Explore topic-wise InterviewSolutions in Current Affairs.

∠APR = ∠PRD (Alternate interior angles)

50º + y = 127º

y = 127º − 50º

y = 77º

Also, ∠APQ = ∠PQR (Alternate interior angles)

50º = x

∴ x = 50º and y = 77º

Let the complement be x then the number = 5x

Now, according to question

x + 5x = 90°

6x = 90°

x = \(\frac{90}{6}\)

x = 30°

Hence, measure of the angle is 30°.

Supplement is defined as the other angle that adds up to 180 degrees. So therefore if you have an acute angle you know by definition that the angle is less than 90 degrees. In order for both of them to be supplementary (add up to 180 degrees) the other angle must be greater than 90 degrees. Angles that are greater than 90 degrees are obtuse angles. So an obtuse angle is the supplement of an acute angle.

It is given that,

Angle = 8 times its compliment

Let x be the measured angle

Angle = 8 (Compliment)

Angle = 8 (90° – x°)

x = 720° – 8x

9x = 720°

x = 80°

Given: Required angle = 8 times of its complement 

Consider a° be one angle then its complementary angle will be equal to (90 – a)°

According to the question; 

a = 8 times of its complement 

a = 8 ( 90 – a ) 

a = 720 – 8a 

a + 8a = 720 

9a = 720 

a = 80

Therefore, the required angle is 80°.

Angle measured will be ‘x’ say

Therefore, its compliment will be (90° – x)

It is given that angle = Compliment – 28°

x = (90° – x) – 28°

2x = 62°

x = 31°

Supplementary angles are in the ratio 4 : 5

Let the angles be 4x and 5x.

It is given that they are supplementary angles

Therefore,

4x + 5x = 180°

x = 20°

Hence, 4x = 80°

5x = 100°

Therefore, angles are 80° and 100°.

Obtuse

If angle is acute angle, then its supplement will be an obtuse angle. As, if we subtract an angle which is less than 90° from 180°, then result will be an angle greater than 90°.

Given perimeter of a rhombus = 28 m

But we know that perimeter of a rhombus = 4 (Side)

4(Side) = 28 m

Side = 28/4

Side = 7m

Now, Area of the rhombus = 28 m²

But we know that area of rhombus = Side x Altitude

(Side x Altitude) = 28 m² 

(7 x Altitude) = 28 m²

Altitude = 28/7 = 4 m

When the sum of two angles is 90 degrees, then the angles are known as complementary angles.

Area of the circular flower garden = 314 sq.m

Radius of the sprinker = 12m

Area of the sprinkler = πr² = 3.14 × 12 × 12

= 452.16 sq.mt

∴ 452.16 > 314

∴ Yes, the sprinkler water the entire garden.

When the sum of two angles is 180°, then the angles are known as supplementary angles.

(i) The sum of an angle and its supplement = 180°. 

Therefore supplement of angle 54° = 180° – 54° = 126° 

(ii) The sum of an angle and its supplement = 180°. 

Therefore supplement of angle 132° = 180° – 132° = 48° 

(iii) The sum of an angle and its supplement = 180°. 

Therefore supplement of angle 138° = 180° – 138° = 42°

Two supplementary angles are in the ratio 4:5.

Let us say, the angles are 4a and 5a

Since angle are supplementary angles; 

Which implies, 4a + 5a = 180° 

9a° = 180° 

a = 20° 

Therefore, 4a = 4 (20°) = 80° and 5a = 5 (20°) = 100° 

Hence, required angles are 80° and 100°.

(i) Given angle is 54°

Since the sum of an angle and its supplement is 180°

Therefore, its compliment will be:

180° – 54° = 126°

(ii) Given angle is 132°

Since the sum of an angle and its supplement is 180°

Therefore, its compliment will be:

180° – 132° = 48°

(iii) Given angle is 138°

Since the sum of an angle and its supplement is 180°

Therefore, its compliment will be:

180° – 138° = 42°

(i) The given angle is 54°. 

