Bank Math

The word PUZZLE has 6 different letters, but ATQ, the vowels should always come together.
Now, let the vowels UE as a single entity.
Therefore, the number of letters is PZZL = 4, and UE = 1
Since the total number of letters = 4+1 = 5

So the arrangement would be in ⁵P₅ = 5!/(5 - 5)!
= 5!/0!
= 5!
= 5 × 4 × 3 × 2 × 1
= 120 ways

Now, the vowels AE can arrange in 2 different ways, i.e., ²P₂ = 2! = 2 × 1 = 2 ways
Hence, the new words that can be formed = 120 × 2 = 240.

Question: The cost of a milk and water mixture in a 7:3 ratio is Tk. 500. Adding 2 liters of water increases profit. Find the profit percentage.

Solution:
Let
Milk = 7 liters
Water = 3 liters
Original mixture quantity = 10 liters

Cost of original mixture = 500 Taka
Cost per liter = 500 / 10 = 50 Taka

Water added = 2 liters
New total water = 3 + 2 = 5 liters
Milk remains = 7 liters
Total mixture = 7 + 5 = 12 liters

Total Selling Price = 12 × 50 = 600 Taka

Profit = SP - CP = 600 - 500 = 100
Profit % = (100/500) ​× 100 = 20%

∴ Profit percent = 20%

Question: The average salary of all the workers in a workshop is Tk. 8,000. The average salary of 7 technicians is Tk. 12,000 and the average salary of the rest is Tk. 6,000. The total number of workers in the workshop is:

Solution:
Let the number of rest workers = x

Now, According to the question,
(7 + x) × 8000 = 12000 × 7 + 6000x
⇒ 56000 + 8000x = 84000 + 6000x
⇒ 2000x = 28000
⇒ x = 14

So the total number of worker
= 14 + 7
= 21

Question: The value of tan 90° is-

Solution:
We know that,
tan⁡θ = sin⁡θ/cos⁡θ
⇒ tan⁡θ = sin⁡90°/cos⁡90°
⇒ tan⁡θ = 1/0 ; [sin⁡90° = 1 ; cos⁡90° = 0 and Division by zero is undefined]
∴ tan⁡θ = undefined

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters
Let us assume that other ‘P’ liters of water are added to the mixture to make water 25% of the new mixture.
So, the total amount of water becomes (30 + P) and
the total volume of the mixture becomes (150 + P)
Thus,
(30 + P) = 25% of (150 + P)
Solving, we get P = 10 liters

(m1×h1×t1)/w1 = (m2×h2×t2)/w2
9 taps × 20 min = t taps × 15 min
So, t = 12 taps

Question: Seats for Chemistry, Bangla and English in a school are in the ratio 5 : 8 : 10. There is a proposal to increase these seats by 20%, 30% and 40% respectively. What will be the ratio of increased seats?

Solution:

Originally, let the number of seats for Chemistry, Bangla and English be 5x, 8x and 10x respectively

Number of increased seats are (120% of 5x), (130% of 8x) and (140% of 10x)
= {(120/100) × 5x}, {(130/100) × 7x} and {(140/100) × 8x}
= 6x, 52x/5 and 14x

∴ The required ratio = 6x : 52x/5 : 14x
= 30 : 52 : 70
=15 : 26 : 35

Let x be the amount of water to be added.
New total amount of water = 4 + x
Total amount of chemical = 4
∴ New total = 4 + 4 + x = 8 + x
Final ratio required (for water) = 5/(5 + 3) = 5/8
Thus, (4 + x)/(8 + x) = 5/8
Or, 32 + 8x = 40 + 5x
Or, 3x = 8
Or, x = 8/3

A completes the work in 12 days at 8 hour per day = 12 x 8 = 96 hours
In 1 hour, A completes (1/96) amount of work

B completes the work in 10 days (2 days less than A) working 10 hours every day (i.e. 2 hours more than A per day) = 10 x 10 = 100 hours
In 1 hour, B completes (1/100) amount of work

Working together, in 1 hour A and B complete (1/96) + (1/100) = (49/2400) amount of work
So they complete the entire work in (2400/49) hours

Since they work for 8 hours per day, they need = (2400/49) × (1/8) = 300/49 days.

