Explore topic-wise InterviewSolutions in .

Correct option is  B) – 4

Correct option is (B) –4

–5 (x + 4) = 0

\(\Rightarrow\) x+4 = \(\frac0{-5}\) = 0

\(\Rightarrow\) x = 0 - 4 = -4.

Correct option is  A) 1

Correct option is (C) 73°

Let x & y are two linear angles.

\(\therefore\) x+y = \(180^\circ\)    ________(1)

Given that difference of these two linear angles is \(34^\circ.\)

\(\therefore\) x - y = \(34^\circ\)    ________(2)

By adding (1) & (2), we get

2x = \(180^\circ\) + \(34^\circ\) = \(214^\circ\)

\(\Rightarrow\) x = \(\frac{214^\circ}2\) = \(107^\circ\)

Put x = \(107^\circ\) in (1), we get

y = \(180^\circ\) - x = \(180^\circ\) - \(107^\circ\) = \(73^\circ.\)

Correct option is  C) 73°

C) ax + 3y + 7z = 0

(d) We know that, the algebraic expression in one variable having the highest power of the variable as 1, is known as the linear expression.

Here, 1 + z is the only linear expression, as the power of the variable z is 1.

(b) 1/15

Given,

1/3 + S = 2/5

S = 2/5 – 1/3

S = (6 – 5)/15

S = 1/15

Given x – 3 = 5

Adding 3 to both sides we get,

x – 3 + 3 = 5 + 3

x = 8

Verification:

Substituting x = 8 in LHS, we get

LHS = x – 3 and RHS = 5

LHS = 8 – 3 = 5 and RHS = 5

LHS = RHS

Hence, verified.

Given x + 9 = 13

Subtracting 9 from both sides i.e. LHS and RHS, we get

x + 9 – 9 = 13 – 9

x = 4

Verification:

Substituting x = 4 on LHS, we get

LHS = 4 + 9 = 13 = RHS

LHS = RHS

Hence, verified.

Given x – (3/5) = (7/5)

Add (3/5) to both sides, we get

x – (3/5) + (3/5) = (7/5) + (3/5)

x = (7/5) + (3/5)

x = (10/5)

x = 2

Verification:

Substitute x = 2 in LHS of given equation, then we get

2 – (3/5) = (7/5)

(10 – 3)/5 = (7/5)

(7/5) = (7/5)

LHS = RHS

Hence, verified

We have,

(5x/(2x – 1)) = 2

By cross multiplication, we get

5x = 2 × (2x – 1)

5x = 4x – 2

Transposing 4x to LHS it becomes – 4x

5x – 4x = -2

x = -2

We have,

((2x – 3)/(4x + 5)) = (1/3)

By cross multiplication, we get

3 × (2x – 3) = 1 × (4x + 5)

6x – 9 = 4x + 5

Transposing -9 to RHS it becomes 9 and 4x to LHS it becomes – 4x.

6x – 4x = 5 + 9

2x = 14

x = 14/2

x = 7

Given 3 (x + 2) = 15

Dividing both sides by 3 we get,

3 (x + 2)/3 = (15/3)

(x + 2) = 5

Now subtracting 2 by both sides, we get

x + 2 -2 = 5 -2

x = 3

Verification:

Substituting x =3 in LHS we get,

3 (3 + 2) = 15

3 (5) = 15

15 = 15

Therefore LHS = RHS

Hence, verified.

Given (x/4) = (7/8)

Multiply both sides by 4 we get,

(x/4) × 4 = (7/8) × 4

x = (7/2)

Verification:

Substituting x = (7/2) in LHS we get,

(7/2)/4 = (7/8)

(7/8) = (7/8)

Therefore LHS = RHS

Hence, verified.

Given 3x = 0

On dividing both sides by 3 we get,

(3x/3) = (0/3)

x = 0

Verification:

Substituting x = 0 in LHS we get

3 (0) = 0

And RHS = 0

Therefore LHS = RHS

Hence, verified.

We have,

(8/x) = (5/(x – 1))

By cross multiplication, we get

8 × (x – 1) = 5 × x

8x – 8 = 5x

Transposing -8 to RHS it becomes 8 and 5x to LHS it becomes – 5x.

8x – 5x = 8

3x = 8

x = 8/3

Given 14 = (7x/10) – 8

Adding 8 to both sides we get,

14 + 8 = (7x/10) – 8 + 8

22 = (7x/10)

Multiply both sides by 10 we get,

220 = 7x

x = (220/7)

Verification:

Substituting x = (220/7) in RHS we get,

14 = (7/10) × (220/7) – 8

14 = 22 -8

14 = 14

Therefore LHS = RHS.

