datavorous/entrance-exam-dataset huggingface dataset:
The mean of 100 observations is 50 and their standard deviation is 5 . Then, the sum of squares of all observations is

Options:
A $252500$
B $250000$
C $255000$
D $50000$

The correct answer is:
$252500$

Solution:
We know, standard deviation,
$\begin{aligned} P^2 & =\frac{\Sigma x^2}{n}-(\bar{x})^2 \\ P & =5, \Sigma x=50, n=100\end{aligned}$
$\begin{aligned} & \Rightarrow \quad 25=\frac{\Sigma x^2}{100}-(50)^2 \\ & \Rightarrow \quad \Sigma x^2=252500\end{aligned}$
Thus, sum of squares of all observation in 252500 .

Topic: Mathematics, Statistics, KCET, KCET 2023
The mean of a group of 100 observations was found to be 20 . Later it was found that four observations were incorrect, which were recorded as $21,21,18$ and 20 . What is the mean if the incorrect observations are omitted? $\quad$

Options:
A 18
B 20
C 21
D 22

The correct answer is:
20

Solution:
Mean of 100 observations $=20$
Required mean $=\frac{20 \times 100-(21+21+18+20)}{96}$
$=\frac{1920}{96}=20$

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 1)
The mean of five numbers is 30 . If one number is excluded their mean becomes 28 . The excluded number is

Options:
A 28
B 35
C 30
D 38

The correct answer is:
38

Solution:
Mean of 5 numbers $=30$ Total sum of 5 numbers $=30 \times 5=150$ After excluded one number
Mean of 4 numbers will be $=28$ Total sum of 4 numbers $=4 \times 28=112$ Thus, excluded number $=$ (sum of 5 numbers $-$ sum of 4 numbers) $=150-112=38$
Option (d) is correct.

Topic: Mathematics, Statistics, NDA, NDA 2015 (Phase 1)
The mean of four observations is 3 . If the sum of the squares of these observations is 48 , then their standard deviation is

Options:
A $\sqrt{7}$
B $\sqrt{2}$
C $\sqrt{3}$
D $\sqrt{5}$

The correct answer is:
$\sqrt{3}$

Solution:
Let four observations are be $x_1, x_2, x_3$ and $x_4$
Given, mean
and
$$
\begin{aligned}
(\bar{x}) & =3 \\
\Sigma x_i^2 & =48
\end{aligned}
$$
$\begin{aligned} \therefore \quad \mathrm{SD} & =\sqrt{\frac{\sum x_i^2}{n}-(\bar{x})^2} \\ =\sqrt{\frac{48}{4}-(3)^2}=\sqrt{12-9} & =\sqrt{3}\end{aligned}$

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2014
The mean of four observations is 3 . If the sum of the squares of these observations is 48 , then their standard deviation is

Options:
A $\sqrt{7}$
B $\sqrt{2}$
C $\sqrt{3}$
D $\sqrt{5}$

The correct answer is:
$\sqrt{3}$

Solution:
Let four observations are be $x_1, x_2, x_3$ and $x_4$
Given, mean
and
$$
\begin{aligned}
(\bar{x}) & =3 \\
\Sigma x_i^2 & =48
\end{aligned}
$$
$\begin{aligned} \therefore \quad \mathrm{SD} & =\sqrt{\frac{\sum x_i^2}{n}-(\bar{x})^2} \\ =\sqrt{\frac{48}{4}-(3)^2}=\sqrt{12-9} & =\sqrt{3}\end{aligned}$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2014
The mean of the series $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{\mathrm{n}}$ is $\overline{\mathrm{X}}$. If $\mathrm{x}_{2}$ is replaced by $\lambda$, then what is the new mean? $\quad$

Options:
A $\overline{\mathrm{X}}-\mathrm{x}_{2}+\lambda$
B $\frac{\overline{\mathrm{X}}-\mathrm{x}_{2}-\lambda}{\mathrm{n}}$
C $\frac{\overline{\mathrm{X}}-\mathrm{x}_{2}+\lambda}{\mathrm{n}}$
D $\frac{\mathrm{n} \overline{\mathrm{X}}-\mathrm{x}_{2}+\lambda}{\mathrm{n}}$

