ব্যাংক নিয়োগ বিষয়ভিত্তিক প্রস্তুতি

Question: Find the equation of the line passing through (2, -3) and parallel to 5x - 2y + 6 = 0.

Solution:
Slope of given line: 
5x - 2y + 6 = 0 ---------(1)
⇒ y = (5/2)x + 3
∴ slope, m = 5/2

If a line has slope = m and passes through a point (x₁ , y₁).
Then the equation of the line is -
∴ y – y₁ = m(x – x₁)

Now, parallel to equation (1), has the same slope and pass through (2, -3).
So, the required line,
y + 3 = (5/2)(x - 2)
⇒ 2y + 6 = 5x - 10
⇒ 5x - 2y - 16 = 0

∴ the required line 5x - 2y - 16 = 0.

Question: The areas of a square and a rhombus are equal. The diagonals of the rhombus are 6 meters and 8 meters, respectively. What is the length of one side of the square?

Solution:
The area of the rhombus = (1/2) × Product of the diagonals
= (1/2) × 6 × 8
= 24 square meters

The area of the square = 24 square meters.
∴ Length of one side of the square = √24 meters
= 2√6 meters

∴ the length of one side of the square is 2√6 meters.

Question: The distance between the points (4, 3) and (1, 7) is -

Solution:
Given points (4, 3) and (1, 7).
Now the formula for the distance between (x₁, y₁) and (x_2, y₂) is
= √[(x₂ - x₁)² + (y₂ - y₁)²]

∴ The distance between the points (4, 3) and (1, 7) is
= √[(1 - 4)² + (7 - 3)²]
= √[9 + 16]
= √25
= 5

Question: From the top of a lighthouse 60 m high above sea level, the angle of depression of a boat is 45°. How far is the boat from the foot of the lighthouse?

Solution:

Let the height of the lighthouse above sea be AC and it is given 60 m.
Angle of depression = 45°

Boat is at point B so the distance between the base of lighthouse A and Boat is AB.

 So, tan 45° = AC / AB
⇒ 1 = 60 / AB
⇒ AB = 60 m

∴ The boat is 60 m away from the foot of the lighthouse.

Question: The midpoint of the line joining (10, 2) and (4, 8) is -

Solution:
The formula for the midpoint of (x₁, y₁) and (x₂, y₂) is
= ((x₁ + x₂)/2 , (y₁ + y₂)/2)

∴ The midpoint of the line joining (10, 2) and (4, 8) is
= (14/2 , 10/2)
= (7, 5)

Question: A chord of length 18 cm is drawn in a circle of radius 12 cm. The distance of the chord from the center of the circle is -

Solution:
Given,
r = 12 cm
and c = 18 cm.
Half of the chord length = 18/2 = 9 cm

∴ Distance = √(12² - 9²)
= √(144 - 81) 
= √63 cm

The chord's distance from the circle's center is √63 cm.

Question: The perimeter of an equilateral triangle is 84√3 cm. Find its height.

Solution:
Given,
The perimeter of the equilateral triangle = 84√3 cm.

∴ Each side of the equilateral triangle
= (84√3)/3 
= 28√3 cm.

We know,
The height of the equilateral triangle = (a√3)/2

∴ The height of the equilateral triangle will be 
= (√3/2) × (28√3)
= 42 cm

Question: The radius of a wheel is 14 cm. How many revolutions will it make in travelling 88 kilometers?

Solution:
আমরা জানি,
চাকার পরিধি = 2πr = 2 × (22/7)​ × 14 = 88 সে. মি.

∴ মোট দূরত্ব = 88 কি. মি.
= 88 × 1000 × 100
= 8800000 সে. মি.