Hence, verified.

Given (1/3) – 2x = 0

Subtract (1/3) from both sides we get,

(1/3) – 2x – (1/3) = 0 – (1/3)

– 2x = – (1/3)

2x = (1/3)

Divide both side by 2 we get,

2x/2 = (1/3)/2

x = (1/6)

Verification:

Substituting x = (1/6) in LHS we get,

(1/3) – 2 (1/6) = 0

(1/3) – (1/3) = 0

0 = 0

Therefore LHS = RHS

Hence, verified.

Given x/2 = 0

Multiplying both sides by 2, we get

(x/2) × 2 = 0 × 2

x = 0

Verification:

Substituting x = 0 in LHS, we get

LHS = 0/2 = 0 and RHS = 0

LHS = 0 and RHS = 0

Therefore LHS = RHS

Hence, verified.

We have,

[(5(1 – x)) + (3(1 + x))/ (1 – 2x)] = 8

By cross multiplication, we get

(5(1 – x)) + (3(1 + x)) = 8 × (1 – 2x)

5 – 5x + 3 + 3x = 8 – 16 x

8 – 2x = 8 – 16x

Transposing 8 to RHS it becomes – 8 and -16x to LHS it becomes 16x.

16x – 2x = 8 – 8

14x = 0

x = 0/14

x = 0

Given 3 (x + 6) = 24

Divide both the sides by 3 we get,

3 (x + 6)/3 = (24/3)

(x + 6) = 8

Now subtract 6 from both sides we get,

x + 6 – 6 = 8 – 6

x = 2

Verification:

Substituting x = 2 in LHS we get,

3 (2 + 6) = 24

3 (8) =24

24 = 24

Therefore LHS =RHS

Hence, verified.

Given x – (1/3) = (2/3)

Adding (1/3) to both sides, we get

x – (1/3) + (1/3) = (2/3) + (1/3)

x = (2 + 1)/3

x = (3/3)

x =1

Verification:

Substituting x = 1 in LHS, we get

1 – (1/3) = (2/3)

(3 – 1)/3 = (2/3)

(2/3) = (2/3)

Therefore LHS = RHS

Hence, verified.

We have,

((0.2x + 5)/ (3.5x – 3)) = (2/5)

By cross multiplication, we get

5 × (0.2x + 5) = 2 × (3.5x – 3)

x + 25 = 7x – 6

Transposing x to RHS it becomes – x and -6 to LHS it becomes 6.

25 + 6 = 7x – x

31 = 6x

x = 31/6

Given 2y – (1/2) = (-1/3)

Adding (1/2) from both the sides, we get

2y – (1/2) + (1/2) = (-1/3) + (1/2)

2y = (-1/3) + (1/2)

2y = (-2 + 3)/6 [LCM of 3 and 2 is 6]

2y = (1/6)

Now divide both the side by 2, we get

y = (1/12)

Verification:

Substituting y = (1/12) in LHS we get

2 (1/12) – (1/2) = (-1/3)

(1/6) – (1/2) = (-1/3)

(2 – 6)/12 = (-1/3) [LCM of 6 and 2 is 12]

(-4/12) = (-1/3)

(-1/3) = (-1/3)

Therefore LHS = RHS

Hence, verified.

Given 3 (x + 2) – 2 (x – 1) = 7

On simplifying the brackets, we get

3 × x + 3 × 2 – 2 × x + 2 × 1 = 7

3x + 6 – 2x + 2 = 7

3x – 2x + 6 + 2 = 7

x + 8 = 7

Subtracting 8 from both sides, we get

x + 8 – 8 = 7 – 8

x = -1

Verification:

Substituting x = -1 in LHS, we get

3 (x + 2) -2(x -1) = 7

3 (-1 + 2) -2(-1-1) = 7

(3×1) – (2×-2) = 7

3 + 4 = 7

Therefore LHS = RHS

Hence, verified.

Given x + (1/2) = (7/2)

Subtracting (1/2) from both sides, we get

x + (1/2) – (1/2) = (7/2) – (1/2)

x = (7 – 1)/2

x = (6/2)

x = 3

Verification:

Substituting x = 3 in LHS we get

3 + (1/2) = (7/2)

(6 + 1)/2 = (7/2)

(7/2) = (7/2)

Therefore LHS = RHS

Hence, verified.

Given 10 – y = 6

Subtracting 10 from both sides, we get

10 – y – 10 = 6 – 10

-y = -4

Multiplying both sides by -1, we get

-y × -1 = – 4 × – 1

y = 4

Verification:

Substituting y = 4 in LHS, we get

10 – y = 10 – 4 = 6 and RHS = 6

Therefore LHS = RHS

Hence, verified.

