ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Let original length = 10
original breadth = 10
Then, original area = 10 × 10 = 100
Length is halved
⇒ New length = 10/2
= 5
breadth is tripled.
⇒ New breadth = 10 × 3
= 30
New area = 5 × 30
= 150
Increase in area
= new area - original area
= 150 - 100
= 50
Percentage increase in area
= {(increase in area/original area) × 100}%
= {(50/100) × 100}%
= 50%.

In Rhombus
Let,
a = length of each side
b = base
h = height
d₁,d₂ are diagonals
Then Perimeter = 4a
= 2√(d₁² + d₂²)

Perimeter = 2√(72² + 30²)
= 156 cm.

In Parallelogram
Let,
b and care sides
b = base
h = height
area = base × height
= b × h
Therefore,
p² - 4 = (p + 4)(p - 3)
p² - 4 = p² + p - 12
p = 8
Hence, actual area
= (p² - 4) = 8² - 4
= 64 - 4
= 60 square units.

Length of the wire fencing
= perimeter
= 2(90+50)
= 280
Two poles are kept 5 metres apart.
Note that the poles are placed along the perimeter of the rectangular plot, not in a single straight line.
Hence, the number of poles required
= 280/5
= 56.

Consider a rectangular solid of length l, width w and height h. Then,
Total Surface Area
= 2lw + 2lh + 2wh
= 2(lw + lh + wh)
Volume = lwh
In this case, l = 25, w = 12 m and h = 6 m and all surfaces needs to be plastered except the top.
Hence, the total area to be plastered
total surface area - an area of the top face
= 2(lw + lh + wh) - lw
= lw + 2lh + 2hw
= (25 × 12) + 2 × (25 × 6) + 2 × (12 × 6)
= 300 + 300 + 144
= 744
Cost of plastering
= 744 × 75
= 55800 Paisa
= Tk. 558

Let, base = x
Then, height = 2x
Area = base × height
= x × 2x
= 2x²
Area is given as 72 cm²
2x² = 72 cm²
⇒ x² = 36 cm²
⇒ x = 6 cm
Hence, the length of the base is 6 cm.

Let original length = 10
original breadth = 10
Then, original area
= 10 × 10
= 100
Length is increased by 20%
⇒ New length = 10 + 2 (2 is 20% of 10)
= 12
Breadth is increased by 20%
⇒ New breadth = 10 + 2 (2 is 20% of 10)
= 12
New area = 12 × 12
= 144
Increase in are =
= new area - original area
= 144 - 100
= 44
Percentage increase in area
= (increase in area/original area) × 100
= {(44/100) × 100}%
= 44%

Let, the area of larger part = x hector
∴ area of smaller part = (700 - x) hectors
The difference between the areas of the two parts
= x - (700 - x)
= 2x - 700
One-fifth of the average of the two areas
(1/5) × (700/2) [total area = 700]
= 70
Given that difference of the areas of the two parts = one-fifth of the average of the two areas
2x−700 = 70
⇒ 2x = 770
⇒ x = 385.

∴ Smaller part of the land is = (700 - 385) = 315 hectares.

Consider the diagram is shown above where QR represents the tree and PQ represents its shadow.
We have, QR = PQ
Let, ∠QPR = θ
tanθ = QR/PQ
= QR/QR [since QR = PQ]
= 1
= tan 45°
⇒ θ = 45°
i.e., the required angle of elevation = 45°

SR = PQ = 2 m
PS = QR = 10√3
tan 30° = TS/PS
1/√3 = TS/10√3
TS = 10√3/√3
= 10
TR = TS + SR
= 10 + 2
= 12 m.

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m
∠ABD = 30°
∠ACD = 60°
Let CD = x, AD = h
From the right ⊿CDA
tan 60° = AD/CD
√3 = h/x ......eq(1)
From the right ⊿BDA
tan30° = AD/BD
1/√3 = h/(70 + x) ........ eq(2)
eq(1)/eq(2)
√3/(1/√3) = (h/x)/{h/(70+x)}
3 = 70 + x/x
2x = 70
x = 35
Substituting this value of x in eq:1, we have
√3 = h/35
h = 35√3
= 60.55 m

ধরি,
AB হচ্ছে দেয়াল এবং BC হচ্ছে মই।
এখানে ∠ACB = 60° এবং AC = 4.6 মিটার
তাহলে, AC/BC = cos 60° = 1/2 [ যেহেতু, cosθ = ভূমি/অতিভুজ]
⇒ BC = 2 × AC
∴ BC = 2 × 4.6
= 9.2m

