Bank Math

Question:  (32)^2/5 + (243)^1/5 = 3k , the value of k is:

Solution:
32^(2/5) = (2⁵)^2/5 = 2² = 4,
243^(1/5) = (3⁵)^1/5 = 3

∴ (32)^2/5 + (243)^1/5 = 3k
 ⇒ 4 + 3 = 3k
⇒ 7 = 3k
⇒ 3k = 7
⇒ k = 7/3

Principle,
= (P.W. of Tk. 882 due 1 year hence) + (P.W of Tk. 882 due 2 years hence)
= [{882/(1 + 5/100)} + {882/(1 + 5/100)²}]
= [{(882 × 20)/21} + {(882 × 400)/441}]
= Tk. 1640.

Question: Sakib can paint 10 desks in 20 minutes. Tamim can paint the same number of desks in only 10 minutes. Sakib paints the first 5 desks, then Tamim paints for 3 minutes, and finally Sakib finishes the remaining desks. How long will it take them to paint 15 desks in total?

Solution:
তামিম 10 মিনিটে রঙ করে 10টি ডেস্ক।
∴ 3 মিনিটে সে রঙ করতে পারে = (10 × 3)/10
= 3টি ডেস্ক

সাকিব প্রথমে রঙ করে 5টি ডেস্ক।
∴ এপর্যন্ত মোট সম্পন্ন কাজ = 5 (সাকিব) + 3 (তামিম) = 8টি ডেস্ক
বাকি থাকে = 15 - 8 = 7টি ডেস্ক

সাকিব 10টি ডেস্ক রঙ করতে সময় নেয় 20 মিনিট।
∴ সাকিবের প্রথম 5টি ডেস্ক রঙ করতে সময় লেগেছে = (20 × 5)/10 = 10 মিনিট
∴ সাকিবের অবশিষ্ট 7টি ডেস্ক রঙ করতে সময় লাগবে = (20 × 7)/10 = 14 মিনিট

∴ মোট সময় = 10 (সাকিব) + 3 (তামিম) + 14 (সাকিব) = 27 মিনিট।

Question: The average of nine numbers is 18. The average of six of these numbers is 16. What is the average of the remaining three numbers?

Solution:
৯ টি সংখ্যার গড় = ১৮
৯ টি সংখ্যার সমষ্টি = ১৮ × ৯ = ১৬২

৬ টি সংখ্যার গড় = ১৬
৬ টি সংখ্যার সমষ্টি = ১৬ × ৬ = ৯৬

∴ বাকী ৩ টি সংখ্যার সমষ্টি = ১৬২ - ৯৬ = ৬৬
∴ ৩ টি সংখ্যার গড় = ৬৬/৩ = ২২

- বৃত্তের কেন্দ্র (center of the circle) থেকে পরিধির (circumference) উপর যে কোন বিন্দুর দুরত্বকে বৃত্তের ব্যাসার্ধ বলে।
- অন্যভাবে বললে, বৃত্তের কেন্দ্র ও পরিধির উপর যে কোন বিন্দুর সংযোজক রেখাংশের দৈর্ঘ্যকে বৃত্তের ব্যাসার্ধ বলে।

Let selling price each oranges is TK 1
so selling price of 36 oranges is Tk. 36
loses on oranges is Tk. 4
so cost price = 36+4=40
therefore, Loss percent= (loss/cost price) × 100
= (4/40) × 100
= 10%

At 5% profit, selling price = 100+5 = 105 tk
At 5% loss, selling price = 100-5 = 95 tk
Difference between selling price = 105-95 = 10 tk
When the difference 10, buying price is 100 tk
∴ When the difference is 15, buying price is (100×15)/10 tk = 150 tk

Question: A technician charges a Tk. 50 service fee plus Tk. 30 per hour for labur. If a customer's total cost is Tk. 200, what is the maximum number of full hours the technician can work?

Solution:
Given that, 
Fixed service fee = Tk. 50
Charge per hour = Tk. 30
Total bill = Tk. 200

Let h = number of full hours the technician works.

Now, total cost equation,
50 + 30h ≤ 200
⇒ 30h ≤ 200 - 50
⇒ 30h ≤ 150
⇒ h ≤ 150 / 30
∴ h ≤ 5

So the technician can work a maximum of 5 full hours.

