Bank Math

Question: What will be the difference between simple and compound interest at 10% on a sum of Tk. 5000 after 3 years?

Solution:
Given that,
Principal, P = Tk. 5000
Rate of interest, r = 10%
Time, n = 3 years

Simple Interest (SI):
SI = (P × r × n) / 100
⇒ SI = (5000 × 10 × 3) / 100
⇒ SI = 150000 / 100
⇒ SI = Tk. 1500

Compound Interest (CI):
CI = P × (1 + r/100)ⁿ 
= 5000 × (1 + 10/100)³
= 5000 × (1.10)³
= 5000 × 1.331 
= 6655

CI = A - P
= 6655 - 5000 = Tk. 1655

∴ Difference between CI and SI:
= CI - SI
= 1655 - 1500
= Tk. 155

Time = 2 year 4 months = 2(4/12) year = 2(1/3) year.
Amount = Tk'. [8000 X (1+(15/100))² X (1+((1/3)×15)/100)]
=Tk. [8000 ×(23/20) × (23/20) ×(21/20)]
= Tk. 11109.
:. C.I. = Tk. (11109 - 8000)
= Tk. 3109.

x + y + z can do 1/6 part of the job in one hour……(i)
x + y can do 1/8 part of the job in one hour……(ii)
and, y + z can do 1/12 part of the job in one hour……(iii)
(ii) + (iii),
x + y + y + z = 1/8 + 1/12
Or, x + z + 2y = 5/24 ........ (iv)
Subtracting (i) from (iv) we get,
y = 5/24 - 1/6 = 1/24
Substituting the value of y in (iv) we get,
x + z + 2/24 = 5/24
x + z = 5/24 - 2/24 = 3/24 = 1/8
Hence x and z will do 1/8 of the job in one hour
So, x and z will do all of the work in 8 hours

Question: There were 800 students in a school in 2025. In 2026, 10% of the male students left the school, while the number of female students increased by 40%. If the total number of students remained unchanged, how many female students were there in the school in 2025?

Solution:
ধরি, 2025 সালে ছাত্রীর সংখ্যা ছিল x জন।
এবং ছাত্রের সংখ্যা ছিল (800 - x) জন।

প্রশ্ন অনুযায়ী, 2026 সালে ছাত্রীর সংখ্যা 40% বৃদ্ধি পায় এবং ছাত্রের সংখ্যা 10% হ্রাস পায়, কিন্তু মোট ছাত্র-ছাত্রীর সংখ্যা 800-ই থাকে।
অর্থাৎ, ছাত্রীর বৃদ্ধির পরিমাণ = ছাত্রের হ্রাসের পরিমাণ।

প্রশ্নমতে,
x এর 40% = (800 - x) এর 10%
⇒ 40x/100 = 10(800 - x)/100
⇒ 40x = 8000 - 10x
⇒ 40x + 10x = 8000
⇒ 50x = 8000
⇒ x = 8000/50
⇒ x = 160

∴ 2025 সালে স্কুলে ছাত্রীর সংখ্যা ছিল 160 জন।

(1500 x R1 x 3)/100 
=> 4500 (R1-R2) = 1350 
=> (R1-R2)= 1350/4500 = 0.3 %

Question: An aeroplane covers a certain distance at a speed of 250 kmph in 4 hours. To cover the same distance in hours, it must travel at a speed of:

Solution:
বিমানটির অতিক্রান্ত  মোট দূরত্ব = গতিবেগ × সময়
= 250 কিমি/ঘন্টা × 4 ঘন্টা
= 1000 কিমি

এখন, একই দূরত্ব ঘন্টায় অতিক্রম করার জন্য প্রয়োজনীয় গতিবেগ নির্ণয় করতে হবে।

ঘন্টা = (1 + 1/4) ঘন্টা = 5/4 ঘন্টা

প্রয়োজনীয় গতিবেগ = মোট দূরত্ব/সময়
= 1000 কিমি/(5/4) ঘন্টা
= (1000 × 4/5) কিমি/ঘন্টা
= 200 × 4 কিমি/ঘন্টা
= 800 কিমি/ঘন্টা

সুতরাং, একই দূরত্ব 5/4 ঘন্টায় অতিক্রম করার জন্য বিমানটিকে 800 কিমি/ঘন্টা গতিবেগে চলতে হবে।

Time taken by Javed =100/{18 × (5/18)} = 20 seconds
Time taken by Naveed = 20 + 5 = 25 seconds
Speed of Naveed = 100/25 × 18/5 = 14.4 km/h

Question: In a business, the ratio of the capitals of Arun and Babul is 2 : 1, that of Babul and Chandan is 4 : 3, and that of Dipu and Chandan is 6 : 5. What is the ratio of the capitals of Arun and Dipu?

