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Class 10 Maths Chapter 1 – Real Numbers

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Exercise 1.1 (Page 7)

Question 1:
1.Use Euclid’s division algorithm to find the HCF of:
i. 135 and 225
ii. 196 and 38220
iii. 867 and 225

Answer 1:
i. 135 and 225
As you can see, from the question 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,
225 = 135 × 1 + 90

Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,
135 = 90 × 1 + 45
Again, 45 ≠ 0, repeating the above step for 45, we get,
90 = 45 × 2 + 0
The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.

Hence, the HCF of 225 and 135 is 45.

ii. 196 and 38220
In this given question, 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as divisor, we get,
38220 = 196 × 195 + 0

We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196.
Hence, the HCF of 196 and 38220 is 196.

iii. 867 and 225
As we know, 867 is greater than 225. Let us apply now Euclid’s division algorithm on 867, to get,
867 = 225 × 3 + 102

Remainder 102 ≠ 0, therefore taking 225 as divisor and applying the division lemma method, we get,
225 = 102 × 2 + 51
Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,
102 = 51 × 2 + 0

The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51, therefore, HCF (867,225) = HCF(225,102) = HCF(102,51) = 51.

Hence, the HCF of 867 and 225 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 2:
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.
Now substituting the value of r, we get,
If r = 0, then a = 6q
Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.
If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 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 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 3:
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 the form 3m or 3m + 1 for some integer m.

Answer 4:
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 = (9q)² 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 5:
Let x be any positive integer and y = 3.
By Euclid’s division algorithm, then,
x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we get,
x = 3q

or
x = 3q + 1

or
x = 3q + 2

Now, by taking the cube of all the three above expressions, we get,

Case (i): When r = 0, then,
x²= (3q)³ = 27q³= 9(3q³)= 9m; where m = 3q³

Case (ii): When r = 1, then,
x³ = (3q+1)³ = (3q)³ +1³+3×3q×1(3q+1) = 27q³+1+27q²+9q

Taking 9 as common factor, we get,
x³ = 9(3q³+3q²+q)+1
Putting = m, we get,
Putting (3q³+3q²⁺q) = m, we get ,

x³ = 9m+1

Case (iii): When r = 2, then,
x³ = (3q+2)³= (3q)³+2³+3×3q×2(3q+2) = 27q³+54q²+36q+8
Taking 9 as common factor, we get,
x³=9(3q³+6q²+4q)+8
Putting (3q³+6q²+4q) = m, we get ,
x³ = 9m+8

Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

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Exercise 1.2 (Page 11)

Question 1:
Express each number as a product of its prime factors:

(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

Answer 1:
(i) 140
By Taking the LCM of 140, we will get the product of its prime factor.

Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2²×5×7

(ii) 156
By Taking the LCM of 156, we will get the product of its prime factor.

Hence, 156 = 2 × 2 × 13 × 3 × 1 = 2²× 13 × 3

(iii) 3825
By Taking the LCM of 3825, we will get the product of its prime factor.

Hence, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 3²×5²×17

(iv) 5005
By Taking the LCM of 5005, we will get the product of its prime factor.

Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

(v) 7429
By Taking the LCM of 7429, we will get the product of its prime factor.

Hence, 7429 = 17 × 19 × 23 × 1 = 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

Answer 2:
(i) 26 = 2 × 13
91 =7 × 13
HCF = 13
LCM =2 × 7 × 13 =182
Product of two numbers 26 × 91 = 2366
Product of HCF and LCM 13 × 182 = 2366
Hence, product of two numbers = product of HCF × LCM

(ii) 510 = 2 × 3 × 5 × 17
92 =2 × 2 × 23
HCF = 2
LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of two numbers 510 × 92 = 46920
Product of HCF and LCM 2 × 23460 = 46920
Hence, product of two numbers = product of HCF × LCM

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
HCF = 2 × 3 = 6
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024
Product of two numbers 336 × 54 =18144
Product of HCF and LCM 6 × 3024 = 18144
Hence, product of two numbers = product of HCF × LCM.

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

Answer 3:
(i) 12, 15 and 21
Writing the product of prime factors for all the three numbers, we get,

12=2×2×3
15=5×3
21=7×3

Therefore,

HCF(12,15,21) = 3
LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29
Writing the product of prime factors for all the three numbers, we get,

17=17×1
23=23×1
29=29×1

Therefore,

HCF(17,23,29) = 1
LCM(17,23,29) = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25
Writing the product of prime factors for all the three numbers, we get,

8=2×2×2×1
9=3×3×1
25=5×5×1

Therefore,

HCF(8,9,25)=1
LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800

Question 4:
Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer 4:
We have the formula that
Product of LCM and HCF = product of number
LCM × 9 = 306 × 657
Divide both side by 9 we get
LCM = (306 × 657) / 9 = 22338

Question 5:
Check whether 6n can end with the digit 0 for any natural number n.

Answer 5:
If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with 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.

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

Answer 6:
7 × 11 × 13 + 13
Taking 13 common, we get
13 (7 x 11 +1 )
13(77 + 1 )
13 (78)
It is product of two numbers and both numbers are more than 1 so it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 common, we get
5(7 × 6 × 4 × 3 × 2 × 1 +1)
5(1008 + 1)
5(1009)
It is product of two numbers and both numbers are more than 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 meet again at the starting point?

