Rd Sharma 2022 _mcqs for Class 10 Maths Chapter 1 - Real Numbers

Answer:

Using the factor tree for prime factorization, we have:

Therefore,

Thus, the exponent of 2 in 144 is 4.

Hence the correct choice is (a).

Answer:

It is given that the LCM of two numbers is 1200.

We know that the HCF of two numbers is always the factor of LCM

Checking all the options:

(a) 600 is the factor of 1200.

So this can be the HCF.

(b) 500 is not the factor of 1200.

So this cannot be the HCF.

(c) 400 is the factor of 1200.

So this can be the HCF.

(d) 200 is the factor of 1200.

So this can be the HCF.

Hence the correct choice is (b).

Answer:

Since, it is given that 

n = $2^3\times 3^4\times 5^4\times 7$

   $$\begin{gathered} = 2^3\times 5^4\times 3^4\times 7 \\ = 2^3\times 5^3\times 5\times 3^4\times 7 \\ = {\left(2\times 5\right)}^3\times 5\times 3^4\times 7 \\ = 5\times 3^4\times 7\times {\left(10\right)}^3 \end{gathered}$$

So, this means the given number n will end with 3 consecutive zeroes.

Answer:

Using the factor tree for prime factorization, we have:

Therefore,

The exponents of 2 and 7 are 2 and 2 respectively.

Thus the sum of the exponents is

Hence the correct choice is (c).

Answer:

We have,

Theorem states: 

Let be a rational number, such that the prime factorization of q is of the form, where m and n are non-negative integers.

Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.

This is given that the prime factorization of the denominator is of the form.

Hence, it has terminating decimal expansion which terminates after places of decimal.

Hence, the correct choice is (b).

Answer:

Two positive integers are expressed as follows:

p and q are prime numbers.

Then, taking the highest powers of p and q in the values for a and b we get:

LCM

Hence the correct choice is (c).

Answer:

Two positive integers are expressed as follows:

p and q are prime numbers.

Then, taking the smallest powers of p and q in the values for a and b we get

HCF

Hence the correct choice is (a).

Answer:

Two positive integers are expressed as follows:

p and q are prime numbers.

Then, taking the smallest powers of p and q in the values for m and n we get

HCF

Hence the correct choice is (b).

Answer:

Using the factor tree for 95, we have:

Using the factor tree for 152, we have:

Therefore,

HCF

Hence the correct choice is (c).

Answer:

HCF  

We have to find the value for LCM

We know that the product of numbers is equal to the product of their HCF and LCM.

Therefore,

Hence the correct choice is (c).

Answer:

(i) We have,

Theorem states: 

Let be a rational number, such that the prime factorization of q is not of the form, where m and n are non-negative integers.

Then, x has a decimal expression which does not have terminating decimal.

(ii) We have,

Theorem states: 

Let be a rational number, such that the prime factorization of q is not of the form, where m and n are non-negative integers.

Then, x has a decimal expression which does not have terminating decimal.

(iii) We have,

Theorem states: 

Let be a rational number, such that the prime factorization of q is not of the form, where m and n are non-negative integers.

Then, x has a decimal expression which does not have terminating decimal.

(iv) We have,

Theorem states: 

Let be a rational number, such that the prime factorization of q is of the form, where m and n are non-negative integers.

Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.

Then, x has a decimal expression which will have terminating decimal after 3 places of decimal.

Hence the (iv) option will have terminating decimal expansion.

There is no correct option.

Answer:

We have, $\frac{14587}{2^1\times 5^5}$.

Theorem states: 

Let  be a rational number such that the prime factorization of q is of the form , where m and n are non-negative integers.

Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. This is given that the prime factorization of the denominator is of the form .

Thus, it has terminating decimal expansion which terminates after 5 places of decimal.

Hence, the correct answer is option (b).

Answer:

Since the denominator is of the form $2^n{5}^m$, the decimal expansion of the given number is terminating.
 
$\frac{31}{2^3\times 5^2}$ = $\frac{31}{200}$
            = 0.155

So, the decimal expansion of $\frac{31}{2^3\times 5^2}$ is terminating.