Solution:
Given,
Arun : Babul = 2 : 1
⇒ Arun/Babul = 2/1

Babul : Chandan = 4 : 3
⇒ Babul/Chandan = 4/3

Dipu : Chandan = 6 : 5
⇒ Chandan/Dipu = 5/6

Now,
Arun/Dipu = (Arun/Babul) × (Babul/Chandan) × (Chandan/D)
= (2/1) × (4/3) × (5/6)
= 20/9

∴ Arun/Dipu = 20 : 9

Question: What is the slope of a line perpendicular to the line whose equation is 3x + 4y = 12?

Solution:
প্রদত্ত সরল রেখার সমীকরণ: 3x + 4y = 12
y = mx + c আকারে লিখি, যেখানে m হলো রেখার ঢাল।

4y = - 3x + 12
⇒ y = (- 3/4)x + 3

অতএব, মূল রেখার ঢাল (m) = - 3/4

আমরা জানি, কোনো রেখার উপর লম্ব রেখার ঢাল m₁ = - 1/m
= - 1/(- 3/4)
= 4/3

∴ লম্ব রেখার ঢাল = 4/3

দেওয়া আছে,
log_x(1/9) = - 2
বা, 1/9 = x - 2
বা,  1/3²  = 1/x²
বা,  1/x = 1/3
বা,  x = 3

Question: If 5 ≥ x ≥ - 1 and y ≥ - 1, which of the following cannot be a value of x - y?

Solution: 
Here, 5 ≥ x ≥ - 1 and y ≥ - 1

Now,
i) If, x = - 1 and y = - 1 then, x - y = - 1 - (- 1) = -1 + 1 = 0 
ii) If, x = 2 and y = 1 then, x - y = 2 - 1 = 1 
iii) If, x = 5 and y = 0 then, x - y = 5 - 0 = 5 
iv) If, x = 5 and y = -1 then, x - y = 5 - (- 1) = 5 + 1 = 6

∴ Any value greater than 6 cannot be a value of x - y.  

Question: A cylindrical tank has a radius of 6 meters and a height of 7 meters. If the metal used to make the cylinder costs Tk. 40 per cubic meter, find the total cost of the metal required.

Solution:
Given,
Radius of the cylinder, r = 6 m
Height of the cylinder, h = 7 m
Cost per cubic metre = Tk. 40

The volume of the cylinder:
V = πr²h = (22/7) × (6)² × 7 = (22/7) × 36 × 7 = 22 × 36 = 792 cubic metres

Total cost = Volume × Cost per cubic metre = 792 × 40 = 31,680

∴ The cost of the cylinder is Tk. 31,680

Here, 2 × 2 + 1 = 5
5 × 2 - 1 = 9
9 × 2 + 1 = 19
19 × 2 - 1 = 37
37 × 2 + 1 = 75

A + B = 70%
B + C = 50%
[∵ (A + B) + (B + C) − (A + B + C) = B]
=> B = 20% ; A = 50% and C = 30%
Hence A is most efficient.

Question: A college has 8 basketball players. A 4 member's team and a captain will be selected out of these 8 players. How many different selections can be made?

Solution: 
We can select the 4 member team out of the 8 in =  ⁸C₄ ways
= 8!/4!(8 - 4)! = 8!/(4! × 4!)
= 70 ways

The captain can be selected from amongst the remaining 4 players in 4 ways.

∴ The total ways the selection of 4 players and a captain can be made = 70 × 4 ways
= 280 ways

So the total number of different selections that can be made is 280.

প্রথম 1000 ডলার এর জন্য ট্যাক্স দিতে হয় না।
বাকি টাকা = (2500 - 1000) = 1500 টাকা

এখানে, (100 - 65)% = 35%

প্রশ্নমতে, 35% = 1500
∴ 1% = 1500/35
∴ 100% = 1500×100/35
= 4285.71 ≅ 4286

∴ নির্ণেয় টাকার পরিমাণ = (4286 +1000) = 5286 টাকা

Question: What will come at the place of the question mark?
1, 16, 81, 256, 625, ?

Solution:
1⁴=1
2⁴= 16
3⁴= 81
4⁴ = 256
5⁴=625
6⁴ = 1296

ধরি, সংখ্যাটি x
প্রশ্নমতে, (2x + 9) × 3 = 75
⇒ 2x + 9 = 25
⇒ 2x = 16
⇒ x = 8

Question: The present ages of P and Q are in the ratio 2 : 7. After 8 years, the ratio of their ages will be 3 : 8. What is the difference in their present ages?

Solution:
Let the present ages be,
P = 2x and Q = 7x

Ages after 8 years,
P = 2x + 8, Q = 7x + 8

According to the problem, the ratio becomes 3:8
(2x + 8)/(7x + 8) = 3/8
⇒ 8(2x + 8) = 3(7x + 8)
⇒ 16x + 64 = 21x + 24
⇒ 64 - 24 = 21x - 16x
⇒ 40 = 5x
⇒ x = 8

P = 2 × 8 = 16 years
Q = 7 × 8 = 56 years

∴ Difference = 56 - 16 = 40 years

Question: A bag costs 20% more than a purse. A wallet costs 30% less than the bag. If the price of the purse is 200 Tk, then by what percentage is the wallet cheaper than the purse?

