Real Numbers Class 10 Mathematics — Important Questions

20 Questions | 58 Marks

Explain why

7 \times 11 \times 13 + 13

is a composite number. 3marks

Express the number

4 \times 10^{3} + 2 \times 10^{2} + 5 \times 10^{1} + 6 \times 10^{0}

as a product of its prime factors. 2marks

Find the HCF and LCM of 6, 72 and 120 using the prime factorisation method. 3marks

Prove that

\sqrt{7}

is an irrational number. 3marks

Prove that

3 + 2 \sqrt{5}

is an irrational number, given that

\sqrt{5}

is an irrational number. 3marks

A seminar is being organised for students of Class X. The number of participants in a dance competition is 105 and the number of participants in a drama competition is 135. If the participants have to be arranged in rows such that each row has the same number of participants and no participant is left out, then what is the minimum number of rows required? 3marks

Find the largest number which divides 650 and 1170, leaving no remainder. 2marks

Prove that

\sqrt{2}

is an irrational number. 3marks

Prove that

15 - \sqrt{7}

is an irrational number, given that

\sqrt{7}

is an irrational number. 3marks

The prime factorisation of a natural number n is

2^{2} \times 3^{1} \times 5^{1}

. Which of the following statements is false?(a) n is divisible by 10.(b) n is divisible by 4.(c) n is divisible by 12.(d) n is divisible by 8. 1mark

Explain why

7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5

is a composite number. 3marks

Find the LCM of 48 and 180 by the prime factorisation method. 2marks

Prove that

\sqrt{13}

is an irrational number. 3marks

Prove that

2 \sqrt{11}

is an irrational number, given that

\sqrt{11}

is an irrational number. 3marks

The decimal expansion of the rational number

\frac{14587}{1250}

will terminate after how many places of decimals? 2marks

A rectangular sheet of paper has dimensions

66 cm \times 30 cm

. The paper is to be cut into identical square pieces of largest possible size. Find the number of square pieces obtained. 3marks

Prove that

5 + \sqrt{3}

is an irrational number. 3marks

The sum of two numbers is 1290 and their HCF is 30. If one of the numbers is 390, find the other number. 2marks

Can the number

4^{n}

, where n is a natural number, end with the digit 0? Give reasons. 2marks

Prove that

\frac{\sqrt{2}}{3}

is an irrational number, given that

\sqrt{2}

is an irrational number. 3marks

Answers

7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78

. Since 78 is also a composite number (

78 = 2 \times 3 \times 13

), the expression

13 \times 78

is a product of primes, thus it is a composite number.

4 \times 10^{3} + 2 \times 10^{2} + 5 \times 10^{1} + 6 \times 10^{0} = 4000 + 200 + 50 + 6 = 4256

. Prime factorisation of 4256 is

2^{5} \times 133

. Further,

133 = 7 \times 19

. So,

4256 = 2^{5} \times 7 \times 19

Prime factorisation of 6 is

2^{1} \times 3^{1}

. Prime factorisation of 72 is

2^{3} \times 3^{2}

. Prime factorisation of 120 is

2^{3} \times 3^{1} \times 5^{1}

.HCF(6, 72, 120) = Product of the smallest power of each common prime factor =

2^{1} \times 3^{1} = 6

.LCM(6, 72, 120) = Product of the greatest power of each prime factor involved =

2^{3} \times 3^{2} \times 5^{1} = 8 \times 9 \times 5 = 360

Proof by contradiction. Assume

\sqrt{7}

is rational. Then

\sqrt{7} = \frac{a}{b}

, where a and b are coprime integers and

b \neq 0

. Squaring both sides,

7 = \frac{a^{2}}{b^{2}} ⟹ 7 b^{2} = a^{2}

. This means

a^{2}

is divisible by 7. Since 7 is a prime number, if 7 divides

a^{2}

, then 7 must divide a. So,

a = 7 k

for some integer k. Substituting this into the equation,

7 b^{2} = (7 k)^{2} = 49 k^{2} ⟹ b^{2} = 7 k^{2}

. This means

b^{2}

is divisible by 7. Since 7 is a prime number, if 7 divides

b^{2}

, then 7 must divide b. So, both a and b are divisible by 7, which contradicts the fact that a and b are coprime. Therefore,

\sqrt{7}

is irrational.