Hence, the correct answer is option (a).

Answer:

We have,
LCM of 2³ × 3² and 2² × 3³  = 2³ × 3³

Hence, the correct answer is option (c).                                                                               

Answer:

Given that,
HCF of two numbers = 18
Product of numbers = 12960

Now, 
Product of two numbers = HCF × LCM
⇒ 12960 = 18 × LCM
$\begin{array}{rcl}& \Rightarrow & \mathrm{LCM}=\frac{12960}{18}\\ & =& 720\end{array}$

Hence, the correct answer is option (c).

 

Answer:

We have,
$$\begin{gathered} 147=3\times 7\times 7 \\ 120=2\times 2\times 2\times 3\times 5 \\ \Rightarrow \frac{147}{120}=\frac{3\times 7\times 7}{2\times 2\times 2\times 3\times 5} \\ =\frac{7\times 7}{2^3\times 5} \end{gathered}$$

The power of 2 in the denominator gives the number of decimal places after which the expansion terminates.
Thus, the places of decimal in the decimal expansion is 3.

Hence, the correct answer is option (c).

Answer:

We know that the factors of any prime number are 1 and the prime number itself.

Hence, the correct answer is option (c).

Answer:

Here,
$12=2^2\times 3$
$21=3\times 7$
$15=3\times 5$
Therefore, HCF(12, 21, 15) = 3 and
LCM(12, 21, 15) = $2^2\times 3\times 5\times 7=420$

Hence, the correct answer is option C.

Answer:

The product of a non-zero rational number and an irrational number is always an irrational number.

Hence, the correct option is (a).

Answer:

It is given that,
$$\begin{gathered} a=x^3{y}^2=x\times x\times x\times y\times y \\ b=x{y}^3=x\times y\times y\times y \\ \mathrm{HCF}\left(a, b\right)=\mathrm{HCF}\left(x^3{y}^2, x{y}^3\right)=x\times y\times y=x{y}^2 \end{gathered}$$
Hence, the correct answer is option B.

Answer:

To find the least number that is divisible by all the numbers between 1 and 10 (both inclusive), find the L.C.M of all the numbers between 1 and 10.

LCM = Product of the highest power of the prime factors 
          = 2³ × 3² × 5 × 7
          = 2520

Hence, the correct answer is option (d).

Answer:

It is given that on dividing 70 by the required number, there is a remainder of 5. This means that 70 − 5 = 65 is exactly divisible by the required number.
Similarly, 125 − 8 = 117 is exactly divisible by the required number.
The required number is HCF of 65 and 117.
By Euclid's division algorithm,
$$\begin{gathered} 117=65\times 1+52 \\ \Rightarrow 65=52\times 1+13 \\ \Rightarrow 52=13\times 4+0 \end{gathered}$$
Here, the remainder is zero. Therefore, the HCF of 65 and 117 is 13.
So, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively.

Hence, the correct answer is option A.

Answer:

LCM   ……(I)

We have to find the value for n

Also

We know that the while evaluating LCM, we take greater exponent of the prime numbers in the factorization of the number.

Therefore, by applying this rule and taking we get the LCM as

LCM ……(II)

On comparing (I) and (II) sides, we get:

Hence the correct choice is (b).

Answer:

We have,

Theorem states: 

Let be a rational number, such that the prime factorization of q is of the form, where m and n are non-negative integers.

Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.

This is given that the prime factorization of the denominator is of the form.

Hence, it has terminating decimal expansion which terminates after places of decimal.

Hence, the correct choice is (d).

Answer:

We have,

Let 

Then,

Subtract these to get

Thus, we can also conclude that all infinite repeating decimals are rational numbers.

Hence, the correct choice is (b).

Answer:

Let the two numbers be a and b.

(a) If we assume that the a and b are prime.

Then, 

HCF

LCM

(b) If we assume that a and b are co-prime.

Then, 

HCF

LCM

(c) If we assume that a and b are composite.

