Answer: Option A. -> 0
:
A
Sol:
$x={16}^3+{17}^3+{18}^3+{19}^3$
$=({16}^3+{19}^3)+({17}^3+{18}^3)$
$=(16+19)({16}^2+16\times 19+{19}^2)+(17+18)({17}^2+17\times 18+{18}^2)$
$=35\times$ (an odd number) $+35\times$ (another odd number)
$=35\times$ (an even number) $=35\times \left(2k\right)..........$ (k is a positive integer)
$x=70k$
Therefore x is divisible by 70. Remainder when x is divided by 70 = 0
Hence, option 1.

Answer: Option B. -> 6
:
B
Option (b):
Let the number be (10y + x)- (10x + y) = 18
$\Rightarrow 9(y-x)=18\Rightarrow y-x=2$
So, the possible pairs of (x, y) are
(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9).
But we want the number other than 13. Thus, there are 6 possible numbers, i.e., 24, 35, 46, 57, 68,79.
So, total number of possible numbers are 6.

Answer: Option D. -> R>1
:
D
$R=\frac{{30}^{65}-{29}^{65}}{{30}^{64}+{29}^{64}}$
$\because a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}{b}^2+.....+b^{n-1})$
$\therefore R=\frac{(30-29)[{30}^{64}+({30}^{63}\times 29)+....+{29}^{64}]}{{30}^{64}+{29}^{64}}$
$\because {30}^{64}+{30}^{63}\times 29+...+{29}^{64}>{30}^{64}+{29}^{64}$
$\therefore R>1$
Hence, option 4.
Alternative approach: Unitary Method
$R=\frac{({30}^{65}-{29}^{65})}{({30}^{64}+{29}^{64})}$
If you take this number to be of the form 'n' and 'n+1' instead of 30 and 29, you can solve it using simple numbers like $1\left(n\right)$ and $2(n+1)$
i.e $\frac{(2^3-1^3)}{(2^2+1^2)}=\frac{7}{5}>1$. Only 1 option satisfies this. Option (d)

Answer: Option D. -> (√15−3)2
:
D
Ans: (d)
Given:y=
y =$\frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+.......}}}}$
Thus y = $\frac{1}{2+\frac{1}{3+y}}$
$y=\frac{(3+y)}{(6+2y+1)}\Rightarrow 2y^2+7y=3+y;$ On solving, $y=\frac{(-3\pm \sqrt{15})}{2};$ since the given fraction is positive, the value of y is
$\frac{(\sqrt{15}-3)}{2}$(CAT 2004)

Answer: Option C. -> 9
:
C
Soln:c.
If $p=\frac{1}{3}$ and $q=\frac{2}{3}$, then $p+q=1$ and $pq=\frac{2}{9}$
$a_1+b_1=p+q=1$
$a_3+b_3=p^{2}q+p{q}^2=pq(p+q)=\frac{2}{9}.$
$a_5+b_5=p^3{q}^2+p^2{q}^3=p^2{q}^2(p+q)=\frac{4}{81}.$
So, in general, for odd 'n' we can write
$a_n+b_n={\left(\frac{2}{9}\right)}^{\left(\frac{n-1}{2}\right)}$ ; we need to find least value of n such that ${\left(\frac{2}{9}\right)}^{\left(\frac{n-1}{2}\right)}<0.01$
Using the options given, if $n=7,a_7+b_7={\left(\frac{2}{9}\right)}^3>0.01;$ if $n=9,a_9+b_9={\left(\frac{2}{9}\right)}^4<0.01$

Answer: Option D. -> 1590
:
D
Ans: (d)
53x - 35x = 540; $\Rightarrow$ 18x= 540 or x = 30; Therefore, new product$=53\times 30=1590.$

Answer: Option D. -> 1
:
D
option (d)
If $P=1!=1$
Then $P+2=3,$ when divided by $2!$ Remainder will be 1.
If $P=1!+2\times 2!=5$
Then, $P+2=7$ when divided by $3!$ Remainder is still 1.
Hence, $P=1!+(2+2!)+(3\times 3!)+.....+(10\times 10!)$
When divided by $11!$ Leaves remainder 1.
Alternative method :
$P=1+2.2!+3.3!+....10.10!$
$=(2-1)1!+(3-1)2!+(4-1)3!+.....+(11-1)10!$
$=2!-1!+3!-2!+...+11!-10!$
$=1+11!$
Hence, the remainder is 1.

Answer: Option D. -> None of these
:
D
Ans: (d)
Total weight of three pieces: $=(\frac{9}{2}+\frac{27}{4}+\frac{36}{5})=\frac{359}{20}=18.45lb.$
Required weight of a single piece is HCF of $(\frac{9}{2}+\frac{27}{4}+\frac{36}{5})=\frac{HCFof(9,7,36)}{LCMof(2,4,5)}=\frac{9}{20}lb.$
Number of guests = $\frac{18.45}{\left(\frac{9}{20}\right)}=14$

Answer: Option C. -> (√13+1)2
:
C
Ans :(c)
$x=\sqrt{4+\sqrt{4-x}}$
$x^2=4+\sqrt{4-x};(x^2-4)=\sqrt{4-x}$
Only option c satisfies this condition.
Shortcut : Reverse Gear
$X=\sqrt{4}+a$ small value ( note the minus signs inbetween )
$\Rightarrow$ the answer is something slightly greater than 2 .
Look for an answer option, which is slightly greater than 2
(a) 3 is much greater than 2
(b) $\frac{(\sqrt{13}-1)\sim }{2}1.1$ (this can never be the answer)
(c) $\frac{(\sqrt{13}+1)\sim }{2}2.2$ (this is a possible answer)
(d) $\sqrt{13}=$ something greater than 3. this is also not possible, as the value is too high

Answer: Option C. -> 781
:
C
option (c) 781
Here, in each step we are adding two number and reducing the sum by 1. So after 39 operations, we will have the sum of all the numbers from 1 to 40 reduced by 39. Hence the final number will be $S_{40}-39=781.$