Chapter 1 - Real Numbers, RD Sharma Solutions - (Part - 9) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 1.58

Q.11. The HCF of 95 and 152, is
(a) 57
(b) 1
(c) 19
(d) 38
Ans. (c)
Solution. Using the factor tree for 95, we have: 
Using the factor tree for 152, we have: 
Therefore,
95 = 5 x 19
152 = 2³ x 19
HCF (95, 152) = 19

Q.12. If HCF (26, 169) = 13, then LCM (26, 169) =
(a) 26
(b) 2
(c) 3
(d) 4
Ans. (c)
Solution. HCF (26, 169) = 13 
We have to find the value for LCM (26, 169)
We know that the product of numbers is equal to the product of their HCF and LCM.
Therefore,
13(LCM) = 26(169)
LCM = 26(169)/13
LCM = 338

Q.13. If a = 2³ ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3ⁿ ✕ 5 and LCM (a, b, c) = 2³ ✕ 3² ✕ 5, then n =
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (b)
Solution. LCM (a, b, c) = 2³ ✕ 3² ✕ 5 ……(I)
We have to find the value for n
Also
a =  2³ ✕ 3
b = 2 x 3 x 5
c = 3ⁿ x 5
We know that the while evaluating LCM, we take greater exponent of the prime numbers in the factorization of the number.
Therefore, by applying this rule and taking n ≥ 1 we get the LCM as
LCM (a, b, c) = 2³ ✕ 3ⁿ ✕ 5 ……(II)
On comparing (I) and (II) sides, we get:
2³ ✕ 3² ✕ 5 = 2³ ✕ 3ⁿ ✕ 5
n = 2

Q.14. The decimal expansion of the rational number 14587/1250 will terminate after 
(a) One decimal place
(b) Two decimal place
(c) Three decimal place
(d) Four decimal place
Ans. We have,
14587/1250 = 14587 / 2¹ x 5⁴
Theorem states:
Let x = p / q be a rational number, such that the prime factorization of q is of the form 2^m x 5ⁿ, where m and n are non-negative integers.
Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.
This is given that the prime factorization of the denominator is of the form 2^m x 5ⁿ.
Hence, it has terminating decimal expansion which terminates after 4 places of decimal.

Q.15. If p and q are co-prime numbers, then p² and q² are
(a) Co-prime
(b) Not co-prime
(c) Even
(d) Odd
Ans. (a)
Solution. We know that the co-prime numbers have no factor in common, or, their HCF is 1.
Thus, p² and q² have the same factors with twice of the exponents of p and q respectively, which again will not have any common factor.
Thus we can conclude that p² and q² are co-prime numbers.

Q.16. Which of the following rational numbers have terminating decimal?
(i) 16/225
(ii) 5/18
(iii) 2/21
(iv) 7/250
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (i) and (iv)
Ans. (d)
Solution. (i) We have,
16/225 = 16/3² x 5²
Theorem states:
Let x = p/q be a rational number, such that the prime factorization of q is not of the form 2^m x 5ⁿ, where m and n are non-negative integers.
Then, x has a decimal expression which does not have terminating decimal.
(ii) We have,
5/18 = 5/2 x 3²
Theorem states:
Let x = p/q be a rational number, such that the prime factorization of q is not of the form 2^m x 5ⁿ, where m and n are non-negative integers.
Then, x has a decimal expression which does not have terminating decimal.
(iii) We have,
2/21 = 2/7 x 3
Theorem states:
Let x = p/q be a rational number, such that the prime factorization of q is not of the form 2^m x 5ⁿ, where m and n are non-negative integers.
Then, x has a decimal expression which does not have terminating decimal.
(iv) We have,
7/250 = 7/2¹ x 5³
Theorem states:
Let x = p/q be a rational number, such that the prime factorization of q is of the form 2^m x 5ⁿ, where m and n are non-negative integers.
Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.
Then, x has a decimal expression which will have terminating decimal after 3 places of decimal.
Hence the (iv) option will have terminating decimal expansion.
There is no correct option.

