Bank Math

Question: Find the value of Cos(3π/4).

Solution:
Cos(3π/4)
= Cos{π - (π/4)}
= - Cos(π/4) [(π - θ) দ্বিতীয় চতুর্ভাগে পড়ে এবং দ্বিতীয় চতুর্ভাগে Cos ঋণাত্মক, তাই Cos(π - θ) = - Cosθ ]
= - Cos(45°)
= - (1/√2)

Question: Find

Solution:

The pattern is followed by
155 - 4 = 151
151 - 7 = 144 {7 = 4 + 3}
144 - 12 = 132 {12 = 7 + 5}
132 - 19 = 113 { 19 = 12 + 7}
113 - 28 = 85 { 28 = 19 + 9}
Hence ? = 85.

Let,
Cost of 1 Notebook = x tk.
ATQ,
300/x = 300/(x + 5) + 10
Or, (300/x) – 300/(x + 5) = 10
Or, (300x + 1500 - 300x)/{x(x + 5)} = 10
Or, x(x + 5)10 = 1500
Or, x(x + 5) = 1500/10 
Or, x² + 5x - 150 = 0
Or, x² + 15x - 10x - 150 = 0
Or, (x + 15)(x - 10) =0
Or, x ≠ - 15, x = 10
As, amount of money can't be negeative, the answer is 10 tk

Question: If one fifth of one sixth of a number is 10, then what is 2/5 of the number?

Solution:
ধরি, সংখ্যাটি = x

প্রশ্নমতে,
(1/5) × (1/6) × (x) = 10
⇒ (1/30)x = 10
⇒ x = 10 × 30
∴ x = 300

∴ সংখ্যাটি = 300

এখন,
সংখ্যাটির 2/5 অংশ = (2/5) × 300
= 600/5
= 120

Question: 

Solution: 

Question: What is the angle between the hour and minute hands of a clock when it is 4 : 20 pm?

Solution:
4টা 20 মিনিট = 4 + (20/60) ঘন্টা
= 4 + 1/3 = 13/3 ঘন্টা

আমরা জানি,
ঘণ্টার কাঁটা 12 ঘণ্টায় 360° ঘোরে।
∴ 1 ঘণ্টায় ঘোরে = 360°/12 = 30°
∴ 13/3 ঘণ্টায় ঘোরে = (30° × 13)/3
= 390°/3 = 130°

আবার,
মিনিটের কাঁটা 60 মিনিটে 360° ঘোরে।
∴ 1 মিনিটে ঘোরে = 360°/60 = 6°
∴ 20 মিনিটে ঘোরে = 20 × 6° = 120°

∴ ঘড়ির কাঁটা দুটির মধ্যবর্তী কোণ = |130° - 120°| = 10°

Given number of boxes = 14
Number of workers = 4
Now, number of whole boxes per worker = 14/4 = 3.5
Hence, number of whole boxes per each coworker = 3

Question: Today is Hasan's 12^th birthday and his father's 40^th birthday. How many years from today will Hasan's father be twice as old as Hasan at that time?

Solution:
Given that,
Today, Hasan is 12 years old
And his father is 40 years old

Age difference = 40 - 12 = 28 years (constant)

Let after x years, father's age = 2 × Hasan's age

Now,
Hasan's age after x years = 12 + x
Father's age after x years = 40 + x

ATQ,
40 + x = 2 × (12 + x)
⇒ 40 + x = 24 + 2x
⇒ x = 40 - 24

∴ x = 16

So, in 16 years, Hasan's father will be twice as old as Hasan.

The population grew from 3600 to 4800 in 3 years.
That is a growth of 1200 on 3600 during a three year span.
Therefore, the rate of growth for three years has been constant.

The rate of growth during the next three years will also be the same.
Therefore, the population will grow from 4800 by 4800 × 1/3 = 1600
Hence, the population three years from now will be 4800 + 1600 = 6400

Question: If the sum of two numbers is 18 and the sum of their squares is 234, then what is the product of the two numbers?

Solution:
সংখ্যা দুটি যথাক্রমে x এবং y
∴ x + y = 18 এবং x² + y² = 234

আমরা জানি,
(x + y)² = x² + y² + 2xy
⇒ (18)² = 234 + 2xy
⇒ 324 = 234 + 2xy
⇒ 2xy = 324 - 234
⇒ 2xy = 90
⇒ xy = 90/2
∴ xy = 45

Question: A boat travels 120 km upstream in 6 hours and the same distance downstream in 4 hours. What is the speed (in km/h) of the stream?

সমাধান:
ধরি, 
​স্থির জলে নৌকাটির গতিবেগ = x কিমি/ঘন্টা
এবং স্রোতের গতিবেগ = y কিমি/ঘন্টা

স্রোতের প্রতিকূলে গতিবেগ = (x - y) কিমি/ঘন্টা
স্রোতের অনুকূলে গতিবেগ = (x + y) কিমি/ঘন্টা

প্রশ্নমতে,
120/(x - y) = 6
⇒ x - y = 120/6
∴ x - y = 20 …………(1)

আবার,
120/(x + y) = 4
⇒ x + y = 120/4
∴ x + y = 30 …………(2)

(2) নং থেকে (1) নং সমীকরণ বিয়োগ করে পাই,
(x + y) - (x - y) = 30 - 20
⇒ x + y - x + y = 10
⇒ 2y = 10
⇒ y = 10/2
∴ y = 5

সুতরাং, স্রোতের গতিবেগ 5 কিমি/ঘন্টা।

Question: The diagonal of a rectangular field is 15 m and its area is 108 sq. m. What will be the total expenditure in fencing the field at the rate of Tk. 5 per metre?

