ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: The wheel of scooter has diameter 140 cm. How many revolutions per minute must the wheel make so that the speed of the scooter is kept at 132 km per hour?

Solution:
Distance travelled by wheel in one revolution = circumference of wheel
= 22/7 × 140 = 440 cm.

Speed of scooter = 132 km/hr = (132 × 1000 × 100)/60 cm/min
= 220,000 cm/min.

The wheel has therefore got to travel 220,000 cm in 1 min i.e. it has to perform 220,000/440 revolution in 1 min = 500 revolutions.

Question: ABCD is a rhombus. AB = 3x - 2, AC = 4x + 4, and BD = 2x. Find x.

Solution:
A rhombus is a quadrilateral with four sides of equal length. Rhombuses have diagonals that bisect each other at right angles.

Thus, we can consider the right triangle AED and use the Pythagorean Theorem to solve for x. From the problem:
AD = 3x - 2
BD = 2x
AC = 4x + 4

Because the diagonals bisect each other, we know:
ED = x
AE = 2x + 2

Using the Pythagorean Theorem,
a² + b² = c²
∴ AE² + ED² = AD²
⇒ (2x + 2)² + (x)² = (3x - 2)²
⇒ 4x² + 8x + 4 + x² = 9x² - 12x + 4
⇒ 0 = 4x² - 20x

Factoring,
0 = x and 0 = 4x - 20
The first solution is nonsensical for this problem. 
0 = 4x - 20
∴ x = 5

Question: The difference between the length and the perimeter of a rectangle is 100 cms. What is the breadth of the rectangle?

Solution:
Let the length of the rectangle be 'x' and breadth of the rectangle be 'y'
According to the question:
2(x + y) - x = 100
⇒ 2x + 2y - x = 100
⇒ x + 2y = 100

From this we cannot find 'y' (breadth), so the given data is inadequate.

Question: If the curved surface area of a sphere is same as the curved surface area of a hemisphere, find the radius of the hemisphere.

Solution:
Let,
R as the radius of the sphere.
r as the radius of the hemisphere.

Curved surface area of a sphere = Curved surface area of a hemisphere
4πR² = 2πr²
⇒ 2R² = r²
⇒ r = √2R

∴ The radius of the hemisphere = √2 times that of the sphere.

Question: Calculate sin(- 585°).

Solution:
sin(- 585°)
= - sin(585°)
= - sin(360° + 225°)
= - sin(2π + 225°)
= - sin225°
= - sin(180° + 45°)
= - sin(π + 45°)
= sin45°
= 1/√2

Question: In the figure AC and BC are radii of circles. The length of AB is 8. If AC = 4, what is BC? (BC is tangent to the circle with center A.)

Solution:
Since BC is tangent to circle with centre A
∴ BC is perpendicular to AC.
ΔABC is right angled triangle.
So,
BC = √(AB² - AC²)
= √(8² - 4²)
= √(64 - 16)
= √48
= √(16 × 3)
= 4√3

Question: If the length of the shorter diagonal is four, what is the length of the longer diagonal of this kite?

Solution:
We can find the longer diagonal by adding together the altitude of the top triangle and the altitude of the bottom triangle. To find these, use Pythagorean Theorem. We can use Pythagorean Theorem because one of the properties of a kite is that the two diagonals are perpendicular.

The top triangle has two sides of length 3 [labeled in the picture], and a base of 4 [provided in the written directions]. To figure out the altitude, split this triangle into 2 right triangles. The two legs are x [the altitude] and 2 [half of the base 4], and the hypotenuse is 3:
x² + 2² = 3²
⇒ x² + 4 = 9
⇒ x² = 5
∴ x = √5

We will do something similar for the bottom triangle. Consider one of the right triangles. It will have a hypotenuse of 7, one leg that we don't know, x [the altitude], and one leg 2 [half the shorter diagonal]. Set up the equation using the Pythagorean Theorem:
x² + 2² = 7²
⇒ x² + 4 = 49
⇒ x² = 45
∴ x = √45 = 3√5

∴ The length of the longer diagonal of this kite = 3√5 + √5 = 4√5

Question: If each side of the square is increased by 50%, what will be the ratio between the new area and the original area of the square?

Solution:
Let,
The side of original square is x
∴ The area of original square is x²

The side of new square is x + 50% of x = x + x/2 = 3x/2
∴ The area of new square is (9x²)/4

∴ The ratio between the new area and the original area of the square = (9x²)/4 : x²
= 9/4 : 1
= 9 : 4

Question: A copper sphere of radius 3 cm is beaten and drawn into a wire of diameter 0.2 cm. The length of the wire is-

Solution:
ব্যাসার্ধ, r = 3 সেমি
গোলকটির আয়তন = (4/3) × π × r³
= (4/3) × π × 3³
= 36π

তারটি ব্যাসার্ধ = 0.2/2 = 0.1 সেমি
তারটির আয়তন = πr²l
= π × (0.1)² × l
= 0.01πl

শর্তমতে,
0.01πl = 36π
⇒ l = 36/0.01
⇒ l = 3600 cm
⇒ l = 36 m

Question: cos(24°) + cos(5°) + cos(175°) + cos(204°) + cos(300°) =

Solution:
cos(175°) = cos(180° - 5°) = - cos(5°)
cos(204°) = cos(180° + 24°) = - cos(24°)
cos(300°) = cos(360° - 60°) = cos(60°)

∴ cos(24°) + cos(5°) + cos(175°) + cos(204°) + cos(300°)
= cos(24°) + cos(5°) - cos(5°) - cos(24°) + cos(60°)
= cos(60°)
= 1/2

Question: The tops of two poles are connected by a wire. The heights of the poles are 10 m and 14 m respectively. If the wire makes a 30° angle with the horizontal, find the length of the wire?

