datavorous/entrance-exam-dataset huggingface dataset:
If the matrix $\mathrm{A}=\left[\begin{array}{rrr}1 & 3 & 1 \\ -1 & 2 & -3 \\ 0 & 1 & 2\end{array}\right]$ then $\operatorname{adj}(\operatorname{adj} A)$ is equal to

Options:
A $\left[\begin{array}{llr}12 & 36 & 12 \\ -12 & 24 & -36 \\ 0 & 12 & 24\end{array}\right]$
B $\left[\begin{array}{lll}12 & 26 & -12 \\ 24 & 36 & -36 \\ 0 & 12 & -24\end{array}\right]$
C $\left[\begin{array}{llr}12 & -12 & 36 \\ 24 & -24 & -36 \\ 0 & 12 & 24\end{array}\right]$
D None of there

The correct answer is:
$\left[\begin{array}{llr}12 & 36 & 12 \\ -12 & 24 & -36 \\ 0 & 12 & 24\end{array}\right]$

Solution:
$A=\left[\begin{array}{ccc}1 & 3 & 1 \\ -1 & 2 & -3 \\ 0 & 1 & 2\end{array}\right]$
$\begin{aligned} \Rightarrow &|\mathrm{A}|=1 .(4+3)-3(-2+0)+1(-1-0) \\ &=7+6-1=12 \\ & \text { So, adj }(\operatorname{adj} \mathrm{A})=|\mathrm{A}|^{\mathrm{n}-2}=\mathrm{A} \\ &=(12)^{3-2} \mathrm{~A}=12 \mathrm{~A} \\ &=12\left[\begin{array}{ccc}1 & 3 & 1 \\ -1 & 2 & -3 \\ 0 & 1 & 2\end{array}\right] \\ &=\left[\begin{array}{ccc}12 & 36 & 12 \\ -12 & 24 & -36 \\ 0 & 12 & 24\end{array}\right] \end{aligned}$

Topic: Mathematics, Matrices, JEE Main
If the matrix A=2513, then the value of A100+A98A20+A18 is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
A=2513=6-5=1
A100+A98A20+A18=A98A2+IA18A2+I=A80=1

Topic: Mathematics, Determinants, JEE Main
If the matrix \( \mathrm{A}=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right] \) satisfies the equation \( \mathrm{A}^{2}+\lambda \mathrm{A}+\mu \mathrm{I}=\mathrm{O} \)
\( \left(\right. \) where,\( I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \) and \( \left.O=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\right) \), then the value of \( \frac{\mu-\lambda}{3} \) is

Options:

The correct answer is:
3

Solution:
This question can be easily done by using the concept of characteristic equation of a matrix.
Consider the equation det (A - kI) = 0 , where k is a scalar.
Now, when we solve the equation A - k I = 0 , then we get a polynomial in k.
Here, if we put k = A(matrix), then the equation obtained is called as the characteristic equation of square matrix A.
The characteristic polynomial of A is A - k I = 0 .
∴ 3 2 7 5 - k 1 0 0 1 = 0
⇒ 3 - k 2 7 5 - k = 0
⇒ 3 - k 2 7 5 - k = 0
⇒ 3 - k 5 - k - 1 4 = 0
⇒ k 2 - 8 k + 1 = 0
∴ A 2 - 8 A + I = 0
On comparing the above equation with the given equation A 2 + λ A + μ I = O , we get
λ = - 8 , μ = 1 ⇒ μ - λ 3 = 3

Topic: Mathematics, Matrices, JEE Main
If the maximum and minimum values of the determinant

1 + sin 2 x cos 2 x sin 2x sin 2 x 1 + cos 2 x sin 2x sin 2 x cos 2 x 1 + sin 2x are Î± and Î² respectively, then Î± + 2 Î² is equal to

Options:

