Bank Math

Question: Given that the diagonal of a square measures 10√6 units, find the area of the square in units.

Solution:
দেয়া আছে,
বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য = 10√6 একক

আমরা জানি,
বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য = √2 × বাহু

প্রশ্নমতে,
√2 × বাহু = 10√6
⇒ বাহু = 10√6/√2
⇒ বাহু = 10√3 একক

এখন, বর্গক্ষেত্রের ক্ষেত্রফল  =  বাহু2
= (10√3)2
= 300 বর্গ একক

∴ বর্গক্ষেত্রের ক্ষেত্রফল 300 বর্গ একক 

Question: If 25% of (A + B) = 50% of (A - B), then find B : A - 

Solution: 
25% of (A + B) = 50% of (A - B)
⇒ (A + B) × 25/100 = (A - B) × 50/100
⇒ (A + B)/4 = (A - B)/2
⇒ 2 (A + B) = 4 (A - B)
⇒ 2A + 2B = 4A - 4B 
⇒ 4A - 2A = 2B + 4B
⇒ 2A = 6B
⇒ A/B = 6/2 = 3/1
∴ B/A = 1/3
= 1 : 3

Question: One diagonal of a rhombus is four times the other diagonal. If its area is 56 sq. cm, find the sum of the diagonals.

Solution:
let, one diagonal is x cm, other is 4x

ATQ,
(1/2) × x × 4x = 56
⇒ 2x2 = 56
⇒ x2 = 28
⇒ x = √28 = 2√7
x = 2√7

sum of diagonals = x + 4x
= 5x
= 5 × 2√7
= 10√7 cm

Area of rectangle without road = 16 × 12 = 192
Area of rectangle with road = (16 + 2 × 3) × (12 + 2 × 3) = 396
Area of road = 396 - 192 = 204.

At 25% discount, 75% = 180
So, 100% = (180×100)/75 = 240

প্রশ্ন: 7 জন ব্যক্তিকে 1 টি গোলটেবিলের চারপাশে কতভাবে বসানো যাবে?

সমাধান: 
7 জন ব্যক্তিকে 1 টি গোলটেবিলের চারপাশে সাজানো যাবে (n - 1)! উপায়ে।
= (7 - 1)! 
= 6!
= 720

At 10% loss, He got = (52000×90) / 100 = 46800 tk
His loss is = 52000 - 46800 = 5200

At 20% profit, He receives = (46800×120)/100 = 56160
Profit, 56160 - 46800 = 9360

∴ His net profit = 9360 - 5200 = 4160

Question: The ratio of the present ages of a father and his son is 5 : 1. Five years ago, the ratio was 9 : 1. What will be the father's age five years from now?
 
Solution:
Let the present age of the son = x years
Then present age of the father = 5x years

Five years ago, Father's age = 5x - 5
And son's age = x - 5

According to the question,
(5x - 5) : (x - 5) = 9 : 1
⇒ (5x - 5)/(x - 5) = 9/1
⇒ 5x - 5 = 9(x - 5)
⇒ 5x - 5 = 9x - 45
⇒ 9x - 5x = 45 - 5
⇒ 4x = 40
⇒ x = 40/4 = 10
∴ x = 10

Therefore, present age of son = 10 years
present age of father = 5 × 10 = 50 years

Five years from now father's age = 50 + 5 = 55 years

So the father's age five years from now will be 55 years.

Let, there was x exams (excluding the makeup exam)
ATQ, 80x + 92 = 84(x + 1)
Or, 80x + 92 = 84x + 84
Or, x = 2
So, total number of exams including the makeup exam = 2 + 1 = 3

Let the number be x and y
Then,
x²−y² = 63 & x−y = 3
On dividing, we get: x + y = 21
Solving x + y = 21 and x - y = 3,
We get: x = 12 and y = 9
∴ Larger number = 12

Question: A fair coin is flipped three times. What is the probability that the coin lands head each time?

Solution:
All possible outcome = {HHH, HHT, HTT, ΗΤΗ, ΤΗΗ, TTH, THT, TTT} = 8

It will be head every time, this occurs 1 time = {HHH}

∴ Probability = 1/8

15% discount - এ T - shirt টির বিক্রয়মূল্য 85 টাকা হলে প্রকৃত মূল্য 100 টাকা
∴ বিক্রয়মূল্য 272 টাকা হলে প্রকৃত মূল্য = (100 × 272) / 85 = 320 টাকা

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।

Question: A student find the average of ten 2 digit numbers, and while copying numbers by mistake, he writes one number with its digit interchanged, as a result of that his average is 3.6 less that the correct answer, then find the difference of the digits of the number in which he made the mistake.

Solution:
The number = 10x + y
After interchanging the digit, the number becomes 10y + x
The difference in average after interchanging the digits = 3.6
As there are 10 numbers,
The difference between the numbers will be = 3.6 × 10 = 36

Now
10x + y - (10y + x) = 36
9x - 9y ⇒ 36
9(x - y) = 36
x - y = 4

Difference in digits, (x - y) = 4

∴ The difference of the digits of the number in which he made the mistake is 4.