The correct answer is:
$\frac{\mathrm{n} \overline{\mathrm{X}}-\mathrm{x}_{2}+\lambda}{\mathrm{n}}$

Solution:
Mean of series $\left(x_{1}, x_{2}, x_{3}, \ldots . x_{n}\right)$
$\bar{x}=\frac{x_{1}+x_{2}+x_{3}+\ldots \ldots x_{n}}{n}$
$\Rightarrow x_{1}+x_{2}+x_{3}+\ldots \ldots x_{n}=n \bar{x}$
Now we will replace $x$, by $\lambda$ so no. of elements in series will not change. New series will include $\lambda$ and exclude $x_{2}$ Hence new series sum
$\left(x_{1}+x_{2}+\ldots . x_{n}\right)-x_{2}+\lambda=n \bar{x}+\lambda-x_{2}$
Now new mean

Topic: Mathematics, Statistics, NDA, NDA 2016 (Phase 1)
The mean weight of 150 students in a certain class is $60 \mathrm{~kg}$. The mean weight of boys in the class is $70 \mathrm{~kg}$ and that of girls is $55 \mathrm{~kg}$. What is the number of boys in the class?

Options:
A 50
B 55
C 60
D 100

The correct answer is:
50

Solution:
Number of students $=150$.
By aligation, ratio $=1: 2$
$\therefore$ No. of boys $=\frac{1}{3} \times 150=50$

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 1)
The mean weight of all the students in a certain class is $60 \mathrm{~kg}$. The mean weight of the boys from the class is $70 \mathrm{~kg}$. while that of the girls is $55 \mathrm{~kg}$. What is the ratio of number of boys to that of girls?

Options:
A $2: 1$
B $1: 2$
C $1: 4$
D $4: 1$

The correct answer is:
$1: 2$

Solution:
Let there be $x$ number of boys and $y$ number of girls. Total students $=\mathrm{x}+\mathrm{y}$
Total weight of the students $=(\mathrm{x}+\mathrm{y}) 60$ Total weight for boys $=x \times 70$ Total weight for girls $=\mathrm{y} \times 55$ Hence, $(\mathrm{x}+\mathrm{y}) 60=70 \mathrm{x}+55 \mathrm{y}$
$60 x+60 y=70 x+55 y$
$5 y=10 x \Rightarrow y=2 x$
$\frac{\mathrm{x}}{\mathrm{y}}=\frac{1}{2} \Rightarrow \mathrm{x}: \mathrm{y}=1: 2$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 1)
The means of two groups of observations A and B are x→, y→ respectively and their standard deviations are respectively 2 and 3. In order that the group A is to be more consistent than the group B, y→x→<

Options:
A 32
B 51
C 23
D 65

The correct answer is:
32

Solution:
We know, coefficient of variation =σx→×100

Given that, the group A is more consistent than the group B,

So, coefficient of variation of A<coefficient of variation of B

Then, Aσx→<Bσy→

⇒y→x→<BσAσ=32

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
The measure of the angle between the lines $x^{2}+2 x y \operatorname{cosec} \alpha+y^{2}=0$ is

Options:
A $\frac{\pi}{2}-\alpha$
B $\frac{\pi}{2}+\alpha$
C $\alpha$
D $\pi-\alpha$

The correct answer is:
$\frac{\pi}{2}-\alpha$

Solution:
We have $x^{2}+2 x y \operatorname{cosec} \alpha+y^{2}=0$
Comparing it with standard form, we get $\mathrm{a}=1, \mathrm{~h}=\operatorname{cosec} \alpha, \mathrm{b}=1$
Let $\theta$ be the angle between the lines.
$$
\begin{aligned}
\tan \theta &=\left|\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \\
&=\left|\frac{2 \sqrt{\operatorname{cosec}^{2} \alpha-1}}{1+1}\right|=\left|\frac{2 \sqrt{\cot ^{2} \alpha}}{2}\right| \\
\tan \theta &=\cot \alpha \Rightarrow \tan \theta=\tan \left(\frac{\pi}{2}-\alpha\right) \\
\therefore \theta=\frac{\pi}{2}-\alpha &
\end{aligned}
$$

