Bank Math

Question: 49 × 49 × 49 × 49 = 7^?

Solution:
49 × 49 × 49 × 49 = 7^?
⇒  7² × 7² × 7² × 7² = 7^?
⇒ 7^{2 + 2 + 2 + 2} = 7^?
⇒ 7⁸ = 7^?
? = 8

Question: What should be the value of "P" so that the expression (9 − 24x + Px²) becomes a perfect square?

Solution:
(9 − 24x + Px²)
= (3)² − 2 × 3 × 4x + (4x)² + Px² - (4x)²
= (3 - 4x)² + Px² - 16x²

∴ the expression becomes a perfect square if,
Px² - 16x² = 0
⇒ Px² = 16x²
∴ P = 16

Question: A cube has a total surface area of 384 square units. What is the volume of the cube?

Solution:
Given, total surface area of the cube, S = 384 square units.
We know, surface area of a cube, S = 6a²

According to the question,
6a² = 384
⇒ a² = 384 / 6
⇒ a² = 64
⇒ a² = 8²
⇒ a = 8

Again, we know, volume of the cube, V = a³
= 8³
= 512

Therefore, the volume of the cube is 512 cubic units.

Speed = 45 km/hr = 45 × 5/18
= 25/2 m/s
Distance travelled = Length of the train + Length of the platform
= 360 + 140
= 500 metre.
Time taken to cross the platform = 500/(25/2)
= 40 seconds

Question: If (a - 18)² + (b - 12)² + (c - 6)² = 0 then, What is the value of (a + b + c)^1/2 = ?
​
​Solution:
​দেওয়া আছে,
​(a - 18)² + (b - 12)² + (c - 6)² = 0
​
​আমরা জানি, যেকোনো বর্গের যোগফল শূন্য হলে প্রতিটি বর্গই শূন্য হবে।
​সুতরাং,
​​(a - 18)² = 0
⇒ ​a - 18 = 0
∴ ​a = 18
​
​আবার,
​(b - 12)² = 0
⇒ ​​b - 12 = 0
∴ ​​b = 12
​
​এবং
​(c - 6)² = 0
⇒ ​​c - 6 = 0
∴ ​c = 6
​
​প্রদত্ত রাশি,
​ (a + b + c)^1/2
​= (18 + 12 + 6)^1/2
​= (36)^1/2
​= 6

Question: If r sinθ = 3, r cosθ = 4, then find the value of (4 tanθ + 1).

Solution:
r sinθ = 3
r cosθ = 4

Now,
(r sinθ)/(r cosθ) = 3/4
⇒ sinθ/cosθ = 3/4
⇒ tanθ = 3/4
⇒ 4 tanθ = 4 × 3/4 = 3
⇒ 4 tanθ + 1 = 3 + 1
∴ 4 tanθ + 1 = 4

Amount = Tk.(30000+4347) = Tk.34347
let the time be n year
Then,
30000(1+7/100)ⁿ = 34347
(107/100)ⁿ = 34347/30000 = 11449/10000 = (107/100)²
n = 2year

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.

Question: By what percentage must 40 be increased to become 70?

Solution: 
Let,
x% should be added.

Therefore,
40 + x% of 40 = 70
⇒ 40 + 40x/100 = 70
⇒ 40x/100 = 30
⇒ 2x/5 = 30
⇒ 2x = 150
⇒ x = 75

Therefore, 75% should be added to 40 to make it 50.

Question: One diagonal of a rhombus is three times the other diagonal. If its area is 54 sq. cm, find the sum of the diagonals.