Given ((x – 3)/5) -2 = -1

Adding 2 to both sides we get,

((x -3)/5) – 2 + 2 = -1 + 2

(x -3)/5 = 1

Multiply both sides by 5 we get

(x – 3)/ 5 × 5 = 1 × 5

x – 3 = 5

Now add 3 to both sides we get,

x – 3 + 3 = 5 + 3

x = 8

Verification:

Substituting x = 8 in LHS we get,

((8 – 3)/5) -2 = -1

(5/5) – 2 = -1

1 -2 = -1

-1 = -1

Therefore LHS = RHS

Hence, verified.

Given 5 (x – 2) + 3 (x +1) = 25

On simplifying the brackets, we get

(5 × x) – (5 × 2) +3 × x + 3× 1 = 25

5x – 10 + 3x + 3 = 25

5x + 3x – 10 + 3 = 25

8x – 7 = 25

Adding 7 to both sides, we get

8x – 7 + 7 = 25 + 7

8x = 32

Dividing both sides by 8, we get

8x/8 = 32/8

x = 4

Verification:

Substituting x = 4 in LHS, we get

5(4 – 2) + 3(4 + 1) = 25

5(2) + 3(5) = 25

10 + 15 = 25

25 = 25

Therefore LHS = RHS

Hence, verified.

Given 5 (2 – 3x) -17 (2x – 5) = 16

On expanding the brackets, we get

(5 × 2) – (5 × 3x) – (17 × 2x) + (17 × 5) = 16

10 – 15x – 34x + 85 = 16

10 + 85 – 34x – 15x = 16

95 – 49x = 16

Subtracting 95 from both sides, we get

– 49x + 95 – 95 = 16 – 95

– 49x = -79

Dividing both sides by – 49, we get

– 49x/ -49 = -79/-49

x = 79/49

Verification:

Substituting x = (79/49) in LHS we get,

5 (2 – 3 × (79/49) – 17 (2 × (79/49) – 5) = 16

(5 × 2) – (5 × 3 × (79/49)) – (17 × 2 × (79/49)) + (17 × 5) = 16

10 – (1185/49) – (2686/49) + 85 = 16

(490 – 1185 – 2686 + 4165)/49 = 16

784/49 = 16

16 = 16

Therefore LHS = RHS

Hence, verified.

Given 6 (1 – 4x) + 7 (2 + 5x) = 53

On simplifying the brackets, we get

(6 ×1) – (6 × 4x) + (7 × 2) + (7 × 5x) = 53

6 – 24x + 14 + 35x = 53

6 + 14 + 35x – 24x = 53

20 + 11x = 53

Subtracting 20 from both sides, we get 20 + 11x – 20 = 53 – 20

11x = 33

Dividing both sides by 11, we get

11x/11 = 33/11

x = 3

Verification:

Substituting x = 3 in LHS, we get

6(1 – 4 × 3) + 7(2 + 5 × 3) = 53

6(1 – 12) + 7(2 + 15) = 53

6(-11) + 7(17) = 53

– 66 + 119 = 53

53 = 53

Therefore LHS = RHS

Hence, verified.

Given 8 (2x – 5) – 6(3x – 7) = 1

On simplifying the brackets, we get

(8 × 2x) – (8 × 5) – (6 × 3x) + (-6) × (-7) = 1

16x – 40 – 18x + 42 = 1

16x – 18x + 42 – 40 = 1

-2x + 2 = 1

Subtracting 2 from both sides, we get

-2x+ 2 – 2 = 1 -2

-2x = -1

Multiplying both sides by -1, we get

-2x × (-1) = -1× (-1)

2x = 1

Dividing both sides by 2, we get

2x/2 = (1/2)

x = (1/2)

Verification:

Substituting x = (1/2) in LHS we get,

(8 × 2 × (1/2)) – (8 × 5) – (6 × 3 × (1/2)) + (-6) × (-7) = 1

8(1 – 5) – 6(32 – 7) = 1

8× (-4) – (6 × 32) + (6 × 7) = 1

– 32 – 9 + 42 = 1

– 41 + 42 = 1

1 = 1

Therefore LHS = RHS

Hence, verified.

Given (4/5) – x = (3/5)

Subtracting (4/5) from both sides, we get

(4/5) – x – (4/5) = (3/5) – (4/5)

– x = (3 -4)/5

– x = (-1/5)

x = (1/5)

Verification:

Substituting x = (1/5) in LHS we get

(4/5) – (1/5) = (3/5)

(4 -1)/5 = (3/5)

(3/5) = (3/5)

Therefore LHS =RHS

Hence, verified.