Topic: Mathematics, Pair of Lines, MHT CET, MHT CET 2020 (20 Oct Shift 1)
The measurement of an unknown resistance $\mathrm{R}$ is to be carried out using. Wheatstones bridge as given in the figure below. Two students perform an experiment in two ways. Thefirst studentstakes $R_2=10 \Omega$ and $R_1=5 \Omega$. The other student takes $\mathrm{R}_2=1000 \Omega$ and $\mathrm{R}_1=500 \Omega$. In the standard arm, both take $\mathrm{R}_3=5 \Omega$.
Both find $\mathrm{R}=\frac{\mathrm{R}_2}{\mathrm{R}_1}, \mathrm{R}_3=10 \Omega$ within errors.

Options:
A
The errors of measurement of the two students are the same

B
Errors of measurement do depend on the accuracy with which $\mathrm{R}_2$ and $\mathrm{R}_1$ can be measured

C
If the student uses large values of $R_2$ and $R_1$ the currents through the arms will be feeble. This will make determination of null point accurately more difficult

D
Wheatstone bridge is a very accurate instrument and has no errors of measurement

The correct answers are:
Errors of measurement do depend on the accuracy with which $\mathrm{R}_2$ and $\mathrm{R}_1$ can be measured
,
If the student uses large values of $R_2$ and $R_1$ the currents through the arms will be feeble. This will make determination of null point accurately more difficult

Solution:
As given that,
Resistance for first student, $\mathrm{R}_1=5 \Omega, \mathrm{R}_2=10 \Omega, \mathrm{R}_3=5 \Omega$
Resistance for second student, $\mathrm{R}_1=500 \Omega$, $\mathrm{R}_2=1000 \Omega, \mathrm{R}_3=5 \Omega$
Now, according to Wheatstone bridge balanced condition, $\frac{\mathrm{R}_2}{\mathrm{R}}=\frac{\mathrm{R}_1}{\mathrm{R}_3} \quad$ or $\quad \mathrm{R}=\mathrm{R}_3 \times \frac{\mathrm{R}_2}{\mathrm{R}_1}$
Now putting all the values in Eq. (i), we get $R=10 \Omega$ for both students. Thus, we can analyse that the Wheatstone bridge is most sensitive and accurate if resistances are of same value. So the results of both students depend on the accuracy of resistances used.
When $R_2$ and $R_1$ are larger, the currents through galvanometer becomes very weak. It can make the determination of null point accurately more complex.

Topic: Physics, Current Electricity
The mechanism of action of a steroidal hormone on target cell is established through the

Options:
A formation of secondary messengers.
B activation of transcriptional apparatus.
C attachment with cytosolic receptor protein.
D regulation of gene expression.

The correct answer is:
regulation of gene expression.

Solution:
The mechanism of action of steroid hormones like estrogen, testosterone, etc. is through regulation of gene expression. These hormones interact with the intercellular receptors. They form a hormone-receptor complex with the genome.
Hormones like FSH do not enter the target cell. They interact with the membrane-bound receptors and generate secondary messengers like cyclic AMP for their action.

Topic: Biology, Chemical Coordination and Integration, NEET
The median AD of a triangle $\mathrm{ABC}$ is bisected at $\mathrm{F}$, and $\mathrm{BF}$ is produced to meet the side $A C$ in $P$. If $A P=\lambda A C$, then what is the value of $\lambda$ ?

Options:
A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{2}{3}$
D $\frac{1}{3}$

