Explore topic-wise InterviewSolutions in Current Affairs.

Correct option is (B) \(\frac{x-3}{2}\)

x – 2y = 3

\(\Rightarrow\) 2y = x - 3

\(\Rightarrow\) y = \(\frac{x-3}{2}\)

Correct option is  B) \(\frac{x-3}{2}\)

Correct option is (B) y = 3x

Linear equation in which constant is not present, always passes through the origin.

\(\therefore\) Among all given lines only y = 3x passes through the origin.

Correct option is  B) y = 3x

Correct option is  A) (1,1)

Correct option is  A) 4

Correct option is (C) (0, 3)

\(\because\) \(0+2\times3\) = 0+6 = 6

\(\therefore\) (0, 3) satisfies the equation x+2y = 6.

Correct option is  C) (0, 3)

The given system of equations is 

x – y = 3 …..(i) 

x/3 + y/2 = 6 ……(ii) 

From (i), write y in terms of x to get 

y = x – 3 

Substituting y = x – 3 in (ii), we get 

x/3 + x −3/ 2 = 6 

⇒2x + 3(x – 3) = 36 

⇒2x + 3x – 9 = 36 

⇒x = 45/ 5 = 9 

Now, substituting x = 9 in (i), we have 

9 – y = 3 

⇒y = 9 – 3 = 6 

Hence, x = 9 and y = 6.

Correct option is (C) 31

\(\because\) (- 3, 2) is a solution of 5x – 8y + k = 0

\(\therefore\) \(5\times-3-8\times2+k=0\)

\(\Rightarrow\) -15 - 16 + k = 0

\(\Rightarrow\) k = 15+16 = 31

Correct option is  C) 31

Correct option is (C) x = 0

Equation of Y – axis is x = 0.

Correct option is  C) x = 0

Correct option is (A) x = 4

A line parallel to Y–axis is x = c, where c is constant. Among all given lines only x = 4 is a line parallel to Y-axis.

Correct option is   A) x = 4

Correct option is (B) y = 0

Equation of X–axis is y = 0.

Correct option is  B) y = 0

Correct option is (C) 3

\(x=\frac{7}{3}y\)

\(\Rightarrow\) 3x = 7y

\(\Rightarrow\) 3x - 7y = 0

By comparing above equation with ax+by+c = 0, we get a = 3.

Correct option is  C) 3

The given system of equation is: 

2x + 3y = 0 ……(i) 

3x + 4y = 5 ……(ii) 

On multiplying (i) by 4 and (ii) by 3, we get: 

8x + 12y = 0 ……(iii) 

9x + 12y = 15 …….(iv) 

On subtracting (iii) from (iv) we get:

x = 15 

On substituting the value of x = 15 in (i), we get: 

30 + 3y = 0 

⇒3y = -30 

⇒y = -10 

Hence, the solution is x = 15 and y = -10. 

Correct option is (D) (0, k)

Line y = k is passing through the point (0, k).

Correct option is  D) (0, k)

Correct option is (C) (k, 0)

Line x = k is passing through the point (k, 0).

Correct option is  C) (k, 0)

Correct option is  B

Correct option is (A) 0

\(x=\frac{9}{4}\)

\(\therefore\) 4x+y = 9

\(\Rightarrow\) \(4\times\frac94+y=9\)

\(\Rightarrow\) 9+y = 9

\(\Rightarrow\) y = 9 - 9 = 0

Correct option is  A) 0

The given system of equation is: 

2x - 3y = 13 ……(i) 

7x - 2y = 20 ……(ii) 

On multiplying (i) by 2 and (ii) by 3, we get: 

4x - 6y = 26 ……(iii) 

21x - 6y = 60 …….(iv) 

On subtracting (iii) from (iv) we get: 

17x = (60 – 26) = 34 

⇒x = 2 

On substituting the value of x = 2 in (i), we get: 

4 – 3y = 13 

⇒3y = (4 – 13) = -9 

⇒y = -3 

Hence, the solution is x = 2 and y = -3.

