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Real Numbers - NCERT Questions
Q 1.

Use Euclid's division algorithm to find the HCF of
(i) 135 and 225 (ii) 196 and 38220
(iii) 867 and 225

SOLUTION:

Soln. : (i) HCF of 135 and 225;
Applying the Euclid's lemma to 225 and 135, (where 225 > 135), we get
225 = (135 × 1) + 90, since 90 ≠ 0, therefore, applying the Euclid's lemma to 135 and 90, we have 135 = (90 × 1) + 45
But 45 ≠ 0
∴ Applying Euclid's lemma to 90 and 45, we get 90 = (45 × 2) + 0
Here, r = 0, so our procedure stops. Since, the divisor at the last step is 45,
∴ HCF of 225 and 135 is 45.
(ii) HCF of 196 and 38220 :
We start dividing the larger number 38220 by 196, we get 38220 = (196 × 195) + 0
Here, r = 0
∴ HCF of 38220 and 196 is 196.
(iii) HCF of 867 and 255
Here, 867 > 255
∴ Applying Euclid's Lemma to 867 and 255, we get
867 = (255 × 3) + 102, 102 ≠ 0
∴ Applying Euclid's Lemma to 255 and 102, we get
255 = (102 × 2) + 51, 51 ≠ 0
∴ Applying Euclid's Lemma to 102 and 51, we get
102 = (51× 2) + 0, r = 0
∴ HCF of 867 and 225 is 51.

Q 2.

Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

SOLUTION:

Soln. : Let us consider a positive odd integer as ′a′.
On dividing ′a′ by 6, let q be the quotient and ′r′ be the remainder.
∴ Using Euclid's lemma, we have : a = 6q + r where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4 or 5 i.e.,
a = 6q + 0 = 6q or a = 6q + 1
or a = 6q + 2 or a = 6q + 3
or a = 6q + 4 or a = 6q + 5
But, a = 6q, a = 6q + 2, a = 6q + 4 are even values of ′a′.
[∵ 6q = 2(3q) = 2m1, 6q + 2 = 2(3q + 1) = 2m2,
6q + 4 = 2(3q + 2) = 2m3]
But ′a′ being an odd integer, we have :
a = 6q + 1, or a = 6q + 3, or a = 6q + 5

Q 3.

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups have to march in the same number of columns. What is the maximum number of columns in which they can march?

SOLUTION:

Soln. : Total number of members = 616
∴ The total number of members are to march behind an army band of 32 members is HCF of 616 and 32.
i.e., HCF of 616 and 32 is equal to the maximum number of columns such that the two groups can march in the same number of columns.
∴ Applying Euclid's lemma to 616 and 32, we get
616 = (32 × 19) + 8 ∵ 8 ≠ 0
∴ Again, applying Euclid's lemma to 32 and 8, we get
32 = (8 × 4) + 0, r = 0
∴ HCF of 616 and 32 is 8
Hence, the required number of maximum columns = 8.

Q 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.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]

SOLUTION:

Soln. : Let us consider an arbitrary positive integer as 'x' such that it is of the form
3q, (3q + 1) or (3q + 2).
For x = 3q, we have x2 = (3q)2
x2 = 9q2 = 3(3q2) = 3m ... (1)
Putting 3q2 = m where m is an integer.
x = 3q + 1,
x2 = (3q + 1)2 = 9q2 + 6q + 1
= 3(3q2 + 2q) + 1 = 3m + 1 ... (2)
Putting 3q2 + 2q = m, where m is an integer.
For x = 3q + 2,
x2 = (3q + 2)2
= 9q2 + 12q + 4 = (9q2 + 12q + 3) + 1
= 3(3q2 + 4q + 1) + 1 = 3m + 1 ... (3)
Putting 3q2 + 4q +1 = m, where m is an integer.
From (1), (2) and (3)
x2 = 3m or 3m + 1
Thus, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Q 5.

Use Euclid's division lemma to show that cube of any positive integer is of the form 9m,
9m + 1 or 9m + 8.

SOLUTION:

Soln. : Let us consider an arbitrary positive integer x such that it is in the form of 3q, (3q + 1) or
(3q + 2).
For x = 3q
x3 = (3q)3= 27q3 = 9(3q3) = 9m ... (1)
Putting 3q3 = m, where m is an integer.
For x = 3q + 1
x3 = (3q + 1)3 = 27q3 + 27q2 + 9q + 1
= 9(3q3 + 3q2 + q) + 1 = 9m + 1 ... (2)
Putting (3q3 + 3q2 + q) = m, where m is an integer.
For x = 3q + 2,
x3 = (3q + 2)3 = 27q3 + 54q2 + 36q + 8
= 9(3q3 + 6q2 + 4q) + 8 = 9m + 8 ... (3)
Putting (3q3 + 6q2 + 4q) = m, where m is an integer.
From (1), (2), (3) we have:
x3 = 9m, (9m + 1) or (9m + 8)
Thus, cube of any positive integer can be in the form 9m, (9m + 1) or (9m + 8) for some integer m.

Q 6.

Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

SOLUTION:

Soln. : (i) Using factor tree method, we have:
ncert
∴ 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7
(ii) Using factor tree method, we have :
ncert
∴ 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13
(iii) Using factor tree method, we have:
ncert
∴ 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17
(iv) Using factor tree method, we have:
ncert
∴ 5005 = 5 × 7 × 11 × 13
(v) Using the factor tree method, we have:
ncert
∴ 7429 = 17 × 19 × 23

Q 7.

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:

Soln. : (i) Using factor-tree method, we have:
ncert
⇒ 26 = 2 × 13 and 91 = 7 × 13
∴ LCM of 26 and 91 = 2 × 7 × 13 = 182
HCF of 26 and 91 = 13
Now, LCM × HCF = 182 × 13 = 2366 and
26 × 91 = 2366
i.e., LCM × HCF = Product of two numbers.
(ii) Using the factor tree method, we have :
ncert
⇒ 510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23
∴ LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23
= 23460
HCF of 510 and 92 = 2
⇒ LCM × HCF = 23460 × 2 = 46920
510 × 92 = 46920
i.e., LCM × HCF = Product of two numbers.
(iii) Using the factor tree method, we have:
ncert
∴ 336 = 2 × 2 × 2 × 2 × 3 × 7
∴ 54 = 2 × 3 × 3 × 3
LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7
= 3024
HCF of 336 and 54 = 2 × 3 = 6
LCM × HCF = 3024 × 6 = 18144
Also 336 × 54 = 18144
Thus, LCM × HCF = Product of two numbers.

Q 8.

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:

Soln. : (i) we have
ncert
⇒ 12 = 2 × 2 × 3, 15 = 3 × 5 and 21 = 3 × 7
∴ HCF of 12, 15 and 21 is 3
LCM of 12, 15 and 21 is 2 × 2 × 3 × 5 × 7
i.e., 420
(ii) we have :
ncert
LCM of 17, 23 and 29 = 17 × 23 × 29 = 11339
(iii) we have :
ncert
⇒ 8 = 2 × 2 × 2 and 9 = 3 × 3 and 25 = 5 × 5
∴ HCF of 8, 9 and 25 is 1
LCM of 8, 9 and 25 is
2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

Q 9.

Given that HCF (306, 657) = 9,
find LCM (306, 657).

SOLUTION:

Soln. : Here, HCF of 306 and 657 = 9 we have :
LCM × HCF = Product of the numbers.
∴ LCM × 9 = 306× 657
⇒ LCM = ncert = 34 × 657 = 22338
Thus, LCM of 306 and 657 is 22338.

Q 10.

Check whether 6n can end with the digit 0 for any natural number n.

SOLUTION:

Soln. : Here, n is a natural number and let 6n end
in 0.
∴ 6n is divisible by 5.
But the prime factors of 6 are 2 and 3.
⇒ 6n = (2 × 3)n
i.e., In the prime factorisation of 6n, there is no factor 5.
So, by the fundamental theorem of Arithmetic, every composite number can be expressed as a product of primes and this factorisation is unique apart from the order in which the prime factorisation occurs.
∴ Our assumption that 6n ends in 0, is wrong.
Thus, there does not exist any natural number n for which 6n ends with zero.

Q 11.

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

SOLUTION:

Soln. : We have 7 × 11 × 13 + 13 = 13[(7 × 11) + 1]
= 13[78]
i.e., 13 × 78 cannot be a prime number because it has a factors 13 and 78.
∴ 13 × 78 is a composite number.
Also 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5[7 × 6 × 4 × 3 × 2 × 1 + 1] which is also not a prime number.
[∵ It has a factor 5]
Thus, '7 × 11 × 13 + 13' and '7 × 6 × 5 × 4 × 3 × 2 × 1 + 5' are composite numbers.

Q 12.

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 meet again at the starting point?

SOLUTION:

Soln. : Time taken by Sonia to drive one round of the field = 18 minutes
Time taken by Ravi to drive one round of the field = 12 minutes
The L.C.M. of 18 and 12 gives the exact number of minutes after which they meet at the starting point again.
We have
ncert
⇒ 18 = 2 × 3 × 3 and 12 = 2 × 2 × 3
∴ LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36
Thus, they will meet again at the starting point after 36 minutes.

Q 13.

Prove that ncert is irrational.

SOLUTION:

Soln. : Let ncert be a rational number.
∴ We can find integers a and b (where, b ≠ 0 and a and b are co-prime) such that ncert
ncert ...(1)
Squaring both sides, we have a2 = 5b2
∴ 5 divides a2 ⇒ 5 divides a ...(2)
a = 5c, where c is an integer.
∴ Putting a = 5c in (1), we have
ncert
⇒ 25c2 = 5b2 ⇒ 5c2 = b2
⇒ 5 divides b2 ⇒ 5 divides b ...(3)
From (2) and (3) a and b have at least 5 as a common factor. i.e., a and b are not co-prime.
∴ Our supposition that ncert is rational is wrong.
Hence, ncert is irrational.

Q 14.

Prove that ncert is irrational.