Then, 

HCFor any other highest common integer

LCM

(d) If we assume that a and b are equal and consider a=b=k.

Then, 

HCF

LCM

Hence the correct choice is (d).

Answer:

Let the HCF be x and the LCM of the two numbers be y.

It is given that the sum of the HCF and LCM is 1260

…… (i)

And, LCM is 900 more than HCF.

…… (ii)

Substituting (ii) in (i), we get:

Substituting in (ii), we get:

We also know that the product the two numbers is equal to the product of their LCM and HCF

Thus the product of the numbers

Hence, the correct choice is (b).

Answer:

We have,
Least composite number = 4
Least Prime Number = 2
$\begin{array}{rcl}\therefore \frac{\mathrm{LCM}\left(4,2\right)}{\mathrm{HCF}\left(4,2\right)}& =& \frac{4}{2}\\ & =& 2:1\end{array}$

Hence, the correct answer is option (b).

 

Answer:

Since 44.123 has a terminating decimal expansion, the prime factors of the denominator will be 2 and 5.

Hence, the correct answer is option (c).

Answer:

Given that, $a^2=\frac{23}{25},$
$\Rightarrow a=\frac{\sqrt{23}}{5}$ which is an irrational number.

Hence, the correct answer is option (b).

Answer:

We have, 
$\mathrm{LCM} \times \mathrm{HCF}=\mathrm{Product} \mathrm{of} \mathrm{two} \mathrm{numbers}$

$$\begin{gathered} \Rightarrow x\times 18=36\times 2 \\ \Rightarrow x=\frac{36\times 2}{18}=4 \end{gathered}$$
Thus, the value of x is 4.

Hence, the correct answer is option (c).

Answer:

Given that, HCF of two numbers = 81

Let the two numbers are 81x and 81y.
$$\begin{gathered} \Rightarrow 81x+81y=1215 \\ \Rightarrow x+y=15 \end{gathered}$$

Thus, the four co-prime numbers are (1, 14), (2, 13), (4, 11), (7, 8).

So,
For x = 1, y = 14, the numbers are 1 × 81 + 14 × 81 = 81 + 1134 = 1215
For x = 7, y = 8, the numbers are 7 × 81 + 8 × 81 = 567 + 648 = 1215
For x = 2, y = 13, the numbers are 2 × 81 + 13 × 81 = 162 + 1053 = 1215
For x = 4, y = 11, the numbers are 4 × 81 + 11 × 81 = 324 + 891 = 1215

Therefore, the numbers of such pairs are 4.

Hence, the correct answer is option (c).

 

Answer:

We have,
LCM of two prime numbers = Product of two numbers
                                              = 13 × 17
Therefore, p = 17 and q = 13.
$\begin{array}{rcl}\therefore 3p-q& =& 3\times 17-13\\ & =& 51-13\\ & =& 38\end{array}$

Hence, the correct answer is option (c).

Answer:

For terminating decimal expansion after one decimal place, the highest power of m and n in should be 1.

Let .

We will get:

$\begin{array}{rcl}\frac{1}{13}\times \frac{13}{100}& =& \frac{1}{100}\\ & =& 0.01\end{array}$

Thus, it is evident that we multiplied it by $\frac{13}{100}$ ​so that its decimal expansion terminates after two decimal places.

Hence, the correct answer is option (a).

Answer:

Consider three numbers a, a + 2 and a + 4.

Case 1: If $a\in \mathrm{even}$, then a, a + 2 and a + 4 are even.
Let a = 2q.
So, a + 2 = 2q + 2 = 2(q + 1) and
a + 4 = 2q + 4 = 2(q + 2)

Out of three consecutive numbers q, q + 1 and q + 2, one of them will always be divisible by 3. Therefore, one out of 2q, 2q + 2 and 2q + 4, one of them will always be divisible by 3.

For a = 2, the three numbers are 2, 4, 6 and only 6 is divisible by 3.
For a = 4, the three numbers are 4, 6, 8 and only 6 is divisible by 3.
For a = 6, the three numbers are 6, 8, 10 and only 6 is divisible by 3 and so on.

