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NCERT Solutions Chapter 1- REAL NUMBERS


EXERCISE 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

Answer: (i) 135 and 225
135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero,the process stops.
Since the divisor at this stage is 196,
Therefore, the HCF of 196 and 38220 is 196.

(iii) 867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51,
Therefore, the HCF of 867 and 255 is 51

Question 2:
Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q is some integer.
Answer:
Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,
a = 6q + r for some integer q ≥ 0 , and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k₁ + 1, where k₁ is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k₂ + 1, where k₂ is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k₃ + 1, where k₃ is an integer
Clearly, 6q + 1, 6q +3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1,
or 6q + 3,
or 6q + 5

Question 3:
An army contingent of 616 members is to march behind an army band
of 32 members in a 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?
Answer:
HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid’s algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.

Question 4:
Use Euclid’s division lemma to show that the square of any positive integer is either of 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.]

Answer:
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
a²=(3q)² or (3q+1)² or (3q+2)²
a²=(9q²)or 9q²+6q+1 or 9q²+12q+4
=3*(3q²) or 3(3q²+2q)+1 or 3(3q²+4q+1)+1
=3k₁ or 3k₂+1 or 3k₃+1
Where k₁, k₂, and k₃ are some positive integers
Hence, it can be said that the square of any positive integer is either of
the form 3m or 3m + 1.

Question 5:
Use Euclid’s division lemma to show that the cube of any positive
integer is of the form 9m, 9m + 1 or 9m + 8.
Answer:
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴a=3q or 3q+1 or 3q+2
Therefore,every number can be represented as these three forms.
There are three cases.
Case 1: When a = 3q,
a³=(3q)³=27q³=9(3q³)=9m,
Where m is an integer such that m = 3q³
Case 2: When a = 3q + 1,
a³ = (3q +1)³
a³ = 27q³ + 27q² + 9q + 1
a³ = 9(3q³ + 3q² + q) + 1
a³ = 9m + 1
Where m is an integer such that m = (3q³ + 3q² + q)
Case 3: When a = 3q + 2,
a³ = (3q +2)³
a³ = 27q³ + 54q² + 36q + 8
a³ = 9(3q³ + 6q² + 4q) + 8
a³ = 9m + 8
Where m is an integer such that m = (3q³ + 6q² + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.


EXERCISE 1.2



Question 1:
Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii)3825 (iv) 5005 (v) 7429

Answer:
(i) 140=2x2x5x7=2²x5x7
(ii) 156=2x2x3x13=2²x3x13
(iii) 3825=3x3x5x5x17=3²x5²x17
(iv) 5005= 5x7x11x13
(v) 7429 =17x19x23

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
Answer:
(i) 26 and 91
26=2x13
91=7x13
HCF=13
LCM=2x7x13=182
Product of two numbers= 26x91=2366
HCFxLCM=13x182=2366
Hence, product of two numbers=HCFxLCM

(ii) 510 and 92
510=2x3x5x17
92=2x2x23
HCF=2
LCM=2x2x3x5x17x23=23460
Product of two numbers= 510x92= 46920
HCFxLCM= 2x23460=46920
Hence, product of two numbers=HCFxLCM

(iii) 336 and 54
336=2x2x2x2x3x7
336=2⁴4x3x7
54=2x3x3x3
54=2x3³
HCF=2x3=6
LCM=2x3³x7=3024
Product of two numbers= 336x54=18144
HCFxLCM=6x3024=18144
Hence, product of two numbers=HCFxLCM


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
(i) 12,15 and 21
12=2²x3
15=3x5
21=3x7
HCF=3
LCM=2²x3x5x7=420

(ii) 17,23 and 29
17=1x17
23=1x23
29=1x29
HCF=1
LCM=17x23x29=11339

(iii)8,9 and 25
8=2x2x2
9=3x3
25=5x5
HCF=1
LCM=2x2x2x3x3x5x5=1800


Question 4:
Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer:
HCF(306,657)=9
We know that, LCMxHCF =Product of two numbers
∴LCMxHCF=306x657
LCM=306x657/HCF = 306x657/9
LCM=22338


Question 5:
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Answer:
If any number ends with the digit 0, it should be divisible by 10 or in
other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of 6ⁿ = (2 ×3)ⁿ
It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6ⁿ will not be divisible by 5.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.


Question 6:
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
Numbers are of two types - prime and composite. Prime numbers can be
divided by 1 and only itself, whereas composite number
have factors other than 1 and itself.
It can be observed that
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)
= 13 × 78
= 13 ×13 × 6
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 ×1009
1009 cannot be factorised further. Therefore, the given expression has
5 and 1009 as its factors. Hence, 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
meet again at the starting point?
Answer:
It can be observed that Ravi takes lesser time than Sonia for
completing 1 round of the circular path. As they are going in the same
direction, they will meet again at the same time when Ravi will have
completed 1 round of that circular path with respect to Sonia.
And the total time taken for completing this 1 round of circular path will be the
LCM of time taken by Sonia and Ravi for completing 1 round of circular
path respectively i.e., LCM of 18 minutes and 12 minutes
18 = 2 × 3 × 3
And, 12 = 2 × 2 × 3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
Therefore, Ravi and Sonia will meet together at the starting pointafter
36 minutes.



