Important Questions for Real Numbers Class 10 Maths
Given below are the Class 10 Maths Important Questions for Real Numbers
a. Concepts questions
b. Calculation problems
c. Multiple choice questions(MCQ)
d. Long answer questions
e. Irrational number proof questions
f. True and false questions
Long answer questions
Question 1. Without actually performing division, state which of these number will terminating decimal expression or non terminating repeating decimal expression
Those rational number which can be expressed in form x/2^{m} X5^{n} 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 Solution
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 Solution
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
Question 6True or False statement
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 6^{n} 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 Solution
True
True
False, as it is written as 6
True ,as any integer can be expressed in the form p/q
True
False,There are finite integer between two integers
False
False
False
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 Solution
Answer is (d)
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 Solution
Answer is (c)
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 Solution
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 Solution
(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 Solution
(d)
Short answer question
Question 12 Find the HCF and LCM of these by factorization technique
a.27,81
b. 120 ,144
c. 29029 ,580 Solution
Question 13. Find all the positive integral values of p for which $p^2 +16$ is a perfect square? Solution
$p^2+16=q^2$
$(q-p)(q+p)=16$
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 $(\sqrt {2} - \sqrt {3}) ( \sqrt {3} + \sqrt {2})$ ? Solution
$(\sqrt {2} - \sqrt {3}) ( \sqrt {3} + \sqrt {2})$
$=(\sqrt {2} - \sqrt {3}) ( \sqrt {2} + \sqrt {3})$
=2-3=-1
So, it is rational number
Question 15 Prove that the sum of a rational number and an irrational number is always irrational. Solution
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 16 Prove that $\sqrt {5}$is an irrational number. Solution
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, p^{2} 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
$5q^2 = 25c^2$
$q^2 = 5c^2$
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 17 Show that $3 + 5 \sqrt {2} $ is an irrational number. Is sum of two irrational numbers always an irrational number? Solution
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.
So,our contradiction is wrong.
$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. Solution
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 p^{2},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. Solution
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. Solution
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 Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. Solution
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 , a^{3}=27q^{3} a^{3} =9m where m=3q^{3}
a=3q+1 , a^{3}=27q^{3} +27q^{2} +9q +1=9(3q^{3}+3q+q) +1 =9m+1 where m=3q^{3}+3q+q
a=3q+2 , a^{2}=27q^{3} +54q^{2} +36q +8=9(3q^{3}+6q+4q) +1 =9m+1 where m=3q^{3}+6q+4q
Question 24 Prove that √2 + 1/√2 is an irrational number Question 25 Prove that for any positive integer n, n^{3} – n is divisible by 6. Solution
n^{3} - n =n(n^{2} -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 divisble 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 27Show 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 Solution
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 Show that 4^{n} can never end with the digit zero for any natural number n. Solution
A number ending with zero must have 2 and 5 as factors like 10=2X5 550=2X5X5X11
Here 4^{n} =(2X 2)^{n}
So it does not have any 5
Therefore we can say that it will not with zero
Question 31The product of a non-zero rational and an irrational number is
(A) always irrational
(B) always rational
(C) rational or irrational
(D) one Solution
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. Solution
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.
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