Solution:
ধরি,
রম্বসের একটি কর্ণ = x সেমি
রম্বসের অপর কর্ণ = 3x সেমি

আমরা জানি,
রম্বসের ক্ষেত্রফল = (1/2) × কর্ণদ্বয়ের গুণফল

প্রশ্নমতে,
(1/2) . x . 3x = 54
⇒ 3x² = 54 × 2
⇒ 3x² = 108
⇒ x² = 108/3
⇒ x² = 36
⇒ x = √36
∴ x = 6

এখন,
একটি কর্ণ = 6 সেমি
∴ অপর কর্ণ = 3 × 6 = 18 সেমি

∴ কর্ণদ্বয়ের সমষ্টি = 6 + 18 = 24 cm

Question: In a class, there are 10 boys and 8 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is:

Solution: 
Total students = 10 + 8 = 18
Let S be the sample space, and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number of ways of selecting 3 students out of 18 = ¹⁸C₃
= (18 × 17 × 16)/(3 × 2 × 1)
= 816

And, n(E) = Number of events of selecting 1 girl and 2 boys = ⁸C₁ × ¹⁰C₂ 
= 8 × [(10 × 9)/2]
= 8 × 45
= 360

∴ Probability = n(E)/n(S)
= 360/816
= 15/34

Question: The least number by which 294 must be multiplied to make it a perfect square is :

Solution:
294 = 7 × 7 × 2 × 3

এখানে
2 এবং 3 জোড়াবিহীন
2 × 3 = 6 দ্বারা গুণ করলে 294 সংখ্যাটি পূর্ণবর্গ সংখ্যা হবে।

Let the total distance covered = x
= Length of journey
According to the question,
x/2 × 1/40 + x/2 × 1/60 = 16
x/80 + x/120 = 160
(3x + 2x)/240 = 16
5x = 16 × 240
x = (16 × 240)/5
x = 16 × 48
x = 768 km

Question: If cosA sinA = 0,then (cosA + sinA)² =?

Solution:
(cosA + sinA)²
= cos²A + 2 cosA sinA + sin²A
= 1 + 2.0 [sin²A + cos²A = 1]
= 1 + 0
= 1

Question: If x = 2y = 4z and xyz = 64, find the value of x.

Solution:
Given,
x = 2y = 4z
So, y = x / 2 and z = x / 4

Now,
xyz = 64
⇒ x × (x/2) × (x/4) = 64
⇒ x³/8 = 64
⇒ x³ = 64 × 8
⇒ x³ = 512
⇒ x = ∛512
∴ x = 8

Let the cost price = 100 units
Marked price = 140 units
Selling price = 140 × 85/100 = 119
Profit % = 19/100×100 = 19%

Question: Rana walked from A to B in the East 10 feet. Then she turned to the right and walked 3 feet. Again, she turned to the right and walked 14 feet. How far is she from A?

Solution: 



∴ Required Distance = AD
= √{3² + (14 - 10)²}
= √(3² + 4²)
= √(9 + 16)
= √25
= 5 ft

Question: An ambulance moving at 33 km/h is 45 m behind a school bus. After 15 s, it is 30 m ahead of the bus. Find the speed of the school bus.

Solution: 
Relative Speed = Total distance/Total time
= (30 + 45)/15
= 75/15 
= 5 m/s
= 5 × (18/5)
= 18 km/h

Now,
Relative Speed = Speed of ambulance - Speed of school bus
∴ Speed of school bus = Speed of ambulance - Relative speed
= 33 - 18
= 15 km/h

Question: The traffic lights at three different road crossings change after every 24 sec, 36 sec, and 72 sec respectively. If they all change simultaneously at 8 : 20 : 00 hrs, then they will again change simultaneously at:

Solution:

Time interval for simultaneous change = L.C.M. of 24, 36, 72 = 72 sec

72 seconds = 1 min 12 sec

Next simultaneous change = 8 : 20 : 00 + 1 min 12 sec = 8 : 21 : 12 hrs

Question: The sum of five consecutive multiples of 6 is 150. What is the second largest number?

