Bank Math

Given: Speed of passenger train = 55 km/hr, length of goods train (P) = 250, length of passenger train (Q)= 200m
Hint:
Time = (P+Q)/(V1+V2) sec
Goods train and the passenger train move in opposite direction. Hence, the relative speed is the addition of two speeds.
Convert 55 km/hr into m/s
55 x (5/18) = 15.277 m/s
Therefore,
10 = (250+200)/(15.27+V2)
V2 = 29.73 m/s 

Let the speed of the stream be x km/hr. Then,
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 - x) km/hr

ATQ, 30/(15 + x) + 30/(15 - x) = 4(1/2)
Or, 900/(225 - X²) = 9/2
Or, 9X² = 225
Or, X²  = 25
∴ X = 5 km/hr

Question: For the word 'MAGIC' how many different types of arrangement are possible so that the vowels are always together?

Solution:
In the Word MAGIC 
There are 2 vowels: A, I 
They can be arranged in 2! = 2 ways

There are three consonants: M, G, C
As the vowels are always together, we consider them as 1 letter.
So, 4 letter can be arranged in 4! = 24 ways

∴ Total number of arrangement is 2 × 24 = 48 words

Let the rate of interest be R%
Amount due in 6 months
= 10 + simple interest on Tk. 10 for six months.
= {10 + 10 × R × (1/2)}/100
= 10 + (R/20)
With the formula mentioned,
3 = 100(10 + 9R/20)/{(100 × 6) + R × 6(6 - 1)}/(2 × 12)
⇒ 3 = (1000 + 5R)/{600 + (5R/4)}
⇒ 1800 + 15R/4 = 1000 + 5R
⇒ 5R/4 = 800
⇒ R = 640.
Hence interest rate is 640%

Let the total capital be Tk. x.
Then, Rahul's share = Tk. x/2
Amin's share = Tk. x/3
Akash's share = [x - {(x/2) + (x/3)}]
= Tk. x/6
∴ Required ratio = x/2 : x/3 : x/6 = 1/2 : 1/3 : 1/6
= 3 : 2 : 1

Question: (x - 35) is divisible by 36, 48, 60. Find the value of x.

Solution:
Find the LCM of 36, 48, and 60.
Prime factorization
36 = 2² × 3²
48 = 2⁴ × 3¹
60 = 2² × 3¹ × 5¹

LCM formula
LCM = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720
Let
x - 35 = 720
x = 720 + 35
x = 755

Total distance = Average speed × total time
So, 2 = 2/3×T
⇒ T = 3 hours
So, the whole journey took 3 hours
Since trip back took half as much time as the trip there, it took 2 hours to reach there and 1 hour to come back.
So, speed of the delivery cart while going to Lark Rise = distance/time = 1/2 mph

Given that,
A takes 20 minutes to fill and B takes 12 minutes to empty
Clearly,
tap B is faster than tap A.
And so, the tank will be emptied.
Half of the tank or 1/2 part of the tank is already filled.
Therefore,
we have to find the time taken to empty that 1/2 part.
Part filled by A in 1 minute = 1/20
Part emptied by B in 1 minute = 1/12.
Part emptied by (A + B) in 1 minute
= (1/12) – (1/20)
= (5 - 3)/60
= 1/30.
Therefore,
The time taken by (A + B) to empty the full tank is 30 minutes.
Time taken to empty 1/2 part of the tank is
= 30/2
= 15 minutes.

Question: Find the smallest positive integer that must be added to 11356 so that it becomes divisible by both 18 and 22.

Solution:
Since the number must be divisible by both 18 and 22, it must be divisible by their Least Common Multiple (LCM).

LCM of 18 and 22:
18 = 2 × 3²
22 = 2 × 11
∴ LCM = 2 × 3² × 11 = 198

Now, dividing 11356 by 198:
11356 = (198 × 57) + 70
Here, the remainder is 70.

Since we need to add a number to make it divisible by 198,
∴ Smallest integer to be added = Divisor - Remainder
= 198 - 70
= 128

Therefore, 128 must be added to 11356 to make it divisible by both 18 and 22.

Question: Karim started a business with Tk. 80,000. After some months, Rahim joined with Tk. 60,000. At the end of the year, the profit was divided in the ratio 8 : 3. For how many months was Rahim in the business?

