Bank Math

Question: ∠A and ∠B are complementary to each other. If ∠A = 30° + 3x and ∠B = 5x, find the value of ∠B.

Solution:
Here,
∠A = 30° + 3x and ∠B = 5x

For complementary angles,
∠A + ∠B = 90°
⇒ (30° + 3x) + 5x = 90°
⇒ 30° + 8x = 90°
⇒ 8x = 90° - 30°
⇒ 8x = 60°
⇒ x = 60°/8 = 7.5°

∴ ∠B = 5x = 5 × 7.5° = 37.5°

Question: The ratio between Karim's age and Rahim's age is 11 : 10. What is the age of Karim as a percentage of Rahim's age?

Solution: 
Given that,
The ratio of Karim's age to Rahim's age = 11 : 10

Let, Karim's age = 11x
and Rahim's age = 10x

Now, we find Karim's age as a percentage of Rahim's age,

∴ Percentage = (Karim’s age/Rahim’s age​) × 100
= (11x/10x) × 100
= 110

So, Karim's age is 110% of Rahim's age.

Question: The graphs of the equations 7x + 11y = 3 and 8x + y = 15 intersect at the point P, which also lies on the graph of the equation.

Solution: 
Given equations
7x + 11y = 3 .....(1)
8x + y = 15 .......(2)

From (1) we get,
8x + y = 15
∴ y = 15 - 8x ......(3)

Substitute into (1) then we get, 
⇒ 7x + 11(15 - 8x) = 3
⇒ 7x + 165 - 88x = 3
⇒ - 81x = - 162
∴ x = 2

Now from (3),
⇒ y = 15 - 8x
⇒ y = 15 - 16
∴ y = - 1

Now check which line also passes through P(2, -1),
A. 2x - y = 1
2(2) - (-1) = 4 + 1 = 5 ≠ 1 ; [Not valid]

B. 3x + 2y = 3
3(2) + 2(-1) = 6 - 2 = 4 ≠ 3 ; [Not valid]

C. 2x + y = 2
2(2) + (-1) = 4 - 1 = 3 ≠ 2 ; [Not valid]

D. 3x + 5y = 1
3(2) + 5(-1) = 6 - 5 = 1 = 1 ; [valid]

So correct answer is D. 3x + 5y = 1

Question: A and B share profits in the ratio 2 : 3. If 20% of the total profit is given to charity and B's share is Tk. 4800, find the total profit.

Solution:
ধরি, মোট লাভ = Tk. x
20% দাতব্য প্রতিষ্ঠানে দেওয়ার পর বাকি থাকে = 100% - 20% = 80% of x
= 80x/100
A এবং B এর লাভের অনুপাত 2 : 3।
অর্থাৎ, B এর অংশ = 80x/100 এর 3/(2 + 3)
= 80x/100 এর 3/5

প্রশ্নমতে,
80x/100 × (3/5) = 4800
⇒ 4x/5 × (3/5) = 4800
⇒ 12x = 4800 × 25
⇒ x = 120000/12
⇒ x = 10000
সুতরাং, মোট লাভ হলো Tk. 10000।

let, the height is x
By applying Pythagoras Theorem, x² + 2² = 4²
⇒ x² = 16 - 4
⇒ x = √12 =  √(4.3)
⇒ x = 2√3 cm
So, the height is 2√3 cm

Question: Fahim is younger than Nabila but older than Ritu. Shanti is older than Fahim. Nabila is the second-oldest person among them. Labib is younger than Ritu. Who is the third-oldest among them?

Solution:
• First statement: Nabila > Fahim > Ritu
• Second statement: Shanti > Fahim
• Third statement: Nabila is the second oldest, meaning one person is older than Nabila. 
Since both Shanti and Nabila are older than Fahim, and Nabila is the second-oldest, Shanti must be the oldest. Therefore, Shanti > Nabila > Fahim
• Fourth statement: Ritu > Labib

• Putting everyone together: Shanti > Nabila > Fahim > Ritu > Labib
∴ The third-oldest person is Fahim.

Question: Silver is 17 times as heavy as water and copper is 7 times as heavy as water. In what ratio should these be mixed to get alloy 13 times as heavy as water?

