Bank Math

Question: If 6,250 Taka is invested at 12% simple interest, how much time will it take for the interest to reach 3,000 Taka?

Answer: 
Given, 
P = 6250 Taka
I = 3000 Taka
r = 12% = 0.12
Time = n

We know,
I = Pnr
n = I/Pr
n = 3000/(6250×0.12)
n = 4 years

Question: There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Tk. 12000 after 3 years at the same rate?

Solution:
Given that,
Increase in amount after 6 years = 60%

We know,
A = P(1 + r/100)n
simple Interest = (principal × rate × time)/100

Now,
Let the amount at the 1^st year be 100x
∴ Increased in amount = 60% of 100x = 60x
⇒ 60x = (100x × 6 × r)/100
⇒ 6r = 60
⇒ r = 60/6 = 10
∴ r = 10% 

∴ Compound Interest after 3 years = P(1 + r/100)ⁿ - P
= 12000(1 + 10/100)³ - 12000
= 12000 × (11/10)³ - 12000
= 12000 × (11/10) × (11/10) × (11/10) - 12000
= 15972 - 12000
= Tk. 3972

∴ The required answer is Tk. 3972

Question: A and B can finish a job together in x days. If A alone can complete the job in x + 3 days, and B alone can complete it in x + 12 days, what is the value of x?

Solution:
A's 1 day's work = 1/(x + 3) part
B's 1 day's work = 1/(x + 12) part
(A + B)'s 1 day's work = 1/x

ATQ,
1/(x + 3) + 1/(x + 12) = 1/x
⇒ (x + 12 + x + 3)/[(x + 3)(x + 12)] = 1/x
⇒ (2x + 15)/(x² + 15x + 36) = 1/x
⇒ 2x² + 15x = x² + 15x + 36
⇒ x² = 36
∴ x = 6

Question: The average of the first 6 numbers out of 7 is 25, and the average of the last 6 numbers is 29. If the first number is 10, what is the 7th number? 

Solution:
Given that,
First number = 10
Average of first 6 = 25
Average of last 6 = 29

Let,
the seven numbers be A₁, A₂, A₃, A₄, A₅, A₆, A₇
A₁ = 10

Sum of first 6 numbers = 6 × 25 = 150 …(1)
Sum of last 6 numbers = 6 × 29 = 174 …(2)

Subtract equation (1) from (2):
(A₂ + A₃ + A₄ + A₅ + A₆ + A₇) - (A₁ + A₂ + A₃ + A₄ + A₅ + A₆) = 174 - 150 
⇒ A₇ - A₁ = 24 
⇒ A₇ - 10 = 24 
⇒ A₇ = 34

∴ The 7th number is 34.

When roots are equal, b² = 4ac
⇒ b² = 4×9×81
⇒ b² = 36×81 = 6 × 6 × 9 × 9
∴ b = ± 54

Let Simple Interest be S
Original sum 'P' and Time 'T' remains the same
Also, 3% increase in rate of interested means, New rate = 15%
Increase of 300 Rs/- in the final amount means, New amount = Tk. 3300/-

So, Amount 2 - Amount 1 = (P + S₁) - (P + S₂) = S₁ - S₂
∴ 3300 - 3000 = S₁ - S₂
∴ S₁ - S₂ = 300

∴ {(P × R × T)/100} - {(P × 12 × T)/100 = 300
3PT/100 = 300
PT = 10000
For R = 15%,

Simple interest = (P × 15 × T)/100
= (10000 × 15)/100
= Tk. 1500

Original sum = Amount - Interest = 3300 - 1500 = Tk. 1800 (Principle amount)

Also, PT = 10000
T = 10000/1800
= 5.55 years.

At, A ; M : W = 20 : 28 = 5 : 7 = 5/12 : 7/12
At, B; M : W = 18 : 30 = 3 : 5 = 3/8 : 5/8
At C ; M : W = 8 : 40 = 1 : 5 = 1/6 : 5/6
M : W = (5/12 + 3/8 + 1/6) : ( 7/12 + 5/8 + 5/6)
= (20 + 18 + 8)/48 : (28 + 30 + 40)/48
= 46 : 98
= 23 : 49

Question: For what value of k (k > 0) will the lines 3x + 4y = 1 and 2x + Ky = 7 be parallel?

