10th Class Real Numbers (6marks/80)
Topics: 1. Euclid's Division Lemma/Algorithm 2. Fundamental Theorem of Arithmetic 3. Irrational Numbers 4. Decimal expression of Rational Number
10th Real Number CBSE Test Papers

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Q. If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number.
Solution:
We know that product of two rational number is always a rational number.
Hence if a is a rational number then
a2 = a x a is a rational number,
a3 = a2 x a is a rational number,
a4 = a3 x a is a rational number,
...
...
an = an1 x a is a rational number.
Q. Write the HCF and LCM of the smallest odd composite number and the smallest odd prime number. If an odd number p divides q2 then will it divide q3 also. Explain
Solution: An odd number p divides q2
=> p is one of the factor of q2 (1)
=>q3 = q2 x q
=>q is one of the factor of q2 (ii)
from (i) and (ii), q2 and q3 has common factor p
Thus p divide q3 also.
Q. Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4
Solution:
Let x=2n+1 and y=2m+1 (Where n and m are positive integers)
Therefore, x 2+y 2 = (2n+1) 2+ (2m+1) 2 =4n 2+4n+1+4m 2+4m+1 =4(n 2+m 2+n+m) +2
Thus, x2 + y2 is even but not divisible by 4
Q. If LCM (p,q,r) = 420, HCF ( p,q,r) = 1, HCF (p,q) = 3 , HCF ( q,r) = 5 and HCF (p,r) = 1,
then find the the product of p,q and r
Solution: We know that : HCF (p, q, r) = [p. q. r. LCM (p, q, r)] / [ LCM(p ,q). LCM(q , r). LCM(p ,r )]
and LCM (p, q, r) = [p. q. r. HCF (p, q, r)] / [ HCF(p ,q). HCF(q , r). HCF(p ,r )]
420 = [pqr x1]/[3x5x1]
pqr = 420 /( 3x5x1) = 28
Q.If d is the HCF of 45 and 27, find x and y satisfying d= 27 x +45 y.
Solution: a = bq + r here a = 45 and b = 27
45 = 27 x 1 + 18  (i)
=> 27 = 18 x 1 + 9  (ii)
=> 18 = 9 x 2 + 0  (iii)
So, HCF is 9
From (ii) 27  18 x 1 = 9
From (i) 27  ( 45  27 x 1) x 1 = 9
27 x 1  45 x 1 + 27 x 1 = 9
27 x 2  45 x 1 = 9
27 x +45 y = 9
x = 2 and y = 1
Q. Prove that one of every three consecutive positive integers is divisible by 3
Solution: Let three consecutive positive integers are n , n+ and n+2 for some integer n
Now, Number divisible 3 are of form 3q , 3q +1 and 3q + 2
For n = 3q
n = 3p, then n is divisible by 3.
n + 1 = 3q + 1 not divisible by 3
and n + 2 = 3q + 2 not divisible by 3
For n = 3q+1 not divisible by 3
n + 1 = 3q +1 + 1 = 3q+2 not divisible by 3
n + 2 = 3q + 1 + 2 = 3(q+1), which is divisible by 3.
For n = 3q+2 not divisible by 3
n + 1 = 3q +2 + 1 = 3(q+1) which is divisible by 3
n + 2 = 3q + 2 + 2 = (3q+3) +1) = 3(q+1) + 1, which is not divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
Q. Is it possible for the hcf and lcm of a number to be 18 and 378 CBSE SA1{2015)
Solution We know that HCF of given numbers must be a factor of their LCM.
378 = 18 x 21 + 0
R = 0
So, HCF of given numbers is a factor of their LCM.
Thus, it possible for the HCF and lcm of a number to be 18 and 378
Q. Show that the square of an odd positive integer is of the form 8m + 1, for some whole number m.
Solution: Let positive odd integer is of the form 2q + 1
Therefore, (2q + 1)2 = 4q2 + 4q + 1 = 4 q (q + 1) + 1 (i)
q and (q + 1) are two consecutive so, their product must be even.
q x (q + 1) = 2m for any integer m it is 2m, where m is a whole number.
Now, (2q + 1)2 = 4 x 2 m + 1 = 8 m + 1. [From (1)]
Q. Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.
[2014]
Solution :
We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.
Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5
Thus we have:
(6 m +1)2 = 36 m2 + 12 m + 1 = 6 (6 m2 + 2 m) + 1 = 6 q + 1, q is an integer
(6 m + 3)2 = 36 m2 + 36 m + 9 = 6 (6 m2 + 6 m + 1) + 3 = 6 q + 3, q is an integer
(6 m + 5)2 = 36 m2 + 60 m + 25 = 6 (6 m2 + 10 m + 4) + 1 = 6 q + 1, q is an integer.
Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.
Q. Prove that if n is odd then n2 – 1 is divisible by 8.
Solution : An odd integer n is either a 4k+1 or a 4k+3.
(4k+1)2 – 1 = 16k2+ 8k + 1  1 =8(2k2 +k), which is a multiple of 8.
(4k+3)2 – 1 = 16k2+ 24k + 9  1 =8(2k2 +3k + 1), which is a multiple of 8.