Case 2: If $a\in \mathrm{odd}$, then a, a + 2 and a + 4 are odd.
Then, out of these consecutive odd numbers, any one of them will always be divisible by 3 as every third natural number is a multiple of 3.

For a = 1, the three numbers are 1, 3, 5 and only 3 is divisible by 3.
For a = 3, the three numbers are 3, 5, 7 and only 3 is divisible by 3.
For a = 5, the three numbers are 5, 7, 9 and only 9 is divisible by 3 and so on.

Thus, in both the cases, any one of numbers a, a + 2 and a + 4 is a multiple of 3.

Hence, the correct answer is option (b).

Answer:

LCM

HCF

We know that the product of numbers is equal to the product of their HCF and LCM.

Therefore,

Hence the correct choice is (c).

Answer:

We know that the co-prime numbers have no factor in common, or, their HCF is 1.

Thus,and have the same factors with twice of the exponents of p and q respectively, which again will not have any common factor.

Thus we can conclude that and are co-prime numbers.

Hence, the correct choice is (a).

Answer:

Since

The least prime factor of has to be 2; unless is a prime number greater than 2.

Suppose is a prime number greater than 2. Then must be an odd number

o one of a or b must be an even number. 

Suppose then that a is even. Then the least prime factor of a is 2; which is not 3 or 7. So a can not be an even number nor can b be an even number. Hence can not be a prime number greater than 2 if the least prime factor of a is 3 and b is 7.

Thus the answer is 2.

Hence the correct choice is (a).

Answer:

Out of the given choices is the only smallest number by which if we multiply we get a rational number.

Hence, the correct choice is (c).

Answer:

For terminating the decimal expansion after one place of decimal, the highest power of m and n in should be 1.

Let

We will get:

Thus, it is evident that we multiplied it by

Hence, the correct choice is (a).

Answer:

We know that is always divisible by both and .

So, is always divisible by both and.

Hence, the correct choice is (c).

Answer:

We know that will end in 6

And will end in 5.

Now, always end with

Hence the correct choice in (a).

Answer:

Any prime number greater than 3 is of the form, where k is a natural number.

Thus,

When, is divided by 6, we get, and remainder as 1.

Hence, the correct choice is (a).

Answer:

It is known that, even integers are ...,−4, −2, 0, 2, 4,...
Observe that, all of the even numbers are the multiple of 2.
So, even numbers can be written as 2m, where, m is an integer.
m can be ...,−2, −1, 0, 1, 2,...
∴ 2m can be ...,−4, −2, 0, 2, 4,...
Hence, the correct answer is option C.

Answer:

We know that, all numbers that are not the multiple of 2 are odd numbers.
Odd integers are ...,−3, −1, 1, 3, 5,...
So, odd numbers can be written as 2m + 1, where m is an integer.
m can be ..., −2, −1, 0, 1, 2,...
∴ 2m + 1 can be ..., −3, −1, 1, 3,...
Hence, the correct answer is option D.

Answer:

From Euclid's lemma, we have
a = 3q + r          
$\therefore 0\le r<3$

Hence, the correct answer is option (a).



 

Answer:

From Euclid's Lemma, for two positive integers a and b, there exist unique integers q and r such that a = bq + r,
Where  0 ≤ r < b.

Hence, the correct answer is option (c).
 

Answer:

The product of a non-zero rational number and an irrational number is always an irrational number.

Hence, the correct option is (a).

Answer:

$\begin{array}{rcl}{25}^{2n}–9^{2n}& =& {\left(5^{2n}\right)}^2-{\left(3^{2n}\right)}^2\\ & =& \left(5^{2n}+3^{2n}\right)\left(5^{2n}-3^{2n}\right)\end{array}$

For n = 1,
$\begin{array}{rcl}{25}^{2n}–9^{2n}& =& \left(5^2+3^2\right)\left(5^2-3^2\right)\\ & =& \left(25+9\right)\left(25-9\right)\\ & =& 34\times 16\end{array}$
Thus, 25²ⁿ – 9²ⁿ is always divisible by 16 and 34 both for n = 1.

