ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: A room measures 7.5 m in length and 3.2 m in width. If the cost of paving is Tk. 750 per square metre, what is the total cost?

Solution:
Given that,
Length = 7.5 m
Width = 3.2 m
And rate = Tk. 750 per square metre

Now,
Area of the room is = Length × Width
= 7.5 × 3.2
= 24 m²

Now, Total cost = Area × Rate
= 24 × 750
= Tk. 18000

Question: In the figure, AOC is the diameter of the circle and arc AXB = (1/2)arc BYC. Find ∠BOC = ?

Solution:
Given that,
arc AXB = (1/2) arc BYC
∴ ∠AOB = (1/2) ∠BOC

We know that,
 ∠AOB + ∠BOC = 180º

Therefore,
(1/2) ∠BOC + ∠BOC = 180º {linear pair since AOC is the diameter}
⇒ (3/2) ∠BOC 180º
⇒ ∠BOC = (2/3) × 180º = 120º
∴  ∠BOC = 120º

Question: The volume V of a right circular cylinder is V = πr²h where r is the radius of the base and h is the height of the cylinder. If the volume of a right circular cylinder is 81π and its height is 9, what is the circumference of its base?

Solution: 
একটি সিলিন্ডারের উচ্চতা h একক ও ব্যাসার্ধ r একক হলে,
উক্ত সিলিন্ডারের আয়তন = πr²h ঘন একক
 
প্রশ্নমতে,
πr² × h = 81π
⇒ πr²  × 9 = 81π
⇒ r² = 9
∴r = 3
 
সুতরাং বৃত্তের পরিধি = 2πr = 2π × 3 = 6π

Question: If the radius of a sphere is 3r, what is its volume?

Solution:
Given that,
Radius of sphere = 3r

We know,
Volume of a sphere = (4/3) × πr³
= (4/3) × π(3r)³
= (4/3) × π × 27 × r³
= 36πr³ 

Question: A rectangular prism has dimensions 10 inches by 12 inches by 15 inches. A cylinder with a radius of inches 5 and a height of 14 inches is placed inside the prism. To the nearest cubic inch, what is the volume of the space in the prism not taken up by the cylinder?

Solution:
Given that,
Rectangular prism are 10 in × 12 in × 15 in
And
Cylinder inside prism radius, r = 5 in and height, h = 14 in

Now, 
Volume of the prism = length × width × height
= 10 × 12 × 15 = 1800 cubic inch

And, Volume of the cylinder = πr²h
= (22/7) × 5² × 14
= 22 × 25 × 2
= 1100 cubic inch

So the volume of the empty space = Volume of the prism - Volume of the cylinder
= 1800 - 1100 = 700 cubic inch

Question: The area of a rectangle and square are equal. The side of the square is 12 cm and the smaller side of the rectangle is one-third that of the square. The length of the other side of the rectangle would be-

Solution:
given that,
Side of the square = 12 cm
Smaller side of the rectangle = one-third of the square’s side = 12/3 = 4 cm
And The area of a rectangle and a square are equal.

Now,
Area of the square = 12² = 144  cm²

∴ Area of rectangle = 144cm² [The area of a rectangle and a square are equal]

Let the other side of rectangle = L
Now,
4 × L = 144
⇒ L = 144/4
∴ L = 36 cm

So the other side of the rectangle is 36 cm.

Question: A trapezium has parallel sides of length 15 m and 35 m. The distance between the sides is 12 m. Calculate the area of the trapezium.

Solution:
Given that,
Parallel sides of trapezium, a = 15 m and b = 35 m
Distance (height) between parallel sides, h = 12 m

We know,
Area of a trapezium =(1/2) × (sum of parallel sides) × height
= (1/2) × (a + b) × h
= (1/2) × (15 + 35) × 12 
= 50 × 6
= 300 m²

So the area of the trapezium is 300 m².

Question: The wheel of a scooter has a diameter of 140 cm. How many revolutions per minute must the wheel make to maintain a speed of 132 km/h?

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

And
Speed of scooter = 132 km/hr = (132 × 1000 × 100)/60 cm/min = 220000 cm/min.

∴ Revolutions per minute = Distance covered per minute/Distance per revolution
= 220000/440 = 500

So the answer is indeed 500 revolutions per minute.

Question: Find the maximum distance between two points on the perimeter of a rectangular garden whose length and breadth are 15 m and 8 m.

Solution:
Given that,
Length of rectangle, L = 15 m
Breadth of rectangle, B = 8 m 

Now, the maximum distance between two points on a rectangle is the diagonal of the rectangle.

∴ Diagonal = √(L² + B²)
=  √(15² + 8²)
= √(225 + 64)
= √289
= 17 m

So the maximum distance between two points on the perimeter is 17 m.

Question: If the difference between the circumference and diameter of a circle is 120 cm, then the diameter of the circle is -

Solution:
ধরি,
বৃত্তের ব্যাসার্ধ = r
বৃত্তের ব্যাস = 2r
বৃত্তের পরিধি = 2πr

প্রশ্নমতে,
2πr - 2r = 120
⇒ 2r(π - 1) = 120
⇒ r = (120/2){(22/7) - 1}
⇒ r = 60/(22 - 7)/7
⇒ r = (60 × 7)/15
∴ r = 28

∴ বৃত্তের ব্যাস = 2r = 2 × 28 = 56 সে.মি.

Question: A rectangular floor of dimensions 18 m × 12 m is to be covered with a carpet 60 cm wide. Calculate how many metres of carpet are required.

Solution:
Given that,
Floor dimensions = 18 m × 12 m
Carpet width = 60 cm = 0.6 m [1m = 100cm]

Now, Area of floor = length × breadth = 18 × 12 = 216m²

And,
Width of carpet = 0.6m Length of carpet required = L m 
Area covered by L m of carpet = 0.6 × L  m²

This must equal the area of the floor, 0.6 × L = 216
L = 216/0.6 = 360 m

So 360 metres of carpet will be required.

Question: Calculate the area of a rhombus if the length of its side is 4 cm and one of its angles A is 120 degrees.

Solution:
Given that,
Side of rhombus, a = 4 cm
And One angle, A = 120°

We know,
Area of a rhombus = a² × sin⁡A [Where a = side of rhombus, A = any interior angle.]
= 4²  × sin⁡120°
= 16 × (√3/2)
= 8√3

So the area of the rhombus is 8√3 cm²

Note:
sin(180∘ - θ) = sinθ,
So sin120° = sin⁡(180° - 60°) = sin⁡60° = √3/2

Question: A circular garden has a radius of 10 feet. If the radius is increased by 10%, what is the new area of the garden?

Solution:
Given that,
Original radius, r = 10 ft
And Increase radius by 10%

∴ New radius, r' = 10 + (10% of 10) = 10 + 1 = 11 ft.

We know, Area of circle = πr²

∴ New area with radius 11 ft
A′ = π(11²) = π × 121 = 121π ft²

Question: Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9), and D(5, 4). What is the shape of the quadrilateral?

Solution:
 
∴ The shape of the quadrilateral is Rhombus.

Question: The length of a rectangle is thrice its breath, and its perimeter is 120 meters. What is its area?

Solution:
Let the breath = x
So, the Length = 3x

Perimeter of a rectangle = 2 (Length + Breadth)
So, 2(3x + x) = 120
⇒ 6x + 2x = 120
⇒ 8x = 120
∴ x = 120/8 = 15

Now, Breadth = 15
so, length = 15 × 3 = 45

So, its area = Length × Breadth = 45 × 15 = 675 sq. m.