ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Given,
(2a + b)/(a + 4b) = 3
2a + b = 3a + 12b
-a = 11b
a = -11b

∴ (a + b)/(a + 2b)
= (-11b + b)/(-11b + 2b)
= -10b/-9b
= 10/9.

x = 1 + √2 + √3 ...........(i)
y = 1 + √2 - √3 .............(ii)

x² + 4xy + y²/(x + y)
= {(x + y)² + 2xy}/(x + y)

From (i) + (ii)
x + y = 2 + 2√2
xy = (1 + √2)² - (√3)²
= 3 + 2√2 - 3
= 2√2

{(x + y)² + 2xy}/(x + y)
= {(2 + 2√2)² + (2 × 2√2)}/(2 + 2√2)
= (12 + 12√2)/(2 + 2√2)
= 12(1 + √2)/2((1 + √2)
= 12/2
= 6.

a² + b² = 5ab
a²/ab + b²/ab = 5
a/b + b/a = 5
Squaring the both sides
(a/b)² + (b/a)² = (5)²
a²/b² + b²/a² + 2 × (a/b) × (b/a)= 25
a²/b² + b²/a² + 2 = 25
a²/b² + b²/a² = 25 -2
a²/b² + b²/a² = 23.

x = a(b - c),
y = b(c - a),
z = c(a - b)

Let,
x/a = b - c = A
y/b = c - a = B
z/c = a - b = C

∴ A + B + C = b - c + c - a + a - b
= 0
A³ + B³ + C³ = 3ABC
= (x/a)³ + (y/b)³ + (z/c)³
= 3 × (x/a) × (y/b) × (z/c)
= 3xyz/abc

let x = 5 and y = 3

Then,
(8x - 5y) : (8x + 5y)
= {(8 × 5) - (5 ×3)} : {(8 × 5} + (5 × 3)}
= (40 - 15) : (40 + 15)
= 25 : 55
= 5 : 11.

If (x - 1) and (x + 1) are the factors y equation then,

x - 1 = 0
x = 1
Put x = 1 we get,
1 + a - 3 + 2 + b = 0
a + b = 0 ........(i)

x + 1 = 0
x = -1
Put x = -1 we get,
1 - a - 3 - 2 + b = 0
b - a = 4 ........(ii)

(i) - (ii) we get,
a + b - b + a = 0 - 4
2a = -4
a = -2
∴ b = 2
a, b = -2, 2.

According to math,
If,
x = y
Then, 1 - q = 2q + 1
⇒ 2q + q = 1 - 1
⇒ 3q = 0
⇒ q = 0.

Let,
x/2 = y/3 = z/4 = m
Then, x = 2m, y = 3m, z = 4m
∴ x/2 = (2x - 3y + 5z)/k = 2m/2
⇒ (2 × 2m - 3 × 3m + 5 × 4m)/k = m
⇒ k = 4 - 9 + 20
= 15.

Paint Area = Total area - Non-Paint area
Subtracting width of the border from all sides we get,
Length = 20 - 4 - 4 = 12
Breadth = 15 - 4 - 4 = 7

∴ Paint Area = (20 x 15) - (12 x 7) = 300 - 84 = 216

Total Cost of painting the border = Rate x Area
= 6 x 216
= Tk. 1296

Let length = x and breadth = y. Then,
2(x + y)= 46 or
x + y = 23 and
x² + y² = (17)²
= 289.

Now, (x + y)² = (23)²
⇒ (x² + y²) + 2xy = 529
⇒ 289 + 2xy = 529
⇒ xy = 120.

∴ Area = xy = 120 cm²

Let the length of the rectangle be x metre.
Then, a breath of the rectangle = (168/x)

∴ √{(x)² + (168/x)²} = 25
⇒ √{(x² + (28224/x²)} = 25
⇒ {(x² + (28224/x²)} = 625
⇒ x⁴ - 625x² + 28224 = 0
⇒ x²(x² - 576) - 49(x² - 576) = 0
⇒ (x² - 576)(x² - 49) = 0
⇒ x² = 576 or x² = 49
⇒ x = 24 or x = 7

Hence, length = 24 cm. and breadth = 7 m.