Question: Solve for x: log₂(x + 3) = 4

Solution: 
Given that, 
log₂(x + 3) = 4
⇒ x + 3 = 2⁴
⇒ x + 3 = 16
⇒ x = 16 - 3
∴ x = 13

Question: Rina is currently four times older than Meena. Six years ago, Rina was ten times older than Meena. What will be the sum of their ages after 5 years?

Solution:
ধরি,
বর্তমানে মিনার বয়স = x বছর
∴ রিনার বয়স = 4x বছর

৬ বছর পূর্বে,
মিনার বয়স (x - 6) বছর.
রিনার বয়স (4x - 6) বছর.

ATC,
4x - 6 = 10(x - 6)
⇒ 4x - 6 = 10x - 60
⇒ 6x = 54
∴ x = 9

∴ মিনার বর্তমান বয়স = 9 বছর
রিনার বর্তমান বয়স = 4 × 9 = 36 বছর

5 বছর পরে,
মিনার বয়স = 9 + 5 = 14 বছর
রিনার বয়স = 36 + 5 = 41 বছর

∴5 বছর পরে উভয়ের বয়সের যোগফল = 14 + 41 = 55 বছর

Question: Find the value of sin(7π/6)

Solution:
sin(7π/6) = sin(π + π/6)
= - sin(π/6) [যেহেতু (π + π/6) তৃতীয় চতুর্ভাগে পড়ে এবং তৃতীয় চতুর্ভাগে sin ঋণাত্মক, তাই sin(π + θ) = - sinθ]
= - sin(30°)
= - 1/2

According to the question,
Let, D is number of days
(639×12×5)/1 road = (30×6×D)/1 road
⇒ D = 213 days

Speed in m/sec = 60×(5/18) = (50/3) m/sec
Time taken to cross the man = 6 secs
Therefore, Length of the train = (Speed×Time)
= (50/3)×6 = 100 metres

Question: What is the solution of the inequality: - 10 < 2x - 4 ≤ 6 = ?

Solution:
Given that,
- 10 < 2x - 4 ≤ 6
⇒ - 10 + 4 < 2x - 4 + 4 ≤ 6 + 4
⇒ - 6 < 2x ≤ 10
⇒ - 6/2 < 2x/2 ≤ 10/2
⇒ - 3 < x ≤ 5

∴ Solution of the inequality: (-3, 5]

Question: If |x - 3| < 4, then for which values of m and n will m < 3x + 5 < n hold?

Solution:
|x - 3| < 4
⇒ - 4 < x - 3 < 4
⇒ - 4 + 3 < x < 4 + 3
⇒ - 1 < x < 7
⇒ - 1 × 3 < 3x < 7 × 3  ; [Now multiply all parts by 3]
⇒ - 3 < 3x < 21
⇒ - 3 + 5 < 3x + 5 < 21 + 5 ; [Now add 5 to all parts]
⇒ 2 < 3x + 5 < 26

Comparing this with m < 3x + 5 < n, then we get,
m = 2 and n = 26.

Question: (cos²θ + 1/cosec²θ) + 17 = x. What is the value of x²?

Solution:
We know,
sin²θ + cos²θ = 1

Given that,
cos²θ + (1/cosec²θ) + 17 = x
⇒ cos²θ + sin²θ + 17 = x   ; [1/cosecθ = sinθ]
⇒ 1 + 17 = x
⇒ x = 18
⇒ x² = 18²
⇒ x² = 324

∴ The value of x² is 324.

Investment = Tk.[20 × (50 − 5)] = Tk. 900.
FaceValue = Tk. (50 × 20) = Tk.1000.
Dividend = Tk. (27/2 × 1000/100) = Tk.135.
Interest Obtained = (135/900 × 100)% = 15%

Question: If the diagonal of a rectangle is 10 cm long and its perimeter is 32 cm. Find the area of the rectangle.

