Bihar Board Class 10th Maths Chapter 1 Real Numbers Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.1

Question 1.
Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255

Solution:

(i) We have:
Dividend = 225 and Divisor = 135
The process can be exhibited as under:

Hence, HCF (135, 225) = 45.

(ii) We have:
Dividend = 38220 and Divisor = 196
38220 = 196 × 195 + 0
Hence, HCF (196, 38220) = 196.

(iii) We have:
Dividend = 867 and Divisor = 255

Hence, HCF (255, 867) = 51.

Question 2.
Show that any positive odd integer is of the form 6q + t or 6q + 3 or 6q + 5, where q is some integer.
Solution:
Let a be any positive integer and b = 6. Then, by Euclid’s Division Lemma a = 6q + r, for some integer q ≥ 0, and where 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 or 5.

That is, a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient. If a = 6q or 6q + 2 or 6q + 4, then a is an even integer.

Also, an integer can be either even or odd. Therefore, any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.

Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
To find the maximum number of columns, we have to find the HCF of 616 and 32.

∴ The HCF of 616 and 32 is 8.
Hence, maximum number of columns is 8.

Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. [CBSE, Foreign, 2008]
Solution:
Let x be any positive integer. Then it is of the form 3q, 3q + 1 or 3q + 2. Now, we have to prove that the square of each of these can be written in the form 3m or 3m + 1.
Now, (3q)2 = 9q2 = 3(3q2)
= 3m,
where in = 3q2
(3q + 1)2 = 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3m+ 1, where m = 3q2 + 2q
and (3q + 2)2 = 9q2 + 12q + 4
= 3(3q2 + 4q + 1) + 1
= 3m + 1, where m = 3q2 + 4q + 1
Hence, the result.

Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is either of the form 9q, 9q + 1 or 9q + 8.
Solution:
Let x be any positive integer. Then it is of the form 3m, 3m + 1 or 3m + 2. Now, we have to prove that the cube of each of these can be rewritten in the form
9q, 9q + 1 or 9q + 8.
Now, (3m)3 = 27m3 = 9(3m3)
= 9q, where q = 3m3
(3m + 1)3 = (3m)3 + 3(3m)2. 1 + 3(3m). 12 + 1
= 27m3 + 27m2 + 9m + 1
= 9(3m3 + 6m2 + 4m) + 8
= 9q + 8, where q = 3m3 + 6m2 + 4m.

BSEB Bihar Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.2

Question 1.
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
(i) We use the division method as shown below:

∴ 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7

(ii) We use the division method as shown below:

∴ 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13

(iii) We use the division method as shown below:

∴ 3825 = 3 × 3 × 5 × 5 × 17
= 32 × 52 × 17

(iv) We use the division method as shown below:

∴ 5005 = 5 × 7 × 11 × 13

(v) We use the division method as shown below:

∴ 7429 = 17 × 19 × 23

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91

∴ LCM of 26 and 91 = 2 × 7 × 13 = 182
and HCF of 26 and 91 = 13
Now, 182 × 13 = 2366 and 26 × 91 = 2366
Thus, 182 × 13 = 26 × 91
Hence verified.

(ii) 510 and 92

510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23
∴ LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23
= 23460
and HCF of 510 and 92 = 2
Now, 23460 × 2 = 46920 and 510 × 92 = 46920
Thus, 23460 × 2 = 510 × 92
Hence verified.

(iii) 336 and 54

336 = 2 × 2 × 2 × 2 × 3 × 7
and 54 = 2 × 3 × 3 × 3
∴ LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7
= 3024
and HCF of 336 and 54 = 2 × 3 = 6
Now, 3024 × 6 = 18144 and 336 × 54 = 18144
Thus, 3024 × 6 = 336 × 54
Hence verified.

Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method:
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) First we write the prime factorisation of each of the given numbers.
12 = 2 × 2 × 3 = 22 × 3, 15 = 3 × 5 and 21 = 3 × 7
∴ LCM = 22 × 3 × 5 × 7 = 420
and HCF = 3

(ii) First we write the prime factorisation of each of the given numbers.
17 = 1 × 17, 1 × 23 and 1 × 29
[Note that here each of the numbers 17, 23 and 29 is a prime]
∴ LCM = 17 × 23 × 29
= 11339
and HCF = 1
[Note that in 17, 23 and 29, there is no common prime factor]

(iii) First we write the prime factorisation of each of the given numbers.
8 = 2 × 2 × 2 = 23, 9 = 3 × 3 = 32, 25 = 5 × 5 = 52
∴ LCM = 23 × 32 × 52 = 8 × 9 × 25 = 1800
and HCF = 1

Question 4.
Given that IICF (306, 657) = 9, find LCM (306, 657).
Solution:
We know that the product of the HCF and the LCM of two numbers is equal to the product of the given numbers.
∴ HCF (306, 657) × LCM (306, 657) = 306 × 657
So, 9 × LCM (306, 657) = 306 × 657
or LCM (306, 657) = 306×6579
= 22338

Question 5.
Check whether 6n can end with the digit 0 for any natural number n.
Solution:
If the number 6n, for any n, ends with the digit 0, then it is divisible by 5. That is, the prime factorisation of 6n contains the prime 5. This is not possible as the only primes in the factorisation of 6n are 2 and 3 and the uniqueness of the Funtlaniental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 6n. So, there is no natural number n for which 6n ends with the digit 0.

Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Now, 7 × 11 × 13 + 13 = 13 × (7 × 11 × 1 + 1)
= 13 × (77 + 1) = 13 × 78
Hence, it is a composite number.
Again, 7 × 6 × 5 × 4 × 3 × 1 × 1 + 5
= 5 × (7 × 6 × 4 × 3 × 1 × 1 + 1)
So, it is a composite number.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they again at the starting point?
Solution:
They will be again at the starting point after common multiples of 18 and 12 minutes. To find the LCM of 18 and 12, we have:

LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36
So, Sonia and Ravi will meet again at the starting point after 36 minutes.

BSEB Bihar Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3

Question 1.
Prove that 5–√ is irrational. [CBSE, Delhi 2009]
Solution:
Let us assume, to the contrary, that 5–√ is rational.
Squaring on both sides, we get
5 = a2b2 or 5b2 = a2 ………………. (1)
This shows that a2 is divisible by 5.
It follows that a is divisible by 5. …………………. (2)
So, a = 5m for some integer m.
Substituting a = 5m in (1), we get
5b2 = (5m)2 = 25m2
or b2 = 5m2 or b2 is divisible by 5
and hence b is divisible by 5.
From (2) and (3), we can conclude that 5 is a common factor of both a and b.
But this contradicts our supposition that a and b are coprime.
Hence, 5–√ is irrational.

Question 2.
Prove that 3 + 25–√ is irrational.
Solution:
Let us assume, to the contrary, that 3 + 25–√ is a rational number.
Now, let 3 + 25–√ = ab, where a and b are coprime and b ≠ 0.
So, 25–√ = ab – 3 or 5–√ = a2b – 32
Since a and b are integers, therefore
a2b – 32 is a rational number.
∴ 5–√ is an irrational number.
But 5–√ is an irrational number.
This shows that our assumption is incorrect.
So, 3 + 25–√ is an irrational number.

Question 3.
Prove that the following are irrationals:

1. 12√
2. 75–√
3. 6 + 2–√

Solution:
1. Let us assume, to the contrary, that 12√ is rational.
That is, we can find co-prime integers p and q (≠ 0) such that

Since p and q are integers, 2pq is rational, and so 2–√ is rational.
But this contradicts the fact that 2–√ is irrational.
So, we conclude that 12√ is irrational.