The correct answer is:
5

Solution:
Given determinant is
Δ = 1 + sin 2 x cos 2 x sin 2x sin 2 x 1 + cos 2 x sin 2x sin 2 x cos 2 x 1 + sin 2x
Applying the operation C 1 → C 1 + C 2 , we get
Δ = 2 cos 2 x sin 2x 2 1 + cos 2 x sin 2x 1 cos 2 x 1 + sin 2x
Applying the operations R 2 → R 2 - R 1 and then R 3 → R 3 - R 1 , we get
Δ = 2 cos 2 x sin 2x 0 1 0 - 1 0 1
Now, expanding the above determinant along R2, we get
Δ = - 0 + ( 2 + sin 2x ) - 0 = 2 + sin 2x
Since, the maximum value of sin 2x is 1 and minimum value of sin 2x is -1.
Therefore, α = maximum value of Δ = 2 + 1 = 3 and β = minimum value of Δ = 2 - 1 = 1 .
⇒ α = 3 and β = 1
∴ α + 2 β = 3 + 2 = 5

Topic: Mathematics, Determinants, JEE Main
If the maximum area bounded by y2=4x and the line y=mx ∀m∈1,3 is k square units, then the smallest prime number greater than 3k is

Options:
A 3
B 5
C 7
D 11

The correct answer is:
11

Solution:
The area covered is maximum if ‘m’ is minimum, i.e. m=1
∴ Required area ∫044x-xdx
= 2⋅ x 3 2 3 2 − x 2 2 0 4
=43x32-x2204
=323-8=83 sq. units
∴k=83
Thus required prime number =11

Topic: Mathematics, Area Under Curves, JEE Main
If the maximum distance of normal to the ellipse x24+y2b2=1,b<2, from the origin is 1 , then the eccentricity of the ellipse is:

Options:
A 12
B 32
C 12
D 34

The correct answer is:
32

Solution:
Given:

x24+y2b2=1

Equation of normal to the ellipse is

2xsecθ-bycosecθ=4-b2

Distance of normal from origin is

d=4-b24sec2θ+b2cosec2θ

For maximum distance, denominator must be minimum.

4sec2θ+b2cosec2θ

=4+4tan2θ+b2+b2cot2θ

Now,

4tan2θ+b2cot2θ2≥4tan2θ×b2cot2θ

⇒4tan2θ+b2cot2θ≥4b

⇒4+b2+4tan2θ+b2cot2θ≥4+b2+4b

⇒4+b2+4tan2θ+b2cot2θ≥2+b2

So, minimum value of 4sec2θ+b2cosec2θ is 2+b2=b+2

So,

d=4-b22+b=1

⇒2-b=1

⇒2-b=±1

⇒b=1, 3

But b<2, so b=1, hence

e=1-14=32

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2023 (31 Jan Shift 1)
If the maximum load carried by an elevator is 1400 kg (600 kg-Passengers +800 kg-elevator) , which is moving up with a uniform speed of 3 m s-1 and the frictional force acting on it is 2000 N, then the maximum power used by the motor is _____________kW. g=10 m s-2

Options:

The correct answer is:
48

Solution:
The maximum force acting is, Fmax=14000+2000 N

= 16000 N

The formula for power is given by, P=Fmaxv.

It is given, v=3 m s-1.

⇒P = 16000 N × 3 m s-1 = 48000 W=48 kW

⇒W = 48 kW

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (10 Apr Shift 2)
If the maximum value of a, for which the function fax=tan-12x-3ax+7 is non-decreasing in -π6,π6, is a¯, then fa¯π8 is equal to

Options:
A 8-9π49+π2
B 8-4π94+π2
C 81+π29+π2
D 8-π4

The correct answer is:
8-9π49+π2

Solution:
Given fax=tan-12x-3ax+7 is non-decreasing function in -π6,π6, so

fa'x=21+4x2-3a≥0

i.e. a≤231+4x2min

Now, maximum value of a is at x=0

i.e. amax=a¯=69+π2

Hence, fa¯π8=tan-1π4-369+π2π8+7

=tan-1π4-9π4π2+9+7=8-9π4π2+9

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (26 Jul Shift 2)
If the maximum value of accelerating potential provided by a radio frequency oscillator is 12kV. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is:

mp=1.67×10-27 kg, e=1.6×10-19 C, Speed of light=3×108 m s-1

Options:

The correct answer is:
543

Solution:
From the conservation of energy,

2neV=12mv2

⇒2n×12×103×1.6×10-19=12×1.67×10-27×5×1072

⇒n=543

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (26 Aug Shift 2)
If the maximum value of the term independent of t in the expansion of t2x15+1-x110t15,x≥0, is K, then 8 K is equal to _____ .