Question: A man lends Tk. 5000 in three parts at different rates of simple interest per annum. If he gets 4% on TK. 1500 and 5% on TK. 2500, what percent must he get for the remainder, if the average rate of simple interest per annum is 5.5%?

Solution:
Total Principal, P = 5000 TK.
Average Interest Rate = 5.5%
Total Interest required = (5000 × 5.5)/100 = 275 TK.

Interest from 1st part (1500 at 4%) = (1500 × 4)/100 = 60 TK.
Interest from 2nd part (2500 at 5%) = (2500 × 5)/100 = 125 TK.

Total interest from these two parts = 60 + 125 = 185 TK.
Remaining Interest needed = 275 - 185 = 90 TK.

Remaining Principal = 5000 - (1500 + 2500) = 1000 TK.

Let the required rate for the remainder be r%.
Using SI = Pnr/100 (where n = 1 year):
⇒ 90 = (1000 × 1 × r)/100
⇒ 90 = 10r
⇒ r = 90/10
⇒ r = 9

So, he must get 9% interest for the remainder.

Question: If sin(θ - 15°) = 1/2, then what is the value of tanθ?

Solution:
Given that,
sin(θ - 15°) = 1/2
⇒ sin(θ - 15°) = sin30°
⇒ (θ - 15°) = 30°
∴  θ = 45°

Now,
tanθ
= tan45°
= 1

Question: If 7 - 3x ≤ 16, then what is the value of x?

Solution:
7 - 3x ≤ 16
⇒ - 3x ≤ 16 - 7
⇒ - 3x ≤ 9
⇒  3x ≥ - 9
⇒ x ≥ - 9/3
⇒ x ≥ - 3

সমাধানটিকে ব্যবধি (Interval) আকারে প্রকাশ করলে হয়: [- 3, ∞)
এখানে তৃতীয় বন্ধনী [ দ্বারা বোঝায় - 3 সমাধান সেটের অন্তর্ভুক্ত, এবং ∞ এর পাশে প্রথম বন্ধনী) বোঝায় যে এটি অসীম পর্যন্ত।

Question: The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Tk. 4,000. The total price of 12 chairs and 3 tables is:

Solution: 
Let the cost of a chair and that of table be Rs. x and Rs. y respectively.
Then,10x = 4y or y = 5x/2
∴15x + 2y = 4000
⇒ 15x + 2 × 5x/2 = 4000
⇒ 20x = 4000
∴ x = 200
So, y = 5 × 200/2 = 500
Hence, the cost of 12 chairs and 3 tables = 12x + 3y = Tk. (2400 + 1500) = Tk.3900

a = 8; b = 9; C = 12
Capacity of a tank
= a × b × c/(c-a)
= (8 × 9 × 12)/4
= 216 Litre.

Question: A sum of Tk 7800 gives a simple interest of Tk. 702 in 2 years and 3 months. The rate of interest per annum is =?

Solution:
Time = 2 years 3 months
= 2 + 3/12
= 2 + 1/4
= (8+1)/4
= 9/4 years

Here,
I = 702
P = 7800
n = 9/4
r = ?

We know,
I = Pnr
⇒ r = I/pn
⇒ r = (702 × 4 × 100)/(7800 × 9)
⇒ r = 4%

Question: An error 4% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

Solution: 

100 cm is read as 104 cm.
∴ A₁ = (100 x 100) sq. cm and A₂ (104 x 104) sq. cm

Now,
(A₂ - A₁) = [(104)² - (100)²]
= (104 + 100) x (104 - 100)
= 816 sq. cm

Percentage error in area = [(A₂ - A₁)/A₁] × 100
= [816/(100 x 100)] × 100%
= 8.16% 

Investment of A for 8 months=Tk 15400
Investment of B for 6 months=Tk. 18200
Investment of C for 8 months= Tk.12600
Ratio of the share of A, B and C
= 15400×8:18200×6:12600×8
= 154×8:182×6:126×8
= 44:39:36
Sum of the terms of ratio
= 44+39+36 = 119
∴ Share of C
= Tk.(36/119 × 28790) ≈ Tk 8710 

Total quantity of petrol consumed in 3 years
= (6460/17 + 6460/19 + 6460/20) litres
= (380 + 340 + 323) litres
= 1043 litres
Total amount spent
= Tk. (3 × 6460)
= Tk. 19380
∴ Average cost
= Tk. (19380/1043)
= Tk. 18.58

Let speed upstream = x
Then,
Speed downstream = 2 x
Speed in still water = (2 x + x) /2 = 3x/2
Speed of the stream = (2 x − x)/2 = x/2
Speed in still water : Speed of the stream
= 3x/2 : x/2
= 3 : 1

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

Solution:
Given that,
Today Ratul 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 × Ratul's age

Now,
Ratul'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, Ratul's father will be twice as old as Ratul.

