datavorous/entrance-exam-dataset huggingface dataset:
Let the length of the focal chord PQ of the parabola $y^2=12 x$ be 15 units. If the distance of $\mathrm{PQ}$ from the origin is $\mathrm{p}$, then $10 \mathrm{p}^2$ is equal to _______

Options:

The correct answer is:
72

Solution:
$\begin{aligned} & \text { length of focal chord }=4 a \operatorname{cosec}^2 \theta=15 \\ & 12 \operatorname{cosec}^2 \theta=15 \\ & \sin ^2 \theta=\frac{4}{5} \\ & \tan ^2 \theta=4 \\ & \tan \theta=2 \\ & \text { equation } \frac{y-0}{x-3}=2 \\ & y=2 x-6 \\ & 2 x-y-6=0 \\ & P=\frac{6}{\sqrt{5}} \\ & 10 p^2=10 \cdot \frac{36}{5}=72\end{aligned}$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let the length of the latus rectum of an ellipse with its major axis along $x$ -axis and centre at the origin, be 8 . If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

Options:
A $(4 \sqrt{2}, 2 \sqrt{2})$
B $(4 \sqrt{3}, 2 \sqrt{2})$
C $(4 \sqrt{3}, 2 \sqrt{3})$
D $(4 \sqrt{2}, 2 \sqrt{3})$

The correct answer is:
$(4 \sqrt{3}, 2 \sqrt{2})$

Solution:
Let the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Then, $\frac{2 b^{2}}{a}=8,2 a e-b^{2}$ and $b^{2}=a^{3}\left(1-e^{2}\right)$

$\Rightarrow a=8, b^{2}=32$

Then, the equation of the ellipse

$\frac{x^{2}}{64}+\frac{y^{2}}{32}=1$

Hence, the point $(4 \sqrt{3}, 2 \sqrt{2})$ lies on the ellipse.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let the lengths of intercepts on x -axis and y -axis made by the circle x2+y2+ax+2ay+c=0, a<0 be 22 and 25, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x+2y=0, is equal to :

Options:
A 11
B 7
C 6
D 10

The correct answer is:
6

Solution:
x2+y2+ax+2ay+c=0

2g2-c=2a24-c=22

⇒ a24-c=2 ...1

2f2-c=2a2-c=25

⇒a2-c=5 ...2

1 & 2

3a24=3⇒a=-2 a<0

∴ c=-1

Circle ⇒x2+y2-2x-4y-1=0

⇒x-12+y-22=6

Given x+2y=0⇒m=-12

mtangent=2

Equation of tangent ⇒y-2=2x-1±61+4

⇒2x-y±30=0

Perpendicular distance from 0,0=±304+1=6

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let the lengths of the altitudes from the vertices A-1,1, B5,2, C3,-1 of ÎABC are p1, p2, p3 units respectively, then the value of 1p12+1p321p22 is equal to

Options:

The correct answer is:
2.5

Solution:
1p12+1p321p22=2Δp12+2Δp322Δp22=BC2+AB2AC2 {where Δ represents the area of triangle ABC}
=4+9+36+116+4=5020=2.5

Topic: Mathematics, Straight Lines, JEE Main
Let the line $2 x+3 y-\mathrm{k}=0, \mathrm{k}>0$, intersect the $x$-axis and $y$-axis at the points $\mathrm{A}$ and $\mathrm{B}$, respectively. If the equation of the circle having the line segment $\mathrm{AB}$ as a diameter is $x^2+y^2-3 x-2 y=0$ and the length of the latus rectum of the ellipse $x^2+9 y^2=k^2$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $2 \mathrm{~m}+\mathrm{n}$ is equal to

Options:
A 11
B 10
C 12
D 13

The correct answer is:
11

Solution:
Centre of the circle $=\left(\frac{3}{2}, 1\right)$
Equation of diameter $=2 \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0$
$\begin{aligned}
& 2\left(\frac{3}{2}\right)+3(1)-\mathrm{k}=0 \\
& \Rightarrow \mathrm{k}=6
\end{aligned}$
Now, Equation of ellipse becomes
$\begin{aligned}
& x^2+9 y^2=36 \\
& \frac{x^2}{6^2}+\frac{y^2}{2^2}=1
\end{aligned}$
$\begin{aligned} & \text { length of } L R=\frac{2 b^2}{a}=\frac{2.2^2}{6}=\frac{8}{6}=\frac{4}{3}=\frac{m}{n} \\ & \therefore 2 m+n=2(4)+3=11\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta=0$. Then $(\alpha, \beta)$ equals