Options:

The correct answer is:
6006

Solution:
Given,

t2x15+1-x110t15

Now rth term is given by,

Tr+1=Cr15t2x1515-r·1-xr10tr

For independent of t,

215-r-r=0

30-2r-r=0

⇒r=10

So, Maximum value of C1015x1-x will be at x=12as dydxx-x2=1-2x & 1-2x=0⇒x=12

=C101512×12=30034

So, K=30034 and 8K=6006

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (25 Jul Shift 1)
If the maximum value of x which satisfies the inequality sin-1â¡sinâ¡xâ¥cos-1â¡sinâ¡x for xâÏ2,2Ï is Î», then 2Î»3 is equal to (take Ï=3.14)

Options:

The correct answer is:
1.57

Solution:
Let t=sinâ¡x
From the graphs of sin-1â¡t and cos-1â¡t, we know that,

sin-1â¡tâ¥cos-1â¡t for tâ12,1

âsinâ¡xâ12,1
Hence, xâÏ2,3Ï4
Maximum value is 3Ï4=Î»

So, 2Î»3=Ï2=1.57

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If the mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, …, x5 and -50 is equal to

Options:
A 582.5
B 507.5
C 509.5
D 586.5

The correct answer is:
507.5

Solution:
We have,

Mean =x1+x2+x3+x4+x55=10

⇒x1+x2+x3+x4+x5=50 ...i
Variance

=S.D2=Σxi25-x¯2=9
⇒x12+x22+x32+x42+x525-100=9
⇒x12+x22+x32+x42+x52=545 ...ii
Now,

x¯new=x1+x2+x3+x4+x5-506=0
New variance

=∑i=16xi26-x¯new2
=x12+x22+x32+x42+x52+25006-0
=545+25006=30456=507.5

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (10 Jan Shift 2)
If the mean and the standard deviation of the data 3,5,7, a, b are 5and 2 respectively, then a and b are the roots of the equation:

Options:
A x2-10x+18=0
B 2x2-20x+19=0
C x2-10x+19=0
D x2-20x+18=0

The correct answer is:
x2-10x+19=0

Solution:
5+3+7+a+b=25⇒a+b=10
S.D. =52+32+72+a2+b25-52=2
=a2+b2+835-25=4⇒a2+b2=62
⇒(a+b)2-2ab=62⇒ab=19
so equation whose roots are a and b is x2-10x+19=0

Topic: Mathematics, Statistics, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If the mean and the variance of a binomial variate X are 2 & 1 respectively, then the probability that X takes a value greater than or equal to one is:

Options:
A 116
B 916
C 34
D 1516

The correct answer is:
1516

Solution:
np=2
npq=1
∴q=12
p=1-q=12
n=4
PX≥1=1- 4C0 124120
1-116=1516

Topic: Mathematics, Probability, JEE Main, JEE Main 2015 (11 Apr Online)
If the mean and the variance of the numbers a, b, 8, 5 and 10 are 6 and 6.8 respectively, then the value of a3+b3 is equal to

Options:
A 58
B 61
C 91
D 89

The correct answer is:
91

Solution:
Mean =6=a+b+8+5+105⇒a+b=7
Variance =6.8=a2+b2+64+25+1005-36
⇒a2+b2=25
2ab=a+b2-a2+b2⇒ab=12
Now, a3+b3=a+b3-3aba+b
=343-3×12×7
=91

Topic: Mathematics, Statistics, JEE Main
If the mean and variance of a binomial variate $X$ are 2 and 1 respectively, then the probability that $X$ takes a value greater than 1 , is

Options:
A ${ }^{\frac{2}{3}}$
B ${ }^{\frac{4}{5}}$
C ${ }^{\frac{7}{8}}$
D ${ }^{\frac{15}{16}}$

