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Correct option is (A) irrational

Number which can’t be expressed in \(\frac pq\) form is called an irrational number.

A) irrational

Correct option is (C) 0.125

\(\frac{1}{2^3}=\frac{5^3}{5^3\times2^3}=\frac{125}{10^3}\)

\(=\frac{125}{1000}\) = 0.125

Correct option is  C) 0.125

(d) 5 

We know, 

 Dividend = Divisor × Quotient + Remainder. 

It is given that: 

Divisor = 143 

Remainder = 13 

So, the given number is in the form of 143x + 31, where x is the quotient. 

∴ 143x + 31 = 13 (11x) + (13 × 2) + 5 = 13 (11x + 2) + 5 

Thus, the remainder will be 5 when the same number is divided by 13.

Any prime number greater than 3 is of the form 6k ± 1, where k is a natural number

and (6k ± 1)² = 36k² ± 12k + 1 = 6k (6k ± 2) + 1

Thus, the remainder is 1.

Correct option is (C) 5

Total possibilities A can assume is 5.

Correct option is (C) 5

Correct option is (D) 1

Any prime number greater than 3 is of the form \(6k\pm1,\) where k is a natural number.

Thus, \((6k\pm1)^2\) \(=36k^2\pm12k+1\)

= 6k (6k \(\pm\) 2) + 1

When \((6k\pm1)^2\) or (6k (6k \(\pm\) 2) + 1) is divided by 6, we get k (6k \(\pm\) 2) as a quotient and 1 as a remainder.

\(\therefore\) When the square of any prime number greater than 3 is divided by 6, we get 1 as a remainder.

Correct option is D) 1

Correct option is (C) either (A) or (B)

Let \(n^2\) is a perfect square.

Then n is a natural number.

\(\because\) n is any arbitrary natural number then for any integers a & b we have n = 3a+b, where \(0\leq b<3.\)  (By division algorithm)

\(\therefore\) Possible values for b is 0, 1 and 2.

Therefore, any natural number is of the form 3m, 3m+1 and 3m+2, where \(m\in N.\)

Case I :- If n = 3m

Then \(n^2=3m^2\)

Thus, \(31n^2\)

\(\therefore\) Remainder of \(n^2\) when divided by 3 is 0.

Case II :- If n = 3m+1

Then \(n^2=(3m+1)^2=9m^2+6m+1\) \(=3(3m^2+2m)+1.\)

Thus, when \(n^2\) is divided by 3 then leaving remainder is 1.

Case III :- If n = 3m+2

Then \(n^2=(3m+2)^2=9m^2+12m+4\) \(=3(3m^2+4m+1)+1.\)

Thus, when \(n^2\) is divided by 3 then leaving remainder is 1.

Hence for all possibilities of n, when \(n^2\) is divided by 3 then leaving remainder is either 0 or 1.

Correct option is (C) either (A) or (B)

Prime factors of 4, 7 and 13

4 = 2×2

7 and 13 are prime numbers.

LCM ( 4, 7, 13) = 364

We know that, the largest 4 digit number is 9999

Step 1: Divide 9999 by 364, we get

9999/364 = 171

Step 2: Subtract 171 from 9999

9999 – 171 = 9828

Step 3: Add 3 to 9828

9828 + 3 = 9831

Therefore 9831 is the number.

(d) 5 

We know, 

Dividend = Divisor × Quotient + Remainder. 

It is given that: 

Divisor = 143 

Remainder = 13 

So, the given number is in the form of 143x + 31, where x is the quotient. 

∴ 143x + 31 = 13 (11x) + (13 × 2) + 5 = 13 (11x + 2) + 5 

Thus, the remainder will be 5 when the same number is divided by 13.

Let a be any positive integer. Let q be the quotient and r be remainder. Then a = bq + r where q and r are also positive integers and 0 ≤ r < b 

Taking b = 3, we get 

a = 3q + r; where 0 ≤ r < 3 

When, r = 0 = ⇒ a = 3q 

When, r = 1 = ⇒ a = 3q + 1 

When, r = 2 = ⇒ a = 3q + 2 

Now, we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m or 3m + 1 for some integer m. 

⇒ Squares of 3q = (3q)² 

= 9q² = 3(3q)² = 3 m where m is some integer. 

Square of 3q + 1 = (3q + 1)² 

= 9q² + 6q + 1 = 3(3q² + 2q) + 1 

= 3m +1, where m is some integer 

Square of 3q + 2 = (3q + 2)² 

= (3q + 2)² 

= 9q² + 12q + 4 

= 9q² + 12q + 3 + 1 

= 3(3q² + 4q + 1) + 1

= 3m + 1 for some integer m.