Since, the sum of an angle and its supplement is 180° . 

its, supplement will be 180° - 54° = 126° 

(ii) The given angle is 132°. 

Since, the sum of an angle and its supplement is 180° .

 its, supplement will be 180° - 132° = 48°

(iii) The given angle is 138°. 

Since, the sum of an angle and its supplement is 180° . 

its, supplement will be 180° -138° = 42°

For ABC to be a line, the sum of the two adjacent angles must be 180° i.e.,x + y = 180°.

Area of sheet (R) = 5 cm

Radius of circle (r) = 4 cm

Area of remaining part = πR² – πr²

= π(R² – r²)

= 3.14(5² – 4²)

= 3.14(25 – 16)

= 3.14 x 9

= 28.26 sq. cm

The first symbol represents the restriction on foot movement and second symbol represents the restriction of smoking.

Third symbol represents the restriction of truck. Fourth symbol represents the restriction on horn. Fifth symbol represents the maximum speed limit of 50 km/hour.

Radius of circle = 21 cm

Circumference of the circle = 2πr ( where π = 22/7)

= 2 x 22/7 x 21 = 2 x 22 x 3

= 44 x 3 = 132 cm.

Area of the circle = πr²

= 22/7 x 21 x 21 = 22 x 3 x 21

= 1386 cm².

No. A triangle cannot have two obtuse angles

Justification:

According to angle sum property,

We know that the sum of all the interior angles of a triangle should be = 180°.

An obtuse angle is one whose value is greater than 90° but less than 180°.

Considering two angles to be equal to the lowest natural number greater than 90°, i.e., 91°.

According to the question,

If the triangle has two obtuse angles, then there are two angles which are at least 91° each.

On adding these two angles,

Sum of the two angles = 91° + 91°

⇒ Sum of the two angles = 182°

The sum of these two angles already exceeds the sum of three angles of the triangle, even without considering the third angle.

Therefore, a triangle cannot have two obtuse angles.

Correct option is C) 80°

Correct option is A) 60°

Correct option is A) 111°

Surface area of cube = 384 m² (Given) 

Let length of each edge of cube = a metre

6a² = 384

a² = \(\frac{384}{6}\) = 64

a = \(\sqrt{64}\) = 8 m

Volume of cube = a³ = (8)³ = 512m³

∴ 8 = 2 x 2 x 2 = 2³

But, 2 + 2 + 2 = 6 ≠ 2 x 2 x 2 

∴ 2³ ≠ 2 + 2 +2 is false and 2³ = 2 x 2 x 2 is true.

(i) \(\{(5^2 + 12^2)^{1/2}\}^3\)

After solving we get,

\(\{(25 + 144)^{1/2}\}^3\) = \(\{(13^2)^{1/2}\}^3\) = {13}³ = 2197

(ii)  \(\{(6^2 + 8^2)^{1/2}\}^3\)

After solving we get;

\(\{(25 + 144)^{1/2}\}^3\) = \(\{(100)^{1/2}\}^3\) = \(\{(10^2)^{1/2}\}^3\) = \({\{10}\}^3\) = \(1000\)

Given one of the angles of a linear pair is obtuse, then the other angle should be acute, because only then their sum will be 180°.

i) Angle formed when a door is opened. 

ii) Angle between two adjacent edges of a blackboard. 

iii) Angle between two adjacent edges of a ruler.

iv) Angle between two adjacent edges of a set square. 

v) Angle formed at elbow.

In Fig. (i) sum of two interior angles 132° + 48° = 180° [∴ equal to 180°]

Here, we see that the sum of two interior angles on the same side of n is 180°, then they are the parallel lines.

In Fig. (ii), the sum of two interior angles 73° + 106° = 179° ≠ 180°. 

