Explore topic-wise InterviewSolutions in Current Affairs.

Correct option is  D) 3/12

Fill in the blanks to make the statements true.

0.5 x 0.7 = 0.35

Correct option is  C) a

 1/5 ÷ 5/6 = 1/5 x 5/6

Correct answer is 1

Correct answer is 741.6 km (approximetly)

Correct answer is 11 days

Correct answer is 0.93 kg

Correct answer is D

Correct answer is  

(a) D, (b) E (c) 3/6 or 1/2 or middle

Let × be the CP of Flower Vase 

CP = x 

SP = (x) × 5/6

= 5x/6 

Loss = CP – SP 

= × – 5x/6 

= x/6 

Loss Percent = (Loss × 100) / CP 

= (x/6 × 100) / x 

= 100/6 

= 16.66% 

So, If Flower vase set is sold at 5/6 price of its CP. Then Loss percent will be 16.66%.

1. 31.997

2. \(\frac{5}{1000}\) (or) 0.005

3. \(\frac{32}{1000}\) (or) 0.032

4. 48

Length of number plate in front of bus = 3 cm

Breadth = 1.6 cm

Perimeter of rectangular size number plate of bus

= 2 (length + breadth)

= 2 (3 + 1.6) cm = 2 (4.6) = 2 x 4.6

= 9.2 cm

Taxi number plate is also in same size as bus so
perimeter of taxi number plate is also 9.2 cm

Length of number plate of private vehicle is = 6.2 cm

Breadth = 1.6 cm

Perimeter of private vehicle number plate is

= 2 (length + breadth) 

= 2(6.2 + 1.6) 2(7.8)

= 15.6 cm

Perimeter of rectangle = 2 (length + breadth)

= 2(10 + 5) 

= 2(15) 

= 30 m

Perimeter of rectangle = 2 (length x breadth)

= 2(25 + 15) 

= 2(40) 

= 2 x 40 

= 80 cm

Perimeter of square = Perimeter of rectangle

4 x side = 80 ⇒ side = 80/4 = 20 cm

⇒ Area of square = side² = 20 x 20 = 400 sq. cm

⇒ Area of rectangle = length x breadth

= 25 x 15 = 375 sq. cm

Area of square is 400 sq. cm and area of rectangle is 375 sq. cm.

Hence area of square has greater area.

Side of square = 12 m

Perimeter of square = 4 x side

= 4 x 12 

= 48 m.

(i) Base of parallelogram = 9 cm

Height of parallelogram = 6 cm

Area of parallelogram = Base x Height

= 9 x 6 

= 54 sq. cm

(ii) Base of parallelogram = 3 cm

Height of parallelogram = 5.5 cm

Area of parallelogram = Base x Height = 3 x 5.5
= 16.5 sq. cm

(iii) Base of triangle = 7 cm

Height of triangle = 4 cm

Area of triangle = 1/2 x Base x Height 1 28

= 1/2 x 7 x 4 

= 28/2 = 14 sq. cm

(iv) Base of triangle = 6.5 cm

Height of triangle = 3 cm

Area of triangle = 1/2 x Base x Height

= 1/2 x 6.5 x 3

= 6.5 x 1.5 

= 9.75 sq. cm.

In this shape we have to find out perimeter, All around of a square, semi circles are joined.

The diameter of every semi circle is 14 cm

Circumference of a circle = πd (∴2r = d)

Half circumference = 1/2 πd

= ( 1/2 x 22/7 x 14) = 22 cm

So, the perimeter of given shape = 4 x 22 = 88 cm.

(i) Sides of triangle: 2 cm, 3 cm and 4 cm

Perimeter of triangle = Sum of all sides

= 2 + 3 + 4 

= 9 cm

(ii) Side of equilateral angle = 8 cm 

Perimeter of equilateral triangle = 3 x side

= 3 x 8

= 24 cm

(iii) Two sides of isosceles triangle = 10 cm, 10 cm Third side = 7 cm

Perimeter of isosceles triangle

= 10 + 10 + 7

= 27 cm

(c) 12.5 cm

Let the sides of the triangle be 12x cm , 14x cm and 25x cm 

Thus, we have Perimeter = 12x +14x + 25x

= 25.5 = 51x

x = 25.5/51 = 0.5

Greatest side of the triangle = 25x = 25 x 0.5 = 12.5 cm

p = perimeter of a rectangle = 130 cm

l = length of a rectangle = l = ?

b = breadth of a rectangle = b = 30 cm

p = 2 (l + b)

130 = 2l + 2 × 30

2l = 130 – 60 = 7

I = \(\frac{70}{2} = 35cms\)

∴ Area of a rectangle = l × b = 30 × 35

= 1050 sq.cms

(c) 180 m²

AE = AB – EB = AB – DC = 16 m – 8 m = 8 m 

∴ In Δ AED,

DE² = AD² – AE² = 17² – 8² = 289 – 64 = 225 

⇒ DE = 15 m 

∴ Area of the trapezium ABCD

= \(\frac12\) x (AB + DC) x DE

= \(\frac12\) x (16 m + 8 m) x 15 m

= \(\frac12\) 24 m x 15 m = 180 m².