Solution:
Two lines are parallel if their slopes are equal.
Rewrite both equations in slope-intercept form (y = mx + c).

Now first line,
3x + 4y = 1
⇒ 4y = - 3x + 1
⇒ y = (- 3/4)x + 1/4
∴ Slope m₁ = - 3/4

And second line, 
2x + Ky = 7
⇒ Ky = - 2x + 7
⇒ y = (- 2/K)x + 7/K     (since k > 0, k ≠ 0)
∴ Slope m₂ = - 2/k

For the lines to be parallel:
m₁ = m₂
⇒ - 3/4 = - 2/k
⇒ 3/4 = 2/k
⇒ 3k = 8
∴ k = 8/3

So the lines will be parallel when k = 8/3.

Question: If x and y are positive real numbers, then (2x⁰ - 5y⁰)⁴ = ?

Solution:
We know that for any positive real number,
x⁰ = 1 and y⁰ = 1

So, (2x⁰ - 5y⁰)⁴
= (2 × 1 - 5 × 1)⁴
= (2 - 5)⁴
= (- 3)⁴
= 81

The value of (2x⁰ - 5y⁰)⁴ is 81

আমরা জানি, সমবাহু ত্রিভুজের শীর্ষবিন্দু হতে ভূমির উপর অঙ্কিত লম্ব ভূমিকে সমদ্বিখণ্ডিত করে
তাই, BD = DC = 1 cm
পীথাগোরাসের সূত্র অনুসারে ΔABD হতে পাই,
⇒ AB² = BD² + AD²
⇒ AD² = AB² - BD² = 2² - 1²
∴ AD = √3

Question: Find the H.C.F. of p(x) = 2x³ - 3x² - 2x + 3 and q(x) = 3x² + 8x + 5.

Solution:
Given that,
p(x) = 2x³ - 3x² - 2x + 3 and q(x) = 3x² + 8x + 5

Now,
The factors of p(x) = 2x³ - 3x² - 2x + 3
⇒ x²(2x - 3) - 1(2x - 3)
⇒ (x² - 1)(2x - 3)
⇒ (x + 1)(x - 1)(2x - 3)

And,
The factors of q(x) = 3x² + 8x + 5
⇒ 3x² + 5x + 3x + 5
⇒ x(3x + 5) + 1(3x + 5)
⇒ (3x + 5)(x + 1)

17% বৃদ্ধিতে বর্তমান মূল্য = 117 টাকা
বর্তমান মূল্য 117 টাকা হলে পূর্ব মূল্য = 100 টাকা
∴ বর্তমান মূল্য 25.74 টাকা হলে পূর্ব মূল্য = (100 × 25.74)/100 টাকা
                                                      = 22 টাকা
∴ প্রকৃত মূল্য বৃদ্ধি = 25.74 - 22 = 3.74

Amount = Tk.(30000+4347) = Tk.34347
let the time be n year
Then,
30000(1+7/100)n = 34347
(107/100)n = 34347/30000 = 11449/10000 = (107/100)2
n = 2year

Given that,
The speed of a boat in still water is 20 km/hr
The speed of the stream is 4 km/hr.
Then,
The speed downstream = (20 + 4) km/hr
= 24 km/hr
The distance travelled in half an hour = 24 x 1/2
= 12 km
Hence the distance travelled by the boat is 12 km.

Question: Find the equation of the vertical line passing through the point (5, - 3).

Solution:
একটি উল্লম্ব রেখা (vertical line) হলো এমন একটি সরলরেখা যা Y-অক্ষের সমান্তরাল। এই ধরনের রেখার একটি বিশেষ বৈশিষ্ট্য হলো, এই রেখার উপর অবস্থিত প্রতিটি বিন্দুর x-স্থানাঙ্ক (x-coordinate) একই থাকে, কিন্তু y-স্থানাঙ্ক (y-coordinate) পরিবর্তিত হতে পারে।

উল্লম্ব রেখার (vertical line) সাধারণ সমীকরণ হলো: x = a 
যেখানে a একটি ধ্রুবক সংখ্যা এবং রেখার প্রতিটি বিন্দুর x এর মান একই থাকে।

