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First option is wrong because second sentence does not contradict the first sentence. Third option is wrong because two sentences are related. Fourth option is wrong because the second sentence does not repeat the first one.
Hence second option is correct which shows the result of the cause.

Possible cases are:

Conclusion-I is incorrect becaue some risk seeker are wealthy.
Conclusion-II is also incorrect because all the entrepreneurs are risk seeker as well as wealthy

"The number of" is singular and it takes singular verb. "A number of " is plural and it takes plural verb.

Option (A) and (B) are weekend by expression 'rhinos have no benefit'. Oxpeckers do not save life of poachers, so option (D) is incorrect.
Hence, option (C) is correct.

The last sentence of the passage is reflecting in option (c) only.

Morning sun is falling from east then the shadow will fall to the west. So west should be on the right side of Ms. X. So Ms. X came out towards south. Hence, her building is facing south.

Hardly/Scarcely/Barely have same sense that is negative and they are used after the verb (could barely and could hardly).

Click here for detail solution by gateoverflow

Click here for detail solution by gateoverflow

Click here for detail solution by gateoverflow

Area=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}

\begin{aligned} \sin 30^{\circ} &=\frac{r}{R} \\ \text { Area ratio } &=\frac{\pi r^{2}}{\pi R^{2}}=\sin ^{2} 30=\frac{1}{4} \end{aligned}

The probability of first ball is blue and second ball is black is given as,
P=\frac{15}{60} \times \frac{45}{59}=\frac{45}{236}

\begin{aligned} \frac{\oplus}{\odot}&=2, \frac{\oplus}{\Delta}=3 \\ \therefore \qquad\frac{\Delta}{\odot}&=\frac{2}{3} &\ldots(1)\\ \odot+\Delta&=5 &\ldots(2)\\ \text{From (1) and (2)}\\ \Delta&=2, \odot=3\\ \text{and}\\ \oplus &=6,2 \times \otimes=10 \\ \otimes &=5 \\ \Rightarrow \quad(\otimes-\oplus)^{2}&=(5-6)^{2} =1 \end{aligned}

LCM of (30 and 32) is 480
480 seconds = 8 minutes
Hence, time will be 10.08 pm

The possible distinct arrangement are
S R P T A, A R P T S, S R T P A
Hence, number of distinct sitting arrangement. = 3

The possible seating arrangements are
QPRST, QTRSP
QPRTS, QTRPS
PQTRS, TQPRS
SPQTR, STQPR
RPQTS, RTQPS
SRPQT, SRTQP
SPRTQ, STRPQ
PSRTQ, TSRPQ
Hence, total seating arrangements are 16

Number of employee in C2 company =\frac{5}{100} \times 10000=500
Number of management degree holder employee in \mathrm{C} 2=\frac{1}{5} \times 500=100
Number of employee in \mathrm{C} 5 company =\frac{20}{100} \times 10000=2000
Number of management degree holder employee in \mathrm{C} 5=\frac{9}{10} \times 2000=1800
Total management degree holder employee =100+1800=1900

In farm P,
Hens =65, Ducks =91, Goats =169
In farm Q,
Hens : Ducks : Goats
5: 14: 13
\begin{aligned} \text { Hens } &=\frac{5}{32} \times 416=65 \\ \text { Ducks }&=\frac{14}{32} \times 416=182\\ \text { Goats }&=\frac{13}{32} \times 416=169 \end{aligned}
\because From farm d, hens, ducks and goats are sent to farm P.
\begin{aligned} \therefore \text{ Total hens }&=65+65=130\\ \text { Total ducks } &=91+182=273 \\ \text { Total goats } &=169+169=338 \\ \text { New ratio } &=130: 273: 338 \\ &=10: 21: 26 \end{aligned}

First option is wrong because second sentence does not contradict the first sentence. Third option is wrong because two sentences are related. Fourth option is wrong because the second sentence does not repeat the first one.
Hence second option is correct which shows the result of the cause.