Options:
A
$(6,-17)$

B
$(-6,7)$

C
$(5,-15)$

D
$(-5,15)$

The correct answer is:
$(-6,7)$

Solution:
Dr's of line $=(3,-5,2)$
Dr's of normal to the plane $=(1,3,-\alpha)$
Line is perpendicular to normal $\Rightarrow 3(1)-5(3)+2(-\alpha)=0 \Rightarrow 3-15-2 \alpha=0 \Rightarrow 2 \alpha=-12 \Rightarrow \alpha=-6$
Also $(2,1,-2)$ lies on the plane
$$
\begin{aligned}
& 2+3+6(-2)+\beta=0 \Rightarrow \beta=7 \\
& \therefore(\alpha, \beta)=(-6,7)
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2009
Let the line L be the projection of the line x-12=y-31=z-42 in the plane x-2y-z=3. If d is the distance of the point (0,0,6) from L, then d2 is equal to

Options:

The correct answer is:
26

Solution:
Let

x-12=y-31=z-42=r

Any point on the line P2r+1, r+3, 2r+4

For the point P2r+1, r+3, 2r+4 to lie on the plane, its co-ordinates should satisfy the equation x-2y-z=3.

⇒2r+1-2r+3-2r+4=3

⇒-2r-9=3

⇒r=-6

∴P-11, -3, -8

Now, let Qh, k, l be the coordinates of the foot of the perpendicular drawn from1, 3, 4 to the plane x-2y-z=3.

⇒h−11=k−3−2=l−4−1=-1-6-4-31+4+1

⇒h−11=k−3−2=l−4−1=2

⇒h=3, k=-1, l=2

∴Qh, k, l=Q3, -1, 2

Therefore, projection line PQ is

L:x−33--11=y+1-1--3=z−22--8

L:x−314=y+12=z−210

L:x−37=y+11=z−25

Let, L:x−37=y+11=z−25=λ …i

Direction ratio of AC are 7λ+3,λ-1,5λ-4

Now, AC is perpendicular to the projection line (i) So,

77λ+3+λ-1+55λ-4=0

⇒49λ+21+λ-1+25λ-20=0

⇒λ=0

So,

∴C≡3,-1,2

Now,

d2=32+-12+2-62

⇒d2=26

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Let the line L pass through the point (0, 1, 2), intersect the line x-12=y-23=z-34 and be parallel to the plane 2x+y-3z=4. Then the distance of the point P(1, –9, 2) from the line L is

Options:
A 74
B 69
C 54
D 9

The correct answer is:
74

Solution:
Given,

The line L pass through the point (0, 1, 2), intersect the line x-12=y-23=z-34 and be parallel to the plane 2x+y-3z=4,

Now plotting the diagram of above data we have,

Now let the direction of line which passes through 0,1,2 be a,b,c

Now the line L & L1 are coplanar,

So coplanar condition we get,

abc2-13-24-3234=0

⇒abc111234=0

⇒a-2b+c=0 .......1

And the line L is perpendicular to the normal vector of the plane,

So by perpendicular condition we get,
2a+b-3c=0 ......2

So, from equation 1 & 2 we get,

a=b=c

So, equation of line L will be,

⇒L=x1=y-11=z-21=λ

So any point on L can be taken as Aλ, 1+λ, 2+λ,

Now AP→ will be perpendicular to the direction ratio of the line L,

So, again by perpendicular condition we get,

AP→·i^+j^+k^=0

⇒λ-1+λ+10+λ=0

⇒3λ+9=0

⇒λ=-3

Hence, A-3, -2, -1 & P1, -9, 2

So, by distance formula we get, AP=74

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Let the line L:x-12=y+1-1=z-31 intersect the plane 2x+y+3z=16 at the point P. Let the point Q be the foot of perpendicular from the point R1,-1,-3 on the line L. If α is the area of triangle PQR. then α2 is equal to _____ .

Options:

The correct answer is:
180

Solution:
We have,

L:x-12=y+1-1=z-31=λ

Any point on L is P2λ+1,-λ-1,λ+3

It lies on plane 2x+y+3z=16, so

⇒22λ+1+-λ-1+3λ+3=16

⇒6λ+10=16⇒λ=1

∴P≡3,-2,4

Let Q2μ+1,-μ-1,μ+3

Direction ratios of QR is

2μ,-μ,μ+6

Direction ratios of L is 2,-1,1.

