Bank Math

d = v × t
= (4 - 3) × 4
= 4 km

Question: A sum of Tk. 4200 is divided among P, Q, and R such that P gets 3/5 of what Q gets and Q gets 1/4 of what R gets. Q’s share is:

Solution:
Let,
R’s share = Tk. x
Then,
Q’s share = Tk. x/4
P’s share = Tk. (3/5) × (x/4) = Tk. 3x/20

∴ P + Q + R = 4200
⇒ 3x/20 + x/4 + x = 4200
⇒ (3x + 5x + 20x)/20 = 4200
⇒ 28x/20 = 4200
⇒ 28x = 4200 × 20
⇒ 28x = 84000
⇒ x = 84000/28
⇒ x = 3000

∴ Q’s share = x/4 = 3000/4 = 750 Tk.

Question: Pipe A fills a tank in 40 seconds, pipe B in 60 seconds, and pipe C empties it in 30 seconds. Initially, A and B are opened, and after 8 seconds, C is also opened. In how much more time will the tank be completely filled?

Solution:
Let the capacity of the tank be LCM of 40, 60, 30 = 120 units.

Efficiency of A = 120/40 = 3 units/second
Efficiency of B = 120/60 = 2 units/second
Efficiency of C = –120/30 = –4 units/second

(A + B open) for first 8 seconds,
Combined efficiency = 3 + 2 = 5 units/second
Work done in 8 seconds = 8 × 5 = 40 units
Remaining work = 120 – 40 = 80 units

Now when all three pipes open,
Combined efficiency = 3 + 2 – 4 = 1 unit/second
More time required = 80 ÷ 1 = 80 seconds

Question: A man deposits Tk. 600 in a bank at 10% interest rate compounded annually. At the end of the second year, what will be the total amount including interest?

Solution:
Given,
Principal, P = 600 Tk
Rate of interest, r = 10% = 10/100 = 1/10
Time, n = 2 years

We know,
Compound Amount = P(1 + r)ⁿ
= 600 × (1 + 1/10)²
= 600 × (11/10)²
= 600 × (11/10) × (11/10)
= 600 × 121/100
= 726 Tk

P's 1 day 1 hrs work = 1/96 part.
Q's 1 day 1 hr's work = 1/80 part.
P & Q together work in 1 hr per day = 1/96 + 1/80 = 11/480 part.
P & Q together completes in 8 hr per day = 11×8/480 = 11/60 part.
So, days required = 60/11 = 5(5/11)

Question: If a half kg of tomato costs 80 taka, how many taka will 300 gm cost?

Solution:
Let the required cost be P TK
Less weight : Less cost (Direct proportion)
500 : 300 : : 80 : P
⇒ 500/300 = 80/P
⇒5/3 = 80/P
⇒ P = (3 × 80)/5
∴ P = 48

ATQ, 14 - x = 10 + x
Or, 2x = 4
Or, x = 2

Area of the rectangle = 10 × 14 = 140 cm²
And, area of the square = (10 + 2)² = 12² = 144 cm²

So, its area increased by 144 - 140 or 4 cm²

Question: Find the value of n, if 27^{{n - (1/3)}} = 243.

Solution:
27^{{n - (1/3)}} = 243
⇒ (3³)^{{n - (1/3)}} = 3⁵
⇒ 3^{(3n - 1)} = 3⁵
⇒ 3n - 1 = 5
⇒ 3n = 5 + 1
⇒ 3n = 6
∴ n = 2

Question: A vendor purchased 200 books for Taka 150 each and sold them for Taka 200 each. Calculate the total profit.

Solution:
Cost Price of 200 books = 200 × Taka 150 = Taka 30,000
Selling Price of 200 books = 200 × Taka 200 = Taka 40,000
∴ Total Profit = Selling Price - Cost Price 
= Taka 40,000 - Taka 30,000
= Taka 10,000

Question:

Solution:
Given that, √{(0.85)² - (0.36)²}
Since, a² - b² = (a - b)(a + b)
= √{(0.85 + 0.36)(0.85 - 0.36)}
= √{(1.21)(0.49)}
= √{(1.1)(1.1) × (0.7)(0.7)}
= 1.1 × 0.7
= 0.77

Question: Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

Solution:
Total outcomes = {TTT, TTH,THT, HTT, THH, HTH, HHT, HHH} = 8
Favorable outcomes = {HHT, HTH, THH, HHH} = 4

So, the probability of getting at least 2 heads = Favorable outcomes/Total outcomes
= 4/8 = 1/2

1 child's 1 day's work = 1/192;
1 adults 1 day's work = 1/96.
Work done in 3 days = (1/96) × 16 × 3
= 1/2
Remaining work = {1 - (1/2)}
= 1/2.
(6 adults + 4 Children)'s 1 day's work = (6/96) + (4/192)
= 1/12
1/12 work is done by them in 1 days
1/2 work is done by them in (12 × (1/2)}
= 6 days.