Solution:
Let the length and breadth of the rectangular field be x metres and y metres respectively.

Given that,
Diagonal, √(x² + y²) = 15
⇒ x² + y² = 225
And area, xy = 108 m²

We know, 
(x + y)² = x² + y² + 2xy
⇒ (x + y)² = 225 + 2 × 108
⇒ (x + y)² = 225 + 216
⇒ (x + y)² = 441
⇒ x + y = √441 = 21
∴ x + y = 21

∴ Perimeter of the field = 2(x + y) = 2(21) = 42 m

∴ Total expenditure for fencing = Perimeter × Rate
= 42 m × Tk. 5 per metre
= Tk. 210

So the total expenditure is Tk. 210.

S.I. for 1 year=Tk.(854−815)=Tk.39
S.I. for 3 year=Tk.(39×3)=Tk. 117
∴Principal=Tk.(854−117)=Tk. 698

Question:

Solution:

After the first bounce it reached to 125 inches.
& there4; After the second bounce it reached to 125 × 2/5 = 50 inches.
& there4; After the third bounce it related to 50 × 2/5 = 20 inches.
& there4; After the fourth bounce it related to 20 × 2/5 = 8 inches.
Answer: 8 inches.

The digits to be used are 0,6 and 9
The required numbers are from 1 to 99999
The numbers are five digit numbers.
Therefore, every place can be filled by 0, 6 and 9 in 3 ways.
Total number of ways = 3 × 3 × 3 × 3 × 3 = 3⁵
But 00000 is also a number formed and has to be excluded.
Total number of numbers,
= 3⁵ - 1
= 243 - 1
= 242

Let, the length of the rectangle is l and breadth is b.
Where the breadth of the rectangular field is 60% of its length.
∴ b = 60l/100
      = 3l/5

Given that, Perimeter of the field = 800 m
⇒ 2(l + b) = 800
⇒ 2{l + (3l/5)} = 800
⇒ l + (3l/5) = 400
⇒ 8l/5 = 400
⇒ l = 250 m.
∴ b = 3l/5
= (3 × 250)/5
= 150 m

∴ Area = lb
= (250 × 150)
= 37500 m²

Question: A water pump fills a tank in 8 hours. If two identical pumps work together, how long will it take to fill the same tank?

Solution: 
One pump fills the tank in 8 hours.
Rate of one pump = 1/8 per hour

If two identical pumps work together, their rates add up.
So, the combined rate is = 1/8 + 1/8
= 1/4 per hour

∴ Time = 1/(1/4) = 4 hours

A:B:C = 35000:45000:55000 = 7:9:11
A's share = (7/27) ×40500 = Tk 10500
B's share = (9/27) ×40500 = Tk 13500
C's share = (11/27)×40500 = Tk 16500

Length of train = x metre (let)
Speed of train
5x + 1250 = 3x + 1500.
⇒ 5x – 3x = 1500 – 1250.
⇒ 2x = 250.
∴ x= 125 metres.

Let,
The amount invested at 12% be Tk. x and that invested at 10% be Tk. y
Then, 12% x + 10% of y = 130
⇒ 12x + 10y = 13000
⇒ 6x + 5y = 6500 .............(i)

And, 10% x + 12% of y = 134
⇒ 10x + 12y = 13400
⇒ 5x + 6y = 6700 .......(ii)

Adding (i) and (ii) we get,
11(x + y) = 13200
⇒ x + y = 1200 ..........(iii)

Subtracting (i) from (ii) we get,
x + y = 200 .........(iv)

Adding (iii) and (iv) we get,
2y = 1400
⇒ y = 700

∴ x = 1200 - 700 = 500

So,
The amount invested at 12% is TK 500
And the amount invested at 10% is TK 700.

According to question,
3a+4b /3a−4b = 3c+4d/ 3c−4d
⇒ 3a/4b = 3c/4d
⇒ad=bc

Question: sin(A + 18°) = √3/2, find the value of A.

Solution:
sin(A + 18°) = √3/2
⇒ sin(A + 18°) = sin60°
⇒ A + 18° = 60°
⇒ A = 60° - 18°
∴ A = 42°

Question: A certain principal amount, invested at simple interest, grows to Tk. 920 after 2 years and Tk. 1010 after 5 years. What is the original principal amount?