∴ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

(d) 3.141141114….. 

3.141141114 is an irrational number because it is a non-repeating and non-terminating decimal.

(c) π is an irrational number 

π is an irrational number because it is a non-repeating and non-terminating decimal.

(i) 0.120120012000120000…

The decimal expansion of this number is non-terminating and non-repeating 

So it cannot be written in the form of p/q

So, this number is not rational number

(ii) 43.123456789 = (3123456789)/(1000000000)

This number is of the form p/q

This is a rational number

q = 1000000000 = (10)⁹ 

= (2 x 5)⁹ = 2⁹ x 5⁹

So, prime factor of q is 2 and 5.

(iii) 27.14\(\over{28}\)57

= 27.142857 142857 142857…

The decimal expansion of this number is non-terminating and recurring.

So, it can be written in the form p/q

This is a rational number.

So, besides 2 and 5, one other prime factor of q other prime positive integers is possible.

Correct option is (C) 2

\(\sqrt{4}+\sqrt{36}\) = 2+6 = 8

Cube root of \(\sqrt{4}+\sqrt{36}\) is \((\sqrt{4}+\sqrt{36})^\frac13\)

\(=(8)^\frac13=(2^3)^\frac13=2\)

Correct option is  C) 2

Correct option is (B) an irrational number

\(\pi\) is an irrational number.

B) an irrational number

Correct option is (C) irrational

Difference of an irrational and a rational number is always an irrational.

\(\because\) x is an irrational number.

\(\therefore\) x - 3 is also an irrational number.

C) irrational

Correct option is (B) 5.3

\(\sqrt{7}\) = 2.65

\(\therefore\) \(\sqrt{28}=\sqrt{4\times7}=2\sqrt7\) \(=2\times2.65=5.3\)

Correct option is  B) 5.3

Correct option is (C) irrational

\(\because\) Sum of a rational number and an irrational number is always an irrational number.

\(\therefore\) Sum of irrational number x and 2 is an irrational number.

\(\Rightarrow\) x + 2 is an irrational number.

∴ The ratio of circumference to the diameter of each circle is \(\sqrt[22]{7}\)

A. always irrational

Product of a non-zero rational and an irrational number is always irrational

For example,√3 x 2 = 2√3 this is irrational number.

i. 3.5, \(\sqrt[22] {7}\)

ii. 7, \(\sqrt[22] {7}\)

iii. 10.5, \(\sqrt[22] {7}\)

∴ The ratio of circumference to the diameter of each circle is \(\sqrt[22] {7}\)

Correct option is (D) None of these

(A) Let both rational numbers are \((\sqrt7+\sqrt3)\;and\;(\sqrt7-\sqrt3).\)

\(\therefore\) \((\sqrt7+\sqrt3)(\sqrt7-\sqrt3)\) = 7 - 3 = 4 is a rational number.

But \(\frac{\sqrt7+\sqrt3}{\sqrt7-\sqrt3}=\frac{(\sqrt7+\sqrt3)(\sqrt7+\sqrt3)}{(\sqrt7-\sqrt3)(\sqrt7+\sqrt3)}\) \(=\frac{7+3+2\sqrt{21}}{7-3}=\frac{5+\sqrt{21}}{2}\) which is an irrational number.

Thus, the ratio of greater and smaller numbers may not be an integer while the product of two irrational numbers is a rational number.

(B) For given center example.

Sum \(=(\sqrt7+\sqrt3)+(\sqrt7-\sqrt3)=2\sqrt7\) which is an irrational number.

Thus, the sum of both irrational numbers may not be a rational number while the product of two irrational numbers is a rational number.

(C) Difference greater irrational number and smaller irrational number \(=(\sqrt7+\sqrt3)-(\sqrt7-\sqrt3)=2\sqrt3\) which is an irrational number.

Here, excess of greater irrational number over the smaller irrational number is \(2\sqrt3\) which is an irrational number.

Thus, excess of the greater irrational number over the smaller irrational number may be irrational while the product of two irrational numbers is a rational number.

Hence, if the product of two irrational numbers is a rational.

Then none of the above can be concluded for sure, it may be happen or may be not.

(D) None of these

i. |15 – 2| = |13| = 13

ii. |4 – 9| = |-5| = 5 

iii. |7| x |- 4| = 7 x 4 = 28

Correct option is (D) None of these

We are disproving all given facts in options by giving a counter example against the fact.