Solution:
Let, Rahim joined for x months.

ATQ,
(80,000 × 12)/(60,000 × x) = 8/3
⇒ (80,000 × 12 × 3) = (60,000 × x × 8)
⇒ 2,880,000 = 480,000x
⇒ x = 2,880,000/480,000
⇒ x = 6

∴ Rahim joined for 6 months.

Question: n(A\B) + n(A ∩ B) = ?

Solution:
We know that,
A\B = {x : x ∈ A and x ∉ B} 

Example: A = {a, b, c} and B = {c, d}, 
then A ∩ B = {a, b, c} ∩ {c, d} = {c}
and A\B = {a, b, c} \ {c, d} = {a, b}

We can say that,
n(A\B) = n(A) - n(A ∩ B)

Now,
n(A\B) + n(A ∩ B) = n(A) - n(A ∩ B) + n(A ∩ B) = n(A)

We are given that the numbers m and n, when divided by 6, leave remainders of 2 and 3, respectively,
Hence, we can represent the numbers m and as 6p +2 and 6q + 3, respectively, where p and q are suitable integers.
Now, m - n = (6p + 2) - (6q + 3) = 6p - 6q - 1 = 6(p-q) - 1.
A remainder must be positive, so let's add 6 to this expression and compensate by subtracting 6 :
6(p - q) - 1 =
6 (p - q) - 6 + 6 - 1 =
6 (p - q) - 6 = 5 =
6 (p - q - 1) + 5
Thus, the remainder is 5,
and the answer is 5

Question: A train 150 meters long passes a signal post in 15 seconds. How long will it take to pass a bridge that is 450 meters long?

Solution:
Train's speed = Distance/Time
= 150/15 = 10 m/s

Total distance to pass the bridge,
= Length of train + Length of bridge
= 150 m + 450 m
= 600 m

∴ Required time = Distance/Speed
= 600/10
= 60 seconds
​= 1 minute

∴ The train will take 60 seconds or 1 minute to pass platform.

এখানে,
3 দ্বারা বিভাজ্য সংখ্যা = 100/3
ভাগফল 33 এবং ভাগশেষ 1
সুতরাং 3 দ্বারা বিভাজ্য সংখ্যা = 33 টি
3 ও 8 এর ল, সা, গু = 24
এখন 100/24 =
ভাগফল 4 এবং ভাগশেষ 4
∴ 3 ও 8 দ্বারা বিভাজ্য সংখ্যা = 4 টি
সুতরাং, (33 - 4) = 29 টি সংখ্যা 3 দ্বারা বিভাজ্য কিন্তু 8 দ্বারা বিভাজ্য নয়।

Question: How many terms are there in the geometric progression,
3, 6, 12, 24, …, 1536

Solution:
First term, a = 3
Common ratio, r = 6/3 = 2

Last term or nth term of GP = ar^{n - 1}
⇒ 1536 = 3 × (2^{n - 1})
⇒ 2^{n - 1} = 1536/3
⇒ 2^{n - 1} = 512
⇒ 2^{n - 1} = 2⁹
So, comparing the power,
Thus, n - 1 = 9
∴ n = 10

∴ Number of terms = 10

Question: How many years will it take for an investment of Tk.1000 to earn Tk. 200 in simple interest rate of 5%?

Solution: 
Given that,
Principal, P = 1000
Simple Interest, SI = 200
Rate of interest, r = 5%
Time, n = ?

We know,
n = I/Pr
= 200/(1000 × 5%)
= (200 × 100)/(1000 × 5)
= 4

So, it will take 4 years for the investment to earn Tk. 200 at 5% simple interest.

Question: Using the digits 2, 5, and 7 exactly once each, three-digit numbers are formed. One number is selected at random. What is the probability that it is divisible by 4?

Solution:
Total numbers = 3! = 6
Numbers are 257, 275, 527, 572, 725, 752

We know, 
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

∴ Check last two digits are 72 and 52 divisible by 4.

∴ Favourable outcomes = 572 and 752 = 2

∴ Probability = 2/6 = 1/3

Question: A man travels from his home to the office at 4 km/hr and reaches his office 5 min late. If the speed had been 5 km/hr he would have reached 10 min early. Find the distance from his home to the office?

