Bank Math

Let,
The distance between A and destination be X
The total time taken by the car to cover X = 8 hours
Since X/2 by 80km/hr and remaining X/2 by 100km/hr
Then by Time = distance/speed, we have
(X/2)/80 + (X/2)/100 = 8
⇒ 1/2 {(X/80 + X/100)} = 8
⇒ X/80 + X/100 = 16
⇒ (5X + 4X)/400 = 16
⇒ 5X + 4X = 6400
⇒ 9X = 6400
⇒ X = 6400/9
⇒ X = 711.11
Hence 711.11 km is the required answer.

Question: Find the value of cos(7π/6).

Solution:
cos(7π/6)
= cos(π + π/6)   [যেহেতু (π + π/6) তৃতীয় চতুর্ভাগে পড়ে এবং তৃতীয় চতুর্ভাগে cos ঋণাত্মক, তাই cos(π + θ) = - cosθ]
= - cos(π/6)
= - cos(30°)
= - √3/2

Question: What is the angle between the hour and minute hand of a clock when it is 9 : 30 pm?

Solution:
9টা 30 মিনিট = 9 + (30/60) ঘন্টা
= 9 + 1/2 = 19/2 ঘন্টা

আমরা জানি,
ঘণ্টার কাঁটা 12 ঘণ্টায় 360° ঘোরে।
∴ 1 ঘণ্টায় ঘোরে = 360°/12 = 30°
∴ 19/2 ঘণ্টায় ঘোরে = (30° × 19)/2
= 15 × 19 = 285°
আবার,

মিনিটের কাঁটা 60 মিনিটে 360° ঘোরে।
∴ 1 মিনিটে ঘোরে = 360°/60 = 6°
∴ 30 মিনিটে ঘোরে = 30 × 6° = 180°

∴ ঘড়ির কাঁটা দুটির মধ্যবর্তী কোণ = |285° - 180°| = 105°

Number of bricks = Volume of the wall/Volume of 1 brick
= (800×600×22.5)/(25×11.25×6)
= 6400

 ৪ ও 18 এর ল.সা.গু = 72
∴ নির্ণেয় পূর্ণ সংখ্যা =10000/72
 = 138.8 ≈ 138 টি

Question: If an exponent or index has base 15 and power zero, then which of the following will be its value?

Solution:
a⁰ = 1 (for any non-zero base a)

Now,
= 15⁰
= 1

Let daughter’s age be x and father’s age be 3x.
Father's age is 3 times more aged than his daughter,
therefore father’s present age = x + 3x = 4x
After 5 years,
The father’s age is 3 times more than his daughter's age.
(4x + 5) = 3 (x + 5)
(4x+5) = 3 (x+5)
(4x + 5) = 3 (x + 5)
x = 10

After 5 years it was (4x + 5),
then after further 5 years, father’s age = (4x +10) and daughter’s age = (x + 10)
(4x + 10)/(x + 10)
= [{(4 × 10) + 10}/(10 + 10)]
= 50/20
= 5/2
= 2.5
After further 5 years, the father will be 2.5 times of daughter’s age.

Let the Cost price of the Chair be y taka.
Selling price = y - 25% of y
                     = y - 25y/100
                     = 75y/100
                     = 3y/4
Therefore 3y/4 = 930
                 or, y = 1240
So, To gain 25%, Selling Price would be
= 1240 + 25% of 1280
= 1240 + 310
= 1550
= 1600 taka

a² + b² = 5ab
a²/ab + b²/ab = 5
a/b + b/a = 5
Squaring the both sides
(a/b)² + (b/a)² = (5)²
a²/b² + b²/a² + 2 × (a/b) × (b/a)= 25
a²/b² + b²/a² + 2 = 25
a²/b² + b²/a² = 25 -2
a²/b² + b²/a² = 23.

Question: The difference between compound interest and simple interest on a sum of Taka 10,000 for 2 years at 10% per annum compounded annually is - 

Solution: 
Simple interest = 10000 × 2 × (10/100)
= 2000 Taka

Compound Interest, 

So, difference = 2100 - 2000 = 100 Taka

Question: A boat travels 24 km downstream in 40 minutes. If the speed of the stream is 6 km/h, what is the speed of the boat in still water?