For n = 2,
$\begin{array}{rcl}{25}^{2n}–9^{2n}& =& \left(5^4+3^4\right)\left(5^4-3^4\right)\\ & =& \left(625+81\right)\left(625-81\right)\\ & =& 706\times 544\\ & =& 34\times 16\times \left(706\right)\end{array}$
Thus, 25²ⁿ – 9²ⁿ is always divisible by 16 and 34 both for n = 2.

So, for any value of n,  25²ⁿ – 9²ⁿ is always divisible by 16 and 34 both.

Hence, the correct answer is option (c).

Answer:

HCF of two numbers is always a factor of their LCM.

Hence, the correct answer is option (b).

Answer:

Given that, $\frac{327}{2^3\times 5}$
The number of terminating decimal places is decided by the number of 2's.

Thus, the given rational number will terminate after 3 decimal places.

Hence, the correct answer is option (c).

Answer:

Given that, (a × 5)ⁿ ends with the digit zero for every natural number n.

For every natural number n, (a × 5)ⁿ can end with zero only if (a × 5) ends with 0. This happens only for even values of a.

Hence, the correct answer is option (b).

Answer:

All decimal numbers are rational numbers.

Hence, the correct answer is option (a).

Answer:

LCM of 4, 7 and 14 = 28
Thus, all the bells rang together again at 6 : 28 AM.

Hence, the correct answer is option (c).

Answer:

Given that,
Numbers of the form 10n +1, where n is a Natural number such that 1 ≤ n <10

Now, put n = 1
The number is 10(1) + 1 = 11

Again, putting n = 2
The number is 10(2) + 1 = 21
and so on till n = 9 as n < 10

The number is 10(9) + 1 = 91
Numbers are 11, 21, 31, 41, 51, 61, 71, 81, 91

Prime numbers are numbers that have exactly 2 factors, 1 and the number itself.

Out of these, the prime numbers are 11, 31, 41, 61, 71.

Thus, 5 prime numbers are of form 10n + 1, where n is a whole number such that 1 ≤ n < 10.

Hence, the correct answer is option (a).

Answer:

We have,
119² – 11² = (119 + 11)(119 − 11)
                  = 130 × 108
Since 130 and 108 both are even numbers. 
Thus 119² – 11² is a composite number.

Hence, the correct answer is option (b).

Answer:

π is an irrational number as it has non-terminating, non-repeating decimal expansion given as follows:
π = 3.1415926535897932...

Hence, the correct answer is option (b).

Answer:

The sum of irrational numbers is an irrational number. Also, the sum of a rational and an irrational number is irrational.

Thus, $\sqrt{5}+\sqrt{3}+2$ is an irrational number because $\sqrt{5} \mathrm{and} \sqrt{3}$ are irrationals.

Hence, the correct answer is option (d).

Answer:

We have,
$\begin{array}{rcl}\sqrt{18}& =& \sqrt{3\times 3\times 2}\\ & =& 3\sqrt{2}\end{array}$

Thus, the smallest number by which $\begin{array}{rcl}& & 3\sqrt{2}\end{array}$ should be multiplied is $\sqrt{2}$.

$\Rightarrow 3\sqrt{2}\times \sqrt{2}=6$ which is a rational number.

Hence, the correct answer is option (c).
 

Answer:

The decimal expansion of π is 3.14159...
Since it is non-terminating and non-repeating.
Thus, π is irrational. 

Hence, the correct answer is option (b).

Answer:

We have, $\frac{441}{2^2\times 5^3\times 7}$.
$\frac{441}{2^2\times 5^3\times 7}=\frac{63}{2^2\times 5^3}$

Since the denominator of this rational number is in the form of 2^m5ⁿ, so it will have a terminating decimal representation.

Hence, the correct answer is option (a).

Answer:

We have, $\frac{17}{8}$ 

Since the denominator of the rational number is of the form 2³ , therefore it will terminate after 3 decimal places.

Hence, the correct answer is option (c).