Question: What will be the simple interest on Tk. 730 at 9% per annum for the period from January 12, 2026 to May 31, 2026?

Solution: 
January 12, 2026 to May 31, 2026 = 140 days

100 টাকার 365 দিনের মুনাফা = 9 টাকা 
1 টাকার 1 দিনের মুনাফা = 9/(365 × 100) টাকা 
730 টাকার 140  দিনের মুনাফা = (9 × 730 × 140)/(365 × 100) টাকা 
= 25.20

Question: The ratio of two numbers is 3 : 4 and their H.C.F is 6. Find their L.C.M.

Solution:
Let the two numbers be 3x and 4x.
∴ H.C.F = x = 6

∴ The two numbers are, 3 × 6 = 18 and 4 × 6 = 24

∴ Product of the two numbers = 18 × 24 = 432
And H.C.F = 6

We know,
L.C.M = (Product of two numbers)/H.C.F
= 432/6
= 72

∴ The L.C.M of the two numbers = 72.

= √(8/3)
= √(8×3)/(3×3)
= √(24/3)
=4.899/3
=1.633

Question: The perimeter of an equilateral triangle is 84√3 cm. Find its height.

Solution:
Given,
The perimeter of the equilateral triangle = 84√3 cm.

∴ Each side of the equilateral triangle
= (84√3)/3 
= 28√3 cm.

We know,
The height of the equilateral triangle = (a√3)/2

∴ The height of the equilateral triangle will be 
= (√3/2) × (28√3)
= 42 cm

Relative speed = (120 + 80) km/hr
= (200×5/18) m/s
= 500/9 m/s
Let, The length of other train be x m
Then,
x + 270 = 500/9 × 9
Or, x = 500 - 270 = 230

Question: The slant height of a right circular cone is 10 m, and its height is 8 m. Find the area of its curved surface.

Solution: 
Here, l = 10 and h = 8
r = √(l² - h²) 
= √(10² - 8²) 
= √(100 - 64) 
= √36
= 6

∴ Curved surface area = πrl 
= π × 6 × 10
= 60π sq. meter

Combined work done by Mamun and Tonmoy in 4 + 6 days (4 initial days and last 6 days)= 10/24 + 10/40
= (50 + 30)/120
= 80/120
= 2/3

∴ Remaining work = 1 - 2/3 = 1/3
Robin works for = (1/3) × 30 = 10.

Question: A can complete a work in 24 days and B in 16 days. They work together for 6 days. How many more days will A take alone to finish the remaining work?

Solution:
A একা কাজটি করতে পারে = 24 দিনে
∴ A এর একদিনের কাজ = 1/24 অংশ
এবং, 
   B একা কাজটি করতে পারে = 16 দিনে
∴ B এর একদিনের কাজ = 1/16 অংশ

∴ A ও B একসাথে একদিনের কাজ = (1/24) + (1/16) = (2 + 3)/48 = 5/48 অংশ
তারা 6 দিনে একসাথে কাজ করে = 6 × (5/48) = 5/8 অংশ

বাকি কাজ = 1 - (5/8) = 3/8 অংশ

অতএব,
A, 1/24 অংশ কাজ করে 1 দিনে 
∴ 3/8  অংশ কাজ করে = (24 × 3)/8 = 9 দিনে 

অতএব, A একা বাকি কাজ শেষ করতে ৯ দিন লাগবে।

Question: The area of a rectangle is equal to the area of a square with side x. If the length of the rectangle is 2x, find its breadth.

Solution:
Given that, 
A square with side x

We know,
Area of square = x²

ATQ,
Area of a rectangle is equal to the area of a square.
∴ Area of a rectangle = x2

And,
Given Length of the rectangle = 2x
Let the breadth of the rectangle = b

We know,
Area of rectangle = length × breadth
2x⋅b = x²
⇒ b = x²/2x
∴ b = x/2

Therefore, the breadth of the rectangle is x/2

Question: Find an equation of the horizontal line containing the point (3, 2).

Solution:
The equation of the horizontal line containing the point (3, 2) is y = 2. 
A horizontal line has a constant y-value for all points on the line.
Since the line must pass through the point (3, 2), its y-coordinate must be 2. 

