ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Let
breadth of a room (b) = x
Then height (h)= x/2
and length (l)= 2x

Now,
Volume of the room =2x × x × x/2 = x³

Volume of the room is given as 512 m³.
⇒ x³ = 512
⇒ x³ = 8³
⇒ x = 8
∴ Length = 2x = 2 × 8 = 16 m

Perimeter of Square = 44 cm 
⇒ Side of square is 11 cm
Area of Square = 11² = 121 cm²

Circumference of Circle = 44 cm
⇒ 2πr = 44
⇒ 2r(22/7) = 44
⇒ r = 7
⇒ Radius of circle is 7 cm
Area of circle =πr²=(22​/7) × 7² = 154 cm²

Circle has larger area by 33 cm² than the area of square.

The longest rod that can be placed in the cuboidal room
= Length of the diagonal 
= √(l² + b² + h^2​) 
= √{(12)² + (9)² + (8)²}
= √{144 + 81 + 64}
= √289
= 17

Let r be the radius of the circle 
Now
Area of circle = circumference of circle 
πr² = 2πr
r²​/r =2π​/π
r = 2 

Diameter of circle = 2r
                             = 2 × 2
                             = 4 units 

 

The angle of elevation of the top of a tower = 30°
The height of the tower = AB = 80√3m 
We have 
tan 30° = AB/PB
⇒ 1/√3 = 80√3/PB
⇒ PB =  80√3 × √3
⇒ PB = 80 × 3 = 240 m

∴ The distance (in m) of point P from the foot of the tower is 240 m

Let r be the radius of circle

Given that,
2πr - r = 74
⇒ r(2π - 1) = 74
⇒ r{(44/7) - 1} = 74 
⇒ r (44 - 7)/7 }=74
⇒ r(37/7) = 74
⇒ r = 74 (7/37)
    r = 14

Wire is in the shape of a cylinder. 
Let the radius of the original cylinder =r cm and height =h cm
Original volume =πr²h

New radius after decrease = r​/4

Let the new height be H.
Now 
πr²h = π(r/4)²H
h = H/16
16h = H

the height becomes 16 times.

Radius of plot = 42/2 ​= 21m
Radius of plot with path = 21 + 3.5 = 24.5m


Area of plot = (22/7)​{(24.5)² - (21)²}
                    = (22/7)​(600.25 - 441)
                    = (22/7)​ × 159.25
                    = 500.5m²

Cost of gravelling = Tk.(500.5 × 6) = Tk. 3,003

Let AB be the wall and AC be the ladder.
Then,
∠ACB =  60° and BC = 5.5 m.
⇒ cos 60° = BC/AC
⇒ 1/2 = BC/AC
⇒ AC = 2 × BC
⇒ AC = (2 × 5.5)m
 AC =11 m

We have
r=3cm
h=4cm

∴Volume=(1/3)πr²h   
        (1/3) × π × 3² × 4)cm³ = 12π cm³

Given that
length = m
Width = m/2

The area of the rectangle is = m(m/2) = m²/2

The ratio of the volume of two cubes = 8 : 27
Let
a and b the sides of the first cube and second cube respectively
Now 
 a³ : b³ = 8 : 27
⇒ a³ : b³ = 2³ : 3³
⇒ a/b = 2/3

 Surface area of cubes =6(side)²

6a²/6b² = a²/b²
              = 2²/3²
               = 4/9
               = 4 : 9

Volume of sphere = 36π cm³ 
Volume of sphere = (4/3) × πr³ 
⇒ (4/3) × πr³ = 36π
⇒ 36π = (4/3) × π × r³
⇒ 9 = (1/3) × r³
⇒ 27 = r³
⇒ r = 3

∴ The radius of sphere is 3 cm.

Radius of the circle r = 14/2 = 7
The circumference of the circle = 2πr = 2 × (22/7) × 7 = 44 cm

Area of rhombus =36cm²
First diagonal =d₁​=8 cm
Second diagonal = d₂​
Now,
Area of rhombus = (1/2)​ × d₁​ × d₂​
36 = (1/2)​ × d₁​ × d₂ 
36 =  (1/2)​ × 8​ × d₂ 
36 = 4 × d₂ 
d₂ = 36/4 
d₂ = 9 cm

The sides of the rectangle are 9 cm and 13 cm.
Perimeter of the rectangle =2(9 + 13)=44 cm
Circumference of circle =44 cm.

Here
2πr = 44
(22/7)r = 22
r/7 = 1
r = 7
Area of circle = πr² = (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm²

Let length and breadth be a and b respectively
⇒ The area before change = ab
The length after change = a + 10% of a
                                         = (a + 10a/100)
                                           = (10a + a)/10
                                           = 11a/10
The breadth after change = b - 10% of b   
                                         = (b - 10b/100)
                                         = b - b/10
                                         = (10b - b)/10
                                          =9b/10
⇒ The area after change = (11a/10)(9b/10)
⇒ The area after change = 99ab/100

⇒ Percentage change = [{ab - (99ab/100) }/ab] × 100%
                                    = (1/100) × 100%
                                     = 1% 
∴ The area of the new rectangle is decreased by 1%.

Let AB be the observer and CD tower
Draw BE perpendicular to CD
Then CE = AB = 1.6 m

And BE = AC = 25√3​​ m
Now 
∴tan 30°=DE/​BE
1/√3  =DE/25√3
DE = 25

CD = CE + DE = 1.6 + 25= 26.6 m

⇒ πr²₁/πr²₂ = 4/1
⇒ r²₁/r²₂ = 4/1
⇒ (r₁/r₂)² = (2/1)²
⇒r₁/r₂ = 2/1
     r₁ : r₂ = 2 : 1

√(l² + b²) = √41.
l² + b² = 41
Also, lb = 20.

(l + b)² = (l² + b²) + 2lb
(l + b)² = 41 + 40 = 81
(l + b) = 9.

 Perimeter = 2(l + b) = 18 cm.

Internal dimensions of the box are : 
Length =(14 - 2)cm =12cm
Breadth =(12 - 2)cm =10cm
Height =(10 - 2)= 8cm
∴ Inner surface area =2×(12 × 10 + 8 × 10 + 8 × 12)cm²
                                  = 2 × (120 + 80 + 96) cm² 
                                  = 592 cm² 

Here, r = 14 cm and h= 21 cm.
Total surface area of cylinder = 2πr(h + r)
                                                = 2π × 14 × (14 + 21)
                                                = 2 × π × 14 × 35
                                                = 980π cm²

Area of square = 576 cm²
Side of squared = √576
                          = 24 cm
Perimeter of square :
= 4a
= (4 × 24 cm
= 96 cm
= diameter of circle

∴ Circumference of circle :
           =πd =96×π=96π cm

Volume of water displaced
= (3.5 × 2 × 0.01) m³
 = 0.07 m³.
 
Mass of man
= Volume of water displaced × Density of water
= (0.07 × 1000) kg
= 70 kg.