(A) Let distinct irrational numbers are \(\sqrt8\;and\;\sqrt2.\)

Then, their product \(=\sqrt8\times\sqrt2=\sqrt{8\times2}=\sqrt{16}=4\) which is a rational number.

\(\therefore\) The product of two distinct irrational numbers is not always irrational.

(B) Let \(1+\sqrt2\) is irrational number.

Then, \(1-\sqrt2\) is one rationalising factor of \((1+\sqrt2).\)

\((\because\) \((1-\sqrt2)(1+\sqrt2)=1^2-(\sqrt2)^2\) = 1 - 2 = -1 (rational number)).

Also, any multiple of \((1-\sqrt2)\) is another rationalising factor of \((1+\sqrt2).\)

For example \(2(1-\sqrt2)(1+\sqrt2)\) = 2 (1-2) = -2 (rational number)

Thus, \(2(1-\sqrt2)\) is another rationalising factor of \((1+\sqrt2).\)

\(\therefore\) The rationalising factor of a number is not always unique.

(C) Let both distinct irrational numbers are \(\sqrt2\;and\;\sqrt3.\)

Their sum \(=\sqrt2+\sqrt3\) which is an irrational number.

Thus, The sum of two distinct irrational numbers is not always rational.

Correct option is (D) None of these

(2) may be a rational or irrational number

Correct option is (B) 6 and 35

Two integers are said to be co-prime if their HCF is 1.

(A) HCF (8, 14) = 2 \(\neq\) 1

\(\therefore\) 8 & 14 are not co-primes.

(B) HCF (6, 35) = 1

\(\therefore\) 6, 35 are co-prime numbers.

(C) HCF (4, 12) = 3

\(\therefore\) 4, 12 are not co-primes.

(D) HCF (9, 18) = 9

\(\therefore\) 9, 18 are not co-primes.

Correct option is (B) 6 and 35

Correct option is (C) 1/3

\(\pi,\sqrt2\;and\;\sqrt3\) are irrational numbers.

While \(\frac13\) is a rational number.

Correct option is (C) 1/3

Correct option is (D) 0.101001000……..

(B) The answer is √5

(3) Every real number is an irrational number

Real numbers contain rationals and irrationals.

Correct option is (A) 1681

(A) 1681 = 1600+80+1 \(=40^2+2\times40\times1+1^2=(40+1)^2=41^2\)

\(\therefore\) 1681 is a perfect square.

Also, its first two digits \(=16=4^2\) is a perfect square.

And its last two digits \(=81=9^2\) is a perfect square.

Thus, 1681 is a type of required number.

(B) 5462 is not a perfect square because any number whose last digit (unit digit) is 2 never forms a perfect square.

Thus, 5462 is not a type of required number.

Similarly 8214 and 7210 are not a perfect square.

Thus, both are not a type of required number.

Correct option is (A) 1681

Correct option is (B) √9

\(\because\) Square root of prime numbers is always an irrational number.

\(\because\) 2, 3 & 7 are prime numbers.

\(\therefore\) \(\sqrt2,\sqrt3\,,\&\,\sqrt7\) are irrational numbers.

Now, \(\sqrt9=\sqrt{3^2}=3\) which is a rational number.

Correct option is (B) √9

(D) 0.101001000…

Since the decimal expansion is neither terminating nor recurring, 0.101001000…. is an irrational number.

(D) Real numbers

Correct option is (A) an odd positive integer

Given that \(log_3\,2,\,log_3\,(2^x-5)\;and\;log_3\,(2^x-\frac{7}{2})\) are in A.P.

\(\therefore\) \(log_3\,(2^x-5)=\frac{log_3\,2+log_3\,(2^x-\frac{7}{2})}2\)

\(\Rightarrow\) \(2\,log_3\,(2^x-5)=log_3\,2+log_3\,(2^x-\frac{7}{2})\)

\(\Rightarrow\) \(log_3\,(2^x-5)^2=log_3\,\left(2(2^x-\frac{7}{2})\right)\)

\(\therefore\) \((2^x-5)^2=2\times2^x-7\)      (By comparing)

\(\Rightarrow\) \(2^{2x}-10.2^x+25=2.2^x-7\)

\(\Rightarrow\) \(2^{2x}-12.2^x+25+7=0\)

\(\Rightarrow\) \(2^{2x}-12.2^x+32=0\)

\(\Rightarrow\) \(2^{2x}-8.2^x-4.2^x+32=0\)

\(\Rightarrow2^x(2^x-8)-4(2^x-8)=0\)

\(\Rightarrow(2^x-8)(2^x-4)=0\)

\(\Rightarrow\) \(2^x\) - 8 = 0 or \(2^x\) - 4 = 0

\(\Rightarrow2^x=8=2^3\) or \(2^x=4=2^2\)

\(\Rightarrow\) x = 3 or x = 2

\(\therefore x\neq2\) because if x = 2 then \(2^x-5=2^2-5\) = 4 - 5 = -1 which can not be a domain for log function.