Answer:

We have, $\frac{63}{72\times 175}$.

$\begin{array}{rcl}& \Rightarrow & \frac{63}{72\times 175}=\frac{7\times 3^2}{2^3\times 3^2\times 5^2\times 7}\\ & =& \frac{1}{2^3\times 5^2}\\ & & \\ & & \end{array}$ 

Since the denominator of this rational number is in the form of 2^m × 5ⁿ, so it will have a terminating decimal representation.

Hence, the correct answer is option (a).

Answer:

(i) By dividing all the options by 2, 3, 4, 5, 6, and 11, we will get that 539 is the only option which leaves the remainder of 1, 2, 3, 4, 5, and 0 respectively.
Dividing by 2, it leaves remainder 1. So, the number should be odd.
Dividing by 5, it leaves remainder 4. So, the number should end with 4 or 9. Option (a) is eliminated.
Dividing by 4, it leaves remainder 1. So, options (b) and (d) are eliminated.

Hence, the correct answer is option (c).

(ii) After removing 3 chairs, the number of chairs left are 536 chairs. Since it is an even number and divisible completely by both 2 and 4, the chairs can be arranged in pairs of 2'^s or 4'^s.

Hence, the correct answers are options (a) and (c).

Disclaimer: There are two correct answers for the question and the explanation is given above.

(iii) If one chair is added to 536 chairs, the number of chairs become 537.
Now, 537 is completely divisible by 3. So, the chairs can be arranged in groups of 3'^s.

Hence, the correct answer is option (b).

Disclaimer: The correct answer for the question is option (b) and the explanation for the same is given above.

(iv) If one chair is added to 539 chairs, the total number of chairs become 540. It is already given in the question that the initial number of chairs is completely divisible by 11.

Therefore, when one more chair is added, the number of chairs left after they are arranged in groups of 11 is 1.

Hence, the correct answer is option (a).

(v) When 539 is divided by 9, the remainder will be 8. Thus, arranging chairs in pairs of 9^'s, 8 chairs will be left.

Hence, the correct answer is option (a).

Answer:

(i) To find the maximum number of guests Mira can invite, find the HCF of the quantities of two fruits which is the highest common factor between the two quantities.

Now, factors of 60 bananas and 36 apples
$$\begin{gathered} 36=2\times 2\times 3\times 3 \\ 60=2\times 2\times 3\times 5 \end{gathered}$$

So, the highest common factor (HCF) between them is 2 × 2 × 3 = 12.
Thus, the maximum number of guests Mira can invite is 12.

Hence, the correct answer is option (c).

(ii) As Mira wants to distribute the apples equally among all the guests and the number of guests is 12. So,
Number of apples each guest will get = $\frac{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{apples} \mathrm{she} \mathrm{has}}{\mathrm{Number} \mathrm{of} \mathrm{guests} \mathrm{she} \mathrm{invites}}$
                                                            $$\begin{gathered} =\frac{36}{12} \\ =3 \end{gathered}$$

Hence, the correct answer is option (a).

(iii) As Mira wants to distribute the bananas equally among all the guests and the number of guests is 12. So,
Number of bananas each guest will get = $\frac{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{bananas} \mathrm{she} \mathrm{has}}{\mathrm{Number} \mathrm{of} \mathrm{guests} \mathrm{she} \mathrm{invites}}$
                                                               $$\begin{gathered} =\frac{60}{12} \\ =5 \end{gathered}$$

Hence, the correct answer is option (d).

(iv) As Mira decides to add 42 mangoes, so we need to find out the HCF of quantities of all three fruits.

Factors 36, 60, and 42 are-
$$\begin{gathered} 36=2\times 2\times 3\times 3 \\ 60=2\times 2\times 3\times 5 \\ 42=2\times 3\times 7 \end{gathered}$$

So, the highest common factor (HCF) between them is 2 × 3 = 6.
Therefore, she can invite a maximum of 6 guests in that case.

Hence, the correct answer is option (b).

