Explore topic-wise InterviewSolutions in Current Affairs.

Before the fire brigade were formed people used to extinguish fire forming human s chain. Everyone was a fireman In that scenario. People used to pass buckets filled with water from a pond or well through each other and the person at the extreme end used to pour it over the flames.

Every year we spend millions of rupees for fighting fires. We spend even large sum of ; money to find out ways to prevent fire from happening. In the process we have learnt to control fire and use it for our betterment.

Non-Cooperation Movement spread in cities across the country:

(i) The movement started with middle class participation in the cities.

(ii) Thousands of students left government controlled schools and colleges.

(iii) Headmasters and teachers resigned and lawyers gave up their legal practices.
(iv) The council elections were boycotted in most provinces except Madras where Justice Party took part in elections.
Effects of Non-Cooperation Movement on the economic front were :
(i) Foreign goods were boycotted.
(ii) Liquor shops were picketed.
(iii) Foreign clothes were burnt in huge bonfires.
(iv) The import of foreign cloth halved between 1921-1922. In value, the drop was from Rs. 102 crore to Rs.57 crore.
(v) In many places, merchants and traders refused to trade in foreign goods or finance foreign trade.
(vi) The people began discarding imported clothes and wore only Indian ones.
(vii) Production of Indian textile mills and handlooms went up tremendously.

From the cities, the Non-Coperation Movement spread to the countryside. It drew into its fold the struggles of peasants and tribals which were developing in different parts in the years after the war.

(i) The movement was primarily against talukdars and landlords who demanded from the peasants exorbitantly high rents. By swaraj they understood that they would not be required to pay any taxes and that land would be redistributed.

(ii) In Awadh, peasants were led by Baba Ramchandra-a sanyasi who had earlier been to fiji as an identured labourer.

(iii) Alluri Sitaram Raju was a tribal peasant leader. During the days of Non-Cooperation Movement, he led the tribal people in the Gudem Hills of Andhra Pradesh.

(a) 7

Let the fraction be \(\frac{X}{y}\)

Given, \(\frac{X+1}{y+2}\)= 2

\(\Rightarrow\) x +1 = 2y + 2

\(\Rightarrow\) x – 2y = 1  ..........(i)

and \(\frac{X-1}{y-1}\) = 3

\(\Rightarrow\) x – 1 = 3y – 3  \(\Rightarrow\)x – 3y = –2  ........(ii)

Now solve eqn (i) and (ii) for x and y.

Let the actual price of the tea and lemon set be Rs.x and Rs.y respectively. 

When gain is Rs.7, then

y/100 × 15 - x/100 × 5 = 7 

⇒ 3y – x = 140 ……..(i) 

When gain is Rs.14, then 

y/100 × 5 + x/100 × 10 = 14 

⇒ y + 2x = 280 ……..(ii) 

Multiplying (i) by 2 and adding with (ii), we have 

7y = 280 + 280 

⇒ y = 560/7 = 80 

Putting y = 80 in (ii), we get 

80 + 2x = 280 

⇒ x = 200/2 = 100 

Hence, actual price of the tea set and lemon set are Rs.100 and Rs.80 respectively.

(b) Rs 80

Let the cost price of the pen and the book be Rs x and Rs y respectively. 

Case I: When the pen is sold at 5% loss and book at 15% gain,

Loss on pen =Rs \(\frac{5X}{100}\) = Rs \(\frac{X}{20}\)

Gain on book = Rs \(\frac{15y}{100}\) = Rs \(\frac{3y}{20}\)

\(\therefore\) Net gain = Rs \(\frac{3y}{20}\)-  Rs \(\frac{X}{20}\)

Given, \(\frac{3y}{20}\) - \(\frac{X}{20}\) = 7

\(\Rightarrow\) 3y -x =140  ..........(i)

Case II: When the pen is sold at 5% gain and book at 10% gain

Gain on pen = Rs \(\frac{5X}{100}\)= Rs \(\frac{X}{20}\)

Gain on book = Rs \(\frac{10y}{100}\) = Rs \(\frac{y}{10}\)

\(\therefore\) Net gain = \(\frac{X}{20}\) + \(\frac{y}{10}\)

Given, \(\frac{X}{20}\) + \(\frac{y}{10}\) = 13 

\(\Rightarrow\) x + 2y = 260 ......…(ii)

Now, solve equation (i) and (ii) for the value of x and y.

