# Class 10 real numbers extra questions

Given below are the important and extra questions for Chapter 1 class 10 maths real numbers
b. True and false questions
c. Multiple choice questions(MCQ)
e. Irrational number proof questions
f. Cross-word Puzzle

Question 1
Without actually performing division, state which of these number will terminating decimal expression or non terminating repeating decimal expression
1. $\frac {7}{25}$
2. $\frac {3}{7}$
3. $\frac {29}{343}$
4. $\frac {6}{15}$
5. $\frac {77}{210}$
6. $\frac {11}{67}$
7. $\frac {15}{27}$
8. $\frac {11}{6}$
9. $\frac {343445}{140}$

Those rational number which can be expressed in form x/2m X5n    are terminating expression and those can not be are non terminating decimal expression
Terminating decimal:  (a), (d)
Non terminating repeating decimal: (b), (c), (e), (f), (g).(h) ,(i)

Question 2
Using Euclid’s theorem to find the HCF between the following numbers
a. 867 and 225
b. 616 and 32

a.
Using Euclid theorem
$867=225 \times 3 +192$
$225=192 \times 1 +33$
$192=33 \times 5+ 27$
$33=27 \times 1+6$
$27=6 \times 4+3$
$6=3 \times 2+0$
So solution is 3
b. 8

Question 3
Write 10 rational number between
a. 4 and 5
b. 1/2 and 1/3
Question 4
Represent in rational form.
a. 1.232323….
b. 1.25
c. 3.67777777
Question 5
a. Prove that 2+√3 is a irrational number
b. Prove that 3√3 a irrational number

a. Let’s take this as rational number
$\frac{a}{b}=2+\sqrt{3}$
Or
$\frac{a-2b}{b}=\sqrt{3}$
Since a rational number can’t be equal to irrational number, our assumption is wrong
b. Let’s take this as rational number
q=3√3
q/3=√3
Since a rational number can’t be equal to irrational number, our assumption is wrong

## True or False statement

Question 6
Mark T/F as appropiate:
a. Number of the form $2n +1$ where n is any positive integer are always odd number
b. Product of two prime number is always equal to their LCM
c. $\sqrt {3} \times \sqrt {12}$ is a irrational number
d. Every integer is a rational number
e. The HCF of two prime number is always 1
f. There are infinite integers between two integers
g. There are finite rational number between 2 and 3
h. √3 Can be expressed in the form √3/1,so it is a rational number
i. The number 6n for n in natural number can end in digit zero
j. Any positive odd integer is of the form 6m+1 or 6m+3 or 6m +5 where q is some integer

1. True
2. True
3. False, as it is written as 6
4. True ,as any integer can be expressed in the form p/q
5. True
6. False,There are finite integer between two integers
7. False
8. False
9. False
10. True

## Multiple choice Questions

Question 7
the HCF (a, b) =2 and LCM (a, b) =27. What is the value $a \times b$
a. 25
b. 9
c. 27
d. 54

LCM X HCF=aXb

Question 8
2+√2 is a
a. Non terminating repeating
b. Terminating
c. Non terminating non repeating
d. None of these

Question 9
if a and b are co primes which of these is true
a. LCM (a, b) =aXb
b. HCF (a, b)= aXb
c. a=br
d. None of these

a

Question 10
A rational number can be expressed as terminating decimal when the factors of the denominator are
a. 2 or 5 only
b. 2 or 3 only
c. 3 or 5 only
d. 3 or 7 only

(a)

Question 11
if $x^2 =3 \;, \; y^2=9 \; ,\; z^3=27$, which of these is true
a. x is a irrational number
b. y is a rational number
c. z is rational number
d.All of the above

(d)

Question 12
Find the HCF and LCM of these by factorization technique
a.27,81
b. 120 ,144
c. 29029 ,580

(a)
$27= 3 \times 3 \times 3$
$81=3 \times 3 \times 3 \times 3$
HCF=27
LCM=81
b.
$120=2 \times 2 \times 3 \times 2 \times 5$
$144=2 \times 2 \times 3 \times 2 \times 2 \times 3$
$HCF=2^3 \times 3=24$
LCM=720
c.
$29029= 29 \times 13 \times 11 \times 7$
$580=29 \times 5 \times 4$
HCF=29
LCM=$29 \times 13 \times 11 \times 7 \times 4 \times 5=580580$

Question 13
Find all the positive integral values of p for which  $p^2 +16$ is a perfect square?