Q.17. If 3 is the least prime factor of number a and 7 is the least prime factor of number b, then the least prime factor of a + b, is 
(a) 2
(b) 3
(c) 5
(d) 10
Ans. (a)
Solution. Since 7 + 3 = 10
The least prime factor of a + b has to be 2; unless a + b is a prime number greater than 2.
Suppose a + b is a prime number greater than 2. Then a + b must be an odd number
o one of a or b must be an even number.
Suppose then that a is even. Then the least prime factor of a is 2; which is not 3 or 7. So a can not be an even number nor can b be an even number. Hence a + b cannot be a prime number greater than 2 if the least prime factor of a is 3 and b is 7.
Thus the answer is 2.

Q.18. is 
(a) An integer
(b) A rational number
(c) A natural number
(d) An irrational number
Ans. (b)
Solution. We have,
= 3.27272727...
Let
x = 3.27272727...
Then,
100x = 327.272727...
10000x = 32727.2727...
Subtract these to get
9900x = 32400
x = 32400/9900
x = 32400/9900
x = 324/99
Thus, we can also conclude that all infinite repeating decimals are rational numbers.

Page No 1.59

Q.19. The smallest number by which √27 should be multiplied so as to get a rational number is 
(a) √27
(b) √33
(c) √3
(d) 3
Ans. (c)
Solution. 
= 3√3
Out of the given choices √3 is the only smallest number by which if we multiply √27 we get a rational number.

Q.20. The smallest rational number by which 1/3 should be multiplied so that its decimal expansion terminates after one place of decimal, is
(a) 3/10
(b) 1/10
(c) 3
(d) 3/100
Ans. (a)
Solution.
For terminating the decimal expansion after one place of decimal, the highest power of m and n in 2^m x 5ⁿ should be 1.
Let m = n = 1
We will get:
1/3 x 3/10 = 1/10
= 0.1
Thus, it is evident that we multiplied it by 3/10.

Q.21. If n is a natural number, then 9²ⁿ − 4²ⁿ is always divisible by
(a) 5
(b) 13
(c) Both 5 and 13
(d) None of these
[Hint : 9²ⁿ − 4²ⁿ is of the form a²ⁿ − b²ⁿ which is divisible by both a − b and a + b. So, 9²ⁿ − 4²ⁿ is divisible by both 9 − 4 = 5 and 9 + 4 = 13.]
Ans. (c)
Solution. We know that a²ⁿ − b²ⁿ is always divisible by both a - b and a + b.
So, 9²ⁿ − 4²ⁿ is always divisible by both 9 - 4 = 5 and 9 + 5 = 13.

Q.22. If n is any natural number, then 6ⁿ − 5ⁿ always ends with 
(a) 1
(b) 3
(c) 5
(d) 7
[Hint: For any n ∈ N, 6ⁿ and 5ⁿ end with 6 and 5 respectively. Therefore, 6ⁿ − 5ⁿ always ends with 6 − 5 = 1.]
Ans. (a)
Solution. We know that 6ⁿ will end in 6
And 5ⁿ will end in 5.
Now, 6ⁿ - 5ⁿ always end with 6 - 5 = 1.

Q.23. The LCM and HCF of two rational numbers are equal, then the numbers must be 
(a) Prime
(b) Co-prime
(c) Composite
(d) Equal
Ans. (d)
Solution. Let the two numbers be a and b.
(a) If we assume that the a and b are prime.
Then,
HCF (a, b) = 1
LCM (a, b) = ab
(b) If we assume that a and b are co-prime.
Then,
(a, b) = 1
LCM (a, b) = ab
(c) If we assume that a and b are composite.
Then,
HCF (a, b) = 1 or any other highest common integer
LCM (a, b) = ab
(d) If we assume that a and b are equal and consider a=b=k.
Then,
HCF (a, b) = k
LCM (a, b) = k
Hence the correct choice is (d).