(b) Rs 2000

Let the C.P. of 1 horse = Rs x and C.P. of 1 cow = Rs y. Then,

4x + 9y = 13400  ..........(i)

Also, 10% of 4x + 20% of 9y = 1880

\(\frac{2}{5}\)x + \(\frac{9}{5}\)y = 1880 \(\Rightarrow\) 2x + 9y = 9400  .........(ii)

Subtracting eqn. (ii) from eqn. (i) we get

(4x + 9y) – (2x + 9y) = 13400 – 9400

\(\Rightarrow\) 2x = 4000  \(\Rightarrow\) x = 2000

Let H be the cost of a horse and C the cost of a cow. 

Then, 4H + 9C = 13400

Also, \(\frac{10}{100}\) x 4H + \(\frac{20}{100}\) x 9C =1880

\(\Rightarrow\)4H + 18C = 18800

Subtracting equation (1) from equation (2), we get

9C = 5400 \(\Rightarrow\) C = \(\frac{5400}{9}\) = 600

Putting C = 600 in (1), we get

4H + 9 × 600 = 13400 \(\Rightarrow\) 4H = 8000 \(\Rightarrow\) H = \(\frac{8000}{4}\) = 2000

\(\therefore\) The cost of a horse is Rs 2000.

We know that ∠BOA + ∠BOC = 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°]

60° + x° = 180°

x° = 180° – 60°

x° = 120°

Let x be the required angle

So, x + 65° = 180°

x = 180° – 65°

x = 115°

The two angles are said to be complementary angles if the sum of those angles is 90° here it is more than 90° therefore the complement of the angle cannot be determined.

False

A linear pair either have both right angles or one acute and one obtuse angle, because angles forming linear pair is 180°.

Given that,

An angle is 14 more than its compliment

Let the angle be x

Compliment = (90° – x)

According to the question,

x – (90° – x) = 14

2x = 90° + 14°

x = 52°

(i) Two adjacent angles are said to be form a linear pair of angles, if their non-common arms are two opposite rays.

Therefore linear pairs are

∠ABD and ∠DBC

∠ABE and ∠EBC

(ii) We know that every linear pair forms supplementary angles, these angles are

∠ABD and ∠DBC

∠ABE and ∠EBC

Given that,

Angle measured is twice its supplement

Let the angle measured be x

Therefore,

Supplement = (180° – x)

According to the question

x° = 2 (180° – x)

3x = 360°

x = 120°

(a) 120^o

When the sum of the measures of two angles is 180^o, the angles are called supplementary angles.

P + Q = 180^o

60^o + Q = 180^o

Q = 180^o – 60^o

Q = 120^o

No, two acute angles cannot form a linear pair because their sum is always less than 180°.

Let p and q be the two supplementary angles that are equal

The two angles are said to be supplementary angles if the sum of those angles is 180°

∠p = ∠q

So,

∠p + ∠q = 180°

∠p + ∠p = 180°

2∠p = 180°

∠p = 180°/2

∠p = 90°

Therefore, ∠p = ∠q = 90°

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

(A) triangles of equal area

Explanation:

The median of a triangle divides it into triangle of equal area.

Hence, option (A) is the correct answer.

No, because each of these will be a right angle only when they form a linear pair.

Given complement of an angle is 28°

Here, let x be the complement of the given angle 28°

Therefore, ∠x + 28° = 90°

∠x = 90° – 28°

= 62°

So, the supplement of the angle = 180° – 62°

= 118°

Given one of the angle of a linear pair is the right angle that is 90°

We know that linear pair angle is 180°

Therefore, the other angle is

180° – 90° = 90°

No, because if the triangle have two obtuse angles i.e., more than 90° angle, then the sum of all three angles of a triangle will not be equal to 180°.

Two adjacent angles are said to be linear pair of angles, if their non-common arms are two opposite rays.

Therefore linear pairs are listed below:

∠1 and ∠2

∠2 and ∠3

∠3 and ∠4

∠1 and ∠4

∠5 and ∠6

∠6 and ∠7

∠7 and ∠8

∠8 and ∠5

∠9 and ∠10

∠10 and ∠11

∠11 and ∠12

∠12 and ∠9

The two angles are said to be vertically opposite angles if the two intersecting lines have no common arms.