ধরি,
ক্ষুদ্রতর বৃত্তের ব্যাসার্ধ r cm
∴ বৃহত্তর বৃত্তের ব্যাসার্ধ = (r + 14) cm
প্রশ্নমতে, Π(r + 14)²-Πr² = 1056
Π{r² + 2b × r × 14 + (14)² - r²} = 1056
r² + 28r + 196 - r² = 1056/Π
28r + 196 = 1056/(22/7)
= 1056 × (7/22)
= 336
28r = 336 - 196
= 140
∴ r = 140/28
= 5
অতএব, ক্ষুদ্রতর বৃত্তের ব্যাসার্ধ 5 cm.

দেওয়া আছে,
বৃত্তকার মাঠের ব্যাসার্ধ r₁ = 50 m
বৃত্তকার মাঠের পরিধি, 2πr₁ = 2 × π × 50
= 100π m.
সাইকেলের চাকার ব্যাসার্ধ, r₂ = 50 cm = 0.5 m.
∴ সাইকেলের চাকার পরিধি, 2πr₂ = 2 × π × 0.5
= π m.
এখন চাকাটি π মিটার গেলে ঘুরবে = 1 বার।
∴ চাকাটি 100π গেলে ঘুরবে = 1 × 100×π
= 100 বার।
এখন, 60 মিনিটে চাকাটি ঘুরবে = 100 বার।
∴ 15 মিনিটে চাকাটি ঘুরবে = (100 × 15)/60
= 25 বার।

দেওয়া আছে,
sin2(A + B + C) = 1
⇒ sin 2(A + B + C) = sin 90°
⇒ 2(A + B + C) = 90°
∴ (A + B + C) = 45°............(i)
আবার, tan (B + C - A) = √3
⇒ tan (B + C - A) = tan60°(tan 60° = √3)
B + C - A = 60°............(ii)
এখন, (i) ও (ii) নং সমীকরণ যোগ করে পাই,
(A + B - C) + (B + C - A) = 45° + 60°
⇒2B = 105°
⇒ B = 52(1/2)°

দেওয়া আছে,
sinA + sin²A = 1
⇒ sinA = 1 - sin²A
⇒ sinA = cos²A
এখন,
cos²A + Cos⁴A
= cos²A + cos²A.cos²A
= cos²A + sinA.sinA [sina = cos²A]
= Cos²A + sin²A
= 1[sin²A + cos²A = 1]

Let,
The edge of the third small cube be x cm
The volume of the cube = (edge)³
According to the question,
6³ + 8³ + x³ = 12³
⇒ 216 + 512 + x³ = 1728
⇒ x³ = 1728 - 728
= 1000
⇒ x = 1000^(1/3)
⇒ x = 10 cm.

Length of the parallel sides of prism = 10 cm and 6 cm
Height of prism = 8 cm
∴ Volume of prism = (1/2){(10 + 6) × 5 × 8}
= (1/2) × 16 × 5 × 8
= 320 cm³

(a - b)² = a² - 2ab + b²
⇒ 6² = (a² + b²) - 2ab
⇒ 6² = 116 - 2ab
⇒ 36 = 116 - 2ab
⇒ 2ab = 116 - 36
⇒ 2ab = 80
⇒ ab = 40.

Given,
a/b + b/a = 2
⇒ (a² + b²)/ab = 2
⇒ (a² + b²) = 2ab
⇒ a² + b² - 2ab = 0
⇒ (a - b)² = 0
⇒ a - b = 0.

Let the missing number be x.
Given,
20 + 8 × 0.5/(20 - x) = 12
⇒ (20 + 4)/(20 - x) = 12
⇒ 24/(20 - x) = 12
⇒ 20 - x = 24/12
⇒ 20 - x = 2
⇒ x = 20 - 2
⇒ x = 18.

As we know,
a³ + b³ + c³ - 3abc = a² + b² + c² - ab - bc - ca(a + b + c)
when (a + b + c) = 0
Then a³ + b³ + c³ - 3abc = 0
When x + y + z = 0
⇒ x³ + y³ + z³ = 3xyz
⇒ x³ + y³ + z³ + 3xyz = 3xyz +3xyz
= 6xyz.