Solution:
ধরি, বাড়ি থেকে অফিসের দূরত্ব = d কিমি।
4 কিমি/ঘণ্টা বেগে সময় লাগে = d/4 ঘণ্টা
5 কিমি/ঘণ্টা বেগে সময় লাগে = d/5 ঘণ্টা

সময়ের পার্থক্য = 5 মিনিট (দেরি) + 10 মিনিট (আগে) = 15 মিনিট
= 15/60 = 1/4 ঘণ্টা

প্রশ্নমতে,
(d/4) - (d/5) = 1/4
⇒ (5d - 4d) / 20 = 1/4
⇒ d / 20 = 1/4
⇒ d = 20 / 4
∴ d = 5

সুতরাং, বাড়ি থেকে অফিসের দূরত্ব 5 কিমি।

Question:

Solution:

Question: In a certain code, AXIOM is written as ZWHNL. How is WAXED written in that code?

সমাধান:
এই কোডিং প্যাটার্নটি হলো প্রতিটি অক্ষরের পূর্ববর্তী অক্ষর ব্যবহার করা।
A ⇒ Z
X ⇒ W
I ⇒ H
O ⇒ N
M ⇒ L
এই নিয়ম অনুযায়ী, WAXED শব্দটির কোড হবে:
W ⇒ V
A ⇒ Z
X ⇒ W
E ⇒ D
D ⇒ C

সুতরাং, WAXED এর কোড হবে: VZWDC

Question: Mahi was facing east. he walked 7 km forward and then after turning to his right walked 5 km. Again he turned to his right and walked 7 km. After this he turned back. Which direction was he facing at that time?

Solution: 
Mahi's Movement-



∴ If Mahi turned back from final position he must be facing in the direction of East.

Question: If 1 < p < 3, then which of the following could be true?
(I) p² < 2p
(II) p² = 2p
(III) p² > 2p

Solution:
Given: 1 < p < 3

Since p > 0,

(I) p² < 2p
⇒ p < 2
True for 1 < p < 2
Example: p = 1.5 gives 2.25 < 3.0
So, condition (I) could be true.

(II) p² = 2p
⇒ p = 2
Since 2 is within the given range (1< 2 < 3), condition (II) could be true.

(III) p² > 2p
⇒ p > 2
Example: p = 2.5, gives 6.25 > 5.0 
So, condition (III) could be true.

Conclusion: Since values of p in the range 1< p < 3 satisfy conditions (I), (II), and (III), the correct choice is: (E) I, II, and III.

Question: A man invests Tk. 8,100 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 = 3780 Tk

Question: At what profit percent must an article be sold so that by selling it at two-thirds of that price, there will be a loss of 20%?

Solution:
Let,
Cost Price be x .
Selling Price be y.

Selling at 2/3 of y, causes 20% loss,
So, 2y/3 = x - 20% of x
⇒ 2y/3 = x - (20x/100)
⇒ 2y/3 = x{(1 - (20/100)}
⇒ 2y/3 = x × (80/100)
∴ y = 6x/5

Profit = Selling Price - Cost Price
= (6x/5) - x
= (6x - 5x)/5
= x/5

∴ Profit Percentage = (Profit/Cost Price) × 100%
= {(x/5)/x} × 100%
= 20%

Question: The monthly incomes of two brothers are in the ratio 11 : 7 and their monthly expenditures are in the ratio 9 : 5. Each of them saves Tk. 4,800 per month. Find the monthly income of the first brother.

Solution: 
Let the monthly income of the two brothers are 11x and 7x
Let the monthly expenses of the two brothers are 9y and 5y
Since each saves Tk. 4800 per month.

Then we get,
11x - 9y = 7x - 5y
⇒ 4x = 4y
∴  x = y

Now, 
11x - 9y = 4800
⇒ 11x - 9x = 4800
⇒ 2x = 4800
⇒ x = 4800/2
∴ x = 2400
∴ y = 2400

Therefore, 
Monthly income of the first brother = 11x = 11 × 2400 = Tk. 26400

So the monthly incomes are Tk. 26400.

