# Chapter 1 Real Numbers Important Questions for CBSE Class 10 Maths Board Exams

**Important Questions for Class 10 Maths Chapter 1 Real Numbers**, which will help the students to prepare for the CBSE Class 10 maths Board exam 2022-23. It help the doing better in their maths paper. Extra questions of Chapter 1 Real Numbers given here which are based on the pattern of CBSE NCERT book. Students will learn about the entire syllabus and learn how to solve problems in preparation for the exams.

## Important Questions for Chapter 1 Real Numbers Class 10 Maths

### Real Numbers Class 10 Maths Important Questions Very Short Answer (1 Mark)

**1. For any two integers, the product of the integers = the product of their HCF and LCM. Is this relation true for three or more integers?**

**Solution**

No

**2. The decimal expansion of the rational number 43/2 ^{4}5^{3} will terminate after how many places of decimals?**

**Solution**

**3. Write one rational and one irrational number lying between 0.25 and 0.32.**

**Solution**

**4. Express 98 as a product of its primes.**

**Solution**

2 × 7^{2}

**5. HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number. (2012)**

**Solution**

**6. If HCF(336, 54) 6 = , find LCM(336, 54).**

**Solution**

**7. a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then calculate the least prime factor of (a+b).**

**Solution**

**8. Calculate the HCF of 3**

^{3}×5 and 3^{2}×5^{2}.**Solution**

^{3}×5 = 3

^{2}×5×3

^{2}×5

^{2}= 3

^{2}×5×5

^{3}×5, 3

^{2}×5

^{2}) = 3

^{2}×5

**9. Calculate 3/8 in the decimal form.**

**Solution**

**10. If HCF (a,b) = 12 and a×b = 1800, then find LCM (a,b).**

**Solution**

**11. Find LCM of numbers whose prime factorisation are expressible as 3 × 52 and 32 × 72. (2014)**

**Solution**

**12. Find the least number that is divisible by all numbers between 1 and 10 (both inclusive).**

**Solution**

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

**13. Find the LCM of 96 and 360 by using fundamental theorem of arithmetic.**

**Solution**

^{5}× 3

^{3}× 3

^{2}× 5

^{5}× 3

^{2}× 5 = 32 × 9 × 5 = 1440

**14. Find the largest number which divides 70 and 125 leaving remainder 5 and 8 respectively.**

**Solution**

^{2}× 13

**15. Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.**

**Solution**

^{2}×101

^{5}×3

^{5}×3 = 9696

**16. Given that HCF (306,1314) = 18. Find LCM 1314**

**Solution**

**17. Complete the following factor tree and find the composite number x.**

**Solution**

**18. Check whether 4 ^{n} can end with the digit 0 for any natural number n.**

**Solution**

If the number 4^{n} for any n, were to end with the digit zero, then it would be divisible by 5 and 2.

That is, the prime factorization of 4^{n} would contain the prime 5 and 2. This is not possible because the only prime in the factorization of 4^{n} = 2^{2n} is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 4^{n}. So, there is no natural number n for which 4^{n} ends with the digit zero. Hence 4^{n} cannot end with the digit zero.

**19. Write the denominator of the rational number 257/500 in the form 2 ^{m} × 5^{n}, where m and n are non-negative integers. Hence write its decimal expansion without actual division.**

**Solution**

500 = 25 × 20

= 52 × 5 × 4

= 53 × 22

Here, denominator is 500 which can be written as 22 × 53.

Now decimal expansion,

**20. Write a rational number between √2 and √3.**

**Solution**

**21. Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons.**

**Solution**

**22. Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number? (2015)**

**Solution**

**…(i)**

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

**23. Prove that 3 + 2√3 is an irrational number.**

**Solution**

**24. Prove that 2+√3/5 is an irrational number, given that √3 is an irrational number.**

**Solution**

**25. Write the smallest number which is divisible by both 306 and 657.**

**Solution**

**26. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together?**

**Solution**

**27. Two tankers contain 850 liters and 680 liters of petrol. Find the maximum capacity of a container which can measure the petrol of each tanker in the exact number of times.**

**Solution**

**28. 144 cartons of Coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and if it equal contain cartons of the same drink, what would be the greatest number of cartons each stack would have?**

**Solution**

The required answer will be HCF of 144 and 90.

144 = 2^{4} × 3^{2}

90 = 2 × 3^{2} × 5

HCF (144, 90) = 2 × 3^{2} = 18

Thus each stack would have 18 cartons.

**29. If p is prime number, then prove that √p is an irrational.**

**Solution**

Let p be a prime number and if possible, let √p be rational

Thus,

where m and n are co-primes and n ≠0 .

Squaring on both sides, we get

or, pn^{2} = m^{2} **...(1)**

Here p divides pn^{2}. Thus p divides m^{2} and in result p also divides m.

Let m = pq for some integer q and putting m = pq in eq. (1), we have

pn^{2} = p^{2}q^{2}

or, n^{2} = pq^{2}

Here, p divides pq^{2}. Thus p divides n^{2} and in result p also divides n.