The given system of equation is: 

3x - 5y - 19 = 0 ……(i) 

-7x + 3y + 1 = 0 ……(ii) 

On multiplying (i) by 3 and (ii) by 5, we get: 

9x - 15y = 57 ……(iii) 

-35x + 15y = -5 …….(iv) 

On subtracting (iii) from (iv) we get: 

-26x = (57 – 5) = 52 

⇒x = -2 

On substituting the value of x = -2 in (i), we get:

–6 – 5y – 19 = 0 

⇒5y = (–6 – 19) = -25 

⇒y = -5 

Hence, the solution is x = -2 and y = -5.

Correct option is (A) 8x + 9y = 0

\(\frac{8}{9}\) x = -y

\(\Rightarrow\) 8x = -9y

\(\Rightarrow\) 8x+9y = 0

A) 8x + 9y = 0

The given system of equation is: 

2x – y + 3 = 0…….(i) 

3x – 7y + 10 = 0 ……(ii) 

From (i), write y in terms of x to get 

y=2x + 3 

Substituting y = 2x + 3 in (ii), we get 

3x – 7(2x + 3) + 10 = 0 

3x – 14x – 21 + 10 = 0 

-7x = 21 – 10 = 11 

x = – 11/7 

Now substituting x = – 11/7 in (i), we have 

– 22/7 – y + 3 = 0 

y = 3 - 22/7 = -1/7 

Hence, x = – 11/ 7 and y = – 1/ 7 .

Correct option is (A) Straight line

Linear equation represents a straight line.

\(\therefore\) 2015 x + 2016 y = 4031 represents a straight line.

A) Straight line

Option : (C)

4x + 3y = 12 

A is (3, 0) B is (0, 4) 

Base of triangle AOB = OA = 3 untis 

Perpendicualr of triangle AOB = OB = 4 units 

Hypotenuse² = perepndicular² + base² 

⇒ Hypotenuse² = 16 + 9 = 25 sq units 

⇒ Hypotenuse = 5 units

Option : (A)

2x – y = 4 

At y = 0, x = 2 

Thus, 

The line cuts the x-axis at (2, 0)

The given system of equations is 

x – y = 3 …..(i)

\(\frac{x}3+\frac{y}2= 6 \)........(ii)

From (i), write y in terms of x to get 

y = x – 3 

Substituting y = x – 3 in (ii), we get

\(\frac{x}3+\frac{x-3}2= 6 \)

⇒ 2x + 3(x – 3) = 36 

⇒ 2x + 3x – 9 = 36

⇒ x = \(\frac{45}5= 9\)

Now, substituting x = 9 in (i), we have 

9 – y = 3 

⇒ y = 9 – 3 = 6 

Hence, x = 9 and y = 6.

Correct option is  B) solution

Let the number of students in the hostel be x

Quantity of rice consumed by each student = 400 gm.

So, daily rice consumption in the hostel mess = 400 (x).

But, daily rice consumption = 50 kg = 50 × 1000 = 50000gm [since 1 kg = 1000gm].

According to the question,

⇒ 400x = 50000

Dividing both sides by 400, we get

⇒ 400 x/400 = 50000/400

⇒ x = 125

Thus, 125 students have their meals in the hostel mess.

Slope of the line that is parallel to y-axis is infinity. 

Eq of the line parallel to y-axis passing through (a, b) is x – a = 0 

(i) (4, 0) 

⇒ x – 4 = 0 

(ii) (-2, 0) 

⇒ x + 2 = 0 

(iii) (3, 5) 

⇒ x – 3 = 0 

(iv) (-4, - 3) 

⇒ x + 4 = 0

Slope of the line parallel to x – axis is 0 

Eq of line parallel to x-axis and passing through (a, b) is y – b = 0

(i) (0, 3) 

⇒ y – 3 = 0 

(ii) (0, 4) 

⇒ y – 4 = 0 

(iii) (2, - 5) 

⇒ y + 5 = 0 

(iv) (-4, -3) 

⇒ y + 3 = 0

Correct option is (D) Parallel to x – axis at a distance 6 units from the origin.

