Bank Math

Question: A factory manufactures products in batches of 16, 24, and 32 units. What is the minimum number of units the factory needs to produce so that each batch can be formed exactly?

Solution:
To find the minimum number of units the factory needs to produce so that each batch size (16, 24, and 32) can be formed exactly, we need to find the least common multiple (LCM) of these batch sizes.

The prime factorization of each batch size is -
16 = 2 × 2 × 2 × 2 = 2⁴
24 = 2 × 2 × 2 × 3 = 2³ × 3
32 = 2 × 2 × 2 × 2 × 2 = 2⁵

Now,
So the highest power of 2 is 2⁵
The highest power of 3 is 3¹

So, the LCM of 16, 24, and 32 is = 2⁵ × 31 = 32 × 3 = 96.

So the minimum number of units the factory needs to produce is 96.

Question: Fifteen distinct points are randomly placed on the circumference of a circle. At most how many triangles can be formed using these points?
​
​Solution:
​Given that,
​Number of distinct points = 15
​Maximum number of triangles = ¹⁵C₃
​= 15!/3!(15 - 3)!
​= (15 × 14 × 13 × 12!)/(3 × 2 × 12!)
= 455
​

Total distance travelled in 12 hours = (35 + 37 + 39 + ..... upto 12 terms)
This is an A.P with first term, a=35, number of terms,
n = 12,
d = 2.
Required distance = n/2 {2a+(n-1)×2}
= 12/2[2 x 35+{12-1) x 2]
= 6(70+22)
= 552 kms.

যেহেতু, x ও y ঋণাত্মক পূর্ণসংখ্যা
এবং x - y = 1 অর্থাৎ x > y
তাহলে, x = -1 এবং y = -2 হলে xy = (-1)(-2) = 2

Question: The average of ten numbers is 25. The average of seven of these numbers is 22. What is the average of the remaining three numbers?

Solution:
১০ টি সংখ্যার গড় = ২৫
১০ টি সংখ্যার সমষ্টি = ২৫ × ১০ = ২৫০

৭ টি সংখ্যার গড় = ২২
৭ টি সংখ্যার সমষ্টি = ২২ × ৭ = ১৫৪

∴ বাকী ৩ টি সংখ্যার সমষ্টি = ২৫০ - ১৫৪ = ৯৬
∴ ৩ টি সংখ্যার গড় = ৯৬/৩ = ৩২.

Question: If the cost price of 15 books is equal to the selling price of 20 books, what is the loss percentage?

Solution:
Let C.P. of each books Tk. 1
Cost Price of 20 books Tk. 20
Selling Price of 20 books Tk. 15

∴ Loss  = TK. (20 - 15)
= Tk.5

∴ Percentage of loss = {(5/20) × 100}%
 = 25%

We know,
Speed of Work / Efficiency ∝ 1/Time needed
So more the efficiency, less time is needed to do the task.

Let us denote Akash by 'A' and Biplob by 'B'
Hence,
if A is 5 times faster than B, then B needs 5 times more time than A.
Let A need 'n' days to complete the work.
So B will need 5n days.
Also, 5n – n = 60 ------------------- Given
∴ n = 15 days = Days needed by A to complete work individually
Days needed by B to complete work individually = 5n = 5 x 15 = 75 days.

Price of 6 mangoes = Price of 9 oranges
⇒ (Price of 1 mango) : (Price of 1 orange) = 9/6 = 3 : 2
Suppose, the price of each mango is Tk. 3x, and the price of each orange is Tk. 2x.
According to the question,
50×3x + 30×2x = 420
⇒ x = 2
Price of each mango = 3×2
= Tk. 6.

Question: The compound interest on Tk. 30,000 at an annual rate of 7% amounts to Tk. 4,347. What is the time period (in years) for which the money was invested? 

Solution:
Amount = 30000 + 4347 = 34347 tk

Let, the time be n years,
Then,
30000{1 + (7/100)}ⁿ = 34347
⇒ (107/100)ⁿ = 34347/30000
⇒ (107/100)ⁿ = 11449/10000
⇒ (107/100)ⁿ = (107/100)²
∴ n = 2 years

Question: Express the following inequality using absolute value notation: 1 < x < 9

Solution:
1 < x < 9
∴ The midpoint = (1 + 9)/2
= 10/2
= 5
Now subtract the midpoint from all parts. then we get,
1 - 5 < x - 5 < 9 - 5
⇒ - 4 < x - 5 < 4
∴ |x - 5| < 4

Question: At a conference, everyone shakes hands with everybody else. If there were 190 handshakes, how many people were at the conference?