The correct answer is:
${ }^{\frac{15}{16}}$

Solution:
We have mean $(X)=n p=2$
and variance $(X)=n p q=1 \Rightarrow q=\frac{1}{2}$ or $p=\frac{1}{2}$ and $n=4$
Thus
$p(X \geq 1)=1-p(X=0)$ $=1-{ }^4 C_0\left(\frac{1}{2}\right)^4=\frac{15}{16}$

Topic: Mathematics, Probability, JEE Main
If the mean and variance of eight numbers 3,7,9,12,13,20,x and y be 10 and 25 respectively, then x·y is equal to

Options:

The correct answer is:
54

Solution:
Mean =x-=3+7+9+12+13+20+x+y8=10⇒x+y=16…i
Variance σ2=∑xi28-x-2=25
9+49+81+144+169+400+x2+y28-100=25
⇒x2+y2=148…ii
x+y2=162=x2+y2+2xy=256⇒xy=54

Topic: Mathematics, Statistics, JEE Main, JEE Main 2020 (07 Jan Shift 2)
If the mean and variance of eight numbers \( 3,7,9,12,13,20, x \) and \( y \) be \( 10 \) and \( 25 \) respectively, then \( x \cdot y \) is equal to

Options:

The correct answer is:
54

Solution:
Mean =x-=3+7+9+12+13+20+x+y8=10⇒x+y=16…i
Variance σ2=∑xi28-x-2=25
9+49+81+144+169+400+x2+y28-100=25
⇒x2+y2=148…ii
x+y2=162=x2+y2+2xy=256⇒xy=54

Topic: Mathematics, Statistics, JEE Main
If the mean and variance of five observations are 245 and 19425 respectively and the mean of first four observations is 72, then the variance of the first four observations in equal to

Options:
A 45
B 7712
C 54
D 1054

The correct answer is:
54

Solution:
Given:

x=245 and variance is 19425

⇒∑i=15xi=245×5

⇒∑i=15xi=24

Also, variance

⇒∑i=15xi25-2452=19425

⇒∑i=15xi2=154

Now, 5th observation =24-sum of first 4 terms

=24-72×4=10

So, new variance =∑i=14xi24-722

=154-1004-494

=54

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (29 Jan Shift 2)
If the mean and variance of six observations 7,10,11,15,a,b are 10 and 203, respectively, then the value of |a-b| is equal to:

Options:
A 9
B 11
C 7
D 1

The correct answer is:
1

Solution:
Given:

Mean=10

⇒7+10+11+15+a+b6=10

⇒a+b=17 …i

And,

Variance=203

⇒72+102+112+152+a2+b26-102=203

⇒495+a2+b26=3203

⇒495+a2+b2=640

⇒a2+b2=145 …ii

Solve i and ii, we get

a=9, b=8 or a=8, b=9

∴a-b=1

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (20 Jul Shift 2)
If the mean and variance of the data 65,68,58,44, 48,45,60,α,β,60 where α>β are 56 and 66.2 respectively, then α2+β2 is equal to

Options:

The correct answer is:
6344

Solution:
Given: x¯=56 and σ2=66.2

We know that, σ2=∑xi2n-x2

⇒652+682+582+442+482+452+602+α2+β2+60210-562=66.2

⇒α2+β2+2567810-(56)2=66.2

⇒α2+β2+2567810=3136+66.2

⇒α2+β2=32022-25678

⇒α2+β2=6344

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If the mean and variance of the following data: 6,10,7,13,a,12,b,12 are 9 and 374 respectively, then a-b2 is equal to:

Options:
A 24
B 12
C 32
D 16

The correct answer is:
16

Solution:
Mean =6+10+7+13+a+12+b+128=9

⇒60+a+b=72

⇒a+b=12 ...1

variance =∑xi2n-∑xin2=374

Here, Σxi2=62+102+72+132+a2+b2+122+122

=a2+b2+642

So, variance is a2+b2+6428-92=374

⇒a2+b28+3214-81=374

⇒a2+b28=81+374-3214

⇒a2+b28=81-71

∴ a2+b2=80

From 1 a2+b2+2ab=144

⇒80+2ab=144 ∴ 2ab=64

So, a-b2=a2+b2-2ab=80-64=16.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (27 Jul Shift 1)
If the mean and variance of the frequency distribution

xi	2	4	6	8	10	12	14	16

fi	4	4	α	15	8	β	4	5

are 9 and 15.08 respectively, then the value of α2+β2-αβ is _____.