Solution:
স্রোতের অনুকূলে 40 মিনিটে যায় 24 কিমি
স্রোতের অনুকূলে 1 মিনিটে যায় 24/40 কিমি
স্রোতের অনুকূলে 1 ঘণ্টা বা 60 মিনিটে যায় (24 × 60)/40 কিমি
= 36 কিমি

∴ স্রোতের অনুকূলে বেগ = 36 কিমি/ঘণ্টা

দেওয়া আছে,
স্রোতের বেগ = 6 কিমি/ঘণ্টা।

∴ স্থির পানিতে নৌকার বেগ = স্রোতের অনুকূলে বেগ - স্রোতের বেগ
= 36 - 6 = 30 কিমি/ঘণ্টা।

Question: If log₄[log₃(log₂x)] = 0, then find the value of x.

Solution:
log₄[log₃(log₂x)] = 0
⇒ log₃(log₂x) = 4⁰  [log_bM = c ⇒ M = b^c]
⇒ log₃(log₂x) = 1
⇒ log₂x = 3¹  [log_bM = c ⇒ M = b^c]
⇒ log₂x = 3
⇒ x = 2³  [log_bM = c ⇒ M = b^c]
∴ x = 8

এখানে দুটি ধারা বিদ্যমান।
বিজোড় অবস্থানের ধারা = 4, 6, 8, 10
জোড় অবস্থানের ধারা = 9, 11, 13, 15

Question: A working partner receives a commission equal to 30% of the profit remaining after paying his commission. If his commission is Tk 12,000, find the total profit.

Solution:
Let,
the total profit be K taka
According to the question,
remaining profit after paying 30% working partners commission = (K - 12000)

∴ (K - 12000) × (30/100) = 12000
⇒ (K - 12000) = 12000 × (100/30)
⇒  (K - 12000) = 40,000
⇒ K = 40,000 + 12,000
∴ K = 52,000

so, Total profit = 52,000 TK

Question: Which of the following is the polynomial equation 7x⁴ + 3x² - 2x + 1 = 0?

Solution:
প্রদত্ত বহুপদী সমীকরণটি x এর সাপেক্ষে।
x এর সর্বোচ্চ ঘাত হলো 4 এবং তাই সমীকরণটির ঘাত 4
∴ এই সমীকরণটির মাত্রা (degree) হলো 4
সুতরাং, এটি একটি চতুর্ঘাত বা দ্বি-দ্বিঘাত সমীকরণ (Biquadratic equation)।

• বহুপদীর সর্বোচ্চ ঘাত 1 হলে তা রৈখিক (linear) সমীকরণ।
• বহুপদীর সর্বোচ্চ ঘাত 2 হলে তা দ্বিঘাত (quadratic) সমীকরণ।
• বহুপদীর সর্বোচ্চ ঘাত 3 হলে তা ত্রিঘাত (cubic) সমীকরণ।
• বহুপদীর সর্বোচ্চ ঘাত 4 হলে তা চতুর্ঘাত বা দ্বি-দ্বিঘাত (biquadratic) সমীকরণ।

Sum = (B.D.×T.D.)/(B.D.−T.D.)
= (120×110) / (120−110)
= 1320

Team won 40 games out of 60 and the remaining games were 30.
Total games = 60 + 30
= 90
70% of 90 = 63
Team has to win 63 games in total.
Team has already won 40.
∴ Games to win = 63 - 40
= 23

Question: In a right triangle, the length of one of the legs is 12 and the length of the hypotenuse is 13. What is the length of the other leg?

Solution:
এখানে, সমকোণী ত্রিভুজের (right triangle) অতিভুজ (hypotenuse) = 13 একক সমকোণ সংলগ্ন এক বাহু = 12 একক
সমকোণ সংলগ্ন অপর বাহু = a একক

প্রশ্নমতে,
a² + 12² = 13²
⇒ a² + 144 = 169
⇒ a² = 169 - 144
⇒ a² = 25
⇒ a = √25
⇒ a = 5

∴ সমকোণ সংলগ্ন অপর বাহুর দৈর্ঘ্য = 5 একক

Let length and speed of the train be x metre and v kmph respectively.
x/9 = (v − 2) × 5/18 ⋯ ( 1 )
x/10 = (v − 4) × 5/18 ⋯ ( 2 )
Dividing (1) by (2) gives,
10/9 = (v − 2)/(v − 4)
⇒ 10v − 40 = 9v − 18
⇒ v = 22
Substituting the value of v in (1)
x/9 = 100/18
⇒ x = 50

Let B's capital be tk. x.
Then, (3500 x 12)/7x = 2/3    
or, 14x = 126000
so, x = 9000 tk

Question: The lengths of the sides of a triangle are in the ratio 2 : 4 : 5. If the perimeter of the triangle is 66 cm, what is the length of the longest side?