Q.Show that 14^n cannot end with digit 0 or 5 for any natural number n. [2014]
Solution:
14^n can end with digit 0 or 5 for any natural number n if 14^n = 2^x x 5 ^y
But 12^n = (2^n x 7^n)
Thus ,14^n cannot end with digit 0 or 5 for any natural number n.
Q. Find LCM of the numbers given below : m, 2m, 3m, 4m and 5m, where m is any positive integer. 2014
Solution: m x 4 x 3 x 5 = 60 m
Q. Determine the values of p and q so that the prime factorisation of 2520 is expressible as 2^3x 3^P x q x 7 [2014]
Solution: 2520 = 2^3 x 3^2 x 5 x 7 = 2^3x 3^P x q x 7
So, p = 2 , q = 5
Q.Pens are sold in pack of 8 and notepads are sold in pack of 12. Find the least number of pack of each type that one should buy so that there are equal number of pen and notepads 2014
Solution: LCM of 8 and 12 is 24
the least number of pack the pack of pen = 22/8 = 3
the least number of pack the pack of note book = pen = 24/12 = 2
Q. If two positive integers x and y are expressible in terms of primes as x =p^2 q^3 and y = p^3 q, what can you say about their LCM and HCF. Is LCM a multiple of HCF ? Explain.
Solution : LCM of x and y = p^3 q^3 and HCF of x and y = p^2 q
Here LCM = pq^2 x HCF So, LCM a multiple of HCF or, HCF of given numbers must be a factor of their LCM.
Q. Find the least positive integer which on adding 1 is exactly divisible by 126 and 600 {2014}
Solution: L CM of 126 and 600 = 2 x 3 x21 x 100 = 12600
the least positive integer which on adding 1 is exactly divisible by 126 and 600 = 12600 – 1 = 12599
Q.A boy with collection of marbles realizes that if he makes a group of 5 or 6 marbles at time there are always two marbles left. Can you explain why the boy can’t have prime numbers of merles?
Solution: Given that, He makes a group of 5 or 6 marbles at time there are always two marbles left.
According to Euclid’s Division lemma : a = bq + r
Numbers of marble = 5 m + 2 or 6 n + 2.
As remainder is 2, when it is divided by 5 or 6.
Thus, number of marbles must be = (the multiple of 5 × 6) + 2
=> 30 m + 2 = 2 (15 m + 1) = even number .
So , it cannot be a prime number
Q. Prove that root 5 is an irrational number.Hence, show that 3 + 2 root 5 is also an irrational number [2014]
Q. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.[2014]
Q.Find the value of (1)^n + (1)^2n + (1)^(2n +1) + (1)^(4n +1) , where n is positive odd integer [2014]
Solution:
We know that product of two rational number is always a rational number.
Hence if a is a rational number then
a2 = a x a is a rational number,
a3 = a2 x a is a rational number,
a4 = a3 x a is a rational number,
...
...
an = an1 x a is a rational number.
Q. Write the HCF and LCM of the smallest odd composite number and the smallest odd prime number. If an odd number p divides q2 then will it divide q3 also. Explain
Solution: An odd number p divides q2
=> p is one of the factor of q2 (1)
=>q3 = q2 x q
=>q is one of the factor of q2 (ii)
from (i) and (ii), q2 and q3 has common factor p
Thus p divide q3 also.
Q. Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4
Solution:
Let x=2n+1 and y=2m+1 (Where n and m are positive integers)
Therefore, x 2+y 2 = (2n+1) 2+ (2m+1) 2 =4n 2+4n+1+4m 2+4m+1 =4(n 2+m 2+n+m) +2
Thus, x2 + y2 is even but not divisible by 4
Q. If LCM (p,q,r) = 420, HCF ( p,q,r) = 1, HCF (p,q) = 3 , HCF ( q,r) = 5 and HCF (p,r) = 1,
then find the the product of p,q and r
Solution: We know that : HCF (p, q, r) = [p. q. r. LCM (p, q, r)] / [ LCM(p ,q). LCM(q , r). LCM(p ,r )]
and LCM (p, q, r) = [p. q. r. HCF (p, q, r)] / [ HCF(p ,q). HCF(q , r). HCF(p ,r )]
420 = [pqr x1]/[3x5x1]
pqr = 420 /( 3x5x1) = 28
Q.If d is the HCF of 45 and 27, find x and y satisfying d= 27 x +45 y.
Solution: a = bq + r here a = 45 and b = 27
45 = 27 x 1 + 18  (i)
=> 27 = 18 x 1 + 9  (ii)
=> 18 = 9 x 2 + 0  (iii)
So, HCF is 9
From (ii) 27  18 x 1 = 9
From (i) 27  ( 45  27 x 1) x 1 = 9
27 x 1  45 x 1 + 27 x 1 = 9
27 x 2  45 x 1 = 9
27 x +45 y = 9
x = 2 and y = 1
Q. Prove that one of every three consecutive positive integers is divisible by 3
Solution: Let three consecutive positive integers are n , n+ and n+2 for some integer n
Now, Number divisible 3 are of form 3q , 3q +1 and 3q + 2
For n = 3q
n = 3p, then n is divisible by 3.
n + 1 = 3q + 1 not divisible by 3
and n + 2 = 3q + 2 not divisible by 3
For n = 3q+1 not divisible by 3
n + 1 = 3q +1 + 1 = 3q+2 not divisible by 3
n + 2 = 3q + 1 + 2 = 3(q+1), which is divisible by 3.