EXERCISE 1.3



Question 1:
Prove that √5 is irrational.
Answer:
Let √5 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that√5= a/b
Let a and b have a common factor other than 1. Then we can divide
them by the common factor, and assume that a and b are co-prime.
a = √5b
a² = 5b²
Therefore, a² is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5k, where k is an integer
(5k)²= 5b²
b² = 5k² This means that b² is divisible by 5 and hence, b is divisible
by 5.
This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
Hence, √5 cannot be expressed as p/q or it can be said that √5 is
irrational.

Question 2:
Prove that 3+2√5 is irrational.
Answer:
Let 3+2√5 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
3+2√5=a/b
2√5=a/b - 3
√5=1/2(a/b-3)
Since a and b are integers,1/2(a/b-3) will also be rational and
therefore,√5 is rational.
This contradicts the fact that is √5 irrational. Hence, our assumption
that 3+2√5 is rational is false. Therefore,3+2√5 is irrational.

Question 3:
Prove that the following are irrationals:
(i) 1/√2 (ii)7√5 (iii)6+√2
Answer:
(i) 1/√2
Let 1/√2 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
1/√2=a/b
√2=b/a
b/a is rational as a and b are integers.
Therefore,√2 is rational which contradicts to the fact that √2 is
irrational.
Hence, our assumption is false and 1/√2 is irrational.

(ii)7√5
Answer:
Let 7√5 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
7√5=a/b for some integers a and b
∴√5=a/7b
a/7b is rational as a and b are integers.
Therefore,√5 should be rational.
This contradicts the fact that √5 is irrational. Therefore, our assumption
that 7√5 is rational is false. Hence,7√5 is irrational.

(iii) 6+√2
Answer:
Let 6+√2 be rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
6+√2=a/b
√2=a/b-6
Since a and b are integers,a/b-6 is also rational and hence,√2 should
be rational. This contradicts the fact that √2 is irrational. Therefore,
our assumption is false and hence,6+√2 is irrational.


EXERCISE 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:
(i)13/3125 (ii)17/8 (iii)64/455 (iv)15/1600
(v)29/343 (vi)23/2³5² (vii)129/2²5⁷ 7⁵ (viii)6/15
(ix)35/50 (x)77/210
Answer:
(i)13/3125
3125=55
The denominator is of the form 5m .
Hence, the decimal expansion of 13/3125 is terminating.

(ii)17/8
8=2³
The denominator is of the form 2m.
Hence, the decimal expansion of 17/8 is terminating.

(iii)64/455
455 = 5 × 7 × 13
Since the denominator is not in the form 2m × 5ⁿ, and it also contains
7 and 13 as its factors, its decimal expansion will be non-terminating
repeating.

(iv)15/1600
1600 = 2⁶× 5²
The denominator is of the form 2m × 5ⁿ.
Hence, the decimal expansion of 15/1600 is terminating.

(v) 23/2³5²
The denominator is of the form 2m × 5ⁿ.
Hence, the decimal expansion of 23/2³5² is terminating.

(vi)23/2²5⁵
Since the denominator is not of the form 2m × 5ⁿ,and it 7 as
its factor, the decimal expansion of 23/2²5⁵ is non-terminating repeating.

(vii) 129/225775
since the denominator is not of the form 2m*5n,and it also has 7 as
its factor ,the decimal expansion of 129/2²5775 is non-terminating.
repeating.

(viii) 6/15=2x3/3x5=2/5
The denominator is of the form 5n.
Hence, the decimal expansion of 6/15 is terminating.

(ix) 35/50=7x5/10x5=7/10
10=2x5
The denominator is of the form 2m × 5n.
Hence, the decimal expansion of 35/50 is terminating.

(x)77÷210=11x7/30x7=11/30
30=2x3x5
Since the denominator is not of the form 2m × 5n, and it also has 3 as
its factors, the decimal expansion of 77/210 is non-terminating repeating.

Question 2:
Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Answer:
(i)13/3125=0.00416 3125⟌








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 , what can you say about the prime factor
of q?
(i) 43.123456789 (ii) 0.120120012000120000… (iii) 43.123456789
Answer:
(i) 43.123456789
Since this number has a terminating decimal expansion, it is a rational
number of the form p/q and q is of the form 2mx5n
i.e., the prime factors of q will be either 2 or 5 or both.

(ii)0.120120012000120000…
The decimal expansion is neither terminating nor recurring. Therefore,
the given number is an irrational number.

(iii)43.123456789 Since the decimal expansion is non-terminating recurring, the given
number is a rational number of the form p/q and q is not of the form
2mx5n i.e., the prime factors of q will also have a factor other than 2 or 5 .

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