$p^2+16=q^2$
$(q-p)(q+p)=16$
So we have
Case 1
$q-p=8$ and $q+p=2$ which gives p=3
Case 2
$q-p=4$ and $q+p=4$ which gives p=0
Case 3
$q-p=2$ and $q+p=8$ which gives p=3
So the answer is 3 only

Question 14
Find the nature of the product $(\sqrt {2} - \sqrt {3}) ( \sqrt {3} + \sqrt {2})$ ?

$(\sqrt {2} - \sqrt {3}) ( \sqrt {3} + \sqrt {2})$
$=(\sqrt {2} - \sqrt {3}) ( \sqrt {2} + \sqrt {3})$
=2-3=-1
So, it is rational number

Question 15
Show that 4n can never end with the digit zero for any natural number n.

A number ending with zero must have 2 and 5 as factors like 10=2X5 550=2X5X5X11
Here 4n =(2X 2)n
So it does not have any 5
Therefore we can say that it will not with zero

Question 16
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

For a and b any two positive integer, we can always find unique integer q and r such that
a=bq + r    ,      0 ≤ r < b
Now on putting b=3  ,we get
a=3q+ r     , 0 ≤ r < 3
a=3q   , a3=27q3  a3 =9m  where m=3q3
a=3q+1  , a3=27q3 +27q2 +9q +1=9(3q3+3q+q) +1 =9m+1 where m=3q3+3q+q
a=3q+2  , a2=27q3 +54q2 +36q +8=9(3q3+6q+4q) +1 =9m+1 where m=3q3+6q+4q

## Irrational number proof questions

Question 17
Show that $3 + 5 \sqrt {2}$ is an irrational number. Is sum of two irrational numbers always an irrational number?

We can solve it using contradiction method.
Let $3 + 5 \sqrt {2}$ is a rational number, then
$3 + 5 \sqrt {2} = \frac {p}{q}$
$3q+ 5q\sqrt {2} = p$
$\sqrt {2} = \frac {p-3q}{5q}$
Clearly,RHS is a rational number ,but LHS is not.
$3 + 5 \sqrt {2}$ is an irrational number

sum of two irrational numbers is not always an irrational number
Example
$3 + 5 \sqrt {2}$ and $3 - 5 \sqrt {2}$ are irrational number, but the sum is rational number
$3 + 5 \sqrt {2} + 3 - 5 \sqrt {2}= 6$

Question 18
Prove that $\sqrt {3}$ is an irrational number and hence show that $2\sqrt {3}$ is also an irrational number.

Since we cannot clearly express in p/q form,it is difficult to say,So lets us assume this is rational number
then
$\sqrt {3}=\frac {p}{q}$

where p and q are co primes.
or
$q\sqrt {3}=p$
squaring both sides
$3q^2=p^2$

So 3 divides p2,from theorem we know that,
2 will divide p also. p=3c

$3q^2=9c^2$
or
$q^2=3c^2$
So q divided by 3 also

So both p and q are divided by 3 which is contradiction from we assumed
So $\sqrt {3}$ is irrational number

Question 19
Prove that $5 - \sqrt {3}$ is an irrational number.

let $5 - \sqrt {3}$ be rational number
then,
$5 - \sqrt {3}=\frac {p}{q}$
$\sqrt {3}= \frac {5q -p}{q}$
Clearly RHS is rational number but LHS is not,
So our assumption is wrong.
$5 - \sqrt {3}$ is an irrational number

Question 20
Prove that $2 \sqrt {5}$ is an irrational number.
Question 21
Show that $(\sqrt {3} + \sqrt {5})^2$ is an irrational number.

Let $(\sqrt {3} + \sqrt {5})^2= 8+ 2 \sqrt {15}$ is a rational number
Then
$8+ 2 \sqrt {15}=\frac {p}{q}$
$\sqrt {15}= \frac {p -8q}{2q}$
Clearly RHS is rational number but LHS is not,
So our assumption is wrong.
$(\sqrt {3} + \sqrt {5})^2$ is an irrational number.

Question 22
Prove that $4 - \sqrt {5}$ is an irrational number.
Question 23
Prove that $\sqrt {5}$ is an irrational number.

