Answer: Option D. -> 50
:
D
The minimum value will occur when $p^2=\frac{625}{p^2}$
$p^4=625\Rightarrow p=5$
Putting the value of p in the equation, min value =$p^4=625\Rightarrow p=5\Rightarrow =25+\frac{625}{25}=25+25=50$

Answer: Option B. -> 84:16
:
B
This question can be solved using alligation. For alligation; remember to always consider the cost price only.
Selling Price (SP)
Cost Price (CP)
SP of rice 1 = 2.70/kg CP (rice 1) = $\frac{2.7}{0.9}$ = 3/kg
SP of rice 2 = 4.5/kg CP (rice 2) = $\frac{4.5}{1.125}$ = 4/kg
SP of mixture = 3.95/kg CP (mixture) = $\frac{3.95}{1.25}$ = 3.16/kg

Required Ratio = 84:16

Answer: Option C. -> 5
:
C

Solving, we get (4x + 3)$\times$3+2 = 71 $⟹$ x=5.

Answer: Option C. -> 9
:
C
Adding the three equations we get, $2(x+y+z)=9.2⟹x+y+z=4.6$.
Subtracting first eq. from the above eq. we get $\left\{y\right\}+\left[z\right]=0.8⟹\left\{y\right\}=0.8$ and $\left[z\right]=0$.
Subtracting second eq. from the above eq. $\left\{x\right\}+\left[y\right]=1.4⟹\left\{x\right\}=0.4$ and $\left[y\right]=1$.
Subtracting third eq. from the above eq. $\left[x\right]+\left\{z\right\}=2.4⟹\left[x\right]=2$ and $\left\{z\right\}=0.4$.
$\therefore x=2.4,y=1.8$ and $z=0.4$.
Hence, choice (c) is the right answer.

:
There are 12 letters, in which 'S' appears 4 times and 3 vowels are A,E,U.
Take all vowels (AUE) as a single object
So number of permutations when vowels occur together =$\frac{10!}{4!}$ $\times$ 3!
Arrangements of 12 letters take all at atime =$\frac{12!}{4!}$
Required arrangements when al vowels do not occur together
=$\frac{12!}{4!}$ - $\frac{10!}{4!}$ $\times$ 3!
=$\frac{10!}{4!}$ (12 $\times$ 11 -6)
=$\frac{10!}{4!}$ $\times$ 126
A = 126

:

The above diagram is self-explanatory. 180-4x+x+y=180
Hence y=3x, k=3.

Answer: Option A. -> 99C2
:
A
Natural number solutions means each of A,B and C can take a value ≥ 1
This is a Similar to Different question, with a lower limit condition. (Refer ebooklet for indepth explanation)
To take care of the condition, that we are dealing with natural numbers,
give A,B &C 1 initially
(A+1)+(B+1)+(C+1)=100
Thus, this reduces to A+B+C=97.
This question is now based on arrangement of 97 zeroes and 2 ones=${}^{99}$C${}_2$
The answer is ${}^{99}$C${}_2$ which is option (a)

:
x : y: z: :2 : 3 : 5
$\therefore$x=$\frac{2y}{3}$ and z=$\frac{5y}{3}$
x+y+z=100
$\frac{2y}{3}+y+\frac{5y}{3}=100$
y= 30 $\&$ x=20
But y= ax-10
30 = 20a - 10
​a=2

Answer: Option B. -> 25
:
B
It is given that A,B and C can take values ≥0
Give values to C from 0 to 3
At C=0, the equation changes to A+B=10. Number of solutions will be based onarrangement of 10 zeroes and 1 one = ${}^{11}$C${}_1$= 11
At C=1, the equation changes to A+B=7. Number of solutions=${}^8$C${}_1$=8
At C=2, the equation changes to A+B=4. Number of solutions=${}^5$C${}_1$=5
At C=3, the equation changes to A+B=1. Number of solutions=${}^2$C${}_1$=1
Total number of solutions = 11+8+5+1 = 25

Answer: Option A. -> abb
:
A
If only 3 digits are involved, the system has to be in base 3 with the digits 0,1,2. We can confirm the base using the examples provided.

Thus, a=1 ,b=2 and c=0 (you can reconfirm using the other examples as well)
17 in base 3 = = abb.