DevSquare Quantitative Aptitude Areas, Shapes, Perimeter Questions

Let breadth = x metres.

Then, length = (x + 20) metres.

Perimeter = ( 5300/26.50 ) m = 200 m.

2[(x + 20) + x] = 200

2x + 20 = 100

2x = 80

x = 40.

Hence, length = x + 20 = 60 m.

We have: l = 20 ft and lb = 680 sq. ft.

So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Area to be plastered =

[2(l+b)×h]+(l×b)

=[2(25+12)×6]+(25×12)

=744sq.m

Cost of plastering = 744×75100 = Rs.558

Area of the park = (60 x 40) m2 = 2400 m².

Area of the lawn = 2109 m².

Area of the crossroads = (2400 - 2109) m² = 291 m².

Let the width of the road be x metres. Then,

60x + 40x - x² = 291

=>x² - 100x + 291 = 0

=>(x - 97)(x - 3) = 0

As, the width of the road cannot exceed the length of rectangular park,

So, x = 3.

Area covered by rabbit = Area of circle(radius 8m)-
Area of Triangle(side 5m)
= π*8*8 - (√(3)/4)*5*5
= 201.06 - 10.82
= 190.24 m²

total no. of points=15
no. of points needed to form a triangle=3
thus total no of triangles formed=^{15}C_{3}=455.

Given the diameter of golf ball = 4.1 cm

=> the radius of golf ball

= diameter/2 = 4.1/2cm

Now surface area of a golf ball without dimples

4 *π r² = 4 * π * (4.1/2)² = 16.81 π cm²

It is given that the shape of dimples is hemispherical and radius of each dimple = 2mm = 0.2 cm

Area of each dimple =

= 2 πr² = 2 * π * (0.2)² = 0.08 π cm²

So area of 150 dimples = 150 x area of each dimple = 150 x 0.8π = 12π cm square

Area of flat surface removed to make 1 dimple

=  πr² = π * (0.2 )² = 0.04 π cm²

Area of flat surface removed to make 150 dimples

= 150 * 0.04 π = 6 π cm²

Now total surface area of golf ball exposed to surroundings = surface area of golf ball without dimples + area of 150 dimples – area of flat surface removed to make 150 dimples

= 16.81 π + 12π - 6π

= 22.81π

= 22.81 x 3.14

= 71.62 cm²

Since the length of the rope is longer than the sides of the triangle it
will eat a bit more at the edges than (330/360)*144π. Hence more than
132π.

We have the rectangle with dimensions 10*11
(10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR
is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle):

^{10}C_{1}\times ^{11}C_{1};

Choose the x coordinate for vertex R (as y coordinate is fixed
by

A): ^{9}C_{1}, (10-1=9 as 1 horizontal dot is already occupied by A);

Choose the Y coordinate for vertex Q (as x coordinate is
fixed by A):

10C1, (11-1=10 as 1 vertical dot is already
occupied by A).

^{10}C_{1}\times ^{11}C_{1}\times ^{9}C_{1}\times ^{10}C_{1}=9900.

1:2
as traingle resides half of parallelogram