Bank Math

Question: The average score of a class of 90 students in an exam was 40. The average score of the students who had passed is 50, and the average score of the students who had failed is 30. How many students failed in the exam?

Solution:
Let
Total number of students who had failed = x
So, the total number of students who had passed = 90 - x

ATQ,
50(90 - x) + 30x = 90 × 40
⇒ 4500 - 50x + 30x = 3600
⇒ 20x = 4500 - 3600
⇒ 20x = 900
∴ x = 45

Question: A man invests Tk. 8100 partly in 14% stock at 294 and partly in 12% stock at 288. If his income from both is the same, find his investment in the 14% stock.

Solution:
Let he invests x at 14% stock.
x Investment at 12% stock = 8100 - x
​
As income is same.
x × (14/100) × (1/294) = (8100 - x) × (12/100) × (1/288)
⇒ x/2100 = (8100 - x)/2400
⇒ x = {(8100 - x)/2400} × 2100
⇒ 24x = 170100 - 21x
⇒ 45x = 170100
∴ x = Tk3780

আমরা জানি,
গ.সা.গু x ল.সা.গু = সংখ্যাদ্বয়ের গুণফল
⇒ 12 x 288 = 96 x অপর সংখ্যা
⇒ অপর সংখ্যা = (12×288)/96
∴ অপর সংখ্যা = 36

Let us assume that he answered x question correctly. Marks scored by him in x question = 2x
Then, wrong answer would be = 60 – x
Marks lost by him in (60 – x) questions = (60 – x)×1
ATQ,
2x – (60 – x) = 39
Or, 3x = 99
∴ x = 33

A's 2 day's work = (1/20) × 2
= 1/10
(A + B + C)'s 1 day's work = (1/20) + (1/30) + (1/60)
= 6/60
= 1/10
(A + B + C) work in 3 days = (1/10) + (1/10)
= 1/5.
Now, 1/5 work is done in 3 days.
∴ Whole work is done in (3 × 5)
= 15 days.

Question: If a shopkeeper sells a bat for Tk. 600, he incurs a loss of 20%. At what price should he sell it to make a profit of 20%?

Solution:
Let the cost price (CP) of the bat be x.
Selling price = x - 20% of x
⇒ 600 = x - (20x/100)
⇒ 600 = 80x/100
⇒ 80x = 60000
⇒ x = 750 (This is the CP)

To gain,
20% = 750 + 20% of 750
= 750 + (20 × 750)/100
= 750 + 150
= Tk. 900

LCM of 4 and 3 is 12. So they meet after every 12 day.
12 day after Monday is Saturday.
So, they will meet on Saturday again after the meeting on Monday.

According to the question,
Cost Price : Marked Price
(100 - Discount) : (100 + Profit)
100 - 10 : 100 + 12
90 : 112
45 : 56

Time taken,
7.25 P.M. - 4.10 P.M. = 3 hr. 15 min.
= 3 × (15/60) hrs.
= 3(1/4) hrs.
= (13/4) hrs.
∴ Required distance = {40 × (13/4)} km.
= 130 km.

Question: In a class of 78 students, 41 are taking French, 22 are taking German, and 9 are taking both courses. How many students are not enrolled in either course?

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:
Total students = 78
Students taking French, n(F) = 41
Students taking German, n(G) = 22
Students taking both French and German, n(F ∩ G) = 9

We know,
n(F ∪ G) = n(F) + n(G) - n(F ∩ G)
n(F ∪ G) = 41 + 22 - 9 = 54

∴ Not enrolled = Total students - n(F ∪ G) = 78 - 54 = 24

Question: A trader sells a bag for Tk. 2,400 at a loss of 20%. If his regular profit is 10%, find the regular selling price.

Solution: 
Selling price = 2,400 Taka
Loss % = 20%

Original cost price = 2400 / 0.8 = 3000 Taka

Regular profit = 10%
Regular Selling price = 3000 × 1.1 = 3300 Taka

∴ Regular Selling price = 3300 Taka

Question: What is the compound amount of Tk. 16000 for 2 years at a rate of interest of 5% per annum?

Solution:
Given,
Principal, P = 16000
Rate, r = 5% = 5/100 = 1/20
Time, n = 2 years

We know,
A = P(1 + r)ⁿ
= 16000 × (1 + 1/20)²
= 16000 × (21/20)²
= (16000 × 21 × 21)/(20 × 20)
= (16000 × 441)/400
= 40 × 441
= 17640

∴ The compound amount is Tk. 17640.

Face Value = Tk.1200
Market Value = Tk.120
Dividend (income) = (1200 × 8)/100
= Tk.96

Question: What would be the value of 20% of m as a percentage of p, if 8% of m = 4% of p?

Solution: 
8% of m = 4% of p
⇒ 8m/100 = 4p/100 
⇒ m = p/2
⇒ 20% of m = (p/2) × 20%
⇒ 20% of m = 10% of p

Question: A solid metal sphere of radius 6 cm is melted and recast into solid cones of radius 4 cm and height 9 cm. How many cones can be made?