Solution: 
let length = x and breadth = y
2(x + y) = 32         
⇒  x + y = 16

x² + y² = 10²
= 100  

now (x + y)² = 16² 
⇒ x² + y² + 2xy = 256
⇒ 100 + 2xy = 256
⇒ 2xy = 256 - 100 = 156
 ⇒ xy = 78 

∴ area = xy = 78 sq.cm

Water content in 40 litres of mixture.
= 40×(10/100) =4 litres
Milk in the mixture.
= 40−4=36 litres
Let x litres of water is mixed
⇒(4+x)/(40+x)=20/100
⇒x=5 litres

Question: A factory produces 480 items in 6 days working 8 hours per day. How many items would it produce in 9 days working 10 hours per day?

Solution: 
480 items in 6 days, working 8 hours per day
Total hours worked = 6 × 8 = 48 hours
So, production rate = (480/48) = 10 items/hour

Total hours for 9 days working 10 hours/day
= 9 × 10 = 90 hours

So, total items = 90 × 10 = 900 items

Work done by P in 1 day = 1/23
Let work done by Q in 1 day = q
q × (130/100) = 1/23
⇒ q = 100/(23 × 130)
= 10/(23 × 13)
Work done by P and Q in 1 day = 1/23 + 10/(23 × 13)
= 23/(23 × 13)
= 1/13
⇒ P and Q together can do the work in 13 days

Question: Rahim weighs 74.5 kg. If he reduces his weight in the ratio 5 : 3, find his reduced weight in kg.

Solution:
মনেকরি,
রহিমের পূর্বের ওজন = 5x
রহিমের বর্তমান ওজন = 3x

প্রশ্নমতে,
⇒ 5x = 74.5
⇒ x = 74.5/5
∴ x = 14.9

∴ ওজন কমে যাওয়ার পর হবে = 3 × 14.9
= 44.70 kg

Question: Solve the inequality: 4(x - 3) > 2(x + 5) + 6 

Solution:
Given that,
4(x - 3) > 2(x + 5) + 6
⇒ 4x - 12 > 2x + 10 + 6
⇒ 4x - 12 > 2x + 16
⇒ 4x - 12 - 2x > 2x + 16 - 2x  ; [Subtract 2x from both sides]
⇒ 2x - 12 > 16
⇒ 2x - 12 + 12 > 16 + 12  ; [Add 12 to both sides]
⇒ 2x > 28
⇒ 2x/2 > 28/2 ; [Divide both sides by 2]
∴ x > 14 

Question: Find the distance between the points (2, - 4) and (- 4, 3).

Solution:
আমরা জানি,
দুটি বিন্দুর মধ্যবর্তী দূরত্ব নির্ণয়ের সূত্র:
দূরত্ব, d = √{(x₂ - x₁)² + (y₂ - y₁)²}

দেওয়া আছে,
দুটি বিন্দু যথাক্রমে (2, - 4) and (- 4, 3)

∴ মধ্যবর্তী দূরত্ব, d = √{(- 4 - 2)² + (3 + 4)²}
= √{(- 6)² + (7)²}
= √(36 + 49)
= √85

In 1 day Z can do = 1/15
In 1 day X + Y can do = 1/10
Given, X = Y + Z
So, X + Y + Z = 1/15 + 1/10 = 1/6
OR, Y + Z + Y + Z = 2(Y + Z) = 1/6
OR, Y + Z = 1/12
OR, Y = 1/12 -1/15 = 1/60
∴ Y alone can do the work in 60 days.

অপশনে একটু মডিফিকেশন করা হয়েছে। 

Question: If sinA + cosA = 1 , then A = ?

Solution:
sinA + cosA = 1
⇒ (sinA + cosA)² = 1²
⇒ sin²A + cos²A + 2sinAcosA = 1
⇒ 1 + 2sinAcosA = 1 
⇒ 2sinAcosA = 1 - 1
⇒ 2sinAcosA = 0
∴ sinAcosA = 0

Here,
sinA = 0
⇒ sinA = sin0°
∴ A = 0°

Or,
cosA = 0
⇒ cosA = cos90°
∴ A = 90°

A = 0°, 90°