The correct answer is:
$\frac{1}{3}$

Solution:
Given, $\mathrm{AD}$ is the median of $\triangle \mathrm{ABC} . \mathrm{F}$ is mid-point of AD and $\mathrm{BF}$ is produced to meet the side $\mathrm{AC}$ in $\mathrm{P}$.
Draw DQ $\| \mathrm{FP}$ In $\Delta \mathrm{ADQ}, \mathrm{F}$ is mid-point of $\mathrm{AD}$ and $\mathrm{FP} \| \mathrm{DQ}$. Pis mid-point of $A Q$ (converse of mid-point theorem) $\Rightarrow \mathrm{AP}=\mathrm{PQ}$ ...(1)
In $\Delta \mathrm{BCP}, \mathrm{D}$ is mid-point of $\mathrm{BC}$ and $\mathrm{DQ} \| \mathrm{BP}$.
$\therefore$ Q is midpoint of $P C$. $\Rightarrow \mathrm{PQ}=\mathrm{QC}$ ...(2)
From $(1),(2)$ we get, $A P=P Q=Q C$ From figure, $\mathrm{AP}+\mathrm{PQ}+\mathrm{QC}=\mathrm{AC}$
$\Rightarrow \mathrm{AP}+\mathrm{AP}+\mathrm{AP}=\mathrm{AC}$
$\Rightarrow 3 \times \mathrm{AP}=\mathrm{AC}$
$\Rightarrow \mathrm{AP}=\frac{1}{3} \times \mathrm{AC} \quad \therefore \quad \lambda=\frac{1}{3}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2006 (Phase 1)
The median of 27 observations of a variable is 18 . Three more observations are made and the values of these observations are 16,18 and 50 . What is the median of these 30 observations ? $\quad$

Options:
A 18
B 19
C $25.5$
D Can not be determined due to insufficient data

The correct answer is:
19

Solution:
Median is middle of data. Observations are 27 and median is $18 .$ So, sum of all the observation are $18 \times 27$ $=486$ Now, 16,18 and 50 are additional three observations. So, Total $=486+16+18+50=570$.
and number of obs. are 30 .
Median $=\frac{570}{30}=19$

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 2)
The member of this Phylum are terrestrial or aquatic having an organ system level organisation bilaterally symmetrical, triploblastic, segmented and eucoelomate animals?

Options:
A Mollusca
B Arthropoda
C Platyhelminthes
D Echinodermata

The correct answer is:
Arthropoda

Solution:
The body od arthropoda is Bilateral, Triploblastic with organ system level of organization. Head is distinct with high degree of caphalization, consists of many fused segments bears well developed sense organ such as eyes, compound eyes, antennae. Arthropoda have various shapes with externally segmented body. Segmentation is either two region head and trunk or cephalothorax and abdomen or three region with head, thorax and abdomen. Some or all segments bear jointed appendages. Hence name Arthro - joints, poda - foot. No internal segmentation like Annelida. Body wall consists of cuticle composed of protein and polysaccharide chitin which is further strengthened by deposition of mineral (calcium phosphate and carbonate).

Topic: Biology, Animal Kingdom, NEET
The meristem that occurs in the mature regions of root and shoots of many plants, particularly those that produce woody axis and appear later than primary meristem is called

Options:
A secondary meristem
B lateral meristem
C both a and b
D permanent tissue

The correct answer is:
both a and b

Solution:
The meristematic cell present on mature root and shoot which produce a woody axis known as lateral meristem. They are cylindrical meristem and are the secondary meristems that induce the growth of the plant with increasing girth.

Topic: Biology, Anatomy of Flowering Plants, NEET
The meristem that occurs in the mature regions of roots and shoots of many plants, particularly those that produce woody axis and appear later than primary meristem, is called

Options:
A Intercalary meristem
B Apical meristem
C Secondary meristem
D All the above

The correct answer is:
Secondary meristem

Solution:
The meristem that occurs in the mature regions of roots and shoots of many plants, particularly those that produce woody axis and appear later than primary meristem, is called secondary or lateral meristem. They are cylindrical meristems and are responsible for producing the secondary tissue.

Topic: Biology, Anatomy of Flowering Plants, NEET
The meristem that occurs in the mature regions of roots and shoots of many plants, particularly those that produce woody axis

Options:
A Primary meristem
B Secondary meristem
C Intercalary meristem
D Both (B) and (C)

The correct answer is:
Secondary meristem

Solution:
The meristem that occurs in the mature regions of roots and shoots of many plants, particularly those that produce woody axis and appear later than primary meristem is called the secondary or lateral meristem. They are cylindrical meristems. Fascicular vascular cambium, interfascicular cambium, and cork-cambium are examples of lateral meristems. These are responsible for producing the secondary tissues.