Since, QR⊥L, so

4μ+μ+μ+6=0

⇒μ=-1

Q≡-1,0,2

Now,

QR→=2i^-j^-5k^ QP→=4i^-2j^+2k^

Hence,

QR→×QP→=i^j^k^2-1-54-22=-12i^-24j^

Therefore,

α=12×144+576

⇒α2=7204=180

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Let the line L:x=1-y-2=z-3λ,λ∈ℝ meet the plane P: x+2 y+3 z=4 at the point (α,β,γ). If the angle between the line L and the plane P is cos-1514, then α+2β+6γ is equal to

Options:

The correct answer is:
11

Solution:
The given equations are,

L:x-01=y-12=z-3λ & P:x+2y+3z=4

So, Vector parallel to line:<1, 2, λ>=b→

Normal vector to plane P:<1, 2, 3>=n→

Angle between plane & line is θ. Then angle between the line and normal to the plane will be 90-θ.
Then, ⇒cos90-θ=i^+2j^+λk^·i^+2j^+3k^12+22+λ2·12+22+32
⇒314=1+4+3λλ2+514⇒λ=23

L1=x-03=y-16=z-32=μ

x,y,z≡3μ,6μ+1,2μ+3

But this point lies on the plane :

3μ+12μ+2+6μ+9=4

⇒μ=-13
Hence, α=3μ=-1, β=6μ+1=-1, γ=2μ+3=73

Now, α+2β+6γ=11

Hence this is the required answer.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let the line passing through the points P2,-1,2 and Q5,3,4 meet the plane x-y+z=4 at the point R. Then the distance of the point R from the plane x+2y+3z+2=0 measured parallel to the line x-72=y+32=z-21 is

Options:
A 61
B 189
C 31
D 3

The correct answer is:
3

Solution:
Given,

The line passing through the points P2,-1,2 and Q5,3,4,

So, equation of line PQ will be,

x-23=y+14=z-22=λ

Now let point R be 3λ+2,4λ-1,2λ+2

Given, R lies on plane x-y+z=4

∴ 3λ+2-4λ+1+2λ+2=4

⇒λ=-1

∴ R-1,-5,0

Now let line SR be :x+12=y+52=z1=k as it is parallel to x-72=y+32=z-21 and passing through -1,-5,0

So, the point S:2k-1,2k-5,k

Now S lies on plane : x+2y+3z+2=0

⇒2k-1+4k-10+3k+2=0

⇒9k-9=0⇒k=1

Hence, S1,-3,1

So, distance SR=4+4+1=3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let the line x1=6-y2=z+85 intersect the lines x-54=y-73=z+21 and x+36=3-y3=z-61 at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane 2x-2y+z=14 is

Options:
A 3
B 113
C 4
D 103

The correct answer is:
4

Solution:
Given,

The line x1=6-y2=z+85 intersect the lines x-54=y-73=z+21 and x+36=3-y3=z-61 at the points A and B respectively,

Now plotting the diagram, we get,

Now solving, x1=y-6-2=z+85=λ and x-54=y-73=z+21=t we get,

A≡(λ,-2λ+6,5λ-8)≡(4t+5,3t+7,t-2)

Now comparing both side and solving we get, t=-1,λ=1

Hence, point A(1, 4, –3)

Now solving, x1=y-6-2=z+85=μ and x+36=y-3-3=z-61=k we get,

For point B≡(μ,-2μ+6,5μ-8) ≡(6k-3,-3k+3,k+6)

Now comparing both side and solving we get, k=1,μ=3

Hence, point B(3, 0, 7)

So, mid-point of AB=M(2, 2, 2)

Now finding, distance of M from the plane 2x – 2y + z – 14 = 0 we get,

4-4+2-1422+22+12=-123=4

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let the line y=mx and the ellipse 2x2+y2=1 intersect at a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at -132,0 and 0,β, then β is equal to

Options:
A 223
B 23
C 23
D 23

The correct answer is:
23

Solution:
Let P be x1,y1

Equation of normal at P is x2x1-yy1=-12

It passes through

-132, 0⇒-162x1=-12⇒x1=132

So, y1=223 (as P lies in Ist quadrant)

So, β=y12=23

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Let the line y=mx and the ellipse 2x2+y2=1 intersect at a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at -132,0 and 0,β, then β is equal to