The amount of money deposited at 5% interest = x
The amount of money deposited at 7% = 1200 - x
A/Q, ( x × 5 × 1)/100 + {(1200 - x) × 7 × 1}/100 = 72
Or, 5x + 8400 - 7x = 7200
Or, 2x = 1200
So, x = 600

Question: A, B, and C are partners in a business. Their investments are in the ratio 1/3 : 1/4 : 1/6 respectively. B withdraws half of his capital after 4 months. The business runs for 12 months. If the total profit is Tk. 6,300, what is C’s share of the profit?

Solution:
Ratio of initial investments = 1/3 : 1/4 : 1/6
= 4 : 3 : 2 (Multiplying by LCM 12)

Let the initial investments be 4x, 3x, and 2x respectively.
A : B : C = (4x × 12) : (3x × 4) + (3x × 1/2 × 8) : (2x × 12) [B withdraws half of his capital after the first 4 months]
= 48x : (12x + 12x) : 24x
= 48x : 24x : 24x
= 2 : 1 : 1

Sum of the ratio = 2 + 1 + 1 = 4
Total profit = Tk. 6,300

∴ C’s share = 6,300 × (1/4)
= 1,575

∴ C’s share of the profit is Tk. 1,575.

Question: A shopkeeper sells 15 notebooks for Tk. 720 and incurs a loss equal to the cost price of 5 notebooks. What is the cost price of one notebook?

Solution:
Let,
Cost price of 1 notebook = Tk. x

∴ Cost price of 15 notebooks = Tk. 15x
∴ Cost price of 5 notebooks = Tk. 5x

We know,
Loss = Cost price − Selling price

According to the question,
5x = 15x − 720
⇒ 15x − 5x = 720
⇒ 10x = 720
⇒ x = 720/10
⇒ x = 72

∴ The cost price of 1 notebook is Tk. 72

Question: A jogger running at 9 kmph alongside a railway track is 240 meters ahead of the engine of a 120 meters long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?

Solution: 
Relative Speed = (45 - 9) kmph
= 36 kmph
= 36 × (5/18)
= 10 m/s

The train must cover the initial distance separating it from the jogger and its own length to fully pass him.
∴ Total Distance = (240 + 120) meters
= 360 meters

Now,
Time = Distance/Speed
= 360/10 seconds
= 36 seconds

∴ The train will take 36 seconds to completely pass the jogger.

Question: If a and b are positive real numbers, then (a⁰ - 3b⁰)⁵ = ?

Solution:
We know that for any positive real number,
a⁰ = 1 and b⁰ = 1

So, (a⁰ - 3b⁰)⁵
= (1 - 3 × 1)⁵
= (1 - 3)⁵
= (- 2)⁵
= - 32

Surface area of the box = 2(lb + bh + hl)
S.A. = 2[(80 × 40) + (40 × 20) + (20 × 80)]
= 2(3200 + 800 + 1600)
= 2 × 5600
= 11200 sq.cm.

Let,
Principal = Tk. 100
Amount = Tk. 200

Alternative Method:
Suppose 100 be the principal.
In 4 years it doubles and becomes 200.
In next 4 years 200 double to 400.
In next 4 years 400 doubles to 800.
Hence the period for 8 times is 12 years.

We know,
Average = Sum of Quantities/Number of Quantities

Here,
Number of quantities = Number of students in each batch

As average marks of students are given, calculate total marks of each batch first. So total marks for
Batch 1 = (40 x 45) = 1800
Batch 2 = (50 x 55) = 2750
Batch 3 = (60 x 70) = 4200

Sum of marks = (1800 + 2750 + 4200) = 8750

Therefore,
Required Average = (Some of works)/(Total number of students in each batch)
= 8750/(40 + 50 + 60)
= 8750/150
= 58.33

In a days P does 1/20 work; And in 1 day Q does 1/25 work
As seen above, Q works alone for 10 days.
In 1 days Q completes (1/25) × 10 = 2/5 work
Remaining work = 1 - (2/5) = 3/5 = Done by P and Q together
Total Work done = Total days x Work done by all in 1 day

Let P and Q work together for total K days.

