Explore topic-wise InterviewSolutions in .

Prime factorization: 

360 = 2³ × 3² × 5

Correct option is (B) 8/3

Let x = \(2.\bar 6\)

i.e., x = 2.666.....   ________(1)

Multiply equation (1) by 10, we get

10x = 26.666.....      ________(2)

Subtract equation (1) from (2), we get

9x = 24

\(\Rightarrow x=\frac{24}9=\frac83\)

Hence, rational number that equals to \(2.\bar 6\) is \(\frac83.\)

Correct option is B) 8/3

Correct option is (B) 6

\(\because\) Cyclicity of 6 is 1.

Thus, last digit of any natural number powers of 6 is 6.

Alternative :-

Last digit of \(6^2\) is 6.  \((\because\) Last digit of \((6^2=36))\)

Last digit of \(6^3\) is 6.   \((\because\) Last digit of \((6^3=216)\) is 6)

Here, we see that 6 to the power any natural numbers ends with 6.

Thus, we conclude that last digit of \(6^{50}\) is 6.

Correct option is B) 6

Correct option is (A) 13

Let n = 1 then \(3^{2n}+2^{2n}=3^2+2^2\) = 9+4 = 13 is not divisible by 15, 17 and 19.

Correct option is (A) 13

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

Correct option is (A) 9/2

\(a=b^2=c^3\)

\(\Rightarrow log_c\,a=log_c\,b^2=log_cc^3\)   (By takin \(log_c)\)

\(\Rightarrow log_c\,a=2\,log_c\,b=3\,log_c\,c\)   \((\because log\,a^n=n\,log\,a)\)

\(\Rightarrow log_c\,a=2\,log_c\,b=3\)    \((\because log_c\,c=1)\) 

\(\therefore log_c\,a=3\,\&\,\,log_c\,b=\frac32\)

Now, \(log_c\,ab=log_c\,a+log_c\,b\)

\(=3+\frac32\) \(=\frac{9}{2}\)

Correct option is A) 9/2

The prime factors of a + b

= 3 + 7 = 10 = 2 × 5

So the least prime factor is 2 .

Correct option is B) -2

Since,

The given number is non - terminating recurring,

It is a rational number.

Given rational number is 64/455

p/q is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

64 = 2⁶

455 = 5 × 7 × 13

64 and 455 have no common factors

Therefore, 64 and 455 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

455 = 5 × 7 × 13

So, denominator is not of the form 2ⁿ5^m

Thus, 64455 is a non-terminating repeating decimal.

Length of the three measuring rods are 64cm, 80cm and 96cm, respectively. 

∴ Length of cloth that can be measured an exact number of times = LCM (64, 80, 96) 

Prime factorization: 

64 = 2⁶ 

80 = 2⁴ × 5 

96 = 2⁵ × 3 

∴ LCM = product of greatest power of each prime factor involved in the numbers = 2⁶ × 3 × 5 = 960cm = 9.6m 

Hence, 

the required length of cloth is 9.6m.

Correct option is (C) 3/8

\(\frac{3}{8}\); denominator = 8 \(=2^3\)

only prime factor is 2.

\(\therefore\frac{3}{8}\) has terminating decimals.

Correct option is C) 3/8

i) 64 = 2^x

We know that 

64 = 2 × 32 

= 2 × 2 × 16 

= 2 × 2 × 2 × 8 

= 2 × 2 × 2 × 2 × 4 

= 2 × 2 × 2 × 2 × 2 × 2 

64 = 2⁶

⇒ x = 6

ii) 100 = 5^b

Here also 100 cannot be written as any power of 5.

i.e., there exists no integer for b such that 5^b = 100

iii) \(\frac{1}{81}=3^c\)

We know that 81 = 3 x 27 

= 3 × 3 × 9 

= 3 × 3 × 3 × 3 

= 34

∴ \(\frac{1}{81}=3^{-4}\) [∵a^-3 = \(\frac{1}{a^m}\)]

∴ c = – 4

iv) 100 = 10²

z = 2

v) \(\frac{1}{256}\)= 4^a

We know that 256 = 4 × 64

= 4 × 4 × 16 

= 4 × 4 × 4 × 4 

∴ \(\frac{1}{256}\) = 4 

∴ a = – 4

Let us consider two numbers 2 + √3 and 2 – √3 which are irrationals

their sum = (2 + √3) + (2 – √3) = 4; Which is a rational numbers.