Options:
A 223
B 23
C 23
D 23

The correct answer is:
23

Solution:
Let P be x1,y1

Equation of normal at P is x2x1-yy1=-12

It passes through

-132, 0⇒-162x1=-12⇒x1=132

So, y1=223 (as P lies in Ist quadrant)

So, β=y12=23

Topic: Mathematics, Ellipse, JEE Main
Let the line y=mx and the ellipse 2x2+y2=1 intersect at a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at -132,0 and 0,β, then β is equal to

Options:
A 223
B 23
C 23
D 23

The correct answer is:
23

Solution:
Let P be x1,y1
Equation of normal at P is x2x1-yy1=-12
It passes through -132, 0⇒-162x1=-12⇒x1=132
So, y1=223 (as P lies in Ist quadrant)
So, β=y12=23

Topic: Mathematics, Ellipse, JEE Main
Let the lines 4x-3y+10=0 and 4x-3y-30=0 make equal intercepts of 6 units with a circle C whose centre lies on 2x+y=0, then the equation of the circle C is

Options:
A x2+y2-2x+4y-20=0
B x2+y2-4x+8y-20=0
C x2+y2+2x-4y-20=0
D x2+y2-4x+8y-5=0

The correct answer is:
x2+y2-2x+4y-20=0

Solution:
Centre of the circle C lies on 2x+y=0 & 4x-3y+10-302=0
⇒ Center of C is 1,-2
Radius of C is 32+42=5

Equation of C is x-12+y+22=52
⇒x2+y2-2x+4y-20=0

Topic: Mathematics, Circle, JEE Main
Let the lines l1 and l2 be normals to y2=4x and tangents to x2=-12y (where l1 and l2 are not x-axis). The absolute value of the difference of slopes of l1 and l2 is

Options:
A 3
B 2
C 1
D 12

The correct answer is:
1

Solution:
The tangent to x2=-12y is kx=y+-3k2
Normal to y2=4x is y=mx-2m-m3
⇒k=m & 2m+m3=-3k2
⇒mm2+3m+2=0⇒m=-1,-2 ( m=0 is rejected)
Hence, the difference of slopes is equal to 1

Topic: Mathematics, Parabola, JEE Main
Let the lines L1:r→=λi^+2j^+3k^, λ∈R and L2:r→=i^+3j^+k^+μ(i^+j^+5k^); μ∈R, intersect at the point S. If a plane ax+by-z+d=0 passes through S and is parallel to the lines L1 and L2, then the value of a+b+d is equal to ______.

Options:

The correct answer is:
5

Solution:
L1≡x=y2=z3=λ passes through 0,0,0 and parallel to i^+2j^+3k^

L2≡x-1=y-3=z-15=μ passes through 1,3,1 and parallel to i^+j^+5k^

i.e. λ=1+μ, 2λ=3+μ⇒μ=1

Both the lines intersect at 2,4,6

Normal to the plane will be i^j^k^123115=7i^-2j^-k^

Since plane is parallel to both the given lines, the equation of the plane will be

7x-2-2y-4-z-6=0

⇒7x-2y-z=0

On comparing we get a=7, b=-2, c=1, d=0

∴ a+b+d=5

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let the lines x-1λ=y-21=z-32 and x+26-2=y+183=z+28λ be coplanar and P be the plane containing these two lines. Then which of the following points does NOT lies on P ?

Options:
A 0,-2,-2
B -5,0,-1
C 3,-1,0
D 0,4,5

The correct answer is:
0,4,5

Solution:
Given, L1:x-1λ=y-21=z-32

and L2:x+26-2=y+183=z+28λ are coplanar

Now using the condition of coplanarity we get,

⇒272031λ12-23λ=0⇒λ=3

Now, normal of plane P, which contains L1 and L2 is given by

i^j^k^312-233=-3i^-13j^+11k^

Now equation of required plane P will be

-3x-1-13y-2+11z-3=0

⇒3x+13y-11z+4=0

Now satisfy all given points from option we get, 0,4,5 does not lie on plane P.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let the lines y-2=m1x-5 and y+4=m2x-3 intersect at right angles at P (where m1 and m2 are parameters). If locus of P is x2+y2+gx+fy+7=0, then g22+f22-7 is equal to