Solution:
Let the number of people at the conference be x.
A handshake occurs between any two people, which can be expressed using combinations.

According to the question,
^xC₂ = 190
⇒ x!/(2!(x - 2)!) = 190
⇒ {x(x - 1)(x - 2)!}/{2 × 1 × (x - 2)!} = 190
⇒ x(x - 1)/2 = 190
⇒ x(x - 1) = 380
⇒ x² - x - 380 = 0
⇒ x² - 20x + 19x - 380 = 0
⇒ x(x - 20) + 19(x - 20) = 0
⇒ (x - 20)(x + 19) = 0

So x - 20 = 0 or x + 19 = 0
∴ x = 20 or x = - 19

Since the number of people cannot be negative, x = 20 

Therefore, there were 20 people at the conference.

Question: A retailer buys a TV from a wholesaler at a 25% discount. He then marks up the price by 50% on the discounted price and offers a 20% discount to the customer. What is the retailer’s profit percentage?

Solution: 
Let the original price of the TV be 1000 Taka.

Wholesaler’s discount = 25%
Retailer’s purchase price:
1000 − 25% of 1000 = 1000−250 = 750 Taka

Retailer’s markup = 50% on Taka 750
New marked price:
750 + 50% of 750 = 750 + 375 = 1125 Taka

Customer’s discount = 20% on Taka 1125
Selling price:
1125 − 20% of 1125 = 1125 − 225 = 900 Taka

So, Profit = {(900 - 750)/750} × 100%
= 20%

Question: What will come at the place of the question mark?
4, 7, 12, 19, 28, ?

Solution:
7 - 4 = 3
12 - 7 = 5
19 - 12 = 7
28 - 19 = 9
Differences increase by 2 each time,
So, next difference = 9 + 2 = 11.
Next term = 28 + 11 = 39

Let the maximum marks be x.
Then,
(40−4)% of x =261
⇒ 36% of x =261
⇒ 36x/100 =261
⇒ x = (261×100/36)
⇒ x = 725

Question: Tanzid has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average by 8 runs. His new average is-

Solution:
Let Tanzid’s average for 9 innings = x
Then total runs in 9 innings = 9x

And,
In the 10th inning, he scores 100 runs
So, total runs after 10 innings = 9x + 100

∴ New average = x + 8  ; (because his average increased by 8)

ATQ,
(9x + 100)/10 = x + 8
⇒ 9x + 100 = 10x + 80
⇒ x = 100 - 80
∴  x = 20

∴ New average = x + 8 = 20 + 8 = 28

Question: The equation of a line that passes through the points (1, 5) and (2, 3) is:

Solution:
We know that the equation of a line passes through two points (x1, y1) and (x2 y2) is
(y - y₁)/(x - x₁) = (y₂ - y₁)/(x₂ - x₁)

Now, substitute the values in the formula, we get
(y - 5)/(x - 1) = (3 - 5)/(2 - 1) ; [(x₁, y₁) = (1, 5) and (x₂, y₂) = (2, 3)]
⇒ (y - 5)/(x - 1) = (- 2)/(1)
⇒ y - 5 = - 2(x - 1)
⇒ y - 5 = - 2x + 2
⇒ 2x + y - 5 - 2 = 0
∴ 2x + y - 7 = 0

Therefore, the equation of a line that passes through the points (1, 5) and (2, 3) is 2x + y - 7 = 0.

Question: A man completes a journey in 8 hours. He travels the first half of the journey at the rate of 40 km/hr and the second half at the rate of 60 km/hr. Find the total journey in km.

Solution:
ধরা যাক, মোট যাত্রার দূরত্ব হলো d কিমি।
তাহলে, যাত্রার প্রথম অর্ধেকের দূরত্ব হবে d/2 কিমি
এবং দ্বিতীয় অর্ধেকের দূরত্বও হবে d/2 কিমি।

প্রথম অর্ধেক যাত্রায়, সময় = দূরত্ব/গতিবেগ
= (d/2)/40 ঘন্টা
= d/80 ঘন্টা

দ্বিতীয় অর্ধেক যাত্রায়, সময় = দূরত্ব/গতিবেগ
= (d/2)/60 ঘন্টা
= d/120 ঘন্টা

প্রশ্নমতে,
(d/80) + (d/120) = 8
⇒ (3d + 2d)/240 = 8
⇒ 5d/240 = 8
⇒ 5d = 8 × 240
⇒ 5d = 1920
⇒ d = 1920/5
⇒ d = 384 কিমি

∴ মোট যাত্রার দূরত্ব 384 কিলোমিটার।

So, the man crosses (10 - 5) = 5 meters east everyday
As he will cross the 380th meter when for the last time he goes 10 m east, remaining distance will take (380 - 10)/5 = 74 days
Total 75 days will take to cross the building which is 380 m east.