Given area of the wall = 3 m x 4 m

= 12 m²

Also given that area of one marble tile = 10 cm x 12 cm

= 120 cm² 

= 0.012 m² [Since 1 m² = 10000 c m²]

Number of tiles required to cover the wall = Area of wall/ Area of one marble tile

= 12/0.012

= 1000 tiles

Cost of one tile = Rs. 2

Total cost = Number of tiles x Cost of one tile

= Rs. (1000 x 2)

= Rs. 2000

Given.

In ΔABC, ∠A = 90°

AB = 5cm; AC = 12cm; AD ⊥ BC; BC = 13cm.

Now area of ΔABC = 12 × base × height

= 12 x AB x AC = 12 × 5 × 12 = 30cm²

Also area of MBC = 12 × BC × AD

30 = 12 × 13 × AD

∴ \(\frac {30\times2}{13} = \frac {60}{13} = 4\frac {8}{13}\) cm

Given length of rectangular shaped wire = 40 cm

Breadth of rectangular shaped wire = 22 cm

Perimeter of the rectangle = 2(Length + Breadth)

= 2(40 cm + 22 cm)

= 124 cm

It is given that the wire which was in the shape of a rectangle is now bent into a square.

Therefore, the perimeter of the square = Perimeter of the rectangle

Perimeter of the square = 124 cm

We know that perimeter of square = 4 x side

Therefore 4 x side = 124 cm

Side = 124/4 = 31 cm

We know that area of rectangle = length x breadth

Now, Area of the rectangle

= 40 cm x 22 cm

= 880 cm²

We know that area of the square = (Side)² 

= (31 cm)² 

= 961 cm².

Therefore, the square-shaped wire encloses more area.

The shape of the wire rectangle.

The perimeter of this rectangle

Total or length of wire =2 (l + b)

length = 40 cm breadth = 22 cm

= 2(40 + 22)

= 2(62) = 124 cm.

If it is reshaped in the form of sq, its side}

= \(\frac{124}{4} = 31cm\)

∴ Its square = side × side = 31 × 31 = 961 sq. cms

Rectangle’s Area = l × b = 40 × 22 = 880 sq.cms.

The area of square encloses more area than the area of rectangle.

Given : In rectangle ABCD,

AB = 8cm, BC = 16 cm, AE = 4cm

Now Area of ΔBAE = 12 x base x height

= 1/2 x \(\overline {BC} \times \overline {AB}\)

= 1/2 × 16 x 8 = 64cm²

= 1/2 × 16 x 8 = 64cm²

 = 1/2 x \(\overline {ED} \times \overline {DC}\) 

= 1/2 × 12 × 8 = 48cm²

⁽\(\overline {ED}\) = AD – AE = 16 – 4 = 12cm and \( \overline {DC} = \overline {AB}\))

Now ΔBAE + ΔCDE = 16 + 48 = 64 = ΔBCE.

Given: PQ = PR = 7.5cm, QR = 9cm PS = 6cm

Area of ΔPQR = 1/2 bh = 1/2 x 9 x 6 = 27cm²

Also area of ΔPQR = 1/2 × PQ × RT

27 = 1/2 × 7.5 × RT [PQ = PR]

∴ RT = \(\frac {2\times27}{7.5} = \frac {2\times27\times2}{15} = \frac {36}{5}\) = 7.2 cm

We know that, π = 22/7 = 3.14 

(a) Shapes 1, 3 and 4 have different areas and different perimeters.

Shapes 1, 3 and 4 have same area and same perimeter.

False

Their corresponding sides and height may be different. So, area cannot be equal.

True

Area of ΔABC : Area of ΔACD = 1/2 x BC x AC : 1/2 x CD x AC

= BC : CD

Circumference ‘C’ of a circle can be found by multiplying diameter ‘d’ with π.

We know that, circumference of circle when radius is considered = 2 πr

When diameter is considered = πd

Let the height of the parallelogram be x cm.

Then the base of the parallelogram is 2x cm. [from given data]

Given that the area of the parallelogram = 512 cm²

We know that area of a parallelogram = Base x Height

512 = (2x) (x)

512 = 2x²

x² = 512/2

= 256 cm²

x² = (16)²

x = 16 cm

Hence height of parallelogram = x = 16 cm

And base of the parallelogram = 2x = 2 x 16 = 32 cm

(a) area of (i) is the same as the area of (ii).

True.

By observing the figure, we can say that area of booth figure same, because number of blocks used in both figures are same.

(b) Perimeter of (ii) is the same as (i).

False.

By observing the figure, we can say that there are 2 new sides are added in figure (ii).