প্রশ্নে বলা হয়েছে রেখাটি (5, - 3) বিন্দুর মধ্য দিয়ে যায়।
যেহেতু, আমরা জানি, একটি উল্লম্ব রেখার প্রতিটি বিন্দুর x-স্থানাঙ্ক একই থাকে, এবং এই বিন্দুর x-স্থানাঙ্ক হলো 5, সুতরাং, রেখাটির সমীকরণ হবে:
x = 5

Let rate of stream be y km/hr, distance be d km
time upstream = 2 (time downstream)
⇒ d/(7.5 − y) = 2d/(7.5 + y)
⇒ 2( 7.5 − y ) = 7.5 + y
⇒ 15 − 2 y = 7.5 + y
⇒ 3 y = 7.5
⇒ y = 2.5

Tank filled or work done by P in 1 hour = 1/22
Tank filled or work done by Q in 1 hour (Q takes 11 hrs more than P) = 1/33

Tank filled or work done by both pipes in 1 hour = 1/22 + 1/33 = 5/66
So the entire tank is full in = 66/5 = 13(1/5) hours.

Question: A pole 90 meters long breaks into two parts without complete separation and makes an angle of 30° with the ground. Find the length of the broken part of the pole.

Solution:

ধরি, খুঁটির মোট দৈর্ঘ্য = 90 m
ভাঙা অংশটির দৈর্ঘ্য, AC = x 
অবশিষ্ট অংশটির দৈর্ঘ্য, AB = (90 - x) 

আমরা জানি, 
sin θ = লম্ব/অতিভুজ 
sin 30°= BC/CD
∴ 1/2 = (90 - x)/x
⇒ x = 2(90 - x)
⇒ x = 180 - 2x
⇒ 3x = 180
∴ x = 60 m

অতএব, খুঁটির ভাঙা অংশটির দৈর্ঘ্য = 60 meters।

Let x be the number of students in the class and y be the average weight of the class
Now according to question,
(xy + 50)/(x + 1) = y + 1
x + y = 49 .......... (i) Again,
(xy + 50 + 50)/(x + 2) = y + 1.5
1.5x + 2y = 97.......... (ii) From equation (i) and (ii), we get y = 47
Answer: 47

Question: Calculate the area of a triangle whose sides are 13 meters, 14 meters, and 15 meters.

Solution:
We know,
Area of a scalene triangle = √{s(s - a)(s - b)(s - c)}
where, s = (a + b + c)/2
= (13 + 14 + 15)/2 = 42/2
= 21 

∴ Area = √{21(21 - 13)(21 - 14)(21 - 15)}
= √(21 × 8 × 7 × 6 )
= √(9  × 4  × 196)
= 3 × 2 × 14
= 84 m²

Question: At a conference, everyone shakes hands with everybody else. If there were 105 handshakes, how many people were at the conference?

Solution:
ধরি, কনফারেন্সে x সংখ্যক লোক ছিল।
দুজন ব্যক্তির মধ্যে একটি হ্যান্ডশেক হয়, যা Combination দিয়ে প্রকাশ করা যায়।

প্রশ্নমতে,
^xC₂ = 105
⇒ x!/(2!(x - 2)!) = 105
⇒ {x(x - 1)(x - 2)!}/{2 × 1 × (x - 2)!} = 105
⇒ x(x - 1)/2 = 105
⇒ x(x - 1) = 210
⇒ x² - x - 210 = 0
⇒ x² - 15x + 14x - 210 = 0
⇒ x(x - 15) + 14(x - 15) = 0
⇒ (x - 15)(x + 14) = 0

হয় x - 15 = 0 অথবা x + 14 = 0
∴ x = 15 অথবা x = -14

যেহেতু লোকসংখ্যা ঋণাত্মক হতে পারে না, তাই x = 15

অতএব, কনফারেন্সে 15 জন লোক ছিল।

We know, the probability of an event = Favorable outcomes / Total outcomes.
In a case of two dices, Total outcomes = 6 × 6 = 36
A.T.Q favorable outcome is a sum of 6,
Ways of obtaining a sum of 6 (1,5), (5,1), (2,4), (4,2) and (3,3).
Thus, there are 5 ways in which 6 can be obtained using two dices.
Therefore, required probability, P = 5/36

Question: The equation cos²θ = (x + y)²/4xy is only possible when ?