Proof by contradiction. Assume

3 + 2 \sqrt{5}

is rational. Then

3 + 2 \sqrt{5} = \frac{a}{b}

, where a and b are coprime integers and

b \neq 0

. Rearranging,

2 \sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3 b}{b}

. So,

\sqrt{5} = \frac{a - 3 b}{2 b}

. Since a, b, 2, and 3 are integers,

\frac{a - 3 b}{2 b}

is a rational number. This implies

\sqrt{5}

is rational, which contradicts the fact that

\sqrt{5}

is irrational. Therefore,

3 + 2 \sqrt{5}

is irrational.

To find the minimum number of rows, we need to find the largest possible size of each square piece, which is the HCF of the dimensions of the rectangular sheet. HCF(66, 30).

66 = 2 \times 3 \times 11

30 = 2 \times 3 \times 5

HCF(66, 30) =

2 \times 3 = 6

. So, the largest possible size of each square piece is 6 cm.Number of square pieces =

\frac{Area of rectangular sheet}{Area of one square piece} = \frac{66 \times 30}{6 \times 6} = \frac{1980}{36} = 55

We need to find the HCF of 650 and 1170.

650 = 2 \times 5^{2} \times 13

1170 = 2 \times 3^{2} \times 5 \times 13

HCF(650, 1170) =

2 \times 5 \times 13 = 130

Proof by contradiction. Assume

\sqrt{2}

is rational. Then

\sqrt{2} = \frac{a}{b}

, where a and b are coprime integers and

b \neq 0

. Squaring both sides,

2 = \frac{a^{2}}{b^{2}} ⟹ 2 b^{2} = a^{2}

. This means

a^{2}

is divisible by 2. Since 2 is a prime number, if 2 divides

a^{2}

, then 2 must divide a. So,

a = 2 k

for some integer k. Substituting this into the equation,

2 b^{2} = (2 k)^{2} = 4 k^{2} ⟹ b^{2} = 2 k^{2}

. This means

b^{2}

is divisible by 2. Since 2 is a prime number, if 2 divides

b^{2}

, then 2 must divide b. So, both a and b are divisible by 2, which contradicts the fact that a and b are coprime. Therefore,

\sqrt{2}

is irrational.

Proof by contradiction. Assume

15 - \sqrt{7}

is rational. Then

15 - \sqrt{7} = \frac{a}{b}

, where a and b are coprime integers and

b \neq 0

. Rearranging,

\sqrt{7} = 15 - \frac{a}{b} = \frac{15 b - a}{b}

. Since a, b, 15 are integers,

\frac{15 b - a}{b}

is a rational number. This implies

\sqrt{7}

is rational, which contradicts the fact that

\sqrt{7}

is irrational. Therefore,

15 - \sqrt{7}

is irrational.

The number n is

2^{2} \times 3^{1} \times 5^{1} = 4 \times 3 \times 5 = 60

.(a) 60 is divisible by 10. (True)(b) 60 is divisible by 4. (True)(c) 60 is divisible by 12. (True)(d) 60 is divisible by 8. (False,

60 / 8 = 7.5

)The false statement is (d).

7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)

.Let

K = 7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1 = 1008 + 1 = 1009

.So the expression is

5 \times 1009

. Since it is a product of two integers (5 and 1009), both greater than 1, it is a composite number.

Prime factorisation of 48 is

2^{4} \times 3^{1}

. Prime factorisation of 180 is

2^{2} \times 3^{2} \times 5^{1}

.LCM(48, 180) = Product of the greatest power of each prime factor involved =

2^{4} \times 3^{2} \times 5^{1} = 16 \times 9 \times 5 = 720

Proof by contradiction. Assume

\sqrt{13}

is rational. Then

\sqrt{13} = \frac{a}{b}

, where a and b are coprime integers and

b \neq 0

. Squaring both sides,

13 = \frac{a^{2}}{b^{2}} ⟹ 13 b^{2} = a^{2}

. This means

a^{2}

is divisible by 13. Since 13 is a prime number, if 13 divides

a^{2}

, then 13 must divide a. So,

a = 13 k

for some integer k. Substituting this into the equation,

13 b^{2} = (13 k)^{2} = 169 k^{2} ⟹ b^{2} = 13 k^{2}

. This means

b^{2}

is divisible by 13. Since 13 is a prime number, if 13 divides

b^{2}

, then 13 must divide b. So, both a and b are divisible by 13, which contradicts the fact that a and b are coprime. Therefore,

\sqrt{13}

is irrational.

