Explore topic-wise InterviewSolutions in Current Affairs.

The author left town to live with his father

The island became a small green paradise.

Correct answer is (c) forest

Correct answer is (c) rocky

The garden became a happy place for the author when his grandfather joined him.

His Grandfather served many years in the Indian Forest Service.

After his retirement, he built a bungalow on the outskirts of Dehradun, planting trees all around.

1. We need trees to keep the desert away. 

2. To attract rain.

3. To prevent banks of rivers being washed away. 

4. For fruits and flowers 

5. For timber

There was a mango tree on the island. So grandfather planted saplings there.

The two reasons the author gives for the plants moving towards grandfather are: 

1. Light & Warmth

2. They liked to be near grandpa.

The thought of a world without trees became a sort of nightmare to the author and so he helped his Grandfather in his tree-planting with greater enthusiasm.

(c) wanted to be near Grandfather. (✓)

(1) The writer points out the fact that simple- hearted workers are carried away by the oratory of shrewd leaders and get befooled. They have to sacrifice everything in their life but sugarcoated speeches of leaders trap them and they lead a miserable life working for them.

(2) According to the writer, in spite of being aware of the fact that their honest sweat, persistent toil and years of struggle bring them nothing on to worth fighting for, the workers foolishly believe that they are a country. This is blind vanity of slaves.

The correct answer is False

(iii) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.

Given (218)² – (217)²

According to the property 8 i.e. for every natural number n, we have [(n+1)²-n²] = [(n+1) + n]

Here n = 217 By applying the above property we get,

=(218)² – (217)² 

= (217+1) + 217 

= 218 + 217 

= 435

When silver nitrate solution is added to sodium chloride solution, sodium nitrate and a white precipitate of silver chloride is formed. This is an example of double displacement reaction. Following is the chemical equation for the reaction

AgNO₃(aq)+NaCl(aq)→NaNO₃(aq)+AgCl(s) ↓

All nitrates are soluble, hence silver nitrate is soluble; and all halides are soluble, EXCEPT for AgX,PbX₂, and Hg₂X₂.

Thus silver nitrate is soluble, but silver chloride precipitates from solution as a curdy white solid.

Given,

Length of field = 200 m

Breadth of field = 150 m

Dimension of plot = 50m × 40m

Depth up to which plot is dug = 7 m

Volume of earth dug out = 50 × 40 × 7 = 14000 m³

Let level of earth rises in field = h meter

Hence, 

= 200 × 150 × h = 14000

\(= h = \cfrac{14000}{200\times150}\) = 0.47 m

Total surface area of one brick = 2(lb + bh + lh)

= [2(22.5 ×10 + 10 × 7.5 + 22.5 × 7.5)] cm²

= 2(225 + 75 + 168.75) cm²

= (2 × 468.75) cm²

= 937.5 cm²

Let n bricks can be painted out by the paint of the container.

Area of n bricks = (n ×937.5) cm² = 937.5n cm²

Area that can be painted by the paint of the container = 9.375 m² = 93750 cm²

∴ 93750 = 937.5n n = 100

Therefore, 100 bricks can be painted out by the paint of the container.

Correct option is (D) 6

Let edge of a cube be a meter.

\(\therefore\) Volume of cube is \(V=a^3m^3\)

But given that volume of cube is \(216\,m^3.\)

\(\therefore\) \(a^3=216=6^3\)

\(\Rightarrow\) a = 6

Therefore, edge of given cube is 6 m.

Correct option is D) 6

Given,

Dimension of a brick = 22.5 cm × 10cm × 7.5 cm

Total surface area = 2(lb + bh + hl)

= 2[22.5× 10 + 10× 7.5 + 22.5 × 7.5] cm²

= 2(225 + 75 + 168.75) cm²

= 2(468.75) cm²

= 937.5 cm²

Area that can be painted = 9.375 m²

As 1 m² = 10000 cm²

9.375 m² = 9.375 × 10000 

= 93750 cm²

Number of bricks = \(\cfrac{total\,area\,that\,get\,paint}{surface\,area\,of\,one\,brick}\) = \(\cfrac{93750}{937.5}\)