Solution: 
Given that,
cos²θ = (x + y)²/4xy
⇒ 1 = (x + y)²/4xy   ; [Max value of cos²θ = 1 ]
⇒ 4xy = (x + y)²
⇒ 4xy = x² + y² + 2xy
⇒ 0 = x² + y² - 2xy
⇒ (x - y)² = 0
⇒ x - y = 0
⇒ x = y

Therefore, the given equation is possible only when x = y.

Required number = (Sum of all numbers) - [(Sum of first 5 numbers) + (Sum of last 3 numbers)]
= 50 x 9 - (54 x 5 + 3 x 52)
= 450 - (270 + 156)
= 450 - 426
= 24

 x + 1/x = 3
Or, (x + 1/x)² = 9
Or, (x - 1/x)² + 4.x.1/x = 9
Or, (x - 1/x)² = 9 - 4 = 5
Or, (x - 1/x) = √5

Question: Anika is taller than Bristi but shorter than Kabir. Rafi is shorter than Bristi. Shawon is taller than Anika but shorter than Kabir. Who is the smallest among them?

Solution: 
From the given statements,
Anika > Bristi
Kabir > Anika
∴ Kabir > Anika > Bristi

Rafi < Bristi
∴ Bristi > Rafi

And
Shawon > Anika
Kabir > Shawon
∴ Kabir > Shawon > Anika

Full order from tallest to smallest,
Kabir > Shawon > Anika > Bristi > Rafi

Therefore, Rafi is the smallest among them.

Question: A train having length 150 m passes a platform of 550 m length. The time taken for it is 56 seconds. In how much time will this train take to pass a platform of 250 m length?

Solution:
We know,
When a train passes a platform, the total distance to be covered is,
Length of train + Length of platform

Case 1:
Distance covered = 150 m + 550 m = 700 m
Time taken = 56 seconds
∴ Speed of the train = Distance/Time
= 700/56
= 12.5 m/s

Case 2: 
Length of new platform = 250 m
∴ Total distance to be covered now = 150 + 250 = 400 m
Time taken to pass the new platform = Distance/Speed
= 400m/12.5 m/s
= 32 seconds

So the train will take 32 seconds to pass the platform of 250 m length.

Question: If the cost of gas on burning 5 burners for 5 hours a day for 7 days is Tk. 525, then how many burners can be used for 10 days at 5 hours a day for Tk. 750?

Solution:
Given that,
5 burners 5 hours a day for 7 days is Tk. 525
Total burner-hours = (5 × 5 × 7) burner-hours
= 175 burner-hours

Cost per burner-hour =(525 ÷ 175)
= 3 Tk

∴ 750 Tk. total burner-hour = (750 ÷ 3) burner-hours
= 250 burner-hours

Let, the number of burners = x

ATQ,
x × 5 × 10 = 250
⇒ x × 50 = 250
⇒ x = 250 ÷ 50
∴ x = 5

There are 9 letters in the given word, out of which 4 are vowels.
In the word ''CANDIDATE'' we treat the vowels ''AIAE'' as one letter.
Thus, we have CNDDT(AIAE).
Now, we have to arrange 6 letters, out of which D occurs twice.
Therefore,
Number of ways of arranging these letters = 6!/2!
= 720/2
= 360 ways.
Now,
AIAE has 4 letters, in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = (1 × 2 × 3 × 4)/2
= 12
Therefore, required number of words = (360 × 12)
= 4320.

Question: If the sum of two numbers is 34 and their H. C. F and L. C. M are 2 and 144 respectively, the sum of the reciprocals of the two numbers is-

Solution:
Let the two numbers are, x and y then
x + y = 34
and xy = H. C. F × L. C. M = 2 × 144 = 288

Sum of their reciprocals = (1/x) + (1/y)
= (x + y)/xy
= 34/288
= 17/144

Question: The average of 15 numbers is 48. If the average of first 7 numbers is 44 and the average of last 7 numbers is 50, what is the middle number?

Solution:
Let
The middle number be x

Total sum of 15 numbers,
= 15 × 48
= 720

Sum of first 7 numbers,
= 7 × 44
= 308

Sum of last 7 numbers,
= 7 × 50
= 350

∴ Sum of first 7 numbers + x + Sum of last 7 numbers = 720
⇒ 308 + x + 350 = 720 
⇒ 658 + x = 720
⇒ x = 720 - 658
∴ x = 62

So the middle number be 62.