Options:
A 1
B 2
C 8
D 10

The correct answer is:
10

Solution:
Since the lines intersect at right angle,

therefore, the locus of the point of intersection of lines is a circle with end points of diameter as 5,2 & 3,-4
x-5x-3+y-2y+4=0
âx2+y2-8x+2y+7=0
Hence, g22+f22-7=16+1-7=10

Topic: Mathematics, Circle, JEE Main
Let the locus of any point Pz in the argand plane is argz-5iz+5i=Ï4. If O is the origin, then the value of max.OP+min.OP2 is

Options:
A 52
B 5+52
C 5+52
D 10-52

The correct answer is:
5+52

Solution:
r2+r2=102⇒r=52
max.OP=OC+ radius =5+52
and min.OP=OA=5
Required value =5+5+522=5+52

Topic: Mathematics, Complex Number, JEE Main
Let the locus of the centre α,β, β>0, of the circle which touches the circle x2+y-12=1 externally and also touches the x-axis be L. Then the area bounded by L and the line y=4 is

Options:
A 3223
B 4023
C 643
D 323

The correct answer is:
643

Solution:
Since the circle with centre α,β touches the circle x2+y-12=1 externally and also touches x-axis,

therefore the distance between the centres will be equal to the sum of the radius of both circles

i.e. α-02+β-12=β+1

⇒α2=4β

Hence the locus of L will be x2=4y

Now, the area bounded by x2=4y & y=4 will be

A=2∫044-x24dx

=24x-x31204=643

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Let the matrix A=1122 and B=A+A2+A3+A4. If B=λA, ∀λ∈R, then the value of λ is equal to

Options:

The correct answer is:
40

Solution:
A2=11221122=3366=3A
Hence, A3=3A2=9A,A4=27A
So, B=1+3+9+27A=40A

Topic: Mathematics, Matrices, JEE Main
Let the matrix A=123012001 and BA=A where B represent 3×3 order matrix. If the total number of 1 in matrix A-1 and matrix B are p and q respectively, then the value of p+q is equal to

Options:
A 3
B 4
C 5
D 7

The correct answer is:
7

Solution:
A-1=1AadjA⇒A-1=1-2101-2001
A=1 and BA=A
BAA-1=AA-1⇒B=I
Total number of 1 in matrix B=p=3 and matrix A-1=q=4

Hence, p+q=7

Topic: Mathematics, Determinants, JEE Main
Let the maximum and minimum value of the expression 2cos2â¡Î¸+cosâ¡Î¸+1 is M and m respectively, then the value of Mm is, (where [.] is the greatest integer function)

Options:

The correct answer is:
4

Solution:
2cos2⁡θ+cos⁡θ+1=2cos2⁡θ+cos⁡θ2+12
2cos⁡θ+142+716
Given expression is maximum when cos⁡θ=1 and minimum when cos⁡θ=-14
⇒M=2542+716=23216=4

and m=2716=78

Hence, Mm=327=4

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
Let the maximum area of the triangle that can be inscribed in the ellipse x2a2+y24=1,a>2, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be 63. Then the eccentricity of the ellipse is:

Options:
A 32
B 12
C 12
D 34

The correct answer is:
32

Solution:
Plotting the diagram as per given information we get,

Now finding area of the triangle we get,

A=12a1-cosθ4sinθ

A=2a1-cosθsinθ

Differentiating to get maxima and minima we get,

dAdθ=2asin2θ+cosθ-cos2θ

dA dθ=0⇒1+cosθ-2cos2θ=0

cosθ=1Reject

OR

cosθ=-12⇒θ=2π3

d2 A dθ2=2a2sin2θ-sinθ

d2 A dθ2<0 for θ=2π3

Now, Amax=332a=63

⇒a=4

Now, e=a2-b2a2=32

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Let the mean and the standard deviation of the probability distribution
$\begin{array}{|c|c|c|c|c|}
\hline \mathrm{X} & \mathrm{\alpha} & \mathrm{1} & 0 & -3 \\
\hline \mathrm{P}(\mathrm{X}) & \frac{1}{3} & \mathrm{~K} & \frac{1}{6} & \frac{1}{4} \\
\hline
\end{array}$
be $\mu$ and $\sigma$, respectively. If $\sigma-\mu=2$, then $\sigma+\mu$ is equal to________

Options:

The correct answer is:
5

Solution:
$\begin{aligned} & \frac{1}{3}+\mathrm{k}+\frac{1}{6}+\frac{1}{4}=1 \quad \Rightarrow \mathrm{k}=\frac{1}{4} \\ & \mu=\frac{\alpha}{3}+\frac{1}{4}-\frac{3}{4} \\ & \mu=\frac{\alpha}{3}-\frac{1}{2} \\ & \sigma=\sqrt{\left(\alpha^2 \frac{1}{3}+\frac{1}{4}+9 \frac{1}{4}\right)-\left(\frac{\alpha}{3}-\frac{1}{2}\right)^2} \\ & \sigma=\sqrt{\frac{2 \alpha^2}{9}+\frac{\alpha}{3}+\frac{9}{4}} \\ & \sigma=\mu+2 \\ & \sigma^2=(\mu+2)^2 \Rightarrow \frac{2 \alpha^2}{9}+\frac{\alpha}{3}+\frac{9}{4}=\frac{\alpha^2}{9}+\frac{9}{4}+\alpha \\ & \frac{\alpha^2}{9}-\frac{2 \alpha}{3}=0 \\ & \alpha=0,(\text { reject }) \text { or } \alpha=6 \\ & (\because \mathrm{x}=0 \text { is already given) } \\ & \Rightarrow \sigma+\mu=2 \mu+2 \\ & =5\end{aligned}$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let the mean and the variance of 6 observation a, b, 68, 44, 48, 60 be 55 and 194, respectively if a>b, then a+3b is

Options:
A 200
B 190
C 180
D 210

The correct answer is:
180

Solution:
Mean, x=a+b+68+44+48+606=55

⇒a+b+220=330

⇒a+b=110 ...i

Variance, σ2=∑xi-x2n

⇒σ2=a-552+b-552+132+112+72+526=194

⇒a-552+b-552+364=1164

⇒110-b-552+b-552=800

⇒255-b2=800

⇒55-b=20

⇒b=35

⇒a=75

⇒a+3b=75+105=180

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let the mean and variance of 8 numbers x, y, 10, 12, 6,12, 4, 8 be 9 and 9.25 respectively. If x>y, then 3x-2y is equal to _______

Options:

The correct answer is:
25

Solution:
Given, the data x, y, 10, 12, 4, 6, 8, 12.

So,

Mean=x+y+10+12+4+6+8+128

⇒9=x+y+528

⇒x+y+52=72

⇒x+y=20

And,

Variance=∑xi2n-∑xin2

⇒9.25=x2+y2+100+144+16+36+64+1448-92

⇒x2+y2=218

⇒x+y2-2xy=218

⇒202-2xy=218

⇒2xy=182

⇒xy=91

Now solving x+y=20 & xy=91 we get,

x=13, y=7⇒3x-2y=25

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let the mean and variance of 12 observations be 92 and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is mn, where m and n are coprime, then m+n are coprime, then m+n is equal to

Options:
A 315
B 316
C 314
D 317

The correct answer is:
317

Solution:
Given,

The mean and variance of 12 observations be 92 and 4 respectively,

So, mean will be,

x¯=x1+x2+……+9+10+…..+x1212=92

⇒x1+x2……+x12+19=54

Now removing the observation 9 & 10 and adding the observation 7 & 14,

So, the new total sum of the observation will be,

x1+x2+……+x12+7+14=54-19+7+14

Now new mean will be,

x1+x2+….+x12+7+1412=5612

⇒x¯new=143
Now using the formula of the variance we get,

x12+x22+……+x122+92+10212-922=4

⇒x12+x22+……+x122+92+102-81×3=4×12

⇒x12+x22+……+x122+92+102=291

Now removing 92 & 102 and adding 72 & 142 we get,

⇒x12+x22+……+x122+72+142=355
New variance =∑xi2N-(x¯)2
=35512-1432
=28136

Now on comparing with mn=28136 we get,

⇒m=281 and n=36

∴ m+n=317

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let the mean and variance of four numbers 3,7, x and y (x>y) be 5 and 10 respectively. Then the mean of four numbers 3+2x, 7+2y, x+y and x-y is ______.

Options:

The correct answer is:
12

Solution:
Mean =5, Variance =10
3+7+x+y4=5,10=∑xi24-∑x2

10+x+y=20, 10=∑xi24-25

x+y=10

y=10-x, 40=∑xi2-100

140=∑xi2

140=9+49+x2+y2
140=58+x2+y2
x2+y2=82
x+y2=x2+y2+2xy
100-82=2xy
18=2xy
xy=9
x10-x=9
⇒x2-10x+9=0
=x=1,9
y=9,1

As x>y

So, x=9, y=1

x¯=3+2x+7+2y+x+y+x-y4

⇒10+4x+2y4

⇒5+2x+y2

⇒5+18+12=242=12

Topic: Mathematics, Statistics, JEE Main
Let the mean and variance of four numbers 3,7, x and y (x>y) be 5 and 10 respectively. Then the mean of four numbers 3+2x, 7+2y, x+y and x-y is ______.