Solution:
Given that,
Density of silver is 17 times as heavy as water
Density of copper is 7 times as heavy as water
Mixture should be 13 times as heavy as water

Let the weights of silver and copper be x and y respectively. Then we get, 
⇒ (17x + 7y)/(x + y) = 13
⇒ 17x + 7y = 13(x + y)
⇒ 17x + 7y = 13x + 13y 
⇒ 17x - 13x = 13y - 7y
⇒ 4x = 6y
⇒ x/y = 6/4 = 3/2
∴ x : y = 3 : 2

So Ratio of silver to copper = 3 : 2

Question: If 4 men or 6 women can complete a work in 20 days, how many days would it take 6 men and 11 women to complete twice the work?

Solution:
এখানে,
4 men = 6 women
∴ 1 man = 6/4 = 3/2 women
∴ 6 men = 6 × 3/2 = 9 women
∴ 6 men and 11 women together = 9 + 11 = 20 women

6 women কাজটি সম্পন্ন করে = 20 দিনে
∴ 1 woman কাজটি সম্পন্ন করে = 20 × 6 = 120 দিনে
∴ 20 women কাজটি সম্পন্ন করে = 120/20 = 6 দিনে

সুতরাং, দ্বিগুণ (twice) কাজ সম্পন্ন করতে সময় লাগবে = 6 × 2 = 12 দিন

Question: If the list price of a mobile phone is Tk. 10000, and a Tk. 1500 discount is offered on the mobile phone, then what is the discount percentage?

Solution:
Marked Price = Tk. 10000
Discount = Tk. 1500

∴ Discount (%) = (Discount/marked Price) × 100%
∴ Discount (%) = (1500/10000) × 100%
= 15%

∴ So the discount percentage is 15%.

Question: In a right-angled triangle, the length of the medians from the vertices of acute angles are 7 cm and 4√6cm. What is the length of the hypotenuse of the triangle (in cm)?

Solution: 

Given that, 
AD = 7 cm
CE = 4√6 cm

Since, 4(AD² + CE²) = 5AC²
⇒ 4{(7)² + (4√6)²} = 5AC²
⇒ 4(49 + 96) = 5AC²
⇒ 4 × 145 = 5AC²
⇒ AC² = (4 × 145)/5
⇒ AC² = 4 × 29
⇒ AC = √(4 × 29)
∴ AC = 2√29 cm

Question: Fahim sold a t-shirt for Tk. 810 and made a gain of 8%. What was the purchase price of the t-shirt?

Solution:
8% লাভে,
ক্রয়মূল্য 100 টাকা হলে বিক্রয়মূল্য = 100 + 8 = 108 টাকা

বিক্রয়মূল্য 108 টাকা হলে ক্রয়মূল্য = 100 টাকা
বিক্রয়মূল্য 1 টাকা হলে ক্রয়মূল্য = 100 ÷ 108 টাকা
বিক্রয়মূল্য 810 টাকা হলে ক্রয়মূল্য = (100 × 810) ÷ 108 টাকা
= 750 টাকা

Question: What will come at the place of question mark?
4, 9, 19, 34, 54, ?

Solution:
9 - 4 = 5
19 - 9 = 10
34 - 19 = 15
54 - 34 = 20

∴ প্রতিবার পার্থক্য 5 করে বৃদ্ধি পাচ্ছে।

∴ পরবর্তী পার্থক্য হবে = 20 + 5 = 25

∴ পরবর্তী সংখ্যা = 54 + 25 = 79

• Shortcut: 4 (+5)→ 9 (+10)→ 19 (+15)→ 34 (+20)→ 54 (+25) → 79.

Question: Let N be the smallest positive integer that is divisible by both 20 and 30. How many distinct prime factors does N have?

Solution:
এখানে, N হলো 20 এবং 30 দ্বারা বিভাজ্য ক্ষুদ্রতম সংখ্যা।
সুতরাং, N হবে 20 এবং 30 এর ল.সা.গু।

এখন, 20 = 2 × 2 × 5 = 2² × 5¹
এবং 30 = 2 × 3 × 5 = 2¹ × 3¹ × 5¹

LCM(20, 30) = 2² × 3¹ × 5¹ = 60
অতএব, N = 60

60 এর মৌলিক উৎপাদক = 2² × 3 × 5

স্বতন্ত্র মৌলিক উৎপাদকগুলি হলো 2, 3 এবং 5।

∴ N এর স্বতন্ত্র মৌলিক উৎপাদকের সংখ্যা হলো 3টি।

ATQ,
x - y = 12 ...... (i)
x + y = 38 ........ (ii)
(i) + (ii), 2x = 50
Or, x = 25
So, y = 13
If 2 is added in both the numbers, then their ratio is:
x+2 / y+2
= 25+2 / 13+2
= 27/15
= 9/5

Question: How many ways the letters of the word 'TEACHER' can be arranged ?