Proof by contradiction. Assume

2 \sqrt{11}

is rational. Then

2 \sqrt{11} = \frac{a}{b}

, where a and b are coprime integers and

b \neq 0

. Rearranging,

\sqrt{11} = \frac{a}{2 b}

. Since a, b, and 2 are integers,

\frac{a}{2 b}

is a rational number. This implies

\sqrt{11}

is rational, which contradicts the fact that

\sqrt{11}

is irrational. Therefore,

2 \sqrt{11}

is irrational.

The given rational number is

\frac{14587}{1250}

. The denominator is

1250 = 125 \times 10 = 5^{3} \times 2 \times 5 = 2 \times 5^{4}

.Since the denominator is in the form

2^{m} \times 5^{n}

, the decimal expansion will terminate. The number of decimal places is the highest power of 2 or 5, which is 4. So, it will terminate after 4 decimal places.

We need to find the HCF of the dimensions of the rectangular sheet to find the largest possible size of the identical square pieces. HCF(66, 30).

66 = 2 \times 3 \times 11

30 = 2 \times 3 \times 5

HCF(66, 30) =

2 \times 3 = 6

. So, the largest possible size of each square piece is 6 cm.Number of square pieces =

\frac{Area of rectangular sheet}{Area of one square piece} = \frac{66 \times 30}{6 \times 6} = \frac{1980}{36} = 55

Proof by contradiction. Assume

5 + \sqrt{3}

is rational. Then

5 + \sqrt{3} = \frac{a}{b}

, where a and b are coprime integers and

b \neq 0

. Rearranging,

\sqrt{3} = \frac{a}{b} - 5 = \frac{a - 5 b}{b}

. Since a, b, and 5 are integers,

\frac{a - 5 b}{b}

is a rational number. This implies

\sqrt{3}

is rational, which contradicts the fact that

\sqrt{3}

is irrational. Therefore,

5 + \sqrt{3}

is irrational.

Let the two numbers be

a

and

b

. We are given HCF(a, b) = 30. We know that HCF(a, b) × LCM(a, b) = a × b.We are given HCF(a, b) = 30 and their sum

a + b = 1290

.Also, we know that for any two numbers

a

and

b

a = 30 x

and

b = 30 y

, where

x

and

y

are coprime integers.So,

30 x + 30 y = 1290 ⟹ 30 (x + y) = 1290 ⟹ x + y = \frac{1290}{30} = 43

.We are given one number is 390. Let

a = 390

. Then

30 x = 390 ⟹ x = \frac{390}{30} = 13

.Now, substitute

x = 13

into

x + y = 43

13 + y = 43 ⟹ y = 43 - 13 = 30

.The other number is

b = 30 y = 30 \times 30 = 900

.Verification: HCF(390, 900).

390 = 2 \times 3 \times 5 \times 13

900 = 2^{2} \times 3^{2} \times 5^{2}

. HCF =

2 \times 3 \times 5 = 30

. Sum =

390 + 900 = 1290

The number is

8^{n}

, where n is a natural number. We need to check if it can end with the digit 0.For a number to end with 0, its prime factorisation must contain both 2 and 5.The prime factorisation of

8^{n}

(2^{3})^{n} = 2^{3 n}

.The only prime factor of

8^{n}

is 2. It does not have a prime factor of 5.Therefore,

8^{n}

cannot end with the digit 0 for any natural number n.

Proof by contradiction. Assume

\frac{\sqrt{2}}{3}

is rational. Then

\frac{\sqrt{2}}{3} = \frac{a}{b}

, where a and b are coprime integers and

b \neq 0

. Rearranging,

\sqrt{2} = \frac{3 a}{b}

. Since a, b, and 3 are integers,

\frac{3 a}{b}

is a rational number. This implies

\sqrt{2}

is rational, which contradicts the fact that

\sqrt{2}

is irrational. Therefore,

\frac{\sqrt{2}}{3}

is irrational.

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Real Numbers — Important Questions

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