Options:

The correct answer is:
25

Solution:
Given,

The mean and variance of the frequency distribution

xi	2	4	6	8	10	12	14	16

fi	4	4	α	15	8	β	4	5

are 9 and 15.08 respectively,

Now mean is given by,

Mean =8+16+120+80+56+80+6α+12β40+α+β=9

⇒360+9α+9β=360+6α+12β

⇒3α-3β=0

⇒α=β ......1

Now variance is given by,

15.08=16+64+36α+960+800+144α+784+128040+2α-92

⇒96.08=3904+180α40+2α

⇒40+2α96.08=3904+180α

⇒3843.20+(192.16)α=3904+180α

⇒(12.16)α=60.80

⇒α=5=β

Hence, α2+β2-αβ=25+25-25=25

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (06 Apr Shift 2)
If the mean deviation about median for the number 3,5,7,2k, 12,16,21,24 arranged in the ascending order, is 6 then the median is

Options:
A 11.5
B 10.5
C 12
D 11

The correct answer is:
11

Solution:
Median is the central value (or middle observation) of statistical data if it is arranged in ascending or descending order.

Thus, if there are n observations (variates) x1, x2, x3, …, xn arranged in ascending or descending order, then

median =n+12th observation, if n is odd

and median =n2 th observation +n2+1 th observation 2, if n is even.

Here we can see there are even number of terms,

So, median =2k+122=k+6

Mean deviation =∑xi-Mn=6

⇒k+3+k+1+k-1+6-k+6-k+10-k+15-k+18-k8=6

⇒ 58-2k8=6

⇒k=5

So, median =2×5+122=11

Topic: Mathematics, Statistics, JEE Main, JEE Main 2022 (25 Jul Shift 2)
If the mean deviation about the mean of the numbers 1,2,3,……,n, where n is odd, is 5n+1n, then n is equal to ______.

Options:

The correct answer is:
21

Solution:
We know that mean deviation about the mean of odd natural number is given by M.D x=1nn2-14

So, now 1nn2-14=5n+1n (given)

⇒n-1=20⇒n=21

Topic: Mathematics, Statistics, JEE Main, JEE Main 2022 (25 Jun Shift 2)
If the mean deviation about the median of the numbers $\mathrm{a}, 2 \mathrm{a}, \ldots, 50 \mathrm{a}$ is 50 , then $|\mathrm{a}|$ equals

Options:
A
3

B
4

C
5

D
2

The correct answer is:
4

Solution:
$\frac{1}{\mathrm{n}} \sum\left|\mathrm{x}_{\mathrm{i}}-\mathrm{A}\right|$
$\mathrm{A}=$ Median $=\frac{25 \mathrm{a}+26 \mathrm{a}}{2}=25.5 \mathrm{a}$
Mean deviation $=\frac{1}{50}\{|\mathrm{a}-25.5 \mathrm{a}|+|2 \mathrm{a}-25.5 \mathrm{a}|\}=\frac{2}{50}\{(24.5 \mathrm{a}+23.5 \mathrm{a})+\ldots(0.5 \mathrm{a})\}$
$=\frac{2}{50}\{312.5 \mathrm{a}\}=50 \quad$ (Given)
$\Rightarrow 625 \mathrm{a}=2500 \Rightarrow \mathrm{a}=4$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2011
If the mean deviation from the mean of the observations $\{1+K \alpha\}$, where $K=0,1,2, \ldots, 100$ and $\alpha>0$ is 255 then $\alpha=$

Options:
A $10$
B $10.1$
C $25$
D $25.5$

The correct answer is:
$10.1$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Statistics, JEE Main
If the mean deviation of number $1,1+d, 1+2 d, \ldots ., 1+100 d$ from their mean is 255 , then the $d$ is equal to