Solution:

Let AD and BE, be the poles of height 10 m and 14 m respectively.
DE is the wire of length = L
DC is parallel to AB so AD = BC = 10 m
So, CE = BE - BC = 14 - 10 = 4 m

In ΔDCE,
sin30° = CE/DE
⇒ 1/2 = 4/L
⇒ L = 8

Solution: A circle of radius 3 cm is drawn inscribed in a right angle triangle ABC, right angled at C. If AC is 10 Find the value of CB.

Solution:

(Tangents from the external point are equal)
(x + 7)² = 10² + (x + 3)²
⇒ x² + 14x + 49 = 100 + x² + 6x + 9
⇒ 8x = 60
∴ x = 7.5cm

∴ CB = x + 3 = 7.5 + 3 = 10.5 cm 

Question: Square ABCD with a perimeter of 48 units. Find length of BD.

Solution:
Each side of the square must be 48/4 = 12 units
∴ 12² + 12² = (BD)²
⇒ BD² = 2 × 144
∴ BD = 12√2

Question: The sides of a rectangular field are in the ratio 3 : 4 and its area is 7500 m². What is the cost of fencing it at Tk. 25 per meter?

Solution:
Ratio between the sides of rectangle = 3 : 4
Let the ratio constant be x then,
Length = 4x and breadth = 3x

ATQ,
7500 = 3x × 4x =12x²
⇒ x² = 625
∴ x = 25

Perimeter = 2(75 + 100) = 2 × 175 = 350 m
Cost of fencing 1 meter = Tk. 25
Cost of fencing 350 m = 350 × 25 = 8750

Question: The length of the longest rod that can be placed in a room 30 m long, 24 m broad and 18 m high, is-

Solution:
Length of room = 30 m
Breadth of room = 24 m
Height of room = 18 m

Length of the longest rod = Diagonal of the room = √(30² + 24² + 18²)
= √(900 + 576 + 324)
= √(1800)
= √(900 × 2)
= 30√2

Question: Calculate sin75°sin15° =

Solution:
sin75°sin15°
= sin(90°-15°).sin15°
= cos15°sin15°
= (1/2) × 2cos15°sin15°
= (1/2) × sin(2 × 15°)
= (1/2) × sin30°
= (1/2) × (1/2)
= 1/4

Question: The top and bottom of a flag on a building subtend angles of 60° and 30° respectively at a point B which is 48 meter away from the building. Find the height of the flag?

Solution:

Let height of building be AC = X and height of flag be CD = h.

In ΔDAB
tan60° = (X + h)/48
⇒ √3 = (X + h)/48
⇒ X + h = 48√3
∴ h = 48√3 - X ..................(1)

In ΔCAB
tan30° = X/48
⇒ 1/√3 = X/48
∴ X = 48/√3

From (1) we get,
h = 48√3 - 48/√3
= (48 × 3 - 48)/√3
= (144 - 48)/√3
= 96/√3
= (32 × 3)/√3
= 32√3

Question: In the given figure, AB is the diameter of the circle with center O. If ∠BOD = 15° & ∠EOA = 85°, then find the value of ∠ECA.

Solution:
∠EOA = 85°, ∠BOD = 15°
∠EOD = 180° - (85° + 15°) = 80°
In ΔOED,
OE = OD (radii)
∠OED = ∠ODE = 50°

In ΔOEC,
∠EOC = 80°+15° = 95°, ∠OEC =50°
∴ ∠ECA = 180°- (95 + 50°) = 35°

Question: In rectangle ABCD, diagonals = 36 unit. Find the value of x, where AE = 2x + 4y and CE = 4x - y.

Solution:
Diagonals = 36 unit
Each diagonal segment = 18.
2x + 4y = 18 ......(1)
and 4x - y = 18 .........(2)

From (1) + (2) × 4 we get,
2x + 4y + 16x - 4y = 18 + 72
⇒ 18x = 90
∴ x = 5

Question: If the size of a tile is 9" by 9", how many tiles are required to cover a 12 ft. wide and 18 ft. long floor?

Solution:
Side of the tile is 9" = 9/12 ft.
∴ Area of the tile is (9/12)² = 81/144 sq. ft.

Area of the floor = 12 × 18 = 216 sq. ft.

Number of tiles = 216/(81/144) = (216 × 144)/81 = 384

Question: The dimensions of a hall are 40 m, 25 m and 20 m. If each person requires 200 cubic meters, find the number of persons who can be accommodated in the hall.

Solution:
Length of the hall = 40 m
Breadth of hall= 25 m
Height of hall = 20 m
Volume of the hall = 40 × 25 × 20 = 20000 m³
Space occupied by each person = 200 m³
Number of person that can accommodate in the hall = 20000/200 = 100

Question: The angles of elevation of the top of a tower from the top and bottom of a tree of height 15 m are 30° and 60° respectively. Find the height of the tower?

Solution:

Let the CE be h meter.
Height of tree be AD = 15m
BE is the height of tower = BC + CE = 15 + h
AB = CD, let it is = X m

From ΔCDE, 
tan30° = EC/CD
⇒ 1/√3 = h/X
∴ X = √3h ..........(1)

From ΔABE, 
tan60° = EB/AB
⇒ √3 = (h + 15)/X
∴ X = (h + 15)/√3 ..............(2)

From (1) and (2) we get,
√3h = (h + 15)/√3
⇒ 3h = h + 15
⇒ 2h = 15
∴ h = 7.5

∴ Height of tower = 15 + 7.5 = 22.5 m

Question: What is the radius of a circle if its perimeter is numerically equal to thrice its area?

Solution:
Let,
The radius be r

ATQ,
2πr = 3πr²
⇒ 3r = 2
⇒ r = 2/3