Important Questions for Class 10 Maths Chapter 1 Real Numbers

Important Questions for Class 10 Maths Chapter 1 Real Numbers

Free PDF Download of CBSE Class 10 Maths Chapter 12 Areas Related to Circles Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Areas Related to Circles MCQs with Answers to know their preparation level.

Important Questions for Class 10 Maths Chapter 1 Real Numbers

Question 1.
The decimal expansion of the rational number will terminate after how many places of decimals? (2013)
Solution:

Question 2.
Write the decimal form of
Solution:
Non-terminating non-repeating.

Question 3.
Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively.
Solution:
Algorithm
398 - 7 = 391, 436 - 11 = 425, 542 - 15 = 527
HCF of 391, 425, 527 = 17

Question 4.
Express 98 as a product of its primes.
Solution:
2 × 7²

Question 5.
If the HCF of 408 and 1032 is expressible in the form 1032 × 2 + 408 × p, then find the value of p.
Solution:
HCF of 408 and 1032 is 24.
1032 × 2 + 408 × (p) = 24
408p = 24 - 2064
p = -5

Real Numbers Class 10 Important Questions Short Answer-I (2 Marks)

Question 6.
HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number. (2012)
Solution:
We know,
1st number × 2nd number = HCF × LCM
⇒ 27 × 2nd number = 9 × 459
⇒ 2nd number = = 153

Question 7.
Find HCF and LCM of 13 and 17 by prime factorisation method. (2013)
Solution:
13 = 1 × 13; 17 = 1 × 17
HCF = 1 and LCM = 13 × 17 = 221

Question 8.
Find LCM of numbers whose prime factorisation are expressible as 3 × 5² and 3² × 7². (2014)
Solution:
LCM (3 × 5², 3² × 7²) = 3² × 5² × 7² = 9 × 25 × 49 = 11025

Question 11.
Find the largest number which divides 70 and 125 leaving remainder 5 and 8 respectively. (2015)
Solution:
It is given that on dividing 70 by the required number, there is a remainder 5.
This means that 70 - 5 = 65 is exactly divisible by the required number.
Similarly, 125 - 8 = 117 is also exactly divisible by the required number.
65 = 5 × 13
117 = 3² × 13
HCF = 13
Required number = 13

Question 16.
Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number? (2015)
Solution:
17 × 5 × 11 × 3 × 2 + 2 × 11 …(i)
= 2 × 11 × (17 × 5 × 3 + 1)
= 2 × 11 × (255 + 1)
= 2 × 11 × 256
Number (i) is divisible by 2, 11 and 256, it has more than 2 prime factors.
Therefore (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number.

Question 17.
Check whether 4n can end with the digit 0 for any natural number n. (2015)
Solution:
4ⁿ = (2²)ⁿ = 2²ⁿ
The only prime in the factorization of 4ⁿ is 2.
There is no other prime in the factorization of 4ⁿ = 2²ⁿ
(By uniqueness of the Fundamental Theorem of Arithmetic).
5 does not occur in the prime factorization of 4ⁿ for any n.
Therefore, 4ⁿ does not end with the digit zero for any natural number n.

Question 18.
Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons. (2017 OD)
Solution:
No, LCM = Product of the highest power of each factor involved in the numbers.
HCF = Product of the smallest power of each common factor.
We can conclude that LCM is always a multiple of HCF, i.e., LCM = k × HCF
We are given that,
LCM = 175 and HCF = 15
175 = k × 15
⇒ 11.67 = k
But in this case, LCM ≠ k × HCF
Therefore, two numbers cannot have LCM as 175 and HCF as 15.

Real Numbers Class 10 Important Questions Short Answer-II (3 Marks)

Question 23.
The length, breadth, and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly. (2015)
Solution:
To find the length of the longest rod that can measure the dimensions of the room exactly, we have to find HCF.
L, Length = 8 m 50 cm = 850 cm = 2¹ × 5² × 17
B, Breadth = 6 m 25 cm = 625 cm = 5⁴
H, Height = 4 m 75 cm = 475 cm = 5² × 19
HCF of L, B and H is 5² = 25 cm
Length of the longest rod = 25 cm

Question 26.
By using Euclid’s algorithm, find the largest number which divides 650 and 1170. (2017 OD)
Solution:
Given numbers are 650 and 1170.
1170 > 650
1170 = 650 × 1 + 520
650 = 520 × 1 + 130
520 = 130 × 4 + 0
HCF = 130
The required largest number is 130.

Question 34.
If two positive integers x and y are expressible in terms of primes as x = p²q³ and y = p³q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain. (2014)
Solution:
x = p²q³ and y = p³q
LCM = p³q³
HCF = p²q …..(i)
Now, LCM = p³q³
⇒ LCM = pq² (p²q)
⇒ LCM = pq² (HCF)
Yes, LCM is a multiple of HCF.
Explanation:
Let a = 12 = 2² × 3
b = 18 = 2 × 3²
HCF = 2 × 3 = 6 …(ii)
LCM = 2² × 3² = 36
LCM = 6 × 6
LCM = 6 (HCF) …[From (ii)]
Here LCM is 6 times HCF.

Question 35.
Show that one and only one out of n, (n + 1) and (n + 2) is divisible by 3, where n is any positive integer. (2015)
Solution:
Let n, n + 1, n + 2 be three consecutive positive integers.
We know that n is of the form 3q, 3q + 1, or 3q + 2.
Case I. When n = 3q,
In this case, n is divisible by 3,
but n + 1 and n + 2 are not divisible by 3.
Case II. When n = 3q + 1,
In this case n + 2 = (3q + 1) + 2
= 3q + 3
= 3(q + 1 ), (n + 2) is divisible by 3,
but n and n + 1 are not divisible by 3.
Case III.
When n = 3q + 2, in this case,
n + 1 = (3q + 2) + 1
= 3q + 3 = 3 (q + 1 ), (n + 1) is divisible by 3,
but n and n + 2 are not divisible by 3.
Hence, one and only one out of n, n + 1 and n + 2 is divisible by 3.

Question 36.
Find the HCF and LCM of 306 and 657 and verify that LCM × HCF = Product of the two numbers. (2016 D)
Solution:
306 = 2 × 3² × 17
657 = 32 × 73
HCF = 3² = 9
LCM = 2 × 3² × 17 × 73 = 22338
L.H.S. = LCM × HCF = 22338 × 9 = 201042
R.H.S. = Product of two numbers = 306 × 657 = 201042
L.H.S. = R.H.S.

Question 37.
Show that any positive odd integer is of the form 41 + 1 or 4q + 3 where q is a positive integer. (2016 OD)
Solution:
Let a be a positive odd integer
By Euclid’s Division algorithm:
a = 4q + r …[where q, r are positive integers and 0 ≤ r < 4]
a = 4q
or 4q + 1
or 4q + 2
or 4q + 3
But 4q and 4q + 2 are both even
a is of the form 4q + 1 or 4q + 3.

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