Therefore, the perimeter of (ii) is not same as (i).

(c) If (ii) is divided into squares of unit length, then its area is 13 unit squares.

False.

We know that, area of square = side × side

Then area of 1 square = 1 × 1

= 1 unit squares

There are 12 squares in figure (ii).

Therefore, area of figure (ii) = 12 × 1 = 12 unit squares

(d) Perimeter of (ii) is 18 units.

True.

The perimeter refers to the total length of the sides or edges of a polygon.

Circumference ‘C’ of a circle is equal to 2π × r.

i) Perimeter =12 cm + 8cm + 12cm + 8 cm = 40 cm 

ii) Perimeter = 3cm + 2cm + 3cm + 2cm + 3cm + 2cm + 3cm + 2cm + 12cm + 8 cm = 40 cm

iii) Perimeter = 5cm + 3cm + 2cm + 3cm + 5cm + 2cm + 5cm + 3cm + 2cm + 3cm + 5cm + 2cm = 40 cm 

By observing the perimeters of the above figures perimeters are same for the different shaped figures.

(b) 2 times

We know that, perimeter of parallelogram = 2 (length + breadth)

If the sides of a parallelogram are increased to twice its original lengths,

= 2 (2length + 2breadth)

= 2 × 2 (length + breadth)

Hence, the perimeter will increase by 2 times the perimeter of the original parallelogram.

Area of triangle = 1/2 x Base x Height.

We know that, 1 cm = 10 mm

∴ 1 cm² = (10)² mm² = 100 mm²

We know that, 1 m = 100 cm

∴                         1 m² = (100)²cm²

⇒                        1m² = 10000 cm²

True.

Perimeter of a closed figure is the distance around it while area is the measure of the part of plane or region enclosed by it.

Hence, perimeter of (ii) is greater than that of (i), but its area is smaller than that of (i).

(c) 3 times

Let us assume x be the side of the rhombus and h be the height of the rhombus.

Then, area of rhombus = x × h = xh

Now, side of a rhombus is doubled

Area of new rhombus = 4xh

(the area of new rhombus after its sides are doubled is 4 times the original rhombus)

Hence, the area will increase by 3 times the area of the original rhombus.

We know that, 1 km = 1000 m

∴ 1 km² = (1000)² m² = 1000000 m²

10 cm² = 0.001 m².

We know that, 1 m = 100 cm

1 m² = 10000 cm²

10 cm² = 10/10000

= 0.001 cm²

Dimensions of wall:

Length = 14 m

Breadth = 10 m

Height = 6.5 m

Dimensions of windows

Length = 1.5 m

Breadth = 1 m

And,

Length of doors = 2.5 m

Breadth of doors = 1.2 m

Cost = ₹35 per m²

Now,

Area of four walls = 2(Length of walls × Height of walls) + 2(Breadth of walls × Height of walls)

= 2(14 × 6.5) + 2(10 × 6.5)

= 182 + 130

= 312

⇨ Area of four walls is 312 m²

Area of two doors = 2(Length of doors × Breadth of doors)

= 2(2.5 × 1.2)

= 6

⇨ Area of two doors is 6 m²

Area of four windows = 4(Length of windows × Breadth of windows)

= 4(1.5 × 1)

= 6

⇨ Area of four windows is 6 m²

Therefore,

Area to be painted = Area of 4 walls – (Area of 2 doors + Area of 4 windows)

= 312 – (6+ 6)

= 300

⇨ Area to be painted is 300 m²

Cost of painting = 300 m² × ₹ 35 per m²

=₹10500

Length of a wall = 12 m

Cost per meter = ₹30

Total cost = ₹ 7560

Cost per meter for floor = ₹ 25

Total cost for floor = ₹ 2700

Let h be the height.

Breadth = (area of the floor) / Length = 108/12 = 9m

Area of the floor = (total cost)/ (cost per meter) = 2700/25 = 108 m²

Again,

Area of walls = (total cost)/ (cost per meter) = 7560/30 = 252 m²

Now,

Area of 4 walls = 2(Length of walls × Height of walls) + 2(Breadth of walls × Height of walls)

252 = 2(12 × h) + 2(9 × h)

252 = 24h + 18h

252 = 42h

h = 6

⇨ Height is 6 m

Therefore dimensions of the room are : 12 m × 9 m × 6 m

Area of a square of side 6 m is equal to the area of 3,60,000 squares of each side 1 cm.

Let us assume the number of squares having side of 1 cm be ‘x’.

As per the condition in the question,

Area of the side 6 m square = Area of side 1 cm square

(6m)² = x(1cm)²

360000 cm² = x cm² … [because 1 m = 100 cm]

x = 360000cm²/1 cm²

x = 36000

(b) 12 squares with side 1m each

Area of square = side × side

Area of square of 1 m side = 1 × 1

= 1 m²

Therefore, area of 12 squares of 1 m side = 12 × 1

= 12 m²