Let us assume that 1/√3 is rational

1/√3 x √3/ √3 = √3/3 is rational

Which is only possible if 1/3 is rational and √3 is rational. As we know that, Product of two rational numbers is rational

But the fact is, √3 is an irrational.

Which is contradictory to our assumption.

which implies 1/√3 is irrational. Hence proved.

We know that if 7 = 2^x then x = log₂ 7 

We want to find the value of 2^log₂ 7;

Now put log₂ 7 = x in the given

∴ 2^log₂ 7 = 2^x =7 (given)

∴ 2^log₂ 7 = 7

Thus  a^log_a N = N

a) If x = 3^log₃ 8 then

log₃ x = log₃ 8

⇒ x = 8 

b) If y = 5^log₅ 10

then log₅ y = log₅ 10

⇒ y = 10

Correct option is (D) 30

Two digits number which are divisible by 3 are 12, 15, 18, ......, 99 or \(3\times4,3\times5,.....,3\times33.\)

\(\therefore\) Total such numbers = 33 - 4 + 1 = 30

Correct option is D) 30

Correct option is (C) distributive law

The relation a (b+c) = ab+ac is a distribution law of multiplication over addition.

Correct option is C) distributive law

Given : 

log₁₀2 = 0.3010 

4²⁰¹³ = (2²)²⁰¹³ = 2⁴⁰²⁶

∴ log₁₀2⁴⁰²⁶ = 4026 log₁₀2 

[∵ log x^m= m log x] 

= 4026 × 0.3010 = 1211.826. 

So 1211 + 1 = 1212 

∴ 4²⁰¹³ has 1212 digits in its expansion. 

(∵ characteristic 1211)

No, log₁₀ 0 doesn’t exist, 

i.e a^x ≠ 0 ∀ a, x ∈ N.

Correct option is (B) (-1, 12)

\(\frac{4+3\sqrt3}{2+\sqrt3}=a+\sqrt b\)

\(\Rightarrow\) \(a+\sqrt b=\frac{4+3\sqrt3}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}\)

\(=\frac{8+6\sqrt3-4\sqrt3-9}{4-3}=-1+2\sqrt3\)

\(=-1+\sqrt{4\times3}=-1+\sqrt{12}\)

\(\therefore\) a = -1, b = 12

Thus, (a, b) = (-1, 12)

Correct option is (B) (-1, 12)

Correct option is (C) 3

\(log_{10}\,125+log_{10}\,8 \) \(=log_{10}\,5^3+log_{10}\,2^3\)

\(=log_{10}\,(5^3\times2^3)\)

\(=log_{10}\,10^3\)

\(=3\,log_{10}\,10\)

= 3

Correct option is C) 3

Let us assume that √3 + √5 is rational

On squaring, we get

(√3 + √5)² is rational

3 + 2√3 x √5 + 5 is rational

8 + 2√15 is rational

Subtract 8 from above result, considering 8 is a rational number.

As we know, Difference of two rational numbers is a rational.

8 + 2√15 – 8 is rational

2√15 is rational

Which is only possible if 2 is rational and √15 is rational.

The fact is √15 is not a rational number.

Our assumption is wrong, and

(√3 + √5) is irrational.

Correct option is (A) real number

(i) If P = 1, Q = 4

Then \(\sqrt P+\sqrt Q=\sqrt1+\sqrt4\) = 1+2 = 3 which is a rational number.

(ii) If P = 1, Q = 2

Then \(\sqrt P+\sqrt Q=\sqrt1+\sqrt2\) \(=1+\sqrt2\) which is an irrational number.

Thus, \(\sqrt P+\sqrt Q\) is rational or irrational. It is depends on fixed value of P and Q. But if P and Q are natural numbers (or positive rational numbers) then \(\sqrt P+\sqrt Q\) always a real number.

Correct option is A) real number

Correct option is (B) 5

For n = 1, \(8^n-3^n=8^1-3^1\)

= 8 - 3 = 5 which is divisible by 5 (not divisible by 3, 8 or 11)

Alternative :-

\(8^n-3^n=(3+5)^n-3^n\)

\(=3^n+\,^nC_1\,3^{n-1}\times5+\,^nC_2\,3^{n-2}\times5^2+....+5^n-3^n\)   (By using binomial expansion)

\(=5\,(^nC_1\times3^{n-1}+\,^nC_2\times3^{n-2}\times5+....+5^{n-1})\)

= 5q, where \(q=\,^nC_1\times3^{n-1}+\,^nC_2\times3^{n-2}\times5+....+5^{n-1}\) \(\in I\,or\,Z.\)

\(\therefore\) \(8^n-3^n\) is a multiple of 5.