Question: There are 5 men and 4 women. Two members are to be selected for a committee. In how many ways can two  people be selected?

Solution:
Total candidates = 5 + 4 = 9

Number of ways to select 2 people = ⁹C₂ 
= 9!/2!(9 - 2)!
= (9 × 8 × 7!)/(2  × 7!)
= 36

Question: If tan(θ - 30°) = √3, then find sinθ.

Solution:
Given,
tan(θ - 30°) = √3
⇒ tan(θ - 30°) = tan60°
⇒ θ - 30° = 60°
∴ θ = 90°

Now,
sinθ = sin90° = 1

30% of 10 = 10% of x
⇒ (30/100) × 10 = (10/100) × x
⇒ 3 = x/10
So, x = 30

তাদের মধ্যবর্তী সিটের সংখ্যা = (39 - 17) - 1 টি
= (22 - 1) টি
= 21 টি।

Question: 20 men can do a piece of work in 16 days. They worked together for 3 days, then 6 men joined. In how many days is the remaining work completed?

Solution:
20 men can complete the work in 16 days.
So, total work = 16 × 20 = 320

∴ Work done by 20 men in 3 days = 20 × 3 = 60

∴ Remaining work = 320 - 60 = 260

After joining 6 men total men = 20 + 6 = 26

So, Time required to finish the remaining work = 260/26=10 days

Question: If 5^{x + y} = 25 and 5^{x - y} = 5, then what are the values of x and y respectively?

Solution:
Given,
5^{x + y} = 25
⇒ 5^{x + y} = 5²
⇒ x + y = 2 .......(1)

Again,
5^{x - y} = 5
⇒ 5^{x - y} = 5¹
⇒ x - y = 1 ........(2)

Now, solving (1) and (2) we get,
x + y + x - y = 2 + 1
⇒ 2x = 3
∴ x = 3/2

Now,
x + y = 2
⇒ 3/2 + y = 2
⇒ y = 2 - (3/2)
​⇒ y = 1/2
​
​(x, y) = (3/2, 1/2)

Question: The range of f(x) = 1/(x + 1) is:

২০২২ সাল ভিত্তিক সমন্বিত ৮ ব্যাংক ও ১ আর্থিক প্রতিষ্ঠান পদের নাম: অফিসার (জেনারেল)
Solution:
দেওয়া আছে,
f(x) = 1/(x + 1)
⇒ y = 1/(x + 1)
⇒1/y = x + 1
⇒ x = (1/y) - 1
⇒ x = (1 - y)/y

∴ f^-1(x) = y = (1 - x)/x
x এর মান 0 ব্যতীত যেকোনো বাস্তব সংখ্যা হবে। কারণ x এর মান 0 হলে ফাংশনটি অসঙ্গায়িত হবে।

অতএব, নির্ণেয় রেঞ্জ: R\{0}

Let the larger number is = a
Then, the other number is = a - 5
ATQ,
a² – (a-5)² = 65
⇒ a² – a² + 10a – 25 = 65
⇒ 10a = 65 + 25 = 90
⇒ a = 90/10 = 9

Question: If x : y = 4 : 5 & y : z = 7 : 9, find x : y : z.

Solution: 
Given that, 
x : y = 4 : 5 = (4 × 7) : (5  × 7) = 28 : 35
∴ x : y = 28 : 35

And,
y : z = 7 : 9 = (7 × 5) : (9 × 5) = 35 : 45
∴ y : z = 35 : 45

∴ x : y : z = 28 : 35 : 45

Question: Tea worth Tk. 240 per kg and Tk. 280 per kg are mixed with a third variety in the ratio 3 : 2 : 5. If the mixture is worth Tk. 300 per kg, the price of the third variety per kg will be:

Solution:
Let the price of the third variety be x Tk. per kg.
The given ratio of the three varieties is 3 : 2 : 5.
For calculation, let the quantities be 3 kg, 2 kg, and 5 kg respectively.

Total weight of the mixture = (3 + 2 + 5) = 10 kg
Total value of the mixture = 10 × 300 = Tk. 3000

According to the question (ATQ),
(3 × 240) + (2 × 280) + (5 × x) = 3000
⇒ 720 + 560 + 5x = 3000
⇒ 1280 + 5x = 3000
⇒ 5x = 3000 − 1280
⇒ 5x = 1720
⇒ x = 1720 / 5
∴ x = 344

∴ The price of the third variety is Tk. 344 per kg.

