Mathematics Test 259

Result

Mathematics Test 259

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Weekly Quiz Competition

Question 1
4 / -1

Solution of the differential equation x cos x (dy/dx) + y(x sin x + cos x) = 1 is

A
xy = sin x + c cosx

B
xy sec x = tan²x + c

C
xy + sin x + c cos x = 0

D
None of these

Solution
Question 2
4 / -1

A
5

B
20

C
10

D
30

Solution
Question 3
4 / -1

A and B toss a die alternatively till one of them gets a six and wins the game. If A begins the game, then the probability of B winning the game is

A
5/11

B
6/11

C
7/11

D
4/11

Solution

Let:

p = 1/6 (probability of getting a 6)

q = 5/6 (probability of not getting a 6)

Since A starts, A wins if:

He gets 6 on the 1st try: p

Or both A and B fail once, and A gets 6 on 3rd throw: q·q·p = q²p

Or both fail twice, and A gets 6 on 5th throw: q⁴p, and so on.

So: P(A wins) = p + q²p + q⁴p + ... = p(1 + q² + q⁴ + ...)

This is a geometric series with first term 1 and ratio q².

Sum of infinite GP = 1 / (1 − q²)

So: P(A wins) = p / (1 − q²)

⇒ (1/6) / (1 − (25/36)) = (1/6) / (11/36) = 6/11

Hence, P(B wins) = 1 − 6/11 = 5/11

Final Answer: Option A: 5/11 is correct.
Question 4
4 / -1

Let X = {x : x = n³+ 2n + 1, n ∈ N} and Y = {x : x = 3n²+ 7, n ∈ N} then

A
X ∩ Y is a subset of {x : x = 3n + 5, n ∈ N}

B
X ∩ Y ⊆ {x : x = n²+ n + 1, n ∈ N}

C
34 ∈ X ∩ Y

D
None of these

Solution

If n³+ 2n + 1 = 3n²+ 7 ⇒ n³− 3n²+ 2n − 6 = 0 ⇒ (n − 3)(n²+ 2) = 0 ⇒ n = 3 as n ∈ N

⇒ n = 3 as n ∈ N So, x =3 × 3²+ 7 = 34 ∈ X ∩ Y.

In (a) and (b) x ≠ 34, for any n ∈ N.
Question 5
4 / -1

For integers m and n, both greater than 1, consider the following three statements:

P : m divides n

Q : m divides n²

R : m is prime

then

A
Q ∧ R → P

B
P ∧ Q → R

C
Q → R

D
Q → P

Solution

(b) 8/4 = 2, 64/4 = 16 ; but 4 is not prime.

Hence P∧Q → R, false

(c) (6)²/12 = 36/12 = 3 ; but 12 is not prime

Hence Q→R, false

(d) (4)²8 = 168 = 2;4/8 is not an integer

Hence Q → P, false
Question 6
4 / -1

The mean of 5 observations is 4.4 and the variance is 8.24. If three of the five observations are 1,2 and 6, the two values are

A
4 and 9

B
3 and 5

C
2 and 6

D
4 and 6

Solution
Question 7
4 / -1

Let a₁, a₂, ... a_n be a given A.P. whose common difference is an integer and S_n = a₁ + a₂ + ... + a_n. If a₁ = 1, a_n = 300 and 15 ≤ n ≤ 50, then the ordered pair (S_{n - 4}, a_{n - 4}) is equal to

A
(2490, 249)

B
(2480, 249)

C
(2490, 248)

D
(2480, 248)

Solution
Question 8
4 / -1

Let S₁ be the sum of first 2n terms of an arithmetic progression. Let S₂ be the sum of first 4n terms of the same arithmetic progression. If (S₂ - S₁) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to:

A
1000

B
7000

C
5000

D
3000

Solution

S₁ = S₂ₙ

= (2n/2) (2a + (2n - 1)d)

= n(2a + 2nd - d)

S₂ = S₄ₙ

= (4n/2) (2a + (4n - 1)d)

= 2n(2a + 4nd - d)

Given S₂ - S₁ = 1000

n(4a + 8nd - 2d - 2a - 2nd + d) = 1000

n(2a + 6nd - d) = 1000

n(2a + (6n - 1)d) = 1000

(6n/2) (2a + (6n - 1)d) = 3(1000)

Sum of the first 6n terms is 3000.
Question 9
4 / -1

then the remainder when K is divided by 6 is

A
1

B
2

C
3

D
5

Solution
Question 10
4 / -1

The greatest positive integer k, for which 49^k + 1 is a factor of the sum 49¹²⁵ + 49¹²⁴ + ... + 49² + 49 + 1, is

A
32

B
60

C
65

D
63

Solution