∴ 3/5 = K × (1/20 + 1/25)
K = 20/3 = 6(2/3) days = Days when P and Q worked together
Thus P leaves after 6(2/3) days.

Question: Solution set of the inequality: 3y + 4 ≥ 2y - 5 is-

Solution:
Given that,
3y + 4 ≥ 2y - 5
⇒ 3y + 4 - 2y ≥ 2y - 5 - 2y
⇒ y + 4 ≥ - 5
⇒ y ≥ - 5 - 4
⇒ y ≥ - 9

∴ Solution set of the inequality is  [- 9,  ∞)

Let the number be x
Then,
⇒ (x+x²)/2 = 5x
⇒ x²− 9x = 0
⇒ x (x−9) = 0
⇒ x = 0 or, x = 9
So, the number is 9.

Question: If tanθ = 4/3, then what is the value of (3sinθ + 2cosθ)/(3sinθ - 2cosθ)?

Solution: 
Given,
tanθ = 4/3

Now,

Question: The present ages of Hasan and Kamal are in the ratio 4 : 5. After 6 years, the ratio of their ages will be 5 : 6. What is the difference in their present ages?

Solution:
Let, their present ages be 4x and 5x.

After 6 years,
Hasan's age = 4x + 6
Kamal's age = 5x + 6

According to the question,
(4x + 6)/(5x + 6) = 5/6
⇒ 6(4x + 6) = 5(5x + 6)
⇒ 24x + 36 = 25x + 30
⇒ 25x − 24x = 36 − 30
⇒ x = 6

Hasan's present age = 4 × 6 = 24 years
Kamal's present age = 5 × 6 = 30 years

∴ Difference = 30 − 24 = 6 years

Question: A clock is started at noon. By 10 minutes past 5, the hour hand has turned through:

Solution: 
Total time from 12:00 to 5:10 is 5 hours 10 minutes
5 hours = 60 × 5 = 300 min
∴ 5 hours 10 min = 300 min + 10 min
= 310 min
= 310 min/60 min 
= 31/6 hours

Now,
Angle traced by the hour hand in 12 hours = 360°

∴ Angle traced by the hour hand in 5 hours 10 min. i.e. (31/6) hours = [(360/12) × (31/6)]°
= 155°

∴ Sum of remaining 2 innings = 1269 - 1050 = 219

Let,
The minimum score be x and the maximum score be x + 157

According to the question,
x + x + 157 = 219
⇒ 2x = 219 - 157
⇒ 2x = 62
∴ x = 31

So, highest score = 157 + 31 = 188

Question: If θ be an acute angle and 7sin²θ + 3cos²θ = 4, then the value of tanθ is?

Solution:
7sin²θ + 3cos²θ = 4
⇒ 7sin²θ + 3(1 - sin²θ) = 4
⇒ 7sin²θ + 3 - 3sin²θ = 4
⇒ 4sin²θ = 1
⇒ sin²θ = 1/4
⇒ sinθ = 1/2
⇒ sinθ = sin30°
∴ θ = 30°

∴ tanθ = tan30° = 1/√3

Let, the number is 3x and 4x
Then 3x + 4x = 420
Or, 7x = 420
Or, x = 60
So, the greater number is 4 × 60 = 240

Question: The greatest value of sin⁴θ + cos⁴θ + 2sin²θcos²θ is?

Solution:
We know,
sin²θ + cos²θ = 1

Squaring both sides,
(sin²θ + cos²θ)² = 1²
⇒ sin⁴θ + cos⁴θ + 2sin²θcos²θ = 1

∴ The greatest value of sin⁴θ + cos⁴θ + 2sin²θcos²θ is 1.

Question: If θ = 45°, then what is the value of (1 - sec²θ)/(1 + sec²θ)?

Solution:
Here, θ = 45°
(1 - sec²θ)/(1 + sec²θ) = {1 - (sec 45°)²}/{1 + (sec 45°)²}
⇒ sec 45° = 1/cos 45° = 1/(1/√2) = √2
⇒ (1 - (√2)²)/(1 + (√2)²) = (1 - 2)/(1 + 2) = - 1/3

∴ The value of (1 - sec²θ)/(1 + sec²θ) = - 1/3