Solution:
ধরি, ৬ এর পাঁচটি ক্রমিক গুণিতক হলো যথাক্রমে (x - 12), (x - 6), x, (x + 6) এবং (x + 12)

প্রশ্নমতে,
(x - 12) + (x - 6) + x + (x + 6) + (x + 12) = 150
⇒ 5x = 150
⇒ x = 150/5
⇒ x = 30

সুতরাং, সংখ্যাগুলো হলো 18, 24, 30, 36, 42।
এদের মধ্যে দ্বিতীয় বৃহত্তম সংখ্যাটি হলো 36।

Let the side of the square be a cm.
Then, its diagonal = √2 a cm.
Now, √2 a = 12√2
⇒ a = 12 cm.
Perimeter of square = 4a = 48 cm.
Perimeter of equilateral triangle = 48 cm.
Each side of the triangle = 16 cm.
Area of the triangle = ((√3/4)×16×16) cm²
= 64√3 cm²

Let C's age be x years.
Then, B's age = 2x years.
A's age = (2x + 2) years.

According to the question,
(2x + 2) + 2x + x = 27
5x = 25
⇒ x = 5
Hence, B's age = 2x = 10 years.

Question: What is the sum of the first 20 terms of the series 6, 11, 16, 21, ...?

Solution:
এটি একটি সমান্তর ধারা (arithmetic series)।
এখানে,
​প্রথম পদ, a = 6
সাধারণ অন্তর, d = 11 - 6 = 5
পদের সংখ্যা, n = 20

সমান্তর ধারার n-সংখ্যক পদের সমষ্টি সূত্র:
S_n = (n/2) [2a + (n - 1)d]

∴ প্রথম 20টি পদের সমষ্টি:
S₂₀ = (20/2) [2 × 6 + (20 - 1) × 5]
= 10 [12 + 19 × 5]
= 10 [12 + 95]
= 10 × 107
= 1070

∴ প্রথম 20টি পদের সমষ্টি = 1070

Question: A train is 100 meter long and is running at the speed of 30 km per hour. Find the time it will take to pass a man standing at a crossing.

Solution: 
Given that, 
Length of train, L = 100 meters
Speed of train, v = 30 km/h
= 30 × (1000/3600) m/s
= 30 × (5/18) m/s
= 150/18 m/s 
= 25/3 m/s

We know, 
Time taken = Distance/Speed
= 100/(25/3) m/s
= 100 × (3/25)
= 300/25
= 12 seconds

So the train takes 12 seconds to pass the man standing at the crossing.

Question: What is the angle between the hour and minute hands of a clock when it is 15 minutes past 3?

Solution:
15 minutes past 3 অর্থাৎ, ৩ টা 15 মিনিট।
= 3 + (15/60) ঘন্টা
= 3 + (1/4)
= 13/4 ঘন্টা

আমরা জানি,
ঘণ্টার কাঁটা 12 ঘণ্টায় 360° ঘোরে।
∴ 1 ঘণ্টায় ঘোরে = 360°/12 = 30°
∴ 13/4 ঘন্টায় ঘোরে = (30° × 13)/4 = 390°/4 = 97.5°

আবার,
মিনিটের কাঁটা 60 মিনিটে 360° ঘোরে।
∴ 1 মিনিটে ঘোরে = 360°/60 = 6°
∴ 15 মিনিটে ঘোরে = 15 × 6° = 90°

∴ ঘড়ির কাঁটা দুটির মধ্যবর্তী কোণ = | 97.5° - 90° |
= 7.5°

Let the son's present age be x years.
Then,
man's present age = (x + 24) years.
(x + 24) + 2 = 2(x + 2)
x + 26 = 2x + 4
x = 22.

Question: If 3 less than twice the number is equal to 2 more than 3 times the number, then 5 less than 5 times the number is:

Solution:
Let the number be x

ATQ,
2x - 3 = 3x + 2
∴ x = - 5

∴ Five times the number = 5 × (- 5)
= - 25
∴ Five less than this = - 25 - 5
= - 30

Question: The ratio of X : Y is 3 : 4, and the ratio of Y : Z is 8 : 9. If X = 18, what is the value of Z?