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ × ⁶C₁)
As remaining person can be any one among three couples left.
Required probability = (⁵C₂ × ⁶C₁)/¹⁰C₅
= (10 x 6)/252
= 5/21

Question: If the fractions 7/13, 2/3, 4/11, 5/9 are arranged in ascending order, then the correct sequence is ?

Solution:
Given that,
(7/13) = 0.538
(2/3) = 0.666
(4/11) = 0.3636
(5/9) = 0.5555

Out of 2/3, 7/13, 4/11, 5/9

2/3 is the largest number followed by 5/9 then 7/13 and the smallest is 4/11.

∴ The ascending order will be 4/11, 7/13, 5/9, 2/3. 

Question: A man walk at a speed of 10 km/h. After every 3 kilometers, he takes a rest of 5 minutes. How much time will he take to cover a distance of 15 km?

সমাধান:
দূরত্ব = 15 কিমি
গতিবেগ = 10 কিমি/ঘন্টা

∴ সময় = দূরত্ব/গতিবেগ
= 15 কিমি/10 কিমি/ঘন্টা
= 1.5 ঘন্টা
= 90 মিনিট

সে প্রতি 3 কিলোমিটারের পরে 5 মিনিট বিশ্রাম নেয়।
15 কিমি দূরত্বে সে (15/3) - 1 = 4 বার বিশ্রাম নেবে (3 কিমি, 6 কিমি, 9 কিমি এবং 12 কিমি অতিক্রম করার পর)।
শেষ কিলোমিটারের পরে তার আর বিশ্রাম নেওয়ার প্রয়োজন নেই।

মোট বিশ্রামের সময় = 4 × 5 মিনিট = 20 মিনিট।

∴ মোট সময় = হাঁটার সময় + বিশ্রামের সময়
= 90 মিনিট + 20 মিনিট
= 110 মিনিট
= 1 ঘন্টা 50 মিনিট

সুতরাং, লোকটি 15 কিমি দূরত্ব অতিক্রম করতে মোট 110 মিনিট বা 1 ঘন্টা 50 মিনিট সময় নেবে।

Question: Which of the following is irrational?

Solution:
​√15 একটি অমূলদ সংখ্যা (irrational number)।
অমূলদ সংখ্যা (irrational number):
- যে সংখ্যাকে p/q আকারে প্রকাশ করা যায় না, যেখানে p ও q পূর্ণসংখ্যা এবং q ≠ 0, সে সংখ্যাকে অমূলদ সংখ্যা বলা হয়।
- পূর্ণবর্গ নয় এরূপ যে কোনো স্বাভাবিক সংখ্যার বর্গমূল কিংবা তার ভগ্নাংশ একটি অমূলদ সংখ্যা। যেমন, √2 = 1.414213..., √6 = 2.229489... ইত্যাদি অমূলদ সংখ্যা।
- কোনো অমূলদ সংখ্যাকে দুইটিপূর্ণ সংখ্যার অনুপাত হিসেবে প্রকাশ করা যায় না।
-অমূলদ সংখ্যাকে একটি মূলদ সংখ্যা দ্বারা গুণ করলে অমূলদ সংখ্যা পাওয়া যায়।
অর্থাৎ, non zero rational number × irrational number = irrational number.

Question: Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice.

Solution: 
Let x be required events and S be the sample space
Total outcomes when two dice are thrown,
6 × 6 = 36 
n(S) = 36
And then x = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (6, 2), (3, 4), (6, 4), (3, 6)}
n(x) = 11

Hence, required probability
= n(x)/n(S) = 11/36

Question: Find the number of triangles that can be formed by joining the angular points of a polygon of 10 sides as vertices.

Solution: 
A triangle needs 3 points.
A polygon of 10 sides has 10 angular points.

Hence, the number of triangles formed = ¹⁰C₃
= (10 × 9 × 8)/(3 × 2 × 1)
= 3 × 5 × 8
= 120

Let the speed of the boat in still water=x km/hr
Speed of the current = 2 km/hr

Then, speed downstream = (x + 2) km/hr
speed upstream = (x - 2) km/hr
Total time taken to travel 10 km upstream and back = 55 minutes
= (55/60) hr
= 11/12 hr.

