Bank Math

3% sugar solution = 50×(3/100) = 1.5 Liters of sugar
This amount must be 5% of a reduced final amount (50 - x)
ATQ,
1.5 = 5% × (50-x)
⇒ 50 - x = 150/5 = 30
∴ x = 20

Relative speed
= (80 + 55) km/hr
= 135 km/hr.
= 135 × (5/18) m/sec
= (75/2) m/sec.

Distance covered = (120 + 90 + 90)
= 300 m.

∴ Required time = {300 × (2/75)}
= 8 sec.

Question: A man rowed 3 miles upstream in 90 minutes. If the river flowed with current of 2 miles per hour, how long did the man's return trip take?

Solution:
Let,
The velocity of the boat is x mph
and the stream be y mph
In upstream,
In 90 minutes, he goes 3 miles
In 60 minutes, he goes (3 × 60)/90 = 2 miles

ATQ,
x - y = 2
⇒ x - 2 = 2 [stream's velocity = 2 mph]
⇒ x = 4

So, velocity in downstream = x + y = 2 + 4 = 6 mph
He goes 6 miles in 1 h
He goes 3 miles in = 3/6
= [(1/2) × 60] minutes
= 30 minutes

Question: 

Solution: 

Question: From a point Q on level ground, the angle of elevation of the top of a building is 45 degrees. If the building is 50 m high, find the distance of point Q from the foot of the building.

Solution:

Height of building AB = 50 m, angle of elevation ∠AQB = 45°

We know, tanθ = opposite/adjacent = AB/AQ
⇒ tan45° = 50/AQ
⇒ 1 = 50/AQ
⇒ AQ = 50 m

Thus, the distance from point Q to the foot of the building is 50 m.

Question: A ladder is leaning against a wall. It makes a 60° angle with the ground. If the length of the ladder is 10 meters, what is the distance between the foot of the ladder and the wall?

Solution:

ধরি, দেয়ালটি হলো AB এবং মইটি হলো AC।
মইটি ভূমির সাথে ∠ACB = 60° কোণ তৈরি করে।
মইয়ের দৈর্ঘ্য, AC = 10 মিটার।
মইয়ের গোড়া থেকে দেয়ালের দূরত্ব হলো BC।

এখন, ΔABC -এ
cos60° = BC/AC
⇒ 1/2 = BC/10
⇒ BC = 10/2
∴ BC = 5
∴ মইয়ের গোড়া থেকে দেয়ালের দূরত্ব 5 মিটার।

Let the speeds of the faster and slower trains be x m/sec and y m/sec respectively.
Then,
240/(x - y) = 60
⇒ x - y = 4 ........(i)

And, 240/(x + y) = 3
⇒ x + y = 80 ........(ii)

Adding (i) and (ii), we get
2x = 84
⇒ x = 42
Putting x = 42 in (i), we get:
42 - y = 4
⇒ y = 38

Hence, speed of the faster train = 42 m/sec.

Speed of train = 650/65 = 10 m/s
Let, speed of the man is x
So, 6:1 = 10:x
∴ x = 10/6 = 5/3
Time required to cross the platform by the man = 240/(5/3) = 144 sec = 2 minutes 24 seconds

Question: A sum of Tk. 2500 amounts to Tk. 2809 in 2 years at compound interest. Find the rate of interest per annum.

Solution:
Here,
Principal, P = 2500 Tk.
Final amount, A = 2809 Tk.
Time, n = 2 years
Interest rate, r = ?

প্রশ্নমতে,
A = P × (1 + r/100)ⁿ
⇒ 2809 = 2500 × (1 + r/100)²
⇒ (1 + r/100)² = 2809/2500
⇒ (1 + r/100) = √(2809/2500)
⇒ 1 + r/100 = 53/50
⇒ r/100 = (53/50) - 1
⇒ r/100 = 3/50
⇒ r = (3/50) × 100
⇒ r = 6

∴ The annual rate of interest is 6%.

let the width of rectangle be x so the length & breath is increased by 2x.

so, new total area along with walkway is (60+2x)×(20+2x)

so, (60+2x)×(20+2x)-60×20 = 516
⇒ (60+2x)×(20+2x) = 1716
⇒ (30+x)×(10+x) = 429 = 33×13
⇒ x = 3

Question: If x² is an odd number, determine the nature of x² - x. 

Solution:
যেহেতু x² বিজোড় তাই x ও বিজোড় হবে।
এখন,
x² - x
= (x - 1)x
= x(x - 1)
∴ (x - 1) এবং x দুইটি ক্রমিক সংখ্যা।

x বিজোড় সংখ্যা হলে (x - 1) অবশ্যই জোড় সংখ্যা হবে।
কারণ দুইটি ক্রমিক সংখ্যার মধ্যে একটি বিজোড় হলে অন্যটি জোড় হবে।

সুতরাং, x ও (x - 1) এর গুনফল = x(x - 1) = x² - x একটি জোড় সংখ্যা।
[জোড় × বিজোড় = জোড়]

Let, 1 man's 1 day's work = x and
1 boy's 1 day's work = y

Then, 12x + 16y = 1/5
and 13x + 24y = 1/4.