Therefore supplement of the angle are listed below:

∠1 and ∠3

∠4 and ∠2

∠5 and ∠7

∠6 and ∠8

∠9 and ∠11

∠10 and ∠12

(i) Vertically opposite angles are

∠1 and ∠4

∠5 and ∠2 + ∠3

(ii)  Linear pairs

∠5 and ∠1

∠4 and ∠5

(i) Radius (r)= 5 cm

Area of circle = πr²

= 3.14 x (5)² 

= 3.14 x 25 

= 78.57 sq. cm

(ii) Diameter (d) = 42 m

Radius (r) = 42/2 = 21 m

Area of circle = πr²

= 22/7 x 21 x 21

= 22 x 3 x 21

= 66 x 21

= 1386 sq. cm

(iii) Radius (r) = 5.6 cm

Area of circle = πr²

= 22/7 x 5.6 x 5.6 

= 22 x 0.8 x 5.6 

= 98.56 sq. cm.

Infinitely many triangles can be drawn having its angles as 53°, 64° and 63°.

Justification:

According to angle sum property,

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

According to the question,

We have the angles 53°, 64°, and 63°.

Sum of these angles = 53° + 64° + 63°

= 180°

Hence, the angles satisfy the angle sum property of a triangle.

Therefore, infinitely many triangles can be drawn having its angles as 53°, 64° and 63°.

(i) The two adjacent angles are said to form a linear pair of angles if their non – common arms are two opposite rays.

∠1 and ∠3

∠1 and ∠2

∠4 and ∠3

∠4 and ∠2

∠5 and ∠6

∠5 and ∠7

∠6 and ∠8

∠7 and ∠8

(ii) The two angles formed by two intersecting lines and have no common arms are called vertically opposite angles.

∠1 and ∠4

∠2 and ∠3

∠5 and ∠8

∠6 and ∠7

(i) 31 

To find unit digit of cube of a number we do the cube of unit digit only 

Here, unit digit of 31 is = 1 

Cube of 1 = 1³ = 1 

Therefore, unit digit of cube of 31 is always be 1. 

(ii) 109 

To find unit digit of cube of a number we do the cube of unit digit only. 

Here, unit digit of 109 is = 9 

Cube of 9 = 9³ = 729 

Therefore, unit digit of cube of 109 is always be 9. 

(iii) 388 

To find unit digit of cube of a number we do the cube of unit digit only. 

Here, unit digit of 388 is = 8 

Cube of 8 = 8³ = 512 

Therefore, unit digit of cube of 388 is always be 2. 

(iv) 4276 

To find unit digit of cube of a number we do the cube of unit digit only. 

Here, unit digit of 4276 is = 6 

Cube of 6 = 6³ = 216 

Therefore, unit digit of cube of 4276 is always be 6. 

(v) 5922 

To find unit digit of cube of a number we do the cube of unit digit only. 

Here, unit digit of 5922 is = 2 

Cube of 2 = 2³ = 8 

Therefore, unit digit of cube of 5922 is always be 8. 

(vi) 77774 

To find unit digit of cube of a number we do the cube of unit digit only. 

Here, unit digit of 77774 is = 4 Cube of 4 = 4³ = 64 

Therefore, unit digit of cube of 77774 is always be 4. 

(vii) 44447 

To find unit digit of cube of a number we do the cube of unit digit only. 

Here, unit digit of 44447 is = 7 

Cube of 7 = 7³ = 343 

Therefore, unit digit of cube of 44447 is always be 3. 

(viii) 125125125 

To find unit digit of cube of a number we do the cube of unit digit only. 

Here, unit digit of 125125125 is = 5 

Cube of 5 = 5³ = 125 

Therefore, unit digit of cube of 125125125 is always be 5.

False.

The cube of 3 .i.e. 4³ = 4 × 4 × 4 = 64

Correct option is D) 120°

No triangle can be drawn having its angles 45°, 64° and 72°.

Justification:

According to angle sum property,

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

But, according to the question,

We have the angles 45°, 64° and 72°.

Sum of these angles = 45° + 64° + 72°

= 181°, which is greater than 180°.

Hence, the angles do not satisfy the angle sum property of a triangle.

Therefore, no triangle can be drawn having its angles 45°, 64° and 72°.

No, because they don’t have common vertex.

Area of remaining sheet = Area of sheet – Area of square

= πr² – 4² 

= 22/7 x 14 x 14 – 16
[∵ r = 14 cm]

= 22 x 2 x 14 – 16

= 616 – 16

= 600 sq. cm

(i) False. The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it.