∴ ঘূর্ণন সংখ্যা = 8800000/88​ = 100000 টি

Question: If tan(3x - 15°) = cot(3y + 15°), then (x + y) is:

Solution:
tan(3x - 15°) = cot(3y + 15°)
⇒ tan (3x - 15°) = tan {90° - (3y + 15°)}
⇒ 3x - 15° = 75° - 3y
⇒ 3x + 3y = 75° + 15°
⇒ 3(x + y) = 90°
⇒ x + y = 30°

∴ x + y = 30°

Question: A square has an area of 25 m². If a circle has a radius equal to the length of the square’s diagonal, what is the area of the circle?

Solution:
Area of square = 25
Side of square = √25 = 5

Diagonal of square = 5√2
So, the radius of the circle is 5√2 m

Area of circle = πr²
= π(5√2)²
= 50π m²

∴ The area of the circle is 50π sq. m.

Question: A lamp post 18 meters tall broke in such a way that the broken part makes a 30-degree angle with the ground. At what height did the lamp post break?

Solution:


Let,
height from ground (A) to broken part (C) = h
rest = 18 - h
The broken part makes a 30-degree angle with the ground at B.
It creates a triangle ABC where,
BC = 18 - h
AC = h

Now,
sin 30° = AC/BC
⇒ 1/2 = h/(18 - h)
⇒ 2h = 18 - h
⇒ 3h = 18
∴ h = 6

∴ the lamp post broke at 6 m height from the ground.

Question: The area of a trapezium is 72 square cm. The lengths of its parallel sides are 12 cm and 6 cm. What is the distance between the parallel sides?

Solution:
We know,
The area of a trapezium = (1/2) × Sum of the lengths of the parallel sides × Distance between the parallel sides.

Let, the distance between the parallel sides be d.
Then,
d = (2 × Area of the trapezium)/Sum of the lengths of the parallel sides
= (2 × 72)/(12 + 6)
= 144/18
= 8 cm

Thus, the distance between the parallel sides 8 cm.

Question: The base of a right-angled triangle is 6 m and the hypotenuse is 10 m. What is its area?

Solution:
Given,
Base = 6 m, Hypotenuse = 10 m

By Pythagoras' Theorem
Height² = Hypotenuse² - Base²
⇒ Height²= 10² - 6²
⇒ Height²= 100 - 36
⇒ Height² = 64
∴ Height = 8

We know,
Area = (1/2) × base × height
= (1/2) × 6 × 8
= 24 sq. meters

Question: A wire can be bent in the form of a circle of radius 21 cm. If it is bent in the form of a square, then its area will be -

Solution:
Given,
radius of the circle r = 21 cm

Circumference of the circle = 2πr 
 = 2 × (22/7) × 21 
= 2 × 22 × 3
= 132 cm 

The length of one side of the square = 132/4 = 33 cm

Area of the ​​square = (33)² cm²
= 1089 cm²

Question: If a 20-meter tall pole creates a shadow of length 20√3 meters, what is the angle of elevation of the sun?

Solution:
 
খুঁটির উচ্চতা AB = 20 m
খুঁটির ছায়ার দৈর্ঘ্য BC =20√3 m

ΔABC হতে পাই,
tanθ = লম্ব/ভূমি
বা, tanθ = AB/BC
বা, tanθ = 20/(20√3)
বা, tanθ = 1/√3
বা, tanθ = tan30°
∴ θ = 30°

Question: If the area of a square is 529 square meters, what is the perimeter of the square? 

Solution:
Given,
The area of the square = 529 square meters.

Therefore,
The length of one side of the square = √529 meters = 23 meters.

We know,
The perimeter of a square = 4 × length of one side
= 23 × 4 meters
= 92 meters

Thus, the perimeter of the square is 92 meters.

Question: If sin 17° = (x/y), then sec 17° is equal to

Solution:
Given,
sin 17° = (x/y)

We know,
sin² θ + cos² θ = 1
∴ cos θ = √(1 - sin θ)

∴ cos 17° = √(1 - sin²17°)
= √{1 - (x²/y²)}
= √{(y² - x²)/y²}
= {√(y² - x²)}/y

∴ sec 17° = 1/cos 17°
= 1/[{√(y² - x²)}/y]
= y/{√(y² - x²)}

Question: What is the value of cos 120°?