Solution: 
মনে করি,
ত্রিভুজের তিন বাহুর দৈর্ঘ্য যথাক্রমে 2x, 4x এবং 5x
∴ ত্রিভুজের পরিসীমা = (2x + 4x + 5x)
= 11x

প্রশ্নমতে, 
⇒ 11x = 66
⇒ x = 66/11
⇒ x = 6

∴ দীর্ঘতম বাহুর দৈর্ঘ্য = (6 × 5) cm
 = 30 cm

Question: If the average of four consecutive odd integers is x, then in terms of x, which is the smallest among the following options?

Solution:
Let the four consecutive odd integers be represented by  n, n + 2, n + 4, n + 6. 
Their average is calculated by summing them and dividing by 4, which equals x. 

Their average is,
x = {(n + (n + 2) + (n + 4) + (n + 6)}/4
⇒ x = (4n + 12)/4 = 4(n + 3)/4
⇒ x = n + 3
∴ n = x - 3

So the smallest integer in terms of (x - 3).

Question: A does a work in 12 days, B in 15 days, and C can do half the work in 10 days. How long will it take them to complete the whole work if they work together?

Solution:
A,
12 দিনে করে কাজটির = 1 অংশ 
∴ 1 দিনে করে কাজটির = 1/12 অংশ

B,
15 দিনে করে কাজটির = 1 অংশ
∴ 1 দিনে করে কাজটির = 1/15 অংশ

C,
10 দিনে করে কাজটির = 1/2 অংশ
∴ 1 দিনে করে কাজটির = 1/20 অংশ

A, B ও C একত্রে করে = (1/12) + (1/15) + (1/20)
= (5 + 4 + 3)/60
= 12/60
= 1/5 অংশ

A, B ও C একত্রে 1/5 অংশ করে = 1 দিনে
∴ 1 বা সম্পূর্ণ অংশ করে = (1 × 5) দিনে
= 5 দিনে

F = (BD×TD)/(BD−TD)
= (200×100)/(200−100)
= 200−100/ 100
= Tk. 200 

April has 30 days. So Protik takes 30 days to build the pavement.
Mahmud is 25% faster than Protik

25% = 25/100 = .25
This means, if Protik is 1, then Mahmud is (1 + 0.25) = 1.25
Protik takes 30 days to do the work.
Mahmud will take = 30/1.24 = 24 days to get the work done.

Question: Arif bought a ticket of a cricket match for Tk. 25 and later sold the ticket to Rafi for Tk. 75. What was the percent increase in the price of the ticket?

Solution:
ক্রয়মূল্য = 25 টাকা
বিক্রয়মূল্য = 75 টাকা
লাভ = 75 - 25 = 50 টাকা

25 টাকায় লাভ হয় = 50 টাকা
1 টাকায় লাভ হয় = 50 / 25 টাকা
100 টাকায় লাভ হয় = (50 × 100)/25 টাকা = 200 টাকা

শতকরা লাভ 200%

 Question: Two pipes can fill a tank in 15 and 20 minutes respectively and a waste pipe can empty 2 gallons per minute. All the three pipes working together can fill the tank in 9 minutes. The capacity of the tank is-

Solution:
Let, the waste pipe empty the tank in x minutes.

According to the question,
⇒ (1/15) + (1/20) - (1/x) = (1/9)
⇒ 1/x = (1/15) + (1/20) - (1/9)
⇒ 1/x = (12 + 9 - 20)/180
⇒ 1/x = 1/180
∴ x = 180

A waste pipe can empty 2 gallons per minute In 180 minutes it can empty = 2 × 180 = 360 gallons.
∴ Capacity of the tank = 360 gallons.

Question: Rafi and Nabil started a business. Rafi invested Tk. 15,000 for 12 months, while Nabil invested Tk. 10,000 for 6 months. After 6 months, Nabil added another Tk. 5,000. If the total profit is Tk. 11,000, what is Nabil’s share of the profit?

Solution:
Rafi's share 15,000 × 12 = 180,000
Nabil's share (10,000 × 6) + (15,000 × 6)
= 60,000 + 90,000
= 150,000

Ratio of Rafi : Nabil = 180,000 : 150,000 = 6 : 5

∴ Nabil's profit = (5/11) × 11,000
= Tk 5,000

Question: What is the minimum percentage increase in the mean of set X {-4, -1, 0, 6, 9} if its two smallest elements are replaced with two different primes?

Solution: 
old mean = - 4 - 1 + 6 + 9 /5 = 2 

-4, -1 will be replaced by 2, 3 

new mean = 2 + 3 + 6 + 9 /5 = 4 

%increase = {(4 - 2)/2} × 100%
= 100%