Mathematics Test 255

Result

Mathematics Test 255

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Question 1
4 / -1

The foot of the perpendicular drawn from the origin on the line 3x + y = λ (λ ≠ 0) is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is:

A
1 : 9

B
1 : 3

C
9 : 1

D
3 : 1

Solution
Question 2
4 / -1

If xlog_e(log_e x) - x² + y² = 4(y > 0), then dy/dx at x = e is equal to:

A

B

C

D

Solution

x * (1 / ln(x)) * (1 / x) + ln(ln(x)) - 2x + 2y * (dy/dx) = 0

At x = e, we get:

1 - 2e + 2y * (dy/dx) = 0

⇒ 2e * (dy/dx) = 2e - 1

⇒ (dy/dx) = (2e - 1) / (2y)

As y(e) = √(4 + e²),

⇒ (dy/dx) = (2e - 1) / (2√(4 + e²))
Question 3
4 / -1

If |z - 3 + 2i| ≤ 4, then the difference between the greatest value and the least value of |z| is:

A
4 + √13

B
2√13

C
√13

D
8

Solution

Given equation represents the circle with center(3,−2) and is of radius (R) = 4

∣z∣ represents the distance of point 'z' from origin

Greatest and least distances occur along the normal through the origin

Normal always passes through center of circle
Question 4
4 / -1

If the system of following linear equations has no solution, then:

x + ay + z = 3

x + 2y + 2z = 6

x + 5y + 3z = b

A
a = 1, b ≠ 9

B
a ≠ -1, b = 9

C
a = -1, b = 9

D
a = -1, b ≠ 9

Solution
Question 5
4 / -1

then the matrix A^-50, when θ = π/12, is equal to:

A

B

C

D

Solution
Question 6
4 / -1

Two different families A and B are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families, so that no child gets more than one ticket. If the probability that all the tickets go to the children of family B is 1/12, then the number of children in each family is

A
6

B
5

C
3

D
4

Solution

Let the number of children in each family be n.

Thus, the total number of children in both the families is 2n.

Now, it is given that tickets are distributed amongst the children of the families.

If all the tickets are distributed to the children in the families, the probability that all the three tickets go to the children in family B is given by:
Question 7
4 / -1

The mean and the standard deviation (sd) of five observations are 9 and 0, respectively. If one of the observations is changed, such that the mean of the new set of five observations becomes 10, then their sd is

A
0

B
2

C
4

D
1

Solution

Since the standard deviation before the change was 0, all the observations were exactly equal to 9.

Since one observation was changed and the new mean is 10, we have the following equation.

(9 + 9 + 9 + 9 + x)/5 = 10

(36 + x)/5 = 10

Solving for x, we have

x = 14

The changed observation is 14.

All the observations are {9, 9, 9, 9, 14}.

Since the mean is 10, the variance is

[(9 - 10)² + (9 - 10)² + (9 - 10)² + (9 - 10)² + (14 - 10)²]/5

= (1 + 1 + 1 + 1 + 16)/5

= 20/5

= 4

Hence, the standard deviation is √4 = 2
Question 8
4 / -1

Consider the experiment of throwing a dice, if a multiple of '3' comes up, throw the dice again and if any other number comes, toss a coin. Find the conditional probability of the event coin shows the tail, given that at least one die shows a '3'.

A
0

B
1

C
2

D
3

Solution
Question 9
4 / -1

A
π/6

B
π/12

C
π/24

D
π/4

Solution
Question 10
4 / -1

A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of the letter is mailed.

A
40611

B
42612

C
43690

D
45656

Solution