Ans: (a)
$\frac{7}{Dx}-\frac{7}{1Dx}=770=(126-\frac{56}{144})x=770=x=1584$

Elimination strategy:

As $\frac{7}{18}$ is slightly less than half of $\frac{7}{8}$,we have to find the answer which is slightly greater than twice of 770, hence check the given conditions for 1584 and 1656, the other options can be easily eliminated. Check manually for 1584. we find that the difference is 770, hence that is the option.

Ans: (d)

Number of students which should be seated in each room is the HCF of 60, 84 and 108 which is 12.

Number of rooms required for subject A, subject B and subject $C=\frac{60}{12}=5$ rooms, $\frac{84}{12}=7$rooms and $\frac{108}{12}=9$ rooms respectively. Hence minimum number of rooms required to satisfy our condition $=(5+7+9)=21.$

This can be solved by Chinese remainder theorem, but as the common difference is constant, it is a special case of Chinese remainder concept.

Required number of the set is calculated by the LCM of (2, 3, 4, 5, 6) - (common difference)

In this case, common difference = (2-1) = (3-2) = (4-3) = (5- 4) = (6 - 5) = 1.

All integers of the set will be given by (60n - 1) ; If n = 1, (60 - 1) = 59; If n = 2,((60*2) - 1) = 119; Since range of the set A is between 0 and 100, hence there will exist only one number i.e., 59.

Ans: (c)

$\frac{(16n^z+7n+6)}{n}=16n+7+\frac{6}{n};$
Since n is an integer, hence for the entire expression to become an integer, $\left(\frac{6}{n}\right)$ should be an integer. And $\left(\frac{6}{n}\right)$ can be integer for $n=1,2,3,6.$ Hence n has 4 values.

Ans: (d)

None of the options (a), (b) and (c) is necessarily true. Hence option (d) is an answer.

As this is a variable based question: the word "ANY" can be used

Let the series of integers $a_1,a_2,.......,a_{50}$ be $1,2,3,4,5,.......,50.$

$S_1=1,2,3,4,.......24,S_2=25,26,\mathrm{27............50}$

Ans: (a)

Let S1 = 1, 2, 3, 4,......., 24,   S2 = 50, 49,........., 25;

New series after interchange S1 = 1, 2, 3, 4........, 25,  S2 = 50, 49,.......24

It is therefore clear that S1 continues to be in ascending order.

As this is a variable based question: the word "ANY" can be used

Let the series of integers $a_1,a_2,.......,a_{50}$ be $1,2,3,4,5,......,50.$

$S_1=1,2,3,4,......24,S_2=25,26,27,.........50$

Ans: (d)

The smallest integer of the series is 1 and greatest integer is 50. If each element of S1 is made greater of equal to every element of S2, then the smallest element 1 should be added to (50 - 1) = 49. Hence option (G-L) is the correct answer.

As this is a variable based question: the word "ANY" can be used

Let the series of integers $a_1,a_2,.......,a_{50}$ be $1,2,3,4,5,.......,50.$

$S_1=1,2,3,4,.........24,S_2=25,26,27,..........50$

Ans: (c)

If $x=30,$

$y=16\times 2=32$

Let Roopa have x flowers with her; then
$\begin{array}{cccc}& Balancebeforeoffering& Flowersoffered& Balanceafteroffering\\ I^{st}Place& 2x& y& 2x-y\\ I{I}^{nd}Place& 4x-2y& y& 4x-3y\\ II{I}^{rd}Place& 8x-6y& y& 8x-7y\\ I{V}^{th}Place& 16x-14y& y& 16x-15y\end{array}$

$16x-15y=0\Rightarrow 16x=15y\Rightarrow y=\frac{16x}{15}$

Ans: (c)

Minimum value for y is available for $x=15;y=x\ast 15\Rightarrow y=16;$

Let Roopa have x flowers with her; then
$\begin{array}{cccc}& Balancebeforeoffering& Flowersoffered& Balanceafteroffering\\ I^{st}Place& 2x& y& 2x-y\\ I{I}^{nd}Place& 4x-2y& y& 4x-3y\\ II{I}^{rd}Place& 8x-6y& y& 8x-7y\\ I{V}^{th}Place& 16x-14y& y& 16x-15y\end{array}$

$16x-15y=0\Rightarrow 16x=15y\Rightarrow y=\frac{16x}{15y}$

Therefore minimum value of x for which y is an integer is 15 hence y = 16.

Ans: (b)

Minimum value of x is available for y = 16. $x=\frac{15}{16xy}=\frac{15}{16}\times 16=15$

Let Roopa have x flowers with her; then
$\begin{array}{cccc}& Balancebeforeoffering& Flowersoffered& Balanceafteroffering\\ I^{st}Place& 2x& y& 2x-y\\ I{I}^{nd}Place& 4x-2y& y& 4x-3y\\ II{I}^{rd}Place& 8x-6y& y& 8x-7y\\ I{V}^{th}Place& 16x-14y& y& 16x-15y\end{array}$

$16x-15y=0\Rightarrow 16x=15y\Rightarrow y=\frac{16x}{15}$