Answer:

(i)

Answer:

(i) From the factor tree, the two factors of x are observed to be 3 and 1275.
So,
$$\begin{gathered} x=3\times 1275 \\ =3825 \end{gathered}$$
Thus, the value of x is 3825.

Hence, the correct answer is option (b).

(ii) From the factor tree, the two factors of 425 are observed to be y and 85.
So,
$$\begin{gathered} 425=y\times 85 \\ \Rightarrow y=\frac{425}{85} \\ \Rightarrow y=5 \end{gathered}$$
Thus, the value of y is 5.

Hence, the correct answer is option (a).

(iii) From the factor tree, the two factors of 85 are observed to be z and 5.
So,
$$\begin{gathered} 85=5\times z \\ \Rightarrow z=\frac{85}{5} \\ \Rightarrow z=17 \end{gathered}$$
Thus, the value of z is 17.

Hence, the correct answer is option (b).

(iv) 
$\begin{array}{rcl}x+y+z& =& 3825+5+17\\ & =& 3847\end{array}$
Thus, the value of x is 3847.

Hence, the correct answer is option (b).

Answer:

Statement-1: HCF and LCM of two natural numbers are 25 and 815.
Now,
HCF of 25 and 815 is 5
LCM of 25 and 815 is (25 × 163) = 4075
Therefore, 25 and 815 are divisible by only 5. So, HCF is 5 and LCM is 4075. Also, the LCM of the two numbers is divisible by their HCF too.

Thus, Statement-1 is true.

Statement-2: LCM of two natural numbers is always divisible by their HCF.

The LCM (least common multiple) or simply, the multiple of two natural numbers is always divisible by their HCF (highest common factor).

Thus, Statement-2 is true and is the correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-1: HCF (234, 47) = 1
234 = 2 × 3 × 3 × 13
47 = 1 × 47

There is no common factor between the two numbers. So, their HCF is 1.
Thus, Statement-1 (Assertion) is true.

Statement-2: HCF of two co-primes is always 1.
If there are two co-prime numbers, then their only common factor is 1. Therefore, HCF of two co-primes is always 1.
Thus, Statement-2 (Reason) is true.

Thus, Statement-1 is true and Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-1: $\sqrt{11}$ is an irrational number.
The given statement is true as $\sqrt{11}$ cannot be written in the form of $\frac{p}{q}, q\ne 0$.

Thus, Statement-1 is true.

Statement- 2: If p is a prime number, then $\sqrt{p}$ is an irrational number.
Let $\sqrt{p}$ be a rational number. Then,
$$\begin{gathered} \sqrt{p}=\frac{a}{b} \\ \Rightarrow p=\frac{a^2}{b^2} \\ \Rightarrow a^2=p{b}^2 \end{gathered}$$
∴ p divides $a^2$.

But when a prime number divides the product of two numbers, it must divide at least one of them. So, let p divides a.
$$\begin{gathered} \Rightarrow a=pk, \mathrm{for} \mathrm{some} k \\ \Rightarrow {\left(pk\right)}^2=p{b}^2 \\ \Rightarrow b^2=p{k}^2 \end{gathered}$$
∴ p divides $b^2$. Thus, p divides b.

Thus, a and b have at least one common multiple p. But it arises the contradiction to our assumption that a and b are coprime.
Thus, our assumption is wrong and p​ is irrational number.

​Thus, Statement-2 is true and it is the correct explanation of Statement-1 is true.

Hence, the correct answer is option (a).

Answer:

Statement-1: HCF of two consecutive natural numbers is 1.
The set of natural numbers is {1, 2, 3, 4, 5, ...}. It is evident that there is no common factor (except 1) between any two consecutive natural numbers. So, the HCF of two consecutive natural numbers is 1.

Thus, Statement-1 is true.

Statement-2: HCF of two co-primes is 1.
There is no common factor between two co-primes numbers other than 1. So, their HCF is 1.