[∵ p is prime and p divides n^{2} ⇒ p divides n]

Thus p is a common factor of m and n but this contradicts the fact that m and n are primes. The contradiction arises by assuming that √p is rational.

Hence, √p is irrational.

**30. Amita, Sneha, and Raghav start preparing cards for all persons of an old age home. In order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together? (2013)**

**Solution**

To find the earliest (least) time, they will start preparing a new card together, we find the LCM of 10, 16 and 20.

10 = 2 × 5

16 = 24

20 = 22 × 5

LCM = 24 × 5 = 16 × 5 = 80 minutes

They will start preparing a new card together after 80 minutes.

**31. 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. (2014)**

**Solution**

x = p^{2}q^{3} and y = p^{3}q

LCM = p^{3}q^{3}

HCF = p^{2}q **…(i)**

Now, LCM = p^{3}q^{3}

⇒ LCM = pq^{2} (p^{2}q)

⇒ LCM = pq^{2} **(HCF)**

Yes, LCM is a multiple of HCF.

Explanation:

Let a = 12 = 2^{2} × 3

b = 18 = 2 × 3^{2}

HCF = 2 × 3 = 6 **…(ii)**

LCM = 2^{2} × 3^{2} = 36

LCM = 6 × 6

LCM = 6 **(HCF) [From (ii)]**

Here LCM is 6 times HCF.

**32. Show that any positive odd integer is of the form 4q + 1 or 4q + 3 where q is a positive integer.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.

### Real Numbers Class 10 Maths Important Questions Long Answer (4 Marks)

**33. Prove that √5 is an irrational number.**

**Solution**

^{2}= 5b

^{2}

^{2}and in result 5 is also a factor of a.

^{2}= 25c

^{2}

^{2}= 5b

^{2}we have,

^{2}= 25c

^{2}

^{2}= 5c

^{2}

^{2}and in result 5 is also a factor of b.

**34. Prove that n**

^{2}- n is divisible by 2 for every positive integer n.**Solution**

We have n^{2} - n = n(n-1)

Thus n^{2} - n is product of two consecutive positive integers.

Any positive integer is of the form 2q or 2q + 1, for some integer q.**Case 1:** n = 2q

If n = 2q we have

n(n-1) = 2q(2q-1)

= 2m

where m = q(2q-1) which is divisible by 2.**Case 2:** n = 2q+1

If n = 2q+1, we have

n(n-1) = (2q+1) (2q+1-1)

= 2q(2q+1)

= 2m

where m = q(2q+1) which is divisible by 2.

Hence, n^{2} - n is divisible by 2 for every positive

integer n.

### Real Numbers Class 10 Maths Case Based Questions

**A. Read the following text and answer the following questions on the basis of the same:**

**1. If p and q are positive integers such that p = ab2 and q = a2b, where a, b are prime numbers, then the LCM (p, q) is**

A. ab

B. a^{3}b^{3}

C. a^{3}b^{2}

D. a^{2}b^{2}

**Solution**

D. a^{2}b^{2}

Given, p = ab^{2} = a × b × b

q = a^{2}b = a × a × b

LCM of (p, q) = a^{2}b^{2}

**2. If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32 , 36) is**

A. 2

B. 4

C. 6

D. 8

**Solution**

B. 4

Prime factor 32 is 2×2×2×2×2

Prime factor 36 is 2×2×3×3

HCF is 2×2 = 4

HCF of 32 and 36 is 4.

**3. 7 × 11 × 13 × 15 + 15 is a**

A. Prime number

B. Composite number

C. Neither prime nor composite

D. None of the above

**Solution**

B. Composite number

7 × 11 × 13 × 15 + 15 is composite number.

Take 15 common

15 ( 7 × 11 × 13 + 1)

So, given number has factor other than 1 and itself so it is a composite number.

**4. What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?**

A. 144

B. 128

C. 288

D. 272

**Solution**

We have to find the LCM of 32 and 36.

LCM(32, 36) = 2^{5} × 9 = 288

Hence, the minimum number of books required to distribute equally among students of section A and section B are 288.

**B. Read the following text and answer the following questions on the basis of the same:**

A garden consists of 135 rose plants planted in certain number of columns. There are another set of 225 marigold plants, which is to be planted in the same number of columns.

**1. Find the sum of exponents of the prime factors of total number of plants.**

A. 2

B. 3

C. 5

D. 6

**Solution**

D. 6

Total number of plants = 135 + 225 = 360

The prime factors of 360 = 2 × 2 × 2 × 3 × 3 × 5

= 23 × 32 × 51

∴ Sum of exponents = 3 + 2 + 1 = 6.

**2. What is total numbers of row in which they can be planted**

A. 3

B. 5

C. 8

D. 15

**Solution**

C. 8

Number of rows of Rose plants = 135/45 = 3

Number of rows of marigold plants = 225/45 = 5

Total number of rows = 3 + 5 = 8

**3. Find the sum of exponents of the prime factors of the maximum number of columns in which they can be planted.**

A. 5

B. 3

C. 4

D. 6

**Solution**

We have proved that the maximum number of columns = 45

So, prime factors of 45

= 3 × 3 × 5 = 32 × 51

∴ Sum of exponents = 2 + 1 = 3.