Solution: 
The volume of a sphere = (4/3)πr³
= (4/3)π(6)³
= (864/3) π
= 288π cm³

The volume of a cone = (1/3)πr²h
= (1/3)π(4)²(9)
= (144/3)π cm^3
= 48π cm³

Number of cones = 288π/48π
= 288/48
= 6

Question: The 3rd term of a geometric sequence is 48, and the 6th term is 384. What is the common ratio is?

Solution:
Given, 
The 3rd term of a geometric sequence, a₃ = 48
The 6th term of the same sequence, a₆ = 384

In a geometric sequence we know,
a_n = ar^{n - 1}

So,
a₃ = ar^{3 - 1}= ar² .......(1)
And
a₆ = ar^{6 - 1}= ar⁵........ (2)

Now (2) ÷ (1), 
ar⁵/ar² = 384/48
⇒ r³ = 8 = 2³
∴ r =  2

So the common ratio is 2.

Area of the base of a cylinder, πr² = 100π
The volume of the cylinder, πr²h = 900π
∴ h = πr²h/πr²
= 900/100
= 9 m

Question: P is 6 times greater than Q then by what percent is Q smaller than P?

Solution:
Let Q = 10.
Then, P = 60.
Q is 50 less than P.
Q, % less than P = (50/60) × 100
= 83.33%

Let AB be the wall and BC be the ladder.
Then, ∠ACB = 60° = AC = 4.6m
AC/BC=cos⁡60∘=1/2
⇒ BC = 2 × AC = 2 × 4.6 = 9.2m

Let,
The C.P (cost price) of 1 kg goods be 1 tk
Then, S.P. of {100 - (25/2)% of 1 kg}
Or, 875 gm goods = .9375 tk
S.P of 1 kg goods = (.9375/875 × 1000) tk
= 1 (1/14) tk
∴ Profit % = (1/14 × 100)% = 7(1/7)%
Answer: 7(1/7)%

Question: Four boys are sitting on a bench to be photographed. Amin is to the left of Bilal. Rafi is to the right of Bilal. Tariq is between Bilal and Rafi. Who would be third from the left in the photograph?

solution:
Amin is to the left of Bilal
→ Amin — Bilal

Rafi is to the right of Bilal
→ Bilal — Rafi

Tariq is between Bilal and Rafi
→ Bilal — Tariq — Rafi

Now place all four:
Amin — Bilal — Tariq — Rafi

∴ Tariq would be third from the left in the photograph.

Average weight of 15 girls = 24 kg
Total weight of 15 girls = 24 x 15 = 360 kg
Average after including a new girl = 24 + 3 = 27 kg
Total weight of 16 girls = 27 x 16 = 432 kg
Weight of the new girl = Total weight of 16 girls - Total weight of 15 girls = 432 - 360 = 72 kg
Hence the required answer is 72 kg.

Question: With an average speed of 50 km/hr, a train reaches its destination in time. If it goes with an average speed of 40 km/hr, it is late by 30 min. The total journey is-

Solution:
Difference between timings = 30 min = 30/60 hr = 1/2 hr.
Let the length of the journey be x km.

Then,
(x/40) - (x/50) = 1/2
(5x - 4x)/200 = 1/2
x/200 = 1/2
x = 200/2
∴ x = 100 km.

∴ The total journey is 100 km.

দুটি একইরকম মুদ্রা ছুড়ে মারা হলে নমুনাক্ষেত্রটি হবে = HH, HT, TH, TT
হেড এবং টেল আসার সম্ভাবনা = 2/4 = 1/2

Question: If the sum of five consecutive odd integers is 255, what is the largest number?

Solution:
ধরি, মাঝের সংখ্যাটি = x
সুতরাং, 5টি ক্রমিক বিজোড় সংখ্যা হবে যথাক্রমে: x - 4, x - 2, x, x + 2 এবং x + 4

প্রশ্নমতে,
(x - 4) + (x - 2) + x + (x + 2) + (x + 4) = 255
⇒ 5x = 255
⇒ x = 51

∴মাঝের সংখ্যা, x = 51 
∴ সবচেয়ে বড় সংখ্যা = x + 4 = 51 + 4 = 55

Question: 605 sweets were distributed equally among children in such a way that the number of sweets received by each child is 20% of the total number of children. How many sweets did each child receive? 

Solution: 
Let Children = X
A/Q,
605/X = 20% of X
⇒ 605/X = X/5 
⇒, X² =5 × 605 
⇒ X² = 5² × 11² 
X = 55 

So each children receive = 605/55
= 11

Given, d = -1.5t² + 120 
As, d =10 
∴ 10 = -1.5t² + 120  
⇒ 1.5t² - 110 = 0 
⇒ 1.5t² = 110 
⇒ t² = 110/1.5 
⇒ t = √(110/1.5)
∴  t = 8.56

As given the questions these numbers are co-primes, so there is only 1 as their common factor.
It is also given that two products have a middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;

So the first number is: 551/29 = 19
Third number = 1073/29 = 37

So sum of these numbers is = (19 + 29 + 37) = 85