Solution:
Given,
the price of the purse = 200 tk

∴ Price of the bag = 200 + (200 × 20%)
= 240 tk

Price of the wallet = 240 − (240 × 30%)
= 240 − 72
= 168 Tk

Difference is = (200 - 168) = 32 tk

∴ Percentage = (32 × 100)/200 = 16%

If he scored a three-point basket, then remaining points are = 26 - 3 = 23, which is not divisible by 2
∴ As, there is no one-point basket, by scoring two three-point basket, the remaining points are: 26 - 3×2 = 20
∴ He scored maximum number of = 20/2 = 10 two-point baskets

C's 1 day's work = 1/3 - (1/6 + 1/8)
= 1/3 - 7/24
= 1/24
C's share (for 3 days) = (3 × (1/24) × 3200)
= 400

Question: Determine the largest number among four multiples of 5 in a row that equal 170 together.

Solution:
Let the four consecutive multiples of 5 be x, (x + 5), (x + 10) and (x + 15)

According to the question,
x + (x + 5) + (x + 10) + (x + 15) = 170
⇒ 4x + 30 = 170
⇒ 4x = 170 - 30
⇒ 4x = 140
⇒ x = 140/4
⇒ x = 35

Therefore, the largest number is = x + 15 = 35 + 15 = 50

Question: A worker union contract specifies a 6% salary increase plus a Tk. 450 bonus for each worker. For a worker, this is equivalent to an 8% salary increase. What was this worker's salary before the new contract?

Solution:
ধরি, কর্মীর পূর্বের বেতন = x টাকা।

6% বৃদ্ধিতে বেতন = x + x এর 6%
= x + (6x/100) = 106x/100

বোনাস হিসেবে 450 টাকা যোগ করলে মোট বেতন = (106x/100) + 450

8% বৃদ্ধিতে বেতন = x + x এর 8%
= x + (8x/100) = (108x/100)

প্রশ্নমতে,
(106x/100) + 450 = (108x/100)
⇒ 450 = (108x/100) - (106x/100)
⇒ 450 = (2x/100)
⇒ x = (450 × 100)/2
∴ x = 22500

অর্থাৎ, কর্মীর পূর্ববর্তী বেতন ছিল 22500 টাকা।

Let, A be the event of the group A pass
Let, B be the event of the group B pass

Then,
A'= Event of the group A's fail and B'= event of the group B's fail.
Therefore, p(A) = 2/7 and p(B) = 2/5,
P(A') = 1 - P(A) = 1- 2/7 = 5/7 and P(B') = 1- P(B) = 1- 2/5 = 3/5

Required probability = P[( A And B') Or (B And A')]
= P[( A And B') Or (B And A')]
= P[( A And B') + (B And A')]
= P[( A And B')] + P[(B And A')]
= p(A) x P(B') + P(A') x P(B)
= (2/7 x 3/5) + (2/5 x 5/7)
= (6/35 + 10/35)
= 16/35

Question: What is the sum of the first 12 terms of an arithmetic progression if the 3^rd term is - 13 and the 6^th term is - 4?

Solution:
In an arithmetic progression. We know
n^th term = a + (n - 1)d
where a = first term,
d = common difference

Given that,
3^rd term = a + 2d = - 13  … (1)
6^th term = a + 5d = - 4   … (2)

Subtract equation (1) from equation (2) then we get,
⇒ (a + 5d) - (a + 2d) = - 4 - (- 13)
⇒ 3d = 9
⇒ d = 9/3
∴ d = 3

Equation (1) we get,
⇒ a + 2(3) = - 13
⇒ a + 6 = - 13
⇒ a = - 13 - 6
∴ a = - 19

Sum of first n terms of an arithmetic progression.
Sₙ = (n/2) × [2a + (n - 1)d]
S₁₂ = (12/2) × [2(- 19) + (12 - 1)3]
= 6 × [- 38 + 11 × 3]
= 6 × [- 38 + 33]
= 6 × (- 5)
= - 30

Speed of aeroplane is 200, 400, 600 and 800 km/h respectively
Let the side of side be LCM of (200, 400, 600 and 800) = 2400
Time taken by aeroplane to travel the side at the speed of 200 km/hr 
⇒ 2400/200 = 12 hours
Time taken by aeroplane to travel the side at the speed of 400 km/hr 
⇒ 2400/400 = 6 hours
Time taken by aeroplane to travel the side at the speed of 600 km/hr 
⇒ 2400/600 = 4 hours
Time taken by aeroplane to travel the side at the speed of 800 km/hr 
⇒ 2400/800 = 3 hours

Average speed = (Total Distance travelled)/(Total time taken)
∴ Average speed = (4×2400)/25 = 384 km/hr

Required number of ways = (⁸C₅ × ¹⁰C₆)
= (⁸C₃ × ¹⁰C₄)
= (8 × 7 × 6)/3! × (10 × 9 × 8 × 7)/4!
= (8 × 7 × 6)/6 ×(10 × 9 × 8 × 7)/(4 × 3 × 2 × 1)
= 11760.