We will try to prove it contradiction method
Let $\sqrt {5}$ be rational
then it must in the form of p/q [q is not equal to 0][p and q are co-prime]
$\sqrt {5}=\frac {p}{q}$
$\sqrt {5}$ \times q = p$squaring on both sides$5q^2 = p^2$So, p2 is divisible by 5 Now from the theorem, p is divisible by 5 let$p = 5c$[c is a positive integer] [squaring on both sides ]$p^2 = 25c^2$So we can obtain below from both the above equation$5q^2 = 25c^2q^2 = 5c^2$So q is divisible by 5 thus q and p have a common factor 5 there is a contradiction so$\sqrt {5}$is an irrational number Question 24 Prove that √2 + 1/√2 is an irrational number Question 25 Prove that for any positive integer n, n3 – n is divisible by 6. Answer n3 - n =n(n2 -1) = n(n-1)(n+1) Any Number can be represented in the form 3q,3q+1 and 3q+2 for n=3q, n(n-1)(n+1)= 3q(3q-1)(3q+1) Divisible by 3 for n=3q+1,n(n-1)(n+1)= (3q+1)(3q)(3q+2) Divisible by 3 for n=3q+2,n(n-1)(n+1)= (3q+2)(3q+1)(3q+3)=3(3q+2)(3q+1)(q+1) Divisible by 3 So product n(n-1)(n+1) is divisible by 3 Similarly Any Number can be represented in the form 2m,2m+1 for n=2m, n(n-1)(n+1)= 2m(2m-1)(2m+1) Divisible by 2 for n=2m+1, n(n-1)(n+1)= (2m+1)(2m)(2m+1) Divisible by 2 So this is divisible by both 2 and 3 =6 Question 26 If n is rational and √m is irrational, then prove that (n + √m) is irrational. Question 27 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 Answer n,n+4,n+8,n+12,n+16 be integers. where n can take the form 5q, 5q+1 ,5q+2 , 5q + 3 , 5q + 4. Case I when n=5q Then n is divisible by 5. but neither of 5q+4 ,5q+8 , 5q + 12 , 5q + 16 is divisible by 5. Case II when n=5q+1 Then n is not divisible by 5. n+4 = 5q+1+4 = 5q+5=5(q +1), which is divisible by 5. but neither of 5q+1 ,5q+9 , 5q + 13 , 5q + 17 is divisible by 5. Case III when n=5q+2 Then n is not divisible by 5. n+8 = 5q+2+8 =5q+10=5(q+2), which is divisible by 5. but neither of 5q+2 ,5q+6 , 5q + 14 , 5q + 18 is divisible by 5. Case IV when n=5q+3 Then n is not divisible by 5. n+12 = 5q+3+12 =5q+15=5(q+3), which is divisible by 5. but neither of 5q+3 ,5q+7 , 5q + 11 , 5q + 19 is divisible by 5. Case V when n=5q+4 Then n is not divisible by 5. n+16 = 5q+4+16 =5q+20=5(q+4), which is divisible by 5. but neither of 5q+4 ,5q+8 , 5q + 12 , 5q + 16 is divisible by 5. Hence, one of n, n+4,n+8,n +12 and n+16 is divisible by 5. Question 28 Prove that √11 is irrational. Question 29 Show that 3√2 is irrational. Question 30 Prove that the sum of a rational number and an irrational number is always irrational. Answer We will try to prove it contradiction method Let z be the irrational number, p/q is the rational number. We assume the sum is a rational number a/b Then z+ p/q = a/b z= a/b - p/q = (aq-bp)/bq Now aq,bp and bq are all integers as a,b,p,q are integers.So (aq-bp)/bq is a rational number but x is irrational number.So our assumption is wrong The sum is a irrational number Question 31 The product of a non-zero rational and an irrational number is (A) always irrational (B) always rational (C) rational or irrational (D) one Answer The product is always irrational We can prove it like below with contradiction method Let y be irrational number and a/b is rational and Product is rational p/q then ya/b = p/q y= bp/aq bp,aq are integers,so ratio is rational but y is irrational so our assumption is wrong. Product is always irrational Question 32 Prove that √p + √q is irrational, where p, q are primes. Question 33 Prove that one of any three consecutive positive integers must be divisible by 3. Answer Let three consecutive positive integers are n,n+1,n+2 where n is any positive integer Any Number can be represented in the form 3q,3q+1 and 3q+2 for n=3q, n=3q So it Divisible by 3 for n=3q+1,n+2=3q+3 Divisible by 3 for n=3q+2,n+1=3q+3 Divisible by 3 So one of any three consecutive positive integers must be divisible by 3. ## Cross-word Puzzle to check your Real number knowledge Across 2. Number which are not divisible by any other number except 1 6. decimal expression can be expressed in the form 1/2m5n Down 1. Number which can be written as product of prime 3. In Euclid division lemma a=bq + r , it is the value r 4. HCF can be found using this division algorithm 5. In Euclid division lemma a=bq + r , it is the value b 7. Numbers of the forms p/q Answer (1) Composite (2) prime (3) remainder (4) Euclid (5) quotient (6) Terminating (7) Rational ## Summary This class 10 real numbers extra questions with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail. Also Read Go back to Class 10 Main Page using below links ### Practice Question Question 1 What is$1 - \sqrt {3}\$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20