2. Let us assume, to the contrary, that 75–√ is rational.
That is, we can find co-prime integers p and q (≠ 0) such that 75–√ = pq.
So, 5–√ = p7q.
Since p and q are integers, p7q is rational and so is 5–√.
But this contradicts the fact that 5–√ is irrational.
So, we conclude that 75–√ is irrational.

3. Let us assume, to the contrary, that 6 + 2–√ is rational.
That is, we can find integers p and q (≠ 0) such that
6 + 2–√ = pq or 6 – pq = 2–√
or 2–√ = 6 – pq
Since p and q are integers, we get 6 – pq is rational, and so 2–√ is rational.
But this contradicts the fact that 2–√ is irrational.
So, we conclude that 6 + 2–√ is irrational.

BSEB Bihar Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.4

Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

1. 133125
2. 178
3. 64455
4. 151600
5. 29343
6. 232352
7. 129225775
8. 615
9. 3550
10. 77210

Solution:
We know that if the denominator of a rational number has no prime factors other than 2 or 5 (or both), then it is expressible as a terminating decimal, otherwise it has non-terminating repeating decimal expansion.
Thus, we will have to check the prime factors of the denominators of each of the given rational numbers.

1. In, the denominator is 3125.53125
We have: 3125 = 5 × 5 × 5 × 5 ×
Thus, 3125 has 5 as the only prime factor.
Hence, 133125 must have a terminating decimal expansion.

2. In the denominator is 8.
We have: 8 = 2 × 2 × 2
Thus, 8 has 2 as the only prime factor.
Hence, 178 must have a terminating decimal expansion.

3. In 64455, denominator is 455.
We have: 455 = 5 × 7 × 13
Clearly, 455 has prime factors ther than 2 and 5. So, it will not have a terminating decimal expansion.
That is, it will have non-terminating repeating decimal expansion.

4. In 151600, the denominator is 1600.
We have:
1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
Thus, 1600 has only 2 and 5 as prime factors.
Hence, 151600 must have a terminating decimal expansion.

5. In 29343, the denominator is 343.
We have:
343 = 7 × 7 × 7
Clearly, 343 has prime factors other than 2 and 5.
So, it will have non-terminating repeating decimal expansion.

6. In 2323⋅52
Clearly, the denominator 23.52 has only 2 and 5 as prime factors.
Hence, 2323⋅52 must have a terminating decimal expansion.

7. In 12922⋅57⋅75
Clearly, the denominator 23.57.75 has prime factors other than 2 and 5. So, it will have non-terminating repeating decimal expansion.

8. In 615, we have 615 = 25
Clearly, denominator 5 has no prime factors other than 2 and 5. So, it will have a terminating decimal expansion.

9. in 3550, we have denominator 50 = 2 × 5 × 5
The denominator has only 2 and 5 as prime factors.
Hence, 3550 must have a terminating decimal expansion.

10. In 77210, we have 77210 = 1130. The denominator is 30.
We have: 30 = 2 × 3 × 5
Clearly, 30 has prime factors other than 2 and 5.
So, it will have non-terminating repeating decimal expansion.

Question 2.
Write down the decimal expansion of those rational numbers in Question 1 above which have terminating decimal expansions.
Solution:
1.

2.

3. Non-terminating repeating.

4.

5. Non-terminating repeating.

6.

7. Non-terminating repeating.

8. 615 = 25
= 410 = 0.4

9. 3550 = 35×250×2 = 70100 = 0.70

10. Non-terminating repeating.

Question 3.
The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form pq, what can you say about the prime factors of q?

1. 43.123456789
2. 0.120120012000120000…
3. 43.123456789¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Solution:
1. 43.123456789 is terminating.
So, it represents a rational number.
Thus, 43.123456789 = 431234567891000000000 = pq. Thus, q = 109.

2. 0.12012001200012000… is non-terminating and non-repeating. So, it is irrational.

3. 43.123456789¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is non-terminating but repeating. So, it is rational.
Thus, 43.123456789¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ = 431234564699999999 = pq.
Thus, q = 999999999.