For n = 3q+2 not divisible by 3
n + 1 = 3q +2 + 1 = 3(q+1) which is divisible by 3
n + 2 = 3q + 2 + 2 = (3q+3) +1) = 3(q+1) + 1, which is not divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
Q. Is it possible for the hcf and lcm of a number to be 18 and 378 CBSE SA1{2015)
Solution We know that HCF of given numbers must be a factor of their LCM.
378 = 18 x 21 + 0
R = 0
So, HCF of given numbers is a factor of their LCM.
Thus, it possible for the HCF and lcm of a number to be 18 and 378
Q. Show that the square of an odd positive integer is of the form 8m + 1, for some whole number m.
Solution: Let positive odd integer is of the form 2q + 1
Therefore, (2q + 1)2 = 4q2 + 4q + 1 = 4 q (q + 1) + 1 (i)
q and (q + 1) are two consecutive so, their product must be even.
q x (q + 1) = 2m for any integer m it is 2m, where m is a whole number.
Now, (2q + 1)2 = 4 x 2 m + 1 = 8 m + 1. [From (1)]
Q. Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.
[2014]
Solution :
We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.
Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5
Thus we have:
(6 m +1)2 = 36 m2 + 12 m + 1 = 6 (6 m2 + 2 m) + 1 = 6 q + 1, q is an integer
(6 m + 3)2 = 36 m2 + 36 m + 9 = 6 (6 m2 + 6 m + 1) + 3 = 6 q + 3, q is an integer
(6 m + 5)2 = 36 m2 + 60 m + 25 = 6 (6 m2 + 10 m + 4) + 1 = 6 q + 1, q is an integer.
Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.
Q. Prove that if n is odd then n2 – 1 is divisible by 8.
Solution : An odd integer n is either a 4k+1 or a 4k+3.
(4k+1)2 – 1 = 16k2+ 8k + 1  1 =8(2k2 +k), which is a multiple of 8.
(4k+3)2 – 1 = 16k2+ 24k + 9  1 =8(2k2 +3k + 1), which is a multiple of 8.
Q.Show that 14^n cannot end with digit 0 or 5 for any natural number n. [2014]
Solution:
14^n can end with digit 0 or 5 for any natural number n if 14^n = 2^x x 5 ^y
But 12^n = (2^n x 7^n)
Thus ,14^n cannot end with digit 0 or 5 for any natural number n.
Q. Find LCM of the numbers given below : m, 2m, 3m, 4m and 5m, where m is any positive integer. 2014
Solution: m x 4 x 3 x 5 = 60 m
Q. Determine the values of p and q so that the prime factorisation of 2520 is expressible as 2^3x 3^P x q x 7 [2014]
Solution: 2520 = 2^3 x 3^2 x 5 x 7 = 2^3x 3^P x q x 7
So, p = 2 , q = 5
Q.Pens are sold in pack of 8 and notepads are sold in pack of 12. Find the least number of pack of each type that one should buy so that there are equal number of pen and notepads 2014
Solution: LCM of 8 and 12 is 24
the least number of pack the pack of pen = 22/8 = 3
the least number of pack the pack of note book = pen = 24/12 = 2
Q. If two positive integers x and y are expressible in terms of primes as x =p^2 q^3 and y = p^3 q, what can you say about their LCM and HCF. Is LCM a multiple of HCF ? Explain.
Solution : LCM of x and y = p^3 q^3 and HCF of x and y = p^2 q
Here LCM = pq^2 x HCF So, LCM a multiple of HCF or, HCF of given numbers must be a factor of their LCM.
Q. Find the least positive integer which on adding 1 is exactly divisible by 126 and 600 {2014}
Solution: L CM of 126 and 600 = 2 x 3 x21 x 100 = 12600
the least positive integer which on adding 1 is exactly divisible by 126 and 600 = 12600 – 1 = 12599
Q.A boy with collection of marbles realizes that if he makes a group of 5 or 6 marbles at time there are always two marbles left. Can you explain why the boy can’t have prime numbers of merles?
Solution: Given that, He makes a group of 5 or 6 marbles at time there are always two marbles left.
According to Euclid’s Division lemma : a = bq + r
Numbers of marble = 5 m + 2 or 6 n + 2.
As remainder is 2, when it is divided by 5 or 6.
Thus, number of marbles must be = (the multiple of 5 × 6) + 2
=> 30 m + 2 = 2 (15 m + 1) = even number .
So , it cannot be a prime number
Q. Prove that root 5 is an irrational number.Hence, show that 3 + 2 root 5 is also an irrational number [2014]
Q. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.[2014]
Q.Find the value of (1)^n + (1)^2n + (1)^(2n +1) + (1)^(4n +1) , where n is positive odd integer [2014]
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