(ii) False. The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.

(iii)True, as the cube of 1 and 2 are 1 and 8 respectively.

(i) The two angles are said to be complementary angles if the sum of those angles is 90°

Complementary angle for given angle is

90° – 35° = 55°

(ii) The two angles are said to be complementary angles if the sum of those angles is 90°

Complementary angle for given angle is

90° – 72° = 18°

(iii) The two angles are said to be complementary angles if the sum of those angles is 90°

Complementary angle for given angle is

90° – 45° = 45°

(iv) The two angles are said to be complementary angles if the sum of those angles is 90°

Complementary angle for given angle is

90° – 85° = 5°

(i) Given angle is 20°

Since, the sum of an angle and its complement is 90° . 

its, complement will be (90° - 20 = 70°) 

(ii) Given angle is 35° 

Since, the sum of an angle and its complement is 90° . 

its, complements will be (90° - 35° = 55°) 

(iii) The given angle is 90° 

Since, the sum of an angle and its complement is 90° . 

its, complement will be (90° - 90°  = 0)  

(iv) The given angle is 77° 

Since, the sum of an angle and its complement is 90° . 

its, complement will be (90° - 77° = 13)  

(v) The given angle is 30° . 

Since, the sum of an angle and its complement is 90° . 

its, complement will be (90° - 30° = 60°)

Let the measure of any angle is ‘ a ‘ degrees.

Thus, its complement will be (90 – a)° 

So, the required angle = Complement of a – 28 

a = ( 90 – a ) – 28

a = 90 - a - 28

a + a = 90 - 28 

2a = 62 

a = 31 

Hence, the angle measured is 31°.

(i) The sum of an angle and its complement = 90° 

Therefore, the complement of 20° = 90° – 20° = 70° 

(ii) The sum of an angle and its complement = 90° 

Therefore, the complement of 35° = 90° – 35° = 55° 

(iii) The sum of an angle and its complement = 90° 

Therefore, the complement of 90° = 90° – 90° = 0° 

(iv) The sum of an angle and its complement = 90° 

Therefore, the complement of 77° = 90° – 77° = 13° 

(v) The sum of an angle and its complement = 90° 

Therefore, the complement of 30° = 90° – 30° = 60°

Let an angle measured by ‘ a ‘ in degrees

Thus, its complement will be (90 – a)° 

Required Angle = 30° + complement/2 

a = 30° + ( 90 – a )°/2 

a + a/2 = 30° + 45°

3a/2 = 75° 

a = 50° 

Therefore, the measure of required angle is 50°.

Two angles are Adjacent when they have a common side and a common vertex.

An obtuse angle

Let the angle be "x"

The, its complement will be (90° – x)

Note: Complementary angles: When the sum of 2 angles is 90°.

It is given that angle = 30° + \(\frac{1}{2}\) Complement

x = 30° + \(\frac{1}{2}\)(90° – x)

x = 30° + 45° - \(\frac{x}{2}\)

x + \(\frac{x}{2}\) = 30° + 45°

\(\frac{3x}{2}\) = 75°

3x = 150°

x = 50°

Thus, the angle is 50°

An acute angle

The complement of an acute angle is an acute angle. Since complementary angles add to 90 degrees, the only angles that add to 90 are acute angles.

Let one of the complement angle be x°

Its complement be = 90° – x

∴ According to the question

x° = 90° – x°

x° + x° = 90°

2x° = 90°

x = 90/2 = 45°

Correct option is (A) 135°

Let the required angle be x.

\(\therefore\) Its supplementary angle is \(180^\circ-x.\)

Now, complementary angle of \(180^\circ-x\) is \(90^\circ-(180^\circ-x)\)

\(=x-90^\circ\)

According to given condition,

\(180^\circ-x=x-90^\circ\)

\(\Rightarrow\) 2x = \(180^\circ+90^\circ=270^\circ\)

\(\Rightarrow\) x = \(\frac{270^\circ}2=135^\circ\)

Correct option is  A) 135°

(i) 130° + 50° = 180°
∴ This pair is supplementary angles.

(ii) 45° + 45° = 90°

∴ This pair is complementary angles.

(iii) 80°+ 10° = 90°

This pair is complementary angles.

If the sum of any two angles is 90°, then the angles are called complementary angles.

A complementary angle of 27° is (90 – 27) = 63°