Answer 7:
They will be meet again after LCM of both values at the starting point.
18 = 2 × 3 × 3
12 = 2 × 2 × 3
LCM = 2 × 2 × 3 × 3 = 36
Therefore, they will meet together at the starting point after 36 minutes.

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Exercise 1.3 (Page 14)

Question 1:
Prove that √5 is irrational.

Answer 1:
Let us assume, that √5 is rational number.
i.e. √5 = x/y (where, x and y are co-primes)

y√5= x
Squaring both the sides, we get,

(y√5)² = x²
⇒5y² = x²……………………………….. (1)

Thus, x² is divisible by 5, so x is also divisible by 5.
Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,

5y² = (5k)²
⇒y² = 5k²
is divisible by 5 it means y is divisible by 5.

Therefore, x and y are co-primes. Since, our assumption about  is rational is incorrect.
Hence, √5 is irrational number.

Question 2:
Prove that 3 + 2√5 + is irrational.

Answer 2:
Let take that 3 + 2√5 is a rational number.
So we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
2√5 = (a-3b)/b
Now divide by 2, we get
√5 = (a-3b)/2b
Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.
Hence, 3 + 2√5 is a irrational number.

Question 3:
Prove that the following are irrationals:

(i) 1/√2
(ii) 7√5
(iii) 6 + √2

Answer 3:
(i) 1/√2

Let us assume 1/√2 is rational.
Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Rearranging, we get,
√2 = y/x

Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.
Hence, we can conclude that 1/√2 is irrational.

(ii) 7√5

Let us assume 7√5 is a rational number.
Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y

Rearranging, we get,
√5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.
Hence, we can conclude that 7√5 is irrational.

(iii) 6 +√2
Let us assume 6 +√2 is a rational number.
Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅
Rearranging, we get,

√2 = (x/y) – 6
Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Hence, we can conclude that 6 +√2 is irrational.

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Exercise 1.4 (Page 17)

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/(2²5⁷7⁵)
(viii) 6/15
(ix) 35/50
(x) 77/210

Answer 1:
(i) 13/3125
Factorize the denominator we get
3125 =5 × 5 × 5 × 5 × 5 = 5⁵So denominator is in form of 5^m so it is terminating .

(ii) 17/8
Factorize the denominator we get
8 =2 × 2 × 2 = 2³
So denominator is in form of 2^m so it is terminating .

(iii) 64/455
Factorize the denominator we get
455 =5 × 7 × 13
There are 7 and 13 also in denominator so denominator is not in form of 2^m × 5ⁿ . so it is not terminating.

(iv) 15/1600
Factorize the denominator we get
1600 =2 × 2 × 2 ×2 × 2 × 2 × 5 × 5 = 2⁶ × 5²
so denominator is in form of 2^m × 5ⁿ
Hence it is terminating.

(v) 29/343
Factorize the denominator we get
343 = 7 × 7 × 7 = 7³
There are 7 also in denominator so denominator is not in form of 2^m × 5ⁿ
Hence it is non-terminating.

(vi) 23/(2³ × 5²)
Denominator is in form of 2^m × 5ⁿ
Hence it is terminating.

(vii) 129/(2² × 5⁷ × 7⁵ )
Denominator has 7 in denominator so denominator is not in form of 2^m × 5ⁿ
Hence it is none terminating.

(viii) 6/15
divide nominator and denominator both by 3 we get 2/5
Denominator is in form of 5^m so it is terminating.

(ix) 35/50 divide denominator and nominator both by 5 we get 7/10
Factorize the denominator we get
10=2 × 5
So denominator is in form of 2^m × 5ⁿ so it is terminating.

(x) 77/210
simplify it by dividing nominator and denominator both by 7 we get 11/30
Factorize the denominator we get
30=2 × 3 × 5
Denominator has 3 also in denominator so denominator is not in form of 2^m × 5ⁿHence it is none terminating.

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

Answer 2:
(i) 13/3125 = 13/5⁵ = 13×2⁵/5⁵×2⁵ = 416/10⁵ = 0.00416
(ii) 17/8 = 17/2³ = 17×5³/2³×5³ = 17×5³/10³ = 2125/10³ = 2.125
(iv) 15/1600 = 15/2⁴×10² = 15×5⁴/2⁴×5⁴×10² = 9375/10⁶ = 0.009375
(vi) 23/2³5² = 23×5³×2²/2³ 5²×5³×2² = 11500/10⁵ = 0.115
(viii) 6/15 = 2/5 = 2×2/5×2 = 4/10 = 0.4
(ix) 35/50 = 7/10 = 0.7.

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, p q what can you say about the prime factors of q?

(i) 43.123456789
(ii) 0.120120012000120000. . .
(iii) 43.123456789

Answer 3:
(i) Since this number has a terminating decimal expansion, it is a rational number of the form p/q, and q is of the form 2^m × 5ⁿ.
(ii) The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.
(iii) 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 2^m × 5ⁿ.

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For More Content related to Class 10 –

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