Q.24. If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is 
(a) 203400
(b) 194400
(c) 198400
(d) 205400
Ans. (b)
Solution. Let the HCF be x and the LCM of the two numbers be y.
It is given that the sum of the HCF and LCM is 1260
x + y = 1260 …… (i)
And, LCM is 900 more than HCF.
y = x + 900 …… (ii)
Substituting (ii) in (i), we get:
x + x + 900 = 1260
2x + 900 = 1260
2x = 1260 - 900
2x = 360
x = 180
Substituting x = 180 in (ii), we get:
y = 180 + 900
y = 1080
We also know that the product the two numbers is equal to the product of their LCM and HCF
Thus the product of the numbers
= 1080(180)
= 194400

Q.25. The remainder when the square of any prime number greater than 3 is divided by 6, is 
(a) 1
(b) 3
(c) 2
(d) 4
[Hint: Any prime number greater than 3 is of the from 6k ± 1, where k is a natural number and (6k ± 1)² = 36k² ± 12k + 1 = 6k(6k ± 2) + 1]
Ans. (a)
Solution. Any prime number greater than 3 is of the form 6k ± 1, where k is a natural number.
Thus,
(6k ± 1)² = 36k² ± 12k + 1
6k(6k ± 2) + 1
When, 6k(6k ± 2) + 1 is divided by 6, we get, k(6k ± 2) + 1 and remainder as 1.

Q.26. For some integer m, every integer is of the form 
(a) m
(b) m + 1
(c) 2m
(d) 2m+ 1
Ans. (c)
Solution. It is known that, even integers are ...,−4, −2, 0, 2, 4,...
Observe that, all of the even numbers are the multiple of 2.
So, even numbers can be written as 2m, where, m is an integer.
m can be ...,−2, −1, 0, 1, 2,... ∴ 2m can be ...,−4, −2, 0, 2, 4,...
Hence, the correct answer is option C.

Q.27. For some integer q , every odd integer is of the form 
(a) q
(b) q + 1
(c) 2q
(d) 2q + 1
Ans. (d)
Solution. We know that, all numbers that are not the multiple of 2 are odd numbers.
Odd integers are ...,−3, −1, 1, 3, 5,...
So, odd numbers can be written as 2m + 1, where m is an integer.
m can be ..., −2, −1, 0, 1, 2,... ∴ 2m + 1 can be ..., −3, −1, 1, 3,...
Hence, the correct answer is option D.

Q.28. n² − 1 is divisible by 8 , if n is 
(a) An integer
(b) A natural number
(c) An odd integer
(d) An even integer
Ans. (c)
Solution. Any odd positive integer is of the form 4m + 1 or 4m + 3 for some integer m.
When n = 4m + 1,
n² −1 = (4m + 1)² − 1 = 16m² + 8m + 1 − 1 = 16m² + 8m = 8m(2m + 1) ⇒ n² − 1 is divisible by 8.
When n = 4m + 3,
n² − 1 = (4m + 3)² − 1 = 16m² + 24m + 9 − 1 = 16m² + 24m + 8 = 8(2m² + 3m + 1) ⇒ n² − 1 is divisible by 8.
Thus, n² – 1 is divisible by 8 if n is an odd positive integer.

Q.29. The decimal expansion of the rational number 33/2⁵ x 5 will terminate after
(a) One decimal place
(b) Two decimal places
(c) Three decimal places
(d) More than 3 decimal places
Ans. (b)
Solution. Consider the rational number 33/2⁵ x 5
Denominator of the above rational number is of the form 2^m × 2ⁿ, so the rational number has terminating decimal expansion.

So, the decimal expansion of the given rational number terminates after two decimal places.

Q.30. If two positive integers a and b are written as a = x³y² and b = xy³ ; x , y  are prime numbers , then HCF(a,b) is
(a) xy
(b) xy²
(c) x³y³
(d) x²y²
Ans. (b)
Solution. It is given that,
a = x³y² = x × x × x × y × y
b = xy³ = x × y × y × y
HCF(a, b) = HCF(x³y², xy³) = x × y × y = xy²