Real Numbers | Maths | Class 10th | Chapter 1 | PYQ Level 1
Real Numbers :- PYQ :- Level 1
Q.1: 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.
Solution: Let x be any positive integer and y = 3.
By Euclid’s division algorithm;
x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3)
Therefore,
x = 3q, 3q + 1 and 3q + 2
As per the given question, if we take the square on both the sides, we get;
x2 = (3q)2 = 9q2 = 3.3q2
Let 3q2 = m
Therefore,
x2 = 3m ………………….(1)
x2 = (3q + 1)2
= (3q)2 + 12 + 2 × 3q × 1
= 9q2 + 1 + 6q
= 3(3q2 + 2q) + 1
Substitute, 3q2+2q = m, to get,
x2 = 3m + 1 ……………………………. (2)
x2 = (3q + 2)2
= (3q)2 + 22 + 2 × 3q × 2
= 9q2 + 4 + 12q
= 3(3q2 + 4q + 1) + 1
Again, substitute, 3q2 + 4q + 1 = m, to get,
x2 = 3m + 1…………………………… (3)
Hence, from eq. 1, 2 and 3, we conclude that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Q.2: Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
(i) 140
Using the division of a number by prime numbers method, we can get the product of prime factors of 140.
Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7
(ii) 156
Using the division of a number by prime numbers method, we can get the product of prime factors of 156.
Hence, 156 = 2 × 2 × 13 × 3 = 22 × 13 × 3
(iii) 3825
Using the division of a number by prime numbers method, we can get the product of prime factors of 3825.
Hence, 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17
(iv) 5005
Using the division of a number by prime numbers method, we can get the product of prime factors of 5005.
Hence, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13
(v) 7429
Using the division of a number by prime numbers method, we can get the product of prime factors of 7429.
Hence, 7429 = 17 × 19 × 23 = 17 × 19 × 23
Q.3: Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution: As we know that,
HCF × LCM = Product of the two given numbers
Therefore,
9 × LCM = 306 × 657
LCM = (306 × 657)/9 = 22338
Hence, LCM(306,657) = 22338
Q.4: Prove that 3 + 2√5 is irrational.
Solution: Let 3 + 2√5 be a rational number.
Then the co-primes x and y of the given rational number where (y ≠ 0) is such that:
3 + 2√5 = x/y
Rearranging, we get,
2√5 = (x/y) – 3
√5 = 1/2[(x/y) – 3]
Since x and y are integers, thus, 1/2[(x/y) – 3] is a rational number.
Therefore, √5 is also a rational number. But this confronts the fact that √5 is irrational.
Thus, our assumption that 3 + 2√5 is a rational number is wrong.
Hence, 3 + 2√5 is irrational.
Q.5: 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:
(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600
Solution:
Note: If the denominator has only factors of 2 and 5 or in the form of 2m × 5n then it has a terminating decimal expansion.
If the denominator has factors other than 2 and 5 then it has a non-terminating repeating decimal expansion.
(i) 13/3125
Factorizing the denominator, we get,
3125 = 5 × 5 × 5 × 5 × 5 = 55
Or
= 20 × 55
Since the denominator is of the form 2m × 5n then, 13/3125 has a terminating decimal expansion.
(ii) 17/8
Factorizing the denominator, we get,
8 = 2× 2 × 2 = 23
Or
= = 23 × 50
Since the denominator is of the form 2m × 5n then, 17/8 has a terminating decimal expansion.
(iii) 64/455
Factorizing the denominator, we get,
455 = 5 × 7 × 13
Since the denominator is not in the form of 2m × 5n, therefore 64/455 has a non-terminating repeating decimal expansion.
(iv) 15/1600
Factorizing the denominator, we get,
1600 = 26 × 52
Since the denominator is in the form of 2m × 5n, 15/1600 has a terminating decimal expansion.
Q.6: 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, p/q what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000. . .
Solution:
(i) 43.123456789
Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.
(ii) 0.120120012000120000. . .
Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number.
Q.7: Check whether 6n can end with the digit 0 for any natural number n.
Solution: If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with a unit place as 0 or 5 is divisible by 5.
Prime factorization of 6n = (2 × 3)n
Therefore, the prime factorization of 6n doesn’t contain prime number 5.
Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.
Extra questions for Class 10 Maths Chapter 1
Q.1: Find three rational numbers lying between 0 and 0.1. Find twenty rational numbers between 0 and 0.1. Give a method to determine any number of rational numbers between 0 and 0.1.
Q.2: Which of the following rational numbers have the terminating decimal representation?
(i) 3/5
(ii) 7/20
(iii) 2/13
(iv) 27/40
(v) 133/125
(vi) 23/7
Q.3: Write the following rational numbers in decimal form:
(i) 42/100
(ii) 27/8
(iii) 1/5
(iv) 2/13
(v) 327/500
(vi) 5/6
(vii) 1/7
(viii) 11/17
Q.4: If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number.
Q.5: Show that 3√6 and 3√3 are not rational numbers.
Q.6: Show that 2 + √2 is not a rational number.
Q.7: Give an example to show that the product of a rational number and an irrational number may be a rational number.
Q.8: Prove that √3 – √2 and √3 + √5 are irrational.
Q.9: Express 7/64, 12/125 and 451/13 in decimal form.
Q.10: Find two irrational numbers lying between √2 and √3.
Q.11: Mention whether the following numbers are rational or irrational:
(i) (√2 + 2)
(ii) (2 – √2) x (2 + √2)
(iii) (√2 + √3)2
(iv) 6/3√2
Thanking You.
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