Solution: 
cos 120°
= cos(90° + 30°)
= - sin 30°
= - 1/2

Question: If each side of the square is increased by 20%, 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 + 20% of x
= x + (x/5)
= 6x/5

∴ The area of new square is
= (6x/5)²
= (36x²)/25

∴ The ratio between the new area and the original area of the square = {(36x²)/25} : x²
= 36/25 : 1
= 36 : 25

Question: If 0 ≤ θ ≤ 90° and 4 cos²θ - 4√3 cosθ + 3 = 0 then the value of θ is

Solution:
4 cos²θ – 4√3 cosθ + 3 = 0
⇒ (2cosθ)² – 2 · 2 cosθ · √3 + (√3)² = 0
⇒ (2cosθ – √3)² = 0
⇒ 2 cosθ – √3 = 0
⇒ 2 cosθ = √3
⇒ cosθ = (√3)/2
⇒ cosθ = cos 30°
∴ θ = 30°

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 a right-angled triangle.

So,
BC = √(AB² - AC²)
= √(8² - 4²)
= √(64 - 16)
= √48
= √(16 × 3)
= 4√3

Question: The area of a square is equal to the area of a parallelogram. If the base of the parallelogram is 45 meters and its height is 5 meters, what is the length of one side of the square?

Solution:
Area of the parallelogram = Base × Height
= 45 × 5
= 225 square meters

Let,
Side of the square = k meters
∴ Area of the square = k² square meters

ATQ,
k² = 225
⇒ k = √225
∴ k = 15
Therefore, the length of one side of the square = 15 meters

Question: A hall measures 40 m in length, 25 m in width, and 20 m in height. If each person needs 200 cubic meters of space, how many people can the hall accommodate?

Solution:
Length of the hall = 40 m
Width 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 people that can be accommodated in the hall
= 20000/200
= 100

Question: A rectangular field has a perimeter of 110 meters. The length of the field is 5 meters less than three times its width. Find the area of the field in square meters.

Solution:
ধরি, আয়তাকার ক্ষেত্রটির প্রস্থ = x মিটার
সুতরাং, ক্ষেত্রটির দৈর্ঘ্য = 3x - 5 মিটার

আয়তক্ষেত্রের পরিসীমা = 2(দৈর্ঘ্য + প্রস্থ)
প্রশ্নমতে,
2((3x - 5) + x) = 110
⇒ 2(4x - 5) = 110
⇒ 4x - 5 = 55
⇒ 4x = 60
⇒ x = 15 মিটার

সুতরাং,
প্রস্থ = 15 মিটার।
দৈর্ঘ্য = 3x - 5
= (3 × 15) - 5 
= 45 - 5
= 40 মিটার।

আয়তক্ষেত্রের ক্ষেত্রফল = দৈর্ঘ্য × প্রস্থ
∴ ক্ষেত্রফল = 40 × 15 
= 600 বর্গ মিটার।

সুতরাং, ক্ষেত্রটির ক্ষেত্রফল হলো 600 বর্গ মিটার।

Question: The volume of a right circular cylinder is 25π cubic units, and its height is 4 units. What is the circumference of its base?

Solution:
আমরা জানি, একটি সিলিন্ডারের আয়তন = πr²h
যেখানে, r হলো ভূমির ব্যাসার্ধ এবং h হলো উচ্চতা।

প্রশ্নমতে,
πr² × 4 = 25π
⇒ 4r² = 25
⇒ r² = 25/4
⇒ r = √(25/4)
⇒ r = 5/2 = 2.5 একক

সিলিন্ডারের ভূমির পরিধি = 2πr
= 2π × 2.5
= 5π একক

∴ সিলিন্ডারটির ভূমির পরিধি হলো 5π একক।