SOLUTION:

Soln. : Let ncert is rational.
∴ We can find two co-prime integers ′a′ and 'b' such that
ncert
ncert
... (1)
a and b are integers,
ncert
ncert
⇒ from (1), ncert is a rational
But this contradicts the fact that ncert is irrational.
∴ Our supposition is wrong.
Hence, ncert is an irrational.

Q 15.

Prove that the following are irrationals.
(i) ncert
(ii) ncert
(iii) ncert

SOLUTION:

Soln. : (i) We have ncert
Let ncert be rational, and ncert
ncert is rational
Let ncert such that a and b are co-prime integers and b ≠ 0.
ncert ... (1)
Since, the division of two integers is rational.
ncert is a rational.
From (1), ncert is a rational number which contradicts the fact that ncert is irrational.
∴ Our assumption is wrong.
Thus, ncert is irrational.
(ii) Let us suppose that ncert is rational.
Let there be two co-prime integers a and b.
such that ncert where b ≠ 0
Now, ncert
ncert
ncert is a rational
This contradicts the fact that ncert is irrational.
∴ We conclude that ncert is irrational.
(iii) Let us suppose that ncert is rational.
∴ We can find two coprime integers a and b (b ≠ 0), such that ncert
ncert
ncert ... (1)
= Rational [ ∵ a and b are integers,]
ncert is a rational number.
From (1), ncert is a rational number, which contradicts the fact that ncert is an irrational number.
∴ Our supposition is wrong.
ncert is an irrational number.

Q 16.

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

(i) ncert

(ii) ncert

(iii) ncert

(iv) ncert

(v) ncert

(vi) ncert

(vii) ncert

(viii) ncert

(ix) ncert

(x) ncert

SOLUTION:

Soln. : (i) ncert
∵ 3125 = 5 × 5 × 5 × 5 × 5 = 55, which is of the form (2m × 5n)
ncert is a terminating decimal.
i.e., ncert will have a terminating decimal expansion.
(ii) ncert
∵ 8 = 2 × 2 × 2 = 23 = 1 × 23
= 50 × 23, which is of the form (2m × 5n)
ncertwill have a terminating decimal expansion.
(iii) ncert
∵ 455 = 5 × 7 × 13, which is not of the form
2m × 5n
ncertwill have a non-terminating repeating decimal expansion.
(iv) ncert
∵ 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
= 26 × 52, which is of the form 2m × 5n]
ncertwill have a terminating decimal expansion.
(v) ncert
∵ 343 = 7 × 7 × 7 = 73, which is not of the form 2m × 5n
ncert will have a non-terminating repeating decimal expansion.
(vi) ncert
ncert i.e., q = 23 × 52, which is of the form 2m × 5n.
ncert will have a terminating decimal expansion.
(vii) ncert
ncerti.e., q = 22 . 57 . 75, which is not of the form 2m × 5n.
ncert will have a non-terminating repeating decimal expansion.
(viii) ncert
ncert which is of the form 2m × 5n.
ncert will have a terminating decimal expansion.
(ix) ncert
∵ 50 = 2 × 5 × 5 = 21 × 52, which is of the form 2m × 5n.
ncert will have a terminating decimal expansion.
(x) ncert
∵ 210 = 2 × 3 × 5 × 7 = 21.31.51.71, which is not of the form 2m × 5n.
ncertwill have a non-terminating repeating decimal expansion.

Q 17.

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

SOLUTION:

Soln. : (i) ncert
ncert
Multiplying and dividing both sides by 25, we have
ncert
(ii) ncert
Multiplying and dividing by 53, we have
ncert
(iii) ncert, represents non-terminating repeating decimal expansion.
(iv) ncert
ncert
Multiplying and dividing both sides, by 54, we have
ncert
(v) ncert, represents non-terminating repeating decimal expansion.
(vi) ncert
Multiplying and dividing by 5, we have
ncert
ncert
(vii) ncert represents non-terminating repeating decimal expansion.
(viii) ncert
ncert
Multiplying and dividing by 2, we have
ncert
ncert
(ix) ncert
ncert
(x) ncert, represents non-terminating repeating decimal expansion.

Q 18.

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 ncert what can you say about the prime factors of q ?
(i) 43.123456789
(ii) 0.120120012000120000
(iii) ncert

SOLUTION:

Soln. : (i) 43.123456789
∵ The given decimal expansion terminates.
∴ It is rational of the form
ncert
ncert
Hence, p = 43123456789 and q = 29 × 59
Prime factors of q are 29 and 59.
(ii) 0.120120012000120000 ...
∵ The given decimal expansion is neither terminating nor non-terminating repeating,
∴ It is irrational number, hence cannot be written in p/q form.
(iii) ncert
∵ The given decimal expansion is non-terminating repeating.
∴ It is rational number.
ncert ... (1)
Multiplying both sides by 1000000000, we have
1000000000x = 43123456789.123456789 ... (2)
Subtracting (1) from (2), we have
(1000000000x) - x = (43123456789.123456789.......)
- 43.123456789
⇒ 999999999x = 43123456746
ncert
Here, p = 4791495194 and q = 111111111, which is not of the form 2m × 5n i.e., the prime factors of q are not of the form 2m × 5n.