Topic: Biology, Anatomy of Flowering Plants, NEET
The meristem which occurs in grasses and regenerates parts removed by the grazing herbivores is

Options:
A Apical meristem
B Intercalary meristem
C Secondary meristem
D Cambium

The correct answer is:
Intercalary meristem

Solution:
The meristem which occurs between mature tissues is known as intercalary meristem. They occur in grasses and regenerate parts removed by the grazing herbivores.

Topic: Biology, Anatomy of Flowering Plants, NEET
The middle point of $\mathrm{A}(1,2)$ and $\mathrm{B}(\mathrm{x}, \mathrm{y})$ is $\mathrm{C}(2,4)$. If BD is perpendicular to $\mathrm{AB}$ such that $\mathrm{CD}=3$ umit, then what is the length BD?

Options:
A $2 \sqrt{2}$ wit
B 2 unit
C 3 unit
D $3 \sqrt{2}$ unit

The correct answer is:
2 unit

Solution:
Given that mid point of $\mathrm{A}(1,2)$ and $\mathrm{B}(\mathrm{x}, \mathrm{y})$ is $\mathrm{C}(2,4)$,
$\frac{1+x}{2}=2$ and $\frac{2+y}{2}=4$
$\Rightarrow \quad x=3$ and $y=6$
So, coordinates of B are $(3,6)$. Given that
$\mathrm{BD} \perp \mathrm{AB}$ and $\mathrm{CD}=3$ unit
$\mathrm{BC}=\sqrt{(2-3)^{2}+(4-6)^{2}}=\sqrt{1+4}=\sqrt{5}$
In right angled $\Delta \mathrm{BCD}, \mathrm{CD}^{2}=\mathrm{BC}^{2}+\mathrm{BD}^{2}$
$\Rightarrow 9=5+\mathrm{BD}^{2} \Rightarrow \mathrm{BD}^{2}=4 \Rightarrow \mathrm{BD}=2 \mathrm{unit}$

Topic: Mathematics, Straight Lines, NDA, NDA 2006 (Phase 2)
The middle point of the segment of the straight line joining the points $(p, q)$ and $(q,-p)$ is $(r / 2, s / 2)$. What is the length of the segment?

Options:
A $\left[\left(s^{2}+r^{2}\right)^{1 / 2}\right] / 2$
B $\left[\left(s^{2}+r^{2}\right)^{1 / 2}\right] / 4$
C $\left(s^{2}+r^{2}\right)^{1 / 2}$
D $s+r$

The correct answer is:
$\left(s^{2}+r^{2}\right)^{1 / 2}$

Solution:
Two joining points are $(p, q)$ and $(q,-p)$ Mid point of $(p, q)$ and $(q,-p)$ is $\left(\frac{p+q}{2}, \frac{q-p}{2}\right)$
But it is given that the mid-point is $\left(\frac{r}{2}, \frac{s}{2}\right)$.
$\therefore \frac{p+q}{2}=\frac{r}{2}$ and $\frac{q-p}{2}=\frac{s}{2}$
$\Rightarrow p+q=r$ and $q-p=s$
Now, length of segment $=\sqrt{(p-q)^{2}+(q+p)^{2}}$ (by distance formula) $=\sqrt{s^{2}+r^{2}}=\left(s^{2}+r^{2}\right)^{1 / 2}$

Topic: Mathematics, Straight Lines, NDA, NDA 2009 (Phase 2)
The minimum distance from the point $(4,2)$ to $y^{2}=8 x$ is equal to