Options:
A
$10.0$

B
$20.0$

C
$10.1$

D
$20.2$

The correct answer is:
$10.1$

Solution:
$$
\begin{aligned}
& \text { Mean }(\bar{x})=\frac{\text { sum of quantities }}{n}=\frac{\frac{n}{2}(a+1)}{\not n}=\frac{1}{2}[1+1+100 d]=1+50 d \\
& \text { M.D. }=\frac{1}{n} \sum\left|x_i-\bar{x}\right| \Rightarrow 255=\frac{1}{101}[50 d+49 d+48 d+\ldots .+d+0+d+\ldots \ldots+50 d]=\frac{2 d}{101}\left[\frac{50 \times 51}{2}\right] \\
& \Rightarrow d=\frac{255 \times 101}{50 \times 51}=10.1
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2009
If the mean deviation of the numbers 1, 1+d,…, 1+100d from their mean is 255, then a value of d is :

Options:
A 10.1
B 5.05
C 20.2
D 10

The correct answer is:
10.1

Solution:
Mean x-=11011+1+d+1+2d….1+100d
=1101 ×1012 1+1+100d=1+50d
Mean deviation from mean
=11011-1+50d+1+d-1+50d….1+100d-1+50d
=2d101 1+2+3….+50

=2d101 50 50+12
=2550101d=255

⇒d=10.1

Topic: Mathematics, Statistics, JEE Main, JEE Main 2016 (09 Apr Online)
If the mean of $4,7,2,8,6$ and a is 7 , then the mean deviation from the median of these observations is

Options:
A
8

B
5

C
1

D
3

The correct answer is:
3

Solution:
Given observations are 4, 7, 2, 8, 6, $a$ and mean is 7.
We know
$$
\begin{aligned}
& \text { Mean }=\frac{4+7+2+8+6+a}{6} \\
& \Rightarrow 7=\frac{4+7+2+8+6+a}{6} \Rightarrow a=15
\end{aligned}
$$
Now, given observations can be written in ascending order which is $2,4,6,7,8,15$ Since, No. of observation is even
$\therefore$ Median
$=\frac{\left(\frac{6}{2}\right) \text { th observation }+\left(\frac{6}{2}+1\right) \text { th observation }}{2}$
$=\frac{\text { 3rd observation }+4 \text { th observation }}{2}$
$=\frac{6+7}{2}=\frac{13}{2}$
$$
\begin{aligned}
& \text { Now, Mean deviation }=\frac{\sum_{i=1}^6\left|x_i-\frac{13}{2}\right|}{6} \\
& =\frac{\left|4-\frac{13}{2}\right|+\left|7-\frac{13}{2}\right|+\left|2-\frac{13}{2}\right|+\left|8-\frac{13}{2}\right|+\left|6-\frac{13}{2}\right|+\left|15-\frac{13}{2}\right|}{6} \\
& =\frac{\frac{5}{2}+\frac{1}{2}+\frac{9}{2}+\frac{3}{2}+\frac{1}{2}+\frac{17}{2}}{6}=\frac{18}{6}=3 \\
&
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2012 (12 May Online)
If the mean of 10 observations is 50 and the sum of the squares of the deviations of observations from the mean is 250, then the coefficient of variation of these observations is

Options:
A 25
B 50
C 10
D 5

The correct answer is:
10

Solution:
Given, x-=50
âxi-x-2=250 Ï2=110Ã250=25
C.V=S.Dx-Ã100=550Ã100=10.

Topic: Mathematics, Statistics, JEE Main
If the mean of 10 observations is 50 and the sum of the squares of the deviations of observations from the mean is 250, then the coefficient of variation of those observations is

Options:
A 25
B 50
C 10
D 5

The correct answer is:
10

Solution:
Given, x-=50
âxi-x-2=250 Ï2=110Ã250=25
C.V=S.Dx-Ã100=550Ã100=10.