Thus, \(8^n-3^n\) is divisible by 5.

Correct option is B) 5

Correct option is (B) 4

\(\because\) \(x=\sqrt{5}+2\)

\(\therefore\) \(\frac1x=\frac1{\sqrt{5}+2}\) \(=\frac{\sqrt{5}-2}{(\sqrt{5}+2)(\sqrt{5}-2)}\)

\(=\frac{\sqrt{5}-2}{5-4}\) \(=\sqrt{5}-2\)

\(\therefore\) \(x-\frac{1}{x}\) \(=(\sqrt{5}+2)-(\sqrt{5}-2)=4\)

Correct option is  A) 2√5

Correct option is (C) 3

Area of rectangular sheet \(=(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})\)

\(=(\sqrt{5})^2-(\sqrt{2})^2\) = 5 - 2

= 3 square units

Correct option is C) 3

Correct option is (D) -5/12

\((\frac{1}{2})^3+(\frac{1}{3})^3-(\frac{5}{6})^3\) \(=\frac18+\frac1{27}-\frac{125}{216}\)

\(=\frac{27+8-125}{216}\) \(=\frac{-90}{216}\)

\(=\frac{-90\div18}{216\div18}\) = \(\frac{-5}{12}\)

Correct option is  D) -5/12

Since a/b is a rational number and it has terminating decimal “b” will in the form 2^m x 5ⁿ where m and n are some non-negative integers.

Correct option is  B) 2

Correct option is (A) \(\frac{1999}{1000}\)

1.999 = \(\frac{1999}{1000}\)

Correct option is  B) 2

Let assume that √p is rational

Therefore it can be expressed in the form of \(\frac{a}{b}\), where a and b are integers and b ≠ 0

Therefore we can write √p = \(\frac{a}{b}\)

(√p)² = (\(\frac{p}{q}\)) ²

P = \(\frac{a^2}{b^2}\)

a ² = pb²

Since a² is divided by b², therefore a is divisible by b.

Let a = kc

(kc)² = pb²

K ²c ² = pb²

Here also b is divided by c, therefore b² is divisible by c².

This contradicts that a and b are co - primes.

Hence √p is an irrational number.

Answer is (B) even number

False

The product of two irrational is sometimes an irrational number.

Example:

If we multiply an irrational number with 0 we will get the product as 0 which is a rational number.

False

The sum of two irrational numbers is sometimes an irrational number.

Example:

If the two irrational numbers have a sum zero, then the sum is a rational number

(i) Prime factorization of 69 and 92 is: 

69 = 3 × 23 

92 = 2² × 23

Therefore, \(\frac{69}{92}\) = \(\frac{3\times23}{2^2\times23}\) = \(\frac{3}{2^2}\) = \(\frac{3}{4}\)

Thus, simplest form of \(\frac{69}{92}\) is \(\frac{3}{4}\)

(ii) Prime factorization of 473 and 645 is: 

473 = 11 × 43 

645 = 3 × 5 × 43

Therefore, \(\frac{473}{645}\) = \(\frac{11\times43}{3\times5\times43}\) = \(\frac{11}{15}\)

Thus, simplest form of \(\frac{473}{645}\) is \(\frac{11}{15}\)

(iii) Prime factorization of 1095 and 1168 is: 

1095 = 3 × 5 × 73 

1168 = 2⁴ × 73

Therefore, \(\frac{1095}{1168}\) = \(\frac{3\times5\times73}{2^4\times73}\) = \(\frac{15}{16}\)

Thus, simplest form of \(\frac{1095}{1168}\) is \(\frac{15}{16}\)

(iv) Prime factorization of 368 and 496 is:

368 = 2⁴ × 23 

496 = 2⁴ × 31 

Therefore, \(\frac{368}{496}\) = \(\frac{2^4\times23}{2^4\times31}\) = \(\frac{23}{31}\)

Thus, simplest form of \(\frac{368}{496}\) is \(\frac{23}{31}\)

No, it is not possible.

Reason:

HCF must be a factor of LCM. Here 18 is not factor of 760.

Correct option is (C) √2

\(\because\) \(\frac{1}{5\sqrt{2}}\times\sqrt{2}=\frac15\) which is an irrational number.

