AP ICET 16th May 2016

Different approaches are possible for this question. 
One of the ways can be to take the square of any option and it should result in the term which is enclosed in the square root sign in the question.

Another approach would be to split the terms of the square root to make it a perfect square.
$$30-12\sqrt{\ 6}=30-2\cdot6\cdot\sqrt{\ 3}\sqrt{\ 2}=30-2\cdot3\cdot2\cdot\sqrt{\ 3}\sqrt{\ 2}$$

It means 30 could be the sum of squares of one of the following possible combinations: $$3\sqrt{\ 3\ }\&\ 2\sqrt{\ 2}$$ OR $$3\sqrt{\ 2\ }\&\ 2\sqrt{\ 3}$$
We can see that it is the sum of the squares of the second combination. 

Hence, $$30-12\sqrt{\ 6}$$ = $$\left(3\sqrt{\ 2\ }-2\sqrt{\ 3}\right)^2$$

Let a = 5x and b = 8x

4a + 5b = 20x + 40x = 60x

5a + 4b = 25x + 32x = 57x

$$\therefore\ $$(4a + 5b) : (5a + 4b) = 60x : 57x = 60 : 57

The largest power of 7 that divides $$(68)!$$ = Sum of the quotients of {$$\frac{68}{7}$$, $$\frac{68}{7^2}$$, $$\frac{68}{7^3}$$, .......}

= Sum of the quotients of {$$\frac{68}{7}$$, $$\frac{68}{49}$$, $$\frac{68}{343}$$, .......}

= 9 + 1 + 0 + 0 + .....

= 10

$$\therefore\ $$n = 10

Hence, the correct answer is Option B

$$\therefore\ $$The binary equivalent of the number 199 is 11000111

Hence, the correct answer is Option C

We are given that the product of 14 and x i.e. 14x when divided by 19, should give a remainder 6. 
One can simply take the different available values of x from the options, multiply it by 14 and then divide by 19 to see whether remainder obtained is 6.

This happens when x = 14 as 14*14 = 196 which when divided by 19 gives us remainder 6. 

Multiplying both numerator and denominator of a by $$\sqrt{\ 5}$$, we get a = $$\frac{\left(\sqrt{\ 5}+1\right)^2}{4}$$

Multiplying both numerator and denominator of b by $$\sqrt{\ 5}-1$$, we get b = $$\frac{\left(\sqrt{\ 5}-1\right)^2}{4}$$

Now, $$a^2=\frac{\left(14+6\sqrt{\ 5}\right)}{4}$$
$$b^2=\frac{\left(14-6\sqrt{\ 5}\right)}{4}$$
$$a\cdot b=1$$

Adding them up, we get 8 as the answer

The largest power of 3 that divides $$(20)!$$ = Sum of quotients of {$$\frac{20}{3}, \frac{20}{3^2}, \frac{20}{3^3},$$....}

= Sum of quotients of $$\frac{20}{3},\frac{20}{9},\frac{20}{27},....$$

= 6 + 2 + 0 + 0 +.....

= 8

$$\therefore\ $$The largest power of 3 that divides $$(20)!$$ is $$3^8$$

Hence, the correct answer is Option D

For a number to be divisible by 33, it must be divisible by both 11 and 3
We can see that all options would be divisible by 3.

However, the number in option B is not divisible by 11 after applying divisibility test of 11 to all the options. 

This question can be resolved by knowing approximate values of the square roots of 3, 5, 7. 
Sqrt(3) = 1.73 approx., sqrt(5) = 2.23 approx., sqrt(7) = 2.6 approx.

Now, we can see that $$10\sqrt{\ 3}$$ will be larger than 17 while $$7\sqrt{\ 5}$$ would be a little over 15. 
While in contrast, $$5\sqrt{\ 6}$$ would not even be 15. 
Also, $$6\sqrt{\ 7}$$ would be greater than 15. 

Hence, we can say that $$10\sqrt{\ 3}$$ is the largest of them all and $$5\sqrt{\ 6}$$ would be the smallest. 
$$6\sqrt{\ 7}\ >\ 7\sqrt{\ 5}\ $$. 
Hence, answer is B

Students who study Physics = 70% of 750 = 525 ... (A)
Students who study Mathematics = 40% of 750 = 300 ... (B)

Total students = (A) + (B) - (Students who study both) 
750 = 825 - (Students who study both)
Students studying both subjects = 75