 = \(\cfrac{937500}{93750}\)

Correct option is (A) a√3

\(\because\) Diagonal of a cuboid \(=\sqrt{l^2+b^2+h^2}\)

For a cube, l = b = h = a

\(\therefore\) Diagonal of a cube \(=\sqrt{a^2+a^2+a^2}\)

\(=\sqrt{3a^2}=\sqrt3\,a\)

Correct option is  A) a√3

Given,

Edge of cube = 9 cm

Dimension of base of rectangular vessel = 15cm × 12cm

Volume of cube = 9 × 9 × 9 = 729 cm³

Area of base of vessel = 15 × 12 = 180 cm²

So, rise of water in vessel = \(\cfrac{volume\,of\,cube}{area\,of\,base}\)

\(=\cfrac{729}{180}\) = 4.05 cm

(b) 5.5 cm

Let the radius of the base of the cylinder be r cm. Then,

\(\frac{22}{7}\times{r}^2\times14=(11)^3\)

⇒ r² = \(\frac{11\times11\times11\times7}{22\times14}\) = \(\frac{121}{4}\)

⇒ r = \(\sqrt{\frac{121}{4}}=\frac{11}{2}\) = 5.5 cm.

Speed of flowing water = 10 km/hr 

In 30 minutes length of flowing water =10 × \(\frac{30}{60}\) km 

= 5 km = 5000 m 

Volume of flowing water in 30 minutes = 5000 × width × depth

 Width of canal = 1.5 m 

Depth of canal = 6 m 

Volume of canal = 5000 × 1.5 × 6 m³ 

= 45000 m³ 

Irrigated area in 30 minutes if 8 cm of flowing water is required 

= \(\frac{45000}{0.08}\)

= 562500 m²

Here, r = 15cm and

θ = 108° = \(\left(108 \times \frac{\pi}{180}\right)^c\) \(=\left(\frac{3\pi}{5}\right)^c\)

Since S = r.θ

S = 15 x 3π/5

= 9π cm.

We know that Sum of angles of a quadrilateral is = 360°

So,

110° + 72° + 55° + x° = 360°

x° = 360° – 237°

x° = 123^o

∴ Value of x is 123^o.

(i) ∠A = ∠C

True, Since ∠A =∠C = 55° [opposite angles are equal in a parallelogram]

(ii) ∠FAB = ½ ∠A

True, Since AF is the angle bisector of ∠A.

(iii) ∠DCE= ½ ∠C

True, Since CE is the angle bisector of angle ∠C.

(iv) ∠CEB= ∠FAB

True,

Since ∠DCE = ∠FAB (opposite angles are equal in a parallelogram).

∠CEB = ∠DCE (alternate angles)

½ ∠C = ½ ∠A [AF and CE are angle bisectors]

(v) CE || AF

True, since one pair of opposite angles are equal, therefore quad. AEFC is a parallelogram.

(i) OB = OD

OB = OD. Since diagonals bisect each other in a parallelogram.

(ii) ∠OBY =∠ODX

∠OBY =∠ODX. Since alternate interior angles are equal in a parallelogram.

(iii) ∠BOY= ∠DOX

∠BOY= ∠DOX. Since vertical opposite angles are equal in a parallelogram.

(iv) ΔBOY ≅ ΔDOX

ΔBOY and ΔDOX. Since OB = OD, where diagonals bisect each other in a parallelogram.

∠OBY =∠ODX [Alternate interior angles are equal]

∠BOY= ∠DOX [Vertically opposite angles are equal]

ΔBOY ≅ΔDOX [by ASA congruence rule]

OX = OY [Corresponding parts of congruent triangles]

∴ XY is bisected at O.