\(\therefore\) \(\sqrt{2}\) is the rationalising factor of \(\frac{1}{5\sqrt{2}}.\)

Correct option is  C) √2

Clearly, 1.732 is a terminating decimal. 

Hence, a rational number. 

Hence, the correct answer is option (b).

7 / 75 = 7 / 3 x 5² 

Since denominator of given rational number is not of form 2 ^m x 5ⁿ

Hence, It is non-terminating decimal expansion

(d) four decimal places

\(\frac{14753}{1250}\) = \(\frac{14753}{5^4\times2^4}\) = \(\frac{14753\times2^3}{5^4\times2^4}\) = \(\frac{118024}{1000}\) = 11.8024 

So, the decimal expansion of the number will terminate after four decimal places.

(c) \(\frac{2027}{625}\) 

\(\frac{124}{165}\) = \(\frac{124}{5\times33'}\). we know 5 and 33 are not the factors of 124. It is in its simplest form and it cannot be expressed as the product of (2^m × 5ⁿ) for some non-negative integers m, n. 

So, it cannot be expressed as a terminating decimal.

\(\frac{131}{30}\) = \(\frac{131}{5\times6'}\). we know 5 and 6 are not the factors of 131. It is in its simplest form and it cannot be expressed as the product of (2^m × 5ⁿ) for some non-negative integers m, n. 

So, it cannot be expressed as a terminating decimal.

\(\frac{2027}{625}\) = \(\frac{2027\times2^4}{5^4\times2^4}\) = \(\frac{32432}{10000}\) = 3.2432; as it is of the form (2^m × 5ⁿ), where m, n are non-negative integers. 

So, it is a terminating decimal

\(\frac{1625}{462}\) = \(\frac{1625}{2\times7\times33'}\). we know 2, 7 and 33 are not the factors of 1625. It is in its simplest form and it cannot be expressed as the product of (2^m × 5ⁿ) for some non-negative integers m, n. 

So, it cannot be expressed as a terminating decimal.

Correct option is (A) an even number

Given that \(P_1\) and \(P_2\) are two odd numbers such that \(P_1>P_2.\)

Let \(P_1\) = 2m+1 and

\(P_2\) = 2n+1

Then m > n \((\because P_1>P_2)\)

Now, \(P_1^2-P_2^2\) \(=(2m+1)^2-(2n+1)^2\)

\(=(4m^2+4m+1)-(4n^2+4n+1)\)

\(=4(m^2-n^2)+4(m-n)\)

= 4 (m-n) (m+n+1)

\(\therefore\) 4 divides \(P_1^2-P_2^2\)

\(\therefore\) 2 divides \(P_1^2-P_2^2\)

Thus, \(P_1^2-P_2^2\) is an even number.

Correct option is A) an even number

(b) two decimal places.

\(\frac{37}{2^5\times5}\) = \(\frac{37\times5}{2^2\times5^2}\) = \(\frac{185}{100}\) = 1.85

So, the decimal expansion of the rational number terminates after two decimal places.

Let x = p/q be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non - negative integers.

Then x has a decimal expansion which terminates.

The maximum power of 2 or 5 in the given rational number is 2.

So, it will terminate after 2 places of decimals.

Let us take p₁ = 5 and p₂ = 3

Then p₁ ² – p₂ ² = 25 – 9 = 16

16 is an even number.

751/2⁴5²

 = 751/ 2² × 5² × 2² 

 =  751 / 10² × 2²

 = 751 × 5² / 10² × 2² × 5²

 =  751 × 25 / 10⁴ 

 = 18,775 / 10⁴

 = 1.8775

Therefore, it will terminate after 4 places of decimal. 

Given:

If p and q are co - prime numbers.

To find:

p² and q² are

Solution:

Two numbers are co - prime if their HCF is 1 i.e they have no number common other than 1.

Let us take p = 4 and q = 5.

As 4 and 5 has no common factor other than 1, p and q are co - prime.

Now p² = 16 and q² = 25,

As 16 and 25 has no common factor other than 1,

So p² and q² are also co - prime.

Let us take the last factors of the tree 3 and 7.

Then 3 × 7 = 21

Now we get the factors in the next level as 21 and 2.

So, 21 × 2 = 42

The missing entries are 42, 21.

The decimal number of expansion of a rational number terminates if the denominator of rational number can be expressed if the denominator of rational number can be expressed as 2^m 5ⁿ where m and n are non-negative integers and p and q both co-primes.

e.g.             3 / 10 = 3 / (2¹ x 5¹) = 0.3