Question: What is the distance of (5, 12) from the origin?

Solution:
Since the distance of the coordinate (5, 12) is taken from the origin, then the coordinates of the origin are (0, 0).
Therefore,
x₁ = 0, y₁ = 0
x₂ = 5, y₂ = 12

We know,
The distance between two points = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(5 - 0)² + (12 - 0)²]
= √[25 + 144] 
= √169
= 13 units

Question: Find the odd number from the given alternatives.

Solution:
256 = 16²
∴ 256 is a perfect square of 16, others are not a perfect square.

Let X be the number which is added to 80
80% of X = 0.8X
Now,
80 + 0.8X = X
0.2X = 80
X = 80/0.2 = 400

Question: [(289)^0.17 × (17)^0.16]² = ? 
(Janata RC 22 এর অনুরূপ)

Solution:
(289)^0.17 × (17)^0.16
= {(17)²}^0.17 × (17)^0.16
= 17^{(2 × 0.17)} × (17)^0.16
= (17)^0.34 × (17)^0.16
= (17)^{0.34 + 0.16}
= (17)^0.50
= (17)^50/100
= (17)^1/2
= √17

Hence, (√17)² = 17

The height of an equilateral triangle is also its median
hence, base gets divided into 2 equal parts

So,
Base = 1cm
Hypotenuse = 2cm
And, Height = p = ?

Now, h² = b² + p²
Or, 2² = 1² + p²
Or, 4 = 1 + p²
Or, p² = 4 - 1
Or, p² = 3
Or, p = 3^1/2

Hence, the height of the ∆ is 3^1/2 = √3 cm

Now, in this numerical, there is no common loss and gain %.
Hence, solve it by making equations.

Let the cost price of the camel be x.
As the cost of camel and carriage = Tk 5000
Cost of carriage = Tk (5000 – x)
After selling the camel he gains 20% and on carriage a loss of 10%. But on the whole, he gains 3%.

Therefore,
20% of x – 10% of (5000 – x) = 3% of 5000
⇒ (20/100) × x - (10/100) × (5000 – x) = (3/100) × 5000
⇒ x/5 - (5000 – x)/10 = 150
⇒ 10x/5 - {(5000 – x) × 10}/10 = 150 × 10
⇒ 10x/5 - (5000 – x) = 1500
⇒ 2x - 5000 + x = 1500
⇒ 3x=1500 + 5000
⇒ x = 2166.67.
The cost of camel = Tk. 2166.67

Question: A committee of 4 members is to be selected from 6 men and 5 women. In how many ways can this be done if exactly 2 men must be selected?

Solution:
Here, to form the committee of 4 members, we need to select 2 men 
and (4 - 2) = 2 women.

Number of ways to select 2 men from 6 men:
⁶C₂ = 6!/[2! (6 - 2)!]
= {6 × 5 × 4!}/[2! × (4)!]
= (6 × 5)/(2 × 1)
= 3 × 5
= 15 ways

Number of ways to select 2 women from 5 women:
⁵C₂ = 5!/[2! (5 - 2)!]
= 5!/[2! × (3)!]
= {5 × 4 × 3!}/[2! × (3)!]
= (5 × 4)/(2 × 1)
= 5 × 2
= 10 ways

So, total number of ways = 15 × 10 = 150.

Question: Solution set of the inequality: p - 5 > 4p + 7 Is
(Janata RC 2022 অনুযায়ী)

Solution:
p - 5 > 4p + 7
⇒ - 5 > 4p - p + 7
⇒ - 5 > 3p + 7
⇒ - 5 - 7 > 3p
⇒ - 12 > 3p
⇒ - 12/3 > 3p/3
⇒ - 4 > p
⇒ p < - 4

∴ নির্ণেয় সমাধান সেট: (- ∞, - 4)

Given, 3√x = 2√3
⇒ √x/√3 = 2/3
⇒ x/3 = 4/9
∴ x = (4×3)/9 = 1.33

Let cost price per item be 1.
cp of 15 items = 15
sp of 15 items = cost price of 20 items= 20

So, profit = 20 - 15 = 5.
∴ required profit % = (5/15) × 100 = 33.3%

Question: If b is equal to 20% of a, then what is b% of 20 equal to?

Solution:
20% of a = b    
⇒ 20a/100 = b

∴ b% of 20
= (b/100) × 20
= (20a/100) × (1/100) × 20
= 4a × (1/100)
= 4% of a

Question: Population of a town increase 2.5% annually but is decreased by 0.5% every year due to migration. What will be the percentage increase in 2 years?

Solution:
Net percentage increase in Population = (2.5 - 0.5) = 2% each year.
Assuming initial population = 100

Population of Town after 1st year = (100 + 2% of 100)
= 100 + 2 = 102

Population of Town after 2nd year = (102 + 2% of 102)
= 102 + 2.04 = 104.04

Increase = 104.04 - 100 = 4.04
∴ Percentage increase = (4.04/100) × 100% = 4.04%

Question: Write an equation of the line with slope 3 and x-intercept (- 2, 0).