Topic: Mathematics, Statistics, JEE Main
If the mean of 10 observations is 50 and the sum of the squares of the deviations of the
observations from the mean is 250, then the coefficient of variation of thoseobservations is

Options:
A 25
B 50
C 10
D 5

The correct answer is:
10

Solution:
Given, number of observation, $n=10$,
Mean, $\bar{x}=50$ and $\sum|x i-\bar{x}|^2=250$
Variance, $\sigma^2=\sum_{i=1}^n \frac{\left|x_i-\bar{x}\right|^2}{n}$
$\begin{gathered}
\Rightarrow \sigma^2=\sum_{i=1}^{10} \frac{\left|x_i-\bar{x}^2\right|}{10} \\
=\frac{250}{10}=25 \Rightarrow \sigma=5
\end{gathered}$
$\therefore$ Cofficient of variation,
$\text { C.V. }=\frac{\sigma}{x} \times 100=\frac{5}{50} \times 100=10$

Topic: Mathematics, Statistics, JEE Main
If the mean of 50 observations is 25 and their standard deviation is 4 and the sum of the squares of all the observations is λ, then λ1000 is

Options:

The correct answer is:
32.05

Solution:
Variance =∑i=150xi2n-x-2
⇒16=∑i=150xi250-252
⇒∑i=150xi2=641×50
∴∑i=150xi2=32050=λ

Topic: Mathematics, Statistics, JEE Main
If the mean of a set of observations x1,x2,x3,....,xn is x¯, then the mean of observations xi+2i, ∀i=1,2,3,....n is equal to

Options:
A x¯+2n+1
B x¯+n+1
C x¯+n+12n
D None of these

The correct answer is:
x¯+n+1

Solution:
Given, nxÂ¯=x1+x2+x3+...+xn
Now, new observations are x1+2,x2+2.2,....xn+2.n
â´ new mean
=x1+2+x2+2.2+x3+2.3+....+xn+2.nn
=x1+x2+x3+...+xnn+21+2+3+...+nn
=xÂ¯+2n+12

Topic: Mathematics, Statistics, JEE Main
If the mean of a set of observations x1,x2,.....,x10 is 40, then the mean of x1+4,x2+8,x3+12,....,x10+40 is

Options:
A 54
B 62
C 38
D 50

The correct answer is:
62

Solution:
Mean=x1+4+x2+8+x3+12+...+x10+4010
=x1+x2+...+x1010+41+2+3+...+1010
=40+22=62

Topic: Mathematics, Statistics, JEE Main
If the mean of the data: $7,8,9,7,8,7, \lambda, 8$ is 8 , then the variance of this data is

Options:
A
$\frac{9}{8}$

B
2

C
$\frac{7}{8}$

D
1

The correct answer is:
1

Solution:
$$
\text { } \begin{aligned}
&\bar{x}=\frac{7+8+9+7+8+7+\lambda+8}{8}=8 \\
&\Rightarrow \frac{54+\lambda}{8}=8 \Rightarrow \lambda=10
\end{aligned}
$$
Now variance $=\sigma^2$
$$
\begin{aligned}
&(7-8)^2+(8-8)^2+(9-8)^2+(7-8)^2+(8-8)^2 \\
&=\frac{+(7-8)^2+(10-8)^2+(8-8)^2}{8} \\
&\Rightarrow \sigma^2=\frac{1+0+1+1+0+1+4+0}{8}=\frac{8}{8}=1 \\
&
\end{aligned}
$$
Hence, the variance is 1 .

Topic: Mathematics, Statistics, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If the mean of the distribution is 2.6 , then the value of $y$ is

Options:
A $24$
B $13$
C $8$
D $3$

The correct answer is:
$8$

Solution:
We know that, Mean $=\frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n f_i}$
i.e., $2.6=\frac{1 \times 4+2 \times 5+3 \times y+4 \times 1+5 \times 2}{4+5+y+1+2}$
or $31.2+2.6 y=28+3 y$ or $0.4 y=3.2 \Rightarrow y=8$.

Topic: Mathematics, Statistics, JEE Main
If the mean of the frequency distribution

Class :	0-10	10-20	20-30	30-40	40-50

Frequency :	2	3	x	5	4

is 28, then its variance is ________ .

Options:

The correct answer is:
151

Solution:
Given that the mean for the frequency distribution is 28.

We know that Mean x=∑fixi∑fi

Now for the class interval 0-10 we get x1=0+102=5.

Similarly we can calculate xi for other class intervals.