Solving these two equations, we get,
x = 1/100 and
y = 1/200
∴ Required ratio = x : y
= 1/100 : 1/200
= 2 : 1.

Total Cost price of 100 mangoes = (100×12) = 1200 tk.
At 10% profit, total Selling price = (1200×110)/100 = 1320 tk.
Total Selling Price of 60 mangoes each 17.40 tk
= (60×17.40) = 1044 tk.
And, total Selling Price of x mangoes each 11.31 tk
= 11.31x tk.

ATQ,
11.31x = 1320-1044
Or, x = 276/11.31
Or, x = 24.40
So the least possible value of x = 25

Question: A bicycle is purchased for Tk. 200 and sold at a 10% loss. Find the selling price and the total loss incurred.

Solution:
Cost Price of the bicycle = Taka 200
Loss Percentage = 10%

∴ Loss Amount = Loss Percentage × Cost Price
= 10% × Taka 200
= Taka 20

∴ Selling Price = Cost Price - Loss Amount
= Taka 200 - Taka 20
= Taka 180

Total Loss Incurred: Tk. 20

মোট বল রয়েছে = 2 + 3 + 2 = 7 টি
নীল বাদে বল আছে = 7 - 2 = 5 টি
∴ বলটি নীল না হবার সম্ভাবনা = ⁵c₂ / ⁷c₂ = 10/21

Question: What will be the number in the question mark?
2, 5, 11, 20, 32, ?

Solution:
প্রদত্ত ধারাটি হলো: 2, 5, 11, 20, 32, ?
ধারার সংখ্যাগুলোর মধ্যে পার্থক্য নির্ণয় করি:
5 - 2 = 3
11 - 5 = 6
20 - 11 = 9
32 - 20 = 12

এখানে, প্রতিবার পার্থক্য 3 করে বৃদ্ধি পাচ্ছে।
∴ পরবর্তী পার্থক্য হবে = 12 + 3 = 15
∴ পরবর্তী সংখ্যাটি হবে = 32 + 15 = 47
অতএব, প্রশ্নবোধক স্থানে 47 বসবে।

Shortcut: 2 (+3)→ 5 (+6)→ 11 (+9)→ 20 (+12)→ 32 (+15) → 47.

Question: A and B can complete a piece of work in 18 days and 12 days respectively. They got a contract to complete the work for TK. 60000. The share of A in the contracted money will be-

Solution:
Ratio of number of days taken by A and B to complete the work = 18 : 12 = 3 : 2
∴ Ratio of efficiency of A and B = 2 : 3

Let their shares is in the ratio of 2x and 3x
Now,
2x + 3x = 60000
or, 5x = 60000
∴ x = 60000/5 = 12000

∴ share of A = 2x = 2 × 12000 = 24000 Taka

Question: What is the original price of a jacket if the sale price after 20% discount is Tk. 480?

Solution: 
Let the original price be x

Discount = 20% of x 
= 0.2x

Selling Price = Original Price - Discount
= x - 0.2x
= 0.8x

Now,
0.8x = 480
⇒ x = 480/0.8
∴ x = 600

∴ The original price of the jacket is Tk. 600

Question: Karim borrowed Tk. 10000 at a certain rate of simple interest for 4 years. If he paid Tk. 2500 as interest, find the rate of interest per annum.

Solution:
Given that, 
Principal, P = Tk. 10,000
Simple Interest, SI = Tk. 2,500
Time, n = 4 years

We know,
SI = (Principal × Rate × Time)/100
⇒ 2500 = (10,000 × r × 4)/100
⇒ 2500 = (40,000 × r)/100
⇒ 2500 = 400 × r
⇒ r = 2500/400
∴ r = 6.25%

Therefore, the rate of interest per annum is 6.25%.

Question: If θ is a positive acute angle satisfying sin²θ + sin⁴θ = 1, then find the value of cot²θ + cot⁴θ.

Solution:
Given that,
sin²θ + sin⁴θ = 1 ..........(1)
⇒ sin⁴θ = 1 - sin²θ
⇒ sin⁴θ = cos²θ
⇒ sin²θ.sin²θ = cos²θ
⇒ sin²θ = cos²θ/sin²θ
⇒ sin²θ = cot²θ 

Now, putting sin²θ = cot²θ ..........(2)
∴  sin⁴θ = cot⁴θ ..........(3)
From equation (2) & (3) we get
sin²θ + sin⁴θ = cot²θ + cot⁴θ
∴ cot²θ + cot⁴θ = 1

Thus, the value of cot²θ + cot⁴θ = 1.