Solution:
Given:
Slope, m = 3
x-intercept = (- 2, 0)

We know the point-slope form of a line:
y - y₁ = m(x - x₁)

Substitute the values (x₁, y₁) = (- 2, 0) and m = 3:
y - 0 = 3(x - (- 2))
⇒ y = 3(x + 2)
⇒ y = 3x + 6

So, the equation of the line is y = 3x + 6.

Let the daily wages of A, B and C be Tk. 5x, Tk. 6x and Tk. 4x respectively.
Then, ratio of their amounts
= (5×6):(6×4):(4x9)
= 30:24:36
= 5:4:6
∴ A's amount
= Tk. (1800 × 5/15)
= Tk. 600

Question: A company increases salary of an officer at 20% per year. In 2025 an employee receives Tk. 43,200. What was his salary in 2023?

Solution:
Let, his salary in 2023 was 100 Tk.

At 20% increment,
Salary of 2024 is = 100(1 + 20/100)
= 120 Tk.

Salary of 2025 is = 120(1 + 20/100)
= 144 Tk.

Now,
144 Tk. corresponds to = 43,200 Tk.
1 Tk. corresponds to = 43,200/144 Tk.
∴ 100 Tk. corresponds to = (43,200 × 100)/144
= 30,000 Tk.

∴ His salary in 2023 was 30,000 Tk.

Question: In how many ways can a garland be made using 6 different flowers?

Solution:
We know,
For a circular garland, the formula for arrangements is (n - 1)!/2
Here, n = 6

∴ The number of ways to make a garland with 6 flowers = (6 - 1)!/2 
 = 5!/2 
= (5 × 4 × 3 × 2)/2 
= 60 

Question: A reservoir can be filled in 4 hours by three taps, P, Q, and R. Tap R is three times as fast as Q, and tap Q is twice as fast as P. How long will tap Q alone take to fill the reservoir?

Solution:
ধরি, নল P একা চৌবাচ্চাটি পূর্ণ করতে x ঘন্টা সময় নেয়।

যেহেতু নল Q, P এর দ্বিগুণ দ্রুত গতিতে পানি সরবরাহ করে, তাই Q এর সময় লাগবে x/2 ঘন্টা।
যেহেতু নল R, Q এর তিনগুণ দ্রুত গতিতে পানি সরবরাহ করে, তাই R এর সময় লাগবে (x/2)/3 = x/6 ঘন্টা।

প্রশ্নমতে, তারা একসাথে 4 ঘন্টায় চৌবাচ্চাটি পূর্ণ করে।
অতএব,
1/x + 1/(x/2) + 1/(x/6) = 1/4
⇒ 1/x + 2/x + 6/x = 1/4
⇒ (1+2+6)/x = 1/4
⇒ 9/x = 1/4
⇒ x = 9 × 4
⇒ x = 36 ঘন্টা।

অতএব, নল P একা চৌবাচ্চাটি পূর্ণ করতে 36 ঘন্টা সময় নেয়।
∴ নল Q একা চৌবাচ্চাটি পূর্ণ করতে সময় নেবে 36/2 = 18 ঘন্টা।

Question: Find the missing term,
3, 12, 27, 48, 75, 108, ?

Solution: 
The terms of the given series are,
3 = 3 × 1²
12 = 3 × 2²
27 = 3 × 3²
48 = 3 × 4²
75 = 3 × 5²
108 = 3 × 6²
So, missing term = 3 × 7² = 3 × 49 = 147

Question: If |2x - 2| ≤ 8, what is the maximum value of x?

Solution:
Given that,
|2x - 2| ≤ 8
⇒ - 8 ≤ 2x - 2 ≤ 8 
⇒ - 8 + 2 ≤ 2x - 2 + 2 ≤ 8 + 2 
⇒ - 6 ≤ 2x ≤ 10
⇒ (- 6/2) ≤ (2x/2) ≤ (10/2)
⇒ - 3 ≤ x ≤ 5

∴ The maximum value of x is 5

Let,
Number of girls = boys = 24 (LCM of 8 and 6)
boys who do not reside in hostel = 24 - (5 × 24/6)
= 24 - 20
= 4
The students consist of boys who do not reside hostel = 4/(2 × 24)
= 1/12.

Question: The average of six consecutive numbers A, B, C, D, E and F is 62. What is the sum of B and F?

Solution:
Let the Numbers A, B, C, D, E, F be x, x + 1, x + 2, x + 3, x + 4, x + 5.
According to question, 
x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 62 × 6
⇒ 6x + 15 = 372
⇒ 6x = 372 - 15 = 357
⇒ 6x = 357
⇒ x = 59.5

∴ B = x + 1 = 60.5 
∴ F = x + 5 = 64.5

∴ B + F = 60.5 + 64.5 = 125

Selling price of egg Tk (M/6) × 12 per dozen
= 2M
profit = 2M - M = M
profit percentage = (M/M) × 100
= 100%.