Now we know that mean =28

⇒2×5+3×15+x×25+5×35+4×4514+x=28

⇒x=6

Also we know that variance σ2=∑fix¯-xi2∑fi

σ2=2×28-52+3×28-152+6×28-252+5×28-352+4×28-45220

σ2=2×232+3×132+6×32+5×72+4×17220

σ2=151

Hence this is the required answer.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
If the mean of two samples of sizes 20 and 30 are 25 and 10 respectively, their variances are 9 and 16 respectively, their combined variance is

Options:
A $67.9$
B $68.2$
C $68.1$
D $67.2$

The correct answer is:
$67.2$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Statistics, JEE Main
If the median and the range of four numbers $\{x, y, 2 x+y, x-y\}$, where $0 < y < x < 2 y$, are 10 and 28 respectively, then the mean of the numbers is :

Options:
A
18

B
10

C
5

D
14

The correct answer is:
14

Solution:
Since $0 < y < x < 2 y$
$$
\begin{aligned}
& \therefore \quad y>\frac{x}{2} \Rightarrow x-y < \frac{x}{2} \\
& \therefore x-y < y < x < 2 x+y
\end{aligned}
$$
Hence median $=\frac{y+x}{2}=10$
$\Rightarrow x+y=20$
And range $=(2 x+y)-(x-y)=x+2 y$
But range $=28$
$$
\therefore x+2 y=28
$$
From equations (i) and (ii),
$$
\begin{aligned}
& x=12, y=8 \\
& \therefore \text { Mean } \\
& =\frac{(x-y)+y+x+(2 x+y)}{4}=\frac{4 x+y}{4}
\end{aligned}
$$
$$
=x+\frac{y}{4}=12+\frac{8}{4}=14
$$

Topic: Mathematics, Statistics, JEE Main
If the median and the range of four numbers $\{x, y, 2 x+y, x-y\}$, where $0 < y < x < 2 y$, are 10 and 28 respectively, then the mean of the numbers is :

Options:
A
18

B
10

C
5

D
14

The correct answer is:
14

Solution:
Since $0 < y < x < 2 y$
$$
\begin{aligned}
& \therefore \quad y>\frac{x}{2} \Rightarrow x-y < \frac{x}{2} \\
& \therefore x-y < y < x < 2 x+y
\end{aligned}
$$
Hence median $=\frac{y+x}{2}=10$
$\Rightarrow x+y=20$
And range $=(2 x+y)-(x-y)=x+2 y$
But range $=28$
$$
\therefore x+2 y=28
$$
From equations (i) and (ii),
$$
\begin{aligned}
& x=12, y=8 \\
& \therefore \text { Mean } \\
& =\frac{(x-y)+y+x+(2 x+y)}{4}=\frac{4 x+y}{4}
\end{aligned}
$$
$$
=x+\frac{y}{4}=12+\frac{8}{4}=14
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2013 (23 Apr Online)
If the median of the data $6,7, x-2, x, 18$ and 21 written in ascending order is 16 , then the variance of that data is

Options:
A $30 \frac{1}{5}$
B $31 \frac{1}{3}$
C $32 \frac{1}{2}$
D $33 \frac{1}{3}$

The correct answer is:
$31 \frac{1}{3}$

Solution:
Since, there are six observations and are written in ascending order then
$\begin{aligned}
& \text { Median }=\left(\frac{n}{2}\right)^{\text {th }} \text { observation }+\left(\frac{n}{2}+1\right)^{\text {th }} \frac{\text { observation }}{2} \\
& \frac{x-2+x}{2}=16 \\
& 2 x-2=32 \\
& 2 x=34 \Rightarrow x=17
\end{aligned}$
So the observations are $6,7,15,17,18$ and 21 .
$\begin{aligned}
& \therefore \quad \text { Mean }=\frac{6+7+15+17+18+21}{6} \\
& \bar{x}=\frac{84}{6}=14
\end{aligned}$
$\therefore \quad$ Variance $=\frac{1}{n} \sum\left(x_i-x\right)^2=\frac{188}{6}=\frac{94}{3}=31 \frac{1}{3}$

Topic: Mathematics, Statistics, JEE Main