Solution:
The word 'TEACHER' has 7 letters
Here, E = 2 times

We know, 
Number of distinct permutations = n!/(p1! × p2!......) 
= 7!/2!
= (7 × 6 × 5 × 4 × 3 × 2!)/2!
= 2520

∴ Distinct permutations 2520

Question:

Solution:

Question: Denominator of a proper fraction is 3 more than the numerator. If the fraction is squared, its denominator will be 51 more than the numerator. The fraction is
(Janata RC 2022 অনুযায়ী)

Solution:
ধরি,
ভগ্নাংশের লব = x 
∴ হর = x + 3

প্রশ্নমতে,
(x + 3)² - x² = 51
⇒ x² + 6x + 9 - x² = 51
⇒ 6x = 51 - 9
⇒ 6x = 42
⇒ x = 30/6
⇒ x = 7

সুতরাং,
ভগ্নাংশটি = x/(x + 3) = 7/(7 + 3) = 7/10

Question: Two pipes A and B together can fill a cistern in 3 hours. If they had been opened separately, B would have taken 8 hours more than A to fill the cistern. How long will A take to fill the cistern separately?

Solution: Let the time taken by A alone be x hours.
Then time taken by B alone = x + 8 hours.
Rate of A = 1/x cistern/hour. Rate of B = 1/(x+8) cistern/hour.
Combined rate = 1/x + 1/(x+8) = 1/3 (since together they fill in 3 hours).

Now,
1/x + 1/(x+8) = 1/3
⇒ (x+8 + x) / [x(x+8)] = 1/3
⇒ (2x + 8) / [x(x+8)] = 1/3

Cross multiply:
3(2x + 8) = x(x+8)
⇒ 6x + 24 = x² + 8x
⇒ x² + 2x - 24 = 0

Factorize:
(x + 6)(x - 4) = 0
So, x = 4 (positive value).

(Other root is negative and discarded.)
Therefore A will take 4 hours alone. 

Question: What should come in place of both n in the equation (n/√162) = (√128/n)?

Solution: 
Here,
n/√162 = √128/n
⇒ n² = √(128 × 162)
⇒ n² = √(64 × 2 × 18 × 9)
⇒ n² = √(64 × 36 × 9)
⇒ n² = √(8² × 6² × 3²)
⇒ n² = 8 × 6 × 3
⇒ n² = 144
⇒ n = √144
∴ n = 12

Let the fixed amount be Rs. x and the cost of each unit be Rs. y.
Then,
540y + x = 1800 ...(i) and
620y + x = 2040 ...(ii)
On subtracting (i) from (ii), we get 80y = 240
⇒ y = 3
Putting y = 3 in (i), we get :
540×3 + x = 1800
⇒ x = (1800 − 1620) = 180
∴ Fixed charges = Tk. 180, and, charge per unit = Tk. 3
Total charges for consuming 500 units = Tk. (180 + 500×3) = Tk. 1680

Question: A can do a piece of work in 30 days. When he had worked for 10 days, B joined him. If the complete work was finished in 24 days, B can alone finish that work in -

 
Solution:

A's 1 day's work = 1/30 part
A's 24 day's work = 24/30 part = 4/5 part

∴ Remaining work = 1 - 4/5 = 1/5 part

This 1/5 part of work was done by B in = (24 - 10) = 14 days

∴ 1 part of work done by B in = 14 × 5 = 70 days

∴ Probability = Possible outcomes/Total outcomes = 8/12 = 2/3

Hence, the required probability is 2/3.

Question: Point A is 10 km west of point B. Point B is 30 km north of point C. Point C is 20 km east of point D. What is the distance between points A and D?

Solution: 


AD = √(30² + 10²)
= √1000 
= 10√10 km

Question: An amount of Tk. 8,000 yields a simple interest of Tk. 1,440 in 3 years. What is the annual rate of interest?

Solution:
Given,
Principal, P = 8000
Simple Interest, SI = 1440
Time, n = 3 years
Rate of interest, r = ?