Options:
A $\sqrt{2}$
B $2 \sqrt{2}$
C 2
D $3 \sqrt{2}$

The correct answer is:
$2 \sqrt{2}$

Solution:
Given equation is $\mathrm{y}^{2}-8 \mathrm{x}$ Take an arbitrary point on this curve. If we take y as $\mathrm{P}$.
then point is $\left(\frac{P^{2}}{8}, P\right)$.
The distance between $\left(\frac{P^{2}}{8}, P\right)$ and $(4,2)$ is
$d^{2}=\left(\frac{P^{2}}{8}-4\right)^{2}+(P-2)^{2}$ ...(1)
$=\frac{1}{64}\left(P^{2}-32\right)^{2}+(P-2)^{2}$
$\Rightarrow 2 d \cdot \frac{d d}{d p}=\frac{1}{64} \times 2\left(P^{2}-32\right) \times 2 P+2(P-2)$
$=\frac{1}{16}\left(P^{2}-32\right) P+2(P-2)$
for minimum distance, $\frac{d d}{d p}=0$ $\Rightarrow \mathrm{P}^{3}-32 \mathrm{P}+32 \mathrm{P}-64=0$
$\Rightarrow \mathrm{P}^{3}=64$
$\Rightarrow \mathrm{P}=4$
$\therefore(1) \Rightarrow d^{2}=\left(\frac{16}{8}-4\right)^{2}+(4-2)^{2}$
$=(-2)^{2}+(2)^{2}=8$
$\Rightarrow d=\sqrt{8}=2 \sqrt{2}$

Topic: Mathematics, Straight Lines, NDA, NDA 2019 (Phase 1)
The minimum value of the function $f(x)=|x-4|$ exists at

Options:
A $x=0$
B $x=2$
C $x=4$
D $x=-4$

The correct answer is:
$x=4$

Solution:
$\quad$ At $x=4, f(x)=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 2)
The modulus and principal argument of the complex number
$\frac{1+2 i}{1-(1-i)^{2}}$ are respectively

Options:
A 1,0
B 1,1
C 2,0
D 2,1

The correct answer is:
1,0

Solution:
We know, if $Z=x+$ iy,$|z|=\sqrt{x^{2}+y^{2}}$
Given, $\frac{1+2 i}{1-(1-i)^{2}}=\frac{1+2 i}{1-\left(1-2 i+i^{2}\right)}=\frac{1+2 i}{1+2 i}=1$
Since, it is purelyreal number, modulus $=1$ and principal argument $=0$

Topic: Mathematics, Complex Number, NDA, NDA 2017 (Phase 1)
The modulus-amplitude form of $\sqrt{3}+\mathrm{i}$, where $\mathrm{i}=\sqrt{-1}$ is

Options:
A $2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)$
B $2\left(\cos \frac{\pi}{6}+\mathrm{i} \sin \frac{\pi}{6}\right)$
C $4\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)$
D $4\left(\cos \frac{\pi}{6}+\mathrm{i} \sin \frac{\pi}{6}\right)$

The correct answer is:
$2\left(\cos \frac{\pi}{6}+\mathrm{i} \sin \frac{\pi}{6}\right)$

Solution:
Let $x+$ iy $=\sqrt{3}+$ i
Comparing real and imaginary parts, $\mathrm{x}=\sqrt{3}, \mathrm{y}=1$ modulus-amplitude of $\mathrm{x}+$ iy is $\mathrm{r}(\cos \theta+\sin \theta)$, where
$r=\sqrt{x^{2}+y^{2}}$ and $\theta=\tan ^{-1}\left(\frac{y}{x}\right)$
$\mathrm{r}=\sqrt{(\sqrt{3})^{2}+(1)^{2}}=\sqrt{4}=2$
$\theta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$
$\therefore$ modulus-amplitude is $2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$.

Topic: Mathematics, Complex Number, NDA, NDA 2018 (Phase 1)
The monthly family expenditure (in percentage) on different items are as follows:
\begin{array}{|c|c|c|c|c|c|}
\hline Food & Rent & Cloth & Transport & Education & Others \\
\hline 38 & 19 & 18 & - & 9 & 6 \\
\hline
\end{array} If the total monthly expenditure is $â‚¹ 9000$, then what is the benditure on transport

Options:
A $â‚¹ 180$
B â‚¹1000
C $\quad$ â‚¹900
D â‚¹ 360

The correct answer is:
$\quad$ â‚¹900

Solution:
Required expenditure $=9000 \times \frac{10}{100}$
$=₹ 900 .$

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 2)
The more positive the value of $E^{\ominus}$, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.
$\mathrm{E}^{\ominus}$ values $: \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 \mathrm{~V} ; \mathrm{I}_2(\mathrm{s}) / \mathrm{I}^{-}=+0.54$; $\mathrm{Cu}^{2+} / \mathrm{Cu}=+0.34 ; \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}$