\(\therefore\) x = 3 which is an odd positive integer.

Correct option is (A) an odd positive integer

Correct option is (B) x² = 49/20

(A) \(\frac{1}{x}=\frac{3}{4}\) \(\Rightarrow\) \(x=\frac{1}{\frac{3}{4}}=\frac{4}{3}\) is a rational number.

(B) \(x^2=\frac{49}{20}\) \(\Rightarrow\) \(x=\sqrt{\frac{49}{20}}=\frac{7}{2\sqrt5}\) is an irrational number.

(C) \(\frac{1}{2}x+\frac{1}{5}=\frac{1}{3}\) \(\Rightarrow\) \(\frac x2=\frac13-\frac{1}{5}=\frac{2}{15}\) \(\Rightarrow x=\frac4{15}\) is a rational number.

(D) \(x^2=\frac{9}{16}\) \(\Rightarrow\) \(x=\sqrt{\frac{9}{16}}=\frac{\sqrt9}{\sqrt{16}}=\frac{3}{4}\) is a rational number.

Correct option is (B) x² = 49/20

(C) The answer is 6

\(\sqrt[3] {\sqrt {5}}\) = \(\sqrt[3 \times 2]{5}\) = \(\sqrt[6] {5}\)

∴ Order = 6

(C) Irrational number

(4) may be rational or irrational

Correct option is (A) 1.121212....

Since, among all given real numbers only 1.121212....... has a repeating decimals.

Thus, 1.121212....... is only rational number among all given real numbers.

Correct option is (B) 1.23874 …....

Correct option is (B) irrational

If n is a natural number other than a perfect square then \(\sqrt{n}\) is an irrational number.

Correct option is  B) irrational

Correct option is (C) 6

Correct option is (C) Irrational number

Correct option is (D) Neither (A) nor (B)

Correct option is (C) \(\sqrt[3]{\sqrt64}\)

It is given that: 

a = (2² × 3³ × 5⁴ ) and b = (2³ × 3² × 5) 

∴HCF (a,b) = Product of smallest power of each common prime factor in the numbers. 

 = 2² × 3² × 5 

 = 180

Step-by-step explanation:

The given numbers are

$a=2^2\times 3^3\times 5^4$

$b=2^3\times 3^2\times 5$

Highest Common Factor of a and b :

It is a largest positive integer that divides both a and b.

H.C.F. = 2^2 \times 3^2 \times 5H.C.F.=22×3<span class="" style="disp

1. b) 12

2. d) 21

3. a) 3780

4. d)45360

5. d) 2² x 3³

a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then least prime factor of (a + b) is 2.

Let, n = 6q + 5, when q is a positive integer
We know that any positive integer is of the form 3k, or 3k + 1, or 3k + 2
∴ q = 3k or 3k + 1, or 3k + 2
If q = 3k, then
n = 6q + 5
= 6(3k) + 5
= 18k + 5
= 18k + 3 + 2
= 3(6k + 1) + 2
= 3m + 2, where m is some integer
If q = 3k + 1, then
n = 6q + 5
= 6(3k + 1) + 5
= 18k + 6 + 5
= 18k + 11
= 3(6k + 3) + 2
= 3m + 2, where m is some integer
If q = 3k + 2, then
n = 6q + 5
= 6(3k + 2) + 5
= 18k + 12 + 5
= 18k + 17
= 3(6k + 5) + 2
= 3m + 2, where m is some integer
Hence, if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q.

Conversely
Let n = 3q + 2
We know that a positive integer can be of the form 6k + 1, 6k + 2, 6k + 3, 6k + 4 or 6k + 5
So, now if q = 6k + 1 then
n = 3(6k + 1) + 2
= 18k + 5
= 6(3k) + 5

= 6m + 5, where m is some integer
So, now if q = 6k + 2 then
n = 3(6k + 2) + 2
= 18k + 8
= 6 (3k + 1) + 2
= 6m + 2, where m is some integer
Now, this is not of the form 6m + 5
Hence, if n is of the form 3q + 2, then it necessarily won’t be of the form 6q + 5 always.

3 / 8 = 3 / 2³

= 3 x 5³ / 2³ x 5³

= 375 / 10³ = 375 / 1, 000

= 0.375