এখানে, আমরা প্রশ্নের option গুলো বিবেচনা করিঃ
a) m যদি বিজোড় হয়, তখন p শুধুমাত্র বিজোড় হলেই (m + p)m জোড় হবে।
b) যদি m বিজোড় হয়, তখন p জোড় হলে কখনই (m + p)m জোড় হবে না।
c + d) যদি m জোড় হয়, তাহলে p জোড় বা বিজোড় যাই হোক না কেন (m + p)m জোড় হবে।
অর্থাৎ, m যদি even হয়, তাহলে প্রশ্নোক্ত সমীকরণ থেকে p জোড় বা বিজোড় যেকোনোটাই হতে পারে যা 'must' শর্তকে মানে না। 
তাই উত্তর হবেঃ If m is odd, then p is odd.

প্রশ্ন: যদি log_x(1/125) = - 3 হয়, তবে x এর মান কত?

সমাধান:
log_x(1/125) = - 3
⇒ x^{- 3} = 1/125    [log_a(b) = c  ⇒  a^c = b]
⇒ 1/x³ = 1/125
⇒ x³ = 125
∴ x = 5

If the remainder is the same in each case and the remainder is not given,
the HCF of the differences of the numbers is the required greatest number.

34 - 24 = 10
34 - 28 = 6
28 - 24 = 4

Hence, the greatest number which divides 24, 28, and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2.

Question: The least number which when divided by 4, 6, 8, 12 and 16 leaves remainder of 2 in each case is = ?

Solution: 
LCM of (4, 6, 8, 12, 16)
⇒ 16 × 3 = 48

∴ The number when divided by (4, 6, 8, 12, 16) leaves remainder 2 is
= 48 + 2
= 50

When Divided by 7,
A = 7x + 4
So, numbers can be: 4, 11, 18, 25, 32, 39, 46, 53…….

Again,
when divided by 11,
A = 11y + 9
So, numbers can be: 9, 20, 31, 42, 53…….

Here, 53 is common.

Now, LCM of 7 & 11 is 77.
So, the numbers pattern is : 77x + 53

Then,
77x ≤ 1000
Or, x ≤ 12.2
x can be 0, 1, 2 ...... 12 = total 13 integers
It can't be x = 13, then the number will be bigger than 1000

Total number of students = 20.
Let S be the sample space.
Then, n(S) = number of ways of three scored first mark
n(S) = ²⁰C₃
= 20 x 19 x 18 / 2 x 3
= 20 x 19 x 3

Let,
E be the event of 1 girl and 2 boys.
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12.
n(E) = ⁸C₁ x ¹²C₂
= 8 x 12 x 11 / 1 x 2
= 8 x 6 x 11.

Now, the required probability = n(E)/n(S)
= (8 x 6 x 11)/(20 x 19 x 3)
= 44/95.

আমরা জানি,
বাক্সের উপরিতলের ক্ষেত্রফল = 2(ab + bc + ca)
= 2(2 × 3 + 3 × 4 + 4 × 2)
= 52 বর্গমিটার
∴ মোট খরচ = 52 × 3 = 156

Question: A bus moves 600 meters in a minute, and a car travels 90 km in 60 minutes. How much faster is the car than the bus? 

Solution:
Speed of the bus = Distance/Time
= (600/1) meters/minute
= (600/1000)/(1/60) km/h
= (600 × 60)/1000 km/h
= 36 km/h

A car travels 90 km in 60 minutes
So, speed of the car = Distance/Time
= 90 km/h [60 minutes = 1 hour]

∴ Difference in speed between the car and the bus = (90 - 36) km/h = 54 km/h.

Question: 2log₁₀5 + log₁₀8 - (1/2)log₁₀4 = ?

Solution:

Question: What is the angle between the hour and minute hands of a clock when it is 20 minutes past 9?

Solution:
20 minutes past 9 অর্থাৎ, 9টা 20 মিনিট।
= 9 + (20/60) ঘন্টা
= 9 + 1/3
= 28/3 ঘন্টা

আমরা জানি,
ঘণ্টার কাঁটা 12 ঘণ্টায় 360° ঘোরে।
∴ 1 ঘণ্টায় ঘোরে = 360°/12 = 30°
∴ 28/3 ঘন্টায় ঘোরে = (30° × 28)/3 = 280°

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

∴ ঘড়ির কাঁটা দুটির মধ্যবর্তী কোণ = | 280° - 120° | = 160°

Question: What is the LCM of 6, 9, and 15?

Solution:
Prime factorization:
6 = 2 × 3
9 = 3 × 3 = 3²
15 = 3 × 5

LCM = take the highest power of each prime:
2¹ × 3² × 5¹ = 2 × 9 × 5 = 90

Question: A’s salary is 25% more than B’s, and B’s salary is 20% less than C’s. If C’s salary is 50,000 Taka, find A’s salary.

Solution:
B’s salary is 20% less than C’s:
B = C × (1 - 20/100) = 50,000 × 0.8 = 40,000 Taka

A’s salary is 25% more than B’s:
A = B × (1 + 25/100) = 40,000 × 1.25 = 50,000 Taka

∴ A’s salary = 50,000 Taka