Area=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}

\begin{aligned} \sin 30^{\circ} &=\frac{r}{R} \\ \text { Area ratio } &=\frac{\pi r^{2}}{\pi R^{2}}=\sin ^{2} 30=\frac{1}{4} \end{aligned}

The probability of first ball is blue and second ball is black is given as,
P=\frac{15}{60} \times \frac{45}{59}=\frac{45}{236}

Possible cases are:

Conclusion-I is incorrect becaue some risk seeker are wealthy.
Conclusion-II is also incorrect because all the entrepreneurs are risk seeker as well as wealthy

\begin{aligned} \frac{\oplus}{\odot}&=2, \frac{\oplus}{\Delta}=3 \\ \therefore \qquad\frac{\Delta}{\odot}&=\frac{2}{3} &\ldots(1)\\ \odot+\Delta&=5 &\ldots(2)\\ \text{From (1) and (2)}\\ \Delta&=2, \odot=3\\ \text{and}\\ \oplus &=6,2 \times \otimes=10 \\ \otimes &=5 \\ \Rightarrow \quad(\otimes-\oplus)^{2}&=(5-6)^{2} =1 \end{aligned}

LCM of (30 and 32) is 480
480 seconds = 8 minutes
Hence, time will be 10.08 pm

"The number of" is singular and it takes singular verb. "A number of " is plural and it takes plural verb.

The possible distinct arrangement are
S R P T A, A R P T S, S R T P A
Hence, number of distinct sitting arrangement. = 3

Given, A P=\alpha^{2} P
By comparison with A X=\lambda X \Rightarrow
\Rightarrow \quad \lambda=\alpha^{2}
Hence, eigen value of A is \alpha^{2}, so eigen value of A^{2} is \alpha^{4}.

By using partial fraction concept.
\begin{aligned} f(t) &=L^{-1}\left[\frac{s+3}{(s+1)(s+2)}\right] \\ &=L^{-1}\left[\frac{2}{s+1}-\frac{1}{s+2}\right] \\ \Rightarrow \qquad f(t) &=2 e^{-t}-e^{-2 t} \\ \text { So, } \qquad f(c)&=2 e^{0}-e^{0}=2-1=1 \end{aligned}

Mean= np
Variance = npq = np(1 - p)

On the basis of nature of carbon present in cast iron, it may be divided into white cast iron and gray cast iron.
In the gray cast iron, carbon is present in free form as graphite. Under very slow rate of cooling during solidification, carbon atoms get sufficient time to separate out in pure form as graphite. In addition, certain elements promote decomposition of cementite. Silicon and nickel are two commonly used graphitizing elements.
In white cast iron, carbon is present in the form of combined form as cementite. In normal conditions, carbon has a tendency to combine with iron to form cementite.

Particle Size, Shape, and Distribution:
Particle size is generally controlled by screening, that is, by passing the metal powder through screens (sieves) of various mesh sizes. Several other methods also are available for particle-size analysis:
1. Sedimentation, which involves measuring the rate at which particles settle in a fluid.
2. Microscopic analysis, which may include the use of transmission and scanning- electron microscopy.
3. Light scattering from a laser that illuminates a sample, consisting of particles suspended in a liquid medium; the particles cause the light to be scattered, and a detector then digitizes the signals and computes the particle-size distribution.
4. Optical methods, such as particles blocking a beam of light, in which the particle is sensed by a photocell.
5. Suspending particles in a liquid and detecting particle size and distribution by electrical sensors.

In contouring systems the machining path is usually constructed from a combination of linear and circular segments. It is only necessary to specify the coordinates of the initial and final points of each segment, and the feed rate. The operation of producing the required shape based on this information is termed interpolation and the corresponding unit is the "interpolator". The interpolator coordinates the motion along the machine axes, which are separately driven, by providing reference positions instant by instant for the position-and velocity control loops, to generate the required machining path. Typical interpolators are capable of generating linear and circular paths.