Here S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting the multiple of 3
Then,
E = {3,6}
P(E) = n(E)/n(S)
= 2/6
= 1/3

Now,
Let the ratio of boys and girls be 7x ∶ 5x

According to the question,
(7x + 35) : (5x + 20) = 3 ∶ 2
⇒ (7x + 35)/(5x + 20) = 3/2
⇒ 14x + 70 = 15x + 60
⇒ 15x - 14x = 70 - 60 
∴ x = 10

∴ The number of girls now = (5 × 10) + 20 = 70

∴ The number of girls in the school now is 70.

Question: If x/a = 4, a/y = 6, a² = 9, and ab² = - 8, then x + 2y =?

Solution: 
Square rooting the given equation a² = 9 yields two solutions: a = 3 and a = - 3
In the equation ab² = - 8, b² is positive since the square of any nonzero number is positive.
Since ab² = - 8 is a negative number, a must be negative.
Hence, keep only negative solutions for a. Thus, we get a = - 3

Substituting this value of a in the given equation x/a = 4 yields
x/(- 3) = 4
∴ x = - 12

Substituting of a = - 3 in the given equation a/y = 6 yields
- 3/y = 6
∴ y = - 3/6 = - 1/2

Hence, x + 2y = - 12 + 2(- 1/2)
= - 12 - 1
= - 13 

A can complete work in 9 days.

So, percentage of work A completed in one day = 100/9 = 11.11%.

B can complete work in 12 days.

B's one day work = 100/12 = 8.33%.
A and B together can complete = 11.11 +8.33 = 19.44% of work in one day.

Now,

Take 2 days = 1 unit of time (one day of A and one of B).
In one unit of time A and B can complete work = 19.44% work.
Total time unit they need to complete whole work = 100/(19.44) = 5.14 time unit
Thus, Total time = 5.14 time unit = 5.14 × 2 = 10.28 days.

Question: If the n^th term of an arithmetic progression is 7n + 1, then what is the common difference?

Solution:
The nth term of an arithmetic progression is Tn = 7n + 1
n = 1 then, T₁ = 7 × 1 + 1 = 8
n = 2 then, T₂ = 7 × 2 + 1 = 15
n = 3 then, T₃ = 7 × 3 + 1 = 22
n = 4 then, T₄ = 7 × 4 + 1 = 29
............................

Common difference,
T₂ - T₁ = 15 - 8 = 7
T₄ - T₃ = 29 - 22 = 7

∴ The common difference is 7.

Question: If a sum of BDT 5,000 becomes BDT 6,728 in 2 years at compound interest, what is the rate of interest per annum?

Solution: Given, 
A = Amount after interest = BDT 6,728
P = Principal amount = BDT 5,000
r = Rate of interest per annum 
n = Time in years = 2

Thus, the rate of interest per annum is 16%.

Question: In a party, there is enough cake for 120 adults or 200 teenagers. If 150 teenagers have already eaten the cake, how many adults can be served with the remaining cake?

Solution:
মোট কেকের পরিমাণ = 200 জন কিশোর (Teenagers)
ইতিমধ্যে কেক খেয়েছে = 150 জন কিশোর
অবশিষ্ট কিশোরদের জন্য কেক আছে = 200 - 150 = 50 জনের

প্রশ্নমতে,
200 জন কিশোরের কেক = 120 জন প্রাপ্তবয়স্কের সমান
∴ 1 জন কিশোরের কেক = 120/200 জন প্রাপ্তবয়স্কের সমান
∴ 50 জন কিশোরের কেক = (120 × 50)/200 জন প্রাপ্তবয়স্কের সমান
= 30 জন

∴ অবশিষ্ট কেক দিয়ে আরও 30 জন প্রাপ্তবয়স্ককে পরিবেশন করা যাবে।

Speed of the boat in still water = 12 km/hr.
Speed of the stream = 4 km/hr.
Speed downstream = (12 + 4)
= 16 km/hr.
Time is taken to travel 68 km downstream
= 68/16
=17/4
= 4.25 hrs.

No. of boundaries = 6
No. of balls = 30
No. of balls without boundaries = 30 – 6 =24
Probability of no boundary = 24/30 = 4/5

Average speed = {(2 × 500 × 700)/(500 + 700)} km/hr
= (1750/3) km/hr
= 583(1/3) km/hr.
Hence, The average speed of the aeroplane for the entire journey is 583(1/3) km/hr

Question: In the first 10 overs of a cricket game, the run rate was only 3.2 runs per over. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?

Solution:
Given that, 
Target = 282 runs
Runs scored in first 10 overs = 10 × 3.2 = 32 runs
∴ Runs remaining = 282 - 32 = 250 runs

∴ Remaining overs = 50 - 10 = 40 overs

∴ Required run rate in the remaining 40 overs
= 250/40
= 6.25 runs per over

(B.D.×T.D)/(B.D.−T.D)
= Tk (72×60/ 72−60)
= Tk. (72×60/12)
= Tk.360