We know,
I = Pnr/100
⇒ r = (I × 100)/(P × n)
⇒ r = (1440 × 100)/(8000 × 3) 
⇒ r = 144000/24000 
∴ r = 6%

So, the annual rate of interest is 6%.

Question: If 3 men or 6 women can plough a field in 42 days, how long will 8 men and 5 women take to plough it?

Solution: 
3 men or 6 women can plough the field in 42 days

3 men = 6 women
1 men = (6/3) women
8 men = {(6/3) × 8} = 16 women

∴ 8 men and 5 women = 16 + 5 = 21 women

6 women can plough field in 42 days
1 women can plough field in (42 × 6) days
∴ 21 women can plough field in (42 × 6)/21 = 12 days

Question: A machine wheel has a circumference of 50 cm and completes 24 rotations in 4 seconds. What is the speed of the wheel in kilometers per hour (km/h)?

Solution:
Total distance covered = (50 × 24) cm
= 1200 cm
= (1200 ÷ 100) m
= 12 m

We know,
Speed = (Total distance ÷ Time)
= (12 ÷ 4) m/sec
= 3 m/sec

Converting into km/h,
= 3 × (18/5) km/h
= 10.8 km/h

∴ The speed of the wheel is 10.8 km/h.

Question: In the figure, AOC is the diameter of the circle and arc AXB = (1/2)arc BYC. Find ∠BOC = ?

Solution:
Given that,
arc AXB = (1/2) arc BYC
∴ ∠AOB = (1/2) ∠BOC

We know that,
 ∠AOB + ∠BOC = 180º

Therefore,
(1/2) ∠BOC + ∠BOC = 180º {linear pair since AOC is the diameter}
⇒ (3/2) ∠BOC = 180º
⇒ ∠BOC = (2/3) × 180º = 120º
∴  ∠BOC = 120º

Question: Walking 3/4 of his normal speed, Rabi is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and office.

Solution:
Let,
Total time = x minutes
So, when it is late then required time = x + 16
If actual speed = d metre/min
Then reduced speed = (3d/4) metre/min

ATQ,
dx = 3d(x + 16)/4
⇒ dx = (3dx + 48d)/4
⇒ 4dx = 3dx + 48d
⇒ 4dx - 3dx = 48d
⇒ dx = 48d
∴ x = 48

∴ Total time = 48 minutes

Question: 94 is divided into two parts such that the fifth part of the first and the eighth part of the second are in the ratio 3 : 4. Find the first part.

Solution:
Let the two parts be x and 94 - x.

According to the problem,
(x/5) : (94 - x)/8 = 3 : 4
⇒ (x/5)/{(94 - x)/8} = 3/4
⇒ 8x/5(94 - x) = 3/4
⇒ 32x = 15(94 - x)
⇒ 32x = 15 × 94 - 15x
⇒ 47x = 15 × 94
⇒ x = (15 × 94)/47
∴ x = 30

∴ First part is 30

Question: If a - (1/a) = √5, what is the value of a³ - (1/a³)?

Solution:
দেওয়া আছে,
a - 1/a = √5

এখন, 
a³ - (1/a³)
= {a - (1/a)}³ + 3 . a . 1/a . {(a - 1/a)}
= (√5)³ + 3
= 5√5 + 3√5
= 8√5

Question: A retailer marked the price of a television at Taka 12000 and gave a discount of 15%. Calculate the selling price and the amount of discount.

Solution:
Marked Price of the television = Taka 12000
Discount Percentage = 15%

∴ Discount Amount = Discount Percentage × Marked Price
= 15% × Taka 12000
= Taka 1800

∴ Selling Price = Marked Price - Discount Amount
= Taka 12000 - Taka 1800 
= Taka 10200

Let AE and BC be the heights of trees.

AE = 28 m
BC = 20 m

Horizontal distance between trees AB = DC

In EDC, EC² = ED² + DC² (Pythagoras theorem)
DC² = EC² - ED²
= 16² - 8²
= 256 - 64
DC² = 192
DC =√192 m.

Question: Find the simple interest on BDT 12000 at 4% per annum for 8 months.

Solution:
Principal, P = 12000 Taka
Time, n = 8 months = 8/12 = 2/3 years
Rate of interest, r = 4% = 4/100

Simple Interest, I = P × n × r
= 12000 × (2/3) × (4/100)
= 40 × 2 × 4
= 320

∴ The simple interest is Tk. 320.