24 men complete work in 10 days.
In 1 day work done by 24 men = 1/10
In 1 day work done by 30 men = ?
∴ ? × 24 = (1/10) × 30
∴ ? = 1/8 = Amount of work done by 30 men in one day.

Thus in this case,
30 men complete the entire work in 8 days.
[If A completes a work in ''n'' days, in 1 days he completes 1/n amount of work; Conversely, if A completes 1/n amount of work in 1 day, he completes the entire work in 'n' days.]

Question: If the side length of a square is increased by 20%, by what percentage is the area of the square increased?

Solution:
ধরা যাক, বর্গক্ষেত্রের মূল বাহুর দৈর্ঘ্য = a একক
সুতরাং, মূল ক্ষেত্রফল = a² বর্গ একক

বাহুর দৈর্ঘ্য ২০% বৃদ্ধির পর নতুন বাহুর দৈর্ঘ্য = a + (a × 20/100)
= a + 0.2a
= 1.2a একক

নতুন ক্ষেত্রফল = (1.2a)²
= 1.44a² বর্গ একক

ক্ষেত্রফল বৃদ্ধি = নতুন ক্ষেত্রফল - মূল ক্ষেত্রফল
= 1.44a² - a²
= 0.44a² বর্গ একক

শতকরা বৃদ্ধির হার = (ক্ষেত্রফল বৃদ্ধি/মূল ক্ষেত্রফল) × 100%
= (0.44a²/a²) × 100%
= 44%

সুতরাং, বর্গক্ষেত্রটির ক্ষেত্রফল 44% বৃদ্ধি পাবে।

Question: A person took a loan of Tk. 2500 with simple interest for as many years as the rate of interest. If they paid Tk. 1600 as interest at the end of the loan period, what was the rate of interest?

Solution:
ধরি, সুদের হার = r%
যেহেতু সময় এবং সুদের হার সমান,
∴ সময়, n = r বছর
আসল (P) = 2500 টাকা
মোট সুদ (I) = 1600 টাকা
আমরা জানি,
সরল সুদ, SI = (P × r × n)/100
⇒ 1600 = (2500 × r × r)/100 [এখানে, n = r]
⇒ 1600 = 25 × r²
⇒ r² = 1600/25
⇒ r² = 64
⇒ r = √64
⇒ r = 8

∴ সুদের হার ছিল 8%।

Question: If 3x/4 = 1, then x is greater than 3/4 approximately by what percent?

Solution: 
Given that, 
3x/4 = 1
⇒ 3x = 4
∴ x = 4/3

∴ Difference = x - (3/4) = (4/3) - (3/4)
= (16 - 9)/12
= 7/12

∴ Percentage increase = (Difference/Original)​ × 100%
= {(7/12)/(3/4)}​ × 100%
= (7/9) ​ × 100%
= 77.78%
= 78%

∴ Rounded to the nearest whole number 78%.  (Correct answer গ)

Question: A and B are two positive integers such that A × B = 96. Which of the following cannot be the value of A + B?

Solution:
Factor pairs of 96:
(1, 96) → A + B = 97
(2, 48) → A + B = 50
(3, 32) → A + B = 35
(4, 24) → A + B = 28
(6, 16) → A + B = 22
(8, 12) → A + B = 20

So, possible values of A + B are:
97, 50, 35, 28, 22, 20
Among options, 49 is not possible.

Question: The difference between two numbers is 5 and the difference between their square is 65. What is the smaller number?

Solution:
Let the numbers be x and y
Then,
x - y = 5.........(i) 
x² - y² = 65
Or (x + y)(x - y) = 105
5(x + y) = 65
x + y = 13 ....... (ii)

by adding (i) and (ii) we get,
2x = 5 + 13
2x = 18
x = 9 

From (ii)  we get
x + y = 13
9 + y = 13
y = 13 - 9
y = 4

Hence, the smaller number is 4

Question: Solve for y, √(2y - 3) = 5 

Solution:
Given that,
⇒ √(2y - 3) = 5
⇒ (√(2y - 3))² = 5²
⇒ 2y - 3 = 25
⇒ 2y = 25 + 3
⇒ 2y = 28
⇒ y = 28/2
∴ y = 14