Thus, Statement-2 is true and it is the correct explanation of Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement I:
Let $f\left(n\right)=n^3-n$
Now, $f\left(1\right)=1^3-1=0$ divisible by 6
Again, $f\left(2\right)=2^3-2=6$ divisible by 6
So the given statement is true for $n=1,2$
Let us take some integer k divisible by 6
$\therefore f\left(k\right)=k^3-k$ divisible by 6
$\Rightarrow f\left(k\right)=6k_1 \left(k_1 \mathrm{is} \mathrm{some} \mathrm{integer}\right)$                    .....(1)

Also,
$\begin{array}{rcl}f\left(k+1\right)& =& {\left(k+1\right)}^3-\left(k+1\right)\\ & =& \left(k^3-k\right)+3k\left(k+1\right) .....\left(2\right)\end{array}$

Now, k(k + 1) is the product of two consecutive integers, which is always divisible by 2.
$\Rightarrow k\left(k+1\right)=2k_2 \left(k_2 \mathrm{is} \mathrm{some} \mathrm{integer}\right)$
From (1) and (2), we have
$f\left(k+1\right)=6\left(k_1+k_2\right)$

So, f(k+1) is divisible by 6.
So, the statement is true for $n=k+1$ if we assume it to be true for $n=k$.

Statement - II
Let the three consecutive positive integers be n, n + 1 and n + 2. When a number is divided by 3, the remainder obtained is either 0, 1 or 2.

Answer:

Statement-1: If HCF (a, b) = 4 and ab = 96 × 404, then LCM (a, b) = 9696.
We have,
$$\begin{gathered} \mathrm{HCF}\times \mathrm{LCM}=a\times b \\ \Rightarrow \mathrm{LCM}=\frac{a\times b}{\mathrm{HCF}} \\ =\frac{96\times 404}{4} \\ =9696 \end{gathered}$$

Thus, Statement-1 is true.

Statement-2 (Reason): LCM of two numbers a and b = HCF (a, b) × ab.
The product of two numbers is equal to the product of their LCM and HCF.
So, Statement-2 is false.
$$\begin{gathered} \mathrm{HCF}\times \mathrm{LCM}=a\times b \\ \Rightarrow \mathrm{LCM}=\frac{a\times b}{\mathrm{HCF}} \end{gathered}$$

Thus, Statement-1 is true and Statement-2 is false.

Hence, the correct answer is option (c).



 

Answer:

Statement-1 (Assertion): 997 is the largest three digit prime number.
Starting from 999, we see that 999 and 998 are not prime numbers. But 997 has only 2 factors, 1 and 997 itself, satisfying all the conditions of a prime number. Therefore, 997 is the largest three-digit prime number in the number system.

 Thus, Statement-1 is true.

Statement-2 (Reason): A positive integer n is a prime number, if no positive integer less than or equal to $\sqrt{n}$ divides n.

Let n be the positive integer such that no prime less than $\sqrt{n}$ divides n. Suppose n is not prime, then we have
$$\begin{gathered} n=ab \mathrm{where} 1<a\le b \\ \Rightarrow a\le \sqrt{n} \mathrm{and} b\ge \sqrt{n} \end{gathered}$$
Let p be a prime factor of a. Then,
$$\begin{gathered} \Rightarrow p\le a\le \sqrt{n} \\ \Rightarrow p|ab \\ \Rightarrow p|n \\ \therefore \mathrm{a} \mathrm{prime} \mathrm{less} \mathrm{than} \sqrt{n} \mathrm{divides} n \end{gathered}$$

This contradicts our assumption that no prime less than $\sqrt{n}$ divides n.
Thus, statement-2 is true but statement-1 is false.

Hence, the correct answer is option (d).
 

Answer:

Statement-1 (Assertion): $\sqrt{2}+\sqrt{3}$ is an irrational number.
The sum of two irrational numbers $\sqrt{2}$ and $\sqrt{3}$ is an irrational number.
Thus, Statement-1 is true.

Statement-2 (Reason): If p and q are prime positive integers, then $\sqrt{p}+\sqrt{q}$ is an irrational number.
The sum of two irrational numbers is an irrational number.

Thus, Statement-2 is true and it is a correct explanation for Statement-1.

Hence, the correct answer is option (a).