Laser beam machining (LBM) is a nonconventional machining process, which broadly refers to the process of material removal, accomplished through the interactions between the laser and target materials. The processes can include laser drilling, cutting, grooving, writing, scribing, ablation, welding, cladding, milling, and so on. LBM is a thermal process, and unlike conventional mechanical processes, LBM removes material without mechanical engagement. In general, the workpiece is heated to melting or boiling point and removed by melt ejection, vaporization, or ablation.

In CPM,
\begin{array}{l} \sigma=\sqrt{\text { sum of variance along critical path }} \\ \sigma=\sqrt{\sigma^{2}+\sigma^{2}+\ldots .+\sigma^{2}} \\ \sigma=\sqrt{9 \sigma^{2}}=\sqrt{9 \times 9}=9 \end{array}

It is minimum clearance or maximum interference. It is the intentional difference between the basic dimensions of the mating parts. The allowance may be positive or negative.

S t=\frac{N u}{R e \times P r}

\begin{aligned} \frac{d y}{d x}+y \cot x&=\text{cosec} x\\ 1.F. \qquad&=e^{\int \cot x d x}=e^{\log \sin x}=\sin x\\ \Rightarrow \quad y(\sin x)&=\int \text{cosec} x \sin x d x+c\\ \Rightarrow \qquad y \sin x&=x+c\\ \Rightarrow \qquad \frac{\pi}{2} \sin \frac{\pi}{2} & =\frac{\pi}{2}+c \\ \Rightarrow \qquad \frac{\pi}{2} & =\frac{\pi}{2}+c \quad \Rightarrow c=0 \\ \Rightarrow \qquad y \sin x & =x\\ \Rightarrow \qquad y \sin \frac{\pi}{6}&=\frac{\pi}{6}\\ \Rightarrow \qquad y\left(\frac{1}{2}\right) &=\frac{\pi}{6} \\ \Rightarrow y &=\frac{\pi}{3} \end{aligned}

\begin{aligned} \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)&=? \;\;\;\;\;\;\left(\frac{0}{0} \text { form }\right) \\ \text { Applying } L \cdot H \text { rule } & =\lim _{x \rightarrow 0} \frac{\sin x}{2 x}\left(\frac{0}{0}\right)=\lim _{x \rightarrow 0} \frac{\cos x}{2}=\frac{1}{2} \end{aligned}

\begin{aligned} \because \qquad \int_{0}^{-} f(t) \delta(t-a) d t&=f(a) \\ \therefore \qquad L\{\delta(t-a)\}&=\int_{0}^{-} e^{-s t} \delta(t-a) d t=e^{-a s} \end{aligned}

\begin{aligned} \frac{y\left(t_{n+1}\right)-y\left(t_{n}\right)}{h} &=-\pi y\left(t_{n}\right) \\ y_{n+1} &=-\pi / y_{n}+y_{n}=(-\pi h+1) y_{n} \end{aligned}
It is recursion relation between y_{n+1} and y_{n}
So solution will be stable if
\begin{aligned} |-\pi h+1| & \lt 1 \\ -1 \lt -\pi h+1 & \lt 1 \\ -2 \lt -\pi h & \lt 0 \\ 0 & \lt \pi h \lt 2 \\ 0 & \lt h \lt \frac{2}{\pi} \end{aligned}
Therefore option (A) is correct.

Elastic strain : Which can be recovered = 0.03 - 0.01 = 0.02
Plastic strain : Permanent strain = 0.01

Drilling: to produce a hole, which then may be followed by boring it to improve its dimensional accuracy and surface finish.

Boring: to enlarge a hole or cylindrical cavity made by a previous process or to produce circular internal grooves.

Reaming: is an operation used to (a) make an existing hole dimensionally more accurate than can br achived by drilling alone and (b) improve its surface finish. The most accurate holes in workpieces generally are produced by the following sequence of operation.

Centering -> Drilling -> Boring -> Reaming.

R_{G}=\frac{\sum_{i=1}^{n} y}{n}=\frac{4}{4}=1