The graph of y = 6 is a line parallel to x–axis at a distance 6 units from the origin.

D) Parallel to x – axis at a distance 6 units from the origin.

Correct option is (D) many

A linear equation has infinitely many solutions.

Correct option is  D) many

The given equations are as follows: 

x – 2y = 0 ………….(i) 

3x + 4y = 20 …………(ii) 

On multiplying (i) by 2, we get: 

2x – 4y = 0 …………(iii) 

On adding (ii) and (iii), we get: 

5x = 20 

⇒ x = 4 

On substituting x = 4 in (i), we get: 

4 – 2y = 0 ⇒ 4 = 2y ⇒ y = 2 

Hence, the required solution is x = 4 and y = 2.

Correct option is (B) (2, 3)

Put x = 2 then 2x + 3y = 13

\(\Rightarrow\) 3y = 13 - 2x = 13 - 2 \(\times\) 2 = 13 - 4 = 9

\(\Rightarrow\) y = \(\frac93\) = 3

\(\therefore\) (2, 3) is a solution of 2x+3y = 13.

Correct option is  B) (2, 3)

The given equations are as follows: 

x – 2y = 0 ………….(i) 

3x + 4y = 20 …………(ii) 

On multiplying (i) by 2, we get: 

2x – 4y = 0 …………(iii) 

On adding (ii) and (iii), we get: 

5x = 20 ⇒ x = 4 

On substituting x = 4 in (i), we get: 

4 – 2y = 0 ⇒ 4 = 2y ⇒ y = 2 

Hence, the required solution is x = 4 and y = 2.

Correct option is (D) many

The number of solutions to 3x – 5y = 8 is infinitely many.

Correct option is  D) many

Correct option is (B) –3

Given that x = 3, y = 3 is a solution of 2x – 3y – k = 0

\(\therefore2\times3-3\times3-k=0\)

\(\Rightarrow\) 6 - 9 - k = 0

\(\Rightarrow\) k = 6 - 9 = - 3

Correct option is  B) – 3

2x + 3y + 1 = 0 ……..(1)

(7 – 4x) /3 = y ……….(2)

Put value of y in (1), we get

2x + 3((7 – 4x) /3 ) + 1 = 0

2x + 7 – 4x + 1 = 0

x = 4

from (2):

(7 – 4(4)) /3 = y

y = -3

Answer: x = 4 and y = -3

Given, (2 α + 1, α – 1 ) is the solution of equation 2x – 3y + 5 = 0. 

Substituting x = 2 α + 1 and y = α – 1 in 2x – 3y + 5 = 0, we get 

2(2 α + 1) – 3(α – 1 ) + 5 = 0 

4 α + 2 – 3 α + 3 + 5 = 0 

α + 10 = 0 

α = – 10 

The value of α is -10.

Given, (-λ, 5/2) is a solution of equation 3x + 4y = k 

Substituting x = – λ and y = 5/2 in x + 4y – 7 = 0, we get 

– λ + 4 (5/2) – 7 = 0 

-λ + 10 – 7 = 0 

λ = 3

Given, 3 x + 4 y = k 

(-1, 2) is the solution of 3x + 4y = k, so it satisfy the equation.

Substituting x = -1 and y = 2 in 3x + 4y = k, we get 

3 (– 1 ) + 4( 2 ) = k 

– 3 + 8 = k 

k = 5 

The value of k is 5.

In case of the sale, the Seller has the right to sell.

Scapigerous inflorescence is seen in Allium cepa.

• Glycogen, poly-B-hydroxybutyrate granules, sulphur granules and gas vesicles.

• The virion is made up of two constituents, a protein coat called capsid and a core called nucleic acid.

Lysozyme is secreted by phage during Penetration. 

• The protein coat of viruses is made up of approximately 2130 identical protein subunits called capsomeres.

• The empty protein coat left outside by the phage after penetrating the host cell is called as ghost.

Herpes virus shows cuboid symmetry.

The base plate of T₄ phage has 6 tail fibres.

• A portion of filament of blue green algae that becomes detatched and reproduces by cell division. Eg : Nostoc.