Mathematics Test

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Mathematics Test - 8

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Question 1
2.5 / -0.83

A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, and scooter or by other means of transport are respectively 0.3, 0.2, 0.1 and 0.4. The probabilities that he will be late are 1/4 , 1/3 and 1/12, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. The probability that he comes by bus is:

A
1/6

B
4/9

C
5/18

D
1/18

Solution

Let E1,E2,E3andE4 be the events that the doctor comes by train, bus, scooter and car respectively. Then,
P(E1)=3/10,P(E2)=1/5,P(E3)=1/10andP(E4)=2/5.
Let E be the event that the doctor is late. Then,
P(E/E1) = probability that the doctor is late, given that he comes by train
=1/4.
P(E/E2)= probability that the doctor is late, given that he comes by bus
=1/3.
P(E/E3)= probability that the doctor is late, given that the comes by scooter
=1/12.
P(E/E4)= probability that the doctor is late, given that that he comes by car
=0.
Probability that he comes by train, given that he is late
P(E2/E)
[P(E2).P(E/E2)]/[P(E1).P(E/E1)+P(E2).P(E/E2)+P(E3).P(E/E3)+P(E4).P(E/E40)][by Bayes 's theorem]
[(1/5 ×1/3)]/(3/10 ×1/4)+(1/5 ×1/3)+(1/10 ×1/12)+(2/5 ×0)
=(3/40 ×120/18)=4/9
Hence, the required probability is 4/9.
Question 2
2.5 / -0.83

If P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find P(A ∩B)

A
0.35

B
0.32

C
0.37

D
0.29

Solution

We have ,
P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4
Question 3
2.5 / -0.83

A random variable X has the following probability distribution:

Then k =?

A
1/8

B
8

C
0

D
4

Solution
Question 4
2.5 / -0.83

The equation of the plane passing through the intersection of the planes and and the point (1, 2, 1) is:

A
18x+6y+14z-23=0

B
18x+7y+14z-46=0

C
9x+3y+7z-23=0

D
18x+7y+14z-38=0

Solution

n1 = 2i + j + k
n2 = 2i + 3j - 4k
p1 = 4, p2 = -6
r.(n1 + λn2) = p1 + λp2
=>r . [2i + j + k + λ(2i + 3j - 4k)] = 4 - 6 λ
=>r . [ i(2 + 2 λ) + j(1 + 3 λ) + k(1 - 4k)] = 4 - 6 λ
Taking r = xi + yj + zk
(2 + 2 λ)x + (1 + 3 λ)y + (1 - 4k)z = 4 - 6 λ
(2x + y + - z - 4) + λ(2x + 3y - 4k + 6) = 0
Given points are (1,2,1)
(2 + 2 - 1 - 4) + λ(2 + 6 - 4 + 6) = 0
-1 + λ(10) = 0
λ= 1/10
Substitute λ= 1/10, we get
18x + 7y + 14z - 46=0
Question 5
2.5 / -0.83

The Cartesian form of the equation of the plane is:

A
2x+3y-z=10

B
2x+3y-z= √14

C
2x+3y-z+√14=0

D
2x+3y-z+10=0

Solution

Let r = xi + yj + zk
(xi + yj + zk) . (2i + 3j - k) = 10
2x + 3y - k = 10
Question 6
2.5 / -0.83

The length of the perpendicular from the origin to the plane 3x + 2y –6z = 21 is:

A
3

B
14

C
21

D
7

Solution

The given point is P(0,0,0) and the given line is 3x+2y-6z-21=0
Let d be the length of the perpendicular from P(0,0) to the line 3x+2y-6z-21=0
Then, d= |3*0 + 2*0 + (-6)*0 -21|/[(3)^2 + (2)² + (-6)² ]^1/2
= |-21|/7
= 3
Question 7
2.5 / -0.83

If l, m, n are the direction cosines of the normal to the plane and p be the perpendicular distance of the plane from the origin, then the equation of the plane is:

A

B

C

D

Solution

Let P(x, y, z) be any point on the plane.
OP = r = xi + yj +zk
Let l,m,n be the direction of cosine of n, then
n - li + mj + nk
As we know that r.n = d
(xi + yj + zk)(li + mj + nk) = d
i.e. lx + my + nz = d
this is the cartesian equation of plane in normal form.
Question 8
2.5 / -0.83

The combined equation to two parabolas, both have their axis along x-axis, is given by y⁴ - y² (4x + 4 - 2 sin² 2 α) + sin² 2 α (4x + 4x + sin² 2 α) = 0. The locus of the point of intersection of tangents, one to each of the parabolas, when they include an angle of 90 °
is

A
another parabola

B
circle

C
a line midway between the tangents at vertices

D
a line midway between their directrices

Solution

The parabolas are y² = 4sin² α(x + sin² α) and y² = 4cos² α(x + cos² α), hence the locus is x + cos² α+ sin² α= 0 ⇒x + 1 = 0
Assume the point as origin and line joining it to the centre as x-axis, the equation to the circle becomes x² + y² - 2x = 0, center is A₁ (1, 0) and the second circle has the equation x² + y² - 2 √2 = 0 center A₀ (0, √2)
Similarly A₃ (-2, 0), A₄ (0, -2 √2) etc.
Question 9
2.5 / -0.83

The straight line x + y = k + 1 touches the parabola y = x(1 –x) if

A
k = -1

B
k = 0

C
k = 1

D
k takes any real value

Solution

Solving x + y = k + 1 and y = x(1 –x) ⇒x² –2x + k + 1 = 0 and Δ= 0
Question 10
2.5 / -0.83

Focus and vertex of the parabola that touches x-axis at (1, 0) and x = y at (1, 1) are (h, k) and (p, q) then the value of 25(p + q +h + k)

A
35

B
37

C
33

D
39

Solution

The x-axis touches at A(1, 0) and x = y touches at B(1, 1). Hence the equation to the curve through these points is given by y(y –x) + k(x –1)² = 0. For this to represent a parabola, 4k = 1. The equation is x² –4xy + 4y² –2x + 1 = 0. Vertex focus