Question: CMM, EOO, GQQ, _____, KUU

Solution:
প্রথম অক্ষর,
C → E → G → ? → K ; [+ 2 করে বাড়ছে]
C + 2 = E, E + 2 = G, G + 2 = I, I + 2 = K)
∴ ফাঁকা জায়গায় প্রথম অক্ষর = I

দ্বিতীয় অক্ষর,
M → O → Q → ? → U  ; [+ 2 করে বাড়ছে]
M + 2 = O, O + 2 = Q, Q + 2 = S, S + 2 = U
∴ ফাঁকা জায়গায় দ্বিতীয় অক্ষর = S

তৃতীয় অক্ষর, দ্বিতীয় অক্ষরের মত
∴ ফাঁকা জায়গায় তৃতীয় অক্ষর = S

সুতরাং ফাঁকা জায়গায় আসবে- ISS

Question: Which of the following is equivalent to the pair of inequalities 2x - 5 ≤ 7 and 3x + 4 > 10?

Solution:
Solve the first inequality,
2x - 5 ≤ 7 
⇒ 2x ≤ 7 + 5
⇒ 2x ≤ 12
∴ x ≤ 6
And,
Solve the second inequality,
3x + 4 > 10 
⇒ 3x > 10 - 4
⇒ 3x > 6
∴ x > 2

∴ We get 2 < x ≤ 6

Question: What decimal of an hour is a second ?

Solution: 
There are 60 seconds in a minute and 60 minutes in an hour.
So, 1 hour = 60 × 60 = 3600 seconds
Therefore, 1 second = 1/3600 = 0.00027 of an hour.

Question: A car is sold at a profit of 25%. Had it been sold for Tk. 15,000 less, there would have been a loss of 15%. What was the cost price?

Solution:
ধরি, গাড়িটির ক্রয়মূল্য = x টাকা

25% লাভে বিক্রয়মূল্য = x + x এর 25%
= x + 25x/100
= x + x/4
= 5x/4

15% ক্ষতিতে বিক্রয়মূল্য = x - x এর 15%
= x - 15x/100
= x - 3x/20
= 17x/20

প্রশ্নমতে,
(5x/4) - (17x/20) = 15000
⇒ (25x/20) - (17x/20) = 15000
⇒ 8x/20 = 15000
⇒ 2x/5 = 15000
⇒ 2x = 15000 × 5
⇒ 2x = 75000
⇒ x = 75000/2
∴ x = 37500

∴ গাড়িটির ক্রয়মূল্য = 37,500 টাকা

Question: What is the least number which, when tripled, becomes exactly divisible by 15, 20, 24, and 30?

Solution:
Let the required number be x.
Then, 3x must be divisible by 15, 20, 24, and 30.

Find the LCM of the numbers:
15 = 3 × 5
20 = 2² × 5
24 = 2³ × 3
30 = 2 × 3 × 5

∴ LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120

∴ 3x = 120
∴ x = 120/3 = 40

Question: The cube shown below has a volume of 64 cubic inches. What is the length of the line segment AB? 



Solution:
Let side length = a

Here,
Volume of the cube, a³ = 64
⇒ a³ = 4³
⇒ a = 4

∴ Length of the line segment (space diagonal) AB = √(a² + a² + a²)
= √(3a²) 
= a√3
= 4√3

Let the quantity of milk purchased be x and quantity of water added be y.
Then, the ratio of water to milk is y : x.
CP = 6.4x
SP = 8(x+y)
Profit per cent = 37.5%
Therefore,
8(x+y) = 6.4x × 1.375
Or, 8x + 8y = 8.8x
Or, 8y = 0.8x
Or, y/x= 0.8/8
∴ y : x = 1 : 10

Question: A sum of money doubles itself in 4 years at a certain rate of simple interest. In how many years will it become four times itself at the same rate of interest?

Solution:
মনে করি, আসল (P) = 100 টাকা
4 বছরে এটি দ্বিগুণ হয়ে হয় (A) = 200 টাকা
∴ 4 বছরের সুদ (I) = 200 - 100 = 100 টাকা

অর্থাৎ, 100 টাকা সুদ হতে সময় লাগে 4 বছর।

এখন, আসল চারগুণ হলে সবৃদ্ধিমূল (A) হবে = 400 টাকা
∴ তখন প্রয়োজনীয় মোট সুদ (I) = 400 - 100 = 300 টাকা

এখন,
100 টাকা সুদ হতে সময় লাগে 4 বছর
1 টাকা সুদ হতে সময় লাগে 4/100 বছর
300 টাকা সুদ হতে সময় লাগে (4 × 300)/100 বছর
= 12 বছর

∴ আসল বা মূলধন 12 বছরে চারগুণ হবে।