Bank Math

We know that,
Two pipes A and B can fill (or empty) a tank in X and Y minutes respectively, while working alone.
If both the pipes are opened together,
then the time taken to fill (or empty) the cistern is given by XY/(X+Y) minutes.
Here,
X = 30 minutes and Y = 40 minutes
Therefore,
the required time = (30 x 40)/(30 + 40)
= 1200/70
= 120/7
= 17(1/7) minutes.
Hence the answer is 17(1/7) minutes.

Multiples of 7 between 11 and 200 are 14, 21, 28, 35, 42, ..........., 189, 196.
T_m = 196
14 + (m - 1) × 7 = 196
⇒ (m - 1) × 7 = 196 - 14
⇒ (m - 1) × 7 = 182
⇒ (m - 1) = 182/7
⇒ (m - 1) = 26
⇒ m = 27.
Multiples of 7 and 3 both, i.e that of 21 are 21, 42, 63, ........, 189
T_n = 189
21 + (n - 1) × 21 = 189
⇒ (n - 1) × 21 = 189 - 21
⇒ (n - 1) × 21 = 168
⇒ (n - 1) = 168/21
⇒ (n - 1) = 8
⇒ n = 9
∴ Required number of terms = 27 - 9
= 18.

Here,
AB = √{(1 – 5)² + (0 – 0)²} = √(-4)² = √16 = 4
BC = √{(5 – 3)² + (0 – 4)²} = √(4 + 16) = √20
AC = √{(1 - 3)² + (0 – 4)²} = √(4 + 16) = √20 

So, AC = BC 
As in a triangle it can be, AC = CA and BC = CB

∴ CA = CB is true

Question: (a² - b² - 2bc - c²)/(a² + b² + 2ab - c²) is equivalent to?

Solution:
(a² - b² - 2bc - c²)/(a² + b² + 2ab - c²)
= {a² - (b² + 2bc + c²)}/{(a² + b² + 2ab) - c²}
= {a² - (b + c)²}/{(a + b)² - c²}
= (a + b + c)(a - b - c)/(a + b + c)(a + b - c)
= (a - b - c)/(a + b - c)

Question: The monthly salaries of A, B and C are in the ratio 1 : 2 : 3. If C’s monthly salary is Tk. 1200 more than that of A, then what is B’s annual salary?

Solution:
Let the monthly salary of A, B and C be x, 2x and 3x
If C’s monthly salary is Tk. 1200 more than that of A, then

ATQ,
3x = x + 1200
⇒ 2x = 1200
⇒ x = 1200/2 = 600
∴ x = 600

Then, B’s monthly salary = 2x = 2 × 600 = 1200

∴ B’s annual salary = 1200 × 12 = Tk. 14400

Question:

Solution: 

Question: Find the average of first 97 natural numbers.

Solution:
We know,
Average = Sum of first n natural numbers​/n

And the sum of first n natural numbers = n(n + 1)/2

∴ Average = {n(n + 1)/2}/n = (n + 1)/2
= (97 + 1)/2  ; [Substitute n = 97]
= 98/2
= 49

2¹²ⁿ−6⁴ⁿ
=(2³)⁴ⁿ−6⁴ⁿ
=8⁴ⁿ−6⁴ⁿ
=(8²)²ⁿ−(6²)²ⁿ
= 64²ⁿ−36²ⁿ (n=1)
= 64²−36²
=(64+36)(64−36)
=100×28 = 2800

Distance covered = (108 + 112) = 220 meter.
Time taken = 6 seconds.
Relative speed = 220/6 = 110/3 m/s.
= (110/3) × (18/5) km/hr
= 132 km/hr.

ATQ, 50 + Speed of second train = 132 km/hr.
⇒ Speed of second train = (132 - 50) = 82 km/hr.

Question: If 1 + tan²θ = 4 and θ < 90°, than what is the value of θ = ?

Solution:
Given that,
1 + tan²θ = 4 and θ < 90°
⇒ sec²θ = 4    ; [sec²θ = 1 + tan²θ]
⇒ (secθ)² = (2)²
⇒ secθ = 2
⇒ secθ = sec60°
⇒ θ = 60°

Question: If a is an even integer and b is an odd integer, which of the following must be an odd integer?

Solution: 
Let, a = 2 and b = 3

Then, 
ক) a/b
= 2/3
So a/b is not an integer, hence cannot be odd integer.

খ) 2(a + b)
= 2 × (2 + 3)
= 10
So 2(a + b) is always even.

গ) ab
= 2 × 3
= 6    ; [even × odd = even]
So ab is always even.

ঘ) 2a + b 
= (2 × 2) + 3
= 4 + 3
= 7
So 2a + b is always odd.

∴ 2a + b This must be an odd integer.

A, B and C are partners of a company. During a particular year A received one-fifth of the profit, B received one-third of the profit and C received the remaining Tk 5600. How much did B receive?

Solution:
Let the total profit be Tk x.

Now, Total profit - A's share - B's Share = C's share
⇒ x - (x/5 + x/3) = 5600
⇒ x - {(3x + 5x)/15} = 5600
⇒ x - (8x/15) = 5600
⇒ (15x - 8x)/15 = 5600
⇒ 7x/15 = 5600
⇒ 7x = 5600 × 15
⇒ 7x = 84000
⇒ x = 84000/7
∴ x = 12000

So, B's share = 12000 × 1/3 = 4000

Question: If tan(θ - 45°) = 1, then what is the value of sinθ?

Solution:
Given that,
tan(θ - 45°) = 1
⇒ tan(θ - 45°) = tan45°
⇒ (θ - 45°) = 45°
∴  θ = 90°

Now,
sinθ
= sin90°
= 1

After down payment and first installment, remaining amount = 3,90,000 - 40,000 - 5,000 = 3,45,000
As the installments will be doubled each month, next installments are 10000, 20000, 40000, 80000, 160000, 320000
As the amounts won't be repaid by 5th installments fully, it will take 6th installments to complete the payment.  

(2a)² + (3b)²
= (2a + 3b)² - 2.2a.3b
= (2a + 3b)² - 12ab
So, we have to add 12ab to make the sum a perfect square

Question:

Solution:

Question: If tan(x - 15°) = 1, what is the value of sin(x + 30°)?

Solution:
দেওয়া আছে,
 tan(x - 15°) = 1
⇒ tan(x - 15°) = tan 45°
⇒ x - 15° = 45°
⇒ x = 45° + 15°
⇒ x = 60°

এখন,
sin(x + 30°)
= sin(60° + 30°)
= sin 90°
= 1

Question: In a class, 25 students play cricket, 25 students play football, and 10 students play both. 10 students play neither cricket nor football. What is the total number of students in the class?

Solution:
Number of students who play cricket, n(C) = 25
Number of students who play football, n(F) = 25
Number of students who play both cricket and football, n(C ∩ F) = 10
Number of students who play neither = 10

n(C ∪ F) = n(C) + n(F) - n(C ∩ F)
= 25 + 25 − 10 = 40

Total students in the class = students who play cricket or football + students who play neither
n(U) = n(C ∪ F) + neither = 40 + 10 = 50

∴ Total 50 students in the class.