Given 7 + 4y = -5

Subtracting 7 from both sides, we get

7 + 4y – 7 = -5 -7

4y = -12

Dividing both sides by 4, we get

y = -12/ 4

y = -3

Verification:

Substituting y = -3 in LHS, we get

7 + 4y = 7 + 4(-3) = 7 – 12 = -5, and RHS = -5

Therefore LHS = RHS

Hence, verified.

We have,

[(y – (4 – 3y))/ (2y – (3 + 4y))] = 1/5

(y – 4 + 3y)/ (2y – 3 – 4y) = 1/5

(-4y – 4)/ (2y – 3) = 1/5

By cross multiplication, we get

5 × (-4y – 4) = 1 × (2y – 3)

20y – 20 = 2y – 3

Transposing – 20 to RHS it becomes 20 and 6y to LHS it becomes -6y.

20y – 2y = 20 – 3

22 y = 17

y = 17/22

Correct option is  D) 7/9

Correct option is (D) 7/9

\(\frac{-4y}{7}=\frac{-4}{9}\)

\(\Rightarrow\) \(y=\frac{-4}9\times\frac{-7}4=\frac79.\)

Correct option is  C) 4

Correct option is (C) 4

2 (a – 3) = 2

\(\Rightarrow\) a - 3 = \(\frac22\) = 1

\(\Rightarrow\) a = 1+3 = 4.

Correct option is  D) none

Correct option is (D) none

\(\frac{1}{3}\) – s = \(\frac{1}{9}\)

\(\Rightarrow\) s = \(\frac{1}{3}\) - \(\frac{1}{9}\) = \(\frac{3-1}{9}\) = \(\frac{2}{9}.\)

2a + 4 = 0 …..(i)

Substituting a = 2 in L.H.S. of equation (i), 

L.H.S. = 2 (2) + 4 

= 4 + 4 

= 8 

R.H.S. = 0 

∴ L.H.S. ≠ R.H.S. 

∴ a = 2 is not the solution of the given equation.

Substituting a = – 2 in L.H.S. of equation (i), 

L.H.S. = 2 (-2)+ 4 

= -4 + 4 

= 0 

R.H.S. = 0 

∴ L.H.S. = R.H.S. 

∴ a = – 2 is the solution of the given equation.

Substituting a = 1 in L.H.S. of equation (i), 

L.H.S. = 2(1)+ 4 

= 2 + 4 

= 6 

R.H.S. = 0 

∴ L.H.S. ≠ R.H.S. 

∴ a = 1 is not the solution of the given equation.

i. x – 4 = 3 ….(i)

Substituting x = – 1 in L.H.S. of equation (i), 

L.H.S. = (-1) – 4

= – 5

R.H.S. = 3 

∴ L.H.S. ≠ R.H.S.

∴ x = – 1 is not the solution of the given equation.

Substituting x = 7 in L.H.S. of equation (i), 

L.H.S. = (7) – 4 

= 3 

R.H.S. = 3

∴ L.H.S. = R.H.S. 

∴ x = 7 is the solution of the given equation.

Substituting x = – 7 in L.H.S. of equation (i), 

L.H.S. = (- 7) – 4

= -11

R.H.S. = 3 

∴ L.H.S. ≠ R.H.S.

∴ x = – 7 is not the solution of the given equation.

9m = 81 …(i)

Substituting m = 3 in L.H.S. of equation (i), 

L.H.S. = 9 × (3)

= 27

R.H.S. = 81 

∴ L.H.S. ≠ R.H.S. 

∴ m = 3 is not the solution of the given equation.

Substituting m = 9 in L.H.S. of equation (i), 

L.H.S. = 9 × (9) 

= 81

R.H.S. = 81 

∴ L.H.S. = R.H.S. 

∴ m = 9 is the solution of the given equation.

Substituting m = – 3 in L.H.S. of equation (i), 

L.H.S. = 9 × (- 3) 

= -27 

R.H.S. = 81 

∴ L.H.S. ≠ R.H.S. 

∴ m = – 3 is not the solution of the given equation.

Correct option is  A) 1

(a) A linear equation in one variable has only one solution.

e.g. Solution of the linear equation ax + b = 0 is unique, i.e. x = -b/a.

(i) a, b are disproportionation reactions.

The correct answer is (i) 2/5

The correct answer is (iii) Tm

HOI < HOBr < HOCl

(iv) Mn²⁺ acts as autocatalyst.

Answer is (A) Haryana

(a) Bryophyllum and money plant.

(b) Bryophyllum and Begonia

Budding  is asexual method of reproduction in yeast. .