Options:

The correct answer is:
12

Solution:
Mean =5, Variance =10
3+7+x+y4=5,10=∑xi24-∑x2

10+x+y=20, 10=∑xi24-25

x+y=10

y=10-x, 40=∑xi2-100

140=∑xi2

140=9+49+x2+y2
140=58+x2+y2
x2+y2=82
x+y2=x2+y2+2xy
100-82=2xy
18=2xy
xy=9
x10-x=9
⇒x2-10x+9=0
=x=1,9
y=9,1

As x>y

So, x=9, y=1

x¯=3+2x+7+2y+x+y+x-y4

⇒10+4x+2y4

⇒5+2x+y2

⇒5+18+12=242=12

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Let the mean of 6 observations 1,2,4,5,x and y be 5 and their variance be 10. Then their mean deviation about the mean is equal to

Options:
A 73
B 3
C 83
D 103

The correct answer is:
83

Solution:
Consider the data 1,2,4,5,x,y.

So,

Mean=1+2+4+5+x+y6

⇒5=x+y+126

⇒x+y=18

And,

Variance=∑xi2n-∑xin2

⇒10=12+22+42+52+x2+y26-52

⇒x2+y2=164

We have x+y=18 and x2+y2=164.

Now solving above equations we get, x=10, y=8,

Now mean deviation about mean will be =∑i=16xi-x6

=1-5+2-5+4-5+5-5+10-5+8-56

=166

=83

Therefore, the required answer is 83

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let the mean of 50 observations is 15 and the standard deviation is 2. However, one observation was wrongly recorded. The sum of the correct and incorrect observations is 70. If the mean of the correct set of observations is 16, then the variance of the correct set is equal to

Options:
A 10
B 36
C 43
D 60

The correct answer is:
43

Solution:
By using the formula of mean we get, =∑xi50=15⇒∑xi=750

Now variance=∑xi250-152=22⇒∑xi2=11450

⇒∑xinew=50×16=800

So, ∑xinew-∑xi=800-750=50

Hence, if wrong observation was x then corrected one is x+50.
⇒x+x+50=70⇒x=10

Therefore, the correct observation =60.

Now, ∑xi2new=11450-102+602=14950

Therefore, variance of the correct set =1495050-162=299-256=43.

Topic: Mathematics, Statistics, JEE Main
Let the mean of 50 observations is 15 and the standard deviation is 2. However, one observation was wrongly recorded. The sum of the correct and incorrect observations is 70. If the mean of the correct set of observations is 16, then the variance of the correct set is equal to

Options:
A 10
B 36
C 43
D 60

The correct answer is:
43

Solution:
By using the formula of mean we get, =∑xi50=15⇒∑xi=750

Now variance=∑xi250-152=22⇒∑xi2=11450

⇒∑xinew=50×16=800

So, ∑xinew-∑xi=800-750=50

Hence, if wrong observation was x then corrected one is x+50.
⇒x+x+50=70⇒x=10

Therefore, the correct observation =60.

Now, ∑xi2new=11450-102+602=14950

Therefore, variance of the correct set =1495050-162=299-256=43.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Let the mean of the data

x	1	3	5	7	9

Frequency f	4	24	28	α	8

be 5. If m and σ2 are respectively the mean deviation about the mean and the variance of the data, then 3αm+σ2 is equal to _______.

Options:

The correct answer is:
8

Solution:
Given,

The mean of the below data is 5

x	1	3	5	7	9

Frequency f	4	24	28	α	8

Now finding mean we get,

4+72+28×5+7α+7264+α=5

⇒α=16

So, total number of observation will be, ∑fi=80

Now finding, mean deviation we get,

M.D=∑fixi-5∑fi

=4×4+24×2+0+16×2+8×480=85

So, m=85

Now finding, variance we get,

σ2=∑fixi-52∑fi

⇒σ2=4×16+24×4+0+16×4+8×1680

⇒σ2=225

Hence, the value of 3αm+σ2=3×1685+225=8

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (13 Apr Shift 1)