ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: The ratio of the length of a rod and its shadow is 1 : √3. The angle of elevation of the sun is:

Solution:
Let AB be rod and BC be its shadow
So, AB : BC = 1 : √3

Let θ be the angle of elevation

∴ tanθ = AB/BC = 1/√3 = tan30°

∴ θ = 30°

Question: If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.

Solution:
Given that arcs AXB and CYD of a circle are congruent,
i.e. arc AXB ≅ arc CYD.

We know that if two arcs of a circle are congruent, their corresponding chords are also equal.
i.e. chord AB = chord CD

Thus, AB/CD = 1

AB/CD = 1/1
∴ AB : CD = 1 : 1

Question: The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

Solution: 
We have,
(l - b) = 23 and
2(l + b) = 206
⇒ (l + b) = 103

Solving the two equations, we get:
l = 63 and b = 40

∴ Area = (l  x  b) = (63  x  40) m² = 2520 m²

Question: A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, then the altitude of the triangle is:

Solution: 
In this question, we will denote ‘b’ as the base h1 and h2 as the altitudes of the triangle and parallelogram respectively.

Then, according to the data in the question:

½ × b ×  h1 = b ×  h2
⇒ h1 = 2 h2
⇒ h1 = 2  × 100 = 200m

∴ The altitude of the triangle is 200m.

Question: If tanθ + cotθ = 5, then tan²θ + cot²θ is?

Solution:
Given,
tanθ + cotθ = 5
⇒ (tanθ + cotθ)² = 5² ( Squaring both sides)
⇒ tan²θ + cot²θ + 2tanθcotθ = 25
⇒ tan²θ + cot²θ = 25 - 2 [∵ tanθ . cotθ = 1]
∴ tan²θ + cot²θ = 23

Question: The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is 30°. The distance of the car from the base of the tower is:

Solution:
AB is a tower and AB = 75 m
From A, the angle of depression of a car C
on the ground is 30°


Let distance BC = x

Now, in ΔACB,
tanθ = AB/BC
⇒ tan30° = 75/x
⇒ 1/√3 = 75/x
⇒ x = 75√3
∴ BC = 75√3 m

Question: The ratio of the surface area of a sphere and the curved surface area of the cylinder circumscribing the sphere is:

Solution: 
Let the radius of the sphere be r
Then, the radius of the cylinder = r
Height of the cylinder = 2r
Surface area of sphere = 4πr²

Surface area of the cylinder = 2πr(2r) = 4πr²

∴ Required ratio 
= 4πr² : 4πr²
= 1 : 1

Question: If the area of the trapezium whose parallel sides are 6 cm and 10 cm is 32 sq. cm, then the distance between the parallel sides is:

Solution: 
Let the required distance be P cm
Then,
½ × (6 + 10) × P = 32
⇒ P = 4 cm

Question: A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

Solution: 
Here,
l = 20 ft and
lb = 680 sq. ft.

So, b = 34 ft.

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

Question: An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:

Solution: 

Let AB be the observer and CD tower
Draw BE perpendicular to CD

Then CE = AB = 1.6 m
And BE = AC =  20√3 m

Then right angle triangle DEB
∴ tan30° = DE/BE
⇒ 1/√3 = DE/20√3
⇒ DE = 20√3m

Then CD = CE + DE = 1.6 + 20 = 21.6 m

Question: The sides of a triangle are consecutive integers. The perimeter of the triangle is 120 cm. Find the length of the greatest side:

Solution:
Let the sides of the triangles be x cm, (x + 1) cm and (x + 2) cm respectively.
Then,
x + (x + 1) + (x + 2) = 120
⇒ 3x + 3 = 120
⇒ 3x = 117
⇒ x = 39

∴ Length of the greatest side
= (39 + 2) cm
= 41 cm

Question: Calculate the area of a rhombus if the length of its side is 2 cm and one of its angles A is 30 degrees.

Solution:
Given,
Side = s = 2 cm
Angle A = 30 degrees
Square of side = 2 × 2 = 4

Area, A = s² × sin (30°)
⇒ A = 4 × (1/2)
∴ A = 2 cm²

Question: A swimming pool 9 m wide and 12 m long and 1 m deep on the shallow side and 4 m deep on the deeper side. Its volume is:

Solution: 
Given, length width of the swimming pool is 9 m and 12 m respectively.

The volume of the swimming pool
= 9 × 12 × {(1 + 4)/2}
= 9 × 12 × (5/2)
= 270 m³

Question: If the volume and surface area of a sphere are numerically the same, then its radius is:

Solution: 
Let, the volume of a sphere = (4/3)πr³
the surface area of a sphere = 4πr²

∴ (4/3)πr³ = 4πr²
⇒ r = 3 units

Question: What is the value of tan240°?

Solution: 
tan240°
= tan(180° + 60°)
= tan(180° + θ)
= tanθ
= tan60°
=√3

Question: The length of a room is 5.5 m and the width is 3.75 m. Find the cost of paving the floor with slabs at the rate of Tk. 800 per square metre.

Solution:
Area of the floor
= (5.5 × 3.75)m²
= 20.625m²

∴ Cost of paying
= (800 × 20.625) Tk
=16500 Tk

Question: A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

Solution: 
Let the side of the square (ABCD) be x metres.
Then, AB + BC = 2x metres

AC = √2x = (1.41x) m
Saving on 2x metres = (0.59x) m

∴ Saving% = {(0.59x/2x).100%}
= 29.5%

Question: How many metres of carpet 63 cm wide will be required to cover the floor of a room 14 metres by 9 metres?

Solution:
Area of the floor = (14 × 9) m² = 126 m²

∴ Length of the carpet = {(126/63) × 100}m = 200 metres

Question: The circumcentre of a triangle ABC is 'O'. If ∠BAC = 85° and ∠BCA = 75°, then the value of ∠OAC is - 

Solution:
According to the question,
Given:
∠BAC = 85°
∠BCA = 75°
∠OAC = ?


∠ABC + ∠BCA + ∠CAB = 180°
⇒ ∠ABC + 75° + 85° = 180°
∴ ∠ABC = 20°

∠COA = 2 × ∠ABC
⇒ ∠COA = 2 × 20 = 40°

In ΔAOC,
We know OC = OA
∴ ∠OAC = ∠OCA
∴ ∠OAC + ∠OCA + ∠COA = 180°
⇒ 2∠OAC = 180° - 40°
⇒ 2∠OAC = 140°
∴ ∠OAC = 70°

Question: If cosθ.cosec23° = 1, the value of θ is:

Solution: 
Here,
cosθ.cosec23° = 1
⇒ 1/cos θ = secθ
⇒ cosec23° = cosec(90° – θ)
⇒ 23° = 90° – θ
⇒ θ = 90° – 23° = 67°

Alternative way,
cosθ.cosec23° = 1
If cosA.cosecB = 1

Then, A + B = 90°
So, θ + 23° = 90°
∴ θ = 67°

Question: The length of a tangent from point A at a distance of 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution: 
Let O be the centre of the circle and AB is the tangent.


OB ⊥ AB then ∆OAB is a right-angled triangle.

OA² = AB² + OB² (by Pythagoras Theorem)
⇒ 5² = 4² + OB²
⇒ OB² = 5² – 4²
⇒ OB² = 25 – 16
⇒ OB² = 9
⇒ OB = √9 = 3

Hence, the radius of the circle = 3 cm.