ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103
Solving the two equations, we get:
l = 63 and b = 40
∴ Area = (l x b) = (63 x 40) m² = 2520 m²

Number of bricks
=Courtyard area/1 brick area
=(2500×1600 / 20×10)=20000

1 hectare = 10,000 m²
So, Area = (1.5 x 10000) m² = 15000 m².
Depth =5/100 m=1/20m.
Volume = (Area x Depth) =15000 x1/20m³= 750 m^3.

In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So, 2(l+b) = 340
As we have to make 1 meter boundary around this,
so Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344
So required cost will be = 344 × 10 = 3440

l = (48 - 16)m = 32 m, [because 8+8 = 16]
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) m cube
= 5120 m cube.

Let the lengths of the line segments be x and x+2 cm
then,
(x+2)²−x²=32
x²+4x+4−x²=32
4x=28
x=7cm

Total volume of water displaced = (4 x 50) m³ = 200 m³.
Rise in water level =200/(40 x 20)m 0.25 m = 25 cm.

We know the product of diagonals is 1/2×(product of diagonals)
Let one diagonal be d₁ and d₂
So as per question
1/2×d₁×d₂=150
1/2×10×d₂=150
d₂=150/5 = 30

Let the thickness of the bottom be x cm.
Then, [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000
320 x 250 x (110 - x) = 8000 x 1000
(110 - x) =8000 x 1000/320 x 250= 100
 x = 10 cm = 1 dm.

Radius of incircle
=a/2√3
=42/ 2√3
=7√3
Area of incircle
=22/7×49×3
=462 cm²

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) cm².
Required number of tiles 
= (1517 x 902) / (41 x 41)= 814.

Number of bricks =Volume of the wall/ Volume of 1 brick
=800 x 600 x 22.5/25 x 11.25 x 6
= 6400.

Area of the park = (60 x 40) m² = 2400 m².
Area of the lawn = 2109 m².
Area of the crossroads = (2400 - 2109) m² = 291 m^2.
Let the width of the road be x metres. Then,
60x + 40x - x² = 291
x² - 100x + 291 = 0
(x - 97)(x - 3) = 0
x = 3.

We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Area to be plastered= [2(l + b) x h] + (l x b)
= {[2(25 + 12) x 6] + (25 x 12)} m²
= (444 + 300) m²
= 744 m².
Cost of plastering = Rs.744 x (75/100)
= Rs. 558

Clearly, l = (48 - 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) m³ = 5120 m³.

√3 এর বদলে 3 দেয়া আছে। উত্তর বাতিল করা হয়েছে।
√3 এর ক্ষেত্রে সমাধান এরকম হবে -

Let AB be the tree and AC be its shadow.
Let Angle ACB =θ
Then,AC/AB = √3
cot θ = √3
θ = 30°.

Let AB be the tower
then∠ACB=30º
AB=100 meter
AB/AC= tan30º
=>100/AC= 1/√3
=>AC= √3/100
=>AC= 1.73×100
=>AC= 173meter

Let AB be the wall and BC be the ladder.
Then, ∠ACB = 60° = AC = 4.6m
AC/BC=cos⁡60∘=1/2
⇒ BC = 2 × AC = 2 × 4.6 = 9.2m

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.
cos⁡60∘=PQ/PR
1/2=12.4/PR
PR=2×12.4=24.8m

100 cm is read as 102 cm.
A₁ = (100 x 100) cm² and A₂ (102 x 102) cm².
(A₂ - A₁) = [(102)² - (100)²]
= (102 + 100) x (102 - 100)
= 404 cm².
Percentage error =404/(100 x 100) x 100%= 4.04%