According to question,
10/(x - 2) + 10/(x + 2) = 11/12
⇒ 120(x + 2)+120(x - 2) = 11(x² - 4)
⇒ 240x = 11x² - 44
⇒ 11x² - 240x - 44 = 0
⇒ 11x(x -22) + 2(x - 22) = 0
⇒ (x -22) (11x + 2) = 0
Since x cannot be negative.
So, x = 22 km/hr.

Hence, the Speed of the motorboat is 22 km/hr.

Question: The slope of the line 4x - 8y = 16 is not the same as the slope of which one of the following lines?

Solution:
প্রথমে, প্রদত্ত রেখাটির ঢাল নির্ণয় করতে হবে। রেখাটির সমীকরণকে y = mx + c আকারে রূপান্তর করতে হবে। এখানে 'm' হলো ঢাল (Slope)।

প্রদত্ত রেখার সমীকরণ:
4x - 8y = 16
⇒ - 8y = - 4x + 16
 ⇒ y = (- 4/- 8)x + (16/- 8)
⇒ y = (1/2)x - 2
∴ এই রেখাটির ঢাল (m) হলো 1/2.

এবার, প্রদত্ত অপশনগুলোর প্রত্যেকটির ঢাল নির্ণয় করি:

ক) x - 2y = 8
⇒ - 2y = - x + 8
⇒ y = (- x/- 2) + (8/- 2)
⇒ y = (1/2)x - 4
∴ ঢাল, m = 1/2

খ) 3x - 6y = 12
⇒ - 6y = - 3x + 12
⇒ y = (- 3/- 6)x + (12/- 6)
⇒ y = (1/2)x - 2
∴ ঢাল, m = 1/2

গ) y = 3x + 5
∴ ঢাল, m = 3

ঘ) y = x/2 + 4
⇒ y = (1/2)x + 4
∴ ঢাল, m = 1/2

সুতরাং, দেখা যাচ্ছে যে শুধুমাত্র অপশন (গ) এর রেখার ঢাল মূল রেখার ঢাল থেকে ভিন্ন।
∴ সঠিক উত্তর: গ) y = 3x + 5

ATQ,
6x - 5 : 4x - 5 = 5 : 3
Or, 20x - 25 = 18x - 15
Or, 2x = 10
Or, x = 5
So, John's age now is = 6×5 = 30 years

Question: If the 11th number in a series of 11 consecutive integers has the value k + 15, what is the 1st number in the series expressed in terms of k?

Solution:
Let the 1st number in the series be x.
Since there are 11 consecutive integers, the series is: x, (x + 1), (x + 2), ..., (x + 10).

According to the question, the 11th number is k + 15.
So,
x + 10 = k + 15
⇒ x = k + 15 - 10
⇒ x = k + 5

∴ The 1st number in the series is k + 5.

Question: A thief committed a crime and escaped from the spot at a speed of 12 m/s. A Security guard started chasing him 20 minutes after the thief started running and caught him in the next 20 minutes. What is the speed (in m/s) of the Security guard?

Solution: 
Given that,
Thief's speed = 12 m/s
Security guard starts 20 minutes or 1200 s later and catches thief in next 20 minutes or 1200 s

Now,
Distance covered by thief before guard starts,
d₁ = speed × time = 12 × 1200 = 14400 m ; [20 minutes = 1200 s]
So, thief has a 14400 m head start.

And
Guard catches thief in 20 minutes = 1200 s
d₂ = 12 × 1200 s = 14400 m 
∴ Guard covers total distance = head start + distance thief runs during chase = 14400 + 14400 = 28800 m
and Time taken by guard = 1200 s

∴ Speed of guard = Distance/Time = 28800/1200 = 24 m

Question: A rectangular hall is 12.5 meters long and 6.4 meters wide. The cost of installing wooden flooring is Tk. 120 per square meter. What is the total cost of flooring the hall?

Solution: 
Given that,
Length of the hall = 12.5 meters
Width of the hall = 6.4 meters
Cost of wooden flooring = Tk. 120 per square meter

We know,
Area = Length × Width
= 12.5 m × 6.4 m
= 80 m²

∴ Total cost = Area × Cost per square meter
= 80 × 120 
= Tk. 9600 

Therefore, the total cost of installing wooden flooring in the hall is Tk. 9600.