Here, we see that the sum of two interior angles on same side of r is not equal to 180°, then they are not the parallel lines.

i) acute 

ii) obtuse. 

iii) right 

iv) acute 

v) obtuse.

Given, diameter of a circular garden = 28 m

Length of the fencing = Circumference of circle

= πd

= 22/7 x 28 = 88 m

Total cost of fencing = 88 x 300 = Rs. 26400

(D)

Explanation:

In figure (D), the parallelograms, PQRA and BQRS are on the same base QR and between the same parallels QR and PS.

Hence, option (D) is the correct answer.

False, because ar (ABD) = ar (ACD) and ar (PBD) = ar (PCD), therefore, ar (ABP) = ar (ACP).

Given that,

Supplement = \(\frac{2}{3}\) of the angle itself

Let the angle be x

Therefore,

Supplement = (180° – x)

According to the question

(180° – x) = \(\frac{2}{3}\)x

540° - 3x = 2x

5x = 540°

x = 108°

Hence, supplement = 72°

Therefore, the angle will be 108° and its supplement will be 72°.

Correct option is (C) 18°

The complemenatry angle of \(72^\circ\) is \(90^\circ-72^\circ=18^\circ\)

Hence, \(18^\circ\) is complemenatry angle of \(72^\circ.\)

Correct option is  C) 18°

We know that ∠LOP + ∠PON + ∠NOM = 180°

[Linear pair: The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180°]

Since, 35° + x° + 60° = 180°

x° = 180° – 35° – 60°

x° = 180° – 95°

x^o = 85°

Let x and (90° – x) be two complimentary angles

According to question,

2x = 3 (90° – x)

2x = 270° – 3x

x = 54°

The angles are:

54° and 90° – 54° = 36°

Thus, smallest angle is 36°

True

e.g. Let one angle = 50°

∴ The other angle = 90 – 50° = 40° < 45°

True

Vertically opposite angles are equal. So, if one angle is acute, then other angle will be acute and if one angle is obtuse, then the other will be obtuse.

We know that ∠POQ + ∠QOR = 180°

[Linear pair: The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180°]

3x° + 2x° = 180°

5x° = 180°

x° = 180°/5

x° = 36°

(a) Rs 62, Rs 34

Let Ram have Rs x and Mohan have Rs y. 

If Ram gives Rs 30 to Mohan, then 

Ram has = Rs (x – 30) 

Mohan has = Rs (y + 30)

 According to the question, y + 30 = 2(x – 30) 

\(\Rightarrow\) y + 30 = 2x – 60 \(\Rightarrow\) 2x – y = 90 …....(i)

If Mohan gives Rs 10 to Ram, then 

Ram has Rs (x + 10) 

Mohan has Rs (y – 10) 

According to the question, (x + 10) = 3(y – 10)

\(\Rightarrow\)x + 10 = 3y – 30 \(\Rightarrow\) x – 3y = – 40 ........(ii)

Now we can easily solve for x and y.

The oxidation state of an element depends on its electronic configuration. 

Zn: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²

When Zn loses two electrons from 4s-orbital thus acquiring +2 oxidation state, it gets a very stable electronic configuration wherein all the electrons in d-orbital are paired.

Zn²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

Since d-orbitals are completely filled with paired electrons, it has a very stable electronic configuration. Therefore, Zn shows only one oxidation state of +2.

i. Scandium (Sc) has electronic configuration, 

Sc: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²

ii. Due to the loss of two electrons from the 4s-orbital, Sc acquires +2 oxidation state. 

Sc²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹

iii. By the loss of one more electron from the 3d-orbital, it acquires +3 oxidation state. 

Sc³⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶

iv. Since Sc³⁺ acquires extra stability of inert element [Ar], it does not form higher oxidation state.

Electronic configuration of Mn²⁺ is [Ar] 3d⁵ which is half filled and hence it is stable. So third ionization enthalpy of manganese is very high i.e., third electron cannot be easily removed. Whereas, in case of Fe²⁺, the electronic configuration is [Ar] 3d⁶ . Therefore, Fe²⁺ can easily lose one electron to form Fe³⁺ (3d⁵) which is more stable as it is half-filled.

