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

Page No 1.60

Q.31. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is 
(a) 10    
(b) 100     
(c) 504      
(d) 2520
Ans. (d)
Solution. We have,
1 = 1
2 = 2
3 = 3
4 = 2 × 2
5 = 5
6 = 2 × 3
7 = 7
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
∴ LCM of numbers from 1 to 10
= LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
= 2 × 2 × 2 × 3 × 3 × 5 × 7
= 2520

Q.32. The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively is 
(a) 13     
(b) 65         
(c) 875       
(d) 1750
Ans. (a)
Solution. It is given that on dividing 70 by the required number, there is a remainder of 5. This means that 70 − 5 = 65 is exactly divisible by the required number.
Similarly, 125 − 8 = 117 is exactly divisible by the required number.
The required number is H.C.F of 65 and 117.
By Euclid's division algorithm,
117 = 65 × 1 + 52
⇒ 65 = 52 × 1 + 13
⇒ 52 = 13 × 4 + 0
Here, the remainder is zero. Therefore, the H.C.F of 65 and 117 is 13.
So, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively.

Q.33. If  the HCF of 65 and 117 is expressible in the form  65m − 117 , then the value of m is 
(a) 4     
(b) 2      
(c) 1    
(d) 3
Ans. (b)
Solution. Use Euclid's algorithm to find the H.C.F of 65 and 117.
By Euclid's algorithm,
b = aq + r, 0 ≤ r < a
⇒ 117 = 65 × 1 + 32
⇒ 65 = 52 × 1 + 13
⇒ 52 = 13 × 4 + 0
∴ H.C.F(65, 117) = 13
It is given that H.C.F(65, 117) = 65m − 117.
⇒ 65m − 117 = 13
⇒ 65m = 130
⇒ m = 2

Q.34. Euclid's division lemma states that for two positive integers a and b , there exist unique integers q and r  such that a = bq + r , where  r must satisfy 
(a) 1 < r <  b   
(b) 0 < r ≤ b  
(c) 0 ≤ r <  b      
(d) 0 < r <  br
Ans. (c)
Solution. Euclid's division lemma states that for two positive integers a and b, there exists unique integers q and r such that a = bq + r, where r must satisfy 0 ≤ r < b.

Q.35. The product of a non-zero rational number and an irrational number is
(a) Always irrational
(b) Always rational
(c) Rational or irrational
(d) One
Ans. (a)
Solution. The product of a non-zero rational number and an irrational number is always an irrational number.

Q.1. If a and b are two positive co-prime integers such that a = 12b, then HCF (a, b) = _________.
Ans. If a and b are two positive co-prime integers
⇒ There is no common factor in a and b except 1.
⇒ H.C.F (a, b) = H.C.F (12b, b)  (∵ a = 12b) = b
Hence, H.C.F (a, b) = b.

Q.2. If two positive integers m and n are expressible in the form m = a²b³ and n = a³b², where a, b are prime numbers, then HCF (m, n) = _______ and LCM (a, b) = _______.
Ans. It is given that, m = a²b³ and n = a³b², where a, b are prime numbers.
H.C.F (m, n) = H.C.F (a²b³, a³b²)
= The lowest of indices of a and b
= a²b² 
Hence, H.C.F (m, n) is a²b².
L.C.M (m, n) = L.C.M (a²b³,  a³b²)
= The lowest of indices of a and b
= a³b³ 
Hence, L.C.M (m, n) is a³b³.
Hence, H.C.F (m, n) =  a²b² and L.C.M (m, n) = a³b³.
Disclaimer: In the question, we need to find LCM (m, n) instead of LCM (a, b).

Q.3. If the HCF and LCM of two positive integers a and b are x and y respectively, then x²y² / a²b² = ______.
Ans. Product of two numbers = L.C.M × H.C.F
⇒ a × b = y × x
⇒ ab = xy ...(1)
Now,
x²y² / a²b² = (xy)² / (ab)²
= (ab)² / (ab)² (From (1))
= 1
Hence, x²y² / a²b² = 1.

Q.4. A positive integer m when divided by 11 gives remainder 6. If 4m + 5 is divided by 11, the remainder is ______. 
Ans. We know that,
Dividend = Divisor × Quotient + Remainder
⇒ m = 11 × q + 6, where m is the dividend and q is the quotient
⇒ 4m = 11 × 4q + 24
⇒ 4m = 11 × 4q + 22 + 2
⇒ 4m = 11 × (4q + 2) + 2
Thus, if 4m is divided by 11, we get 2 as the remainder.
Therefore, If 4m + 5 is divided by 11, the remainder is 2 + 5 = 7.
Hence, If 4m + 5 is divided by 11, the remainder is  7.