The measure of interior angle A of a polygon of n sides is given by A = [(n-2) ×180°]/n

Angle of quadrilateral is 160°

160° = [(n-2) ×180°]/n

160°n = (n-2) ×180°

160°n = 180°n – 360°

180°n – 160° = 360°

20°n = 360°

n = 360°/20

= 18

∴ Number of sides are 18

We know that Sum of angles of a quadrilateral is = 360°

Let each angle be x°

So,

x° + 2x° + 4x° + 5x° = 360°

12x° = 360°

x° = 360°/12

= 30°

Value of angles are

x = 30°

2x = 2 × 30 = 60°

4x = 4 × 30 = 120°

5x = 5 × 30 = 150°

∴ Value of angles are 30°, 60°, 120°, 150°

By using the formula,

The sum of interior angles of a polygon = (n – 2) × 180°, (where n = number of sides of polygon.)

We know, a hexagon has 6 sides. So,

The sum of interior angles of a hexagon = (6 – 2) × 180° = 4 × 180° = 720°

x°+ (x-5)°+ (x-5)°+ (2x-5)°+ (2x-5)°+ (2x+20)° = 720°

x°+ x°- 5°+ x° – 5°+ 2x° – 5°+ 2x° – 5°+ 2x° + 20° = 720°

9x° = 720°

x = 720°/9

= 80°

∴ Value of x is 80°^.

We know that Sum of angles of a quadrilateral is = 360°

Let each angle be x°

So,

3x° + 5x° + 7x° + 9x° = 360°

24x° = 360°

x° = 360°/24

= 15°

Value of angles are

3x = 3 × 15 = 45°

5x = 5 × 15 = 75°

7x = 7 × 15 = 105°

9x = 9 × 15 = 135°

∴ Value of angles are 45°, 75°, 105°, 135°

The measure of interior angle A of a polygon of n sides is given by A = [(n-2) ×180^o]/n

Angle of quadrilateral is 162°

162^o = [(n-2) ×180^o]/n

162^on = (n-2) ×180^o

162^on = 180^on – 360^o

180^on – 162^o = 360^o

18^on = 360^o

n = 360^o/18

= 20

∴ Number of sides are 20.

We know that Sum of angles of a quadrilateral is = 360°

Let the sum of two angles be 180°

Let angle be x°

So,

180° + x° = 360°

x° = 360° – 180°

x° = 180°

∴ Sum of remaining two angles is 180°

By using the formulas,

The sum of interior angles of a polygon = (n – 2) × 180° ……..(i)

The Sum of exterior angle of a polygon is 360°

We know that Sum of exterior angles/Sum of interior angles = 1/5…..(ii)

So, equating (i) and (ii) we get

360^o/(n – 2) × 180° = 1/5

On cross multiplication,

(n – 2) × 180° = 360^o × 5

(n – 2) × 180° = 1800^o

(n – 2) = 1800^o/180^o

(n – 2) = 10

n = 10 + 2

= 12

∴ Numbers of sides of a polygon is 12.

By using the formulas,

The sum of interior angles of a polygon = (n – 2) × 180° …..(i)

The Sum of exterior angle of a polygon is 360°

So,

Sum of interior angles = 3 × sum of exterior angles

= 3 × 360° = 1080°…..(ii)

Now by equating (i) and (ii) we get,

(n – 2) × 180° = 1080°

n – 2 = 1080^o/180^o

n – 2 = 6

n = 6 + 2

= 8

∴ Number of sides of a polygon is 8.

By using the formulas,

The sum of interior angles of a polygon = (n – 2) × 180°

The sum of interior angles of a hexagon = (6 – 2) × 180° = 4 × 180° = 720°

The Sum of exterior angle of a polygon is 360°

∴ Sum of interior angles of a hexagon = twice the sum of interior angles.

Hence proved.

We know that the sum of exterior angles of a polygon is 360°

Measure of each exterior angle of a polygon is = 360°/n , where n is the number of sides

We know that number of sides in a pentagon is 5

Measure of each exterior angle of a pentagon is = 360°/5 = 72°

∴ Measure of each exterior angle of a pentagon is 72°

We know that Sum of angles of a quadrilateral is = 360°

In the quadrilateral MPNO

∠NOP = 45°, ∠OMP = ∠PNO = 90°

Let angle ∠MPN is x°

∠NOP + ∠OMP + ∠PNO + ∠MPN = 360°

45° + 90° + 90° + x° = 360°

x° = 360° – 225°

x° = 135°

∴ Measure of ∠MPN is 135°

(i) A quadrilateral has four sides.