Thus, manganese is more stable in the +2 state than in the +3 state and the reverse is true for iron.

Iron (Fe) is placed in the 4^th period and group 8 of the modern periodic table.

Electronic configuration of Fe²+ : 1s² 2s² 2p⁶ 3s² 3p6 3d6

Electronic configuration of Fe³⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

Due to the presence of half filled ‘d’ orbital, Fe³⁺ is more stable than Fe²⁺.

Oxidation states of Mn: +2, +3, +4, +5, +6 and +7.

Electronic configuration of Mn:1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²

Due to the presence of half filled ‘d’ orbital, the +2 oxidation state of manganese is more stable.

Ionization enthalpy is defined as the amount of energy required to remove the outermost electron completely from a gaseous atom in its ground state.

(b) 43

Let the larger number be x and the smaller one be y. 

We know that, 

Dividend = (Divisor × Quotient) + Remainder 

By the first condition,

3x = 4y + 3  \(\Rightarrow\) 3x – 4y = 3 ........(i)

By the second condition,

7y = 5x + 1  \(\Rightarrow\) 5x – 7y = –1  .........(ii)

Now we can easily get x and y.

i. The thermodynamic stability of transition metal compounds can be predicted on the basis of their ionization enthalpy value. 

ii. When the sum of the ionization enthalpies required to attain a particular oxidation state of transition metal ions is small, the thermodynamic stability of the compounds of the metal in that oxidation state is high.

eg. 

a. Compounds containing Ni (II) are more stable than compounds containing Pt (II). Less amount of energy is required for the ionization of Ni to Ni²⁺, than the energy required for the ionization of Pt to Pt²⁺. This is because the sum of first and second ionization enthalpies (IE₁ + IE₂) for nickel is lesser as compared to that of platinum.

Ni → Ni²⁺ (IE₁ + IE₂ = 2.49 x 10³ kJ mol^-1)

Pt → Pt²⁺ (IE₁ + IE₂ = 2.66 x 10³ kJ mol^-1)

b. Compounds containing Pt (IV) are more stable than compounds containing Ni (IV). Less amount of energy is required for the ionization of Pt to Pt⁴⁺ than the energy required for the ionization of Ni to Ni⁴⁺. This is because the sum of first four ionization enthalpies (IE₁ + IE₂ + IE₃ + IE₄) for platinum is lesser as compared to that of nickel.

Ni → Ni⁴⁺ (IE₁ + IE₂ + IE₃ + IE₄ = 11.29 x 10³ kJ mol^-1)

Pt → Pt⁴⁺ (IE₁ + IE₂ + IE₃ + IE₄ = 9.36 x 10³ kJ mol^-1)

Note: K₂PtCl₆ is a well known compound of Pt (IV). The corresponding compound of nickel is not known

i. In the beginning of 3d transition series, Sc²⁺ is virtually not known or in other words it is not stable in comparison to Sc³⁺. Ti²⁺, V²⁺, Cr²⁺ are known but less stable in comparison to their most common oxidation state of +3.

ii. In the middle of the 3d transition series, Mn²⁺, Fe²⁺, Co²⁺ are known and quite common. Mn2+ and Mn⁷⁺ are most stable in Mn. Fe²⁺ is less stable in comparison to Fe³⁺, due to fact that Fe²⁺ tends to lose one electron to acquire d⁵ structure, which has extra stability.

iii. Co²⁺ is less stable as compared to Co³⁺. Ni²⁺ is most common and most stable among its +2, +3 and +4 states. Cu+ is more stable and is most common species as compared to Cu²⁺.

iv. At the end of the 3d transition series, Zn forms only Zn²⁺ which is highly stable as it has 3d¹⁰ configuration.