Q.5. If HCF (306, 657) = 9, then LCM (306, 657) ______.
Ans. we know that,
Product of two numbers = LCM × HCF
⇒ 306 × 657 = LCM × 9
⇒ LCM × 9 = 306 × 657
⇒ LCM = 306 × 657 / 9
⇒ LCM = 34 × 657
⇒ LCM = 22338
Hence, LCM (306, 657) is 22,338.

Q.6. The LCM of the smallest prime number and the smallest odd composite number is ______.
Ans. Smallest prime number is 2.
Smallest odd composite number is 9.
Therefore, LCM (2, 9) is 18.
Hence, The LCM of the smallest prime number and the smallest odd composite number is 18.

Page No 1.61

Q.7. The sum of the exponents of prime factors in the prime factorisation of 250 is ______.
Ans. The prime factorisation of 250 are:
250 = 2 × 5 × 5 × 5
= 2¹ × 5³ 
The sum of exponents is 1 + 3 = 4.
Hence, The sum of the exponents of prime factors in the prime factorisation of 250 is 4.

Q.8. The LCM of the smallest prime number and the smallest composite number is ______.
Ans. Smallest prime number is 2.
Smallest composite number is 4.
Therefore, LCM (2, 4) = 4
Hence, The LCM of the smallest prime number and the smallest composite number is 4.

Q.9. 6ⁿ can not end with digit 0 for _________ value of n.
Ans. The prime factors of 6 are 2 and 3.
⇒ 6ⁿ = 2ⁿ × 3ⁿ 
Since, 5 is not the factor of 6ⁿ 
Therefore, for any value of n, 6ⁿ will not be divisible by 5.
Hence, 6ⁿ can not end with digit 0 for any value of n.

Q.10. The ratio between the HCF and LCM of 5, 15 and 20 is ______.
Ans. Prime factorisation of 5, 15 and 20 is
5 = 5
15 = 3 × 5
20 = 2 × 2 × 5
Thus, LCM of 5, 15 and 20 is 2 × 2 × 3 × 5 = 60
and HCF of 5, 15 and 20 is 5.
Therefore,
HCF / LCM
= 5 / 60
= 1 / 12
Hence, The ratio between the HCF and LCM of 5, 15 and 20 is 1 : 12.

Q.11. The HCF and LCM of two numbers are 33 and 264 respectively. When the first number is completely divided by 2 the quotient is 33. The other number is _______.
Ans. It is given that the first number is completely divided by 2 the quotient is 33,
∴ First number = 2 × 33 + 0 = 66
Let the other number be x.
Now,
Product of two numbers = LCM × HCF
⇒ 66 × x = 264 × 33
⇒ x = 264 × 33 / 66
⇒ x = 264/2
⇒ x = 132
Hence, the other number is 132.

Q.12. If the prime factorisation of a natural number n is 2⁴ × 3⁴ × 5³ × 7, then the number of consecutive zeros in n, is ______.
Ans. To calculate the number of consecutive zeros, we need to find the number of pairs of 2 and 5 to form a product of 10.
Here, n = 2⁴ × 3⁴ × 5³ × 7
= 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 × 7
= 2 × 5 × 2 × 5 × 2 × 5 × 2 × 3 × 3 × 3 × 3 × 7
= 10 × 10 × 10 × 2 × 3 × 3 × 3 × 3 × 7
= 2 × 3⁴ × 7 × 10³ 
Hence, the number of consecutive zeros in n, is 3.

Q.13. If 2520 = 2^a × 3^b × 5^c × 7^d, then a + b – 2c – 3d = ______.
Ans. Prime factorisation of 2520 is
2520 = 2³ × 3² × 5¹ × 7¹ 
Therefore, a = 3, b = 2, c = 1 and d = 1.
a + b – 2c – 3d = 3 + 2 – 2(1) – 3(1)
= 5 – 2 – 3
= 5 – 5
= 0
Hence, a + b – 2c – 3d = 0.

Q.14. If n is a natural number, then the number of consecutive zeros in 7ⁿ, is _______. 
Ans. To calculate the number of consecutive zeros, we need to find the number of pairs of 2 and 5 to form a product of 10.
But, 2 and 5 both are not the factors of 7ⁿ.
Hence, the number of consecutive zeros in 7ⁿ, is 0.

Q.15. Two numbers are in the ratio 21 : 17. If their HCF is 5, the numbers are ______ and ______.
Ans. Let the two numbers be 21x and 17x.
LCM of (21x, 17x) = 17 × 21 × x
We know that,
Product of two numbers = LCM × HCF
⇒ 21x × 17x = 17 × 21 × x × 5
⇒ 21 × 17x² = 17 × 21 × 5x
⇒ 21 × 17x = 17 × 21 × 5
⇒ x = (17 × 21 × 5)/21 × 17
⇒ x = 5
Therefore, first number is 21 × 5 = 105 and second number is 17 × 5 = 85.
Hence, the numbers are 105 and 85.