AP ICET 2014

Firstly, we should know the identities:
$$\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc$$
$$\left(a-b-c\right)^2=a^2+b^2+c^2-2ab-2ac+2bc$$

In the question, the first expression inside the square root can be rewritten as:

$$10+2\sqrt{\ 6}+2\sqrt{\ 10}+2\sqrt{\ 15}=2+3+5+2\sqrt{\ 2}\sqrt{\ 3}+2\sqrt{\ 2}\sqrt{\ 5}+2\sqrt{\ 3}\sqrt{\ 5}$$=$$\left(\sqrt{\ 2}+\sqrt{\ 3}+\sqrt{\ 5}\right)^2$$
Hence, the first modulus will return the net value $$=\sqrt{\ 2}+\sqrt{\ 3}+\sqrt{\ 5}^{ }$$

Similarly, the expression in the second square root can be simplified as $$=\left(\sqrt{\ 2}-\sqrt{\ 3}-\sqrt{\ 5}\right)^2$$
As sqrt(3) and sqrt(5) are greater than sqrt(2), the modulus will change the sign of the expression finally and it will becomes $$=-\sqrt{\ 2}+\sqrt{\ 3}+\sqrt{\ 5}$$

Hence, the resultant value will be option A

315 x k = 3 x 3 x 5 x 7 x k
To make the LHS a perfect square, k should definitely have at least one power of 5 and 7. 
For the least value of k, we should have only the minimum requirements i.e. k = 5 x 7 
hence, value of k must be 35. 

The best approach in these questions is to evaluate the options and arrive at the required answer by some trial and error. 
As we can see, the desired number has to be an odd number because 2 cannot divide it completely. Hence, options A and B get eliminated quickly. 
The required conditions get satisfied with option C.

The theoretical approach for such questions is long and not advisable when options are present. 

Given,  LCM = 144

GCD = HCF = 12

One of the integer = 36

Let the other integer = a

We know that,

Product of 2 integers = HCF x LCM

$$\Rightarrow$$  36 x a = 144 x 12

$$\Rightarrow$$  a = 48

$$\therefore\ $$The other integer = 48

Hence, the correct answer is Option C

The question asks for a number which will leave remainder 4 whether it is divided by 9 or 11 or 13. 
The number has to be less than 2580 and largest possible in this range.

This means that the desired number will be of the form = k*(LCM of 9,11,13) + 4
where K means any multiple of the LCM of 9,11,13 and should result in a number less than 2580
LCM if 9, 11, 13 is 1287. For k = 2, we get the desired number as = 2574 + 4 =  2578
Hence, 2 must be subtracted from 2580 to get desired number. 

Let the 3 numbers be a, 2a, 3a respectively.

According to the problem,

$$a^2+4a^2+9a^2=504$$

$$\Rightarrow$$  $$14a^2=504$$

$$\Rightarrow$$  $$a^2=36$$

$$\Rightarrow$$  $$a = 6$$

$$\therefore\ $$The largest of the numbers = 3a = 18

Hence, the correct answer is Option C

By observation, we can see that (5/7) is less than (11/14) as 5/7 is equal to (10/14)
Comparing 5/7 to 6/8 we can see that (5/7) is less than (6/8). 

Lastly, we can compare (9/11) with (11/14). The approx value of the first is 0.81 while the value of the second one is approx. <0.8 
Hence, the correct ascending order is the one given in the answer option. 

A gets (3/7)th of the property means that he gets (6/14)th of the property.
B gets (5/14)th of the property. This leaves (3/14)th of the property of C and hence, least amount is received by C. 

We raise all the terms to an equal power. Let's take the 12th power of each term. 
We get $$6^4,\ 10^3\ and\ 3^6$$

The value of the 4th power of 6 is easily over 1000, which is nothing but cube of 10. And lastly 6th power of 3 is 729.
Hence, the descending order of the terms would be the answer option. 

Let total votes polled be V.
The winner got 65% votes and won by margin of 12000 votes. Means the rival got 35% votes.
hence, 0.65V - 0.35V = 12000 which gives us V = 12000/0.3 = 40,000 i.e. 0.4 lakh