Answer: Option A. -> 53
:
A
Step 1 : Underline the keywords
Henko is renting a car. The rental charge is $\$17.50$ per day plus $\$0.16$ per mile. Henko can spend at most $\$33$ for the cost of the car rental. If Henko rents the car for one day, which of the following is one possible number of miles Henko can drive the rental car?
Step 2 : Recognize the variable
Number of miles ,is what the question is looking for.
Let number of miles be x
Step 3 : Decipher the relation
Total cost = no. of days rented $\times$ 17.50 + number of miles $\times$ 0.16
As henko rented the car for only one day
Total cost $=17.50+x\times 0.16$
Step 4 : Decide the inequality
Look for cues : Henko can spend at most $33\$$ for the car rent. Therefore, we get
$33\ge 17.50+x\times 0.16$
As, the question is looking for possible value of x, simplify the given inequality
$96.87\ge x$
Only, option A satisfies the given inequality.

Answer: Option C. -> Quadrant IV
:
C

Given above is the graphical representation of the solution region. It can be seen that the system of linear inequalities have no solution in the fourth quadrant.

Answer: Option D. -> x≥−14
:
D
$\frac{3+12x}{4}\ge \frac{2(4x+1)}{5}$
$(3+12x)\times 5\ge 2(4x+1)\times 4$
$15+60x\ge 32x+8$
$60x-32x\ge 8-15$
$28x\ge -7$
$x\ge -\frac{7}{28}$
$x\ge -\frac{1}{4}$

Answer: Option B. -> (2, 7)
:
B
$-5<p<-2$ and $7<q<9$
Adding two inequalities with same natured inequalities will preserve the sign.
So adding the two inequalities:
$-5+7<p+q<-2+9$
$\Rightarrow 2<p+q<7$

Answer: Option D. -> All of these
:
D
Let y =${\log }_{25}36$
=$\frac{{\log }_{e}36}{{\log }_{e}25}=\frac{{\log }_e{6}^2}{{\log }_e{5}^2}=\frac{2{\log }_{e}6}{2{\log }_{e}5}=\frac{{\log }_{e}6}{{\log }_{e}5}$
= ${\log }_{5}6$
${\log }_{25}36$ can be written as,
${\log }_{5^2}{6}^2=2{\log }_{5^2}6$
or $\frac{1}{2}{\log }_5{6}^2$

Answer: Option A. -> {1,2,3}
:
A
$-1<2x+3\le 10$
Subtract 3 from each sides
$-1-3<2x+3-3\le 10-3$
$-4<2x\le 7$
Divide by 2 each side
$-\frac{4}{2}<\frac{2x}{2}\le \frac{7}{2}$
$-2<x\le 3.5$
From here we can say that x will that all the value between -2 to 3.5 [including 3.5 because $\le$ sign].
Now lets look at options
A) {1,2,3}
B) {1,2,3,4}$\to$ not possible because 4 is not in the range
C) {2,3,4}$\to$ same reason
D) {2,3,4,5} $\to$ Same reason

Answer: Option D. -> 12ab−1
:
D
$ab=lo{g}_{4}5.lo{g}_{5}6=lo{g}_{4}6=\frac{1}{2}lo{g}_{2}6$
$ab=\frac{1}{2}(1+lo{g}_{2}3)\Rightarrow 2ab-1=lo{g}_{2}3$
$\therefore lo{g}_{3}2=\frac{1}{2ab-1}$

Answer: Option A. -> x≥−52
:
A
$-2x+5\le 10$
Subtract 5 on both side
$-2x+5-5\le 10-5$
$-2x\le 5$
Divide by -2 on both side
$-\frac{2x}{-2}\ge \frac{5}{-2}$
$x\ge -\frac{5}{2}$
The correct answer is option A

Answer: Option D. -> p≥115.3
:
D
Step 1 : Underline the key words.
To pass a bill in the Lower house, the Indian government requires at least $\frac{2}{3}$ majority in the house. There are 545 seats in the lower house. 248 members are in favor of passing a particular bill. Which of the following inequality represents the number of additional lower house members, p, who would support the bill, to achieve the required majority.
Step 2 : Recognize the variable
Additional members required to support the bill
Step 3 : Decipher the relation
Given, there are 248 members already supporting the bill and we require 'p' members more.
Thus, 248 + p is required to pass the bill
Step 4 : decide the inequality
Look out for cues :at least $\frac{2}{3}$ majority. We also, know there are 545 seats.
Therefore, $248+p\ge \left(\frac{2}{3}\right)$ 545.
Since, the answer options are simplified. Simplify the given inequation :
$p\ge 115.333$
Hence, option D is correct.

Answer: Option A. -> 40x+65≤2400;x+y≤45
:
A
The first constraint is on the weight of the boxes. The total weight of the boxes cannot exceed 2400. Since there are x 40 pound boxes and y 65 pound boxes, total weight is (40x + 65 y). So,
$40x+65y\le 2400$
The total number of boxes cannot exceed 45. Total number of boxes is (x + y). So,
$x+y\le 45$
Hence, A is the correct choice.