Area of the park = (60 x 40) m² = 2400 m²
Area of the lawn = 2109 m²
∴ Area of the crossroads = (2400 - 2109) m² = 291 m²

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 m

Area of the plot = (110 × 65)m²
= 7150m²
Area of the plot excluding the path = [(110 - 5) × (65 - 5)]m²
= 6300 m²

∴ Area of the path = (7150 - 6300) m²
= 850 m²

Cost of graveling the path = Tk. {850 × (80/100)}
= Tk. 680

The perimeter of the square = 160 m.
Side of square = (160/4) m
= 40 m.
Area of square = (40 × 40) m² = 1600 m²
Area of rectangle = (1600 - 100) m² = 1500 m²
Let the length and breadth of the rectangle be 'l' and 'b' respectively.
Then, 2(l + b) = 160
⇒ (l + b) = 80
⇒ b = 80 - l.

∴ lb = 1500
⇒ l(80 - l) = 1500
⇒ 80l - l² = 1500
⇒ l² - 80 l + 1500 = 0
⇒ (l - 50)(l - 30) = 0
⇒ l = 50. or l = 30

Hence, length = 50 m, breadth = 30 m.

Given that,
The area of the field = 680 sq. feet
⇒ lb = 680 sq. feet
Length(l) = 20 feet
⇒ 20 × b = 680
⇒ b = 680/20
= 34 feet

∴ Required length of the fencing = l + 2b
= 20 + (2 × 34)
= 88 feet

Let the breadth be x metre,
Then, length = 4x metre
∴ Volume of the room = (4x × 2x × x) m³
= (8x³) m³

8x³ = 512
x³ = 64
x = 4.

The length of the room is (4 × 4) = 16 m.

Let the common factor be K

∴ Length = 6K;
Breadth = 5K and
Height = 3K

We know,
Total Surface area of cuboid = 2(LB + BH + LH)
L = Length; B = Breadth/width; H = Height

Whole surface area of the rectangular plank = 2{(6K x 5K) + (5K x 3K) + (6K x 3K)}
∴ 504 = 126K²
∴ K = 2
∴ Breadth = 5K = 10 m

Let original edge = a,
Then, original volume = a³

New edge = (150/100)a
= 3a/2
New volume = (3a/2)³
= 27a³/8

Increase in volume = (27a³)/8 - (a³)
= 19a³/8

∴ Increase% = {(19a³/8) × (1/a³) × 100}%
= 237.5%

Volume of the block = (6 × 12 × 15) cm³
= 1080 cm³

Side of the largest cube = H.C.F of 6 cm, 12 cm, 15 cm
= 3 cm.

Volume of this cube = (3 × 3 × 3) cm³
= 27 cm³

Number of cubes = 1080/27
= 40.

External dimensions,
l = 50 cm,
b = 40 cm,
h = 23 cm

Internal dimension,
l' = 50 - (2 × 3) = 44 cm
b' = 40 - (2 × 3) = 34 cm
h' = 23 - 3 = 20 cm

The volume of the metal used in the box = External Volume - Internal Volume
= [( 50 × 40 × 23) - (44 × 34 × 20)] cm³
= 16080 cm³.

∴ Weight of the metal =
(16080 × 0.5)/1000 kg
= 8.04 kg

Total volume of the body :
= Volume of the cylinder + Volume of the cone + Volume of the hemisphere
= πr².r + (1/3)πr².r + (2/3)πr³
= 2πr³
= 2π × (2)³
= 16π.

Let BA be the ladder and AC be the wall as shown above.
Then the distance of the foot of the ladder from the wall = BC

Given that BA = 10 m , BAC = 60°
sin 60° = BC/BA
√3/2 = BC/10
BC = 10√3/2
= 5√3.
= 5 × 1.71
= 8.65 m.

Length of the tower AB = h meter.
∠DAC = ∠ACB = 60°
BC = 70 metre

In ABC,
tan 60° = AB/BC
⇒ √3 = h/70
⇒ h = 70√3 meter.

Let AB be the tower and C and D be the objects.
Then, AB = 150 m,
∠ACB = 45° and
∠ADB = 60°

AB/AD = tan 60° = √3
AD = AB/√3
= 150/√3 m.

AB/AC = tan 45° = 1
AC = AB = 150 m.

∴CD = (AC - AD)
= {150 - (150/√3)} m
= [{150(√3 - 1)/√3} × {(√3)/(√3)}] m
= 50(3 - √3) m
= (50 × 1.27) m
= 63.5 m.

Given, BC = 20 m
∠ACB = 30°

The total height of the telegraph post is (AB + CA) = ?
In ABC, tan 30° = AB/BC
1/√3 = AB 20

∴ AB = 20/√3m

Now, cos 30° = BC/AC
√3/2 = 20/AC
∴ AC = 40/√3 m

So, AB + CA
= (20/√3) + (40/√3)
= (60/√3)
= 20√3 m