(ii) A quadrilateral has four angles.

(iii) A quadrilateral has four, no three of which are collinear.

(iv) A quadrilateral has two diagonals.

(v) The number of pairs of adjacent angles of a quadrilateral is four.

(vi) The number of pairs of opposite angles of a quadrilateral is two.

(vii) The sum of the angles of a quadrilateral is 360⁰.

(viii) A diagonal of a quadrilateral is a line segment that joins two opposite vertices of the quadrilateral.

(ix) The sum of the angles of a quadrilateral is four right angles.

(x) The measure of each angle of a convex quadrilateral is less than 180°.

(xi) In a quadrilateral the point of intersection of the diagonals lies in interior of the quadrilateral.

(xii) A point is in the interior of a convex quadrilateral, if it is in the interiors of its two opposite angles.

(xiii) A quadrilateral is convex if for each side, the remaining vertices lie on the same side of the line containing the side.

We know that

∠POH + 70° = 180° [Linear pair]

∠POH = 180°-70°

∠POH = 110°

∠POH = ∠x = 110° [opposite angles are equal in a parallelogram]

∠x + ∠z + 40° = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]

110° + ∠z + 40° = 180°

∠z = 180° – 150°

∠z = 30°

∠z +∠y = 70°

∠y + 30° = 70°

∠y = 70°- 30°

∠y = 40°

Let us consider one of the adjacent angle as x°

Other adjacent angle is = 2x^o/3

We know that sum of adjacent angles = 180°

So,

x° + 2x°/3 = 180°

(3x° + 2x°)/3 = 180°

5x°/3 = 180°

x° = 180°×3/5

= 108°

Other angle is = 180° – 108° = 72°

∴ Angles of a parallelogram are 72°, 72°, 108°, 108°

Let us consider one of the adjacent angle as x°

Other adjacent angle = 2x°

We know that sum of adjacent angles = 180°

So,

x° + 2x° = 180°

3x° = 180°

x° = 180°/3

= 60°

So other angle is 2x = 2×60 = 120°

∴ Measures of the remaining angles are 60°, 60°, 120° and 120°

Let us consider each angle of a parallelogram as x°

We know that sum of angles = 360°

x° + x° + x° + x° = 360°

4 x° = 360°

x° = 360°/4

= 90°

∴ Measure of each angle is 90°

Yes, this quadrilateral is a parallelogram.

Since each angle of a parallelogram is equal to 90°, so it is a rectangle.

We know that opposite angles of a parallelogram are equal.

So, (3x – 2)° = (50 – x)°

3x° – 2° = 50° – x°

3x° + x° = 50° + 2°

4x° = 52°

x° = 52°/4

= 13°

Measure of opposite angles are,

(3x – 2)° = 3×13 – 2 = 37°

(50 – x)° = 50 – 13 = 37°

We know that Sum of adjacent angles = 180°

Other two angles are 180° – 37° = 143°

∴ Measure of each angle is 37°, 143°, 37°, 143°

Let us consider one of the adjacent angle as x°

Other adjacent angle = 70°

We know that sum of adjacent angles = 180°

So,

x° + 70° = 180°

x° = 180° – 70°

= 110°

∴ Measures of the remaining angles are 70°, 70°, 110° and 110°

Let, f (x) = x³ - 3x² - 10x + 24 be the given polynomial.

In order to prove that (x – 2) (x + 3) (x – 4) are the factors of f (x), it is sufficient to show that f (2) = 0, f (-3) = 0 and f (4) = 0 respectively.

Now,

f (x) = x³ - 3x² - 10x + 24

f (2) = (2)³ – 3 (2)² – 10 (2) + 24

= 8 – 12 – 20 + 24

= 0

f (-3) = (-3)³ – 3 (-3)² – 10 (-3) + 24

= -27 – 27 + 30 + 24

= 0

f (4) = (4)³ – 3 (4)² – 10 (4) + 24

= 64 – 48 – 40 + 24

= 0

Hence, 

(x – 2), (x + 3) and (x – 4) are the factors of the given polynomial.