//cbse.gov) .in, cbseresults.nic.in

Thank you

Sir, did you get any problem in result
I have also done this same mistake in board 2023 exam

Hello SIR CBSE board 9th class admission 2022 Kara Raha hu entrance ki problem hai please show the Entrance for Baal vidya mandir school sambhal

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

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 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

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Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

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##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

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Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
##### Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
##### Videographer

Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.

2 Jobs Available
##### SEO Analyst

An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available
##### Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available
##### Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available
##### Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available