Solution:
Given,
Amount after 2 years = Tk. 920
Amount after 5 years = Tk. 1010

∴ Interest for (5 - 2) = 3 years = 1010 - 920 
= Tk. 90
∴ Interest for 1 year = 90/3 = Tk. 30
∴ Interest for 2 years = 30 × 2 = Tk. 60

∴ Principal = 920 - 60 = Tk. 860

Question: A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm³, then the weight of the pipe is:
(একটি ফাঁপা লোহার পাইপের দৈর্ঘ্য ২১ সেন্টিমিটার এবং এর বাইরের ব্যাস ৮ সেন্টিমিটার। যদি পাইপের পুরুত্ব ১ সেন্টিমিটার হয় এবং লোহার ঘনত্ব ৮ গ্রাম/সেন্টিমিটার^৩ হয়, তাহলে পাইপের ওজন কত হবে?)

Solution: 
পাইপটির দৈর্ঘ্য (L) = 21 cm
বাহ্যিক ব্যাস (D) = 8 cm
পাইপের প্রস্থ (t) = 1 cm

বাহ্যিক ব্যাসার্ধ R_external​ হল বাহ্যিক ব্যাসের অর্ধেক: 8/2 = 4 cm
অভ্যন্তরীণ ব্যাসার্ধ R_internal​ হল বাহ্যিক ব্যাসার্ধ থেকে প্রস্থ বিয়োগ:
R_internal = R_{external - t} = 4 - 1= 3 cm

লোহার আয়তন = π (4² - 3²) 21 [πr²h সূত্রানুসারে]
=  π (16 - 9) 21
= π 7 × 21
= 462 cm³

পাইপের ওজন = 462 × 8 g
= 3696 g
= 3.696 kg

Question: If 40 workers can construct a road in 30 days, how many workers are needed to construct the same road in 20 days?

Solution:
Here,
M₁ = 40, M₂ = ?, D₁ = 30, D₂ = 20

∴ (M₁ × D₁) = (M₂ × D₂) 
⇒ (40 × 30) = (20 × M₂)
⇒ M₂ = (40 × 30)/20 
⇒ M₂ = 60

So 60 workers are needed to construct the same road in 20 days.

We are to choose 11 players including 1 wicket keeper and 4 bowlers or 1 wicket keeper and 5 bowlers.
Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in ²C₁ × ⁵C₄ × ⁹C₆ = 840
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in ²C₁ × ⁵C₅ × ⁹C₅ = 252
Total number of ways of selecting the team = 840 + 252 = 1092

Here,
The selling price of 10 m cloth is obtained as profit.
Profit of 10 m cloth = (S.P. of 30 m cloth) – (C.P. of 30 m cloth)
The selling price of 20 m cloth = Selling Price of 30 m of cloth

Let the cost of each metre be Tk. 100.

Therefore,
the cost price of 20 m cloth = 20 × 100 = Tk. 2000 and
S.P. of 20 m cloth = Tk. 3000

Profit% = (10/20) × 100
= 50%

Question: Four girls are sitting on a bench to be photographed. Where Nila is to the right of Sara. Maya is to the left of Sara. And Rima is between Sara and Nila. Who would be second from the left in the photograph?
 
Solution:
Nila is to the right of Sara
∴ Sara — Nila

Maya is to the left of Sara
∴ Maya — Sara

Rima is between Sara and Nila
∴ Sara — Rima — Nila

Now place all four together:
∴ Maya — Sara — Rima — Nila

∴ the second from the left is Sara.

Question: A gardener planted trees in rows and columns such that the number of rows is five more than the number of columns. If the total number of rows and columns is 105, find the number of trees.

Solution:
Let the number of columns = x.
Then, number of rows = x + 5

According to the question: x + (x + 5) = 105
⇒ 2x + 5 = 105
⇒ 2x = 100
⇒ x = 50

Number of rows = x + 5 = 55

Total number of trees = rows × columns = 55 × 50 = 2750

Question: In a circle, if the inscribed angle on an arc is 35°, what is the measure of the central angle subtended by the same arc?
(কোন বৃত্তের একই চাপের উপর দণ্ডায়মান বৃত্তস্থ কোণ 35° হলে, কেন্দ্রস্থ কোণের পরিমাণ কত?)

Solution:
দেয়া আছে,
একই চাপের উপর দণ্ডায়মান বৃত্তস্থ কোণ = 35°
আমরা জানি,
কোন বৃত্তের একই চাপের উপর দণ্ডায়মান কেন্দ্রস্থ কোণ বৃত্তস্থ কোণের দ্বিগুণ।

∴ কেন্দ্রস্থ কোণ = 2 × বৃত্তস্থ কোণ
=2 × 35°
=70°

অতএব, কেন্দ্রস্থ কোণের পরিমাপ 70°।

There are 12 triangles in the figure.
These are: ABF, AEF, BCF, DEF, CDF, ACF, ADF, ACE, ABD, BCD, CDE, ACD

Question: The complement of an angle is 4 times the angle. Find the angle.

Solution:
Let the angle be x degrees.
The complement of an angle = 90° - x

According to the question,
90° - x = 4x
⇒ 90° = 4x + x
⇒ 90° = 5x
⇒ x = 90°/5
∴ x = 18°

∴ The angle is 18°

As the toys are distinct and not identical,
For each of the 8 toys, we have three choices as to which child will receive the toy.
Therefore, there are 5⁸ ways to distribute the toys.
Hence, it is 5⁸ and not 8⁵.