Solution:
দেওয়া আছে,
X : Y = 3 : 4
Y : Z = 8 : 9
এবং X = 18

এখন,
X : Y = 3 : 4
⇒ X/Y = 3/4
⇒ 3Y = 4X
⇒ Y = (4 × 18) / 3 ; [X = 18]
∴ Y = 24

আবার,
Y : Z = 8 : 9
⇒ Y/Z = 8/9
⇒ 8Z = 9Y
⇒ Z = (9 × 24) / 8
∴ Z = 27

সুতরাং, Z এর মান হলো 27

Question: If a right-angled isosceles triangle has base 4 cm, then height is:

5 Combine Banks (২০২২ সাল ভিত্তিক) Post Name: Officer Cash/Officer Teller (১০ম গ্রেড) Exam Date: 11.07.2025 Faculty of Business Studies (FBS), DU

Solution:
(Right-angled isosceles triangle) সমকোণী সমদ্বিবাহু ত্রিভুজ এর ভূমি = 4 cm.
সমকোণী সমদ্বিবাহু ত্রিভুজ এর ভূমি ও উচ্চতা সমান।
ভূমি = উচ্চতা = 4 cm.
∴ উচ্চতা = 4 cm

Question: A can finish a work in 15 days, and B can do it in 25 days. They work together for 5 days. What fraction of the work remains unfinished?

Solution:
Work done by A in 1 day = 1/15
Work done by B in 1 day = 1/25

Combined work in 1 day = 1/15 + 1/25
= (5 + 3) / 75
= 8/75

Work done in 5 days = 5 × (8/75) = 40/75 = 8/15

Fraction of work left = 1 - 8/15 = 7/15 

Question: A group of 10 boxes has their average weight increased by 4 kg after replacing a 60 kg box with a new one. What is the weight of the new box?

Solution:
Let the weight of the new box be x kg.
Let the total weight of the original 10 boxes = W
Original average = W/10

After replacing the 60 kg box with x,
The total weight becomes: W - 60 + x
The new average = (W - 60 + x) / 10

Accordingly:
New average = Old average + 4
(W - 60 + x) / 10 = W/10 + 4
⇒ W - 60 + x = W + 40
⇒ x = 40 + 60
⇒ x = 100

Question: Alam sold two vehicles for Tk. 46,000 each. If he gained 10% on the first and lost 10% on another, then what is his percentage profit or loss in this transaction?

Solution:
In the first case he gains 10%
∴ C. P. of the first car = (100 × 46000)/110 = 41,818.18

In the second case he loses 10%
∴ C. P. of the second car = (100 × 46000)/90 = 51,111.11

∴ Total cost of both cars = 41818.18 + 51111.11 = Tk 92,929.29
∴ Total S. P. of both cars = 2 × 46000 = Tk.92,000

Total loss = 92,929.29 - 92,000 = Tk. 929.29

%loss = (929.29/92,929.29) ×100 ​= 0.99% ≈ 1%

Let,
The required number of days be x.
Less Cows, More days[Indirect proportion]
Less Bags, Less days [Direct proportion]
Cows(1 : 40), bags(40 : 1)}::40 : x
∴ 1 × 40 × x = 40 × 1 × 40
= 40.

Given, ar⁰ = 1/2
Or, a = 1/2
And, ar⁵ = 1/64
Or, r⁵ = 1/32
Or, r⁵ = (1/2)⁵
∴ Ratio, r = 1/2

Given,
a - (1/a) = 2
On cubing both of the sides,
a³ - (1/a³) - 3 × a × (1/a) (a - 1/a) = 2³
⇒ a³ - (1/a³) - 3(a - 1/a) = 8
⇒ a³ - (1/a³) - 3 × 2 = 8
⇒ a³ - (1/a³) = 8 + 6
∴ a³ - (1/a³) = 14.