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Correct option is (D) infinite

Infinitely many rational numbers are exist between any two distinct rational numbers.

Correct option is D) infinite

Correct option is (D) 105^1/35

\(\sqrt[35]{105}\) = \((105)^\frac1{35}\)

Correct option is  D) 105^1/35 

Correct option is (A) 3 x 1.732

\(\sqrt{27}=\sqrt{3^3}=\sqrt{3^2\times3}\)

\(=\sqrt{3^2}\times3=3\times\sqrt3\)

= \(3\times1.732\)

A) 3 x 1.732

Correct option is  A) 1,3, 4, 2

Correct option is (D) Odd number

Sum of three odd numbers is odd.

Sum of an odd number and an even number is odd.

\(\therefore\) Sum of three odd numbers and one even number is odd.

D) Odd number.

Correct option is (C) \(\frac{10}{14}\)

\(\frac{5}{7}=\frac{5}{7}\times\frac22\) \(=\frac{5\times2}{7\times2}\)

= \(\frac{10}{14}\)

Correct option is   C) \(\frac{10}{14}\)

Correct option is (C) \(\sqrt{125}\)

Given that \(\sqrt{5}\) = 2.236

Now, 11.18 = 2.236 \(\times\) 5

\(=5\sqrt5=\sqrt{5^2\times5}\)

\(=\sqrt{125}\)

Correct option is   C) \(\sqrt{125}\)

Correct option is  A) 2√2 

Correct option is (B) 3^2/5

\(\sqrt[5]{3^2}=(3^2)^\frac15=3^{2\times\frac15}\)

\(=3^{\frac25}\)

Correct option is  B) 3^2/5

Correct option is (A) \(\sqrt[3]{3}\)

\(\sqrt[3]{9}=(9)^\frac13=(3^2)^\frac13=(3)^\frac23\)

\(\because\) \((3)^\frac23\times(3)^\frac13=3^{(\frac23+\frac13)}\) \(=3^1=3\) which is a rational number.

i.e., \(\sqrt[3]{9}\times\sqrt[3]{3}=3\) a rational number.

It implies \(\sqrt[3]{3}\) is a rationalising factor of \(\sqrt[3]{9}.\)

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

Correct option is (C) 2√3

\(\because\) \(x=2+\sqrt{3}\)

\(\therefore\) \(\frac1x=\frac1{2+\sqrt{3}}\) \(=\frac1{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}\)

\(=\frac{2-\sqrt{3}}{4-3}\) \(=2-\sqrt{3}\)

\(\therefore\) \(x-\frac1x=2+\sqrt{3}-(2-\sqrt{3})=2\sqrt{3}\)

Correct option is  B) √3

Correct option is (B) 3.142

\(\frac{22}{7}\) = 3.142  (upto three decimals)

Correct option is  B) 3.142

Correct option is  C) 0.3162

Correct option is (C) 10

x+y \(=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)

\(=\frac{(\sqrt{3}+\sqrt{2})^2+(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}\)

\(=\frac{(3+2+2\sqrt6)+(3+2-2\sqrt6)}{3-2}\)

\(=\frac{5+5}1=10\)

Correct option is  C) 10

Correct option is (C) 0.3162

\(\because\) \(\sqrt{10}=3.162\)

\(\therefore\) \(\frac1{\sqrt{10}}=\frac{\sqrt{10}}{\sqrt{10}\times\sqrt{10}}=\frac{\sqrt{10}}{10}\) \(=\frac{3.162}{10}=0.3162\)

Correct option is  C) 0.3162

Correct option is (A) 6.324

\(\because\) \(\sqrt{10}\) = 3.162

\(\therefore\) \(\sqrt{40}=\sqrt{4\times10}\) \(=2\sqrt{10}=2\times3.162\)

= 6.324

Correct option is  A) 6.324

Six bells toll together at intervals of 2,4, 6, 8, 10 and 12 minutes, respectively. 

Prime factorization: 

2 = 2 

4 = 2 × 2 

6 = 2 × 3 

8 = 2 × 2 × 2 

10 = 2 × 5 

12 = 2 × 2 × 3 

∴ LCM (2, 4, 6, 8, 10, 12) = 2³ × 3 × 5 = 120 

Hence, 

after every 120minutes (i.e. 2 hours), they will toll together

∴ Required number of times = \((\frac{30}{2}+1)\) = 16

Six bells toll together at intervals of 2,4, 6, 8, 10 and 12 minutes, respectively. 

Prime factorization: 

2 = 2 

4 = 2 × 2 

6 = 2 × 3 

8 = 2 × 2 × 2 

10 = 2 × 5 

12 = 2 × 2 × 3 

∴ LCM (2, 4, 6, 8, 10, 12) = 2³ × 3 × 5 = 120 

Hence, after every 120 minutes (i.e. 2 hours), they will toll together.

∴ Required number of times = (30/2 +1) = 16

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively.

Find LCM using prime factorization:

2 = 2 (prime number)

4 = 2 × 2

6 = 2 × 3

8 = 2 × 2 × 2

10 = 2 × 5

12 = 2 × 2 × 3

Therefore, LCM (2, 4, 6, 8, 10, 12) = 2 x 2 x 2 × 3 × 5 = 120

After every 120 minutes = 2 hours, bells will toll together.

So, required number of times = (30/2 + 1) = 16 times.

Correct option is  C) √3 – √2

Correct option is (B) 12

To find:

The number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21

Solution:

To be exactly divisible by each of 8, 15 and 21,the required number must be divisible by the LCM of 8, 15 and 21 i.e.by 840.

Now on dividing 110000 by 840 we get 800 as remainder.

Therefore the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21 is 110000 – Remainder ⇒ 110000 – 800 = 109200

Now 109200 gives remainder 0 when divided by 8, 15 and 21 as it is completely divisible by their LCM.

We know that, the greatest 6 digit number is 999999. 

Let’s assume that 999999 is divisible by 24, 15 and 36 exactly. 

Then, the LCM (24, 15 and 36) should also divide 999999 exactly.

Finding the prime factors of 24, 15, and 36, we get 

24 = 2 × 2 × 2 × 3 

15 = 3 × 5 

36 = 2 × 2 × 3 × 3 

⇒ L.C.M of 24, 15 and 36 = 360 

Since, \(\frac{(999999)}{360}\) = 2777 × 360 + 279 

Here, the remainder is 279. 

So, the greatest number which is divisible by all three should be = 999999 – 279 = 9997201 

∴ 9997201 is the greatest 6 digit number which is exactly divisible by 24, 15 and 36.

Minimum number of rooms required =Total number of participants /HCF (60,84,108) 

Prime factorization of 60, 84 and 108 is: 

60 = 2² × 3 × 5 

84 = 2² × 3 × 7 

108 = 2² × 3³

HCF = product of smallest power of each common prime factor in the numbers = 2² × 3 = 12 

Total number of participants = 60 + 84 + 108 = 252 

Therefore, minimum number of rooms required = 252/12 = 21 

Thus, minimum number of rooms required is 21

(c) 0 ≤ r ˂ b

Euclid’s division lemma, states that for any positive integers a and b, there exist unique integers q and r, such that a = bq + r 

where r must satisfy 0 ≤ r ˂ b

Correct option is (A) 25