Options:
A
$\mathrm{Fe}^{3+}$

B
$\mathrm{I}_2(\mathrm{s})$

C
$\mathrm{Cu}^{2+}$

D
$\mathrm{Ag}^{+}$

The correct answer is:
$\mathrm{Ag}^{+}$

Solution:
Since the redox couple $\mathrm{Ag}^{+} / \mathrm{Ag}$ has the highest positive value of $\mathrm{E}^0$ i.e., $0.80 \mathrm{~V}$, therefore, $\mathrm{Ag}^{+}$is the strongest oxidising agent.

Topic: Chemistry, Redox Reactions
The morphological nature of the edible part of coconut is

Options:
A Perisperm
B Cotyledon
C Endosperm
D Pericarp

The correct answer is:
Endosperm

Solution:
The edible part of coconut is endosperm. Coconut water is free nuclear endosperm and the white kernel is the cellular endosperm.
Initially, the coconut is a nuclear liquid endosperm. it is a coenocytic liquid tissue with lots of rapidly dividing nuclei. At some point, gelification of the soil surrounding the nuclei occurs and cellular boundaries start to appear. These cells and nuclei start to settle down on the endothelium (inner lining of the integuments) and the cells divide to produce a columnar sort of a covering over it. This is what forms the characteristic of coconut meat. Development of the cellular endosperm is by coalescence of these deposited cells and nuclei at the antipodal end and continues towards the top. The cells right above the endothelium stay meristematic for a long time while the columnar cells on top of this meristematic layer grow radially inward.

Topic: Biology, Plant Kingdom, NEET, NEET 2017
The most abundant antibody in colostrum is

Options:
A Immunoglobulin A
B Immunoglobulin G
C Immunoglobulin M
D Immunoglobulin E

The correct answer is:
Immunoglobulin A

Solution:
Colostrum contains large numbers of antibodies called "secretory immunoglobulin" IgA that help protect the mucous membranes in the throat, lungs, and intestines of the infant.

Topic: Biology, Human Health and Diseases, NEET
The most abundant chemical in living organisms could be

Options:
A
protein

B
water

C
sugar

D
nucleic acid.

The correct answer is:
water

Solution:
Approx $70 \%$ of living orgamisms'body weight is water.

Topic: Biology, Biomolecules
The most abundant component of living organisms is

Options:
A Protein
B Water
C Sugar
D Nucleic acid

The correct answer is:
Water

Solution:
The most abundant component of living organisms is water.

Topic: Biology, Biomolecules (B), NEET
The most abundant element present in the plants is:

Options:
A carbon
B nitrogen
C manganese
D iron.

The correct answer is:
carbon

Solution:
Carbon is the most abundant element in living matter.
Related Theory
The six most common elements of life on Earth (including more than $97 \%$ of the mass of $a$ human body) are carbon, hydrogen, nitrogen, oxygen, sulphur and phosphorus.

Topic: Biology, Mineral Nutrition, NEET, NEET 2004
The most abundant enzyme in the biosphere is

Options:
A Collagen
B RuBisCO
C Trypsin
D Insulin

The correct answer is:
RuBisCO

Solution:
The world's most abundant and most important enzyme is RuBisCo. It's the most abundant because it's present in relatively large quantities in every photosynthetic organism on the planet - from microscopic cyanobacteria and phytoplankton in the oceans to the leaves of giant-sized trees in the tropics.

Topic: Chemistry, Biomolecules, NEET
The most abundant intracellular cation is

Options:
A $\mathrm{Na}^{+}$
B $\mathrm{Ca}^{2+}$
C $\mathrm{H}^{+}$
D $\mathrm{K}^{+}$

The correct answer is:
$\mathrm{K}^{+}$

Solution:
Major intracellular cation is $\mathrm{K}^{+}$.
Major extracellular cation is $\mathrm{Na}^{+}$.
$\mathrm{Ca}^{+}$play major role in signal transduction pathways.
$\mathrm{H}^{+}$is found in aqueous solution of all acids.

Topic: Biology, Biomolecules (B), NEET, NEET 2013 (All India)